LIBRARY OF THE UNIVERSITY OF CALIFORNIA. Class APPLIED MECHANICS A Treatise for the use of Students who have time to work Experimental, Numerical, and Graphical Exercises illustrating the subject BY JOHN PERRY, M.E., D.Sc., F.R.S. WHIT. SCH., Assoc. M.INST.C.E. PROFESSOR OP MECHANICS AND MATHEMATICS IN THE ROYAL COLLEGE OF SCIENCK, ' SOUT1I KENSINGTON ; PRESIDENT OF THE INSTITUTION OF ELECTRICAL ENGINEERS WITH 372 ILLUSTRATIONS CASSELL AND COMPANY, LIMITED LONDON, PARIS, NEW YORK & MELBOURNE. MCMV ALL RIGHTS RESERVED ERAL 6 t. First Edition 1897. Revise^ Edition 1898, Reprinted 1899, 1 9<>3 I 9S- PREFACE. THIS book describes what has for many years been the course of instruction in Applied Mechanics at the Finsbury Technical College. All mechanical and electrical engineering students in their first year had two lectures a week ; the substance of these lectures is here printed in the larger type. Mechanical engineers had three lectures a week in their second year ; the substance of these lectures is here printed in small type. As I found our arrangement of hours per week to work fairly well, I give it here : Mechanical Engineers. Electrical Engineers. 1st year. 2nd year. 1st year. 2nd year. Mathematics 4 4 4 4 Graphics and Machine Drawing Mechanics: Lectures ... 7 2 13 3 4 2 3 Mechanical Laboratory 3 2 2 Mechanics : Numerical Exercises 2 1 2 Mechanism 1 Wood and Iron Workshops ... 4 7 3 6 Chemists and building trade students also attended in the mechanical department, and the mechanical engineering students had courses of study in the physics and chemistry departments of the College. The Mechanics course included work on the steam and gas engine not given in this book. When, after much experience in teaching at an English public 160010 iv PREFACE. school, in the Imperial College of Engineering, Japan, and other places, I ventured sixteen years ago to publish my method of teaching mechanics, it was met with some ridicule. Even without encouragement I was prepared to pursue the course which I had tested and found to be good, but I met with a great deal of encouragement from thoughtful men of about my own age. I had myself to scheme out and make drawings for every piece of apparatus for the laboratory. I knew of no collection of numerical and graphical exercises which were suitable for students, but gradually a collectior was made from those given out in the lectures which seemed to me to be less objectionable, less academic, less misleading, than those hitherto available. I had difficulty in getting clever assistants trained in academic ways to sympathise with me. It was found in time that students took very eagerly to the quantitative experimental work, and that the whole system, faithfully followed, produced men whose knowledge was always ready for use in practical problems, and who knew the limits of usefulness of their knowledge. I am glad to say that more than twenty complete sets of the apparatus have been made and sent to various institutions by my workshop assistant, Mr. Shepherd. It would have made this book too large if I had included in it, as I should have liked to do, copies of the instructions which each student receives when he begins on a new piece of apparatus. Professor (now Sir Robert) Ball, at the Ptoyal College of Science, Dublin, started quantitative experimental mechanical work. He used the well-known frame of the late Professor Willis, which was taken to pieces and built up in new forms for fresh experiments. What I have done has been to carry out Professor Ball's idea, using a distinct piece of apparatus for each fresh kind of experiment. A student measures things for himself ; illustrates mechanical principles ; finds the limits to which the notions of the books as to friction and properties of materials are correct ; learns the use of squared paper, and PREFACE. the accuracy of graphical methods of calculation ; and, above all, really learns to think for himself. Professor Ewing, at Cambridge ; has developed the ideas of Professor Ball to a far greater extent than what I have had opportunity for, and I know of no place in which a better engineering education can be obtained at the present time than at Cambridge. I am glad to think that a system begun under the Sciei ce and Art Department by Sir Robert Ball is now likely to be adopted generally in science classes. I am under great obligations to my assistant, Mr. G. A. Baxandall, who has been to great trouble in adding to the exercises, verifying answers, and correcting proofs. Professor Willis, D.Sc., has been kind enough to read through the proofs, and consequently I feel that there can be no im- portant mistake anywhere. I should like to think that, before a student begins the part in small type, he has worked through Thomson and Tait's small book on " Natural Philosophy," and that he has read the early part of my book on " The Calculus for Engineers." JOHN PERRY. 16th July, 1897. Royal College of Science, London S.W. CONTENTS. ; J CHAP. PAGE I. INTRODUCTORY 1 II. VECTORS : RELATIVE MOTION 29 III. WORK AND ENERGY .38 IV. FRICTION 56 V. EFFICIENCY . . 88 VI. MACHINES: SPECIAL CASES 98 VII. ELEMENTARY ANALYTICAL AND GRAPHICAL METHODS . . 120 VIII. EXAMPLES IN GRAPHICAL STATICS 151 IX. HYDRAULIC MACHINES 170 X. MACHINERY IN GENERAL 220 XL KINETIC ENERGY 242 XII. MATERIALS USED IN CONSTRUCTION 275 XIII. TENSION AND COMPRESSION 290 XIV. SHEAR AND TWIST 323 XV. MORE DIFFICULT THEORY 361 XVI. BENDING . . 381 XVII. STRENGTH AT ANY SECTION OF A BEAM . . . .393 XVIII. SOME WELL-KNOWN RULES ABOUT BEAMS . . . .410 XIX. DIAGRAMS OF BENDING MOMENT AND SHEARING FORCE . 427 XX. MORE DIFFICULT CASES OF BENDING OF BEAMS . . . 441 XXI. BENDING AND CRUSHING 457 XXII. METAL ARCHES 478 XXIII. MEASUREMENT OF A BLOW ....... 4S'> XXIV. FLUIDS IN MOTION 505 XXV. PERIODIC MOTION . . . . ."' . . . 546 XXVI. MECHANISM 570 XXVII. CENTRIFUGAL FORCE . . . . . . . .597 XXVIII. SPRINGS 613 XXIX. RESILIENCE OF SPRINGS . ' , . . . . . . " . 641 XXX. CARRIAGE SPRINGS . . .....>.. , . 646 APPENDIX . . ... . .<. - . . 654 INDEX . ...,_ .-:. ; . G<>7 LIST OF TABLES. PAGB I. Normal Pressure of Wind against Koofs . . . t . 157 II. Moments of Inertia and the M and k of Rotating Bodies . . 251 III. Young's Modulus (Wertheim) . . ... . . . 302 IV. Factors of Safety for Different Materials and Loading . . 305 V. Ultimate Stresses for Loads Repeated a Great Number of Times 310 VI Moment of Inertia and Strength Modulus of Various Sections . 398-400 VII. Strength and Stiffness of Rectangular Beams Supported at the Ends and Loaded in the Middle 411 VIII. Diagrams of Bending Moment, with Strength and Stiffness, of a Uniform Beam when Supported or Fixed and Loaded in Various Ways "... 414-16 IX. Breaking Stress of Iron and Timber Struts - . . . .468 X. Values of the Constant B for Different Lengths of Struts and Different Materials . . ... . . . . ., 468 XI. Values of the Constant n for Struts of Different Sections . 469 XII. Relative Amplitudes for Different Frequencies in a Case of Forced Vibration 616 XIII. Spring Materials Subjected to Bending . . . - ., . 621 XIV. Materials for Cylindric Spiral Springs . . . . .622 XV. Values of Torsional and Flexural Rigidity and Axial Load for Various Sections of Wire used in Springs .... 635 XVI. Relative Elongations of Various Springs for Same Axial Loads. 638 XVII. Proof Load and Resilience of Various Springs . . . .640 XVIII. Resilience of Spring Materials for Different Kinds of Loading . 642 XIX. Useful Constants ... c .... 654-6 XX. Moduli of Rigidity 656 XXI. Moduli of Compressibility . 657 XXII. Melting Points, Specific Gravity, Strength, etc., of Materials 658-61 XXIII. Logarithms 662-3 XXIV. Antilogarithms . .' . . . . . . . 664-5 XXV. Angles in Degrees and Radians, Sines, Tangents, etc. . . 666 : ?^ UNIVERSITY OF APPLIED MECHANICS. CHAPTER I. INTRODUCTO BY. 1. THE student of Applied Mechanics is supposed to have some acquaintance already with the principles of mechanics ; to be able to multiply and divide numbers and to use logarithms ; to have done a little practical geometry; to know a little algebra and the definitions of sine, cosine, and tangent of an angle ; and to have used squared paper. He is supposed to be working many numerical and graphical exercises \ to be spending four hours a week at least in a mechanical laboratory ; to be learning about materials and tools in an iron and wood workshop; and to be getting acquainted with gearing and engineering appliances in a drawing-office and elsewhere. Unfortunately, many students are deceived as to their fitness to begin the study of A.pplied Mechanics, and we think it necessary in this introductory chapter to suggest some pre- paratory exercise work, and also to state certain definitions and facts which will be afterwards referred to, perhaps as if they were still unknown. When we think of what goes on under the name of teaching we can almost forgive a man who uses a method of his own, however unscientific it may seem to be. Nevertheless, it is not easy to forgive men who, because they have found a study interest- ing themselves, make their students waste a term upon it, when only a few exercises are wanted on what is sometimes called the scientific study of arithmetic, for example, or of mensuration. In our own subject of Applied Mechanics there are teachers who spend most of the time on graphical statics, or the graph- ing of functions on squared paper, or the cursory examination of thousands of models of mechanical contrivances. One teacher seems to think that applied mechanics is simply the study of kinematics and mechanisms ; another, that it is simply exercise work on pure mechanics ; another, that it is the breaking of specimens on a large testing machine ; another, that it is the trying to do in a school or college what can only be done in real engineering works ; another, that it is mere graphics ; another, that it is all calculus and" no APPLIED MECHANICS. graphics; another, that it is all shading and colouring and the production of pretty pictures without centre lines or dimensions. Probably the greatest mistake is that of wasting time in a school in giving the information that one cannot help picking up in one's ordinary practical work after leaving school. We believe that the principles which an engineer really recollects and keeps ready for mental use are very few. By means of lectures, models, drawing-office and laboratory, and numerical exercise work, we show a man how these simple principles enter, in curiously different-looking shapes, into his engineering practice. We give him the use of all the necessary methods of study, and we send him out into practical life pre- pared to study things for himself. We ought to recognise the fact that his real study of his profession is not at school or college. We ought to teach him how to learn for himself. Any child can state Newton's second law of motion, and the other half-dozen all-important principles of mechanics, so as to get full marks in an examination paper; the engineer knows that the phenomena he deals with are exceedingly complex, and that only a long ex- perience will enable him to utilise the so easily stated principles. Schools and colleges are the places in which men ought to learn the uses of all mental tools ; they are sure to specialise afterwards, but in the meantime we ought to give them plenty of tools to choose from. The average student cannot take in more than the elementary principles , the best students need not take in more. 2. The most important lesson for a beginner, however he may have studied mathematics and mechanics, and however able he may be as a mathematician, is this that he must not go on merely assuming that he knows how to do things ; he must know things by actual trial and not mere hearsay. He must actually calculate certain numerical results ; he must actually illustrate principles with laboratory apparatus ; and, if there is a school workshop, he must get to know the properties of materials by chipping and filing and paring and planing and turning. It is just the same as in one's after school work. There is no great mechanical engineer who has not himself worked like a workman with other workmen, and got to understand men and things by actual contact with them. The man who shirks the following exercises and laboratory work will lose a great deal more than he is aware of. Teachers will notice that things requiring even a little pre- paration more than other things will gradually become neglected. Therefore, let it be part of the daily work for every student to use logaritlims, drawing instruments in graphical exercise work, and squared paper. In the drawing-office, blue prints ought to bo made by some student or other every day ; the planimeter ought APPLIED MECHANIC'S. 3 to be used every day, and some student ought to be resetting his drawing-pens and other instruments every day. It ought to be a rule that all apparatus must always be ready for use, and that it is always in use. Teachers can arrange their own work in such a way that they cannot help seeing every day how the practical work of students is being done. When we find our system to be going with clockwork regularity and we feel no worry, we ought to believe that some change is necessary. If we find that the students are not absorbed in their work, we must understand that we teachers are in fault. 3. Students cannot spend too much time in multiplying, factoring, and simplifying algebraical and trigonometrical expressions. These are our tools, and we must get familiar with them. We may easily spend too much time in studying roots of equations, permutations and combinations, etc., and in the solution of triangles ; and therefore, if it is possible, we try to learn all our mathematics, mechanics, physics, and chemistry from teachers who are engineers. What acquaint- ance with these subjects we have, ought to be a real knowledge, .not the glib pretence which suffices for examinations ; it must not be something apart from our life and work. To effect this object we must work many numerical and graphical exercises, and try to conquer our contempt for simple laboratory experi- ments, and illustrate the forty-seventh proposition of the Fir.jt Book of Euclid by actually drawing some right-angled triangles and measuring their sides to illustrate rules about triangles by actual measurement. In this way we learn much more than the ordinary geometrician knows ; among other things, we obtain a valuable knowledge of the errors we are likely to make in graphical calculation. 4. Every student is supposed to be able to calculate the values of any algebraical or trigonometrical expression when numerical values are given. He ought to be able, when given at random such an expression as w = a 1 * b~ ll 2 4- v/w 2 + n* a log e b . cos. 9, to be able to calculate w when he knows the values of a, b, m, n, and 0. He ought to be able to use logarithms in multiplica- tion and division and extraction of roots ; to know all the usual mathematical symbols, and how it is sometimes convenient to use A/ 'a and sometimes ai It is pedantic to say that a man must not use a formula unless he is able to prove its truth. It is usually of great help in learning to prove a 4 APPLIED MECHANICS. formula to have previously used the formula and know the value and meaning of what we are to prove. A living Northern professor of great eminence has declared that a boy ought not to be allowed to use logarithms until he is able to calculate them ; he has not said that a boy ought not to use a watch or wear a coat until he is able to make them. EXERCISES. 1. Find 4-326 x 0-003457 to four significant figures, leaving out all unnecessary figures in the work. Find 0*01584 4- 2-104 to four signifi- cant figures. Also do these using logarithms. Find log e 7. Calculate 5 2 ' 43 , 3-' 246 , -042 ' 476 , /\/246-3, 31-01* x 0-02641T . Ans., 0-01495, 0-007529, 1-94591, 49'95, 0-7632, 0-2211, 3-008, 5'872. 2. If m = (a 3 + 2 a?b + s - '345)1 -=- (a 2 i 2 ) 5 find m if a = 0-504, b = 0'309, s = 1-667. Ana., 1-453 x 10 8 . 3. What errors are there in assuming (1 + a )n = I -f no, to be true in (1-001) 3 = 1-003, (1-01)* = 1-0033. (0-99) 2 = (1 - -Ol) 2 = 1 -02 = -98. !_ = 1 = (1 - -01) ~ l = 1 -f -01 = 1-01. 99 1 - -01 V - = (1 + -01) ""*=(! -0033) = -9967. Vl-01 ^99 = v/100^0" ~ '01) = 10 (1 - -01)* = 10 (1 - -005). = 9-95 ? The above answers are very nearly correct ; the student is expected to find the correct answers. 4. How much error is there in the assumptions 1 + = 1 + a - 0, (1 + a) (1 + 0) = 1 + a + |8, when a = '01 j8 = -01, a = - '003 = - '005 ? Ans. t No error : -01 per cent., -004 per cent., -0015 per cent. 5. If d is the diameter of the bore or the " calibre " of a gun, it is usually assumed that the weight of the gun is proportional to d 3 , and that the thickness of armour which its projectile will pierce is propor- tional to d. If an 8-inch gun weighs 14 tons and can pierce 11 inches of armour, what thickness will be pierced by a 10-inch gun, and what is the weight of the gun? Ans., 13*75 inches, 27*34 tons. 5. The linear expansion of bodies by heat is practically proportional to the rise of temperature. The values of a, the co-efficient for linear expansion (the fractional increase in length for a rise in temperature of 1 Centigrade), are the APPLIED MECHANICS. 5 following numbers divided by 10 5 : Aluminium, 2-34 ; copper, 1-79; gold, 1-45; iron, 1-2; lead, 2-95 ; platinum, -9 ; silver, 1-94; tin, 2-27 ; zinc, 2-9; brass (71 copper to 29 zinc), 1-87; bronze (86 copper to 10 tin to 4 zinc), 1-8; German silver, 1-8; steel, '1-11 ; brick, 0*5; glass, 0'9; granite, 0*9; sand- stone, 1-2; slate, 1'04; boxwood (across the fibre), 61; box- wood (along the fibre), O3; oak (across), 5*4; oak (along), 0'5 ; pine (across), 3 - 4 ; pine (along), 0'5. The co-efficient, k, of cubical expansion is three times the co-efficient of linear expansion, because (1 -f a) 3 = 1 -f 3 a is practically correct for these small values of a. The average values of k between and 100 C. are the following numbers divided by 10 3 : Alcohol, 1-26 ; mercury, 0-18 ; olive oil, 0-8; petroleum, 1-04 ; pure water, CH3 ; sea-water, 0'5. The student is supposed to have worked many exercises like the following ones : 1. Steel rails of C. have an aggregate length of 1 mile. What is the length at 33 C. ? Ans., 1 mile 24-2 inches. 2. A ring of wrought iron has an inside diameter of 5 feet when at a temperature of 970 C. What is the diameter at C. ? Ans., 4-9 feet. 3. A cylindric plug of copper just fits into a hole 4 inches diameter in a piece of cast iron. After heating the mass to 1,240 C., by how much is the diameter of the hole too small for the plug? Ans.^ -0293 inch. 4. A bar of iron is 70 centimetres long at C. What is its length in boiling water (100 C.) ? What is its length at 50 C. ? Ans., 70-079 centimetres, 70-039 centimetres. 5. Two rods one of copper, the other of iron measure 98 centimetres each at C. What is the difference in their lengths at 57 C. ? Ans., -027 centimetre. 6. Bars of wrought iron, each 3 '4 metres long, are laid down at a temperature of 10 C. What space is left between every two if they are intended to close up completely at 40 C. ? Ans., 1'26 mi Hi metres. 7. A wrought-iron connecting-rod is 12 feet long at 10 C. What is the increase of length at 80 C. ? Ans., 0-121 inch. 8. A wrought-iron Cornish boiler is 33 feet long; the shell is at C., the flue at 100 C. What would the difference of the lengths be if the flua were not prevented from expansion ? Arts., 0*475 inch. 9. A steel pump rod is 1,000 feet long. What is its change of length for a change of 10 C. Ans., 1'44 inch. 10. In a thermometer *01 cubic inch of mercury at 10 C. is raised to 15 C., and rises 1 inch in the tube. What is the cross-section of the tube ? Ans., 9 x 10~ 6 square inch. 11. The volume of a lump of iron being 5 cubic feet at 10 C., find its volume at 80 C. Ans., 5-0126 cubic feet. 6. A student's knowledge of mathematics ought to be such that he can work out for himself all the rules given in such an APPLIED MECHANICS. excellent book on mensuration as that of Professor A. Lodge. The thorough study of such a book is one of several ways which may be recommended of getting familiar with mathe- matical principles. But nobody's life is long enough to use all these ways, and, besides, unnecessary study leads to dul- ness. Hence, if a student has taken some other way, he need not be alarmed at his ignorance of the more complex rules in mensuration ; he may feel absolutely certain that he can work out such rules for himself, given time and necessity. He will study the more complex rules, such as prismoidal formulae, if he needs to use them practically, not otherwise. The following rules are in constant use and must be familiar to the student, whether or not he knows the reasons for them. If he is familiar with the rules and does not anxiously search for the reasons for them, he lacks the necessary spirit of the practical engineer. RULES IN MENSURATION. An area is found in square inches if all the dimensions are given in inches. It is found in square feet if all the dimen- sions are given in feet. Area of a parallelogram. Multiply the length of one side by the perpendicular distance from the opposite side. The centre of gravity of a parallelogram is at the point of intersection of its diagonals. Draw a right-angled triangle ; measure very accurately the lengths of the sides. You will find that, no matter what scale of measurement you use, the square of the length of the hypothenuse is equal to the sum of the squares of the lengths of the other two sides. Area of a triangle. Any side multiplied by its perpendicular distance from the oppo- site corner and divided by two. The centre of gravity, or, rather, the centre Fig. i. of area, of a triangle is found by joining (Fig. 1) any corner, A, with the middle point, D, of the opposite side, B c, and making D G one-third of D A. G is the centre of gravity. Area of an irregular figure. Divide into triangles, and add the areas of the triangles together. Circumference of a circle. Multiply the diameter by 3-1416. APPLIED MECHANICS. 7 Arc of a circle. From eight times the chord of half the arc subtract the chord of the whole arc ; one-third of the remainder will give the length of the arc, nearly. Area of a trapezium. Half the sum of the parallel sides multiplied by the perpendicular distance between them. Area of a circle. Square the radius, and multiply by 3'1416 ; or square the diameter, and multiply by 0-7854. Area of a sector of a circle. Multiply half the length of the arc by the radius of the circle. Area of a segment of a circle. Find the area of the sector having the same arc, and the area of the triangle formed by the chord of the segment and the two radii of the sector. Take the sum or difference of these areas as the segment is greater gr less than a semicircle. Otherwise, for an approximate answer : Divide the cube of the height of the segment by twice the chord, and add the quotient to two-thirds of the product of the chord and height of the segment. When the segment is greater than a semi- circle, subtract the area of the remaining segment from the area of the circle. The areas of curves may be found by Simpson's rule. Divide the area into any even number of parts by an odd number of equidistant parallel lines or ordinates, the first and last touching the bounding curve. Take the sum of the extreme ordinates (in many cases each of the extreme ordinates is of no length), four times the sum of the even ordinates, and twice the sum of the odd ordinates (omitting the first and last) ; multiply the total sum by one-third of the distance between any two successive ordinates. The ordinary rule for an indicator diagram is : Draw lines at right angles to the atmospheric line, touching the extreme ends of the diagram. Divide the distance between them into ten equal parts (a parallel ruler with ten pieces is sometimes supplied), and at the middle of each part dra'w a line at right angles to the atmospheric line. Measure the breadth of the diagram on each of these ten lines, and take one-tenth of their sum. This givea the average breadth, and represents the average pressure to scale. The better plan is to find the area by a planimeter. The earnest student will practise the use of the planimeter, finding its error by tests on rectangles and circles. The average breadth of a diagram is used in manjr ways. 8 APPLIED MECHANICS. Exercise for Advanced Students. Prove that Simpson's rule gives the area correctly if the arcs of curve between the odd ordinates follow, each of them, any such law as y = a -f bx + ex"* . Surface of a sphere. Multiply the diameter by the circumference. Surface of a cylinder. Multiply the circumference by the length, and add the areas of the two ends. Surface of a right circular cone. Multiply half the circumference of the base by the slant side, and add the area of the base. Lateral surface of the frustum of a right cone. Multiply the slant side by the circumference of the section equidistant from its parallel faces. Area of an ellipse. Multiply the product of the major and minor axes by '7854. The areas of two similar figures are as the squares of their like dimensions. The volumes are as the cubes of their like dimensions. The cubic content of a body is calculated in cubic inches if all the dimensions are given in inches ; in cubic feet if all the dimensions are given in feet. Cubic content of a plate. Multiply area of plate by its thickness. Cubic content of a sphere. Cube the diameter, and multiply by -5236. Cubic content of the segment of a sphere. Subtract twice A Fig. 2. APPLIED MECHANICS. the height of the segment from three times the diameter of the sphere ; multiply the remainder by the square of the height, and this product by -5236. The cubic content and surface of a sphere are each two- thirds of that of the cylindric vessel which just encloses it. Cubic content of any prismatic body (Fig. 2). Multiply the area of the base by the perpendicular height. This will give the same product as, Multiply the area of cross section by length along the axis of the prism. (The axis of a prismatic body goes from centre of gravity of base to centre of gravity of top. ) The centre of gravity of a prismatic body is half-way along the axis. Cubic content of any pyramidal or conical body (Fig. 3). Fig. 3. Multiply the area of the base by one- third of the perpen- dicular height. Centre of gravity is one-quarter of the way along the axis from the base. (The axis of any such body joins the centre of area of base with the vertex.) The cubic content of the rim of a wheel is found by multi- plying the area of a cross section by the circumference of the circle which passes through the centres of gravity of the cross sections. The weight of a cubic inch of each of the following mate- rials is given in Ibs. : Cast iron, -26; wrought iron, '28; steel, 28 ; brass, -3 ; copper, -32 ; bronze, -3 ; lead, 4; tin, -27 ; zinc, 26. Hence to find the weight of a body of cast iron or any other of these substances, find the volume in cubic inches and multiply by one of the above numbers. Very often it is only the approximate weight that is wanted, so that a moulder may know how much metal to melt, or for other purposes. Now, suppose we want the approxi- mate weight of a cast-iron beam. Find roughly the average section and get its area in square inches, multiply by the length 10 , APPLIED MECHANICS. in inches, add to this the cubic content of any little gusset plates or other excrescences, multiply by '26 and we have the weight in pounds. The specific gravity of a substance means its weight as compared with the weight of the same bulk of water. Now, it is known that a cubic foot of water weighs very nearly 1,000 ounces, or rather 62 - 3 Ibs. The specific gravity of a brick varies from 2 to 2-167, and therefore the weight of a cubic foot of brick varies from 2x1,000 or 2,000 to 2-167x1,000, or 2,167 ounces. We see, then, that from a table of specific gravities we can get the weight of a cubic foot of a substance, and therefore if we know the cubic content of a body formed of this substance we can calculate its weight. Various plans for saving labour in calculation suggest themselves to people working at any particular trade. For instance, if a pattern has no prints for cores, the weight of the pattern bears nearly the same proportion to the weight of the casting as the weight of a cubic inch of the wood bears to the weight of a cubic inch of cast iron. This is not always a con- venient rule, because the pattern is a little larger than the casting, and the density of wood alters as it dries. The area of an irregular figure may be obtained approxi- mately by cutting it out of a uniform sheet of cardboard and weighing it. Now cut out a rectangle or square whose area it is easy to calculate. Weigh this also. The areas are in the same proportion as the weights. The area of cross section of a fine wire in square inches can be determined with some accuracy by weighing a considerable length of the wire, dividing by the weight of the material per cubic inch, and dividing by the length of the wire in inches. EXERCISER 1. Find the number of revolutions per mile made by a rolling wheel 4| feet diameter. _ Ans., 373. 2. Find the area and circumference of a circle of 4 inches radius. Find the circumference and diameter of a circle whose area is 20 square inches. Ans., 50-27 square inches, 25-13 inches; 15-85 inches, 5-046 inches. 3. Find the area of a parallelogram whose adjacent sides are 50 and 30 feet and the angle hetween them 65. Ans., 1359-45 square feet. 4. Find the area of a sector of a circle, radius 4 mches, angle 50. Ans., 6-982 square inches. 5. Find the area of the segment of a circle, chord 20 inches, height 3 inches. If this were a parabolic segment, its area would ibe |rds of chord multiplied by height. Ans., 40'6 square inches. APPLIED MECHANICS. 11 6. Ordinates of a curve, 1-5 inches apart, are 2*30, 2-35, 2-46, 2-57, 2-42, 2-21, 2-10. Find the area between the first and last by Simpson's rule. Test your answer by drawing the curve and using a planimeter. Ana., 21-34. 7. Find the area of the surface of a sphere, its radius being 8 inches. Ans., 804-2 square inches. 8. A boiler has 300 tubes 8 feet long, 3 inches diameter. What is the total cross-sectional area ? What is the area of tube-heating surface ? Ans., 2,121 square inches, 1,885 square feet. 9. Eight circular cone, base 3 inches radius, height 8 inches ; find its curbed surface and volume. Ans., 80-52 square inches, 75-41 cubic inches. 10. Segment of sphere, height 4 inches, diameter of base 15 inches : find volume. Ans. , 309-9 cubic inches. 11. Find by Simpson's rule the number of cubic feet in a log of timber 36 feet long, the cross- sections at intervals of 6 feet being 8-20, 5-68, 4-04, 2-92, 2-16, 1-54, 1-02 square feet. Ans., 124-36. 12. A railway cutting is mile long; the thirteen cross-sectional areas, measured at equal intervals of 220 feet, are respectively 280, 462, 594, 685, 757, 742, 500, 346, 320, 418, 512, 626, 560 square feet. Find the volume by Simpson's rule. Ans., 52,480 cubic yards. 13. Two models of terrestrial globes ; the areas of Africa are in the ratio 3 to 2. What is the ratio of the diameters of the globes ? What is the ratio of their volumes ? Ans., 1*225, 1-837. 14. Find the surface and volume of a sphere of 2-361 inches radius. If it is of cast iron, what is its weight ? Find the weight of a segment of this sphere 1 inch in height. Ans., 70'1 square inches, 55-12 cubic inches ; 14-44 Ibs., 1*67 lb. 15. Find the volume and weight of the rim of a cast-iron wheel ; section circular, outside and inside radii 20 feet and 18 feet 6 inches. Ans., 213-7 cubic feet, 42-87 tons. 16. Find the circumference and area of a circle whose radius is 5 inches. This is the base of a cylinder of 11 inches height. What is its volume ? Find its curved surface. If made of cast iron, what is its weight? What is the area of a section making 70 with the axis? (A x cos. 20 is the area of the circular base which is its projection.) Ans., 31-42 inches, 78 -54 square inches, 863*9 cubic inches, 2'4 square feet, 226 -4 Ibs., 83'58 square inches. 17. A cone on an elliptic base, whose principal diameters are 12 inches and 8 inches, is 20 inches in vertical height. What is its volume ? What is its weight if made of cast iron ? Ans., 5027 cubic inches, 131 '7 Ibs. 18. A plate of wrought iron ^ inch thick weighs 0'6 lb. What is its area ? Ans., 34-3 square inches. 19. A disc of zinc 10 inches diameter, with a hole of 2 inches diameter in it, weighs 0-2 lb. Find its thickness. Ans., 0*0106 inch. 20. 4 Ibs. of copper is drawn into wire 0'14 inch diameter. Find its length. Ans., 6 7 '66 feet. 21. A piece of round copper wire 100 feet long weighs 5 Ibs. What is its diameter ? Ans., *128 inch. 22. A spherical shell 10 inches outside and 8 inches inside diameter, of cast iron : what is its weight ? Ant., 66-45 Ibs. 12 APPLIED MECHANICS. 23. A rolled girder of wrought iron is 12 inches outside depth, flanges are 6 inches "by \ inch, web is \ inch thick: what is its weight if 18 feet lon S ? Am., b95'6 ihs. 24. Cylindric boiler 30 feet long, 7 feet diameter. Two cylindric Sues each 2 feet 6 inches diameter: what is the area of the plates? Plate everywhere -i? inch thick : what is its weight ? When in this boiler water covers the flues and its level is a quarter of the diameter of the shell from the top, what are the volumes of water and steam ? Am ^ 1>189 gquare feetj 1Q . 7 tongj 2 . g to ^ 25. A circular plate of lead 2 inches thick, 8 inches diameter, is con- verted into spherical shot of the same density, each of '075 inch radius. How many shot does it make ? Ans., 56,889. 26. Walking from the centre towards the end of one span of a lattice girder railway bridge, I count 10 wrought iron bars of rectangular section, each 14 feet long ; the cross section of the first is 5 inches x $ inch. If the widths increase by ^ inch and the thicknesses by -^ inch, find the total weight of bars. ' Ans., 1-08 ton. 7. The idea of velocity involves two things the direction and the speed. When the direction does not alter, we speak of the speed as if it were the whole idea. Find the time in seconds taken by a body to traverse a certain distance measured in feet. This distance divided by the time is called the average velocity. Thus, if a railway train moves through 200 feet in 4 seconds, its average velocity during this time is 200 -f- 4, or 50 feet per second. If we find, with careful measuring instruments, that it moves through 20 feet in -4 second, or through 2 feet in '04 second, the velocity is 20 -f- *4, or 2 '04, or 50 feet per second. It is important to remember that the velocity may be always changing during an interval of time, however short. To get the velocity at any instant, we must make very exact measurements of the time taken to pass over a very short distance, and even this will only give us the average velocity during this short time. But if we make a number of measurements, using shorter and shorter periods of time, the average velocity becomes more and more nearly the velocity which we want. Thus, at 10 o'clock, a man in a railway train making a careful measurement finds that the train passes over 200 feet in the next 4 seconds. He finds the average speed for 4 seconds after 10 o'clock to be 200 -=- 4, or 50 feet per second. Another man finds that it passes over 100-4 feet in the two seconds after 10 o'clock, and finds during these two seconds an average velocity of 1004 -f- 2, or 50'2 feet per second. Another man finds 50-25 feet passed over in APPLIED MECHANICS. 13 one second after 10 o'clock, which shows a velocity of 50-25 feet per second. Another man finds 25-132 feet passed over in half a second after 10 o'clock, and finds 25-132 -4- 0-5, or 50-264 feet per second. Another man finds 12-567 feet in a quarter second after 10 o'clock, and his observation gives 50-268 feet per second, and so on. It is evident that the values given by these various observations are approaching the real value of the velocity at 10 o'clock. Tabulating these results, we have : Intervals of Time in Seconds after 10 o'clock. Average Velocity in Feet per Second. 4 50-00 2 50-20 1 50-25 i 50-264 * 50-268 Plot the two sets of numbers on squared paper, and draw a curve through the points so found. Produce the curve, and we have the means of finding the average velocity for an infinitely small interval of time after 10 o'clock. This is ths required velocity. 8. Acceleration. This is the time rate of change of th& velocity of a body. Thus it is known that the velocity of o body falling freely in London At the end of one second is 32-2 feet per second. two seconds is 64*4 three 96-6 . four ^128-8 and we see that there is an increase to the velocity of 32-2 every second. The acceleration in this case is always of the same amount hence we call it uniform acceleration, and say it is 32-2 feet per second per second. EXERCISES. 1. One mile per hour; also one knot; convert each, of these into feet per minute and feet per second. Ans., 88, 1-467 ; 101-3, 1-689. 2. A torpedo-catcher travels at 32 knots ; convert this into English miles per hour. Ans., 36-85. 3. Prove that 60 miles per lu,ur means -0268 kilometres * per second. * See tables on page 654. 14 APPLIED MECHANICS, 4. An eccentric disc is 10 inches diameter; the shaft makes 300 revolutions per minute. What is the rubbing velocity on the straps in feet per second ? Ans., 13'09. 5. In running a race of 1 mile long, A beats B by 100 yards, and B beats C by 90 yards. By how many does A beat C ? Ans., 185 yards, nearly. 6. Ten miles per hour; state this in feet per second and in centimetres per second. Ans., 14 ; 447. 7. An acceleration of 32-2 feet per second per second ; state this in miles per hour per second ; state it in centimetres per second per second. Ans., 21-96; 981-4. 8. 200 gallons of water per minute ; how many pounds per second ? How many cubic feet per second ? Ans., 33-3 ; -535. 9. A round pipe 6 inches diameter has 30 gallons per second flowing through it. What is the velocity? If the diameter becomes 10 inches, what is the velocity ? Calculate in the two cases the kinetic energy of one pound of water, this being the square of the velocity divided by 64 '4. Ans., 24'5 feet per second ; 8'8 feet per second ; 9'3 foot-pounds ; 1*2 foot-pound. 9. Example. Two fine wires are 10 feet apart; a bullet breaks them both. The breaking of each wire causes an electric spark to make a mark underneath a fixed platinum pointer on a revolving drum. If the drum is 4 feet in diameter, and revolves at 300 revolutions per minute, and when the drum is at rest the spark-marks are found to be 1-32 feet asunder on the curved surface, assuming that the intervals of time between the breaking of the wires and making the marks were the same, find the time between the breaking of the wires, and find the velocity of the bullet. The surface velocity is ^ , or 62-83 feet per second; 1-32 divided by this gives '02101 seconds ; dividing this into 10 feet gives 476 feet per second as the velocity of the bullet. Exercise. In some gun experiments screens 150 feet apart were cut by a bullet at the following times (in seconds), counting from the time of cutting the first screen : 0, 0-0666, 0-1343, 0-2031, 0-2729, 0-3439, 0-4159. Find the average velocity between every two successive screens. Ans., 2,252, 2,216, 2,180, 2,149, 2,113, and 2,083 feet per second. 10. Speed or velocity is a rate the rate of increase of space with regard to time. If a body has passed through the space * at the time t, and if it goes over the additional space Ss in the additional time of 8*, then -^ is the average velocity. Observe that $s is one symbol; APPLIED MECHANICS. 15 it does not mean a quantity called 8 multiplied by a quantity called s. When we imagine 82 to get smaller and smaller without limit, the average velocity becomes what we call the real velocity, and we indicate it by the symbol -j- . When, instead of space and time, we have other quantities, we generally use y for the dependent and x for the independent variable, and the symbol - means " the rate of increase of y with regard to x." Thus, if y is the ordinate of a curve and x is the abscissa at the point p (Fig. 4), and o, K =: 5#, P n = 8#, then 8y/8# is the tangent of the angle Q v x. As 8# and 8y become smaller and smaller, so that $y/8x is to be written dyjdx, then dy\dx is evidently tangent of the angle which the tangent at p makes with the axis of x. We usually call this the slope of the curve at the point P. If y is any quantity that depends- upon another, x t and if we plot the values of x and y to scale on a sheet of squared ' paper, the curve shows by its slope everywhere the rate of increase of y with regard to x. If we know the alge- braic law connecting y and x, we can find this rate by certain easy rules. Thus, if y = Ao;n + B (1), - 1 (2), whatever kind of number n tfig- 4. may be. 11. In teaching beginners it is well to start on the assumption that students already possess the notions of the differential and integral calculus, and it is a teacher's duty to put before them the symbols used in the calculus at once. It is surely much better to do this than to evade the calculus in the fifty usual methods which we sometimes see adopted. Unfortunately many readers of this book are likely to be preparing for examinations in which only academic methods of elementary study are recognised, so we must keep our calculus symbols for the smaller print paragraphs. We may say, however, that we think even beginners in this subject will be able to understand the author's book on the calculus ; and if so, they will find the study of applied mechanics very greatly simplified. The language of the calculus is the natural, easy, simple language of the engineer ; but it is in writing, whereas most engineers only speak of rates and integrals. If y is known as a function of x, we can find , which we ax may call u. Conversely, if u is known as a function of x, we can 16 APPLIED MECHANICS. find y. The two symbols are J, called a rate, or " the differential ax co-efficient of y with regard to x " ; and I u . dx, called "the in- tegral of u with regard to ar." J 12. The following formulae will suffice for nearly all engineering calculations : y dy dx H I-" c Car 1 T5 A. x^ A log. x A a; Aa;- 1 A log, x . ^ Ke ax I A,- a A log. (x 4 a] A A A log. (x 4 ) x 4 a; 4 a A sin. (ax + V} A cos. (ax 4 4) A sin. (ax.-\- b} cos. (&x 4 <>) a A cos. (# 4- b) a A sin. (ax 4 ^) A cos. (ax 4 i) ^. sin. (ax 4 fy and all the important part of the author's book on the calculus is devoted to illustrating the use of them. 13. For example, if x is the abscissa o B, and y is the ordinate t p B of the curve shown in Fig.- 5 ; if the area between the curve and o x and p B, and any other ordinate nearer o, say D s, be called A ; and if the area to a T be called A + S A, o T being called x + 8#, the area SA = p Q T B, and as 5x is made smaller and smaller this area gets more and more nearly s Q T is y + 5y. I I V B T -H^-IW Fig. 6. Thus APPLIED MECHANICS. 17 . and as S.r gets smaller and smaller this in the limit gets to be dx ~ Hence, if we know y in terms of x, we can find A by integration. If we know A, we can find y by differentiation. Thus, if we have the parabolic curve, y = ax* (3) A = 1 aaM- c (4), a constant is added, because the rate of change of a constant is 0, and we wish our answer to be as general as possible. The value of c is fixed as soon as we settle from which ordinate A is to be calculated. Thus, if A is when # is 0, c is 0. If r (x) is the general integral of u, r j A dx tells us to use x 2 for x in the general integral, then to use x^ for x and subtract, or u . dx = F (a? 2 ) - F (a?!) .... (5). If u is the ordinate of a curve, (5) means the area between the curve, the axis of x, and the two ordinates at x\ and x 2 . Any summation which may be indicated by an area can be effected in this way by integration. It is very easy to recollect the rule that the rate of change of x 1 the integral of z m is x m + l . is- nx n ~ *, and that This rule suffices for most engineering problems. There is one case in which the rule for integrating x m is useless to us namely, when m is 1 : in this particular case -l 14. Integrations. In a great number of problems we have the ordinates like AP, called the y, of every point of a curve, given us, the abscissa o A of the point being called x. If we are given a list of corresponding values of y and x, we can draw the curve. Now, if A Q represents to scale the area of A P R o, and if we find many points like Q Fig. 6. 18 APPLIED MECHANICS. and join them, we get a new curve, which is said to represent the integral or area of the first. The areas may he found by means of a planimeter. I always use the following method myself. It is inaccurate to this extent, that the curve R P is taken to be a many- sided polygon instead of a curve. But it leads to a very quick solution of complicated-looking problems. Thus the following values of y are given for the corresponding values of x. ~x y JV* 0-5 1-869 1 1-869 1-5 1-756 2 3-625 2-5 1-657 3 5-282 3-5 1-571 4 6-853 4-5 1-497 5 8-350 55 1-432 6 9-782 6-5 1-375 7 11-157 7-5 1-326 8 12-483 8-5 1-247 9 13-730 It will be seen that we assume the ordinate at any point like x =3 5 to be the average height of the curve from x = 3 to x =4. Notice that to get, say 6 853, we add 1-571 times 5x, which is 1, to 5-282. We always take the values of y at equal intervals in the values of x. If the above values of x had been 0, 0-05, O'lO, 0-15, etc., beginning with the first, we should have had to divide all the numbers in the last column by 10. Exercise. i^or values of x = Q'5, 1'5, 2-5, ..., calculate various values of y fromy = 2 + 3 aj+0'5 x 2 , and find the integrals. The true value of the integrals is A = 2 x -f 1 '5 x 1 + % x*. Calculate this for x=0, 1, 2, ..., and note the errors in our numerical method. 15. The symbol ~ 2 means the rate of change of -j- with regard to x ; -j means the rate of change of -^ w ^ regard to x. It will be found that much of the difficulty which some APPLIED MECHANICS. 19 students find in using the calculus is due to their not being able to differentiate t n with regard to t, although they know how to differentiate x n with regard to x the mere use of another letter than the one to which they have been accustomed causing the difficulty. It is well, therefore, from the beginning to get used to many other letters, such as t, v, s, p, w, etc. If s is space and t time, ~ is velocity v, and ^ = ^ is acceleration. Newton's symbols were: * for space (length), s for velocity, s for acceleration. Example. If acceleration ^ = a, a constant ; integrate with regard to t, and we have ^r^= velocity at + b . . . . (6), where I is the constant which we always add to make the answer general. We note the meaning of b to be the velocity when t = ; and perhaps we had better use Vo instead of b, so that -r = at + Vo Integrate again, and we have * \at* i + v t -f- c . . . . (7). Evi- dently c means the value of s when t = Q. (6) and (7) are well- known laws of uniformly accelerated motion. 16. Example. A chain of length x, hanging vertically down- wards, is being lowered from a capstan. Its weight per foot of its length is w. When the weight wx is lowered through the distance Sx, the work done is wx 8x, and the whole work done by it from the time it is of no length, till it is of length I, is I wx dx, or | wo? , or, or its total weight multiplied by half its vertical length. Suppose the chain to get thicker as more of it is let down ; say that w = a + b x, then the above summation becomes I (a + bx] x dx = I (ax + bx*) dx = f ax* + \bx* = |fl/ 2 + $ bl 3 . These two answers represent also the work done in lifting the chain. 17. Example. Fluid expanding through the volume dv at the pressure p does work p. $v, and f Jv p . tiv is the work done in expanding from volume v^ to volume % Thus, suppose that expanding fluid follows the law p 20 APPLIED MECHANICS. where k and c are constants, we have to integrate cv " * witt regard to v, and according to our rule the answer is Putting in the limits, our work is and this may he put in other shapes. When the law w pv=:c t the work is c log e - 2 . v i 18. The Compound Interest Law. If we are told that the rate of increase or diminution of y with regard to x is proportional to y } say that _? = ay, then we know that y = A eo#, where A is any constant. 19. The Harmonic Law. If y = A sin. (ax + b), we find that - = A a cos. (0# -f b). It will be found that this includes the case : If y = A cos. (ax -f b), d -l = - A a sin. (ax + J). & Thus if y = -A sin. (a* + i), ^ = A a cos. (a* + >), ax g = - A a 2 sin. (s + J) = - V- With this knowledge a student can put in a few words all the theory of Chapter XXV. He will see that simple harmonic motion is stated hy x == a sin. (qt -f e) ____ (1), where a is the amplitude, q is 2 TT/ or _ , where / is the frequency or r is the periodic time, and e is the lead, a quantity introduced "because x is not when t = ; that is, we give generality to (1) hy assuming that we may begin to count time from any position of the body. Observe that the motion may be angular, x and a being angles. ^ wiU then' be angular velocity, and ^-? will be angular at at* acceleration. Differentiating twice we get velocity = = aq cos. (qt + (2), at acceleration = ^L - of sin. (qt + *) ---- (3). dt 2 Notice the sign of C-jL and see how it agrees with the statements of Chapter XXV. If we only think of the numerical value of ^ APPLIED MECHANICS. 21 we notice at once that #*-== - .... (4), and the square root of this is T/2 IT, which is the rule found in another way in Chapter XXV. 20. Students will do well to graph on squared paper some curves like the following : 1. If y = 0-1 x 2 + 3, take x = 0, x 1, x = 2, etc., and in each case calculate y. Plot the values of x and y as co-ordinates of points on squared paper, and draw the curve passing through these points. 2. Graph y = a + bx. 1st, when a = 0, b 1. 2nd, when a = 1, b = 1. 3rd, when a zz: 1, b = 1*5. 4th, when a = 1, b T5. 5th, when = 1, b = 1. 3. Graph y = -1 **. 4. Graph y = 10 sin. ( =- x + jV \ 3/ 5. Graph y = 120 x ~ l . 6. Graph y = 120 # ~ 1>3 . 7. Graph y = 120 x " 7 . 8. Graph y = 10 a; *. 9. Graph y = Hh -v/25 a: 2 , and also y 4^ 1-3 */ 25 a?. 21. The following observed numbers are known to follow a law like y ~ a + bx; but there are errors of observation. Find by the use oi squared paper the most probable values of a and b. X 2 3 4i 6 7 9 12 13 y 5-6 6-85 9-27 11-65 12-75 16-32 20-25 22-33 Ans. t y = 2'5 -f 1-5 .r. 22. The following observed numbers are known to follow a law like y:=ax/ (I +'*): Find by plotting the values of y/ and y on squared paper that these follow in a law y\x +sy = a, and so find the most probable values of a and *. X 5 1 2 3 1-4 2-5 V 78 97 1-22 55 1-1 1-24 Ans., y = Zx I (1 + 2*}. 22 APPLIED MECHANICS. 23. Vertical Line. A line showing the direction in which that force which we call the resultant force of gravity acts. It is a line at right angles to the surface of still water or mercury. 24. Level Surface. A surface like that of a still lake, everywhere at the same level, and everywhere at right angles to the force of gravity or other volumetric force which is acting upon matter. It is not a plane surface. 25. Curvature. For any curve we can find at any place what circle will best coincide with the curve just there. The radius of this circle is called the radius of curvature at the place. But since we say, for instance, that a railway line curves much, when we mean that the radius is small, the name curvature is always given to the reciprocal of the radius. Thus, if the radius is 8 feet, we say that the curvature is J. If at another place the curvature is . t the change of curva- ture in going from the one place to the other is the difference between these two fractions. Curvature may also be defined as the angle turned through by a tangent to the curve per unit of its length. A student ought to see for himself that the two definitions are the same. If in Fig. 5 the distance along the curve between p and o. is called 8s, and if 50 is the angle which the tangent at a makes with the tangent at p, then the average curvature between p and a is 50/Ss. As p and Q become closer and closer, without limit, the curvature is -=-. as If a curve is defined by its x and y co-ordinates, the curvature T \dx, Example. Find the curvature,^ where x = 0, of the parabola y = ax" 1 . Here -% = 2 #a? and - , = 2 a, so t'hat the curvature anywhere is 2 -r J 1 + 4 % 2 j *> and at tne vertex where a; = the curvature is 2 a. When, as in 'ordinary beams, -~- is small, it is evident that the curvature may be taken to be -y-f . This is what we take to be true in the discussion of beams and struts. Example. In making 100 steps round a curve, my compass, showing the direction of motion, changes from N. to N.E. What is the average curvature? Answer from N.to N.E. is 45 degrees, or 0'7854 radians, and this divided by 100 steps, or "007854 radians per step, is the average curvature. The APPLIED MECHANICS. 23 reciprocal of this, or 127 '3 steps, is the radius of curvature, if the curvature is constant that is, if the curve is an arc of a circle. Many unpractical rules will be found in books, requiring us to draw a tangent to a given curve at a given point, or to find its curvature there by trial. .These are only academic suggestions. If half a dozen students get tracings of the same curve, and two points to measure the angle between the tangents there, they will obtain six very different answers. EXERCISES. 1. Through what angle must a rail 10 feet long be bent to fit a curve of half a mile radius? Ans., 0'22 degrees. 2. Arc s is 10 feet long, radius r half a mile ; find the versed sine x that is, the greatest deflection from straightness. Prove that, practically, 3 8 rx ; so that in this case x = -0047 feet. 3. Find the radius of curvature of the catenary y=. a the vertex. at 4. Find the radius of curvature of the ellipse -- -f- TO = 1. (1) At the end of the major axis ; (2) at the end of the minor axis. J2 ffi Oi 26. Angle. An angle can be drawn : First, if we know its magnitude in degrees ; a right angle has 90 degrees. Second, if we know its magnitude in radians J a right angle contains 1-5708 radians. Two right angles contain 3'1416 radians. One radian is equal to 57'2958 degrees. One radian has an arc, B c, equal in length to the radius A B or A c. It some- times gets the clumsy name " a unit of circular measure." Third, we can draw an angle if we know either its sine cosine, or tangent, etc. Draw any angle, BAC (Fig. 7). Take any point, P in AB, and draw PQ at right angles to ^ A c. Then measure P Q, A P and A Q Fig. 7. in inches and decimals of an inch. p Q ^ A P is called the sine of the angle. A Q ~ A P is called the cosine of the angle. P Q -f- A Q is called the tangent of the angle. Calculate each of these for any angle we may draw, and measure with a protractor the number of degrees in 24 APPLIED MECHANICS. the angle. We shall find from a book of mathematical tables whether our three answers are exactly the sine, cosine, and tangent of the angle. This exercise will impress on our memory the meaning of the three terms. It will also impress upon us the fact that if we know the angle in degrees, we can find, by nreans of a book of Cables, its sine, or cosine, or tangent ; and if we know any one of the sides A P, or P Q, or A Q, of the right-angled triangle A P Q and the angle A, we can find the other sides. .Divide the number of degrees in an angle by 57*2958, and we find the number of radians. Suppose we know the number of radians in the angle BAG, and we know the radius A B or A C, then the arc B c is A B x number of radians in the angle. Given, then, a radius to find the arc, or given an arc to find the radius, are very easy problems. A student becomes accustomed, on seeing an angle drawn on paper, to judge from a mere glance how. many degrees the an^le contains. It would be an advantage to acquire the habit of judging how many radians there are in the angle. What we mean is, that he ought to be as ready to think in radians as in degrees, and to do this he requires to be familiar with the size of a radian. EXERCISES. 1. Draw an angle of 35. Find by measurement the sine, cosine, tangent, cotangent, secant, and cosecant of the angle, and compare with the numbers in a "book of tables. Calculate the number of radians. Try if sin. 2 35 + cos. 2 35 = 1 ; if sin. 35 -i- cos. 35 =r tan. 35 ; if tan. 2 35 -f 1 = sec. 2 35 ; and if cot. 2 35 + 1 = cosec. 2 35. 2. If a = 55, = 20, illustrate the following important formula) by numerical calculation : Sin. (o -f j8) = sin. a cos. -f cos. a sin. /3. Sin. (a /3) sin. a cos. j8 cos. a sin. ;3. Cos. (a 4- j8) cos. a cos. /3 sin. a sin. . Cos (a &} = cos. a cos. + sin. a sin. j3. Sin. a cos. = ^ { sin. (a + j8) + sin. (a 0) Cos. a cos. j8 = G O S - (a + ]8) + cos. (a /3) Sin. a sin. 6 = ^ { cos. (a 0) cos. (o + j8) } . Cos. 2a = 2 cos. 2 o 1 = 12 sin. -a. APPLIED MECHANICS. 25 3. What are sin. 150, cos. 130, tan. 170, cos. 240, sin. 220, tan. 218, sin. 290, cos. 310, tan. 320? Express aU these angles in radians. Ann., -5, - -6428, - -1763, - -o, - -6428, -7813, - -9397, '6428, - '8391; 2-6180, 2-2089, 2-9671, 4'1888, 3-8397, 3-8048, 5-0614, 5-4105, 5-5851. 4. The sine of an angle is 0'25; find its cosine, tangont, cotangent, secant, and cosecant. Find the angle by actual drawing. How many radians? Ana., '9683, -2582, 3-875, 1-033, 4; 145 ; -2528. 5. What are the sine, tangent, and radians of 1^ degrees ? Ans., Each '0262. 6. If in Fig. 7 A is 47, and A p 5-23 feet, find A a and p ct. Ans., A Q = 3-567, P Q = 3'824. 27. Angular Velocity. If a wheel makes 90 turns per minute, this means that it makes 1'5 turns per second. But in making one turn any radial line moves through the angle ot 360 degrees, which is 6*2832 radians; so that 1-5 turns per second means 6*2832 x 1'5, or 9'4248 radians per second. This is the common scientific way in which the angular velocity of a wheel is measured so many radians per second. If a wheel makes 30 turns per minute, its angular velocity is 3-1416 radians per second; n turns per minute mean 2irn radians per minute, or 2vn -f- 60 radians per second. One turn is the angular space traversed in one revolution. Exercise. Show that the linear speed in feet per second of a point in a wheel is equal to the angular velocity of the wheel multiplied by the distance in feet of the point from the axis. 28. Angular Acceleration. The increase of angular velocity per second. If a wheel starts from rest, and has an angular velocity of 1 radian per second at the end of the first second, its average angular acceleration during this time is 1 radian per second per second. EXEECISES. 1. A shaft revolves at 800 revolutions per minute. What is its angular A^elocity in radians per second ? Ans., 83'79. 2. A point is 3,000 miles from the earth's axis, and revolves once in 23 hours 56 minutes 4 seconds. What is its velocity in miles per hour ? Ans., 787-5. 3. The average radius of the rim of a fly-wheel is 10 feet. When the wheel makes 150 revolutions per minute what is the average velocity of the ^im ? Ans., 157'1 feet per second. 4. An acceleration of 1 turn per minute every second ; how much is this in radians per second per second ? Ans., -1047. 5. A wheel is revolving at the rate of 90 turns a minute. What is Us angular velocity ? 26 APPLIED MECHANICS. A point on the wheel is 6 feet from the axis ; what is its linear specil ? If its distance from the centre be increased by 50 per cent., what does its speed become ? If at the same time the speed of the wheel increases 50 per cent., what is now the linear speed of the point ? Ans., 9-425 radians per second ; 56-55 feet per second; 84 '82 feet per second ; 127 '2 feet per second. 6. There is a lever, A o, 30 inches long, works about an axis at o. The lever is made to turn by applying a force at a point B in A o, 15 inches from o, so that B receives a velocity of 2 feet per second. What is the angular velocity of the lever ? If the same velocity had been given to the point A instead of B, what would the angular velocity have been ? Ans., 1-6 radians per second ; 0-8 radian per second. 29. Length of Belt. Let D and d be the diameters of pulleys, c the distance between their centres, L the length of belt, let D + d .be called s. Prove that for a crossed belt s + 2* cos. 0....(1), where sin. 6 =. s -f- 20. Example. Find the length of a crossed belt for pulleys of 20 and 15 inches diameter, the distance of their centres apart being 120 inches. First find 0, if sin. = 35 -f- 240 = -1458. So that 9 = 8 23', or -1463 radians. Also cos. = -9893. Hence we have L (1-5708 + -1463) 35 + 240 x '9893 = 297'5 inches. Notice from the formula (1) that as L depends only on c and s, if c and s are the same, the same length of belt will do. Thus the same crossed belt does for any two corresponding steps in stepped cones if D + d is the same. It is exceedingly easy to calculate the sizes of these steps when we know the speeds. Thus the above- mentioned pulleys were 20 and 15 inches, and their speeds were ^ as 3 to 4. If there were two steps, and we want another pair to give a speed ratio, say, of 1 to 2 using the same belt, we know that if their diameters are D and d T> + d = 35, T> = 2d. Hence, 3d=35 and *=llf, D = 23. Exercise. The centres of two pulleys, 3 and 1 feet in diameter respectively, are 10 feet apart. Find tbe length of crossed belt required. Ans., 28-48 feet. If the belt is an open belt, let the student prove that the length L is L =| (D + d] + e (D - d] + 2c cos. 0, where sin. = (D rf) + 20. It is easy to show that the following answer,, which is much easier to deal with later, ia practically correct : 1 = * (D + rf) + 2o + l(^... .(2). The length of belt depends upon D -f d, c, and D d. APPLIED MECHANICS. 27 Exercise. Pulleys of 20 and 15 inches diameter, whose axea are 120 inches apart, are connected by an open belt. Find its length. = 295-03". The student ought to find the correct answer, and see how small is the error in the use of our approximate formula. Suppose we have a pair of pulleys DJ and d v and we want another pair D 2 and rf 2 on the same shafts to work with the same length of belt, the ratio of D 2 to d 2 being known. Putting the two sxpressions for L equal, we have * + ^ + 2V, (D * - d ^ = D > + ^ + 27, (Dl - d ^ "" (3) * The right-hand side is known, and we know D 2 in terms of rf 2 , so that it is easy to calculate. Example. In the above example of D X = 20, d^ = 15, c = 120, let us calculate D 2 and d 2 , if D 2 = 2 d 2 . Our equation (3) becomes 36 + -^4i2o w = 3 * + anriro w *- The solution of this quadratic gives us d z 33 ; and,_ therefore, D 2 = 66. In practice we usually calculate D 2 and d% as if the belt were crossed. These are nearly the right answers. Then, taking the D 2 d 2 so found, we find from (3) a corrected value of D 2 + ^2> an d we use ^his, knowing the ratio of D 2 to 2 = | d 2 . Hence \d^ + <, = 30, 2^ d^ = 30, d 2 = 12 inches, D 2 = 18 inches. In the same way we find d 3 = 15, D 3 = 15 ; d^ = 17}, D 4 = 12f ; rf 5 = 20, -D 5 = 10. 30. If a line, AB, makes an angle with the horizontal, the projection of its length on the horizontal, is A B cos. 0. Its projection on a vertical line is A B sin. 0. Exercise. Draw two lines ox, o Y at right angles to each other. Now draw lines o P, o Q, o R, o s, of lengths 3, 5, 2|, and 4 inches, and making angles of 35, 72, 130, and 220 with ox." Find the projection of each line on o x and on o Y, and the sum of the projections on o x and on o Y. An., 2-457; 1-545; -1-607; -3-064 ; 1721 ; 4-755 ; 1-915; -2-571; -0-669; 5-820. If a plane area of A square inches is inclined at an angle with the horizontal, its area, as projected on the horizontal, is A cos. square inches. 28 APPLIED MECHANICS. Try to prove that this must be so by dividing the area into strips by horizontal lines. EXERCISES 1. A plane area of 35 square feet is inclined at 20 to the hori- zontal : find its horizontal and vertical projections. Am., 32*89 square feet; 11-97 square feet. 2. The cross-section (a cross-section always means a section by a plane at right angles to the axis or line of centres of sections) of a cylinder is a circle of 0'7 inch radius. Find the areas of sections which make angles of 25 and 45 with the cross-section. Note that the cross- section is a projection of any other section. Ans., 1-699, 2-177 square inches. 3. The above cylinder is a tie bar of wrought iron. The total tensile load is 12,000 Ib. ; how much is this per square inch of the cross-section? How much is it per square inch of either of the other sections ? ^ Ans., 7794 Ibs., 7063 Ibs., 5512 Ibs. 4. The cross-section of a pipe is a circle of 15 inches diameter; what is the area in square feet? If 13 gallons flow per second, what is the velocity Vo ? What is the area of a section at 28 to the cross- section? What is the velocity v, normal to this section, if normal velocity x area = cubic feet per second ? Show that v is the resolved part of VQ normal to the section. Ans., 1-228, 17 feet per second ; 1-39, 1-5 feet per second. 5. Part of a roof, shown in plan as 4,000 square feet, is inclined at 24 to the horizontal ; what is its area ? Ans., 4378*7 square feet. 6. A tie bar or short strut of 2 square inches cross- section ; what is the area of a section making 45 with the cross-section? If the total tensile or compressive load is 20,000 Ibs., how much is this per square inch on each of the sections? Resolve the total load normal to and tangential to the oblique section, and find how much it is per square inch ea;h way. Ans., 2-828 square inches; 10,000 Ibs., 7,070 Ibs., 5,000 Ibs. 29 CHAPTER II. * VECTORS. RELATIVE MOTION. 31. ANY quantity which is directive is called a vector quan- tity for example, a velocity or a force. It can be represented by a line. Its amount can be represented to some scale by the length of the line. The clinure of the line and an arrow-head represent the clinure and sense of the vector. Vector quanti- ties are distinguished from scalar quantities, such as a sum of money, the mass of a body, energy, temperature, etc. The resolved part of a vector in any new direction is represented by the projection of ^ its representative length in the new IL direction. Thus in Fig. 8, if OP represents to scale a velocity or a ^ force, its resolved part in the direc- tion o x is o A, the amount of which ^ is o P cos. A o P, and its resolved part / )- jr~ X in the direction o Y is o B, the Fio . g amount of which is o P cos. BOP. Thus, if a ship is going at 9 knots north-eastward, the northerly component of its velocity is 9 cos. 45, or 6 -3 63 knots, and its easterly component is 6-363 knots. If a body has an acceleration of 20 feet per second per second in the direction 25 east of north, the northerly com- ponent of this is 20 cos. 25, or 18*13 feet per second per second ; and the easterly component is 20 sin. 25, or 8*452 feet per second per second. If a force of 30 Ibs. is in a northerly direction, its com- ponent in a north-easterly direction is 30 cos. 45, or 21*21 Ibs. 32. The resultant of two or more iorces is a force which might be substituted for them without changing the effect. If two strings pull a small body with forces of 5 Ibs. and 7 Ibs. (Fig. 9), and if the angle between them is 30, draw o P equal in length to 5 inches, and make the angle QOP equal to 30. Make the length of o Q 7 inches. Com- Fig. 9. plete the parallelogram Q o P R, and draw the diagonal o R. Measure o R in inches ; we find it to be 11 *6 inches, so that the resultant of the two forces 30 APPLIED MECHANICS. is 11 '6 Ibs. One string acting in the direction OR with a pull of 11 '6 Ibs. will produce the same effect at o as the two strings did. In using this construction, take care that the arrow-heads are confluent that is, that they all point away from o, or they all point towards o. Suppose when the two strings were acting we had found by experiment that a third p string o E (Fig. 10) would just prevent the two strings from causing motion at o, then o experiment would also show that the force Fig. 10. in o E, which may be called the equilibrant of o p and o Q, is exactly equal and opposite to the resultant of o P and o Q. We see that when OQ and OP are given, and the angle between them, we may use the above principle, called the parallelogram of forces ; or, what comes to the same thing, the triangle of forces. Draw o Q one of the forces, draw Q R the other, and let their arrow-heads be circuital ; then the non- circuital force OR is the resultant; or a circuital force RO would be the equilibrant. It is in this way that we find the resultant or vector sum of any two vectors. In vector language, o Q + QR = o R, or OQ = OR-QR. This principle is very easy to express ; to be able to apply it implies a considerable experience. We mention it now merely to introduce a few exercises. It is very easy to solve problems graphically. The student must work some numerical exercises, and test his answers graphically. Example. Two forces, o P and o Q, of 5 Ibs. and 7 Ibs. respectively, act a point, at an angle of 56 ; find their resultant by calculation. In Fig. 11 OP and o Q represent, to scale, the two forces, the angle P o Q being 56. Find the resolved part of o P (Art. 31) in the direction o x, say ; it is o A = o P cos. 56 = 5 x '5592 = 2-796. Now obtain the resolved part of o P in the direction o Y, taken at right angles to o x ; it is o B = o P . cos. P o B = 5 sin. 56 = 5 x -829 = 4-145. In- stead of the given forces we now have o A and OQ acting along ox, and OB along o Y. Draw OH (Fig. 11) equal to o A + OQ, that is, 9796, and ov = 4-145. The resultant of these is o R, and we have OR 2 = on 2 + ov 2 = (9-796) 2 + (4-145) 2 = 24-996; . . o R=5 very nearly APPLIED MECHANICS, '6\ To obtain the direction of o R, we have 4-145 , EH tan. a = = = -4231 OH 9-796 . . a = 23 nearly. The student should test this result by finding o R graphi- cally by the method explained above. He should also observe that if the angle P o Q had been, say, 130, OA would have been directly opposed to Q, in which case o H would have been obtained by subtracting o A from o Q. It is of the utmost importance that he should work many exercises similar to this one, so as to become familiar with the method. It will be shown later (Art. 94) that the same method is employed for calculating the resultant of any number of forces. In every case we resolve the forces along any line o x, and let o H represent their resultant ; then resolve along o Y taken at right angles to o x, and let ov represent the resultant of these resolved parts, o R and a are then easily calculated. EXERCISES. 1. Force of 20 Ibs. at an angle of 72 with the horizontal; what are its horizontal and vertical components? Am., 6 - 18 Ibs., 19-02 Ibs. 2. A man walks towards the north-north-east at 4 miles per hour ; at what rate is he getting towards the east ? And at what rate towards the north? Am., T53 miles an hour; 3-69 miles an hour. 3. Forces o p of 10 Ibs. and o Q of 7 Ibs. at an angle o, o p of 35 ; find the resultant. Test your answer by working the problem graph- ically. Ans., 16-24 Ibs. inclined 14 18' with the force 10 Ibs. 4. On a horizontal surface there is a normal pressure of 4 tons per square inch and a tangential force of 3 tons per square inch from N.E. to S.W. What is the total force per square inch ? Am., An oblique pressure of 5 tons per square inch, making an angle tan ~ 1 1 with the N.E. to S.W. direction in a vertical plane. 32 APPLIED MECHANICS. 5. An anvil is carried by three ropes, which make angles 20, 30, 25 with the vertical ; the tensions in the ropes are known to be 1,000 Ibs., 700 Ibs., and 1,200 Ibs. What is the weight of the anvil ? If the hori- zontal components of these three forces are drawn as balancing one another, find the azimuthal angles which the vertical planes through the ropes make with each other that is, find the angles in the plan of the ropes. Ans., 2,633 Ibs. ; 85 47', 137 43', 136 30'. 33. If the magnitude of o p (Fig. 1 1 ) is called p, of o Q, is Q, of o R is R, and if the angle o. o p is 6, it can be proved, that R = \/p 2 4- a' 2 + 2 P Q cos. Q. It will be seen from Fig. 1 1 that K H OB P sin. 9 tan. o = = = . OH OA + OQ P. COS. 6 + Q If 6 is a right angle, P and Q are the resolved parts of R in their two directions R = V p 2 + a 2 , and tan. R o Q == P/Q. EXERCISES. 1. P = 3, Q = 4, = 90. Then R = 5 and a is an angle whoso tangent is 0'75. [The neat way of making this statement, is to write a = tan.- 1 075.] 2. P = 3-045, a = 7-462, 6 = 37 ; find R, a. An*., R = 10-06. o= 1030'. 3. p = 3-045, a = 7-462, = 143; find R, a. Ans., R= 5-353. a = 20. 4. p = 12-06, a = 1-002, 6 = 184 ; find R, o. Ans., R = 13-05. a = 187 42'. 34. A river flows at 1 mile per hour ; a swimmer has a velocity of 2 miles per hour relatively to the water. What is his velocity relatively to the bank 1 Ihis will depend upon the direction in which he swims. Make AB represent 1 mile per hour down the river (as some students cannot get out of their heads the wrong notion that these lines represent distance, imagine the drawing to be infinitely ^small but greatly magnified merely that it may be >c examined) ; make B c represent the velocity of swimming to scale, the direction and sense being correct. Take care that the Fig. 12. arrow-heads are circuital. Then A c is the sum of the two velocities possessed by the swimmer, and is therefore his velocity relatively to the bank. Let the student draw B c in all sorts of directions and reflect upon his answers. If he has ever swam in a broad river or has watched a swimming dog trying to reach his master, he will understand his answers more readily. APPLIED MECHANICS. 33 EXERCISES. 1 . Given the above velocities and that the stream flows due south, if the absolute motion of the swimmer is to be south-east, in what direction ought he to swim ? Am., 24 18' S. of E. 2. A steamer moves westward at 10 feet per second ; a boy throws a ball across the deck northwards at 4 feet per second. What is the velocity of the ball relatively to the water ? Ans., 10-77 feet per second, 21 48' N. of W. 3. A steamer has a velocity of 14 knots due west ; the wind blows with a velocity of 7 knots from the north. What will be the apparent velocity of the wind to a person on board the steamer ? Am., 15-7 knots from W. 26 34' N. 4. Velocity of a ship westwards 10 feet per second ; velocity of a ball on deck 5 feet per second north-east relatively to the ship. What is the total velocity of the ball ? Ans., 1 "37 feet per second, 28 40' N. of W. 5. If the total velocity of the ball is 12 feet per second northwards, what is its velocity relatively to the ship ? Ans., 15-62 feet per second, 50 12' N. of E. 6. A railway train is going at 30 feet per second ; how must a man throw a stone from the window so that it shall leave the train laterally at 1 foot per second, but have no velocity in the direction of the train's motion ? Ans., 30-02 feet per second ; at an angle of 178 6' with direction of motion of train. 35. A bicyclist is ordered to travel so that he shall be more to the north at the rate of 3 miles every hour, and he must keep to roads. Notice that if he is on a north and south road his task is very easy. If his road is directed N.W., he must travel at 4-242 miles per hour. If his road is W.N.W., he must travel at 7*839 miles per hour. If his road is due west, his task is an impossible one. If his road makes an angle 6 with due north, he must travel at the rate of 3 -=- cos. miles per hour. 36. Water flowing from the inner to the outer part of a motionless wheel of a centrifugal pump is guided by vanes to follow a curved path. Suppose its radial velocity v r known ; show that its real velocity anywhere is v r -f- cos. 6, if 6 is the angle which the vane there makes with the radial direction. A student will find this an excellent graphical exercise. Draw circles of 1 and 2 feet radii to represent the inner and outer cylindric surfaces of the wheel of a pump. Draw any shape of vane connect- ing these, but you had better take a shape from an actual wheel (see Art. 427). Let the angle at A with the tangent / to the circle there (Fig. 13) be 18. Now imagine a particle of water travell- "x ing out radially at O'l foot per second. p . J3 Imagine it to take 10 seconds to get to o,. Mark its successive positions along the vane. You had better also trace its positions backward for a few seconds towards A from s. Now imagine the wheel to revolve about its centre so that A moves C 34 APPLIED MECHANICS. at 0-3 foot per second. Map out the real positions of the points which you found on the vane, and therefore the real path of a particle of water. Try to follow it after it has left Q, assuming plenty of space outside ; hut this is a prohlem which you perhaps may return to later. 37. Let P (Fig. 14) he a point on the rim of the wheel of a centrifugal pump. Suppose that we know the velocity of P, and represent it hy the distance P B ; also that we know the velocity of the water along the vane of the wheel at p, and represent it hy the distance A p. Now, a particle of water at P has hoth these velocities ; A p relatively to the wheel, together with p B "because the wheel is in motion. Hence the total velocity of the particle of water is repre- sented in direction, amount, and sense hy the vector sum A p -f P B, and we can either use the parallelogram method or the triangle method to find it. Pig. 14. Exercise. The rim of the wheel of a centrifugal pump goes at 30 feet per second ; water flows radially at 5 feet per second ; the vanes are inclined backward at an angle of 35 to the rim. What is the absolute velocity of the water? What is the component of this parallel to the rim? " Am., 23-4 feet per second; 22-8 feet per second. When we know the velocity of water before it enters a wheel and the velocity of the wheel at the place, and we wish the water to enter without shock, this simply means that the total velocity of the water the instant after it enters the wheel shall be exactly the same as before it entered. Thus in Fig. 15, if the velocity, c P, of water is known before it enters the turbine wheel, and P B represents the velocity of the wheel, find the velocity which, added to P B, will have c P for a resultant. It is p A if P Q represents the velocity c p to scale. Make, then, p A the direction of the vane at p. The water will flow in the direction p A c relatively to the wheel, and it has also the x x velocity p B because it moves with the wheel; its total velocity is the same as before it entered, and it has entered without shock. We don't much care how the vane curves afterwards so long as it curves gradually ; it is its direction at p that is important. Exercise. The inner circumference of a centrifugal pump wheel goes at 15 feet per second ; water approaches it radially at 5 feet per second. What is the angle of the vanes if the water is to enter without shock ? Ans., 18 26'. Pig. 15. Note that the angle chosen by us for the vane at A (Fig. 13) enabled the water to enter the wheel without shock. We may at once say that it is only the angles at A and at a that are of real APPLIED MECHANICS. 35 importance in the design of a pump. The actual shape of the vane is unimportant so long as the curve is fairly direct. But the exercise ought to "be worked carefully throughout. We made the tangent of the angle at A equal to the radial velocity !, divided by the velocity of the wheel at A (see Art. 33). We always endeavour to have this sort of relation true at the inner side of the wheel of a centrifugal pump or turbine. 38. Usually we consider the earth and frame of a machine to be fixed. The student will find it very instructive to think of some link or wheel as fixed (which he usually thinks of as moving) and now note what the motions are. One simple example of this is given in Art. 467, and we there show that to understand the relative motions in a four-link mechanism is to understand the motions in a very great number of other mechanisms. Train A is passing train B. Looked at by an observer on the ground, how very different is their appearance from what it is to an observer in either train ! The mathematics of the subject is quite easy. It is simply this : If a, b, c, etc., are lines drawn representing what may be called the absolute displacements of points A, B, c, etc., or the rotations of bars about axes, then their displacements relatively to a frame F, whose own absolute displacement or rotation is /, are a /, b f, c f, etc., the sign meaning vector subtraction. But mathematics does not satisfy us ; we want the instinct of comprehending easily these relative motions. That we do not possess it is evidenced by the fact that all the mechanisms of Art. 467 do really seem to us different, and that we need to give the name epicyclic train to a train of wheels when the framework which connects their centres is allowed to move instead of remaining at rest, as it does in our usual way of studying things. Notice that this, which ought to be an easy subject, follows in smaller printing. 39. Thus, for example, in Fig. 16 we have a train of three wheels (the student ought to take two or four or more). When the frame F is at rest (that is, we take speeds relatively to the frame) let A, B, c have the angular velocities o, ba, ca. (evidently in the figure b is negative). Now let F have an absolute angular velo- city, /, and the absolute velocities of the Fi - 16> wheels are a + /, ba -\- /, ca + f. 36 APPLIED MECHANICS. 1. Suppose A is at rest ; a + /= 0, a = /. B'S velocity is (_ J + I)/; c's velocity is (- e + I)/. Thus, let c have the same number of teeth as A ; so. that c = 1, c's velocity is 0. If c has 100 teeth, and alongside it on what is practically the same spindle, let there he wheels of 99 and 101 teeth also gearing with B. Evidently the absolute speeds of the three will be (since e has the te ee TO lues and ) a, - to to ^ Wien the arm goes round and the motions of the three wheels are observed, we call this Ferguson's paradox. If we could, like flies, move with the arm, there would be nothing paradoxical about it. Notice that A, B, and c may be bevel wheels, as shown in Fig. 17. 2. Simpler Case. Wheels A and B connected by arm F ; arm Fig. 17. rotates. The absolute rotational velocity of B is 0. What is A'S velocity ? Here la + / = 0. Hence a= f/b ; so that A'S velocity is _ fib +f or / (l -\ Thus, let b = 1, as it is in Watts' sun and planet motion ; A'S velocity is 2/. In this case, however, because of the angularity of the connecting- rod, B has some angular velocity, fluctuating between, sav> -f j8 and j8. Hence (since b = 1), a +/ + 0, and A'S velocity is 2/ + 0. 3. In Fig. 17 we have A, B, and c, three bevel wheels. B may be carried round the axes of the others on a spur wheel D. Suppose we are looking from the point D at every wheel, and the numbers of teeth on A and c are as 1 to c, then relatively to D the speeds of A and c are a and ae. If D rotates at speed /, the absolute velocities of A and c are a + / an( i - ac + f. Example. Let e =. 1 ; then, if A'S speed a +/ is called a, so that a = a - /, c's speed is - a + / or - a + 2/. Thus, APPLIED MECHANICS. 37 suppose A goes at 20 revolutions per minute or a = 20, then c's speed is 20 + 2/; so that if / is gradually changed in the following way, we get the following speeds for c shown in the table : Again, we may imagine the speed of F to keep constant and that of A to vary, and we get the same result. By means of a cam we may gradually change from negative to positive velocities, but here we have a very much better means by ordinary gearing. In Fig. 17, if the shaft of A is rotated by coned pulleys, by which it may be given varying speeds, and if A is kept rotating at a fixed speed, c gets speeds which may be negative or positive or zero. Speed of F. Speed of C. 4 - 12 6 - 8 8 4 10 12 4 14 8 16 12 18 16 20 20 Exercise. An epicyclic gear consists of an annular wheel A of 72 teeth, a pinion B, and a spur wheel c of 40 teeth concentric with A. The arm which carries the axis of B makes 30 revolutions per minute. (1) If A be a dead wheel, find the revolutions of c. (2) If c be a dead wheel, find the revolutions of A. Am., (1) 84 ; (2) 46|. Again, if A makes 4 revolutions and c 6 revolutions in the same direction, find the revolutions of the arm. Am. , 4|. Exercise. In a horse-gear for driving a chaff-cutter, the bracket that holds the pole supports also a short horizontal shaft carrying a bevel wheel of 31 teeth and a bevel pinion of 16 teeth. The pinion gears into a horizontal bevel ring of 80 teeth that is stationary, and forms part of the framing. The bevel wheel of 31 teeth also gears with a bevel pinion of 22 teeth which is loose on the central vertical axis, and this pinion carries with it a bevel wheel of 60 teeth that gears with a pinion of 16 teeth on the high speed horizontal shaft. Find the number of revolu- tions of the high speed shaft for each circuit of the horse. ,.,., 30 I- 58 CHAPTER III. WORK A^D ENERGY. 40. Work. To do work it is necessary to exert a force through a certain distance in the direction of the force. Thus, if we exert a force of 20 Ibs. through a distance of 6 feet, we do 20 x 6, or 1 20 foot-pounds of work. If a body of 5 Ibs. weight changes its level by the amount of 10 feet, whether it does this by a direct vertical fall or rise, or is moved up or down an inclined plane or curved surface, so long as there is no friction, the amount of work given out by the body in falling or given to it to make it rise is always the same, 5 x 10, or 50 foot- pounds. Example. The weight in a certain clock is 20 Ibs., and after being wound up it can fall through a distance of 40 feet. Suppose we wish to alter this height, making it 10 feet ; what weight must we use 1 Evidently the work given out by the new weight in falling 10 feet must be equal to that given out by the old weight, or 800 foot-pounds. In fact, the new weight must be 80 Ibs. Of course we must apply this weight to the clock by means of a block and pulleys, or we must reduce the diameter of the drum proportionately ; and if in applying it we introduce more friction than there used to be in the clock, we must further increase the weight, so as to be able to overcome this friction. The work done by a force is well illustrated by the pulling of a tra.mcar. If the pulling force P Ibs. is not directly along the track, but makes an angle 6 with it, the effective force, or the resolved part of P in the direction of motion is P cos. 0, and this, multiplied by the distance moved through in feet, is the work done in foot-pounds. When a body is pulled up a curve the work done in over coming the force of gravity (we are neglecting work spent in overcoming friction) is simply the weight of the body multiplied by the difference in level. Thus in Fig. 18 the work done in moving a body of weight w from A to B along the curve is simply w y, where y is the difference in level of A and B. Proof : Let the co-ordinates of any point P be x, y ; and of Q, 9 point indefinitely near to P, x -\- 8#, y -f- 8y. APPLIED MECHANICS. 39 The weight w resolved in the direction of the tangent at p is w sill. 0, and this multiplied by the distance p Q, (we are supposing that the tangent at o, is parallel to the tangent at p, as Q is supposed to be indefinitely near to P), or p Q . w sin. Q is the work done against graAdty in pulling the body from p to Q. Therefore the whole work done in pulling the body from A to B is the sum A N Fig. 18. of all such terms as P Q, . w sin. 0. But p a sin. is Q R, or in the limit dy. Therefore the whole work done is co- J! Jo wdy or Gravity does this work when the body falls. But, both when a body is moved up and down, energy is wasted or con- verted into heat in overcoming friction. Hence, when a weight, w Ibs. , is lifted in level h feet, the useful work done ia wh, but there is energy wasted. Again, when w falls, gravity does the work wh, but there is energy wasted. If we are depending upon the total store wA to drive machinery, the useful work done is less than wh, the difference being wasted, or rather converted into heat, by friction. 41. Horse-power. One horse-power is the work of 33,000 foot-pounds done in one minute. Power means not merely work, but work done in a certain time ; the time rate of doing work. The work done in one minute by any agent divided by 33,000 is the horse-power of that agent. In a steam-engine the piston travels four times the length of the crank in one revolution, and all this time it is being acted upon by the pressure of steam. If the mean or average pressure urging the piston is 60 Ibs. per square inch, and the area of the piston is 150 square inches, then the total average force urging the piston is 150 x 60, or 9,000 Ibs. If the crank, whose length is 0*9 foot, makes 70 revolutions per 40 APPLIED MECHANICS. minute, then the piston travels 4 times 0*9 x 70, or 252 feet per minute, so that the work done in one minute is 9,000 X 252, or 2,268.000 foot-pounds. Dividing this by 33,000, we find the horse-power of the steam-engine to be 68'7. The mean pressure is best found by the use of an indicator which draws for us an jndicator diagram. Measuring the pressures at ten equidistant places on this diagram, adding them together, and dividing by ten, gives the average pressure. Or, measure the area by means of a planimeter in square inches ; divide by the extreme length of the diagram parallel to the atmospheric line; this gives the average breadth, and therefore the average pressure to scale. As the pressure of steam is usually given per square inch, it is usual to take the diameter of the cylinder in inches, but distances passed through by the piston are evidently to be measured in feet. Example. We find by a spring balance that some horses or a steam-engine have been pulling a carriage with an average pull of 120 Ibs. during one minute, the space passed over in the minute being 500 feet ; what is the horse-power expended on the carriage ? Here 120 Ibs. act through the distance of 500 feet, and the work done in one minute is evidently 500 x 120, or 60,000. Dividing by 33,000, we find the horse-power to be 1-818. 42. Energy is the capability of doing work. When a weight is able to fall, it possesses potential energy equal to the weight in Ibs. multiplied by the change of level in feet through which it can fall. When a body is in motion, it possesses kinetic energy equal to half its mass (its weight in London in pounds divided by 32-2 is its inertia, which is usually, but we think unwisely, called its mass), multiplied by the square of its velocity in feet per second. (See Art. 190.) Example. A body of 60 Ibs. is 100 feet above the ground, and has a velocity of 150 feet per second; what is its total amount of mechanical energy 1 that is, what energy can it give out before it reaches the ground, and becomes motionless? Here the potential energy is 60 x 100, or 6,000 foot-pounds. Its kinetic energy is 60x 150 x 150-=-64'4, or 20,963 foot- pounds. So that the total amount is 26,963 foot-pounds. Suppose this body to lose no energy through friction with the air, and suppose that, after a time, it is at a height of 20 feet above the ground ; find its velocity. Answer : Its potential \ APPLIED MECHANICS. 41 energy is now 60x20, or 1,200 foot-pounds, therefore its kinetic energy must be 25,763; and evidently this, multiplied by 64-4 and divided by 60, gives 27,652-29, the square of the new velocity. Its velocity is therefore 166-3 feet per second. In such a question we are not concerned with the direction in which the body is moving. It may be a cannon-ball, or a falling or rising stone, or the bob of a pendulum. Given its velocity and height at any instant, we can find for any other height what its velocity must be, or for any other velocity what its height must be. On a Switchback railway, the loss of energy (roughly pro- portional to distance travelled) every journey is represented by the lift of a few feet which is effected by the attendant at each end before making a fresh start. Neglecting this loss, it is easy to calculate the velocity at any pla,ce if we know the vertical depth of the place below the starting point. The deepest places on the line are places of greatest velocity; the highest places are those of least velocity. In a bob of a pendulum we see a continual conversion of kinetic into potential and of potential into kinetic energy, the total store remaining constant, except in so far as friction is converting mechanical energy into heat. (See Art. 192.) The energy of a strained body for example, a strained spring is another store of mechanical energy. It is an excellent laboratory experiment to measure how much energy a spring can store without getting permanently hurt in shape or broken. This store of energy is called its resilience. As the elongation of a spring is proportional to the load, if we gradually increase the load from to w Ibs., the elongation increases from to x feet, so that the average force for the whole length being w, the energy stored is -| w x foot-pounds. A weight vibrating vertically at the end of a spiral spring gives a good example of the continual conversion of the three kinds of mechanical energy into one another. The total store gradually diminishes, partly by friction in the atmosphere, but greatly also by an internal frictional resistance or viscosity in the material of the spring. The student will find it interesting to compare the behaviour of a spring of steel and a spring of indiarubber in this respect ; the viscosity of the indiarubber damps the vibrations with great rapidity. Many hours may be well spent in studying these phenomena, making quantita- tive measurements. 42 APPLIED MECHANICS. Air and other fluids in a compressed condition contain stores of energy. 43. A student may employ his leisure in calculating the possible stores of energy in 1 lb. of various materials : 1 Ib. of hydrogen, 4'8 x 10 7 foot-pounds; kerosene, 2 x 10 7 foot-pounds; coal, 12x10 foot-pounds (the weight of oxygen or air for combustion is not counted in) ; 1 lb. of cast iron in rim of pulley at highest speed to produce greatest working stress, 1,000 foot-pounds ; steel spring of the best kind, 270 foot- pounds ; usual spiral spring of round wire, 135 foot-pounds. The heat given to 1 lb. of water to raise it from to 1 c C. or from 99 to 100 C. in temperature, is nearly the same. This is a unit of heat energy. Joule showed that this is equivalent to 1,400 foot-pounds.* When energy is in the heat form, a heat-engine may be used to convert part of it into mechanical energy ; unfortunately the amount convertible, depends upon how high is the temperature of the stuff contain- ing the heat energy above the temperature of the refrigerator or exhaust. Hence it is that in the very best steam-engines we seldom convert more than one-seventh of the heat energy of the steam into mechanical energy, the other six-sevenths being degraded in temperature, and going to the refrigerator useless for our purposes. In discussing heat-engines we express all energy in foot-pounds. The energy store is most intense as to volume (we do not include the weight of the air needed for combustion) in a pound .of kerosene, being about half as much again as in a pound of coal. We look forward to the time when no heat-engine will be needed in the conversion, and we may then be able to convert over 90 per cent, of the energy of a pound of kerosene into the mechanical form, as at the present time the chemical energy of zinc is convertible in a battery or electric motor, or the chemical energy of oats and other food is convertible in the animal machine, which is probably a gas battery and electric motor. 44. When the engine of the Finsbury College is working mainly for the electric light and possibly one electric motor, the students have sometimes, during a long-continued measurement, been able to trace what becomes of the energy of one pound * The latest determination for the mean specific heat of water from C. to 100 C. is 1399 foot-pounds, or for 1 gramme it is 0'995 calorie ; 1 calorie, the heat to raise 1 gramme of water from 10 C. to 11 C. is 4 '2 Joules or 4-2 X 10? ergs. The heat from 20 C. to 21 C. is l-30th of 1 per cent. leas. APPLIED MECHANICS. 43 Of coal. They had previously measured the chemical energy in a pound of coal, and tested all sorts of instruments and ideas used in their measurements. The energy of 1 Ib.of coal is, say, 12,000,000 foot-pounds; of this 4,000,000 go up the chimney or get wasted by radiation as heat, and 8,000,000 foot pounds of heat energy reach the engine in steam ; of this only one-thirteenth, or 600,000 foot- pounds is converted into mechanical energy and given to the piston ; the other twelve-thirteenths go off to the condenser and are wasted. Only 500,000 are given out by the engine to the long shaft which drives the dynamo machine, about 400,000 foot-pounds leave the dynamo as electric energy, and part is wasted by conversion into heat in the conductors to the lamps. At the lamps we have about 370,000 foot-pounds of electric energy converted into heat and light. If any of the electric energy or all of it is given to an electric motor, perhaps more than 90 per cent, of it will be converted into mechanical power. Men who make measurements of this kind get to have very clear ideas as to the various forms of energy and the fact that it is indestructible, and cannot be wasted, although it may alter into forms in which it may not be available for use ; and therefore we say that it is wasted. The very best steam-engines use more than 1 J Ibs. of coal per hour for each horse-power given out. We cannot hope for much improvement; this is about one useful for nine total. Gas-engines using Dowson gas already give out one horse- power for 1 Ib. of coal consumed per hour ; we hope for con- siderable improvement. Oil-engines give out one horse-power for less than 0*9 Ib. of kerosene per hour ; we hope for very considerable improvement. 45. Students ought to work many numerical exercises on me- chanical energy : w Ib. of water raised vertically h feet, the energy is wh foot-pounds. If time is given, find the work done per minute and divide by 33,000; this is useful horse-power. Divide this by the efficiency of a pump, and we have the actual horse- power which must be supplied to the pump. Or, if the w Ibs. of water fall per minute through a turbine, the head available being h feet, we have wh -- 33,000 as the total horse-power, and the turbine will probably give out 70 per cent, of this usefully, 0-70 or 70 per cent, being the efficiency of a good turbine. If there were no friction, a waggon of weight w requires to 44 APPLIED MECHANICS. be pulled with a force of w-i-s up a road which rises 1 foot in every s feet of its length. The fractional resistance to motion of a vehicle on a level road is usually stated as so many pounds per ton weight of the vehicle. If there is also an incline wo add the two tractive forces together one for the incline without friction, the other to overcome friction on the level. The resistance in pounds per ton of a moving train (including engine) on the level is found roughly by adding 2 to one- quarter of the speed in miles per hour. This is for speeds greater than 20 miles per hour. At less speeds there is quite a different law, which may for some trains and permanent ways be indicated by the following figures : Speed in miles per | hour. j Resistance in pounds } per ton. I 20 1* 10 2 7 6 5 10 6 A curved line adds 1 2 per cent, id the resistance on the average English railways. The tractive force of heavy waggons on macadamised roads may be taken as 50 Ibs. per ton, on paved roads 30 Ibs. per ton, and on gravel roads as 150 Ibs. per ton. These rules are good enough for academic exercise work. If the pull on a tramcar is recorded as the ordinate of a diagram, of which the abscissa represents distance along the track, the area of the diagram represents to some scale the work done. The average height of the diagram represents the average pull P. This average pull p, multiplied by the whole length of the track, is the whole work done. In most practical cases the average can only be obtained by making the diagram, finding its area, and dividing by its length, just as we do with an indicator diagram of an engine. But there are cases where we can calculate an average. Example. A chain of length I and weight wl, with a weight w at the end of it, is to be wound up by a capstan ; what work will be done ? Obviously the average pull is w-j- half the weight of the chain, or w-f J wl, and the total distance is I, so that the work done is wl + ^ wl*. As a rule, it is wise to plot the varying pull as the ordinate of a curve on squared paper, as, without the aid of the calculus, one is apt to state what is the average pull without due thought. Thus, if the abovn- APPLIED MECHANICS. 45 mentioned chain varies in heaviness per foot, the average pull due to it is not half its weight. Observe that we ought to call the above the space average of a force. The space average of a force, multiplied by the distance, is the work done. A little consideration will show a student that a time average is a very different thing. 46. According to the results of some experiments by Prof. E. H. Smith on the cutting of metal in the lathe without water or oil, the force on the tool is not much affected by For both thin and moderately thick shavings at all speeds, feeds, and depths of cut, we may roughly take it that forged steel takes twice as much power to cut it as does cast iron; wrought iron takes one to one and a half times as much as cast iron. For broad thin shavings, cast iron required more cutting force than wrought iron. The force is neither proportional to the breadth of the shaving nor the depth, but it is more nearly proportional to depth than breadth. It is interesting to note that before these experiments it was usual in books to follow Weisbach in saying that for iron P =fb d where f was 50,000 Ibs. per square inch. Smith's P experiments show that this rule is not true, and that varies bd from 92,000 to 239,000. Possibly when it is discovered that there are other things to be done in college engineering labora- tories than to break endless numbers of specimens of metal with a 100- or 200-ton testing-machine, we may have further experimental results on which practical engineers may rely. Professor Smith tried various depths, d inches, and breadths, 6 inches, of shaving from cast iron, wrought iron, arid forged steel, in every case measuring P, the pressure on the tool in pounds, at the cutting edge. In almost every case we find from his numbers that the energy E usefully spent per cubic inch of metal removed, diminished about 30 per cent, as b was increased from -03 to '05, being about its minimum value when b was -05 ; but probably it does not increase much for greater values of b. The minimum values of E for various depths of cut were as follow : APPLIED MECHANICS. d * p E 05 135 056 056 288 690 8,700 7,700 - Cast iron. 03 06 14 056 06 056 215 570 810 10,700 13,000 8,700 ) Wrought v iron. 02 056 250 18,700 \ 04 056 480 18,000 ( Steel. 06 056 705 17,700 With cast iron if b '05, and probably for greater values of b, E is much the same for cuts of the depths '05 and '135 inch, being about 9,000 foot-pounds per cubic inch of metal removed. With steel, if b=-Q5, and probably for greater values of b, E is much the same for cuts of the depths '02, '04, and 06 inch, being a')out 18,000 foot-pounds per cubic inch of metal removed. With steel, if b ='05, and probably for greater values of b, E is 10,700 footpounds for a cut of -03 inch depth, 13,000 foot-pounds for a cut of '06 inch depth, and is practically the same as cast iron for a cut of "14 inch depth. EXERCISES. 1. Two tons of rock can fall to a depth of 320 feet; find the work which it may do. Ans., 1,433,600 foot-lbs. 2. In lifting an anchor of \\ tons from a depth of 15 fathoms in six minutes, what is the useful man-power, if a man-power is defined as 3,500 foot-lbs. per minute? Am., 14-4. 3. The pull on a tramcar is 200 Ibs. at an angle of 25 with the track ; what is the component in the direction of the track ? "What work is done in a distance of 10 feet along the track? If the speed is 4 feet per second, what is the usefully expended horse-power ? Ans., 181-26 Ibs. ; 1812-6 foot-lbs ; 1-318. 4. What horse-power is involved in lowering by 2 feet the level of the surface of a lake 2 square miles in area in 300 hours, the water being lifted to an average height of 5 feet ? Ans., 58'5. 5. Taking the average power of a man as T Vth of a horse-power, and the efficiency of the pump used as 0-4, in what time wiU ten men empty a tank of 50 feet x 30 feet x 6 feet filled with water, the lift being an average height of 30 feet ? Ans., 21 hours 14 minutes. 6. The diameter of a steam-engine cylinder is 9 inches, the length of crank 9 inches, the number of revolutions per minute 110, and mean effective pressure of the steam 35 Ibs. per square inch; find the indicated horse-power. An*., 22-3. APPLIED MECHANICS. . 47 7. One gas-engine uses 24 cubic feet of coal-gas, and another 98 cubic feet of Dowson gas per hour per useful horse-power ; what are their efficiencies ? The calorific powers of coal-gas and of Dowson gas per cubic foot are 520,000 and 123,000 foot-pounds respectively. Am., 0'159; 0-164. 8. What would be the indicated horse-power of an Otto gas engine which has a piston 12 inches in diameter and a crank 8 inches long? The engine works at 150 revolutions a minute, and there is an explosion every 2 revolutions, the mean effective pressure in the cylinder during a cycle being 62 Ibs. per square inch. Ans., 21. 9. The average breadth of an indicator diagram for one side of the piston is 1-58 inches, and for the other side it is T42 inches, and 1 inch represents 32 Ibs. per square inch. Piston, 12 inches diameter; crank, 1 foot ; 110 revolutions per minute. What is the indicated horse-power ? Ans., 72-38. 10. What must be the effective horse-power of a locomotive which moves at the steady speed of 35 miles an hour on level rails, the weight of engine and train being 120 tons, and the resistances 16 Ibs. per ton? What additional horse-power would be necessary if the rails were laid along a gradient of 1 in 112 ? Ans., 179-2 ; 224. 11. In 10 find in each case how far the train would move after steam was shut off, assuming the above constant resistance and neglecting rota- tory motions. Find also the speed of the train after the latter had moved over a distance of 1,000 feet from the point where steam was shut off. Ans., 5,728 feet; 2,545-8 feet ; 3 1-8 miles per hour; 27 '3 miles per hour. 12. A flywheel weighs 1\ tons, and its mean rim has a velocity of 40 feet per second ; what is its kinetic energy ? If the velocity be reduced 3 per cent., what is the reduction in the kinetic energy? If the kinetic energy be reduced by 10,000 foot-lbs., by how much is the velocity reduced ? In estimating the latter, why would it be wrong to subtract from 40 feet per second the velocity which corresponds to 10,000 foot-lbs. of energy in this flywheel ? Ans. 139,130 foot-lbs. ; 130,900 foot-lbs. ; 1-5 ft. per second. 13. A machine discharges n projectiles per minute, each of w Ibs. moving with the velocity of v feet per second ; what is the actual horse- power ? If the efficiency of the machine is e, and it is driven by a steam- engine which uses w Ibs. of steam per hour per horse-power given out by it, what is the total steam per hour? If the engine is governed by throttling, and if the total steam per hour follows the rule: steam per hourr=0 -f- b x brake horse-power (where a and b are known to us), and if for half an hour n^ projectiles are discharged per minute, and if for the next half hour w 2 projectiles are discharged per minute, how much steam is used during the hour ? . wwv 2 e wnvw 2 . b^ + >? 2 )wt? 2 '' ^x 66000 ' 66000*7* ' ' 1 32000 ^ 14. Town refuse is about \ a ton per unit of the population per year. In the careful burning of 1 Ib. of refuse, 0-5 Ib. of water at 212 F. may be converted into steam at 212 F. If 1 Ib. of coal is able to evaporate in a good boiler 10 Ibs. of water, how many tons of coal per year would pro- duce the same amount of steam as the refuse from 5,000,000 inhabitants? Ans., 5,000,000 x \ x 0-4 -f- 10, or 100,000 tons of coal per year. If we get one actual horse-power for 40 Ibs. of refuse per hour ; if the engines drive pumps of 90 per cent, efficiency, pumping water to a reservoir ; if the water drives motors in the city with a total efficiency of 30 per cent., 48 APPLIED MECHANICS. working on an average 2 hours per working day ; what is the average horse-power supplied to each house if there are 500,000 houses in the city ? Am., 5,000,000x1^x2240-7-40 or 42xl0 7 is the actual energy in horse-power hours. The pumped energy is 3*78 x 10 8 horse-power hours. The supplied energy is 1 134 x 10 8 horse-power hours. The average total horse-power (2 hours a day for 313 days) is 1/8 x 10 5 , or 180,000, and per house it is 0'36 horse-power. 15. The section of a stream is 12 square feet, the average velocity of the water is 2 feet per second ; there is an available fall of 25 feet ; what is the horse-power available ? A turbine drives a dynamo machine which sends electric power to a motor at a distance. The efficiency of the turbine is 70 per cent.; of the dynamo, 87 per cent.; ten per cent, of the energy from the dynamo is wasted in transmission, and the efficiency of the motor is 72 per cent. ; how much power is given out by the motor ? The voltage at the dynamo is 102 ; what is the current in amperes ? An*., 68; 26'8 ; 303. 16. Electric lamps giving 1 candle-power for 4 watts; how many 10- or how many 16 -candle lamps maybe worked per electric horse-power? The combined efficiency of engine, dynamo, and gearing being 70 per cent., what is the candle-power available for every indicated horse- power? Ans., 18; 11; 130'55. 17. On a switchback the carriage is 9G6 Ibs., neglecting friction ; find its kinetic energies when it is 5, 10, and 15 feet below its starting-point. And if the starting-point is 20 feet above datum level, write out in two columns its two* kinds of energy at each point. If the above points are 0-1, 0-4, and 0-7 of the distance along the track, and if loss of energy by friction is proportional merely to distance along the track, and if the carriage has to be lifted 1-6 feet at the end of each journey, find the correction in the kinetic energy at each place. Ans., Potential energies, 14,490, 9,660, 4,830 foot-lbs. ; kinetic energies, 4,830, 9,660, 14,490 foot-lbs. ; corrected kinetic energies, 4,675-4, 9,041-6, 13,407 foot-lbs. ] 8. The calorific powers of 1 Ib. of each of the following fuels are given in centigrade-pound heat units ; convert into foot-lbs. if 1 heat unit = 1,400 foot-lbs. : charcoal, 7,000; coke, 7,000; coal, 8,800 to 7,330; wood, 4,200; and kerosene, 12,200. 19. The pull on a tramcar was registered when the car was at the following distances along the track: 0, 200 Ibs.; 10 feet, 150 Ibs.; 25 feet, 160 Ibs.; 32 feet, 156 Ibs.; 41 feet, 163 Ibs.; 56 feet, 170 Ibs.; 60 feet, 165 Ibs. ; 73 feet, 160 Ibs. ; what is the average (space) pull on the car, and what is the effective work done in pulling the car through the distance of 73 feet? Ans., 161 Ibs., 11,800 foot-lbs. about. 20. A chain hanging vertically 520 feet long, weighing 20 Ibs. per foot, is wound up ; what work is done ? Ans., 2, 704, QUO foot-lbs. 21. Four cwt. of material are drawn from a depth of 80 fathoms by a rope weighing 1-15 Ibs. per linear foot ; how much work is done altogether, and how much per cent, is done, in lifting the rope ? What horse-power we nld be required to raise the material in four and a half minutes ? Ans., 347,520 foot-lbs ; 38 ; 2'34. 22. (a) A cut of -06 inch depth is being made on a 4-inch wrought- iron shaft revolving at 10 revolutions per minute ; the traverse feed is 0'03 inch per revolution ; the pressure on the tool is found to be 435 Ibs. . APPLIED MECHANICS. 49 what is the horse-power expended at the tool ? How much metal is removed per hour per horse-power ? Ans., -1381 ; 98-28 cuhic inches. (b] When the traverse feed is -06 inch per revolution, the pressure on the tool is found to he 570 Ibs. ; find the horse-power, and the metal removed per hour per horse-power. Ans. y '181 ; 150 cuhic inches. (c) If the above horse-power is called the useful power TI, and it is found that the actual horse-power given to the lathe is 0-1 + 1-5 u, find actual horse-power and metal removed per hour per actual horse-power. ^Ans., -3071, 44-2 cubic inches; -3715, 73*1 cubic inches. 23. What is the kinetic energy of a tramcar moving at 6 miles per hour, laden with 36 passengers, each of the average weight of 11 stones ? Weight of car, 2^ tons. What is its momentum ? If stopped in 2 sees., what is the average force ? If the force is constant, this must also be the space average force ; find the distance of stopping if the force is constant. Ans., 13,400 foot-lbs; 3045-565; 1522-8 Ibs. ; 8 -8 feet. 24. A ball weighing 5 ounces, and moving at 1,000 feet per second, pierces a shield, and moves on with a velocity of 400 feet per second ; what energy is lost in piercing the shield ? Ans., 4,076 foot-lbs. 25. A fire-engine pump is provided with a nozzle, the section area of which is 1 square inch, and the water is projected through the nozzle with an average normal velocity of 130 feet per second; find (1) the number of cubic feet discharged per second, (2) the weight of water discharged per minute, (3) the kinetic energy of each pound of water as it leaves the nozzle, (4) the horse-power of the engine required to drive the pump, assuming the efficiency to be 70 per cent. Ans., (1) -9 cubic feet ; (2) 1-51 tons; (3) 262-3 foot-lbs. ; (4) 38-3. 47. Bicycle problems. When a man says that his bicycle is geared to D inches, he means that he advances IT D inches for one turn of his pedals. Let the diameter of the pedal circle be d inches. Let W be weight in Ibs. of rider, W Q of machine, W + w =w. Let w be the uniform vertical force which the rider applies to each pedal alternately ; if w is negative, it means that he is back-pedalling. Let F be the force in Ibs. which would pull the bicycle along at the velocity of v miles per hour ; 2 w d is the work done in inch-pounds by the rider in one revolution, and this is equal to IT D F. r, or FV -r 375, the horse-power usefully expended.... (2) 60 X 33,000' 12 x v x 5,280 = 3360 = number of revolutions of a pedal per 7TD X 60 D minute.... (3). [It is useful to remember that a machine geared to 56 inches goes at 10 miles an hour when the pedals make one turn per second.] 50 APPLIED MECHANICS. The vertical force is not by any means constant in practice, nor indeed ought it to be ; it is and ought to be greatest when the crank is horizontal. With proper ankle action the force is always somewhat in the direction of the circular path of the pedal, but for exercise work there is no harm in assuming a uniform vertical force w in the down-stroke and no force in the up-stroke. In practice it is difficult to avoid pressing on the pedal in its up-stroke. If F is the value of P on a level road, then on a rising slope of 1 in s we have F = F + ? .... (4). If it is a descending slope, is negative. P may usually be taken as proportional to w. The following data were roughly measured by myself. They are good enough for academic problems. They suit my own bicycle on a good road. Students ought to obtain data of their own by careful measurement. A rider weighing 127 Ibs. on a cycle weighing 33 Ibs. (or w=160 Ibs.) finds that on a descent of 1 in 80, with his feet off the pedals, he is just able to get on very slowly but steadily ; on a long slope of 1 in 40 his steady speed (feet off pedals) was 9 J miles per hour ; on a long slope of 1 in 20 his steady speed (feet off pedals) was 20 miles per hour. In these three cases the values of F O were : - or 2, - or 4, - or 8. 80 '40 20 We need careful experiments ; and it is not of much use speculating on the probable law of resistance of a safety bicycle with pneumatic tyres on a certain kind of road. A constant term for quasi solid friction, and a term (the most important at high speeds) proportional to the square of the speed relatively to that of the atmosphere for air resistance : these we ought to have. Resistance due to unevenness of the ground would be constant if a certain kind of unevenness were to repeat itself at intervals so far apart on the road that the vibration due to each had time to die away before the next ; but speculation is vague, especially as the kind of vibration will often depend on the velocity. In our present state of knowledge we cannot be far wrong in assuming F W = w (a + bv 4- cv-). APPLIED MECHANICS. 51 If the above values of F O are correct at the given speeds, =0, 9|, and 20 miles per hour, we find as a formula which represents our experimental data well enough for exercise purposes. Hence, going up a slope of 1 in s we have - - 80 20 200 for this bicycle. When w= 160 Ibs. 7 = 2 + v . + *L + *22 . . (6). r 10^ 100 ^ * Example 1. What horse-power is expended in going at 12 miles an hour on a level road ] Here -- =0. Th ) speed is 88 v or l/)56 feet per minute, and the horse-power Example 2. At 12 miles an hour, going up or down au incline of 1 in 60, what is the useful horse-power ? F = 4-64 + ^ =; 7-31 or 1-97 Ib. Horse-power up = 1>056 x 7 ' 31 = Q-234. 33,000 Horse-power down = 1>056 x 1-97 = 0-033. 33,000 EXERCISES. 1. In Examples 1 and 2, what force, w, does the rider exert upon his pedal if his bicycle is geared to D = 60 inches and the diameter of the pedal circle is 13 inches'? Ans., On the level, w = 33-6 Ibs. ; going up, w = 63 Ibs. ; going down, w =.- 14-3 Ibs. 2. On what slope downward would the velocity of 12 miles an hour be steadily maintained, feet off pedals ? Ans. , 1 in 34. 3. Going down a slope of 1 in 30 at 8 miles per hour, what is the force on the pedal? Ans., -12-30 Ibs. The minus sign means that the rider must back-pedal. 4. If a rider whose weight is 157 Ibs., back-pedals on this same bicycle with a force of only 10 Ibs., down a slope of 1 in 25, what is his velocity P Ans. i 13'6 miles per hour. B2 APPLIED MECHANICS. The following example is for students who can integrate ! The resistance to motion being r a -f bv + cv 2 on a road which does not alter in character for a whole journey, compare the work done in going over a certain distance, I first, at a constant speed, v ; second, at the same average speed, but varying according to the law v = v -}- /sin. qt. In both cases the time is the same if the journey is just so long as to be finished by the rider in the same state as to speed, etc., with which he starts. 9 + bv 2 + cv*) dt t the work done. If we calculate this for a time, T, where q = 27TT, it is just as good as for any number of such periods. The value divided by T gives the average work per second, or whereas, if / is and the speed is constant, we have the average work per second av + bv, the angle of friction, is such that sin. ^> = '015. Re- member that tan. = /t Ans., 2-16 APPLIED MECHANICS. 71 2. Find the horse -power absorbed in overcoming the friction of a foot-step bearing 4" diameter, the total load being 1J tons, the number of revolutions 100 per minute, and the average co-efficient of friction -07. Ans., 0-5 nearly. 3. If it be assumed that the power wasted between the end of a flat pivot and its step is proportional at each point to the product of the velocity and pressure, what horse-power will be absorbed by such a pivot, 3" diameter, when running at 120 revolutions per minute, the load on the pivot being 2| tons, and the average co-efficient of friction -06 ? Inte- gration gives the total energy wasted per second as f a p w 11, where a is the angular velocity in radians per second, w is the total load, it is outside radius of pivot, and p. is the co- efficient of friction. Ans. t -639. 4. The length of a journal is 9" and its diameter 6"; it carries a load of 3 tons. What horse-power is absorbed in friction when making 100 revolutions per minute, the average co- efficient of friction being '015? What number of thermal units per minute will be conducted away per square inch of the brass ? Ans., 0'48 ; 0'2 centigrade heat units. 5. A shaft makes 50 revolutions per minute. If the load on the bear- ing be 8 tons, and the dia- meter of the bearing 7 inches, at what rate is heat being generated, the average co- efficient of friction being 05? If 3 thermal units escape per minute when the temper- ature of the bearing is 1 C . higher than that of surround- ing objects, what will be the increase in temperature caused by the heat produced at the bearing? Ans., 19 0< 6 C. 59. Friction and Speed You will find it instructive to experi- ment with such a piece of apparatus as is re- presented in Fig. 28, designed to measure the friction between sliders of different materials and the cast-iron sur- face P. Here we have a pulley with a broad, smooth outer surface. On this surface lies a slide made slightly concave, to fit the rim of the pulley. On this slide we can hang different loads w by the arrangement shown in the figure, and the slide can only move a small distance in 72 APPLIED MECHANICS. any direction on account of stops. There is a fly-wheel to give steadiness of motion when the apparatus is worked hy hand. Sup- pose, now, that p rotates in the direction of the arrow. Friction causes the slide to move in the direction of motion until it is brought up by a stop. Now let weights be placed in the scale- Pig. 28. pan w until the slide is held in a position half way between the stops. Evidently the force of friction between the slide and p is just balanced by the weight in the scale-pan. With this appar- atus you can not only find the co-efficient of friction for two rubbing surfaces easily at any speed, but you can very quickly vary your experiments, s is a speed counter. To what extent I ought to be ashamed of the following facts I don't know. They are instructive. Successive generations of fitudents at Finsbury obtained results from an apparatus like Fig. 28. APPLIED MECHANICS. 73 It was arranged to be driven by the college engine at many different speeds ; and there was a speed counter. The slide had a much longer arc of contact than is shown in the figure, because a rocking motion, instead of sliding, was apt to be set up. The load was applied at a single point in the centre of the top of the slider. In every case when the load was kept constant the friction was greatest at low speeds ; it got less and less as the speed increased, and reached a minimum value ; after which it increased steadily as the speed increased, the rate of increase of friction with speed getting less towards our highest speeds. I thought these curves obtained by students well worthy of special study, and several times projected an investigation of my own, and urged others to take it up ; but I was too busy with other matters to give it much attention. Now, from Prof. Osborne Reynolds's explanation of the results of Mr. Beauchamp Tower's experiments one sees how very different the phenomenon is from what occurs between flat surfaces. Air is being pumped by friction into the space between the slider and pulley, and the pressure underneath the slider is greater than the atmospheric pressure, and varies from point to point, as we have proved by inserting little pressure-gauges. We are about to use now another piece of apparatus (Fig. 29), the slider being flat, and lying on a circular, flat horizontal plate, which may be kept rotating at any speed. The slider is prevented from moving much by stops, and friction is balanced, as in the above case, by a scale-pan. The apparatus is only now (July, 1896) being run for the first time. What sorts of results will be obtained from it I do not know. 60. Although, on the whole, in any machine the average forces of friction do not seem to depend much upon speett, and they increase in proportion to the load, and in other ways seem to follow the laws set down in Art. 55, when we make experiments on the friction at any one place in a machine we obtain inconsistent results. So long as our theory of an action is wrong, our experi- ments give rise to what are called inconsistent results. On the hypothesis of Art. 55 the mathematicians have built a science, and thousands of examples and exercises have been invented to illus- trate it. The exercises and examples are valuable to the engineer ; but he must remember that they have been invented by mathema- ticians for the training of mathematicians, and he must exercise caution in using the results. I give some examples in Art. 58. Two substances, when they really touch, get welded together ; and this seizing seems to occur in some journals and footsteps under heavy loads. Bodies said to touch or rub on one another are really separated by a layer of air or other fluid. A slider like c (Fig. 24) is separated from the table AB by a layer of air; and the greater the load, the less is the thickness of air. Students will find it very interesting to study the friction between two scraped surfaces in the workshop. If one plate is laid down on the other, there is usually very little friction, because there is a thick layer of air separating the surfaces. By putting on a load, and giving small sliding motions, we can make the layer of air very thin so thin, indeed, that when the toj> -?* 74 APPLIED MECIIANK plate is lifted the bottom one sticks to it, and lifts also, partly because there is a partial vacuum between them, partly also, prob- ably, because of molecular attraction seeing that it occurs even in a good vacuum. Now, when, through the one plate having lain on the other a considerable time, or through pressure, we get the layer of air very thin, it is found that there is considerable resistance to sliding. In fact in this case, where we might expect to find the phenomena of friction assuming their simplest form, we find what seem to be the most inconsistent results. When c (Fig. 24) slides, it seems as if fresh air were being carried into the space between the surfaces, keeping them apart, and that the greater the velocity Fig. 29. of rubbing the more air is carried in, so that the separation between the surfaces is proportional to the velocity of rubbing. When we come to discuss fluid friction, and reflect that all the friction we know of is really fluid friction, we shall not be aston- ished that the laboratory results from the rubbing of solids are sometimes inconsistent-looking we shall wonder greatly that the above-mentioned law should be even approximately true. But I cannot think the effect merely one of the fluid, there is also molecular attraction. In any lubricated bearing which is not kept flooded with oil, the rules called the laws of solid friction are found to be approximately true that is, we may take the load on the bearing in pounds, multiplied by a coefficient p., as representing a force of friction ; and this, multiplied by the distance of rubbing in^ feet, is the mechanical energy converted into heat by friction, /u is less as the temperature is greater, partly because the viscosity APPLIED MECHANICS. 75 of a fluid diminishes with temperature (see Art. 63) ; "but it is not only because of this, for the body of the lubricant also alters with temperature that is, it tends more to get squeezed out of place. Of course the friction depends greatly upon the nature of the lubricant, and the phenomena are really so very complicated that in the present state of our knowledge the reader must perforce be satisfied with the rough general law that I have mentioned. That the supply of oil, however small, shall be continuous, and not inter- mittent, is regarded as the most important condition in the lubrica- tion of bearings. The lubricant ought to suit the nature of the load. Thus great body is necessary when the loads are great, so that the oil may not be squeezed out; and greases and solid lubricants, such as soapstone and plumbago, must be used for very heavy loads. It is also usual to cast plugs of white metal and other soft alloys in recesses of the step, and in some cases to line the whole step with such a soft alloy. As for the detailed construction of pedestals, hangers, A frames, and other supports for shafting, this is to be learnt in the drawing- office and shops, and it would be useless to refer to it here. 61. As it has been found that with some kinds of material the statical friction that is, the friction which resists motion from rest is somewhat greater than the friction of the surfaces when actually moving, experiments have been made to deter- mine whether, at very small velocities indeed, with such materials, there is not a gradual increase in the friction. It is known that at ordinary velocities the friction is much the same as at a velocity of '01 foot per second. We have reason to believe that with metals on metals and air between, there is the same friction at all velocities, even down to one-five- thousandth of a foot per second, whereas with metals on wood the friction increases gradually as the velocity diminishes, until when the velocity is 0, the friction is what we call static friction. Again, at very high velocities it has been found that there is a very decided diminution of the coefficient of friction between a cast iron railway brake and the wrought iron tyre of a wheel. The coefficient was '33 for very slow motion, '19 for a speed of 29 feet per second, and *127 for a speed of 66 feet per second. It has also been observed in these railway brake experiments that when a certain pressure is applied for a short space of time the friction diminishes. All such results as these, however interesting they may be to the railway engineer, tell us nothing about what I have hitherto called friction, because I have supposed the rubbing surfaces to remain unaltered, whereas these railway brakes are rapidly worn away, and the effects of abrasion and polishing are of an 76 APPLIED MECHANICS. utterly different kind from the effects of friction of which I have hitherto been speaking. 62. We must remember that although friction leads to waste of energy, all the energy spent in overcoming friction being converted into another form of energy called heat, still the force of friction is very useful. The weight resting on the driving-wheels of a locomotive engine multiplied by the co- efficient of friction between the wheels and rails represents the greatest pull which the engine can exert upon a train. Suppose the weight on the driving-wheels to be 15 tons,. and that the co- efficient of friction of wrought iron on wrought iron is about -2, the greatest pull which the locomotive can exert is 15 x 0*2, or 3 tons. If the train, including the locomotive itself, resists with a greater force than this, the driving-wheels must slip ; if the train resists with a less force than this, there is no slipping, the wheels simply roll on the rails. Again, it is the friction between the soles of our feet and the ground that enables us to walk ; friction enables us to handle objects ; friction enables a nail to remain in wood ; friction keeps mountains from rolling down. 63. Fluid Friction. I have been considering the friction between solid bodies only. The friction between liquids and solids or between liquids and liquids is of a very different kind. If a man attempts to dive into water unskilfully, and falls prone, you know that the water offers a very considerable resistance to a change of shape. Now this is mainly the resistance that any body offers to being rapidly set in motion. If you came colliding against the end of the most frictionless carriage, you would also experience its resistance to being suddenly set in motion; whereas the constant steady resistance to motion which the carriage experiences when moving with a uniform velocity is called friction. What I wish rather to refer to is the resistance to the motion of water in a pipe, the resistance to the steady motion of a ship. In nearly all ordinary cases the motion is complicated and difficult to study. The simplest motion is in plane parallel layers. Imagine two infinite plane parallel boundaries with the fluid between : one of the boundaries at rest, the other moving with uniform velocity v in its own plane. Imagine the fluid to stick to each boundary. If b is the distance between them, the tangential force per unit area required to keep up the motion is p v 4- b if p is the coefficient of viscosity. Theory shows that p ought to be constant if the APPLIED MECHANICS. 77 motion is truly in plane layers. As we cannot experiment with infinite surfaces, I thought that I could approach the condition most nearly with the apparatus shown in Fig. 30. F is a hollow cylindric body supported so that it cannot move sidewise, and yet so that its only resistance to turning is clue to the twist it would give the c suspension wire, A. c c is water or other liquid fill- ing the annular space be- tween the cylindric sur- faces D D and E E, and wetting both sides of F. When the vessel D D, E E is rotated, the water mov- ing past the surfaces of F tends to make F turn IJ round, and this frictional Fig. so. torque is resisted by the twist which is given to the wire. The amount of twist in the wire gives us, then, a measurement of the viscosity of liquids, and investigations may be made under very different conditions. The above apparatus was designed and partly constructed in Japan in 1876. Experiments made with it by Finsbury students on olive oil are described in the Proceedings of the Physical Society of London, March, 1893. At constant temperature below a certain critical speed, I found that the friction was proportional to the velocity, so that p could be found. At that critical speed I found that there was a sudden change in the law, and above that speed the friction is pro- portional to a higher power of the speed than 1. We know that above the critical speed the plane motion which I described above would become unstable, and eddies would be formed. From a theoretical point of view it is curious* that * Some experiments of Mr. D. Baxandall, not yet published, show that the forces of friction that is, the resultant forces applied to the solid bodies which form the boundaries of a mass of fluid to maintain relative motion are strictly proportional to the relative velocity at small speeds, and there is always a critical speed above which the friction is proportional to a higher power of the velocity. We have tried surfaces arranged like those of churns, with paddles and curiously shaped vanes, and the law is always true. With mixtures of glycerine and water, the higher power of the velocity above referred to depends on the proportions of glycerine and water. 78 APPLIED MECHANICS. the phenomenon should have been so marked between my cylindric surfaces, because even at slow speeds cylindric motion ought to be unstable inside a fixed cylindric surface that is, in the inner part of my trough. At speeds below the critical, I measured fi at many different temperatures, and noted the rapid decrease in it as the temperature increased. Very interesting observations may be made at small speeds by immersing similar and equal heavy discs of brass in air, water, and oil, suspending them by fine steel wires. (See Fig. 30.) When the suspension wires are twisted and let go, the bodies vibrate like the balance of a watch. But it is only the one which vibrates in air that goes on vibrating for a long time ; the one in water keeps up its motion longer, however, than the one in oil, showing that there is more f fictional resistance in oil than in water, and more in water than in air. The rates of diminu- tion of swing or the stilling of the vibrations tell us the relative viscosities of the fluids. If, by means of a pointer or mirror attached to the wire, you observe the various angular displace- ments, noting the time for each, and then plot your observa- tions on squared paper (as in Art. 53), you will find what is very nearly a curve of sines for the vibrations in air ; and for the different liquids damping curves, which show the effect of friction in the liquids. Similarly, the rate of diminution of SAving of the vibrating fluids in U tubes, one containing water and the other oil, tells us about the relative co-efficients of viscosity of the liquids. 64. The motions in these cases are not so simple as in the case which I considered experimentally. The question of the resist- ance to the passage of fluids through pipes is one which has attracted much attention, and the results of experiments seemed very inconsistent until Professor Osborne Reynolds considered the problem. It was known that the pressure difference at the ends of a level uniform pipe necessary to produce a certain flow was proportional to the length of the pipe, and it was usual to say that the force of friction was proportional, as in all other cases of fluid friction, to the wetted area ; it is quite independent of the pressure ; it is proportional to the velocity of the water when the velocity is small, but at high speeds it increases much more quickly than the speed. Thus, as I said in the first edition of this book, of water flowing in a certain pipe, " at the velocities of 1,2, 3, etc. inches per second, the APPLIED MECHANICS. 79 friction is proportional to the numbers 1, 2, 3, etc., whereas at the velocities of 1, 2, 3 yards per second the friction is pro- portional to the numbers 1, 4, 9, etc." At small velocities, three times the speed means three times the friction ; \vhereas at great velocities, such as those of ships, three times the speed means nine or more times the friction. We see, then, that friction in fluids is proportional to the speed when the speed is small, to the square of the speed when the speed is greater, and at still greater speeds the friction increases more rapidly than the square of the speed. The resistance to motion of a rifle bullet is proportional to the square root of the fifth power of the speed ; that is, a bullet going at four times the velocity meets with thirty-two times the frictional resistance from the atmosphere. (See Art. 68.) Again, it has been found that the friction is much the same whatever be the pressure. Thus it is found that when the disc and liquid apparatus is placed in a partial vacuum or under considerable pressure, there is exactly the same stilling of the vibrations. This fact is illustrated by the apparatus, Fig. 31. Water tends to flow from vessel A to vessel B, through the long tube. Whether the tube is in the position shown in Fig. 31, or in the position Fig. 32, or is acting as a syphon, we find the same flow through it ; the same quantity of water passes through it per second, although the pressure of the water in the tube in Fig. 81. the position Fig. 32 is very much greater than in the position Fig. 31, or again when the tube is a syphon. In the apparatus actually used by Fig> 32< me, there is a stopcock in the middle of the tube, and by nearly closing it one is sure that the friction occurs at the place where the pressure is 80 APPLIED MECHANICS. greatest in Fig. 32. The comparison is most readily made by observing how long it takes for a certain change of levels to take place in the two vessels, repeating this several times with the tube in various positions, beginning and ending each experiment with the same difference of levels. Again, fluid friction, for even considerable velocities, does not seem to depend much on the roughness of the solid boundary. This seems to be due to the fact that a layer of fluid adheres to the solid surface and moves with it. Even when the disc of Art. 63 is indented, or when large grooves are cut in it we find practically the same frictional resistance. Comparison of the Laws of Fluid and Solid Friction. Friction between Solids. Fluid Friction. 1. The force of friction does not much depend on the velocity, but is certainly greatest at slow speeds. 2. The force of friction is pro- portional to the total pressure between two surfaces. 3. The force of friction is in- dependent of the areas of the rubbing surfaces. 4. The force of friction depends very much on the nature of the rubbing surfaces, their roughness, etc. 1. The force of friction very much depends on the velocity, and is indefinitely small when the speed is very slow. 2. The force of friction does not depend on the pressure. 3. The force of friction is pro- portional to the area of the wetted surface. 4. The force of friction at moderate speeds does not much depend on the nature of the wetted surfaces. 65. Molecular theory gives us the cause of fluid friction in such a fluid as air. Layers of fluid at different velocities are continually interchanging molecules by ordinary diffusion ; consequently, the relative motion is being destroyed, the rate of loss of momentum by one layer and gain of it by the other enabling us to state that the force required to maintain the motion is proportional to the surface of contact and to the relative velocity. In regard to friction in gases, the explanation is complete, the greater diff usivity at higher temperatures causing the viscosity to be greater also. Indeed, viscosity is proportional to the square of the absolute tem- perature. In the same way, if two trains were passing one another and the same number of passengers jumped from each train to the other, the trains would become more equal in speed ; there would seem to be a mutual frictional force between them proportional to the APPLIED MECHANICS. 81 rate of loss or gain of momentum per second. But in liquids there is less viscosity at higher temperatures, although there is greater diffusivity. This is probably due to the fact that mere diffusivity is all-important in gases, the molecules of which exert no forces upon one another except by collision ; wheroas in liquids, although the greater diffusivity at higher temperatures would tend to make them behave like gases, in regard to viscosity, forces are ^always acting between the molecules which resist shearing strain, and these forces get less as the temperature increases. Now, in any case of relative motion between the bounding surfaces of a fluid, beyond a certain velocity, motion in plane layers becomes unstable and sinuous motion sets in. This means that the surfaces across which interchange of momentum by diffusion may take place become greater in area ; so that above a certain critical speed I take it that we may expect almost any law connecting friction and speed. Theory shows that for any given shape of surface the critical speed will be less as the density of the fluid is greater, and it is less as /* is less. A very friction- less fluid is very unstable. I believe that all friction said to be between solid surfaces is to be regarded as taking place in the fluid which always separates such surfaces. This statement seems a mere truism. It is like many another yet to be made by discoverers in applied physics. As a matter of fact, the above table showing the utter difference in character between the phenomena of solid and fluid friction quite hid from everybody's view the fact that all friction must be a fluid friction, until Professor 0. Reynolds opened our eyes. He has given us in his lectures at the Royal Institution and in his paper published in the Transactions of the Royal Society the suggestion that it is to some extent in the solution of hydrodynamic problems we must look for an explanation of the curious phenomena of solid friction. I have already mentioned a curious phenomenon often brought to my notice in connection with the use of the apparatus shown in Fig. 27. Let me now describe some experiments made on the friction of journals. Probably everybody has been occa- sionally interested in curious results obtained when testing oils with the Thurston oil-tester (Fig. 33). Many of these will be found published in Mr. Thurston's book on "The Materials of Engi- neering," Part I. ; but every mechanical laboratory ought to be provided with the apparatus, that students may study the phenomena for themselves. In Hirn's experiments, made in 1855, he found that the force of friction was proportional to the square root of the product of load and velocity. In the experiments of Mr. Beauchamp Tower upon a steel journal with a gun-metal cap only, the cap being loaded, the lubrication being practically an oil batli, the friction was found to be practically independent of the load for loads so excessive as from 100 to 520 Ibs. to the square inch (the diameter being 4 inches and length 6 inches, the pressure is taken as the whole load divided by 24), and in all cases to be practically proportional to the square root of the velocity. If, instead of such excessive lubrication as we have in an oil,bath, there was only the lubrication due to an oily pad pressed APPLIED MECHANICS. Fig. 83. the journal below, the ordinary law assumed for the friction oi solids was found to he approximately followed. In collar hearings, as in the thrust hearings of a propeller shaft, he again found that the ordinary laws of solid friction are fairly well followed, and that only very much less pressures (75 Ihs. per square inch at high speeds and 90 Ihs. per square inch at low speeds) were possible without seizing. These curious phenomena have "been completely explained hy Professor O. Reynolds. They depend upon the APPLIED MECHANICS. 83 carrying of the lubricant into the space between the step and the journal, and if there is not an oil bath and the motion is all in one direction, the lubricant leaves the place where it is most wanted and the journal seizes at comparatively low pressures ; whereas if there is such an irregularity of motion or reversal of motion as helps the lubricant to maintain its place, very great pressures may be employed. Thus in crank pins and railway axles pressures as high as 400 Ibs. per square inch with sperm oil and 600 Ibs. with mineral grease have been used ; and, indeed, in slow-moving steam engines nearly double these pressures have been used. It is well to remember that at the place where the journal most nearly approaches the step the pressure in the oil becomes very great, and if there is an opening there the oil is forced out. It is now quite common to employ a force-pump to pump oil at great pressure into these parts of the bearings, and in consequence much higher loads on bearings are possible than used to be the case. In any case, we must try to understand the distribution of pressure in the bearing, so that in our endeavours to utilise an ordinary syphon-lubricator for example, we shall not be attempting impossible things. 66. It is in the drawing-office and shops that students will become acquainted with the methods in actual use for supporting horizontal shafts. Footsteps for vertical shafts, which give endless trouble in many high factories, are now in many cases water- or oil-borne, being converted into the rams of hydraulic presses having only a very small range of vertical motion, or, rather, so arranged that the lifting force of the fluid shall always be less than the weight of the shaft. 67. It is when great forces have to be overcome slowly, and particularly with a long translational motion, that water-pressure machinery shows itself most greatly superior to other machinery, for friction seems to be nearly independent of pressure. But if in any place the water is set in rapid motion there is internal friction and ^waste of energy. Where fluids move so slowly that the friction is proportional to the velocity, we seldom consider it in pur engineering work. At the valves of pumps and in pipes it is usually, on the whole, economical to let energy be wasted in water friction. There is a certain relationship between velocity, v, and diameter, d, of pipe for a particular fluid which causes a certain v to be critical. Below that value of v the water flows in straight streams ; at that critical value the beautiful straight lines which Professor Eeynolds shows coloured with aniline dye suddenly break up into confused, smoke-like eddying cloud. Below the critical velocity the total pressure difference or friction, as we may call it required to keep up the flow is proportional to the velocity. Above that critical velocity the friction is proportional to a power of the velocity which varies from 17 to 2, depending upon the nature of the material of the pipe. 68. The mathematical investigation of the resistance to the pas- sage of a body through a viscous fluid is so difficult that we have almost no results which may be relied upon. Without viscosity 84 APPLIED MECHANICS. there would be no resistance to steady motion, whatever tho shape of the object. It is difficult to imagine that there would be no propelling force on a sailing-boat if the air were frictionless, and yet this is so. Even in the case of a ship, experiments on which have been going on continuously since ships were first built, our knowledge is very incomplete. Roughly, we may take it that resistance is generally proportional to square of speed. In the case of shot this law holds, projably up to speeds of 300 feet per second ; from 400 to 1,000 feet per second the resistance is possibly proportional to. the 2^ power of the speed. Beyond 1,100 feet per second we may take, F being in pounds, d the diameter of a shot in feet, v the velocity in feet per second, r =fd 2 (v 800), where f 3 for spherical and 2 for elongated shots with ogee-shaped heads. The velocity is greater than that of sound, and probably it is to this that the change of law is due. The fact that even in the steadiest winds there is pulsation, causes scientific speculation about wind pressure to be difficult. 69. Reynolds has deduced from hydrodynamics the rational formula LB TC P 2 ~ n D"- 3 v n / A . . . . (1) as the loss of energy per pound of fluid passing through a pipe of length L feet and diameter D feet at v feet per second. (I have reduced his numbers to suit the foot us the unit of length.) The index n is 1 for velocities below the critical velocity, v c '039 P/D, and n varies from 1'7 to 2 at higher velocities than the critical. p is proportional to the co-efficient of viscosity, which changes with temperature. In the case of water he takes p - 1 -r- (1 - -0336 6 + -000221 2 ) .... (2), where 6 is temperature Centigrade. A = 1-917 x 10 6 ; B = 36-8. Note that the critical velocity depends upon the temperature and size of pipe. Thus, for a tube T \jth of an inch in diameter, the critical velocity is 4-65 feet per second ; for a pipe 1 inch in diameter the critical velocity is -465 feet per second; for a 6-inch pipe (d 0-5) the critical velocity is -077 feet per second. In all practical hydraulic cases the critical velocity is exceeded, and for a general rule, with cast-iron pipes in actual use, we usually take n = 2. In this case, in (1), the influence of p, the temperature term, is unfelt ; that is, in practical hydraulic work, temperature has no important influence. The formula now becomes LB 2 t> 2 /AD or '0007 Lfl 2 /D .... (3). As a mnemonic for this simple formula, let the student imagine that a solid prism of water of length L is moved along a pipe rubbing all round its perimeter, the friction being proportional to the square of the velocity and to the area of the rubbing surface. Thus, if * is the wetted perimeter, LS is the area and the force of friction is Lsv 2 ; that is, if p is the pressure difference which produces the motion and A is the area of cross-section, pA. ocLSfl 2 . The loss of energy per pound being proportional to p, this oc LV S . Now, A/* is called APPLIED MECHANICS. 85 the hydraulic mean depth, m, of any channel, and we find loss of energy per pound = cLv^/m .... (4). In the case of a round channel full of water Comparing (4) with (3), the formula of Osborne Reynolds corre- sponds to c = -000175. The constant in the formula (3) agrees with that of D'Arcy for small pipes. Reynolds has compared (1) with D'Arcy's experi- ments as well as with his own, from the smallest sizes of pipes to 20 inches diameter, and finds that there is practical agree- ment. D'Arcy's pipes had joints which somewhat vitiated the results. Reynolds gives for n the values : Lead- jointed pipes, 1-79; varnished, T82 ; glass, T79 ; new cast-iron, 1-88; incrusted pipe, 2-0 ; cleaned pipe, T91. This formula of Reynolds is rational, and suits every imaginable size of pipe, and I prefer it. But that of D'Arcy is more commonly used for pipes of from 3 inches to 2 feet in diameter, and it is often used in academic exercises, in which, indeed, almost any loss of energy per pound of water (usually called loss of head} is expressed as / x the kinetic energy per pound of water, or / x . D'Arcy gives / = -02 ( \ = -02 l +_ 12D for a straight pipe of length L feet and diameter D feet. 70. Resistance to Rolling. When one wheel or cylindric body rolls upon another there is some conversion of mechanical energy into heat. The power lost seems, roughly, to be propor- tional to the force pressing the two bodies together, to the velocity of rolling, and to the curvature of the smaller of the two. We have very little experimental knowledge of the subject. In all probability the power wasted is proportional to the strain energy per second stored in the material, the waste being due to viscosity. When the velocity is very great, as at the driving-wheels of locomotives, secondary effects are pro- duced, waves of compression and extension travelling in the rim of the wheel and in the rail, with very curious results. In some experiments which I have made with great pressures between hard cast iron wheels rolling upon one anothei, there seems to have been much local heating just at the surfaces. At the end of some months of work a quantity of black dust had been produced, and each particle, when examined by the microscope, looked like a piece of slag. Besides energy wasted by changing strain in the material, there is slipping at the surfaces in contact. A student who 86 APPLIED MECHANICS. remembers that when a strut is compressed it swells, and when a tie bar is lengthened it gets thinner, can study the " creep " which occurs both here and in belting, for himself. Imagine points one inch apart upon the rim of an iron wheel, and another set upon an unstrained plane indiarubber surface. Now draw the wheel as it indents the surface. As in Fig. 34, points 1, 2, 3, and 4 are further apart, and points 6, 7, 8, 9 are nearer together than in the unstrained con- dition, and hence the metal and indiarnb- ber surfaces slide upon one another. No ordinary material F i gi 34. coating seems to have much effect in pre- venting the sliding. A cast iron wheel on planes of cast iron, boxwood, and on indiarubber, seems to have frictional resistances to rolling in the proportion of 1 : 2 : 8. The loss of energy in belting is partly due to this, partly due to energy wasted in bending and unbending the bolt. Both the lessened distance of rolling and the slip of a belt seem to be proportional to the power transmitted. M. Raffard has actually used a dynamometer on this principle. He transmits his power, to be measured, by means of a thick indiarubber belt through two equal pulleys; the difference of speed of these pulleys is taken to be a measure of the power. The slip is quite noticeable when speed cones are used in driving machines at various speeds with variable power, for the actual speed has to be carefully measured ; calculation from the known sizes of the steps of the cones giving in- accurate results. When pressures are not too great, as in the ball bear- ings of cycles and some machine tools, there can be no doubt whatever of the ease of running. Fig. 35 shows the ordinary adjustable ball bearing used in bicycles. D is the fork and H the hub of a wheel. The spindle A is fixed to the fork D. One of the hard steel cones c is tight against a shoulder v ; the other c' is tightened just enough to let the wheel revolve easily, and then it is locked by the lock-nut K. The linings of the shaped APPLIED MECHANICS. 87 ends of c are hardened steel, and a number of hard steel balls are placed between. Fig. 36 is an enlarged drawing of the Fig. 35. Fig. 36. ball and the linings, showing that the radii of curvature of the ball, cone, and cup are different; the friction of the bearing will be much less than if the radii of curvature were nearly the same. A very little oil getting inside the hub finds its way to the balls. Some experiments (Proc. I. C. JE., Vol. 119, p. 456) on rollers between flat cast-iron plates, give as the resistance in pounds to rolling c/ \/r where r is radius in inches and c=-0063 for cast-iron, 0120 for wrought iron, 0073 for steel. These are 13 per cent, greater for wrought iron plates and 13 per cent, less for steel plates. The crushing load in pounds on a wrought iron roller seems to be 444 r per inch of its length. Several experimenters are now engaged in procuring for us more exact information on rolling friction. In using roller bearings on carriages, it has been found that there is a diminution of from 23 (on gradients of 1 in 20) to 60 per cent, (on gradients of 1 in 140) of the tractive effort required with ordinary bearings. In ordinary machinery, the loss of energy by friction has been found (in one experiment) to be less than one- third of what it is with good ordinary bearings. Oil is only needed to prevent rusting. 88 CHAPTER V. EFFICIENCY. 71. Mechanical Advantage. In books on mechanics you will usually find that when simple machines are described, they are only considered in relation to their Mechanical Advantage. That is, suppose a small weight E, now usually called the effort, is able by means of the mechanism to cause a larger weight, R, usually called the resistance, to rise, the ratio of R to E is called the mechanical advantage. Now, in nearly all cases you will find that, when there is a mathematical investigation of a machine, the assumption is made that there is no friction. I have already shown you that the problem of taking friction into account is a very difficult one. But, as we have seen, a practical man can experi- ment on the effect of friction; and, happily for us, he obtains re- sults which are gener- ally very simple. Let the reader make a few experiments himself, or let him by means of squared paper find the relation between E and R from the following results, taken from a crane, Fig. 37, whose gearing was well oiled, Pig. $7. and whose handle was replaced by a grooved wheel, round which was a cord supporting E : APPLIED MECHANICS. 89 R. Resistanc Overcoi 10011 200 300 400 500 600 700 800 E. 3 just Effort jus ne. Overcome 1 IM, '' ' * '" '' 8-5 t able to lesistanco. Ibs. > y > M u 12-8 17-0 21-4 25-6 29-9 34-2 ,, 38-5 We found that E fell forty times as rapidily as R rose, and you may have imagined that the mechanical advantage was forty, or that a weight, E, could lift a weight, R, forty times as great as itself. This would be true if there were 110 friction ; but we see that in practice it is not the case. Plot the above values of E and R on squared paper, and you will find that, if the weight R is increased 1 lb., E must be increased '0429 Ib. ; and also that when R is 0, an effort E of 4-21 Ibs. is needed to cause a slow motion of the crane ; so that the law is E = 4-21 + -0429 R. Namely, multiply the resistance R in pounds by the fraction 0429, and add 4'21 : the answer is the effort required to lift R. When you have worked out this rule, employ it in finding how much effort, E, is required to lift a ton with such a crane. Answer, 100 -3 Ibs. The word power is generally, but very unscientifically, used as the name of the force which I have called E, the effort. Power may, of course, be used in many senses by newspaper writers, but when used by the engineer it is a technical term, meaning the rate of doing work. If a weight of 1,000 Ibs. falls 100 feet in two minutes, it does 1,000 x 100 or 100,000 foot-pounds of work in two minutes, or 50,000 foot-pounds of work in one minute. Now, 33,000 foot-pounds of work done in one minute is called a horse-power, and hence our falling weight gives out 50,000 -f 33,000 or 1-5 horse-power. Ten horse-power means ten times 33,000 foot-pounds of work done in one minute. The idea, then, of power is an idea of work done in a certain time. 72. Economical Efficiency. Take any pair of numbers from the above table, say E = 8*5 Ibs., when R = 100 Ibs. Let us suppose that E is moving at the rate of forty feet per second, then we know that R is rising at the rate of one foot per 90 APPLIED MECHANICS. second. E is giving out the power 8 -5 x 40, or 340 foot-pounds per second ; R is receiving 100 foot-pounds per second. The ratio of the power usefully employed to the power given to the machine is called the efficiency of the machine, so that our crane has an efficiency 100^-340, or *294. Sometimes the efficiency is put in the form of a fraction ; sometimes we say that it is 2 9 '4 per cent., meaning that the machine employs usefully 2 9 -4 per cent, of the energy given to it. Now take another pair of numbers, say E = 38 -5, R = 800, and let E fall forty feet in one second, a^ before. We now get as our answer '519 that is, more than half, or 51*9 per cent, of the power given to the crane is usefully employed. We see, then, that as the power given to the crane is greater, the efficiency is also greater. This arises from the fact that the friction of the unloaded crane is always entering into the calculation ; and if we take the case where no resist- ance, R, is being overcome, and E must be 4 -21 Ibs., we shall find an efficiency, 0, because work is being given to the crane, and none is coming out usefully. You will always find that the power usefully given out is a certain fixed fraction of the total power given to the machine, minus the power required to drive the crane at the given speed when it is unloaded. Choose some speed, say that E falls forty feet per second ; find the total power or 40 E ; find the usefully employed power 1 x R for every case of the above table. Plot your answers on squared paper, and you will find this rule : if P is the power required to drive the crane at the same speed when unloaded, if u is the useful power, and T is the total power supplied : T = l-761 U + P O 73. In some machines there have been attempts to ap- portion the loss of power to various parts. As a rule, these are very speculative. Roughly, we may say thac the frictional loss in a steam-engine may be divided in the following propor- tions : Crank-shaft bearings and eccentric sleeves, 1 ; valve, if unbalanced, '6 ; valve, if balanced, -05 ; piston and rod, -4 ; crosshead and slides, *2 ; crank pin, -14 ; total loss because of air pump, 0-3 to 0'5. It is fairly obvious that on account of the great weight of the fly-wheel and other parts much of the energy loss in a steam-engine is the same whether the engine is giving out much or little power, and in many cases for purposes of calculation this assumption may be made. APPLIED MECHANICS. 91 74. The work of this and the succeeding two pages is more suggestive than any other work in this boo_k. From the follow- ing results of experiment with a gas-engine, show, by plotting on squared paper and correcting for errors of observation, that if I is the indicated horse-power, B the brake horse- power, G the cubic feet of gas per hour, including what is used for ignition, then G = 20-3 i + 8, G = 20-4 B -f 45, B =.i 1-8. I B G G/I O/B Efficiency. 13-4 11*6 280 20-9 24-1 166 10-2 8-4 216 21-2 257 155 7-3 5-4 156 21-4 28-9 123 4-6 2-9 104 22-6 37-9 112 1-8 45 25-0 ~ When we plot the values of G and I and of G and B on squared paper, we find points lying very nearly in straight lines. Assuming that they ought to lie in straight lines, we find the above laws satisfied. Let the student fill in the columns showing G/I and G/B. Also from the brake horse-power, and knowing that one cubic foot of gas per hour means an actual supply of energy of a quarter horse-power, let him fill in the column of efficiencies. [The calorific power of one cubic foot of average coal gas may be taken to be from 500,000 to 540,000 foot-pounds; of Dovvson gas, it is about 124,000 foot-pounds.] Again, taking the following experimental results from an oilengine (one pound of oil being taken to give out 11,700 centigrade heat units in burning), i being the indicated, B the brake horse-power, and o the pounds of oil used per hour. I B o O/I O/B Efficiency. 7-41 6-77 6-4 86 95 128 8-33 6-88 6-8 82 99 122 4-71 3-62 5-0 1-06 1-38 088 0-89 3-1 3-48 92 APPLIED MECHANICS. Let the student show that o = 0*505 I + 2-62, o = 0'52 B + 31, B = 0-98 i - -89. Let him also fill in the columns o/i and O/B. Prove that 1 Ib. of oil per hour means 8 '27 horse- power actually supplied, and fill in the column of efficiency. 75. Accept the following measurements. A steam-engine employed in driving a dynamo machine delivering electric energy to customers, each load being kept steady for four hours, each measurement being the average of the results obtained during the four hours. I is indicated horse-power ; B the brake horse-power measured by a transmission dynamo- meter ; the electrical horse-power E is obtained by multiplying amperes and volts to get the power in watts and dividing by 746 ; c is the coal used per hour ; and w is the weight of steam used by the engine per hour. The governor acted upon the throttle-valve, and not upon the cut-off. B Amperes. Volts. E w c 190 163 1,050 100 143 4805 730 142 115 730 100 96 S770 544 108 86 506 100 69 3080 387 65 43 219 100 29 2155 218 19 1220 First plot the values of I and w, I and B, E and B, I and c on squared paper. It will be found that there is approxi- mately a linear law in every case. See if you get some such laws as w = 800 + 21 1 B = 95 i - 18 B = -93 B - 10 c = 4-2 i - 62 Now produce a few more columns of numbers and study them. Give w -j- i and c -f- i. Give w -i- B and c -f- B. Give w -4- E and c -i- E. Also give w -f- c. 76. Students may compare the above results with the following average measurements made at an electric supply station in 1891 : APPLIED MECHANICS. 93 E w c Average for 7 hours 11 a.m to 6 p.m. 80-3 57-1 3,268 552 Average for 6 hours. 6 p.m. to midnight. 2277 163-2 7,122 742 Average for 1 1 hours. Midnight to 11 a.m. 37 23-64 2,143 232 24 hours. 11 a.m, to 11 a.m. 97-3 68-3 3,718 453 Here it will be found that although the load was constantly varying even when the averages for the 24 hours are taken with the others, we have linear laws between i, E and w, w=1150 + 26-25 i, and E =-72 i 2. But c does not follow a linear law with the others. The reason lies in the fact that a spare boiler was used during part of the time, and there is consequently a greater consumption of fuel than if one or two boilers had been used the whole time. Since we have considered fuel consumption in the above exercises, it may not be out of place to introduce here some figures from the testing of a water- tube boiler : Water evaporated per Steam per hour from and at 100 C. per Ib. of coal. Coal per sq. ft of grate per hour. sq. ft. of total toiler heating surface per hour. This is not re- w. duced to 100 C. 13-40 7'74 1-24 103 12-48 18-6 3-20 233 12-00 29-8 4-70 357 10-29 66-8 8-50 686 If w = steam per hour per square foot of grate, f = fuel per hour per square foot of grate. Plotting w and /"on squared paper, we find a fair approach to a linear law, w = 45 + 9-78/ or -3- = jr 94 APPLIED MECHANICS. 77. If c is the total cost per hour when the useful horse-power p is being sent out by an hydraulic or electric or other supply com- pany (c includes interest and depreciation on first cost, rent, taxes, repairs, wages, stores, coal, water, office expenses and management), and if it is found that there is a simple law like c = a p + b, where a and b are constants, prove that the average cost per hour is calculated from the average power in exactly the same way as the real cost c in any hour is from the P during that hour. For if t is time in hours, then the cost during the year is, if T is the number of hours in a year, I G.dt = I OP + V) dt = a\ p Jo Jo Jo is- 1 p.dt T Jo p dt + b T. Hence the average cost per hour is Now - I P .dt is the average power, call it p m , and we see that if, T Jo the average cost is a ~e m + b. The average power delivered in a day or year, divided by the maximum power, is called the daily or yearly load factor. If / is the load factor and PJ is the maximum power, then average cost per hour = a p a -f- / + b. 78. An electrical company has arranged for a maximum output of 1,000 horse-power. It is found that the total cost per hour in pence c is c = 0-8 P + 350. If ju pence is charged for every horse-power hour sent out, what is the yearly profit when the average power sent out day and night is P TO ? Ans. Subtract from p P, M the average cost per hour to the company which is 0-8 p m + 350, and the profit per hour is (p - 0'8) r m - 350. The profit per annum in pounds is therefore this divided by 36-5. Exercise. What charge per horse-power hour will give just no profit ? Ans. x = 0-8 + 350/p TO . Thus if P TO has the following values, we have the values of x Fro 1,000 500 200 100 X 1-15 1-50 2-55 4-30 79. We have seen that if a force B lb., acting through e feet, overcomes a force n lb., acting through r feet, instead of the non- frictional law E= . . . . (1), we find experimentally some such law as It is evident that the constant A is a frictional resistance from the mere weight of the parts of the machine ; a is always found to be APPLIED MECHANICS. 95 greater than -. The work done by B in the ^distance e feet ie B e or ea R -f A e , and the useful work is r R, so that the r Efficiency x = ean-\- A.e i _i_ ^_ ~*~R It will be observed that the larger R is, the smaller is > or a& the more insignificant is the term due to weight of parts of machine, and the more nearly does the denominator approach unity. However great R may become, the efficiency cannot exceed . ; and as a is always greater than -, the efficiency is always less than unity, as was to be expected. Denote a by , * > where k is always less than 1. Then the K B efficiency can never be greater than k. If e and r are feet per second, then ^-^ and ^-^- represent the horse-powers. Let us call ^v^ the total horse-power T, and ^- the useful horse-power ooO ooO u, and we shall call T u the lost or wasted horse-power L. Then as 80. It is interesting to know, from such examples as we have studied in Arts. 73-76, that if T is the indicated horse-power of a steam-engine, and u is the brake horse-power, or what is given out usefully by the crank shaft, and if this shaft drives a dynamo machine, and E is the electrical horse-power given out ; or if the shaft drives a pump, and E is the effective horse-power of the pump, we always find (probably with only approximate accuracy) a linear law connecting T and u, and u and E, and therefore also connecting T and E. Furthermore, in a steam- engine which does not vary its period of cut-off that is, the regulation is by throttling or the boiler-pressure changes if w is the poundage of steam per hour, there is a linear law connecting any two of the quantities w, T, u, and E. If we write (1) as v = ki a, orT= r u + r a, fC K> it is curious that in so many of our machines in which the transmission of power is by mere shafting k should be so much less than unity a/3 it is. This occurs because a me v e torque is very seldom applied to a shaft when power is being given or taken. When the power is supplied or taken off by a belt, the power is proportional to the difference of pulls in the belt, whereas the loads on the bearings and the friction are nearly constant. Such kinds of driving tend rather to increase a than to diminish k. All spur or bevel gearings 96 * APPLIED MECHANICS. tend to diminish k, and only by their mere weight do they tend to increase o. The poor efficiency of ordinary lines of shafting is a matter to which few people seem to have paid any attention ; and this is, I think, altogether due to the absence of dynamometer couplings or other means of forcing upon the attention the actual amounts of power which are being transmitted at each place. Electrical engineers have very accurate methods of measuring the power given out by their dynamos, and hence they are concerned as to the actual mechanical power supplied to them. It was almost altogether due to this that direct driving of dynamos from engines took the place of indirect methods; and, indeed, I may go further, and say that it was due to this that the great improve- ments have taken place in steam-engine manufacture and working during the last fifteen years. If even the practical engineer is beginning, therefore, to think of economy, it is greatly due to the fact that electricity is paid for at so much per unit of energy; although, no doubt, much is also due to the fact that economy of coal is very important in ocean-going steamships. 81. In any machine in which a small effort, E, overcomes a great resistance, R, we usually find that the frictional loss of energy in the machine is almost altogether dependent upon a. Thus in the inclined plane (Fig. 25), with E parallel to the plane, the friction is independent of E. When in the inclined plane of small inclina- tion E is horizontal, the friction is almost altogether dependent on the load to be lifted ; and it is so in a screw-jack and in the differential pulley-block. It is not so much the case in cranes, unless such gearing as worm-gearing is employed. Let us, then, suppose that the loss of energy in the machine is altogether due to R. Then in the direct use of the machine that is, E overcoming R if the efficiency is only 50 per cent., the loss of energy is equal to R r (if R is lifted r feet), R r being also the useful energy. Consequently, if we try by increasing R to R 1 to make the machine reverse, the loss is always nV; or the whole energy would be wasted if the machine could be supposed to move. If, then, the direct efficiency is equal to or less than 50 per cent., the machine cannot reverse or overhaul. In any machine there is some loss due to E, and also to the weight of the parts of the machine, as well as to R, and conse- quently the direct efficiency must always be less than 50 per cent. if the machine is not to overhaul. On the assumption that the lost energy, L, is where A is a constant, due to the weight of parts of the machine, and m and n are fractions, we have, in direct working, E 1 n Hence the direct efficiency r is (l-*)Br . ( + 1) ar + A APPLIED MECHANICS. 97 or i~ n ~l . Again* i 11 reversed working R V _ E e = L, R V E e = m RV -^ w E -j- A, H'r(l-m)-A l+ The reversed efficiency , = = ^f^ - - . (2). This is or negative when 5+* | ** ar > 1 - As R 1 may be made very great, our condition of non-reversibility must be m = 1 or > 1, Hence the direct efficiency must be equal to or less than ~ * . We therefore see that there is no general & -J- A/R f rule such as many people seem to believe in the result of mis- leading mathematics. We can hardly call it a general rule to say that, if m is equal to 1, R cannot overcome E, because m = l means that all R'S energy would bo wasted. 98 CHAPTER VI. MACHINES. SPECIAL CASES. 82. Blocks and Tackle. It is very good to have a general law telling us about machines in which there is no friction. That law vou now know. The mechan- ical work given to a machine is equal to the work given out by it^ unless it is stored up in the machine itself by the coiling of a spring or in some other way. But, besides knowing the law itself, it is well to know what it leads to in certain special cases. Take, for instance, a pulley-block, Fig. 38. It is evident here that if we have three pulleys in the block B, if the effort P acts through six inches, w will only rise one inch, and therefore P will balance six times its weight at w if there is no friction. The mechanical advantage is therefore six. This is a case in which it is less dif- ficult than usual to trace the loss of energy due to friction. If, when the pull of a cord is P O on the overcoming side of a pulley, the tension on the other side is given by PI = a P O b, where a is less than 1, and b does not depend much on the weight of a sheave, out rather on the vis- cosity of the rope, we may take it that Fig. 88. P 2 =rapj b, and so on. I shall leave it to students to work out an algebraic expression connecting w, the sum of p p P 2 , Pg, etc. , and P O . w will include the weight of the lower block. But it is good to take a numerical example. For instance, let P! = 0-9 P O - 3, the forces being in pounds. Then P 2 = 0-81 P O - 2-7 -3 = 0-81 F O 57, P 3 = 0-729 P O - 5-13 -3 = 0729 r - 8'13, P 4 = -6561 P O - 7-317 - 3 = -6561 r - 10'32, P 6 = -5905 P O 9-288 3 = '5905 P O 12*29, P 6 = '6315 P O 11-06 -3= -5315 P O - 14-06; Hence w = 4'217l P O - 53-50, or P O = 0'237 w + 12 7. APPLIED MECHANICS. 99 The efficiency x 167 . 237 w 1 + Fig. 89. Hence, however great w may be, the efficiency is less than 70'5 per cent. When w = 53 '6 the efficiency is only half this. 83. This is a fairly good example of the cumulative effect of friction. If we give power P O to a machine whose efficiency is e , tf this machine gives power to another whose efficiency is e^ and so on, the power given out by the last of a series of machines is P O x e x e l x e 2 x etc. If, as in transmitting power for a great dis- tance by means of ropes, all the contrivances are the same, we have the compound _ interest law of diminution of power. (See Exercise, page 232.) Students may find it interesting to study a machine or method of transmission of power in a manner allied to what follows in Art. 91 of this chapter. 84. Inclined Plane. Again, take the inclined plane, Fig. 39 ; w is a weight which may roll down the plane without friction, let us suppose ; P is the pull in a cord which just prevents w from falling. The cord is parallel to the plane. Evidently when w rises from level A to level B the cord is pulled the distance A B ; that is, w multiplied by the height of the plane is equal to P multiplied by the length of the plane. Thus, if w is 1,000 Ibs., and the length of the plane 10 feet for a rise of 2 feet, then ten times P is equal to 2,000, or p is 200 Ibs. Barrels and boilers are often raised along an inclined plane, ropes or chains held fast at the top of the 'plane passing round the cylindric object and back towards the top of the plane where force is applied to them. In this case P is evidently only half what is stated above. As the weight rolls on the double rope or chain there is not much friction. In Art. 90 we shall consider the effect of friction in the inclined plane. 85. The Screw. Again, suppose there is no friction in the 100 APPLIED MECHANICS. screw A B, Fig. 40 ; if it rises it lifts a weight say of 3,000 Ibs. Now, if the screw make one turn it rises by a distance equal to its pitch; that is, the distance (measured parallel to the axis) between two threads. Say that the pitch is '02 foot, then when the screw makes one turn it does work on the weight 3,000 x '02, or 60 foot-pounds. But to do this, P must fall through a distance equal to ; Fig. 40. the of the pulley A, about which I sup- pose the cord to be wound. Suppose the circumference of the pulley to be 6 feet, then p multiplied by 6 must be 60, or P is 10 Ibs. The rule, then, for a screw is this effort multiplied by circum- ference of the pulley equals resistance multiplied by pitch of screw. It is not usual to have a pulley and a cord working a screw ; it is more usual to have a handle, and to push or pull at right angles to the handle. Instead of the circum- ference of the pulley, we should take, then, the circumference of the circle described by the point where the effort is applied to the handle. Example. A. steam-engine gives to a propeller shaft in one revolution 60,000 foot-pounds of work ; the pitch of the screw is 12 feet. What is the resistance to the motion of the vessel? Answer: The resistance in pounds multiplied by 12 gives the work done in overcoming this resistance, and this work must be equal to 60,000 foot-pounds ; hence the resistance to the motion of the vessel is 5,000 Ibs. [It would be more correct to say that this is the work done per foot-travol of the vessel, assuming no slip of the screw.] Screws are used for many purposes. When used, as in bolts to fasten things, the threads are triangular in section. The Whitworth thread is shown in Fig. 41. A B is the pitch ; G n APPLIED MECHANIC*. 101 1 8 is -96 of the pitch, the angle H B j being 55 ; the corners are rounded, so that i B is '64 of the pitch. The Sellers thread used in America has an angle of 60. Fig. 42 shows a square thread. There is less friction and less wear with this form of screw, and it is used when accuracy of motion is important. As there is only half the amount of material resisting shearing, the square thread has only half the strength of a triangular thread. The Buttress thread of Fig. 43 has strength and accuracy a^ to motion; it is used when the important motion is in one direction only. A student has plenty of opportunity of examin- ing examples of the use of screws. To fasten things together we have many kinds of bolts and many forms of heads and nuts, and many ways of locking nuts. Other fastenings also, such as pins and keys and cottars, come I before his eyes every day Fig. 42. Fig. 43. in the workshops, and he must become familiar with them and their usual proportions both in the shops and the drawing-office. He must make a very careful examination of such a tool as a good screw- cutting lathe, and especially of the mechanism of the slide rest. The calculation of the proper change wheels for the cutting of a particular screw is a very simple matter ; but all so simple as it is, he must work out some examples. He must note the shapes of threads of screws for wood, for sand or mud, and for water; the use of screw piles as the supports of structures, the forms of screw-propeller for steam- ships and wind-mills, and screws used as fans for blowing air. 86. Wheel and Axle. If A and B, Fig. 44, are two pulleys or drums on the -same axis and having cords round them, a small weight, p, hung from A, will balance a larger weight, w, hung from B. For, suppose that one complete turn is given to the axis, P falls a distance equal to the circumference of A whilst w is rising a distance equal to the circumference of B. Hence Fig. 41. 102 APPLIED MECHANICS. p x circumference of A = w x circumference of B, or, what really comes to the same thing, p x diameter of A = w x diameter of B .... (1), or p x radius of A = w x radius of B. In practice A is often a handle and B a sort of barrel on which a chain may be wound. Or the axle may be compound, consisting of two parts, the diameters D and d being nearly equal, the rope being coiled round them in opposite directions so as to form a loop, upon which hangs a pulley. In this case (1) becomes D d. p x diameter or A = w x ^~ Or, again, A may not be on the same shaft as the barrel B, but may gear with it through spur and bevel-gearing. Thus, in a hand crane, the handle may turn many times for one turn of the barrel. Also there may be a snatch-block, so that when two feet of chain are coiled on the barrel, the weight rises only one foot. There is no mystery about tooth gearing, and anyone who has looked at a crane knows how by merely measuring the length of the handle, or, rather, the circum- ference of the circle described by the handle, and the amount of chain coiled on in one revolution of the barrel (this is larger than the circumference of the barrel itself), and counting the number of teeth of the driving and driven wheels, what is the velocity ratio, and, therefore, the hypothetical mechanical advantage. I write for men who go about among machinery, and the most illiterate workman knows well how speeds of shafts depend upon numbers of teeth. Also, all my readers have seen com- binations of barrels, snatch-blocks, blocks and tackle and crab- winches. A worm and worm wheel, or other kinds of screw gearing may also be imagined. Example. The handle of a crab or crane is 18 inches long; 20 inches of chain are wound on in one revolution of the barrel. The barrel is driven from the handle by a train of wheels driver 15 teeth, follower 64 teeth; driver 16 teeth, follower 60 teeth (the value of this train is said to be Fig. 44 APPLIED MECHANICS. 103 ^70 ~ ); so that the handle makes 16 revolutions for one of the barrel ; the chain lifts the weight through the agency of a block and tackle, with 3 sheaves below. When the barrel turns once, 20 inches of chain are coiled 20 on, and therefore the weight rises -~ ^ inches or 3J inches. The handle turns 16 times, and the hand moves through 16 x 36 TT inches. Hence the velocity ratio of hand to weight is ^ -- = 543. Often we have a means *jg provided of disengaging the spur wheels, driver 16, follower 60, and so we can de- j n crease the velocity ratio to 543 x or 60 about 145. 87. A differential pulley-block is shown in Fig. 45. When the chain E is pulled, it turns the two pulleys, or rather one pulley with two grooves, B and c. Now c is a little smaller than B, so that, although at D the chain is lifted, it is lowered at F. If the circumference of B is 2 feet and that of c is 1-95 feet, then, when E is pulled 2 feet, D is lifted 2 feet, but F is lowered 1'95 feet, so that there is 0*05 foot of chain less than before in the parts D and F, and the pulley G rises the half of this, or -025 foot If R is 2,000 Ibs., then 2,000 x '025, or 50, must be equal to the pull E multiplied by 2, hence E is 25 Ibs., or an effort of 25 Ibs. is able to overcome a resistance of 2,000 Ibs. The general rule, then, for the differential pulley-block is, effort E multi- plied by circumference of larger groove B is equal to resistance R multiplied by half the difference between the circumference of the two grooves B and c. You will find that this rule comes to the same thing effort multiplied by diameter of B is equal to resistance multiplied by half the difference between the diameters Fig. 45. 104 APPLIED MECHANICS. of B and c. The grooves are furnished with ridges to catch the links of the chain, so that there shall be no slipping. It is only when we experimentally measure the effort E which will slowly overcome the resistance R in this pulley -block as actually made, that we see how great is the f rictional waste of energy. It is in consequence of this that, however great the re- sistance R may be, it will not fall, even when there is no force exerted at E. This property of not "overhauling" makes the differential pulley- block the very useful implement which we know it to be in a machine shop. It is the characteristic of .any machine which has a very great velocity ratio that if its Fig. 46 Fig 48. efficiency is less than half, it will not overhaul (see Art. 81), and lifting machines which do not overhaul are often very convenient. 88. Equilibrium in one Position. In all the machines which we have hitherto considered, we could give motion without altering the balance of the forces, but there are many machines in which the mechanical advantage alters when motion is given. In such cases you will employ your general principle, but you must make your calculation from a very small motion indeed. For instance, in the inclined plane, if the cord which prevents the weight from falling is not parallel to the plane say that it is like M, Fig. 46 you will find that the necessary pull depends on the angle the cord makes with the plane. Now, suppose that the cord pulls the carriage from b to c, evidently the angle of the cord alters. The question is, what is P, that it may support w in the position shown in the figure 1 We APPLIED MECHANICS. 105 know that it will be different after a little motion, but what is it now ? Imagine such a very small motion from b to c to occur that the angle of the cord does not alter perceptibly, and now make a magnified drawing, Fig. 47. P has not fallen so much as the distance b c, it has only fallen the distance b a (c a is perpendicular to b a). In the meantime the weight w has been lifted the distance k c. Hence, w x k c ought to be equal to P x 6 a. Thus, if you measure k c and b a on your magnified drawing to any scale you will find the relation between p and w. Another way of finding the same relationship is this. "We know that the weight of w acting downwards, the pull in the cord, and a force acting at right angles to the plane, are the three forces which keep w where it is. Draw a triangle whose three sides are parallel to the directions of these three forces, Fig. 48, with arrow-heads circuital; then x and y are in the propor- tion of w and p. Here we have used the principle called " the triangle of forces " to find p. EXERCISES. 1. Find the force parallel to the plane required to draw a weight of 2 cwt. up a smooth inclined plane. Height of plane, 3 ; length, 5. Ans., 1-2 cwts. 2. In a screw-jack the pitch of the screw is f inch : radius of circle described by hand, 19 inches ; find the velocity ratio. It is found that a force A at the handle of 30 Ibs. will OA r ercome a weight of 2,300 Ibs., and one of 10 Ibs. will overcome a weight of 500 Ibs. ; what law connects A and w ? When w is 3,000 Ibs., what is A ? What is the efficiency ? Am., 318-47 ; A = j? -f 4| ; 37 j Ibs. ; 25 per cent. 3. The handle of a lifting-jack measures 24 inches in length; the pitch of the screw is f inch ; what force applied at the end of the handle would be required to raise a load of 22 cwt., the effect of friction being neglected? Ans., 6-125 Ibs. 4. Pitch of screw-propeller, 18 feet; slip, 10 per cent.; speed of ship, 15 knots ; find the revolutions per minute. What is the thrust if the actual horse-power spent by the propeller is 2,000, and the waste by surface friction is 30 per cent ? What is the torque in the shaft ? Ans., 93-83, 13-57 tons; 50 ton-feet. 5. The British Association rules for the pitch and diameter and index number of screw-threads for instrument work are Taking the values ofO, 1,2,3,... .12 for s, calculate p and d, and keep for reference in a table. Try to what extent the rule p = -08 d -+- -04 for the pitch and outside diameter of triangular screws agrees with the well- known Whitworth table. 106 APPLIED MECHANICS. 6. What must be the difference in the diameters of a compound wheel and axle so that the velocity of E may be eighty times that of R, the length of the handle being 2 feet? Ans., 1-2 inches. 7. The weight on a crane is carried by a snatch-block ; the chain goes to a barrel on which 17 inches are wrapped in one revolution. The barrel is driven by wheels, which give it 1 revolution for 15 revolutions of the handle, and the hand describes a circle of 21 inches radius ; what is the velocity ratio? Am., 232-8. 8. In a single-purchase crab the pinion has 12 teeth and the wheel has 78 teeth, the diameter of the barrel (or rather of the chain on the barrel) being 7 inches, and the length of the lever -handle 14 inches. It is found that the application of a force of 15 Ibs. at the end of the handle suffices to raise a weight of 280 Ibs. ; find the efficiency of the machine. Ans., 0-718. 9. In the differential pulley, if the weight is to be raised at the rate of 5 feet per minute, and the diameters of the pulleys of the compound sheave are 7 and 8 inches, at what rate must the chain be hauled ? Ans., 80 feet per minute. 10. In a differential pulley in which the velocity ratio of fall to lift is 30, a pull of 7 Ibs. will just raise a load of 24 Ibs., and a pull of 25 Ibs. a load of 240 Ibs.. Find the pull required to lift 600 Ibs., and the efficiency of the machine when such a weight is being raised. Ans., 55 Ibs. ; 0-36. 89. "When one body touches another, and there is equilibrium, if there is no friction, this means that there is no tangential force at the surface. The force with which either body acts on the other has no tangential component. If, however, there is friction, and we assume that the friction is /j. R where R is the force normal to the rubbing surface, it is evident that the total force at the place makes an angle with the normal if tan. = /j.. If the total force makes an angle less than q>, its tangential component is less than the friction, and there can be no motion. If, then, two bodies A o B and DOC touch at o, and N o N 1 is their com- mon normal at o, so long as the direction of the force acting between the bodies lies inside the cone Q o P, whose axis is o N, the angle PON or Q o N being $, there will be no sliding motion or rubbing at o. Thus, when the block lies en the inclined plane (Fig. 50), the total force trans- mitted between block and plane is w, the vertical weight of the block. So long as the angle is less than , there is no sliding. But the angle is the angle of the plane ; hence sliding is about to begin when we are increasing the angle of the plane, and it has become as much as . If the body DOC is at rest, and A o B moves, touching the first, the motion is one of sliding, rolling, or spinning, or combinations of these three. If there is no sliding, so that o is momentarily at test, tho instantaneous motion of A o B must be an angular velocity APPLIED MECHANICS. 107 y about some axis through o. If this axis makes an angle 6 with the tangent plane, there is an angular velocity of rolling 7 cos. 6 Fig. 50. about an axis in the tangent plane, and an angular velocity of spinning 7 sin. 6 about the common normal to the two planes. If in Fig. 49 the surfaces are cylindric and the rolling angular velocity is ta, and if r is the radius of curvature of A o B at o, and if R is the radius of curvature at o of the fixed surface B o c, and if v is the linear velocity of the point of contact, it is easy to show that ft>= Vr + R/ * The centres of curvature are supposed to be on different sides of the tangent plane. If in Fig. 50 there is friction between w and the plane, and motion up the plane is steadily taking place, draw R, making an angle N o R = <|>, with o N the normal. Let p A show the direction of the pulling force, and w of the weight. Knowing only w to calculate P and R, we have simply to use the triangle of forces. We are given one side and the angles to find the other sides. Thus draw AB repre- senting w to scale; and draw BC an I AC parallel to the two unknown forces, with the arrow-heads going circuitally. Measure B c and c A, and the forces p and R are known. Analytically. Resolve all the forces horizon- tally and vertically, and we have (see Art. 31), if the angle p o B is o, as p makes an angle a + with the horizontal, and R makes an angle R o H, which is the complement of ( + 0), p cos. (a + 0) = R sin. (0 + 0) .... (1) p sin. (a + 6) + R cos. ^ + 0) = w . . . . (2). As we really do not wish to know R, use its value from (1) in (2), Fig. 61. 108 APPLIED MECHANICS. and we find P sin. (a + 0) + cos. (ft + 0) ^ ^ + w sin. (ft + 0) cos. (ft + 9) cos. (a + 0) -f sin. (ft + 0) sin. (a + 0) w sin. (ft + 0) = w sin. (0 + ft) If p is the force required to just prevent the tody sliding down the plane, it is necessary to draw the angle NOB. upwards, instead of downwards, and to take ft for ft in the formula. Then p _ w sin, (0 - ft) (4) cos. (a+ft)" Example 1. If a = o, P = w for upward motion ; F = w for downward motion, p is for downward COS. ft motion when = ft. Under these circumstances the body will just be about sliding down under the action of its own weight. When = ft, it takes a force p = 2 w sin. ft to drag the body up the plane. Notice that when the body is pulled up the plane, if 6 is a small angle, expanding sin. (ft + 6), and dividing, we find P = w (tan. ft cos. 8 + sin. 9). Taking cos. = 1, we have, since tan. ft = /*, p = ft w + \v sin. 0. Now /*w is the force that must be exerted if we have no inclination 0, and w sin. is the force that must be exerted if we have inclination but no friction. Hence the rule usually employed in calculating the pulling force on a vehicle (Art. 47) namely, the total pull when the inclination is 1 in n ( or sin. = - j is equal to the pull necessary on a level road plus - th of the weight. II It is only true when is small. Example 2. If a == 0, so that p acts horizontally, P = w sin. (ft + 0) sin. (0 ft) cos. (ft + 0) = W ** < e + *> f r upward motlon ' P = W cos. (0- ft] = w. tan. (0 ft) for downward motion. Example 3. If ft = o, so that there is no friction, p = w - . Example 4. Find a so that p for upward motion may be a minimum. That is, what value of a will cause -4 to be cos. (a ft) a minimum ? That is, what value of o will make cos. (a - ft) a maximum ? Evidently it is a maximum when a ft = or o = ft, and then p = w sin. (ft -f 0). This important result neems to be completely ignored by men who deal practically with traction problems. Example 5. II there is no force P, what is the acceleration APPLIED MECHANICS. 109 down the plane ? We must first find what value of p acting parallel to the plane would just "be overcome. This is given in Example 1 as w ~i-J "~ ^ . Now, this force acts upon the mass cos. , and acceleration is force -r- mass ; so that the acceleration is g sm-J -_4>)^ QJ, course, when d> = 0, this becomes the well- cos. known g sin. 0. Example 6. When the nut of a square-threaded screw is turned, the surface of the thread is like an inclined plane, the tangent of whose inclination, 6, is -^- . if d is the mean diameter of the thread (or the diameter of the pitch cylinder of the screw). The nut presses upon the inclined plane exactly as in Example 2, and hence P = w tan. (0 -f ) . . . . (1) if P over- comes w, the total weight to be lifted when the nut is turned ; or P = w tan. (0 <>) . . . . (2) if w overcomes P. If the length of the handle to which the real force A is applied be I, then / A = P d or A = | y ; so that (1) and (2) become A = i|wtan. (0 + *)....(!), A = I j w tan. (0 - >) . . . . (2). (1) and (2) are equal, of course, if is 0, and then In this case, as there is no friction, the efficiency is 1, and the mechanical advantage is = ----- (3) 4 The mechanical A d tan. = A d tan advantage from (1) when there is friction is ^^ an (0-fdT) ---- ( 3 )- Hence the efficiency when there is friction is e = - ' : .... (4). tan.^a + cp) When w overcomes A the efficiency is similarly e = -^ - .. (5). Both (4) and (5) become 1 when there is no friction. It is an easy exercise in the calculus to show that (4) is a maximum when = ? -$.... (6), and that (5) is a maximum when = | + i ---- (7). If, then, a screw is to work as much in one way as the other, it seems reasonable to use = J- or 45 as the angle of its mean spiral. Note that when is as small as , w cannot overcome A, and the screw will not overhaul or reverse. In this case, if A over- 110 APPLIED MECHANICS. comes w, e = 2 '-f- . When a lubricant is used we cannot be tan. 2 sure that the here assumed law of friction holds ; that is, /t is. not a constant. When a lubricant is not used it is safe to say that in no actual screw-jack does one find 6 approaching- in value. Note that when there is no friction the mechanical advantage . w 2 1 pitch IB = , where tan. 6 = = ; so that A d tan. 6' ird W = 2 fad = 2*1 A. d x pitch pitch' or circumference described by the end of the handle -f- pitch, the rule given already. I have not considered the very considerable loss of efficiency due to friction because the load which is lifted is being kept from turning. Even the carefully-constructed and well-lubricated square- threaded screw-jack in my laboratory has an efficiency of only about '25 even at the highest loads. The ordinary jack used in workshops has a very much smaller efficiency than this. When the screw has a triangular thread we may assume that with the same kinds of rubbing surface the co-efficient of friction has greatly increased ; in reality it is the normal pressure on the thread which has increased. 90. When & pulling force F, making an angle with the normal, is applied to move a block which is being pressed against a surface, the tangential component which overcomes friction Fig. 52. is F sin. 0. The normal component F cos. diminishes R, so that the fric- tion is /A (R F cos. 0). Hence, when sliding occurs, F sin. = p. (R F cos. 0), r F = sin.0+*cos.0 (1) ' For any given value of /* it is easy to find the value of $ which will make F a minimum. In fact, the denominator of (1) is a maximum when cos. = /* sin. 0, or ju, tan. = 1, or is the complement of , the angle of repose. But if F is a pushing force, we have F sin. = /* (R + F cos. 0) ; so that F = - ^ R . In this case F becomes infinity sin. /3 n cos. if /3 is less than the complement of 0. For an angle (3 which is less than 90 the pushing F must be greater than the pulling F. When one piece of machinery drives another at a sliding contact this great distinction between pushing and pulling must be remembered. EXERCISES. 1. A weight of 5 cwt. resting on a horizontal plane requires a horizontal force of 100 Ibs. to move it against friction. What, in that case, is the value of the co-efficient of fiiction P Ans., 0-18. APPLIED MECHANICS. ni 2. A weight of 50 Ibs. is supported by friction alone on an inclined plane ; what is the force of friction ? Angle of plane, sin -f . Am., 20 Ibs. 3. A body placed on a horizontal plane requires a horizontal force equal to one-half its own weight to overcome the friction. If the plane be gradually tilted, at what angle will the body begin to slide ? Ans., 26 34'. 4. Find the force parallel to the plane required to draw a weight of 40 lbs.*up a rough inclined plane if n = %, the inclination of the plane being such that a force of 12 Ibs. acting at an angle of 15 to the plane would support the weight if the plane were smooth. Ans., 24'35 Ibs. 5. On a rough inclined plane it is found that a body is just supported on it by a horizontal force equal to three-quarters the weight of the body. Find the co-efficient of friction. Angle of plane, sin -f. Ans., ^. 6. Two unequal weights, Wj and w 2 , of the same material on a rough inclined plane are connected by a string which passes over a fixed pulley in plane. Find inclination of plane when the system is in equilibrium. Ans., tan + W 2 ) 7. Two rough bodies, Wj and w 2 , rest on an inclined plane and are connected by a string, the inclination of string to the horizontal being the same as the plane. If the co-efficients of friction are m and // 2 respectively, find the greatest inclination of the plane when the system is in equilibrium. Ans., tan 8. A plane is inclined to the horizontal at 24; there is a load of 1,200 Ibs. on it, the co-efficient of friction being '18. Determine (1) the force required to just draw the load up the plane ; (2) the force required to just prevent it slipping down ; (3) the force required to support it if there be no friction, the direction of the line of action of the force being in each case 23 with the plane, and above it. Determine the correspond- ing results if the direction of the force be parallel to the plane, and again if the force be horizontal. Obtain also the least force, in magnitude and direction, which is necessary in each of the three cases first referred to ; that is, for raising, lowering, and supporting without friction. There are twelve results in all. Ans., Effort E in pounds. E making 23 with plane. K parallel to plane. E horizontal. E a mini- mum. r>74 Upward motion. 692 685 816 Downward motion. 342 291 295 286 No friction. 530 488 534 488 112 APPLIED MECHAtffOS. 9. A load of 1,200 Ibs. has to be pulled along a horizontal plane Determine in magnitude and direction the least force necessary to do this, the co-efficient of friction being taken 0'4. Ans. y 446 ; 21 48 '. 10. Determine the efficiency in the case of a screw 2^ inches diameter, in which there are four threads to the inch, taking the co- efficient of friction -04. Ans., 0-442. 11. Taking the mean diameter of the threads of a 1 inch bolt to be 92 inch, number of threads per inch 8, the co-efficient of friction reduced to an equivalent square thread '17, find the turning couple required to overcome an axial force of 2^ tons. Ans., -20 ; 46'1 pound-feet. JOINTS WITH FRICTION. 91. When we know the resultant of the loads or forces on a pin W journal, and the law of friction, we can calculate the resultant force on the eye or step. Thus, if RO re- presents the resultant load w upon the journal s P, and there is also a twisting couple turning the journal in the direction of the arrow T, the journal rides up in the step till P is the point of contact, and until o P Q, = is the angle of repose, P o, being a vertical force equal to w, and this is the resultant of the forces with which the step acts on the journal; /u, the coefficient of friction, is tan. o P Q. It is usually stated in books that the resul- tant force between an eye and pin must act through some point such as P in the surface of contact. To show that this is a mistake, imagine A B and B c two pieces with a pin joint at B. These letters indicate the centres of pins at A, v f and Fig. 53. 0. If we neglect the weights of the pieces, imagine frictionless pins at A and c, then any force communicated from A to c through the frictional joint B is in the straight line A c ; for the resultant force of the pin A upon the piece A B acts through the centre A, and the resultant force of the piece B c on the pin c acts through the centre c. Imagine the pin at B to be rigidly fixed to B c. The force between A B and c B at the joint B must be in the direction A c ; and we can have so much friction at B that A c does not Fig, 5&. necessarily act through any point of the ALLIED MECHANICS. H3 surface of contact. We cannot, however, imagine this^ happening if the pin and eye at B meet only at one point or line at right angles to the paper ; and it is therefore important that pins and eyes should he f slack fit, and we shall in future imagine them to he so. We avoid unnecessary complication if we imagine that when two pieces act through a pin, the pin is rigidly attached to one of them. Let c n Fig. 56. and E F he two fractional pins acting on the piece o H. Let A and B he their centres. Let the coefficients of friction he a and , or the angles of repose and (pi 1 . Let the radii of the pins he A and B. Describe about A a circle with the radius A sin. #, or a b A /-, ;, and about B a circle with the radius B /-, :~. v 1 + a" V 1 + o* These may be called the friction circles of the pins. Ef there were no friction, the equal and opposite forces at A and B acting on the pin would be in the direction A B. They are really in the direction of a common tangent to the two circles. There are four positions of this common tangent depending upon the direction of relative sliding of pin and piece at each place. There are certain cases in which we see the direction without trouble. Thus : if A (Fig. 57) is the centre of the crosshead of a Fig. 57. steam-engine, and c of the crank shaft. If the resultant force due to the piston-rod, etc., is F at A, with no fiiction, we know that F sec. B A c is the pushing force in the connecting-rod A B, and the moment of this about c is F sec. B A c x CD cos. B A c = F x c D. If, then, we had no friction at the slide, but only in the pins A and B, as we see that both frictions diminish the turning moment, the line of resistance touches the two friction circles drawn at A and B in the manner shown in the figure, and the turning moment on the crank shaft is F x c E. Happily, in steam-engines that are well attended to the friction circles are so small that we may practically neglect them, and consider the force between a rod and pin to be truly through the centre of the pin. Should we, however, as in the case of the hinged joint of an arch, fear friction at the joint, it is worth while remembering that whatever be the loads on a piece, IU APPLIED MECHANICS. the necessary equilibrating forces at the pins are to be foond tangential to the friction circles there. Thus, if A and B (Fig. 58) are the centres of the hinges of the arch A c B, draw the friction circles at A and B. It is not enough to find the line of resistance A c B, but also other pos- sible ones not passing through the centres A and B, but touching the friction circles there. If a is the inclination of the centre line of the arch at A, and r is the radius of the friction circle there, there will be no great error in adopting this rule ; The line of resistance may Fig. 58. start from any point between A' and A" in the vertical through A, if A'A = A A* = r/sin. a. The smaller, therefore, the radius of the cylindric surface or hinge of a hinged arch the better. When an arch ring abuts at a cylindric surface of large radius, I cannot see much distinction between it and an arch fixed at the ends, except that there may be no tension in the joint, or rather that the line of resistance may not pass outside the joint. 92. Body turning about an Axis. In Fig. 59 we have a body which can move about an axis. It is acted on by a number of cords exerting forces which just balance one another. Now, if you make this ex- periment you will find that you must keep your finger on the body, because it is in such a state that a very small motion either way causes the forces to no longer balance. g Jgjpgl - = Suppose, however, ^**^ you were to let the Fig. 59. cord A be wound on a pulley whose radius is equal to the distance A o ; the cord B on a pulley whose radius is equal to B o, and so on, you would have the arrange- ment shown in Fig. 60, which differs from Fig. 59 in that a gmall motion has no effect on the balance. Now what is the APPLIED MECHANICS. 115 Fig. 60. condition of balance in this case 1 ? Suppose one complete turn given to the axis, every cord shortens or lengthens by a distance equal to the circumference of the pulley on which it is wound. Let A and B lengthen, and let C shorten, then we know that the work done by A and B must be equal to the work done against c. Hence, Pull in A x cir- cumference of A'S pulley, together with pull in B x circumference of B'S pulley, must be equal to pull in C x circumference of c's pulley. We might, however, use the diameters or radii of the pulleys, and so we see that in Fig. 60 there is balance if Pull in A x A o, together with pull in B x B o, equals pull in C X CO. The pull in A x A o is really the tendency of A to turn the body about the axis, and in books on mechanics it is called the moment of the force in A about the axis o. The law is then, if a number of forces try to turn a body and are just able to balance one another, the sum of the moments of the forces tending to turn the body against the hands of a watch must be equal to the sum of the moments of the forces tending to turn the body with the hands of a watch. We sometimes say : " The algebraic sum of the moments of all the forces is zero" That is, we regard one kind of moment as positive and the other as negative. Another way of putting the proof is this : If a much magnified drawing be made showing a very small motion through the angle 0, it will be seen that the work done by c being, either c x motion of point of application resolved in direction of c, or motion of point x resolved part of c in direction of motion; in either case this work is equal to X C x 6, that is, moment of c x 0. Hence, total work = 116 APPLIED MECHANICS. algebraic sum of moments x 0, and if total work is 0, then the algebraic sum of the moments is 0. When work is done upon a body in turning it, observe that the work in foot-pounds is equal to the moment in pound-feet multiplied by the angle in radians turned through. If a constant moment M acts upon a body rotating at a radians per second, Ma is the work done per second. If n is the number of revolutions per minute, 2 TT n is the angle per minute, and hence M 2 TT n -^ 33,000 is the horse-power. [No doubt the student has already become quite familiar with the idea expressed by "work = force x distance." It has just been shown how, by the very simplest transformation, this expression becomes, in the case where a force is producing turning about an axis, " work done = moment x angle turned through in radians." Although not required for the study of the portion of the subject now under consideration, yet we may remind the student that at a later stage it will be an advantage to him if he has accustomed himself to calculating work done when the turning moment (called the torque) and angle turned through are given.] 93. The Lever. Thus, for example, a lever is a body such as I have spoken about, capable of turning about an axis. You will find that our general rule of work and this rule of moments will give the same result. . If two forces act on a lever, they will balance when their moments about the axis are equal; that is, when p, multiplied by the shortest distance from the fulcrum or axis to the line in which P acts, is equal to w multiplied by the distance of the fulcrum from the line in which w acts. If a number of forces balance when acting on a lever, tJie sum of the moments tending to turn the lever against the hands of a watch must be equal to the sum of the moments tending to turn the lever with the hands of a watch. It must be remembered that if the body acted upon has its centre of gravity somewhere else than in its axis, then we must consider that the weight of the body is a force acting vertically through its centre of gravity. Example. The safety valve, Fig. 61, must open when the pressure on the valve is just 100 Ibs. per square inch. The mean area of the valve A, on which we assume that the pres- sure acts, is 3 square inches; CD is 2 inches, E is 50 Ibs., the weight of the lever is 6 Ibs., and its centre of gravity is 6 inches APPLIED MECHANICS. 117 from D j where must E be placed 1 All distances are measured horizontally. Here, the upward force is 100 x 3, or 300 Ibs , and its moment about D is 300 x 2, or 600. The moment of the weight of the lever is 6 x 6, or 36. The moment of the weight E is 50 x the required distance from D. Hence, 600 36, or 564 divided by 50, is the answer; 11-28 inches from D. Thus we find where the mark 100 ought to be placed. Let the student repeat this for pressures of 90, 80, and 70 Ibs. per square inch, stating in each case the distance of the weight from D. What are the distances apart of the marks ? This is not the place to consider the various forms of safety Pig. 61. valve employed by engineers. The force which fluid at rest can exert against a valve is not the same as when the valve is open and the fluid is moving ; but it is very important that a valve should stay open, allowing the fluid to escape. This is effected partly by using a peculiar shape of valve, but more usually by letting E come closer to the fulcrum when the valve opens. In the above figure the valve seat is conical ; in practice, a flat seat is now much more commonly used, the breadth of the seat being very small. Exercise. A weighbridge consists of three levers whose mechanical advantages help each other ; I mean, the short arm of each supports the long arm of the next. Suppose that the weights of all parts are arranged so as just to be balanced when no load is on the bridge, and that the mechanical advantages of the three levers are 8, 10 and 12, what load will be balanced by a weight of 15 Ibs. ? Answer 14,400 Ibs. Suppose that it is the first of these levers that is alterable 118 APPLIED MECHANICS. (that is, there is a sliding weight), what is its mechanical advantage altered to when the load is 16,000 Ibs. 1 Answer It was 8 ; it now becomes increased in the proportion of 16,000 to 14,400, so that it becomes 8 -8889. Show that the graduation of the lever with the sliding weight is in equal divisions for equal alterations of the load. Do this by finding the position of the sliding weight for various loads. Friction is greatly got rid of in weighing-machines by using steel knife edges as fulcrums of levers resting on hard steel or agate plates. Students must examine actual specimens. The common chemical balance must be examined. There is one thing about it which may trouble the student. " Why are shortarmed balances now preferred to the older forms'?" The short-armed balance has less moment of inertia, and this causes it to be quicker in its motions, so that time is saved. (See Art. 453.) But there is much more to be said about it than this. Indeed, in this as in all other cases, to thoroughly understand one machine requires a knowledge of the whole of applied mechanics and applied physics. I am not now dis- cussing any one machine exhaustively. I was strongly tempted to take up the thorough consideration of one machine and call this the study of applied mechanics ; and if I had a student with a particular interest in one machine this would be the very best way to put before him the study of applied mechanics. The method I have adopted in this book is to illustrate each principle by means of a machine in which that principle happens to appear most important. The defect of the method arises from its causing a student to think that he knows all about a machine when he only knows the most important principle of applied mechanics which is illustrated by it. The cure for this academic training defect comes when a student is compelled to take a special interest in some one machine, and it is then, in practical work, that he really is learning applied mechanics. We can only partially correct the defect by our numerical exercise and laboratory work and experience in the workshop. Students looking at weighing-machines ought particularly to notice that when objects to be weighed are placed not on swinging scale-pans, but on fairly firm platforms, the construc- tion of the balance must be such as to make the measurement to be independent of the position of the object upon the platform. APPLIED MECHANICS. 119 EXERCISES. 1. A uniform straight bar, 2 feet long, weighs 5 Ibs. ; it is used as a lever, and an 8 Ibs. weight is suspended at one end ; find the position of the fulcrum where there is equilibrium. Am., 4-5 inches. 2. A lever safety-valve has the following dimensions : Mean diameter of valve, 3 inches ; weight of valve, 8 Ibs. ; distance of centre of valve from fulcrum, 2 inches ; weight of lever, 16 Ibs. ; distance of its centre of gravity from fulcrum, 13 inches. Find where a weight of 35 Ibs. must be hung from the lever, so that the steam may blow off at 85 Ibs. per square inch. Ans., 36-4 inches from fulcrum. 3. A B c D is a rectangle, and A c a diagonal. In A c take a point o, such that A o is a'third of A c. Forces of 30, 10, and 5 act from A to B, c to B, and D to c respectively. If A B = 3 inches, and B c = 4 inches, write down the moment about o of each force, with its proper sign, and find their algebraic sum. . 40 20 3 3 4. The diameter of the safety-valve of a steam-boiler is 4 inches ; the weight on the lever is 90 Ibs., and distance from centre of the valve to the fulcrum is 2'5 inches ; what must be the distance of the point of suspension of the weight from the fulcrum in order that the valve will just lift when the pressure of steam in the boiler is 80 Ibs. per square inch ? Length of lever, 30 inches ; weight, 12 Ibs. ; distance of centre of gravity from fulcrum, 14 inches. Ans., 26*059 inches. 5. A ball weighing 4 Ibs. is fixed to one end of a bar hanging vertical, the other end of which can turn about a fixed axis. If the ball be pulled out by a force which acts horizontally through the centre of the ball, what will be the amount of this force necessary to keep the bar at rest in a position such that the bar makes 30 with the vertical ? If the length of the bar be 30 inches, what would be the force if applied at a point on the bar 10 inches from the centre of the ball ? 4 ^-,^/3; 2v/3. 120 CHAPTER VII. ELEMENTARY ANALYTICAL AND GRAPHICAL METHODS. 94. IN Fig. 62 we have another example of the fact that when there is a displacement of a point, o, in the direction, o T, and a force, P, acts in the direction o A, the angle A o Q being called a, the work done is the displacement multiplied by the com- ponent of P in the direction o T. This component is P cos. a, or it may be defined as the projection of o A, the line which represents the force, upon o T. The student must note what we mean by the projection of a line. The projection of A upon the line o T is o Q. If the arrow-head is not put upon o Q, the order of the letters will indicate the sense of the action. Thus 1 shall use Q o to mean a note that if we have lines, I call the sum of Fig. 62. sense opposite to o Q. Now o A, A B, B c, c D, and D E (Fig. 63), what of their projections upon any line is really the projection o E. If I have a closed polygon, the sum of the projections of its sides upon any line is zero, if the sense of every side is circuital round the polygon with the sense of all the rest. If the sides are parallel to, and proportional to forces, this means that the sum of all the components of all the forces in any direction whatsoever is zero. Now suppose we have a body acted upon by any number of forces in all sorts of directions (we can only illustrate this by strings and weights) ; and if the weight of every portion of the body . itself be con- sidered also acting ; and if the body is in equilibrium, Fig. es. a small translational the principle of work tells us that if the body receives displacement, , in any direction APPLIED MECHANICS. 121 whatsoever (translation means that every point moves parallel to the motion of every other point, and through the same distance), then the total work done by all the forces is zero. But the work done by any force is its component in the direc- tion of the motion multiplied by the displacement ; hence, we see that the sum of the components in any direction whatsoever must be zero. This is a condition which must hold when any body is in equilibrium. But it is evident that we have the same rule when we say that the forces are proportional to the sides of a closed polygon, the senses of the forces being cir- cuital round the polygon. If the forces are not in one plane, the polygon will be what is called a gauche polygon, but if the amounts and directions of the forces are given in plan and elevation, the polygon may be drawn in plan and elevation. To give the analytical rule more simply : Take three lines mutually at right angles to one another, o x, o Y, o z ; project all the forces upon o x, they balance (that is, the algebraic sum of their projections is zero) ; project all the forces upon o Y, they balance ; project all the forces upon o z, they balance. These three analytical conditions are the same as the graphical rule, " the force polygon is closed." Perhaps, however, we had better confine our attention to forces all in one plane. The analytical rule is : The algebraic sum of all the horizontal components is zero, or the horizontal components balance ; the algebraic sum of all the vertical components is zero, or the vertical components balance. These two conditions are the same as " the force polygon is closed." EXERCISE. Given the [following forces, find their equilibrant Force o x of 50 Ibs., force o A of 20 Ibs., the angle, x o A, being 33; force M o of 56 Ibs., the angle XOM being 150; force o G of 100 Ibs., the angle x o G (all angles are measured counter-clockwise) being 217 ; force H o of 70 Ibs., the angle x o H being 315. Now let a student draw these on paper. I have assumed them all to be in lines through a point o, but this was for ease in setting the question. The forces need not act through one point. Let him, by drawing, or by using a table of sines and cosines, find the components of all in the direction o x and o Y, if o Y is perpendicular to o x (Art. 31). I make these projec- tions to be 122 APPLIED MECHANICS. In the direction ox. 50 cos. = 60 20 cos. 33 = 16-773 -|- 56 cos. 30 = 48-497 - 100 cos. 37= - 79-863 70 cos. 45= - 49-497 Algebraic sum = 2 x = 14-09 In the direction OY 50 sin. = 20 sin. 33 = 10-893 - 56 sin. 30 = - 28 - 100 sin. 37 = - 60-181 + 70 sin. 45 = 49'497 Algebraic sum = 2 Y = 27 '791 The Greek letter 2 is used to mean " the sum of all such terms as." x means any P cos a, and Y any p sin a. We have found 2x and SY ; call them x and Y respectively. Draw o c = - 14*09, o B= - 27-79, and complete the rectangle o B R c ; then o R is the resultant and R o the equilibrant. It is evident that o R 2 = \/x2 + Y*=3M6. Also tan R o C=Y/X= 1-971 ; therefore ROC=63 7', and x o R=243 7'. The student must get accustomed to symbols which so greatly shorten our use of language. Our rule is: If P I? P 2 , etc., are forces in a plane, and if they make angles a 1? a 2 , etc., with any line, say a horizontal line, then P l cos. a^ = Xj, P 2 cos. a 2 = X 2 , etc. , are the horizontal components of the forces. The directions of arrow-heads must be noted, and the algebraic sum of the horizontal components is de- noted by x = 2P cos. a, or 2x. In the same way Y= SP sin. a, or SY, denotes the algebraic sum of the vertical components Y I = P I sin. a } , Y 2 =P 2 sin. ct 2 , etc. Then the re- sultant, R, of all the forces is the resultant of x horizontally and Y vertically, and R 2 = x 2 + Y 3 , and if R makes the angle a, with the horizontal, then tan. a = Y -^ x. We have only found the amount, clinure and sense of R. We do not yet know where it acts. (The word clinure was, I believe, invented by the late Prof. Thomson ; the word direction implies more than what we try to express by clinure.) Fig. 63A. APPLIED MECHANICS. 123 Similarly, if the forces are not in the same plane, and if each p makes the angles a, ft, and 7, with three standard lines at right angles to one another (usually called the axes of x, y, and z), then if p cos. a is called x and SP cos. a x, if p cos. is called Y and 2ji j cos. /3 = "Y, and if P cos. 7 = z and SP cos. 7 ="z, then R 2 = x 2 f x 2 + Y 2 and the cosines of the angles which K makes with the axes are x/ R , Y/ R , and z/ B . 95. Any physical quantity which is directional is called a vector (such as a displacement, a velocity, an acceleration, a force, a stress, the flow of a fluid, etc.), and may be represented by a straight line. The length of the line represents the quantity to some scale of measurement ; the line's clinure represents the clinure of the vector, and the barb of an arrow represents its sense. It is easy by actually drawing lines and measuring their lengths to solve problems which would otherwise require a good deal of mathematical knowledge. This sort of graphi- cal calculation having proved useful, it has attracted the attention of men who have leisure enough to make an elaborate study of its methods. It has, unfortunately, become a com- plicated weapon with which these men can attack all sorts of problems which are much more easily solved in other ways. We shall only use graphical methods where they happen to be the best methods. Now a force is a vector ; it has magni- tude, clinure, and sense, but it is more than a vector ; it has a fourth quality not possessed by ordinary vectors namely, actual position in space. Settling any one point in its line of application settles its position when its clinure is known. But if we are told that forces all act at a point, they are added exactly as all mere vector quantities are added. When, then, I speak of finding the resultant or equilibrant of forces at a point, I may be said to speak of any vectors. 96. Forces acting at a Point. The line A B (Fig. 64) represents a force in clinure by its own clinure, in amount by its length to any scale we please, and in sense by its arrow-head, which shows that the action of the force is from A to B. It would not be correct to call this the force B A, because this is opposed to the sense of the arrow-head. The forces A o, o B, o c, o D (Fig. G5), all act upon a small body, o, or their lines of action when produced all pass through 124 APPLIED MECHANICS. Fig. 65. a point, o, in a large rigid body. The amount of each force is shown by the length of the line, representing it to some scale. Now to add these forces together in the most perfect manner that is, to find a force called their resultant, which shall be exactly equivalent in its effects to all the above forces acting together we draw a polygon (Fig. 66). ^Each side of this polygon is parallel to, and proportional to a force in Fig. 65 j thus, the side A corresponds to the forco A o, and the ai row-heads agree, and lastly the action indicated by the arrow-heads is circuital. Fig. 66 is always called the force polygon. When it is unclosed, as it is in the present case, we know that the forces A o, etc. (Fig. 65), are not in equili- brium. To keep A o, o B, etc., in equilibrium, a new force, called the equilibrant, must be introduced corresponding to the side E (shown dotted), which will close the polygon, its arrow-head being circuital with the others. Now if we want the resultant of A o, o B, o c, o D, it evidently acts through o and corresponds to E, Fig. 66, but with arrow- head reversed. The resultant of a number of forces is equal and opposite to their equilibrant. Prove now the following statements by actual drawing : 1st. The resultant of any number of forces does not depend on the order in which they are drawn as sides of the polygon. 2nd. Any lines or forces whatever which form a closed polygon in any given order will form a closed polygon if drawn in any other order. 3rd. In adding forces we may first find the resultant of some of the forces, and then add together this resultant and all the rest of the forces. The answer will always be the same, however we may group the forces before adding them. When the forces are not all in one plane the polygon must be drawn by descriptive geometry, and to draw it, and so find the resultant or equilibrant as the closing side, is an excellent graphical exercise. APPLIED MECHANICS. 125 In working exercises we recollect the fact that the resultant of all the forces due to gravity is called the weight of the body and acts through its centre of gravity (seepage 136). This one force replaces the millions which are due to gravity. When we observe that we have only three forces acting upon a body at rest, we know that they must be in a plane and act through a point unless they are parallel. In the ordinary books on mechanics there are many problems which are easily solved if we remember this fact. Four forces in equilibrium, if not all in one plane, must meet in one point. When a body touches a smooth surface we know that the force which there acts upon the body is normal to the smooth surface. When a body touches a rough surface we know that the force which there acts upon the body makes an angle with the normal, and the limiting value of this angle when sliding is about to occur is what is called the angle of repose, or the angle whose tangent is p, the co-efficient of friction there. It is astonishing what a number of exercises are easily worked out if one will only recollect these few general principles. The student will at this place work again the exercises of page 111. It ought to be getting clear to him that the most difficult analytical work has really nothing in it more com- plicated than these exercises have. The exercises are usually called easy, certainly students get easily into the way of working them, and I am always sorry to notice too great an ease of this kind. It often indicates shallowness of comprehension. EXERCISES. 1. Forces o A = 30 Ibs., o B = 50 Ibs., co = 15 Ibs., D o = 80 Ibs., o E = 150 Ibs. ; the angles are B o A = 45, c o A = 90, D o A = 135, E o A = 270. Find the resultant analytically and graphically. Ana., 223 Ibs. at an angle of 303" with o A. 2. Sheer legs each 50 feet long, 30 feet apart on horizontal ground, meet at point c, which is 45 feet vertically ahove the ground ; stay from c is inclined at 40 to the horizontal ; load of 10 tons hanging from c. Find the force in each leg and in stay. dns., 7'8 tons ; 6'4 tons. 97. In many engineering problems, when forces A, B, c, D, etc., are given, it is sometimes important to be able to show graphically the resultant of A and B, the resultant of A, B, and c, the resultant of A, B, c, D, and so on. Thus (Fig. 67) A, B, c, D, etc., are given forces. Draw the unclosed force polygon (Fig. 68). Join the point with each corner of the force polygon. From the point B' 126 APPLIED MECHANICS. where A and B meet (Fig. 67) draw a line B' c' parallel to the line o B c (Fig. 68) (o B c is the line from o to the corner where B and c meet) ; from c' draw c' D' parallel to o c D, and so on. Then B'C' represents the position and direction, Pig. 67. and B c represents to scale the resultant of the given forces A and B. Similarly D'E' represents the position and direction, and ODE represents to scale the resultant of the given forces A, B, c, and D. Note the arrow-heads of the resultants we have found. The line A B' c' D' E' F' (Fig. 67) is usually called a line of resistance. 98. We have in Art. 96 confined our attention to the forces acting upon a small body, or forces which all pass through one point if they act on a large body. But in Fig. 67 and in our description we assumed a large body, and our forces were any forces whatsoever. We gave it a small motion of translation, and obtained an important result from the consideration that, on the whole, no work was done upon the body. Now, let us assume that any point o in the body is fixed, or, rather, that an axis o is fixed, the axis being at right angles to the plane in which all the forces act ; about this axis we assume that the body may rotate. Consider the work done by all the forces during any small rotation Q ; it is zero. But the work done by any force is, as we have already seen (Art. 92) the moment of the force multiplied by 0. Hence the sum of all the moments of all the forces about is zero if there is equilibrium. But any -small plane motion Fig. 68. APPLIED MECHANICS. 127 of the body whatsoever we know to be resolvable into a motion of translation and a rotation about some axis o at right angles to the plane. Hence the law of work tells us that if any system of forces is in equilibrium their compo- nents in any direction balance one another and theii moments about any axis balance one another. In the numerical exercise Art. 94 let the respective forces (only given in amount, clinure, and sense as yet) be at the following distances from a certain point, which I shall call s. The sign -f means that a force tends to turn the body against the hands of a watch, ox is at the distance -f 5 feet, o A is at 2 feet, MO is at 7 feet, o G is at + 3 feet, H o is at 4 feet. Let the student now draw these forces in their proper positions relatively to s. The sum of their moments is 50 x 5 - 20 x 2 - 56 x 7 + 100 x 3 - 70 x 4, or 162 pound-feet. This is the moment of their resultant which is 31 '158 Ibs. ; so that its distance from s is 5-2 feet. 99. If the clinure and sense of a force p be given, it is also neces- sary to give some point through which its direction passes. Thus, let all the forces be in one plane ; let P make an angle a with the horizontal ; let the co-ordinates of the given point be x and y referred to the axes of x and y. If P cos. o is x and p sin. o is Y, then as the moment of p about any axis is equal to the sum of the moments of x and Y, taking moments about the origin, the moment of p is Y x - x y. If R is the resultant, its components are 2 x and 2 Y ; call them X and Y. Also, if ~x and y are the co-ordinates of a point in R, XY y x = 2 (YSC xy). For equilibrium we must have 2 x = . . (1), 2 Y = . (2) 2(Y - xy) = 0..(3). Notice that we may have (1) and (2) true without (3) being true. In this case the system of forces reduces to a mere couple whose moment about any point is 2 (Y a? x y} ; such a system is called a torque. If we choose any point in the plane, we can replace any system of forces by a single force through this point^ together with a couple whose moment is the sum of the moments of the system of forces about this point. This is often an exceedingly important fact to remember. (See Art. 100.) The student ought to work many numerical exercises graphically and analytically. Example 1. A beam, ABODE, is supported at A and E by forces x and y. The load at u is 3 tons, and A B = 4 feet ; load at c is 2 tons, and A c is 7 feet ; load at D is 2i tons, and A D is 9 feet ; A Bis 12 feet. Here # -f- y = 3 + 2-J-2| = 7J ton's. Taking moments about A., the moments with the hands of a watch are 3x4 + 2x7 + 2 x 9 = 48 ton-feet. 128 APPLIED MECHANICS. Die moment against the hands of a watch is y x 12, and x has no moment, because it acts at A. Hence our second equation is 12 y = 48, y = 4-0417 tons, and therefore x is 3*4583 tons. Example 2. We neglected the weight of the beam itself in Example 1. If its centre of gravity is 4 feet from A, and if the weight of the beam is half a ton, and if x' and y' are the additional supporting forces, v 1 + V' = i, i X 4 = y' x 12. Hence, y 1 = i ton, x' = i ton. Example 3. In Fig. 69 o A is part of a beam. Considering only the portion of beam to the right of the section at o, let the loads downward, P 15 P 2 , and P 3 , and the supporting force upward, Q, be given. Let the perpendicular A distances from o be a lt 2 , 3 , and q. Find a force through o, and a couple to balance the given forces. [CaU the force s. If it acts downward at o, its amount Fig. 69. must be s = Q P! K P 2 P 3 . If the couple is called M and it acts tending to turn the body o A round o, with the hands of a watch M = Q q - PJ ^ - P 2 2 - P 3 3 . When we come to consider beams we shall call s the shearing force and M the bending moment at the section o. The student will at this point work the Exercises 1, 2, 3, of pages 134-5, as well as the following : 1. A uniform beam, 20 feet long and supported at its ends, has weights of 1, 3, 2, and 4 cwts. placed at distances of 2, 8, 12, and 15 feet respect- ively from one end. Taking the weight of the beam to be 5 cwts., find the reactions at each of the supports. Ans., 7 cwts. ; 8 cwts. 2. Draw any line o x, and lines o p, o Q, o R, o s, o T, making angles of 28, 62, 118, 220, and 305 respectively with o x. Consider that forces act along these lines, their amounts being 25, 34, 14, 42, and 18 Ibs. Find the amount and direction of the force which balances these. In doing this first determine the components of each force in the directions o x and o Y, and arrange these in columns as shown in Art. 94. Ans., 15-67 Ibs. ; 232-2 with o x. 3. A trap-door of uniform thickness, 5 feet long and 3 feet wide, and weighing 5 cwts., is held open at angle of 35 with the horizontal by means of a chain. One end of the chain is fixed to a hook placed 4 feet vertically over the middle point of the edge on which the hinges are, the other end being fixed to the middle point of the opposite edge. Deter- mine the force in the chain and the force at each hinge. Ans., 2*65 cwts. ; 2'5 cwts. 4. A uniform beam, weighing 2 cwts., is suspended by means of two chains fastened one at each end of the beam. When the beam is at rest it is found that the chains make angles of 100 and 115 with the beam; find the tensions in the chains. Ans., 1 cwt. ; I'l cwt. APPLIED MECHANICS. 129 5. AB is a horizontal uniform bar 1| feet long, and F a point in it 1Q inches from A. Suppose that A B is a lever that turns on a fulcrum under p, and carries a weight of 40 Ibs. at B ; weight of lever, 4 Ihs. If it is kept horizontal by a fixed pin above the rod, 7 inches from p and 3 inches from A, find the pressure on the fulcrum and on the fixed pin. Ana., 89-14 Ibs. ; 45'14 Ibs. 100. When the forces are not in one plane, let the force P make angles a, /3, and y with the three axes of co-ordinates. Let a point in the direction of P be x, y, z. If p cos. o, or x, p cos. /3, or Y, and p cos. 7, or z be the three components of p, we can use x, Y, and z instead of P for all purposes. Positive directions are the direc- tions of increasing x, y or z. Thus the moment of p about any axis is equal to the sum of the moments of x, Y, and z about the axis. Attention, must be paid to the sense of each force, and whether it tends to turn the body against or with the hands of a clock. The student ought to spend time in fixing clearly in his mind the truth of the following statements : The moment of p about the axis of x is z y Y z ; the moment of p about the axis of y is x z z x ; the moment of P about the axis of z is Y a? x y. Hence we see that if R is the resultant, its components are 2 x, , 2 Y, 2 z, and the sum of their squares is R 2 , which is therefore easily calculated ; also each of them, divided by R, is a direction cosine of R. Again, if x, y, and z be the co-ordinates of a point in R, then y2z-s2Y=2(zy - Y z), 2 2 x i2z = 2(xz z a?), W 2 Y - y 2 x = 2 (Y x - x y). The value of R and its clinure (the angles which it makes with the axes) having already been found, these equations enable the posi- tion of a point in the resultant to be found ; so that R is completely determined. For equilibrium we must have 2x = o, SY = O, 2z = o, 2 (z y Y z) = o, 2 (x z z a?) = o, 2 (Y x x y) = o. Given a set of forces, it is evident that we can always sum them into a resultant force acting at any point we please to choose, together with a couple about some axis. If we are not given the point, it is always possible to reduce any system of forces to a resultant, and a couple whose axis is the resultant force. 101. The Link Polygon. We shall now consider graphical methods of dealing with forces which do not necessarily act through one point. Take, for example, the forces of 1, 2, 3, 4 (Fig. 70). Draw the unclosed force polygon 1', 2', 3', 4' (Fig. 71), with its sides parallel to and proportional to the forces, and the arrow-heads circuital. Now the dotted line b a, with its arrow non-circuital with the rest, is parallel to and proportional to the resultant of all the given forces. But this does not tell us where the resultant force is situated, although it tells us its direction and amount. 130 APPLIED MECHANICS. From any point, o (Fig. 71), draw a line to the junction of 1' and 2' (it is easier to say draw the line o 1' 2'), o 2' 3', etc., to all the angles of the force polygon. Now construct a new unclosed polygon, with its corners on 1, 2, 3, 4 (Fig. 70), and its sides parallel to o 1' 2', o 2' 3' etc. (Fig. 71), its last side being parallel to oa, and its first parallel to 06. We have now found the point, 5 (Fig. 70), where the first and last sides of the link polygon meet. The resultant of the forces Fig. 71. 1, 2, 3, 4, passes through this point, 5, and corresponds to the closing side, ba, in direction and magnitude. The new polygon is called the link polygon of the forces relative to the pole, o. The position of A, the point at which we start to draw the link polygon, may be chosen anywhere on 1, and hence there may be any number we please of link polygons for a given position of the pole, o. Again, there are any number we please of link polygons corresponding to any other posi- tion of o, and we can choose o where we please. Any student who studies this in the light of what he did in Art. 96, will see that the link polygon really consists of a system of links which would be in equilibrium under the given set of forces and the force we have found ; and since the mere links only introduce forces which are of themselves in balance, being equal and opposite in each link, the system of forces acting at the joints must balance. Suppose we find that when we are given the forces 1, 2, 3, and 4 (Fig. 72), and we draw the force polygon (Fig. 73), and any link polygon (Fig. 72), that these are both closed, let us prove that the forces are in equilibrium; A system of forces acting on a rigid body is not affected by introducing any number of forces which separately halance one another. Now let a force represented hy the length^ of the line o ]/ 2' act at the point A in the direction BA, its sense* "being shown by the arrow-head near A, and let an equal force act at B in the APPLIED MECHANICS. 131 direction AB, its sense "being- opposite to that of the force at A. These two forces are in equilibrium with one another, and they cannot therefore affect the original system of forces in any way. P Fig. 72. Fig. 73. Similarly, the forces shown by the arrow-heads in B c, c D, D A are introduced, every pair balancing one another. Now we see that the three forces at the point A are in equi- librium with one another, because they are parallel to and pro- portional in amount to the sides of the triangle omn (Fig. 73), * and corresponding arrow-heads would run right round the triangle. Similarly, there is equilibrium at every other corner of the link polygon A B c D ; hence all the forces are in equilibrium, and hence the forces, 1, 2, 3, 4, taken by themselves, must be in equilibrium. The theorems which we wish students to prove by construction can be proved to be generally true, reasoning from the fact that a number of forces acting at a point can only have one resultant. 102. We see, then, that the force polygon alone is sufficient to find the resultant of any number of forces if the forces meet at a point, but we need also the link polygon if the forces do not meet at a point. The link polygon really shows that the sum of the turning moments of the forces 1, 2, 3, 4 (Fig. 70) about any point is equal to the moment of the resultant about the same point. The force polygon pays no regard to turning moments of forces ; it merely tells us about the resultant of the forces, supposing that they all passi d. through the same point. 103. You ought by actual drawing to illustrate the truth of the following four ways of putting one statement. If you use coloured inks your drawings will be more instructive. 1st. The resultant of any number of forces is independent of the order in which we draw them in the force polygon, and draw between them the sides of the link polygon. 2nd. In adding forces we may first find the resultant of some of the forces, and then add together this resultant and all the other forces. The answer will always be the same, however we may group the forces before adding them. 132 APPLIED MECHANICS. 3rd. If the force polygon of a number of forces is closed, and if we can draw a closed link polygon, then all the link polygons we may draw will also be closed. 4th. If any other pole be taken in Fig. 71, and another link polygon be drawn and a new point 5 (Fig. 70) is found, both of the points so found lie in a straight line parallel to b a of Fig. 71. You will also find, and it is easy to prove, that the locus of the point in which any two sides of the link polygon meet is parallel to the line which closes the corresponding portion of the force polygon. Again, if b is taken as pole instead of o, the last side of the link polygon is found to be in the direction of the resultant of the forces 1, 2, 3, 4; and, generally, any side of the link polygon is in the direction of the resultant of the corresponding number of the given forces. Thus, if b is taken as pole, 4, 5 becomes the resultant of the forces 1, 2, 3, 4, and 34 becomes the resultant of the forces 1, 2, 3. It is evident from this that the direction of the resultant of any two forces, or of any number of forces, which meet at a point passes through their point of intersection. A system of forces may not reduce to a resultant force, but be equivalent to a couple. When this is the case the force polygon is closed, and the first and last sides of any link polygon that may be drawn are parallel to one another. You may also find it worth your while to prove by construction this statement : if two link polygons are drawn for two positions of the pole o, the correspond- ing sides of the two polygons meet in points which lie in a straight line parallel to the line joining the two positions of the pole o. If you have been able to make a few drawings such as I have been speaking about, and so take an interest in this easy and instructive method of working mechanical exercises, you ought to work by means of it some such exercises as the following : 104. 1. Exercises. In Exercise 2 of page 128, draw the force polygon, taking the forces in the order OP, OR, o Q, o T, o s, and observe that the resultant and equilibrant are the same as before. Obtain also the resultant of o P, o B, o T, and the resultant of o Q, o s, and show that these have a resultant equal to the resultant of the five forces. 2. Draw a line H K 4*7 inches long. On H K take points A, B, c, distant from H 0-4 inch, 1'5 inch, and 3 -3 inches respectively. Now draw A p, B Q, c R inclined at angles of 72, 57, and 37 with H K. Suppose that a force of 214 Ibs. acts from K to H, one of 576 Ibs. from R to c, one of 132 Ibs. from Q to B, and one of 237 Ibs. from p to A. (1) Draw a force polygon to deter- mine the amount, clinure, and sense of the resultant. (2) Take any pole o and draw a link polygon to determine the lateral position of the result- ant. In giving your answer say what are the perpendicular distances of H and K. from the resultant. (3) Draw a new force polygon, taking the forces in a different order, and observe that the resultant is the same as before as regards amount, clinure, and sense. With respect to a new pole o, now draw a second link polygon, and observe that the lateral position of the resultant agrees with that obtained before. (4) Take the first force polygon and choose a new pole, and with respect to it draw a APPLIED MECHANICS. 133 new link polygon. Observe that this gives rise to the same resultant. That is, the closing sides of each link polygon meet at points, all of which should lie on the same straight line. (5) Calculate the above answers according to the rules of Art. 99. Ans., (1) 1,066 Ibs. ; 397 with HK; sense same as force RC; (2) 1-312 inches; 17 inches. 105. Example. Given a set of forces, find two forces, one given in position, and clinure the other to pass through a given point P, such that these two forces will balance the given set. The ingenious student will find out how to do this himself; other students will benefit by the following instructions. Draw the force polygon with one corner unknown ; choose a pole and begin drawing a link polygon from the point p ; the side closing the link polygon enables the missing corner of the force polygon to be found. This problem is one of the very commonest to come before the engineering student. Thus, let any structure (a roof principal or a railway girder, for example) have given loads. Let it be supported at a hinge joint at p, and upon rollers at Q, to allow for expansion. The direction of the supporting force at Q is known, and one point, p, in the other supporting force is known. 106. A student who.sees the essential ideas underlying our methods of working will have pleasure in working curious problems, such as the following exercise : Given a set of forces and given three points, A, B, and c. Draw a link polygon with three of its sides passing respectively through A, B and c. Hint. You must first find the resultant of the three given forces, and observe where its line of action meets Y Y ; this point of intersection, i, joined to x, gives the line of action of the balancing force through x. Three forces now act at i. Their lines of action are known, and the magnitude of one force ; hence the amounts and senses of the other two can be found. 107. Exercise. Draw a rectangle A B c D, A B = 5 inches, B c = 1-8 inch. From A, along A B, measure off lengths, A E, A F, A G 175, 2-8, 57 inches respectively. From D along D c measure off D H, D K, D L = 1-4, 2 g 4, 2*85 inches respectively. Suppose that forces of the following amounts act on 'a rigid body viz. , 1,460 Ibs. from A to H, 1,085 Ibs. from E to K, 808 Ibs. from F to L. These are balanced by two others, one of which acts through x, a point in D A distance 0'6 inch from D, and the other has for its line of action Y Y, the line passing through c and G. Find the magnitudes of the balancing forces and the angle between them. (The rectangle is intro- duced merely as a convenient way of settling, without a figure, the given forces.) Scale i inch to 100 Ibs. Ans., 2,550 Ibs. ; 1,140 Ibs. ; 60. 108. Example. Given a set of forces, A, B, c, j> ; given also three lines, x, Y, z, as the positions of three forces which are to balance our given set. Find these three. The method here 134 APPLIED MECHANICS. is not so obvious as the method of the last example. Draw A', B', c', D', known sides of the force polygon. Draw, also, after D', x' parallel to x unlimited in length. In it choose o the pole and join to all known corners of the force polygon. Note that o x' is itself two radiating lines from o, because there are two corners of the force polygon in it. Now let the point where D and x meet be called a side of the link polygon ; the intercept on x till it meets Y is another side, and it will now be found that we have sufficient information for the completion of both force and link polygons. As an example, Fig. 74 represents a ladder whose centre of gravity is at G, and weight 300 Ibs. A string fastened to it at c in the direction c o keeps it in equilibrium, its end A resting on the smooth wall o A, and its end B on the smooth floor o B. Find the pull in the string and the reactions at A and B. The forces acting on the ladder are shown by the arrows. Draw Y x (Fig. 75) vertically to represent the weight of the ladder. Draw w Y horizontally of unknown length. Draw Fig. 74. Fig. 75. x o parallel to o c, and take o anywhere in this line. Use as pole of the force polygon. Join o Y. Now the link polygon is M N P M, and drawing o w parallel to P M, and w z parallel to B M, we find that Y x z w is the force polygon. The lengths of x z, z w, w Y represent the forces at c, at B, and at A. The student will find it very instructive now to introduce friction into this problem, and thus create two new problems. (1) If the pull in the cord is just sufficient to prevent the ladder from moving. Here the angle OEM ought to be 90 0j where tan. 0j is the coefficient of friction at B, and GAP ought to be 90 + 2 if tan.

e ^c., from a line in the plane, and if A is j + a 2 + etc., the whole area, and if x is the distance of the common centre of area (often called the centre of gravity of the area) from the line, then oS = (j x l + a z x z + etc.) -j- A. 110. We can use a graphical statics method of finding the centre of mass, or area G. Thus, let there be masses or areas, w l5 w 2 , etc., whose centres are at the points 1, 2, etc. (Fig. 76). Draw parallel forces 11, 22, etc., in any direction proportional to m lt m 2 , etc., and find the resultant by the above method. Suppose M N to be the direction of the resultant. Now repeat the process, taking the parallel forces of the same magnitudes as before, but in a different direction, and let MP be the direction of the resultant. Evidently M, where these lines meet, is the centre of gravity of the masses or areas. This method may often prove useful, for areas especially. Thus, to find the centre of gravity of any given area, divide it into any suitable number of parts, so that the centre of gravity and area of each part may be found easily. If we divide the area by parallel lines, these lines may be drawn equidistant, and the area of each part is approximately given by the length of the line which separates it from either of its neighbours. A repetition of the process has been employed to determine the moment of inertia of the area about any given line. 111. I do not advise students to adopt this link polygon method of finding centres of gravity or of calculating moment of inertia. A practical engineer will always apply the ordinary formula to find the centre of gravity of an area. Thus, if you want the centre of gravity of the figure M N o P (Fig. 77), draw two parallel lines, G H, K o, touching the figure at two opposite sides. Draw a line, K G, at right angles to G H, and APPLIED MECHANICS. 137 divide it into any number of parts, each equal to d. Draw the lines A B, c D, etc., parallel to G H, so that they are at the distance d apart, the distance from A B to G H, or from Y z to K o being J d. It is evident that if N x p is a line parallel to G H through the centre of gravity, then approxi- mately _j ~~ 2 ' etc. -7- Fig. 77. AB + CD + EF + etC. We have thus obtained one line through the centre of gravity, and in a similar way we may obtain another such line, and their point of intersec- tion is the centre of gravity required. Sometimes we choose as our line of reference, a line, G H, which cuts through the area ; in this case distances on one side of the line are to be considered negative ; and if G H happens to pass through the centre of gravity, of course the sum % a x is 0. We often cut the shape of an irregular area from sheet zinc and balance it in two positions on a straight edge to find the centre of gravity of the area.' Exercises. 1. Masses whose centres are in a straight line at A, B, c, D are 4 Ibs., 8 Ibs., 7 Ibs., 6 Ibs. ; where is the common centre of mass if A B = 0-5 feet, A c = 2 feet, and A D = 2 feet ? Ans., A a = 1-32 ft. 2. A disc 8 inches diameter, 2 inches thick, with a hole 4 inches 8.86" Fig. 78. <-. 3" --> X ... f_\* 1 r, <- 5.64" 3. 71" *' 1 v diameter ; centre of hole, 1 inch from centre of disc ; where is the centre of mass ? Ans. , inch from centre. 3. ABC is an equilateral triangle, each side being 3 inches long ; particles whose masses are 1, 2, 3 are placed at A, B, c respectively; find their centre of gravity by construction, and note its distance from A. An$., 2-18 138 APPLIED MECHANICS. 4. ABC D is a square lamina of uniform density; E, F are the middle points of A B and B c. If the corner of the square is turned down along the line E F, so that B comes on to the diagonal A c, find the centre of gravity of the lamina under the new circumstances. Ans., $ of the diagonal from the centre. 5. A circular disc, 8 inches in diameter, has a hole 2 inches in diameter punched out of it, the centre of the hole being 3 inches from the circumference of the disc. Find the centre of gravity of the remaining portion. Ans., 0-0667 inch from centre of larger circle. 6. A thin plate of metal is in the shape of a square and equilateral triangle, having one side common; the side of the square is 12 inches long. Find the centre of gravity of the plate. Ans., 2-86 inches from centre of square. 7. Find the centres of area of the areas in Fig. 78. 112. To find the moment of inertia, i, of a great number of little masses about an axis, multiply each mass by the square of its distance from the axis, and add up. If the whole mass is M, we often write M k 2 = i ; and when i and M are known we can calculate &, which we call the radius of gyration of the mass about the axis. To find what is called the moment of inertia, i, of a great number of little areas about a line in their plane, multiply each by the square of its distance from the line, and add up. If A is the whole area, and A & 2 = i, we call k the radius of gyration of the area about the line. 113. Just as we found centres of gravity, so we may obtain the moment of inertia, i, of any area about any line; or, as is often the case, suppose we wish to find the moment of inertia of M N o p (Fig. 77) about NXP, a line which passes through the centre of gravity. Evidently the moment of inertia about G H is approximately i = ^3(AB + 9cD + 25EF + etc.)/ 4 . Students ought to practise this method first upon a rectangle and a circle whose moments of inertia have been worked out for them by the calculus. Now, it is well known that the moment of inertia of an area about any line is equal to its moment of inertia about a parallel line through its centre of gravity, together with the product of the area into the square of the distance between the two lines. Hence, the moment of inertia about NXP is i = i ox 2 . d (AB + CD + EP + etc.). It will be found in practice that this easy way of carrying out, simple ideas is better than the compli- cated use of the link polygon method for finding moment of inertia. If the area may be divided into rectangles D ! whose sides are parallel to and per- pendicular to the axis, we need not subdivide these rectangles. It must be remembered that the moment of inertia .- i . of any given area such as a rectangle ' about any axis is equal to the area multiplied by the square of the dis- tance of its centre of gravity from the APPLIED MECHANICS. 139 axis, plus the moment of inertia of the area about a parallel line through its centre of gravity. Thus the rectangle A B c D (Fig. 79) is known to have, about N M N, the moment of inertia A. B B C^ j-~ - so that the moment of inertia of the rectangle about ofoo'ia A B . B C 3 /B C 2 , ,\ r-^ I-AB.BC.MO 3 , OrAB.BCI -TT^- + M O 8 I 12 \ 12 / It is mainly by the use of this rule that we have found the moments of inertia of the various sections shown in Table VI., and all these ought to be worked out as exercises by students. The student will find it good exercise to take a few sections of angle-iron, T-iron, rails, and other specimens of rolled iron, and find by the above graphical method the position of centre of gravity of each section, and the moment of inertia of each area about any line through the centre of gravity. The exact forms ought to be taken from real specimens. If the area is symmetrical, one line through the centre of gravity can always be found by mere inspection. In using this or any other graphical method, it is well to know what is the error involved in having each strip of width d, instead of being infinitely narrower. We ought to add to the sum in (2) the moment of every strip about its own central line that is, d 3 (AB + CD + EF, ete.)/ 12 or A.d 2 /^, if A represents the total area. It is easy to see that in a rectangle of depth D, if we divide into strips of breadth d, the fractional error is an d if the greatest moment of inertia is about o B, and is -^ then A B A' B' being an ellipse whose major and minor axes are A A' and B B', the moment of inertia about an axis, o c, is -^ 115. To know the principal moments of inertia of an area is important for many purposes. It is specially important in regard to struts. A strut will bend in such a way, that the axis through the centre of gravity of a section, at right angles to the plane of bend- ing, is the axis about which there is least moment of inertia of the section. The ellipse lets us see the moments about all axes. To draw it for any particular section, if the section is symmetrical, as in sections of Fig. 80, we know that the axis of symmetry and the axis at right angles to it are the principal axes of inertia. If the section is not uo APPLIED MECHANICS. symmetrical, as in the case of an angle-iron, we find the moment of inertia about three axes as o P, o Q, o R of Fig. 81, and we set off the distances o P, o Q, o R to represent the reciprocals of the radii of gyration. We now have the graphical problem: given three points P a R of .an ellipse, and its centre o to draw the ellipse. Mr. Harrison thinks the following solu- tion better than any other. With centre o, radius o Q, de- scribe a circle, and through H, where P R cuts o Q, draw the chord pr such that j?H:Hr = PH:HR. Draw the radius os perpendicular to o Q, ; through s draw * M, * N Pig. 80. parallel respectively to p o, or; andi through N and M draw lines parallel to R o, o P, intersecting in g ; then o s is conjugate to o Q. fig. 81. For proof, suppose the ellipse orthogonally projected into a circle, and let the smaller figure be similar to this projection. This figure can be drcwn, remembering that parallel lines project into parallel lines the mutual ratios of whose lengths remain APPLIED MECHANICS. 141 unaltered ; also that conjugate diameters project into perpendicular diameters. The solution given consists in drawing this figure with oq coinciding with o o,, and then locating s. To determine the principal axes, through s draw a line (not shown) perpendicular to o Q, and on it take two points D, E, opposite ways from s, such that SD = SE = OQ. Then the axes of the ellipse are respectively equal to the sum and difference of o D and c E, and the major axis bisects the angle DOE. MOMENT OF INERTIA OF A RECTANGLE. 116. The moment of inertia of a rectangle about the line o o through its centre, parallel to one side. Let A^ = b, B c = d. Consider the strip of area between A - 23 o P = y and o o, = y + Sy. Its area is b . 5y, and its moment of inertia about o o is b . y 2 . Sy ; so that the moment of inertia of the whole rectangle is J * The moment of inertia of the area ABC about an axis o at right angles to the area is equal to the sum of the moments of inertia i x and ly, about o x and o y, axes at right angles to one another Fig- 82. in the area. For if a, is a portion of area at P, ix is the sum of all such terms as a . P R 2 , ly is the sum of all such terms as a . p o, 2 ; and the sum of each such pair of terms is a term a . o P 2 . 117. Moment of inertia of a circle about its centre. Consider the ring of area between the radii r and r + 5r. Its area is Z-n-r . 5r more and more nearly as Sr is made smaller and smaller, and its moment of inertia is 2 irr B . 8r. The integral of this is ^TrR 4 for a circle of radius R. The square of the radius of gyration is ^R 2 . Now, in this case i x = iy, each being half of |?rR 4 ; so that the moment of inertia of a circle about its diameter is -jTTR 4 , or V* 71 " 04 if D is the Fig. 88. diameter. 118. The student ought to be able to prove the propositions referred to in Art. 112 : 1. As to mass or inertia. To prove that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the centre of gravity together with the whole mass multiplied by the square of the distance between the two axes. Thus, let the plane of the paper be at right angles to the axes. 142 APPLIED MECHANICS. Let there be a little mass w at P in the plane of the paper. Let o be the axis through the centre of gravity of the whole mass, and o' be the other axis. We want the sum of all such terms as TO . (o' p) 2 . . Now, (o' p) 3 = (o' o) 2 + o P 2 + 2 . o o 7 . o Q, where Q is the foot of a perpendicular from p upon o o 7 , the plane containing the two axes. Then, calling 2 m . (o' p) 2 by the name i, calling 2 m . o p 2 by the name I , the moment of inertia about the axis o through the centre of gravity of the whole mass, then i = (o' o) 2 2 m + I + 2 . o o' . 2 m . o a. But S m . o Q means that each portion of mass m is multiplied by its distance from a plane at right angles to the paper through the centre of gravity, and this is zero. So that ^e proposition is proved. Or, letting 2 m be called M, the whole mass, i = i + M.(o'o) 2 . 2. To prove the proposition about areas. Let o o (Fig. 85) be the axis through the centre of gravity, and o' o' a parallel axis. We want i, the sum of all such terms as a (o' p) 2 , and this is the same as 2 a . o P 2 + 22a . OP . oo'+ 2a. oo' 2 . But 22.op.oo'= 2.oo'2a.op, and this is from our definition of centre of gravity ; so that i = I + o o' 2 a. EXERCISES. 1. A fly- wheel has a rim of rectangular section, the outside and inside radii being 8 feet and 7 feet. What is the error in assuming the radius of gyration to be 7 '5 feet? Ans. The true radius of gyration is 7 '5 16 feet, and hence the assumed radius of gyration is '21 per cent, wrong. It would give a moment of - inertia -4 per cent, wrong. 2. Find the radii of gyration of the sphere of Table II., p. 251, about a line touching its surface ; the solid cylinder about a line touching its - outside, parallel to the axis ; the rod about a line at right angles to it at one end. Ans., -5916 rf; -514 rf; '5773*. 3. What error is introduced in Table II., by neglecting the size of the section of the prismatic rod ? 4. Two homogeneous spheres of weights 12 and 20 Ibs., radii 0'2 and 0'3 feet, their centres 5 feet apart ; find the distance of G, the centre of gravity, from o, the centre of the smaller ; find i , the moment of inertia about an axis through o at right angles to o G ; find the moment of inertia about o G itself ; find the moment of inertia about any line G A if th and x, and the tensile forces in the links are proportional to the lengths of the lines om,on,op, and o q. For it is evident that the three forces at E, keeping the joint in equilibrium as they do, must be proportional to the sides of the triangle omn. If you put arrow-heads on om and on circuital with the one already on m n, you will see that the bars B E and D E do not push the joint E ; they pull it and are tie-bars. Thus, then, the lengths of the lines in Fig 105 represent, to some scale, all the forces acting at the joints, c, D, and E. All so easy as it is to prove that the above proposition is correct, it is well to illustrate its truth in the laboratory. Let A, c, D, E, B be a string fastened to a vertical board at A and B, with loads cc, y, and z applied at c and D by means of strings passing over pulleys. The string, instead of being fastened at A and B, may there pass over pulleys with balancing weights. Let the vertical board be covered with paper ; with a pin prick points on the paper showing the directions of the string everywhere, and transferring the paper to another board, find what is the pull in A c, c D, D E, and A B. We have no Pig. 105.. APPLIED MECHANICS. 161 actual test of the pulls in c D and D E, unless an ingenious student can introduce very light spring balances whose own weights are negligible. The student may compare this construction with exactly the same construction in Art. 97. He will see that in Fig. 104 the resultant of the forces E B and z is in the direction D E ; the resultant of the forces E B, z and y is in the direction c D ; the resultant of the forces E B, , y and x is in the direction A c. Fig. 104 shows the/Mfttton*, and Fig. 105 shows the amounts of these successive resultants. B E D c A is what we called a line of resistance. 126. Loaded Chain. If we want to find the pull in every part of one chain of a suspension bridge, and to draw the shape of the chain, it is first necessary to know the weight of the bridge at every place. This weight is probably supported by two chains, so, as we have only one chain to deal with, we only take half the weight of the bridge. We shall suppose that there is no long girder or other support for the bridge but the chain. It is usual to suspend the supporting beams of the roadway from the chain by vertical iron rods, placed at equal horizontal distances from one another. We may imagine the roadway to be as heavy at one place as another, so that the pulls in all the rods will be the same. Suppose there are ten rods, and in each a pull of 20 tons. Draw ten equidistant vertical lines (Fig. 106) to repre- sent the rods. We must get another condition before we can draw the chain. Let it be this, that the chain in the middle, where it is horizontal, shall be capable of with- standing a pull of 20D tons. Now draw o H horizontally (Fig. 107), and make its length on any scale represent 200 tons. Make H A and H B on the same scale represent each 100 tons (if your chain is to be symmetrical), and divide them up, so that each portion represents 20 tons that is, the vertical load communicated to the chain by each tie-rod. Now join o with each point of division in A B. Suppose Fig. 106. Fig. 107. \' 162 APPLIED MECHANICS. that P (Fig. 106) is one point of support of the chain, draw p a (Fig. 106) parallel to o A (Fig. 107), a c parallel to o c, c d parallel to o D, and so on till you reach the point Q, which I suppose to be on the same level as p. Of course, the points of support, P and Q, may be anywhere on the lines a P and m Q. It is quite evident from what you have already learnt that the pull in any part of the chain is represented by the length of the line from o, which is parallel to it in Fig. 107, and it is also evident that the chain will take this shape without any tendency to alter. Note that when all the loads on a chain are vertical, the horizontal component of any of the sloping forces is the same as that of any other, being o H. This is always called the horizontal pull of the chain. 127. We began by assuming a pull of 200 tons in the part/A, where the chain is horizontal. We might have assumed a pull of 300 tons in fh ; this would have caused the chain to hang in a flatter curve. Assuming a pull of 100 tons infh, we should have obtained a greater difference of level between p and h. It will be found that in the present case, where the load is supposed to be uniformly distributed along the horizontal, the links would just circumscribe the curve called a parabola. With any other distribution of load they will fit some other curve than a parabola, but in any case you know now how to draw the shape of such a chain, and to determine the pull in any part of it. 128. Arched Rib. If instead of a hanging chain you wanted to use a thin arched rib to support your roadway, then if you have numerous vertical rods by which to hang your load to the rib, and if the distribution of the load is known, you can draw the curve of the rib in exactly the same way, but it will now be convex upwards, of course. With uniform horizontal distribu- tion of your load you will get a parabolic rib. The difference between the two cases is this : a slight inequality in your loads or a temporary alteration will only cause the chain to take a slightly different position for the time, and it will get back to its old shape when the old loading is returned to ; whereas the arch is in a state of unstable equilibrium, and as it is very thin, so that it cannot resist any bending, a slight change of loading will very materially alter its shape and it will get destroyed. Such a rib or series of struts is either stayed with numerous diagonal pieces or else it is made very massive, so that should the line like P h Q (inverted), which APPLIED MECHANICS. 163 is supposed to pass everywhere along its axis, deviate a lifetle from this position, the rib may resist alteration of shape by refusing to bend. 129. The load carried by an arch may either be hung from it by means of tie-rods, or else it may rest on the top of the arch, the weight being carried from the different parts by means of struts or pillars of iron, stone, or brick, or the arch may be levelled up to the roadway by means of a solid mass of masonry, or merely by one or two pillars of masonry, the roadway being carried on little arches from one to the other ; or there may be a filling- in of earth. It is rather difficult in a stone or brick bridge to say exactly what is the load on every portion of the arch, but it is guessed at, and a curve or line of resistance, such as P h Q, Fig. 106 (inverted), drawn. It is shown in Art. 368, that in a stone or brick arch it is dangerous to have the arch so thin that the line P A Q (inverted) passes anywhere outside the middle third of the arch ring. Thus, in Fig. 108 we have a section of a stone arch, the various stones or voussoirs, as they are called, being separ- ated by joints of mortar or cement. Now divide each joint into three equal parts and draw two polygons, mmm and n n n, marking out the middle third of every joint. Let us suppose we know the weight which each voussoir supports, including its own weight (it is usual to consider the arch as one foot deep at right angles to the paper), and let these weights be the weights w lt Wfr etc., shown in Fig. 108. I have taken a case in which these loads are symmetrical to right and left of the crown. Now draw the force polygon, Fig. 109 ; it happens to be all in one vertical line, the forces being all vertical. And now we come to the drawing of our line of resistance, but we are stopped at the outset by not knowing what is the thrust Fig. 109. 164 APPLIED MECHANICS. at the crown of the arch. The pull at the middle of our suspension bridge chain was quite definite, but the thrust at the crown of the arch may be what we please, and the arch will remain stable if the link polygon which we draw never passes outside the middle third of any of the joints.* Suppose we draw any symmetrical link polygon to begin with, by bisecting a k in H (Fig. 109), draw H o horizontal, and take o anywhere we please, o H will be the thrust in the crown of our arch, if this link polygon is the correct one. Join o a, o 1 2, o 2 3, etc. Start from any con- venient point in w^ Fig. 108, say E, within the space which contains the middle thirds of all the joints. Draw E D, Fig. 108, Fig. 110. parallel to o 4 5, Fig. 109 ; draw D c, c B, B A, A P, in succes- sion parallel to the corresponding lines in Fig. 109, and so also for E F, etc., to K Q. If any of the lines so drawn passes out- side the space mm, n n, we must choose some other point E to begin at, and if we find that no choice of E will allow the link polygon to lie altogether within the space mm, nn, then we must choose another pole, o, in Fig. 109, until at length we find, as in the figure, a link polygon, P E Q, which cuts within the middle third of every joint. The lengths of the lines in Fig. 109 tell us the forces acting at the joints of Fig. 108. Thus o a, Fig. 109, is the force PA, Fig. 108, the resistance of the abutment of the bridge. Again, the length of o 4 5 is the force acting in the direction E D between the stones E and D. 130. Professor Fuller has made the work of drawing such a link polygon very easy. In case the loads are all parallel to one * It is obvious also that the link polygon, wherever it crosses a joint, must make an angle so near a right angle with the joint that there can be no slipping or rupture by shearing there. APPLIED MECHANICS. 165 another, it can be shown that if a number of link polygons are drawn in Fig. 108 for different lengths, o H, Fig. 109, then the vertical distances between the points A, B, c, D, etc., are in the same proportion in all the link polygons.* Beginning with the first load w lt draw (Fig. 110) A 1' 2' 3' 4' 5' E, the half of any link polygon corresponding to P A B c D E in Fig. 108. Divide the horizontal A s into any number of equal parts ; I choose six. Erect perpendiculars at A, 1, 2, 3, 4, 5, s. Draw horizontal lines AST and from 1', 2', 3', etc. ; draw any inclined straight line of convenient length, ET; draw vertical lines from T, 1", 2", 3", etc. From the points where the verticals A A', 1 1', 22', etc., cut m m and n n, draw horizontals to cut the correspond- ing verticals from T 1", 2", 3", etc. Join the points so found by the curves m m" and nn" ; then, just as the straight line ET represents a link polygon, m m" n n" represents the area bound- ing the middle third of all the joints, and any link polygon will be represented on the right hand side by a straight line. Now draw a straight line lying altogether within the space m m" n n". If you can draw several, then draw that one which is steepest, in this case c T. Project this over to the left hand side, and you will find that you have the link polygon, which supposes the least thrust at the keystone. The corresponding force polygon has its o H less than the o H of A E in the proportion so to s' E. The proof of this is easy. In the figure the possible line A' s' if is so close to A s T that it is not shown. The proposition that the line of resistance must lie within the region of middle thirds will be dealt with in Art. 368. Fig. 109 shows the amounts of thrust between the voussoirs, and it is worth while comparing these forces with the areas of the joints. In Art. 368 the strength of such joints is more particularly studied It is also worth while to consider whether the line of resistance may not be made to pass outside the region of middle thirds by a possible unsymmetrical load. But we cannot give a proof of the proposition on which the whole work depends. If any line of resistance may be drawn to pass inside the region of middle thirds, >the arch is stable, and the steepest of such lines is the actual line. We have a lecture model in which a number of roiinded blocks of wood CDEFGHIJ represent voussoirs, the fixed parts A and B being abutments. We can load these voussoirs in all sorts of ways. When the loading is changed, the voussoirs will be seen to roll on one another into new * See Art. 349. Proof. Each link polygon is really a diagram of bending moment, supposing the loads were acting on a horizontal beam, and the scalo of each diagram is proportional to tho length of o n. 166 APPLIED MECHANICS. ' positions ; and if the points of contact are joined, we have the line of resistance. When the loading is symmetrical, loads on the haunches cause the line to rise at the haunches and get lower at the crown, and a load on the crown reverses this effect. A student may see on this model how change of loading causes such a yielding in the arch as produces pressures at the abutments just suited to equilibrium, provided only that a line of resistance cuts all the joints. But we must confess that the sudden step in the reasoning from this to the mere statement of the above proposition does not satisfy us. At the actual plane joints of an arch, changes occur when the load is changed, but these changes are very different from the changes that occur in the model; and it seems to us that the only legitimate method of study is that of assuming that a masonry arch behaves exactly like an arched rib of iron fixed at the ends, and Fig. m. this is studied in Art. 380. The line of resistance being found aa for an iron arch, the masonry arch must be so designed that this line is kept within the middle third. In a few cases in Germany an attempt has been made to build masonry arches hinged at the ends, and, indeed, also with a hinge at the crown. This can be done by bedding the masonry at these places upon iron. A quasi hinge has also been employed by introducing plates of lead at the two joints at the springings, and one at the crown, the lead only extending over the middle thirds of the joints. 131. Buttresses. To find the force which acts from one stone to another in a buttress, it is necessary to know the force acting on every stone from the outside, and also the weight of the stone. Find the resultant of these two forces for each stone, and draw the link polygon whose first side is the force on the top stone. In Fig. H'JpABKpis the link polygon so APPLIED MECHANICS 167 drawn. Each side of it shows the resultant of the forces acting at every joint, and the length of the corresponding line in Fig. 113 shows its amount. Thus the resultant of the forces acting at the joint s T is shown by the position of the line B K, and its amount is shown by the length of the line o t in Fig. 113. If we see that any of the sides of the link polygon passes outside the middle third of the corresponding joint between two stones, we know that part of that joint will be subjected Fig. 112. Fig. 113. to tension, a condition to which we suppose that a common masonry joint ought not to be subjected.* The student ought to work examples in which, besides the force F acting on the top stone, there are forces acting on the other stones, due, say, to the pressure of water (Art. 173), or to the pressure of earth (Art. 293). 132. To return to the hanging chain. If the total weight of a chain and all the vertical loads upon it is w, and if its ends are sup- ported at two points, A and B, and at these points it makes angles a and with the horizontal, the tensions there are represented by the two sides of a triangle, which are parallel to the directions there, the .third side, representing w, being a vertical line. Let the student sketch examples. * In many cases it will be found well to magnify all the horizontal components of all the forces, magnifying the horizontal dimensions of all the stones in the same proportion. In this way the points in which each side of the link polygon cuts each joint may be found more accurately. 168 APPLIED MECHANICS. The tension T at A is evidently w cos. /8/sin. (a -f j8), and the tension at B is w cos. a/sin, (a + 0). If at B the chain is horizontal, then T at A is w/sin. a , and the tension at B is w cot. o. Call this T O . If # is horizontal distance of any point in the chain from B (where the chain is horizontal) and y is vertical height above B ; if the load on the chain "between B and any point A is wx where w is the load per unit length of horizontal projection of the chain [here I assume that the load is uniformly spread horizontally, and this is nearly the case in any very flat chain or telegraph wire], then, as "before, if o is the angle of inclination of the chain at A, we have TO = wx cot. a ; or, in the language of the calculus, T O ~~- = wx ; and it follows that the shape of the chain is y = % a? 2 .... (1), T o a parabola with vertical axis whose vertex is at B, the lowest point. Given the heights of the two points of support of such a chain ahove its lowest point, and also the horizontal distance between them and w, it is easy to calculate the value of T O and everything eise. EXERCISE. In a suspension bridge of 800 feet span and 80 feet dip, the weight of the platform, chains, and rods, etc., is 2| tons per foot run, what will be the horizontal tension in each of the two chains, and the tension in each at the point of support at the piers 1 Ans., 1125 tons ; 1212 tons. 133. In a telegraph wire of span I, the points of support being at the same level, if the dip (supposed to be small compared with 1} is a, the weight of wire being w or wl nearly, it is easy to show that the tension, which I shall here call p, is wP/8a or vrl/8a at the lowest point, and the tension elsewhere is not much greater. Now sup- pose s, the whole length of a wire, to be b (1 + kQ + j8p) where b is its length at the lowest temperature, and this we shall call the zero of temperature, under no tension, being the temperature Centigrade at any other time above the zero, k the coefficient of expansion, and j3 a constant easily determined when one knows Young's modulus for the material and the cross-section of the wire, just as w is known from the material and cross-section. If a is the dip, find the tension p, and find the length of the wire. The length of a parabolic curve is easily found by integration, but x + 2 y 2 /3 x gives the length from the origin to any point with sufficient accuracy for such purposes as the present. [Check this statement when you have leisure.] Sp. ;md p P Q = u>h. T APPLIED MECHANICS. 219 Example 2. Take w to be variable, say w = cp, a rule that holds for gases at constant temperature, then (3) becomes ~ = cp, or = an p c . dh, or log. p = ch + c where c is a constant. Let p = p Q where h = o, then c = log. jt? , and we have log. P/Po = eA . . . . (4). The actual value of c depends upon the constant temperature supposed to be maintained. Example 3. Take w to be variable, say w = cp l lv where y = 1-414 for air, being the ratio of the specific heats. This is the law which is much more likely to hold in a mass of gas than the constant temperature law. Then = cp l h, oTp~ l h . dp = c . dh, or p 1 ~ l n/(l - i/ 7 ) = eh + c. It is easy to show that this leads to the result that the temperature increases in proportion to the increase of depth in the atmosphere. Example 4. In our whirling fluid it is easy, since F = w V 1 + a 4 a? 2 /y 2 and ~ = V 1 + # 2 /aV 2 , to find from (2) the law of y, if one knows how to integrate. Take the simpler case, in which x is so great that the lines of force may be regarded as horizontal, and the level surfaces vertical circular cylinders. Then letting 5s be called 5#, as being the radius of the path in which a particle revolves, w 9 , w F = o 2 #, and a?x . dx dp. g g (1) If w is constant, -^ a? 2 p + c where c is some constant. o Let p=p where x = a? , and we have p-p Q = -- (# 2 -> 2 ) .... (2) If w = cp k , then ~ = - * "* T a ^ 2 *y If we take k= where y=l - 414 for air, and the easily found value of c for any given conditions, we can find the increase of pressure due to centrifugal force in a mass of whirling gas, as in the wheel of a fan when it is not delivering much fluid. 220 CHAPTER X. MACHINERY IN GENERAL. 176. Mechanism. -When the power of a steam-engine is dis- tributed through a factory, the distribution is performed by means of shafts, spur and bevil wheels, belts and pulleys, and other kinds of gearing. As we are writing for men who have observed such transmission of energy, it is no part of our object to describe here what can be seen in any workshop. Perhaps no study is more useless from books alone than the study of mechanism ; whereas it is very useful and easy if we examine the actual thing, or make a skeleton model or a skeleton drawing. At the same time it is necessary to read books. The present book deals with the kinetics of mechanism ; but there is another part, a preliminary part, and students must read some book giving descriptions of contrivances and the mere kine- matics of the subject. We give a short sketch of certain important principles here, and in Chap. XXVI., because they are necessary for the proper understanding of our own division of the subject. 177. Velocity Ratio. In any machinery the velocity of a point may be calculated when the velocity of any other point is known. The number of revolutions per minute made by a shaft tells us the velocity of any point on any wheel or pulley fixed on the shaft ; the circumference of the circle described by such a point, multiplied by the number of revolutions per minute, is evidently the distance moved through by the point in one minute. Now, when one shaft drives another by means of spur or bevil wheels, or by two pulleys and a strap, it is evident that the number of revolutions per minute made by one of the shafts, multiplied by the number of teeth of the wheel, or by the circumference or diameter of the wheel or pulley, is equal to the number of revolutions per minute made by the other shaft, multiplied by the number of teeth, or by the cir- cumference or diameter of the other wheel or pulley. This is evidently true, supposing that the strap does not slip on the pulley. Hence the rule to find the speed of a shaft, driven from another by means of any number of wheels and pulleys, multiply the speed of the driving shaft by the product of the APPLIED MECHANICS. 221 diameters or numbers of teeth in all the driving wheels or pulleys, and divide by the product of tJie diameters or numbers of teeth in all the driven wheels or pulleys. By the diameter of a spur wheel we mean the diameter of its pitch circle. Two spur wheels enter some distance into one another, and the circle on one which touches the circle on the other (the dia- meters of these circles being proportional to the numbers of teeth on the wheels), is called the pitch circle. The circum- ference of the pitch circle, divided by the number of teeth, gives the pitch of the teeth. 178. Shapes of Wheel Teeth. We know that if two spur wheels gear together, however badly their teeth are formed, so long as a tooth in one drives past the line of centres of a tooth in the other, their average speeds follow the above rule. But if we want the speed ratio at any instant to be the same as at any other instant, it is necessary to form the teeth in a certain way. The curved sides of teeth ought to be cycloidal curves. The proof of this is not very difficult ; it is given in Art. 462. It is not usual to employ these cycloidal curves, for it is found that certain arcs of circles approximate very closely to the proper curves. The method of drawing rapidly the curved tooth of a wheel you will find taught by every teacher of mechanical drawing, you will find described in a great number of books, and you will see it in use in the workshop.* You must remember that no study of books, and I may also say, no fitter's or turner's work that you may engage in, will make up for want of the experience which you would gain by actually drawing to scale a spur or bevil wheel, a bracket or pedestal with brasses, and a few other contrivances used in machinery. A worm and worm-wheel, that is, a screw, every revolution of which causes one tooth of a wheel to be driven forward, is sometimes used when we wish to drive a shaft with a very slow speed. If the worm-wheel has 30 teeth, it evidently makes one-thirtieth of the number of revolutions of the driving shaft. 179. Skeleton Drawings. When we, consider the relative motions of, say, a piston and the crank which it drives, we come to something which it is not so easy to state without some little knowledge of mathematics. It is the same with all sorts of combinations of link work, and with cams. Even a good knowledge of mathematics is only sufficient to give one a rough general idea of the relative motion in such cases ; and for tho * Consult Professor Un win's " Machine Design " ^n the teoth of wheels. 222 APPLIED MECHANICS. study of any special case there is nothing so good as a skeleton drawing or a model. I give one example of the use of skeleton drawings a crank and connecting- rod. Let A and B (Fig. 146) be the ends of a connecting-rod. As A moves from ! to c and back again, B de- scribes the complete circle, b l d b. Set off equal distances, b 5 2 , 6 2 B, B > 4 , etc., and make 6 2 2 , 6 4 a 4 , etc., equal to the length of the connecting-rod. Then the points a^ a^ etc., and 6j > 2 , etc., show in a very good way the relative motions of A and B. When you have finished this exercise, work others in which, with the same length of crank, you have longer or shorter connecting-rods. You will get some such results as are shown in the upper part of the figure. In every case, if we imagine the crank to revolve uni- formly, the motion of A, the end of the connecting-rod, is shown the distance from any one point to the next is passed over by A in the same in- terval of time. Simple Harmonic Motion (see Chap. XXV.) is the name given to the motion of the piston- rod, when we imagine the connecting- rod to be infinitely long ] or rather, as we make the connecting-rod longer and longer, we get more and more nearly to this sort of motion. You see, then, that by skeleton drawings I mean drawings which show successive posi- tions of the different parts of a mechanism whose motions we want to U> ^ ^ not> a mere numlt)er > but we ma y it to be the mere number ~ multiplied by 7 - 9 In truth, h J (seconds). we find out in various ways that b had better be written 2 g where g is an acceleration of 32 -2 feet per second per second in London. We have, then, the rule for bodies falling freely from rest, v = y^ .... (i). Now, h is v^l 2 g, and hence, when a body has fallen h feet, and its velocity is v, the potential energy, wh, lost by it, may be written w or, as it is more usually written, Iw x v 2 ... (2). w is the gravitational force or weight of the body in pounds, and g is 3 2 -2 feet per second per second in London. 191. It was by experiments on falling bodies that Galileo was led to formulate the law that at any place a body's w/g, which we call the mass or inertia of it, is constant ; and we have been led by astronomical observations to believe that the mass or inertia of a body is a constant everywhere. The gravitational force on a body, w, may vary, and g, the acceleration due to gravitational force, may vary; but the ratio between them seems everywhere to be constant. We generally denote the mass or inertia w/g by the letter m. A body kept in London, denned by law to have a weight of 1 Ib. if weighed in vacuo, has a mass 1 ~- 32-2 in the absolute units which are most con- venient for engineering purposes. Any body whose weight in London is w Ib. has a mass numerically expressed by w -f- 32 -2. Any body whose weight in any kind of units is 10 at a place where the gravitational acceleration in any units is g, is said to have a mass w/g. If we wish to be very exact, we notice that if we speak of a force as w Ibs., the w is a mere number ; if we speak of a force as w, then w is much more than a mere number. A body whose weight in London is 32 -2 Ibs. has a mass 1, 192. We have now seen that half the mass of a body, multi- plied by the square of its speed in feet per second, is its kinetic energy. When the bob, Fig. 162, is at B, let us say that its total store of energy is kinetic. When it is at A, the energy APPLIED MECHANICS. 245 is all potential. When it is anywhere between, its total amount of energy is exactly the same as before, but part is potential and part is kinetic. During the swinging of a pendulum there is a constant change going on, potential energy changing into kinetic or kinetic into potential, and the sum of these two would always remain the same only that friction is constantly reducing this sum by converting part of it into energy of another order namely, heat. Exercise. Imagine no frictional resistance to the motion of a projectile. A projectile of 100 Ibs., with a muzzle velocity of 1,000 feet per second. What is its kinetic energy when it is at heights of 10, 50, and 100 feet? When its whole velocity is 707 feet horizontally per second, what is its height 1 This being the horizontal component of the velocity through- out, what was the angle of elevation of the gun 1 Ans., 1,551,000, 1,547,000 and 1,542,000ft. Ibs.; 8,241 ft.; 45. 1$ is to be noticed that we assume that the potential energy of a body is w, multiplied by the vertical height through which it can fall (that is, to some fixed datum level), so that at a certain height we know its potential energy ; and if we know its total energy, the remainder is kinetic, and the kinetic being always m- v 3 , the velocity is known when the height is known. '2i 193. Test of our Law. We now have our rule to calculate the energy stored up in a moving body, every part of which is moving with the same velocity, and we can test it in the following way. Get a pulley (Fig. 163) as light and frictionless as possible, because we must, at the beginning, neglect both the energy stored up in the pulley itself and the loss by friction. Fasten the pulley at a considerable height above the floor. Let two equal weights, A and B, balance one another at the ends of a long silk cord, passing over the pulley ; and let there be a wooden scale, close alongside which A passes as it ascends and descends. Let us be able to fix to this scale, at any place, a plate which will suddenly stop A, and, above this, a ring which will just allow A to pass through. You will find such an arrangement as I speak of in almost every little collection of apparatus in the kingdom, and it is called an Attwood's Machine. Now let A be as high as possible at the beginning ; place on it a little weight w, such as will be lifted off when A passes through a ring ; and place a ring so that it will lift the little weight off A when A has fallen, say, 3 feet. You know 246 APPLIED MECHANICS. that, so long as the little weight lies on A, the speed of A down wards and B upwards must become greater and greater. In fact, the potential energy lost by the little weight becomes converted into kinetic energy of the whole arrangement. Now, as soon as the little weight is stopped, A and B move with a steady motion; and if the table is placed by trial so that one second after A passes tho ring it is suddenly stopped by the table, the distance between the ring and table shows the velocity which A, B, and the little weight had when the little weight was removed. In one experiment A being 1 Ib. and B the same, and the little weight 0'25 Ib. the velocity was measured after A had fallen 3 feet, and was found to be about 4*5 feet per second. Now, the potential energy lost by the little weight was 3 x '25, or *75 foot-pound. The kinetic energy was stored up in 2*25 pounds, moving with the velocity of 4 '5 feet per second, and, according to the above rule, its amount is 2-25 ^ 64-4 x 4-5 x 4-5, or -71 foot-pound, or -04 foot-pound too little. If we consider that there was some friction, that the pulley retained some kinetic energy, and that it was difficult to fix the table, so that exactly one second elapsed from A'S passing through the ring until it was stopped, we see that the experiment is a fairly good illustration of the rule. You ought, with your own hands, to make a number of such experiments. In exercise work we shall call A = B = w Ib. , and we shall call the little extra weight w. In Art. 50 we found the sort of corrections which had to be introduced by friction at the wheel pivots and by the bend- ing of the cord. Their study led us to the great subject of friction. But in a well-made Att wood's machine, a far more important matter to consider is the energy (so far neglected) in the pulley. The experimental study of this leads us to ft consideration of angular motion. Fig. 163. APPLIED MECHANICS. 247 194. Energy in a Rotating Body. Suppose now that the pulley is so massive that its kinetic energy is considerable, and may not be neglected, is there any way of finding from its speed how much energy it has stored up in it 1 We can easily calculate the energy in any little portion of a wheel if we know its velocity and mass, but those portions near the centre are moving more slowly than portions near the circumference, so that we have to calculate the energy in each little portion separately, and add all the results together. There is one thing which all portions of a wheel have in common they all go round the centre the same number of times per minute. Suppose now that the number of revolutions of a wheel is doubled, the real velocity of every point in the wheel is doubled, whether that point be near the axis or not, so that the kinetic energy of the whole wheel is quadrupled ; in fact, then, we find that the kinetic energy stored up in a wheel depends on the square of the number of revolutions which it makes per minute, so that the energy must be equal to a constant number multiplied by the square of the number of revolutions per minute. 195. To find experimentally how much energy is possessed by a wheel when it is rotating, let the wheel be mounted on an axle supported on very frictionless bearings. If the centre of gravity of the wheel is not exactly in the axis, then it is better to place the wheel as in Fig. 164. Now let a cord with a loop from the pin B be wound round the axle and pulled by a weight w. Suppose the weight to be 1,000 Ibs., and that we only allow it to fall 8 feet from rest before the cord drops off the pin, so that when it has fallen this distance it no longer acts on the wheel, which will then rotate with a constant speed. Roughly speaking, the wheel possesses 1,000 x 8, or 8,000 foot-pounds of energy stored up in it. This is not quite true, because the weight itself possessed a certain amount of energy of motion which must be subtracted. Suppose that at the instant before being stopped the weight was moving with a velocity of 1 -5 foot per second, then we must subtract ^? x 1-5 x 1-5, or about 35 foot-pounds. 64-4 If there were no friction, and we find that a speed of 10 revolutions per minute has been given to the fly-wheel, we know that we have to find a constant number, M, which, when multiplied by the square of 10 or 100, will give 7,965 foot- pounds. Evidently M = 79*65, and hence, if ever we find this 248 APPLIED MECHANICS. fly-wheel rotating, we know that it has stored up in it the amount of energy in foot-pounds 7 9 -65 x square of number of revolutions per minute. 196. In the above calculation we have neglected friction ; but, as a matter of fact, in experiments the friction never is negligible. As for the friction of the wheel itself, on a cord similar to that which you have already used, hang a small weight such as will merely overcome friction, so that when you Fig. 164. give the wheel a jerk for the purpose of starting the motion, this weight will just suffice to prevent friction reducing the speed. Suppose this weight to be 5 Ibs., then it is quite evident that 5 Ibs. of the original 1,000 were really employed in over- coming friction and not in storage. Hence our calculation gives 995 x 8 - 35, or 7,925 foot-pounds as the total storage. This is at ten revolutions per minute. When it makes one revolution per minute the storage is 79-25 foot-pounds, and at any other speed we multiply 79 '25 by the square of the number of revolutions per minute ; 79 -25 is called the M of the wheel. As for the friction of the pulley P and the energy waited in bending the rope, the pulley must be tested like the pulley APPLIED MECHANICS. 249 of Art. 50, and a correction made on account of the two parts of the rope being at right angles to one another. Or two pulleys may be tested at the same time, the cord going over both of them to two weights. 197. It is obvious that we must be pretty quick in count- ing the number of revolutions of the wheel produced by the falling of the weight. Indeed, we ought to observe, if possible, the time taken in part of one revolution, using some special form of time-measurer, because the speed will now continually decrease on account of friction. But there is another way in which it is easy to find the speed at the instant when the weight ceases to act. Find the total number of revolutions made by the wheel till the cord drops off its pin, and let someone observe this time in minutes. Then, as we know that the speed increases uniformly during this interval of time, the mean speed is just half the speed at the end of the interval ; that is, divide the number of revolutions by the number of minutes in which they were per- formed, and twice the quotient will give the number of revolutions per minute made by the wheel when the weight just ceases to act. You can test your result by counting the number of revolu- tions from the time the cord drops off until the wheel is stopped by its own friction and dividing by the time which elapses; twice this quotient ought also to be the speed you want to know. 198. It is not necessary even to measure the friction directly, for we found that 7,965 foot-pounds were given out by the weight in falling; now if we count the total number of revolu- tions made by the wheel from the time of starting until stopped by its own friction, and divide 7,965 by the total number, we shall find the loss of energy due to friction during one revolution, since there is just as much energy wasted by friction in any one revolution as in any other [a statement to be tested by the student]. Ten times this must be the same amount of energy as 5 x 8, or 40 foot-pounds, for we .measured the friction during 10 revolutions of the wheel as equivalent to 5 Ibs. falling 8 feet. This, then, is the method we ought to employ. I know of no exercise in my laboratory which is so useful as this. A thoughtful student revives his knowledge of nearly all the important dynamical principles in making the corrections, which enable him to get nearer and nearer to the correct answer for M. 250 APPLIED MECHANICS. 199. You see that M is a number which ought to be known for every fly-wheel ; it is just as important to know the M of a fly-wheel as to know the weight of an ordinary body. We have only to multiply the M by the square of the number of revolu- tions per minute, and we find at once the energy in foot-pounds stored up in the wheel. I have shown you how to find the M of a fly-wheel by experiment ; I will now give you an idea of its value in different cases. Imagine a grindstone whose diameter is 4'5 feet, whose breadth is 1*4 foot, the weight of its material per cubic foot being 132 Ibs. ; then we can calculate its M by first finding 132 x 1-4 x 4-5 x 4-5 x 4-5x4-5, and dividing this answer by 59,800. For any rotating object of cylindrical shape, the shape of a grindstone, this rule will always find M. Multiply the weight of the material per cubic foot by the breadth or width ; multiply this by tJie fourth power of the diameter, and divide by the constant number 59,800. Whether the material is wood or stone or metal, this will give M, and this multiplied by the square of the number of revolu- tions per minute will give the energy in foot-pounds stored up in the rotating body. For the above grindstone, on calculating out, you will find the M to be 1'26. So that when it makes 1 revolution per minute, there is stored up in it 1-26 foot- pound of energy ; when it makes 2 revolutions per minute, there is stored up in it 1-26 x 4, or 5-04 foot-pounds. Foot-pounds. At 3 revolutions, 2-53 x 9, or 22-68 20 2-53 x 400, 1,012 50 2-53 x 2,500, 6,325 100 2-53 x 10,000, 25,300 200. If we fix a small weight of 20 Ibs. on a wheel, at 12 feet from the axis, this adds to the M of the wheel the amount 20 x 12 x 12^5,873, or 0-49; or the weight multiplied by the square of its distance from the axis, divided by 5,873. If we add a very thin rim to a wheel, the addition to M ia found by multiplying the weight of the rim by the square of its average radius, and dividing by 5,873. Note. In Table II. all dimensions are supposed to be in feet See Art. 203. APPLIED MECHANICS. TABLE II. 251 Nature of Rotating Body. I M k Sphere of diam-^ eter d, rotating I about diameter i as axis. . . / wd 5 X -001626 wd*+ 112, 166 3162l(d*-di*) -r 59,814 { |N x -00305 | 3535y* +rf , 1 t *S| > .. i i i < 4, > Thin rim, mean} radius r of > weight W . . J wr 2 -j- 32-2 wr 2 -5- 5,873 r Thin rod, of N length Z, rotat- ing about axis th rough its middle point, at right angles to its length. Weight of rod W/ wZ 2 '00258 w? 2 ^ 70,474 288U Thin rectangular s plate, rotating about axis through its cen- tre parallel to side 6, the side d being at right angles to axis. Weight of plate w . : . . ., W(P -00258 wd 2 -i- 70,474 '2882d i 252 APPLIED MECHANICS. It will be found that if a fly-wheel has light arms and a heavy rim, as we often see on such wheels, a fairly good approximation to its M is found by multiplying the weight of the wheel by the square of the mean diameter of the rim, and dividing by 23,000. Example. The rim of a fly-wheel weighs 15 tons; its mean diameter is 20 feet. Calculate approximately what energy is stored up in it when it makes 60 revolutions per minute. Here you will find the M of the fly-wheel to be about 584, and hence the stored energy is 584 x 60 x 60, or 2,102,400 foot-pounds. The mathematicians do not use our M. They use a quantity called I, the moment of inertia of a rotating body, which is numerically equal to twice the energy stored in the body when it has an angular velocity of one radian per second. See Art. 203. I give the values of I as well as of M in the above table. In the case of any rotating body, if we imagine a fly-wheel made of the same weight and same M or I, with an exceed- ingly thin rim, the average radius of this rim is called the radius of gyration of the body. In fact, the mass of a body multiplied upon the square of its radius of gyration is its moment of inertia. The radius of gyration of a fly-wheel rim is usually taken to be the average radius of the rim. 201. Steadiness of Machines. A fly-wheel is put upon a riveting- or shearing-machine, or other machine, because the supply of energy to the machine is not given regularly, or else because the demand for energy from the machine is irregular. The fly-wheel enables the machine to maintain a more constant speed. In calculating the proper size of a fly-wheel for any machine we must know two things : first, what is the greatest alteration of speed allowable in the case ; and secondly, the greatest fluctuation of the demand and supply of energy. Thus, suppose we wish never to have the speed of the fly-wheel more than 51 nor less than 49 revolutions per minute, and that during some interval of time the fly-wheel has to give out 20,000 foot-pounds more than it receives during that time ; then, although the fly-wheel will afterwards have this deficiency made up to it by some steady supply, it is obvious that its speed must diminish. We wish its speed to diminish only from 51 revolutions to 49 revolutions per minute in this interval of time. Now, when the fly-wheel runs at 51 revolutions, it has stored up an amount of. energy equal to APPLIED MECHANICS. 253 its M X 51 x 51 ; and when it runs at 49 revolutions, its store is M x 49 x 49, and the difference between these two ought to be 20,000. Hence, subtracting 49 x 49, or 2,401, from 51 x 51, or 2,601, we get 200; and dividing 20,000 by 200, we find 100 as the required value for M. Subtract, then, the square of the least speed from the square of the greatest, and divide the greatest excess of demand or supply by this remainder ; the quotient is the M of the fly-wheel. Having found M, the question is, how can you tell from it the size, and weight of the wheel 1 ? Find the M of any wheel of the same shape and material as that which you want to use. It is obvious that the diameters of the wheels are as the fifth roots of their M's.* We want a wheel whose M is 100. Suppose I find a wheel of the shape I wish to use whose outer diameter is 8 feet, and I calculate its M, and find it to be 11 ; then The fifth root of 11 : fifth root of 100 :: 8 : answer. Log. 11 = 1-0414; divided by 5 it is 0-2083, which is the logarithm of 1-615. Log. 100 = 2-0; divided by 5 it is 0'4, which is the logarithm of 2 -5 12. Hence 1-615 : 2-512 :: 8 : answer. This is an easy exercise in simple proportion. I find my answer to be 12-44 feet, or 12 feet 5J inches, the diameter of the required fly-wheel, which is to be similar in form to the smaller specimen used by me for calculation. 202. The total kinetic energy stored up in any machine is found by calculating the energy in every wheel and in every * If we have any two similar wheels, or other rotating bodies of the same material ; if we consider any similar small portions of them ; it it evident that their weights are proportional to their cubic contents, or to the cubes of any similar linear measurements. Hence, if one is, say, twice the diameter of the other, as every dimension of the one is twice that of the other, the weight of one must be 2 X 2 X 2, or eight times that of the other. Now, the M of any rotating body depends not merely on the weight of each portion of the body, but on the square of its distance from the axis. so v that the M of one must be 8 X 2 X 2, or thirty-two times the M of the other. Similarly, if the linear dimensions were as 3 to 1, the values of M would be as 243 to 1 for a pair of similar wheels. Example. "We want a wheel which will have a store of 1,000 foot-pounds when rotating at twenty revolutions per minute, and it is to' be of the same shape as that of an already existing wheel, which is 4 feet in diameter, and which contains a store of 1,350 foot-pounds when running at 30 revolu- tions. Evidently the M of this second wheel is 1,350 -5- 900, or 1'5, and the M of the first wheel is to be 2 "5. Using logarithms, we find that the fifth root of l - 5 is to the fifth root of 2 "5 as 4 feet is to 4 '4 feet, the answer. 254 APPLIED MECHANICS. moving part, and adding all together. But suppose that in the machine there is some shafb of more importance than any other, it is usual to give the speed of this shaft only, because if its speed be doubled, the speed of every other is doubled. Thus, in a steam-engine we state the number of revolutions per minute of the crank shaft, and this tells us the speed of every part of the engine. Let, then, the number of revolutions of some such principal axle of a machine be found. If this number of revolutions is doubled, the kinetic energy stored up in the machine is quadrupled ; and, in fact, the kinetic energy stored up is equal to a certain number which can be found for the machine, and which we shall call its M, multiplied by the square of the number of revolutions of this particular axle per minute. The M of any machine may be experimentally deter- mined in exactly the same way as we have shown above. If we know the M of any machine, then the M of any other machine made to the same drawings, and of the same materials, but with all its dimensions twice as great, is thirty- two times as great, because the M's of the two machines are proportional to the fifth powers of their corresponding dimensions. 203. The energy stored up in a rotating body is equal to ia 2 , where I is moment of inertia about the axis ; that is, the sum of all such terms as mass of a little portion multiplied by the square of its distance from the axis, a is angular velocity in radians per second. Hence, as a = -^ , if n is number of revolutions per minute, and IT is 3*1416, the energy is i n v . ; so that our M is I ' v . In 1,800 1,800 Table II. we give values "both of M and of I. In both cases w is in Ibs. per cubic foot, and all dimensions are supposed to be in feet. We ask for the advice of students in an important matter. Is it good to use the idea of our M, the energy stored in a body when it makes one revolution per minute ? To the weaker vessel, the beginner, or the man who dislikes algebra, we know it to be useful. Indeed, to any practical engineer, however mathematical he may be, it is convenient to have such a quantity. But is its convenience overpowered by the inconvenience of having a new quantity to think of ? We do not know, and we should like to receive advice. As for the idea of moment of inertia, it comes in in this way. If we have a small mass m moving round an axis at the distance r feet with angular velocity a, its linear velocity is ra feet per second ; its kinetic energy is fyn)- 2 a? foot-pounds. Now, if we have many small masses and make this calculation, we see that every little term has a 2 in it, common to them all Hence we multiply APPLIED MECHANICS. 255 every little m by its r 2 and add up, calling the sum I, or, as we say mathematically, Swr* = i, and evidently ^ia 2 is the kinetic energy. In any book on the calculus, exercises will be found on the calculation of the various values of I given in Table II. Every student who knows how to do so ought to calculate all these, and also the value of I for various other bodies in such shapes as may be defined mathematically. EXERCISES. 1. A fly-wheel of cast iron, whose rim is 10 feet mean diameter and section 8" x 10", has a volume 8 x 10 x lOn- x 12 cubic inches, and as a cubic inch weighs 0'26 Ib. the weight is 7,840 Ibs. Its moment of inertia is, approximately, 7,840 x 5 2 -~ 32-2, or 6,087. Find its M also. The M of a fly-wheel is its kinetic energy when going at one revo- lution per minute. This is an angular velocity a = 2ir 60, or 0*1047 radians per second; so that M is \ x 6,087 X (*1047) 2 = 33'48. 2. What energy will this fly-wheel store in changing from 98 to 102 revolutions per minute? Ans. , M (102 2 98 2 ), or 26,782*8 foot-pounds. 3. A gas engine has 6 indicated horse-power at 150 revolutions per minute ; what is the indicated work of one cycle (or two revolutions) ? Ans., 6 x 33,000 -*- 75, or 2,640 foot-pounds. 4. If one-half of this is stored in changing speed from 149 to 151 revolutions per nJnute, what is the M of the fly-wheel ? What is its moment of inertia ? If it is made to the same drawings as the fly-wheel of Exercise 1, what is the average radius of its rim ? What is its weight ? Ans., 2*2 ; 400 ; 2-9 ft. 1,534 Ibs. 5. If the gas-engine of Exercise 3 goes at 200 revolutions per minute, and if there is a complete cycle in every revolution and half of its energy is stored, what is the M of the fly-wheel if fluctuation of speed is to be the same as in (4) ? Ana., 0-46. 6. In Attwood's machine each w = 1*5 Ib., w = 0'3 Ib. If w is applied for a height of 2 feet, what ought to be the velocity ? Ans., Total mass is 3*3 -*- 32-2, or 0-1025, and its kinetic energy is X *1025 x v 2 . This is equal to 0'3 x 2, or 0*6 foot-pound. Hence v a = 1-2 -4- 1,025, or v = 3*42 feet per second. 7. If in Exercise 6 the wheel weighs 0*17 Ib., with a radius of gyration which happens to be equal to the radius of the circle of the centre of the cord as it goes round, we have simply to imagine the moving mass to have 0-17 Ib., or rather 0'17 -r 32-2, or -0053, added to it. What is the corrected value of v at the time when w is lifted off ? Ans., The whole mass is now *1078, and | x *1078 v 2 = 0*6 ; so that v = 3-34 feet per second. 8. If the radius of the cord circle on the pulley is 0*24 foot and the radius of gyration of the pulley is 0-21 foot, , imagine the pulley of 0'17 Ib. and radius of gyration 0*21 replaced by another of x Ibs. and radius of gyration 0*24; so that x x (0'24) 2 = 0-17 x (0*21) 2 , or x = 0-13. We therefore imagine a mass 0*13 + 32*2, or -0040, added on to the original weight. In this case find t*. Ans., v = 3-36 feet per second. It is this sort of calculation, when experimenting with an Attwood's machine, which gives a man a practical knowledge of mechanics. 9. A homogeneous cylinder of mass m and radius r rolls down an 256 APPLIED MECHANICS. inclined plane. If the linear speed of its centre Is v, show that its angular velocity a about the centre is v/r. What is its kinetic energy ? Ans., |M? a + io 2 . Table I. shows that its radius of gyration, k, is -3535^; and as i = mk 2 , we have the kinetic energy fyn (v 2 + A 2 a 2 ), or \m (> 4. 2 ^\ t or % mv * (l + ^ ) Notice that in a cylinder & 2 /r 2 ^ . go that the total kinetic energy is 3wv 2 / 4 . If the cylinder is cast iron, r = 0-1 7 foot, its length 0-3 foot; then its weight is 12-257 Ibs., its mass is 0-3808, and its kinetic energy is -2855V 2 . If this cylinder has rolled without friction along a switchback path through a vertical height of 1 '5 foot, show that v 8'02. If the path was an inclined plane, such that for a vertical fall of 1-5 feet the distance traversed is 20 feet, and if the average velocity was half the final velocity, find the time taken to reach this point. Ans., 4 -99 seconds. On a plane at the angle a with the horizontal, show that the time taken for a sloping distance I feet, being I -f- %v, is V3l/y sin a. 10. A riveting-machine needs 3 horse-power; a fly-wheel upon it fluctuates in speed between 80 and 120 revolutions per minute; an operation occurs every two seconds, and this requires fths of all the energy supply for two seconds. Find the M and I of the fly-wheel. The energy supply for one minute is 3 x 33,000, and |ths of the supply for two seconds is 3 x 33,000 x x ^ x 2, or 2,875 foot- pounds. This is equal to M (120 2 80 2 ) ; so that M is 0'36, I is 36 x 1,800 -T- 7T 2 , or 65 -6. 11. A machine (consisting of a grindstone and the shafting of a shop) has an M = 12 '13. Suppose the loss of energy by friction in one revolution to be always 1,550 foot-pounds, whatever the speed. If the machine (the grindstone spindle) is making 140 revolutions per minute and ceases to receive energy, in how many revolutions will it come to rest? Let the answer be x ; then x x 1,550 = 12-13 x 140 2 , so that so = 154 revolutions. As the average speed will be 70 revolutions per minute, the stoppage will take 2-2 minutes. 12. Find the M and I of a fly-wheel which stores 25,000 foot-pounds of energy when changing in speed from 50 to 70 revolutions per minute. It is to be similar, and of the same material, to an existing fly-wheel whose M is 0-047. What is the ratio of their sizes ? ^ws.,10-4; 189-7; 2-93: 1. 13. If the earth described a circular path of 92 x 10 6 miles radius in 365 days, if it were a sphere of 8,000 miles diameter, of uniform density 6-6 times that of water, what is its mass in engineer's units ? What is its kinetic energy because of the motion of its centre? What is its kinetic energy because it rotates on its axis once in 23 hours 56 minutes 4 seconds? 14. An engine is running at 200 revolutions per minute. Suppose that after the steam is shut off and the load removed the engine made 250 revolutions before coming to rest. Assume a constant moment of resistance, and calculate its amount if the moving parts are capable of APPLIED MECHANICS. 257 storing as much energy as a fly-wheel weighing 1 ton and having a radius of gyration of 2| feet. Ans., 73 '6 Ibs. 15. The rim of a fly-wheel is 9 inches broad and 4| inches deep, the external diameter being 9 feet. Find the moment of inertia of the rim. If an engine be supposed to drive this wheel (neglecting other resistances), how many revolutions will have been made before the speed acquired from rest by the wheel is 120 revolutions per minute, the diameter of the cylinder being 18 inches, the stroke 2 feet 6 inches, and the mean effective pressure 15 Ibs. per square inch? What time would be required ? Ans., 2134 ; 17'6 revs.; t=i8'8. 16. The fly-wheel of an engine ef 4 horse-power, running at 75 revolu- tions per minute, is equivalent to a heavy rim, 2 feet 9 inches mean diameter, weighing 500 Ibs. Estimate the ratio of the kinetic energy in the fly-wheel to^the energy developed in a revolution, and determine the maximum and minimum speeds of rotation when the fluctuation of energy is one-fourth the energy of a revolution. Ans., 1 : 1-94 ; 75'2, 74*8. 17. A fly-wheel is made to rotate by means of a weight of 5 Ibs. attached to the end of a cord passing round the shaft, the diameter of which is, together with thickness of cord, 1 inches. Find the moment of inertia of the wheel if the weight falls a distance of 8 feet from rest in 1 minute, friction being neglected. Ans., 4-35. 204. Exercise. In the Attwood's machine, if the pulley of moment of inertia I is supported on four friction wheels eacn of moment of inertia Ij about its own axis, and if the spindle of the pulley has a diameter one- twentieth of the diameter of the rim of each friction wheel, show that the pulley has a virtual moment of inertia I -f '01 Ij. In a good Attwood's machine calculate carefully the virtual moment of inertia of the pulley. Find experimentally for any values of A and B the lessening of the preponderating force which represents the friction and stiffness of the cord. If you can work with a discarded old machine with massive wheels and much friction and hemp-cords, you will find it far more instructive than an expensive, nearly frictionless contrivance. In measuring time, exercise a little ingenuity of your own in getting accuracy ; but if your teacher has made everything very easy for accurate measurement, you will learn nothing. If you have read too much about other people's methods, you will learn very little. You will learn most when you try to get accuracy of measurement with very rough apparatus. Teachers who do not think will introduce roughness and inaccuracy in the wrong places ; they will give you a bad timekeeper, perhaps, or an inaccurate scale for measuring distances. After having worked with the Attwood's machine, in which the velocity ratio is 1, you will find it very interesting (if you have time enough) to experiment with the apparatus shown in Fig. 22, in which the velocity ratio may have any value. In the experiments described in Art. 51 you aimed at keeping speed constant. Now let there be an excess weight B, and measure the kinetic energy as with the Attwood machine. We need not here describe more fully the obviously interesting and instructive experiments which may be made. 205. In a Bull engine (see Fig. 16 6) steam-pressure in the space B, p Ib. per unit area (in excess of the pressure from above the piston)acts on a piston of area A with a total upward force p A. 258 APPLIED MECHANICS. This causes a total weight w to be lifted (consisting of steacn- piston, rod, pump-plunger, etc.). If p varies when the piston rises through a height h, the total work E done by the steam must be calculated for small changes of level, and the results added up. Thus in Table HA. we give the pressure of the steam as the piston rises, the pressure being measured by an indicator. The area of the piston is 900 square inches, so that 900 p is the lifting force. It is therefore evident that the fourth column shows at each place (approximately) the work already done upon the piston when w (w = 20 tons, or 44,800 Ibs.) has risen h feet. If it is rising with the velocity v feet per w second, it possesses wh foot-pounds of potential and J 1> 2 foot- pounds of kinetic energy. But we know the total E ; and if we subtract w^, we know the kinetic energy, and we can therefore calculate v. TABLE HA. p Pounds E, the total w fcFeet. ser Square A p Pounds. Work done in wA. V 2 - v. Inch. Foot-pounds. y 100 90,000 0-5 100 90,000 1-0 100 90,000 90,000 44,800 45,200 8-06 1-5 100 90,000 2-0 100 90,000 180,000 89,600 90,400 11*4 2-6 80 72,000 3-0 66-7 60,030 252,000 134,400 117,600 13-0 3-5 57-1 51,390 4-0 50-0 45,000 303,390 179,200 124,190 13-36 4-5 44-4 39,960 5-0 40-0 36,000 343,350 224,000 119,350 13-1 5-5 36-4 32,760 6-0 33-3 29,970 376,110 268,800 107,310 12-41 6-5 30-8 27,720 7-0 28-6 25,740 403,830 313,600 90,230 11-38 7-5 26-7 24,030 8-0 25 22,500 427,860 358,400 69,460 9-98 8-5 23-5 21,150 9-0 22-2 19,980 449,010 403,200 45,810 8-10 9-5 21 18,900 10-0 20 18,000 467,910 448,000 19,910 5-33 10-25 19-5 17,550 10-5 19 17,100 10-75 18-6 16,740 481,600 481,600 APPLIED MECHANICS. 259 206. The earnest student will work out the numbers in the above table, and draw curves showing h and p and h and v. In Fig. 165 o G H u shows p the pressure, o P being the distance travelled by the piston and weight. The ordinates of o P N repre- sent the velocity at every instant. Notice that if we divide the c c B B w Pig. 165. Fig. 166. force p A and the weight w by A, we get jt? and w/A to compare. The straight line K M represents w/A. The area of o K M N is equal to the area o G H u N ; that is, w/A is the average value of the pressure until h = ON. The student must see clearly that the force increasing the velocity is represented to scale by the ordinate of KGHLIJMK, and this force becomes negative after L, so that the point L tells us where v reaches a maxi- mum and begins to diminish. The positive area K G H L K is equal to the negative area L u M L. It is worth while spending a good deal of time over such curves and such an investigation as this. The student ought now to assume that besides the force w there is a constant force of friction (say 2,000 Ibs.) to be overcome. He will therefore subtract 2,000 h foot-pounds TIT from E, as well as w A, to find the kinetic energy, \ - v 2 , and he will calculate v, again obtaining a new curve. In this engine, when the steam is allowed to escape, the weight w falls, and performs the actual pumping operation. 207. Should the advanced student care to take up a problem that more nearly approaches the actual case (namely, assume that the force of friction r is not constant, but is some function of the velocity), he must find out for himself some graphical method of working. Our own plan is this : We calculate v at the end of an interval from a rough notion of the velocity, and therefore of P, the 360 APPLIED MECHANICS. force of friction daring the interval. We can now approximate more closely to the actual friction and calculate more exactly. We advise the student to proceed in this way in the above case, taking F as following the law F = 500 + 100 v. When a torque M turns a body through the angle radians, it does work of the amount M 0. Thus, if a turning moment of M pound-inches is acting on a shaft which revolves at n revolutions per minute, or 2?ra radians per minute, it does work M (pound x inch) 2irn per minute, or vn M 6 foot-pounds per minute. If the twisting moment on a shaft M is proportional to 6 radians, the angle through which a certain length of shaft is twisted, then |M is the strain energy. If bending moment M acting on a certain length of a beam between two cross-sections causes an angular change 0, and is proportional to M, then ^M is the strain energy produced by the action of M. These two propositions enable us to calculate the resilience of many springs ; that is, the energy which it is possible to store up in them. The force F, acting through a little distance Sx, does work F . 5#. If F varies, a curve must be drawn showing its value for each value of a?, and the work done in any distance is shown as an area or an integral. Thus, if F = ax, it is easy to show that the work done from x = to any other value is \ax^ ; or from x = x l to x = a? 2 the work done is \a (o? 2 2 x-?). Or we may put it that the work done is x multiplied by the average value of F, the average value being the half sum of the two extreme values. Thus if the gradually applied load w on a beam produces the deflection y where y is proportional to w, the energy stored in the loaded beam is \w], But if there is no loss of the energy due to loading, any method of loading which during the deflection y is calculated to do the same work will cause the same deflection. One such method is to suddenly apply half the final load or \w. Here, as in structures generally, we have assumed that deformation is pro- portional to load. If the law of yielding is force =f(y] whore there is any curious law, and if the integral of f(y) with regard to y is f(y) ; or if in case the deformation is through an angle 6 the moment M = /(0), then the energy stored in any configuration is F(*/) or r(0). If a force or moment which, gradually applied, would only produce a yielding y or 9 is suddenly applied of its full amount, the yielding is y 1 or 1 where y l f(y) = ?(y l ) or Thus if the righting moment M of a ship is a known function of the heeling angle 8 (say M = A sin. aO), if the moment M due to the wind produces the steady heeling angle 9, and the energy ri A A -cos. ae,dO=- (1 cos. 0), if this a a moment were suddenly applied, and the vessel heeled over under a gust of wind from o to lt the work done is t M or 0jA sin. a6 ; and APPLIED MECHANICS. 261 as this is all stored as - (l - cos. a^), we can calculate 6 lt knowing Q from 61 A sin. a6 = graphically. If the curve o P R v repre- sents the righting moment (QP) corre- sponding to the in- clination 6 (or TP), then a steady mo- ment P Q produces the heel o s if the rectan- gular area o T P u s o is equal to the area of OPRSO. If the law is M = A sin. 40, the vessel cannot The work is easily done Fig. 167. right herself if Q is more than 45 degrees. But if the student studies the matter, he will see that if a steady moment gets the vessel over to anything approaching 45 degrees, she must heel farther than 45 degrees. We are neglecting friction. There is the same moment at 2 as at 6 when sin. 2 = sin. ; and if 2 sin. a0 = - (1 - cos. a0 2 ), the steady wind which would keep the ship at will, as a steady gust acting from = 0, heel the vessel to 2 and beyond. Thus, taking the ahove example a = 4, sin. 40 2 = sin. 40, or 40 2 = TT 40, or 2 = - 0, 2 sin. 40 = (1 - cos. 40 2 ), it cos. 40 2 = cos. (TT 40) = cos. 40, (Z - 0) sin. 40 = | (1 -f cos. 40), cos. 40 = 2 cos. 2 20 - 1, /, 4fl - 1 + cos - 4 * I*." sin. 40 ' ^_- ' sin. 20 v 40 cot. 20 = 0. Calling this (0)=0, 1 find that =*llf degrees. Hence, if a wind is such tiiat it would maintain a steady heel of llf degrees ; if it caught the vessel sud- denly and acted steadily, and we ne- glect friction, the heel would he as much as 33^ degrees ; and if the wind still continued to act with the same moment, the righting moment being now less than the moment due to the wind, the vessel must go on her beam-ends. She will recover from a gust of wind less strong than this. 0deg. *() 18 16 3142 2793 0-5083 0-4241 12 11 2095 1920 0576 - -1015 262 APPLIED MECHANICS. 208. So long as a constant force acts it produces a uniform acceleration in the direction in which it acts. We may experi- ment with Attwood's machine, or simply use, as the body acted on, a small carriage running very freely on a very smooth level table; and tJie force acting, the pull in a string passing over a pulley on the edge of the table, and having weights in a scale-pan at its end, Fig. 168. Here, however, friction will be greater than in Attwood's machine, We can illustrate the following rule. If a force of 2 Ibs. acts on a body whose weight is 50 Ibs. at London (the 50 Ibs. includes the weight of everything which is set in motion, so that if we use a little weight of 2 Ibs. for the purpose of exerting the force, we must remember that this little weight of 2 Ibs. is included in the 50 Ibs. ; we may take 48 Ibs. as the body acted upon, and the pull in the string, which is less than 2 Ibs., as the force), then the acceleration or increase of the velocity every second is equal to the force divided by the whole mass moved. In this case the mass is 50 -f 32*2, or 1-553, so that we have 2-?- 1-553, or 1-223 feet per second per second as the acceleration. Thus, if the body started from rest, the velocity would be 1-223 x 5, or 6-115 feet per second at the end of 5 seconds. And now comes the question, how far will the body move from rest in five seconds ? Evidently its average velocity during this time is half its terminal velocity, or 3-058, so that 5 x 3-058, or 15-388 feet is the distance. It is evident that to get the space passed over we have multiplied half the acceleration by the square of the time. I do not suggest that the apparatus of Fig. 168 is the most suitable apparatus for use in the laboratory ; there is too much friction, and it seems difficult to measure the velocity. Attwood's machine is used for this illustration, and is described MECHANICS. in Art. 193. In both pieces of apparatus corrections must be made for the motion of wheels. When a body falls freely, its own weight is acting on its own mass. For instance, say the weight is 2 Ibs., then the mass is 2 -f- 32 '2, and weight divided by mass is acceleration, which we find to be in every case 32-2 feet per second per second at London. Anywhere else than in London its weight is w Ibs., and its mass is, as before, 2 ^ 32-2 ; and consequently o its acceleration is w 4- -^ But we call this acceleration g, 32*2 and hence w = 2 x '-f ~ We see that anywhere is the 6'2'2. g same as anywhere else. Keeping to London, where g = 32 - 2 feet per second per second, the velocity at the end of any number of seconds is 32-2 multiplied by this number ; and the space fallen through in any number of seconds is half 32*2, or 16-1 multiplied by the square of the number of seconds. You can check these rules by the rules given you for potential and kinetic energy, and you will find them quite consistent with one another. 209, Momentum. If a body's weight is 2 Ibs., its mass is 2 -f 32-2, or '0621. Now, if the body is moving with a velocity of 20 feet per second, its momentum is -0621 x 20, or 1*242. If this momentum is created or destroyed by a force acting for only one second, the force must be 1-242 Ibs. ; if it is created or destroyed by a force acting for 5 seconds, the force must be 1-242 -7- 5, or -2485 Ib. The mass of a body multiplied by its velocity represents its momentum. Momentum is sometimes defined as the quantity of motion possessed by a body. It represents the constant force which, acting for one second, would stop the body. Suppose a certain amount of momentum is produced by a force of 1 Ib. acting on a body for one second, the same amount of momentum would be produced by a force of 2 Ibs. acting for half a second, or by 1,000 Ibs. acting for the thousandth of a second, or by "001 Ib. acting for 1,000 seconds. Example. A bullet of 2 ounces, or -125 Ib., at 500 feet per second, directed towards the centre of mass of a body of 200 Ibs. at rest, in which it lodges and which is free to move ; what is the velocity after the impact ? The momentum before impact is *125 200*125 - x 500. The momentum after impact is __ _ __ x the 62''2 6'2' 2 264 APPLIED MECHANICS. 125 x 500 required velocity. Hence, the answer is 900-125 OI 0*312 feet per second. Exercise. Chipping hammer, 1J Ibs. ; a velocity of 30 feet per second is destroyed in the one five hundredth of a second. What is the average force of the blow 1 Ans., 700 Ibs. (nearly). Example. One hundred and twenty cubic feet of water leave the rim of the wheel of a centrifugal pump every minute ; its com- ponent velocity in the direction of motion of the rim is 25 feet per second. What is the retarding force on the vanes at the rim of the wheel 1 Ans., Two cubic feet of water per second have the mass ; this, multiplied by 25, gives the momentum lost by the wheel per second, which is force, and amounts to 97 Ibs. If the velocity of the rim is 30 feet per second, the work done per second is 30 x 97 foot-pounds ; the work done per minute is 30 x 97 X 60 ; dividing by 33,000, we find 5-56 horse-power. Assuming that there is no force at the inside of the wheel, the water entering radially and without shock, this is the power given to the water. If we neglect friction inside the wheel and also out- side it retarding its motion, this is the total power given to the wheel itself. What is the work done per pound of water? There are 2 x 6 2 '4 Ibs. of water per second, and the work done per second is 30 x 97 foot-pounds, so that the work done per pound of water 30 x 97 is 9775 or 23*3 foot-pounds, or energy sufficient to lift the water to a height of 23-3 feet. 210. If the velocity of a body has been produced or destroyed by various forces, each acting for a certain time, multiply each force by the time during which it acted (each of these products is called an impulse), and the sum must be equal to the whole momentum generated or destroyed. When we know the time during which a certain force has acted on a body giving to it motion, we generally determine the motion by calculating the momentum of the body. When we know the distance through which a force has acted on a body giving to it motion, we generally first find the kinetic energy of the body. 211. Knowing the force F acting at any instant on the mass APPLIED MECHANICS. 265 M, the acceleration a is F -f- M. Thus, suppose the following values of F to be given ; the varying force acting on the mass of a body whose weight is 6 4 -4 Ibs. in London. In engineers' units its mass is 64'4-r 32*2, or 2. Suppose that at time the body has a velocity v=3Q feet per second. Time, in seconds. F, in pounds. a, in feet per second per second. V, in feet per second. *> in feet. 20 10 30 1 20 10 31 3-05 2 19 9-5 31-975 6-199 3 18 9 32-900 9-443 4 16 8 33-750 12-775 5 14 7 34-500 16-188 6 11 5-5 35-125 19-669 7 8 4 35-600 23-205 8 5 2-5 35-925 26-781 9 2 1-0 36-100 30-382 1-0 -1 -0-5 36-125 33-993 1-1 -3 -1-5 36-025 37601 1-2 -4 -2-0 35-040 41-163 To obtain the numbers in column 4, take an example. Suppose we know that when t = *4 second from the beginning v= 33*750 feet per second. Now, in the next O'l second, the average acceleration is approximately J (8 + 7), or 7*5 ; and in the time O'l the actual increase of velocity is 7'5 x O'l, or '75 ; and this is what we add to 33-750 to get 34 '500 the velocity at the end of the little interval. We warn beginners that there is no easier way than this, of getting several very important, essential ideas, and every number of such a table ought to be calculated. Now, notice that s, the space passed over, is made up by multiplying an interval of time O'l by the average velocity during tha,t interval. Thus, at t = -4, s = 12 '7 7 5 feet is the distance passed through since the time t = 0. During the interval from t = -4 to t = -5 the average velocity is f (33 -75 + 34-50), or 34-125 feet per second, and the space passed through in this O'l second is 34-125 x O'l, or 3-4125; and this, added to 12-775, gives s= 16-1875, or, rejecting the last figure, $=16-188 at the r* APPLIED MECHANICS. time =0 '5 seconds from the beginning. Note that in approximate calculations of this kind we cannot be certain of our last figures in each number. It will assist the student now to illustrate this work by curves. Plot a and t so that the ordinate of B c D is a and the abscissa is t. Anyone who thinks a little must see that the area of B c E o re- presents (to scale) the total increase of velo- city at the time repre- sented by o E. Add this on (taking care about the scale of measurement) to 30, the velocity at o, and we have the true velocity at the time o E. Let the velocities be found and plotted as o G H u. In the same way the area of the v curve must give the space s curve ; that is, the area of o G H E o represents to some scale the space s passed over since the time o, and we can now show on a curve s at every instant. This is given as o K L. To repeat : E c represents to scale the acceleration at the time o E ; E H represents to scale the velocity at the time o E ; it is the area of o B c E o added to the velocity at the time o, care being taken as to the scale of the diagrams ; E K represents 8 ; it is measured as the area of o G H E o. A student who works such an exercise as this carefully is getting all sorts of valuable notions, not merely of mechanics but of practical mathematics. Unfortunately, twenty academic exercises can be worked out without thought or trouble to teacher or student, and by the rules of the game this is sufficient for the passing of examinations. For the present, therefore, my advice will be followed by a few earnest students only the men who want to know, the men who are not merely in search of examination tips, the men who find academic exercises difficult because they think about what they do. If such exercises as the above ever become obligatory on all examination candidates, of course their academic friends will discover ways of doing such exercises without the trouble of thinking about them. 212. When the resultant force F acting on a body of mass M is constant, the acceleration a=F -r M is constant. There is no such APPLIED MECHANICS. 267 case in Nature, but it is commonly studied. When a body falls in a vacuum in an ordinary laboratory, and there are no magnetic or electric or frictional effects, we may for all practical purposes assume that the force, the weight of the body w, is constant. The mass is w -r gr, and the constant acceleration is g, or 32-2 feet per second per second in London. If the student treats this case in a table like that of p. 265, or by means of curves like Fig. 169, he will see that v = g t, s = | g fi, and hence that v 2 = 2 g s. Or if v is the velocity downwards at the time t = o and s is the vertical height through which the body falls from t = o to any other time t, then v = V Q + g t, s = v Q t + and 2gs = v'* v ( f. If a body w Ibs. falls without friction down an inclined plane, making an angle a with the horizontal, the constant tmr force is w sin. a, the constant acceleration is w sin. a -f- - r g sin. a. In any case when the acceleration a is constant, v = v -j- a t 213. I have described the units of measurement employed practically, not merely in calculation but in thought, by English-speaking people. In some parts of our work we find it necessary to calculate in a system based upon other units the centimetre as the unit of length, the inertia or mass of the amount of stuff called one gramme as the unit of mass or inertia, and the second as the unit of time. This is called the C.G.S. system. Its advantages lie in its being used by scientific men of all countries. One of its disadvantages lies in this, that all answers to problems must be multiplied by coefficients to bring them into practical language (see p. 655). EXERCISES. 1. A bullet takes 2 seconds to fall to the bottom of a well. What are tbe depth and the velocity at the bottom ? Assume no resistance of the atmosphere. Ans., Depth s = \g (2) 2 ; and taking g = 32'2, s = 100-6 feet. The velocity is 2^, or 79 '5 feet per second. 2. The bullet of (1) leaves the hand with a velocity of 20 feet per second downwards. What are the two answers ? s = 20 x 2 + \g (2i) 2 = 150'6 feet ; also v = 1\g + 20 = 99-5 feet per second. 3. The bullet of (1) leaves the hand with a velocity of 20 feet per second upwards. What are the two answers ? = - 20 x 2 -f \g (2i) 2 = 50 6 feet; v = 2|y - 20 = 595 feet 268 APPLIED MECHANICS. 4. How high did the "bullet reach in (3) ? Ans., v 2 = 2 ff h, or 20 2 -5- 2g = h = 6'21 feet. In the above exercises time and space are measured from leaving the hand. 5. A bullet leaves o, a point on the ground, with an upward velocity of 300 feet per second. Find y, the vertical height of it, at the times t = 0, t !, t = '2, etc., seconds. 6. A bullet leaves o with a horizontal velocity of 400 feet per second, and no force acts upon it to alter its horizontal velocity. Find x, its horizontal distance from o, at the times 0, !, -2, '3, etc., seconds. 7. If a bullet has both the velocities of (5) and (6) when leaving o, plot its positions on a sheet of squared paper at the times 0, !, *2, etc., and show that the path is parabolic. 8. With different horizontal and vertical components, but the same total velocity (500 feet per second), let the bullet of (7) leave o and again plot the path. Do this in several cases. If you knew a little mathe- matics, you could prove that an angle of elevation of 45 degrees will give the greatest range on a horizontal plane. 9. A force acts on a body of 8 ozs. for 6'9125 minutes, and produces a velocity of 10 feet per second. Find the force. Express it in dynes (see p. 655). Ans., 0-000372 lb., or 165'6 dynes. 10. How far will a lateral force of 1 oz. move 100 Ibs. on a smooth horizontal plane in 5 minutes ? *Ans., 900 feet. 11. In Attwood's machine, where the weights are 17 ozs. and 16 ozs., find the acceleration and the tension of the cord. Ans., 0'976 foot per second per second ; 1 lb. 12. How long must a force of 14 Ibs. act on a body of 1,000 tons to give it a velocity of 1 foot per second? Ans., 5,000 seconds. 13. A rifle 5 feet above a lake discharges a bullet horizontally, which strikes the water 400 feet away. What was the velocity of the bullet ? Ans., 720 feet per second. 14. A man weighing 168 Ibs. is standing on the floor of a lift. What force does he exert on it (1) when the lift is stationary? (2) when it is falling freely that is, with an acceleration of 32-2 feet per second per second? (3) when it is descending with an acceleration of 12 feet per second per second ? (4) if it is ascending with the latter acceleration ? In each case indicate by a figure what are the forces which act on the man, and give the resultant force in the direction of motion. Ans., (3) 105 Ibs., (4) 231 Ibs. 15. A jet of water having a sectional area of 12 square inches, and a velocity of 16 feet per second, impinges normally on a fixed plane surface. What is the mass of the water which comes on the plane per second ? What is the momentum of this quantity before impact ? What is the force on the plane ? If the jet impinges normally on a plane surface which has a velocity of 6 feet per second in the direction of the jet, what is the velocity of the water relatively to the surface ? And what is the force exerted on the surface ? Find the amount of work per second necessary to maintain the jet, and the work done by it per second ; and find the efficiency. Ans., 2-58 ; 41-28 ; 41-28 Ibs. ; 10 feetjper second ; 16-1 Ibs. ; 332 ft. Ibs. ; 96-7 ft. Ibs. ; -29. 16. The head of a steam-hammer weighs 50 cwts. j steam is admitted APPLIED MECHANICS. 269 on the under side for lifting only, and there is a drop of 5 feet. What will be the velocity and momentum of the head the instant before the blow is given if the fall is without resistance ? If the time during which the compression of the iron takes place be -^ second, find the average force of the blow. Ans., 17*95 feet per second ; 3,121 ; 111-5 tons. 17. A body has its velocity diminished by one-third. By how much are its kinetic energy and momentum diminished? If this diminution was brought about by a certain constant force acting on the body through a distance of 5 feet, through what further distance would this force have to act in order to bring the body to rest? If, on the other hand, the diminution of velocity had taken place in five seconds, what additional time would be required to bring the body to rest, the same constant force still acting ? Ans., 4 feet ; 10 seconds. 213. Force may be defined as the space-rate at which work is done or any form of energy converted into another, or it may be defined as the time-rate of transference of momentum. We would advise students to make two sets of curves from Table HA., p. 265. The first is given in Fig. 169. The second set Distance. 1 Pressure. P Acceleration. a Area = V 3 2 V ft* t 100 32-5 5 100 32-5 2481 1 100 32-5 32-5 8- 6 2481 1-6 . 100 32-5 1027 2 100 32-5 65-0 11-4 3508 2-5 80 19-5 082 3 66-7 10-9 84-5 13-0 4328 3-5 57-1 4-7 0758 4 50 1 89-2 13-36 5086 4-5 44-4 - 3-5 0756 5 40 - 6-3 85-7 13-1 5842 5-5 36-4 - 8-7 0784 6 33-3 -10-7 77-0 12-41 6626 6-5 30-8 -12-3 0841 7 28'6 -13-7 64-7 11-38 7467 7-5 26-7 -14-9 0936 8 25 -16-0 49-8 9-98 8403 8-5 23-5 -17-0 1106 9 22-2 -17-8 32-8 8'1 9509 9-5 21 -18-6 1490 10 10-25 10-5 20 19-5 19 -19-3) -19-6J-9-8 14-2 4-4 5-33 2-97 1217 1-0999 1-2216 10-75 18-6 -20 ) ~ 4 ' 9 -0-5 1700 1-3916 1-3916 270 APPLIED MECHANICS. ought to have * for the horizontal co-ordinates or abscissae. We hava seen that it was a most instructive problem, when given the a, t curve, to find the v, t and s, t curves. Now, suppose we have the a, s curve and we wish to find the v, s and the t, s curves. As a matter of fact, we have already worked out a v, s curve when given an a, * curve in Art. 211. But let us look at it from our new point of view. In the Bull engine of Art. 205 the force causing upward motion of w is Ap w ; the mass is w/y, and so the acceleration is o= (A^ - w) -7- -, or a = (p - -\ 4- . This accelera- g \ A/ Ag tion is given in Column 3 of the table on p. 269 ; Column 1 shows s, and from * and a we can draw our *, a curve, which is really shown as GHLiJMKof Fig. 165, a being the ordinate measured upwards from KM. Now, a = * = ^ . ^ = v ~ t and hence a . ds = v . dv, or dt ds dt ds I- . ds = v 2 , or twice the area of the a , s diagram up to any place is the square of the velocity. We have, in our usual way, worked the integral of a . ds numerically in the table, and we give the values of v. Note that as v = -j-, or 8t = 5s/v, we get the intervals of time by dividing the intervals of space by the average velocity during the interval. Thus the interval of time from * = 4 to s = 5 feet is 1 foot 4- (13-36 + 13-1), or 0-0756 second; and if we had already determined that t = -5086 for * = 4, we know that t = -5086 + -0756 for s = 5. In this way the numbers of column 6 were obtained. If the v, s diagram is given, and we are asked to find the a, a diagram, we notice that _ dv dv ds dv Therefore o is represented by the sub-normal to the v diagram. We advise no student to use the measurement of the normal of a curve for any useful purpose. It is practically too inaccurate. But from a table of the values of v and s, taken from a curve which will correct errors of observation, values of 8v/8s, and therefore of v Sv/Ss or 0, may be calculated with no great error. The values of t may be obtained as in the last example, and the whole motion is known. Of course if of the variables t, s, v, . any one is known as an algebraic function of the other, it is an exercise in the calculus to find any or all the others in terms of any one ; as also the work done, or the kinetic energy, and other things. Example. A body has reached the earth at London from space, no other force than the earth's gravitation having acted upon it ; what is its velocity ? If r feet is the distance of London from the earth's centre, and at any other place reached by the body if the distance is r, then g, the acceleration at r, is ^ V Let 2^ r, 2 be called i 2 , then APPLIED MECHANICS. 271 J,et = when r = oo, and we need to add no constant. Hence at = -, and therefore r^dr = b . dt. "We take the sign as we imagine the body falling. Integrating again, we have | r I = bt + c. Let t = when r r , so that all our times before reaching the earth will be negative. It is (2) that we asked for. Note that a body of w Ibs., when it reaches r from space, has wi 2 a kinetic energy ^ -- , or w r. "We are therefore prompted to study the problem from the energy rather than the momentum point of view. Imagine a body of 1 Ib. leaving the earth (say at London). At the distance r feet from the earth's centre the weight of the body is ~ if r is the distance of the earth's surface from the centre. The potential energy here being v, and being v + 8v at r + 8r, 5v = weight x 5r = -~ . 8r. Hence v = -- + const. If v is the potential energy of 1 Ib. (we are really in all this neglecting the fact that the earth moves relatively to the object) at y ^ the earth's surface, V = 4- const., so that our constant is V + r . Hence v = - -4- + v + r . The potential energy when r = oo is v + *V Hence we have the easily remembered fact : A body of w Ibs. lifted infinitely high from the earth's surface would receive a store of w r foot-pounds of energy if r is the radius of the earth in feet ; and if we imagine all this converted into kinetic energy, we see that the velocity of the body coming from rest in space would be J 2^ r when g Q = 32-2. If g is the gravitational acceleration at the place r, the 272 APPLIED MECHANICS. body's speed on reaching this place would be -v/ Igr and if we remember that g ^ r oV r2 > calculation is easy, v = V 2^r r 2 /r. If we write v= where t is time, and if V 2y r 2 is called J, then r\ .dr = b . dt, or - f r a = & -f const. Thus, if we count time from the time of reaching the earth's surface, so that it is always negative, let t = when r r . The constant is - fr i, and so t = ^ (r c l - ri). Also, as vr$ = b, r$ = -^ and hence t = (r 5 -- 3), or 125. Force F is rate of change of momentum M. If force F acts for time 8t, it increases the momentum of a body by the amount 5 M ; so we can say either that F = -3-, or that \ F . dt = whole gain ) o of momentum if T is the time during which the gain occurs. Of course if F is constant, we have F T = gain in momentum ; or the whole momentum gained is the force multiplied by the time. But if F varies, we can only say that the whole gain of momentum, divided by the time, is the average force during the time. Here we have a time average. , work done Space average value of force = -,- ; time average . momentum gained value of force = , . ,. -- -. whole time Continually in dynamics we are considering the two great ideas of energy and momentum. On any system if a force acts from outside bodies it gives energy, and it gives momentum. If no force acts from the outside, the momentum and moment of momentum remain unaltered ; and the total energy would remain unaltered were it not that other forms of energy become changed to heat, and a system loses heat by radiation. If the earth-moon system were alone in space, we have to consider that its moment of momentum remains constant, whereas its total store of mechanical energy is diminishing. Professor Purser pointed out that this idea gives us the past and future history of the earth-moon system, and Professor Darwin has worked it out for us. One of the most instructive of laboratory experiments is that in which two bodies, A and B, are suspended so that they may collide, their motions before and after collision being measurable. The whole momentum after impact is the same as before impact ; and it is very interesting to notice that when A strikes B at rest the extent of swing of the two in combination after impact is not affected by the initial rebounding and chattering and the many little circumstances which cause the energy after impact to be quite different from what it was before. Men who do not make experiments of this kind have no clear notions of dynamical APPLIED MECHANICS. 273 phenomena, unless they are very exceptional (that is, men of genius). When the momentum H, as of a pile-driver, is destroyed in the time T, M/T is the time average force of the blow; we have a perfectly definite idea of what is meant. But when we are told that the whole energy of the falling pile-driver, divided by the distance through which the pile is forced into the ground, gives us the resistance of the ground to the pile, we get a misleading academic statement. The "blow is a complicated phenomenon ; and even in the much simpler case of the history of the collision of two hilliard-halls we are only now beginning to see how and where the total energy is converted into heat (see Art. 404). Indicator diagrams of engines are space diagrams of force. Their average heights enable the work done to be calculated. t p a - a8t = - Sv v f& 8s = v5t * p.S* 8-8 1 49 141 028 8-756 1-760 86 2 97-9 281 112 8-772 1-76 4 195-8 562 8-687 3-475 680 G 293-8 844 337 5-24 8 391-7 1-125 8-350 3-334 1,310 1-0 489-6 1-406 562 8-53 1-2 587-5 1-687 7'787 3-115 1,830 1-4 678-4 1-968 787 11-69 1-6 783-4 2-250 7-000 2-800 2,194 1-8 881-3 2-540 1-016 14-49 2-0 979-2 2-812 5-984 2-393 2,350 2-2 1077-1 3-193 1-277 16-93 2-4 1175-0 3-374 4-707 1-883 2,220 2-5 1224-0 3-519 2-6 1175-0 3-374 1-350 18-85 2-8 1077-1 3-193 3-357 1-353 1,460 3-0 979-2 2-812 1-125 20-22 3-2 881-3 2-540 2-233 893 790 3-4 783-4 2-250 900 21-16 3-6 678-4 1-968 1-333 533 362 3-8 587-5 1-687 675 *, 21-74 4-0 489-6 1-406 658 263 129 4-2 391-7 1-125 450 22-04 4-4 293-8 844 208 083 24-5 4-6 195-8 562 225 -017 22-16 4-8 97-9 281 028 055 -007 - -7 5-0 -045 22-19 8-844 21-892 13,635 274 APPLIED MECHANICS. When a mass is vibrating at the end of a spiral spring, the space diagram of the force exerted by the spring upon the body is a straight line. The space average of the force (neglecting the weight of the body) between the end of a swing and the mid position is half what the force was at the end of the swing; whereas the time average of the force in this interval is the o fraction - of the force at the end. jr 216. Example. A body of 5 tons moving at 6 miles per hour ; what are its momentum and kinetic energy? Find the time average of the force which will stop it in 5 seconds. Ans., The mass is 347-8 ; momentum, 1*97 x 10 5 ; the kinetic energy is 13466-8 ; time average .of force = 612 Ibs. If the force increases for 2| seconds, and then diminishes again, in both cases uniformly with time, draw curves showing the velocity with time, and also with distance ; also of force with distance. What is the space average of the stopping force ? Draw a curve showing acceleration a and t (a is negative) ; the integral of this shows v and , the ordinate at t = o being 6 miles per hour, or 8' 8 feet per second. The integral of the v, t curve shows * and t where * is distance from where the force began to act. Now a x mass represents force. Hence the values of a represent force to some scale. Now plot a new curve showing force and s. The whole area of it is the kinetic energy. The average height of it is the space average of the force. Ans., 621 Ibs. We have performed the integrations numerically, and have shown the results in the table on p. 273, and we have shown the various curves in Fig. 169A. Fig. 169A. o A A A v. shows a and t, t parallel to o D. o c B E B shows v and t, t parallel to o D. o 1 1 J B shows * and t, t parallel to o D. o H H H G shows F and *, * parallel to o D. o c v v o shows v and , * parallel to o D. 275 CHAPTER XII. MATERIALS USED IN CONSTRUCTION. 217. Mere reading will give to no student a knowledge of the properties of materials. I insist on the necessity for work in a pattern-shop and a fitting-shop and forge ; and setting work in machine tools. Workshops at a college or school are not intended for the teaching of trades, which can only be done in real shops beside real workmen. A student learns facts about materials which are necessary for his study of mechanics ; it is a secondary matter that he also acquires some skill which enables him to learn his trade quickly in a real shop afterwards. Town boys buy their toys and never come in contact with nature ; country boys are always making things and learning much besides the properties of materials. 218. Stone. The rocks which have once been melted, and have cooled slowly, are usually hard, compact, strong, and dur- able. They are most easily worked when regard is paid to the fact that they naturally divide up into certain regular shapes. They are all more or less crystalline in texture. Stratified rocks are those which have been deposited at the bottom of a sea or river; they are often easily divided in a direction parallel to the layers of which they are built up, but sometimes there are lines of easy cleavage in other directions. These rocks vary very much in appearance, according to the method of their formation, and to the heat and pressure to which they have been subjected, sometimes being very crystalline, strong, and durable, like marble. Slaty rocks may be hard and durable, or soft and perishable ; they are not much used in construction, except as roof covering. Sandstones are hardened sand of very different degrees of compactness, porosity, strength, and durability. There are limestones whose particles seem to form one continuous mass, and which, when they have been subjected to great heat and pressure, become marbles; there are also limestones which are composed of distinct grains cemented together, and which may vary very much in compactness, strength, and durability. Besides these there are conglomerates, in which fragments of older rocks are imbedded. A little knowledge of geology is necessary in order to understand the properties of rocks. Stones are preserved by coating them 276 APPLIED MECHANICS. with some material such as coal-tar, various kinds of oil and paint, or soluble glass, which fills their pores and prevents the entrance of moisture. An artificial stone, which can be made in blocks of any required size and shape, is obtained by turning out of moulds and afterwards saturating with a solution of chloride of calcium, a mixture of clean sharp sand and silicate of soda. The chloride of calcium and silicate of soda produce silicate of lime, which cements the sand together, and thus gradually consolidates the whole mass. 219. Bricks. Bricks are made of tempered (that is, freed from pebbles, saturated with water, and well ground and mixed) clay, moulded, dried gently, then raised to and kept at a white heat in a kiln for some days, and cooled gradually. Sand in the clay prevents too much contraction and helps vitrification. Bricks should have plane parallel surfaces and sharp right- angled edge 3, should give a clear ringing sound when struck, should be compact, uniform, and somewhat glassy when broken, free from cracks, and able to absorb not more than one-fifteenth of their weight of water. They ought to require at least half a ton per square inch to crush them. The published tests are sometimes much more than a ton per square inch. Probably half the published strengths are the true strengths of bricks or of brickwork. The standard brick is 8f x 4$ x 2| inches. The average weight of brickwork is 116 Ibs. per cubic foot. A bricklayer lays from 100 to 200 bricks per hour. 220. Limestone, when burnt in kilns, gives off carbonic acid. If pure it forms quick-lime, which combines readily with water, becoming larger in volume. Mixed with clean sand this forms mortar, which, in the course of time, hardens by losing its water and combining with carbonic acid from the air. If the burnt limestone were not pure, but contained certain kinds of clayey materials, iron, &c., it would not combine with much water, but when ground up fine, water enables its particles to combine chemically with one another, forming compound sili- cates with greater or less rapidity, depending on its composition. Such cement first sets, acquiring a large degree of firmness, and then more slowly and without much expansion becomes as hard as many limestones. These natural hydraulic limestones are not much used now. Nearly pure limestone or chalk is mixed with about one-third of its volume of blue clay to produce -when ground and mixed in plenty of water, then drained and dried, then burnt to incipient vitrification and ground up again APPLIED MECHANICS. 277 very finely indeed an artificial cement, which is equal, if not superior, to the natural cement. This is the Portland cement now in use. Fineness in the particles is exceedingly important. Sand in mortar saves expense, and prevents the cracking ot the mortar in drying; coarse sand seems better than fine. Two measures of sand to 1 of slaked lime or 3 to 7 of sand to 1 of cement are the average allowances, but every person who uses mortar ought to test a particular lime or cement to see how much sand it will bear to have mixed with it. Concre'e is a mixture of 6 of gravel or broken stones and 1 of cement. From the time of setting the tensile strength of cement increases at first rapidly and gradually more slowly. The French standard is 280 Ibs. per square inch at the end of seven days, 500 Ibs. in 28 days, 640 Ibs. in 84 days. The initial strengths of neat cement, 1 of cement to 2 of sand, 1 of cement to 5 of sand being about 1 : J : f- ; the gains of strength in a year are about as 5 : 4 : 3 . Using too much water weakens cement ; water about ^ to \ of the weight of the cement is found to give the best results in testing. The true crushing stress of cement is probably about six times the tensile stress. 221. Timber. A tree is made up of a great number of little tubes and cells arranged roughly in concentric circles, one circle for each year of growth, because the sap which circulates out- side is checked every winter. The process of seasoning consists in uniformly drying the timber. As each little portion dries, it contracts and becomes more rigid, and it contracts much more readily in the direction of the circular arrangement of the tubes than it does towards the centre of the tree, and least easily in the direction along the tree. It is obvious, then, that if the tree is dried whole, there will be a tendency to splitting radially. If the tree is cut up before drying we can tell the way in which the planks will warp if we remember the above facts. Firwoods are easily wrought, and possess straightness in fibre and great resistance to direct pull and transverse load, and are largely used because of their cheapness. They differ greatly in strength, but their weak point is their inability to resist shearing. The best of these is the red pine or Memel timber from Russia, which can be had in large scantlings, and thus used without trussing. The white fir or Norway spruce is suitable for planking and light framing, and is imported 278 APPLIED MECHANICS. from Christiania in " deals," " battens," and " planks." Larch is a very strong timber, hard to work, and has a tendency to warp in drying, and is therefore not suitable for framing, but is largely used for railway-sleepers and fences, because of its durability when exposed to the weather. Cedar lasts long in roofs, but is deficient in strength. The English oak is the strongest and most durable of all woods grown in temperate climates, but is very slow-growing and expensive. Its great durability when exposed to the weather seems to be due to the presence of gallic acid, which, however, in any wood corrodes iron fastenings ; trenails or wooden spikes should be used instead. Teak, which is grown in the East, is the finest of all woods for the engineer. It is very uniform and compact in texture, and contains an oily matter which contributes greatly to its durability. It is used specially in ship-building and railway carriages. Mahogany is unsuitable for exposure to the weather, but it has a fine appear- ance and is not likely to warp much in drying. It is chiefly used for furniture and ornamental purposes, and to some extent in pattern-making. Ash is noted for its toughness and flexibility, and a capability of resisting sudden stresses of all kinds, which make it specially adapted for handles of tools and shafts of carriages. It is very durable when kept dry. It is not obtainable in large scantlings, and is sometimes very difficult to work. Elm is valuable for its durability when constantly wet, which makes it useful for piles or foundations under water. It is noted for its toughness, though inferior to oak in this respect, as also in its strength and stiffness. It. is very liable to warp. Beech is smooth and close in its grain. It is nearly as strong as oak, but is durable only when kept either very dry or constantly wet. It is very tough, but not so stiff as oak. (See also Table VII.) The best time for felling timber is when the tree has reached its maturity, and in autumn when the sap is not circulating. We want to have as little sap in the timber as possible, and in order to harden the sapwood, some foresters are of opinion that the bark should be taken off in the spring before felling. After timber is felled, it is well to square it by taking off the outer slabs. Timber is for the most part dried by putting it into hot- air chambers, from ono to ten weeks according to the thickness. Even when kept quite dry, ventilation is necessary to prevent APPLIED MECHANICS. 279 dry rot. The circumstances least favourable to the durability of timber are alternate wetting and drying, as in the case of timber between high and low water mark, whereas good seasoning and ventilation are most favourable conditions. The most effective means adopted for preserving timber is by saturating it with a black oily liquid called creosote. The timber is placed in an air-tight vessel, and the air and moisture extracted from its pores as far as possible. The warm creosote is then forced into these pores at a pressure of 170 Ibs. per square inch. In this way timber may be made to absorb from a tenth to a twelfth of its weight of creosote. I shall give few numbers here for the strength of timber. Tests of small specimens are not to be relied upon. The time of felling, the duration of drying, the part of the tree from which the specimen is cut, and many other circumstances affect the strength. A beam will sometimes break with a long con- tinued load only about half of what will fracture it in the ordinary way. The ultimate shearing stress along the grain of ash is about 600, oak 850, pine and spruce 300 Ibs. per square inch. For bending, the average value of yin (1) of Art. 341 seems to be for spruce 5,000, yellow pine 7,300, oak 6,000, and white pine 5,000 Ibs. per square inch ; their Young's moduli being respectively 14 x 10 5 , 17 x 10 5 , 13 x 10 5 , and 11 x 10 5 Ibs. per square inch. The crushing strength of timber may roughly be taken as from 4,000 to 3,000 Ibs. per square inch, and its tensile strength as about 10,000 Ibs. per square inch. 222. Glass. Glass is a combined silicate of potassium or sodium, or both, with silicates of calcium, aluminium, iron, lead, and other chemical substances. Certain mixtures of flint and chemicals are melted in crucibles, formed when hot into the required shapes, and cooled as slowly as possible. This may be called the devitrification of glass by slow cooling, giving rise to crystallisation. The more slowly and more uniformly the cooling is effected, the more likely is it that the glass will be without internal strains. When glass js suddenly cooled, as when a melted drop falls into water, the outside is suddenly contracted, becomes hard and brittle, and there are such internal strains that if the tapering part be broken or scratched at the point the whole drop crumbles into a state of dust. A blow or scratch on the thick part produces no such effect. Heating and gradual cooling destroy this property. Many peculiarities in the behaviour of metals when heated and cooled 280 APPLIED MECHANICS. seem to be caricatured in glass, possibly because they are due to the fact that all the portions of matter which are about to form one crystal must be at the same temperature, and when the substance is a bad conductor of heat there is great variation in temperature. Pure metals are good conductors, but the admixture of small quantities of carbon and of gases hurts their conductivity. Toughened glass is the name wrongly given to the hardened glass produced by plunging glass, in a nearly melting state, into a rather hot oily bath. This* glass is somewhat in the condition of the glass in a Rupert's drop. It is so hard that it is difficult to cut it with a diamond, but if the diamond cuts too deep the whole mass breaks up into little pieces. Objects made of it may be thrown violently on the floor without breaking. 223. Cast Iron. Certain chemical changes occur when tho ores of iron, generally oxides, are smelted with coke ; the iron ceases to be in combination with the oxygen, and appears in metallic form, associated, however, with carbon derived from the fuel. There are usually other impurities besides, derived from the same source or from the ores. When the carbon is all combined with the iron the cast iron is "white," and is very hard and brittle. When only a little is combined, and most of its particles crystallise separately, the cast iron is grey in colour ; it is weaker and more fusible. Using the common names for the different varieties, No. 1 is darkest in colour, and from No. 4 to No. 1 there is a gradual darkening in colour. In the cupola of the foundry a little purification is effected, and it is found that the composition of a casting is from 97 to 95 per cent, of iron, the remainder being nearly pure carbon, often very largely in the combined form owing to the elimination of one of the impurities namely, silicon. Nos. 2, 3, 4 are com- monly used in the foundry, mixtures being made of them in various proportions according to circumstances. A greater proportion of No. 3 or No. 4 gives greater strength, whereas a greater proportion of No. 1 gives greater fluidity and a better power of expanding at the moment when the metal solidifies, so that the sharp corners of the mould are better filled. Higher numbers than 4, as 8, 7, 6, and 5 the white varieties are seldom used in the foundry, but they may be converted into grey varieties by cooling from a very high temperature at a slow rate, but much more easily and immediately by the addition of certain brands of cast iron containing special APPLIED MECHANICS. 281 impurities such as " siliconeisen " or "glazed pig."^ A special degree of fluidity and resistance to action of acids is conferred upon cast iron by the presence of a little phosphorus, but this impurity renders iron fragile at low temperatures, just as the presence of much silicon will render it weak and liable to fracture from shock. To soften a hard casting, it is heated in a mixture of bone-ash and coal-dust or sand, and allowed to cool there slowly. The density of oast iron varies from 6 '8 in dark grey foundry iron to 7 - 6 in white iron. Of late years cast iron has greatly improved in strength; this is due probably to our better knowledge. Contracts are sometimes undertaken to deliver iron of nearly 50 per cent, greater strength than the average number of our table, p. 411. The crushing fracture usually makes an angle of 56 degrees with the axis of a test column. The strengths of little round columns of lengths equal to from one to three diameters are much the same, but shorter columns are very much stronger, and longer columns are very much weaker. The reason for this is given in Art. 256. The specified test for cast iron is often this that a bar 3 feet between supports, section 2 inches deep and 1 inch broad, should carry a middle load of 25 to 35 cwt., and will deflect before fracture 0'2 to 0'5 inch. The average ultimate shearing stress is 15 tons per square inch as tested by torsion. Kernel ting im- proves the strength, but not beyond a certain number of times, a tenacity of 6 tons per square inch in the pig becoming 9 tons after the second melting, and 12 tons after the fifth. This seems connected with the change of the iron from grey to white by increase of the combined carbon and decrease of silicon. Mr. Turner has arrived at the following percentages as giving the following qualities in the highest degrees : Combined Carbon. Graphitic Carbon. Silicon. Softness 0-15 3-1 2-5 Hardness , Under 0-8 General strength 0-50 2-8 1-42 Stiffness 1-0 Tensile strength 1-8 Crushing strength Over 1-0 Under 2*6 About 0-8 * 282 APPLIED MECHANICS. 224. Patterns of objects are usually made in yellow pine (sometimes of metal if many castings are wanted), about one- eighth of an inch per foot in every direction larger than the object is to be, because the iron object contracts to this extent in cooling. Gun-metal contracts about one- eleventh of an inch per foot, and brass about twice as much. A thoughtful pattern-maker can often greatly dimmish the labour of moulding. Prints are excrescences made on the patterns to show in the mould where certain cores are to be placed. These cores are made of loam or core-sand in core-boxes, which the pattern-maker supplies ; they represent the spaces in the object where the melted metal is not to flow. They are coated with a wash of charcoal dust and clay. Common casting is green-sand ; there is the more elaborate dry-sand for such objects as pipes, and there is the most expensive loam moulding, in which the mould is built up without a pattern. You must see for yourself in a foundry what are the usual methods of preparing a mould : How the pattern is made so as to draw out easily ; how the surface of the sand is blackened ; how the moulder arranges his vents to let gases escape from the more compact parts of the sand ; how he places his gates to let the metal run into the mould with just enough rapidity and yet without hurt to the mould. You must also see for yourself, taking sketches in your note- book and making a drawing of the cupola, how the pig iron is melted and poured into the moulds ; how the moulder stands moving an iron rod up and down in one of the gates, producing just so much circulation and eddying motion in the melted iron as is likely to remove bubbles of gas which may otherwise be unable to escape from the sides and corners of the mould, as well as to prevent the formation of cavities by shrinkage or "piping"; how in some castings he exposes to the air certain parts which would otherwise cool too slowly for the rest of the object } how next morning he screens his sand and wets it. You ought to observe the appearance of the castings before and after they are cleaned up next morning. 225. The Cooling of Castings. The most important matter in connection with moulding is that there shall be the same amount of contraction at the same time in every portion of the mass of metal as it cools ; otherwise, when finished, there may be internal strains, which very much weaken the object, and often produce fracture. In designing the shape of an object which is to be cast, care is taken that when a thin portion joins a thick APPLIED MECHANICS. 283 one it shall do so by getting gradually thicker, and not by an abrupt change of size. The thin piece exposes more surface, and cooling is effected through the surface. The thin rim of a pulley cools sooner than the arms, and becomes rigid sooner ; when the arms cool they contract so much as sometimes to produce fracture near the junction. In a thick cylindric object the outer portion becomes rigid first; now when the inner portion contracts it tends to make the outer portion contract too much, and the outer portion prevents the inner from contracting so much as it ought to, so that the outer portion retains a compressive strain, and the inner a tensile strain. When a hollow cylinder is cast, and is required to withstand a great bursting pressure that is, all the metal is required to withstand tensile tresses it is usual to cool it from the inside by means of a metal core, in which cold water circulates. The inside now becomes rigid sooner, the outer portions as they solidify contract, and tend to make the inner portion con- tract more than it naturally would, and there is a per- manent state of compressive strain inside and tensile strain outside in the object, which materially helps it to resist a bursting pressure. This inequality of contraction and production of internal strains in objects cause them to vary in their total bulk as compared with that of their patterns, but it is probable that some of this variation is due to the fact that the contraction of grey oast iron is only 1 per cent, of its linear dimension, whereas white cast iron contracts 2 to 2J per cent. The fractional difference between size of pattern and the finished object varies from one-twenty-fifth of an inch per foot in small, thin objects to one-eighth of an inch per foot in heavy pipe castings and girders. Small castings seem to be considerably stronger than large ones. As there is always great inequality in the rate of cooling of a casting near a sharp corner, internal strains may be expected here, and also an inequality in the nature of the cast iron, since the grey variety gets whiter the more rapidly it iscooled. In nearly all bodies a re-entrant corner is a place of weakness (see Art. 303), and is specially to be guarded against in castings. Crystals of cast iron and other metals group themselves along lines of flow of heat. When a plate or wire of iron or steel is rolled or pulled, the crystals become more longitudinal, and the wire or plate becomes stronger, whereas annealing allows the crystals to arrange themselves laterally, and the material is weakened 284 APPLIED MECHANICS. It is said that time gradually reduces some internal strains. Castings which have been rapidly cooled by being cast in an iron mould (painted on its inside with loam) are white, and very hard in those parts which lie nearest the mould, whereas they are grey and strong inside. These, like the hollow cylinder above mentioned, are called chilled castings. When a cleaned casting, preferably of white iron, is put in a box, surrounded with oxide of iron (hematite iron ore or rolling mill scale), and kept at a white heat for a length of time (say a week), its surface, to a depth dependent on the time, loses its carbon and becomes pure or wrought iron, which is much tougher than cast iron. The teeth of wheels are sometimes heated in this way. Such are malleable castings. Malleable cast iron seems to be of 50 per cent, greater tensile strength than cast iron, with a contraction of 8 per cent, before fracture ; it stands about 60 tons per square inch crushing stress. Melted cast iron possesses the property of dissolving pieces of wrought iron, and is then said to be toug-hened cast iron. When sufficient iron is so added it becomes an inferior variety of cast steel. 226. Wrought Iron. Cast iron is exposed to the air in a melted state for a long time, and the carbon is burnt out of it. The pig-iron really undergoes two processes, one called refining, the other puddling. It is then hammered and rolled, when hot, into bars of various shapes. The quality of wrought-iron bars as bought in the market varies greatly. We have common iron, used for rails, ships, and bridges ; best, double best, and treble best Staffordshire iron, used for boilers and forgings generally ; Lowmoor, Bowling, and other good irons for the most difficult forgings ; and, lastly, charcoal iron, which is nearly pure. Iron is softer and more ductile the purer it is. Traces of sulphur make it red-short, difficult to work hot. Phosphorus has effects like carbon, but also makes the iron cold-short, or treacherous eold. Maganese and silicon seem also to act like carbon, but we are still afraid of them. There is usually \ to \ of 1 per cent, of manganese present, and one-third of these amounts of silicon if castings are wanted. Up to the tempera- tures of ordinary boilers, the tensile strength of iron is not much diminished by heating, but at a red heat it is very much less than in the cold state. Repeated forging increases the strength of wrought iron up to a certain number of times, after which it diminishes the strength. This is why small rods and small forgings and the outside of large forgings are respectively of APPLIED MECHANICS. 285 stronger material than large rods and large forgings and the inside of large forgings. Cuts of metal in certain directions from heavy forgings seem surprisingly weak. By rolling and hammering when hot, iron gets a fibrous texture, and becomes more tenacious. By hammering when cold, or by long con- tinued strains of a vibratory kind, wrought iron changes its fibrous and tough for a crystallised and more brittle condition. This brittle condition may be removed by heating and slowly cooling (annealing). Iron wire is stronger the thinner it is. Bar iron is generally stronger than angle or T-iron, and this again than plate iron. The toughness of an iron bar is best shown by the contraction it undergoes before it breaks. The section of a very tough bar may contract as much as 45 per cent, in area. Case hardening of a wrought-iron object is effected by heating it in a box with bone dust and horn shavings. The iron absorbs cai'bon, and is partially converted into steel. 227. Steel. Steel contains less carbon and impurities than cast iron, and thus lies intermediate bet ween cast iron and wrought iron. Expensive steel is produced by giving carbon to wrought iron (the best Swedish charcoal iron), keeping the iron heated for some days in contact with powdered charcoal, and then hammering it, whilst hot, till it is homogeneous (shear steel), or else (and this is the most usual practice) casting it when melted into ingots. .v , Cheap steel is produced by taking only a portion of the carbon from very pure varieties of cast iron by a puddling process such as is employed in the production of wrought iron, or by the Bessemer process. In the Bessemer process, air is forced into the melted cast iron for a time, and very pure white cast iron is then added to help in removing bubbles of gas. In Art. 235 I shall speak about the tempering of steel. Many varieties of soft steel do not harden when suddenly cooled. Some of it is like pure iron almost, and some of it has half as much carbon as the hardest cast steel. v It differs from wrought iron in having been melted, and so there are no minute streaks of slag giving heterogeneousness ; and so a plate is nearly as strong one way as another. The sudden hardening of steel when rapidly cooled seems greater the more carbon there is, up to a certain limit. It is more fusible than wrought iron, and much success has been met with in the production of steel castings in spite of the fact that unless certain precautions are taken, and the 286 APPLIED MECHANICS. addition of silicon, aluminium, etc., there is a tendency to con- tain cavities. This steel has about twice the strength of cast iron. Annealing is necessary after casting. The strength of crucible cast steel is greater than that of any other material, and is greater as it contains more carbon and is harder. The pro- perties of steel depend so much on so many seemingly small things small impurities, a little too much heating or variation in the rate of cooling at different places that great care must be taken in working it. By the Bessemer, Basic, and Siemens processes great quantities of steel are prod need cheaply, contain- ing small percentages of carbon. This steel has largely replaced wrought iron in its use in locomotive rails, bridges, and ships. Taking it that crucible steel has elastic limits of stress 26 to 20 tons per square inch, and ultimate stress 52 to 34 tons, and contraction of section 5 to 20 per cent, before fracture, these are for Bessemer steel 17 tons, 34 tons, and 20 to 45 per cent. Steels to be easily welded together ought to be of the same kind. As steel cools from a welding heat it passes through the " blue heat " stage (about 300 0.), where it is brittle. If hammered or bent in this state it is permanently injured, and if the work was local in a large plate the plate becomes treacherous. Soft steel and iron seem to get a little stronger at very low temperatures. The amount of carbon present in steel does not seem to affect much the Young's modulus, which in all kinds of steel and wrought iron seems to be about 3 x 10 7 , as the modulus of rigidity is not very different from 12 x 10 6 . Various numbers are given in Table III., injdeference to custom. The carbon produces other effects, shown in the following table : % Carbon. Elastic limit stress. Breaking stress. Contraction of area. Ultimate shearing stress. 14 18 28 50 22 51 22 36 25 . 26 96 31 63 10 37 Stresses are in tons per square inch. A formula has been given : Strength = 19 -5 + 114 C 2 + 30 2 + 114 Mn + 9'5 P, and elongation per cent. = 42 - 36 0. 5 -5 Mn - 6 Si, where APPLIED MECHANICS. 287 C, Mn, P, and Si represent the fractional amounts of carbon, manganese, phosphorus, and silicon in the steel. The oxide and silicate skin on cast iron is less liable to corrosion than clean iron, but it is advisable to paint or varnish all iron exposed to oxidation. Sometimes, as for water-pipes, the iron is heated to 150 0., and placed in pitch with some oil in it at 100 C. Mr. Barff keeps the iron surface exposed to superheated steam at a high temperature, and this coats it with a film of protecting oxide. Iron is " galvanised " by putting it in a bath of melted zinc. Boilers are sometimes protected inside by the contact of blocks of zinc. Some alloys of iron and manganese are very strong, and so hard as to prevent their being readily tooled; They can, how- ever, be both cast and forged. They are specially important to electrical engineers, as they are not magnetic. The power of a small trace of manganese to destroy all the magnetic properties of iron is remarkable. Certain alloys of iron and nickel are not quite so hard, but they are extremely tough and strong. Steels containing a little tungsten or chromium have the special properties that fit them for self -hardening tools or projectiles. Mitis castings are of wrought iron, to which 0'5 to 1 per cent, of aluminium has been added to lower the fusing point. The tenacity seems to be 20 per cent, greater than that of wrought iron, and the ductility about equal to that of wrought iron. 228. Copper is noted for its malleability and ductility when both hot and cold, so that it is readily hammered into any shape, rolled into plates, and drawn into wires. When cast it usually contains much oxide and many cavities, but when pure it may be worked up by hammering into a state of great strength and toughness, whereas slight traces of carbon, sulphur, and other impurities necessitate its being refined to do away with its brittleness. The brittleness produced by hammering when cold is very different, as it is removable by annealing. Phosphorus is sometimes used to assist casting, and the strength is greater with more phosphorus, whose function seems that of reducing the oxides. Copper is an expensive metal, and is only used now for pipes which require to be bent cold, for bolts and plates in places where iron would be more readily corroded, and for electrical purposes. Its tensile strength is more reduced by heating than that of iron. Boron seems to affect copper as carbon does iron, the wire 288 APPLIED MECHANICS. alloy standing 22 to 27 tons per square inch without loss of conductivity. 229. Brass consists of about two parts by weight of copper to one of zinc, with a little tin and lead. The copper is first melted and the zinc added, not long before casting. It is used chiefly on account of its fine appearance and the ease with which -it can be worked. Cheap brass things have more zinc usually. The small amount of lead added in melting makes it much softer. Muntz metal contains more zinc than ordinary brass 3 copper to 2 zinc, or 2 to 1 with a little lead. Like copper, it is not much weakened by heating up to 260 C. ; it can be rolled hot. It is used for sheathing ships and for the tubes of boilers. 230. Sterro and Delta metal are brasses to which iron has been added, the latter name being given to the metal after it has also received special mechanical treatment, the rods being made by being forced tp flow under great pressure through dies. Delta metal can be worked hot or cold, and may be brazed. Bronze and gun-metal are alloys of copper and tin in varying proportions, more tin giving greater hardness. Twelve of copper to 1 of tin seems best for guns. Five of copper to 1 of tin is the hardest alloy used by the engineer in bear- ings, but bell metal is 3 -3 to 1 A slight addition of zinc helps in casting, and increases the malleability. A great many experiments have been made on bronze. Its strength depends very much upon the care taken in mixing and melting the metals, for it is easily injured by oxidation. Gun-metal is a good material for castings, which, however, should be quickly cooled to give more uniformity, density, strength, and toughness ; and hence they are sometimes made in cast-iron moulds. Hard bronze is much used for the bearings of shafts. There are also various soft alloys of copper with lead, zinc, tin, and antimony, which are used for this purpose. Phosphor bronze is an alloy of copper and tin to which some phosphorus has been added. It bears re-melting better than gun-metal, and its properties may be varied at will. It may be either strong and hard, or weaker but very tough. The phosphorus appears to act by removal of the oxide of tin, and to tend to prevent segregation in cooling ; and, indeed, we may say that when care is taken against oxidation the difficulties in the way of re-melting gun- metal vanish. Hard wire has broken with from 100 to 150 tons per square inch, and after annealing has stood half these APPLIED MECHANICS. 289 amounts. It has been used successfully in railway axle and crank shaft bearings. It is good in resisting shocks, and has been used instead of steel for chisels in powder factories. Gun-metal slowly decreases in strength as it is heated, until at a certain temperature its strength is suddenly halved, and there is almost no ductility. Phosphor bronze is less affected. 231. Manganese bronze and Silicon bronze are special alloys of copper with manganese and with silicon, the former made by adding ferro-manganese to bronze or brass, and used where unusual strength and power (as for screw propeller blades) of resisting sea-water are required. It may be cast and also forged. The manganese seems to act like phosphorus in clearing off the oxide. Some containing zinc can be forged and rolled hot. Silicon bronze has fair electric conductivity, and resists atmospheric corrosion when used as telephone wire, and is of great strength. As tested for tension in the condi- tion of wire for telephonic purposes, it seems to stand from 30 to 50 tons per square inch ; the drawn copper wire for the same purpose standing about 29 tons to the square inch; phosphor bronze, 44 to 70 ; brass, 25 ; German silver, 30 ; iron, 57 ; Martin steel, 86 ; and crucible steel, 102 tons per square inch. Pianoforte wire sometimes stands 150 tons per square inch. 232. Aluminium bronze is formed of 9 parts of copper and 1 of aluminium, and has a tenacity of 43 tons per square inch. The 5 per cent, alloy stands 30 tons per square inch. Copper with only 2 to 3 per cent, of aluminium is stronger than brass. The usual alloy is of the colour of gold, but, like all aluminium alloys, must be prepared from materials free from iron. When ad\antage is to be taken of the lightness of aluminium, there is an alloy of 32 parts of aluminium and 1 of nickel that can be employed, and to which 1 part of copper may also be added. Specimens have broken with 45 and 50 tons per square inch tensile stress, the first with no elongation and the second with 33 per cent, elongation. The aluminiunrand silicon bronzes and alloys of silver and other metals with aluminium are pro- duced in an electric furnace, the ores being mixed with retort carbon and an electric current passed. About 7 per cent, each of copper and antimony being added to tin, we have white metal or Babbitt's metal, which fuse easily and may be cast inside a bracket, round a journal, as the step of a bearing if not too larga 290 CHAPTER XTTI. TENSION AND COMPRESSION. 233. THE following chapter on the behaviour of material when subjected to tension and to compression has so much to do with the physical properties of matter, that students ought, before reading it, to refresh their memories in regard to the sim- pler principles of chemistry and physics and the notions which laboratory work in these subjects gives to us in regard to the probable molecular condition of matter, and also of that very much larger coarse-grained ness which we sometimes call hetero- geneity. Besides giving us general notions, chemistry gives us useful facts as to the changes which occur in the manufacture of metals, the cause of the rusting of metal, the burning of fuel, etc. A little knowledge of electricity enables us to have clear ideas as to the action by which when two metals touch, and are also connected by liquid, one of them rapidly corrodes and the other does not, and how it is that oil preserves a polished metal surface. A little knowledge of heat tells us how friction wastes mechanical energy ; how heat energy is measured ; when a body is heated how much it expands ; the laws of expansion of gases ; the properties of steam ; the laws of flow of heat in conduction and radiation, and other pheno- mena which continually influence our mere mechanical work. 234. It will be found by students who read what is in the smaller printing in this chapter that the usual statements on which we base our mathematical calculations are very incom- plete. The manufacturer depends on many curious properties of materials, many of which are familiar to and helpful to inarticulate workmen and unknown in the laboratory as yet. Some rude processes and shop beliefs, which sometimes seem to be no more worthy of attention than superstitions, have suggested scientific experiments and new industries. Some phenomena that seem curious at first sight are really easily ex- plained thebehaviour of James Thomson's overtwisted shaft, for example, explains many curious things ; and our knowledge of such things as how initial strains are induced in castings by unequal cooling has helped us greatly in systematising our notions. 235. There is one phenomenon which has been known almost APPLIED MECHANICS. 291 since iron was discovered by man namely, that steel hardens with sudden cooling, and we seem to be almost as far from understanding it as our ancestors were. There has been some advance, for we have no such notions about incantations and the virtues of particular kinds of water for quenching the heat as were held in the Middle Ages. Mild steel is almost like pure iron, and does not harden, but with more carbon (over 0'4 per cent.), such as there is in cast steel, the more rapidly the steel is cooled the harder it gets. So that, for example, thin pieces of steel are apt to be harder than thick pieces. The more carbon a steel contains, the less need it be heated before being suddenly cooled to acquire a particular hardness. Re-heating diminishes the hardness, or, as it is called, tempers the steel ; and the higher the temperature of re-heating, the softer does the steel become. It may roughly be taken that sudden cool- ing in oil doubles the proof tensile strength of the material that is, the stress which would permanently hurt the material. In every workshop the common method adopted for tempering a fitter's chisel is as follows : Heat the chisel to a dull red colour, put the edge in water to a distance of say half an inch, so that it may become very hard; quickly brighten the edge with pumice or a file ; watch it till, as it heats by con- duction from the thicker portion, you know that a certain temperature has been reached by seeing a certain colour (purplish-yellow for a chisel) of oxide of iron making its appearance. When this colour appears, plunge the whole chisel into water. Thus the steel is first made extremely hard at its edge, and is then brought back to the required degree of hardness by re-heating up to a certain temperature and then cooling. This simple process is in common use. In tempering other objects sometimes much greater care must be taken, since it is often necessary that every portion of the object shall be of the same hardness, and in such cases the whole may be cooled at first and then re-heated in a bath of oil, mercury, or other melted metal whose temperature is definitely known. The effect is of the same kind, however, whether the process is the rough one which I have described or a more careful one. There must be no attempt to make large objects glass-hard : they would cool very unequally and might fly to pieces or develop flaws ; a less rapid cooling in hot oil or melted lead tempers such objects in one process. It is interesting to see that one or other of the above two principles is carried out in 292 APPLIED MECHANICS. all sorts of industries, but in a great number of different ways. A certain size of watchmaker's drill is stuck when at a red heat into sealing-wax. This gives the right temper. Another smaller drill is merely waved about in the atmosphere to cool it. The tempering colours of steel, beginning with the hardest, are : Straw colour to yellow, (this is about 220C) for light turn- ing tools, milling cutters, screw-cutting dies and taps, punches, chasers ; straw to purplish-yellow, rimers, wood-chisels, plane- irons, twist-drills ; light purple to dark blue, augers, chisels for steel, axes, chisels for cast iron, chisels for wrought iron, saws for metal ; less dark blue (this is about 320C), screw-drivers and springs. These colours are supposed to indicate fairly exactly the temperature, irrespective of time ; but we cannot say that there is conclusive evidence yet that time produces no effect on the thickness of the film of oxide. If true, it is a very curious phenomenon. But surely it cannot be true ! 236. The volume gets greater in hardening. Curiously enough I have a note saying that steel "becomes denser by hardening, but its authority is unknown. Eepeated hardening and annealing seem to strengthen the steel. It is usual to explain the hardening by saying that in sudden cooling the particles of iron and carbon have not had time to get into their natural positions when cold, and that they jam one another somehow, getting into positions of instability. As regards the influence of impurities, of gases from the atmosphere which may possibly be suddenly imprisoned among the particles, very little is known. _ It may help towards an explanation to say that Abel found that in annealed steel the carbon is in the form of a chemical carbide Fe 3 C mixed in the mass. In hardening, the formation of the carbide is prevented (just as suddenly-cooled gases remain dissociated). At various tempers we have various pro- portions of the carbide, but it is always the same kind of carbide. Speculation as to the molecular constitution of iron does not yet seem to have sufficient facts to go upon. It is sometimes assumed that there are two kinds of iron mixed together, the soft a particles and the harder )8 particles. With slow cooling from a high temperature, when the mass is soft, although the particles may be hard, the /3 particles (practically all are of the kind at a high temperature) change to a. That there is some great molecular change even in the purest iron is evidenced by recalescence and other allied phenomena. In sudden cooling most of the /3 particles have no time to change. Any effect due to the carbon is produced at a much lower temperature than that at which the change from ft to a occurs in slow cooling ; and although the presence of carbon seems necessary to the hardening of steel, changes in its mode of existence are not of much importance. The a particles are changed to j3 in the plastic condition of iron in the ordinary testing opera- tions at low temperatures. APPLIED MECHANICS. 293 237. How a pull is exerted. Howis it that a cord transmits force from my hand to an object when I pull the object by means of a string 1 If you study this matter, you will see that every particle of the string coheres to the next ; and although the refusal of one particle to come away from its neighbour might easily be overcome, there are so many of them to be separated at any particular section of the string that it requires a considerable pull to perform this operation. When a string is pulled it really lengthens a little, and it lengthens more the more force is applied, although it may not break. A string is not so easy to experiment with as a wire of metal, because we find that it differs more in its quality at different sections, and it is affected by dampness arid many other circumstances. No doubt it is also dif- ficult to obtain a metal wire which shall just be as willing to break at one place as another that is, which shall be exactly of the same material everywhere ; but metal wire is certainly more uniform than string. 238. Strain. Take, then, a steel wire, A B (Fig. 170), fastened near the ceiling at A, between two pieces of wood, screwed together firmly so that there may be no tendency for the wire to break just at the fastening. Similarly fasten at B a scale- pan arrangement, and first place just so much weight in the pan as keeps the wire taut. Let there be two light little pointers stuck or tied on at a and 6, and let there be a vertical scale on the wall. Now read off the distance between a and b on the scale, and note the weight. Add more weight, and again read the distance, and continue doing this until the wire breaks. You will prove by means of squared paper that the amount of the extension of a wire is nearly proportional to the weight which produces the extension. 239. My students axe in the habit of using a more exact method Tig. 170. 294 APPLIED MECHANICS. of measurement, a scale hanging from A and two verniers being attached to the wire at a and b. They also use a method in which a vernier on a is brought close to a scale on J, the wire being passed over pulleys. In some cases they use micrometer methods of measurement for greater accuracy, and they also experiment with larger specimens, loading them by means of levers or wheels and screws; and advanced students may be allowed to use machines in which loads up to 100 tons are applied, and arrange- ments may be employed for making automatic records of the load and the extension. If, however, the apparatus is elaborate and imposing as compared with the specimen, a beginner cannot readily pick up the essential idea of an experiment, and hence he had better begin with visible specimens loaded with visible weights. He nray proceed to use such a machine as Bailey's wire-testing machine, and after- wards make a few tests with large commercial testing machines. Bailey's latest form of machine is shown in Fig. 171. The specimen, say of ^-inch wire, is shown at D, being gripped at c and B. By turning the handle, A, we turn a worm driving a worm-wheel, turn- ing a screw whose nut is part of the frame, and so the Fig. 170A. gripping piece B pulls on the specimen D. The other gripping piece, c, tilts the weight, F, and the amount of tilting which measures the pulling force is indicated by a pointer on the dial, E. In some English engineering laboratories experimenting with a steam-engine and testing specimens in tension by means of a large testing-machine are supposed to be the only experimental exercises in which students ought to engage. Tests of specimens in com- pression, bending, and twisting are, however, sometimes made. Consequently a description of all the kinds of large testing-machines which have ever been constructed forms a large part of the college instruction in mechanical engineering. It seems to me, however, somewhat out of place in a text-book, as almost every stu-dent has opportunities of examining some such machine for himself. Complete information will be found in a paper by Prof. Kennedy read before the Inst. of Civ. Eng. (Proceedings I.C.E., vol. Ixxxvi'ii., 1889). The machines in most common use apply tensile load to the lower end of a test-piece by means of an hydraulic press. The upper end is pulled by means of a lever (whose fulcrum is a knife- edge), over which a weight may be rolled by machinery into such APPLIED MECHANICS. 295 Fig. 171. positions that the lever is kept horizontal ; the position of the weight measures the pull. Instruments have heen designed which register on a sheet of paper (as the pencil of a steam-engine indicator does) the load pulling a rod, and the extension which it produces. A little brass cylinder covered with paper is touched hy a pencil on the end of the rod. The amount of rotation of the barrel is regulated so that it is proportional to the load. By this means curves like that of Fig. 172 may rapidly be drawn as the load on the rod is gradually made to increase till the rod breaks (see Art. 244). When Young's modulus for a material is wanted, my advanced students employ Prof. Swing's extensometer. A half -inch bar of iron being given, even a load of 40 lus. causes sufficient extension 296 APPLIED MECHANICS. of the distance between two marks about a foot asunder to be measured with sufficient accuracy for the determination of Young's modulus by this beautiful instrument, shown in Fig. 170A. 240. When we speak of the tensile strain in the wire, and 808 APPLIED MECHANICS. producing permanent set) on a tie-bar of length I and cross-section A, the stress being W/A, the strain is W/AE, the elongation is W//AE, and the work done as the load is increased from to w is |w 2 /AE. This is called the resilience of the bar. Now, the volume of the bar is IA, and hence the resilience per unit volume is -iw 2 /A 2 E. Replacing W/A by/, the proof stress which the material will stand, we have the resilience per cubic inch to be / 2 /2E. If/ is the proof stress in compression, we have the same expression for struts. When a torque M produces an angular change 80, the work done is M . 50 ; and when a twisting moment M has been gradually increased from to M, the twist of a shaft increasing from to 9, the work done is -|M0. When there is a shear stress / and a corresponding shear strain //N, the shear strain energy per unit volume is / 2 /2N ; and if / is the proof shear stress, / 2 /2N is called the resilience of the material per unit volume. The resilience is the strain energy which material may store before permanent set takes place. It is evidently / 2 -f- 2E per cubic inch in tie-bars and struts, if / is the tensile or compressive stress which would produce permanent set and E is Young'a modulus. In shear the resilience is /2 ^ 2w, if / is the limiting elastic shear stress and N the modulus of rigidity. When the stress is not uniform, as in beams and shafts, the average resilience per cubic inch is, of course, less. Shocks due to blows, as of falling weights, will often cause the strain energy to exceed the amount which the material will stand, and local set and plastic yielding may take place. Much depends upon the rate at which strain energy is carried off to the rest of the material. 261. Let us consider why a chisel cuts into an iron plate. When I strike the head of a chisel with a hammer I give to the chisel in a very short period of time a certain amount of energy. This energy is transmitted very quickly to the plate through the edge of the chisel. The shorter and more rigid the chisel, the more quickly portion of material just edge is very great, and the material is fractured there. As the energy of strain is proportional to the product of stress and strain or to the square of either stress or strain, the possibility of fracture for a material is represented by the square root of the strain energy it contains per cubic inch. If a material is brittle, there is a sort of instability which causes fracture at one place to extend to all neig'h bouring places. And hence, if we deliver with great rapidity to a small portion of such a material a moderate supply of energy, it is sufficient to produce a large fracture. As our material becomes less and less brittle, we must have, over a larger and larger part of the volume in which we want fracture to occur, a sufficient supply of strain energy delivered. Hence, in cutting wood, we use a wooden mallet and a more or less lengthened wooden-headed chisel. The mallet and chisel act as a reservoir for the energy of the blow which is delivered to the wood from the edge of the chisel with APPLIED MECHANICS. 309 comparative slowness and just in sufficient quantity to cause rupture in front of the edge. If the wood, without gaining in strength, became more rigid so as to be able to carry off more rapidly the energy given to it by the chisel's edge, it would be necessary to make the supply more rapid by using a more rigid chisel and mallet, and as we do this we must take care that the chisel itself near the edge is strong enough to resist fracture. This is one way of considering the effect of a blow. The exact mathe- matical consideration of what occurs in the impact of elastic bodies is not easy even for spheres and cylinders and other bodies of simple shape (see Art. 404). 262. As we have just seen, the extra stresses due to loads suddenly applied are easy enough to understand. It is not so easy to comprehend why quietly varying loads which pro- duce no visible vibrations should produce what we call fatigue. The ordinary kinds of test as to strength under statical load, ductility, or elongation before fracture, applied to old rails, tyres, and other well-used material, show no great difference from what we obtain with new non-ductile material. Sometimes flaws are found, and it may be that fatigue somehow acts in pro- ducing minute flaws. The nature of the fracture in "Wohler's experiments is the same as that observed in old tyres and axles ; it shows no signs of ductility, and is as if the material were brittle. 263. A piston-rod is subjected to tensile and compressive stresses, often repeated. It is found that its breaking strength is not 45,000 Ibs. per square inch, which, let us say, it would be for a steady pull or push, but 15,000 Ibs. per square inch. If, instead of such an action, we have a tensile stress which varies frequently from 30,000 Ibs. per square inch to zero, the rod will break after a time. In the same way, steel which will bear a steady stress of 84,600 Ibs. per square inch will only bear 46,500 Ibs. per square inch if the stress varies between this and zero, but is always of the same kind ; whereas it will only bear 25,000 Ibs. per square inch if the stress is sometimes a pull of this amount and is sometimes a push of the same amount. The above statement, the outcome of Wohler's experiments begun twenty-five years ago, and Fairbairn's experiments made very much earlier, was made fifteen years ago in the first edition of this book. The experiments made since give results of much the same kind. In the following table we have the most important results arrived at up to the present time, being the stresses in tons which require from five to ten millions or an indefinitely large number of applications of the load to cause fracture : 310 APPLIED MECHANICS. TABLE V. Similar Stresses. One Stress Zero. Opposite Stresses. Material. |J g> HI Least. Greatest. Least. Greatest. Least. Greatest. Wrought Iron. 22-8 12-0 20-5 15-3 - 8- + 8-6 Krupp's Axle Steel 52-0 17-5 37-8 26-5 -14-1 + 14-1 a Untempered |O Spring Steel . 57-5 12-5 34-8 25-5 -13-4 + 13-4 Iron Plate ... 22-8 11-4 19-2 13-1 - 7-2 + 7-2 Bar Iron 26'6 13-3 22-0 14-4 - 7-9 + 7-9 Bar Iron 26-4 13-2 21-9 15-8 - 8-7 -r- 8-7 V Bessemer Mild Steel Plate ... 28-6 14-3 23-8 15-7 - 8-6 + 8-6 J Steel Axle ... 40-0 20-0 32-1 19-7 -10-5 + 10-5 I Steel Eail ... 39'0 19-5 30-9 184 - 9-7 + 9-7 2 Mild Steel n Boiler Plate . 26-6 13-3 22-6 15-8 - 8-7 + 8-7 This exceedingly great weakening in material due to fatigue seems almost as if it had "been vaguely known to English engineers from the beginning, and justifies the larger factors of safety which were wisely used in this country fifty years ago in railway bridges. Professor James Thomson showed that the two elastic limits ought to he called inferior and superior, as they are not necessarily equal positive and negative, hut might even both be on the same side of the zero if the overstraining were great enough, and that variable displacements outside these limits would produce a 'destructive succession of sets. This theoretical deduction from the considera- tion of his overtwisted shaft has been completely verified by the experiments of Bauschinger, and may be said to completely explain the phenomena of fatigue. Bauschinger found that the elastic range does not alter much, although either end of it may be altered, even, indeed, nearly to the ordinary breaking stress. This change of either limit also is not very permanent, altering with hammering and other violent treatment. 264. We can only refer to a few of the ways in which our principles are applied. When bolts tightly screwed up fasten two things the flanges of a cylinder cover, for example the bolts get lengthened and other things (let us call them the cushion) get shortened. If extra forces are now applied it may APPLIED MECHANICS. 311 bo that for a comparatively small extra lengthening of the bolts the cushion is nearly relieved and does not add its force as it did previously ; whereas in other cases, when the cushion is more springy, there may be almost no relief, and the old forces due to the cushion may act along with the new forces. Very often this initial tightening up of bolts is too great, and we have rule of thumb methods of designing sizes based upon an experience of the carelessness of workmen which it is difficult to express algebraically. One rule based on experience is that the length of a spanner shall never much exceed fifteen times the diameter of the bolt. Another, that, in certain kinds of machinery, bolts of less than a certain size shall not be used. Students ought to make careful sketches of the various kinds of bolts and nuts, including the usual forms of lock nuts and ways of locking, and also other kinds of fastening. In study- ing some of the proportions of these important details of machinery our theories are useful in suggestion; in some hands the theory is only a snare. Every true engineer must respect the proportions which have been arrived at by the fit and try, the failure and success methods of generations of engineers. When a very novel thing has to be made the good engineer sighs for practical guidance, and he is very cautious in using his theory. 265. It is found then that when a rod is pulled, with however small a force, not only does it get longer, but its diameter gets less. When, for example, a rod of glass is pulled so that its length increases by the one- thousandth of itself, it is found that its diameter gets less by the one three-thousandth of itself. When a strut shortens, it also swells laterally. The ratio of the lateral to the axial strains in compression or in tension is called Poisson's ratio. It is of great importance in the theory of structures. The nature of the strain in a wire which is being extended, or in a column which is being compressed, cannot be said to be simple. If all lines in one direction, and in one direction only, became shorter or longer, the strain would be called simple, but it needs rather a complicated system of external compression or extension to produce this effect. No^ matter how a body is strained, if we consider a small portion of it we shall find that, besides angular changes, any strain simply consists of extensions and compressions in different directions. In fact, imagine a very small spherical portion of the body before it is strained. The effect of strain is to convert the little sphere into a figure called an ellipsoid (that is, a figure every section of which is an ellipse) or a circle. Eemember that every section of a sphere is a circle. It may be proved that there were three diameters of the sphere at right angles to one another, which remain at right angles to one 312 APPLIED MECHANICS. another in the ellipsoid, and are known as the principal axes of the ellipsoid. These directions are now called the principal axes of the strain existing at that part of the strained body. Along one of these directions the contraction or extension is less, and in another greater, than in any other direction whatever. Example. Thus if M'N' (Fig. 173) is part of a long wire subjected to a pull, the portion of matter which was enclosed in the very small imaginary spherical surface A B c D before the pull was applied is now enclosed in the ellipsoidal spherical surface A' c' B' D'. The sphere has become an ellipsoid of revolution; A B becomes A! B', c B becomes c' D'. The strain in the direction A' B' A B A B is - and this is AB , equal to the pull in the wire Fig- 173. per sqiiare inch divided by Young's modulus of elasticity, B. As, however, it is often more convenient to use a multiplier than a divisor, we are in the habit of using the reciprocal of E, and denoting it by the letter a. Thus, if the pull per square inch is p pounds, it produces a strain of the amount pa. in the direction AB; the lateral contraction of the material is , and CD in this case is usually denoted by p&, the ratio ft/a being Poisson's ratio. 266. In the following exercises on struts, the load is sup- posed to be carefully applied so that there is the same stress at every point of the section : 1. Find the area of the base of a sandstone column carrying a dead load of 6 tons. Take the ultimate crushing stress as 3,600 Ibs. per square inch, and use a factor of safety of 20. Am., 75 square inches nearly. 2. Find the least safe sectional area of a short cast-iron strut to bear a load of 12 tons. Take the safe stress as 15,000 Ibs. per square inch. Ans., 1'8 square inch. 3. The external and internal diameters of a short hollow cast-iron column are 10 inches and 8 inches respectively. If the safe working stress be 15,500 Ibs. per square inch, find what load it will safely bear. If the external diameter had been 7 inches, what ought the thickness to have been to bear a load of 100 tons? Ans., 196 tons ; -73 inch. 4. A tie-rod made from f-inch wrought-iron plate has to sustain a load of 15 tons. What should be its width, allowing a working stress of 7,000 Ibs. per square inch ? Ans., 6-4 inches. 5. The steam pressure in a locomotive boiler is 175 Ibs. per square inch ; the stay-bolts which connect the flat sides of the firebox with the end plate of the boiler are placed 4 inches from centre to centre, vertically and horizontally. What is the tensile force in each stay-bolt, and what APPLIED MECHANICS. 313 must be the diameter of each, the metal not being subjected to a greater stress than 5 tons per square inch of the section of the bolts? Ana., 2,800 Ibs. ; 0'54 inch. 6. What are the working and breaking loads Li pianoforte wire T V inch diameter ? And if the wire hangs vertically 5 miles, what weight at the end will, together with the weight of the wire, produce the working stress at the top ? What is now the average load and the average stress in the wire ? By how much will the wire lengthen when subjected to it ? Ans., 297 Ibs. ; 1,040 Ibs.; 27 Ibs. ; 162 Ibs. ; 16,460 Ibs. per sq. in. ; 20 yds. 7. A strut 100 feet long is made up of four angle irons of wrought, iron (3| + 3|) f , which are prevented from bending. Find the ultimate proof and working loads. How much shortening occurs under the working load? Am., working load 91,000 Ibs. ; 0*28 inch. 8. A piston-rod of mild steel 4 inches diamejter, 6 feet long, the piston 30 inches diameter; what maximum pressure of steam may be used (1) if the engine is double acting; (2) if the engine is single acting? (See Art. 263.) What lengthening and shortening occur under these pressures? Ans., 53 Ibs. ; 160 Ibs. ; (1) 0-08 inch ; (2) 0*12 inch. 9. Columns about 8 feet high of brickwork, sandstone, and granite, 14 inches square ; what working loads will they carry ? If they are 6 feet apart, and carry a wall of brickwork 28 inches thick, to what heights may it be carried? The brick columns being replaced by cast-iron cylinders 6 inches diameter, what ought to be their thickness? How much do they shorten under the load ? Ans., 15| tons each ; 30 ft. ; 0' 12 inch. 10. The tight side of a gearing- chain taking a pull p, let it act at 5 inches from the centre of the wheel, and transmit 20 horse-power at 100 revolutions per minute ; find p. Each of a pair of links takes this pull. If the thickness of a link is one-third of the diameter of the pin, and if its breadth is two and a half times the diameter of the pin, find the section of each link. It is a good exercise to design the chain. Ans. t 5,044 Ibs. 11. A single (a little over 0'2 inch thick) leather belt will stand an average pull of 1,000 Ibs. per inch of its width. The weakness of the average fastener reduces this to about 300 Ibs. per inch of width, and it is usual to take 66 Ibs. as a working load. The pull on the tight side of a belt is two and a half times that on the slack side. The pulley is 30 inches diameter, 150 revolutions per minute ; find the breadth of a single belt to transmit 5 horse-power. (See also the exercises Art. 185). Am., 3 '5 inches. 12. Columns of different material, constrained to keep of the same length, of cross-sections A I? A 2 , A 3 , etc, of coefficients of expansion a 1} 03, o 3 , etc., unstressed, at a particular temperature^ are raised t degrees in temperature. What is tho fractional elongation? And what are the etresses in them ? Ans., If unconstrained, their fractional elongations would be ta v ta%, etc. If a; is the real elongation, each has a compressive strain t^ x, ta% x, etc. Their compressive stresses are E! (ta-^ %}, E 2 (to^ a?), etc., if B I , E 2 , etc., are their Young's moduli, and the total push in each is AjEi (tai - x], A 2 E 2 (ta% so], etc. The Bum of these pushes is o (some of them being negative ; that is, tensile forces), or AJ E, (fa, - x) + A^ E (ta^ on) + etc. = o, or 314 APPLIED MECHANICS. ____ A! Ej + A 2 E 2 -f- 6TC. etc., can be found when a? is known. Thus, for example, suppose there are columns equal in section, two of them of brass, one of cast iron. Then AJ = 2 AO, ttl = 19 x 10~ 6 , 03= 11 x lO- 6 for centigrade degrees; E X = 9-2 xlO 6 , E 2 =17x 10 6 , 2 x 9-2 x 10 6 x 19 x 10- 6 + 17 x 10 s x 11 x 10~ 6 2 x 9-2 x 10 6 + 17 X 10 = t x 15-15 x 10~ 6 . The compressive stress in the brass is 9'2 x 10 fi (l9xlO- 6 - 15-5 x 10~ 6 )=:x 30-2 Ibs. per square inch. The stress in the cast iron is 17 x 10 s (t 11 x 10- 6 - t 15'5 x 10~ 6 ) = Thus, if t is 30 centigrade degrees, there is a compressive stress of 900 Ibs. per square inch in the brass, and a tensile stress of 2,300 Ibs. per square inch in the iron. 13. A bar of wrought iron 25 feet long and 1'75 inch diameter is heated to 180 0. While in this condition it is made to connect (by means of nuts screwed on the ends) the two side walls of a building which have fallen outwards from the perpendicular. If the walls do not yield against the tendency of the bar to contract, find the pull between them when the bar has cooled down to 80 C. Take the mean coefficient of linear expansion of wrought iron as '0000124 for 1 0. An*., 92,350 Ibs. 14. Two bars of copper with a bar of wrought iron between them, all of the same section and length, have their ends rigidly connected together. If the bars are heated from 15 0. to 100 C., find the stresses in the bars, and their fractional elongation, the coefficients of expansion for copper and wrought iron being taken at -0000172 and -0000124 respectively. Ans., copper, 2,940 ; iron, 6,000 Ibs. ; 0-00125. 15. A tie-rod 100 feet long and 2 square inches in sectional area carries a load of 32,000 Ibs., by which it is stretched | inch. Find the stress, strain, and E. Ans., 16,000 Ibs. per square inch ; -000625 ; 25,600,000 Ibs. per square inch. 16. A wooden tie 40 feet long, 12 inches broad, 7 inches thick, was tested with a pull of 130 tons, which stretched it 1-28 inches. Find the value of E for the timber. Ans., 1,300,000 Ibs. per square inch. 17. A vertical wrought-iron tie-rod 200 feet long has to lift a weight of 2 tons. Find the area of the section and the diameter if the greatest strain is -0005 and E = 30,000,000. Neglect the weight of the rod. Ans., 0-298 square inch; 0-616 inch. 18. Steam at a pressure of 200 Ibs. per square inch is suddenly admitted upon a piston 18 inches in diameter. If the piston-rod be 3 inches in diameter and 7 feet long, what is the compression and strain energy in the rod at maximum compression? E = 30,000,000. Find, also, the maximum stress in the rod. Ans., -04 inch; 171 foot-pounds ; 14,400 Ibs. per square inch. 19. A ship is moored by two cables of 90 feet and 100 feet in length, APPLIED MECHANICS. 315 respectively. The first cable stretches 2f inches, and the second stretches 3 inches under the pull of the ship. Find the strain of each cable. Ans., -0024 ; -0025. The yireight of a rope in pounds per foot is taken as ag"- where g is its girth in inches, and its breaking weight in pounds is kg* where a and b have about the following values : a 6 Tarred Hemp White Hemp ' ... Iron AVire . . . 04 . -04 13 900 1,300 4,000 Steel Wire 13 6,700 20. What lengths of themselves will each of these kinds of rope carry ? Ans., 22,500, 32,500, 30,770, 51,540 ft. 21. Compare the weights and strengths of iron- and steel- wire ropes with iron and steel wires of the same circumference. Ans., *49, '4, '49, '65. The Admiralty rule for the proof -lostd p in tons of the ordinary close-link chain of welded iron is 12d 2 , where d is the diameter of the iron in inches ; this means nearly 8 tons per square inch in the iron. For studded chain- cables it is p =. ISd 2 , which means 11 tons per square inch in the iron. The working load is from half to one-quarter of this, depending on circumstances. The weight of either chain is about 10^ 2 Ibs. per foot. Hemp rope of girth g is taken as being of about the same strength as a chain if q = 10d to lid. 22. Two close-link chains, each making an angle of 50 degrees with the vertical, are to support a working load of 10 tons ; what is the proper size of the iron ? Ans., 0'8 inch diameter. 23. A ship of 2,000 tons (take it that one-quarter as much mass of water moves as the ship moves) is moving at 0*1 knots, and is brought to rest in three seconds, the law of the motion during stoppage being v = 0-1 cos. Jet, where t is time and k a constant, and v is in knots. The pull comes directly on a studded chain. If the chain gets its proof load, what is the diameter of the iron ? When three seconds elapse, cos. kt is 0, or cos. 3& = cos- -, or k = . In feet-second units, v = 0-1689 cos. - 1, because 1 knot = 6Q ' x , or 1-6889 feet per second, and acceleration is -1689 x sin. Jrf, being numerically -1689 x , or 0-5305 feet U v) Q per second per second where greatest. The mass is 2,500 x 2,240 -:- 32-2, or 173,900 in engineer's units. Hence the greatest force is 92,300 Ibs. 24. Find the work which may be stored up in a pound of hard spring 316 APPLIED MECHANICS. steel when stretched to its elastic limit, taking the modulus of elasticity as 35 x 10 6 Ibs. per square inch; the elastic limit, 100,000 Ibs. per square inch; and the weight of one cubic inch, *29 Ib. Ans., 410 '5 ft.-lbs. 25. A cylindrical rod of copper inch diameter and 4 feet long, and one of wrought iron |- inch diameter and 3 feet long, ar to be stretched the same amount. Compare the forces necessary to do this, the values of E for copper and wrought iron being 17,000,000 Ibs. per square inch and 29,000,000 Ibs. per square inch respectively. Compare also the amounts of work expended in each case. Am., 0*28 : 1, 0-28 : 1. 26. What would be the resilience of a steel tie-bar 1 inch in diameter and 4 feet long if the bar becomes permanently stretched under a load of 10 tons, the modulus of elasticity being 32,000,000 Ibs. per square inch? Ans., 479 inch-lbs.. 267. Exercise. Find every number in the following table. Values of tensile (/* 2 /2E) ot compressive (/c 3 / 2 ^) resiliences. These numbers express tlie relative values of the following materials for the making of springs in which elongation, compression, or bending occurs. In bending, the smaller of the two values, or possibly an intermediate value, must be taken. The numbers ///2N, or the shear resiliences per cubic inch, express the relative values of the following materials for the making of springs, such as spiral springs, in which shearing is most important. The numbers in each case show the amount of energy (in inch-pounds) which it is possible to store in each cubic inch of the material in the most carefully con- structed springs. In Art. 518 we give a statement showing how much less these stores usually are in ordinary springs. The work actually done upon ductile materials before fracture is often 1,000 times as great as the resilience, and in hard steel it is 150 times, in cast iron being twenty times the resilience. * * /.*. Cast Iron 3 12 5 Wrought Iron 10 . 10 19 Mild Steel (Hardened) Best Hard Steel 83 600 83 128 810 Copper (Rolled or Drawn) . 0-5 0-5 0-73 Fir ... 4 Oak 6 ... ... 268. The diminution in bulk of a substance when it is sub- jected to pressure uniform all round, as, for instance, when it is surrounded by water in an hydraulic press, or sunk in the sea, APPLIED MECHANICS. 317 has been experimented upon. The lessening in the bulk per cubic inch is called the cubical strain of the substance. The pressure in pounds per square inch all over its surface represents the stress, and it is found that the strain is proportional to the stress. In fact, in any substance the stress is equal to the strain multiplied by a certain number, for which the letter K is usually employed, called the Modulus of Elasticity of bulk. If an increase of pressure Sp causes the volume v to become v + 8v (8v is usually negative), the stress being Sp and the strain Sv/v, the elasticity K is denned by 8p = K 8v/v, or K = v . dp/dv. In solids it is found that, whether the change takes place quickly (at constant entropy, as it is called in thermodynamics) or slowly (at constant temperature), there is no very great differ- ence ; but in v gases and liquids it is very important to specify under what circumstances the rate of relative change of p and v is measured. The ratio is in air 1-41; water, 1-004; alcohol, 1-22; ether, 1-58; mercury, 1-38; flint-glass, 1*004; drawn brass, 1-028; iron, 1-019; copper, 1-043.* The table, page 657, of slow moduli of elasticity of bulk is in pounds per square inch. 269. A cube 1 inch in each edge (Fig. 174), subjected to a uniform compressive force of 1 Ib. per square inch on the opposite faces, A D E F and B c L G. Evidently the edges AB, CD, LE, and GF, become 1 a inch in length, a being the reciprocal of Young's modulus used above. Also the edges AD, B c, G L, and F E get the length 1 + inch. If now we give to the faces ABCD and E F G L of this cube compressive forces 1 Ib. per square inch, it is the edges AF, etc., which shorten, and the edges AB, etc., Fig. 174. which lengthen. Again, give the com- pressive forces to the third pair of opposite faces, A B G F and c D E L, and we have the edges AD, etc., shortening and B G, etc., lengthening. If, now, all three sets of - compressive forces act at the sam.e time (that is, the cube gets on every face a pressure of 1 Ib. per square inch), as the compressions and extensions are exceedingly small, each edge shortens by the amount a and lengthens by the amount 2j8. Hence the edge which used to be 1 inch is now I a + 2 j8 inch. The cubic contents used to be 1 cubic inch ; it is now 1 3 (a 2 ) with great exactitude. Hence 3 (a - 2)8) is- the amount of cubical strain produced by 1 Ib. per square inch. That is, the Modulus of Elasticity of bulk, 1 K== 3 (a- 2)8)' and if we know a and /3 it may be calculated. * Lord Kelvin (article on elasticity) gives the above, and also the follow- ing numbers. The ratios of the quick to the slow Young's moduli arc : zinc, 1-008; tin, T0036 ; silver, 1 '00315; copper, 1 '00325; lead, 1 '00310; glass, 1'0006; iron, 1'0026; platinum, I'OOIS. 318 APPLIED MECHANICS. Even very porcms bodies, such as cork, have some elasticity oi bulk. Fluids and homogeneous solids, such as crystals, are probably perfectly elastic as to bulk even at enormous pressures. Manufactured metals are generally porous, and alter (not necessarily increasing) in density after they have been hammered or drawn. Experiments on metals with great negative stress in all directions are wanting. Liquids do not seem capable of resisting great negative pressures, and the contrast between them and solids in this respect is remarkable. 270. Students now know enough to make calculations on the stiff- ness and strength of ties and struts [only when the struts are kept from bending]. Before going on to other structures, even such simple structures as boilers and pipes, I wish to make a few remarks on the application of theory. The engineer must have some sort of theory to work upon. I shall give the theory recognised by men like Rankine, and also by the most successful practical engineers. It is simple as I shall five it, and fits fairly well a great number of practical conditions, t is founded on the assumption that materials are homogeneous and not loaded beyond elastic limits, and yet it gives us knowledge which the judicious man finds useful beyond elastic limits. It is also founded on the assumption that certain things, too difficult to calculate, are negligible, and hence its mathematical results ought to be tested by experiment when this is possible. To give an example. Our theory of bending is founded on the assumption that the plane cross-section of a beam remains plane after bending. Every mathematical result seems to agree with every experiment made on beams, seeming discrepancies being always explainable by the tests not having been confined between the elastic limits of the materials. Again, the more elaborate theory of St. Venant (Art. 311), involving fewer hypotheses, gives results that are practically in agreement with us. Hence we regard this easy theory which I shall give as one which may be depended upon in long beams, in spite of the fact that a plane section of a beam does not remain exactly plane. When it is applied in new cases not yet tested by St. Yenant's theory or by experiment, we use it only as a fairly trustworthy guide in our practical work. Another example. In the great twisting, which might occur without fracture in indiarubber shafts, it was known that the plane section of a square shaft did not remain plane. Nevertheless, the warping in ordinary shafts being small, a simple theory was adopted, in which there was the assumption of no warping. Results of experiments on round shafts agreed with the simple theory. Results from other shafts did not agree, and St. Venant has shown us why there is a discrepancy. Although his investiga- tion is difficult to follow mathematically, his results are easy enough to comprehend. I find it necessary, therefore, to give not merely the simple theory of the engineer, but an account of St. Tenant's results, and also, for advanced students, a short account of St. Yenant's theory of beams and shafts. It is to be remembered that APPLIED MECHANICS. 319 our theory assumes perfect elasticity of the material. We shall see that at the bottom edge of a key-way in a shaft we have a place where the stress becomes very great indeed. I have made experiments on such a shaft with a key- way, and I find that it is by no means such a great source of weakness as the theory supposes, and this leads us to consider how the results of the theory are modified by the flow of the material, instead of its fracture, under 271. In thin-shelled vessels, such as boilers and pipes, sub- jected to fluid pressure^ inside, we assume that the tensile stress f is the same throughout the thickness ; so that if a is the area of metal cut through at any plane section of the boiler, ay is the resistance of the metal to the bursting of the boiler at that section. Now the equal and opposite force due to the fluid is A p if A is the whole area of this plane section of the boiler. Hence Ap af and p = af/A ... (1) gives us the bursting or working pressure if / is the ultimate or working stress. To prove this : In Fig. 175 let F B c D E be part of a boiler whose separation from the rest by a plane section at F E we are now studying. Arrow-heads are drawn showing the forces with which the fluid every where acts nor- mally on the shell. We want to know the resultant of these forces. Imagine a boiler made with the part F n c D E and a rigid flat- plate F E closing it. If we neglect the weight of the fluid, all the pressure forces on the shell balance one another. This is Newton's law of motion. The mutual forces of parts of a system cannot affect the motion of the centre of inertia of the whole system. (See Art. 482.) If the above boiler were placed upon a truck with frictionless wheels, there will be no more tendency to Fig. 175. move on a level road when there is great pressure inside than when there is little. The force due to pressure on any one little portion of the surface balances the forces on all the rest of the surface. Hence it is that if we make a hole there is a want of balance and our truck will move. When we make the hole the pressure everywhere changes because of the motion of the fluid, and hence we can only calculate the unbalanced force by knowing the momentum of the fluid which leaves the vessel per second. In the closed 320 APPLIED MECHANICS. vessel of Fig. 175 we know that the resultant pressure on the flat surface F E is A/>, therefore the equal and opposite force on F BCD E is also A p. As an example of this we saw in Art. 122 that the resultant force axially on the ram of an hydraulic press is exactly the same whatever be the shape of the end of it. Example. Spherical boiler of diameter D. Any plane sec- tion is a circle. If we use the above rule for any such section we find that less pressure will burst the boiler if the section is diametral. The area of such a section is A = - c? 2 , and the area of metal laid bare is a = IT d t. Hence (1) becomes p = IT d tj -f Z<& or 4 tf/d . . . (2). 272. In a long cylindric boiler or pipe it is easy to show that (neglecting the strength of the ends) the tendency to burst laterally is twice as great as the tendency to burst endwise. Endwise, the area of a circular cross-section is A = -j d 2 , and the metal laid bare is ir d t, so that (1) becomes p = just as in a spherical boiler. But laterally, at a plane passing through the axis of the boiler, A = I d if Us the length, and a = 2 It if we neglect the metal at the ends; and hence (1) becomes p = %ft/d . . . (3). The bursting pressure endwise being twice as great as this, we always take (3) as the formula for the strength of a cylindric pipe or boiler. Headers will now understand why in cylindric boilers the longitudinal seams are always much stronger than the girth seams. When the boiler has riveted joints we must, of course, regard the material as weaker than if it could resist tensile stress everywhere like a continuous boiler-plate. The working /for copper ought not to be taken greater than 2,400 Ibs. per square inch for steam pipes. In cast-iron pipes and in steam- engine cylinders it has to be remembered that the difficulty in getting castings which are of the same thickness everywhere, and the allowance that must be made for tendency to cross- breaking when the pipes are handled, as well as the great allow- ance that must be made in steam-engine cylinders for stiffness, the difficulty of casting, and boring out, cause such calculations as the above to be somewhat useless. Thus it will usually be found that, whereas a large cast-iron water-pipe is not much thicker than the above calculation would lead us to expect, APPLIED MECHANICS. 321 because it is usually carefully moulded in loam, yet a thin cast- iron pipe is often of more than twice such a thickness on the average, and it is our rule never to attempt casting a 9-foot length of pipe of less than f inch thick. In these cases the maximum /for cast iron is taken as 1,500 Ibs. per square inch, whereas for large pipes we usually take 3,000. EXERCISES. 1. What must be the thickness of the plates used in the construction of a boiler 10 feet in diameter working under a pressure of 120 Ibs. per square inch, taking the efficiency of the joints to be 70 per cent, and the safe stress at 10,000 Ibs. per square inch? Ans., T03 inch. 2. A copper pipe is 4 inches diameter and f inch thick. What is the working pressure? Take/ = 2,000 Ibs. per square inch. Ans., 375 Ibs. per sq. in. 3. A vertical cast-iron pipe is 4 inches in internal diameter. The pressure at a certain place is 50 Ibs. per square inch. At this place, and at places 100, 200, 300, etc., feet lower in level, find the proper thickness of the metal if the working stress /is taken as 3,000 Ibs. per square inch. Ans., -033 inch; -062 inch; -091 inch; -149 inch, etc. 4. The rule used for loam-moulded cast-iron water mains is t = ^ + hd -T- 13,000 where h is head of water in feet, d diameter of pipe in inches, t the thickness in inches. A pipe 3 feet diameter, 1 inch thick ; find h. Find the corresponding pressure in pounds per square inch. Find what vahie of / the working stress will cause the ordinary rule for thin cylinders to give the same answer. Ans., 316 ft., 136-7 Ibs. per sq. in., 2,460 Ibs. per sq. in. 5. A wrought-iron pipe 2 feet diameter, | inch thick ; its working stress is 5 tons to the square inch, but strength of plate is diminished 30 per cent, because of riveted joint. What is the working pressure ? Ans., 326 Ibs. per sq. in. 6. A cylindrical boiler 12 feet diameter is constructed of f f inch steel plate. The test pressure applied is 245 Ibs. per square inch. Find the stress produced in the plate, and hence deduce the stress in the plate between the rivet holes, the sectional area being there reduced to -77 of the solid. Ans., 19,500 and 25,300 Ibs. per square inch. 273. Storage of Energy in Fluids. The volume of a cylin- dric vessel being v and the safe pressure being p } we may take vp as proportional to the energy which may be stored. If the diameter is d and thickness t and length I, the volume is v = j- dH. Assume what is known to be true (see Art. 272), that the safe pressure for a long cylindric vessel is (neglecting the strength of the ends) p = T, where f is the safe stress which the material will stand. The weight of the metal, neglecting the ends, is w = irdllw, if w is the weight of unit fc 322 APPLIED MECHANICS. volume of the material. The surface of the vessel is s = irdl In all cases we neglect the ends. Then the storage capacity of energy per unit weight of vessel is ~<&l ~T -*- irdtlw, or jp. So we see that it i\ independent of the diameter. In water-tube boilers, therefore, which must store energy in this way, and where it is of importance that there should be great surface, we must consider surface -h vp. This is %/tf or \\pd. Hence the thinner the tubes are, and, if the pressure is fixed, the smaller they are, the more surface they have as compared with their storage capacity for energy. The same considerations cause us to use thin tubes for surface condensation and other purposes, and there is the further consideration that accidents are less likely to be serious. In cases where energy is stored in hot water and steam, the rate of waste of energy is proportional to the surface, and this requires just the opposite conditions. EXERCISES. 1. Sixty tubes of wrought iron 4 inches inside diameter, 10 feet long, \ inch thick. Find volume, weight, internal area in square inches, and working pressure if working / = 10,000 Ibs. per square inch. Neglect the ends. How many tubes 22 inches diameter, 10 feet long, will have the same volume? And find the thickness suitable for the working pressure. Find also the area and weight. An*., 52-36 cub. ft., 6,729 Ibs. wt., 12-54 sq. in., 1,250 Ibs. persq. in., 2, 1-375 inches, 380 sq. in., 6,525 Ibs. 2. Cylindric boiler of mild steel 5 feet 6 inches in diameter, at elastic limit; pressure 300 Ibs. per square inch. What is the thickness? Joints supposed to be of 60 per cent, of strength of unhurt plate/. Replace this boiler with tubes 5 inches diameter of same length. How many tubes are needed to make up the same volume ? What will be their thickness (no seams) ? Their weight ? Now replace with 3-inch tubes, finding thickness and weight. Ans., -47 inch ; 174 ; -021 inch ; 193 Ibs. per foot length ; '013 inch ; 1 99 Ibs. per foot. 274. When a belt or rope of weight w Ibs. per inch of its length is moving with a velocity of v feet per second in a curved path of radius r, the centrifugal force on a small length of it, ? . SO, is w.r.SO' Now, if a sketch be made showing how this force is balanced by tensile forces T at the ends of the small length r . 86, it will be seen that the centrifugal force is equal to T . 80 if 9 is very small ; so that T = wv*/ff, being independent of the radius. This, then, is the tensile force which acts in a belt or rope when in motion an addition to the tensile force which acts when the rope APPLIED MECHANICS. 323 is at rest and it must "be taken into account in considering the strength of a belt. Notice also that if a is the section of the rim of a pulley of wrought iron, the weight of it is -28 Ib. per inch of its length. Hence the tensile force is '28av 2 /ff, or '28v 2 /y Ib. per square inch is the tensile stress induced in the rim by centrifugal force when it moves at v feet per second. Taking the working tensile stress in wrought iron of a pulley as 6,000 Ibs. per square inch, the rim speed of a wrought-iron pulley ought not to exceed 850 feet per second. The usual limiting speed of cast-iron pulley rims is 80 feet per second. Arms, if numerous, serve to diminish this action. If an arm of uniform cross-section moves at n turns per second, the limiting length L of it is A / /, if w is the weight per cubic inch. Thus, if we take /= 6,000, w = -28, 916 L = . Thus, if n = 50 revolutions per second, L is about 18 inches. If such arm has a section a at the distance r from the centre, it is easy to show that if each section has simply to with- stand a pull due to the centrifugal force of the part outside it a = e-^, where, if r is in inches, b = 7r 2 w;wV 2 /193/. The condition as to strain of a rotating disc has been investi- gated by Dr. Chree (see Art. 307). As the tensile force in a perfectly flexible rope due to its motion is independent of the shape of the path in which a particle is moving, a very rapidly-moving rope, if no external force such as those due to gravity act upon it, has no tendency to any alteration of shape ; each particle follows the path of every other, like particles of water in a stream-line in a state of steady motion. We can impulsively alter the shape of such a rope at any place, and then the new shape will be retained. Vibrations will be transmitted by such a rope as if along a naturally straight rope not moving in the direction of its length in which there is the same tension as is here produced by motion. A thinking student will from these facts readily see that there is a quasi-rigidity produced in the rope by the rapid motion. EXEKCISES. 1. Two pulleys, 3 feet 6 inches in diameter, running at 150 revolu- tions per minute, are connected by a leather belt weighing 0'6 Ib. per foot in length. Taking /A = '3, find the greatest tension in belt when transmitting 7^ horse-power. Ans., 260 Ibs. 2. In a travelling crane the driving rope * runs at 5,000 feet per minute. Find the tension due to centrifugal action, having given that a rope 1 inch diameter weighs 0'28 Ib. per foot of length. Am., 60-4 Ibs. 275. Strength of Thick Cylinders. Let the inside and outside radii be i\ and r , and the inside and outside fluid pressures be j! and p . Consider an elementary ring of metal, 1 inch parallel to the axis, inside radius r, outside radius r + 5r. Imagine a compressive stress p inside it and p -\- 8p outside (5p is usually negative in our examples), and a compressive stress q in the 324 APPLIED MECHANICS. material (q is usually negative or the stress is tensile) at right angles to the radius, p x 2r is a force tending to fracture this ring at a diametrical plane; 2 (p + 8p) (r -}- br) 2q . 8r is the force tending to prevent fracture. Note that there is a possible shear stress on the sides of this strip that we are neglecting, and a careful student will give thought to the matter. ur j ustification for neglecting it lies in this, that the strength of our cylinder cannot be imagined to depend upon its length ; and if we consider a very long cylindric strip, end effects are negligible. Balancing the forces, we have the well-known rule for the strength of a thin cylinder. Divide by the thickness Sr, and imagine Sr smaller and smaller, and we find p + r -j- = q . . . . (1). As the material is subjected to crushing stresses p and q in two directions at right angles to one another in a cross-sectional plane, the dimensions parallel to the axis of the cylinder elongate by an amount which is proportional to^? + q. We must imagine this to remain constant if a plane cross- section is to remain a plane, and we make this reasonable assumption. Hence (1) has to be combined with p + q = 2#...(2), where 2a is a constant. Substituting q from (2), in (1) we get ^ = - ^ . . . . (3), and we find by trial that the solution dr r T is p = a -\ g- and q = a - -^- . . . . (4), where the constants a and b are to be found by the conditions of pny problem. Thus, in the case of a gun or hydraulic press, let the pressure inside be p^ where r r v and let p n = where r = r . If we insert these conditions in (4), we find (q is everywhere negative, and I shall use/, the tensile stress, to replace q] _ / is greatest at r = r lt and is then / x = p v 2 L .... (6). ^*o **i The student is to note that the circular compressive strain at any place is qa -p&. This is the fractional diminution of the radius r. A. student ought to take an example such as : An hydraulic press has an inside radius r x = 4 inches ; the stress is not to exceed 5,000 Ibs. per square inch; find the greatest possible pressure pi, first, if the thickness is 1 inch, then if the thickness is 2 inches, and so on. Note how little gain there is by increasing the thickness more than a certain amount ; and it may be well to write out a list of numbers for various thicknesses, showing among other things the gain in weight. EXERCISES. 1. Tn a certain kind of work either one cylindric hydraulic press of 24 inches diameter or four are needed, of same aggregate area and same material, to stand the same pressure. Compare the square area on which the two arrangements will stand. Observe that the ratios of internal to external radii will be the same in the small and large presses. APPLIED MECHANICS, 325 If the student figures it out, lie will find that the four will just occupy the same square space as the single press, and they will weigh just the same. 2. Thin Cylinder. Take r l = $d and ? = \& + t, where d is the inside diameter of a cylindric boiler and t is its thickness, and assume that t is small. Then (6) becomes A first approximation which is generally used, and has been given above, fe/= ^ . . . . (7). A second is/=|^ +&.... (8)- 3. A gun of 12 inches internal and 24 inches external diameter is subjected to a maximum internal pressure of 40,000 Ibs. per square inch. Find the stress produced at r = 6, 7|, 9, 10 J, and 12 inches. Find what was the initial stress everywhere if it was just sufficient to cause the final stress everywhere to be the mean of the stress produced at r 6 and r = 12. Now make diagrams showing the state of stress when Pi = 60,000, 50,000, 40,000, 30,000, 20,000, and 10,000 Ibs. per square inch. Ans., 66,666 ; 47,466 ; 37,037 ; 30,748 ; 4,444 Ibs. per square inch. 4. Pipes of a water-pressure supply company are to withstand a possible pressure of 1,000 Ibs. per square inch; they are of 6 inches internal diameter. What is the outside diameter, the safe tensile stress oi the metal being 3,000 Ibs. per square inch? Ans., 8'485 inches. 5. Pipes are to withstand a working pressure of 1,000 Ibs. per square inch. If their internal diameters d are 2, 3, 4, o, and 6 inches in each case, find the thickness. Find in each case the weight w per foot length. Draw a curve showing the relative values of w and d. Am., thicknesses 0-41, 0-62, 0-83, 1-04 inch. 6. A cast-iron water-main is 30 inches internal diameter and 1 inch thick. What is the greatest head of water that it ought to be subjected to ? Safe tensile stress, 3,000 Ibs. per square inch. Numbers to recollect are : 34 feet of head represent 1 atmosphere, or head in feet + 2-3 = pressure in Ibs. per square inch. If the pipes are wrought iron, what ought their thickness to be if safe /= 10,000 Ibs. per square inch, and if the longi- tudinal seams are of 60 per cent, of the strength of the unhurt plate ? Ans., 460 ft.; -5 inch. 276. In the above theory we have considered the material initially unstrained ; or, rather, the stresses and strains calculated by us are additional to any initial stresses and strains in the material. The student will see why the outer material of a thick cylinder is comparatively useless if he shows in a curve /for various values of r, calculating from (5), for /decreases as the inverse square of r. If, when p n =Pi = 0, there are already strains and stresses in the material, the stresses given in (5) are algebraically added to those already existing at any place. Hence, in casting an hydraulic press, wo chill it internally, cold water circulating in a metal core painted with loam ; and in making a gun we build it of tubes, en oh of which squeezes those inside it. So that there is considerable compressive stress where r = r l and considerable tensile stress where r = r before any pressure comes on inside. We try to 326 APPLIED MECHANICS. create such initial stresses that when there is the maximum pressure p l the material has about the same tensile stress in it everywhere. Much knowledge is needed to produce this result in guns. Exercise. Thick spherical shell subjected to internal fluid pressure. If p is the radial compressive stress at a point at the distance r from the centre, and q is the tangential tensile stress there, show after the manner of Art. 275 that p = A -f- 2B/r 3 , q = A B/r 3 , where A and B are constants, which may he found if the internal pressure and the inner and outer radii are stated. 277. The above theory of the strength of thick cylinders seems to agree with our practical experience for such ratios of r Q and r a as we find in the pipes and presses used by engineers. But all the rules given above show that a flat plate has no strength. The neglected terms in our theory become important in this case. In truth, the mathematical theory of a shell is so troublesome that wo cannot say there is yet a satisfactory treatment of it. The strength of a flat plate has, however, been investigated in a number of cases, and we are led to the following results : For a circular plate of thickness t and diameter d supported all round its edge, with a normal load of p Ibs. per square inch, if / is the greatest stress in the material, /= 5r 2 p/6fi. If the plate is faced all round its edge, /= 2r 2 jj/3 2 . A square plate of side s, fixed at the edges, /= s^p/^t . A rectangular plate of length I and breadth -b, fixed round the edges, f=l 4 Pp/'2tf(l* + i 4 ). A round plate with a load w in the middle, supported at the edges, /= w/ir< 2 . For stays in square formation, distance asunder being * ; each stay has a load ps 2 , and the greatest stress in the plate of thickness t is 278. When the pressure is greatest outside a thin shell, its strength to resist collapse ought evidently to follow the law (1), which becomes (3) if the vessel is cylindric ; but it is in the very way in which a strut may be relied upon if the slight lateral restraints are provided which prevent bending. So also the thin shell of a boiler flue must be provided with certain restraints against buckling ; and just as we have (Art. 372) a theory of laterally unsupported struts, so we have a theory of long boiler flues. Beyond a certain length for a given pressure there is instability, and hence flues are either corrugated or furnished with a number of rings. The most important practical outcome of the theory is that the distance between two rings divided by Vdi must not exceed a certain limit. The rules usually followed by boiler- makers are given in the new edition of my book on Steam. Exercise. In a corrugated flue of 3 feet mean diameter the plate is i inch thick, but the corrugations, being longer than the axial length, make it virtually f inch thick. What is the working pressure if the working compressive stress (we allow for corrosion, etc.) in the material is 3,000 Ibs. per square inch ? Ans., 125 Ibs. per square inch. APPLIED MECHANICS. 327 MORE DIFFICULT EXERCISES ON THICK CYLINDERS. 1. A tube of wrought iron, inside radius 3 inches, outside 4 inches, outside pressure 0. What is the inside pressure r to produce a maximum tensile stress of 15,000 Ibs. per square inch ? Find the fractional increase in size of the inside radius. Here p = where r = 4 ; p = p, and q = - 15,000 where r = 3. Inserting these values in (4) Art. 275, we find b 0=a+ lQ - P - a + - q = a - b/r' 2 - 15,000 = a - - 144 b = 86,400 -5,400 p = - 5,400 + 9,600 = 4,200 Ibs. per square inch. Let the student calculate q for several values of r from 3 to 4, and plot his results on squared paper. r 3 3-25 3-5 3-75 4 -<1 15,000 13,580 12,453 11,544 10,800 The fractional diminution in size of any radius .is q a - p p, or (a - j8) a - - 2 (a + j8). Taking a = -i x 10 ~ 7 , ,8 = 1 10 ~ 7 , the frac- r o 14 tional diminution of the 3 inch radius is * 10 -7a-| ,- 10~ 7 = - 5,350 x 10~ 7 ' That is, the 3 inch radiiis increases, becoming 3 x 5,350 x 10~ 7 , or 0*00161 inch larger. 2. A tube of wrought iron, inside radius 2 inches, outside 3 inches, no pressure inside ; pressure ^=4,200 Ibs. per square inch outside. Find the circular compressive stress everywhere, and also the diminution oi the outer radius. Heroin p = a + ^ > ? = - ^ . . . (1). Fractional diminution i b of the 3 inch radius J [ = ( ~ ^) a - 9 ( a + ' ( 2 ) We have p = 4,200 where r = 3; p = o where r = 2. 4,200 = a + | - 30,240, = 7,660 828 APPLIED MECHANICS. Hence q = 7,560 -f 30,240/r 2 . Thus for the following values of r we have r 2 2-25 2-5 2-75 3-0 15,120 13,533 12,400 11,560 10,920 And the 3-inch radius decr< ^ i Ann 5ases by the x 10~7 + it the 3 in fraction of ] 30,240 5 , tself 0-7, becomes '00099 inch or 3-29 x 10 ~ 4 , so thf 9 12 ] ch radius smaller. 3. A tube of radius 4 inches outside, and radius 2-9984 inside, is squeezed in some way upon a tube 2 inches inside radius and 3-00098 inches outside radius. Find the compressive circular stress at all points in both tubes. Ans. t It will be seen that I have taken just the sizes necessary to produce the states of Exercises 1 and 2. It is evident that we may take the outside radius of the inner tube as 3 inches, and the inside radius of the outer tube '2 '9974 inches. Observe that as the coefficient of ex- pansion of iron IB 1*2 X 10 , .to produce a fractional increase in size 0026 -j- 3, the outer tube must be raised in temperature more than K. -5- (1-2 x 10 ~ 5 ), or 72 Centigrade degrees, before it will slip over the other. 4. A tube of wrought iron, 2 inches radius inside, 4 inches outside, is subjected to a fluid pressure of 50,000 Ib. per square inch inside and no pressure outside. Find the tensile stress everywhere in the material or the values of - q. Here in p = a + -, q = a -^ , insert p = 50,000 where r = 2, p = where r = 4, and so find - q = -q is the tensile stress 266,667 + 16,667. Of course r 2 H 3 8i 4 -f 83,333 59,333 46,300 38,440 33,333 5. To sum up our results. The built-up tube of Exercise 3, with initial tensile stress/ or - q of Err. 3 taken from the tables of Exercises 1 and 2, is subjected to the internal fluid pressure of 50,000 Ibs. per square inch of Exercise 4, there being no pressure outside. We have seen that/' or - q of Ex. 4 shows the tensile stress produced by the fluid pressure, and hence/", which is/' +/, is the real tensile stress everywhere in the compound tube. Students ought to draw curves showing/,/', and/". APPLIED MECHANICS. 329 r 2 2* 2* 2| 3 3 3i Si 3J 4 / - 15,120 - 3,533 - 12,400 -11,560 - 10,920 15,000 13,580 12,453 11,544 10,800 / 83,333 69,337 59,333 51,937 46,300 46,300 41,317 38,440 35,727 33,333 r 68,213 55,804 46,933 40,377 35,380 61,300 54,897 50,893 47,271 44,133 279. These exercises will enable the student to understand the usual calculations of shrinkage which must be made in building up a gun. He will have no great difficulty in working out all the necessary formulae himself, but a man interested in gun-making ought to refer to three articles published in Nature about August, 1890, by Prof. Grcenhill. It will be noticed that when a gun is built of tubes there is a sudden change in the tensile stress at the common surface of two tubes. To get a more uniform stress throughout, instead of using many thin tubes we now use an inner tube of steel strong enough to withstand the longitudinal forces, then a thick layer of steel wire wound on with varying tension, and then a covering tube, which is almost unstrained initially. Probably the hydraulic presses of the future will also be built-up in this way. Prof. Greenhill says : " Mr. Longridge's principle of strengthen- ing a tube with wire wound with appropriately varying tension will be found useful in peace and in war. He can claim credit that a gun strengthened on this principle (the 9-2 inch wire gun) was chosen, from its great strength, to test the extreme range of modern artillery in 1888, with what were called the "Jubilee rounds," when, with an elevation of about 40 degrees, a range of 21,000 yards, or 12 miles, was attained, the projectile weighing 380 Ibs., and the muzzle velocity being about 2,360 feet per second." 280. The following example will illustrate how calculations may be made : Example. A steel tube 4 inches internal and 5 inches external diameter has steel strip wound on it to the external diameter of 10 inches, under a constant tensile winding stress w Ibs. per square inch. Find the radial and circumferential stress anywhere. Consider the place r in the wire. When the wire has already been wound to an outside radius p, let a new layer of thickness 5p be wound on. This produces a pressure w . 8p/p. It is easy to see, just as in Example 2, that a radial pressure w . 8p/p at p, there being no radial pressure at r l = 2", produces at r an increase 8p of radial compressive stress, and an increase $q of circumferential compressive stress where L* 330 APPLIED MECHANICS. _ w . 5p/r i + i) - (2)- These effects are produced by winding on a new thickness 8 j of strip. Integrating these from p = r to p r , we find where # stands for ?- 2 /ri 2 and # for r & 2 /ri 2 . The tensile stress in the strip or wire was w Ibs. per square inch, and we see that this is now reduced to w - q =f where . .(5). I have used (3) and (5) in calculating the following numbers : r 2-5 3 3-5 4 4-5 5 P 402 w 399 ' 315 w 210; 103 w f - -833 w - -036 w 382 w 65 w 847 w w At r = 2 inches, the outside of the steel tube and inside of the wire, there is a discontinuity. Calculating from (3) and (5), the tensile stress there is /= 0-832 w> and p = 0-402 w. Now the metal of the solid tube is subjected to external pressure 402 w and no internal pressure, and we find that in it the compressive circumferential stress q = 4-7 tv (% +;s)----( G )> and the radial stress iap = 4-47 w (* ~ ^V . . . (7). I have calculated the values of p and g, or rather of - q, which I call/, from (6) and (7) for the solid tube : r 2 2* 21 -*=/ 2-235 w - 2w - 1-83 w P 234 w 402 w APPLIED MECHANICS. 331 We know now the internal condition of this coiled structure, which we may call a gun. Let the gun, considered as homogeneous, r = 2, r = 5, be subjected to fluid pressure 3 w inside and no pressure outside. It is easy to see that this fluid pressure would produce the tensile stress everywhere ( '57 + ^- \ w, and as before, leaving out the t0, the numbers marked /' in the following table show its value. In this table the initial radial compressive stress p and circular tensile stress / are copied from the previous tables. The actual tensile stress, therefore, in the gun is/' + /, shown as/". r 2 2 ai H 3 H 4 *i 5 P 234 402 . -402 399 315 210 103 f 2-235 - 2 1-83 -0-833 -036 + -382 + 65 + 847 1 f 4-14 2-85 2-85 2-16 1-74 1-46 14 f" 1-90 1-02 2-02 2-12 2-12 2-11 2-14 It will be observed in the above table that the actual tensile stress in the wire is nearly constant. This is a happy accident. We usually assume that f" shall be a fixed constant in the wire ; /' is known, so we may calculate the necessary values of /, the initial tensile stress everywhere, in the finished gun. /is evidently of the shape f=m- h/r 2 . To produce this particular /, it is necessary to alter the winding tensile stress w in the wire as the winding proceeds. Thus in (1) and (2), if, instead of w being constant, we have w a function of p ; remembering that w is to be looked upon either as a function of r or p, integrating (2) and writing what is equivalent to (5), we have Differentiating with regard to r, we see that if 6 be written for w -f 'and if x stands for r 2 /?-j 2 , then dO *2 + 4*-l *+l dx n * r J ' ~- It is not difficult to solve this and so find the way in which w, the winding tension, ought to vary. 332 CHAPTER XIV. SHEAR AND TWIST. 281. LET c D (Fig. 176) be the top of a firm table, F H a long prism of indiarubber glued to the table, A B a flat piece of wood glued along the upper side of the indiarubber. We try in this way to apply a horizontal force to the whole upper surface of the indiarubber, so that if, for instance, the pull in the cord is 20 Ibs., and the upper surface of the indiarubber is 10 square inches in area, there will be a force of 2 Ibs. per square inch acting at every part of the surface, and this force will be transmitted through the indiarubber to the table. When the length of the prism is great compared with z F, we may suppose that the bending in it is very small, and in this case we say that the indiarubber is being subjected to a simple shear strain ; the force per square inch acting on its surface is also acting from each horizontal layer to the next, and is called the shear stress. If you had drawn vertical lines like Y' x before the cord was pulled, you would now find them sloping like Y x. Thus, making a magnified drawing of Y x in Fig. 176, the point Y' has gone to Y, and any point like M has gone to N. Points touching the table cannot move, but the farther a point is away from this fixed part the farther it can move. Now suppose that Y' Y is 0*01 inch, and we know that x Y' is 2 inches, what is the amount of motion of M if 11 x is 1 '7 inch *? Evidently APPLIED MECHANICS. 333 Y' Y is greater than M N just in the proportion of Y' x to M x, or 2 to 1'7 ; hence MN is 4 0085 inch. Thus the motion of any point is simply proportional to its distance above the fixed plane, and if we know the amount of motion at, say, a distance of 1 inch, we can calculate what it must be anywhere else. The amount of motion at one inch above the fixed plane is called the shear strain. The motion is small, and it is evident that the shear strain is the angle Y' x Y in radians. In this case we have supposed the force on F G to be 2 Ibs. per square inch. This is said to be the amount of the shear stress, and it pro- duces or is produced by a shear strain whose amount is '005 inch per inch. If the shear stress were 4 Ibs. per square inch, you would find the strain to be '01; if the stress were 8 Ibs. per square inch the strain would be '02. In fact, we find experi- mentally that the stress and strain are proportional to one another. Thus if, instead of indiarubber, we had a block of tempered steel, we should find that the force in pounds per square inch is equal to 13,000,000 times the strain. This number is called the modulus of rigidity for steel ; it is given in Table XX. It seems a pity that the name shearing elasticity is not given to this number. 282. However long we may make our "block of indiarubber in Fig. 176, we shall still have some bending in it ; that is, the stress will not he uniformly distributed over each horizontal layer (see Chapter XXI.). To prevent this "bending effect, and to produce a really simple shear strain, we ought to have force distributed over the ends F z and o H of the same amount per square inch as we have now acting ^ p > over F G and z H. These are shown in Fig. 177. Where P is the pull in the cord ^ p of Fig. 176, p' is the equal and opposite force exerted by the p| m table on the glued underside of the india- rubber, and F and F' are equal and opposite forces distributed over the ends, such that the couple FF is able to balance the couple P P'. There can now be no bending moment at any place. As F multiplied by the length of the prism is the moment of the couple F F', and is equal to p multiplied by the vertical dimension, we see that p distributed over the horizontal surface is the same stress per square inch as F distributed over the 334 APPLIED MECHANICS, ends. From, such a material then, if we cut a cubical block A (Fig. 178), its horizontal faces Y y and xa? are acted upon by equal and opposite tangential forces, and its faces Y x y x are acted upon by forces of exactly the same amount. The faces parallel to the paper have no forces acting on them. This will give you the best idea of simple shear stress. The material in Fig. 176 near the ends of the block does not get a simple shear ; but if the block is very long, then at the middle there is a nearly simple shear acting. In Fig. 178 the cube x Y' y' as has become x Y y x. Suppose the side of this cube to be 1 inch, then Y' Y is the shear strain, which I shall call *. The tangential force distributed over Y y is p Ibs., let us say. Then, if we denote by the letter N the modulus of the rigidity of the material, p = N *. 283. Example. A beam of steel has one end fixed, and at the other is a weight of 20 tons. The cross section of the beam is 2 square inches in area, and the length of the beam is 5 inches. Besides the deflection of this beam due to bending, there is a certain deflection due to shearing ; how much is it 1 Answer : the shear stress is 10 tons, or 22,400 Ibs. per square inch. This produces a shear strain of 22,400 -r 13,000,000, or -00172. This is the amount of yielding at 1 inch from the fixed end, and at 5 inches the yielding must be 5 x -00172, or -0086 inch. In a short beam like this, or in one 20 inches long, if we con- sider, for example, that it is 1 inch broad and 2 inches deep, we may calculate the deflections due to bending and to shear, and reflect upon the fact that in very short beams the yield due to shear is much more important than the yield to bending ; whereas in long beams we may, and indeed always do, neglect altogether the deflection due to shear. These reflections are in the main correct, but the actual distribution of shear stress over the sections of a beam not short and not long is unknown to us. Its distribution in the section of a long beam will be given in Art. 369, and it is obvious that the calculation of the deflection due to shear is not so simple as in very short beams. 284. The shear stress which will produce rupture is not well known for any substance except cast- andwrought-iron, but the shear stress which will produce permanent set is fairly well known, and we are also agreed as to the ordinary working shear stress of materials. For wrought iron and mild steel it is usually regarded as from 75 to 85 per cent, of the tensile stress ; but in a single-riveted lap joint in boiler-plates, as the holes are usually punched (and this weakens the metal), and as rivet iron APPLIED MECHANICS. 335 is usually of a better quality than plate, the cross section of the iron which is left, which is resisting pull, is made to have the same area as the cross sections of all the rivets, which, of course, resist shearing. Besides breaking by either a tensile or a shear stress, a riveted joint may give way by the rivet crushing or being crushed by the side of its hole. Again, in many riveted joints, when the rivets are long, as they tend to contract in cooling and are prevented by the plates, so much tension may remain permanently in them that they are greatly weakened. In bolts there is usually some bending, and consequently a want of perfectly, uniform distribution of the shear stress, and they are made larger than rivets in the same positions. 285. In the punching of rivet-holes it may be taken that a shearing force V acts on the material ; the area of the curved side of the hole, multiplied by the breaking shear stress of the material per square inch, represents the force with which the punch 'must be pressed down on the plate. The punch must be able to resist this force as a compressive stress on its own material. Experiments made on punching-machines show that about 24 tons per square inch is the average shearing force required. This pressure has to be exerted through a very short distance. In shearing-machines, if the entire edges of the shears coincided with the plate as soon as they touched anywhere there would be the same sort of effect produced ; but by inclining the edges the shearing action does not occur instantaneously at every place, and the rupture being more gradual than in punching, the shearing resistance is usually from 10 to 30 per cent. less. It is very probable that the power lost in punching- and shearing-machines is wasted rather in the f riction of the heavy parts of the mechanism than in the almost instantaneous effort of cutting the material. Tl>3 effort required seems rather that of an impact than of the more gradual action to be found in most existing machines. At the same time we must remember that M. Tresca's experiments indicated a flowing of the metal. Machines such as hydraulic bears and shears may be uneconomical as to mere energy, but they produce the higher economy due to convenience and certainty. A man can stop the motion of the hydraulic punch very rapidly if he sees that the plate is not quite right in posi- tion. In the fly presses used for hand punching, and used largely in coining, the idea of an impact is already in use ; it will come much more into use in large machines when engineers become 336 APPLIED MECHANICS. better acquainted with the distinction between force and energy. 286. Riveted Joints. Students are supposed to have made sketches of a number of well-proportioned joints. In applying the results of experiment arid our imperfect theory to actual structures we must remember that our practical conditions some- times closely agree with our hypothetical conditions, and then our calculations may be fairly exact ; whereas in other cases our theory is only an imperfect sort of guide in our workshop systems of seeking for success and avoiding failure. It is only Fig. 179. a very capable engineer who knows exactly what weight ought to be given to his theory in every case. If we had a perfect theory we need only consider the various ways in which a riveted joint may fracture. Then we should state the algebraic conditions that the joint shall be equally ready to fracture in these various ways, and we have at once the right proportions. Thus, take the strip which illustrates the single riveted lap joint of a boiler plate, Fig. 179 (1), or the single riveted butt, joint of Fig. 179 (5). A B = p, the pitch or distance from centre to centre of rivets ; t is thickness of plate ; \ is D c, the overlap APPLIED MECHANICS. 337 minus radius of rivet ; f s the shearing stress which the rivet will stand ; f t the tensile stress which the plate will stand. If fracture occurs as in (1), the tensile force P which the joint will stand is P = ?<% .... (1). If in double shear, as in (5), P = \&f. . . . . (5). If, as in (2), P=(p -d)tf t .... (2). If as in (3), when the iron of the plate is crushed by the rivet or the rivet is crushed by the iron, even the most absurd person will hesitate when he puts ?=f c td .... (3),/ c being some kind of surface crushing resistance per square inch, and td being assumed somehow to represent the surface at which crushing may take place. If, as in (4), we assume that the part E G F breaks like a beam fixed at the ends and loaded in the middle, we have a much wilder assumption, giving p = \ 2 / t -f- \d . . . . (4). Now we cannot assume that P has any of the above values. There is always great friction at the joint ; the rivet is in a state of unknown tension, and we have no inform- ation as to f s under such circumstances, and in a lap joint there is evidently a bending action due to the plates not being in line. Nevertheless, all the above formulae and similar formulae easily made out for all joints may be made use of to guide us in obtaining information when we vary proportions and make tests. For example, tests may be made upon X and d, and it has been found that \ from the edge of the hole to the edge of the plate must be at least equal to d, and considerations of economy and workmanship guide us in adding about a quarter of an inch when only half-inch rivets are used. Now if we put (4) equal to (3), or \*tf t -f- Id =f c td ---- (6), we have \ = . ft It is dangerous to follow this any further ; we have reached a rule (by taking convenient values for the stresses) that agrees so well with the practice of the best makers that we shall be apt to think our theory more valuable than it is. It will bo noticed that if we write (3) equal to (1) we get a rule which also agrees fairly well with practice ; but if we write (4) equal tc (5), which we have just as good a right to do, we find that \ ought to be v/2 times as great when the rivets are in double shear, and this is certainly riot in agreement with good practice. If we imagine that instead of (4) we use the idea of a beam 338 APPLIED MECHANICS breaking by shearing, we again find that X ought to be propor- tional to d.~~ But the result of a careful consideration of this and many other points in machine design is that possibly such formulae as (4) are more likely to be misleading than useful. A complete theory to replace (4) is perfectly possible, but it has not yet had the services of the necessary good mathematician. With (1) and (2) we feel much safer than with the others. Putting them equal, j d z j a = (p - d)tf t .... (7). We need some other equation evidently. Now, in all design we try to obtain maximum advantage of some kind. To get maximum economy of material we were obviously right in trying to have a joint equally ready to break in various ways. But there is a more general kind of economy, not at all easy to express algebraically (see my Calculus for Engineers, Art. 37), which tells us to punch holes in thin plates up to f " thick and drill them in thick plates. For thin plates, then, we have such a rule as d oc t, or taking into account cost of riveting, say d = \t + J if we are not to have too much expense in the fracture of punches [compressive strength of punch more than equal to shearing resistance of plate], whereas we have only one of these things to consider in drilling holes. We may take the rule d = 1.2 \/ t . . . . (8) as probably agreeing with the best practice from t = \ to t = 1 inch. If this rule were followed, then, using (7) and (8), we find l'44:f s = (p d)f t , or p d = '36vf s /f t = A, say. The strength of the joint is the strength of the unhurt plate multiplied by ^ , or by P i .... (9) where A stands for p d. For a rivet in double A -T- a shear we use (2), (5), and (8), and find p - d='72irf s /f tt and we may call this A in the formula. It is just twice the last. Now for complicated kinds of joint we must make assump- tions which are less likely than the above. It is usual to calculate, instead of p d in (2), a width of strip w equivalent in tensile strength to one rivet in single shear, or f0 = -j d 2 f s /tf t . Draw round each rivet a circle of diameter d + w, and let lines come to the circles dividing the plate up into strips of the breadth w ; thus we allot a strip of plate to each rivet. Students ought APPLIED MECHANICS. 339 to scheme for themselves examples such as are figured in books on " Machine Design." In such books they will also find sketches of many riveted joints which are well worth study in case there are no actual specimens to look at. The result of the calculation is a formula like (9). The results of actual tests show that instead of A in the numerator of (9), we have a different number k A, and students who have read Chap. XIII. carefully, and also what I have here given, will know fairly well why k is not equal to unity. We have, then, d = 1-2 ^ this off from M to D'. Hence the pushing forces on A D and B c cause D to move to D'. Again, the pulling forces on D c and A B further lengthen A D by the distance pa, which we set off from D' to L, and shorten D c by the distance pfi, which we set off from L 4. // Hence the motion of D due to the pulls and pushes acting Fiy. .81. tOD" together is D D", and we see that this is (D M + L D") A/2 or (a -|- /3) p V2~? But s, the amount of shear, is D D" ~ D o, and as D o = = inch, v 2 as A D is 1 inch, we have * = (o + )8) p A/2 -r- ~j= or 1p (a + 0). That is, shear strain shear stress multiplied by 2 (a + 0). So that the reciprocal of 2 (o + ft} is what we called N, the modulus of rigidity of the material. 289. General Results. Referring back to Arts. 269 and 288, you will see that we have Modulus of rigidity N = Modulus of elasticity of bulk ... K. = Young's modulus of elasticity ... E = and you will also see that if we know two of these for any material, we can find the third. These results are so important that we put them also thus: shapes : APPLIED MECHANICS. 343 = l ^=JL + JL B=L-1 H= ^- / 3 ...(3), where has the value given in (1). p. 2 is supposed to be less than Pi. If p 2 is a tensile stress, we have only to give it a negative sign in (3). (3) may be written mp\ np}>/ 9 .... (4), where m and n are constants for the material. Now let p v p%, and p be tensile stresses, and reverse the arrowhead on q. We have the same numerical relations between P> 1i P\i Pv an( i fracture occurs by shearing when f s has the value q + pp, so that fracture occurs when p t (cos.0 sin.0 + /* cos. 2 0) p 2 (cos.0 sin.0 ft sin. 2 0) > f & if is taken of such a value as to make q + ftp a maximum; that is, if = 45 + f s , where m l and n l are constants for the material. Let p 2 0. If p l is compressive stress, = 45 + ^, and it will be found that ?-C! = sec.0 - tan. .... (4), If ^j is a tensile stress, 9 = 45 - , and it will be found that S2 = se , tan. .... (6), APPLIED MECHANICS. 347 using f c and ft as the usual crushing and tearing stresses of the material and eliminating between. Given compressive stress p l and the value of / to find the least value of p z for resistance to fracture "by shearing ; take = 45 + | as giving the plane on which the tendency to shear is greatest, whatever p 2 may he, and we find _ 2fs . cos. (/> + 2h ( sin. ) ^ 2 1 + sin. Thus, for example, taking cast iron, let us suppose that ^ = 28 and /j. - 53, as already found, if the ultimate f c = 95,000, and /, = 28,500, then sin. = -47, p* = 34,000 + . 36 p l (that is, p l increases 27,800 Ibs. per square inch for every increase of 10,000 Ibs. per square inch in p 2 ). If for any material / c , /*, f s are the numerical values (treated all as positive quantities) of the three stresses which the material will stand, 2f s =f c (sec. tan.) =f t (sec.^> + tan.^>). Thus, for example, = } + ^. ft 1 sin.< Now in cast iron the proof-stresses given in Table XXII. are f c = 21,000, /,= 10,500, /,= 8,000, = 2=iT, and i = 19, tan.^ = -35, O c = 54|, Ot = With this value of , f g ought to be -356/ c , whereas it really is 38/ c . This is a discrepancy very allowable, and we may take it as some sort of verification of the theory. These were the first numbers I tried, but I have since found that other published numbers are less satisfactory in their support. For example, f c ought to be greater than f t for all materials if the theory is correct. It is evident that special experiments are required as a test. Tresca and Darwin have propounded the hypothesis that Pi Pz is constant; and it ought to be easy, by experiment, to decide whether this is constant, or whether, as by my theory, it increases as p l and p z increase. For example, I make p l p 2 equal to 94,000 when p 2 = in cast iron, and equal to 272,000 when p<2 = 100,000. These are from the ultimate values. As to the permanent set strength of cast iron, I makejOj jt> 2 == 21,000 when p 2 = 0, and 121,000 when p 2 = 100,000. In fact, instead of making p\ p z = 94,000, I make p\ 2'78 p 9 94,000. The hypothesis of Poncelet and St. Venant is that the material is fractured when the strain exceeds a certain amount, and this hypothesis is often used to give ultimate strength conditions when the strains have been investigated mathematically. But in a wrought-iron bar we have sometimes before fracture strains which are several hundreds of times as great as the greatest strains to which the calculations apply, and it seems to me, therefore, that Poncelet' s theory has no probability of correctness in its favour. If we assume that in any kind of earth there is a coefficient of 348 APPLIED MECHANICS. friction n between layers, but no permanent resistance to true shear- ing (that is, f s = 0), we get Rankine's theory of earth pressure from the above theory. The lateral pressure p z necessary to prevent a direct pressure p l from causing motion or fracture is given by 1 - ^ ^ the gtatic coefficieilt of friction in certain kind of earth is '9, so that <|> is 42, then pjpi is .figor- or, say, . Example. The weight of a building is 10 7 Ibs. The area of the concrete bottom of the foundations is 2,000 square feet. Afc what depth ought it to be below the level of the soil, if the soil is such that = 42 ? Ans., the pressure pi Ibs. per square foot is 5,000 Ibs. To resist this, a horizontal pressure p z of 1,000 Ibs. is needed. Regarding the horizontal pressure of 1,000 Ibs. as a new p v it needs a vertical pressure of 200 Ibs. per square foot, due to the weight of outside earth, to balance it. If the earth weighs 100 Ibs. per ctibic foot,- the depth of the foundation must be at least 2 feet. Assuming that the theory is right, the difficulty in carrying out a rule of this kind is that we do not know <, the angle of repose of a particular kind of earth. Rankine assumed that a long mound of earth would, after much weathering and rest, get to have a natural slope . If the natural surface of earth is hori- zontal, it is easy to find on the above theory the stress on any interface when motion is about to take place, and particularly on a vertical interface, and so the reason for part of the following rule of Rankine's is known. The study of the stresses when the ground is sloping must be left as an exercise for students. 293. Rankine's Rule for Earth. Draw an angle x o R, Fig. 184, to represent the static permanent angle of repose of the kind of earth. Describe Y R x, a semicircle touching on. If A B, Fig. 185, is the vertical face of a wall sustaining a bank of this earth, whose slope is A c, make the angle x o P equal to the inclination of A c to the horizon. Find B D = o Q.A B/O P. Then Fig. 184. Fig. 185. A B D is a wedge of earth whose weight represents the total pressure acting on A B. The pressures act in directions parallel APPLIED MECHANICS. 849 fco A c, and the resultant force, representing the total pressure, acts a third of the way up from B to A. This rule may be compared with the rule of Art. 173 for water pressure against a vertical wall. Rankine neglects fric tion against the wall face. Students of this subject are directed to a paper by Sir B. Baker, Proc. Inst. C.E., Vol. 65, and its discussion. When grain is stored in vertical prismatic cylinders, the average pressure in pounds per square foot on the flat bottom is c d w where d is the diameter in feet, w the weight of a cubic foot, and c is 0'84 for wheat, 0*96 for peas. At greater heights than three times the breadth of section of a bin the pressure on the sides is constant, being about 50 Ibs. per square foot for all kinds of grain until we get near the surface. 294. Twisting. In Fig. 186, AB represents a wire held firmly at A. At B there is a pulley fixed firmly to the wire, and this pulley is acted upon by two cords, which tend to turn it without moving its centre sideways. In fact, they act on the pulley with a turning mo- ment merely. But the pulley can only turn by giving a twist to the wire, and the amount of motion it gets tells us how much the twist is. A little pointer fastened at c moves over a cardboard dial, and tells us accurately how much twist is given to the wire. The angle turned through by the pointer is called the total angle of twist at c. If we had a pointer at each of the places G, H, and c, and if A, G, H, and c were one foot apart from one another, we should Fig. 186. find that the 350 APPLIED MECHANICS. angles of twist at.G, H, and c are as 1 : 2 : 3 ; in fact, the angle, of twist is proportional to the length oftvire tivisted. You will find that if a twisting moment of 10 pound-feet produces a twist of 4, then a twisting moment of 20 pound- feet produces a twist of 8, and, in fact, the twist is proportional to the twisting moment which is applied. You will also find that if you try different sizes of wire of the same material, say wire whose diameters are in the proportion of 1, 2, 3, &c., and to each of them you apply the same twisting moment, the amount of twist produced in them will be in the proportion 1, , , &c. ; that is, inversely as the fourth power of the diameter 16 81 of the wire. Lastly, taking wires all of the same diameters and lengths, but of different materials, and applying to them the same twisting moment, the amount of twist will be inversely proportional to the number which we call the modulus of rigidity of the material. The exact rules are given in (1) and (2), and the values of N given in Table XX. may be relied upon in such calculations, because they have all been determined from experiments on the twisting of wires and shafts. Exercise. A brass wire 20 inches long, O'l inch diameter, twists through a total angle of 130 degrees when a twisting moment of 4 pound-inches is applied. Find N for the material. Answer : 3 '6 X 10 6 Ibs. per square inch. Exercise. What would be the twist of a shaft of the same material with a twisting moment of 600 pound-inches, 20 feet long, 1*2 inch diameter? Answer: 7 '8 degrees. It is, however, well to notice that the drawn brass will probably have a diflerent value for its N than the brass of a much larger shaft. It will be seen that the strain is a shear strain. Con- sider M H G (Fig. 187) to be a cross-section of the wire ; then a point which is at H before the twist occurs is found to be at G when there is a twist in the wire, and a point such as p' moves to P, but a point o in the centre of the wire does not move. Now there is no such motion at the fixed place A, Fig. 186, and in each .section there is more of this motion the farther it is away from A ; in fact, the motion is just as it was in the Pig. is?. indiarubber of Fig. 176, only that it varies APPLIED MECHANICS. 351 in the section, the motion being greatest at the outside of the wire, and nothing at the centre. The material breaks when the shear stress at the surface becomes too great, and the rule found by experiment is that for any material, whatever the length of the wire, the twisting moment which will cause rupture is proportional to the cube of the diameter. It is well known that when a shaft is transmitting power, the horse-power transmitted is proportional to the twisting moment or torque in the shaft multiplied by the number of revolutions made by it per minute. The rule used by engineers is this \d= 3-3 \/H/n .... (1), as giving the safe diameter of a wroughtiron shaft at n revolutions per minute, if it is only subjected to torsion. Observe that if we double the speed, the shaft is strong enough for double the power. Instead of 3*3 we use 2 -9 for a mild steel shaft and 4 for cast iron, 294a. Shafts usually carry pulleys, and are otherwise loaded as beams, as by the pull of belts, and therefore, for reasons given in Art. 379, we take 1J to 1^ of the above size for mill shafting, and for crank shafts and shafts subjected to shocks we sometimes add 50 per cent, to the diameter as given by (1). We have some explanation in our theory of Art. 263 for this increase ; some of it is due to the variation in stress, and therefore to fatigue ; some of it is due to the fact that in crank shafts the maximum torque is often double the average torque. In a long line of shafting, if the power is given off at various places with some irregularity, it may even become evident to the eye that the shaft is perpetually twisting and untwisting, for of course the twist is proportional to the horse- power transmitted if the speed is constant. When this is the case, although the shaft may seem to be strong enough, it is weak because it is not stiff enough. A very long shaft some times gets into a state of torsional vibration just in the same way that the cage-rope of a coal-mine gets into a state ot longitudinal vibration. The nature of this vibration will depend on accidental causes, and should the impulses that give rise to it happen to repeat themselves at proper intervals, the vibration may go on increasing until the torsion at some place may be sufficient to produce rupture. In the same way a number of men walking from side to side of a large ship, just taking as much time in going from one side to the other as the ship takes to make a vibration, may make the rolling dangerous. 352 APPLIED MECIIAMC3. It is for this reason that we endeavour to make the period of oscillation of a ship differ greatly from the probable period of waves which she may experience (see Art. 489). In very small shafting this vibration often occurs, and it is usual to add vaguely f to f inch to the diameters found by the above rule a sacrifice to the Goddess of Chance. 295. Consider a little prism, r B (Fig. 188), whose ends lie in two cross-sections of a shaft near together, o being the centre of one of the sections, and o' the centre of the other. The twisting strain to move to B', regarding r as fixed. (The motion is, of course, usually very much less than I have here shown it). There must, then, he shearing forces acting on the ends in opposite directions. If a is the angle of twist of the shaft per inch of its length, then B o' B' is a multiplied by o o' ; and if o P or o' B is ?', then B B' is r oo o', where a is an angle measured in radians. The shear strain in the little prism is BB' divided -! o 1 by P B or o o', so that it is ra ; hence the shear stress is N r a (see Art. 282). If a is the area of the end of the little prism in square inches, the shear force acting on it is Nra, and as this acts in the direction at Fte- 188t right angles to the radius, its moment about oo' is Nr^aa. But we have a similar moment for every such little area into which the cross-section may be divided, and to find the total torque we must take the sum of all such terms. Now, N and a are the same every- where, so that in taking such a sum our only difficulty is with thu factors r 2 a. But the sum of all such terms as r"*a is called the moment of inertia of the section about the axis o o', and it has been calculated for us. Thus, if D is-'the diameter of a round shaft, the moment of inertia of its section about an axis through its centre at right angles to the section is ir n 4 -f- 32, and for a hollow shaft whose outside diameter is D and inside diameter d, the moment of inertia is TT (D 4 d*) -f- 32 ; and hence we see that the twisting moment T necessary to produce a twist of a radians per inch in a round shaft of diameter D is T = TTN a D 4 /32 .... (1), and for a hollow shaft it is T TT N a (D* d 4 )/32 .... (2). The torsional rigidity of a shaft is defined as A if o T/A. The values of A will be found in Table XV., Art. 532, for various sections of shaft. The verification of these rules is an excellent laboratory exercise. 296. The strength of a shaft is to be calculated on the assump- tion that rupture occurs when the shear stress N r a mentioned APPLIED MECHANICS. 353 above exceeds the greatest shearing stress to which the material ought to be subjected ; and as the stress is greatest when r is the outer radius of the shaft or | D, so that / = | N D a, and as from equation (1) (Art. 295) we find that Na is 32 T -f- irD 4 , we know that/= ^D x 32T 4- WE 4 , and this is the condition of strength of a cylindric shaft. It is more compactly put in the form IT D 3 / T = fi for solid cylindric shafts .... (1), and in the same way we get T = - Y* f r hollow cylindric shafts .... (2), / being the breaking shear stress of the material in pounds per square inch, T the twisting moment in pound-inches which will cause rupture, D the outer diameter, and d the inner diameter (if the shaft is hollow) in inches. We see, then, that the strength of a solid shaft depends on the cube of its diameter, whereas its stiffness depends on the fourth power of its diameter. As to the practical rule given in (1) (Art. 294), we saw in Art. 182 that torque in pound-//BC 2 + Ac 2 ), or p . (B c + A B), a rule which is perhaps a little un- Fig. 190. expected. 301. Ordinary shafts are of wrought iron or mild steel, materials such that their resistances to compression, tension, and shearing are not very different, and therefore when they are subjected to T and M we at once calculate (6), and say that the strength of the shaft is the same as if it were subjected to this twisting moment only. The practical rule (Art^294a) is worked on the idea of a twist- ing moment only, or d ex \/ T\ If there is also a bending moment M, which is of the value k T, then the rule becomes evidently ~.v d = 3-3 - 3.3 The extreme values of k for many kinds of shafting have beori worked out, and the consequent changd in the multiplier of (1), by means of shrewd guessing and calculation; but to my mind it is a better recommendation of the rules given in Art. 294a that they are consistent with theory and with the best practice of engineers. I have given here the usual theory of shafts subjected to twisting and bending ; it assumes that the strength is determined by the maximum stress. A true theory would take account of the considerations of Art. 294a. if a tensile stress p and a shearing stress / act on the same 356 APPLIED MECHANICS. interface, and if p l and p* are the principal stresses, we see from (3) of Art. 298 that The theory of Arts. 290 and 292 tells us that fracture will take place if nflpi n l p z > f s , where m 1 , n 1 , and f s are constants which ought to be known for the material. If p is a compressive stress, we have fracture if mpi np 2 > f s . The probable values of m, n, w 1 , n l for cast iron are given in Art. 292. For wrought iron and mild steel it is possible that of Art. 292 is 0, and hence m = ^, n = , m 1 = , n l = . Hence we have fracture either on the compressive or tensile side of a shaft that is, whether p is compressive or tensile stress il 2 -v/ p 2 + / 2 >/, Hence, if a shaft is subjected to the twisting moment T and the bending moment M, we calculate an equivalent twisting moment v/T 2 + M 2 , and assume that the shaft is subjected to this alone. For materials in general, and probably even in the case of wrought iron and mild steel, the equivalent twisting moment ought to be calculated from h M + k x/x 2 + M 2 , where h and yfcare constants, depend- ing upon the nature of the material, which are not known at the present time. Fig. 191. Kg. 192. 302. The demonstration of Art. 296 is found to agree with experi- ment, but its results must not be applied except to shafts which are circular in section. Our assumption, which experience warranted, was that when such a shaft as A u (Fig. 191) is fixed at B, and when Fi. 101 APPLIED MECHANICS, 357 to an arm, CD, a twisting couple is applied, every straight line in a section remains straight, and moves through the same angle as every other line. But it can be shown that this is not the case for a shaft of any other than a circular section. Thus, let o (Fig; 192) be the centre of gravity of the section PQS, and let us suppose that a shaft of this section is subjected to the sort of strain I have described. The shear strain at the point r is in the direction P K, perpen- dicular to o P. Let its amount, which we know to be o P X angle of twist, be represented by the length of P K. ^ It is easy to show that this is just the same as a shear strain P N in the direction P N, normal to the surface of the shaft at P, together with a shear strain in the direction P T, tangential to the shaft at p. But shear strain in any direction is al- ways accompanied by a similar strain in a plane at right angles to this direction (see Art. 282), so that since we have the shear PN, we must also have a shear parallel to the axis of the prism along the surface at P, and this cannot be produced merely by a twisting moment. We must imagine that along with the twisting moment there is a force distributed over the sur- face of the shaft to produce the above effects. The result of an exact investigation (Art. 313) is that a twisting couple produces a greater twist than might appear from what I have said in Art. 295, and it also produces a warping of the naturally plane sections of the shaft. Thus Fig. 193 is the shape assumed by each section of an elliptic shaft, and Fig. 194 of a square shaft. Imagine a section to be distinguishable, say, in a glass shaft by a thin layer of a different colour "from the rest. Deeper shading indicates greater distance from the observer who is looking towards the fixed end of the shaft. The arrows show the direction of the twisting moment. In the fol- lowing three sections, instead of the torque for a twist of one radian being equal to N times the moment of inertia of the cross-section, it is only 84 times this for a square section (Fig. 195), - 54 times it for the section Fig. 196, and -6 times it for the section Fig. 197. Indeed, Fig. 195. Fig. 196. Pig. 197. 358 APPLIED MECHANICS. the square section has only -88 times the torsional rigidity of a cylindric shaft of the same sectional area; Fig. 196 has -67 times, and Fig. 197 has -73 times the torsional rigidity of a cylindric shaft of the same sectional area. The numbers in the column headed w, Table XV. , express the rela- tive strengths to resist twisting of the various sections there figured. The torsional rigidity of an elliptic section whose principal semi-diameters are a and b is N7r# 3 3 /(a 2 + b 2 ). If M is the twisting moment, the shear stress at a point x, y (the axis of x being a) is 2 M t/bW + aV/ira 3 3 . This is greatest at the end of the minor axis, being 2 ufirab 2 . The torsional rigidity of a rectangle, if its length is two or more times its breadth, is, with some accuracy, the same as that of the inscribed ellipse multiplied by the ratio of their polar moments of inertia. The greatest shear stress occurs at the middle of the longer side of the boundary, and is 3 M (a 2 + # 2 )/8 cflb z , if M is the twisting moment and a is half the longer, b half the shorter, side of the rectangle. 303. A very interesting result of the investigation is that there is always greatest distortion at that part of the surface of a shaft where the surface is nearest the axis. Thus in an elliptic shaft the substance is most strained at the ends of the shorter diameter of the section. Imagine a very light box to be made so as to contain frictionless liquid exactly of the shape of a shaft. If we give a sudden turn to the box about the axis, the liquid will be left behind if the box is circular in sec- tion, but it will have motions relatively to the box which can very readily be imagined if the shaft is not circular in section. Now the actual velocity of the liquid at any place re- latively to the box is in the same direction as, and is proportional to, the shear in a similar shaft when it is twisted. This has been proved by Lord Kelvin. You will see from this that there is very little strain at the projecting ribs of the shaft, whose section is shown in Fig. 196, and just at the projecting angles of Figs. 195 and 197. This reminds me of a general remark which I have to make, and which I must leave without proof. A solid of any elastic substance cannot experience any finite stress or strain in the neighbourhood of a project- ing point, unless acted on by outside forces just at the point. In the neighbourhood of an edge it may have strain only in the direc- tion of the edge, and generally there will be exceedingly great strain and stresses at any ro-entrant edges or angles. An important application of the last part of the statement is the well-known practical rule that every re-entering edge or angle ought to be rounded to prevent risk of rupture in solid pieces o Fig. 199. Fig. 200. APPLIED MECHANICS. 359 designed to bear stress. An illustration of the principle is the stress at the centre of the circular outline in the three sections of shafts (Figs. 198, 199, and 200). In Fig. 198, at o, there is no stress when the shaft is twisted; in Fig. 199 the stress may he calculated ; in Fig. 200 the stress is exceedingly great for even the smallest twist (see Art. 302). EXERCISES. N.B. A most usual error of students is to forget that moments in pound-inches are not numerically the same as in pound-feet. Beginners had better leave the exercises involving bending moment until they have studied bending. 1. A shaft 1 inch in diameter can safely transmit a torque of 2,400 pound-inches. "What diameter of shaft would be required for transmitting 15 H.P. at 200 revolutions per minute ? Ans., 1 inch. 2. Find the horse-power which may be transmitted by a shaft 4 inches in diameter when running at 150 revolutions per minute, if the stress due to twisting be limited to 9,000 Ibs. per square inch. Ans., 269. 3. A line of steel shafting is 80 feet long ; if a twisting moment of 4,000 pound-inches bfe applied at one end, what will be the total angle of twist, the diameter of the shaft being 2J inches ? What horse-power will this transmit at 220 revolutions per minute ? Am., 4*7 ; 14 H.P. 4. A solid wrought-iron shaft is to be replaced by a hollow steel shaft of the same diameter. If the material of the latter is 30 per cent, stronger than that of the former, what must be the ratio of internal to external diameter ? What is the percentage saving in weight ? Ans., -693 ; 48 per cent. 5. The amount of twist in a solid shaft is to be limited to 1 for each 10 feet of length. Find the diameter for a twisting moment of 50 ton- inches, the modulus of torsional rigidity being 10,000,000 Ibs. per square inch. Ans., 5'2 inches. 6. A wrought-iron shaft is subjected simultaneously to a bending moment of 8,000 pound-inches, and to a twisting moment of 15,000 pound-inches. Find the twisting moment equivalent to these two and the least safe diameter of the shaft, the safe shear stress being taken at 8,000 Ibs. per square inch. Ans., 25,000 pound-inches; 2*52 inches. 7. Find the diameter of a shaft for a winding drum which works under the following conditions : the load lifted is 1 ton ; diameter of drum, 5 feet ; width of face of drum, 26 inches ; distance from inner face of drum to the middle of the bearing of shaft, 13 inches; maximum stress, 7,000 Ibs. per square inch. . Ans., 5-44 inches. 8. A wrought-iron shaft 3 inches in diameter, and making 140 revolu- tions per minute, is supported at wall brackets 16 feet apart. There is a pulley on the shaft, midway between the bearings; if ihe resultant side pull due to the weight of the pulley and the pull of the belt be 210 Ibs., what is the greatest horse-power the shaft will transmit with safety ? Safe shear stress, 7,800 Ibs. per square inch. Ans., 66. 9. A bar of iron is at the same time under a direct tensile stress of 6,000 Ibs. per square inch, and to a shearing stress of 3,500 Ibs. per square inch. What would be the resultant equivalent tensile stress on the material ? Ans., 6,801 Ibs. square inch. 360 APPLIED MECHANICS. 10. Taking the safe tensile stress of wrought-iron to be 10,000 Ibs. per square inch, determine whether it would he safe to subject a piece of wrought-iron to a tensile stress of 3^ tons per square inch, together with a shear stress of 3 tons per square inch. Ans. Unsafe; max. stress = 11,700 Ibs. square inch. 11. A wrought-iron shaft is subjected to a twisting moment of 36,000 pound-inches and a bending moment of 18,000 pound-inches; find the diameter when the maximum shear stress is 8,000 Ibs. per square inch. Find also the twisting moment which alone would produce a shear stress of the same numerical value. Ans., 3'3"; 58,200 pound-inches. 12. A screw propeller-shaft 10 inches in diameter is subjected to a twisting moment of 35 ton-feet, and to a bending moment of 10 ton-feet, due to the weight of the shaft and the pitching of the ship. What is the maximum compressive stress if the thrust of the screw be 10 tons ? Am., 2-9 tons. 13. A shaft 12 inches diameter, transmitting a twisting moment of 100 ton-feet, is also subject to a bending moment of 20 ton-feet. Find the maximum stress induced. Ans., 4*3 tons per square inch. 14. Find the diameter of a wrought-iron shaft to transmit 90 horse- power at 130 revolutions per minute. If there is a bending moment equal to the twisting moment, what ought to be the diameter ? Ans., 2-7 inches ; 4-1 inches. 15. A steam-engine crank is 12 inches long, and the greatest force which is transmitted through the connecting-rod is 9,000 Ibs. Find the diameter of the wrought-iron crank- shaft, taking the safe shear stress at 9,000, the distance of the centre of the crank-pin from the centre of the bearing nearest it being 10 inches, measured horizontally. Ans., 5'07 inches. 16. A round shaft 3 inches diameter, find the sizes of equivalent shafts of square, elliptic and rectangular sections if the breadth and thickness of each of these latter are as 1 to 2. If these shafts are 20 feet long, and they are transmitting 20 H.P. at 100 revolutions, what is the total twist on each of them. N = 10,500,000. Ans., Eectangle, 2-15 x 4-3 inches; ellijee, minor axis, 2-38 inches; square, 2-73 inches side. Twist on circular shaft =2 -07 3, on square shaft, 1-78; elliptical shaft, 1-05; rectangular shaft -93. 361 CHAPTER XV. MORE DIFFICULT THEORY. 304. MATHEMATICIANS endeavour to help engineers (including in this term all men who apply the principles of natural science) by investigations concerning ideal elastic material shaped like actual beams and shafts. The mathematical analysis is exact and difficult, and only a few problems have yet been solved, and it is only by leaving out terms that seem insignificant that we are able to apply the results to actual problems. The engineer recognises from the beginning that his problems are too complicated for any exact mathematical investigation. He therefore leaves out his apparently insignificant terms rather at the beginning ^ than ^ the end ; but indeed he leaves them out in any part of his investiga- tion if they are likely to give trouble, for he recognises from the beginning that his theory is only to guide him, and that the final appeal must be to experiment. The engineer looks upon the phenomena involved in the loading of the tie-bar as simple because experiment is easy; whereas the mathematician, seeing that a lateral contraction accompanies an axial elongation, regards it as complicated. The engineer ought to study, and develop, and correct, by experimental observations, his usual method of investigation aa described in this book, but he ought also to study the mathema- tician's treatment of the subject, and let it assist him as he lets experiment assist him. The following very short sketch needs much time for its proper comprehension ; it is from the mathe- matician's point of view. Students may pursue the subject in Mr. Love's treatise on elasticity, or Thomson and Tait's treatise on natural philosophy. 305. The consideration of homogeneous strain in general (when any portion of stuff in tho shape of a sphere is changed into what is called the strain ellipsoid, any three co-orthogonal diameters of the sphere becoming conjugate diameters of the ellipsoid, planes parallel to one another remain parallel ; all parallel lines get the same fractional changes in length) is not so difficult as it is tedious. Unfortunately the authors of books insist on its being studied. For 2 our purposes we have only to deal with infinitesimal strains, and this is easy. Suppose that a point as, y, z is dis- placed to x + u,.y + v, z + w where u, v, w are very small, then _ du f dv _dw . . are the tensile strains of the stuff in directions parallel to a?, y, z. In Fig. o 201 let the axis of a* be at right angles Fig. M* 363 APPLIED MECHANICS. to the plane of the paper. Let the traces of two planes, c B and c A, parallel to o Y and o z, and perpendicular to the paper, "become changed to C 7 B', c' A'. The angle hy which A' c' B' is less than the original right angle is evidently a shear strain. I shall call the amount of it a ; and as it is a shear of planes normal to Y, parallel to z, or a shear of planes normal to z, parallel to Y, there is no great harm in calling it a shear about the axis of x. Now the angle turned through hy c A, clockwise, is really the horizontal motion of A minus the horizontal motion of c, divided hy AC, since the angle is very small. But this is dv/dz. Let the student he quite sure of this fact. Similarly the angle, anti- clock wise, turned through by c B, is dw/dy, and hence a = ^ h -^ ( 2 )- Similarly if b and c are to y and z what a is to a 1 , then du dw dv du b = Tz + di> c --dx + dy--M' A student may keep these in mind by means of the mnemonic a b c } x y }.... (2). it ] = 2 = w 3 = are evidently the conditions that there is a function (called the strain potential), such that dd> d

, u = -,~, v = - , w = ~ ; and we see that, in case there is no dx dy' dz cubical dilatation, (4) becomes At any point o of a body let there be three planes of reference meeting at the three mutually perpendicular axes of a?, of y, and of z. Let tensile stress be called positive. Across the plane of yz (usually APPLIED MECHANICS. 363 called the plane x, because the axis of a? is normal to it) let the stress be resolved into its three components- xz, parallel to ox; YZ, parallel to oy ; and z x , parallel to oz. Across the planes y and z let the component stresses he xy, yy, z y> and Xz, \z, z z . To find the stress components F, G, H in the three directions across any plane whose direction cosines (or the direction cosines of its normal) are I, m, n, consider the equilibrium of the tetrahedron formed by the four planes under the action of the tractions acting from material outside it. The area of the sloping face being A, the areas of the other faces are I A, m A, and n A. The resultant force on each area is stress multiplied by each area. We hare, then, resolving parallel to x, FAzz: and two Other equations. Hence, H = zx + m zy-T-n zz. In a fluid the static stress is always normal to any interface. Hence F, G, and H are in this case the components of a normal stress p on the new plane, and all the tangential stresses vanish. Hence, F =. I P =z I Xa; or P = Xj G =m'pi=.m'yy or P = Y^ H~wp:rzwz2 or p := z 3 ; so that the stress is the same across any interface whatsoever. 306. Now consider in the general case the equilibrium of a paral- lelopiped whose corner is at o (Fig. 202), the co-ordinates of which are x, y, z, and let the co-ordinates of the opposite corner o' be x + 8#, y + 8z/, z + 8z, and let us consider the equilibrium of all the surface tractions acting on the faces from the outside. The re- sultants of the normal forces meet in the centre, and if we neglect volumetric forces, they are in equilibrium. They are not shown in Fig. 202. We show the tangential forces per unit area with which outside stuff acts on inside on three of the faces. The other three faces have similar forces, the arrows being in opposite senses to these.* Hence their moments are just the same as the moments of these about axes through the centre. * It is easy to take into account the fact that there may be volumetric forces, and that z a; on the x plane through o' is not equal to Z on the x plane through o, but is rather Zx-}-5x z x . We do this later, and it is easy to see that these extra terms vanish in the present problem. Fig. 202. 364 APPLIED MECHANICS. Taking moments of the figured forces about the axis through o parallel to a?, we find zy . 8x . Sz . % 8y = YZ 5a? . Sy . f 5z; so that Zy YZ. This is Cauchy's theorem. We call each of them s. In the same way we have the other two relations here given : We also give the names p to x-c Q to Yy } B, to z g . s, T, u are the shear stresses, and p, Q, and n are the normal tensile stresses in the material. Hence our equation may bo written : e Itr + *t Q + MB >> . . . (1). H= IT + ms -j- n) Exercise. Across what interfaces at a point are the stresses normal ? That is, find the principal stresses and their directions. If p, G, and H are the components of a force B normal to the plane I m n, then B/ = P, BW = G, BW = H; and if we sub- stitute these in (1), remembering that I 2 + w 2 + 2 = 1, we can find Imn and B. The answer tells us that there are always three directions and amounts of principal stress. In our most general state of stress, Fig. 203 shows the surface tractions acting on the element Sx Sy Szfrom the outside. Imagine equal and opposite forces on the other three faces. Now let us consider volumetric forces and the rates of variation of the stress. On the faces meeting at o we have P Q B s T and u. But the forces on the other faces are, as regards normal forces, dv , ^ do. ' dx Q+ Si dR Fig. 203. On Sy . Sz we have also u + Sx , and T + 82? . -T- ; on Sz . Sx we v dv , ds have u + Sy .-5- and 8 + 8z . -7- ; dy dz on Sx . Sy we have T + Sz ^ and s + Sz . ^ . Now we know that the parts of these forces p and p, s and s, etc., balance in- dependently ; therefore we need only consider the extra ones as shown in Fig. 204. If the volumetric force in the direction of increasing x is p . x . So? . Sy . &z, then APPLIED MECHANICS. 365 fi~3L.Sx.Sy. + "5? We have similar equations for the other directions, and these reduce to the following equations of equilibrium : J_^J_ ft I f 2^ dx dy dz II U Uw ffo/ A^ ' ^ + ^ + ^ =0 "--( 5 )- The surface conditions (3) become IT + ms = . . . . (6). He also assumes that the origin is fixed, and that particles in the directions of the axes of z and y just there are fixed. This means that at the origin The following purely mathematical work is tedious, but quite easy. The student will do it carefully for himself, following our directions. [Inserting (4) in (2), so as to calculate the strains, we find that du dv dw ... where a is Poisson's ratio, or /8/o, or means that + N). Also u = dy dx dT . ... d 2 u , d^w -- = means that -y s- + ~s - 5 dz dz 2 dz . dx ds .. . d*w d* = means that -= - = f- -=- dz dz . dy d and the last condition of (5) gives (d 2 u d 2 w d 2 w d^o \ ' dx . dz + ^2 + ~df~r dy . dz) + = 0, dy . The first and second of (10) and (8) give (11) and (8) give, after simplification, Differentiate (13) with respect to z, and (10) with respect tc x and y, and use (8) to eliminate u and v, and we get Differentiate (10) with respect to y and a?, add and use (9), and we get APPLIED MECHANICS. 373 "Differentiate (10) with respect to y and a?, and use (8) and (14), and we get Hence -,- is linear in z, and linear in x and y separately. Hence az ^ = a + ai0 + 2?y + (/8 + J8^ + j^y) .... (17). [a and )8 are here any constants, and not the> coefficients of formula (2)]. Now use (8) with (17), and also (9), and we have u = - ff (ax + I ai* + oaya + )8za? -f ^yzx + -f M! + where Wj and w are functions of y. Similarly, where v and v l are functions of x. Now try these in (9), and find values of U Q , v v v , v v and get w = z (a + a^x + o 2 7/) + ^ 2 2 (j8 -f )3i + Atf) + <^ (*i ^) = - Using this in (13), we see that we must solve To get the complete solution, we find the general solution when the right hand side is zero, and add to it a particular solution. For the particular solution, try = 2 (a (fty + |z 2 (ft* + -r = at all points of any cross-section, and the condition + m = j8 (m* - iy) + - cr) ---- (19) at all points of the cylindric boundary. The stresses at any point are P = 0, Q = 0, u = 0. - 4% .... (20). K = E [a + o^ + OsT/ + z (fta; + By giving any values to a, a 15 o 2 , /8 , ft, /8 2 we get all the possible solutions. It must be remembered that a point x y z has the new position x + u, y + v, w + z. Notice that if , ft, 2 are 0, then < = 0, because s = N and T = N -i, and each of these is at the boundary, and the only conditions to be satisfied by d> are that l- + dx - = at the boundary. Hence may be a constant, or rather a function of z ; but it is where z = 0, and hence = everywhere. 310. Example 1. Let all the arbitrary constants vanish except a. We must remember that a is any arbitrary constant, and is not the a of Art. 289. Then u = - (rax, v = - (x, y}, where T is a constant and w is a function of x and y only. We find that P = such that -~ + -y-| = all over the section, with the boundary condition l~- + m-^- = r (ly - mx). To do ctx cky this, it is well, as in many other such problems, to find, first> a conjugate function tl/, such that -^ + - = 0, and ~ = -^, and ax* ay* ax ay -^ = - -jr. So that the boundary condition becomes l^. - m^- = r (ly - mx). dy dx Now we see by trial that if // = T (x 2 + y z ) + c, the boundary ^ = ry, and -^ y ax then, that if we can find a solution of -^ + -5-^ = 0, subject to the condition ^ = \ r (x 2 + y*) + c at all points of the boundary, we have the correct answer. # 2 v 2 Example. Prism of elliptic section, -^+- = 1> gi yin g the shape of the boundary. It is found by guessing that ;J/ = A (a; 2 - t,- 2 ) is a solution, if A is properly evaluated to satisfy the boundary condition. So that A (a; 2 - y 2 ) = \ r (x 2 + y 2 ) + const, at the b 2 boundary that is, where y z = -^ (a 2 - x 2 ). We substitute this, and find that A is \ r (a 2 - b 2 )/(a 2 + l> 2 ). So that ^ = A (x 2 - y 2 }, and = - 2 A xy. Evidently the curves in the section along which is constant are rectangular hyperbolas, with the principal diameters as axes. The total twisting couple is APPLIED MECHANICS. 377 The shears are now easily calculated. 6 = N (- 2 A.? + rx), T - N (- 2 Ay - T y), 8 - N * ( T - T #T?) = NT * (^Vp T = Ny _ T _ s is greatest when x = #, then The maximum stress anywhere in the section is the value of T when y = b, viz. N T 2 a , a . For / is a maximum when b 4 x z + V is greatest. Now, whatever be the value of x for which the maximum occurs, for that value of x the maximum will be when y is greatest namely, on the boundary. Put x = a cos.0, y = b sin.0, then * 4 OJ 2 + a 4 / becomes 2 6 2 [A 2 + ( 2 - & 2 ) sin. 2 0], which is obviously greatest when = 90. Hence the stress is a maximum at the extremities of the minor axis. Example. Torsion of a Shaft whose Section is an Equilateral Triangle. Let 3 a be the vertical height of _the triangle, thenthe equation of the boundary is. (x - a) (x - y V 3 + 2 a) (x + y V 3 + 2 a) = 0, or a; 2 - 3 a;?/ 2 + 3 (x 2 + y 2 ) - 4 a 3 = 0. The function i|/ = A (a; 3 - 3 #y 2 ) satisfies the differential equation, and A (x* - 3 xf) _ i T (#2 + y2) ^-QI k e constant over the boundary if A = - -. Hence = - (y 3 - 3a%) ---- (1). From this it is easy to find the strains and stresses. It will be found that the shear is in the corners, and is a maximum at the middle of the sides. 313. Approximate Formulae. St. Tenant worked out and the strains and stresses for a number of sections, and he found that if the section of a shaft is not too different in any two of its dimensions across the centre, the torsional rigidity (or twisting 6 4 N moment per unit angle of twist) is A = 'in where s is the area of the section and i its moment of inertia about its centre oi gravity ; N is the modulus of rigidity, and m is a number which does not greatly differ in different cases, m = -02533 for an ellipse; and in the sections examined its lowest value was '023, and its highest was '026. Consequently, if we take A = j- s%/i in all such cases, there is no great error. 378 APPLIED MECHANICS. 314. Non-uniform Flexure. In Art. 309 let all the constants be zero except fa. The displacements are where d> satisfies -^ + = at all points of a normal section ; dx 2 dy 2 and at the "boundary the condition The stresses at any point of a section z = constant, are S = N(-^- (2 + " K i dy E I write out the shear strains a and b in terms of , etc. = 0. Now in cases that have been considered in which x and y are small compared with I - z, it is found that < is of the third degree in x and y. So that a and b are small compared with e, /, and g. For the same reason, the term 2 xy 2 is not important in w above. And the engineer's theory based on the assumption that a plane section remains plane, may be taken as correct. When x and y are not small compared with I - z, we must find , and this is difficult. As w contains , although the tensile strain is pro- portional to x, the plane section does not remain plane if beams are short. St. Venant's solution assumes that w is distributed in a particular manner over the end section. But by the principle of equipollent loads the actual distribution of w is of very little consequence, except close to the end section itself, and hence is of no practical importance except in beams that are not long in comparison with the values of x and y in their sections. 315. Vertical loads are often applied to beams on their hori- zontal top surfaces. We know from the Cciple of equipollent .s that the actual distribution has little effect except in the neighbourhood of the surfaces to which the loads are applied. We can obtain a fairly clear notion of the effect by thinking of the load on a plane surface bounding an infinite elastic solid. Vig. 205. 380 APPLIED MECHANICS. M. Boussinesq has given the solution.* The following brief memorandum may be useful, o o being the plane surface bounding the infinite isotropic solid. M is a point within, situated at a distance MN = z below the surface. K is any element of the surface, situated at the distance K M / from the point M, and subject to a given exterior surface pressure up = p per unit area, having the component K^? 1 = p 1 per unit area along KM. The pressure which a plane ele- ment E E 1 taken through M parallel to the surface o o will support per unit area in consequence of the pressure p will be found directed along K M produced, and is equal to M F = - ^ * t If, as a particular case, the pressure KP =p be normal, then p 1 = p cos. NMK. =p-, and MF = 3^z 2 /2 in- 4 . If we want the vertical component of M F, we have 3pz 3 /2 Trr 5 . Generally. Let Wi be the normal force per unit area at any point on the plane bounding surface at the point x = x 1 , y = y 1 , z = 0, the axes of x and y being in the plane, and the axis of z being the normal to the plane drawn into the material. Let (/> = I \w l . rdx 1 . dy l , and x = I I w- log. (z -\- r) dx l . dy l , where r is the distance from x l y l O to xyz. Then u, v, and w being the displacements at x, y, z, 1 dJi 1 d? 4 v (\ + N) dy 4 IT N dz . dy' _ 1 dm 1 d^<> " __ _ \ + N) ~dz ~ 4^N ' ~dz* 47TN (x + N) ' From M, v, and w all the strains and stresses may be calculated at any point. * The interested student may refer to a paper read by Prof. Carus Wilson before the Physical Society of London (June, 1891) on " The Influence of Surface Loading on the Flexure of Beams," and to another by Dr. Chrco. 381 CHAPTER XVI. BENDING. 316 IN Fig. 206, c D is a beam carrying a weight w. We know that the beam transmits the weight to the walls, and that in doing so the beam is kept in a strained condition ; we must consider what is the state of strain in the beam. To observe this it will be well to take a beam which is very visibly strained, a beam of indiarubber. A B is its appearance when lying on the table; draw upon it a number of parallel lines in chalk or pencil, a b, c d, ef t etc. Now if we support the beam at its two ends, and load it, we find that the lines A ff c g B a b, c d, etc., remain E |__4.__i__j__-} j.j-|~ | |~j~j--j 1 |F straight, but they are no longer parallel. We find the distance a c' to be less than a c, but b' d' is greater than b d. In fact, a' c' is compressed, b' d' is extended. We find also that along the line E' P' there is neither compression nor extension. E' p' remains of its old length, although it is no longer straight. If we consider each cross section of such a beam we see that the upper part of it is in compression, the lower part of it is in exten- sion, and there is a straight line in the middle where there is neither compression nor extension. This line is called the neutral axis of the cross section, and all these axes lie in a surface called the neutral surface of which E' p' is an edge view. Fig. 207 is a magnified drawing of the small portion of the beam between two cross sections, c e f d shows its original shape, c' e' f d' its shape when strained. Evidently there is more compression at c' & than at I' m. Fig. 206. 382 APPLIED MECHANICS. The compression becomes less and less as we come nearer H'J', then the extension begins, and becomes greater and greater as we get farther away from H' j' until we get to d'f, where it is greatest. If the material is likely to break in compression it will be most likely to break at c' e'. If it is likely to break in tension it will be most likely to break at d'f. 317. If we know the compression or extension at any place, we can calculate what it is at any other place, for the strain is evidently proportional to the distance from the middle. Thus if at e' there is a compressive strain of -002, that is, there is a compression of *002 foot for every foot in length, then at m', half-way between j' and e', there is only a strain of '001. There is the same strain at t' the same distance below j', but it is now an extension. The material resists being strained in this way, and the pushing and pulling forces which it exerts at the section e'f, Fig. 207, are just the forces required to balance all the other forces acting on the part e' D Tf. As e' D Tjf is a body kept at rest by forces, and which is no 1", v* X gf Fig. 208. longer altering in shape, it is to be regarded as a rigid body.* Now what is the condition under which it is kept at rest ? 318. The beams used by us are almost never deformed so much as the beam shown in Fig. 206, and indeed our theories are only true on the assumption of exceedingly small changes of shape. Let, then, the part e' D if be drawn less deformed as E D T F in Fig. 208, and consider its equilibrium. We had better consider more weights than one, loading it. The forces w x , W 2 , and W 3 represent loads, and P is the * In boobs on mechanics you may have read much about rigid bodies and the laws of their equilibrium, and you may have thought that such bodies had no existence ; but you must remember that we can regard a quantity of water, or a piece of steel spring, or a rope, as a rigid body for the time being, if it is being acted on by forces, and is no longer changing its shape. APPLIED MECHANICS. 383 supporting force at the end T of the beam. In Art. 99 we saw that if all the loads on a structure are given, the supporting forces may be calculated. 319. We are now considering only horizontal beams on which the loads are only vertical and the supporting forces vertical. Students had better work again here a few exercises, to find the supporting forces when the loads are given. Either neglect the weight of E D T F itself or imagine it represented by W 2 . Suppose, then, p to be known. The molecular forces acting on the surface E F (by the material to the left upon the material to the right of E F) balance all the external forces acting upon ED T F, namely, w l5 w 2 , w 3 , and P. Art. 98 gave the following three as the conditions of equilibrium. I. The upward tangential resultant of the molecular forces, which I shall call s, the shearing force at the section, is equal to Wj -j- W 2 + W 3 P. The student ought now to work a number of exercises. Given loads, find supporting forces ; find s at many sections ; show all answers in one diagram and call it the diagram of shearing force. Observe that we assert nothing as to how this shearing force is distributed over the section. We shall find in Art. 369 that it is most intensely dis- tributed about the middle parts near the neutral axis o o. II. As the loads are all vertical there is no horizontal component in the resultant of the mole- cular forces; in fact, all the pushing forces on E F balance all the pulling forces. Figs. 209, 210 show E F magnified, its actual shape and side elevation also. At H, a point in E F at the distance y from o o the neutral axis, the pushing force per square inch or the compressive stress being proportional to y, let us call it c y, when c is some constant ; if a point is at H' (Fig. 209), Kg 210 we shall call o H' a negative value of y, so that a pulling force is regarded as the negative of a pushing force. Now at H let there be an exceedingly small portion of area a, the force on this is c y a, and we must have the sum of all such terms as c y a for the whole area to be zero. This is only another way of saying that the pushing and pulling forces are equal. 384 APPLIED MECHANICS. All the terms c y a have the same multiplier c ; hence what we state is that if every little portion of area a be multi- plied by its 2/, the sum is zero. When this is so, Art. 109 tells us that o o, the neutral axis, must pass through the centre of gravity of the area. This is why we are always so anxious to find the centre of gravity of the section of a beam. The rules for finding the centre of gravity of the area are given in Art. 111. We are now about to find the value of c. Notice that if we know c we know the stress at any point in the section, and we particularly want to know c. o E or c. o F, the greatest stress. III. The moments of all the molecular forces about any axis balance the moments of W 1? W 2 , W 3 and p about the same axis. Now as these are all vertical forces, if we choose an axis at right angles to the plane of bending (the plane of the paper) in the plane E P, notice that about any such axis these moments will be the same ; hence we speak of the moments of W 1? W 2 , W 3 and p about any such axis as the bending moment M about the section E F. When it tends to make the beam convex upwards I call it positive. Hence M= -P-TF +w 1 'iE + w 2 'JE + w 3 -LE The student ought to practise the calculation of M for many sections of a beam and then show his answers in a diagram^ Fig. 211. We have an easy graphical method of drawing such a diagram (see Art. 349). We call it a diagram of bending moment. 320. Very well, then, the sum of the moments of all such molecular forces as c y a must be equal to M. Take the moments about o 0, the neutral axis ; the moment of c y a is c y a x y or c y 2 a. Now, when every little portion a of an area is multi- plied by the square of its distance from a line and the sum taken, it is called the moment of Inertia I. of the whole area about the line, and it is easy to find I. for any area about the axis o. Part of Chap. VIL is devoted to this subject of moments of inertia. We have, then, c I M, or c = M/I ; and hence the compressive stress p at points y inches from the neutral axis IF APPLIED MECHANICS. 385 p M y/I . . . (1). As y is greatest at points like E or P, we have the greatest compressive and tensile stresses at these points. In fact, the greatest compressive stress in the section is at E, and its amount is o E. M/I . . . (2) ; the greatest tensile stress is at P, and its amount is F. M/I . . . (3), and these two expressions give us the great laws of strength of beams. If we know the stresses which the material will stand, we know whether the section E F will withstand the bending moment M. 321. If the material is cast iron it is advisable to have o E= about 4J times O F, because cast iron will stand about 4 J times as much compressive as tensile stress. Hence the usual economical cast-iron sections are as shown in Fig. 211, with centres of gravity near the bottom boundaries of the sections. Whereas in wrought iron and mild steel and other materials the resistance to compressive stress is much the same as the resistance to tensile stress, and consequently o E is made equal to F, and the usual economical sections are as shown in Fig. 221. 322. In the model, Fig. 212, which shows a beam fixed at one end and loaded at the other, part of the material has been removed, ind instead of it we have inserted a chain or link A, which is only capable of pull, and a rod B, which is only capable of push. It is found that forces acting merely horizontally on M N o F are not sufficient to keep it at rest ; we also need an upward force at M F, which is equal to the weight w, together with the weight of M N o F itself. We see then that at such a section as M F of a beam we need pulling and pushing forces, but also to satisfy the first condition given above we need the shearing force at M p. In fact, an upward force w' must be exerted at M F equal to the weight w and also to the weight of M N o F. At M F the bending moment is W X o p, together with the weight of M N o F x the distance of its centre of gravity from M F. This is to be balanced by the pull in the chain A or the push in the rod B, for these are equal, multiplied Fig. 212. exerting exerting 386 APPLIED MECHANICS. by the distance between their lines of action. If a beam ia long, the shearing force exerted by the material at a section of the beam is usually not so important to consider as the pushing and pulling forces, and in many cases it is neglected. When a beam is very short the shearing force becomes more important to consider. 323. We shall now take a casein which there is only bending moment to be balanced by the material at a section. Let A B (Fig. 21 3) be a strip of wood or metal originally straight, whose weight we shall neglect. Fix or solder to the ends stout pieces of metal, and by means of cords and weights, or in any other Fig. 213. Fig. 214. way, exert couples on these ends as shown. Consider now the equilibrium of any portion say, c D B (Fig. 214). At the sec- tion c D we know that pulling and pushing forces must be exerted by the material which exists at the left of c D on the material which exists at the right of C D, and the moments of these just balance the moment of the forces F and F, and this is evidently the same at any section of the strip. The bending moment at any section is then the moment of the couple or torque M acting at either end. Magnifying the section E F, as in Fig. 210, and representing the amounts of the pulling and pushing stresses by arrows, we see as before that as the sum of all the forces one way must be equal to the sum of all the forces acting the other way, and as the stress at each place is proportional to distance from o, the part where there is no stress or the neutral axis is as before, a line through o at right angles to the paper, and this must pass through the centre of gravity of the section. We see also that the stresses at all points of the section are given by (1), and if we particularly desire to know the greatest stresses they are given us by (2) and (3). 324. Unsymmetrical Bending. If the bending moment M in a section is about an axis o x 1 through o the centre of gravity of the section, and if this is not a principal axis (see Art. 114), but makes APPLIED MECHANICS. 387 the angle x v ox = a with a principal axis ox (about which tha moment of inertia is ij), the other being OY (about which the moment of inertia is I 2 ), then M may be resolved into M cos. a about o x, and M sin. a about o Y. The stresses and strains due to these separately have now to be combined to obtain the actual stresses and strains due to M. Thus at a point whose co- ordinates referred to the principal axes are x and y, we have the total stress . __ M cos. a . y M sin, a . x /= II 12 If / is put equal to 0, we find the position of the neutral line. It evi- dently is like o Q, and makes the angle /8 = Q, o x with o x, such that tan. = - 1 tan. a. It is now easy to find the points which are at the greatest distance from o a, and these are the points at which there is greatest stress. Fig. 215, Exercise. In a beam of rectangular section, if the axis about which the bending moment takes place is at right angles to a diagonal, show that the greatest stress is at corners, and is of the amount 6 M ~- bd D if b, d, and D are the breadth, depth, and diagonal of the rectangle. 325. The line which passes through the centre of gravity of every cross section, being neither extended nor compressed, is of the original length of the strip. When the beam is bent as in the figure, A B becomes longer than this, and a b shorter, yet their ends are in the same planes A a and B b. Thus the strip may be considered as a bundle of fibres lying in arcs of circles which have the same centre and subtend the same angle at that centre. If we know their relative lengths we can tell where the centre of the circle is. Now we know the stress per square inch on a certain fibre, and we know its original length, hence we can calculate its present length (see Arts. 241 and 265), and its length is to the length of the neutral fibre as its radius is to that of the neutral fibre. In this Way we find that the radius of the neutral fibre is numerically equal to the modulus of elasticity of the material multiplied by the moment of inertia of the cross section, and divided by the bending moment at the 1 M section, or the curvature - ... (4). To put it in another way : The curvature in Fig. 207 being 388 APPLIED MECHANICS. angular change per unit length, is the angle between c' $ and e' f if H' j' is unity. But this angle is, strain at y -r- y, or p -r- fy . . . (5) and by (1) this becomes (4). The form (5) is sometimes useful in indicating the greatest curvature which may be given to a beam without hurt, by making p the greatest stress which the material will stand, y being the greatest distance of any point in the section from the neutral axis on the com- pression or tension side. In the same way it is easy to see that if a beam was already bent, having a curvature , the change r o of curvature -==--.... (6). r o ^ L Example. A straight strip of tempered steel, 07 inch broad, O'l inch thick (this represents the depth of a beam), is subjected to a bending moment of 100 pound inches : find its radius of curvature. Answer : The moment of inertia of the section is 0'7 x -1 x '1 X -1 -f-12, or -0000583. The modulus of elasticity of steel is, say 36,000,000, and 36,000,000 x 0000583 -f- 100 gives 21 inches for the radius of curvature of the bent strip. The curvature is ~T. If the strip was already bent before the load was applied and had a radius of curvature 50 inches, then the change of curvature - FQ 22 5. important web, and in all mechanical engineering calculations we keep to the correct rule. I therefore give here a list of the values of I and of z for various forms of section. Exercise. Show that the centre of gravity o of the area in Fig. 225 is 2 inches above the bottom. Take it as x inches. The middle of the bottom rectangle is x - \ from o and the middle of the top one is 3 \ - x from o. Hence (x f ) 5 = (3J x) 5 or #=2 inches. The moment of inertia of the top 5 x I 3 rectangle about axis o o is -^ +5 (l^) 2 - 21f. The moment 398 APPLIED MECHANICS. of inertia of the bottom rectangle about axis oo is ' + 5 (]J) 2 =11|; and the sum isi=33-J-. z l for the top is 1-^4=81 Z 2 for the bottom is I -f- 2 = 1 6f . TABLE VI. Moment of Inertia of Section. Strength Modulus of Section. '2 8 9 BD - QD W + APPLIED MECHANICS. 399 Moment of Inertia of Section. z = i/#. Strength Modulus of Section. (B n 2 - 6A 2 )* - 4 B H bh (H - h)'* 12 (BH - bh) 36 * [*<-- 2 - 6 (B H 2 + M 2 - 2 * H A) _ ' 12(2* / ,_ 12(6 z = - or Z'rr 400 APPLIED MECHANICS. APPLIED MECHANICS. 401 Exercise. Find the greatest load that may be uniformly distributed on a cast-iron girder having top and bottom flanges united by a web, of the following dimensions : Width of upper flange, 3 inches; of lower flange, 9 inches; total depth, 12 inches; thickness of each flange and of the web being 1 inch; distance between the points of supports, 10 feet; when the greatest admissible stress in the compression flange is 6 tons per square inch, and that in the tension flange 1^ tons per square inch. Am., 9;8 tons. 336. Proportion of Depth to Length in Railway Girders. It is usually assumed that maximum economy of weight in booms and diagonal pieces leads to a most economical ratio of depth d to length Z, but we must confess that we feel dissatisfied with the easy mathematical statements sometimes deduced on this subject from incomplete data. Take it that in girders of the same style the diagonal pieces make some known angle with the horizontal. Let us take 45, for example. Then each bar of length dJ 2 and cross-section a withstands a shearing force s = of x V 2, where/ is the shear stress, and has a weight d*/ 2 . a ('28), or the weight is d V 2 ('28) -~. The weight of the corresponding pieces of boom is Id A (-28), where t^fd = M, where / is also the tensile stress. Total weight of a bay is therefore or the weight in pounds per inch run of the girder is "We see, therefore, that if the inclination of the diagonal pieces is fixed, the greater d is the better ; and there is no most economical depth of a girder derivable at all events, from these simple considerations. It is true, however, that making the depth great, if the inclination is constant, means that we are increasing the length of the unsupported part of the compression boom and struts, and they may need more lateral bracing ; or, as it is better to put it, the cost of the compression members per pound will increase. Again, the weight of the platform will also increase. These are, however, questions of a different kind, difficult to settle by elementary mathematical equations when systems are so different. Writing down such general expressions as we may, there is evidence that there is greater economy in weight, whatever there may or may not be in cost, in letting the depth get less where the bending moment is less, instead of keeping it constant. The most important matter, how the natural period of vibration of a bridge ought to come in, seems never to be brought forward in these calculations. Professor Milne finds that the horizontal trans-, verse deflection is the most serious motion of a railway bridge. It begins when the train is perhaps 200 feet or more away ; it becomes accentuated with every passing carriage, and when the 402 APPLIED MECHANICS. whole train has passed there is the natural vibration of the bridge, which continues for some time. These vibrations are due to lurch- ing of carriages and impact of wheel-flanges against the rails Sometimes a light waggon seems to produce much larger effects than the heavy loccmotive, and there is some speed of train with which the vibration is much more serious than with quicker or slower speeds. Bridges have not yet been studied from this point of view, and engineers must for the present rely upon their large factors of safety. We are sure that more attention ought to be paid to these lateral vibrations, which seem to be greatly accen- tuated when gusts of wind are acting laterally. The Board of Trade rule is 5 tons per square inch on wrought iron and 6 on steel. This is not sufficiently safe. It is well to say 5 in tension on iron, 4 in compression, and steel, say, ^ greater. In some large bridges it is estimated that stresses due to wind are greater than those due to rolling loads. The force acting on the rails in the direction of their length is sometimes as much as ^-th of the weight of a train if the train is quickly stopped. As for the vertical motion, its consideration will probably lead to some rule connecting maximum deflection y^ under static load, and length of girder I. There is a vague sort of understanding that yi shall be something between ^tn,th and yJ^^th of I. If beams are of uniform strength and depth, d, the curvature is constant, being 2//E<# (see Art. 362), / being the greatest stress in every section; and hence in girders supported at the ends the deflection y 1 is equal to P//4<*B. If, now, we take ^ to be J/1,200, ~ = /J- TaHng E / /as 3 ' 000 ' 5 = 10 & It is not quite fair to say that the calculation of the probable loading of railway bridges is more scientific in America than in Britain. British engineers differ much more in their assumed loads than American, but in neither case can it be said that there is a scientific basis for the rules in use. It is very important that the student should know this, because there is sometimes a pre- tence of accuracy of treatment in bridge calculations which is quite misleading. The following rules are more common than others, and may well be used in academic problems. In ordinary railway bridges the greatest possible rolling load may be taken as if it were a static load of w 1 tons per foot-run for a double line, where w l = 3 + 176/1, if I is the span in feet. This is really on some such assumption as that a rolling load must be multiplied by If to convert it into the equivalent static load. The diminution with length is due to the fact that the engine weight is more intense per foot than other parts of the train. The weight of the platform may be taken as w^ = 0-7 + -0072 I tons per foot of the span for a double line. w 2 increases with the span because of greater wind-bracing and the greater width of larger spans necessary for lateral stability. APPLIED MECHANICS. 403 Half of every term may be taken for a single line, and used even as low as for 15-feet spans. A railway girder is usually built with so much negative deflection or " camber," as it is called, that it will become just about level when loaded. From (2) of Art. 336, taking M as proportional to wP, if w is the total load per foot run, and s as proportional to wl, the whole weight, M? 3 , of the girders per foot run is equal to ( a ~ -- bw \ I, where a and I are two constants. If we take it, as is usual, that l(d is nearly constant (say 10, as above mentioned), this becomes w 3 = lw{c, where c is some constant. Now, CW 3 W = W l + W . d$ = M, a constant. r J Differentiating with regard to 0, we find p sin. ; and as this is true for all values of 0, we must have p = everywhere. That is, the shape can only be maintained by couples applied at the two ends. It is quite a usual thing to see workmen beating such a ring out of shape near the joint, after it has been manufactured, because it is known that, so far from pressing uniformly all round, it does not even fit the cylinder, but concentrates its pressure at certain points. 404 APPLIED MECHANICS. B The following method of making a piston ring produces uni- form pressure all round, even if the section of the ring varies very greatly. A ring is cast which is larger than what is required ; a piece is cut from it, and the two free ends are brought together by a clamp. This clamp pulls on both ends, and the more n-early the resultant pull is tan- gential to the mean cylindric surface at the end the more nearly perfect will the ring be. The ring is now turned up to the size of the cylinder and finished. It is undamped, and may be sprung into place. To prove that the pressure must be uniform all round. In Fig. 226 let BPA be part of a piston ring constrained to be of the size of the cylinder, of radius r. It may either be kept in its present shape by the equal and opposite forces F at its ends, Fig. 226. or by pressures p Ibs. per inch of its length. Let us suppose at first that p may not be the same all round, being some function of the angle A o p = 0. Draw p Q, perpendicular to o A. r with- out p produced bending moments at all the sections of the ring ; the pressures p without F must produce the same bending moments. Let A o R = <, where P and R are any two places on the ring, and A is one end. Bending moment at p due to F is M = F . A Q, or M = Fr (1 - cos.0) .... (1). The pressure at R on the element r . 80 is prty, and the bending moment due to this at P ispr 2 . 80 . sin (Q ), and we require pr z sin.(0 0) . d$ = F r (1 cos.0) .... (2). r. In the integral p is a function of . Now we know that in general (* J and hence equation (2) is p sin.^) . r. = vr(\ cos.0) .... (3). Differentiating with regard to 0, we have r^sin.0 = Fr sin.0, or p = F/r, a constant .... (4). APPLIED MECHANICS. 405 It will be observed that the pressure p is uniform all the way round, even if the ring varies greatly in section. It is quite true that the proof assumes the thickness to be everywhere incon- siderable. If, however, we assume a uniform thickness, it is easy to see that if the resultant force F exercised by the clamp acts on each end exactly at the mean radius, there is absolutely a constant pressure per inch all the way round. In Japan, twenty years ago, Professor R. H. Smith told me that the above proposition was true. I worked out the proof very easily. I do not think that it has been published before. The usual method of making the rings is very much more mischievous than it may appear to be on a hurried examination. It seems to me that if the correct method is adopted, care being- taken as to the proper method of applying the clamp, piston rings may be made in this way for the very largest cylinders, and it is evident that there must be a very great reduction in the cost of large pistons in consequence. 338. Curvature. The curvature of a circle is the reciprocal of its radius ; and of any curve, it is the curvature of the circle which best agrees with the curve. The curvature of a curve is better given as " the angular change (in radians) of the direction of the curve per unit length." Now draw a very flat curve, with very little slope *. Observe that the change in i or jj- in going from a point P to a point Q is almost exactly a change of angle change in -J- is really a change in i, the tangent of an angle ; but when an angle is very small, the angle, its sine and its tangent, ore all equal . Hence, the increase in -=- from p to Q, divided by the length of the curve P Q, is the average curvature from p to Q ; and as p o, is less and less, we get more and more nearly the curvature at P. But the curve being very flat, the length of the arc P Q, is really 8x, and the change in ^ divided by 8x, as Sx gets less and less, is the rate of change of ~- with regard to #, and the symbol for this is ~y~. Hence we may take 33 as the curvatura of a curve at any place, when it is everywhere nearly horizontal. If the beam was not straight originally, and if ]/ was its small d ' 2 i/' deflection from straightness at any point, then ^4 was its original curvature. We may generalise the following work for beams not straight to begin with by using -j- 2 (y - y') instead of -^ everywhere. It is easy to show that a beam of uniform strength that is, a beam in which the maximum stress / (if compressive, positive ; if 406 APPLIED MECHANICS. tensile, negative) in every section is the same has the same curvature everywhere, if its depth is constant. If d is the depth, the condition for constant strength is that M M - . i d = + /, a constant. But - = E x curvature ; hence curva- Example. In. a beam of constant strength, if d jjz = j ( + to). Integrating, we find ^ . ~ = c + ax + and again, . y e + c-x f | ax 2 + . lx*, where e and c must be determined by some given conditions. Thus, if the beam is fixed at the end, where x = 0, and = there, and also y = there, then c = and e = 0. 339. In a beam originally straight, we know now that if x is distance measured from any place along the beam to a section, and if y is the deflection of the beam at the section and i is the moment of inertia of the section, then d 2 y _ M m dx2 " El ' ' ' ()t where M is the bending moment at the section, and E is Young's modulus for the material. We give to ~J the sign which will make it positive if M is positive. If M would make a beam convex upwards and y is measured downwards, then (1) is correct. Again, (1) would W _ P _ -Q 1-x --* Fig. 227. bo right if M would make a beam concave upwards and y is measured upwards. Example 1. Uniform beam of length I fixed at one end, loaded with weight w at the other. Let x be the distance of a section from the fixed end of the beam. Then M = w (I - x} ; so that (1) becomes A.PPLLED MECHANICS. 407 Integrating, we have, as E and i are constants, From this we can calculate the slope everywhere. To find c we must know the slope at some one place. Now, we know that there is no slope at the fixed end, and hence ~ = 0, where x = ; hence c = 0. Integrating again, To find c, we know that y = when x 0, and hence c = ; so that we have for the shape of the beam that is, the equation giving us /, the deflection for any point of the beam- s' = ^ (^* 2 - '!*).... (3). We usually want to know y when x I, and this value of $ is called D, the maximum deflection of the beam ; so that Example 2. A beam of length I loaded with w at the middle and supported at the ends. Observe that I if half of this beam w in its loaded condition | has a casting of cement made round it so that i i , I. it is rigidly held, the Wt- ---- .......... }- ........ - ..... 4W other half is simply 2 } a beam of length I, &S. 223. fixed at one end and loaded at the other with J- w, and, according to the last example, its maximum deflection is The student ought to make a sketch to illustrate this method of solving the problem. Example 3. Beam fixed at one end with load w per unit length spread over it. The load on the part PO,iswxpQort0 (I - x). The resultant of the load acts at midway between p and Q, so, multiplying by \ (I - x}, we find M at p, or M = \w (I - x) 2 ____ (6). Using this in (1), we have --- -~ = I 2 - 2 to + a? 2 . Integrating, we have - = Px - Ix* + i^ 3 + e. This gives us the slope 408 APPLIED MECHANICS. everywhere. Now ^ = where x = 0, because the beam is fixed there. Hence c = 0. Again integrating, Q . X- . M - l_ X -.._. Fig. 229. and as y = where x = 0, c = 0, and hence the shape of the beam is 4i* + *).... (7). y is greatest at the end where x = I, so that the maximum deflection is 24BI ' 8 EI ' if w = wl, the whole load on the beam. Example 4. Beam of length I loaded uniformly with w per unit length, supported at the ends. Each of the supporting forces is half the total load. The moment about p of \wl, at the distance PQ, is against the hands of a watch, and I call this direction negative ; the moment of the load w (^l - x] at the average distance p Q is therefore positive, and hence the bending moment at p is ~ *) - i (V - *) 2 } , or - [>^ _ |^2 J m . um (9)> so that, from (1), EI -j^ = - j $wP - %wx 2 } . Integrating, we CvUs \, J Fig. 230. cty have E i = - 4 w?x + | wsfl + c, a formula which enables us to find the slope everywhere, c is determined by our knowledge that = where x = 0, and hence e = 0. Integrating again, APPLIED MECHANICS. 409 Ely = ^wPx 2 + -fWX* + c, and c = 0, "because y = where X = 0. Hence the shape of the beam is y is greatest where x = \ I, and is what is usually called the maximum deflection D of the beam, or i> = ^-j if w = lw> the oo4 E I total load. Example 5. In any beam, whether supported at the ends or not, if w is constant, integrating (4) of Art. 357, we find = b + wx, and M = a + bx + ^wx? .... (5). In any problem we have data to determine a and b. Take the case of a uniform beam uniformly loaded, and merely supported at the ends. Measure y upwards from the middle, and x from the middle. Then M = where x = \l and - \l, = a + \ll + %wP, and = a - \U + wP. Hence b = 0, a = - %wP, and (5) becomes M = - lwP + %WX* .... (6), which is exactly what we used in Example 4, where we afterwards divided M by E i, and integrated twice to find y. With regard to the following important practical problem -.When the sizes of an angle iron are given to find the least radius of gyration of its section about a line through its centre of gravity, I give students a number of angle irons every year ; for each of them they find the least radius of gyration. From the tabulation of these results I hope to get an empirical formula. I had hoped to be able to publish this now, but T have not yet obtained a sufficient number of results. 410 CHAPTER XVIII. SOME WELL-KNOWN RULES ABOUT BEAMS. 340. WHEN beams have the same section everywhere we look for the place where the bending moment is greatest, as that is the place where fracture tends most to take place, and we find the cross section to withstand this greatest bending moment. We shall now consider such a uniform beam loaded and supported in various ways. Thus, the load may be hung from one* end of the beam, the other end being rigidly fixed, say by being built into a wall. When we say that the end of a beam is fixed, we mean that it is rigidly held in position, whereas when we say that a beam is supported at its ends, we mean that it is merely held up there. In Table VIII. six ways are shown in which the same length of beam is supposed to be loaded. The total load is supposed to be the same in every case, and the length from A to B is supposed to be the same. Then, we see that when the beam is fixed at both ends, and the load spread over it, it is twelve times as strong as when one end is fixed, and the whole load hung from the other end. This means that if, with the beam fixed at one end, a load of one ton, hung at the other end, breaks the beam, then, when fixed at both ends, and the load spread uniformly over it, the same sized beam will carry 12 tons. Hence, if experiments are made on the strength of the beam when loaded in any of these ways we know what its strength ought to be when loaded in any of the other ways. Now a great many experi- ments have been made upon beams of rectangular section, supported at both ends and loaded in the middle, the third case given in the Table ; and from these experiments we know how to find the load which such a beam will carry. Having found this, we know that when loaded and supported in a different way, the beam will carry more or less according to the numbers in the column headed "Strength." 341. If M is the bending moment and z the strength modulus of the section, and f the stress which the material will stand, M = z/. . . . (1). Let us take as an example beams of rectangular section, APPLIED MECHANICS. 411 breadth 6, depth d; the strength modulus is b d^/6, so that M fb d^/Q (2). Now our theory is based on the idea of perfect elasticity ; we cannot, therefore, assume that M is the bending moment which will break a beam if /is the ultimate tensile or compres- sive stress, because our theory cannot hold beyond the elastic limit, bat we find by experiment that the breaking, bending moment is proportional to b d 2 . Thus if rectangular beams of the same material, but of different lengths (I feet), breadths (b inches), and depth (d inches) are supported at the ends and loaded in the middle, we find that the breaking load w Ib. follows with fair accuracy the rule w = c b d^/l (3) where c is given in the following table and stands for 1/18 of our old/. It is well to try to remember that doubling the breadth of a beam doubles its strength, but doubling its depth gives four times the strength. TABLE VII. BEAMS SUPPORTED AT THE ENDS AND LOADED IN THE MIDDLE. Nature of Material. c to Calculate Strength. e to Calculate Deflection. Teak 820 00018 Oak 450 to 600 00044 to -00020 English Oak 557 0003 Ash. 675 00026 Beech 518 00031 Pitch Pino 544 00035 Eed Pine 450 00023 Fir. 370 0005 to -0002 Larch 284 00041 Deal GOO 00023 Elm 337 00061 Cast Iron 2,540 000024 Wrought Iron 3,470 000016 Hammered Steel 6,400 000013 Marble . 150 Good Sandstone 50 to 80 " The numbers given in this table are merely the average values found by various experimenters. You may wish, how- ever, to find for yourself whether they are correct or not. You are designing a beam of pitch-pine, say ; then take a rod of pitch-pine, 1 foot long, 1 inch broad, 1 inch deep ; support it at the ends, and load in the middle till it breaks : the Table says that the load will be 544 Ibs., but you may find it to be more 412 APPLIED MECHANICS. or less than this. Remember also that it is near the middle that your beam is likely to break ; this, then, ought to be the soundest and most evenly grained part of the timber if possible, and the specimens which you try ought to be as nearly as possible the same kind of timber. If your beam is loaded or supported in any of the other five ways described in Table VIIL, you will multiply the breaking load which you have found by the number called strength in Table VIIL The reason is obvious. 342. In certain standard cases we like to state algebraically the amount of bending moment at any section of a beam. We shall do this in the six standard cases, so well known to all carpen- ters, shown in pages 413 to 415. I. Beam of length I inches fixed at one end A, loaded only with w at the other end B. At a section x inches from B the bending moment is evidently M = w x. The shearing force is s = w. The diagram, Case L, shows M. Notice that M is greatest at A, and is then w L II. Beam fixed at A (Case II.,), load w Ib. per inch of its length or total load w spread uniformly. Note that the load on p B, if P is a section x inches from B, would be w x, and the resultant of this acts at the distance f x from the section, so that its moment is J x X w x or J w x*. It is greatest at A, being there J w I 2 , s the shearing force at p is w x. Note that w I =w, so that the bending moment anywhere is -J a; 3 and is w I at the end A. The shearing force at the end A is numerically greater than anywhere else, being w. III. Beam of length I, load w in middle supported at the ends. The supporting forces are each | w. At a section x inches from either end the . bending moment is J w x, being - j AAAAA/uMMAMWk yf I a t the middle, and the shearing force s is - J w from A to the middle, when it suddenly changes Fig. 231. and becomes -f J w from the middle to B. IV. Beam of length I supported at the ends, load w Ib. per inch of its length or total load w spread uniformly (see APPLIED MECHANICS 413 Fig. 231). The supporting forces at A and B are each J w or \ w 1. At P if O P = x, the bending moment is the moment of the supporting force about P minus the moment of the load on p B, or, since o B = 1 1 and P B = J I - cc, and this simplifies to M = - -J- w (Z 2 4 # 2 ). It is numerically greatest at the middle where x is o, being there w I 2 or - % w I. The shearing force at p is J w I - w ( I x), or w x. The student ought himself to draw all the diagrams of M (bending movement) and s (shearing force). The diagrams of M for cases Y. and VI. are shown in the figures. (Case VI. will be worked out in Art. 360). They are simply the diagrams of III. and IV. with the average value of M subtracted from every ordinate that is, the whole diagrams are lowered by this amount. The diagrams of s are exactly the same whether a beam is merely supported at the ends or is fixed, if the loading is symmetrical (see Art. 362) that is, the fixing does not alter the actual supporting forces at the ends. We have, in fact, the following rule for finding the bending moment diagram for a uniform beam symmetrically loaded, fixed at the ends. Find the diagram of bending moment as if the beam were merely supported at the ends : raise it by a distance equal to its average height. "We now have the diagram of the bending moment when the ends of the beam are fixed. The shearing force diagram is not altered by fixing the ends. If for any two kinds of loading we have the diagrams of M and s, then for the two kinds of loading applied at the same time we simply add algebraically the ordinates of the separate diagrams. 414 APPLIED MECHANICS. ,*i-l < nj eg J|-a oc APPLIED MECHANICS, 415 s > i , ^ ^ 4oH|C,0 15 , i SD' S ' -a ^ g rf^-ag-a g^ I ; 1 li W 416 APPLIED MECHANICS, 2IJ*3 Jllli i ill is !fS|*a o o > ,0 W ft O P fl*i ^11! r s l' B fl ^ O S rt W h + 1 * -9 -g g ^ ^ a + H fi " H * .-|ot>l APPLIED MECHANICS. 417 343. At the Imperial College of Engineering, in Japan, wo had a testing-machine with which I made a great many experiments with my students. It increased the load on a beam at a uniform rate, and registered the load and deflection of the beam at every instant that is, it drew a curve, each point of which showed the deflection and the load which pro- duced it. Mr. George Cawley, instructor in mechanical engineering at the college, lithographed a number of these curves, taken by himself ; and although the experiments were made on Japanese wood, so that the actual amounts of load and deflection are not of general interest, yet the shapes of the curves are so interesting as to be worthy of publication. With only one exception, two beams were broken and two curves taken for each kind of wood. The mean of these, two curves has been given in Fig. 232 that is, a curve lying between the two. The specimens were all free from knots. They were all 28 inches long and If inch square. The distance ow repre- sents one ton, and the distance o D represents a deflection of 2 inches, so that the scale of the diagram is known. The load was in each case added to at a uniform rate, beginning with o, and the rate at which it increased was one ton in two minutes, and we see from the figure that practically only in three cases did the breaking of the beam take more than two minutes. The end of each curve shows where the specimen broke ; it is easy to see where the curve ceases to be a straight line that is, where the law, " Deflection is proportional to load," ceases to be true ; and this point is therefore the elastic limit. In some cases the load corresponding to the elastic limit is less than half the breaking load, and in some "cases greater than this, but usually it may be seen that it is about one-half. 344. What about beams that are not rectangular in section 1 Suppose we have a beam of the same section everywhere, whose strength and stiffness we know, and suppose we want to know the strength and stiffness of another beam 'which has the same form of section that is, suppose the new section is such that all the old lateral dimensions are increased in a certain ratio then the strength and stiffness increase in this ratio ; if all the old vertical dimensions are increased in a certain ratio, then the strength increases as the square of this ratio, and the stiffness increases as the cube of this ratio. The effect of change of length is just the same as it was with rectangular beams, and 418 APPLIED MECHANICS. Note. Students will do well to calculate for each of these materials (1) Young's modulus, (2) c and e of Table VII., and insert in an extra page in their copy of this book. APPLIED MECHANICS. 419 we know the effect produced by different methods of supporting and loading the beam from Table VIII. From Arts. 341 and 342 it is evident that the load which a beam will carry without breaking is proportional to the strength modulus of its section divided by the length of the beam. The deflection of the beam is proportional to the load multiplied by the cube of the length, divided by the moment of inertia of the cross section,, 345. Beams of uniform strength are those in which the nature of the loading is exactly known, and every section is made just of such a shape and size as to be equally ready to break with all the other sections. There is no difficulty in making z, the strength modulus, exactly proportional to M. Thus taking up the four well-known cases already described ; let us design beams of rectangular section everywhere of breadth b or depth d, or of circular section of diameter D, which shall be of uniform strength. Note that for a rectangular section z a bcl*, and for a circular section z a D 3 . In the case of the circular section, the plan and elevation of the beam are of the same shape. Case 1. M = war, so that bd 2 oc #. Keep b constant, the elevation showing d is a parabola. Keep d constant, the plan showing b is a triangle. D 3 or #, so that plan and elevation are what is sometimes called the cuhic parabola ; anyhow, it is easy to draw. Case 2. M a x 2 , so that bd 2 oc x 2 . Keep b constant, the elevation showing d is a triangle. Keep d constant, the plan showing b is a parabola. D 3 oc x 2 ; the plan and elevation are easily drawn. Case 3. From the middle to each end, this beam is the same as the beam of Case 1. Case 4. M oc (P - 4 z 2 ), so that bd 2 (P - 4 a; 2 ). Keep b constant, the elevation showing d is an ellipse. Keep d con- stant, the plan showing b is two parabolas. D 3 a (J 2 - 4 a; 2 ), so that the plan and elevation are easily drawn. We cannot treat Cases 5 and 6 in the same way, because we only know M in these cases on the assumption that the beams are of the same section everywhere (see Art. 362). EXERCISES. Rolled girder section, or two equal flanges and web, like A, Fig. 223. In Table VI. we see that the moment of inertia is i = j bd* - (b - 40 d{ \ + 12, and the strength modulus z is i divided by d, if d is the depth over all, d\ the depth between the flanges, b breadth of either flange, &i the thickness of web. 420 APPLIED MECHANICS. 1. I^et the student calculate i and z in all the cases of the following Table, page 421. 2. Taking / = 20 tons per square inch for iron, and 30 tons for steel, find /z in each case. This is the greatest bending moment in inch-tons which each section will stand. 3. Show that a beam I feet long, supported at the ends and loaded in the middle, will break when the load in tons is /z -^ 3 1, Calculate this for I = 10 feet in each case of the table. 4. Beams 10 feet long. For one ton at the middle the deflection in 2 240 x 120 3 inches is D = -jL . ^ ^ , or 2-688 4- i. Find this in each case. I take E the same for iron and steel namely, 30 x 10 6 Ibs. per square inch. 5. An iron beam of the rolled girder section of the table, 18 inches deep, 25 feet long, fixed at the ends. What is the breaking load if spread uniformly ? Ans., The load for a 10-foot beam in the table is 89 tons ; for a 25-foot 89 beam, supported at the ends, and loaded in the middle, it is ^ x 10, or 2o 35 -6 tons. Table VIII. shows that fixing at the ends and loading all over allows us three times as much breaking load, or 106-8 tons. 6. What is the mid-deflection of the beam of Exercise 5, fixed at the ends, when the load spread all over is 20 tons ? Ans., For a 10-foot beam it would be -00223 x -125, according to the " deflection " column of Table VIII.; and for a 25-foot beam we multiply by 25 3 -T- 10 3 , and the answer is -0044 inch. 7. Compare the strength to resist bending of a wrought-iron I section when it is placed like this : I, and like this : M. The flanges of the beam are each 6 inches wide and 1 inch thick, and the web is f inch thick, and measures 8 inches between the flanges. Ans., 4-57 ; 1 . 8. What is the greatest stress in a bar which is subject to a bending moment of 4,000 inch-pounds (1) if the section is a circle of f inch radius; (2) if of I form, 2 inches deep and 1 inch wide, the web and flanges each being f inch thick. Ans., 5'4 tons; 3'1 tons. 9. The dimensions of the section of a cast-iron girder are the follow- ing: top flange, 4 by 1| inches; bottom flange, 12 by If inches; web, 16 by 1^ inches. Determine the position of the neutral axis, and calculate the moment of inertia of the section. Find, also, the moment of resistance, the greatest permissible tensile and compressive stresses being 1\ and 1\ tons per square inch respectively. If the girder be 20 feet long, and is supported at its two ends, find also the greatest safe load which it will carry when uniformly distributed along its length. Ans., 7| inch from bottom ; 2,280*5 inch-units ; 800 ton-inches ; 26| tons. 10. Find the moment of resistance to bending of a beam of wrought iron which has a section like that of the third figure in Table VI., taking b = 4 inches ; depth, 5 inches ; the thickness of metal everywhere, \ inch; and /= 20 tons per square inch. What is the greatest load, placed at the centre, which a 10-foot beam of this section will stand when supported at both ends ? Ans,, 60-45 inch-tons; 2 tons. 11. In a wrought-iron girder, supported at both ends, the section is that shown in the third figure of Table VI. B = 8 inches, b = 7 inches, d = 10 inches, D = 12 inches. The length is 24 feet. Find the greatest APPLIED MECHANICS. 421 uox [ aoj uicea mi-Ol jo S9HOUI ui uorpayoQ Bending Moment o Section i Inch-Ton -5 CO The sup- required i . tween A and B. porting forces by this are p = 2 -5 and Q = 2-5 tons. The bend- ing moment diagram is a parabolic curve A E B, where Fig. 244. 434 APPLIED MECHANICS. D E = - 9-375 ton-feet. The shearing force diagram is B o D H A B (Fig. 244), where BG = 2'5 tons, AH = - 2*5 tons. Let the student work v out all these diagrams and add all the ordinates together. Now let him obtain the same result by one graphical construction. The continuous loading he will break up into a number of detached loads, and the following example will show him how to work the problem. Let the beam K L N r- be loaded and supported as shown in Fig. 246. Use Bow's lettering. 355. In Fig. 246a the loads are given, half -barbs being used for / A B c r^P K Fig. 245. Fig. 246. the arrow-heads ; the missing corner of the force polygon ABCDEFGHI is i. Choose a pole and join with ABC, etc. We can- not yet draw 1. Now Fig 246a. start at K, or any point in the line oi load AB (Fig. 246), and draw a line through the space A parallel to A (Fig. 246a) to meet (produced) the force A i in Q. Also draw K x through the space B parallel to B 'to meet the load-line B c in x. Draw x w through c parallel to c, and so on to R. Now join B, and d. This is parallel to the missing line i (Fig. 24 6a), which may now be drawn and i found. HI and IA (Fig. 246a) are the amounts of the supporting forces. The diagram QAXWVUTSRQ is the diagram of bending moment, APPLIED MECHANICS. 435 vertical distances such, as VY representing the "bending moment at each corresponding section. The scale of measurement may be computed by Art. 349. We have positive bending moment Fig. 247. (tending to make the beam convex upwards) between K and z and between M' and p, and we have negative bending moment between z' and M'. z' and M' are places of no bending moment, where the beam has no curvature; they are called points of inflexion. The shear force diagram 1, 2, 3, 4, 5 presents no difliculty. It is usual, when we finish the work, to draw the diagram of bending moment as in Fig. 247, the ordinates being measured, not from a broken line as in Fig. 246, but from a horizontal line. Travelling Loads. 356. I. Suppose the load w (Fig. 248), to travel over the beam from A to B. When w is at any point c, between A and D, the s^\ Fig. 248. shearing force at D (say S D ) is positive, and equal to the reaction Q. This positive shear at D increases with Q as the load approaches D, and in the limit, when w is very near to D, it has the value x. At the instant the load passes i>, the shearing force at D diminishes by the amount w, and becomes x - w, or - ^ (L - x) L L thus becoming negative, and equal to the reaction at p ; and as the load moves on towards B, the negative shear at D diminishes 436 AJPPLIED MECHANICS. numerically. We thus see that the greatest positive shearing force at any place occurs when the load is just to the left of the place, and the greatest negative shear when the load is just to the right. Expressed algebraically : Max. + S D = x ; max. - S D = (L - x). L L These are the equations to straight lines, and the corresponding diagrams are set out in Fig. 248 ; that for maximum positive shear is the triangle ABE, and that for maximum negative shear the triangle B A F. Next consider the maximum bending moment at D (say max. M D ). Pig. 249, As the load advances from A towards D, M D increases, since its value is Q (L - x), and a increases. When the load passes D, M H diminishes, since its value is now r x, and P diminishes. Therefore the maximum value of M D occurs when the load is at D, and its value is given by the equation = P x or Q, (L - x) = L - x This is the equation to a parabola passing through A and B, axis vertical, the ordinate of the vertex (see Fig. 248) being w L/4. II. Two travelling loads w l and w 2 at a fixed distance I apart ; draw the diagrams of maximum positive and negative shearing force and maximum bending moment. As the front of the first load ^ approaches D, + s D increases, APPLIED MECHANICS. 437 since its value = Q; when vf l passes D, S D undergoes a sudden diminution by the amount Wi ; and as Wj moves from D towards B the value of + s D increases, on account both of the advance of W! towards B and the approach towards D of the other load w 2 . This increase goes on until w 2 crosses D, when there is a second sudden drop in the value of + 8 D , followed by a gradual increase until w 2 arrives at B. We see that the maximum positive or negative s7> occurs when the front or back respectively of Wi or w 2 is at n. To draw the diagram : Draw the diagrams of s for the front and back of each of the loads Wj and w 2 in its passage across the beam ; the boundary line is the diagram required. The diagram for front of Wi will be a straight line through A, the equation of which is + s = Q = Wj -, until w 2 comes on, when the slope L suddenly alters, the s at front of Wj then being + s = a = Wl ? + w 2 ^=-^ = ( Wl + w 2 ) ^ - w 2 ^. The slope now is - ' 2 , and the diagram of s in front of Wj continues a straight line in this direction. Next, for s at back of Wi. This is equal to s at front - Wi for all positions, and the diagram is a broken line parallel to the one just drawn at a distance below it, equal to Wi. Next, for s at front of w 2 . The s between Wi and w 2 is the same at all points ; that is, when w 2 is just coming on, the s in front of it is the same as the s at back of Wj. Through K draw the horizontal line k m, then in is the starting-point for the 8 diagram in front of w 2 . The initial slope is 1 Wg ; this extends 3J to a point distant horizontally I from B, when Wi passes off the beam, and the slope diminishes to 3. The diagram for the rear of W 2 is parallel to the line just drawn and at a distance below it, equal to w 2 . As a test of accuracy, it is to be noted that this broken line should end at B. Consider now the maximum bending moment. Let w l + w at t be the resultant of Wj at a and w 2 at b. Let ta = a and ib = 0. Then ^ = 2 , that position being shown in Fig. 249 P w i in which the loads Wi and w 2 are one on each side of D. Let the distance of line of load w x from D be x." The maximum M D evi- dently occurs either when Wi or w 2 is at D, or for some intermediate position. Consider an intermediate position as shown. M D = Q (L - X) - Wl * = (W X + W 2 ) * + *-* (L - X) - Wl * = ( Wl + W 2 j ^^-(^ - x) + x { (w x + w 2 ) ^^ - Wj I . Itwillbeseen that so long as the expression in the last bracket is positive, M D is a maximum when x is greatest that is, when x I, or, that if, when 438 APPLIED MECHANICS. loadw a isatD. And | (Wj + w 2 ) - - w l j is positive so long as When W 2 . L - x L - X is > w i + W 2 > the expression in the last "bracket is negative, and the value of M D is therefore a maxi mum when x is zero that is, when load AVj is at D. A T "W To draw the diagram, take point T in A B such that = * B T >A | Let A T be called the field of w 2 , and B T the field of Wi; then M D ia a maximum for any point D in the field A T when the load w , which governs that field, is directly over point D, and for any point of the field B T when w x is at the point, provided in each case the load can completely traverse its field without the other load leaving the beam, which condition requires I to be not greater than the smaller of the two values L_- L, or _-. L. Curve for field B T. Putting x o in the expression for M D , we have ma^iTmnn M O = - - (x - o) (L - x), which is the equa- tion to a parabola passing through B and a point N distant a from A; and by symmetry the curve for field AT is a parabola passing through A and a point K distant /3 from B ; the two parabolas will Pig. 250. be found to intersect at i directly under T. The distance x of the mid point of B N from A is a + ^-- = . Putting this value in the equation to the curve, we have the ordinate H Y = - - r*+i - .1 |~L - L -^l = =^(* - -)'; the correspond- L J I -yy I -^y ing value for G x, or depth of other parabola A K, is a x = (L - ) 2 . When the distance I is greater than the shorter of the APPLIED MECHANICS. 439 two fields, there is then a third parabola through A and B corre- sponding to the greater of the two loads taken alone by method shown in Case I. III. A load as of a travelling train, w Ib. per unit length, comes upon a girder A B from the left, covering it from end to end, and then leaving it. Show that the greatest positive shearing force at a section D occurs when the front of the train reaches D, and that the greatest negative shearing force occurs when the rear of the train leaves D. Also find the maximum bending moment at D. We have seen that any load produces positive s if it is to the left of D, and negative s if it is to the right of D. Hence, to produce the greatest positive s at D, there ought to be no load to the right of D, and for greatest negative s at D there ought to be no load to the left of D, so the proposition is proved. When the train covers A D, s at D = Q, w x 2 or .... (1), being shown by DO in Fig. 250. When the w train covers DB, s at D is - p, or - 5 (L - (2), being shown at D H in Fig. 250. B E and F A are numerically equal to half the load when it covers the span. The curves A E and B F are parabolic, the equations of which are given in (1) and (2) respect- ively. The student ought now to add to the ordinates of Fig. 250. the shearing force due to a uniformly distributed constant load, Pig. 251. and write out in words actually what s is as a train rolls on, covers, and rolls off the bridge. Next consider the maximum M at D. Any load anywhere on a girder increases the bending moment anywhere. Hence where load due to a travelling^train comes upon a girder, covers the girder, and leaves it, the bending moment at any place is never greate* 440 APPLIED MECHANICS. than what it is when the whole girder is covered. Therefore, the maximum bending moment will occur when the girder is fully loaded. A curve to this is known to be a parabola through A and B, the depth of vertex being - . o IV. Travelling load of uniform intensity w Ib. per unit length, of length I less than the span L. The maximum positive s occurs swhen the front of the load is at D, and maximum negative when the rear of the load is at D ; so that while the load is only partially on the beam the diagram of maximum + s will be a parabola similar to that of Case III., the equation being for a distance I from A maximum + s = -= . When the load comes wholly on the beam, as 2, L in Fig.251, the diagram alters ; maximum + s = Q = ( x - ) or The equation to a straight line slope , intersecting L A B at a distance ^ from A, and therefore tangential to the parabola. The diagram of positive s is most readily set out by first drawing tht line E c, as shown in figure, then drawing the parabola to touch it at E, and the line A B at A. A similar curve set out downwards from B will be the diagram of maximum negative s. Maximum bending moment at D. It is easily seen that the maximum M D occurs either when the front a or the rear b of the load is at D, or for some intermediate position, as in Fig. 251. To find value of M D for an intermediate position let the condition of the co-ordinates be as in Fig. 251. Bn . a(l _ x) _f *r/_. v ,1 *+ dis. For a maximum, -3 = (x being constant), .. maximum M D occurs when (L - x) - wx = 0, or = . ; i.e. when = , L x I - x AD D or the maximum M D occurs for such a position of the load that T> divides it and the beam into segments, the ratio of which are respectively equal. Putting this value of x, namely x = I. , into the expres- L sion for M D , we have maximum M D equation to a parabola passing through A and B, the depth of the vertex being 441 CHAPTER XX. MORE DIFFICULT CASES OF BENDING OF BEAMS 357. WE do not usually trouble ourselves as to whether we call the tending moment which makes a beam convex upwards positive or negative, representing it by upward-drawn or by down ward- drawn ordinates. There is the same sort of choice in regard to shearing force. But for the sake of having plus signs in the following expressions (1) and (2), we had better adhere to the following definitions. A section which is being sheared is supposed to be at the positive distance x to the right of the zero point. Loads are positive forces ; supporting forces are negative. Positive shearing force s means that the material to the left of a section acts with downward force on the material to the right of the section. Positive bending moment M causes a beam to be convex upwards ; positive y at a place is a downward dist)lacement. When this is the case we know that m di _. d?y _ M m ' (1) ' dx~d^-Ti""^' In the same way we can show that To prove (3) and (4), consider the equilibrium of the portion of beam between the sections A B and CD. At A B there is the bending moment M and shearing force s, and at c D there are M + 8 M and s -f 8 P, and 0' is 8 x. The forces acting on this portion of beam are shown in Fig. 252. The load being w per unit length, the resultant load here +_ is w . 8x. Hence, considering the vertical forces, 8s = w . Sa; or d s/dx = w . . . . (4). Taking moments about 0', M -j- s . Sa? + J w (8a?) 2 = M + __ 8 M or 8 M/8a? = s + % w 8a? ; and in the limit, as 5x gets smaller and d M smaller, = s . . . . (3). 358. We saw (Art. 349) that if we have a diagram of w we can find easily, graphically, the diagram of M. We now see that if the value of M at Fig. 252. every section be divided by the value of E i there, and if we treat this diagram, showing M/E i every- where, exactly as we treated the w diagram, we obtain y. 442 APPLIED MECHANICS. This graphical method of working is quick and accurate. Ef we only possessed an accurate mechanical integrator, such that when given a curve showing x and v, v being a function of #, we could at once draw another curve whose ordinate for a particular value of a?i represented the area of the v curve up to that place from some datum value of x, we could easily solve more difficult problems. I have often done this by counting squares on squared paper ; also I have worked to obtain a number of points in the new curve by using a planimeter a number of times. We see now that if we know w, the integral of w shows s, the integral of 8 shows M, the integral of shows i, and the integral of i shows /, the shape of the beam. When we integrate, however, we must settle the start- ing value of the new ordinate, and this is what usually gives trouble. Thus the starting value of s (when we integrate w) is not zero, but depends upon the supporting forces. The student must see very clearly that the change in i, in going from one value of x to another, is equal to the area of the M/-E i curve between those places. 359. I have found the arithmetical method of Art. 214 very satis- factory. Its accuracy depends, of course, on the number of ordinates taken. A student ought to test for himself the accuracy of the method on some such exercise as the following, of which he knows the answer. A beam of rectangular section 1 J inch broad, 2 inches deep, and of length 15 inches, is fixed at one end and is loaded uniformly with 10 Ib. per inch. Its E is 25 x 10, find M everywhere and y. Here i = 2-25 x 4 3 -7- 12, or i = 12. x is distance in inches measured from the free end. Imagine the free end to be on our left and the rest of the beam on our right. The table gives the whole work, and needs almost no explanation. The student who thinks will have no difficulty in working out a problem of this kind. To find such a number as M f or x = 9, for example, we add the average value of 130 and 140 to the previous M, and so get 855. We know that i is when x = 15, and so we subtract 3,752 x 10 ~ 8 from all the found values of i + c. In the same way to get the last column we add 39,801 to every number of the previous column. Let us compare our results with the true answers, which the student can easily work out as in Art. 339 : s =50 + 10 x, M = 50 119,531-11,250*+ *+*-*{ o 6 J Thus when x = 15, M = 1,875, as in the table. When x 0, i or dyjdj- -= - 3,750 x 10 ~ 8 , whereas the table gives -3,752 x 10~ 8 . APPLIED MECHANICS. 443 Again, when x 0, y 39,844 x 10 > whereas the table gives 39,801 x 10 8 , which is the same for all practical purposes. After having worked this example a student must feel confidence in using this method of integration which gives us answers so readily. X w 8 M M El xio' i + c xio' i xio" i y + c x 10* V x 10 s 50 -3,752 39,801 10 1 60 55 18 9 -3,743 -3,747 36,054 10 2 70 120 40 38 -3,714 -7,476 32,325 10 3 80 195 65 91 -3,661 -11,163 28,638 10 4 90 280 93 170 -3,582 -14,785 25,016 10 5 100 375 125 279 -3,473 -18,312 21,489 10 6 110 480 160 421 -3,331 -21,714 18,087 10 7 120 595 198 600 -3,152 -24,956 14,845 10 8 130 720 240 819 -2,933 -27,998 11,803 10 9 140 855 285 1,082 -2,670 -30,800 9,001 10 10 150 1,000 333 1,391 -2,361 -33,315 6,486 10 11 160 1,155 385 1,750 - 2,002 -35,497 3,304 10 12 170 1,320 440 2,162 -1,590 -37,293 2,508 10 13 180 1,495 498 2,631 -1,121 -38,648 1,153 10 14 190 1,680 560 3,160 -592 -39,505 296 10 15 200 1,875 625 3,753 -39,801 360. Beams Fixed at the Ends. If the loading on a symmetrical beam is symmetrical so that we find, graphically or any other way, the diagram of bending moment m as if it were supported at the ends, we know that equal and opposite torques are needed to fix the ends. Thus, if A B is, say, a beam of uniform section and has a loading of any kind whatsoever which will produce (the beam being only supported at the ends) a diagram of bending momojit m 444 APPLIED MECHANICS. such as is shown in A c D E B (Fig. 253), and if equal and opposite torques are applied to fix the ends such as alone would produce the diagram of bending moment shown to scale in B G F A, then the algebraic sum of the two (for the A c D E B is negative and ABGFA positive) is shown in A F c' D' E' G B A, being positive from A to c' and from E' to B, so that in these parts the beam is convex upwards ; and being nega- tive from c' to E', where the beam is concave upwards. Fig. 258. We want, then, to know exactly how much G B or AF must be to fix the ends of the beam. Now, the difference of slope, of beam between A and B is nothing if the ends are both fixed, and therefore the total area of the bending moment diagram must be zero (since EI is constant, we say M instead of M/EI). Hence, the area of the portion c' D' E' being negative, must be numerically equal to the sum of the areas A c' F + B E' a. In fact, the average ordinate of the F curve must be zero, and therefore we raise the m curve A c D E B A by its average height to get the M curve. Example 1. Beam of length I, with load w in the middle, fixed at the ends. The diagram of m is A D B A (Fig. 254), where D D presents 4 w? ; raise it therefore by the amount ^ AF c' D' E' G B A. Fig. 254. re- and we find Evidently A F = - D" D' = GB = | wJ. The beam is convex at the ends, concave in the middle, equally ready to break at ends and middle, and the points of inflexion are half- way between ends and middle (see Art. 342) Example 2. Beam A B of F G- length I, with total load w spread uniformly. A D B shows the m curve, a parabola, the diagram of bending mo- ment if the beam were merely supported at the ends, D D" being wj. The average or- dinate of A D B A is |D D". Eaising the diagram by this amount, we find the true diagram AFC' D'E' G B A of M for the beam fixed at the ends. There is & -wl at each end and -_i_ w ; a t the middle (see Art. 342). The points of inflexion are at c' and E', no longer exactly half-way between ends and middle. 361. Students will do well to work at least one complicated APPLIED MECHANICS. 445 example of a uniform beam fixed at the ends with symmetrical loading. If the beam varies in section, but is symmetrical that is, if at two points equally distant from the two ends the sections are the same, and if the loading is symmetrical, first obtain the in curve graphi- cally and measure m at a number of equi-distant points. Thus, for some beam of 20 feet long, let us suppose that the values of m as measured are given in the following table. We need not use E, as we will suppose it to be constant. Distance m Values of from end in feet. in ton-feet. in inches to the 4th power. T M. 1 16 500 03 002 17-85 3 25 300 0833 00333 7-85 5 35 250 140 004 - 2-15 7 40 320 125 003125 - 7-15 9 44 360 122 002778 -11-15 11 44 360 122 002778 -11-15 13 40 320 125 003125 - 7-15 15 35 250 140 004 - 2-15 17 25 300 0833 00333 + 7-85 19 15 500 03 002 + 17-85 Total 1-0006 03046 Average 0-10006 003046 M We see that the area of the curve is to be zero, and if m l is the unknown bending moment applied at each end to fix it, M = m + m^ ; so that the average value of - 1 must be zero, or the average value of must be equal to mi multiplied by the average value of . Hence m^ is equal to the average value of -^ divided by the average value of -. In the above case this is 0-10006 -7- 003046, so that wj = 32-85 ton-feet; and M = 32 -85 + m (algebraical sum) is the true bending moment everywhere. 362. The shearing force diagram in the symmetrical cases is the same whether a beam is fixed or onVv supported at the ends ; and a8 ^ is not altered bv the nxin g the shearing force and the deflec- tion due to shear are everywhere the same in the beam. (See Art. 369.) 446 APPLIED MECHANICS. The solution just given is applicable to a beam of which the I of every cross-section is settled beforehand in any arbitrary manner, so long as i and the loading are symmetrical on the two sides of the middle. Let us give to i such a value that the beam shall be of M uniform strength everywhere; that is, that - z =/.... (2), where z is the greatest distance of any point in the section from the neutral axis on the compression or tension side, and / is the constant maximum stress in compression or tension to which the material is subjected in every section. Taking z = d, where d is the depth of the beam, (2) becomes - d = + 2/ ____ (3), the + sign being iaken over parts of the beam where M is positive, the sign when M is negative. As the area of the curve from end to end of the beam is i to be zero, and - =. 2f/d, we see that the area of a curve show- ing everywhere the value of +, l[d ought to be zero, the positive sign being taken from the ends of the beam to the points of inflexion, and the negative sign being taken between the two points of inflexion. We see, then, that to satisfy (4) we have only to solve the following problem. In the figure, EATUCGEis a diagram whose ordi- nates represent the values of - or the reciprocal of the depth of the beam which may be arbitrarily fixed, care being taken, however, that d is the same A c at points which are at the same distance from the centre. E F G E is a diagram of the values of what the bending moment m would be if the beam were merely supported at its ends. We are required to find a point p such that the area of E p T A = area of POO' T, where o is in the middle of the beam. When found, this point p is a point of inflexion, and p R is what we have called m^ That is, m - p R is the real negative bending moment M at every place, or the diagram E F a must be raised vertically till R is at p to obtain the diagram of M. Knowing M and d, it is easy to find i through (3). It is evident that if such a beam of uniform strength is also of uniform depth, the points of inflexion are half-way between the middle and the fixed ends. 363. In the most general way of loading, the bending moments required at the ends to fix them are different from one another. APPLIED MECHANICS. 447 Thus in Fig. 257 let A F c G B A be what the bending moment m would be if the beam were merely supported at its ends ; let fixing moments mi = A H and w 2 = E B be applied at the ends, producing of themselves (see Art. 354) a bending moment diagram shown by A H E B A, or, if A D is x, then K D, the bending moment produced by the end couples is m x + m * ~ m * x, if Us the length A B. Let us call this * + te. Hence M or R D - D s is M = m, + bx - m . . . . (1), Fig. 257. Now, as i is known, let AJ 1 J 2 J 3 J 4 BA be drawn to represent the value of everywhere. Let it be integrated ; that is, let such an ordinate as D K 1 (call it Y) represent the area of A J 1 J 2 D, and so obtain A K! K 2 K 3 B A, and let Y! be the value of this when x = A B. Also integrate - , and say that the ordinate of the resulting curve Fig. 258. is x, and let x x be the value of this when x = A B. Integrate also the curve whose ordinate is everywhere, calling 448 APPLIED MECHANICS. the answer /t, and /ij its value for x A B ; then (2) becomes, since -~ is the same at both ends, dx = wj YI + * Xl - to ---- (3). Again integrating Y, x, and p (but it is only necessary to get the areas over the whole length at once), calling the answers Y I} X 1? and M I} we see that, since at the two ends y is 0, = W^Y! + iXj - M! . . . (4). The two unknowns, m l and b, can now be found from (3) and (4).* We give in Art. 365 an example completely worked out. The column headed m represents the bending moment in ton-feet due to a given set of loads, if the beam were merely supported at its ends. These values may be found, of course, by the graphical method of Art. 349. The values of i are supposed to be given us, and they are in inches to the fourth power. 384. We shall now consider a beam fixed at one end, B, and merely supported at the other, A, which is on the same level as B. If, as before, m is the bending moment at any place, D, which would exist if the beam were supported at each end, and if m^ is the fixing couple, the true bending moment is Take, first, a simple case, a uniform beam uniformly loaded with w Ib. per inch. It is easy to prove, as in Art. 339, that m = - \wlx + %wx*, and Hence j*, '.... (3), * We have used the symbols M, x, Y, j*i, x l5 Y b M, x, Y, MI, x b Y b fearing that students are still a little unfamiliar with the symbols of the calculus. Perhaps it would have been better to put the investigation in its proper form, and asked the student to make himself familiar with the usual symbol, instead of dragging in fresh symbols. After (3) above, write as follows : Again integrating between limits, The integrations indicated in (4) and (5) being performed, the unknowns m\ and 6 can be calculated and used in (1). The student must settle for himself which is the better course to take to use the formidable-looking but really easily understood symbols of this note, or to introduce the letters whose meaning one is always forgetting. APPLIED MECHANICS. 449 and ^c We need not add a constant here, because y is when a; is 0. In 3) and (4) we insert the conditions that , = o when x * x \ W> Pig. 259. y o when x and we find J ) ,_ x { ( 5 /> I J From these we find that c = T ' 5 i0P, w 2 = $ wJ 2 , so that the true bending moment everywhere is given by (2), and the shape of the beam by (4). The slope of the beam at the supported place is wl s /4S E i. If the loading is of any kind whatsoever, and if the section varies in any way, it is necessary to be able to integrate a function when given as the ordinate of a curve. We have, as before, in (1 Let the integral of - be called x, and the integral of be called /*, and let curves be drawn showing the values of x and /t. Let their values when x I be x x and /*j. Then We can now integrate (9) again, and obtain E^ ; but if we do not actually wish to find the shape of the beam, we need only use the two ideas first, that y = when x = 0, and this shows that no new constant needs to be added ; secondly, that y = when x = /; and hence, if the areas of the whole x and p curves from x = to v I are x 2 and M I} we have = ??x, +*, + **. ..(10), APPLIED MECHANICS. Also, using in (9) the statement that -~ is zero when a, L we ax have = ^ Xl +/*! + *.... (11). From (10) and (11) we can find c and ^. Using the value of /^ so found, we find the "bending moment everywhere given in (1).* If in the above exercise we imagine the supported end at A to be raised a distance a above its original position, or that B has settled downwards by this amount, find (supposing the beam, fixed at one end, B, to be unloaded) what upward force at A (call it P) will cause it to rise through the distance a. We have only to assume that there is an additional supporting force of this amount, and that the bending moment due to it acts as well as the other- in fact, instead of (1), we have as the bending moment 365. It is very important that the student should work out care- fully such an exercise as the following. A beam is given with loads, and we know m and i at every place. The integrations to be performed are much the same whether it is a case of a beam fixed at the ends or fixed at one end and merely supported at the other, and therefore we give both. The two results are stated imme- diately after the table. "Without using the letters /t, x, /*!, Xi, etc., the above investigation 13 Integrating (9) between the limits and I, and recollecting that y is zero at both limits, Also using in (9) the statement that is zero when x = I, we have The integrations in (10) and (11) being performed, the unknowns -- and c can be calculated. The true bending moment everywhere is what WP started , W&2 with, m + --x. APPLIED MECHANICS. 451 X HI i + 1 i + f + i + x = JV + M I M = J"* M Beam Fixed at Ends. M Beam Fixed at one End, Supported at the Other. '5 8 500 00200 00100 01600 + 22-30 - 7-13 00200 00100 01600 1-5 lo 450 00222 00333 03330 + 14-73 - 12-38 00422 00433 04930 2-5 21 400 00250 00625 05250 + 8-16 - 16-64 00672 01058 10180 3-5 25 350 00286 01001 07150 + 3-59 - 18-90 00958 02059 1733 4-5 28 320 00313 01404 08740 + -03 - 20-15 01271 03467 2607 5-5 32 300 00333 01826 1066 - 4-54 - 22-41 01604 05289 3673 6-5 34 300 00333 02159 1132 - 7-11 - 22-66 01937 07448 4805 7-5 3G 320 00313 02340 1123 - 9-67 - 22-92 02250 09788 5928 8-5 37 350 00286 02431 1058 - 11-24 - 22-18 02536 1222 6986 9-5 35 400 00250 02375 0875 - 9-81 - 18-43 02786 1459 7861 10-5 32 400 00250 02625 0800 - 7-38 - 13-69 03036 1722 8661 11-5 31 400 00250 02875 0775 - 6-95 - 10-94 03286 2009 9436 12-5 30 380 00263 03288 0789 - 6-51 - 8-20 03549 2338 1-0225 13-5 28 360 00278 03753 0774 - 5-08 - 6-46 03827 2713 1-0999 14-5 26 330 00303 04393 0788 - 3-65 - -71 04130 3153 1-7187 15-5 24 300 00333 05161 0799 - 2-22 + 3-03 04463 3669 1-2586 16-5 18 330 00303 05000 0545 + 3-22 + 10-78 04766 4169 1-3131 17-5 12 380 00263 04602 0316 + 8-65 + 18-52 05029 4629 1-3447 18-5 ;') 400 00250 04625 0125 + 15-08 + 27-26 05279 5092 1-3572 195 4 500 00200 03900 0080 + 15-52 + 30.01 05479 5482 1-3652 *i = 5748 Xl = 4-062 MI = 15-276 452 APPLIED MECHANICS. Beam Fixed at the Ends. /*i + w x YJ + p Xj = 0, where m^ is the bending moment where x = 0, m. 2 where x = I, p = } *> or - 1-365 + -0548 m^ + -548 p = ; also - M, + ^ y l + p Xi = 0, or - 15-276 + -5748 m l + 4-062 p = 0. Hence, m 1 = 30-58, P = - -5673, m a = 19-23. Adding p a; to m, we find the true bending moment M given in the tenth column of the table. It will be found that if every value of M be divided by the corresponding value of i, the algebraic sum for the whole beam is - -0013 instead of a very close approximation to accuracy. The student may easily proceed to find the shape of the beam. Beam Fixed at One End. Where x = I and the bending moment is w 2 , supported merely at the other, where x = 0, - ft + PXj + e = 0, where c is E / at x = 0, or - 1-365 + -5482 p + c = 0; ax - M] + P . Xi + el = 0, or - 15-276 + 4 062 p + 20 c = 0. Hence p = 2 = 1'744, or w a = 34-88, and c = -4091. Adding p a; to m everywhere, we find the true bending moment M given in the eleventh column of the table. The student may easily proceed to find the shape of the beam. 366. As we have proved (Art. 357), since dy = i di _ M. <*M If = 10 dx '' dx EI' dx ' dx we have a succession of curves which may be obtained from knowing the shape of the beam y by differentiation, or which may be obtained frcm knowing w, the loading of the beam, by integra- tion. Knowing w, there is an easy graphical rule for finding M ; r knowing , we have the same graphical rule for finding y. E I Some rules that are obviously true in the w to M construction and need DO mathematical proof may at once be used without ma the- If matical proof, in applying the analogous rule from to y. Thus M the area of the curve between the ordinates x\ and x 2 is the increase of i from x^ to x 2 , and tangents to the curve showing the shape of the beam at x l and x z meet at a point which is vertically in a line with the centre of gravity of the portion of area of M the curve in question. The whole area of the curve in a I span H j is equal to the increase in -^ from one end of the span to the other, and the tangents to the beam at its ends H j meet in a APPLIED MECHANICS. 453 point r, which is in the same vertical as the centre of gravity of the M whole curve. These two rules may be taken as the starting- I point for a treatment of the most difficult problems in beams by graphical methods. If the vertical from this centre of gravity is at the horizontal distance HG from H and GJ from j, then p is higher than H by the amount H a x i H , the symbol i H being used to mean the slope at H; j is higher than p by the amount GJ x i at J. Hence j is higher than H by the amount H o . t H + GJ.J a relation which may be useful when conditions as to the relative heights of the supports are given, as in continuous beam problems. 367. Theorem of Three Moments. For some time railway engi- neers, instead of using separate girders for the spans of a bridge, fastened together contiguous ends to prevent their tilting up, and so made use of what are called continuous girders. It is easy to show that if we can be absolutely certain of the positions of the points of support, continuous girders are much cheaper than separate girders. Unfortunately a comparatively small settlement of one of the supports alters completely the condition of things. In many other parts of applied mechanics we have the same difficulty in deciding between cheapness with some uncertainty and a greater expense with certainty. Thus there is much greater uncertainty as to the nature of the forces acting at riveted joints than at hinged joints, and therefore a structure with hinged joints is preferred to the other, although, if we could be absolutely certain of our conditions, an equally strong riveted structure might be made which would be much cheaper. Students interested in the theory of continuous girders will do well to read a paper published in the " Proceedings of the Royal Society," cxcix., 1879, where they will find a graphical method of solving the most general problems. We shall take here, as a good example of the use of the calculus, a uniform girder resting on supports at the same level, with a uniform load distribution on each span. Let ABC be the centre line of two spans, the girder originally straight, supported at A, B, and c. The distance from A to B is li and from B to c is 7 2 , and there are any kinds of loading in the two spans. Let A, B, and c be the bending moments at A, B, and c respectively, counted positive if the beam is convex upwards. At the section at p at the distance x from A let m be what the bending moment would have been if the girder on each span were quite separate from the rest. We have already seen that by introducing couples m 2 and m at A and B (tending to make the beam convex upwards at A and B) we made the bending moment at p really become what is given in Art. 363. Our w 2 = A, Wj = B, and hence the bending moment at p is 454 APPLIED MECHANICS. where m would be the bending moment if the beam were merely supported at the ends ; and the supporting force at A is lessened by the amount Assume E i constant and integrate with regard to x, and we have fm. APPLIED MECHANICS. 455 The equality of (7) and (9) is A*! + 2B(f, + Z 2 ) + Cl 2 = 6 - ! - . . . . (10), an equation connecting A, B, and c, the hending moments at three consecutive supports. If we have any number of supports, and at the end ones we have the bending moments because the girder is merely supported there, or if we have two conditions given which will enable us to find them in case the girder is fixed or partly fixed, note that by writing down (10) for every three consecutive supports we have a sufficient number of equations to determine all the bending moments at the supports. Example. Let the loads be w 1 and w 2 per unit length over two consecutive spans of lengths ^ and l z . Then - \wlx + %wx z , I m . dx = - - * If, \m.dx.dx = - and and /MI = Hence ^ + ^ - & becomes - ^ wj* - If + or - A (W + and hence the theorem becomes in this case A?J + 2 B (/! + J a ) + c 1 2 = J (w 2 l* + wjf) ---- riO). If the spans are similar and similarly loaded, then A + 4B + c = %wP ---- (11). Case 1. A uniform and uniformly loaded beam rests on three equidistant supports. Here A = c = and B = + ^wP. m = - ^w (Ix - # 2 ), and hence the bending moment at a point p distant x from Ais - \w (Ix - a?} + - - ^ wP. The sup- porting force at A is lessened from what it would be if the part of the beam AB were distinct by the amount shown in (2) ^ ? or \ wl. It would have been i wl, so now it is really f wl at each of the end supports, and as the total load is 2 wl, there remains ^wl for the middle support. (See also Example on next page.) Case 2. A uniform and uniformly loaded beam rests on four equidistant supports, and the bending moments at these supports are A, B, c, D. Now, A = D = 0, and from symmetry B = c. Thus (11) gives us + 5s = \wP or B = c = ^wP. If the span A B had been distinct, the first support would have had the load | wl ; it now has \wl - ^ wl or ^ wl. The supporting force at D is also fawl. The other two supports divide between them the remainder of the total load, which is altogether 3 wl, and 4:56 APPLIED MECHANICS. so each receives ^ wl. The supporting forces are then ^ wl, Exercise. If a beam ABC has any kind of loading, and varies in section in any way, and is supported at three places, A and c, on the same level, B, m a level b inches below A or c, first find the diagram of bending A C Fig. 260. moment and the deflection y everywhere of the beam, assuming the support B not to exist. Find, in particular, y b the deflection at the point B. Now consider a new problem the same beam supported at two places, and with only an upward force at B. Find what the force B must be to cause a deflection upwards, y - B, at B, and what the upward deflection z everywhere is. This force B is evidently the supporting force at B in the real problem, and the deflection in the real problem everywhere is y - z. Example. Uniform loading w per inch on A B and on B c, each of length I; beam of uniform section, the supports all on same level. (1) If prop B is absent, the deflection at B would be (Art. 339) ^r-7 . Call this b. Each of the supporting forces is 384 EI wl. (2) Beam of length 2 1, supported at the ends. If an upward force B produces an upward deflection, we know that o = -^ (2 1^. Hence we have -g = ~r , and B = | wl that is, when a uniform and uniformly loaded beam is on three equidistant supports, the supporting forces are $wl, ^wl, and $tvl. Mr. George Wilson (Proc. Royal Soc., Nov., 1897) describes a method of sol\ ing the most general problems in continuous beams which is simpler than any other. Let there be supports at A B c D E. (1) Imagine no supports except at A and E, and find the deflections at B c and D. Now assume only an upward load of any amount at B, and find upward deflections at B c and D. Do the same for c and D. These equations enable us to calculate required upward loads at B c and D, which will just bring these points to their proper levels. 457 CHAPTER XXI. BENDING AND CRUSHING. 368. Stress over a Section. When any portion of a column or beam or arch on one side of a section, B c, is acted upon by loads and supporting forces, we can generally find one force, representing the resultant of the stresses at the section, which will balance them all. If, instead of a force, we merely get a couple, then the section is exposed solely to bending moment, and we know now how to find the effect of this. If the force is parallel to the section, then we know that the section is either exposed to mere shearing strain or shearing and bend- ing, as in a horizontal beam with vertical loads ; but if the force is inclined to the section, there will usually be shearing and bending, and besides this a uniform distribution of com- pression or extension all over the section. In practice we generally find that compression and bending alone have to be considered. Thus, if BC (Fig. 261) is the edge view of the section of a structure, O being a line at right angles to the paper through its centre of gravity, and it' F is the resultant of all the forces supposed in the plane of the paper which act on the structure to the right of the section, let p and s be the resolved parts of F normal to and tangential to the section ; then s is balanced by an equal and opposite shear- ing force which must be exerted by the pj g . 201. material to the left, p is a compressive load which is spread uniformly over the section, producing a compressive stress P/'A if A is its area ; but besides this we have in the section the varying compressive stress on the B side of o and the tensile stress on the c side of p, which the bending moment p . o D produces. In fact, the compressive stress at any place which is at the distance y from o on the . p P . o D compression side is - + - y . . . . (1). A I The student ought to draw a slice between two parallel cross-sections like B c near one another^ and draw the change p* 458 APPLIF.D MECHANICS. of shape, first making it uniformly thinner because of P/A, and then making it wedge-shaped. We have, in fact, the wedge- shape c e'f d' of Fig. 207, where H' j' = H J, and in addition we have to imagine e'f moved parallel to itself in the direc- tion c d' ; but students must draw this for themselves. They will see that the result may be compression everywhere, but much more at B than at c ; or compression everywhere and just no stress at c j or compression at B, tension at c, and a neutral line somewhere between. Most people have the habit of calling the line through o the neutral line of the section, although it is the neutral line only when the com- ponent P is zero. The proof of the above statement is this : Assume that a plane section remains plane, it follows, as it did in Art. 319, that there is a neutral line, say at o', at right angles to the paper. Let any point H be at the distance z from o' ; the compressive stress there is cz say, where c is some constant. The force on a small area a there, is cza. Then P = c 2 za . . . (2) and P . O D = 02*0 . ff . . . (3), because, if OH = y, za.y is the moment of za about o. Now z = y + o o', so that equation (2) is p = c.oo'20 + c?,y . a, c.oo'.A... . (4), because 2 ya = o ; and p . o D = c 2 y 2 a + c'2,oo / .y.a=ci.... (5) if i is the moment of inertia of the area about o. We see that cy from (5) is y and c . o o' from (4) is - ; but the sum of I A these is cz, the compressive stress, and so we have proved (1). If the section is rectangular, the dimension at right angles to the paper being 1, and BC being d, then i = 1 x *? 3 /12 and A = 1 x d. The compressive stress is least at c where y = - J d, or by (1), the compressive stress at c is -= - 4j .... (6). As o D gets greater, the compressive stress at c becomes less until there is a value of o D which just causes c to have no stress ; a greater value of o D than this would create tensile stress at c. We usually take it that in a masonry joint there ought to be no tensile stress, and hence in a masonry joint the limiting value of o D is given by putting (6) equal to zero ; that is, 6 . o D = d. Hence D must fall within the middle third of the masonry joint, B c, if there is to be stability. This is the fundamental rule in the design of arches and buttresses. Another condition of stability for a masonry joint is that s shall not exceed the f rictional resistance, or the angle J D L must be less than the angle of repose (see Art. 96) for the materials. The above rulo is very generally useful in machine design, APPLIED MECHANICS. 459 but we need not give many examples. Fig. 262 is a crane-hook. The section at B c is not usually elliptic ; rather like an ellipse, with the end at B blunter than that at c. If o is a line at right angles to the paper through the centre of grav- ity of the section, and w is the load, the stress at any place is that due to a bending moment w . o D, together with a tensile stress due to w being spread uniformly over the section. When a weight w hangs from a bracket as in Fig. 263 the strength at any section, such as o, is merely calculated from the bending moment w . D B, because when the distance D B is con- siderable the stresses due to the bend- ing are usu- ally much greater than those due to w, divided by the area of cross-section. 369. Shear Stress in Beams. Let the distance measured from any section of a beam, say at o (Fig. 264), to the section at A be x, and let ov = x + 8x. Let the bending moment at c' A c be M, and at D' B D be M + SM. Let A c be the com- pressive side of c c'. Let o AB (Fig. 264) and A A (Fig. 265) repre- sent the neutral surface. We want to know the tangential or shear stress f s at E on the plane c A c'. Now it is known that this is the same Pig 261 w Fig. 26a C D Fig. 264. Fig. 266. as the tangential stress in the direction E F on the plane E F, which is at right angles to the paper, and parallel to the neutral surface at A B. Consider the equilibrium of the piece of beam E c D F, shown 460 APPLIED MECHANICS. in Fig. 265 as E c E, and also shown magnified in Fig. 266. We have indicated only the forces which are parallel to the neutral surface, or at right angles to the sections. The tctal pushing forces on D F are greater than the total pushing forces on CE, the tangential forces on E F making up for the difference. We have only to state this mathematically, and we have solved our problem. At a place like H in the plane c A c', at a distance y from the neutral surface, the compressive stress is known to be j? = y ; and if b is the breadth of the section there, shown as H H (Fig. 265), the total pushing force on the area E c E is f A c fA c p = I b-y dy, or P = T fy.dy ---- (1). JAE I VAE Observe that if b varies, we must know it as a function of y before we can integrate in (1). Suppose we call this total pushing force on E c by the name P, then the total pushing force on D F will be p + Sx . -j-. The tangential force on E F is f 8 x area of E F, 01 f t . Sz . E E, and hence Example. Beam of uniform rectangular section, of constant breadth b and constant depth d. Then and hence so that f 8 is known as soon as M is known. As ~ is the shearing cttx/ force over the whole section, we may regard (3) as telling us what fraction of the total s there is on every square inch. As to M, let us choose a case say the case of a beam supported at the ends, and loaded uniformly with w Ib. per unit length of the beam. We saw that in this case, x being distance from the ATJL middle, - M = %wP - %wx 2 . Hence -^- = wz, so that (3) is - -fir at the section, there is nothing very extra- ordinary in finding that the actual shear stress anywhere in the section depends upon its value. Deflection of Beams due to Shear. If a bending moment M acts at a section of a beam, the part of length far gets the strain-energy ^ M 2 SXJE i, because M 5a?/E i is the angular change, and therefore the whole strain-energy in a beam due to bending moment is 2E JL i*-*.. i* If / is a shear stress, the shear strain-energy per unit volume 462 APPLIED MECHANICS. is / 2 /2N .... (7), and by adding we can therefore find its total amount for the whole beam. By equating the strain-energy to the loads multiplied by half the displacements produced by them we obtain interesting rela- tions. Thus in the case of a beam of length I, of rectangular section, fixed at one end and loaded at the other with a load w ; at the distance x_ from the end, M = w#, and the energy due to bending is 1 fW _ 2i "T-*^ +> o =w 2 flj 6 E i (8). The above expression (5) gives for the shearing stress The shear strain-energy in the elementary volume b. 8x. 8y is b. 8x. 8y. / 2 /2 N. Integrating this with regard to y from - \d to + \d, we find the energy in the slice between two sections to be 3 w2. 8#/5 N b d, so that the shear strain -energy in the beam is 3 w 2 ;/5 N b d ____ (10). If now the load w produces the deflection z at the end of the beam, the work done is m . . . . (11). Equating (11) to the sum of (8) and (10) we find ._+ wl SET o N bd Note that the first part of this due to bending is the deflection as calculated in Art. 339, Example 1. We believe that the other part due to shearing has nevor before been calculated. 370. Springs which Bend. We consider the bending in springs of regular shape, such as spiral springs, later, in Art. 521. But it seems natural to consider certain irregularly shaped springs here. Let Fig. 267 show the centre line of a spring fixed at A, loaded at B with a small load w in the direction Fig. 267. shown. To find the amount of yielding at B, the load and the deflection are supposed to be very email. Consider the piece of spring bounded by cross-sections at P and Q. Let p Q = 8s, the length of the spring between B and v being called . APPLIED MECHANICS. 163 The bending moment at P is w . P R or w . #, if x is the length of the perpendicular from P upon the direction of w. Let B R he called y. Consider first that part of the motion of B which is due to the change of shape of QP alone; that is, imagine A Q to be perfectly rigid and p B a rigid pointer. The section at Q, -being fixed, the section at p gets an angular change equal to 5s x the change of curvature there, or 5s or - .... (1), where E is Young's modulus and i is the moment of inertia of the cross- section. The motion of B due to this is just the same as if p B were a straight pointer ; in fact, the pointer p B gets this angular motion, and the motion of B is this angle multiplied by the straight distance p B, or * ' Wa? . PB . . , . (2). Now how much of B'S EI PR motion is in the direction of w ? It is its whole motion j<: P B or x , and hence B'S motion in the direction of w is - - . . . (3). PB E i Similarly B'S motion at right angles to the direction of w is In the most general cases it is easy to work out the integrals of (3) and (4) numerically. We usually divide the whole length .of the spring from B to A into a large number of equal parts so as to ' have all the values of 5s the same, and then we may say (s being the whole length of the spring) that we have to multiply - y.2 %y upon the average values of and for each part. In a well- made spring, if b is the breadth of a strip at right angles to the paper and t its thickness, so that i = ^ bt 3 , we usually have the spring equally ready to break everywhere, or -- = /, a constant. When this is the case (3) and (4) become - . and E I *.: . y.. And if the strip is constant in thickness, varying in 2 f ft? 2 f S breadth in proportion to x, then - - J ' . x is (3) and -^- . y E t K is (4). If aTand y are the x and y of the centre of gravity of the curve (see Art. 110), -^ is the total yielding parallel to w, and this is what we generally desire to know. ' is the total yielding at right angles to w. 371. Struts. As we have already said, when a very short column is loaded (as it usually is) in such a way that the material is prevented from swelling laterally, it will withstand exceedingly great loads without fracture. It is only when the 464 APPLIED MECHANICS. length of a square or round prism is three or four times its diameter that the material gets a chance of showing how it behaves under compressive stress ; up to a length of about ten times its diameter its breaking load is now nearly the same, and it remains the same for any length if slight lateral restraints against bending are provided. In a tie-rod, want of uniformity of stress causes yielding, and this is the same in a strut ; but in a tie the effect of yielding is to remedy the defect, the tie gets straighter, the stress better distributed, whereas in a strut local failure causes bending, instability, and a worse distribu- tion of stress. In Art. 243 we have discussed the behaviour of material under uniform axial compressive stress merely, and also in columns which are too short for mere axial compressive stress. We now know that if B c, Fig. 261, is a cross section of a strut, and if the resultant load on the part of the strut on one side of B c is F or p, not only have we the usually assumed stress P/A over the section, but also the bending moment stress. What this may amount to de- pends upon o D ; and this depends on two things first, the want of accuracy in applying the load at the ends of the strut, a matter which cannot be taken into account in calculation unless we know how much error there is ; secondly, the bending of the strut. 372. Bending of Struts. Consider a strut per- fectly prismatic, of homogeneous material, its own weight neglected, the resultant force F at each end passing through the centre of each end. Let A c B, Fig. 268, show the centre line of the bent strut. Let p Q = y be the deflection at p where o Q = x. Let o A = o B = I ; y is supposed everywhere small in com- parison with the length 21 of the strut. Let E be Young's modulus for the material, and I the least moment of inertia of the cross-section everywhere, Fig 268 about a line through the centre of gravity of the section. Then the result of the following theory is that the load F, which will produce bending, is E 1 7r 2 /4 2 . F t/ F y is the "bending moment' at p and is the curvature there. Then as in Art. 339 the curvature being - | we have *y &y . /n H = ~ 5 (1) ' APPLIED MECHANICS. 465 Notice that when we choose to call ~ the curvature of a curve, if the. expression to which we put it equal is essentially positive, we must give such a sign to -^ as will make it also positive. Now if the slope of the curve of Fig. 268 be studied as we studied the curve of Art. 338, we shall find that j-| is negative from x = to a; = o A, and as y is positive so that is positive, we must use ffi'ii - ~ 2 on the right-hand side. If the student tries he will find that y = a cos. x satisfies (1) whatever value a may have. When x = we see that y = a, so that the meaning of a, is known to us ; it is the deflection of the strut in the middle. The student is instructed to follow carefully the next step in our argument. When x = I, y = 0. Hence How can this be true ? Either a = 0, or the cosine is 0. Hence, if bending occurs, so that a has some value, the cosine must be 0. Now if the cosine of an angle is 0, the angle must be - or or _^ etc. If we confine our attention to ^, the condition that bonding occurs is /~V~ >* T? T ir2 (4) is the load which will produce bending. This is called Euler's law of strength. It is easy to see why we confine our attention to ^ as it gives the least value of w. The meaning of the other cases is that y is assumed to be one or more times between x = and x = /, so that the strut has points of inflexion. It will be found that the complete solution of any such equation as (1), which may be written -r^ + M 2 y = 0, is y = a cos. nx + b sin. nx where a and b are arbitrary constants, a and b are chosen to suit the particular problem which is being solved. In the present case it is evident that as y when x = I and also when x = - I = a cos. nl + b sin. nl, = a cos. nl - b sin. nl, so that b is 0. 466 APPLIED MECHANICS. Now let us consider a strut fixed at the ends. The resultant force F at an end no longer acts through the centre of gravity of the end section. The solution is as before y = a cos. nx + b sin. nx . . . . (5). Differentiating we have dy -j- = - na sin. nx + nb cos. nx . . . . (6). Now j- = for x 0, x = I, x = - I, and hence = nb, so that b = or y = a cos. nx . . . . (7). Also = - na sin. nl = na sin. nl (8). Now (3) tells us that a is the deflection at c, where x is 0, and if a has any value, (4) tells us that sin. nl is that is, nl = IT, or J /v /r7^i = ,r,orF = ^ (9). Hence if a strut is fixed at both ends the load which it will stand before bending is the same as for a strut of half the length hinged at the ends. In fact, if G c H is a strut fixed at the ends, its strength is like that of a strut K L hinged at IK and L, or again like that of a strut G K L fixed at G and hinged at L. The strength to resist bending of : G- a strut of length 2 I hinged at the ends may be cal- culated from (4) ; this is the strength of a strut of length 3 I, hinged at one end and fixed at the other, or of a strut of length 4 1 fixed at both ends. It is only necessary, then, to remember the rules for j _ struts hinged at the ends. The load given by (4) will produce either very little or very much bending equally well, and we may take it that F given by (4) is 'the load which will break a strut if it breaks by bending. If /is the compressive stress which will produce rupture and A is the area of cross section, the load /A will break the strut by direct crushing, and we must take the smaller of the two answers. In fact, we see Fig. 26J> that /A is to be taken for short struts or for struts which are artificially protected from bending, and (4) is to be taken for long struts. Now, even when great care is taken we find that struts are neither quite straight nor homo- geneous, nor is it easy to load them in the specified manner. Consequently, when loaded they deflect with even small loads. APPLIED MECHANICS. 467 and they break with loads less than either /A or that given by (4). Curiously enough, however, when struts of the same section but of different lengths are tested, their breaking loads follow, with a rough approximation to accuracy, some rule as to length. Let us assume that as F =/A for short struts and what is given in f k) for long struts, then the formula . . . (10) EI7T2 may be taken to be true for struts of all lengths because it is true both for short and for long ones. For if I is great we may neglect 1 in the denominator and our (10) is really (4) , again, when I is small, we may regard the denominator as only 1, and so we have w =/A. We get in this way an empirical formula which is found to be fairly right for all struts. To put it in its usual form, let I = A & 3 , k being the least radius of gyration of the section, then where a is 4//E ?r 2 . If F does not act truly at the centre of each end, but at the distance h from it, our end condition is that y = h when x = I. This will be found to explain why struts not perfectly truly loaded break with a load less than what is given in (4). Students who wish to pursue the subject are referred to pages 464 and 513 of the Engineer for 1886, where initial want of straightness of struts is also taken account of. When we consider, therefore, how the rule (11) has been arrived at, it is evident that it needs to be tested by practical experiment, the constants a for various materials and k 2 for various kinds of section being also determined by experiment. This has been done, and on the whole we feel fairly well satisfied when the rule is put in the following form : For a strut whose ends are hinged, or a column whose ends are not fixed, the breaking load in pounds is equal to the breaking stress per square inch given in Table IX. multiplied by the area of cross section in square inches, and divided by 1 + nE where n is given in Table XL, and B is given in Table X. 468 APPLIED MECHANICS. Exercise for Advanced Students. Test in the six cases of Table XL to what extent this rule agrees with (11). TABLE IX. Breaking Stress, in pounds per square inch. Cast Iron ... 80,000 36,000 Timber 7,200 TABLE X. Value of B for struts of the sections shown in Table XI. The first column gives the length of the strut divided by its least lateral dimension. Length divided by Lateral Dimension d. B for Cast Iron. B for Wrought Iron. B for Strong Dry Timber. 10 0748 0-132 1-0 15 1-68 0-300 3-6 20 3-00 0-532 6-4 25 4-64 0-832 10-0 30 6-76 1-200 14-4 35 9-20 1-632 19-6 40 12-00 2-132 25-6 45 18-72 3-332 40-0 EXERCISES. 1. Find the breaking load for a cylindrical strut of wrought iron 3 inches diameter and 10 feet long, supposing it to be not fixed at the ends. Ans., 30-1 tons. 2. A hollow cast-iron pillar 10 feet long, and fixed at the ends, has an external diameter of 6 inches ; what should be the thickness of the metal to carry a load of 30 tons, allowing a factor of safety of 8 ? Ans., -42 in. 3. What is the safe load for an angle iron, least breadth 3 inches and 7 feet long, acting as a strut, firmly fixed at both ends ? Factor of safety, 6. Ans., 3*51 tons. 4. The diagonal brace of a Warren girder is 10 feet long, and is composed of two tee-bars 6" x 3" x ", placed back to back and riveted together. Find the maximum compressive working load which may be applied to it when the ends are firmly riveted to the boom. Use a factor of safety of 4. Ans., 17'5 tons. 5. Section 4, Table VI., has flanges 4*02 inches broad, -56 inch thick, depth over all 8 inches, web '42 inch thick. The greatest moment of inertia about a line through its centre of gravity is 74 -2, according to the rule of the table. Its least moment of inertia is about a line at right - . , . '56 x (4-02) 3 _ 6-88 x (-42) 3 angles to the first, and is '- x 2 + L, or 6-1. APPLIED MECHANICS. 469 The area of the section is 7 '39 square inch. The ultimate stress being taken as 45,000 Ibs. per square inch, and the Young's modulus as 30 x 10 6 Ibs. per square inch, the following are the breaking loads, according to Euler's theory, if it is the section of a strut fixed at the ends of the following lengths. For these loads to be withstood, it is necessary to carefully adjust the application of load at the ends so as to have absolutely no bending until the breaking loads are reached, w = A/ or 4 K i Tr 2 /L 2 , if L is whole length of strut fixed at the ends ; the lesser answer of the two to be taken. L in inches ... 96 120 144 180 216 240 288 Breaking load in tons 148-5 148-5 148-5 99-5 69-1 56 38-9 TABLE XI. Values of n for struts and pillars of the following sections : Value ofn. Square of side d, or rectangle with smallest side d ., Hollow rectangle, or square with thin sides ... Circle, diameter d Thin ring, external diameter d Angle iron, smallest side d 1-00 2-00 Cruciform, smallest breadth d ... ... ^ 2-00 _i Jl If we want the breaking load for a strut whose ends are not hinged, it is necessary to find in what way it tends to bend, and to use the above rule regarding the strut as hinged at two points of contrary flexure. Thus in Fig. 270 the strut or column B is as strong as a strut hinged or rounded at both ends, whose length is only a b. The rule becomes For a strut fixed at both 470 APPLIED MECHANICS. ends, calculate by the above rule, but take n one fourth of what I have given in Table XI. For a strut one end of which is fixed and the other is only hinged ; calculate the breaking load as if both its ends were hinged \ then calculate as if both its ends were fixed, and take the mean value of the two answers. 373. Struts with Lateral Loads. If the lateral loads are such that by themselves, and the necessary lateral supporting forces, they produce a bending moment which we shall call < (#), then (1) Art. 372 becomes vy + (x) =. - EI^. When we know the lateral loads we know (x), and it is quite easy to integrate. Thus let a strut be uniformly loaded laterally, as by centrifugal force or its own weight, and then (x) = \w (I - #) 2 if w is the lateral load per unit length. We find it slightly more convenient to take $ (x) = %-wl cos. ^- x, where w is the total lateral load. This is not a very different law. Hence We find here that cos. * . . . . (2). Observe that when F = this gives the shape of the beam. The deflection in the middle is and the greatest bending moment ju is p. = F y l + w I, or If ^ = 0, and if fj. has any value whatever, the denominator of (4) must be 0. Putting it equal to 0, we have Euler's law for the strength of struts, which are so long that they bend before break- ing. If Euler's value of F be called u, or u = E 1 7^/4 P, (4) becomes If z c is the greatest distance of a point in the section from the neutral line on the compressive side, or if i -r z c = z, the least strength modulus of the section, and A is the area of cross-section, and if f is the maximum compressive stress to which any part of the strut is subjected, + - =/. Using this expression, if Z A stands for - (that is, Euler's breaking load per square inch of (( UNIVERSITY ) \ of / I a^^g_,Sj|tt^*B. 471 section), and if w stands for - (the true breaking load per inch of . This formula is not difficult to remember. From it w may be found. EXERCISES. 1. Every point in an iron or steel coupling rod of length 2b inches describes a radius of r inches. Its section is rectangular, d inches in the plane of the motion, and b at right angles to this. "We may take w = Ibdrn 2 -r- 62,940 in pounds where n = number of revolutions per minute. Take it as a strut hinged at both ends for both directions in which it may break. (1) For bending in the direction in which there is . -r, , . . no centrifugal force where i is - .- Euler's rule gives -j-- Now we shall take this as the endlong load, which will cause the strut to break in the other way of bending also, so as to have it equally ready to break both ways. (2) Bending in the direction in which bending is helped by centrifugal force. Our w is the above quantity divided by bd, or, taking b 2 E = 3 x 10 7 , w = 6-17 x -p x 10 6 . Taking the proof stress / for the steel used as 20,000 Ibs. per square inch (remember to keep/ low, because of reversals of stress), and recollecting the fact that i in this other bd 3 direction is , we have (6) becoming 8-4 x 108 ^ _ 308 j!) (i _ ^ = n *Fr - d . . . . (7). Thus, for example, if b = 1, I = 30, r 12, the following depths d inches are right for the following speeds. It is well to assume d, and calculate n from (7). d 1-5 n 205 277 327 2 2-5 3 4 G 437 545 2. A round bar of steel, 1 inch in diameter, 8 feet long, or I = 48 inches. Show that an endlong load only sufficient of itself to produce a stress of 1,910 Ibs. per square inch, and a bending moment which by itself would only produce a stress of 816 Ibs. per square inch, if both act together produce a stress of 23,190 Ibs. per square inch. Students will find that this subject will well repay further study. The effect of a small lateral load on a strut is sometimes very striking. Again, in tie-bars it is very important to consider the effect of a lateral load. The subject is treated more fully in a paper in the Philosophical Magazine, March, 1892. 374. Beams without Compression. In Art. 368 we have seen that a tensile load applied to extend a beam may not only diminish the greatest compressive stress, but also the tensile stress. Again, there are many cases of beams or infinitely 472 APPLIED MECHANICS. flat arches in which there is no tensile stress anywhere. In such a case, of course, the earth takes the necessary tensile load. When the pneumatic wheel tyre was invented, Pro- fessor Fitzgerald pointed out to me that columns to support loads, and military bridges easy to pack and unpack, might bo made of inflated tubes, the solid material being everywhere in tension. I consider this a notion of great importance. In a thin straight tube of circular section if the greatest bending moment is M and R is the radius, t the small thickness of the material, the compressive stress anywhere due to bending is - y where y is the distance from the diameter which is the TT R 8 1 neutral line of the section on the compressive side. The greatest compressive stress is M/TT R 2 t. Now imagine the tube to be subjected to internal fluid pressure p above that of the atmosphere ; there is a tensile endlong stress PirR 2 -r27rRor p R/2 t, and hence the greatest compressive stress is M/TT R 2 1 P R/2 t. This is just o when P = 2 M/TT R 3 . The greatest tensile endlong stress is then, of course, P n/t ; but this is equal to the lateral tensile stress which the mere internal pressure produces. When, therefore, the internal pressure is just sufficient to remove all compressive stress in the material, the tensile stress, where it is greatest, is the same in all directions and is 2 M/TT R 2 1. We see, therefore, that great loads may be carried by inflated tubes of thin material if they are only large enough in diameter, or by a bundle of small tubes. Shearing forces might be taken up by side frameworks like lazy- tongs. I make no attempt here at exact theory. What I give is sufficiently correct to show the general value of the suggestion. One may go far in speculation on this idea rigidity gained by using thin material and subjecting it to internal fluid pressure, so that there shall be no compressive stress. The great ships of the future may owe their stiffness and strength to the general use of fluid pressure in those parts of them where cargo is stored, and the same pressure which gives strength may serve to keep out the sea in case of a leak. It is the means by which the leaves of plants are made rigid. Similarly, large flat areas might be made of considerable size by fastening together two plane sheets by means of many connecting ties so that they may not balloon out, and then inflating them like an air cushion. Aeroplanes of sufficient size to support a man by Lilienthal's method can be made with APPLIED MECHANICS. 473 comparatively small internal fluid pressures and are not liable to make splinters when they fall to the ground, these splinters being a cause of considerable risk with aeroplanes made with sticks as stiffeners. Kites much larger than those suggested for military purposes might be made, in which the whole kite might be like an air cushion, or thin tubes with compressed air might take the place of the present bamboo framework. The inflation might be maintained automatically. 375. Inflated Columns. Again, a thin tube of radius Rand thickness t has to act as a column carrying a load w, and this is the load which is carried when there is no axial tensile stress. The pressure of the fluid inside being p, we have if R 3 p=w . . . . (1). Also the lateral tensile stress produced in TTT the material is P nit or ;, so that great loads may be sup- 7T R t ported by inflated tubes of thin material if they are large enough in diameter. Thus, for example, I find that a tower of thin steel 1,000 feet high would have in it a lateral tensile stress of only 3 tons to the square inch, due to its own weight and the necessary fluid pressure. Being all in tension there is no danger of instability such as exists in ordinary pillars. If large in diameter, the hemispherical top cap becomes of importance as a load. Any moderate diameter like 20 feet would bear many tons on the top in addition to the weight of the structure itself. Thus, 1,000 feet high and 20 feet in diameter and -01 foot thick would itself weigh about 125 tons. Its hemispherical cap would weigh 6 -3 tons, and it would support 325 tons on its top. The internal pressure would be 23 Ibs. per square inch and the tensile stress 10 tons per square inch. There would be no compressive stress. Neglecting lateral stiffness, whether we assume the adiabatic or constant temperature law for increasing pressure down- wards in the fluid, we are led to rules as to the relationship of radius of column to thickness of metal if the column is to have the same stress in its material at all levels. In fact, Fitzgerald's idea gives rise to a number of easy and interesting problems which I have worked out, but for which I have no space. In all probability it would be found cheaper to use long stays from the top, and possibly from several places' at different heights to resist lateral motion due to wind pressure, etc., than to stiffen as with lazy tongs the sides of the tube itself. It is certain that the thing is practical. 474 APPLIED MECHANICS. 376. Tapering Column of Circular Section. If a; is the distance downwards to any section where t is the thickness and r the radius, p then being the internal pressure due to fluid and q the external pressure due to the atmosphere ; if f l is the compressive stress in the material and w l the weight per unit volume of the metal and w of the inside fluid, 8# . 2 irrt . w l + 7TT 2 . 8x . w + 2 irr.'br.q = Sx . (pirr 2 + 2 irrtf 1 ). It follows from this, if w cp l ly, as w 1 is constant and q is a known function of #, that if a = 1 + 2/ 1 //, where / is the greatest stress in the material ; p =p (I - bx) s , q =. q Q (1 - cx) s i where s = 2L_ } taking the same fluid inside and outside. The result is, if q is taken to be small, JOL x + ^ "*" a ) s log. (1 - bx) + 2 log. r = constant. G,f d If q is not small, the integration seems troublesome. 377. The limiting length of a vertical prism supporting its own weight. h being measured downwards to any place from the centre of gravity of the uppermost section, let y be the deflection. Let the load per unit length be w, and let w == i w . dh, the total w I w . dh, load above any section. Considering bending moment M and M -f SM at the sections at h and h -f- SA, we see that w.Sy + i w.Sh.Sy = 5 M, In the most general case, where w and i vary, and where there is a load r on the top, (f + Let i and w be constant, and let F be 0, an equation whose solution is useful in other problems. If H is the total length of the column whose end is fixed in the ground, lot x = A/H and (2) becomes EI The solution of (3) in series may be indicated by y= A*r(X> + B* 2 /O) ---- (4). If we put ~ = when x = at top, we find u = 0, and the APPLIED MECHANICS. 475 other part of (4) is -= 2.3.42.3.5. 6.7 2.3.5.6.8.9.10 nd In this, if we put -- = when x = 1, we find dx ~ 278 2 . 3 6 . 6 2 . 3 . 5 6 . 8 . 9 " and hy trial we find that this is satisfied hy m = 7'85.* Hence if bending occurs, m H S /E i = 7'85 ; so that this gives us the limiting height H of a uniform column. Thus we find H in feet = 8 x 10 6 /L 2 for thin steel tubes, and H = 4 x 10/L 2 for solid rods of steel, where L is the ratio of length to diameter. Thus, for example, the maximum height of a tube 8 feet in diameter is 800 feet. Let A k z = i : a = diameter in inches. Greenhill give? p is j-. First put^> = #% then put a? = r 2 ; we get 2 <2 2 2 , d* , This is r 2^f , r & dr 2 dr 4 w Bessel's equation, where k* = g^F' ^ = *' Solution of ( 3 ) ^ z == A J n (^rr) + B j _ n (^r). Consequently L = when x = 0, makes A = ; dx ... p = B */~ x j _ i (kx ) ____ (4). Put j? = when x = h, the lowest point. J _ i (^ *) == . If c is the least root of j__^ (0) == 0, then c = A A *. Greenhill gives the expansion of j _ ^ (c), and finds root by trial. 378. Stability of Shafts. Eankine considered the effect of cen- trifugal force on shafting in 1869. Professor Greenhill has investi- gated the stability of a shaft between bearings (see his Paper to the Institution of Mechanical Engineers, 1883). He took into account * Prof. Maurice Fitzgerald published curves showing the values of y and -jt when x = 1 for various values of m (Proc. Phys. Soc. of London, Oct., 1892). Prof. Greenhill had previously given the solution, not only for uniform solid cylinders, but also for cones and paraboloids of revolution (Proc. Camb. Phil. Soc., 1881). 476 APPLIED MECHANICS. an end thrust, F, a twisting moment, T, and centrifugal force. He found in practical examples of propeller shafts that the twisting moment was not nearly so important as the thrust in determining the maximum^ length of shaft for stability ; in fact, that the shaft might he designed, merely like a strut, by Euler's formula. I consider, therefore, that such a shaft ought to be studied under the rules of Art. 372. 379. Whirling Loaded Shaft. If a uniform shaft, originally straight, is w Ib. per unit length, and has an angular velocity of a radians per second ; if B is its flexural rigidity, or E i ; if the deflection from straightness is y at a point distant x from the centre, then the load due to centrifugal force is o 2 y per unit length. If we also take into account the effect of gravity, there is one position in a revolution when the centrifugal force and the weight produce their greatest effects. We may approximately take it that the path of every point is a circle, and that the weight is a radial force always acting in the same direction as the centri- fugal force. If we take into account also an endlong thrust, r, we are led to the equation where y is to have the same values for equal positive and negative values of x, so that A] = 0, A 3 = A 4 . Putting y = where x = I and con- fining our attention to shafts whose bearings do not constrain them in direction at the ends, so that dPyfdx 2 =. where x = I we find A! = 0, A.J = -y*g/a? (-f + 2 ) cos. 01, A 3 = A 4 = )8 2 y/2a 2 (72 + 2 ) cosh yl. The bending moment anywhere is E i dPyfdx'*, and hence the bending moment in the middle, where it is evidently greatest, is M =:Ei(-)8 2 A 2 +2A37 2 ), or / and to the solution y = A! sin. fix + A 2 cos. /Bar + A 3 e 7 X + A 4 ~*X x y/a 2 .... (2) fl2 _ JL , , /J 2 , wa ? - 2B + ^v 4 i 2 + TIT The greatest stress at the middle is where z is the strength modulus of the section, and a is the area of the section ; or if R is outside radios of a circular shaft we have '/..-. (6). APPLIED MECHANICS. 477 Exercise. Show that in the case of a propeller shaft 13-4 inches diameter, of length 98 feet between its bearings, with endlong thrust 50,000 Ibs., if a = 2 TT, or GO revolutions per minute, tho important terms in the above calculation are F 2 /4B 2 =30 x 10~~ , and wo 2 /^B = 89 x 10 . So that the centrifugal force is 300 times as important as the endlong thrust. If we altogether neglect twisting moment and endlong thrust and write n* for w o?\g B, we have to solve Our old = y = w, and the greatest stress is (_! 1 Vcosh nl cos. wZ ---- (8). This is infinite if cos. nl is 0, that is if nl = - ; that is, if This limiting length may be obtained very simply by leaving out the constant term in (1), as Rankine does. Mr. Bunkerley (Phil. Trans, for 1894) discusses the stability under centrifugal force of shafts loaded with pulleys, and he has illustrated his results experimentally. EXERCISES. 1. If the critical length of a shaft 13-4 inches in diameter, subjected to endlong thrust alone, is equal to the critical length when subjected to centrifugal force alone, show that F = 68,500 o. Show that if the length is 98 feet, the critical F is 327,600, and the critical a is nearly 4'1 radians per second, or 46 revolutions per minute. 2. Professor Greenhill's result is that if F is the endlong force, and T the twisting moment, the limiting length of shaft being 2 I, we have TT* _ F T^ 4Z2- EI T 4E 2 l2 - The propeller shaft of the Cunard steamer Servia is of wrought iron, 22| inches diameter. The pitch of screw is 35^ feet. At 53 revolutions per minute the indicated horse-power was 10,350. Assuming that all this power is really utilised an assumption which is, of course, quite wrong prove that F = 181,530 Ibs., and that T = 12 '3 x 10 6 pound- inches. Take E = 29 x 10", and show that tHe above formula becomes jJ = 4-98 x 10~ 7 + 2-8 x 10~ 10 , so that the twisting moment term is quite inconsiderable compared with the thrust term. Show that 2 I = 4,454 inches. Consider this shaft of 22^ inches diameter, 4,454 inches long between bearings, subjected to no thrust or twisting 1 moment, and not even to its own weight ; prove that il cannot be rotated at a higher speed than a = '14, or 1 revolution per minute, without fracture by centrifugal force. 478 CHAPTER XXII. METAL ARCHES. 880. THE student must examine drawings and actual specimens of bridges to see how the weight of a roadway is brought to bear upon the arched ribs whose function is to carry weight, trans- mitting it with certain horizontal forces to the abutments. Our problem is, given the distribution of vertical loads on an arch-ring of which the shape of the centre line and the shape of cross-section everywhere are known, to find the stress everywhere. If K. z o is the centre line of the rib shown in Fig. 270, then any cross-section B c is supposed by our theory to remain plane. The resultant of Fig. 270. all the loads acting to tho right of B c, together with the resultant force R at the abutment, is a force whose direction is shown at D by the direction of the line of resistance there, and the force polygon shows its amount. We saw that in Fig. 261 this force, which we there called F, produces a bending moment M at the section whose amount is p^. o D. But evidently this is the same as F . oj (Fig. 271) if GJ is perpendicular to F. A much more important thing for our present purpose is to know that it is also equal to the horizontal component H of the force F multiplied by the vertical distance o L. This the student will easily prove for himself. Now if all the loads are vertical, we know from Art. 349 that (1), the resultant force F at every joint has the same horizontal component H as at any otb.3r. This is represented on the force polygon (Fig. 236) by OH. It is, of course, also the horizontal APPLIED MECHANICS. 479 Fig. 271. thrust on each of the abutments, and is generally called the horizontal thrust of the arch. Of course the force polygon shows all the forces F at all the sections. (2) If two vertical lines he drawn through K and G, as in Fig. 270, and if any line of resistance he drawn whose ends K' and G' are in these lines, and if E/ and G' are joined, the vertical height anywhere,B T,of this link polygon is inversely proportional to the o H of the force polygon by means of which it is drawn ; so that if two such figures as X'Z'G' are drawn, the vertical ordrnates DT are all in the same proportion. We see, therefore, that since the banding moment at B c is the horizontal thrust multiplied by o L, the vertical distance from the centre line anywhere to the line of resistance there represents to some scale the bending moment at the section there. At v and u the lines cross. These are points of no bending. From K to u, and from G to v the bending moment tends to make the arch more convex upwards. From v to TJ the bending moment tends to make the arch less convex upwards. If we can only find the true line of resistance K' Z'G', we know by Art. 129 the stress in every section. The true po- sition of K' z' G' is determined by these conditions : 1. In an arch hinged at the ends, if K and G are the centres of the hinges, we know that the line of resistance must pass through them. We only need one other condition, and that is given by the statement : the yielding everywhere of the arch must be such that the distance K a remains constant. 2. In an arch fixed at the ends, if the fixing is perfect, we have the above condition that the distance K G remains constant, and also the condition that the inclinations of the centre line at K and G remain constant. Just as in the case of beams fixed at the ends, it is exceedingly difficult to fix the arch at the ends so perfectly that we can rely upon this condition being fulfilled ; and hence, on account of our uncertainty, we prefer to use arches hinged at the ends. 3. The simplest case of all is that in which there are three hinges in the arch, one at each abutment K and G, and another at the crown. Even when the loads are not vertical it is easy to find the line of resistance, because two corners and a point in it are given, and we have merely the exercise of Art. 106. Any system of oblique loads is given, acting upon an arch whose end hinges are K and G and whose crown hinge is L. Find the resultant of all the loads from K to L, and let it be A B Fig. 272. 480 APPLIED MECHANICS. (Fig. 273). Find the resultant of all the loads from L to o, and let it he BC. Draw AB and B c in the force polygon (Fig. 274). Choose any pole o. Join o A, OB, and o c. Draw K P, p Q, and Q u parallel to o A, o B, and o c. The resultant of A B and B c passes through s. Now join o s and draw the new diagram, c o' parallel 'C Fig. 274. Fig. 273. to Q s, and join o' B ; so that K p, p Q', Q' G correspond to o' A, o' B, and o'c. We have then a link polygon. But it does not pass through L. Produce P Q' to meet K G in T. Join T L and let it meet the given forces in p" and a". Join K p" and Q" G, and K p" Q" o is the link polygon or line of resistance required. I set the problem as an exercise for students, and one of them, Mr. Stansfield, gave me this solution. If the loads are vertical, and K, G, and z are the given hinges, Fig. 272, draw any line of resistance K z' G', its corners z' and G' being in the verticals from z and G. K z" G' is a straight line. Draw a vertical 2.' z x z" through z. Construct a figure on the base K G such that if B' is any point in it, B' c is in the same proportion to B D that z x is to z' z". This will evidently be the true line of resistance, Pigr. 275. and the vertical ordinates between it and the centre lines of the portions of the arch will be the bending moments. One example of each of these must be worked out by each student. 381. Arch Hinged at the Ends. Imagine the small sliw of APPLIED MECHANICS. 481 the arch between two sections B o c and u' o' (/, at the small distance 8s asunder, to yield, and study the effect of the yielding of this portion alone, exactly as we did with our spring in Art. 370. That is, we imagine the part o' a (Fig. 275) to he absolutely rigid and fixed ; the part o K to be rigid, hut to move ahout o as centre, as if o K were merely a pointer, the motion K K" being the angular motion of B c relatively to B' c', multiplied by the straight distance o K. The angular change from o to o', if o o' is 5s, is 8s, if M is the bending moment s i the moment of inertia of B c about o through its centre of gravity, and E is Young's modulus ; so that K K" is 5s . o K. The horizontal component of this is K K."' or K K" cos. K" K K'". But this angle is the same as K o N, whose cosine is o N/O K. Hence the horizontal motion of K due to the yielding of the little slice is ON or (1). EI OK El And as x does not yield at all, we make this sum zero. We beg to point out that we cannot in the same way state the vertical displacement of K. If o N is called y, and if K N is called x, then (1) is 5 y . 8* = ; and we might in the same way think that, M as x . 8s is the horizontal displacement of K due to the yielding 30 1 of the slice ; therefore the sum of all these terms ought also to be 0. If the end G is fixed, this is true, and it enables us to calculate the fixing moment. But in truth 5 x . 8s is not if G is hinged ; E I * it is equal to iilie angular movement at o multiplied by K G. Note that for the horizontal displacements the angular yielding at o produces no effect. 482 APPLIED MECHANICS. Since M is represented to scale by the distances OL, (1) becomes oi" ON. 5 = o .... (2). In Fig. .776 K z G is the centre line of the arch of which K and G are the hinges. K z' G is the line of resistance which we want to find. If a student wishes to keep (2) in his memory, let him imagine that the line K z G has positive and negative density, represented by M/B i per unit length. Then (1) or (?) tells us that if the whole positive part of the linear mass ia m Vi and if y p is the distance of its centre of gravity from K G, and if the whole negative part of the linear mass is m n> and if y n is the distance of its centre of gravity from K G, then m p y p = m n y n . To find K z' G, let K z" G be any -line of resistance through K. Q drawn at random. Then L N = k . p N, where Jc is an unknown constant; and if we know k, knowing p N, we can find L N. Divide the given centre line K o z G into any number of equal parts, and draw an ordinate N o p at the middle of each of them. All values of 5* are now ON.OL ON (LN-ON) equal, and (2) is 2 - - - = .... (3), or 2 ' = ; so that, as L N is k . p N, we have k 2 ON ' PN = 2 .... (4), and Jc is evidently 2 -- -f- 2 .... (5) This is easily calculated. Thus, for example, take K. z G, an arc of a circle, span 200 feet, rise 30 feet. For a given system of unsymmetrical loads, K P z" G was drawn. Dividing K z G into sixteen equal parts and raising perpendiculars at the middles of the parts, we found the following values. The curves as drawn were measured in inches, no attention being paid to scale. The values of i are in inches to the fourth power. We see that k = -03014 -f- -07447 or 0-405. Multiplying each value of P N by this, we find the values of L N. The horizontal thrust corresponding to this true line of resistance is greater than the one for K. z" G in the ratio 1 : -405, and its force polygon may be drawn. The student needs no further hints for the completion of the calculation. The loads taken were as follows : APPLIED MECHANICS 483 s g M O COOO5<>t^O500*0 i-l'OO I! 8 * H* i&DQ f-i * **< Oft-O O Cft * <* t-l QD >C v-4 OOOt I i i i (i IC^C^" li ii li IOOO OOOOOOOOOOOOOOOO OOOOOO OOOO CD 00 . sJ OOOOOOOOOOOOOOOO oooooooooooooooo CO CO O CO t~- i i CO GO O5 O5 O5 O O O O O O i I o o o o CO .-I 00 CO O CO CO O O O5 O5 O5 CO CO i-H r-l O O O O O o o o o o o o T C -^ A (A\ . m l5 - t . w. (5) becomes We also now find another condition by stating that the sum of the vertical displacements of K is when G is held fixed that is, 2 K N . 5s = . . . . (6), and this gives us . 2 and the summations indicated in (4), (5), and (6) being effected, we have no difficulty in solving for k, tn v and P, and so finding the true line of resistance. The table shows these summations. It ia for the same arch ring, with same loads, as in Art. 381, and therefore some of the columns are the same. The student will note that if y 1 is the ordinate of the centre line of the unloaded arch at any place, and if y is the ordinate of the loaded arch, and if y l - y, the downward motion of the point, be called z, then, unless in some case there is very great slope indeed, we may take -* I 1 + \~r) j as the change of curvature. The proof of this is easy. Equating this to M/E i, we see that we have only to imagine i at every place divided by -j- ) [ a there, calling it by the new name i 1 , and we take -r4 = M / E jl Having a diagram showing everywhere th-e value of M/E i, it is easy to obtain the diagram of z by graphical statics, just as in Fig. 217 we obtained the shape of a beam from a diagram Of M../EI. 486 CHAPTER XXIII. MEASUREMENT 0*F A BLOW 383. IN Art. 46 we considered what occurs when a chisel is cutting metal. We shall now consider the same subject from a point of view which at first seems different. Consider how it is that a blow of a hand hammer will indent a steel surface, whilst a steady force applied to the same hammer-head would require to be very great to produce any indentation. The pressure between the hammer and the steel is very great, and it must be all the greater because the time of contact is very short. Indeed, if a hammer weighs 2 Ibs., so that its mass is 2 ^ 32'2 or -0621, and if, just before touching the steel, its velocity was 10 feet per second, then we know that its momentum was 10 x -0621, or -621. Now, if one-ten-thousandth of a second elapses from the time of actual contact until the hammer's motion there is destroyed that is, until the elasticity of the steel is just about to send the hammer back again a little the momentum -621 is destroyed in '0001 second ; hence the average force (notice that it is a time average) acting between hammer and steel during this short time must have been '621 -^ -0001, or 6,210 Ibs. It is certain, however, that this average force is less than what the force actually was for some very small portion of the time. We observe, then, that we cannot tell the average force of an impact unless we know two things the momentum and the time in which it was destroyed. Now the duration of an impact depends greatly upon the nature of the objects which strike one another, and we see that the average force of a blow is less as the time is greater. Sometimes, instead of a great force acting for a very short time, what we require is a smaller force acting for a longer time. For instance, when cutting wood we obtain this result by using a wooden mallet and a chisel >vith a long wooden handle, because the force required to make the chisel enter the wood is not very great, and we wish this force to act for some time, so that much wood may be cut at one blow. In chipping, we have the time short, because considerable force is required to cause the chisel to enter metal. The duration of an impact depends on the APPLIED MECHANICS. 487 shapes of the bodies and their masses, and on the elasticity of their materials. 384. Why is it that in driving a nail into wood our blows seem to be of no effect unless the wood is thick and rigid, or un- less it is backed up by a piece of metal or stone ? It is because the wood yields quite readily, and so prevents the hammer losing its momentum rapidly. There are few subjects in which people are so apt to have erroneous ideas as in this one of impact. Thus a man will speak of the force produced by a weight falling through a height without having any idea of the time during which the motion of that weight is being stopped in fact, without considering what time the weight is allowed for delivering up its momentum. Now, a little consideration will show that the mean force of the blow will be quite different according as the weight falls on a long and yielding bar or on a short and more rigid one. If we could imagine bodies to be formed of perfectly unyielding materials, then the slightest jar of one against the other would produce an infinitely great pressure between them ; and in the blow produced by a falling body there may be every gradation from exceedingly great pressures to very small ones, depending on the yielding power of the body that is struck. Everybody is acquainted with the sensation produced by suddenly placing one's foot on a level floor when one was preparing for a step downwards. The downward momentum of the body is suddenly destroyed, and there are great pressures in all the bones of the body. Carriages are hung on springs for the purpose of preventing their losing or gaining momentum with too great rapidity when the carriage wheels pass over obstacles. When we are sitting on a hard seat in a third-class railway compartment, -and the carriage gets a slight jerk upwards, momentum is given much too rapidly to our bodies for perfect comfort, and to sit on cushioned seats is preferable. A cannon- ball is safely, because comparatively slowly, stopped by sand- bags or bales of cotton. 385. Example. A pile driver of 300 Ibs. falls through a height of 20 feet, and is stopped during O'l second. What average force does it exert upon the pile 1 A body which has fallen freely through a height of 20 feet has acquired a velocity equal to the square root of 64-4 x 20, or 35-89 feet per second. Its momentum is 35 -89 x 300 -=- 32-2, or 334-4, and this divided by O'l gives 3,344 Ibs., which, together with the 488 APPLIED MECHANICS. weight 300 Ibs., is 3,644 Ibs., the answer. From the instant when the motion of the driver ceases to diminish, the force exerted by it is its own weight. We have considered the time average of the force. The average force of friction in pounds between the pile and the ground, multiplied by the distance in feet through which the pile descends during the stroke, is equal to 300 x 20, or 6,000 foot-pounds, if we neglect the loss due to vibrations of the body and the energy carried off to the ground to be wasted in earth vibration, and if we also ignore the fact that the weight really descends a little farther than 20 feet. The neglected energy may be the whole of the energy if, for example, the blow does not make the pile move further into the ground. We can only be certain about a time average force, and even in this case we must assume that there is an instant at which one of the bodies has the same average velocity in all its parts. 385a. Example. A column of water in a pipe 6 inches diameter, 30 feet long, was moving behind a piston at 15 feet per second. The piston's motion is stopped in 0-1 second. What is the time average of the pressure due to the stoppage ? Ans. The area of the circular section of the column of water being 28*274 square inches, the quantity of water in motion is 30 x 28-274 -=- 144 or 5-89 cubic feet, or 5-89 x 62-3 or 366*9 Ibs. ; and on the assumption that all the water has exactly the same motion, its momentum is its mass 366*9-^32 -2 multiplied by 15, or 1-70 units. The time average of the force required to stop the motion in O'l second is therefore (62-3 x 30 x 28-274 x 15) -r (144 x 32-2 x 0-1) Ibs., which is equivalent to a pressure of 60-5 Ibs. per square inch over the area of the piston. 386. Suppose a body A to strike another B, and that we can neglect the actions of outside bodies upon them both. If A loses momentum, B must gain the same amount because their mutual pressures are equal and opposite during the time of impact. It is our knowledge of this fact that enables us to calculate the motions of bodies after they strike one another. Again, for the same reason, if from any internal cause the parts of a body separate from one another, either violently or gently, the total momentum remains as it was ; it is only the relative momentum which alters. Hence, when a shell bursts in the air, some parts move in the same direction more rapidly than before, but others less rapidly ; one part may double its APPLIED MECHANICS. 489 velocity and another may drop nearly vertically, its forward motion being stopped, but, on the whole, the total forward momentum is what it was originally.* 387. Examples. If a cannon were perfectly free to move backward when the shot leaves it, the backward momentum of the cannon would be exactly equal to the forward momentum of the shot. Thus, if a shot of 20 Ibs. leaves a cannon whose weight is 2,240 Ibs. with a velocity of 1,000 feet per second, the velocity of the cannon backward would be 1,000 x 20 -f- 2,240, or about 9 feet per second, neglecting the fact that the gases leave the gun also with a certain momentum. When a ship fires her broadside, each gun runs back, communicating, as it is stopped, its momentum to the ship, which heels over in consequence. A gun firing the above shot of 20 Ibs. directly astern from a ship whose total weight is 600 tons gives to the ship (neglecting the momentum of water moving with the ship) so much momentum that its speed is increased (neglecting her friction with the water) 9 -4- 600, or '015 foot per second. We see, then, that a ship might propel herself by means of her guns. The steamship Waterwitch had power- ful steam pumps, wherewith she brought a great quantity of water in nearly vertically, and sent it out backwards on the two sides below water level. The momentum given to the water backwards was equal to the momentum given in the other direction to the ship. It is on this principle that Hero's steam-.engine and Barker's mill work, the momentum given to jets of fluid passing out of certain pipes being equal to the momentum given in an opposite direction to the vessel from which the fluid passed. In all such cases the propelling force in pounds is numerically equal to the momentum of the fluid which passes out in a second. Thus, if from a vessel moving with a velocity of 14 feet per second water comes through orifices of 4 square feet in area with a velocity (relative to the orifices) of 20 feet per second, then the quantity passing out in one second is 4 x 20, or 80 cubic feet that is, 80 x 6 2 '3, or 4,984 Ibs. Now, recollecting that this water was first brought in and is now sent out, what is the velocity which we have really impressed upon it in the process 1 At the beginning it * Of course the kinetic energies of the parts of the shell added together are greater than they were before the shell burst ; we are now merely speaking of the momentum. The total momentum of two equal bodies going in opposite directions with the same velocity is nothing^ whereas their total kinetic energy is double that of one of them. <** 490 APPLIED MECHANICS. was motionless with respect to the sea ; it now has a velocity of 2014, or 6 feet per second with respect to the sea, so that the momentum given to it is its mass, 4,984-4-3 2 -2 multiplied by 6, or 928*7 : hence, as this momentum is given every second, 9 28 '7 Ibs. is the propelling force exerted on the ship. In one second the ship moves through 14 feet, so that the useful mechanical work done is 14 x 928 '7, or 13,002 foot- pounds. We have given to 4,984 Ibs. of water a velocity of 6 feet per second, the kinetic energy of this water is wasted, and this kinetic energy is J of 4,984^32-2x6x6, or 2,786 foot-pounds. In fact, we have altogether spent 15,788 foot- pounds, and 13,002 of this have been usefully employed, so that the efficiency of the method is 13,002 -r 15,788, or -824, or 82 -4 per cent. As a matter of fact, however, the friction in pumps and pipes usually causes a third of the actual horse- power given out by the engine to be wasted, so that the true efficiency of this method of propulsion is two-thirds of the above, or 0'55, or 55 per cent., neglecting the friction of the engine itself. You will remember a fact which has come in casually here : if the water leaves any turbine, water-wheel, or any propeller of a vessel with a velocity re]ative to the still water into which it passes, or if it has any other form of energy, this energy has been wasted. 388. By calculation you will find that, when two free and inelastic bodies strike, the momentum communicated from one to the other is their relative velocity multiplied by the product of their masses and divided by the sum of their masses, and this quotient divided by the time of the impact gives the mean pressure. This pressure acts equally on both, of course, but it may not hurt both equally. If the bodies are surrounded by water, like ships, they can no longer be regarded as free bodies, and it is not easy to say in a few words how much mass we must add to the bodies to represent the mass of the water, which has also to undergo change of motion. In the case of a ship, the mass of water to be moved broadside on is much greater than when the ship is struck stem 011. 389. A body falling into a liquid sets it in motion, and this motion appears at distant places more and more nearly instantaneously as the liquid becomes more and more incom- pressible. The nature of this motion is known to us if we know the velocity of yielding at the place of contact, and from this the total momentum given to the liquid. This APPLIED MECHANICS. 491 represents a very considerable pressure applied at the place of contact, and this pressure becomes greater as the velocity of the body, before it touches the liquid, increases. Hence a cannon-ball fired at sea rebounds from the water as from a rigid body. Hence also a man diving unskilfully, as he falls prone on the water, gets a very unpleasant shock, whereas a skilful diver enters in such a way as to make the momentum of the moving water as small as possible, and to make the creation of this momentum gradual. 390. If a body of inertia or mass M with velocity v overtakes a body of mass M 1 with velocity v 1 , the motion being in the same direction, there is an instant when they move at the same speed, and then they usually separate. The equal and opposite forces equalise their velocities until they are both moving with the velocity v, such that (M + M 1 ) v = M v + M 1 v 1 . . . . (1), or MV + Miyi 9 IFT^ (2)< There has been a communication of the momentum, M (v v) t or M 1 (v v 1 ), and this is the amount of the impact or the time integral of the force from either on the other. We usually imagine the contact surfaces to be normal to the direction of motion.* During the impact the total kinetic energy, EI = M v2 + jMiyi 2 (3) becomes lessened to B = J (M + Ml) . . . . (4), the amount EJ E being stored as strain energy. In truth, much of it travels off and vibrations take place (see Art. 486), but let us speak of E l E as the stored energy. Assume that the energy of the bodies when free after collision * If not normal we may speculate on the connection between the ideas of force as rate of communication of momentum, the direction of the force being the same as that of the momentum communicated, and yet the direction of communication being different from both. Students who have leisure will find three quite neglected letters in Nature for 1878, by Prof. R. H. Smith, which will give them novel ideas on the subject ideas likely to be of use to engineers 492 APPLIED MECHANICS. O C) E 2 = | (M u + M 1 u 1 ) is less than B X by an amount which is a fraction of the stored energy, 01 E l - E 2 = k ( El - E) ---- (5), where k is a constant depending on the nature of the materials and the shapes of the bodies. We know also that . .. M + M 1 From (2), (5), and (6) we can calculate u, u 1 and v in terms of the initial velocities, and we find that Or, " the relative velocity of separation is v/ 1 k times the previous relative velocity of approach." This was Newton's assumption. It is usual to denote the ratio J \ k by the letter e, and to call e the " elasticity," but this is unscientific. 391. Newton found by experiment that e = f for com- pressed wool, iron nearly the same, ^J- for glass. Hence for these substances our k has the values -^ and -J. That is, in the compressed wool or iron -^ of the stored energy is wasted ; in glass only -J of the stored energy is wasted. It is very difficult for us to imagine how this difference between iron and glass should occur ; and there are no recent experiments. Note that k = 1 or e = indicates the case of maximum loss of energy ; in fact, there is maximum loss of energy when the two bodies continue to move together as the bullet and bob of a ballistic pendulum do. In this case the kinetic energy before collision is J M v 2 , and after collision it is I (M + Mi) ' or i (M y + M i yi)V(M + M'). Suppose v 1 = 0, or we have M impinging on M 1 at rest, the energy remaining is \ M v 2 III -\ -- j. Hence, the greater M 1 is in comparison with M the greater the loss. In fact, lost energy _ Mj remaining energy M So that the larger the stationary body before collision, the greater the loss. 392. Example. Thirty gallons of water per second enter a wheel in a direction A B from a horizontal pipe 4 inches in APPLIED MECHANICS. 493 diameter, the shortest distance from the axis of pipe to vertical axis of wheel being 1*3 feet. The water leaves the wheel in a horizontal direction, c D, with a velocity of 3 feet per second, the shortest distance between c D and the axis of wheel being 0-8 feet. What is the turning moment? And what is the power if the wheel makes 150 revolutions per minute? Ans. The 30 gallons, or 300 Ibs., or -^^ cubic feet, have 300 144 an initial velocity of ^3 42 x -7854 ^ eet P er secon( *> an( ^ 300 mass ojTo. '^ ie product is the momentum per second, and is force. Multiply by 1 '3, and we have the moment of the force due to the entering water, or 668-2 pound-feet. The same mass multiplied by 3 and by 0-8 gives 22 -36 pound-feet; and subtracting this from the former, we have the resultant moment (or moment of momentum per second), 645 -8 pound- feet. Multiplying this by 150 x 2 IT radians per minute, and dividing by 33,000, we find 18-4, the horse power. 393. Example. A ball of 4 Ibs., moving to the north on a smooth, level table with a velocity of 6 feet per second, strikes another ball, and after the collision is found to be moving at 5 feet per second to the east. What was the amount of the impulse? If the collision lasted 0-002 second, what was the average force of the blow ? Ans. Subtracting the second velocity (in the way vectors are subtracted) from the first, we find that the sudden gain of velocity was \/~61 -=- ' 002 f b -> or 485 Ibs. EXERCISES. 1. Compare the amounts of momentum in a pillow of 20 Ibs. which has fallen from a height of 1 foot, and an ounce bullet moving at 200 feet per second. Ans., 12-8 to 1. 2. A ball of 56 Ibs. is projected with a velocity of 1,000 feet per second from a gun weighing 8 tons. What is the maximum velocity of rocoil of the gun ? Ans., 3-1 feet per second 494 APPLIED MECHANICS. 3. There are two bodies whose masses are in the xatio of 2 to 3, and their velocities in the ratio of 21 to 16. What is the ratio of their momenta ? If their momenta are due to constant forces acting on the bodies' respectives for times which are in the ratio of 3 to 4, what is the ratio of these forces? Ans., 7 to 8 ; 7 to 6. 4. A body weighing 10 Ibs. impinges on a fixed plane with a velocity of 20 feet per second. If the coefficient of restitution e is 0-5, find the velocity of rebound, and how many foot-pounds of energy are wasted in the collision. Ans., 10 feet per second; 93-17. 5. A ball of 6 ounces strikes a bat with a velocity of 10 feet per second, and returns with a velocity of 30 feet per second. If the duration of the blow be ^V second, find the average (time) force exerted by the striker. Ans., 9-32 Ibs. 6. A body weighing 50 Ibs. moving at the rate of 10 feet per second overtakes another body of 25 Ibs. moving at the rate of 6 feet per second. If both masses be perfectly elastic, find their velocities after the shock. Ans., 7 3 and 11^ feet per second. 7. A hammer-head of 2| Ibs., moving with a velocity of 50 feet per second, is stopped in 0-001 second. What is the average force of the blow ? Ans., 3,882 Ibs. 8. A shell bursts into two fragments, whose weights are 12 and 20 Ibs. The former travels onward with a velocity of 700 feet per second, and the latter with a velocity of 380 feet per second. What was the momentum of the shell when the explosion took place ? Ans. t 496-8. 9. A shell weighing 20 Ibs. explodes when in motion with a velocity of 600 feet per second. At the moment of explosion one-third of the shell is reduced to rest. Find the momentum of the other two-thirds. Ans., 372-7. 10. Water is flowing through a service pipe at the rate of 60 feet per second. If the water be brought to rest uniformly in one-tenth second by closing the stop-valve, what will be the increase of the pressure of the water near the valve, the pipe being taken as 50 feet long, the resistance of the pipe and the compressibility of the water being neglected ? Ans., 403 Ibs. per sq. in. 394. As for the way in which the vibrations take place in two colliding bodies, if mathematically treated it is generally difficult; but very good working notions for the engineer are derivable from the results of experimental study, the subject being taken up in books on physics under the head " Acoustics." The best mathematical treatise is Lord Ray- leigh's " Sound." The time of an impact is known in several cases in which it has been calculated from theory. (See Art. 404.) 395. Let us consider what takes place when two ivory balls come together. There is a certain instant after they first touch when their centres move together just as if they were composed of soft clay then they act on each other with their greatest pressure ; they are in their most strained condition, APPLIED MECHANICS. 495 and supposing no loss by internal friction, the strain in the balls represents an amount of stored-up energy (see Arts. 259, 267) equal to the kinetic energy which the bodies have lost. It is very important to remember this fact, that if bodies are to return to their old states after the collision, we must suppose that during the collision there is a storage of kinetic energy in the form of strain. All the kinetic energy will not be given out again, nor can we say that it is all stored, because there is a sort of internal friction causing part of the strain energy to be converted into heat when any change occurs. Now, if the whole of the stored-up energy is confined to one portion of the body, the strain may be too great. Thus, a steel rod 1 square inch in section, 1 foot long, will store up 167 foot-pounds of strain energy in its stretched condition before it breaks. For suppose breaking stress to be 100,000 Ibs. per square inch. This will occur when there is a length ening or shortening of -0033 foot, so that the energy stored up is the work done by a force whose average amount is 50,000 Ibs. acting through -0033 foot, or 167 foot-pounds. If 2 feet of the same rod stored up this same amount of energy, there would only be 83 foot-pounds in each foot of its length ; and it is easy to see that the stress is no longer the breaking stress of 100,000 Ibs. per square inch, but only 70,700 Ibs. per square inch. As we store the same amount of energy in smaller and smaller portions of a body, it is evident that we must approach a condition of fracture. 396. We see, then, that at the place where contact occurs, two bodies, A and B, are strained ; but if A is of some very elastic material, such as tempered steel, the strain energy is conveyed very rapidly to every part of the body ; whereas if B is a feebly elastic body, the strain accumulates at one place, leaving the rest of the body unstrained, whilst at this place the strain may produce fracture. This slowness to communi- cate strain to the rest of the body may also be produced by the shape of the body. For instance, a rod struck sidewise or a thin plate struck in the middle does not so immediately com- municate its strain to the remote parts as a rod struck end- wise. Again, the nature of the parts of A and B in contact may be such that not only does the strain energy leave this part of A rapidly, but immediately in the neighbourhood of the place of contact there is a greater capacity to bear strain energy without rupture than is the case with B. Thus, when 496 APPLIED MECHANICS. ship A rams the broadside of ship B, the side of B is bent inwards and the strain energy produced is accumulated near the place of contact till fracture occurs ; whereas, not only is A'S stem able to transmit to all parts of A with great rapidity the strain energy which must be stored up in the whole mass, but at the stem itself the material of A is capable of with- standing greater stresses than the material of B'S side. Suppose, however, that A'S stem is not of steel, still B'S iron or wooden side will be perforated if A has enough velocity ; A'S stem may also be damaged in the impact in such a case. 397. A candle may be fired, it is said, through a thin deal board with very little injury to its shape, and the usual explanation of this phenomenon given in books is that the candle has not time to get broken. This explanation is not satisfactory ; it is a little too vague. If we had the board in rapid motion, and striking the candle in the same relative position, the candle having previously been at rest, would the candle perforate the board ? There cannot be any doubt that it would. Hence it is not the body struck which must in every case get hurt ; the pressure on one is equal to that on the other. Suppose ship A rushes at ship B when B is broad- side on, and rams her, B will probably be sunk, even if she is a much larger and better ship than A. But suppose that B is able to meet her adversary stem to stem, if they are equally strong they will equally injure one another, and if B is the stronger A will suffer the most. This case differs very much from that of the candle, because we can assume greater strength even for slowly applied pressures from stem to stern of the ship A than from side to side of B ; whereas the strength of the candle for slowly applied pressures cannot be compared with that of the wood which it punches from the board. What is meant by the usual explanation, " The candle has not time to get deformed " ? Why has not the soft candle time to get broken, and yet the wood has time to get torn asunder 1 The fact is, the wood, if it were slowly pressed, would communicate its strain energy to every part of the board and its supports; but this communication takes an appreciable interval of time, however suddenly the pressure may be applied, or however great it may be. As the strain energy is rapidly produced it becomes accumulated near the place of contact to such an extent as to produce fracture of the wood. Now the point of the candle is subjected to the same APPLIED MECHANICS. 497 pressure as the wood, and begins to get spoiled in shape that is, it is compressed and this compression produces a lateral spreading. In the meantime, however, the compressive strain energy is communicated very rapidly backwards along the candle, and the spreading and spoiling goes on along its entire length, but is small at any point, since it is distributed over the whole mass. Practically, therefore, the spoiling occurs only at the point of the candle, since time is needed for fracture of the material. 398. An earthquake, when it acts on a house, usually tends to move it through a distance of probably a very small fraction of an inch, but it does this in a very short time that is, the house gets a considerable velocity. The mass of the house multiplied by the greatest velocity, and divided by the short time during which the momentum is being communi- cated, gives the pressure which the foundations of the house are subjected to. Now, when the foundations are not very rigidly connected with the ground, the time of communication of the momentum is lengthened, and the pressure is conse- quently diminished. This is the usual Japanese plan of providing for earthquake effects. Unfortunately, the very means taken to dimmish the pressure on the foundations also diminishes their capability of withstanding forces, and it has not yet been decided what sort of a house is best fitted to with- stand destructive earthquakes. Want of rigidity, combined with strength or toughness in the materials, and especially the quality of internal friction in the materials, so that vibrations may rapidly die away these are the qualities needed. They are found in steel, wrought iron, and wood, and especially in wicker-work, in a less degree in east iron and in brick or stone set in cement, and less still in brick and stone set in bad mortar. 399. When you in some way understand the possibility of a candle perforating a board, you will be able to comprehend how sand, when blown in air against tempered steel, is able to abrade it how the emery wheel and grindstone going at great velocities are able to cut into hard metals ; and how in Cali- fornia a jet of water going with very great velocity is used for mining purposes instead of iron tools. 400. Quasi-Rigid ity Produced by Rapid Motion. A top when not spinning can with difficulty be balanced on its point, and if left to itself it almost instantly falls ; whereas when it 4:98 APPLIED MECHANICS. is spinning the effect of slightly tilting it out of the perpen- dicular is not to make it fall, but to make it take a slow precessional motion. There is a piece of apparatus called a gyrostat, which is, in a more or less perfect form, to be found in every mechanical laboratory, and the student ought to experiment for himself with this apparatus on the curious effects of quasi- rigidity which manifest themselves in tops and other spinning bodies. If he has a slight acquaintance with astronomy he will be interested in tracing the connection between the behaviour of a tilted top and the ^precession of the earth's equinoxes. When a circular sheet of drawing-paper is mounted like a very thin grindstone on an axis, and is gradually made to rotate rapidly, it is found to have become quite rigid that is, it greatly resists bending as if it were made of steel. In the same way a long loop of rope, hanging round a high pulley, which gives it a quick motion, takes a certain form which it is very difficult to alter, as may be shown by striking it with the hand or with a stick : it resembles more a rigid rod than a flexible rope. Again, in the well-known lecture-experiments on smoke rings, we see that these little whirlpools of air have many properties in common with elastic solid bodies on account of the partial rigidity which is due to their rapid motion. In objects which spin and rub on a level surface, like tops, we have the interesting general rule, "Positions which are stable when there is no spin, are unstable when there is spin, and vice-versd." Students of .statics insist on a low centre of gravity in vehicles ; students of dynamics sometimes insist on a high centre of gravity, as giving greater stability when there is rapid motion. It would be beyond the scope of a book like this to explain these curious phenomena, and I merely direct the attention of students to these instances in order to incite them to make experiments, and to seek for the explanation of what they observe. My popular lecture on Spinning Tops may be worth reading. 401. Motion Produced by a Blow. When a body sub- jected to a blow is quite free to move in any way, unless the blow acts through its centre of gravity, the body will not merely move as a whole, but it will revolve. When the blow acts in a direction through the centre of gravity there is no APPLIED MECHANICS. 499 rotation produced. It is usual in such a case to consider the motion of the centre of gravity of the body, and the motion of the body about an axis through its centre of gravity, for it is known that any motion whatsoever of the body consists of a combination of two such motions. It is found that the kinetic energy communicated to a body by means of a blow i best calculated in the following way. if we know the nature of the motion ; First, find the "kinetic energy, a* if every portion of the body had the motion of the centre of gravity. Secondly, find the kinetic energy of rotation as if the axis of rotation through the centre of gravity were fixed in space. Add these two re- sults together. We may regard from another point of view the instan- taneous motion of a body when it is struck namely, as a rotation about some axis which does not itself move. This is only the case for an instant; immediately afterwards we must re- gard the body as moving about another fixed axis. If a body is hinged so that it can only move about a fixed axis, it is always possible to find the point at which the body may be struck, and the direction of the blow, which will tend to produce an instan- taneous rotation about this particular axis, and therefore to produce no pressure at the hinge. Thus the ballistic pendulum of Fig. 278 is always struck in such a way, and the point in which it is struck is called the " centre of percussion." An easy way to find the centre of percussion is as follows : Make the body vibrate like a pendulum, about its axis of suspension, under the action of gravity. Now find the length of the equivalent simple pen- dulum. This is the distance of the centre of percussion from Fig. 278. 500 APPLIED MECHANICS. the axis. In a tilt hammer all blows ought to be delivered from this centre of percussion if we wish to have no pressure on the bearings. A cricket-bat or a rod of iron tingles the hand when we strike a blow with it, unless we happen to strike at the centre of percussion. For a rod of iron free to move about one end, the centre of percussion is at two-thirds of the way towards the other end. (See Art. 454.) 402. The Ballistic Pendulum of Fig. 278 is a contrivance which enables us to measure the velocity of a bullet. It con- sists of a mass of wood, A, forming part of a pendulum. The bullet is fired into it, and the wood swings backwards in con- sequence. The bullet is fired in such a way that it will cause no jar to be given to the pivot B. The momentum existing in the bullet before it enters the wood belongs now to the whole mass of which it becomes a part, c is a silk ribbon which is pulled through a moderately tight hole or over the edge of a table by the swing of the pendulum, and the length of ribbon pulled through is found to be proportional to the momentum of the bullet before entering the wood. If w is the weight of the bullet, its horizontal velocity in the plane of swinging being v, its momentum is v. The moment of momentum x - v (if x is the perpendicular distance from B to the axis of the bullet) before impact is equal to the moment of momentum after impact ; and if i is the moment of inertia of A and the bullet ahout B after collision, and a is its initial angular velocity, then i a = x v .... (I). A certain part of the whole kinetic energy has "been conyerted into heat ; the mechanical energy now in the system, 3 i 2 , will be converted into potential energy, w being the total weight, the centre of gravity will be raised through the distance h such that w A = -| i 2 . . . . (2) ; so that, as a is known, h is known. Knowing h, we can either find 0, the total swing, or we can calculate or find experimentally the length of the ribbon which may be drawn up, and from the length of the ribbon we can calculate v. In the short time when the bullet is being lodged it is exercising horizontal force at each instant, and there may be horizontal force at H in the same direction. We may say that these external forces, acting at the centre of gravity, arc balanced by the mass of the whole body multiplied by the acceleration of the centre of gravity at each instant, and hence their integral effects balance , that is, then, the impulse v -f p (where P is the horizontal impulse from APPLIED MECHANICS. 501 the point of support in the same direction) = - a . B G if G is the centre of gravity, hecause a . B G is the velocity produced in G, and w is the weight of the pendulum and bullet Hence w w p = - a . BG - -v . , . . (3), 9 9 This is 0, therefore, if W..BG = >;.... (4). But by (1), (4) means that w . a B G = w# 2 0/a;, or x & 2 /B G. Now this is the rule (Art. 454) by which we find the length of the equivalent simple pendulum or the distance to the point p of oscillation (Art. 454). It also for this reason gets the name " point of percussion." APPENDIX TO CHAPTER XXIII. 403. In the Example, Art. 385a, we had an example of stop- page of water in a pipe as if all the water had exactly the same motion. It was an interesting academic exercise. The following exercise is also interesting, and will give us more information. Sudden Stoppage of Water in a Pipe of Uniform Section. Suppose the pipe to be infinitely rigid. It will be found that the effects are independent of whether the pipe is vertical or horizontal, but we shall consider it to be vertical. It will be found on closer examination that if we know the pressure we need only con- sider, in our problem, p, the pressure above the normal pressure, and this is what we shall do. Let v be the axial velocity of a particle of water, w the weight per unit volume, so that w\g is its mass per unit volume, often called p; if K is the cubical elasticity (during quick changes of volume) and ^ t is time; if x is distance to. a point o, measured along the pipe in the direction of motion, and if the pipe is taken of unit cross- sectional area ; if x is the total distance already travelled by a particle at Q, so that at x + 1, x becomes x + ~ t fluid which once occupied length 1 of the pipe now occupies length 1 ^ so that its increased volume is . Hence, as pressure- dx dx producing volumetric strain = K x compressive strain, Now velocity v = -, , and as the mass between x and x + Sx is p . 8x and the pressures are p and p + tip, we have the force - Sp 502 APPLIED MECHANICS. . causing acceleration --y in the mass p . tiz, and hence d? x dp d?JL * = , . te . -jj, or - f = , . gr . . . . (2), rf 2 X ^X or, from (1), K ^ = p ^. If K/p or ~K.ff/w be called a 2 , we have and we recognise this as the eqxiation of wave propagation with the velocity a. Differentiate (3) with regard to t and use v for ,,d?v d?v dx/dt, and we have a -T- 2 = ,7/2 * Let our problem be this. When water is flowing with a uniform velocity - v , let an. infinitely rigid, thin diaphragm suddenly produce a stoppage at x = ; what will happen ? Con- sider the pipe only for positive values of x. We have v = - v everywhere at t = 0, and at x = we have v such a function of. the time that it was - v till t = 0, and ever afterwards v there is 0. To make things simpler, let v 1 be a new variable such that v l v + v , and therefore the conditions are that v l every- where at t = 0, and at x we have v 1 such a function of the time that it was till t = 0, and ever afterwards v 1 = v . We see that If we let - be denoted by & in the usual symbolical way, (4) is Solving this as the simplest linear equation would be solved if 6 were a quantity independent of x, we have if A and B are functions of the time corresponding to x 0. As x may be infinite, B = ; so that where / (t) is till t = 0. and th.an. is v^, and remains of the value V Q ; so that v = - v +f(t - */) ---- (6).* We see now what occurs. Until t - xja = or t = x/a, v - v at any place ; and after t = sc/a, v 0. Hence we have from = a state of no velocity spreading along the pipe with the * The symbolic c??/) means f(t + 6) APPLIED MECHANICS. 603 velocity a to infinity. Since v = -?-, if we integrate (6) in regard to time, we have x = - V t + F (t - xfa) ---- (7), where F is such that = / (y), whatever y may be. Hence, using (l),p = f(t - x/a); or, since we have This tells us thatj? = till t - x/a = 0, and then is A , and remains of this value. This is the state of pressure produced every- where with perfect rest accompanying it, at the velocity a from the place of the sudden stoppage if the pipe is infinitely rigid. If at x = I there is a place where the pressure is kept constant (we say p = there), at zero, our wave is reflected as a wave of velocity + v and no pressure, till on reaching x = it is again reflected as a wave of no velocity, and so on. As the pipe is not infinitely rigid, and as there is friction, the wave really diminishes in its values of velocity and pressure as it travels, but we may take a pressure approaching the value v */K.w\g as being instantly pro- duced by a sudden stoppage. Experiments with suddenly closed valves give measured pressures in the laboratory considerably less, partly because the stoppage is not one of infinite suddenness, partly because of the inertia of the pressure-measuring apparatus. In the case of water the pressure in pounds per square inch is 20 v if v is the velocity in feet per second. This great pressure produced by stoppage is taken advantage of in the hydraulic ram. If the bottom of the sea were smooth, and a sea with a translational velocity v were suddenly stopped on its motion by a vertical wall, the pressure on the wall would be what we have given above, for a very short time, if the wall were per- fectly rigid ; what would occur subsequently I do not know, be- cause there is atmospheric pressure at the surface of the sea. We know enough, however, to see the necessity for some springiness at the wall surface. It is for the same sort of reason that heavy seas produce so much damage sometimes when objects are struck by them on board ship. Curious stories are told by sailors of half-incl bars of iron being bent and broken by seas coming over the bul- warks, and it is just possible that they may be true, although it is more probable that such fractures are due to blows from passing wreckage. Anyone who has seen the ruined breakwater at Wick will believe in the greatness of the forces due to blows from ocean waves. * 404. Students will find it an excellent easy mathematical exercise to assume that in an infinite length of pipe filled with water there is a piston whose displacement is a pure function of the time. The 504 APPLIED MECHANICS. problem is identical with the simplest problem in Telephonic Signalling. If the pipe is supposed to yield and to be leaky, and if the viscosity of the liquid is considered, we have the same problem as that of the Philosophical Magazine, page 223, August, 1893. When a prismatic bar of length ^ and velocity v l in the direc- tion of the common axis overtakes a bar of length ? 2 (greater than J ; ), moving with a velocity v%, the bars being of the same material and cross section, Mr. Love says that the ends at the junction move with a common velocity \ (vi + v 2 ), and a compressive strain i ( v i ~ v %}l a i 8 produced, a is the velocity of a longitudinal wave of sound, or E = a?p where E is Young's modulus and p is the mass of the bar per unit volume, or our w/g above. Waves of compres- sion run from the junction along both bars, and each element of either bar, as the wave passes over it, takes suddenly the velocity (v l + v z ] and the compression \ (v\ - v 2 )/a. When the wave reaches the free end of the shorter bar it is reflected as a wave of extension ; each element of the bar as the wave passes over it takes suddenly the velocity which initially belonged to the longer bar and zero extension. After a time, equal to twice that required by a wave of compression to travel over the shorter bar, this bar has uniform velocity, equal to that which originally belonged to the longer bar, and no strain. The impact now ceases, and there is in the longer bar a wave of compression of length equal to twice that of the shorter bar. The wave at this instant leaves the junction, and the junction end of the longer bar takes a velocity equal to that which it had before the impact. The ends, therefore, remain in contact without pres- sure. This state of things continues until the wave returns reflected from the further end of the longer bar. When the time from the beginning of the impact is equal to twice that required by a wave of compression to travel over the longer bar, the junction end of the latter suddenly acquires a velocity equal to that origin- ally possessed by the shorter bar and the bars separate. The shorter bar rebounds without strain and with the velocity of the longer, and the longer bar rebounds vibrating. 605 CHAPTER XXIV. FLUIDS IN MOTION. 405. WE tried in Art. 145 to give exact notions on the subject of pressure in fluids not in motion. When we supposed that the weight of the fluid itself was insignificant, we found that the pressure on each square inch of surface touched by the water, and on each square inch of interface separating two portions of the fluid, was everywhere the same. One of the best methods of observing the pressure, any- where, is by inserting a pressure gauge. The gauge indicates the pressure per square inch at the part of the liquid to which a communicating little gauge tube penetrates. It always does so when there is no motion of the fluid at the point in ques- tion ; but unfortunately, when there is motion there, the introduction of the tube alters the motion there and more or less falsifies our measuremeht. We are now about to consider pressure in fluids in motion, and we approach our subject by first speaking of pumps. 406. A pump is a machine which gives energy to water that is, it can raise water to a height, giving it potential energy. It can force water into a vessel under great pressure, moving pistons that is, it can give to water pressure energy. It can set water in motion that is, give it kinetic energy. Reciprocating-pumps are either lift-pumps, or force-pumps, or combinations of both. In Art. 137 we described the force-pump used with hydrau lie presses. In principle this is the same as all other force- pumps. The feed and other pumps of steam-engines, and the pumping parts of pumping engines, are described in detail in works on those subjects. As to Lifting Pumps, in Fig. 279 we have a diagrammatic representation of ' the common village or house pump. The rod R, usually worked by means of a handle or lever, pulls the bucket, B, up and down. The bucket has openings arranged with flaps or other valves in various ways, so that fluid may pass upwards but not down- wards. At w there is another valve which opens upwards. First suppose there is only air between B and the well, j. Sometimes in the morning it is necessary to throw in some 506 APPLIED MECHANICS. water on the top of B to make it air-tight. As B descends, much of the air between B and w passes through B. As B is lifted it produces a partial vacuum below it air passes from H to the barrel, and water rises in H. As the pumping pro- ceeds there is more of a vacuum until water fills H and the barrel, in case B is not nearly 34 feet above the level in the well. 25 feet is probably the maximum height in house pumps. After this it is water that pasaes through B in every down stroke, and is lifted in every up stroke. When a higher lift is needed, we sometimes place a valve in the upper part of the barrel, or in the delivery pipe, to prevent the return of the lifted water. Fig. 280 is a dia- grammatic repre- sentation of a force-pump. ^jr^'lE:- Sometimes apiston kK ~~ " is used instead of Fig. 279. the plunger, P. A is an air-vessel. As more and more water enters at B, the pressure of the air becomes great enough to force the water up the delivery pipe, E. In this case the stream is not so intermittent. Un- fortunately the air is apt to dissolve in the water and get carried off at a rate which depends upon how much air is already in the water. Fig. 281 shows a double-acting force-pump. When the piston B moves to the right, water passes through the valves, A, to the delivery pipe, D, and enters the barrel through D from the suction pipe, s, and well. When B moves to the left, c is the delivery valve, and E is the suction valve. Fig. 2&2 shows a combined force and lift pump whose action is very easy to understand. The delivery valve is not shown. Fig. 280. APPLIED MECHANICS; 507 407. A pump must be efficient that is, it must do nearly as much work on water as the pump itself receives from an engine or labourers. But it must be remembered that the best pumps used for different purposes have very differ- ent efficiencies. Forty per cent, would be regarded as a reasonable efficiency for reciprocating pumps of low lift, Fig. 282. | S 1 Fig. 281. whereas it would be regarded as rathei ==. poor for pumps of high lift. Example. (1) Suppose that when a pump is delivering a certain quantity Q = 1,000 gallons per minute, there is a loss of 35 foot-pounds per pound of water in the pump and horizontal pipes, and 0-05 foot-pound per pound for each foot of vertical pipe. Calculate the total loss per pound in each of the following cases, the vertical height of pipes and of delivery being called h feet : Useful work h per pound of Lost energy. Total energy. Efficiency. water. 20 20 36 >56 0-36 50 50 37'5 87-5 0-57 100 100 40 140 0-71 200 200 45 245 0-82 300 300 50 350 0-8G 400 400 55 455 0-88 (2) If the loss of the pump is 20 + 15 x 10~ 6 Q per pound of water, and of the vertical pipe is5 x 10 ~ 8 a 2 per pound, o, being 508 APPLIED MECHANICS. gallons per minute, work out two more tables like the above one when the delivery is 500, and the other when it is 1,500 gallons per minute. 408. The peculiarity of reciprocating-pumps is that, when there is no slip, there is the same quantity of water passed through the pump at every stroke. If we know the size of everything, then the speed of the pump tells how much water is delivered, and if we know the height to which it is delivered, we know the work done. Now, in a centrifugal - pump things are somewhat different ; with a given speed of pump we may have very different quantities of water passing. We may have the pump running at a certain speed, and no water being delivered, and very little work being done, only frictional work, in fact. Now a slight increase of speed may cause an abundant flow of water, and a tremendous increase in the work usefully done. Strange to say, if Fig. 283. this speed be now diminished to what it was at first, it does not follow that the water will cease to flow. In any centrifugal-pump such as Fig. 283 we observe that there is a central wheel, A B, with vanes, which can be rotated very rapidly. Water can enter the wheel on both sides, c c, at its centre from two supply pipes, D D, which meet in one pipe, E, below. The water fills the wheel or space bet ween the vanes, and, being whirled round with great velocity, tries to get away at the circumfer- ence of the wheel, because of the cen- trifugal force, and it flows out into the casing, F F, which gradually becomes the discharge-pipe of the pump, as shown in the small scale drawing, Fig. 284. This is a popular Fig. 284. APPLIED MECHANICS. 509 explanation of what occurs, but we must examine the matter more carefully. 409. What occurred when the pressure in the pump of the hydraulic press became greater than the pressure of fluid in the press ? There occurred a flow of fluid. The fluid was set in motion from pump to press. A difference of pressure between two places which communicate with one another usually means a tendency to produce motion. In the hydraulic press the flow is intermittent. Why? Because the pressure is intermittent. We may be sure that when we have a certain difference of pressure between two places, and this is always the same, the flow of fluid is per- fectly steady ; and we saw in Art. 408 that work is then done on the fluid with perfect uniformity. We also saw how to tell what work was done. The work done on a fluid in a minute is simply the difference of pressure per square foot which causes the flow, multiplied into the number of cubic feet of water which flow per minute. This gives the answer in foot-pounds. Suppose. that with pumps, or in any other way, we establish i difference of pressure between a place A and a place B, and suppose we know that our pumps and other arrangements will not break down in fact, that the difference of pressure between A and B is really a fixed thing on which we can depend we know that the flow of water from A to B will be the same at all times, and that the same amount of work will be done upon it every minute. If, now, we leave out of our minds all consideration of how that constant difference of pressure has been produced merely think of the two vessels A and B we know that this difference of pressure which has been established is really a store of energy. What enables us to call it a store of energy ? The fact that we know it will not be suddenly destroyed. Suppose we know that a man has a certain income paid, say, by Government, and suppose we are perfectly certain that this income is constant, we can regard the certainty of the man's income as a store. To say that a man makes a sovereign in a day is not of much importance, but to say that the man has a regular income of one pound a day makes him a respect- able member of society, and a store of social energy. A man may be sitting in Parliament, but this in itself does not make him a store of political energy ; whereas if we know that he is certain to sit there for a length of time that a 510 APPLIED MECHANICS. general election or general vote of the House is unlikely to unseat him we can regard him as possessing a store of political energy. Similarly,, a pound of water in the vessel A, at rest, possesses more energy than a pound of water in the vessel B, at rest. (We must remember that there are some means of keeping the pressures in A and B what they were.) How much more energy has it ] If the difference of pressure is p Ibs. per square inch, then it has 2'3 p foot-pounds of energy ; not in virtue of its own intrinsic worth, but because it is where it is, and because we know that if it flows into the vessel B, nothing will alter in the pressure conditions of the two vessels till it gets into the vessel B. We understand, then, that a pound of water, subjected to a pressure of p Ibs. per square inch, may be said to have a store of 2-3 p foot-pounds of energy, if we know that the motion which is occurring in the water is steady, and is not altering capriciously. Thus a pound of water at the pressure of the atmosphere, 14-73 pounds per square inch, possesses, in virtue, of this pres- sure, 2*3 x 14'73 or 34 foot-pounds of energy. It would possess the same energy if it were at the pressure o, the pressure in a vacuum, but were 34 feet higher than it is in position. Suppose that A is a closed box, and that it is filled with water, under great pressure. Now, suppose we open a valve, and let this water escape. Although there was a great pres- sure, p, just for an instant, and therefore a rapid flow of water just for an instant, this almost instantly dies away, because the pressure in A is almost instantly diminished. Every pound of the water did not then have a store of 2' 3 p foot- pounds of energy, and yet it was at the pressure of p pounds per square inch. It is the certainty that the state of pressure in A will continue constant that gives to pressure its signifi- cance, and gives to us the liberty of regarding pressure as a store of energy. 410. Suppose that we have, anywhere, steady motion of water. Consider a pound of the water. What is its total store of energy ? 1. It is A feet above some datum level. Then if one pound of water ever were allowed to fall to the datum level, it would do h foot-pounds of work in falling. The mechanical energy stored up in a miller's dam is simply the weight of the APPLIED MECHANICS. 511 water, multiplied by the height through which it can fall. Of course, if any other volumetric force acts on the pound of water, this will constitute another store of potential energy. We are supposing that the weight of the water is the only volumetric force. 2. As the motion is steady, if the water is at a place where pressure is p pounds per square inch, it possesses, in virtue of the fact that the motion is of a steady character, the store 2 '3 p foot-pounds of energy. 3. As the water is in motion, if v is its velocity in feet per second, as its mass (the mass of one pound) is 1-4-32-2, we know that its kinetic energy, or energy of motion, is the square of the velocity divided by 64-4. Exercise. Calculate the numbers in the following table, which shows the relative values of h and p and v, if we wish to convert one of these forms of energy into another. Thus, bhe energy due to a difference of level of 2 '3 feet is equiva- lent to that due to a difference of pressure of 1 Ib. per square inch, or to that due to a velocity of 12-18 feet per second. Difference of level. Pressure. Velocity. 2-3 feet. 16-1 34 64-4 1 Ib. per square in. 14-73 ", 28 12-18 feet per second 32-2 45-7 64-4 411. Now, we shall not suppose that the pound of water has any other stores of energy than these. We know that, as it is compressible to some extent, it -may have a store as the mainspring of a clock has. This store must always be taken into account when the fluid is air or any other gas. It may also be electrified or at a high temperature, or it may have other stores of energy which we are neglecting. Merely think of these three stores : Potential energy due to height above a datum level ; pressure energy due to the unchangeableness of things ; kinetic energy due to its actual motion. What we must remember carefully is the fact that this pound of water retains all of this energy except what it loses in friction. Suppose that a man has capital in the shape of gold, capital in the shape of shares which never alter 512 APPLIED MECHANICS. in price, and capital in the shape of a pet manufactory, which wastes money just in proportion to the amount of capital invested in it. He may buy more shares or sell them out, in- vest more or less money in the pet manufactory, but all the time his only loss is the loss from the manufactory. He may have no gold, or no shares, or very little money in the factory, but all the time his total capital is unchanging, except that he loses in proportion to the value of his factory. It is only when water has part of its energy in the shape of kinetic energy, only when it is in motion, that it loses any part of its total store. 412. Consider a lake of water at rest. Consider a point A, and somewhat below its level another point B. A pound of water at A has the same store of energy as if it were at B. In neither case is there any energy of motion. The store of energy at A is merely due to height above some datum and the pressuro 285. p er square inch at A. Now, if a pound of water <* ets to A from B, it loses potential energy h foot-pounds, if the difference of the level is h feet, and it ought to gain an equiva- lent of pressure energy, and the gain of pressure is, as we have already seen, simply ^-= pounds per square inch. Thus, if h is 34 feet, and p is bhe gain of pressure, then there is a gain of pressure energy of 34 foot-pounds that is, there is a pressure at B of 34-f-2-3, or 14-73 Ibs. per square inch greater than the pressure at A. This is an increase of pressure called one atmosphere. In still water there is an increase of pressure of one atmosphere for every 34 feet of descent. (See Art. 173.) 413. To further familiarise us with the idea, consider the flow of water from an orifice. In Fig. 286 we see the stream lines along which water flows out of the orifice. The shapes of these stream lines will depend 2g6 on the shape and position of the orifice.. Now, a pound of water at the upper still surface A is at atmospheric pressure, and when it reaches c it is also at atmospheric pressure, so that its pressure energy remains the same. But at c it has fallen h feet ; it has lost h foot-pounds of potential energy, and it must therefore have gained h foot-pounds APPLIED MECHANICS. 513 Fig. 286. of kinetic energy. If v is its velocity, then v 2 -f- 64-4 must be equal to A, so that v may be calculated; hence its velocity is just the same as that of a stone which had fallen freely from A to c. To illustrate this, a high vessel may be used, from which at different levels tubes come out, ending in nozzles, throwing jets vertically upwards. These jets do not rise to the same height, because there is a loss of energy due to friction, and this friction occurs principally at the nozzles. If the jet reaches within a dis- tance A! of the level of still water inside, and if the nozzle is at the depth h below still water level, then if we may say that all the friction occurs at the nozzles, h^h expresses the loss as a frac- tion of the whole kinetic energy at the nozzle. When the measurements and calculations are made for different levels of the water in the vessel, we find nearly the same result in every case, showing that the frictional loss of energy seems to be proportional to the kinetic energy there. There is very little friction in the case we are considering in Fig. 286, where the orifice is sharp-edged, and we have a very simple statement of the velocity at c. Can we say the same about the velocity at B ? Certainly not. If we knew the pressure energy at B, we could say how much of the lost potential energy has been invested in this shape, and therefore how much has been invested in the shape of kinetic energy ; but without knowing the pressure at B, we cannot tell what is the velocity there. 414. In this subject of flowing water there are more mislead- ing hypotheses, due to perverted ingenuity, than in almost any other ; and, unfortunately, the logical conclusions drawn from these hypotheses, when known to be untrue; are said to be the statements of theory as opposed to practice. We often hear the statement, "the theoretical velocity at an orifice is the velocity which the water would have acquired if it had fallen freely, as in a vacuum, from still water level ; " whereas it is evident that we cannot tell the velocity at any point in the jet unless we know the pressure there, and we only know the pressure on the very outside of the jet. 514 APPLIED MECHANICS. Now, although we do not know the pressure or velocity at every point of water flowing from an orifice, the studies of Professor James Thomson enable us to make certain im- portant statements which agree with experiment. One of these is this : When frictionless liquid flows from two similar vessels through similar orifices similarly situated with regard to free water level, the lines of flow are exactly of similar shape ; the velocities at similar points are exactly as the square roots of the dimensions of the vessels, and the total quantities of liquid which flow are proportional to the square roots of the fifth powers of the dimensions. Thus, if we have water flowing similarly from three similar vessels, all made from the same drawings, but to different scales say one 1 foot, another 4 feet, and another 9 feet deep the velocities at similar points are as 1 to 2 to 3, the sections of stream tubes are as 1 to 16 to 81, and the quantities of liquid flowing from the three vessels are as 1 to 32 to 243 ; thus we are quite sure that 243 times as much liquid flows from the third vessel as from the first. Hence, suppose we want to know how much water is flow- ing in a small stream, we dam the water up somewhere, and let it flow out of our dam through a notch like Fig. 287 (a right-angled isosceles notch) in a wooden board with sharp edges. Indeed, we prefer to have the edges of the notch made of metal, so that we shall be sure that they are straight and sharp. Measure the height of the still water in the dam above the lowest point of the notch (D). A graduated post, rising from the bottom of the dam some feet away, its zero being on the level of D, is very convenient for this. This one measurement tells us how much water is tumbling over. For we know that, sup- pose there is rather a drought one day, and a shows the appearance of the notch ; and on another, a rainy day, A shows its appearance, we observe that the orifices through which the water is flowing are similar and similarly situated with regard to the still water level ; and Thomson's theory enables us to say that, if on one day the height is 1 foot, and on another ife Fig. 287. APPLIED MECHANICS. 515 is 4 feet, then 32 times as much water comes over on the second day as on the first. The flow of water is exactly proportional to the square root of the fifth power of the height D A. Now, Thomson measured very accurately how much water is flowing when the height is 1 foot, and he found it to be 2 -635 cubic feet per second. Hence we have the rule : measure the vertical height D A, at any instant, in feet ; raise this to the fifth power, and extract the square root, and multiply by 2-635, and we know how much water is flowing in cubic feet per second. If we know the cubic feet of water flowing per second, we know the weight of the water, since a cubic foot of water weighs 62 -3 Ibs., and the weight of water, multiplied by the number of feet through which we can let it fall, tells us the foot-pounds per second the available power of the stream. The foot-pounds per minute, divided by 33,000, is the horse- power of the stream. 415. Gauge notch observations, made from day to day on a stream, enable a person to make very exact calculations as to the power available for the driving of mills by turbines, for the working of hoists, cranes, and lifts, and for hundreds of other purposes. How Thomson used his theory in proving that the famous Lowell empirical formula, for rectangular gauge notches, is really a rational one, will be found in Art. 436. Students ought to treasure anything published by Thomson on fluid motion. 416. Many rather abstruse-looking questions are easily an- swerable when we fully grasp the significance of the energy law. The fundamental fact which makes any hydraulic problem clear to you is this. If we may neglect friction, then a pound of water at any place has its total energy in three shapes. It has h foot-pounds of energy, because it is h feet above a datum level. It has 2 -3 p foot-pounds of energy, because its pressure is p pounds per square inch ; and it has v^-i- 64 '4 foot-pounds of energy, because its velocity is v feet per second. To take another example : Suppose we have a pipe which is in the main horizontal, so that we may neglect differences of level. Then we have to remember that the pressure energy, plus the kinetic energy, of a pound of water does not alter. When water flows along a pipe which is full, there must be the same quantity flowing 516 APPLIED MECHANICS. everywhere. We are sure, 'therefore, that there must be greater velocity wherever the pipe is contracted. But greater velocity meats greater kinetic energy, and if this water invests 3 K IL M ^^^L^E 5 - 58" V o 7>-^ - _J ~-T-^~ a s E . l-B j. _____ Fig. 288. more of its energy kinetically, it must have less in the shape of pressure energy. That is, the pressure of the water at B (Fig. 288) is less than at A or at c. The pressure at B may become very small indeed. It is easy, in this way, to reduce the pressure to much less than the atmospheric pressure merely contracting the cross section of the pipe is sufficient. We cannot make the pressure as small as that of a vacuum, for before that limit is reached vapour forms. 417- Suppose a conduit of this kind were carried over our fields, and that a quantity of water lay in our fields, at not too great a depth below the conduit. If we bring a pipe to the point B from the field- water, this becomes a suction-pipe, and we get our fields drained at the expense of the conduit owners. We spoil their water, if it is clean, but at all events we get our fields drained. In this lies the theory of jet pumps, and much of the theory of injectors, etc. The jet pump of Professor James Fig. 289. Thomson simply consists of a small pipe, A (Fig. 289), which ends in a nozzle, Through this, let us suppose, we have a APPLIED MECHANICS. 517 small supply of water flowing from some pretty high reservoir. Suppose that this water flows into the atmosphere at c. Evidently the pressure at a is much less than the pressure at c. It is less then than the atmospheric pressure, and hence the neighbourhood of a is a partially vacuous space, so that the pipe B E becomes a suction-pipe, and water tends to flow from a point at E, to B, and on to c. Thus, if we have a small supply of water from a high reservoir, we are able to drain a marsh with it. 418. There is not space here to speak of the hundred other ways in which our principle comes in to simplify all sorts of puzzling phenomena. We may refer to Mr. James Perry's siphon for the discharge of flood- waters at the weirs in rivers. It has no moving parts in the water, being simply a wide, open pipe of varying sectional area, through which an object as large as a bullock might pass, without injury to the sluice. It has been found, by actual trial, that the quantity of water passing per minute through such a sluice is independent of the fall, so long as there is sufficient fall to balance the waste by friction in the pipe, a few inches being enough in the case of wide pipes ; and a velocity of 45 feet per second, at the smallest section, may be calculated upon when the sluice is working full power. The cross-section of the siphon is like the letter D at the tail ; the masonry, or concrete basin, is actually a portion of the siphon, and the horse-shoe shaped space between the lip of the basin and the iron edge of the siphon proper is the actual opening. The quantity of water passing at any time is regulated by the admission of air to, or its exclusion from the siphon. There is a throttle-valve arrangement by means of which the siphon may be adjusted at any time, so as to vary what may be called the normal water level in the reach above the sluice, and to vary the opening through which a constant stream of water falls on a ridge-shaped portion of the siphon. This stream of water is needed to exhaust the siphon of air, so that action may be set up at any time, even when the water is not passing over the top of the throttle- valve. The manipulation of such sluices is perfectly simple they may be made self-acting ; but it is proposed in important places to work all the sluices on a river from a single station electrically. Example. There is a circular sharp-edged orifice in a tank in which there is liquid kept standing to a certain height. A 518 APPLIED MECHANICS. man is told that, without interfering with the actual edge of the hole, he is allowed to do what he pleases to increase the flow. How may he do so? Evidently by fitting on a tube which quickly but gradually gets to be of much larger diameter. He takes care that the tube shall run full. Example. A number of mill-owners receive water from the same lake. Each has a rectangular opening the depth of which below the lake and its breadth are supposed to nx his supply. Show that if a mill-owner is allowed to do anything he pleases on his own side of the opening, he may procure a very much greater supply of water. 419. We know that loss of energy per pound by friction, in water, is proportional to the square of the velocity, at such speeds as are common in pumps. In Art. 48 we gave the rules by which we are able to calculate the loss of energy which a pound of water experiences in going along pipes. But we wish to impress on you the fact that this loss always be- comes very great when the flow of the water has to occur ab- normally. In the pipe (Fig. 288) the wide and narrow parts gradually change into one another by continuous curves. It is practically impossible for a liquid to flow in a discontinuous curve. Suppose we try, then, to make it flow along the pipe shown in Fig. 290. What the water does is this : when it comes to the corner Fig. 290. it produces for itself wheels, little eddies or whirlpools, as we might put rollers under a log of wood that we wanted to get along easily; and there is great loss of energy due to this, for the eddies have not only to be in the corners, but there are smaller eddies carried along by the water itself, maintained so long as they are needed. We have been speaking of actual discontinuity in the flow. Bnt there is a fact which many hydraulic engineers seem to be quite ignorant of namely, that a liquid cannot flow along a path which suddenly changes in curvature. A liquid cannot, for example, flow along this path (Fig. 291). At A it changes from a straight line suddenly to the arc of a circle, and, con- sequently, the water digresses at A, and creates little castors, APPLIED MECHANICS. 519 little eddies to carry it by a path of continuous change of curvature from B to c. Now this analogy can be shown to be true experimentally. Thus it has been found that if water flows along the bend (Fig. 291) it loses a certain amount of energy on account of the bend ; but if we make the pipe bend as much again in the same direction as in A B (Fig. 293), we do not get again the same loss ; indeed, there is comparatively very kittle loss at the second bend. But if we bend the pipe in the opposite direction, as in c D (Fig. 292), there is as much loss at the second bend as at the first. The little wheels, or castors, or eddies which the fluid creates for itself to Fig. 292. Fig. 293. carry it through the bend A are available when the water needs them again at B, whereps the wheels or eddies produced at c have to be destroyed, and new ones created, rotating just in the opposite direction to carry the fluid through the bend D. We see now how necessary it is that all curved vanes or other surfaces along which water flows should be drawn, not with a pair of compasses, but rather with a batten, a thin Strip of wood, which bends gradually. 420. We regard a purnp as a contrivance which gives to every pound of water passing through it an additional store of energy. From the pond to the entrance to the pump, every pound of water has just the energy it has in the pond, barring frictional loss. From pump to cistern, every pound of water has an additional store of energy, and it is the pump which gives it this store. Suppose we have a centrifugal pump going at a regular speed, and discharging a regular quantity of water. In the supply-pipe a pound of water has a certain total amount of energy which we know, if we know its height above datum, its pressure, and its velocity. But, in passing through the wheel, it receives a supply of energy. The total energy of a pound of water in the discharge-pipe is greater than what it is in the supply-pipe. We can make all sorts of calculations, if we know what is the clear gain, the clear gift of energy it gets in passing through the pump. 421. Suppose a man jumps into an American railway train 520 APPLIED MECHANICS. anywhere, and after wandering about, fore and aft, jumps out again. Find the man's momentum in the direction of the train's motion just before he alights on the train. Find his momentum in the same direction when he has just sprung from the train, the difference of these is the total impulse with which he acts on the train. It is the momentum which he gives to the train. Suppose that a Fig. 294. number of people could perform this acrobatic feat every second with the greatest regularity, then the momentum given in one second to the train could be calculated. But momentum given per second is what we call force ; hence we have found the force acting on the train due to these jumping individuals, and this force, multiplied into the space passed through by the train in one second, gives the propelling work done upon the train per second. We have nothing to do with whether it is a pro- pelling force or a retarding force. In the one case, the acrobats give momentum to the train ; in the other, the train gives momentum to them. The loss of momentum per second in a regular stream of people, going on and off the train, is a force which is applied to the train. We only have to do with their momentum in the direction of the train's motion. Now suppose that, instead of its being a train, it were a sort of circular turntable, or a merry-go-round, and that a regular stream of people jumped on and off. In this case, the place where a man jumps on may be going at a different speed from the place where a man jumps off ; but our rule is not very different. Find how much is added per second to the momentum of the wheel at the point where people leap on, and regard this as a force. Multiply by the speed of the wheel there, and this is the work done by the mere leaping on, and staying on the wheel. Now, find how much per second is taken from the momentum of the wheel at the place where leaping off occurs, and regard this as a force opposing the motion. Multiplied into speed at the leaping-off place, we have the work taken by the stream of people from the wheel, per second, because they leap off. The difference in these two things is, of course, the work done in the jumping on and off. Now, the water moves from the centre towards the vanes APPLIED MECHANICS. 521 of the centrifugal pump, merely radially, and hence the water entering the vane cannot add to or diminish the momentum of the wheel just there. It had no momentum of its own previously in this direction. What occurs inside the wheel now we have nothing to do with, excepting that we know that frictional loss occurs there. We are only concerned with how the water leaves the wheel. The way in which it is made to leave the wheel determines how much energy it takes from the wheel. Take the simplest, case. Suppose the vanes to be radial at B (Fig. 294). That is, besides moving outwards radially'at B, the water leaves the vane with the same tangential velocity as the vane at B has. Suppose this tangential velocity to be v. Then w Ib. of water leaving B per second leaves with a tan- gential momentum, wv -4- 32 '2, and retards the wheel with a force of this amount acting at B. This force x v is the energy which it receives per second from wheel, or wv 3 -f- 3 2 -2. One pound of water, therefore, receives the energy v 2 -f- 32-2 from the wheel in passing through it. We understand, then, that a pound of water in the dis- charge-pipe of the centrifugal pump has this greater store of energy than a pound of water in the supply-pipe, except for frictional losses. If we make the water go out from the wheel as backward bent vanes make it go (Fig. 296), and as it would be dangerous for us to do from a railway train, less work has been done upon it by the wheel. If it goes out in the direction of motion relatively to the wheel, more work is done upon it than we are now supposing. 422. The wheel with radial vanes gives v 3 -4-32 > 2 foot- pounds of energy to one pound of water. If we know how- many pounds of water pass through the wheel, we know then the total amount of work done by the wheel. The water gets this energy to squander or store as it pleases, and it does squander it in friction to a large extent. But suppose it squandered none of it, but converted it all into potential energy in lifting itself up to a cistern, it would lift itself v* -T- 32-2 feet high; that is, it would lift itself above the pond to twice the height due to the velocity of the rim of the wheel. Suppose the rim of the wheel has a velocity of 45*7 feet per second, a stone would have to fall freely 34 feet to acquire this velocity, and hence the total rise of water would be 68 feet> twice the height due to the velocity of the 522 APPLIED MECHANICS. rim of the wheel. In this case we should say that the pump was perfect. The wheel itself receives energy from the engine, else it could not give energy to the water. It gives out all the energy that leaves the engine, except what is wasted in bear- ings everywhere, and what is wasted in friction with the water. The energy given out by the engine per pound of water, divided into v* -f- 32-2, is the efficiency of the shafting, belt- ing, and wheel. Again, the real height to which water is lifted by the pump, divided by the ideal height, -y 2 -r- 32 -2, is the efficiency of the water passages from pond to wheel and from wheel to cistern. The loss in these places is due to friction. Make the supply-pipe wide, bell-mouthed at the bottom, where water enters it, so that it may enter by gradual curves ; make the approach to the wheel as gradual as possible ; let the vanes of the wheel make the calculable angle with the central circle, which will reduce the shock there (see Art. 428) ; make the discharge-pipe wide, and let the velocity with which the water enters the upper cistern be as small as possible, and we greatly reduce the waste of energy. But there is one particular place where there is usually much greater waste than anywhere else, and that is the chamber outside the wheel. 423. Just when the water leaves the wheel a large portion of its energy is kinetic. It is in rapid motion. Now, in the large discharge-pipe there may be as little kinetic energy as we please. Hence, from the time the water leaves the wheel till it enters the discharge-pipe there ought to be great care taken in allowing the kinetic energy to become converted into pressure energy. Professor James Thomson discovered here the efficiency of a whirlpool cham- ber. When we let water escape from a wash-basin, we know that the surface of the water takes a shape like this (Fig. 295). The velocity of the water is greater the nearer it is to the centre. The pres- sure is greater the farther away from the centre. The spiral motion which we ob- serve in this case is the only steady motion of water which ajlows a constant radial discharge without the water making APPIJED MECHANICS. 523 objections, setting up little eddies of its own, and thus wasting energy in friction. Hence, in Thomson's pump, after the water is discharged, it circulates in this cylindric whirlpool chamber, which he made of twice the diameter of his wheel. When a pound of water reaches this place, in consequence of its radial and circular motions, it retains more nearly the whole of its total energy than if we let it discharge in any other way. It has lost much of its velocity, but has gained in pressure. This whirlpool chamber of Thomson's, then, did for the water what gradual curves do for the water in a pipe; it enables the water to convert its kinetic into pressure energy, with a mini- mum of waste in friction. It has, however, to be remembered that even here, at the outside of the whirlpool chamber, the water retains a very considerable amount of kinetic energy, even when the whirlpool chamber is made very large, and much of this is wasted afterwards. And, although no one believes more firmly in the reasoning of Thomson than I do, I feel that perhaps the whole problem admits of a better common- sense solution. Let the whirlpool chamber get wider or broader, as well as larger in diameter. Let the wheel have larger orifices on its outer circumference than on its inner circumference. The water will lose its kinetic energy far more rapidly as its passage widens more rapidly. The result of this will be that although in this rapid change there is more friction, yet when the water is only a short distance out, it is not moving much faster than it will do in the discharge-pipe, and there is much less loss in entering the discharge-pipe. We have always thought that since Thomson's chamber is expensively large, we ought to sub- mit to a modification of his con- ditions even from the place where water leaves the wheel. But when we begin to consider a much smaller chamber, we see that possibly the vanes ought rather to slope back- wards than to be radial. Let the radial velocity at M N be v r . The velocity relatively (see Art. 36) to Fig. 296. the vane is v r -r- sin. 9, if M N P is 0. 424. Let v be the tangential velocity of the wheel at N. 524 APPLIED MECHANICS. Using a very easily understood graphical method of working, let N represent the radial velocity at N to scale. Let M N and N p represent the direction of the vane and rim at N. Let N P represent v, the tangential velocity, and if o M is drawn at right angles to ON, MN represents the velocity of the water relatively to the vane. Hence, as the resultant of M N and N P is M P, M P is the total velocity of the water leaving the wheel. It has the tangential component M Q, and the radial component Q P which it had in the wheel. Suppose we call M Q by the letter v. Now, every pound of water leaves with the tangential momentum v -7- 32 -2, or say v/g. Every pound per second therefore represents a tangential retarding force, v/g, acting on the rim of the wheel, and this multiplied by v, or v v/g, is the work done usefully per pound of water. If we take it that in all cases there is a loss of the fraction s of the kinetic energy due to tangential motion, only s v^/g is wasted ; v v/g is total energy ; (v v s v 2 )/g is useful energy, or height to which the water is lifted. We may say roughly that the efficiency is 1 s - ; s depends on the size of the chamber. What the value of v, and therefore the angle d, ought to be, is therefore a question of minimum total waste of value, in interest on plant and waste of energy, etc. 425. In the case of pumps we saw that each pound of water gets an increased store of energy, which may be in the shape of pressure energy, or kinetic energy, or both, but which mainly becomes potential. Now, in water-wheels, turbines, water- pressure engines, including hoists and lifts, we take part of the store of energy from each pound of water, giving it to machinery. As a simple case of the abstraction of energy from water, and as an illustration of the acrobat and railway-train principle, con- sider the vessel (Fig. 298) from which the water is flowing. Water leaves this vessel horizontally from an orifice, taking away with Fig. 29* it momentum. The quantity of momentum APPLIED MECHANICS. 525 it takes away per second is simply the force acting on the vessel. You see that there is a force acting, for I have arranged the vessel as the bob of a pendulum. 426. If we let the water flow from an orifice through which it comes in parallel streams, it is easy to show that the force acting on the vessel is twice the total pressure which would act on this little sluice when it closes the orifice, and no water is flowing. For if a little area a (square feet) of the orifice is h feet below still water level, the pressure at a being atmo- spheric, the velocity v= -. This is the only case in which we can make any easy attempt to calculate the area of the contracted vein. Note that with this orifice, whatever be the shape of the vessel, the total pressure of 534 APPLIED MECHANICS. the fluid upon it in the direction opposite to the arrow at c is known to us, being A w H if H is the depth of the centre of A below water-level, because the velocities, and therefore the pressures, everywhere, except near MN, are the same as if the orifice were closed; and near MN there are no pressure-forces parallel to the arrow at c. Hence A w H is the momentum leaving the vessel per second. It is only when the orifice is small that we can be sure that the average velocity at c is *J 2 g H, and the volume flowing per second is a, ^ 2g H. The mass per second is this multiplied by w/ff, and the momentum per second is this multiplied by ^/ 2 g H ; and hence A w H = 2 aw HJ or A = 2 a. 434. For all other sharp-edged orifices than that shown in Fig. 305, we rely upon experiment. Q the quantity in cubic feet per second flowing from a sharp-edged orifice of area A square feet, the centre of the orifice being H feet below still water level, Q = k v/ 2 g H. In Fig. 305 k is J, as we have seen. For circular orifices, k is 0-62 very nearly, even when the upper edge of the orifice is comparatively near the upper level, so long as the orifice keeps filled. Again, for square and rectangular orifices K is very nearly 0-62. We shall not give the great table of numbers, varying from 0'61 to 0'63, which have been experimentally determined for various sizes and positions of rectangular orifices, because we do not think it more accurate than the statement that 0'62 is very nearly correct for all. 435. The shape of the stream coming from a rectangular orifice is very interesting ; and a student must meditate, when looking at such a stream, upon the way in which its component parts collide to cause the curious palpitating change of shape M Fig. 304. Fig. 305. and section which is going on. Fig. 306 gives some notion of these changes. 436. Experiment has shown that in the case of sharp-edged orifices there is no practical difference in the actual flowing of 536 APPLIED MECHANICS. water from what the flow would be if the liquid were fdction- less. It can be shown that when liquid is frictionless the stream lines from similar and similarly placed orifices in similar vessels with the same kind of liquid at similar heights are similar, the corresponding velocities being proportional to the square roots of the dimensions, and therefore the volumes A D Fig. 307. flowing being proportional to the two and a half powers of the dimensions. If, then, water flows from a pond over a sharp- edged notch shaped like a right angled isosceles triangle, each of the edges making 45 with the horizontal, as in Fig. 287, and if the difference of level from B to A is H, the quantity Q flowing in cubic feet per second is proportional to H^. Prof. James Thomson gave us the above principle and this method of measuring water. By careful measurement, he found that, H being in feet, Q = 2'635H| . . . (1). The rectangular notch is more convenient. Professor Jas. Thomson showed that the empirical formula of Mr. Francis, of Lowell, arrived at with great care and at great expense, is a rational one. If L is the length of the notch in feet, H being vertical height in feet from sill B to still-water level, for a given H there is a certain value of L beyond which increase in L means that the increase in Q is proportional to the increase in L. In fact, we distinguish the flow through the two ends of length m H at one side and m H at the other, and the flow through the middle part L - 2 m H, where all the lines of flow may be regarded as in vertical planes. We have good reason to take m to be constant. Imagine an orifice of length 2 m H. The flow through it is ki H f , where Jc\ is some constant. The flow through a square orifice of height H, the lines of flow being in vertical planes, is # 2 H f , where & 2 is some constant, and therefore the middle flow is (L ~ 2 WH)/H times this, or , 5 T L 2 m H 5 Q = ki H 7 + & 2 H H This will be found to reduce to a = b (L - c H) H f . If there is only one end contraction, c is evidently halved ; and if there are no end contractions, c is 0. The experiments of Mr. Francis give us the values of 6 and c, so that Q = 3'33(L- H)H*....(2), APPLIED MECHANICS. 537 where n is 2 or 1 or 0, according as we have all the edges sharp or we have the edge B c a smooth vertical guiding plane, or both B c and D C smooth vertical guiding planes. 437. In a gas, we have w oc p, where p is the pressure in pounds per square foot, if the temperature could be kept con- stant, or we have the rule for adiabatic flow w oc p f , where y is the well-known ratio of the specific heats. In either of these cases it is easy to find I -2- and write out the law. This law is of universal use in all cases where viscosity may be neglected, and is a great guide to the hydraulic engineer. Thus in the case of adiabatic flow, w = cp f , the integral of is - l - I y i. 1 i *<*-; ji* Y > v 2 17 1-- and hence h + + = p 7 = constant .... (4). In a great many problems, changes of level are insignificant, and we often use t> 2 H - ^ -, p t = constant . . . . (4) for gases. Thus, if p Q is the pressure and W Q the weight of a cubic foot of gas inside a vessel at places where there is no velocity, and if outside an orifice the pressure is p, the constant in (4) is evidently + =__ p ~, and hence outside the orifice and as c is #> p ?, it is easy to make all sorts of calculations on the quantity of gas flowing per second. Observe that if p is very little less than p Q and if we use the approximation (1 + a) n = l+na, when a is small, we find v 2 = (p - p} .... (6), a simple rule which it is well to remember in fan and windmill problems. In a Thomson water turbine the velocity of the rim of the wheel is the velocity due to half the total aA r ailable pressure ; so in an air turbine, when there is no great difference of pressure, the velocity of the rim of the wheel is the velocity due to half the pressure difference. Thus, if p of the supply is 7,000 Ibs. per square foot, and if p of the exhaust is 6,800 Ibs. per square foot, and if we take w = 0-28 Ib. per cubic foot, the velocity of the rim v is, since tho difference of pressure is 200 Ibs. per square foot, ~| (100) = 151 feet per second. 538 APPLIED MECHANICS. Returning to (6) : Neglecting friction, if there is an orifice of area A near which the flow is guided so that the streams of air are parallel, Q, the volume flowing per second, is o, = v A ; and if the pressure is p, the weight of stuff flowing per second is w = v A w or v A cpi . Using v from (5), and letting p/p be called a, we have after simplification Problem. Find p, the outside pressure, so that for a given inside pressure there may be the maximum flow. It is obvious that as p is diminished more and more, v, the velocity, increases more and more, and so does Q. But a large o, does not necessarily mean a large quantity of gas. We want w to be large. When is w a maximum ? That is, what value of o in all will make ?/ Ill\ * 1 + - 1 a 7 1 1 - a 7 l or a 7 - a 7 a maximum r Differentiating with regard to a, and equating to 0, we have Dividing by a ? , we find a = ( 7 * V - or In the case of air y = 1-41, and we find p = -527 p . That is, there is a maximum quantity leaving the vessel per second when the outside pressure is a little greater than half the inside pressure. Problem. When p is indefinitely diminished what is v ? Answer : v = A / 2 ff? -^i V y - 1 WQ This is greater than the velocity of sound in the ratio being 2*21 for air. That is, the limiting velocity in the case of air is 2,413 feet per second x A / J_, where t is the V 273 absolute temperature inside the vessel, and there is a vacuum outside. This involves the idea of the jet creating such intense cold as to be at the absolute zero of temperature. Returning to equations (3) and (4), we assumed h to be of little importance in many gaseous problems of the mechanical engineer. But there are many physical problems in which it is necessary to take account of changes in level. For example, if (3) is integrated on the assumption of constant temperature, and we assume v to APPLIED MECHANICS. 539 keep constant, we find that p diminishes as h increases, according to the compound interest law. Again, under the same condition as to v, but with the adiabatic law for w, we find that p diminishes with h according to a law which may be changed into " the rate of diminution of temperature with h is constant. " A great number of interesting examples of the use of (2) might be given. It enables us to understand the flow of fluid from orifices, the action of jet-pumps, the attraction of light bodies caused by vibrating timing-forks, why some valves are actually sucked up more against their seats instead of being forced away by the issuing stream of fluid, and many other phenomena which are thought to be very curious. 438. Example. Particles of water in a basin, flowing very slowly towards a hole in the centre, move in nearly circular paths, so that the velocity v is inversely proportional to the distance from the centre. Take~v = -, where a is some constant and x is the radius a 2 p or distance from the axis. Then (3) becomes h + - + = c. Now at the surface of the water p is constant, being the pressure a 2 of the atmosphere ; so that there h = - ^ g> ancl tnis S* ves us the shape of the curved surface. Assume c and a any values, and it is easy to calculate h for any value of x and so plot the curve. This curve rotated about the axis gives the shape of the surface, which is a surface of revolution. A student who depends upon a text-book to give him complete information is not learning to become an engineer. Meditation when looking at water flowing from a basin ought to greatly add to what he will obtain from such a book as this. Perhaps he will begin to notice that it is centrifugal force due to whirling motion which maintains pressure at a place. What will be the effect of friction at the solid surface of the basin ? It will diminish velocity and diminish pressure. Water will flow, therefore, down the surface of the basin and towards the hole. If this is well under- stood, the student will understand how it is that at a bend of a river the earth from the outer bank is dragged along the bottom and deposited on the inner bank, and hence that a river through an alluvial plain is always tending to get more crooked, until at length it cuts off a bend and so straightens itself in a new channel. In the basin problem we have also James Thomson's explanation of the phenomena of great forest fires, and also of the prevailing wind system of the earth. (See Phil. Trans.) 439. If water flowing spirally in a horizontal plane follows the law v = -, where r is distance from a central point ; note that p = c - | -- 2- ^ ne ingenious student ought to study how p and v vary at right angles to stream lines. He has only to consider the equilibrium of an elementary portion of fluid, c 11 APPLIED MECHANICS. (Fig. 303), subjected to pressures, centrifugal force, and its own weight in a direction normal to the stream. He will find that if dp means the rate at which p varies in the direction of the radius of curvature away from the centre of curvature, and if a is the angle DOE (Fig. 303), the stream being in the plane of the paper, which is vertical, and if r is the radius of curvature, dp w v 2 ~ = --- w sin. a . . . . (1). dn g r v ' If the stream lines are all in horizontal planes, ^ = ^-....(2) dn g r Stream lines all circular and in horizontal planes in a liquid, so that h is constant. If v = -, where b is a constant, dp-~w IP -7 = . -y dr g v* p = - - f + constant ---- (3). We see, therefore, that the fall of pressure as we go outward is exactly the same as in the exercise at the end of last article. Example. Liquid rotates about an axis as if it were a rigid body, so that v = br, then d = -Pr, p = 1 -#* + c. This approximately shows the law of increase of pressure in the wheel of a centrifugal pump when full, but when delivering no water. It is the answer already obtained (Art. 175). EXEECISES. 1. The pressure at the inside of the wheel of a centrifugal pump is 2,116 Ibs. per square foot; the inside radius is 0'5 foot; the outside radius 1 foot. The angular velocity of the wheel is b = 30 radians per second. Draw a curve showing the law of p and r from inside to outside when very little water is being delivered. If the water leaves the wheel by a spiral path, the velocity everywhere outside being inversely proportional to r, draw also the curve showing the law of p in the whirlpool chamber outside. 2. The expression 2 i _ + - + *.. ...<4), which remains constant all along a stream line, may be called the total store of energy of 1 Ib. of water in the stream if the motion is steady. ^E 1 dv 1 dp dh , Now = - H T- + T- becomes from equation (1), an g an w an an n being in a normal direction away from the centre of curvature and r the radius of curvature. This expression -=(- + ~j is called the APPLIED MECHANICS. 5-41 -T- = x rotation .... (6). an 'average angular velocity" or " the rotation" of the liquid. Hence 2_t> g 3. Show that the law (1) for a gas under adiabatic conditions, being 4 of Art. 437, the above laws (5) and (6) hold for a gas as well as a liquid. 4. Circular Stream Lines. What is v as a function of r, the radius, if the flow is to be irrotational that is, if a pound of water or of gas has the same energy in one stream line as in another ? Here v/r + dv/dr = 0, or dr/r + dv/v 0, or log. r + log. v c, a constant, or v ' ^ 1\ energy per second per unit volume, or 2 ri z m z /j. l~ 1. Total loss = | 4 irr^V (^2 ~ l ) dr (~~ ~ = 4WlV (- 2ri MISCELLANEOUS EXERCISES. 1. In a force pump used for feeding a boiler the ram has a diameter of 2 inches and a stroke of 24 inches. How many gallons of water (neglecting leakages) would be forced into the boiler for each 1,000 double strokes (one forward and one backward) of the pump ? Ans., 272 gallons. 2. How many gallons of water will be delivered per hour by a single- acting pump (diameter of plunger, 4 inches; length of stroke, 12 inches) making 24 strokes per minute, the slip being 15 per cent. ? How long would it take this pump to fill a tank 10 feet by 6 feet to a depth of 3 feet ? Ans., 665 gallons ; 1 hour, 41 mins. 3. Certain machinery worked from an accumulator requires 20-horse power for a quarter of an hour every hour. The pressure in the accumulator is 700 Ibs. per square inch. If 25 per cent, be allowed for frictional losses, find the size of a single-acting pump which, driven for thirty-five minutes every hour by a donkey engine, making 50 strokes a minute, will just have the ram at the top of its stroke at the beginning of each hour. Assume a slip in the pump of 25 per cent. Ans., -094 cubic foot. 4. In a double-acting force-pump the diameter of piston is 12 inches and the stroke 2 feet 6 inches. The distance from the pump to the well is 15 feet, and from the pump to the place of delivery is 35 feet. Find the horse-power required to work the pump if 30 per cent, is wasted in friction and the number of strokes be 40 per minute. Ans., 21 16. 5. The diameter of a pump bucket being 6 inches, and the vertical lift from the well to the point of delivery being 40 feet, find the load on the bucket. What horse-power will be necessary if the stroke be 15 inches and there are 20 double strokes per minute? Allow 30 per cent, for all losses. , Ans., 489 Ibs. ; -53. 6. The discharge from a pipe is 12 gallons a second. At a point 110 feet above datum level the diameter is 5 inches, and the pressure 2,050 Ibs. per square foot. Find the total energy of each pound of water. If at a point where the pipe is 2 inches diameter and 10 feet above datum the pressure is 1,000 Ibs. per square foot, find the loss of energy between the two points mentioned. Ans., 146 ft. Ibs. ; 70'4 ft. Ibs. 7. The volume of water passing along a pipe running full is 10 cubic feet per second. At one section the area is 2 square feet; at another 544 APPLIED MECHANICS. place, 12 feet below the level of the former, the area of the cross- section is 1J square feet. Find the difference of pressure at the two sections, friction being neglected. Am., 710 Ibs. per sq. ft. 8. At a certain point, 15 feet above datum, in a stream flowing from a reservoir, the still surface of which is 150 feet above datum, the velocity is 20 feet per second. What is the pressure at this point, assuming no loss by friction? Ans., 70 '4 Ibs. per sq. in. 9. Find the power of a waterfall where 2,000 cubic feet of water pass per minute, the height of the fall being 30 feet. In a waterfall, 20 tons of water fall from a height of 36 feet in each minute, and are employed to turn a turbine which transforms six-tenths of the energy of the water into useful work. Find the horse-power of the turbine. Ans., 113-3 h.p. ; 29-32. 10. The vertical distance from still- water level to the lip of a rectangular notch was observed to be '3 feet during an interval of six hours, -65 feet during the next twelve hours, and - 4 feet during the next six hours, the width of the notch being 2| feet. Find the number of gallons passing through during the twenty-four hours. Ans., 1,567,000 gallons. 11. The flow of water in a certain stream is measured by employing a Thomson's V-shaped weir gauge. The vertical distance from still- water level to the lowest point of the notch is observed to be 1-4 feet. The water which passes through has a fall of 20 feet, and is employed to drive a, water-wheel having an efficiency of 60 per cent. Find the horse-power which may be obtained from the wheel. Ans., 8-33. 12. A waterfall is to be utilised for electric lighting. The engineer sent to inspect the place finds out the following data The water at one place flows in a straight rectangular channel 4| feet wide, 2J feet deep, with an average velocity of 3 feet per second. The available fall is 20 feet, and the water-wheels to be used have an efficiency of 62 per cent., the dynamo efficiency being 80 per cent. Neglecting all other losses of energy, find approximately how many 60- watt incandescent lamps may be supplied. Ans., 471. 13. The rim of the wheel of a centrifugal pump goes at 30 feet per second. Water flows radially at 5 feet per second. The vanes are inclined backwards at an angle of 35 to the rim. What is the absolute velocity of the water ? What is the component of this parallel to the rim? If 120 cubic feet of water leave the rim every minute, find the tangential retarding force at the rim. What is the work done usefully per pound of water? Am., 23-4 ft. per sec. at 12-3 with direction of rim ; 22-86 feet per sec. ; 88 -45 Ibs. ; 21-3 ft. Ibs. 14. Suppose water to flow in a steady stream with a constant total head of 100 feet, reckoned from the datum plane and from zero pressure. Determine the discharge into the atmosphere in gallons per minute from a pipe 2 inches diameter at a point 5 feet above the datum. Ans., 638. 15. An orifice 1 square inch area is made in the side of a large tank, at a depth of 4 feet below the surface of the water, and the issuing jet is horizontal. If the jet falls vertically through If feet in a horizontal APPLIED MECHANICS. 545 motion of 5 feet, and the discharge be 16 gallons per minute, find the coefficients of velocity, contraction, and discharge. Am., -97 ; -64 ; -62. 16. Find the discharge in gallons per minute from an orifice 2 inches in diameter in the side of a tank under a constant head of 6 feet, measured from the centre of the orifice. The coefficient of discharge may be taken at '6. Am., 96. 17. In an inward flow wheel the velocity of flow is 8 feet per second, the internal diameter 9 inches, and the revolutions 10 per second. Find the angle of the vanes at exit, so that the water may leave the wheel radially. Ans., 18-75. 18. Determine the velocity with which the water enters an inward flow turbine under a head of 36 feet, the speed of the periphery of the \vheel being 32 feet per second. The vanes of the wheel are radial at entrance, the velocity of flow is constant, and the water leaves the wheoi with no tangential velocity. Ans., 32-56 ft. per sec. 19. What horse-power is required to drive a radial- vaned pump of 15 feet diameter at 50 revolutions per minute when delivering 1 15,000 gallons of water per minute ? What is the efficiency if the lift is 22 feet ? Am., 219; -46. 20. A stream of water, the volume of flow of which is 3,000 gallons per minute, has a velocity of 20 feet per second. It impinges on a succession of curved vanes moving with a velocity of 8 feet per second in a direction inclined at 45 to the direction of the stream. Determine the direction of the tangent to the vane at entrance, so that the water may impinge without shock. If the vanes are circular arcs of 90, find the resultant pressure on the vanes, and the component force in the direction of motion. Am., 21'5 with direction of jet. 21. Find the horse-power developed in a Thomson turbine which is supplied with 15 tons of water per minute, with a forward tangential velocity of 40 feet per second, equal to the speed of the periphery of the wheel, the diameter of which is 2 feet. The water leaves the turbine at a radius of 6 inches, with a backward tangential velocity of 10 feet per second. Ans., 57-27. 22. The diameters of the inner and outer circumferences of an inward flow turbine are 2 feet and 4 feet respectively. The direction of the vanes at their outer ends is radial. Determine the angle at which the inner ends are arranged, supposing that velocity of flow through the turbine is one-eighth the velocity due to the total head, and that of the outer ends is that due to half the head. ^Ans., 19-6 with rim. 23. A turbine with radial vanes receives 50 gallons per minute with an effective head of 28 feet. Find what should be the total area of the inlet passages, and the velocity of the lips of the vanes for maximum efficiency. Ans., 1-51 sq. ft. ; 28 ft. per sec. 24. The wheel of a centrifugal pump is -6 feet in diameter; the turning moment on the spindle is 12 pound-feet. If 160 gallons of water are raised per minute, find the mean velocity with which the water leaves the wheel, assuming that on entering it has no velocity of whirl. Ans., 24 ! ft. per sec. 546 CHAPTER XXV. PERIODIC MOTION. 441. WHEN, after a certain interval of time, a body is found to have returned to an old position, and to be there moving in exactly the same way as it did before, the motion is said to be periodic, and the interval of time that has elapsed is said to be the periodic .time of the motion. Thus, if a body moves uniformly round in a circle, the time which it takes to make one complete revolution is called its periodic time. 442. When a body moves uniformly in a circle, as, for instance, the bob of a conical pendulum, if we look at it from a point in the plane of its circle, it seems merely to swing backwards and forwards in a straight line. Thus, it is known that Jupiter's satellites go round the planet in paths which are nearly circular, but a person on our earth sees them move backwards and forwards almost in straight lines. Now, if we were a very great distance away from the bob of a conical pendulum in the plane of its motion, we should imagine it to be moving in a straight line, and the motion which it would appear to have slow at the ends of its path, quick in the middle would be a simple harmonic mo- tion. To get an exact idea of the nature of this motion in fact, to define what I mean by simple harmonic motion draw a circle, * o' L o" (Fig. 308), and divide its circum- ference into any even number of equal parts. Draw the perpendicu- Fig. 308. lars B' B, c' c, etc., to r \ \ '"J ) V"" c -c >. < C H > >--.< / APPLIED MECHANICS. 547 any diameter. Now, if we suppose a body to go back- wards and forwards along AOL, and if it takes just the same time to go from A to B as from B to c, or from any point to the next, then its motion is said to be a simple harmonic motion. This sort of motion is nearly what we observe in Jupiter's satellites ; it is almost exactly the motion of the bob of any long pendulum or the cross-head of a steam- engine; it is the motion of a point in a tuning-fork, or a stretched fiddle-string when it is plucked aside and set free ; of the weight hung from a spring balance when it is vibrating ; of the up and down motion of a cork floating on the waves in water ; and of the free end of a rod of metal when the other end is fixed in a vice and the rod is set in vibration ; it tells us in all these cases the nature of the motion, when such motion is of its simplest kind. Thus, for example, a cork floating on water may really have a very complicated motion, but if the wave in the water is of its simplest kind, the cork goes up and down with a simple harmonic motion. If you study the figure which you have drawn, and then watch the vibration of a very long pendulum, you will learn about this kind of motion what cannot be learnt by reading. 443. Now let me suppose that the body takes one second to go from A to B, or from B to c, or from any point to the next in Fig. 308. Then the length of AB in inches represents the average velocity between the points A and B, and in the same way we get the average velocity anywhere else. Thus, in the figure from which the woodcut is drawn I find A B c D E F a H i j K Velocity from to to to to to to to to to to to to B C D E F G H I j K L 8 in inches per "1 second J 0-34 1-00 1-59 2-07 2-41 2-59 2-59 2-41 2-07 1-59 1-00 0-34 We observe that the velocity increases as the body ap- proaches the middle of the path, and diminishes again as it goes away from the middle. Now the increase in the velocity of a body every second is called its acceleration, and so we can observe what is the acceleration at every place. You see that the velocity changes from '34 to I'OO near B in one second that is, the acceleration near B is -66 inch per second 548 APPLIED MECHANICS. per second. Similarly subtracting l-OO from 1'59 we find the acceleration at c to be 0'59, and so on. Make a table of these values, and place opposite them the distances of the points B, c, etc., from the centre. In this way we find from the figure the following Table of Values : Distance from o to Acceleration at Displacement divided by Acceleration. B is 9-66 B is 0-66 14-6 C is 8-66 o is 0-59 14-7 D is 7-07 D is 0-48 14-7 BIS 5-00 E is 0-34 14-4 pis 2-59 p is 0-18 14-4 o is o o is o G is 2-59 a is 0-18 14-4 H is 5-00 H is 0-34 14-4 i is 7-07 i is 0-48 14-7 j is 8-66 j is 0-59 14-7 K is 9-66 K is 0-66 14-6 From this it is evident that when the distance of a point from the centre is divided by the acceleration at the point, we get about 14 '6 in every case that is, if we worked more exactly we should have the exact law that the acceleration at a place is proportional to the distance from the centre. This curious property is characteristic of the kind of motion which we are describing. If, again, we draw a number of figures, such as Fig. 308, and divide the circles into very different numbers of equal parts, we shall find that in every case the following law is true : The periodic time of a simple har- monic motion that is, the time which elapses from the moment when the body is in a certain condition until it gets into exactly the same condition again is equal to 6-2832 multiplied by the square root of the ratio of displacement to acceleration given in the third column of the above Table. Thus, in the Table we find the mean value of the ratio (adding all the quotients and dividing by their number we get 14*56) to be, let us say, 14'6. Now the square root of 14'6 is 3-82, and this multiplied by 6-2832 is 24 seconds, which we see by inspection is the periodic time in Fig. 308. 444. We see, then, that if the force acting on a body and causing it to move is always proportional to the distance of APPLIED MECHANICS. 549 the body from a certain point, and acts towards that point, the body gets a simple harmonic motion, and we have a rule for finding the periodic time. The acceleration is always towards the middle point that is, whilst a "body is leaving the middle its velocity is being lessened ; when it is approaching the middle its velocity is being increased. The velocity at the middle is equal to the uniform velocity in the circle from which we imagine the harmonic motion to be derived that is, the velocity in the middle is equal to 3-1416 times the distance A L divided by the periodic time. Suppose the body to be at Q, Fig. 309, moving with a harmonic motion in the path AOL. Describe the circle, draw QP perpendicular to AL, then P is the position of a body which has corresponding uniform circular motion. Let the time t seconds have elapsed since the point a was at o. The corre- sponding point P was then at c. Let the uniform speed of P be v feet per second ; then c P = vt. Let o P = r ; let the angle c o P = vt/r, so that the radius o P moves with the angular velocity v/r. The time T of revolution of P or of complete oscilla- tion of Q is 2 TT r -s- v, or 2 v -5- the angular velocity, which we shall call p. Then o a = o P . sin. o P Q = o P . sin. COP, or, if we call o Q = x, x = r sin. pt. By Art. 19, dx velocity of Q = ^ = rp cos. pt, acceleration of Q = -^ = - rp z sin. pt. Hence the displacement #, divided by the value of the acceleration, 4 7T 2 if we neglect the - sign, is l/p 2 or % (since T = Notice that the acceleration - rp 2 sin.pt is the resolved part, in the direction x of the acceleration of P. The speed of P is v, a constant. If, then, it has an acceleration, it cannot be, nor any part of it, in the direc- tion of its path ; it is, then, in the direction of the radius. Let the acce- leration of P in the direction o P be a, then the resolved part id the direction o L is a cos. P o L = acceleration of Q, or a sm.pt = rp 2 sin. pt, or a= - rp z . The acceleration of P ia, then, centripetal from P towards o, and this amount is rp z . Thus p = -, so that a = - v z /r, the ordinary formula for centrifugal acceleration. Here we have obtained a knowledge of the centripetal accelera- tion mathematically from knowing what is meant by linear acceleration. Conversely, suppose we know that the centripetal acceleration of p in the direction p o is v 2 /o p, then the acceleration of a towards the centre is this multiplied by cos, POL; that is, by 550 APPLIED MECHANICS. o a/o P, and it is therefore equal to . or o a . v*lo p 2 , BO o P o P that it is proportional to o Q. Also^o o, divided by the acceleration (towards the centre) is o p 2 /v 2 or 4 7r 2 /T 2 (since 2 -ir . o ?/v = T). We are therefore led in many wax s to the rule Periodic time = (See also Art. 19.) Displacement Acceleration" 445. JSxample.In Fig. 310, A is a ball of lead weighing 20 Ibs. carried by means of a spiral spring whose own weight may be neglected, let us suppose.* Find by experiment how much the spring lengthens when we add 1 Ib. to the weight of A, or shortens when we subtract 1 Ib. from the weight of A. Let it lengthen or shorten 0-01 foot. Evidently, if ever A is 001 foot upwards or downwards from its position of rest, it is being acted upon by a force of 1 Ib. tending to bring it to its position of rest. We know also that if A is - 02 foot or 0'03 foot above or below its place of rest, there is a force of 2 or 3 Ibs. trying to bring it back. "We see, then, that the up and down motion of A must be simple harmonic. When the dis- placement is, say 0*02 foot, the force acting on A is 2 Ibs., and the acceleration of A is force 2 -f- mass of A ; and as the mass of A is 20 -r- 32-2, or 0*621, the acceleration of A is 3 '22 feet per second per second when it is displaced 0-02 foot from its middle position. Now, employing the rule given above, divide 0*02 by 3 '2 2 and extract the square root, then multiply by 6-2832, and we get 0495 second, or about half a second as the periodic time of the swinging ball. When we make experi- ments we find that, unless the coils of the spring are flat, and the rigid support of A exactly in the axis, the ball has a tendency to turn and vibrate laterally, which disturbs observa- tions if we make careful measurements of the length of swing. 446. Example. The Simple Pendulum. A simple pen- dulum consists in an exceedingly small but heavy body suspended by means of a long inextensible thread, whose * We really assume that one-third of the mass of the spring is added to A. APPLIED MECHANICS. 551 weight may be neglected, capable of swinging backwards and forwards in short arcs. If the arcs are not too long, the time of one swing is always the same. Thus, in Fig. 311, s is the point of suspension, s P a silk thread, p a small ball of lead. p will move backward and forward along the path AOL with a motion which is simple harmonic, provided the thread is so long and A L so short that the force acting on the ball at any time in the direction of its motion is proportional to the distance of the ball from o. To show that this is so, resolve the verti- cally acting weight of the ball in the direction of its motion along A o. We find that it is not quite proportional to AO unless AO is very short, but if this slight' discrep- ancy is neglected the force urging the ball towards o is the weight of the ball multiplied by o A and divided by s A, the distance from the point of support to the centre of gravity of the ball. As a matter of fact, the nature of the vibration does not depend on the weight of the ball ; but, to fix our ideas, let us suppose that the weight is 2 Ibs., then the mass of the ball is 2 -f- 32*2, and acceleration along A o is the force -4- mass, or divided by r|^ ^h< 2 X AO the force, Fig. an. so acceleration is 32-2 x A o -4- s A. Now, our rule is to divide A o by the acceleration at A, and this gives j- } extract the square root, and multiply by 6 '2832 for the periodic time of oscillation of the pendulum. The general rule for a simple pendulum swinging in short arcs is then : Time of a complete oscillation = 6'2832v/ *^ and the displacement divided by acceleration is x - or ^ ; so that the periodic time is 2 TT /L _i V 2 ' g- 449. It will be observed that in all these cases of vibration of bodies there is a continual conversion going on of one kind of energy into another. At each end of a swing the body has no motion; all the energy is therefore potential, whether it is the potential energy of a lifted weight or the potential energy of strained material. In the middle of the swing the body is going at its greatest speed, and its energy is kinetic. At any intermediate place the energy is partly potential and partly kinetic,, but the sum of the two remains. a * 554 APPLIED MECHANICS. always the same, excepting in so far as friction is wasting the total store. Now, in time-keepers the office of the mainspring is to give just such supplies of energy to the balance as are necessary to replace the loss by friction ; and we have to ask the question At what part of the swing of a pendulum or balance can we give to it an impulse which shall increase its store of energy without disturbing its time of oscillation? The answer is this. If a blow is given to the bob of a pendu- lum when it is just at its lowest point, energy is given to the pendulum ; we give it power to make a greater swing, but the time which it will take to make this greater swing is just the same as the time it would have taken for a smaller swing. This middle point is the only point at which we can give an impulse to the bob without altering the time of its swing. In the lever escapement, and in other detached escapements of watches, the impulse is always given just at the middle of the swing. EXEBCISES. 1. A point describes a S.H. motion of 1 foot amplitude in a period of -fth of a second. Find its maximum velocity and maximum accelera- tion. Ans., 21,-jr feet per second ; 294 Tr 2 feet per second. 2. A piston with rod and crosshead weigh 350 Ibs. If they have a S.H. motion with amplitude I'l foot, and if the maximum accelerating force is equal to that produced by a pressure of 15 Ibs. per square inch on a piston 14 inches diameter, what is the periodic time ? Ans., 0-37 sec. OTHER EXAMPLES OF PERIODIC MOTION. 450. When the periodic motion of a body is not simple har- monic, we find that by imagining the body to have two or more kinds of simple harmonic motion at the same time we can get the same result. Thus, it is known that a float, em- ployed to measure the rise and fall of the tide by marking on a moving sheet of paper with a pencil, has a motion which is periodic and not simple harmonic. If horizontal distances represent the motion of the paper (unwound from a barrel by means of clockwork), and ^^^ therefore represent time, ^-^ and if vertical distances mean the rise or fall of Fig. 314. water-level in feet, we get such a curve as is shown in Fig. 314. Now this is not a simple harmonic motion. The difference becomes evident if you plot on squared paper the APPLIED MECHANICS. 555 Fig. 315. distances o A, OB, o c, o D, o E, OF, etc. (Fig. 308), for equal intervals of time, for you will get a curve like Fig. 315, which is easily recognised, and is called a curve of sines or cosines. But it has been found that if we take certain curves of sines whose periodic times are 1, the semi-lunar day ; 2, the semi-solar day, and some others, their amplitudes and epochs being properly chosen and draw them on squared paper, and add their ordinates together, we get the curve which shows the real rise and fall of the tide. In the very same way we can combine simple harmonic motions to arrive at any periodic motion. A good way of combining simple harmonic motions experimentally is to let a body hang from a string which passes over two or more movable, and the same number of fixed, pulleys. These pulleys are pivoted on crank pins, and their pivots are made to revolve at any desired rela- tive speeds, and each gives to the body a purely simple har- monic motion by its action on the string. The body gets a motion compounded of the mo- tions of the pulleys, and if it is an ink-bottle or pencil press- ing on the paper on a revolv- ing paper roller, we get a time curve of the periodic motion. This is the principle of the construction of Lord Kelvin's Tide-Predicting Machine. 451. When a body can swing east and west under the influence of forces which have Fig. 310. no tendency to move it except in a direction due east and west, and if forces acting due north and south can make it swing in their direction, then both sets of forces acting together on the body will give it a motion 556 APPLIED MECHANICS. compounded of the two simpler motions. Thus, a ball A (Fig. 316) is suspended by a string, PA, which is knotted at p to two other strings, P s and P s', equal in length, and fastened at s and s'. The ball may swing in the direction E o w as if it were the bob of a pendulum hung directly from the ceiling at P', but it may also swing in the direction N o s at righb angles to E o w, and if it does so it swings as if the point p were the fixed end of the pendulum A p. When it swings under the influence of the two sets of forces tending to make it move both ways at once, the motion of A is compounded of the other two simpler motions. If p A is one-quarter of the length o P', then the east and west swing takes twice as long as the north and south M swing. If p A is one-ninth of o P', then the east and west swing takes three times as long as the north and south swing. The motion of A is sometimes very beauti- ful, and the experiment is easily arranged. 452. The motion is quite easily represented on paper. Thus, in Fig. 3 17, A' M is the north and south direction, and A M, at right angles to it, is the east and west di- rection. Let the points 0, 1, 2, etc., in each of these lines be found as in Fig. 308. Let the bob be supposed to go from to 1 in A' M in the same time as it goes from to 1 in A M. Notice that we have twice as many points in AM as in A' M, showing a slower oscillation in the direction A M. We can begin to number our points anywhere, remembering that when the bob completes its range it comes back again in the opposite direction. Now put marks where the east and west lines meet the north and south ones, drawn through corresponding points. It is evident that the curve drawn through these successive marks is the real path traced out by the ball when acted upon simultaneously by the two sets of forces urging it in a north and south, and an east and west direction. If we have the same number of points in A' M as in A M \ Fig. 317. APPLIED MECHANICS. 557 we get a circle, ellipse, or straight line, as in c, B, A, Fig. 318. This represents the motion of a conical pendulum free to swing in every direction. Again, D, E, F, and many other curves that might be drawn, represent the case which we took up in Fig. 317, where one vibration is twice as quick as the other. If the time of vibration in A M is to the time of /*./. Fig. 318. vibration in A' M as 2 to 3, we get curved paths like a, I, j, and so on. In experimenting with the pendulum, Fig. 316. it will usually be found that slight inaccuracies in the lengths of the cords will cause a continual change to go on in the shape of the path traced out by the ball. We can produce these motions by spiral springs, and in other ways. Thus, for example, if we use instead of the strip 558 APPLIED MECHANICS. of steel, in Fig. 312, a combination of two strips, B and B', as in Fig. 319, so that the heavy bright bead A is capable of vibrating in two directions at the same time, we get the same combinations of simple harmonic motions, depending on the point at which B is held in the vice c. 453. When a body has a periodic rotational motion about an axis like the balance of a watch or a rigid pendulum, we must no longer speak of the force causing motion, and the mass of the body, and the distance of displacement; but if we sub- stitute for these terms, moments of forces, moment of inertia Fig. 819. of the body and angle of displacement, we have exactly the same rule for finding the periodic time of oscillation. The periodic time is 6'2832 times the square root of the an^lar displacement of the body at any instant, divided by the angular acceleration at that instant. And we know that angular acceleration may be calculated by dividing the turning moment acting on a body by the moment of inertia of the body. A point in the balance of a watch swings in circular arcs, but if ,we only take account of the distances which it passes through, and suppose it moved in a straight line instead of in the arc of a circle, the motion is very nearly simple harmonic. If there were no friction or other forces acting on the balance except the turning moment of the balance spring (see Arts. 521-2-3), and if the moment of the spring were always exactly proportional to the angular displacement of the balance, the motion would be simple harmonic. We shall see in Art. 522 that the turning moment due to the spring is -r-^-j 9, if E is the modulus of elasticity of the spring, b its i 2i i breadth, t its thickness, and I its length, and if is the angular displacement in radians. Angular acceleration is this moment divided by moment of inertia i of the balance, or =-=-, . Hence l L l i 1 2 l\ angular displacement 9 divided by angular acceleration is . 3 so that the periodic time of the balance is T = 6-2832 ./ APPLIED MECHANICS. 559 Increasing the moment of inertia of the balance or the length of the spring makes the vibration slow. Increasing the breadth and, what is still more important, increasing the thickness of the spring makes the vibration quick. As we shall see in Arts. 521-3 that our calculation of the turning moment of the spring is not quite right ; that the dimensions of the balance and spring alter with temperature, and that, above all, the elasticity of the steel alters with temperature, and with its own state of fatigue, the rule is not perfectly true, nor can any balance be regarded as taking exactly the same time for its oscillation in different lengths of arc. At the same time it is of great help to the watchmaker to know that with con- siderable, although not with perfect accuracy, the time of vibration of a balance is proportional to the square root of the length of the spring, and so on. For example, suppose the spring is 3 inches long, and the balance makes one swing in 0*251 second, now if he wishes it to make a swing in 0'25 second, he must shorten it in the ratio of '251 x '251 to 25 x -25, or in the ratio -063001 to 0625, so that the length of his spring ought to be 3 x '0625 + -063001, or 2-976 inches that is, it ought to be shortened -024 inch. In the same way he can calculate the effect of adding little masses at any distances from the centre of the balance, so that its moment of inertia may be increased, and the balance made slower in its swing. The same law tells him how he can compensate the balance, so that when in summer the steel of the spring loses its elasticity, some of the mass of the balance will come nearer the centre, in order that the moment of inertia may diminish in the same proportion. 454. Compound Pendulum. The simplependulum described in Art. 446 is not like the pendulums used in practice. In these the bob is not so small that we can consider it as a point ; the long part is not a thread but a stiff rod of metal or wood, and there is usually a knife-edge for support, about which it can turn with little friction. In common clocks, however, the top end of the pendulum is a thin strip of steel held firmly in the chops, but the easy bending of this strip is such that we may imagine an equivalent pendulum to move freely about an axis. Employing our general rule of Art. 453, we find how to calculate the time of vibration. This compound pendulum vibrates in the same time as a certain simple pendulum, called the equi- valent simple pendulum^ whose length we find by experiment. 560 APPLIED MECHANICS. In Fig. 320 let 8 be the axis of suspension, a the centre of gravity, and P a point in the continuation of the line s G such that s P is the length of the equivalent simple pendulum. Then P is called the centre of oscillation, and it is also known to be the centre of percussion of the pendulum (see Art. 402). It can be proved that if the pendulum be inverted and made to vibrate about a parallel axis through P, it will vibrate in exactly the same time as it does about 8 \ and it was in this way, by in- verting a pendulum which had two knife- edges, and adjusting these until the pendulum took the same time to vibrate about one as about the other, and then measuring the distance between them, that Captain Kater found the length of the simple pen- dulum which vibrates in a given time. This method is still employed in gravitation experiments everywhere to find the value of g, which is 32-2 feet per second per second at London. It is an excellent laboratory exercise. If s is the axis of suspension, G the centre of gravity, w the weight of the pendulum, then the moment with which gravity urges the pendulum to return to its position of rest is w x G N ; but if the angle G 8 o be measured in radians, and if it is very small, this moment is almost exactly equal to w x s G x angle G s o. The angular acceleration is obtained by dividing this by i, the moment of inertia of the pendulum about s, and our rule becomes or 6-2832 T = 6-2832 angle o s N . s G . angle G s N -j- I* V. .... (1). . SG When we examine this formula we see that it may be put in another form. Find a point K. such that if all the mass of the pendulum were gathered there its moment of inertia about s would ry be the same as at present ; in fact, such that ^77-7,, the mass of the o/'Z pendulum x s K 2 , would be equal to i. The distance s K is called the radius of gyration of the pendulum (see Art. 112), and our rule now becomes ' T = 6-2832 A / 8 V ff ---- (2),. APPLIED MECHANICS. 561 .where g is 32 '2. In the simple pendulum SK and s G are equal, and (2) gives the same rule which is given in Art. 446. However, in an ordinary pendulum, s K and s G are not equal, but s K 2 -5- so is equal to some length such as s P, and our rule becomes i = 6-2832 Evidently s P is the length of the imaginary simple pendulum which mould vibrate in the same time as our real pendulum. The imaginary point P has been called the centre of oscillation, because when the pendulum is inverted and made to vibrate about an axis through P it vibrates in the same time as before. To prove this it is necessary to return to equation (1). We know that i is equal to the moment of inertia of the body calculated as if all its mass existed at G, together with the moment of inertia of the body as it is at present, but calculated about an axis through o parallel to the present axis ; that is, I = W SG 2 + ^ 9 9 where k is some length unknown to us just now, being the radius of gyration about the axis through the centre of gravity. Rule (1) becomes G-2832 or T = 6-2832 V V . S G That is, the length of the simple pendulum which will vibrate in the same time is so + , and we have already found it to be s F 7i 2 in equation (3) ; so that GP = , or G P x s G = & 2 . But in the S G very same way, if we considered the pendulum as vibrating about p, we should find the length of the equivalent simple pendulum to be greater than a P by an amount equal to , and we know that s G is equal to this amount ; so that s P would, as before, be the length of the equivalent simple pendulum. The axes of oscillation and suspension are therefore interchangeable. 455. Examples. The bar of Fig. 321 with two adjust- able masses may be fixed to one end of a wire, the other end of which is fixed to the ceiling. By twisting and untwisting the wire the bar will oscillate with a motion which is much more nearly simple harmonic than that of the balance of a watch. Students who experiment with such a bar can adjust 562 APPLIED MECHANICS. the weights A and B at any distance from the axis (there ought to be an engraved scale on the har), so that the moment of inertia can be varied. They can fasten the bar at the end of a wire, or they can use it as in Fig. 321, with a flat spiral Fig. 821. spring, or as in Fig. 322, with a cylindric spiral spring ; and the rate of its vibration gives one of the best ways of in- vestigating the twisting moments of wires and such springs when strained through given angles. In the case of a wire the twist always tends to bring the bar to its position of rest with a moment which is proportional to the angle of displacement from this position it is this property which causes the motion to be simple harmonic. This moment is also proportional to the fourth power of the APPLIED MECHANICS. 563 diameter of the wire, and it becomes less as the length of the wire is increased. By means of a circular scale and a pointer we can measure the extent of each swing, and this is found to decrease gradually, due to friction with the air and the in- ternal friction or viscosity of the metal. The amount of Fig. 822. diminution of swing gives us a means of determining the viscosity, and the apparatus can so easily be fitted up that no person who wishes to understand the properties of materials can be excused from making these experiments. This is a common method of finding N, the modulus of rigidity of a material. If the length, of the wire is I inches, its diameter d, and if N is its modulus of rigidity (see Table XX.), then from Art. 295 we seo that the moment with which the wire acts on the bar. when its 564 APPLIED MECHANICS. angle is from the position of rest, is ~ N - d*. If the moment of inertia of the bar is i (we axe neglecting the fact that the wire itself has some mass which has to be set in motion), then the moment, divided by i, is the angular acceleration ; and using this quotient as denominator, and 6 as numerator, extracting the square root, and multiplying by 6-2832, or 2 IT, by the general rule of Art. 453, we find the square of the period of a complete oscillation to be x 8 = N ^ 4 -. If we are in doubt as to our calculated i, we can find it experimentally ; and this is very necessary in many magnetic experiments. "We add a known moment of inertia i (say two equal small masses at equal distances from the axis), and find the new T (call it TI ). Then T^/T 2 = (i + i)/i, and i may be found. When motion is slow, the friction in fluids is proportional to the velocity, and any friction which follows this law is called fluid friction. A great many vibrating bodies tend to come to rest by the action of such friction as this; and it is found that if the friction is numerically / times the angular velocity, then the logarithm of the ratio of the length of one complete swing to the next is nearly equal to k times the periodic time. Hence this logarithmic decrement, as it is called, is proportional to the friction co-efficient. If we observe twenty-one elongations on one side of the middle position, then one-twentieth of the logarithm of the first elongation divided by the last is k times the periodic time of oscillation. EXERCISES. 1. Bifilar Suspension. In many measuring instruments a body is suspended by two thin wires nearly vertical. If the vertical length of each of these is /, the distance between their ends at the top a, and at the bottom b, and the weight of the body w, it is easy to show that for a small angular displacement 6, the moment tending to bring the body to its position of rest is very nearly (neglecting . torsion of the wires them- selves) %-:-wQ. Find the time of vibration of such a body when its moment of inertia is known. In truth, the constraining moment is proportional to sin. 6. 2. A magnet, turning on a frictionless pivot at its centre of gravity, is subjected to a turning moment H sin. 6, or very nearly H 9, due to the earth's magnetic action, if it makes only a small angle 6 with its position of rest. Find the time of a vibration if the moment of inertia is known, and show that the square of the time of vibration of the magnet in different places is inversely proportional to H. Find what is the effect of adding a known moment of inertia, and show that the observations on the APPLIED MECHANICS. 565 two times of vibration enable us to calculate the original moment of inertia if it was unknown. 3. Prove that the time of complete oscillation of a ship is 2-jrkj */gd, where d is the distance from the centre of gravity to the metacentre. 456. Stilling of Vibrations. When a simple harmonic motion is represented on paper in the manner described in Art. 450, we have a curve of sines. The curve may be ob- tained by producing the lines B B', c c', etc., of Fig. 308, cutting them at right angles by equidistant horizontal lines, and joining the successive points of intersection so found. It may also be drawn by finding from a book of tables the sines of 0, 10, 20, etc., and plotting and sin. 0, 1 and sin. 10, 2 and sin. 20, etc., on a sheet of squared paper. A curve of sines expresses the fact that, if d represents the displacement of a vibrating body from its middle position after an interval of t seconds since it was at the middle of its course, then d = a sin. pt where a is the greatest displacement of the body from its middle position. This displacement is usually called the amplitude of the vibration. If T is the time of a complete vibration, it is easy to see that the equation is d = a sin. ~t t or a sin. 2irf . t if/ is the frequency or number of complete oscillations per second. If we make the bob of a pendulum terminate below, in a tube which can act as a pencil-holder, and in which a well- fitting pencil can slide freely, and if we move a sheet of paper at a uniform rate underneath this pencil at right angles to the direction of motion of the pencil, a curve of sines will be traced out, if the pendulum swings without friction. But in practice we always find that, what with the friction at the point of support, friction with the atmosphere, etc., a pendulum's swings get smaller and smaller that is, the amplitude of the vibration gets less and less as time goes on, until the pendulum at length comes to rest. This motion is not a simple harmonic motion, but, within certain limits, each swing may be regarded as very nearly a simple harmonic motion. Practical men who deal with oscillating bodies, such as pendulums, ships, tuning forks, magnetic needles, and suspended coils of wire, usually assume that the motion during each swing is a simple harmonic motion. The frictional resistance to motion of any ordinary vibrating body in a fluid medium, or of a magnetic needle vibrating near any body capable of conducting electricity, is almost always su-ch that the quicker the motion th,e 566 APPLIED MECHANICS. greater the friction (see Art. 64) that is, fricticnal resistance is proportional to speed ; and in this case it is not difficult to show that, instead of the law d a sin. t . . . . (1), we have the law d = ae~ at sin. t . . . . (2). That is, if the strength of the spring, or other governor of vibration, and the character of the vibrating body are such that without friction the law would be (1), then, when the vibration is damped by frictional resistance of the above character, the law of the motion becomes that given by equation (2). Here o is a constant which depends on the character of the friction. Thus a is greater when a pendulum swings in water than when it swings in air. Also, T I} the periodic time of the vibration, is no longer the same T as it was for undamped vibrations, and the 11 2 rektion between T and TI is -^ = ^ + -f-^ .... (3) ; or, if / is the undamped frequency, and /i the damped frequency, then o 2 y 2 =/i 2 + 7 g .... (4). In order to get exact ideas on this subject of ?& damping of vibrations, the student ought to plot on squared puper a curve such as o A' B' c' D' E' F' G' H' i, Fig. 323, which corresponds with equation (2). Thus, let us suppose that a body undamped in its vibrations gets an impulse which sends it from its position of rest in such a way that its amplitude is 10 inches, and let the time of a complete oscillation be 1-6 second. 6*2832 Then the law of its motion would be x = 10 sin. t, or 1*6 x = 10 sin. 3-927 1 .... (5) where x is in inches, t in seconds, and the angle 3-927* in radians.* If, now, the friction is such that a = 07, we find from (3) that the time of an oscillation is practically unchanged. Find, there- fore, the original curve of sines by calculating the second column of the following table. The numbers of the first two columns plotted on squared paper would represent the undamped vibrations. But for damped vibrations the numbers of the second column have all to be multiplied by ~'^*; and if we denote this multiplier by the letter z, we see that z being 6-' rf , or log. z = - 0-304 1. We have calculated z for the various values of , and placed the results in the third column. Multiplying, therefore, the respective numbers of the second and third columns together, we get the fourth column of numbers ; and plotting the numbers of the first and fourth columns on squared paper, we find the curve which shows the nature of the damped vibrations. * We may write (5) in the form x 10 sin. 225$. In this case the angle 2251 is expressed in degrees. 568 APPLIED MECHANICS. t in seconds. 10 sin. 3*927 1, or 10 sin. 225 1, if angle is taken in degrees. f -OT 6 - ' 1 10 sin. 3-927 f 1 0-2 7-07 869 6-14 0-4 10 756 7'56 0-6 7-07 657 4-65 0-8 571 1-0 - 7-07 497 - 3-51 1-2 - 10 432 - 4-32 1-4 - 7-07 375 - 2-65 1-6 326 2-0 10 247 2-47 2-4 186 2-8 - 10 141 1-41 3-2 106 3-6 10 080 8 4-0 061 4-4 - 10 046 - -46 4-8 035 5-2 10 026 26 5-6 020 6-0 - 10 015 - -15 6-4 Oil 4:57. Some students may find it as instructive to first draw a curve of sines, then draw the logarithmic curve, correspond- ing to column three, on the same sheet of squared paper, and multiply the ordinate of one curve by that of the other to get the ordinate of the real curve which exhibits the damped vibrational motion. This is what has been done in Fig. 323 ; OABCDEFGHI is the curve of sines, L p Q is the logarithmic curve, showing how rapidly the amplitude of the vibration diminishes, and o A' B' c' D' E' F' G' H' I is the curve which repre- sents the actual motion of the vibrating body. In this figure the logarithmic curve is drawn to such a scale as seemed con- venient for showing its properties distinctly. It would be very easy to dilate on the nature of the resulting curve o A' B', etc., but this book is written to "help students who are earnest enough to calculate the above numbers and plot the curve, and when they perform these operations they will have very clear notions about the motion we have been investigating. Exercise. A heavy disc, suspended by a wire, vibrates in each of a number of fluid media, its periodic time of vibration in all APPLIED MECHANICS. 569 being sensibly the same, or 1'5 second. The ratio of the amplitudes of two successive swings in one direction being 0'9 in one fluid that is, the second swing being only nine-tenths of the first, and the third being only nine-tenths of the second, and so on and 0-8 in another fluid, and 0*7 in another, what numbers will express the relative viscosities of these fluids ? Here we have, taking common logarithms, 09 = -l'5a f or ^ e first fluid, so that - log. 0'9 = 1-5 a log. e, or a = ~ g ' 9 , that is, o = 0-07. In the same way a = 0-15 and a = 0-24 for the other fluids, and hence 7, 15, and 24 are the required numbers expressing the relative viscosities as measured by the vibrating disc method. A very slowly swinging disc and pointer will enable us to plot the complete curve from actual observations. The nature of the motion when the friction is that of solids rubbing on solids is studied in my book on the Calculus. 570 CHAPTER XXVI. MECHANISM. 458. IN Art. 179 we pointed out how skeleton drawings and models may be made useful incases where velocity ratios vary greatly. We shall now give a sketch or outline of a general theory from which students may find benefit if they fill it in for themselves. We cannot say that there is as yet any satisfactory treatise on this subject. The most interesting part of it (to us) is that which concerns the mechanism of steam and other engines, and this will be found in the author's book on the Steam Engine. 459. We have already referred to spur and bevil gearing used to drive one shaft from another at uniform velocity ratios. Consider at any instant two teeth in contact ; each of them is a rigid piece rotating about a fixed centre and acting on the other by rubbing contact. We shall now briefly refer to this way of transmitting motion. Let A w be a body moving in the plane of the paper about the axis A, let B v be another body moving about the axis B, and let these two bodies keep in contact, as we see them in contact at I. Now, this keeping in contact, what does it mean ? It is that, whatever may be the rela- tive tangential or sliding motion at the point of con- tact P, they have the s.ame velocity in the direction of their common normal there at the instant when we study them. Consider their motions during an exceed- ingly short interval of time St. There are two points P to be considered. One is on the body A w, and it moves to Q, where P Q is an arc of a circle about A as centre ; the other P is on the body B v, and it moves to n, where P R is an arc of a circle about B as centre. The angular velocity of A w being a, the angle P A Q is a . St ; the angular velocity of B v being b, the angle P B R is b . St. As St is con- sidered' smaller and smaller, P a and PR may be considered more and more nearly short straight lines. P Q = A P . a . St and p R = B P . b . St. Now observe that if T P s is the direction of APPLIED MECHANICS. 571 the common normal to the two bodies at P, the normal component p s of the motion p Q must be the same as that of the motion P R ; in fact, the straight line Q, s R must be at right angles to the normal T p s. Now draw T E parallel to A p, and observe that the two triangles T E p and a p R have their sides Q R and T p, K p and p E, p Q, and E T respectively at right angles to one another, and hence they are similar ; that is, QR : RP : PQ = TP:PE:ET. ...(1), TT Hence or = . P Q E T B P . PE b A P . # ET' a BP.ET But because TE is a line drawn parallel to AP, a side of the triangle A p B, we have AP AB P E AT. E T ~ B T an B P ~~ A B ' therefore AF ' PE = ^- T ET . BP BT Hence * = il ____ (2). a BT Hence the ratio of the angular velocities is inversely as the segments into which the common normal at the point of contact divides the line of centres. Hence, if the ratio of the angular velocities is constant, the common normal at the point of contact p passes always through the same point T in the line of centres A B. Notice also that ^ _ !?. But p Q ~ a.AP.S*, TE = AP P Q T E A B and - - j. Hence it will "be found that AB a + b 0, R = P T (a + b) St. That is, the slipping speed at P is the speed at the end of a radius T P when the radius revolves at the angular velocity a + b. Hence there is always slipping at p, unless P is on the line of centres and the speed of slipping is proportional to the distance P T. The student will find that when there is friction, if DPT = D' p T = angle of repose, then p D is the direction of the mutual force when AW rotates as shown, with the hands of a watch; whereas P D' is the direction of the mutual force when A w rotates in the opposite direction, so that B v is the driver. If we have given the shape of B v P, and we desire to find the shape of a piece A w p which will gear with it at a constant angular velocity ratio, make a template of B v P, and arrange that when this template is moved about the fixed centre B a sheet of paper shall move about A through the proper angular distances. If for each position of the two the curved shape of B' p v' be drawn on the paper, the pencil marks will show the proper shape of A w P if it is to touch B p v. In fact, A w p will be the envelope of the shapes of B' p V. When B' p v' is circular with B as centre, A w p will also be circular. True rolling will be possible, for r will be at T, and as T will be constant in position the angular velocity 572 APPLIED MECHANICS. ratio will be constant. When B' p v' is the arc of an ellipse whose focus is at B, and if the distance B A is equal to the major axis of the ellipse, A w P will be a similar ellipse ; there will be true rolling, with changing angular velocity ratio. The other foci may be connected by a link. If B' p v' is shaped like an equiangular spiral whose pole is at B, A w P will be a similar spiral, and there will be true rolling between them, but with changing angulai velocity ratio ; in this way lobed wheels are formed to gear together. 460 Discs or cylinders touching each other, their axes parallel, are used for friction gearing. If the horse-power H is to be trans- mitted, and v is the common circumferential speed in feet per minute, P being the necessary tangential force, P = 33,000 H/V. Slipping is to be impossible, and therefore p -f- /j. is the force necessary to press the cylinders together. If this force acts through the bearings of the two shafts, it is usually found in practice that, unless at very high speeds and with small power, there is so much practical difficulty that the gear is never used. Compressed paper and leather have been used to work with iron. Sometimes nest-gears are used to produce the necessary pressure, tout when the necessary pressure has been produced it has led to disintegration of the surface, or such local elastic changes of shape as produce annoying sound. In one case, where the driven pulley is very heavy and the pressure is produced by its own weight and that of the spindle and part of the weight of the rotating armature of a little dynamo machine, the gear has been used satisfactorily. Wedge-shaped grooves and projections have been cut in the rime of the pulleys, and sufficient grip has been produced in this way, but there is no longer true rolling. 461. When we attempt by using teeth to get the necessary driv- ing forces we introduce sliding contact, using spur, bevil, and skew bevil wheels ; the names pitch circles, pitch cones, and pitch hyperboloids being used for the friction gear, which would run with the same velocity ratios. When the axes are not in one plane, frusta of hyperboloids, generated by the rotation of the same straight line round both axes, will gear with one another, always touching along a straight line. But there will not be simple rolling ; there is sliding along the line of contact. Every student ought to study the shapes of spur-wheel teeth ; it is easy to apply one's knowledge to other kinds of teeth and rubbing and rolling gear. Nothing illustrates the fact that we do not really think, so well as this, that all the principles for the proper construction of worm-wheel teeth and chain gearing were to be found in books many years before there existed any good worm-wheel teeth or chain gearing. The subject, like all other parts of machine design, is best studied when one draws things to scale, and there are now many books to assist the student in machine design. Perhaps it will be well to neglect almost all the mathematical parts of such books on strength. 462. Suppose we have two curved rollers, v T v moving about the axis B and w T w about the axis A, and suppose that these are capable of rolling on one another as they rotate ; they touch at T, and the angular velocities about A and B are as B T to A T. Now APPLIED MECHANICS. 573 suppose we wish wheels, with teeth centred at B and A, to have exactly the same angular velocity ratios as the rollers ; if B' p V and A' P w' are the shapes of the teeth in contact at P, it is necessary that during the motion the common normal at p should pass through the pitch point T. To effect this object it is necessary that B' P v' and A' P V should be two troch- oidal curves, generated by the rolling of a curve inside the curve T v and outside the curve T w. Thus in Fig. 325 imagine the curve v and the curve w to move about B and A, rolling on one an- other and keeping in contact at T in the straight line B A, and imagine the curve p P'T to roll also, keeping in contact with both v and w at T. It is really rolling inside v and outside w. If any point of P P' T, such as P, is always on the contours of the two teeth, the straight line PT is al- ways normal to these teeth at P, their point of contact, and we have ensured that the com- mon normal to both passes through the pitch point T. w and v may be ellipses, but they are generally circles, A and B being their centres. If P P' T is a circle, rolling inside v and outside, w, the trochoidal curves are called hypo- and epi-cycloids. When a number of wheels are to gear any one with any other, we usually choose one rolling circle for the insides and outsides of all the pitch circles ; it is taken of a diameter equal to the radius of the smallest pitch circle. When it rolls inside this smallest pitch circle the hypocycloid is a* radial line. A rack may be regarded as part of a wheel of infinite diameter. Sometimes the trochoidal curves are involutes of circles, the rolling curve being really a straight line or an infinite circle. Let w and v (Fig. 326) be the pitch circles. Draw any line c D through T, the pitch point, and describe circles with B and A as centres touching this line at D and c respectively. Draw through T' o T H, the involute of the circle D o, and F T E, the involute of the circle OF. If G H and E F be the contours of teeth rotating about B and A, their common normal remains always CD. Thus at any point T' in c D, if we draw 325. 574 APPLIED MECHANICS. involutes to the two circles, they must touch there, because a tangent to a circle cuts all the involutes at right angles. It is evident that a pair of wheels with involute teeth may have less or more distance between their centres without any alteration of their velocity ratio, this being B T : A T or B D : A c. If there is no friction, c D is the direction of the driving force. It is usually more oblique to the line of centres in wheels with involute teeth than in cycloidal teeth. For the most accurate work the actual trochoidal curves are drawn by the rolling of templates, but there are well-known drawing-office rules by which we approximate to the true best curves by means of arcs of circles. 463. In Art. 462 we showed how, when given the shape of a tooth, to find the shape of another tooth to gear with it with constant velocity ratio. A good example of the application of this rule is found when designing the shape of the tooth of a worm wheel. The shape of the thread of the worm being chosen, any section of the worm thread by a plane at right angles to the axis of the worm wheel may be drawn by elementary practical geometry. Eegard it as the tooth of a rack. The section there of the tooth of the worm wheel must be the shape of a tooth to gear with the given APPLIED MECHANICS. 575 rack tooth, and must be drawn by some such rule as we have given in Art. 462. 464:. Sometimes pieces centred at A and B, acting upon one another, do not act directly as shown in Fig. 324, but through the Fig. 327. Fig. 328. agency of a third body. Thus, if pins on v and w are connected by a link w v, and if the pins are frictionless, we know that the direction of the force at any instant must be in the direction of the centres of the pins or in the line w v. It is easy to show as before that if a is the angular velocity of A w and b the angular velocity of B v, then -, = if T is where w v cuts the line of centres A B. The b TA' complete study of the relative motion of four links such as AW, w v, v B, and B A is of course a very complicated business if we imagine them to be of all sorts of lengths. We may, if we please, imagine any one of them fixed and consider the motions of the others. Thus in Fig. 328 consider A B to be fixed. Think that A and B are merely pins in some fixed object of any shape whatso- ever. Now consider the other pieces to be of any curious shapes and lengths. The motion is taken to be in the plane of the paper, w v is only a straight line joining the centres of the two pins w and v ; but imagine w v to represent any curiously shaped body we might wish to know the motion of any point in this body. The student's great aim is to get a correct mental picture, and if he recollects that w is moving about A as an instantaneous centre or at right angles to A w, and v is moving about B, the whole body w v must just at the present instant be moving about the point c as a centre. We have produced w A and v B to meet at c, the instantaneous centre of motion of the whole body w v. One has then a mental picture of the motion of any point whatsoever in the body vr v. 465. General motion of a body parallel to a plane. Let us gimply say a plane figure or body in its own plane. If we consider 576 APPLIED MECHANICS. the motion of any two points, this settles the motion of every other point, so we shall speak only of two points. It is easy to prove that when we consider two positions of the body there is one point of the body whose position is the same. Thus, if wv is one position and w' v' is another position, bisect w w' by a line at right angles to it, and let it meet the rectangular bisector of v v' in c. Evidently c is a point in the body which has not altered its position; it is the instantaneous centre of the motion. Notice Fig. 329. Fig. 330. that if w w' is parallel to v v', c is at an in finite distance ; we then say that the motion is one of translation merely. In Fig. 329 we have a piece, w v, whose ends move in the paths V w and v' v. At any instant, if we draw w c and v c, normals to the paths, we find c the instantaneous centre. Consider any point p in w y, and join p p. P traces out some path during the motion. Notice that p c is the normal to this path, and a line through p at right angles to p c is the tangent to it. If we want to consider the envelope of the straight line w v, notice that the foot of the perpendicular from c upon w v is a point in that envelope, because it is the only point of w v which moves in the direction of the length of w v. The student must not imagine that because at the instant every point of a body is moving about c, therefore c is the centre of curvature of the path of w, or of v, or of p. c is only for the instant the centre of motion, and such lines as we, v c, and p c are the directions of the normals to the actual paths of these three points. As the figure moves in its own plane from one position to others successively, let Ci c 2 c 3 , etc., be the successive points of the figure about which the rotations take place, and let c\ c 2 c 3 , etc., be the positions of these points on the fixed plane when each is the instantaneous centre of rotation. Then the figure rotates about Ci (or Ci, which coincides with it) till c 2 coincides with c 2 ; then about c z till c 3 coincides with c 3 , and so on. Hence, if we join Ci CaCg, etc., in the plane of the figure, and c\ c 2 c s , etc., in the fixed plane, the motion will be the same as if the polygon Ci c 2 c 3 , etc., rolled upon the fixed polygon c\c z c^ etc. By supposing the successive displacements smaller and smaller, we have curves, and hence any motion whatever of a plane figure in its own plane may be imagined as produced by the rolling of a curve fixed to the figure upon a curve fixed to the plane. It immediately follows that any displacement of a rigid solid parallel to a plane may be produced by the rolling of a cylinder fixed in the solid on another APPLIED MECHANICS. 577 cylinder fixed in space, the generating lines of the cylinders being at right angles to the plane. 466. What we have said about motion in a plane is true about motion of figures shaped to fit a spherical surface ; straight lines in the one case corresponding to great circles of the sphere in the other case. It is easy to have a mental image of this. Now MR. 381. imagine that any point of the superficial spherical figure is joined to the centre of the sphere, and we see that (1) if a rigid body has one point fixed (imagine this the centre of an imaginary spherical surface), however it may move, in any t~ffo positions of the body there is one line of the body which is common to the two positions. (2) Any motion may be regarded as due to the rolling of a cone fixed in the body upon a cone fixed in space. What we have said, therefore, about the pieces AW and B v of Fig. 327 moving about axes at right angles to the plane of motion may be at once applied to pieces moving in a spherical surface about axes meeting in the centre of the sphere. Or the cylindric pieces A w and B v may be imagined to be conical pieces moving about axes A and B, meeting at the vertex of the cones. Hence all that we have said about spur wheels is at once applicable to bevil wheels, which are suitable for shafts whose centre lines or axes meet at a point. 467. Fig. 329 is worth a very great deal of consideration. There is a link w v, and motion of a point w in it is known. The shape of the path of v is known, to find the motion of the whole link, andjof any point in it, at every instant. When the paths of w and v are arcs of circles, we have the four-bar kinematic chain of Fig. 327. When w (a crank pin) moves in the arc of a circle, and v is a block moving in a straight slot, one particular case is called a slider crank chain (see Fig. 332). When both w and v are blocks moving in straight slots at right angles to one another, we have the ordinary trammels for describing an ellipse. We have the mathe- matical basis of the elliptic chuck, for we may imagine any part of a mechanism to be at rest, and tnen the relative motions of the others become absolute motions. The theory, then, of Art. 464 is the same for a very great many mechanisms. 578 APPLIED MECHANICS. 468. In Fig. 327, if we imagine B A to be fixed, B in the position in which it is in the figure, w A and B A infinitely long, so that w's arc of a circle is really a straight line, we have the usual crank B v and connecting rod v w, which we can indicate in its simplest form by Fig. 332. Now take this as it stands. A Fig. 332. piece B A x along which w slides, and the con- necting-rod w v and the crank v B ; think merely of the relative motion of each piece to the rest. You may imagine any one of the pieces fixed, and you may consider the motion of any point in any piece. If BX is fixed, v describes a circle, w a straight line, and any point in w v describes a curve which is some- what like an ellipse, only blunter at one end than the other. Now imagine BV = vw: v will not make a rotation. Imagine w v extended beyond v as far again : that point will describe a straight line, and we have a parallel motion. A student must try these things with models. Now imagine B v fixed, and let vw turn uniformly. [Draw the fixed part of any curious shape, the frame of a machine, v and B here being two pins.] We get a quick return mechanism for shaping and other machines. Imagine v w fixed, and we have the mechanism of oscillating cylinder engines and of other machines as well. Now imagine w fixed, and guiding B x, so that it shall only move in the direction of its length, and we have a well-known pump mechanism. When we consider that, even from Fig. 332, with the motion of w in a line with B, we have obtained a number of different-looking mechanisms, and that these can be varied very curiously by taking various lengths of the parts, it will be seen that there is a nearly endless variety of forms derivable from the mechanism shown in Fig. 327. Now imagine a pin Y in the piece w v linked to z, a pin in an arm z B, and we have a much more complicated problem to study in the relative motion of six pieces. Any one of them may be imagined fixed, and they may be of all sorts of lengths. 469. The following example is of geometrical interest. In the Peaucellier cell (Fig. 333), AB and AW of Fig. 327 are made Fig. 833. Fig. 334. equal in length, and so are B v and v w. Two pieces, w c and OB } equal to B v and v w, are added. It is easy to see that A c v is always a straight line. Also the positions of A, c, and v are such APPLIED MECHANICS. 579 that AC . AV remains constant. For if ACV and WB be joined, and they meet in o, the centre of the rhombus, A w 2 = A o 2 -f- o w 2 , and v w 2 = v o 2 + o w 2 . Hence A0 2 _ V0 2 _ ( A0 + ye) (AO - VO) = AV. AC. Hence, if A is fixed, and c traces out any curve, v will trace out the reciprocal curve. If c is constrained by a link o' c to move in a circle, and if o' A = o' c, v will describe the reciprocal curve, which in this case is a straight line. If o' c is not equal to o' A, v will describe the arc of a circle of much greater radius than c's. 470. When the four links of Fig. 327 form a parallelogiam, as Fig. 835. Fig. 388. A w v B of Fig. 334 , then, if p is a point anywhere rigidly part of the link w v, but in the line w v, and if at any time it is in the same straight line with A, and a point Q, which is rigidly a part of the link BV, and in the line BV, then p and Q will always be in a straight line with A. We can imagine now that either Q or p or A is a fixed point, and the other two will follow paths which are similar to one another. Thus, for example, if AWVB (Fig. 335) is the four-bar mechanism of Fig. 327 (A B need not be drawn because we imagine it fixed), and if A w D is one piece, v P and P D being links equal to w D and w v respectively, then the pin p will travel in a path similar to that of Q if QW : A w : : PD : AD. If a travels very nearly in a straight line (if BV : A w : : QW : Q v, we have a near approach to straight-line motion of Q, so long as B v and A w do not make too great an angle with one another), then p will also travel very nearly in a straight line. 471. Again, if A w v B (Fig. 336) is a parallelogram, and if p is a point rigidly part of the link B v, and if Q is a point rigidly part of the link w v, and if p and Q are so placed that the ratios p v : v B : B P are tho same as the ratios v a : Q w : w v, then it can be shown that A a and if P and Q are so placed that the ratios p v : Y B : B p are the same as the ratios v Q : Q w : w v, then it can be shown that A Q and A P always make the same angles with one another, and that the ratio of the lengths. A Q and AP keeps constant. In the triangles p v B and v Q w, the angles at P and v, at v and Q, and at B and w are respectively equal. Hence the triangles QWA and A B P are similar, and it follows that the angle Q A P keeps constant in any motion of the mechanism, and the ratio of the lengths A Q 580 APPLIED MECHANICS. and A P keeps constant. It follows, therefore, that if A is fixed and p is allowed to follow any curved path, Q will follow a similar path. 472. Note that Fig. 334, where P and o, are in the straight lines connecting w v and B v, is only a particular case of Fig. 336. Note also that if A w and w v are a pair of links guided at A and v along any paths, and if A B and B v are another pair of links whose A and v follow identical paths with the first pair's A and v, then for each point in the one mechanism there will be a corresponding point in the other which follows a similar path. It is interesting to prove that if the links A w, w v, A B, B v, forming a parallelogram as in Fig. 336, be joined up as in Fig. 337, then any four points PQRS in a line parallel to AV remain in a line parallel to AV, however the mechanism may alter in shape ; and the distances P Q, a R, B, s are always such that P u . P o, is con- Fig. 837. stant. Also p Q = us. 473. We study mechanisms, 1st, Because in designing new machinery we ought to have a general fairly exact knowledge of the sort of motion which each piece has in existing machines. For this we watch existing machinery, and work with rude models whose dimensions may be varied. 2nd, Because we wish to perfect an existing mechanism. For this we need a more exact study of the motion of every part, and we use skeleton drawings, algebraic and trigonometric analysis, or graphical methods such as I have described. 3rd, Because in these days of increasing speed of machinery, the forces necessary to produce the accelerations of all the parts have become important, both for the strength and stiffness of all the parts, and also for the effects of vibration. 474. In Fig. 329 we saw that, knowing the directions of motion of w and v, we can find the instantaneous centre. Now let w v get any motion whatsoever in the plane of the paper. Let the velocities of w and v be w and v in the directions marked. If we draw perpen- diculars to w and v, meeting at c, the angular velocity of w about A I* Pig. 338. c must be the same as that of v, and this must be the same as that of any other point K, rigidly attached to w, about c. Hence, join any point K to c, the velocity of iv is at right angles to CK, and is represented in amount by the length of c K. Hence, as soon as we find the instantaneous centrej we have a diagram of velocities of all points in the body. Notice that if we have the instantaneous APPLIED MECHANICS. 581 Fig. 839. centre c and the velocity of any one point, we have the whole diagram to scale. If c D is a perpendicular to v w, as the length of every line now in the diagram represents the amount of a velocity which is at right angles to its direction, c D is the com- ponent or common velocity of all points in the straight line w v in the direction w v. D w is the velocity at right angles to the direc- tion w v of the point w, and D v is the same for v, hence w v represents v's velocity in excess of w's in this direc- tion. Hence, if w v, v x, x Y are given links, and if we know the directions of the * velocities of each joint at the instant, say the directions of the dotted lines, we can find the velocity of every point in any body which is rigidly a part of any of the links if we know the actual speed of any one point. Choose a pole o. Draw lines from o at right angles to the velocities of all the joints o w, o v, etc. Let the distance along any one of them represent the velocity of that joint to scale, and now draw the lines tw, vx, xy parallel to the real links. Then the lengths o w, ov, ox, o y re- present the velocities of the joints. To prove this, drop a perpendicular o D from o upon any of the link directions in our diagram ; we choose vw. If o w is the amount to scale and is at right angles to the velocity of w, then o D repre- sents the velocity of w in the direction of the link w v ; hut this ought to be the same for v. We have, then, the amount o D of a component of v's velocity, and we have the direction of its whole velocity, and our construction is the very one which we should adopt to obtain v's velocity. If w and Y are fixed, the diagram wvxy is a closed polygon, in the present case a triangle. Notice that in any case, if from o we drop a perpendicular o D on a side of the diagram, say wv, then D w and D v represent the velocities of w and v at right angles to the length of D w. Hence the difference tvv, divided by the actual length of w v, represents the angular velocity of wv. Thus, in Fig. 343, take AWVB as four links, A and B having no velocities. From the point o, which we may also call a or b, draw lines parallel to A w and B v, because we know that these are at right angles to the velocities of w and v, and draw vw parallel to vw. Then bv -5- B v, aw ^ AW, and vro ~- v w represent the angular velocities of the three links. 475. In Fig. 339, if K is a point rigidly attached to the link v x Pig. 340. 582 APPLIED MECHANICS. (in the diagram find k so that vkx is a triangle similar to v x x), then o k is at right angles to and represents to scale the velocity oi K. Professor li. H. Smith, who was, I "believe, the inventor of this Fig. 842. method, takes his radiating lines parallel to, instead of at right angles to, the actual velocities (Trans. R.S.E., Jan., 1885). He gives a similar construction for accelerations. If a point B is fixed and a body rotates in the plane of the paper round it, the accelerations of any points A and K are in the directions A B and K B, and they are proportional to the distances A B and K B. Let A K B "be a rigid body with a motion in the plane of the paper. Let the actual amounts of the accelerations of A and B be known ; represent these in amount and direction by the lines o a and o b. Now make the figure akb exactly similar in shape to the real body A K B. The true acceleration of any point K is represented in amount and direction by ok. To prove this : The acceleration of A is the vector sum, acceleration of B + accelera- tion of A relatively to B. Hence ba repre- sents this acceleration of A relatively to B. But the motion of A and the whole body relatively to B is a rotation, and hence Ik represents the acceleration of K. relatively to B to the same scale. The vector sum o b + bk = o k represents therefore K'S ac- celeration. Notice that the true acceleration of any point A is the vector sum of two accelera- tions the first in the direction of motion, the second the centripetal acceleration if the path of A is curyed. 476. Let a known force in any direction be applied to A, a point in the body A K B which has a known motion in the plane of the paper. 1st. Imagine the body divided into small equal masses. The construction of Fig. 342 enables us to find the forces with which these masses resist their accelerations. They may be combined Fig. 343. APPLIED' MECHANICS. 583 with, the weights of the parts to get what may be called the load diagram of the body. Let it be known that there is another force of unknown amount, but known in direction, acting at A (for example, the guiding force, with or without friction, at the crosshead of a steam engine). It is easy to see that, by Art. 475, we can find the amount of this guiding force at A, and also the total force which must be acting at a given point B. 2nd. In Art. 475 we have a construction which enables us to find the stress at any point of any section of the structure A K B, if it may be imagined to be any quasi-prismatic structure like a beam or connecting-rod, or even if it be shaped like part of a metal arch. Tn such a case as that of the connecting-rod, the steam-engine maker ought to study the motion from many points of view, ov and v w being the crank and connecting-rod in any position. Produce w v to meet o B in B. o B is at right angles to the line of centres, o w. Then the lines o v and o B are at right angles to the velocities of v and w, and v B is in the direction of the link w v ; consequently, o v B is (Art. 475) a diagram of velocities, o v representing to scale the constant velocity of v, OB represents to the same scale the velocity of w ; also the distance v B represents the angular velocity of the connecting-rod. It follows from this at once that a force P at w in the direction w o would, if there were no friction and the connecting-rod were massless, produce a turning moment F x o B on the crank shaft. It can be shown that if we draw B A at right angles to w B, join T A, draw B c parallel to v A, let w be in w o as far from A as c is, but on the opposite side of A, and join w v, then o w v is an acceleration diagram such that if K is any point in the connecting-rod, and we let the points TV, k, and v be relatively to each other as w, K, and v, (k is evidently in the same horizontal as K), then ko represents the acceleration of K. in direction and magnitude to the same scale to which vo represents the centripetal acceleration of v. In fact, o w &v is a diagram of accelerations. 477. When one point of a body is fixed, and we may neglect centripetal accelerations, we may take up the problem as a particular case of the general problem already considered, or we may attack it analytically as follows : If p P is the straight centre line of an arm, o being fixed, and if a is its angular acceleration ; neglecting forces parallel to OP, let there be a force P acting at p; the acceleration at any point Q (neglecting radial acceleration) is #a, if OO is x. If m is the mass per unit length, the load due to 584 APPLIED MECHANICS. acceleration per unit length is max; and if u is the bending moment at the cross- section (see Art. 357), Assume m to be some function of x, and let s be the shearing force at a section. Let \mx.dx = v, I v . dx = , and let the values of v and u become v\ and tt lt when x = I. Then + a + .... (3), i + c .'. i) + * . '. e = a (Jt'i - Wi) - F e. If A is the area of cross-section and p is the mass per unit volume, A p = m. If we wish to have the same maximum stress f in every cross-section, and if z is the strength modulus of the section, then M= Z/, ^ = L/:... ( 4). m A p Dividing, therefore, (3) by m, we know the value of z/A every- where, if the arm is to be of uniform strength. Thus, for example, let the section be rectangular, of breadth z and depth (in the plane of motion) y. Then A = zy, z = \ mf-; so that z/A is \y. Hence from (4) we have = %r y = .... (5). Since pzy = m, z = , so that when we know y we know ; Example. Let m = a - bx, v = 478. In discussing the reciprocating motion of a point, I have found in practice much simplification in my ideas when I have reduced the motion to some such shape as x = a sin. (qt + e{) + b sin. (2 qt + e 2 ) 4- etc. where x is the displacement of the point from some fixed point in its path, and q is 2 TT multiplied by the frequency, or 2 ir divided by the periodic time. For many purposes we find the first term sufficient. In most valve motions, such as link motions and radial APPLIED MECHANICS. 585 gear, I find that only two terms are ever needed even for a very exact definition of the motion. One interest attaching to this method of working is in its showing how the b term becomes twice as important in the velocity, and four times as important in the acceleration as the fundamental term. Another important matter is this : in modern machinery vibration is becoming very important, and the above de- scription of a motion is the one that lends itself most easily to a discussion of the vibra- tions which want of balance gives rise to. (See the author's books on the Calculus and on the Steam Engine.) 479. The following proposition is the foundation of most calculations on motion communicated through links. The points A c B are in a straight line. They move, keep- ing in a straight line, and at the same dis- tances from one another. (In engineer's language ACS is a moving link, and c is a point in it.) Prove that the motion of o A C is the vector sum of the fraction of B'S motion, and the fraction A B B C of A'S motion. For let A A', B B', c c' be any displacements of A, B, A B and c. Join A' B. Draw c c" parallel to A A', and join c" c'. Now = = ^7, so that c" c' is parallel to B B'. Hence, as c c A B A B A B is the vector sum of c c" and c" c', we have proved the proposition. It is easy to extend our reasoning to motion which is not parallel to one plane. The motions of any three points A B c of the body, not in one line, define the whole motion ; and when we are given the accelerations of A, B, and c, it is easy for anyone who knows descriptive geometry to make a diagram showing the accelera- tion of any point of the body. 480. Newton's great one law of motion for any, however complex, system of bodies is this : Look upon the rate of change of momentum of any small portion of matter of a system as a force in the opposite direction. All such forces are in equilibrium with the forces which act on the system from the outside. In most cases it is this most general way of stating the law that is most useful to engineers. Given their diagram of accelerations, they really have a force or load diagram, and the problems to be dealt with are now merely worked out by graphical statics. 481. Most men are led to their study of this subject through what is called D'Alembert's principle. This is a principle which served a very useful purpose. At a time when English mathematicians were stagnating, being academically learned as to Newton's methods, but being really ignorant of them, the French mathema- ticians were developing kinetics practically independently of Newton's methods. Newton's third law was quite misunderstood, . 586 APPLIED MECHANICS. and D'Alembert discovered a principle which gave the power of solving dynamical problems with certainty. It will not astonish anyone who knows academic methods to be informed that, although D'Alembert gave what is now seen to be merely a rather cumbrous explanation (easily misunderstood, because of certain technical terms which may be confounded with one another) of Newton's third law, it is through the clumsy explanation that the subject is nearly always approached. We believe that this is one of the greatest reasons why many engineers are so disgusted with higher studies in dynamics. There seems to be as much absence of common sense now in academic persons as there was in the time of Erasmus. He, the greatest scholar of the fifteenth century, wrote : " They are a proud, susceptible race. They will smother me under six hundred dogmas. They will call me heretic, and bring thunderbolts out of their arsenals, where they keep whole magazines of them for their enemies. Still, they are Folly's servants, though they disown their mistress. They live in the third heaven, adoring their own persons, and disdaining the poor crawlers upon earth. They are surrounded with a bodyguard of definitions, conclusions, corollaries, propositions explicit, and propositions implicit." 482. Let the engineer take Newton's law in its very simplest form, as above expressed, and he will have no difficulty in attacking the most complicated problems, for the dynamical becomes a static problem on the equilibrium of forces. It is usual to express part of the result analytically in the following way : If in any direc-^ tion, which we may call x, the small mass m has the acceleration z, then the resultant force acting from outside the system in that direction is equal to the sum of all such terms as mx. State this as being true in any three directions, and of course it is true in any direction whatsoever. Now if the x of the centre of gravity of the whole system is x t we know that x 5 m = 2 mx. Differentiate with regard to time once, and again, and we see that the whole mass 2 m, multiplied by the acceleration of the centre of gravity, is equal to the sum of all the masses multiplied by their accelerations in the direction x. It follows, therefore, that the motion of the centre of gravity of a system is the same as if the mass of the system were collected there, and all the forces acting from outside on the system acted there. The other part of Newton's law is, of course : The resultant moment of all the outside forces about any axis is equal to the rate of change of moment of momentum of the whole system about the same axis. We very often choose as our axis an axis through the centre of gravity, but it is well to notice that this is not necessary. 483. Angular Motion. We know (Art. 92) that when a rigid body can only rotate about an axis, if the sum of the moments of the forces acting on the body is M, when the body moves through the angle 86 radians the work done is M . 80. If any little portion in of the mass of the body is at the distance r from the axis, and the APPLIED MECHANICS. 587 body is rotating with the angular velocity a, so that ar is the velocity at the place, the energy stored up in the little mass is %tna 2 r 2 . If this is summed up for the whole body, we see that every little term contains 2 ; so that if we know the sum of all such terms as mr 2 (called i, the moment of inertia of the body about the axis), we can say that the kinetic energy is %ia 2 . When a force r acts on a body through the distance 5s, the work done being F . 5s, and the kinetic energy of the body being \ mv z , by assuming that work done is equal to gain of kinetic energy, we are led to the law F = m x acceleration. "We have an exactly analogous set of terms for angular motion. The work M . 59 corresponds with ? .5s; the energy %ia 2 corresponds with \mv 2 . The proof, therefore, is exactly as in Art. 497 namely, if a moment M acts, through the angle 80, on a body moving with the angular velocity a, and whose moment of inertia is i, so that its kinetic energy is \ ia 2 , and if a + 5a is its new angular velocity, then H . 56 = i (a + 5a) 2 - $ia 2 = i . a5a + i . (5a) a , 5a , 5 a . As 50, and therefore 5a, become smaller and smaller, we have 5a 56 5a or da M = I ^ =Itt if a is used for , or the angular acceleration. This is exactly analogous with the force law in linear motion, i a is called the moment of momentum of the body, and i a is the rate of change of the moment of momentum. 484. To arrive directly at the moment of momentum of a rigid body about any axis, consider a portion of mass in at the distance r from the axis ; r making an angle with a fixed plane through the axis ' x = r cos. e,y = r sin. 0, dx d0 dy - dx dx The moment of the momentum m -=7 about the axis is - my . in at * the direction of increasing 0, and the moment of momentum of m -j- is mx . _f , and the sum of these is at at m (T* sin j? + r 2 cos. 2 0^), or mr 2 . Q, and the sum of all such terms is 1-7-. or ia. The rate of change of at thi 8 i s ! _| } or i a. This also may be obtained by differentiating 588 APPLIED MECHANICS. (1), and so finding -^ and -~. Taking the moments of the virtual forces - m-^ and - m^f about the axis, we arrive at dt 2 dt 2 j-2a - i-T-2, or - i o, which, together with the moments due to the outside forces, produce equilibrium; or, with signs changed, are equal to them, as above. Students ought to think of other ways of reaching these results concerning rigid bodies. 485. Kinetic Energy of any System. Let wbe_the mass of a por- tion of the system in the position x, y, z. Let x, y, z at this instant be the position of the centre of gravity. The whole kinetic energy is the sum of such terms as Let x = x + x l , y y + y 1 , z = z + z 1 ; so that x 1 , y 1 , z 1 show the position of m relatively to the centre of gravity. Note that dx dx l \* /dx 2 dx dx 1 But 2m . 2^ . ^ = 2^ 2m . ^, and 2m 1 is because it ii at at at at at the rate of change of 2 m x 1 and 2mx l = 0. Hence The first of these terms is the whole mass multiplied by the square of the x component of the velocity of the centre of gravity. It follows, therefore, that the kinetic energy of any system is equal to the kinetic energy which the system would have if it were all moving with the velocity of the centre of gravity + the kinetic energy due to the motions of all the parts relatively to the centre of gravity. Now, the motion of a rigid body relatively to' any point of it can only be a rotation. Hence for rigid bodies we have the rule : If the velocity of the centre of gravity is v, and if there is a rotation of angular velocity 9 about some axis, and if I is the ' moment of inertia of the body about a parallel axis through the centre of gravity, and M is the whole mass, the kinetic energy E is $ M v 2 + \ i (0) 2 . If io = M& 2 , then E = \ M | w 2 + & 2 (0) 2 } . It is only in the case of a rigid body (in all this by rigid body I mean an infinitely rigid body) that we have this simple rule. In the case also of a rigid body we can find law (Art. 482) from the law of energy, as we do in Art. 495 ; but we cannot do this so easily for a system having internal relative motion, because there is internal potential energy, which may alter. The virial and other laws, which may easily be arrived at, do not concern engineering applications of mechanics. APPLIED MECHANICS. 589 486. In any ordinary elastic body the internal motions are vibra- tioiial. The momentum and moment of momentum of the system are therefore practically the same as if the body were quite rigid. This is not the same in regard to the energy. We say that it disappears or is changed to heat and other forms ; this means merely that it has become molecular, and equations which regard the body as if it were rigid are quite inapplicable. Thus it is that the momentum equations are what we rely upon, because when we study the motions of bodies momentum cannot be hidden as energy may be. 487. When we work exercises on rigid dynamics we ought to apply Newton's law always in the shape given in Art. 480. When we work a new exercise let it be regarded as a new illustration, to be worked to making our knowledge of the fundamental principle clearer. Most details of one's theory and artifices for the solution of problems will disappear from view in one's professional work. Let the fundamental notion be so well fixed that it cannot disappear. 488. Exercise. In the compound pendulum of Fig. 346 find the force act- irg at s, the point of suspension. In Fig. 347 let a be the centre of gravity. Fig. 346. Pig. 347. Let x and Y, as shown, be the horizontal and vertical components of the force at s. Let G s y be 0, and let the body be moving so that increases. If s G be called r, the velocity of & in its path is rQ. It has an accelera- tion rQ in its path, and r(0) 2 , a centripetal acceleration in the direction G s. I. Eegard now the whole system of forces, which are supposed to balance if they were to act at one point, G. Let M be the mass or w/y, we have - M rQ in the direction G A, - M r(0) 2 centripetal in the direction G s, w downwards (the weight of the body), x horizontally, and Y down- wards. Resolve these in any direction whatsoever, they are to balance. Thus, resolve them horizontally and vertically, x - Mr0 cos.0 + M r(0) 2 sin.0 = .... (1), Y + w + Mr0 sin.0 + M>-(0) 2 cos.0 = .... (2). Observe that the forces in G A and GS are virtual forces, or forces equal and opposite to mass multiplied by acceleration, II. Now take moments about any axis. The most convenient seems to be s. w is the only externally applied force that has any moment about s, and so W sin.0 4-10 = 0.... (3), 590 APPLIED MECHANICS. if i is the moment of inertia about s. From these equations x and Y, in terms of 9 and 9, may be calculated. We first express 9 in terms of 6 from (3). If i = ^ & 2 , (C) gives = - r cos.9(- 2T &m.9\ - - r sin.9 . (0) a , 9 \ & ) g = - w - - r sin.0 ( - 9 L sin.0) - - r (0) 2 cos.0, will depend on the limit of the swing, and may have any value. 489. Vibration Indicator. Fig. 348 shows an instrument which has been used for indicating quick vertical vibration of the ground. The mass OP a is supported at p by a knife-edge, or by friction- wheels. The centre of gravity o is in a horizontal line with p and Q. Let p G = a, GQ, = b, -po, = a + b = L The vertical spring A R and thread R Q, support the body at Q. As a matter of fact, A R is an Ayrton- Perry spring, which shows by the rotation of the pointer R the relative motion of A and Q. Let us neglect its inertia now, and consider that the pointer faithfully record^ relative motion of A and Q. It would shorten the work to only consider the forces at p and a in excess of what they are when in equilibrium ; but, for clearness, we shall take the total forces. When a body gets motion in any direction parallel to the plane of the paper, we get one equation by stating that the resultant force is equal (numerically) to the mass multiplied by the linear acceleration of the centre of gravity in the direction of the resultant force. We get another equation by stating that the resultant moment of force about an axis at right angles to the paper through the centre of gravity is equal to the angular APPLIED MECHANICS. 591 acceleration, multiplied by moment of inertia about this axis through the centre of gravity. I shall use x, x, and x to mean displacement, velocity, and acceleration, or x, , and -^. Let P and A hare a displacement x l downward. Let Q be displaced^ downward. Let the pull in the spring be o, = Q O + c (x - Xi) where c is a known constant (c is the reciprocal of the h used in Art. 514). Let w be the weight of the body. Then if P O and Q O be the upward forces at the noints marked P and Q, in the position f equilibrium, 00 (a + b) = w a and P O + % = w. Hence + Q = Q + C (X - Xi). Now G is displaced downwards T#I + - ?x. so that & + a -j- The body has an angular displacement clockwise about its centre of mass, of th inertia about o, of mass, of the amount : j- So ^ a ^ if i is its moment of Hence (2) and (3) give us, if M stands for , and if i = M & 2 where Ic is the radius of gyration about a, If &! is the radius of gyration about P, we find that (4) simplifies to 'x + n*x = e*xi + ri*Xi (6) if n stands for j- / = 2 v x natural frequency and ^ stands for 1 - ra- Call x - x\ by the letter y, because it is really y that 592 APPLIED MECHANICS an observer will note, if the framework and room and observer have the motion x\. Then, as y = x - x\, or x = y + x\, y + Xi + n 2 (y + *,) = &x\ + ri l x\. So that y + ny = (e* - 1) *, ---- (6), or al .. o (7). Thus let x l = A sin. qt. "VVe are neglecting friction tor ease in understanding our results, and yet we are assuming that there is enough friction to destroy the natural vibration of the body. We find that if we assume y = a sin. qt, then al a* a = 7-9 o - O A< &i 2 ri 2 - q* That is, the apparent motion y (and this is what the pointer of an Ayrton- Perry spring will show ; or a light mirror may be used to throw a spot of light upon a screen) is ^ ~ 27T /2lT \ v or a = A cos. f t + e j, * or 41? A "body moves backwards and forwards under the action of a variable force, which is always proportional to the distance of the body from its middle position, and which always acts towards this position; and if the force at a distance of 1 foot is 5 Ibs., then the time of vibration is 2 TT times the square root of the quotient of the mass of the body divided by 5. If m is the mass ; if the force of friction is b times the velocity; if the constraining force is n times distance from centre, If a body vibrates about a fixed axis under the action of the torque (say from a spiral spring or twisted wire), so that it is always propor- tional to e, the angular displace- ment of the body from its mean position, and if the torque is 5 pound-feet when the body is 1 radian from the mean position, then the time of a vibration is 2 IT times the square root of the quotient of the moment of fnertia divided by 5. If i is moment of inertia, fric- tional torque is b times the angular velocity. If the constraining torque is n times the angular distance from the mean position, * or 6 = A e~. As we imagine ds, and therefore dv, to be smaller and smaller without limit, the last term in (1) gets to be nearer and nearer 0, and v gets to be better and better represented by ds/8t. Hence Ss dv dv * = *,-._,,.-, or force is equal to mass multiplied by acceleration, or force is the rate oi change of momentum. 498. 8. So long as we deal with force in the direction of the motion of a body, there is no difficulty in showing that the law of work or the conservation of mechanical energy leads to the rule, " Force is rate of change of momentum." When a point is moving in a curved path, we can say that the component of the force in the direction of motion is equal to the rate of change of momentum in the direction of motion. But what of the other component at right angles to this ? Here it is necessary to observe that mere change of direction of motion, and not merely of speed, indicates that force is acting on a body. It is a matter of common observation that a centripetal force is necessary to keep a body APPLIED MECHANICS. 603 moving in a circle, and that the body exercises a centrifugal force against those constraints which compel it to move circularly. When a curved surface gradually changes the direction of a moving body, these two forces act at the point of contact.* To get dear ideas, consider the conical pendulum of Fig. 354. The body p revolves in a circle at constant speed. To keep it at the constant distance o, P = r from the centre a, we know that, if it were at rest, a force F must act. But it is in motion, and yet keeps out at the distance r. There is a centri- petal force whose amount is known to us, w tan. o, and it is evidently balanced by what we can only call a centrifugal force of this amount which is due to the motion. Observe that we here have a case of centri- petal force acting on a body, creating no increase of kinetic energy, and creating no change of any kind of energy, for when p comes round to the same position again everything is just as before, although the force has been acting for a whole round. Mathematically there is no great difficulty. We assume that we have proved that forces in directions at right angles to one another may be studied independently and may be combined as vectors. Consider circular motion in the plane of the paper. Constant speed v of P means a velocity v cos. 9 in the direction Q A and - v sin. 8 in the direction a o. The two accelerations are v sin. 9 . a and - v cos. 9 . a, a being , the angular velocity, or -. If m is the mass of the body, we know that there are two forces acting upon the body : - mv sin. 9 . a in the direction a A, - mv sin. 9 . a in the direction QC. Forces always compound accord- ing to the vector law, as may be proved from the law of work (see Art. 495), and hence we see that the force acting upon P in the direction p a is a centripetal force mvrv or m , and the acceleration of p is a cen- tripetal acceleration v 2 /r. We have seen (Art. 493) another way of arriving at the amount of the acceleration. In the very same way it is easy to show that when a small body Fig. 355. moves, not merely in a circle, but in * People may have notions of force that seeta to be quite different from a metaphysical point of view, but which are really the same mathematically. Thus one man puts it that there is no sucli thing as force ; we have only mass X acceleration. Another says "Yes, we have a centripetal force ; say, acting on a body which is whirled round at the end of a string, but it is not right to speak of the equal and opposite force, sometimes called centrifugal force. " Now, if we only think of the body, it may be enough to speak only of centripetal force and centripetal acceleration ; but a string in tension is really acted on by equal and opposite forces. If one of these is exerted towards the centre, the other is exerted outwards by the body, and we call it a centrifugal force. 604 APPLIED MECHANICS. any curved path, the force acting upon it has two components one in the direction of the path, equal to its mass, multiplied by the rate of change of its mere speed ; the other at right angles to the path, equal to its mass, multiplied by its linear velocity, multiplied by its angular velocity. 499. It is only when we have looked at the subject from the energy point of view, and in many others in which it will strike a thought- ful student to make experiments, that the beauty of Newton's generalisation comes home to us and we see how all the results of observation, experiment, and speculation are given to us in his statements. In fact, we gradually get to know that, whether acceleration of a small body is along the path or at right angles to it, force is equal to the vector rate of change of momentum. Now, this is Newton's definition of force, and we may begin our study of mechanics from this point of view. But speculation of the above nature is very far from being useless. Taking up our subject from the easiest point of view, accelera- tion is the time rate of change of velocity, and force is the time rate of change of momentum. We first consider acceleration in the direction of motion, and then, if a body changes the direction of its velocity, the lateral rate of change of velocity or lateral acceleration must be considered. A velocity is a vector quantity, and velocities are added as all vectors are added. A velocity of 5 feet per second eastward, added to a velocity of 5 feet per second southward, are equal to a velocity of 7'070 feet per second south- eastward. 500. When I speak of the motion of a body, I usually mean the motion of its centre of mass. If a body is at p moving along a curved path with the constant speed of v feet per second, when it is at Q Fig. 357. its velocity is in a direction making, let us say, an angle 89 with its old direction. In Fig. 357 let OB represent to scale the velocity at P ; and let o c, equal in length, but making an angle c o B = 80, with o B, represent the velocity at Q. Now, in vector addition, OB + B C = C ; so that if dt is the time taken by the body to move from p to Q, in that time there has been the lateral change of velocity BC. B c =: o B . 50 more and more nearly as 80 is made smaller and smaller, or B c = v . SO. Hence the lateral accelera- tion, which is lateral change of velocity divided by the time of the The academic person may be quite right to stick to his conventions as to force, because he never has to think of the medium (string or other) through which forces are exerted when he is working his problems. The engineer is compelled to deal with the larger question ; he very wisely converts all liis dynamical problems into statical problems, and all his forces are always balanced. APPLIED MECHANICS. 605 50 59 8s 9 50 change, is v . or v . - = v z - = v z x curvature, because our definition of curvature is -T. In a circle the curvature dOfds happens to be the reciprocal of the radius. Hence we speak of this acceleration towards the concave side of a body's path as if it were moving in a circle (curiously enough, the fact seems always to be forgotten that the path is usually not a circle, nor even the very smallest arc of a circle) of radius r with a centripetal acceleration v 2 jr.* The centripetal force causing the change of motion is the mass multiplied by this. The engineer is usually concerned with the equal and opposite force which the body exerts upon the constraints, and calls it centrifugal force. We see that it is mv^/r or mcPr or wr 2 /2937, where m is the mass of the body or w is its weight in pounds at London, r the radius of curvature of the path in feet, v the velocity in feet per second, a the angular velocity in radians per second, n the number of revolutions per minute. Observe that if masses m^ and w 2 are attached to the same shaft at distances TI and r 2 from the axis, their centrifugal forces are in the proportion of m^i to w 2 r 2 ; if we have half the mass at twice the distance, we have the same centrifugal force. If a body changes in its speed v, and so has an acceleration in the direction of its path, its total acceleration is the vector sum of the two accelerations, and the resultant force acting on the body at any instant is the resultant of m along the path and m at T at right angles to and in the plane of the path. Example. A body of \v Ibs. is moving along a curve with a velocity v and an acceleration a in the direction of motion ; the radius of curvature of the curve is r. What is the total accelera- 9 tion and the force causing it ? Here we have and o two accelera- tions at right angles to one another. The answer is, an acceleration V-TJ + a 2 in the direction making an angle with the direction of motion, where tan. 6 = Multiply by the mass w -j- 32% and we have the total force. 501. Centrifugal Force in Belts or Ropes. In Fig. 356 let p a be a small portion of a flexible body of w . 8s Ibs. ; its centrifugal force * To keep in our minds the fact that all motion is relative, we ought to remember that, relatively to the body, other bodies have a centrifugal acceleration. The words "centripetal" and "centrifugal" are technical terms now ; their origins seem to have had something to do with the notion that a curve has millions of centres. It is all very well for the mathe- maticians to speak of a small part of a curve as being an arc of a circle, but the engineer knows that it is only in this matter of curvature or rate of change of with s that it is like an arc of a circle, for he knows that complete information about the very, very smallest portion of any curve implies a complete knowledge of the whole curve. 606 APPLIED MECHANICS. is - Ss . -, or - Ss . ?> 2 . ^, or - v 2 . 50. If the tension is T Ibs., r g Ss g we know from Fig. 357 that T . 80 = - v 2 . 59, so that T = tf. g g If a is the section in square feet, and / is the tensile stress per square foot, and w is the weight of 1 cubic foot of the material, n = an> , T = /; so that/= -$ v* being independent of the radius of the path. Example. What velocity will produce a tensile stress of 3,000 Ibs. per square inch in the thin rim of a cast-iron pulley ? Here/ = 3,000 x 144, and w = -26 x 1,728. Hence v = 175 feet per second; or, in engineers' Janguage, a velocity of 10,500 feet per minute will produce what is usually taken as the working tensile stress in cast iron. 502. If the plane of the path alters, if the plane rotates about the tangent to the path through the angle 8 in the distance 8s, then d/ds is called the tortuosity of the path. When a body moves in a tortuous curve it has acceleration dv/dt along the path and v 2 /r, a centripetal acceleration in the plane of the path, or the plane of curvature, as it is called. And if a model made of three very short pieces of wire, o s, o R, and o N, be made, the angles between os, OR, and o N being right angles, and if we conceive o s to keep parallel to the path 8*, if o R keeps pointing to the centre of curva- ture, then the angles turned through about the axes o N and o R per unit length of the path represent the curvature and the tortuosity. A student ought to make a model of a curve with wire and let a little frame like this slide along it, and study the matter for himself. A spiral path in which the curvature and tortuosity are constant is particularly interesting. If we refer the position of a particle to three axes of reference, its total acceleration at any instant is compounded of the three d?z/dt 2 , d?y/dt 2 , d?z/dt 2 . The three components of the resultant of all the forces which are acting are m times these, if m is the mass or inertia of the particle. THE BALANCING OF MACHINES. 503. If a wheel is fixed eccentrically on its shaft, or if to a shaft there is attached any object whose centre of gravity is not exactly in the axis when the shaft rotates, centrifugal force causes pressures on the bearings of the shaft which .are always in the direction of the centre of gravity of the rotating mass. In this case there is said to be a want of balance. If you wish to observe the effect produced by such want of balance, mount an axle to which a wheel is keyed on APPLIED MECHANICS. 607 any support which is not very firm ; fix a small weight on one of the arms of the wheel, and rotate it rapidly. You will find that, even if the weight is small, surprising effects are produced, and show themselves in a shaking of the supports ; and the evil effects are four times as great at 200 revolutions per minute as at 100 revolutions per minute. Centrifugal force is proportional to the mass of a rotating part multiplied by the distance of its centre of gravity from the axis of rotation, multiplied by the square of the number of revolutions per minute. 504. If a number of bodies are attached to a shaft and are whirling round with it, each of them at any instant exerts a force on the shaft which can be calculated, and the resultant effect on the two bearings may easily be determined, just as easily as in the static problem of Art. 99. If the axis of rotation passes through the centre of gravity of all the rotating parts, the pressure on one bearing is equal and opposite to the pressure on the other; and by properly placing the masses, the pressure on either bearing may be reduced to nothing. Thus it is evident that when two masses are directly opposite to one another on a shaft, their cen- trifugal forces may be made to balance one another. When not opposite they cannot be made -to balance, but two masses may balance one which is directly opposed to the resultant force of the two. When there is no pressure on either of the bearings, so that there is no tendency to change the direc- tion of the axis, it is said to be the permanent axis of the rotating masses. All axes of rotation in machines ought to be permanent axes. When this is the case in a rotating machine, and it is suspended by ropes and made to work, there are no visible oscillations. 505. The balancing of a machine consists in adding masses in such positions, or re-arranging the positions of the existing masses so that the centrifugal forces due to their rotation are just able to balance the * otherwise unbalanced forces which act on the various shafts. The student will find that the study of one problem in balancing will make him familiar enough with the method of calculation for its applica- tion to almost any other case which is likely to occur in practice. The most usual case for the student to take up is that of the locomotive engine, because want of balance in the locomotive is capable of producing very serious effects indeed. 608 APPLIED MECHANICS. Fig. 358 shows an electromotor driving a shaft on which a number of discs are keyed. Weights may be fastened on these discs ; the want of balance is evident when the shaft Pig. 358. rotates, and students will find it easy to illustrate how the shaft may be balanced by other weights properly placed. They will see that when motions are merely rotatory we can always have a perfect balancing of machinery. APPLIED MECHANICS. 609 506. Example. It has been shown by experiment that the appli- cation of suitable balance weights is attended by a sensible reduction of resistance on railways at high speeds. Locomotive engines unbalanced cannot attain as high speeds as when balanced, with the same consumption of fuel. There are two separate sets of unbalanced forces acting on the crank shaft of a locomotive.. (1) The centrifugal force of the crank, crank-pin, and as much of the connecting-rod as may be supposed, roughly, to follow the path of the crank-pin (say one-half of it). The mass or weight of each of these multiplied by the distance of its centre of gravity from the axis, divided by the length of the crank, gives the mass which, on the crank-pin, would produce the same centrifugal force. Let this weight be called w Ibs. In designing engines, we consider half the connecting-rod to act as if collected at the crank-pin, the other half to be moving with the piston. At the end of the stroke, when the horizontal component of the centrifugal force is greatest and the vertical component vanishes, the horizontal pressure on the axle caused by the centrifugal force is w v- w /27rim\ 2 i 32^' R or 32^2 (-go") R <*-- 2937, K being the length of the crank in feet, and n the number of revolutions per minute. (2) We have the force due to the mo- mentum of the reciprocating mass, including piston, piston-rod, slide, and the second half of the connecting-rod. The loss of momentum is most rapid just at the end of the stroke ; and as loss of momentum per second is what we call force, the force acting on the axle at the end of the stroke due to this cause is easily found, and proves to be where w is the weight of the total reciprocating mass. Now a weight w\, or weights whose sum is w\, may be placed on the driving-wheel or wheels at a distance r from the axis, such that the centrifugal force of w\ may be equal to the sum of the above forces. This leads to w\r = w R + w R ; and if we assume any distance, r, we can calculate the balance weight or weights, w\. 607. Now, for the axis to be permanent in inside-cylinder engines, w\ must be divided into two parts, one for each wheel, inversely proportional to the distances of the wheels from the crank. For outside-cylinder engines we get balance weights for the two wheels whose difference is w\, and -they are, as before, inversely proportional to the distances from the wheels to the crank in question. Hence, a consideration of each cylinder gives two balance weights, one usually much smaller than the other. As the cranks are at right angles, the balance weights ought to bo 90 apart on each wheel. Instead of using these two, we can use one weight placed between their positions, so that its centrifugal force is the resultant ol theirs. Thus, if we found 20 Ibs. and 6 Ibs. for the two placed at the same distance from the axis but 90 apart, make o A equal 20, 610 APPLIED MECHANICS. and o B, at right angles to o A, equal to 6 according to any scale ; complete the parallelogram, and o c represents on the same scale the weight which will replace them. It ought to be placed at just the same distance from the axis as they were supposed to be placed; and in position it makes the angle AOC with the larger weight. In this case it will be found that 20-88 Ibs. placed 18-3 from the position which the weight of 20 Ibs. might have occupied will be required to replace the two. 508. It often happens in outside- cylinder engines that the distance from one wheel, or rather from the centre of gravity of a balance weight, to the crank, is so little that the corresponding weight for the other wheel is very small, and may even be neglected. In inside-cylinder engines it will be found that, whereas the cranks are at right angles to one another, the balance weights on the two wheels on the opposite side of the axis to the cranks are often only 50 apart. In inside- cylinder engines with coupled wheels the outside coupling rods and cranks are usually made to balance the inside moving parts. These engines work very smoothly indeed. Outside- cylinder engines with coupled wheels are very unstable, from the use of small wheels requiring very rapid revolution of the crank axle; from the cylinders being farther apart than usual, so that the coupling-rods may have room, and from the number of reciprocating parts being increased. The conditions seem to admit of no remedy for these defects. The balance weight ought to be distributed over two or three of the spaces of the wheel, that the tire may not be unduly strained. 509. We have, then, the following easy, approximately correct rules for locomotives : If R is length of crank, r the distance of centres of gravity of every balance weight from centres of wheels, e the distance apart of the centre lines of cylinders, d the distance apart of the wheels or centres of gravity of the balance weights, w the total weight of crank (referred to the pin), pin, connecting- rod, piston, slide, and piston-rod, A the angle which the position of centre of gravity of balance weight makes with near crank : (1) Inside-cylinder engines with uncoupled wheels. Each balance weight = -^j- >/ 2 d 2 + 2eP, tan. A = ; so that A is easily obtained from a book of tables. (2) Outside -cylinder single engines with uncoupled wheels. Each balance weight = ^, A = 180 ; so that in this case the balance weight is placed exactly opposite to the crank. (3) Inside-cylinder engines with wheels coupled. Find by rule (1) if the weight of the coupling-rods, etc., is too great. If so, let counter weights equal to the difference be placed opposite the outside cranks. If too small, the difference must be made up with balance weights, as in rule (1). The positions of the outside cranks are found by rule (1). (4) Outside -cylinder coupled engines. Find revolving weight of coupling-rods, etc., for each wheel. Also find sum of the weight APPLIED MECHANICS. 611 of the piston, rod, slide, and half connecting-rod. Divide this latter among the wheels, adding the given revolving weight already on them. Let this be used on each wheel according to rule (2). 510. We have dwelt upon these practical rules for balancing in locomotives, because they give good illustrations of centrifugal force. But the student ought clearly to see that it is only when a body rotates about an axis that we can exactly balance the forces. A body with reciprocating motion can only be balanced by another body with a reciprocating motion ; and hence it is that, after much expense and quarrelling with persons complaining about the vibration of their houses, many electric lighting companies have discarded reciprocating steam-engines, replacing them with steam turbines. In Arts. 427 and 429 we give the principles on which the subject may be studied. I give practical examples of their use in my book on steam, gas, and oil engines. In that book I give the more exact constructions, which are so commonly taught now to advanced students, to find the forces at the ends of a connecting- rod for any position. It seems never to strike a student or his teacher that such elaborate calculations may possibly not give a very different result from what they may obtain by the simple assumption like what I have stated above. One of my students has made the comparison. His three weeks' constructions to find the forces acting on the frame of a Willans engine, the turning moment on the crank shaft, etc., nowhere differ appreciably in their results from those obtained by assuming that half the mass of the connecting-rod has the motion of the cross-head, and the other half that of the crank pin. The very much more important matter, the effect of the motion of the rod upon its strength as a laterally loaded strut, seems never to engage much attention. EXERCISES. 1. A crank pin 4 inches in diameter and 6 inches long has to be balanced. If the length of crank be 9 inches, and the balance weights are placed directly opposite each crank arm, find the weights, the centre of gravity of each being 6 inches from the centre of the crank shaft. Ans., 16-24 Ibs. 2. A shaft is in balance under the action of three weights, one of 300 Ibs., at a distance of 12 inches from the axis; another of 100 Ibs., at a distance of 20 inches from the axis, on the opposite side, and 30 inches to the right of the first. How much must the third weight be, and where must be its position along the shaft, if its distance from the axis is 28 inches ? Ans., 57'1 Ibs.; 37.5 inches to left of first weight. 3. In a locomotive the distance between the centre lines of the cylinders is 27 inches. Balance weights are fixed at a horizontal distance apart of 59 inches, the centre of gravity of each describing a circle of 55 inches diameter. If the weight of the reciprocating masses for each cylinder be 400 Ibs., and the stroke be 25 inches, find the position and magnitude of balance weights to counteract the horizontal and alternating force and couple. Ans., At 160 with near crank; 142 Ibs. each. 612 APPLIED MECHANICS. 4. In a steam-engine the piston at the beginning of its stroke is exposed to a total effective steam pressure of 2,000 Ibs., but the inertia of the piston is such that the thrust of the piston-rod is only 1,600 Ibs. The speed of the engine is now raised until it becomes half as great again as before, while the steam-pressure is unchanged. What is the thrust of the piston-rod? Ans., 1,100 Ibs. 5. An engine is making 150 revolutions per minute. What is the acceleration of the piston at the commencement of each stroke, the connecting-rod being 4 feet long and the crank 9 inches ? Ans., 223 ; 153. 6. The weight of the reciprocating parts of a steam-engine is equivalent to 3 Ibs. per square inch of the area of the piston. If the length of crank be 9 inches, find how much the initial effective pressure is reduced by the inertia of the reciprocating parts when the crank makes 70 revolutions per minute, the obliquity of the connecting-rod being neglected. Am. 3-8 Ibs. per square inch. 613 CHAPTER XXVIII SPRINGS. 511. ANY contrivance which can store energy as strain energy, and give it out again readily, is a spring. Hence, any tie-bar or strut, any beam in fact, any object whatsoever is a spring. The term is, however, generally applied only to such objects as can be changed in shape very much without fracture. A tie-rod of indiarubber can be stretched to eight times its old length, again and again, without hurt ; whereas a tie-rod of the best steel can only be stretched to g-Vth ot its old length, again and again, without hurt. Hence the indiarubber tie-rod or a strut may be called a spring, just like the spiral spring ; but it would not be right to speak of a tie-rod or strut of steel as a spring. The differ- ence is, however, only one of degree, and indeed, a mine cage suspended by a steel rope half a mile long vibrates up and down just as if hung from any ordinary spring. 512. Springs are almost always used as reservoirs of energy that is, as hydraulic accumulators are used, or fly-wheels of steam-engines, or cisterns of water, or electric accumulators. The mainspring of a clock or watch takes a store of energy in winding-up, and gives it out gradually for about twenty- four hours in unwinding. A bow gets gradually a store of energy, which it gives out rapidly when the arrow is set free. A buffer-stop spring stores up all the kinetic energy of a train, and the stiffer it is the more quickly will it store the energy, and therefore the more suddenly will the train be brought to rest. Ordinary buffer springs are continually storing up and giving out energy, equalising only gradually the velocities of the two railway carriages, so that if one gets a sudden change of velocity, the other shall. only be affected gradually. In the same way, when any two objects have a springy connection, if one of the objects alters its velocity suddenly, the other alters its velocity in consequence only gradually. These are cases in which a spring is used to prevent shocks, or, as we may put it, a spring is used to lengthen the time of a blow, and there- fore to diminish the average force of a blow. 513. Now there are several distinct cases here to consider. 614 APPLIED MECHANICS. I. A body of mass B, moving with velocity v^ overtakes a body A, moving with velocity v . The buffer between them is strained until they move with the same velocity. If the common velocity then is v, B(v b -v)=A (V - V a ) = A + B The energy now stored up in the spring is In case the bodies B and A are themselves elastic, they them- selves, to a greater or less extent, act as buffers, and less energy is stored in the buffer itself. Also as changes of shape are usually accompanied by friction, some of the energy is wasted. If no energy were wasted, the two objects would keep going faster and slower relatively to one another, as the spring was compressed and extended, and to some extent this does go on after the blow ; but these vibrations are usually rapidly stilled as the energy is wasted in friction, partly, as we have already said, in the buffer spring itself, and partly by friction opposing generally the motion of the bodies. II. If one of the bodies, A, is fixed to the earth, then the mass of A may be regarded as infinite ; v and v in the above calculation become 0, but the same reasoning applies as before. III. Two carriages, A and B, are at rest connected by a spring. A is made suddenly to move through a distance a. Now if the spring were infinitely stiff, B would just as suddenly move through the same distance a. As the spring is less and less stiff, B moves over the distance a with less and less suddenness, because the kinetic energy which must eventually be given to B is suddenly stored in the spring, but is only gradually given to B by the spring. If there is no friction, B will be left vibrating. If there is no friction opposing B'S motion, but there is friction due to change of shape of the spring, B will vibrate and gradually come to rest. If there is friction opposing B'S motion, but there is no friction in the spring, the store of energy in the spring is greater than before. B must, of course, come to rest. With friction in the spring, B will more rapidly come to rest. APPLIED MECHANICS. 615 In all these cases, B comes to rest at the distance a from its old position. IV. As in the last case but A moves suddenly through a distance , and after a short time T is moved suddenly back to its old position. (1) If T is exceedingly small, B does not move. (2) If T is great, B moves as described in III., and repeats its motion in the reversed direction. (3) If T is neither too great nor too small, B has a motion intermediate between the (1) and (2) described motions. (4) If A'S motion is quickly vibrating through the ampli- tude a, B gets a vibratory motion of the same period; but superimposed on this is the natural vibratory motion of a period which depends on the stiffness of the spring and the mass of B; the natural vibrations will die away if there is friction. 514. It is well worth while for a student to illustrate this last case by means of a model. The crank Q, Fig. 359, is turned round regularly ; if this is done by hand, a fly-wheel ought to be used to give steadiness of motion. By means of a connecting-rod, P gets an up and down motion which, diminished in the ratio OA/OP, is given to A ; this creates a forced vibration in B. The natural vibration of B, when A is not moving, ought previ- Fig. 359. ously to have been studied (1), when there is very little friction, and B goes on vibrating with nearly the same amplitude for a long time ; (2) when there is fluid friction, B vibrating in a vessel of water. In this case there is really a change of inertia difficult to calculate; the water being set in motion. The amplitude of B'S motion gra- dually diminishes because of friction. When A suddenly begins to vibrate, B may have" a large natural vibration, but this gradually gets destroyed by friction, just as if there were no forced vibration. We shall now speak of the forced vibration only, and assume no friction. Let A vibrate with f times the frequency of the natural vibrations, and let A'S amplitude be 1. Then the amplitude of B'S motion will be as follows : when there is no friction B moves synchronously with A, so that B 616 APPLIED MECHANICS. is at the top when A is at the top, or else (and this is indicated with a sign) B is at the bottom when A is at the top. To prove this, let w be the weight in Ibs. of 15, so that its mass is \fjg. As before, a better approximation to accuracy is obtained by letting w include one- third* of the weight of the spring. Let the spring be such that a force of 1 Ib. elongates it h feet. Then - x the accelera- 9 tion is the force in the spring ; and if x is the distance in feet of the body below its position of equilibrium, - h j7% is the extra elongation of the spring when there is this acceleration. ' But if B is # feet below, and if A is y feet below their positions of equilibrium, then x - y is this extra elonga- tion, and hence or Letting n 2 = -^, we see that n divided by 2 TT _is the natural frequency of vibration. As in Art. 19, if we introduce our force of friction equal to b times the velocity, using 2 F for bff/w, we have -8 1 t By putting y = a sin. qt, it is easy to find the forced vibration. For simplicity, if we let P = o, then Table XII. is obtained by letting qfn be called/. Note in the model and from the tabled numbers that when the forced frequency is a small fraction of the natural, the forced vibration of B is a faithful copy of the motion of the point of support A ; the spring * Let the student prove that i of the weight of the spring is to be taken and not of it. APPLIED MECHANICS. 617 and B move like a rigid body. When the forced frequency is increased, the motion of B is a faithful magnification of A'S motion. As the forced gets nearly equal to the natural, the motion of B is an enormous magnification of A'S motion. There is always some friction, and hence the vibration cannot become infinite. When the forced frequency is greater than the natural, B is always a half-period behind A, being at the top of its path when A is at the bottom. When the forced is many times the natural, the motion of B gets to be very small ; it is nearly at rest. 515. Men who design Earthquake recorders try to find a steady point which does not move when everything else is moving. For up and down motion, observe that in the last case just mentioned B is like a steady point. When the forced and natural frequencies are nearly equal we have the state of things which gives rise to resonance in acoustic instruments ; which causes us to fear for suspension bridges or rolling ships. It is obvious, then, that the simple statement of the problem, " How do springs prevent shocks ? " presents, when we consider it very carefully, many quite different problems. It is worth while observing carefully the up and down motion of the body B of a waggon on the street when the wheels A go up and down over the stones. But we can study the sub- ject better perhaps in the model. We see that although a spring connection between A and B does prevent shocks, the motion of B may be dangerously great. Thus, for example, when an earthquake occurs it does not always do to merely have an elastic connection between the ground and the house, as the earthquake leaves a Japanese house vibrating sometimes so much as to give quite a sea-sick feeling to the inhabitants. We see that in all cases, unless there is friction opposing vibration of the body B whether this friction exists in the parts of the spring itself, or more directly opposes the motion of B there will be vibrations, sometimes dangerously large in amplitude. When by means of the spring connection we seek to diminish shocks, friction may be introduced by some dash- pot arrangement, which may consist of a porous piston moving in a cylinder filled with water or oil ; or the piston may be solid, and there may be a pipe and cock connection between the opposite sides of it. Here the friction is mostly fluid friction. If it were all fluid friction, there would be no 618 APPLIED MECHANICS. opposition to motion there would, in fact, be no friction if the motion were slow ; the friction is greater and greater as the motion is quicker and quicker. The friction between the plates of a carriage spring, which rub together every time the spring is changed in shape, is solid friction, which, if anything, is probably greater for slow motions than for quick motions. In all cases when vibrations are to be stilled it is better that the friction should be of the nature of fluid friction, but this is not always convenient. Its effect in stilling vibrations may very readily be studied in the laboratory. It will be found that by adjusting the cock of this dash-pot we can vary at will the rate of stilling of the vibrations of even very large masses, so that after a shock through the spring the body B may vibrate for a long time before it comes to rest, or it may come to rest after one slow lurching motion only. Now, it is necessary to understand that a dash-pot arrangement, or any other arrangement for introducing fluid friction, will not affect the static law of the spring in any way. It may be introduced on spring balances which are required to measure forces accurately, for example. But the sort of friction we find in carriage springs is very different. Here the plates of which the spring is made rub on one another, and there is solid friction, and such a spring as this cannot be used for spring balances. The law of the spring is altered ; we cannot depend upon the spring for measuring forces if there is any place where rubbing of solids takes place. Such springs are never used for purposes of measurement ; they are only used for preventing shocks. 516. Springs are of many different forms ; they are used as small as balance springs in watches, and they are sometimes so large as we see them in some buffer and locomotive springs. The following list is not given as by any means an exhaustive one: I. Cylindric spiral springs, subjected to axial loading. In these it will be found that the wire, whether round or nearly square in section, is twisted like an ordinary revolving shaft transmitting power, and the strain is one of torsion in the material. Examples. Most forms of spring balance and dynamo- meters ; the springs of indicators ; many railway carriage and tram-car springs ; safety-valve springs for marine and loco- motive boilers. APPLIED MECHANICS. 619 II. Cylindric spiral springs, subjected to a torque about the axis : In these it will be found that the material is sub- jected everywhere to bending. Examples. The balance springs of the best chronometers ; the springs used as elastic joints between two lengths of small shafting. III. Flat spiral springs, subjected to a torque about the axis. In these it will be found that the material is subjected to bending. Examples. The main and balance springs of watches and many clocks ; the springs used in nearly all contrivances which require to be wound up, or wherever a large reservoir of energy is required in a small space. IV. Nearly straight strips of material subjected to bend- ing. These are used in a great variety of cases, sometimes in one piece, as in the limb of a tuning-fork. V. More or less flat, or corrugated, circular, or of other outline, fixed or only supported at three or more points at the edges. VI. Indiarubber springs, usually subjected to merely ten- sion or compression, now being used largely for tram-cars. In all the above cases there is an absence of solid friction. VII. The Ayrton-Perry spring, used in indicating the amount of a force by a large relative rotation of a pointer. It will be found that the material in these springs is subjected to a combination of bending and twisting strains. The Ayrton- Perry twisted strip, in which a very small elongation is accom- panied by great relative rotation. VIII. More or less straight strips of material subjected to bending (like IV.), a number of pieces being used in one spring, these pieces rubbing on one another. When bending occurs, this introduces solid friction. Examples. Locomotive, waggon, and carriage springs, and nearly all the large springs used for minimising shocks, as buffer-stop springs. IX. Gases in closed vessels, the volume of which may be altered, as in the air-chamber of some force-pumps ; the cylinder and piston air-spring. It is, however, obvious that ^he functions of springs cannot be specified in a few words. Springiness comes in 620 APPLIED MECHANICS. usefully in the packing-rings of pistons, and in packing generally, to produce a good fit between pieces which rub on one another ; in spring split rings, as washers, which prevent a nut from becoming loose ; when a springy piece of material is used between two more rigid pieces which are bolted together, to give a uniform bearing without undue strains in the nuts. When, in fact, we discuss the elasticity of material generally, we see that everything in nature is a spring, and performs most of its functions in nature by means of this elastic property. 517. The main uses to which springs are put are these : 1. Lengthening impacts, so as to diminish the forces of blows, and therefore absorbing, and in these cases usually also dissipating energy. 2. Regulating motion that is, preventing large fluctua- tions in speed in driven pieces of machinery. This is not a common use of springs, because of the waste of energy. 3. As reservoirs of energy. 4. Regulating, as in watches and clocks. 5. As measurers of force. 6. As measurers of distance. 7. As measurers of angles. 518. The Best Materials to Use in Springs. A spring's usefulness depends primarily on its being a reservoir of energy. In the first two cases of the preceding table this capacity for storage of energy, and of course cheapness and ease of manu- facture, ought to settle for us the material of which a spring should be made. In the other cases we must also consider the question of perfect or imperfect elasticity and viscosity of the material. First, then, as to the energy which may be stored. The energy which must be given to distort a spring before it takes a permanent set is called its resilience. Now, it will be shown in Art. 535 that in all springs sub- jected to bending, as springs of classes 2, 3, 4, 5, 7, and 8, the resilience per unit volume of the material depends upon the resilience per cubic inch of the material when subjected to compressive or tensile force, and this is f 2 -r 2 E in inch- pounds, where f is the greatest tensile or compressive stress which the material will .stand without taking a set, and E is APPLIED MECHANICS. 621 Young's modulus of elasticity. The following table shows the value of this constant for various materials : TABLE XIII. SPUING MATERIALS SUBJECTED TO BENDING. / E In Millions of Pounds /2-J.2B per Square Inch. Wrought Iron 24,000 29 10 Mild Steel 35,000 30 20 Mild Steel, Hardened 70,500 30 83 Cast Steel, Unhardened 80,000 30 107 Cast Steel, Hardened 190,000 36 50J Copper 4,300 15 0-62 Brass 6,950 9-2 2-62 Gun Metal ... 6,200 9-9 2-00 Phosphor Bronze 19,700 14 13-85 Glass 4,500 8 1-26 The numbers in Tables XIII. and XIV. are subjected to very considerable variations, especially in the cases of copper, brass, gun metal, phosphor bronze, and glass. Indeed, in our opinion, definite statements as to the values of f 2 -f- 2 E, or f* -f- 2 N, ought not to be made, until careful experiments have been made on such varieties of these materials as are actually used in spring-making, and this has not yet, we be- lieve, been done. We have taken the values of f, E, and N from Table XXII. In the best spiral steel springs, for example, the value of J[ (the proof shear stress) has been found to be rather 60,000 or 70,000 Ibs. per square inch than the 145,000 given in the table. It will be shown in Art. 535 that in all springs where the material is subjected to twisting merely, as springs of Class 1 the most important class, probably, for the use of mechanical engineers and instrument makers the resilience per unit volume of the material depends upon the value of where J^ is the greatest shear stress which the material will stand without taking a set, and N is the modulus of rigidity of the material. The following table shows the value of this constant for various materials : 622 APPLIED MECHANICS. TABLE XIV. MATERIALS FOR CYLINDRIC SPIRAL SPRINGS. /i N Iii Millions of Pounds /1 2 -T-2N per Square Inch. Wrought Iron 20,000 10-5 19 Mild Steel 26,500 11 32 Mild Steel, Hardened 53,000 11 128 Cast Steel, Unhardened ... 64,000 11 186 Cast Steel, Hardened 145,000 13 809 Copper 2,900 5-6 0-75 Brass 5,200 3-4 4-00 Gun Metal 4,150 3-7 2-33 Phosphor Bronze ... 14,500 5-25 20 The numbers given in Tables XIII. and XIV. are supposed to express, then, the actual relative values of the various materials for spring-making. It will be observed that hardened cast steel is very much better than any other material for spring-making ; hardening makes it five times more valuable. It is about 35 to 40 times more valuable than phosphor bronze; more than 40 times more valuable than wrought iron (which is not so good as phosphor bronze). The fact that phosphor bronze makes probably the best non-magnetic material for springs has been known to me for fifteen years. I tested this result by a great deal of experimenting with various materials. But this is not the only virtue of phosphor bronze. In quite a remarkable degree it is free from many of the vices of other metals sub-permanent set after small, loads, effects due to fatigue, etc. It is worth while to mention that by the nature of the process of manufacture, the material may have initial strains in it ; before applying it in an instrument it ought to receive a considerable set in the direction in which it will most usually be strained. Phosphor bronze spring's in my electrical instruments receive a set from a load which is six times as great as the greatest load ever applied to the spring when it is in use. The numbers in the tables give us guidance, but we must also consider special conditions. The hardening and temper- ing of steel require great care ; so great is this that we may almost say that there is only one steel spring maker in the APPLIED MECHANICS. 623 whole of England. Now, phosphor bronze and brass and copper receive their greatest hardness by drawing through dies or rolling. They can, in fact, be hardened very uni formly in the cold state quite readily, and springs of them are easy to make. Again, although tempered steel has usually an oxide of iron to protect it, and a soft iron spring of any kind can also be given such an oxide by Barff's process as a cover- ing, yet, on the whole, steel and more especially iron springs are much more subject to rust when exposed to a damp atmo- sphere than copper, brass, or phosphor bronze. Again, in certain electrical instruments where springs are used, steel and iron must not be used because of their magnetic pro- perties, and in other measuring instruments the properties catalogued in the tables may not always be all-important. 519. The following property is not of importance in springs used to prevent shocks, as in buffer and carriage springs : When a spring is loaded with even a small load it may continue to lengthen axially slowly if the load is kept on ; and afterwards, when the load is taken off, it may not immediately shorten to its original length, but needs time. This is usually called sub-permanent set, and is greater with greater loads. When such a spring is unloaded after it has experienced loads of various amounts and various periods of rest, it will not usually go back to its old length, but will slowly undergo slight shortenings and lengthenings of various amounts de- pending on its previous experiences. I sometimes call this property the " creeping " property of the material. Professor Ayrton and I have written a paper concerning this property. A material possessing much of it is quite unsuitable for the springs of measuring instruments. 520. Spiral Springs. A spiral spring is a wire, or rod, or strip of any constant or varying section (we shall always speak of it as a wire, of whatever size or shape its section may be), coiled so that the centre line of the wire lies everywhere on some surface of revolution. In most cases the wire is wound on a cylindric surface, the winding being perfectly regular that is, the angle made by the centre line of the wire, Fig. 860. 624 APPLIED MECHANICS. with a plane at right angles to the axis of the spiral, is constant. Many cylindric spiral springs in use have wire of square or of elliptic section. In another form which is used as a buffer spring, the diameter of the coils varies as if the wire had been wound on the half of a very bulging barrel-shaped mandril ; and we see that under pressure this spring can get very short axially without the coils coming into contact. In the form Fig. ""* 360, the mandril was of a coni- cal shape. Lastly, in Fig. 361, we have the flat spiral spring what the conical form Avould be- Fig. 861. come if it were squeezed until all its coils lay one plane. 521. As an example of the bending of a strip of material, which might have been considered after Art. 327, let us take the case of a flat spiral spring, such as the main or balance spring in watches. Let N P M (Fig. 361) be such a spring, fastened to a case at N, and to an arbor or axle at M. When no forces are acting on the spring it has a spiral shape. Suppose that in this case, at a point P, the radius of curvature is r , and that when the spring is partly wound up there is at P a radius of curvature r, then is the change of curvature at P, and we know that the bending moment which produced this change of curvature is equal to E i ( - ), where \?' 7*Q / E is the modulus of elasticity of the material and i is the moment of inertia of the cross-section. (Thus, taking E at 36,000,000, if the breadth of the spring is 0-2 inch and its thickness 0'03, then EI is 16-2.) Now suppose the arbor to have turned through the angle x o G (which we shall call A) from the unstrained condition. What are the forces acting on P M and the arbor ? Whatever these forces may be, they must be in equilibrium. If these forces were changed, there would be an alteration in the shape ; but so long as these forces do not change, the shape and position of things do not alter. This is why we can apply to the spring, P M, and the arbor the laws of forces acting on rigid bodies. So long as P M o does not alter in shape, it obeys the laws of rigid bodies. 522. Now, the forces acting on the arbor may be very numerous pressure of the pivots, pull of the f uzee chain, or pressure of teeth of wheels ; but whatever they may be, we know that they can be represented by one force acting at o, the centre, together with a couple, c. If the spring is not in contact with the top or bottom of its case, and if the coils are not in contact with one another, no other forces act on the spring, M P, except at p. The particles of APPLIED MECHANICS. 625 steel on one side of the section at p are acting on the particles on the other side ; but whatever the forces at each of the particles may be, we know that the total effect at P is the same as that of one force and one couple. We cannot easily say what the force is, but if r is the radius of curvature at p, and if r was the radius of curvature at P when the spring was unstrained, then the couple at p is what we have already called the bending moment, E_fo3 /l l\ "12" \r ~ V' Let us suppose, for simplicity, that the spring is everywhere of the same breadth and thickness, and let us use the letter e instead E bt 3 of -y^-, which is now, of course, the same everywhere. The couple at p is then e ( - \ The only forces acting on PMO are: \7* TQ/ A force at o, of amount H, in the direction o H, say ; a couple at o whose moment is - L ; a force at p ; a couple at p whose moment is given above and is positive. Now, we know that the sum of the moments of all the forces about any point must be nothing. Take all the moments about the point p. The force at P has, then, no moment, and is to be neglected, and we have /I 1 \ A - H X P H L + . We shall 634 APPLIED MECHANICS. neglect the weight of the spring itself as it will be quite easy afterwards to correct for this. Let r be the radius of the coils that is, the distance of the centre of the wire everywhere from the axis. Let the whole length of wire be I, the angle of the spiral a. Let B be the flexural rigidity of the wire in the osculating plane of the spiral. B = E i where E is Young's modulus of elasticity, and i is the moment of inertia of the section of the wire about the line through its centre of gravity which touches the cylindric surface, and is at right angles to osculating plane. The bending moment, divided by the flexural rigidity, gives the change of curvature produced by bending. Let A be the torsional rigidity of the wire. The twisting moment applied to a wire divided by A gives the angle of twist produced per unit length of wire. In Table XV. a number of values of A and B are given for sections of wire which are in common use in springs. A is the' torsional rigidity of the wire, being the twisting moment required to produce unit angle of twist per unit length. B is the flexural rigidity of the wire in the osculating plane of the spiral, being the bending moment required to produce unit change of curvature in that plane. The line P Q represents the axis of the spiral relatively to the wire. N is the modulus of rigidity, and E the Young's modulus for the material. In the last two cases t is supposed to be small in comparison with b. Notice that A in the elliptic sections becomes N7rD^ 3 /16 when d is small in comparison with D, and it becomes N bt 3 /3 in the rectangular sections when t is small. Coulomb's wrong assumption was that A = N i when i is the moment of inertia of a section about its centre. Now, for the elliptic section (o 3 ^ + D^ 3 ), and hence o2 Coulomb's A _ D 3 d + 2 correct A if x is T/d. Coulomb's value is correct when the section is circular, and we see that it is more and more wrong as the section gets natter and flatter. The true value of A is always easy to calculate APPLIED MECHANICS. TABLE XV. 635 ^5 I 1 1 A. Torsional rigidity. 32 16 D 2 + N 7T Drf 16 b 2 + 0- 14058 N* 4 N W 3 iP~VP B. Flexural rigidity. 64 ; 64 12 12 I? w. Axial load when /is greatest shear stress. (r = radius of coils) 636 APPLIED MECHANICS. in the case of an elliptic section. Its value for a rectangular section is calculated from an infinite series, and it is therefore very important to know that Cauchy has proved that the torsional rigidity of a rectangle bears (approximately) to the torsional rigidity of an inscribed ellipse the proportion of their moments of inertia in the case when D is several times d (see Art. 313). Consider the portion of spring below any cross-section p. This portion is in equilibrium. Hence the molecular forces exerted on it at the section p must balance F and L, and these are the only conditions which we find it necessary to consider. Let Fig. 366 represent the elevation of a portion of the wire below the section p. Let T T be the elevation of the axis of the spring. Then about the axis PM we know that the moment due to the force F must be balanced by a torque of molecular forces whose amount is p r ; and about the axis PT there must be a torque of molecular force of amount L. Resolving these in the directions PS and PU in the plane of the paper, which is tangential to the cylindiic surface at p, we have about p s the torque y r sin. a - L cos. a, and about p u we have r cos. a + L sin. a. Now the moment about p s means a bending moment which produces an angular change per unit length whose amount is T?r sin. a + L cos. a B and the moment about p u means a twisting moment which pro- duces an angle of twist per unit length of the amount F r cos. o + L sin. a . . j - .... (2). Let us now consider how these two angular changes in unit length of the wire cause elongation and rotation at the bottom end of the spring. If the spring is sufficiently long, it is obvious that an axial elongation x and a rotation will occur at the free end of the spring, for we can then imagine that the lateral motions due to all the elements of the wire exactly counteract one another. It is therefore only necessary for us to obtain from the above expressions those elements which produce x and . Now any rotation of the body below p about any axis can be resolved into equivalent rotations about other axes, according to the laws for the resolution of vectors generally. The rotation (1) about the axis PS is equivalent to (1) multiplied by cos. o about p T, and to (1) multiplied by sin. o about p M. The rotation (2) about the axis p u is equivalent to (2) multiplied by sin. a about the axis P T, and to (2) multiplied by cos. a about p M. Adding, we get, on the one hand, the rotation about p T, produced by the flexural and torsional strains in unit length of the wire ; and this multiplied by I, the total length of the wire, gives ^>. Adding, we get, on the other hand, the rotation about p M, produced by these same flexural and torsional strains ; and it is obvious that this rotation causes a point on the axis of any portion of the spring APPLIED MECHANICS. 637 below F to be lowered by a distance which is obtained by multiply- ing the rotation by r, the radius of the spiral. Multiplying by I, we obtain the whole axial lengthening of the spring that is, putting our answer in its simplest shape, and x being the rotation and elongation produced in a spiral spring by an axial force F and a couple L acting together. In the theory < and x are assumed to be very small. If we use 5 (J> and 5 x for them, using < for 2 ir n where n is the number of turns, and x for the axial length of the spring, so that x\n is what we sometimes call the pitch of the spiral ; then - -j- 2 tr r = tan. . . . . (5) and sin. = x\l . . . . (6), and loy means of integration we can find accurately the effect of alteration in r and a as the spring changes in shape. Taking (3) and (4) as they stand, however, I find that young students can have no better easy mathematical exercises than are to be obtained by working problems on them. For example, let I, r, a, A and B for a spiral spring be given, we can calculate x and when given F and L, or if given x and we can calculate F and L. The very common case of a small, say a = 0, leads to z = Tlr*l'A. .'. . (8). Thus if the spirals are very flat, whatever the nature of the section of the wire may be, the spring when subjected to axial force has no tendency to rotate at its end, and the spring when subjected to a couple merely has no tendency to alter in axial length. Again, in any given spring, suppose we are informed that the rotation of the end is prevented, and that a force F acts, putting = we find L in terms of F from (3*), and using this in (4) we have x in terms of F. Again, as in chronometer springs, if the axial elongation is prevented and only a torque L acts, put x in (4) and find F in terms of L ; use this in (3) to find in terms of L. Let o = ^ or 45. (3) and (4) become If we apply to (9) and (10) the first condition given above, that B ~ A is, - In estimating for general purposes the effect of altering the angle a ; or the effect of constraint, such as preventing rotation of the end of a spring when an axial force is applied ; or the effect of change of shape of section, etc., it is useful to remember that for most substances we may take it that approximately E = 2-5 N. It is very interesting to study the formulae (8), (10), (12), using the Table XV. for the values of A and B for various forms of section. We shall leave this study to each student for himself working out only the following interesting examples. Students will please work carefully the following fifteen exercises : Take N = 10-2 x 10 6 , E = 25-5 x 10 6 . Apply an axial load of 1 Ib. and find the elongation in each case. There are 10 springs of wire, 100 inches long ; diameter of coils 2 inches. The first has round wire of 0-1 inch diameter. All the others are of such sizes of section that there is the same total volume of material. Half the spirals have an angle of 45. When the exercises are finished, divide all the elongations by that of the first spring, and enter the results on such a table as the following : TABLE XVI. RELATIVE ELONGATIONS FOR SAME AXIAL LOADS. 0=0 a =45 Whether the end is allowed freedom to rotate or not. End free to rotate. End restrained from rotating. Solid round wire. 1 0-9 0-889 Hollow circular section, thickness of tube ^th of outside diameter. 0-196 0-197 0-195 Square wire. 1-132 0-948 0-912 Rectangular wire, as in Fig. 6, Table XV. Breadth 4 times thick- 2-029 1-108 0-349 Rectangular wire, as in Fig. 7. Breadth 4 times thickness. 2-029 2-544 2-440 APPLIED MECHANICS. 639 In the theory of the spiral spring I considered the bend and twist given to each small portion of a spring, and assumed that the resultant action at the bottom of the spring was a rotation and an axial elongation that is, that the lateral motions produced on the bottom by all angular motions of portions of the spring exactly balanced one another. As a matter of fact, however, it is only when the spring is very long that these lateral actions balance. To take the lateral actions into account is of no practical importance in such springs as we have been considering (cylindric spiral springs) ; but when the spires of a spring rapidly change their character, particu- larly in the case of flat spiral springs, the lateral actions are of importance (see Art. 522). ' 533. If any important object were to be served, I would calculate the relative strengths of all the springs under the various condi- tions here mentioned. It is only necessary to know the load F, which, when producing a bending moment F r sin.a and a twisting moment Frcos.a in a wire of the particular section at the same time, shall be capable of just producing permanent set. For a circular section this is well known that is, it is well known that for a circular section the above two moments, acting together, are equivalent to a twisting moment vr sin.a + N /rV 2 sin. 2 a + F 2 r 2 cos. 2 a. or vr (1 + sin. a) if the material has equal strength to resist shearing and tensile stresses. This is merely (3) of Art. 298. We see, then, that for a round wire the breaking-load for a spring of angle a is less than for one of the same size of wire and length of wire and diameter of coils, but quite flat in its spirals, in the ratio 1 : (1 + sin.a.) Thus a spring with a = 45 has a strength only a little over one-half of a spring with flat coils. This is the reason why, when a weight has been hung from a spring which produces permanent sot, the spring so very rapidly gets completely spoilt, oven although there is a counteracting influence due to the coils becoming smaller in diameter. It is sufficient for most practical purposes, then, to say that if the load to produce permanent set in a round wire spring is 1 when the coils are flat, the load which will produce permanent set when the coils are not flat is 1 -T- (1 -f sin. a). Again, in springs with flat coils, all of radius 1", the loads to produce permanent set are simply equal to the twisting moments which, when applied to 'the wire, produce permanent set; and hence the following table is very useful. w, in the last column of Table XV., gives the load which will produce the maximum shear stress /in each of the sections of wire in the case of cylindric spiral springs, the angle of the spiral a being 0. Hence, if / is the proof shear stress of the material, w is the proof load. The student will calculate the numbers of Table XV. as an exercise 640 APPLIED MECHANICS. In the following table the load to produce permanent set and the resilience are given for cylindric spiral springs, so that students who are in the habit of only calculating the strength and stiffness of springs with round wire may be able to arrive easily at the strength and stiffness of springs made with other lands of wire. Comparison of the loads which will produce permanent set in springs of the same diameter of coils and same area of cross-section of wire : in all of them o = that is, the coils are supposed to be as flat as possible. This is really a comparison of the torsional strength moduli of sections of the same area. Let students calculate the various numbers here given. TABLE XVII. Load to Produce Permanent Set. Resilience per Cubic Inch. Circular ... 1 1 Hollow circular, thickness of } tube J^ of outside diameter } 2-733 ( More and more nearly 2, as tube is thinner ( and thinner. Square . . 828 938 Rectangular wire, as in Fig. 6. ) Breadth, 4 times thickness ... \ 110 224 Rectangular, as in Fig. 7. | Breadth, 4 times thickness ... \ 110 224 641 CHAPTER XXIX. RESILIENCE OF SPRINGS. 534 An examination of our formulae will show that for nearly round wire, even when a is considerable, the rotation depends almost altogether on the couple L, and the elongation x depends almost altogether on the axial force F. But when the section of the wire is very different from circular, the angle a enters very materially into the calculation. Many examples of great interest may be taken up. Nearly flat coils are presumed. 1. If r alters in a spring, and if we want the resilience of the material to be the same everywhere ; that is, the wire being round, if we want the material to be equally ready to break everywhere, we must make constant. That is, if the coils get twice as large, T the diameter of the wire becomes 2 or 1-26 times as large. 2. Eound wire all of same diameter d, but the coils capable of lying just within one another if the spring is compressed by axial force. Let n be the number of coils to any point starting from the end of the wire at the small coil side, where the radius is r . We may take r = r Q + nd, I = r 2 irn + 4 iftfid ; and adding together the elongations produced on each elementary length into which we may imagine the wire divided, we find The strength of this spring is to be calculated as if all the coils were equal in size to the largest. Some students may perhaps be interested in working the problem : Find the cylindric spiral spring which, made of the same wire with the same number of turns, will receive the same axial elongation or compression from the same axial force. The answer is, if n is the total number of turns, R, is the radius of coils in the new spring, r 2 nd + r nd* Thus, as a numerical example, if the first and last coils were 4 and 9 inches in radius, d = 1 inch, n = 5 turns, R is 6'8 inches. 535. Resilience. The average resilience per cubic inch of a spring is the whole resilience divided by the volume. It is only in the case of uniform stress and strain in the material everywhere that we have maximum resilience for the whole spring. In actual cases this only occurs in tie-rods and struts, and in spiral springs made of a wire which is a thin tube, 642 APPLIED MECHANICS. | M & is the resilience per unit length of a wire, if M is the proof twisting moment and is the proof angle of twist, or il M is the proof bending moment and is the proof change of curvature produced. It is easy, therefore^ to work out the following table of values, and every student ought to do this as an exercise : TABLE XVIII. RESILIENCE PER UNIT VOLUME IN INCH-POUNDS. Simple compression or extension. Only convenient in j , / 2 ^QQ springs made of indiarubber ... ... ... ) ~E ~ Bending if the maximum stress is reached in every ^ section, and section is rectangular. As in beams of | /- 2 uniform strength, in well-made carriage springs, and } = 133 in spiral springs subjected to a torque only ; also { C-springs. Art. 370 J Bending. Uniform strip fastened at one end and loaded \ at the, other, or supported at the ends and loaded in > ^ = 44 middle; ... ... ' .;. ) E Simple shearing. Possible in springs of indiarubber; j ^2 also possible in spiral springs made of thin tubes > =1,000 circular in section ... ... ) N Torsion. As in spiral springs of round wire subjected to j j / 2 ^QQ axial loads } * ^ Torsion. As in spiral springs of square wire subjected j .^^f^__ 469 to axial forces. ) ' N Torsion. As in spiral springs of oval or rectangular wire subjected to axial forces, anything less than for circular depending on ratio of diameters or of breadth to thickness. 536. The student may not yet have been sufficiently im- pressed with the importance of knowing the resilience per unit volume of the material in all the springs. The resilience per unit volume tells us the value of a particular shape of spring. Suppose that a spiral spring of any given material is to be used for any purpose. The greatest load is stated, and the elongation or com- pression due to that load. Then whatever other conditions may be given as to the coils being of the different sizes, etc., we know that the spring may be made of any shaped section of wire, but that the thin tubular circular section is the very best and the solid circular section is half as good, and any other sections are not half as good, and that the spiral spring form is better than any other. In any spring, if a force r produces a known motion x in APPLIED MECHANICS. 643 its own direction, then the resilience per unit volume, multi- plied by the volume of material, gives the proof load P' multi- plied by half the motion x' produced by it. This principle enables easy calculations to be made. Again, if in any spring a torque L produces a known angular motion in its own direction, then the resilience per unit volume, multiplied by the volume of material, gives the proof load L' multiplied by half the angular motion ^''produced by it. In nearly any case we consider, it will be found that the important thing looked for when we choose a particular shape of spring is total resilience total energy stored up. Thus for example : We want a spring to exert a force P' and only to alter, say, w lb., for a change of shape indicated by a motion in the direction of P' by, say, b inches. Evidently here the law of stiffness of the spring is b a w, say b = k w, where k is a known number. And, therefore, if x' is the greatest motion, x' = F' ; and the total resilience is -~ P' x', or -^ - P' 2 . That is, we are given P', w, and b, and therefore the total resilience of the spring, and if we know the type of the spring we can find the volume of the substance required and therefore its weight. What people generally mean by the "springi- ness " of a spring means that it shall not change much in the force with which it acts, for a considerable amount of alteration in shape. Now b is the change of shape and 7 is the fractional change of shape, so that p' represents the springiness of a spring. But the value of a spring also depends upon the greatest force it can exert. That is, its value depends on F'& j that is, on its resilience. Looked at from almost any point of view, we see that the value of a particular form of spring is represented mainly by its resilience. Of course, its shape and cost of manufacture are also of importance. We may know that a tubular spiral spring may be the best for weight, and yet a C-Spring shape may be best suited to the use to which it is to be put, and we put up 644 APPLIED MECHANICS. with the disadvantage of using from twenty- two to twenty-eight times the weight of metal because of some other convenience. 537. If the angle of the spiral is a, the length of each turn, instead of being 2irr, as we take it in approximate calculations, is really 2 IT rf sin. a. If the axial length of a spring is x, the angle a is such that cos. a = x /I. If the axial length changes to x + x, the angle changes to a 1 , such that cos. a 1 = (x + #)//. We have not taken these matters into account, assuming that our elongations were small or that our calculations were to be only approximate. The student who knows a little calculus may use x and 5

in Art. 532, and by integration obtain general and accurate expressions. A conical spiral spring of round wire of diameter d, whose. spires vary^ gradually from greatest radius r r to smallest r ; the proof load is evidently to be calculated from ^ or w 1 = The elongation for a given load w ought to be calculated for short lengths and added up. Mathematicall ly it is evident that ~7. 32 w r- .//i *.,;., (2). Taking it that if s is distance from the end of the wire where r r=sr + s ^ ?, and it is easy to show that In fact, we see that in a conical spring, instead of ?' 2 for a cylindnc spring we take EXERCISES. 1. If a conical spring is formed of round wire 0'2 inch diameter, 40 inches long, the coils varying from r^ = 2 inches to r = 1 inch, what is the proof load, and the axial shortening with this load ? Ans., 47 Ibs. if/= 60,000; 0-4 inch. 2. An Ayrton-Perry spring is made of strip section, as in Table XV., which nearly covers a cylindric surface, and a =_45, so that 16 =. 2 irr x if x is the axial length, and x = I sin; o, or I/ \/2. Hence lb 2 irrl/ A/2. b = TT A/2>. If, then, r = 0-1 inch, b = -44 inch. If. t = -001 inch, and we have been able to obtain very broad strips of steel of this thinness, A = -0019 B = -00132 according to Table XV. Hence a load F will pro- duce an elongation x = 12'8 F^ and a rotation

Hence the force w s at its end, producing- the P 8 n + st "W A bending moment w A, must be such that produces this change of curvature. Hence We must, therefore, write out equation (1) for every strip to find the A r alues of p s for each, in terms of pi, which is known, and also equation (2), thus calculating every value of w a , the unknown t\ being in every term. Now we make the statement that the sum of all the values of w s is 0, and this enables n to be found. Thus (2) and (1) give and we must find i\ by inserting 1, 2, 3, etc., for in the equation S If st is small compared with r\ and KI, we find that, if there are strips altogether, if + 5", - 5?-5 ( " - 1( ) r ' 2 - n + 4 M - 1) = o . . . . (5), a quadratic to find r^ It is not necessary to pursue the subject farther. Student? have the means of working all sorts of practical problems. APPLIED MKC;I.\:;ICS. 649 541. Tempering Carriage Springs. In tho last part of the work on the strips, a plate being red hot, by an interesting manipulation between several pairs of smiths' tongs held by two workmen it is fitted to lie everywhere close to a curved piece of metal, so that it receives- a definite shape. It is now a dark red, and the workman dips the plate into water, letting the two ends go in simultaneously and the middle of the plate last. The plate is now placed in an air furnace, where it gradually heats ; it is taken out once or twice and rubbed with a piece of partly charred wood. The experienced workman can tell by the nature of the smoke coming from the wood whether the temperature of the plate is high enough ; when it has been heated to a sufficiently high temperature it is withdrawn from the air furnace and allowed to cool, lying on a metal table with others which have preceded it. 542. Tempering Spiral Springs. The processes employed by Messrs. Salter are, unfortunately, unknown to me. In all cases, however, they probably consist in winding wire or rod in the red hot state on a smooth iron mandrel, shaping the end parts of the spring, if the wire is small, by pliers ; if large, by special tools which can readily be designed for special cases. The formed spring is now heated to a dull red heat in an air furnace and plunged into hot oil, just of such a temperature as will produce the blue colour peculiar to spring steel. It is in this dipping of the hot steel into the oil bath that any trade secret can exist. If all the steel could from the same instant and at the same rate lose its heat, so that the hardened steel would be uniformly hard everywhere, the process would be perfect. Large spiral springs for safety-valves are dipped so that their axes remain vertical. -To lay them sideways into the bath might cause the lower half of each coil to cool more quickly than the upper half. It is quite possible that good spring makers may not only dip their springs axially but also give to them a rotatory motion round their axes as they enter the bath. 543. Very small spiral -springs for the balances of chrono- meters and watches require special care, whether they are cylin- dric or flat spiral springs. This is on account of the thinness of the material and the rapidity with which it may cool. They are usually enclosed in a box very small springs are enclosed in a platinum box either on their mandrels if cylindric, or coiled up together with others if flat, surrounded by powdered 650 APPLIED MECHANICS. charcoal. The box is heated to redness for a sufficient time to ensure the springs inside being all red hot, and the box is then immersed in water. The little box is manipulated by means of a long handle. It is now placed upon an iron pan placed over a flame, and lying beside the box on the pan is a piece of brightened steel. As the pan gets heated the brightened piece of steel takes different colours, and it is presumed that the steel springs have about the same temperature. When the dark blue colour is reached, which is so characteristic of spring steel, the box is removed from the pan and allowed to cool. 544. Tempering Flat Spiral Springs. In many cases these springs are not shaped before tempering. Great lengths of steel strip are passed through an air furnace from one set of rollers to another, at a rate which depends upon the particular purpose for which the strip is designed, upon its breadth and thickness, upon whether it is Bessemer or crucible steel. Just when in the red hot state and leaving the air furnace, the strip is passed through a vessel through which cold water is kept circulating, and is consequently made very hard. It then passes between pieces of cotton waste kept well soaked in oil, and then over a flame which keeps the oil ignited ; and on leaving this region the strip is gradually allowed to cool before being wound upon a roller. If allowed to move too slowly, the strip and the oil covering its surface are for too long a time subjected to the heating effects of the second furnace, and the steel becomes softer. I have seen strips which were said to be continuous and a mile long which had been tempered in this way. The very fine strip steel which is used in the Waterbury watch is tempered in this way. Great quantities of cheap Bessemer steel strip for use in ladies' corsets and numerous other purposes are also tempered in this way. 545. I have never seen this process carried out by electrical heating as a substitute for the air furnace, but it is obviously' quite easy to make the substitution. The strip or wire to bo tempered receives an electrical current from a certain roller A, over which it moves, and another roller B at a short distance from the first ; and at B, or even before it reaches B, the steel passes into an oil bath. In this case we have no sudden great hardening and then a softening process ; the tempering is all done in one operation, a cooling from red heat to the tempera- ture of hot oil. As the wire or strip is still kept heated when APPLIED MECHANICS. 651 in the oil, the cooling is gradual, and obviously the softness of the Steel will depend 011 how much of it is kept heated in the oil and on the average temperature of the oil. 546. In well-made carriage springs the resilience is per cubic inch, O E and in designing a spring a knowledge of this fact enables us somewhat to shorten the work. Thus, suppose a spring is wanted to take a proof load of 3 tons, with a deflection of 3 inches just making it straight. The total resilience is J x 3 x 2,240 x 3, or 10,080 inch pounds. If/' = 30,000 and E = 30,000,000, the resilience per cubic inch may be 5 inch pounds, so that the volume of steel required for the spring is 2,016 cubic inches. Now if there are n plates, this volume may be taken as n b tl (I being half the length of the longest strip), or nbtl = 2,016 .... (1), which is one equation towards the solution of some problem. EXERCISES. 1. What ought the initial curvature of a -inch plate to be if the proof stress of the material is 30,000 Ibs. per square inch (exists when the plate is straight) and E is 30,000,000 ? Here, by formula of Art. 540, 30,000 = 15,000,000 x \w, w = '004. The radius of curvature is _-^ =250 inches. 2. If the above plate is 36 inches long, what is its initial dip, oc, in Fig-. 367 ? Using the approximate formula oc = P -j- 2 r, we have 3. If the spring is formed of six plates, the longest being 36 inches, the overlap on one side of plate "on its neighbour is 36 -r 12, or 3 inches. 4. If the breadth of each plate is 3 inches, D = -000259 w. 5. The load w' which will produce the deflection 0'64 inch that is, which will straighten the spring is " 652 APPLIED MECHANICS. 6. How many plates of f inch thick and 3 inches broad, the longest being 30 inches in length, of the ahove-mentioned steel will be required for a spring whose proof load is to be 1 ton ? o 40 2* x 3 x & x 30,000 2}24 3-x 30 ' or n = 8. Hence the spring ought to be made of eight plates. 547. The answers to the above exercises guide us as to the best way of constructing carriage springs so that the strips throughout shall have the strength of the overlapping parts. The following rules are based upon the dimensions of these overlapping parts, which are merely little cantilevers. If X is the overlap, Fig. 367, and I is the half-length of the longest strip, so that n being the number of strips n \ = I , then t being thickness of each strip, b its breadth, d the deflec- tion of the spring, f the maximum stress, E Young's modulus, w the load at A and at B, there being a load 2 w at the middle, it is evident that /= 6wZ -=- nb&, D = QwP/nEbtf. Iff is the proof stress, the resilience in a well-made spring 6 E inch pounds. Carriage springs are usually made of Bessemer steel, and we may take f = 30,000 Ibs. per square inch and E = 30,000,000. Examples. 1. If the overlap X is 2 inches, and there are 10 strips each f inch thick, so that tfie half-length of the whole spring is I = 20 inches, t = f , and if b = 2 J inches ; find the load which will deflect the spring 2 inches. Here D = 2 2 = 6 w 20 3 -r- 10 x 3 x 107 x 2J x (f) 3 , and hence w = 5-926 x 105 Ibs. If the plates were not of the same initial curvature that is, if the spring was, in this respect only, badly made, then the law for deflection is still true for the badly made spring ; but the rules for strength are untrue. We only calculate the increase in f due to a load w. 548. In the above theory I have assumed no friction between the plates. This friction makes the bending to be less for a given load if the load has been increasing, but if the load has been diminishing the bending will be greater than the formula (1) gives, on account of friction. When we test a carriage spring, the effect of friction in causing the deflections to be greater with diminishing than with increasing loads is very noticeable. I have already pointed out the unsuitableness (on this account) APPLIED MECHANICS. 653 of these springs for all measuring purposes, and how suitable they are for carriages and under other conditions where vibra- tions must be rapidly stilled by frictional forces. 549. In a spring such as Fig. 367, if 2 w is the upward load Ili! L Fig. 368. applied at the middle, the downward supporting forces at A and B are each w, and the end of each plate may be regarded as applying to the plate above it this force w. Thus in Fig. 368, if A B is the overlap of one plate on another, every part of the plate except the overlap part A B at each end is subjected to a bending moment w . A B everywhere, so that the change of curvature everywhere is the same. Also the part A B ought to be fashioned like a beam of uniform strength and curvature fixed at B and loaded at A. We find from Exercise 2, Art. 540, that we can make it of I Fig. 370. Pig. 371. uniform strength and uniform curvature at the same time by varying the breadth of the plate but not its thickness, the proper shape being shown in Fig. 370. This is roughly approximated to in many springs where the ends are shaped as in Fig. 368. If we keep the breadth constant, as in Fig. 369, but alter the thickness so as to have constant curvature everywhere (so that the end part of one plate may fit properly against another plate), we see from Exercises 3 and 4 that the thick- ness ought to vary as the cube root of the distance from A ; but it is not possible to have also uniform strength in this part of the plate. APPENDIX. TABLE XIX. USEFUL CONSTANTS. Time. One sidereal day = 86,164 seconds. Mean solar day = 86,400 seconds. One year = 365-24224 mean solar days. Length. British standard, the yard = 3 feet = 36 inches. 1 chain = 66 feet =100 links = 4 perches. 1 mile = 1,760 yards = 5,280 feet = 80 chains = 8 furlongs. 1 nautical mile = 6, 080 feet (average). = 10 cables = 1,000 fathoms (nautical). Telegraph poles 220 feet apart. 1 fathom = 6 feet. French standard, the metre = 39*37 inches = 3-2809 feet. 1 kilometre = '6214 miles. 1 inch = -0254 metre = 2-54 centimetres. 1 foot = 30-48 centimetres. 1 yard = -9144 metres. 1 mile = 1,609 metres = 1-609 kilometres. Surface. 1 square inch = 6 -451 square centimetres. 1 acre =4,840 square yards. 1 square mile = 640 acres = 2-59 square kilometres. Volume. 1 cubic inch = 16-387 cubic centimetres. 1 cubic foot = -02832 cubic metre = 28-31 litres. 1 litre = -22 gallon. For many practical calculations it is sufficiently correct to take : Length. 1 inch = 25 millimetres, but 25-4 is more correct. 1 metre = 3 feet 3 inches and f ths of an inch. 10 metres =11 yards. 20 metres = 1 chain. 8 kilometres = 5 miles. Surface. 1 square inch = 6^ square centimetres. 6 square yards = 5 square metres. 1 acre = 4,000 square metres. Volume, 4 cubic yards = 3 cubic metres. 1 gallon = 4J litres. 4 litres = 7 pints. 100 litres = 22 gallons. 1 cubic foot =28-3 litres. Weight. 1 gramme = 15.| grains. 10 kilogrammes = 22 pounds. 50 kilogrammes = 1 hundredweight. 1,000 kilogrammes = 1 English ton. Speed. 1 knot = l nautical mile per hour =1-16 miles per hour =1-7 ft. per sec. = 101 ft. per min. =51-5 centimetres per sec. APPLIED MECHANICS. 655 60 miles per hour =88 feet per second. Weight and Force. British engineer's unit of force = the weight of the standard pound in London. Weight of 1 Ib. = 16 ozs. = 7,000 grains = 453-6 grammes =: 445,000 dynes = '4536 kilogrammes. 1 oz. = 28-35 grammes. Weight of 1 grain = 63-57 dynes. 1 kilogramme = 2-2046 Ibs. 1 ton = 2,240 Ibs. = 1,016 kilogrammes. 1 ,000 kilogrammes = 1 ton (metric) = 2,205 Ibs. = -9842 ton (British). g = 981 centimetres per second per second. = 32-2 feet Value of g at London = 32-182 feet per second per second. Equator = 32-088 the Poles = 32-253 * Inertia or mass of a body = weight in Ibs. at London + 32-2. 1 gallon of water at 62 F. weighs 10 Ibs. by English law. 1 cubic foot of water weighs 62 '3 Ibs. 1 cubic foot of air at C. and 1 atmosphere weighs -0807 Ib. 1 cubic foot of hydrogen ,, '00557 Ib. For academic calculations we usually take the weights in pounds of 1 cubic foot of each of the following to be : Brickwork, 112; concrete, 150; grindstone and Portland stone, 131 ; granite and marble, 164 ; dry oak, 58; dry fir, 47; cork, 15; coal, 80; clay, 120. Weights per cubic inch of each of the following materials, in pounds cast iron, -26 ; wrought iron, '28 ; steel, '28 ; brass, '3 ; copper, '32 ; bronze, '3 ; lead, '4 ; tin, '27 ; zinc, '26. Work, Energy. 1 erg 1 dyne x 1 centimetre. 1 gramme- centimetre = 981 ergs. 1 foot pound = 1 '356 x 10 7 ergs. 1 kilogramme metre = 7 '2 3 3 foot pounds. 1 horse-power =33,000 ft. Ibs. per min. = 550 ft. Ibs. per sec. = 7 '46 x 10 9 ergs per sec. = 746 watts (watts = volts x amperes). 1 kilo-watt = 1,000 watts = 1 -34 horse power. 1 force-de-cheval = '9863 horse-power. Energy obtainable from 1 Ib. of coal = 8,500 Centigrade heat units = 15,000 Fahrenheit heat units = 12 x 10 6 foot pounds. Joule's equivalent 1 pound Fahrenheit unit of heat = 780 foot pounds. 1 pound Centigrade unit of heat =1,400 foot pounds. See note, p. 42. 1 horse-power hour = 1 '98 x 10 6 = 2 x 10 6 ft. -Ibs. nearly. 1 Board of Trade electric unit = 1,000 watts for 1 hour = ^ horse-power hour = 2-654 x 10 ft.-lbs. The base of the Napierian logarithms is B = 2 7 1 8. To convert com- mon logarithms into Napierian logarithms, multiply by 2-303. 656 APPLIED MECHANICS. TABLE XX. MODULI OF RIGIDITY. Substance. N = Modulus of Rigidity in Million? of Pounds per Square Inch. Steel Plates, \ per cent. Carbon 2 5> Steel Bo'iler*Plates ' ... ... Cast Steel (tempered) ... (untempered) Soft (unhardened) (hardened) Cast Iron Iron Boiler Plates "Wrought Iron Bars Plates ... Soft Iron Brass Copper ... ... Lead ... Zinc Tin ... ... ... ' ... Gold Silver Platinum Aluminium Delta Metal (rolled) Gun ,.., Phosphor Bronze Glass Wood Granite ... Marble Slate 13 13-5 14-0 12-0 11-0 11-0 5-0 to 7-6 14-0 10-5 9-5 10-8 to 11 5 to 5-5 5-6 to 6-7 27 5-1 to 5-4 2-2 4-0 to 5-6 3-8 8-9 to 9-4 3-4 to 4-8 6-25 3-7 5-2 3-3 to 3-9 1 to -17 1-8 1-7 3-2 APPLIED MECHANICS. 657 TABLE XXI. MODULI OF COMPRESSIBILITY. Substance. Modulus of Compressibility in Pounds per Square Inch. Temperature. Authority. Steel A.. .I. 21-6 x 10 6 Amagat. Steel ... ... 26-7 x 10 6 Thomson's "Elasticity." Iron 21-1 x 10 6 > Brass 15-3 x 10 6 > > Copper 17-1 x 10 6 Buchanan. Copper 24-4 x 10 6 Thomson's " Elasticity." Delta Metal ... 14-4 x 10 6 Amagat. Lead ..V "... 5-3 x 10 6 n Glass 5-8 x 10 6 Distilled Water... 3-2 x 10 s 15 C. Paglianni. Alcohol ... ,5,. 1-76 x 10 5 0C. Amaury and Deschampa Alcohol 1-62 x 10 5 15 C. Ether u. ''" ... 1-35 x 105 C. Ether -.;. 1-15 x 10 5 15 C. Carbon Bisulphide. 2-32 x 10 5 14 C. 'n Glycerine.,. 5-85 x 10 5 20-5 C. Qiiincke. Patroleum 2-11 x 10 5 16-5. C. Martini. Mercury ... 7-85 x 10 6 15 C. Amaury and Deschamps. Mercury ... 4-35 x 10 6 C. Colladon and Sturm. Mercury ... 3-74 x 10 6 0C. Amagat. 658 APPLIED MECHANICS. TABLE Breaking Stress, Weight in Ibs. per MATERIAL. Melting Point Specific of One Cubic Foot square inch. (Fahr.). ravi y. in [Itfl Pounds. Tensile. Soft Steel (unhardened) } Soft Steel (hardened) I 2,400 to 7'85 490 60, 000 to 100, 000 120,000 Oast Steel (untempered) ( Cast Steel (tempered) ) 3,000 90, 000 to 150,000 Steel Plates 60,000 to 80,000 Steel Bars 100, 000 to 130, 000 Cast Steel (drawn) 77 480 120,000 Steel Wire (English), drawn... Steel Wire (Common), tern- ) pered blue ... ) 7'4 460 120,000 to 140,000 330,000 Steel Pianoforte Wire, English 7-73 480 340,000 Manganese Steel, Cast 85,000 Nickel Steel, unhardened 75,000 Nickel Steel, hardened 190,000 Wrought Iron Bars and Bolts ^ 55,000 to 70,000 Wrought Iron Plates, with fibre Wrought Iron Plates (across | fibre) ... ... 3,000 to 3,300 77 480 51,000 46,000 Wrought Iron Plates (mean)J 48,500 Cast Iron < 2,000 to 2,500 7'2 450 14,000 to 30,000 Aluminium 1,300 2-6 162 33,000 Aluminium (annealed) 13,500 Brass, Yellow ) 1,700 7-8 490 17,500 Brass, Sheet ... ... j to 1,850 to 8-4 to 525 30,000 Brass, Tube 80,000 to 100,000 Brass, Cast 8-0 500 20,000 Brass, Wire 50,000 Copper, Wrought 2,000 8-8 550 33,000 Copper, Cast 20,000 Copper Wire, hard-drawn ... 8-9 555 58,000 Copper Wire, annealed 47,000 Zinc, Cast 750 7'0 436 7,500 Zinc, Sheet 7'2 450 30,000 Lead 615 11-4 712 1,900 Lead, Cast 11-2 700 3,000 Tin 450 7-4 462 4,600 Platinum Wire... 3,300 21-5 1,340 50,000 Gold (drawn) 2,200 19-2 1,200 38,000 to 41,000 Silver (drawn) ... 1,850 10-4 650 42,000 Phosphor Bronze (cast ) 55,000 Phosphor Bronze Wire (hard)... 100, 000 to 150, 000 Phosphor Bronze Wire (an- ) nealed) ) 50,000 to 60,000 APPLIED MECHANICS. 659 XXII. Breaking Stress, in Ibs. per square inch. P ./ t Stress which produces Permanent Set, in Ibs. per square inch. Safe Limit of Stress, in Ibs. per square inch. Young's Modulus of Elas- ticity, in 1 Com- pressive. Shear. Tensile. Com- jressive. Sheat Tensile. Com- )ressive. Shear. millions of Ibs. per square inch. 35,000 26,500 ' 17,700 17,700 13,000 30 Qfk olF 30 36 29 to 42 27 28 26 29 45,000 35,000 56,000 50,000 50,000 24,000 24,000 20,000 10,000 10,000 7,800 29 25 27 20,000 20,000 15,000 10,000 10,000 7,800 2b i 50,000 to j 120,000 28,500 10,500 21,000 8,000 3,500 10,500 2,700 14 to 23 10,500 7,000 5,000 3,600 2,700 9 92 14-2 58,000 4,300 4,000 3,000 3,600 3,200 2,300 15 3,200 7,300 1,500 72 6 24 12 11 20,000 14,500 10,000 7,400 14 660 APPLIED MECHANICS. TABLE Weight Breaking Stress, in Ibs. per MATERIAL. Melting Point Specific Gravity. of One Cubic Foot square inch. (Fahr.). in Pounds. Tensile. Aluminium Bronze, \ Cu95A15/ 8-25 515 60,000 Aluminium Bronze, ^ Cu90A110J 77 480 100,000 Manganese Bronze Delta Metal (cast) 65,000 to 85,000 47,000 Delta Metal (rolled) 74,000 Muntz Metal 45,000 Sterro Metal 60,000 . Gun Metal 1,900 8-6 536 25,000 to 50,000 Ebony 17 73 Oak, European 93 58 14,500 Mahogany, Spanish 85 53 15,000 Ash 8 50 17,500 Pitch Pine 7 44 12,000 Red Pine 10,500 Birch 54 34 14,500 Beech 7 44 11,500 Fir, Larch 53 33 11,000 Fir, Spruce 54 34 12,500 Hornbeam 76 47 15,000 Teak, Indian 78 49 15,000 Lancewood 95 59 20,000 Elm, British 55 34 14,000 Lignum Vitse ... ... -j 65 to 1-33 41 to 83 16,000 Sycamore 13,000 Cedar of Lebanon 59 37 11,400 Granite ... 27 170 Marble 2-8 175 700 Limestone 2-8 175 Sandstone 2'3 144 800 Slate 2-8 175 11,000 Brick, Red \ Brick, Fire ] 2 to 2-2 25 to 137 300 Brickwork 1-8 112 Concrete 1-9 to 2'2 19 to 137 Leather ... 4,000 Hemp Rope (in ordinary state) Glass, Plate ... 1-3 27 170 10,000 2,700 Ice 917 57 Quartz Fibre (Professor Boys') 140,000 APPLIED MECHANICS. 661 XXII. (continued). Breaking Stress, in Ibs. per Stress which produces Permanent Set, Safe Limit of Stress, Young's Modulus square inch. in Ibs. per square inch. in Ibs. per square inch. of Elas- ticity, in Com- pressive. Shear. Tensile. Com- pressive. Shear. Tensile. Com- pressive. Shear. millions of Ibs. pei square inch. 17,000 12 50,000 13 10 18,000 10,000 2,300 1-5 i 1-4 1-6 6,000 650 1-4 1-65 1-35 11 1-6 12,000 2-4 7,000 10,000 1-04 48 17,000 7 6,000 6 5,000 5,000 2 15,000 13 800 2,000 500 2,000 025 26,000 8 8-5 662 APPLIED MECHANICS. LOGARITHMS. TABLE 1 1 2 3 4 5 6 7 8 9 123 456 7 8 9 10 0000 0414 0792 1139 0043 0453 0828 1173 0086 0492 0864 1206 0128 0531 0899 1239 0170 0569 0934 1271 0212 0607 0969 1303 0253 0294 0334 0719 1072 1399 0374 4 8 12 17 21 25 29 33 37 11 12 13 0645 1004 1335 0682 1038 1367 0755 1106 1430 4 8 11 3 7 10 3 6 10 15 19 23 14 17 21 13 16 19 26 30 34 24 28 31 23 26 29 14 15 16 17 18 19 1461 1761 2041 1492 1790 2068 2330 2577 2810 3032 1523 1818 2095 2355 2601 2833 3054 1553 1847 2122 1584 1875 2148 1614 1903 2175 1644 1931 2201 1673 1959 2227 1703 1987 2253 1732 2014 2279 369 368 358 12 15 18 11 14 17 11 13 16 21 24 27 20 22 25 18 21 24 2304 2553 2788 3010 2380 2625 '2856 3075 2405 2648 2878 2430 2672 2900 2455 2695 2923 2480 2718 2945 2504 2742 2967 2529 2765 2989 257 257 247 10 12 15 9 12 14 9 11 13 17 20 22 16 19 21 16 18 20 20 21 22 23 3096 3118 3139 3160 3181 3201 246 246 246 246 8 11 18 15 17 19 3222 3424 3617 3243 3444 3636 3820 3997 4166 4330 4487 4639 3263 3464 3655 3838 4014 4183 4346 4502 4654 3284 3483 3674 3856 4031 4200 4362 4518 4669 3304 3502 3692 3324 3522 3711 3345 3541 3729 3365 3560 3747 3385 3579 3766 3404 3598 3784 8 10 12 8 10 12 7 9 11 14 16 18 14 15 17 13 15 17 24 25 26 3802 3979 4150 3874 4048 4216 3892 4065 4232 4393 4548 4698 4843 3909 4082 4249 4409 4564 4713 3927 4099 4265 4425 4579 4728 4871 3945 4116 4281 4440 4594 4742 3962 4133 4298 245 235 235 7 9 11 7 9 10 7 8 10 12 14 16 12 14 15 11 13 J5 27 28 29 4314 4472 4624 4378 4533 4683 4829 4969 5105 5237 4456 4609 4757 235 235 134 689 689 679 11 13 14 11 12 14 10 12 13 30 4771 4914 5051 5185 4786 4800 4814 4955 5092 5224 4857 4997 5132 5263 4886 4900 134 679 10 11 13 31 32 33 4928 5065 5198 4942 5079 5211 4983 5119 5250 5011 5145 5276 5024 5159 5289 5038 5172 5302 134 134 134 678 578 568 10 11 12 9 11 12 9 10 12 34 35 36 5315 5441 5563 5328 5453 5575 5340 5465 5587 5353 5478 5599 5366 5490 5611 5378 5502 5623 5391 5514 5635 5403 5527 5647 5416 5539 5658 5428 5551 5670 134 124 124 568 567 567 9 10 11 9 10 11 8 10 11 37 38 39 5682 5798 5911 6021 5694 5809 5922 5705 5821 5933 6042 5717 5832 5944 6053 5729 5843 5955 5740 5855 5966 5752 5866 5977 5763 5877 5988 5775 5888 5999 5786 5899 6010 123 123 123 567 567 457 8 9 10 8 9 10 8 9 10 40 6031 6064 6075 6085 6096 6107 6117 123 456 8 9 10 4.1 42 43 44 45 46 47 48 49 6128 6232 6335 6435 6532 6628 6721 6812 6902 6990 7076" 7160 7243 6138 6243 6345 6149 6253 6355 6160 6263 6365 6170 6274 6375 6180 6284 6385 6191 6294 6395 6201 6304 6405 6212 6314 6415 6222 6325 6425 123 123 123 456 5 6 5 6 789 789 789 6444 6542 6637 6454 6551 6646 6464 6561 6656 6474 6571 6665 6484 6580 6675 6493 6590 6684 6503 6599 6693 6513 6609 6702 6794 6884 6972 6522 6618 6712 123 123 123 5 6 5 6 5 6 5 5 4 5 4 5 345 345 345 345 345 789 789 778 6730 6821 6911 6739 6830 6920 6749 6839 6928 6758 6848 6937 7024 6767 6857 6946 7033 6776 6866 6955 6785 6875 6964 6803 6893 6981 123 123 123 678 678 678 678 50 IT 52 53 54 6998 7084 7168 7251 7007 7016 7042 7050 7059 7143 7226 7308 7067 123 123 122 122 7093 7177 7259 7101 7185 7267 7110 7193 7275 7356 7118 7202 7284 7126 7210 7292 7135 7218 7300 7152 7236 7316 678 677 667 7324 7332 7340 7348 7364 7372 7380 7388 7396 122 6 6 7 XXIII. APPLIED MECHANICS. LOGARITHMS. 6G3 1 1 2 3 4 5 6 7 8 9 123 456 789 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 122 345 567 56 57 58 7482 7559 7634 7490 756(5 7642 7497 7574 7649 7505 7582 7657 7513 75S9 7664 7520 7597 7672 7528 7604 7679 7536 7612 7686 7543 7619 7694 7551 7027 7701 122 122 112 345 345 344 567 567 567 59 60 61 7709 7782 7853 7716 7789 7SCO 7723 7796 7868 7731 7803 7875 773S 7810 7882 7745 7818 7889 7752 7825 7896 7760 7832 7903 7767 7839 7910 7774 7846 7917 112 112 112 344 344 344 567 566 566 62 63 64 7924 7993 8062 7931 8000 8069 7938 8007 8075 7945 8014 8082 7952 8021 8089 7959 8028 8096 7966 8035 8102 7973 8041 8109 7980 8048 8116 7987 8055 8122 112 112 112 334 334 334 566 556 556 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 112 334 556 66 67 68 8195 8261 8325 8202 8267 8331 8209 8274 8338 8215 8280 8344 S222 8287 8351 8228 8293 8357 8235 8299 8363 8241 8306 8370 8248 8312 8376 8254 8319 8382 112 112 112 334 334 334 556 5 5 6 456 69 70 71 83S8 8451 8513 8395 8457 8519 8401 8463 8525 8407 8470 8531 8414 8476 8537 8420 8482 8543 8426 S4S8 8549 8432 8494 8555 8439 8500 8561 8445 8506 8567 112 112 112 234 234 234 456 456 455 72 73 74 8573 8633 8692 8579 8639 8698 8585 8645 8704 8591 8651 8710 8597 8657 8716 8603 8663 8722 8609 8669 8727 8615 8675 8733 8621 8681 8739 8627 8686 8745 112 112 112 234 234 234 455 455 455 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 SS02 112 233 455 76 77 78 8808 8865 8921 8814 887 1 8927 8820 8876 8932 8825 8882 8938 8831 8887 8943 8837 8893 8949 8842 8899 8954 8848 8904 8960 8354 8910 8965 8S59 8915 8971 112 112 112 233 233 233 455 445 445 79 80 81 8976 9031 9085 8982 9036 9090 8987 9042 9096 8993 9047 9101 8998 9053 9106 9004 9058 9112 9009 9063 9117 9015 9069 9122 9020 9074 9128 9025 9079 9133 112 1 1 2 112 233 233 233 445 445 445 82 83 84 9138 9191 9243 9143 9196 9248 9149 9201 9253 9154 9206 9258 9159 9212 9263 9165 9217 9269 9170 9222 9274 9175 9227 9279 9180 9232 9284 9186 9238 9289 112 112 112 233 233 233 445 445 445 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 112 233 445 86 87 88 9345 9395 9145 9350 9400 9450 9355 9405 9455 9360 9410 9460 9365 9415 9465 9370 9420 9469 9375 9425 9474 93SO 9430 9479 9385 9435 9484 9390 9440 9489 1 1 2 Oil 1 1 233 223 223 445 344 344 89 90 91 9194 9542 9590 9499 9547 9595 9504 9552 9600 9509 9557 9605 9513 9562 9609 9518 9566 9614 9523 9571 j)619 9528 9576 9624 9533 9581 9628 9538 9586 9633 1 1 Oil Oil 223 223 223 344 344 344 92 93 94 9638 9685 9731 9643 9GS9 9726 9647 9694 9741 9652 9699 9745 9657 9703 9750 9661 970S 9754 9666 9713 9759 9671 9717 9763 9675 9722 9768 9680 9727 9773 Oil 1 1 Oil 223 223 223 344 344 344 95 9777 9782 9786 9791 9795 9800 9805 9S09 9814 9818 Oil 223 344 96 97 98 9S23 98GS 9912 9827 9S72 9917 9832 9877 9921 9836 9881 9926 9S41 9SS6 9930 9S45 9S90 9934 9850 9894 9939 9854 9899 9943 9859 9903 9948 9S63 9908 9952 1 1 Oil 1 1 223 223 223 344 344 344 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 1 1 223 334 664 APPLIED MECHANICS. ANTILOGARITHMS. TABLE 1 2 3 4 5 6 7 8 9 1 2 3 456 789 oo 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 001 111 222 01 02 03 1023 1047 1072 1026 1050 1074 1028 1052 1076 1030 1054 1079 1033 1057 1081 1035 1059 1084 1038 1062 1086 1010 1064 1089 1042 1067 1091 1045 1069 1094 001 001 001 111 1 1 1 111 222 222 222 04 05 06 1096 1122 1148 1099 1125 1151 1102 1127 1153 1104 1130 1156 1107 1132 1159 1109 1135 1161 1112 1138 1164 1114 1140 11 V 67 1117 1143 1169 1119 1146 1172 1 1 Oil 1 1 1 1 2 112 112 222 222 222 07 08 09 1175 1202 1230 1178 1205 1233 1180 1208 1236 1183 1211 1239 1186 1213 1242 1189 1216 1245 1191 1219 1247 1194 1222 1250 1197 1225 1253 1199 1227 1256 1 1 Oil Oil 112 1 1 2 1 1 2 222 223 223 10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 Oil 112 223 11 12 13 1288 1318 1349 1291 1321 1352 1294 1324 1355 1297 1327 1358 1300 1330 1361 1303 1334 1365 1306 1337 1368 1309 1340 1371 1312 1343 1374 1315 1346 1377 1 1 1 I 01-1 2 2 2 2 2 2 223 223 233 14 15 16 1380 1413 1445 1384 1416 1449 1387 1419 1452 1390 1422 1455 1393 1426 1459 1396 1429 1462 1400 1432 1466 1403 1435 1469 1406 1439 1472 1409 1442 1476 1 1 Oil Oil 2 2 2 2 2 2 233 233 233 17 18 19 1479 1514 1549 1483 1517 1552 1486 1521 1556 1489 1524 1560 1493 1528 1563 1496 1531 1567 1500 1535 1570 1503 1538 1574 1507 1542 1578 1510 1545 1581 Oil Oil Oil 122 122 1 2 2 233 283 383 20 1585 1589 1592 1596 1600 1603 1607 1611 1614 1618 Oil 1 2 2 333 21 22 23 1622 1660 1698 1626 1663 1702 1629 1667 1706 1633 1671 1710 1637 1675 1714 1641 1679 1718 1644 1683 1722 1648 1687 1726 1652 1690 1730 1656 1694 1734 Oil Oil 1 1 222 222 222 333 333 334 24 25 26 1738 1778 1820 1742 1782 1824 1746 1786 1828 1750 1791 1832 1754 1795 1837 1758 1799 1841 1762 1803 1845 1766 1807 1849 1770 1811 1854 1774 1816 1858 Oil Oil Oil 222 222 223 384 334 334 27 28 29 1862 1905 1950 1866 1910 1954 1871 1914 1959 1875 1919 1963 1879 1923 1968 1884 1928 1972 1888 1932 1977 1892 1936 1982 1897 1941 1986 1901 1945 1991 Oil Oil Oil 223 223 223 334 344 344 30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 Oil 223 344 31 32 33 2042 2089 2138 2046 2094 2143 2051 2099 2148 2056 2104 2153 2061 2109 2158 2065 2113 2163 2070 2118 2168 2075 2123 2173 2080 2128 2178 2084 2133 2183 Oil Oil Oil 223 223 223 344 344 344 34 35 36 2188 2239 2291 2193 2244 2296 2198 2249 2301 2203 2254 2307 2208 2259 2312 2213 2265 2317 2218 2270 2323 2223 2275 2328 2228 2280 2333 2234 2286 2339 112 112 112 233 233 233 445 445 445 37 38 39 2344 2399 2455 2350 2404 2460 2355 2410 2466 2360 2415 2472 2366 2421 2477 2371 2427 2483 2377 2432 2489 2382 2438 2495 2388 2443 2500 2393 2449 2506 112 112 112 233 233 233 445 445 455 40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 112 234 455 41 42 43 2570 2630 2692 2576 2636 2698 2582 2642 2704 2588 2649 2710 2594 2655 2716 2600 2661 2723 2606 2667 2729 2612 2673 2735 2618 2679 2742 2624 2685 2748 112 1 1 2 112 234 234 334 455 456 456 44 45 46 2754 2818 2884 2761 2825 2891 2767 2831 2897 2773 2838 2904 2780 2844 2911 2786 2851 2917 2793 2858 2924 2799 2864 2931 2805 2871 2938 2812 2877 2944 112 [ j 2 112 334 334 334 4 5 f. 5 5 5 5 ('. 47 '48 49 2951 3020 3090 2958 3027 3097 2965 3034 3105 2972 3041 3112 2979 3048 3119 2985 3055 3126 2992 3062 3133 2999 3069 3141 3006 3076 3148 3013 3083 3155 1 1 2^ 112 112 334 344 344 5 5 f, 560 5 6 APPLIED MECHANICS. 665 XXIV. ANTILOGARITHMS. 50 51 52 53 i 3170 3243 3319 3396 2 3177 3251 3327 3404 3 4 5 8l9fl 6 7 8 3221 9 123 456 789 3162 3236 3311 3388 3184 B192 3206 3214 3228 112 344 567 3258 3334 3412 320(5 3342 3420 T2.-3 3350 3428 3281 3357 3436 3289 3365 3443 3296 3373 3451 3304 3381 3459 122 122 122 345 345 345 5 6 7 567 667 54 55 58 3467 3548 3631 3475 3556 3639 3483 3565 3648 3491 3573 3656 3499 3581 3664 3508 3589 3673 3516 3597 3681 3524 3606 3690 3532 3614 3698 3540 3622 3707 122 122 123 345 345 345 667 677 678 57 58 59 3715 3802 3890 3724 3811 3899 3733 3819 3908 3741 3828 3917 3750 3837 3926 3758 3846 3936 3767 3855 3945 3776 3864 3954 3784 3873 3963 3793 3882 3972 123 1 2 3 123 345 445 455 678 678 678 60 3981 3990 3999 4000 4018 4027 4036 4046 4055 4064 123 456 678 61 62 63 4074 4169 4266 4083 4178 4276 4093 4188 4285 4385 4487 4592 4102 4198 4295 4395 4498 4603 4111 4207 4305 4121 4217 4315 4130 4227 4325 4140 4236 .4335 4150 4246 4345 4159 4256 4355 123 123 1 2 456 456 456 456 456 456 789 789 789 64 65 66 4365 4467 4571 4375 4477 4581 4406 4508 4613 4416 4519 4624 4426 4529 4634 4436 4539 4645 4446 4550 4656 4457 4560 4667 123 123 123 789 789 7 9 10 67 68 69 4677 4786 4898 4688 4797 4909 4699 4808 4920 4710 4819 4932 4721 4881 4943 4732 4842 4955 4742 4853 4966 4753 4864 4977 4764 4875 4989 4775 4887 5000 123 1 2 3 1 9 3 457 467 567 8 9 10 8 9 10 8 9 10 70 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 124 567 8 9 11 71 72 73 5129 5248 5370 5140 5260 5383 5152 5272 5395 5164 5284 5408 5176 5297 5420 5188 5309 5433 5559 5689 5821 5200 5321 5445 5212 5333 5458 5224 5346 5470 5236 5358 5483 124 I 2 4 1 3 4 567 567 568 8 10 11 9 10 11 9 10 11 74 75 76 5495 5623 5754 5888 6026 6166 5508 5636 5768 5521 5649 5781 5916 6053 6194 5534 5662 5794 5929 6067 6209 5546 5675 5808 5572 5702 5834 5585 5715 5848 5598 5728 5861 5998 6138 6281 5610 5741 5875 6012 6152 6295 134 134 134 568 578 578 9 10 12 9 10 12 9 11 12 77 78 79 5902 6039 6180 5943 6081 6223 5957 6095 6237 5970 6109 6252 5984 6124 6266 134 1 3 4 134 578 678 679 10 11 12 10 11 13 10 11 13 80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 134 679 10 12 13 81 82 83 84 85 83 6457 6607 6761 6471 6622 6776 6486 6637 6792 6501 6653 6808 6516 6668 6823 6531 6683 6839 6546 6099 6855 6561 6714 6871 6577 6730 6887 6592 6745 6902 235 235 235 689 689 6 8-9 11 12 14 11 12 14 11 13 14 6918 7079 7244 6934 7096 7261 6950 7112 7278 6966 7129 7295 6982 7145 7311 6998 7161 7328 7015 7178 7345 7031 7194 7362 7047 7211 7379 7063 7228 7396 235 235 235 6 8 10 7 8 10 7 8 10 11 13 15 12 13 15 12 13 15 87 88 89 T413 7586 7762 7430 7603 7780 7447 7621 7798 7980 7464 7638 7816 7482 7656 7834 7499 7674 7852 8035 7516 7691 7870 8054 8241 8433 8630 7534 7709 7889 7551 7727 7907 7568 7745 7925 235 245 245 7 9 10 7 9 11 7 9 11 12 14 16 12 14 16 13 14 16 90 7943 7962 8147 8337 8531 7998 8017 8072 8091 8110 246 7 9 11 13 15 17 91 92 93 8128 8318 8511 8166 8356 8551 8185 8375 8570 8204 8395 8590 8222 8414 8610 8810 9016 9226 8260 8453 8650 8279 8472 8670 8299 8492 8690 246 246 246 8 9 11 8 10 12 8 10 12 13 ]5 17 14 15 17 14 16 18 94 95 96 8710 8913 9120 8730 8933 9141 8750 8954 9162 8770 8974 9183 8790 8995 9204 8831 9036 9247 8851 9057 9268 8872 9078 9290 8892 9099 9311 246 246 246 8 10 12 8 10 12 8 11 13 14 16 18 15 17 19 15 17 19 97 '98 99 9333 9550 9772 9354 9572 9795 9376 9594 9817 9397 9616 9840 9419 9638 9863 9441 9661 9886 9462 9683 9908 9484 9705 9931 9506 9727 9954 9528 9750 9977 247 247 2 5 7 9 11 13 9 11 13 P 11 14 15 17 20 16 18 20 16 18 20 666 APPLIED MECHANICS. TABLE XXV. Angle. Radians. Sine. Tangent, Co-tangent. Cosine, oc 1-0000 1-5708 90 1 0175 0175 0175 57-2900 9998 1-5533 89 2 0349 0349 0349 28-6363 9994 1-5359 88 3 0524 0523 0524 19-0811 9986 1-5184 87 4 0698 0698 0699 14-3006 9976 1-5010 86 5 0873 0872 0875 11-4301 9962 1-4835 85 6 1047 ' -1045 1051 9-5144 9945 1-4661 84 7 1222 1219 1228 8-1443 9925 1-4486 83 8 1396 1392 1405 7-1154 9903 1-4312 82 9 1571 1564 1584 6-3138 9877 1-4137 81 10 1745 1736 1763 5-6713 9848 1-3963 80 11 1920 1908 1944 5-1446 9816 1-3788 79 12 2094 2079 2126 4-7046 9781 1-3614 78 13 2269 2250 2309 4-3315 9744 1-3439 77 14 2443 2419 2493 4-0108 9703 1-3265 76 15 2618 2588 2679 3-7321 9659 1-3090 75 16 2793 2756 2867 3-4874 9613 1-2915 74 17 2967 2924 3057 3-2709 9563 1-2741 73 18 3142 3090 3249 3-0777 .-9511 1-2566 72 19 3316 3256 3443 2-9042 9455 1-2392 71 20 3491 3420 3640 2-7475 9397 1-2217 70 21 3665 3584 3839 2-6051 9336 1-2043 69 22 3840 3746 4040 2-4751 9272 1-1868 68 23 4014 3907 4245 2-3559 9205 1-1694 67 24 4189 4067 4452 2-2460 9135 1-1519 66 25 4363 4226 4663 2-1445 9063 1-1345 65 26 4538 4384 4877 2-0503 8988 1-1170 64 27 4712 4540 5095 1-9626 8910 1-0996 63* 28 4887 4695 5317 1-8807 8830 1-0821 62 29 5061 4848 5543. 1-8040 8746 1-0647 61 30 5236 5000 5774 1-7321 8660 1-047-2 60 31 5411 5150 6009 1-6643 8572 1-0297 59 32 5585 5299 6249 1-6003 8480 1-0123 58 33 5760 5446 6494 1-5399 8387 9948 57 34 5934 5592 6745 1-4826 8290 9774 56 35 6109 5736 7002 1-4281 8192 9599 55 36 6283 5878 7265 1-3764 8090 9425 54 37 6458 6018 7536 1-3270 7986 9250 53 38 6632 6157 7813 1-2799 7880 9076 52 39 6807 6293 8098 1-2349 7771 8901 51 40 6981 6428 8391 1-1918 7660 8727 50 41 7156 6561 8693 1-1504 7547 8552 49 42 7330 6691 9004 1-1106 7431 8378 48 43 7505 6820 9325 1-0724 7314 8203 47 44 7679 6947 9657 1-035 7193 8029 46 45 7854 7071 1-0000 1-0000 7071 7854 45 Cosine Co-tangent Tangent Sine Radians Anglb INDEX. The References are to Pages. A Absorption Dynamometers, 233 Acceleration, 13, 262 Angular, 25 Linear, 13 in Mechanism, 583 in S.H.M., 548 Accumulator, 187 Advantage, Mechanical, 59, 88 Air-vessel, 506 Algebra, 1 Alloys of Copper, 288 of Iron and Manganese, 287 of Iron and Nickel, 287 Aluminium, 285, 289 Bronze, 289 Amperes, 92 Amplitude, 20 of Vibration, 565 Analogies, Linear and Angular Motion, 593 Angle, Measurement in Degrees, Radians, 23 of Repose, 110, 112, 125 Sine, Cosine, Tangent of, 23 of Twist, 225,349 Angular Acceleration. 25 Motion, 586 Ratio, 571 Velocity, 25 Aqueduct, 208 Arch, 113, 391, 457 Least Thrust at Keystone of, 165 Line of Resistance of, 163 > Load Carried by, 163 Masonry, 166 Metal, 478 Middle Third of Section, 163 Rib of Iron, 166 Thrust at Crown of, 163 Arched Rib, 162 Area, Centre of, 136 of Curve by Integration, 17' of Curve by Planimeter, 7 of Irregular Figure, 10 Moment of Inertia of, 136 - Projection of, 27 Areas of Figures, 6 Armstrong's Hydraulic Crane, 194 Ash, 278 Attwood's Machine, 64, 245 Average Curvature, 22 Space, Time, 272, 274 Velocity, 12 The References are to Pages. Axial Flow Turbine, 531 Axis, Neutral, 381 of Suspension, 560 Axles, Railway, 83 Ayrton- Perry Coupling, 239 Ayrton and Perry's Dynamometer, 225 Babbit's Metal, 289 Bailey's Testing Machine, 294 Balance, Chemical, 118 Compensation of, 559 of Watch, 558 Spring, 40 Balancing of Hoists, 201, 203, 204, 205 of Locomotive, 609 of Machines, 606 of Reciprocating Body, 611 Ball-Bearings, 69, 86 Ballistic Pendulum, 499 Barge, 208 Barker's Mill, 486 Basin, Water in, 539 Bauschinger's Experiments, 310 Beam, 127 Bending Moment Diagrams, Six Cases, 415, 428 Cast Iron, Weight of, 9 Curvature of a, 386 Fixed at the Ends, 413, 443 Not Rectangular in Section, 417 Shear Stress in, 459 Shearing Force in, 414, 427 Stiffness of, 431 of Uniform Depth, 446 of Uniform Strength, 419 Well-known Rules about, 410 Without Compression, 471 Bear, Punching, 179 Bearing, Diameter of, 68 Friction at, 68 Bearings, Ball, 69, 86 of Fans, Centrifugal Pumps, Dy- namos, 70 Beech, 278 Bell-mouthed Pipe, 522 Belt, Centrifugal Force on, 827 Length of Crossed, 26 Length of Open, 26 Slip of, 86 Belting, Creep in, 86 Strength of, 231 668 APPLIED MECHANICS. The References are to Pages. Belts, Difference of Pulls in, 226 Bending, 380 and Crushing, 457 Moment, 128, 383 Moment in Beams (more difficult portion), 441 Moment Diagrams, Six Cases, 414, 427 of Prism 369 of Struts, 464 Theory of, 318 , Unsyrn metrical, 387 Bends in Pipes, 519i Bevel Gearing, 36, 95, 102 Bicycle Problems, 49 Resistance to Motion of, 50 Bicyclist, 83 Bifilar Suspension, 564 Blue Prints, 2 Block, Pulley, 98 Snatch, 102 Blocks and Tackle, 98 Board of Trade Rules for Railway Girders, 305, 402 Body Turning about an Axis, 114 Boiler, Bursting of, 181 Cylindrical, 320 Spherical, 320 Strength of, 319 Testing of Water-tube, 93 Bolts, 100 : Tightening of, 311 Boundary Condition, 372 Boussinesq, M., 379 Boy's, Prof., Quartz Fibres, 299 Braced Triangular Pier, 148 Bracket, 459 Brake Block, 240 Horse-power, 91, 95 Knot, 236 Brass, 288 Weight of, 9 Bricks, 276 Bridge, Forth, 299 Suspension, 161 Bronze, 288 Brotherhood's Water-pressure Engine, 191 Bulk. Diminution in, 316 Modulus of Elasticity of, 316 Bull Engine, important example, 257, 269 Bullet, 14, 79, 500 Bursting of Boiler, 181 Buttress Thread, 101 Buttresses, 166, 458 Calculus, Differential and Integral, 15 Calorific Power of Coal-Gas, 91 Cam, 37 Canal, 170, 206 Candle, 496 The References are to Pagu. Cannon, Momentum of, 486 Carbon, 280, 281, 285 Case-Hardening of Wrought Iroii^ 2S5 Castings, Chilled, 183, 284 Cooling of, 282 Internal Strains in, 282 Malleable, 284 Softening of Hard, 281 Strain in, 183 Cast Iron, 280 Density of, 281 Shell, 173 Toughening of, 284 Weight of, 9 Cast Steel, 183 Catenary, 169 Cauchy's Theorem, 363 Cedar, 278 Cement, 276 Centre of Area, 136 of Fluid Pressure, 215 of Gravity, 6, 136 of Inertia, 136 Instantaneous, 575 of Mass, 135 of Oscillation, 560 of Percussion, 499, 560 of Pyramid or Cone, 9 Centrifugal Force, 597, 606 - Force rs Belts or Ropes, 327, 605 Pump, 33, 264, 540 Centripetal Force, 603 Acceleration, 603 C.G.S. System of Units, 267 Chain Gearing, 232, 572 Hanging, 167 Horizontal Pull of, 152 Loaded, 161 Chains, Admiralty Rule for, 315 Change Wheels, 101 Charcoal Iron, 284 Chemical Balance, 118 Energy, 42 Chemistry, 290 Chilled Castings, 183 Chipping, 488 and Filing, 56 Hammer, 264 Chisel, Tempering of, 291 Chree, Dr., 371, 380 Chromium, 287 Chronometer, 649 Chuck, Elliptic, 577 Circle, Circumference of, 6 Circuital, 30 Circular Stream Lines, 541 Clinure, 29, 122 Clock, 38 Closed Polygon, 120 Coal, Energy of 1 Ib. of, 42, 43 Co-efficient of Friction, 65 of Friction, Alteration of, 75, 236 of Viscosity, 76 Collars, Leather, 170, 176 Colours in Tempering, 291, 292 INDEX. 669 The References are to Pages. Columns, 473 Combination of S.H.M.'s, 555 Combined Bending and Twisting, 353 Commercial Tests, 306 Compound Interest Law, 20, 99, 230 Pendulum, 559 Wheel and Axle, 102 Concrete, 183, 277 Conduit, 516 Cone. Area of Surface of, 8 Rolling of a, 577 Surface of, 8 Coned Pulley, 37 Cones, Stepped, 26 Conical Pendulum, 546 Confluent, 30 Connecting Rod and Crank, 222 Construction of Gun, 330 Materials Used in, 275 Contracted Vein, 533 Contraction of Section, 306 Cooling of Castings, 282 Copper, 287 Wire, 289 Cores, 10, 282 Cork, Floating, 547 Corrugated Flue, 326 Cosine of an Angle, 23 Cottars, 101 Cotton Press, 184 Couple, 127 Coupling, 225 Ayrton-Perry, 239 Dynamometer, 96 Flange, 227 Rod, 471 Crab, 102 Crane, 88 Double-Powered, 195 Hook, 458 Hydraulic, 193 Crank and Connecting-Rod, 222 Overhung, 355 Creep in Belting, 86 Creeping, 299 Cricket Bat, 500 Critical Velocity, 83 Crosshead of Engine, 113, 547 Crushing and Bending, 457 Cupola, 282 Curvature of a Beam, 386, 405 Deflnition of, 22 Radius of, 22 of Steel Spring, 388 Curve of Cosines, 555 Elastic, 390 Integral of, 18 .Logarithmic, 568 of Sines, 78, 555, 565, 568 Curved Rollers, 522 Curves, Damping, 78 Trochoidal, 573 Cutting of Metals, 45 Cylinder, Rotating, 369 Thick, 183, 323 The References'are to Pages. D Damping Curves, 78 - of Vibration, 566 D'Arcy's Experiments, 85 Dash-pots, 239 Deadloads, 305 Definition of a Radian, 23 Definitions of Trigonometrical Ratios, 23 Deflection due to Shearing, 334, 460 - in Beams, 407, 429, 431, 460 Delta Metal, 288 Descriptive Geomotry, 124 Diagonal Pieces, 162 - of Velocities, 583 Diagram of. Accelerations, 583 - of Bending Moment, six cases, 413- 427 Diagrams of Shearing Force, 427 Differential Co-efficient, 16 - Pulley Block, Efficiency of, 104 - Pulley Block, 103 Diminution of Swing, 563 '- Displacement of a Ship, 55 - Relative and Absolute 3 85 Distance of Rubbing, 68 Draining of Fields, 516 Drawing Office, 1 Drill, Watchmaker's, 291 Dunkerley, Mr., Stability of Shafts, 477 Dynamo Machine, 43, 92 Dynamometer Coupling, 96 - Froude's, Thorneycroft, 234 - Hefner- Alteneck, 235, 240 - Raflard's, 86 - Transmission, 92 - Transmission and Absorption, 233 Earth, Pressure of, 167 Rankine's Rule for,, 349 Earthquake, 497-617 Eccentric, 222 Economical Efficiency, 89 Economy, General, 339 j Eddies, 518 Effect of Friction, 60] Efficiency, 95 Economical, 89 Hydraulic, 529 of Screw Jack, 110 of Turbine, 522 Elastic Curve, 390 Elasticity, More Difficult Theory of, 361 . of Bulk, Modulus of, 317 Shearing, 333 Young's Modulus of, 299 670 APPLIED MECHANICS. The References are to Pages. Electric Energy, 92 Lamps, 43 Motor, 42 Supply Station, 92, 94 Electrical Horse-power, 92, 95 Electricity, 290 Ellipse, Area of, 8 Elliptic Trammel, 57T Chuck, 577 Elm, 278 Energy, Chemical, 42 Indestructibility of, 59 Kinetic, Potential, 40, 242, 505, 511 Loss of, due to Friction, 67 Loss of, in Change of Strain, 303 Lost", Fluid in Pipe, 53, 84 of 1 Ib. of Coal, 42-43 Pressure, 505-511 in a Rotating Body, 247, 254 Storage of, in Fluids, 326 Store of, 188 Stored up in any Machine, 253 Strain, 85 of Strained Spring, 41 Transmission of, Hydrauli rally, 170 Waste of, in Conductors, 43 of Waterfall, 59 Engine, Bull, 257 Hero's, 486 Steam, Gas, Oil, 43 Water Pressure, 189-191 Willans', 611 English, Colonel, 55 Epicyclic Train, 35 Epicycloids, 573 Equilibrant, 30 of Forces in One Plane, 122 Equilibrium, Conditions of, 121 in One Position, 104 Equi-pollent Loads, 366 Equi-potential Surfaces, 218 Errors of Observation, 62 Escapements, 554 E wing's, Prof., Exteusometer, 294 Expansion, Cubical, Go-efficieuts of, 5 Linear, Co-efficients of, 4 of Girder, 133 Experiments by Bauschinger, 310 by D'Arcy, 85 by Fairbairn, 309 by M. Tresca, 301 by Reynolds, Prof. O., 81 by Tower, M. B., 73, 81 by Wohler, 305, 309 on Attwood's Machine, 245 on Balancing, 608 on Crane, 88 on Deflection of Beams, 431 on Fluid Friction, 77, 79 on Forces at a Pin Joint, 151, 152 on Friction, 60 on Friction of Cords, Belts, 228 on Friction of a Machine, 63 on Friction and Speed, 71, 72 on Guns, 14 The References are to Pages. Experiment* on Hydraulic Jack, 175 on Inclined Plane, 104 on Jack, 63 on Loaded Links, 160 on Moments, 115 on Shearing, 332 on Strain, 293 on Triangle of Forces, 56 on Twisting of Wires, 349 to Find Co-efficient of Friction, 67 to Find Energy in Rotating Body, 247 to Find Modulus of Rigidity, 563 with Discs in Fluids, 78 with Gas-engine, 91 with Oil-engine, 91 with Pulley Block, 103 with Small Carriage, 262 Extensometer, Prof. E wing's, 294 Factor, Load, 94 of Safety, 304 Fairbairn's Experiments, 309 Falling Weight, Work Done by, 39 Fatigue, 559 Ferguson's Paradox, 36 Fir, White, 277 Firwoods, 277 Flat Plate, 330 Flexible Pipe, 197 Flexui-al Rigidity, 634 Flexure, Non-uniform, 378 Flow of Gas from Orifice, 537 of Metals, 335 of Water, Sudden Stoppage of, 192 of Water through Notch, 514, 536 Flue, Corrugated, 331 Fluid Friction, 76 Friction, Cause of, 80 Fluids in Motion, 505 Steady Motion in, 532 Storage of Energy in, 326 Whirling, 216 Force Acting on a Body, important example, 265 Centrifugal, 597, 606 Constant, 267 Due to Pressure of Fluids, 214 Lines of, 217 Moment of, 115 of a Blow, 490 of a Blow, Time Average of, 273 of Friction, 64 Polygon, 124 Shearing, 383, 385 Space Average of a, 45 Time Average of a, 45 Forces Acting at a Point, 123 Graphical Representation of, 30 in One Plane, Analytical Rule, 121 in Space, Examples on, 143, 144, 145 in Structures, Determination of, 156 INDEX. 671 The References are to Pages. Forces Not All in One Plane, 124 Parallelogram of, 30 Resultant of, 29 Triangle of, 30, 56 Forging, Hydraulic, 170 Forth Bridge, 299 Four-Bar Kinematic Chain, 577 Fracture Surface, 306 Frame of a Machine, 35 Framed Structures, 143 Francis, Mr., 536 Frequency, 20 Friction and Speed, 71 at Bearing, 68 at Different Speeds, 73 at Joints, 112 at Leather Collar, 171, 176 Between Cord and Pulley, 228 Circles, 113 Co-efficient of, 65 Cumulative Effect of, 99 Effect of, 60 Experiments on, 60 Fluid, 76 Fluid, Experiments of, 77, 78 Force of, 64 Internal, 495 Laws of Fluid and Solid, 80 Never Negligible, 248 of a Pivot, (58,71 of Pulley, 58, 248 of Railway Brake, 75 of Ships, 56 on Inclined Plane, 107, 108 Quasi-Solid, 177 Skin, 54 Statical, 75 Wheels, 69 Frictional Loss, Reduction of, 68 Frictionless Hinges, 151 Froude, 54 Froude's Dynamometer, 234, 238 Fuel Consumption, 93 Fuller, Prof., 164 Functions, Graphing of, 21 G Gulileo, 244 Gas-Engine, 43 Experiment with a, 91 Gas Flowing from Orifice, 537 Gauche Polygon, 121, 143 Gauge Pressure, 505 Gearing Chain, 232, 572 Spur, Bevel, 95, 102 General Plane Motion, 675 Geology, 275 Geometry, Practical, 1 German Silver, 289 Girder, 393 Bending Moment in, 393, 896 Board of Trade Rules for, 305 Calculation of Forces in, 151 The References are to Pages. Girder, Diagonal Braces of, 393, 396 Flanges of, 393, 396 Railway, 401 Rolled, 419, 421 Shearing Force in, 393, 396 Web of, 393, 397 Glass, 279 Sudden Cooling of, 279 Toughened, 280 Graphical Statics, 1, 149, 151 Graphing of Functions, 21 Greenhill, Prof., on Stability of Shafts, 475 Grid, Hydraulic, 212 Guide Blades, 529 Gun Experiments, 14 Lifting of, 199 Making of, 325 Metal, 288 Tensions in, 183 Wire, 325 Gyration, Radius of, 138, 580 Gyrostat, 498 Hammer, Chipping, 264 Tilt, 500 Hammering of Iron, 285 . Hanging Chain, 167 Hardening, Case, 285 of Steel, 290 Harmonic Motion, 20 Law, 20 Heat, 290 Energy, Unit of, 42 Heavy Disc, Vibration of, 568 Hemp- Packing, 177 Hero's Engine, 486 Hinge, Quasi, 166 Hinged Arch, 114 Hoists, Balanced, 178 Balancing of, 201, 203, 204, 205 Hotel, Mill, Warehouse, 200 Horse-power, 39, 89 Brake, 91 Electrical, 92 Indicated, 92, 95 Hydraulic Crane, 193 Efficiency of Turbine, 529 Forging, 170 Grid, 212 Intensifies 186 Jack, 63, 178 Jack, Efficiency of, 175 Limestones, 276 Mean Depth, 85 Power Company, 94, 187 Power, Loss of, 53 Press, 170 Ram, 501 Transmission, 52 Hydraulics, 170 Hypocyeloids, 578 672 APPLIED MECHANICS. (The References are to Pages. I I, for Various Sections, 251 Idea of Law of Dependence of two Vari- able Things, 63 of a Rate, 14 of Slope, 15 of Velocity, 12 Impact, 487, 505 Impulse, 264 Inclined Plane, 99, 104 Friction on, 107 Indiarubber, 305 Cup, 177 Shaft, 318 Spring, 41 Indicated Horse-power, 95 Indicator Diagram, 40, 274 Centre of, 149 Inertia, 40 . Moment of, 135 Principal Moments or, 137 Rule for, 7 Instantaneous Centre, 575 Integral of a Curve, 18 Integrals, Table of, 16 Intensifier, 186, 188 Involute of Circle, 573 Inward Radial Flow Turbine, 531 Iron, Alloys of, 287 Cast, 280 Charcoal, 284 Galvanised, 287 Pig, 284 Rolling and Hammering, 285 Wrought, 284 Irregular Figure, Approximate Area of, 10 Jack, Hydraulic or^Screw, 63, 178 Jet Pump, 516 Joint, Masonry, 458 Joints, Riveted, 336 - with Friction, 112 Joule, -42 Journal, Friction at, 68 Journals and Footsteps, 78 Rater's Pendulum, 560 Kelvin, Lord, 802 Kelvin's Analogy, 358 - Tide-Predicting Machine, 555 Kerosene, Energy of, 42 Keys, 101 Keystone of Arch, 165 Kinematics, 1 Kinetic Energy, 242, 505-511 - of Any System, 588 Kinetics of Mechanism, 220 The References are to Page*. Kirchhoff, 366 Knife Edges, 118 Knot Brake, 236 Kohlrausch, 303 Laboratory, Mechanical, 1 Ladder, Example on, 134 Lake of Water, 512 Landing Stage, 206 Larch, 278 Lathe, Screw-Cutting, 101 Law Connecting Variable Things, 63 of Crane, 89 of Friction for a Machine, 63 of Moments, 115 of Worth, 58 Laws of Fluid and Solid Friction, 80 Leather Collars, 170, 176 Friction at, 171 Level Surface, 22 Lever, 116 Limestones, 275 Natural Hydraulic, 276 Line of Resistance, 113, 126, 161, 163 Projection of, 2* Vertical, 22 Lines of Force, 217 Link Motions, 584 Link Polygon, 129 Links, Loaded, 160 Live Load, 305 Load Factor, 94 Rolling, 402 Loaded Chain, 161 Links, 160 Loads, Equi-pollent, 366 Suddenly Applied, 306 Loam, 282 Local Strengthening, 306 Lock Nuts, 311 Locomotive, Pull of, 76 Logarithmic Curve, 568 Logarithms, 1 Loss of Energy, by Impact, 488 of Energy due to Friction, 67 of Energy due to Fluid in Pipe, 84. of Energy in Change of Strain, 303 of Energy per Ib. of Water, 518 of Power at Different Parts of Engine, 90 Love's Treatise on Elasticity, 3G1 Lowell Formula for Rectangular Notch ; 515, 536 Lubricant, 66 Body of, 75 M M, for Various Sections, 251 of a Wheel, 247, 248, 254 Machine, Bailey's Testing, 294 INDEX. 673 The Rejerences are to Pages. Ma ihine, Cotton Baling, 186 Design, 70 Kinetic Energy Stored up in, 253 Riveting, 170 Machinery, Quick Speed, 530 Water Pressure, 83 Machines, Balancing of, 606 Forging, Welding, Panelling, Stamp- ing, Shearing, 196, 198 Shearing, 335 Steadiness of, 252 Magnet, 564 Mahogany, 278 Mainspring of Timekeeper, 554 Malleable Castings, 284 Manganese, 284 Bronze, 289 Marbles, 275 Masonry Joint, 459 Mass, 40 Centre of, 135 Materials, Behaviour of, 290 Used in Construction, 275 Weights of, 9 Mathematical Tables, 24 Mean Depth, Hydraulic, 85 Mechanical Advantage, 59, 88 Hypothetical, 102 of Hydraulic Jack, 174 Mechanism, 220, 570 \ Four-Link, 35 Kinetics of, 220 Quick Return, 578 Mechanisms, Acceleration of, 583 Velocity of, 583 Memel, 277 Mensuration, 1, 6 Metal, 288 Arches, 478 Babbit's, 289 Delta, 289 Flow of, 298, 301, 335 Gun, 289 Muntz, 289 Sterro, 289 White, 289 Mills, Driving of, 515 Milne, Prof., 401 Mitis Castings, 287 Modulus of Elasticity of Bulk, 317 of Rigidity, 333, 563 of Section, 398 Young's, 299, 305 Moment, Bending, 128, 384 Law of, 115 of a Force, 115 of Inertia, 135, 136 of Inertia of a Circle, 141 of Inertia of Area, 149 of Inertia of Rectangle, 141 of Inertia of Sections, 398 of Momentum, 500, 528, 587 Work done by, 115 Momental Ellipse, 142 Momentum, 263 W The References are to Pages. Momentum, Important Example, 274 of Cannon, 456 of Shot, 486 Tangential, 524 Mortar, 276 Motion, Fluids in, 505 of Rotation, 127, 592 of Translation, 121, 592 Periodic, 546 Produced by Blow, 498 Motions, Link Value, 584 Moulder, 9 Moulding, 282 Muntz Metal, 88 N Neutral Axis, 381 Surface, 381 Newton's Second Law, 602, 604 Nickel, 287, 289 Non-Redundant Frame, Criterion for, 148 Norway Spruce, 277 Notch, 514 Gauge, 515, 536 Rectangular, 536 Triangular, 514, 536 Nozzles, 513 Nuts, 311 Oak, English, 278 Oil Engines, Experiments wltn, fil Engines, 43 Sperm, 66 Tester, Thurston, 81 Orifice, 512 Gas Flowing from, 537 Rectangular, 535 Sharp-Edged, 513, 534 Triangular, 535 Water from, 512 Oscillating Cylinder Engine, 578 Outward Radial flow Turbine, 531 Overhauling, 96, 104, 109 Overhung Crank, 355 Packing Hay, 185 Parabola, 162, 168 Parabolic Rib, 162 Paraboloids of Revolution, 218 Paradox, Ferguson's, 36 Parallel Motion, Watt's, C9 Parallelogram of Forces, 30 Patterns, 10, 282 Peaucellier Cell, 578 Pendulum, Ballistic, 499 500 Compound, 559 Conical, 546 - .V.-i 674 APPLIED MECHANICS. The References are to Pages. Pendulum, Equivalent Simple, 559 Impulse given to, 554 Simple, 242 Simple, Time of Swing, 551 Percussion, Centre of, 499 Periodic Motion, 546, 592 NotS.H.M., 554 Time, 20, 546 ~ Time of Balance, 558 Permanent Axes, 607 . Set in Wire, 300 Perry's, Mr. James, Syphon, 517 Phosphor Bronze, 288 Phosphorus, 284, 287 Physics, 290 Pianoforte Wire, 289 Piezometer, 173 Pig-iron, Puddling and Refining of, 284 Pile Driver, 273, 489 Pin, Resultant Force at, 112 Pine, Red, 277 Pins, 101 Pipe, Bell-Mouthed, 522 Flexible, 197 Resistance to Motion of Fluid in a, 76 Strength of, 182, 319 Suddenlv Enlarged, 5 Wooden," 183 Pipes, Bends in, 519 Flow of Water in, 53 Piston-Rod, 309 Pitch of Screw, 100 Plane Motion, 576 Planimeter, 2, 7 Plastic Elongations, 301 Plasticity, 301 Plate, Flat, 330 Platform, Weight of, 402 Plotting on Squared Paper, 92, 229 Poisson's Ratio, 343 Polygon, Closed, 120 Gauche, 121 Link, 129 of Forces, 12-4 Unclosed, 124 Portland Cement, 277 Potential Energy, 40, 242, 505, 511 Poundage of Steam, 95 Power, Loss of, at Different Parts of Engine, 90 Misuse of this Expression, 89 of a Stream, 515 Transmission by Shafts, 224 Precession of Top, 596 Press, Cotton, 184 for Packing Hay, 185 for Warehouses, 185 Hand, 185 Hydraulic, 170 Pressure Energy, 505, 511 Gauge, 505 < of a Fluid, 215 of Earth, 167, 848 02, The References are to Pages. Pressure of Water, 167 on Immersed Surface, 215 Principal Moment of Inertia, 139 Stress, 354 Stresses, 364 Principle of Work, 120 Prints, 10, 282 Prism, Centre of Gravity of, 9 of Elliptic Section, 376 Twisting or Bending of, 369 Volume of, 9 Prismatic Body, Volume of, 9 Projectile, 245 Projection of Area, 27 of Line, 27 Propeller Shaft, 82, 100 Propulsion of Ships, 52, 525 Protractor, 23 Puddling of Pig-iron, 284 Pull, How it is Exerted, 292 Pulley Block, 98 Block, Efficiency of, 98 Coned, 37 Friction of, 58 Rim of, 283 Pump, 505 Centrifugal, 33, 264, 503 Double-Acting Force, 50d Efficiency of, 43 Force, 506 Jet, 516 Lifting, 505 Mechanism, 578 Punching, 335 Bear, 179 Q Quantity, Scalar, 29 Vector, 29 Quartz Fibres, 299 Quasi-Hinge, 166 Rigidity, 497 Solid Friction, 177 Quicklime, 276 Quick Return Mechanism, 578 E Rack, 573 Radial Gears, 584 Radian, Definition of, 23 Radius of Curvature, 22, 388 of Gyration, 136, 251, 560 Rafford's Dynamometer, 80 Railway Axles, S3 Brake, Friction of, 75 Girder, 133, 402 Uam, Hydraulic, 503 Rankine, 319 Rankine's Rule for Earth, 349 Rate, Algebraical Representation zt, IS Ratio, Velocity, 59 INDEX 675 The References are to Pages. Rectangular Notch, 536 - Orifice, 534 Red Pine, 277 Redundant Bars, 395 Refining of Pig-iron, 284 Regulation of Turbine, 529 Relative Velocity, 32, 34, 35 - Viscosities, 569 Resilience, 41, 308 - Compressive, 316 - Shear, 316 - Tensile, 316 Resistance, Line of, 113, 114, 161, 430 - of a Moving Train, 44 - to Motion of Fluid in a Pipe, 76 - to Rolling, 85 - "Wave, 54 Resolved Part of a Vector, 29 Eesultant Force, 29, 129 - on Forces in One Plane, 122 Reynolds, Prof. O., 73, 78, 81-4, 541 Rib Arched, 162 - Parabolic, 162 Rigidity, Flexural, 634 - Modulus of, 333, 563 - Quasi, 497 - Torsional, 634 Rim of Wheel, Volume of, 9 River Weaver, 206 Riveted Joints, 336 - Work, 297 Riveting Machine, 170 Rocks, Slaty, Stratified, 275 Rolled Girder Section, 419, 421 Rollers, Curved, 572 - for Girder, 133 Rolling, 106 - and Hammering of Iron, 285 - of a Cone, 577 - Load, 403 - Resistance to, 85 - True, 571 Roof, Calculation of Forces in, 151, 153 - Principal, 133 Rope, Centrifugal Force in, 322 - Weight of, 315 Ropes, Wire, 232 Rotating Body, Energy in, 247 - Cylinder, 368 Rotation, Motion of, 127 Rubbing, Distance of, 68 Rules for Beams, 410 - for Deflection of Beams, 431 - for Flow of Water through Notch. 515, 536 - . for Periodic Time in S.H.M. , 549 ' -- for Simple Pendulum, 551 - for Struts, 466 - for Strength of Pipe, 182 Rupert's Drop, 183, 280 Safety, Factor of, The References are to Pages. Safety-valve, 116 Sandstones, 275 Green, Dry, 282 Scalar Quantity, 29 Scraped Surfaces, 73 Screw, The, 99 Cutting Lathe, 101 Jack, 63 Jack, Efficiency of, 110 Piles, 101 Pitch of, 100 Propeller, 101 Propeller Blades, 289 Square-Threaded, 109 Thread, Whitworth, Sellers, Buttress, 101 Seasoning of Timber, 101 Section, Change of Shape of, 374 Sections of Structures, 159 Seizing, 73 Sellers' Thread, 101 Sense of Vector, 29 Shaft, Hollow Round, 354 Indiarubber, 318 Overtwisted, 290 Section, an Equilateral Triangle, 377 Shafts of Various Sections, 357 Stability of, 475 Strength of, 351, 352 Subjected to Twisting and Bending, 353 Whirling of Loaded, 476 Shape of Stream from Orifice, 584 of Teeth, 572 of Worm Thread, 574 Sharp-edged Notch, 536 Orifices, 534 Shear and Twist, 332 Strain, 332, 341, 350 Stress, 332 Stress in Beams, 460 Shearing Force, 128, 383, 385, 631 Force in Beams, 414, 427 Machines, 335 Ship, 496 Displacement of a, 55 Heeling Angle of, 260 Righting Moment of, 200 Ships, Friction of, 56 Horse-power of, 54 Models of, 55 Propulsion of, 52, 525 Resistance to Motion of, 54 Speed of, 53 Shot, Momentum of, 486 Silicon, 284 Bronze, 289 Silver, Alloys of, 289 Similar Figures, Area of, 8 Structures, Similarly Loaded, 424 Simple Harmpnic Alotion, 20, 222, 546 Harmonic Motion, Amplitude of, 20 Harmonic Motion, Frequency of, 20 Harmonic Motion, Periodic Time of, 20 676 APPLIED MECHANICS. The References are to Pages. Simple Harmonic Motions, Combination of, 555 Simpson's Rule, 7 Sine of an Angle, 23 Sines, Curve of, 78 Skeleton Drawings, 221, 570 Skew Bevel Wheels, 572 Slider Crank Chain, 577 Sliding, 106 Contact, 571 Slip of Belt, 86, 232 Slope of a Curve, 15 Smith, Prof. R. H., 45, 5S2 Snatch Block, 102 Soapy Water, Use of, 237 Space Average of a Force, 45 Average, Time Average, 272, 274 Rate, Time Rate, 269 Specific Gravity, 10 Speed of Bullet, 14 of Commercial Shies. 53 of Train, 12 Sperm Oil, 66 Sphere, Surface of, 8 Volume of, 8 Spherical Shell, Thick, 325 Spinning, 106 Tops, 498 Spiral Flow of Water, 539, 540 Spring, Vibration of, 550 Spring Balance, 40 of Indiarub ber, 41 of Steel, 41 Spiral, Vibration of, 550 Springs, 613 - B uffer Stop, 613 - C, 643 - Carriage, 618, 646 - Clock, 613 - Cylindrical Spiral, 624, 638 - Different Forms of, 618 - - Elongation of, 630 - Flat Spiral, 624, 627, 628, 629 - Hardening and Tempering of, 622, 649 - Materials Used in, 620 - Phosphor Bronze, 622 - - Resilience of, 620, 641 . - Spiral, 622 . -- Tubular Spiral, 642, 643 . - Uses of, 620 - Vibration, 615 - which Bend, 462 Spruce, Norway, 277 Spur Gearing, 95, 102 - Wheel, 36 Squared Paper, 1, 13, 15, 21, 44, 60, 89, 90, 229, 293, 300, 565 Square-threaded Screw, Efficiency of, 109 Statical Friction, 75 Steadiness of Machines, 252 Steady Motion in Fluids, 532 Steam Engine, 43 - Turbine, 531 Steel, 285 - Basic, 286 The References are to Pages Steel, Bessemer, 285 Cast, 183 Castings, Annealing of, 2a6 Crucible, 286 Hardening of, 290 Ingots, 285 Martin, 289 Rails, 286 Shear, 285 Siemens', 286 Spring, 41 Strength of, 2Sft Tempering of, 28 Weight of, 9 Step, 68 Stepped Cones, 26 Stiffness of Beams, 432 Stilling of Vibrations, 565 Stone, 275 Artificial, 276 Stoppage (Sudden) of Water in a Pipe, 192, 501 Storage of Energy in Fluids, 321 Stores of Energy, 42, 509 Straight Line Law, 62 Strain, 293, 296, 312 Energy, 85, 307, 489 in Castings, 183 Nature of, 311 Potential, 362 Shear, 331, 341 Strained Spring, Energy of, 41 Stream Lines, Circular, 541 Strength of Boiler, 319 of Chains, 315 Modulus of Sections, 8:> of Pipe, 182, 319 of Ropes, 315 Stress, 296 Principal, 354 Shear, 331 Structures, Determination of Forces in, 156 Method of Sections, 159 Struts, 151, 463 Bending of, 464 Euler's Formula, 465 Rules for, 467 Shortening of, 298 with Lateral Loads, 470 St. Venant, 296, 318, 348, 366, 370, 379 Suddenly Applied Load, 306 Sulphur, 284 Sun and Planet Motion, 86 Surface, Level, 22 of Fracture, 306 Scraped, 73 Surfaces, Equi-potential, 218 Suspension, Axis of, 560, 561 Bifilar, 564 Bridge, 161 Switchback Railway, 41 Syphon, 79 Lubricator, 83 Mr. James Perry's, 517 INDEX. 677 The References are to Pages. T Table of Constants, 654 I. of Sections, 397 of Strength Moduli, 397 Tables, Mathematical, 24 Tangent of an Angle, 23 Tangential and Normal Components, 106 Teak, 278 Teeth of Spur-wheel, 572 of Wheels, 423 of Worm-wheel, 572 Telegraph Wire, Dip of, 168 Tempering Colours, 291, 292 of Steel, 290 Template, 218 Testing Machine, 1 Machine, Bailey's, 294 of Water-tube Boiler, 93 Theorem of Three Moments, 454 Theory of Bending, 318 of Fluid Friction, 80 Thick Cylinder, 183, 321, 328, 367 Spherical Shell, 325 Thin Cylinder, 324 Thomson's Jet Pump, 516 Prof. J., Dynamometer, 235 Prof. J., Overtwisted Shaft, 290 Triangular Notch, 514, 536 Turbine, 526 Whirlpool Chamber, 522 Thorneycroft's Dynamometer, 234 Thread, Whitworth, 100 Threads, Shapes of, 101 Thurston Oil Tester, 81 Tide-predicting Machine, Kelvin, 555 Tide, Rise and Fall of, 554 Tie Rod, 151 Tilt Hammer, 500 Timber, Felling of, 278 Preserving of, 279 Seasoning of, 277 Strength of, 279 Time Average of a Force, 45 Time Average, Space Average, 272, 274 Time, Periodic, 546 Time Rate, Space Rate. 269 Tin, 289 Top, Precession of, 596 Torque, 77, 95, 127, 260, 351 Torsional Rigidity, 634 of a Shaft, 352 Tortuosity of Path, 606 Total Energy of 1 Ib. of Water, 515, 533 Toughened Cast Iron, 284 Glass, 280 Tower, Mr. Beauchamp, Experiments of, 73,81 Train, Moving, 44 Epicyclic, 35 Speed of, 12 Tramcar, Pull on, 44 Work Done on, 38 Trammels, Elliptic, 577 Translation, Motion of, 121 TJie References are to Pages. Transmission Dynamometers, 92, 233 of Power by Chains, 233 of Power by Shafts, 224 Travelling Loads, 436 Tresca, M., Experiments by, 301 Triangle of Forces, 56 Triangular Orifice, 534 Trochoidal Curves, 573 Tube, Built up, 323 Strengthening of, 324 Tungsten, 287 Turbines, 515, 526 Arranging of, 530 Axial Flow, 531 Efficiency of, 43 Guide Blades, 529 Hydraulic Efficiency of, 529 Inward Radial Flow, 531 Outward Radial Flow, 531 Regulation of, 529 -^ Steam, 531 Tweddell's Hydraulic Tools, 199 Twist, Angle of, 349-352 and Shear, 332 Twisting, 349 and Bending in Shafts, 353 in Wire, 349 Moment, 350, 352, 353 of Prism, 369 Unclosed Polygon, 124, 129 Unit of Heat Energy, 142 Useful Work, 39 U-Tube, Motion in, 553 Valve Motions, 584 Vanes, Radial, 521 of Centrifugal Pump, 508 Vector, 29, 123 Clinure of, 29 Resolved Part of, 29 Sense of, 29 Vectors, Difference of, 30 Sum of, 30 Vehicle, Pulling Force on, 10S Velocity, 12, 14 Angular, 25 at Any Instant, 12 Average, 12 Critical, 83 Ratio, 59, 103, 220 Relative, 32, 34, 35 of Rubbing, 74 Venant, St., 296, 318, 348, 366, 370, 379 Vertical Line, 22 Vibrating Adjustable Masses, 561 Weight, 41 Vibration, Amplitude of, 565 Damping of, 566 678 APPLIED MECHANICS. The References are to Pages. Vibration Indicator, 590 of Heavy Disc, 568 of Spiral Spring, 550 of Strip of Steel, 552 Stilling of, 565 Viscosities, Relative, 569 Viscosity, 41 Co-effieient of, 76 Volts, 92 Volumes of Solids, 88 Voussoirs, 163, 165 w Waste of Hydraulic Power, 53 Watchmaker's Drill, 291 Water Flowing Spirally, 539 Lake of, 512 Pressure Engine, 189, 191 Pressure Machinery, 83 Pressure of, 167 Waterfall, Energy of, 59 " Waterwitch" Steamship, 486 Watt's Parallel Motion, 69 Sun and Planet Motion, 36 Wave Propagation, 367 Weighbridge, 117 Graduation of Lever of. 118 Weights of Materials, 9 Wertheim's Experiments, 302 Wheels and Axle, 101 and Axle, Compound, 102 Friction, 69 Teeth of, 423 Teeth, Shapes of, 221 Value of Train of, 102 Worm, 102 Whirling Fluids, 216 Loaded Shaft, 476 Whirlpools, 518 Chamber, Thomson's, 522 White Fir, 277 Metal, 75, 289 Whitworth Thread. 100 The References are to Pages. Willans' Engine, 611 Wind Pressure, 156 Windmills, 101 Wire, Area of Section, 10 Copper, 289 Drawing Through Die, 298 Experiment on, 293, 300 Gun, 325 Pianoforte, 289 Ropes, 232 Telegraph, 168 Telephone, 289 Wohler's Experiments, 305, 309 Wooden Pipe, 183 Work, 38 by Expanding Fluid, 19 Cutting of Metals, 46 Done by a Moment, 115 Done by Turbine, 521 Law of, 58 Lost in Friction, 67 Principle of, 120 Useful, 39 Workshop, 1 Worm and Worm Wheel, 102, 22 i Thread, 574 Wrought Iron, 284 Case-hardening of, 285 Charcoal, 284 Forging of, 284 Lowmoor, 284 Red-short, 284 Staffordshire, 284 Weight of, 9 Yield Point, 300 Young's Modulus, 299, 805 Zinc, 287, 238 >*^ "*" m. /v^^vv ,/ OF THE ' X 1 UNIVERSITY 1) OF PAINTED BY CABSELL & COMPANY. I.IMITKD, LA BELLE SAUVAQE, LONDON, E.O. 20.1034 THIS BOOK u DUE ON THE STAMPED BELOW AN NOW; 5 o or LD 21-100W-7/33 160010 THE UNIVERSITY OF CALIFORNIA LIBRARY