LIBRARY 
 
 OF THE 
 
 UNIVERSITY OF CALIFORNIA. 
 
 Class 
 
A TEXT-BOOK 
 
 OP 
 
 ENGINEERING 
 
A TEXT-BOOK 
 
 OF 
 
 ENGINEERING 
 
 FOR 
 
 SECONDARY 
 TECHNICAL SCHOOLS 
 
 BY 
 
 WILLIAM R. KING, U. S. N., Retired 
 
 Principal and Head of Department of Engineering 
 Baltimore Polytechnic Institute 
 
 PUBLISHED BY 
 THE FRIEDENWALD COMPANY 
 
 BALTIMORE, MD. 
 1906 
 

 COIERAL 
 
 COPYRIGHT, 1906 
 
 BY 
 WILLIAM R. KING 
 
 jfrufcennrnffc 
 
 BALTIMORE, MD., U. S. A. 
 
PREFACE 
 
 The preparation of this text-book is the result of four years' 
 experience in the class-room in giving students of a secondary 
 technical school of the highest grade the maximum of engineering 
 instruction admissible, considering the time at disposal and the 
 maturity of the students. 
 
 It is believed that the simplicity employed in the presentation of 
 the elements of the different subjects gives an excellent ground- 
 work upon which to build, should the graduate fail to pursue to 
 completion his technical education at a university. In the event 
 of his entry into a university, it is established by precedent that he 
 should be able to obtain a degree in two years. 
 
 It is intended that the first six chapters of Part L, supplemented 
 by some elementary text on the constructive details of steam 
 machinery such as that by Spangler, Greene, and Marshall shall 
 constitute the course for the third year students; the remainder of 
 the book is to be covered during the fourth year. 
 
 Correlative with the subject-matter of the book, it will be found 
 that there is necessity for a course in mathematics, including the 
 Calculus, and one in Demonstrative Mechanics, together with facili- 
 ties for experimental work in a fairly equipped mechanical 
 laboratory. 
 
 The article in the Appendix on the distribution of steam in a 
 cylinder and the formation of the indicator diagram was contributed 
 by Mr. W. L. DeBaufre of the class of 1903, Baltimore Polytechnic 
 Institute, now a senior at Lehigh University. 
 
 Due acknowledgment is made to Mr. S. P. Platt, Instructor in 
 charge of Mechanical Drawing at the Baltimore Polytechnic Insti- 
 tute, for assistance in making the drawings. 
 
iv PREFACE 
 
 Much of the matter on Iron and Steel in Chapter I., Part III., 
 has been taken from Durand's Practical Marine Engineering, by 
 permission of the author. 
 
 Works on engineering subjects have been consulted, including 
 those of Sennett, Seaton, Holmes, Merriman, Goodman, and 
 Jamieson. 
 
 WILLIAM E. KING, 
 'Passed Assistant Engineer, U. S. N. f Retired. 
 
 BALTIMORE POLYTECHNIC INSTITUTE, 
 January, 1906. 
 
CONTENTS 
 
 PART I 
 
 I. INTRODUCTORY 
 
 II. THE APPLICATION OF HEAT TO WATER n 
 
 III. FUELS AND COMBUSTION 
 
 ERRATA 
 
 A few errors escaped notice while passing through the press. On 
 page 3, last line of Art. 5, read on the Fahrenheit scale, the tem- 
 perature of the original volume heing that of melting ice. On 
 page 141, second line of last paragraph, read AB instead of AD. 
 On page 146, last line, read stroke of the piston instead of travel 
 of the valve. On page 170, fifth line from the bottom, read that 
 instead of than. On page 189, sixth line from the top, read steam 
 instead of stream. On page 288, third line from bottom read W 3 
 instead of W. On page 290, in the value of M w ?, read W s (dj + d 2 ) 
 instead of W^ X d 2 . On page 322, fourth line from bottom, read 
 K 2 instead of I. On page 325, fifth line from bottom, read 7,OOOD :5 
 instead of 7,OOOD. In Example V, page 326, the factor 10,000 
 has been omitted under the radical. On page 332, in the value of 
 Ely, read Lx 3 instead of wx 3 . On page 334, in the value of M, 
 read w instead of W. On page 338, third line from bottom, read 
 SL 2 instead of SL 3 . In Art. 36, page 366, read 2n 2 instead of 
 n 2. On page 371, seventh line from bottom, read a instead of a. 
 
 VII. RESILIENCE ... ** 
 
 337 
 
 PART V 
 
 I. Bow's SYSTEM OF LETTERING. FORCE DIAGRAM. FUNICULAR 
 .POLYGON 
 
 II. FRAMED STRUCTURES. RECIPROCAL DIGRAM " 
 APPENDIX 
 
 INDEX 38 ^ 
 
 395 
 
iv PREFACE 
 
 Much of the matter on Iron and Steel in Chapter I., Part III., 
 has been taken from Durand's Practical Marine Engineering, by 
 permission of the author. 
 
 Works on engineering subjects have been consulted, including 
 those of Sennett, Seaton, Holmes, Merriman, Goodman, and 
 Jamieson. 
 
CONTENTS 
 
 CHAP. PART I PAGE. 
 
 I. INTRODUCTORY 3 
 
 II. THE APPLICATION OF HEAT TO WATER 11 
 
 III. FUELS AND COMBUSTION 22 
 
 IV. EFFICIENCY 33 
 
 V. THE VALVE AND ITS MOTION 45 
 
 VI. THE CONVERSION OF MOTION. ACTION OF THE CRANK AND 
 
 CONNECTING-ROD 64 
 
 VII. STAGE EXPANSION ENGINES 70 
 
 VIII. THE INDICATOR AND ITS DIAGRAM 75 
 
 IX. MEAN PRESSURE OF A GAS EXPANDING WITHIN A CYLINDER. ... 96 
 
 X. BOILER AND ENGINE EFFICIENCY 103 
 
 XL ENGINE DESIGN 121 
 
 XII. THE ZEUNER VALVE DIAGRAM 137 
 
 XIII. ECONOMY OF THE STEAM ENGINE AND BOILER 150 
 
 XIV. STEAM BOILERS 179 
 
 XV. THE STEAM TURBINE 190 
 
 PART II 
 
 I. BELTING 205 
 
 II. WHEELS IN TRAIN 217 
 
 PART III 
 
 I. MATERIALS 231 
 
 II. TESTING MATERIALS 255 
 
 PART IV 
 
 I. MOMENTS. CENTER OF GRAVITY 267 
 
 II. BENDING MOMENT. SHEAR. BENDING-MOMENT DIAGRAM. SHEAR 
 
 DIAGRAM 282 
 
 III. MOMENT OF INERTIA. RADIUS OF GYRATION 304 
 
 IV. THE THEORY OF BEAMS 311 
 
 V. COLUMNS. SHAFTS 320 
 
 VI. THE DEFLECTION OF BEAMS 329 
 
 VII. RESILIENCE 337 
 
 PART V 
 
 I. Bow's SYSTEM OF LETTERING. FORCE DIAGRAM. FUNICULAR 
 
 POLYGON 345 
 
 II. FRAMED STRUCTURES. RECIPROCAL DIAGRAM 362 
 
 APPENDIX 383 
 
 INDEX . 395 
 
PART I 
 
 THE ELEMENTS OF STEAM 
 ENGINEERING 
 
JNIVERSITY 
 
 CHAPTEE I. 
 INTRODUCTORY. 
 
 1. A gas may be defined as a fluid of such nature that if a certain 
 volume of it be admitted into a vessel, the volume admitted will 
 distribute itself throughout the vessel, whatever the volume of the 
 vessel may be. 
 
 2. The Kinetic Theory of Gases is that the molecules of a gas 
 are in a state of rapid motion in straight lines, and that as a result 
 of their collision with each other and with the walls of the contain- 
 ing vessel, pressure is exerted. 
 
 3. The two laws of gaseous expansion which are of wide applica- 
 tion are those of Boyle and of Charles. 
 
 4. Boyle's Law. The law of Boyle may be stated as follows : 
 The pressure of a mass of gas at constant temperature varies 
 
 inversely as its volume. 
 
 This law may also be expressed as follows : 
 
 The product of the pressure and volume of a mass of gas is a 
 constant quantity as long as the temperature is constant. 
 The algebraic expression for Boyle's law is 
 
 PV = C 
 
 in which P is the pressure in any units, usually in pounds per sq. 
 foot, V the volume in cubic feet, and C a constant quantity to be 
 determined experimentally for a particular gas. 
 
 5. Charles's Law, known also as the law of Gay Lussac. This 
 law asserts that all gases have the same coefficient of expansion, and 
 this coefficient is the same whatever the pressure supported by the 
 gas. Hence for each degree of rise or fall of temperature of a gas 
 its volume will be increased or diminished by a fixed fraction of 
 its original volume. This fraction has been computed to be 
 
 2^ 0.0036625 on the Centigrade scale, and Tcrr-g = 0.0020347 
 on the Fahrenheit scale. 
 
 6. As in the case of the air thermometer, if a column of air be 
 inclosed in a tube of uniform bore, and separated from the atmos- 
 
4 THE. ELEMENTS OF STEAM ENGINEERING 
 
 phere by an air-tight piston which is free to move in the tube ; and 
 if when the temperature of the confined air is that of melting ice 
 the height of the column is unity, then, if the tube be exposed to 
 the steam of boiling water, the pressure remaining constant, the 
 temperature of the inclosed air will rise to 100 C., or to 212 F. 
 The volume of the air will then be, by Charles's law, 
 
 1 + (100 X .0036625) = 1.36625 
 or 1 + (212 32) .0020347 = 1.36625 . 
 
 This unit volume of air having expanded the fraction 0.36625 of 
 itself by raising its temperature 100 C., or 180 F., it follows that 
 its volume will be doubled when the temperature is raised through 
 x , thus: 
 
 0.36625 : 1 : : 100 : x, whence x = 273 C. 
 or 0.36625 : 1 : : 180 : x, whence x = 491.6 F. 
 
 It follows, by the law of Charles, that if the unit volume had been 
 cooled through 273 C., or through 491.6 F., there would be no 
 existing volume. 
 
 This point is called the absolute zero of temperature, and tem- 
 peratures reckoned from this point are called absolute tempera- 
 tures. 
 
 The absolute zero is then 273 below the zero of the Centigrade 
 scale, and 491.6 32 = 459.6 below the zero of the Fahrenheit 
 scale. To convert temperatures measured on the C. or F. scales to 
 absolute temperatures, we have only to add 273, or 459.6, as the 
 case may be. Absolute temperatures will be denoted by the capital 
 letter T. 
 
 It must be observed that if the volume of a gas be kept constant 
 and its temperature raised, the pressure will be increased or dimin- 
 ished in the same ratio that the volume was increased or diminished 
 when the pressure was constant. For let P and V be the pressure 
 and volume of a portion of gas at 32 F., and 1.366257 the volume 
 when heated to 212 F. under the constant pressure P. If the 
 temperature of the gas be now kept constant at 212 while it is 
 being compressed back to its original volume 7, it will then have 
 a certain pressure P', and we will have, by Boyle's law, 
 
 P'V = P X 1.366257, whence P' 1.36625P. 
 
 In other words, if a volume 7 of a gas at 32 and pressure P be 
 heated to 212, the volume remaining at 7, its pressure will be 
 1.36625P. 
 
INTRODUCTORY 5 
 
 We may now regard this absolute zero from another viewpoint. 
 Let us consider a mass of perfect gas of volume v , pressure p , and 
 a temperature C., or 32 F. Imagine the volume kept constant 
 while its temperature is raised or lowered t, the resulting pressure 
 p will be, by Charles's law, p = p p at = p (l at}, where a 
 is the coefficient of expansion. If the gas now be cooled until its 
 
 1 
 temperature is reduced to - - , we will then have : 
 
 1 
 That is, at a temperature - - the gas would exert no pressure 
 
 on the walls of the containing vessel. If the kinetic theory of gases, 
 or the theory that heat is a form of energy, or of motion, be 
 accepted, then when the point is reached where molecular motion 
 ceases, all heat and pressure must be non-existent; and as it is 
 impossible to imagine a body colder than one devoid of heat, that 
 
 1 
 is, one at a temperature , this temperature is called the abso- 
 
 lute zero. 
 
 7. Combination of the Laws of Boyle and Charles. A very use- 
 ful equation in the solution of questions concerning the pressure, 
 volume, and temperature of gases may be obtained from a com- 
 bination of the laws of Boyle and Charles. 
 
 Let v be the volume and p the pressure of a portion of gas at 
 temperature. Let v -denote its volume when the temperature 
 is raised to if, the pressure remaining at p , and let v' be the volume 
 under the pressure p', the temperature remaining if. Then by 
 Boyle's law we shall have p'v' = p v = C. By Charles's law we 
 have v = v (l + at'). Hence : 
 
 But = T = the absolute zero of temperature ; hence a = -^ , 
 and 1 + f = T ; therefore p'v' = (&} T< , or # = 
 
 In other words, the product of the volume and pressure of a gas is 
 proportional to the absolute temperature. 
 
 By Boyle's law we have, PV = a constant in foot-pounds, where 
 P = pressure in pounds per square foot, and V volume in cubic 
 
6 THE ELEMENTS OF STEAM ENGINEERING 
 
 feet. The constants for different gases have been calculated by 
 Kegnault with great accuracy. 
 
 The constant for air may be found as follows : It is known that 
 water is 773 times as heavy as air, the temperature of which is 
 32 F., or C., and the pressure 14.7 Ibs., or 760 mm.; and since 
 
 773 
 a cubic foot of water weighs 62.4 Ibs., it will take 12-387 
 
 cubic feet of air to weigh one pound. This pound of air may be 
 assumed to be contained in a c}dinder 1 sq. ft. in area and 12.387 
 feet in height. The pressure of the air being 14.7 Ibs. per sq. inch, 
 we may assume at the bottom of the cylinder a piston weighing 
 144 X 14.7 pounds P. The movement of this piston through 
 the height of the cylinder will displace 12.387 cu. ft. = V, and 
 perform 144 X 14.7 X 12.387 foot-pounds of work. Hence, PV = 
 Constant 144 X 14.7 X 12.387 = 26,220 ft. Ibs. 
 
 We have shown that the pressure multiplied by the volume of a 
 portion of gas is proportional to the absolute temperature; so that 
 if T be the absolute zero on the Fahrenheit scale (491.6 below 
 32) corresponding to P V , then for any other pressure and vol- 
 xume, as P and V, we shall have : 
 
 -> 
 
 wlience FT' = 53.3T'. 
 
 This equation for a perfect gas is usually written PV = RT , 
 where R is a constant depending on the density of the gas. For 
 air we have found it to be 53.3. For superheated steam it is 85.5. 
 
 8. The Caloric or Material Theory of Heat was advanced by Black 
 in 1798, but in 1802 it w T as shown that heat could be produced to an 
 unlimited extent by friction, thus proving its identity with motion. 
 
 9. The Kinetic Theory of Heat teaches that heat is a form of 
 motion consisting of the agitation of the molecules of matter, either 
 iby rotation or by vibration, and the hottest bodies are those whose 
 molecules are agitated at the greatest velocities. 
 
 Another statement of the kinetic theory is that heat is a form 
 of energy, or that it is energy in the invisible form of molecular 
 motion. 
 
 Since the motion of the molecules of a body is invisible, the exact 
 nature of molecular motion is unknown, but the application of the 
 laws of motion and energy to phenomena produced by heat has 
 
INTRODUCTORY 7 
 
 produced the knowledge upon which is founded the science of 
 Thermodynamics, or the Mechanical Theory of Heat. 
 
 10. Energy is defined as the capacity to do work. A force acting 
 through a distance exerts energy, and the resistance overcome is the 
 work done. 
 
 Energy is of two kinds Kinetic and Potential. The energy 
 of a body due to its motion is kinetic, and is measured by the work 
 done by the body in being brought to rest. The work done in 
 raising a body of weight w through a height h is ivh, and if v be 
 the velocity acquired by the body in falling a distance In, then 
 
 7 wv z mv 2 i i ,1 
 v 2 = 2gh, whence ivli - -3 = w~, which is the expression for 
 
 the kinetic, or actual, energy of a body of weight w, moving at a 
 velocity v. It is in fact the work stored up in a body of weight w, 
 moving with a velocity v. 
 
 If a body be raised a distance above the surface of the earth a 
 certain amount of work is performed in overcoming the attraction 
 of gravity, and this work, stored in the body, gives to it the power 
 of doing work. Energy thus possessed by a body, due to its posi- 
 tion and not its velocity, is called potential energy. A body may 
 possess potential energy due to the action of forces other than 
 gravity, as, for example, in winding a watch, work is done on the 
 spring by which its energy is increased, giving to it potential 
 energy, which causes a.reappearance of the work in driving the watch 
 while the spring unbends. 
 
 11. Temperature. Temperature refers to the condition of a body 
 as regards its sensible heat, and this condition is measured by the 
 thermometer. Two bodies are said to be of equal temperature when 
 there is no tendency of heat to pass from one to the other. 
 
 12. Thermometer. A thermometer is an instrument for measur- 
 ing temperatures. The lower fixed point on thermometric scales is 
 the temperature of melting ice under atmospheric pressure. The 
 upper fixed point is the temperature of boiling water under the nor- 
 mal pressure of the atmosphere. The scale universally used for 
 scientific work is that known as Centigrade, but in England and in 
 the United States the Fahrenheit scale is in general use. In Eussia 
 the Eeaumur scale is used. 
 
 The temperature of melting ice is marked on the Centigrade 
 and Eeaumur scales, but on the Fahrenheit scale it is marked at 32 t 
 
8 THE ELEMENTS OF STEAM ENGINEERING 
 
 The temperature of boiling water is marked 80 on the Keaumur, 
 100 on the Centigrade, and 212 on the Fahrenheit scales. Denot- 
 ing the three scales by (7, F, and R, we shall have C = 32 F = 
 R . The zero of the Fahrenheit scale is then 32 below the tem- 
 perature of melting ice, so if 32 be deducted from the Fahrenheit 
 reading we shall get the number of degrees on each scale by which 
 any given temperature differs from that of melting ice. Let the 
 given temperature be that of boiling water : then 
 
 C : F 32 : R : : 100 : 180 : 80 : : 5 : 9 : 4 
 
 5 9 ~ 4 * 
 
 By means of these equations temperatures on one scale may readily 
 be converted into either of the others. 
 
 13. Calorimetry. In order to measure the quantity of heat 
 which a body absorbs or yields in passing through a given range of 
 temperature, or in changing its state, we adopt as a unit that quan- 
 tity of heat which, acting on a given weight of water, changes its 
 temperature a definite amount. The apparatus in which the meas- 
 urement is made is called a calorimeter. The specific heats, the 
 latent heats of vaporization, and the heats of combustion of sub- 
 stances are determined by the calorimeter. 
 
 14. Thermal Unit. The unit of Heat, or the British Thermal 
 Unit, is the quantity of heat required to raise the temperature of 
 one pound of water through one degree Fahrenheit, when at or near 
 its temperature of greatest density, 39. More recently the meas- 
 ure of the unit is taken when the water is at a temperature of 62. 
 
 15. Specific Heat. The specific heat of a substance, or its capac- 
 ity for heat, is the ratio of the quantity of heat required to raise 
 the temperature of a given weight of the substance through 1 F., 
 to the quantity required to raise the temperature of an equal weight 
 of water at 39 F. through 1 F. 
 
 16. Thermodynamics. The science which treats of the mechani- 
 cal or dynamical theory of heat, or of the relation between heat and 
 work, is called thermodynamics. 
 
 17. First Law of Thermodynamics. Heat and mechanical energy 
 are mutually convertible, and heat requires for its production, or 
 produces by its disappearance, mechanical energy equivalent to 778 
 foot-pounds for each thermal unit. 
 
INTRODUCTORY 9 
 
 18. Mechanical Equivalent of Heat. One thermal unit is equiv- 
 alent to 778 foot-pounds of mechanical work. 
 
 19. The Conservation of Energy. The first law of thermody- 
 namics expresses the principle of the conservation of energy that 
 the total energy of the universe is constant, that energy can neither 
 be created nor destroyed by any process of nature, and that every 
 gain in one form of energy corresponds exactly to a loss in some 
 other form. While no heat is absolutely lost in the transformation 
 of energy of one form into that of another, a considerable amount is 
 lost as available energy, being transformed into useless heat and 
 dissipated, or diffused among surrounding bodies and thus becoming 
 unavailable. It is only when a body is hotter than surrounding 
 objects that its heat energy can be utilized in doing work. If all 
 bodies were of the same temperature no work of any kind would be 
 possible. 
 
 A general statement of the conservation of energy is : 
 
 Kinetic Energy + Potential Energy Constant, 
 but the practical statement is : In any machine we have : The total 
 energy deposited = Useful work given out -f- Work lost by resist- 
 ances. This disposes of the possibility of "perpetual motion," 
 since a machine cannot deliver more work than was deposited. 
 
 20. Second Law of Thermodynamics. Heat cannot pass from a 
 cold body to a hot one by a self-acting process unaided by external 
 agency. 
 
 It follows from this law that no engine can convert into work 
 the whole of the heat supplied to it; for as soon as the heat sup- 
 plied has fallen in temperature to that of the surrounding atmos- 
 phere the heat remaining is no longer available for doing useful 
 work. It follows also from this law that it is impossible to convert 
 any part of the heat of a body into mechanical work, except by 
 allowing it to pass to another body at a lower temperature. 
 
 21. Definition. Force is that which moves or tends to move a 
 body, or which changes or tends to change the motion of a body. 
 
 The unit of a force is a pound, and forces are expressed as being 
 equal to so many pounds. 
 
 22. Definition. Work is the overcoming of resistance through 
 space, and its measure is the product of the resistance and the space 
 through which it is overcome, the space being usually reckoned in 
 feet. 
 
10 THE ELEMENTS OF STEAM ENGINEERING 
 
 23. Unit of Work. Since the unit of force is a pound, and the 
 unit of distance a foot, the product formed by multiplying a foot 
 by a pound is the unit of work, and is called a foot-pound. A foot- 
 pound represents the work done in raising the weight of a pound 
 through the distance of one foot. If 10 pounds be raised 5 feet, 
 or 5 pounds be raised 10 feet, the work done in either case is rep- 
 resented by 50 foot-pounds. 
 
 Other units of work are sometimes used as a matter of con- 
 venience. For example, an inch-ton represents the work performed 
 in raising 2240 pounds one inch, and it is equal to 2240 inch- 
 pounds, or to ..Q foot-pounds. A foot-ton represents an expendi- 
 ture of work equal to 2240 foot-pounds. It is thus seen that the 
 different units of work are readily converted one to another. 
 
 24. Definition. Power is the rate of doing work, and is meas- 
 ured by the number of foot-pounds of work done in a unit of time. 
 
 25. Definition. Horse-power is the unit employed to represent 
 the rate of work of a steam engine, and its measure is the quantity 
 of work equivalent to the raising of 33,000 pounds through one foot 
 in one minute. The performance of 33,000 foot-pounds of work 
 in one minute is then a horse-power, and an engine performing this 
 amount of work in a second w r ould be working at the rate of 60 
 horse-power. 
 
 26. Definition. Brake Horse-power is the power which the en- 
 gine transmits for useful work ; that is, it is the total, or Indicated 
 Horse-power, minus the pow r er absorbed in driving the engine itself. 
 
 PEOBLEMS. 
 
 1. The ratio of the volumes of air under the same pressure at 
 temperatures of C. and of 100 C., or at 32 F. and at 212 F., 
 is 1 to 1.36625. Find the absolute zero of temperature on the 
 Centigrade and Fahrenheit scales. 
 
 2. A volume of air at temperature of 32 F., and pressure of 14.7 
 Ibs. per sq. inch, is heated to a temperature of 100 F., the volume 
 remaining the same. What will be its pressure? Ans. 16.73 Ibs. 
 
 3. A pound of steam at temperature of 232.8 and absolute 
 pressure of 22 Ibs. per sq. inch has a volume of 18 cubic feet. What 
 will be its volume at 341 and pressure of 120 Ibs.? 
 
 Ans. 3.812 cu. ft. 
 
CHAPTEK II. 
 THE APPLICATION OF HEAT TO WATER. 
 
 27. The effect of the application of heat to water is of the great- 
 est importance in the study of the steam engine. 
 
 Eadiant heat is given off from hot bodies in straight lines, and 
 the rays of heat are subject to the same laws as the rays of light. 
 As far as the generation of steam in a boiler is concerned,, the useful 
 radiation is confined to the furnace, the crowns and sides of which, 
 intercepting the rays of heat from the burning fuel, become them- 
 selves heated, and the heat passes through them to the water in the 
 boiler. A considerable amount of heat is given off by radiation 
 from burning coal, and it is very important to intercept this, and 
 to insure, as far as possible, that the whole of the heat diffused in 
 this way should be transmitted, directly or indirectly, to the water 
 in the boiler, and not wasted on the external air. 
 
 Eadiation is an important item to be considered with reference 
 to the economical employment of steam, for it always causes a cer- 
 tain loss of heat, and unless proper precautions are taken this loss 
 may become very considerable. The surfaces of the boilers, steam 
 pipes, cylinders, etc., when the engines are at work, are very much 
 hotter than the surrounding bodies, and in order that the loss of 
 heat by radiation may be avoided as far as possible, all these sur- 
 faces should be clothed, or lagged, with some non-conducting ma- 
 terial. 
 
 The transfer of heat by convection is the manner in which liquids 
 and gases are heated. When heat is applied to the bottom of a 
 vessel containing a fluid, liquid or gaseous, the particles in contact 
 with the bottom are first heated, and, becoming less dense, they 
 therefore rise, allowing cooler particles to take their places, which 
 become themselves heated, rise and circulate through the mass in a 
 similar manner. 
 
 It is essential that circulation and mixture of all the particles of 
 a fluid should take place to cause the temperature to be uniform 
 throughout the mass. In order that heat may be efficiently trans- 
 mitted through boiler plates and flues, each of the fluids in contact 
 
12 THE ELEMENTS OF STEAM ENGINEERING 
 
 with them, viz., the water on one side, and the heated gases on the 
 other, should have free circulation, so that the particles in contact 
 with the plates should not be considerably different in temperature 
 from those at some distance from the plate. Boilers are some- 
 times fitted with circulating plates to set up currents in the water, 
 and with bafflers in the flues to break up the currents of hot gases 
 and form eddies, in order to promote circulation and mixture in the 
 respective fluids. 
 
 In order to render most efficient the transfer of heat from one 
 fluid to another through a plate, the general movement of the two 
 fluids should be in opposite directions, so that the hottest parts of 
 the two fluids are opposed to each other and the difference of tem- 
 peratures between them a minimum. Consequently, in a boiler,, 
 in order that the best results may be obtained, the feed pipes and 
 circulating plates should be so arranged that the general movement 
 given to the water should be as far as possible in the opposite direc- 
 tion to that of the hot gases on their way to the funnel. For the 
 same reason, the action of surface condensers is most efficient when 
 the cold water for condensing the steam enters the condenser at 
 the end at which the condensed water leaves it, so that the entering 
 steam will be opposed to the water which has been heated to some 
 extent by its passage through the condenser. 
 
 Upon the application of heat to the water in the boiler the tem- 
 perature is at first raised. The particles of water in contact with 
 the heating surface which in a marine boiler consists of the fur- 
 nace plates, combustion chamber,, and tubes become heated, and 
 rise and circulate through the mass of the water, their places being 
 taken by cooler particles, till at length the whole of the water is 
 raised in temperature to the boiling point by the process of con- 
 vection. 
 
 28. Sensible Heat. The heat added to the water up to the tem- 
 perature at which boiling occurs is called sensible heat, its effect 
 being simply to change the temperature, and not the state, of the 
 water, and its amount or quantity may easily be calculated; for it 
 must be understood that the quantity of heat of a body, or the 
 amount of heat energy which a body gains or loses in passing 
 through a given range of temperature, is measured in thermal units, 
 commonly denoted by B. T. U. (British Thermal Units), and is 
 given by the continued product of its weight, range of temperature, 
 and specific heat. 
 
THE APPLICATION OF HEAT TO WATER 13 
 
 29. Latent Heat. After the temperature of boiling is reached, 
 and in order to convert the boiling water into steam, a large quan- 
 tity of heat has to be expended which does not produce any increase 
 in the temperature. This heat is known as latent heat. It was 
 supposed at an early period that heat was a substance, and that the 
 heat thus expended became hidden or latent during the change 
 from the liquid to the gaseous state, and that it again became sen- 
 sible on the reverse process being performed. The science of ther- 
 modynamics, however, has shown that heat is not a substance, but 
 a form of mechanical energy, and that this heat is expended in per- 
 forming the internal work of overcoming the molecular cohesion 
 of the particles of water which resist the change of state, and also 
 performing the external work of overcoming the resistance of ex- 
 ternal bodies to the change of volume which ensues. Work is 
 therefore done on changing the water from a liquid into a gas, and 
 this is stored up as mechanical energy, which can be yielded back 
 again either in work, or in heat, when the gas returns to its original 
 state of water. 
 
 It is now seen that this latent heat of evaporation performs the 
 work which is necessary in the interior of the water and exterior 
 to it, the external work being a measure of the useful effect directly 
 obtained. 
 
 In the case of the boiler and the steam engine, the external work 
 of evaporation is represented by the work performed by the piston 
 while the cylinder is in communication with the boiler. It will be 
 seen later on that this external work is the equivalent of but a 
 small part of the latent heat, the greater part being expended in 
 internal work, and is represented by the energy acquired by the 
 water during the evaporation. 
 
 The term latent heat has been retained for convenience, but it 
 must be understood as an expression that means simply the quan- 
 tity of heat that must be added to or subtracted from a body in a 
 given state, to change it into another state without altering its tem- 
 perature. 
 
 30. Boiling Point. The boiling point, or the temperature of 
 ebullition, of any liquid may be defined .as that stage in the addition 
 of heat to the liquid at which the pressure on it is just overcome 
 by the pressure of vapor due to the temperature. 
 
 The temperature of the boiling point depends on the pressure 
 under which the liquid is evaporated. The greater the pressure, 
 
14 THE ELEMENTS OP STEAM ENGINEERING 
 
 the higher is the temperature at which the liquid boils. The tem- 
 perature depends upon the total, or absolute, pressure, i. e., the pres- 
 sure including that of the atmosphere, so that in ascertaining the 
 temperature corresponding to any given pressure, as shown on the 
 ordinary pressure gauge, the atmospheric pressure must be added. 
 The boiling point of a liquid is affected by its density. Ordin- 
 ary sea-water contains about ^ of its weight in salt, and this raises 
 the temperature of the boiling point by 1.2 F., so that the boiling 
 point of sea-water under the atmospheric pressure is 213.2 F. 
 
 31. The Generation of Steam. The generation of steam under 
 constant pressure and at constant volume will best be understood 
 by the consideration of the two following examples, respectively. 
 These examples are, substantially, as given in Northcott's " The 
 Steam Engine," and in Sennett's " The Marine Steam Engine." 
 
 Example I. The Generation of Steam under Constant Pres- 
 sure. Imagine one pound of water at a temperature of 32 F., or 
 C., contained in an upright tube of indefinite height, and 
 1 sq. ft., or 144 sq. inches, cross sectional area. The one pound 
 
 i r/oft 
 of water has a volume of -Q^-J = 27.7 cubic inches nearly, and it 
 
 27 7 
 will consequently occupy ^r 0.1923 inch of the height of the 
 
 tube, or the water will expose an area of surface equal to one square 
 foot and will be 0.1923 inch deep. The tube is to be opened to the 
 atmosphere, so that each square inch of the water will be pressed 
 upon by the average atmospheric pressure of 14.7 Ibs., equal to 
 14.7 X 144 = 2116.8 Ibs. per sq. ft. On heat being applied to 
 this water, the temperature rises until 212 F. is reached, when 
 vaporization commences, and there is no further rise in temperature 
 until the whole of the water is converted into steam, the tempera- 
 ture of which will also stand at 212 F. 
 
 Immediately previous to the commencement of vaporization a 
 quantity of heat will have been imparted to the water equal to 
 1(212 32)1 = 180 thermal units, or to 180X778 = 140,040 
 foot-pounds. When the pound of water is all converted into steam 
 a further quantity of heat will have been imparted equal to 965.7 
 thermal units, or to 965.7 X 778 = 751,315 foot-pounds; and yet, 
 notwithstanding this large absorption of heat, the temperature 
 remains at 212 F. Since this heat has changed the sitate of the 
 water from that of a liquid to that of a gas without raising its 
 
THE APPLICATION OF HEAT TO WATER 15 
 
 temperature, it is, from our definition, latent. We know, however, 
 that the- only manner in which heat energy can disappear as heat, 
 is by its being transformed into some other form of energy or in 
 the performance of work; and, in order to understand what has 
 become of this large quantity of heat, we must inquire as to what 
 work has been done. 
 
 In the first place 180 units were absorbed in raising the tempera- 
 ture of the water from 32 to 212, and this we know as the sensible 
 heat. This change of temperature is practically the only change 
 effected up to that point of the process, as the expansion of the 
 water in rising from 32 F. to 212 F., is extremely small and may 
 be disregarded. The effect of the heat abstracted from the source of 
 heat and communicated to the water up to this point has been only 
 to raise the temperature of the water. 
 
 Steam of a pressure of 14.7 Ibs. per sq. inch has been found to 
 occupy a volume 1644 times the water from which it was gener- 
 ated; that is, a cubic inch of water is converted into 1644 cubic 
 inches of steam at atmospheric pressure of 14.7 Ibs. per sq. inch. 
 The steam from one pound of water will thus have a volume of 
 
 1644 X 27 7 
 
 ..yog = 26.36 cubic feet. In expanding to this volume 
 
 under atmospheric pressure, it will necessarily have performed ex- 
 ternal work to the extent of 26.36 X 2116.8 = 55,798 foot-pounds. 
 This, perhaps, may be more clearly seen if we disregard the pres- 
 sure of the atmosphere, and assume that the steam expands in the 
 tube under a piston weighing 14.7 X 144 = 2116.8 Ibs., which it 
 will evidently have to lift in the tube through a height of 26.36 
 feet. 
 
 The heat required to convert the water at 212 F. into steam of 
 the same temperature has been found by experiment to be 965.7 
 units, or 965.7 X 778 = 751,315 foot-pounds, and of this, 55,798 
 foot-pounds are accounted for in the external work as shown above, 
 leaving 751,315 55,798 = 695,517 foot-pounds as the mechani- 
 cal equivalent of the internal work, or of the heat absorbed in 
 effecting the expansion against the internal or molecular forces. 
 
 Of the total quantity of heat energy imparted to the water 
 
 ipjo 180 thermal units have been expended in increasing the 
 
 temperature of the water by 180 ; ~~ = 893.98 thermal units 
 have been expended in the internal work of expanding the steam 
 
16 THE ELEMENTS OF STEAM ENGINEERING 
 
 55 798 
 against molecular forces, and ^ =71.72 thermal units have 
 
 been expended in the external work of overcoming the resistance of 
 the atmosphere which opposed the expansion of volume. The sum 
 of these three quantities of heat, that is, the sum of the units due 
 to the sensible heat, the internal work, and the external work, is 
 known as the total heat of steam, or the total heat of vaporization, 
 and is, in this case, 180 + 893.98 + 71.72 = 1145.7 thermal units, 
 which is the total heat of one pound of steam at temperature of 
 212 F. ; that is, it is the units of heat required to evaporate one 
 pound of water from 32 F. into steam at 212 F. 
 
 The results of very careful experiments have shown that the total 
 heat of steam increases but slightly as the temperature increases, 
 the increment of increase being 0.305 of a thermal unit for each 
 degree of increase in temperature. This fact enables us to deduce 
 a formula for ascertaining the total heat of steam of any tempera- 
 ture. Thus, knowing the total heat of steam at 212 F. to be 
 (accurately) 1146.6 units, we will have for the total heat, H T , at 
 any other temperature t, reckoning from 32, the formula: 
 
 H T 1146.6 + .305(^ 212). 
 
 It will be convenient to change the form of this expression, as 
 follows : 
 
 #2, =,1146.6 -f .305( 212) = 1146.6 + .305 [t (32 + 180)] 
 = 1091.7 + .305(J 32). 
 
 The sum of the two quantities of heat that disappears, represent- 
 ing the internal and external work, is equal to the total heat minus 
 the sensible heat, above 32 F., and is known as the latent heat of 
 vaporization, or the latent heat of steam under the given pressure, 
 and equals in this case 893.98 + 71.72 = 965.7 thermal units. 
 
 It has been experimentally determined that the increment of 
 decrease in the latent heat of steam is 0.7 of a thermal unit for 
 each degree of increase of temperature. We can, therefore, deduce 
 a formula for ascertaining the latent heat of steam at any tem- 
 perature. 
 
 Thus, knowing the latent heat of steam at 212 F. to be 965.7 
 units we will have for the latent heat, H L , at any temperature t, 
 reckoning from 32 F., H L = 965.7 .7 (t 212). We may 
 change the form of this formula in order to make it very similar to 
 that for the total heat, and consequently more easily remembered. 
 
THE APPLICATION OF HEAT TO WATER 
 
 17 
 
 Thus, H L = 965.7 .7(* 212) = 965.7 .7 [t (32 + 180) ] 
 
 = 1091.7 .7(^ 32). 
 
 An important modification in the conditions under which the 
 steam is generated in the above example will now be considered. 
 
 Example II. The Generation of Steam at Constant Volume. 
 In Example I, the piston was free to move and allowed the steam 
 to gradually expand in volume as the water evaporated. We will 
 now suppose this not to be the case, and the piston is fixed in the 
 cylinder at a certain distance above the water as shown in Fig. 1. 
 Suppose further that only water is contained in the space A B, so 
 that there is no pressure therein. If heat be now applied to the 
 water, instead of the formation of steam taking place only when a 
 certain temperature is reached, as in our first case, steam imme- 
 
 1. 
 
 diately commences to form, and both the temperature and pres- 
 sure gradually increase as more of the water is evaporated, although 
 the temperature and pressure are still connected by the same law as 
 before. 
 
 The pressure and temperature of the steam when all the water has 
 just been evaporated can be ascertained from our knowledge of the 
 weight of water, and the known volume of the steam; for the 
 weight of the steam is the same as that of the water from which it 
 was formed, so that its volume per pound can be calculated and its 
 pressure and temperature then ascertained from the tables. 
 
 It will be seen that this can occur in practice when steam is being 
 raised in a boiler from cold water with the stop valve and other out- 
 lets closed, except that instead of there not being any pressure on 
 the water prior to the application of heat, there is the pressure of 
 
18 THE ELEMENTS OF STEAM 'ENGINEERING 
 
 the atmosphere. In this case the temperature of the water will 
 gradually rise, but no steam will be formed, and the pressure will 
 not increase, until the temperature reaches that corresponding to 
 the atmospheric pressure, viz., 212 F., when steam commences to 
 be given off, and both temperature and pressure rise. It is thus 
 seen that the absolute pressure of boiler steam is in excess of the 
 gauge pressure by an amount equal to the atmospheric pressure. In 
 our practical illustration, also, the evaporation of the water does 
 not continue until all the water is evaporated, as the amount of 
 water is so large compared with the volume of the boiler from 
 65 per cent to 75 per cent of the volume of the boiler that the 
 desired pressure is obtained when only a small portion of the water 
 has been evaporated. 
 
 31a. Formula Connecting Pressure and Temperature. The rela- 
 tions between the pressure and temperature of steam are very com- 
 plicated, and for practical purposes these relations are best ascer- 
 tained from tables giving the properties of saturated steam. Many 
 formulas have been devised to represent the relations, but most of 
 them are of only theoretical interest, and those of logarithmic form 
 the most accurate. 
 
 Eankine gives the following equation connecting the temperature 
 and pressure of saturated steam, which gives fairly accurate results : 
 
 B C 
 
 log p = < 4 7, 7p 2 , 
 
 in which T = absolute temperature = t -\- 461.2. For a value of 
 p in pounds per square inch the values of A, B, and C are : A = 
 6.1007, log B = 3.43642, and log C = 5.59873. 
 
 EXAMPLE. Find the pressure per gauge of steam having a tem- 
 perature of 285 F. 
 
 Solution: log p A ^ ~ 2 . T 285 -f 461.2 = 746.2. 
 
 746.2 log 2.87286 
 746.2 log 2.87286 
 
 log 5.74572 
 C log 5.59873 
 
 -^=0.71287 log 9.85301 10 
 
THE APPLICATION OF HEAT TO WATER 19 
 
 B log 3.43642 
 746.2 log 2.87286 
 
 B = 3.66067 log 0.56356 
 1 A = 6.10070 
 
 :P + ^_ 2 3.66067 + 0.71287 = 4.37354 
 
 p 53.35 log 1.72716 
 Pressure per gauge = 53.35 14.7 = 38.65 Ibs. per sq. inch. 
 
 EXAMPLE. Find the temperature on the Fahrenheit scale of 
 steam of 80 Ibs. absolute pressure per square inch. 
 Solution: Solving for T, we have : 
 T _B 
 
 A 6.10070 
 p = 80 log 1.90309 
 
 A log p 4.19761 log 0.62300 
 C log 5.59873 m 
 4 log 0.60206 ' 
 
 40 (A log p)= 6664900 log 6.82379 
 
 B log 3.43642 
 B log 3.43642 
 
 B 2 7461700 log 6.87284 
 
 14126600 log 7.15003, log 3.57502 antilog = 3758.6 
 
 Therefore, ^C(A logp)+B 2 = 3758.6 
 
 B =2731.6 log 3.43642 
 
 6490.2 log 3.81225 
 
 A logp 4.19761 log 0.62300 colog, 9.37700 10 
 2 log 0.30103 colog 9.69897 10 
 
 T 773.1 log 2.88822 
 2 = 773.1 461.2 = 311.9. 
 
 32. Formula Connecting Pressure and Volume. Various formu- 
 las have been devised to show the relation between the pressure and 
 volume of saturated steam. 
 
 The following formula gives results that are quite accurate within 
 the range of pressures of recent practice, viz., 100 Ibs. to 280 Ibs. 
 per sq. inch : 
 
 . pvU =482, 
 
20 THE ELEMENTS OF STEAM ENGINEERING 
 
 in which p = absolute pressure in pounds per sq. inch, and v = vol- 
 ume of one pound of steam in cubic feet at pressure p. 
 
 EXAMPLE. Find volume of a pound of steam at 280 pounds ab- 
 solute pressure. 
 
 AR9 
 
 Solution : pv& = 482, whence i; = S 
 
 /ooO 
 
 U log v = log 482 -f colog 280 
 482 log- 2.68305 
 280 colog 7.55284 10 
 
 log 0.23589 Hog 9.37271 10 
 
 16 log 1.20412 
 17 colog 8.76955 10 
 
 v 1.6675 log 0.22207 Hog 9.34638 10. 
 
 EXAMPLE. Find volume of a pound of steam at 500 Ibs. abso- 
 lute pressure. 
 
 Solution: pv& = 482, whence -J-J log v = log 482 log 500 
 482 log 2.68305 
 500 log 2.69897 
 
 log 9.98408 10 
 log ( ) 0.01592 Hog ( ) 8.20194 10 
 
 16 log 1.20412 
 
 17 colog 8.76955 10. 
 
 - 0.01498 Hog ( ) 8.17561 10 
 v =, 0.9661 log 9.98502 10. 
 
 PEOBLEMS. 
 
 4. The specific heat of mercury is 0.033. How many pounds of 
 it at a temperature of 240 will be necessary to raise the tempera- 
 ture of 12 pounds of water from 50 to 58 ? Ans. 16 Ibs. 
 
 5. Work at the rate of a horse-power is expended for an hour in 
 creating friction, the heat from which is all communicated to 10 
 cubic feet of water contained in a non-conducting tank.. The orig- 
 inal temperature of the water was 60. What will be its tempera- 
 ture at the end of the hour? Ans. 64.1. 
 
 6. The formula connecting the temperature and pressure of steam 
 
 B C 
 is log p = A -ff, -fpi , in which p = absolute pressure in pounds 
 
THE APPLICATION OF HEAT TO WATER 21 
 
 per square inch, 4 = 6.1007, log B = 3.43642, log C = 5.59873, 
 and T = the absolute temperature. Find the temperature of steam 
 of 120 Ibs. absolute pressure. Ans. 341.69. 
 
 7. The formula connecting the pressure and volume of steam is 
 pv = 482, in which p = absolute pressure in pounds per square 
 inch, and v = volume in cubic feet of a pound of steam at pressure 
 p. Find the volume of one pound of steam at pressure of 515 Ibs. 
 
 Ans. 0.9396 cubic feet. 
 
CHAPTEE III. 
 FUELS AND COMBUSTION. 
 
 33. The fuels commonly used for generating steam are coal, 
 coke, wood, and mineral oil. 
 
 Coal is an insoluble combustible mineral widely distributed over 
 the earth, and of incalculable value in the application of the 
 sciences. More than 1,000,000,000 tons <are mined yearly, and it 
 is estimated that there is enough in the crust of the earth to make 
 a layer 3 feet thick over the land surface. 
 
 Coal consists essentially of carbon, together with hydrogen, oxy- 
 gen, nitrogen, and sulphur, and its formation is the result of a dis- 
 tillation of organic matter in the earth, the process being conducted 
 on an immense scale under the influence of heat and pressure, and 
 for an unlimited time. If the distillation is complete the result is 
 pure carbon in the shape of graphite. If less complete the result is 
 hard coal, called anthracite, which may contain more than 90 per 
 cent of carbon. If still less complete, the result is soft, or bitu- 
 minous, coal, containing upwards of 60 per cent of carbon. 
 
 34. Anthracite Coal is hard and lustrous, makes an intense fire,, 
 and when dry burns without smoke and with very little flame. It 
 contains but a small percentage of volatile matter. It is much 
 more expensive and far less abundant than soft coal. 
 
 35. Bituminous or Soft Coal contains less carbon than anthracite, 
 but is richer in hydrogen and has a larger percentage of volatile 
 matter which is driven off when the coal is ignited. Bituminous 
 coal contains no bitumen, its name being due to the fact that its 
 behavior when heated is similar to that of bitumen that is, it 
 swells more or less, fuses, and burns with a bright flame and con- 
 siderable dense smoke. 
 
 Varieties of coal known as semi-anthracite and semi-bituminous 
 are so called from their near approach in character to anthracite or 
 bituminous coal. 
 
 36. Coke is the solid product resulting from the destructive dis- 
 tillation of bituminous coal, and has more value for its use in the 
 metallurgic process of reducing iron than for the purpose of gen- 
 
FUELS AND COMBUSTION 23 
 
 erating steam. Coke is carbonized bituminous coal, the volatile 
 constituents having been driven off by the application of heat, and 
 it bears to coal the same relation that charcoal bears to wood. 
 
 37. Wood is only used for generating steam in regions where it is 
 abundant, or where coal is scarce or difficult to obtain. The rela- 
 tive value of the heat producing qualities of wood and coal is ex- 
 pressed by assigning to a cord of hard wood the equivalence of a 
 ton of anthracite coal. 
 
 38. Mineral Oil as a fuel is destined in the future for extensive 
 use, but at present the imperfection of the method of firing it, its 
 unequal distribution over the earth, and the danger attending its 
 storage and carriage, has limited its application for steam generating 
 purposes. The oil used for this purpose is crude petroleum divested 
 by distillation of its most volatile and dangerous parts. The usual 
 arrangement for firing is to spray the oil into the furnace by means 
 of a steam jet, which also induces a sufficient supply of air for com- 
 bustion. An ordinary fire is first kindled on the grate to produce 
 steam of sufficient pressure for the jet, after which oil only is sup- 
 plied as fuel. The richness in hydrogen of this crude petroleum 
 gives it an evaporative effect of 1.5 times that of coal. 
 
 39. Owing to its extreme abundance coal is the fuel generally used 
 in the furnaces of boilers, and the processes through which it passes 
 during combustion should be studied in order that precautions may 
 be taken to insure economy in its use. 
 
 40. Carbon and hydrogen are the heat-producing elements of coal, 
 the small quantity of sulphur ever present contributing very little 
 heat. The presence of oxygen in a fuel really detracts from the 
 total heat of combustion, for it is known to combine with one-eighth 
 of its weight of hydrogen to form water existing as such in, the 
 fuel and only such hydrogen in excess of one-eighth of the oxygen 
 may be reckoned in estimating the total heat of combustion of a 
 fuel. 
 
 41. Combustion in a furnace is a rapid oxidation caused by the 
 chemical union of the oxygen of the air with the hydrogen and 
 carbon of the coal, heat being produced in the process. It has been 
 seen that coal consists largely of carbon, part of which is free, while 
 the other part is in combination with hydrogen, forming hydro- 
 carbons. These hydro-carbons are solid at ordinary temperatures, 
 
24 THE ELEMENTS OF STEAM ENGINEERING 
 
 but being volatile they take the form of gas when heated to tem- 
 peratures below that necessary for combustion. The gases thus 
 formed consist of compounds of hydrogen and carbon which readily 
 decompose under the action of heat, the carbon becoming more dis- 
 tinctly separated from the compounds as the temperature rises. 
 When under the influence of temperature sufficiently high, which 
 not infrequently occurs in furnaces, the hydrogen and carbon be- 
 come entirely separated, and as each of these elements has a stronger 
 affinity for oxygen than for each other they will, if the supply of air 
 be just sufficient, combine with oxygen separately and burning en- 
 sues. Under such ideal conditions combustion would be perfect, 
 with no formations of soot or smoke, the results being carbonic acid 
 (C0 2 ), superheated steam in process of falling to water, and free 
 nitrogen. The earthy impurities, amounting to from 2 per cent to 
 10 per cent, would form the residue of ash and clinker. 
 
 If the supply of air through the ash-pit be insufficient for com- 
 plete combustion, or if it is not perfectly mixed with the gases the 
 result of the union of the carbon and oxygen will be carbonic oxide 
 (CO), which produces less than one-third the heat due the produc- 
 tion of C0 2 ; and unless provision is made for the admission of air 
 above the fuel to supply oxygen to convert this CO into C0 2 , much 
 of the carbon escapes unconsumed, resulting in a loss of fuel, the 
 production of smoke, and the deposit of soot in the tubes and up- 
 takes. The prevention of smoke is only possible when the supply 
 of air to the gases is sufficient when they are at the temperature of 
 ignition, and as the gases after formation pass off very quickly 
 from the furnace, it has been found to be practically necessary to 
 supply one and a half times the quantity of air theoretically neces- 
 sary for combustion if the draft be artificial, and twice the theoreti- 
 cal amount if the draft be natural. 
 
 42. The quantity of air necessary for the combustion of a pound 
 of carbon may be estimated as follows: In the formation of CO, 
 the C and combine by weight in the proportion of 12 to 32, or as 
 1 to 2f ; it follows that every pound of C requires 2f Ibs. of 0. 
 This must come from the air containing oxygen and nitrogen in 
 the proportion of 23 parts of and 77 parts of N" in every 100 parts 
 by weight. Since 1 cubic foot of air weighs very nearly 0.08 lb., 
 it contains 0.23 X 0.08 = 0.0184 lb. of 0. To produce 2f Ibs. 0, 
 
 the amount required for the combustion of one pound of carbon, we 
 
 02 
 must, therefore, have -foT 145 cubic feet of air. 
 
FUELS AND COMBUSTION 25 
 
 Eankine gives the following formula for the weight of air theo- 
 retically necessary for the complete combustion of a pound of fuel : 
 
 where A =. pounds of air required per pound of the fuel, and (7, H, 
 and are the fractions of carbon, hydrogen, and oxygen, respect- 
 ively, contained in a pound of the fuel. 
 
 This formula is obtained from the following considerations: 
 The average composition of air is, by weight, as we have seen, 77 
 parts of nitrogen and 23 parts of oxygen. For the complete com- 
 bustion of one pound of carbon (C0 2 ) there will be required 
 
 DO 
 
 - = 2f pounds oxygen, and since each pound of air contains, but 
 
 0.23 pound oxygen, there will be required for the combustion of 
 
 2 2 
 
 one pound of carbon ~ Jo =11.6 pounds air. For the complete 
 
 combustion of one pound of H there are required 8 pounds 0. 
 
 This must be abstracted from the air, and there will be required, 
 
 g 
 therefore, ^-^ =35 pounds air to furnish it, which is three times 
 
 as much as was required for the combustion of the pound of carbon. 
 In estimating the thermal value of a fuel only such H as is unbal- 
 anced by is taken as available for the generation of heat, and 
 
 therefore H -^- is the expression for this free H. The H in the 
 
 coal is diminished by one-eighth of the because in the formation 
 of the coal by natural processes that amount has already combined 
 with to form water existing as such in the fuel and is there- 
 fore no longer available for generating heat. In the formula given, 
 11.6 is replaced by the approximation 12, and all within the large 
 brackets is expressed in terms of carbon. 
 
 The amounts of heat liberated by the combustion of a given 
 quantity of carbon and of hydrogen have been determined experi- 
 mentally, and the results are as follows : 
 
 A pound of H on entering into combination with 8 pounds 0, 
 liberates a quantity of heat equal to about 62,000 thermal units. 
 
 A pound of carbon on combining with 2f pounds of oxygen, and 
 falling to CO 2 , liberates about 14,500 thermal units. 
 
 A pound of carbon, combining with 1J pounds of gaseous oxygen 
 to form CO, generates only about 4500 thermal units. 
 
26 THE ELEMENTS OF STEAM ENGINEERING 
 
 43. Total Heat of Combustion of Fuels. There are two methods 
 of ascertaining the calorific value of fuels : First by burning samples 
 of them in a calorimeter; and secondly by chemically analyzing 
 them and calculating the heating value of the separate components. 
 The first plan is now the more accurate and reliable. 
 
 To make a test by means of the bomb calorimeter a known weight 
 of the fuel is powdered and burned in pure oxygen under pressure, 
 the containing vessel being of iron and submerged in water. The 
 ignition is caused electrically, and the rise in temperature of the 
 water gives the heating value of the fuel. 
 
 The method based on chemical analysis is not very reliable be- 
 cause of the assumption made that the value of the elements as 
 existing in the fuel is the same as if they existed in the separate 
 and uncombined condition, and that no heat is -absorbed in the 
 separation of the carbon and hydrogen of the hydro-carbons. In 
 the absence of exact formula?, the calorific value of a fuel may be 
 very approximately determined by the following formula : 
 
 h =. 14,500 ["0 + 4.2S(H- \] , 
 L V o / J 
 
 where h = the total heat of combustion of a pound of the fuel in 
 thermal units, and (7, H, and are the fractional weights of car- 
 bon, hydrogen, and oxygen in the fuel. As already explained, the 
 
 expression f H - - -g-J is the free hydrogen, or the H not combined 
 
 nn f)00 
 
 with to form H 2 0. The thermal value of this free H is J^QQ 
 
 = ! 4.28 times the value of an equal weight of carbon, therefore the 
 whole quantity within the large brackets is the equivalent weight 
 of the pound of fuel in pure carbon, the coefficient 14,500 being the 
 thermal value of one pound of carbon as it exists in fuel. 
 
 44. Evaporative Power of Fuel. To convert one pound of water 
 at a temperature of 212 F. into steam of the same temperature 
 requires, as was shown, 965.7 thermal units, or in round numbers 
 966 units; therefore, by dividing the total heat of combustion of a 
 pound of fuel by 966, the quotient will be the number of pounds of 
 water the pound of fuel is theoretically capable of converting into 
 steam of 212 from water of 212, and is equal to 
 
FUELS AND COMBUSTION 2T 
 
 EXAMPLE I. A variety of good steaming coal contains 90 per 
 cent by weight of C, 3.5 per cent of H, 6 per cent of 0. Find the 
 total heat of combustion, and the evaporative power of a pound of 
 the coal. 
 
 Solution: 
 
 Total heat of combustion = 14,500 fo.9 + 4.28 (0.055 - ^ ) 1 
 
 14,757 units, 
 therefore the evaporative power is ',. = 15.27 pounds water. 
 
 We infer from this that the complete combustion of a pound of 
 good coal should evaporate about 15 pounds of water, but owing to 
 several sources of waste an evaporation of from 10 to 11 pounds 
 from and at 212 F. is a result which can only be obtained under 
 the most favorable circumstances. The principal losses which con- 
 tribute to this result are the waste of unburned fuel, both in the 
 solid and gaseous state, the loss of heat in the gases escaping through 
 the smoke pipe, and the loss from radiation. 
 
 Example in the Combustion of Coal. Coal containing 90 per cent 
 of carbon, 4 per cent of hydrogen, 4 per cent of oxygen, and 2 per 
 cent of ash is burned in a furnace with a supply of air of 24 pounds 
 per pound of coal. Temperature of air supplied to furnace 60 , 
 and temperature of chimney gases 600. The specific heats of 
 N, 0, C0 2 , and superheated steam are respectively 0.244, 0.218, 
 0.217, and 0.475. Find the temperature of the furnace. 
 
 The combining weights of and N in air are as 23 to 77, there- 
 fore in 24 Ibs. of air there are 24 X 0.23 = 5.52 Ibs. of 0, leaving 
 24 5.52 = 18.48 Ibs. of N free. There are then 5.52 + 0.04 
 5.56 Ibs. of in all. 
 
 C and combine to form C0 2 in the proportion by weight of 
 12 to 32, or as 1 to 2f ; therefore, 2f X 0.9 = 2.4 Ibs. required to 
 unite with the 0.9 Ib. C in the coal to form 2.4 -f 0.9 = 3.3 Ibs. 
 C0 2 . H and combine to form water in the proportion of 2 to 16, 
 or as 1 to 8 ; therefore, 0.04 X 8 = 0.32 Ib. unites with the 0.04 
 Ib. of H in the coal to form 0.04 -j- 0.32 = 0.36 Ib. superheated 
 steam. 
 
 There are 2.4 + 0.32 = 2.72 Ibs. of used altogether, leaving 
 5.56 _ 2.72 = 2.84 Ibs. of free 0. 
 
28 THE ELEMENTS OF STEAM ENGINEERING 
 
 The gases to be heated are : The gases put in the furnace were : 
 
 18.48 Ibs. N 24.00 Ibs. air 
 
 3.30 " C0 2 0.90 " C 
 
 0.36 " H 2 0.04 " H 
 
 2.84 " 0.04 " 
 
 24.98 Ibs. gases. 24.98 Ibs. gases.' 
 
 Since the amount of heat the gases gained in having their 
 temperature raised is the continued product of their temperature, 
 weights, and specific heats, we have : 
 
 18.48 X 0.244 = 4.509 units 
 
 3.30 X 0.217 = 0.716 " 
 
 0.36 X 0.475 = 0.171 " 
 
 2.84 X 0.218 = 0.619 " 
 
 6.015 units 
 required to raise the temperature of the gases 1. 
 
 The total heat of combustion of a pound of the coal is 
 
 14,500[~0.9 + 4.28 f 0.04 - - ^ \1 15^22 thermal units. There- 
 fore, the elevation of temperature of the gases above their initial 
 
 15 222 
 temperature is ' ^- =. 2530, and the temperature of the furnace 
 
 is 2530 + 60 = 2590. 
 
 EXAMPLE II. Suppose, in the above example, it were required to 
 find the number of thermal units available for generating steam. 
 We would then proceed as follows : For each degree of temperature 
 of the discharged gases there are 6.015 units of heat carried off, or 
 a total of (600 60)6.015 = 3248 units carried off to waste, leav- 
 ing 15,222 3248 = 11,974 units available per pound of coal for 
 generating steam. 
 
 It has been shown that the expression 1091.7 + 0.305(2 32) 
 gives the total heat of a pound of steam at temperature t, or, in 
 other words, it is the total heat required to evaporate a pound of 
 water from a temperature of 32 into steam at temperature t. A' 
 feed water temperature of 32 is taken as an origin, and the 
 evaporation is said to be from 32 and at t. 
 
 If our feed water be taken at any other temperature denoted by 
 tf 9 we will have in every pound of feed water l(t f 32)1 thermal 
 units which will not be supplied by the fuel. (The two factors of 
 
FUELS AND COMBUSTION 29 
 
 unity represent, respectively, the weight of water considered and 
 its specific heat, but hereafter they will be neglected.) 
 
 Obviously, when the temperature of the feed is t f) an amount 
 of heat equal to tf 32 units must be subtracted from the amount 
 given by the formula in order to obtain the heat units to be sup- 
 plied by the fuel to evaporate a pound of water of temperature if 
 into steam of temperature t. 
 
 If, then, we denote by H w the units of heat necessary to evapo- 
 rate a pound of water from a temperature tf, and at a temperature 
 t, we will have the expression, general ^in its application, 
 
 H w = 1091.7 + 0.305(^ 32) (t f 32). 
 
 If t and tf each equal 212, we shall have: 
 
 H w = 1091.7 + 0.305(212 32) (212 32) = 966.6 units, 
 or more accurately by experiment, 965.7 units, which, we have 
 already seen, is the amount of heat required to convert a pound of 
 water, having a temperature of 212, into steam at the same tem- 
 perature. 
 
 In the example above considered we have 11,974 units available 
 per pound of coal for generating steam. Suppose the feed water to 
 be at a temperature of 32, and the water converted into steam of 
 212 temperature; how many pounds of water would a pound of 
 this coal evaporate ? 
 
 Here we have H w = 1091.7 + 0.305(212 32) (32 32) = 
 
 1146.6 units required for each pound evaporated; hence ^ ' ' = 
 10.44 pounds of water evaporated per pound of coal. 
 
 This result might have been obtained as follows : Latent heat of 
 steam at 212 = 1091.7 0.7 (212 32) = 965.7 units, and 212 
 32 180 units required to raise each pound of water from 32 to 
 212. 965.7 + 180 = 1145.7 units required from fuel per pound 
 
 11 974 
 
 of water evaporated ; hence ^^ ^ = 10.44 pounds water evapo- 
 rated per pound of coal, as before. 
 
 Suppose, in the above example, the temperature of the steam to be 
 350, and that of the feed water 110. How many pounds of water 
 would a pound of the coal evaporate under those conditions ? 
 
 We have: H w =1091.7 + 0.305(350 32) (110 32) = 
 1110.69 units required per pound of water evaporated; hence 
 11,974 
 1110139 = l^'^ pounds of water evaporated per pound of coal. 
 
30 THE ELEMENTS OF STEAM ENGINEERING 
 
 45. At their meeting in 1899, The American Society of Mechani- 
 cal Engineers revised the code for conducting steam boiler trials. 
 
 By this code the " unit of evaporation " is stated to be, " one 
 pound of water at 212 F. evaporated into dry steam at the same 
 temperature." This unit is equivalent to 965.7 B. T. IT. 
 
 The code recommends that the capacity of a boiler be expressed 
 in terms of the "number of pounds of water evaporated per hour 
 from and at 212." It does not, however, abandon the widely 
 recognized measure of capacity of stationary boilers expressed in 
 terms of " boiler horse-power." 
 
 46. The unit of commercial horse-power developed by a boiler was 
 adopted by a Commission of the Centennial Exhibition in 1876, 
 and is taken to be " an evaporation of 30 pounds of water per hour 
 from a feed water temperature of 100 F. into steam at 70 pounds 
 pressure per gauge." 
 
 47. It is the common practice to reduce the results of boiler tests 
 to the common standard of weight of water evaporated by the unit 
 weight of the combustible portion of the fuel, the evaporation being 
 considered to have taken place at mean atmospheric pressure, and 
 at the temperature due that pressure, the feed water being also 
 assumed to have been supplied at that temperature. This is, in tech- 
 nical language, said to be the equivalent evaporation from and at 
 atmospheric pressure, or "from and at 212 F." 
 
 48. In determining the commercial rating, or horse-power of 
 boilers from tests, the evaporative results obtained under the con- 
 ditions of the trials must be brought to the uniform standard of 
 comparison of "from and at 212 F." This can readily be done 
 by means of a " factor of evaporation " which is immediately de- 
 ducible from the results of any trial test. 
 
 In the example last given the steam was generated from feed 
 water of 110 temperature into steam of 350, and the units re- 
 quired for each pound of water evaporated, or II w , equalled 1110.69. 
 It has been shown that if feed water at 212 be evaporated into 
 steam at 212 there are required only 965.7 units, therefore the 
 conditions of the example would produce an evaporating result 
 
 r =1.15 times greater from and at 212 ; or the equivalent 
 
 evaporation from and at 212 per pound of coal would be 
 10.78 X 1.15 = 12.4 pounds. 
 
FUELS AND COMBUSTION 31 
 
 TT 
 
 In any case, then, the factor of evaporation is equal to gg^ ; 
 
 and the evaporative results obtained by the tests of the boiler, mul- 
 tiplied by this factor, gives the equivalent evaporation from and 
 at 212. 
 
 49. The unit of commercial boiler horse-power has been stated 
 to be the evaporation of 30 pounds of water per hour from a feed 
 water temperature of 100 F. into steam at 70 pounds pressure per 
 gauge, or an absolute pressure of 84.7 pounds. The temperature 
 of steam at 84.7 pounds is found by calculation to be 316. We 
 have, under these conditions, 
 
 H w 1091.7 + 0.305(316 32) (100 32) .== 1110.32 units. 
 
 Then the factor of evaporation = ^^ = 1 ^^ 2 = 1.1498. 
 
 Therefore, an evaporation of 30 pounds water from 100 into steam 
 at 316 is equivalent to the evaporation of 30 X 1.1498 = 34.5 
 pounds from and at 212. 
 
 50. The practical method of determining the commercial horse- 
 power of a boiler is to ascertain by test the number of pounds of 
 water evaporated per hour under the working conditions, and then, 
 by means of the factor of evaporation, find the equivalent number of 
 pounds from and at 212. This latter quantity divided by 34.5 
 will give the horse-power of the boiler. 
 
 PKOBLEMS. 
 
 8. The average composition of air by weight is 77 parts of N" 
 and 23 parts of 0. How many pounds of air are theoretically 
 necessary for the complete combustion of one pound of carbon? 
 
 Ans. 11.6 Ibs. 
 
 9. Find the number of pounds of air necessary for the complete 
 combustion of one pound of hydrogen. Ans. 35 Ibs. 
 
 10. Write the formula for the total heat of combustion of a pound 
 of fuel, and explain its derivation. 
 
 11. Coal containing 92 per cent of C, 4 per cent of H, 3 per cent 
 of 0, and 1 per cent of H 2 0, is burned in a furnace with a supply of 
 air of 22.5 pounds per pound of coal. Temperature of the entering 
 air 50, temperature of discharged gases 600, and temperature of 
 the feed water 112. The specific heats of N, C0 2 , superheated 
 
32 THE ELEMENTS OF STEAM ENGINEERING 
 
 steam, and 0, are 0.244, 0.217, 0.475, and 0.218, respectively. Find : 
 (a) Temperature of furnace. (&) Heat available for generating 
 steam per pound of coal, (c) The number of pounds of coal neces- 
 sary to evaporate 50 pounds of water into steam of 270 tempera- 
 ture. Ans. (a) 2810. (&) 12,474 units, (c) 4.35 pounds. 
 
 12. An analysis of coal shows its composition to be 91.5 per cent 
 C, 3.5 per cent H, and 2.6 per cent 0. How much air will be theo- 
 retically necessary per pound to burn it ? Ans. 11.714 pounds. 
 
 13. (a) How many pounds of water will a pound of the coal of 
 Example 12 evaporate from feed water of 120 temperature into 
 steam of 373 temperature? (b) What is the evaporative power of 
 this coal from and at 212 ? 
 
 Ans. (a) 13.75 pounds. (6) 15.80 pounds. 
 
 14. Assuming the whole of the 14,500 thermal units of a pound 
 of carbon were available for useful work, how many pounds per 
 I. H. P. per hour would be required? Ans. 0.176 pound. 
 
CHAPTER IV. 
 EFFICIENCY. 
 
 51. The total work performed by any machine is not usefully 
 employed, as there are always causes which occasion a waste of 
 work, notably that expended in overcoming the friction of the 
 machine. 
 
 The ratio of the useful work performed to the total energy ex- 
 pended is called the efficiency of the machine, or 
 
 Useful work performed 
 ~~ Total energy expended 
 
 52. Before any further consideration of the question of efficiency, 
 a brief reference will be made to Carnot's Reversible Cycle or Per- 
 fect Heat Engine. 
 
 Any series of operations which returns upon itself is termed a 
 cycle, and in physics a cycle is a series of operations by which a 
 substance is brought back to its initial state. If a gas in a working 
 process passes through a series of heat changes and finally returns 
 to its initial condition, the changes constitute a cycle. 
 
 The perfect heat engine of Carnot is imaginary in conception, as 
 it is supposed to be built of materials which have no practical ex- 
 istence, such as are perfectly conducting and perfectly non-conduct- 
 ing. The study of the results of the perfect heat engine is of great 
 importance, however, from the fact that it enables us to point out 
 the imperfections of existing engines, and to assign to each its 
 share of the responsibility for the waste of heat. 
 
 A full discussion of the operations of the Carnot cycle may be 
 found in any treatise on the thermodynamics of the steam engine, 
 and only a reference incidental to an understanding of engine effi- 
 ciency is here attempted. 
 
 The gas assumed to be used in the Carnot cycle is air, and during 
 the changes of state which it undergoes heat is expended in the 
 performance of external work of raising a piston, and on the inter- 
 nal work of changing the temperature or the molecular condition 
 of the gas. During the process, by which the gas is returned to its 
 initial state, work is performed upon the gas and the heat expended 
 3 
 
34 THE ELEMENTS OF STEAM ENGINEERING 
 
 in the internal work will have been returned or rejected. The final 
 result is the performance of external work equal to the total heat 
 expended, diminished by the heat of the internal work. 
 
 It will be shown later on that no engine can be devised that will 
 
 rp rp 
 
 show a greater efficiency than - J - 2 , in which T is the absolute 
 
 * i 
 
 temperature of the source of heat, and T 2 the absolute temperature 
 of the source of cold. The engine producing this efficiency is known 
 as the ideal engine. 
 
 T T 
 
 The expression ^ l for the efficiency would have its maxi- 
 mum value of unity when T 2 = 0, or when the source of cold has 
 the absolute zero of temperature, a result practically unattainable. 
 The best we can do is to make the efficiency of the engine approach 
 unity by making T^ T 2 as nearly as possible qual to T 19 which is 
 practically accomplished by making 2\ as large and T 2 as small as 
 possible. Unfortunately, the practical range of temperature in the 
 cylinder of a steam engine is very limited and the efficiency, there- 
 fore, very low. The maximum temperature which could possibly 
 be available is that of the products of combustion of the fuel, and 
 if there were practical means of utilizing the temperature of furnace 
 gases, some material other than iron would have to be used in the 
 construction of engines. We are, therefore, limited for the maxi- 
 mum temperature to that of the steam in the boiler, and for the 
 minimum, a condenser temperature of about 126 F., corresponding 
 to an absolute pressure of 2 pounds. 
 
 The object in a steam engine is to obtain external work by an 
 expenditure of heat; and while the ideal efficiency cannot be ob- 
 tained in practice, yet it is the aim of engineers to approach it as 
 nearly as possible. 
 
 53, Losses Affecting Efficiency. The principal causes which re- 
 duce the efficiency of the steam engine below that of the ideal engine 
 may be enumerated as follows : 
 
 1. Steam is far from being a perfect gas, and its employment as 
 the medium is primarily disadvantageous. It is impossible to com- 
 press the condensed steam to its initial state, necessitating a reheat- 
 ing in the boiler. 
 
 2. Steam is not supplied to the cylinder at the temperature of the 
 furnace, the source of heat. 
 
EFFICIENCY 35 
 
 3. Steam is not rejected to the condenser, the source of cold, at 
 the condenser temperature and pressure, but suffers a fall in both 
 particulars, in consequence of incomplete expansion. 
 
 4. Owing to inevitable leakages an equivalence in water to 
 the steam supplied to the cylinder is not returned to the boiler, 
 necessitating an addition to the feed of water of much lower tem- 
 perature which must be raised to the temperature of the boiler 
 steam. 
 
 5. Initial condensation of steam in the cylinder occasions a loss, 
 though this is partially recompensed by re-evaporation during ex- 
 pansion. 
 
 6. The unavoidable, yet practically desirable, clearance space in 
 the cylinder must be filled with steam at every stroke, and this steam 
 does no work except during the period of expansion. 
 
 7. " Priming " in the boiler occasions an admixture of steam and 
 water, by which particles of water are carried into the cylinder and 
 thence to the condenser without doing any work, and abstracts heat 
 from the cylinder in their efforts to evaporate. 
 
 8. Heat is lost by radiation, and by wire drawing of the steam. 
 
 9. Free and unrestricted expansion of the steam, resulting in the 
 performance of no external work, occasions a loss, as instanced by 
 the " drop " from cylinder to cylinder in multiple-expansion engines. 
 
 10. Loss due to friction of the working parts and of friction of 
 the steam in pipes. 
 
 54. Condensation and Re-evaporation of Steam in Cylinders and 
 Methods of Reducing Losses Therefrom. When, at the beginning 
 of the stroke, fresh boiler steam enters the cylinder, which, has just 
 been cooled by exhaust connection with the condenser, initial con- 
 densation ensues which is not shown by the indicator diagram. If 
 the cut-off is earlier than half stroke it is probable that a further 
 condensation, due to the work done and the difference in tempera- 
 tures of the steam and cylinder, occurs during the earlier part of 
 the expansion, resulting in the lowering of the expansion curve. As 
 the stroke proceeds, the latent heat of the liquefied steam raises the 
 temperature of the cylinder walls, and the reduction of pressure 
 due to the expansion lowers the temperature of the steam, with the 
 result that abstraction of heat from the cylinder ensues, causing a 
 
36 THE ELEMENTS OF STEAM ENGINEERING 
 
 partial re-evaporation of the water and the consequent raising of 
 the expansion curve. This re-evaporation does not ever make up 
 for the loss due to condensation, and there is, therefore, always a 
 quantity of water rejected at release, and owing to the low pressure 
 during exhaust much of it is re-evaporated, resulting in the crea- 
 tion of back pressure. 
 
 There are four methods of reducing the losses from cylinder con- 
 densation and re-evaporation, viz. : 1. The use of superheated steam, 
 2. Steam jacketing the cylinder. 3. Quick running. 4. Com- 
 pounding, or allowing the steam to expand in two or more cylinders, 
 thus reducing the range of temperature in each. 
 
 55. Superheated Steam. Steam in the condition in which it is 
 ordinarily produced and used is said to be saturated,, the name im- 
 plying that it is a saturated vapor having the maximum density 
 and hence the smallest volume per pound consistent with its pres- 
 sure or with its temperature; it is steam at the point of condensa- 
 tion, and any reduction in the temperature will cause liquefaction. 
 If saturated steam be heated so that its temperature rises above that 
 of the saturation point it is said to be superheated, and the employ- 
 ment of such steam in cylinders is the most effective means of reduc- 
 ing the loss from liquefaction. The behavior of superheated steam 
 approaches much nearer that of a perfect gas than does that of 
 saturated steam, and having a specific heat at constant pressure of 
 0.48, it requires less than half a thermal unit to raise the tempera- 
 ture of a pound to the extent of one degree. 
 
 It must be understood that the addition of heat to saturated steam 
 will raise the temperature and leave the pressure constant, provided 
 the steam be allowed to expand as the heat is added. The practical 
 method of superheating steam is to cause it to pass, on its way from 
 boiler to cylinder, through coils surrounded by hot gases from the 
 furnace. The steam absorbs heat from the gases with a consequent 
 rise in temperature, the pressure remaining constant from the fact 
 that the steam is used almost as fast as it is generated, and the dis- 
 placement of the piston in the cylinder causes a virtual extension 
 of the volume of the boiler, thus allowing the steam to expand and 
 the pressure to remain constant. 
 
 The total heat of superheated steam is equal to the total heat of 
 the steam in the saturated condition, plus the quantity of heat 
 0.48 (t s t) required to raise the temperature of the steam from 
 
EFFICIENCY 37 
 
 that due its saturated condition to the temperature t s of its super- 
 heat, the pressure remaining constant; or, 
 
 H a = H T +QAS(t 8 t) = 1091.7 + 0.305(2 32) -\-QAS(t s t). 
 
 Steam entering the cylinder in a superheated condition is able to 
 give up to the cylinder walls the whole of its superheat without 
 undergoing any liquefaction, and if the superheat be sufficiently 
 high the loss from liquefaction may be entirely prevented. 
 
 The amount of superheat necessary to prevent liquefaction in any 
 particular case is not susceptible of exact calculation, but experi- 
 ments made by Mr. Ripper on the behavior of superheated steam in 
 a small engine indicated that for each 1 per cent of wetness at 
 cut-off 7.5 F. of superheat must be present in the steam on admis- 
 sion to the cylinder in order to render the steam dry at cut-off. 
 
 For example, if a small simple engine using saturated steam at 
 90 Ibs. pressure shows a condensation of 20 per cent of the steam 
 up to cut-off, there would be required 7.5 X 20 = 150 of super- 
 heat to secure dry steam at cut-off. As the temperature of saturated 
 steam of 90 Ibs. pressure is 320, the temperature of the super- 
 heated steam supplied to the cylinder would have to be 320 + 
 150 = 470. 
 
 The loss from initial condensation in small engines is greater than 
 in large ones, for the reason that the smaller the diameter of the 
 cylinder the greater the ratio of the interior surface to the volume. 
 That is, the smaller the cylinder the greater the surface to be alter- 
 nately heated and cooled per pound of steam used, and hence the 
 greater the liquefaction. 
 
 56. Steam Jacketing. The jacket of a cylinder is usually formed 
 by an annular space between the working barrel of the cylinder and 
 the cylinder proper. Steam of boiler pressure is circulated through 
 the jacket, the object being to supply heat to the interior surface of 
 the cylinder, and thus reduce condensation and promote re-evapora- 
 tion. This transfer of heat from the jacket to the walls of the 
 cylinder causes a certain amount of the jacket steam to liquefy, but 
 this is drained and returned to the boiler without material waste. 
 The virtue of the jacket lies in the fact that the latent heat due to 
 the condensation of its steam is delivered to the steam within the 
 cylinder, with the result that the working steam is kept compara- 
 tively dry, and being a poor conductor and radiator of heat when in 
 that condition, the cylinder walls are unable to give heat to and 
 
38 THE ELEMENTS OF STEAM ENGINEERING 
 
 receive it from the steam, and are thereby deprived of the power of 
 taking up the extremes of temperature. The greatest benefit to be 
 derived from jacketing is with slow speed simple engines, its value 
 being less with multiple expansion engines, and almost valueless 
 with engines of high rotational speeds. 
 
 57. Quick Running. It is readily seen that with engines making 
 a large nurnber of revolutions per minute, there is insufficient time 
 to make the transfer of heat to and from the cylinder walls, and 
 condensation is therefore diminished. 
 
 58. Compounding. The introduction of higher steam pressures, 
 with increased rates of expansion, produced such wide ranges of 
 temperature in a single cylinder that the remedies already consid- 
 ered for cylinder condensation failed in their application. A di- 
 vision of this wide range of temperature was then resorted to by 
 allowing the steam to expand successively through two, three, and 
 sometimes four cylinders, thus introducing the double, the triple, 
 and the quadruple expansion engines. 
 
 For example, the range of temperature in a triple expansion 
 engine using steam of initial pressure of 160 Ibs. would be from 
 363 to about 152, corresponding to a back pressure in the con- 
 denser of 4 Ibs. ; and a ratio of expansion of 13 could easily be 
 attained. If such an expansion were attempted in a single cylinder 
 the fall in temperature would be 363 152 = 211, a range 
 which would cause excessive condensation, besides other evils which 
 will not be noticed now. With the triple expansion engine this 
 range of temperature could be about equally divided between the 
 three cylinders, and the initial condensation much reduced. 
 
 59. Efficiencies of Engines and Boilers. There are a number of 
 ways of expressing the efficiencies of engines and boilers, and as 
 such expressions are the means of estimating and comparing the 
 economies of engine and boiler systems, they should be carefully 
 studied in order that there should be no confusion in their appli- 
 cation. 
 
 60. Pounds of Coal per I. H. P. per Hour. This is a very com- 
 mon standard of economy of an engine and boiler when considered 
 as one machine. Its value may be averaged as 3 for non-condensing 
 engines, 2.25 to 2.5 for condensing engines, and 1.5 to 2 in multiple- 
 expansion engines. 
 
EFFICIENCY 39 
 
 61. Pounds of Steam per I, H. P. per Hour. This standard is a 
 measure of the performance of an engine, independently of the 
 boiler. The value of this standard varies from 12.5 to 16 for 
 multiple-expansion engines, and from 24 to 30 for non-condensing 
 engines. 
 
 62. Thermal Units per I. H. P. per Minute. The number of ther- 
 mal units expended per I. H. P. per minute is a standard of engine 
 performance much used in preference to the standard of weight of 
 steam per I. H. P. per hour. 
 
 63. Thermal Efficiency. Knowing the number of thermal units 
 per I. H. P. per minute, the standard of thermal efficiency may be 
 expressed as of the engine alone, or it may be made to include both 
 boiler and engine. 
 
 The development of one horse-power is equivalent to the expen- 
 
 33 000 
 diture of ' = 42.42 thermal units, so that the 
 
 I / O 
 
 Thermal Efficiency of Engine 
 
 Thermal Efficiency of Eng. and Boiler = 
 
 Thermal units in steam per I. H. P. per min. 
 42.42 
 
 Thermal units from coal per I.H. P. per min. 
 
 As an example of the first efficiency take the case of a non-con- 
 densing engine working between the temperatures of 324 and 220, 
 corresponding to an initial pressure of 95 Ibs. and a back pressure 
 of 17 Ibs., and using 24 Ibs. steam per I. H. P. per hour. If the 
 feed water were taken at a temperature of 32, the heat expended 
 per pound of steam would be the total heat of formation of steam 
 at 324, but if the feed water be heated to the temperature of the 
 exhaust, we must subtract 220 32 = 188 units from that amount; 
 hence : 
 
 4.9 A.O v fift 
 
 Thermal Efficiency of Engine = = 0.107. 
 
 24 [1091 .7 + 0.305(324 - 32) - (220 - 32) J 
 
 If the example be that of a triple-expansion engine working be- 
 tween the temperatures of 366 and 126, corresponding to an initial 
 pressure of 165 Ibs., and a condenser pressure of 2 Ibs. ; and if the 
 consumption of steam be 12.5 Ibs. per I. H. P. per hour, we will 
 have: 
 
 Thermal Efficiency of Engine = = 0.195. 
 
 12.5[1091.7 + 0.305(366 - 32) - (126 - 32)] 
 
 It is seen from this that 10.7 per cent and 19.5 per cent are good 
 values for the efficiency of the engine alone. 
 
40 THE ELEMENTS OF STEAM ENGINEERING 
 
 As an example of the combined engine and boiler efficiency, we 
 will suppose the expenditure of coal to be 2.4 Ibs. per I. H. P. per 
 hour, and the thermal value of the coal to be 14,200 units per pound. 
 Then, 
 
 Thermal Efficiency of Engine and Boiler =^'^ * ^ = 0.0747. 
 
 14:j^UU /\ /C.4- 
 
 Good values for the thermal efficiency of the engine and boiler 
 combined are from 7 per cent to 10 per cent. 
 
 64. Relative Engine Efficiency. If the thermal efficiency be 
 compared with the efficiency of the ideal engine of the same range 
 of temperature, the result is the standard of relative efficiency. 
 
 The ideal efficiency of the triple-expansion engine above con- 
 sidered is 
 
 T, T 2 _ (366 + 461) (126 + 461) _ 9q 
 T, 366 + 461 
 
 195 
 and the relative efficiency is Q ^ 0.673, or 67.3 per cent of the 
 
 ideal efficiency. 
 
 65. Mechanical Efficiency of an Engine. Brake horse-power is 
 the power developed by an engine independently of the power ab- 
 sorbed in friction in driving the engine itself. Indicated horse- 
 power includes this frictional work; hence, 
 
 Brake Horse-power 
 Mechanical Efficiency == Indicated Horse-power 
 
 and its value should vary from 85 per cent to 90 per cent. 
 
 66. Boiler Efficiency. A standard of boiler efficiency in general 
 use is that of the number of pounds of water it evaporates per 
 pound of coal, corresponding to the standard of engine efficiency 
 expressed by the number of pounds of steam it consumes in the de- 
 velopment of an indicated horse-power. 
 
 67. The real efficiency of a boiler is the product of two efficien- 
 cies, that of the heating surface and that of the combustion. 
 
 a , Heat absorbed by water per Ib. of fuel 
 
 Efficiency of Heating Surface = . 
 
 Heat available per Ib. of fuel 
 
 Efficiency of Combustion = Heat available per Ib. of fuel ^ therefore> 
 Heat contained in a Ib. of fuel 
 
 Boiler Efficiency = Heating Surface Efficiency x Combustion Efficiency 
 _ Heat absorbed by water per Ib. of coal 
 Heat contained in a Ib. of coal 
 
EFFICIENCY 41 
 
 This may be better understood from a numerical example. 
 
 EXAMPLE I. A boiler using coal containing 90 per cent C, 3 per 
 cent H, and 6 per cent 0, evaporates 10 Ibs. of water per pound of 
 coal into steam at 341 from feed at 122. 3130 units are wasted 
 through the smoke pipe per pound of coal used. It is required to 
 find: (a) The efficiency of the heating surface. (&) The efficiency 
 of the combustion, (c) The efficiency of the boiler. 
 
 Solution: 
 
 h = 14,500 [0.9 -f 4.28 ( 0.03 - ^ \~\ = 14,446 units con- 
 
 tained in a pound of the fuel. 
 H w 1091.7 + 0.305(341 32) (122 32) =1096 units 
 
 required to evaporate one pound of water. 
 
 1096 X 10 = 10,960 units absorbed by the water per pound of coal. 
 14,446 3130 = 11,316 units available per pound of coal. 
 
 Efficiency of heating surface = = 0.9686. 
 
 Efficiency of combustion = jg = 0.7835. 
 
 Efficiency of boiler = 0.9686 X 0.7835 = 0.7588. 
 
 Or, we might have proceeded as follows : 
 Heat available per pound of coal = 1^316 = Ia8g ]bg 
 
 MW 
 
 should have been evaporated per pound of coal from the available 
 heat. 
 
 Efficiency of heating surface = T-QQ = 0.9686. 
 
 Heat units in pound of coal = 146 
 
 be evaporated if all the heat in the coal were utilized. 
 
 10 33 
 
 Efficiency of combustion = -..,' = 0.7835. 
 
 68. Total Efficiency of System. The percentage of indicated work 
 performed by the engine from a given weight of coal is expressed by 
 the product, 
 
 Efficiency of Boiler X Thermal Efficiency of Engine, 
 
 and the percentage of the work performed that is applied to the 
 shaft is : 
 
 Boiler Efficiency x Thermal Engine Efficiency x Mechanical Efficiency. 
 
42 THE ELEMENTS OF STEAM ENGINEERING 
 
 69. To illustrate the imperfection of the steam engine we will 
 consider its efficiency by means of a numerical example from the 
 conditions of actual practice. 
 
 EXAMPLE II. A high-speed non-condensing engine works with 
 an initial absolute pressure of 95 Ibs., the temperature of which is 
 324. The feed water has a temperature of 180. By a measurement 
 of the feed water it is found that 25 Ibs. of steam are used per 
 I. H. P., and a test of the boiler shows that 10 Ibs. of water are 
 evaporated per pound of coal. The thermal value of the coal used is 
 14,320 units per pound. It is required to find the efficiency of the 
 system. 
 
 The heat required for the production of one pound of the steam is 
 H w = 1091.7 + 0.305(324 32) (180 32) = 1032.76 units, 
 
 therefore one pound of the coal should evaporate i QQO 75 13.88 
 pounds of water. 
 
 Efficiency of Boiler = |Jj 0.72. 
 
 lO.oo 
 
 1032 76 ^ 25 
 
 ^- - = 430.3 thermal units required per I. H. P. per 
 
 minute. 
 The development of one horse-power is equivalent to the expendi- 
 
 qq ooo 
 
 ture of ~r = 4 ^.42 thermal units; hence: 
 
 Thermal efficiency of engine = ^-^ = 0.098. 
 
 The efficiency of the boiler is 0.72, and as only 9.8 per cent of the 
 heat supplied to the engine is converted into work, the efficiency of 
 the system is 0.098 X 0.72 = 0.07 ; or only 7 per cent of the heat of 
 the fuel is converted into mechanical work; and if the mechanical 
 efficiency of the engine is 87 per cent, we would have only 
 0.07 X 0.87 X 100 = 6.09 per cent of the heat energy of the fuel 
 applied to the shaft. 
 
 PEOBLEMS. 
 
 15. It is claimed that the best modern engines develop one 
 I. H. P. by the expenditure of 1.6 Ibs. of coal. Comparing this 
 result with that obtained from carbon in problem 14, determine the 
 efficiency of the modern engine. Arts. 11 per cent. 
 
EFFICIENCY 43 
 
 16. Diameter of cylinder 24", stroke 30", cut-off f stroke, revo- 
 lutions per minute 125. Thermal value of the fuel 15,000 units 
 per pound, of which 4000 are wasted. Temperature of feed water 
 102, and of steam 327.5, corresponding to an absolute pressure 
 of 100 Ibs. Eelative volumes of steam and water at 100 Ibs. pres- 
 sure are 271 : 1. Eequired: (a) The pounds of coal used per 
 hour, (b) The evaporative* power of the coal from and at 212. 
 
 Ans. (a) 1030.26 Ibs. (b) 11.39 Ibs. 
 
 17. The efficiency of a given boiler is 70 per cent, and .of the 
 engine 10 per cent. The composition of the fuel used is 90 per 
 cent C, 6 per cent H, and 4 per cent 0. How much coal must be 
 burned per hour to develop 1000 I. H. P. ? Ans. 2208 Ibs. 
 
 18. A boiler is to evaporate 45 cu. ft. of water per hour into 
 steam of 300 temperature from feed water of 100. Average rate 
 of transmission per sq. ft. of heating surface per hour is 3000 units. 
 Efficiency of boiler 70 per cent, and thermal value of one pound of 
 the coal used is 15,000 units. Eequired: (a) Heating surface. 
 (b) The evaporative power of the coal from and at 212. 
 
 Ans. (a) 1036.55 sq. ft. (6) 10.87 Ibs. 
 
 19. A boiler using coal containing 88 per cent C, 5 per cent H, 
 and 4 per cent 0, evaporates 9.5 Ibs. of water per pound of coal into 
 steam of 366 from feed water of 152. 4550 units are wasted per 
 pound of coal used. Find: (a) Efficiency of heating surface, (b) 
 Efficiency of combustion, (c) Efficiency of the boiler. 
 
 Ans. (a) 92.75 per cent, (b) 70.73 per cent, (c) 65.60 per cent. 
 
 20. A non-condensing engine working with an initial absolute 
 steam pressure of 100 Ibs., the temperature of which is 328, uses 
 26 Ibs. steam per I. H. P. per hour. The boiler evaporates 9.5 Ibs. 
 of water per pound of coal, the temperature of the feed being 202. 
 The thermal value of the coal used is 14,400 units per pound, and 
 it is found that 6 per cent of the heat energy of the fuel is applied 
 to the shaft. Find: (a) Efficiency of the boiler, (b) Thermal 
 efficiency of the engine, (c) Mechanical efficiency of the engine. 
 
 Ans. (a) 66.76 per cent, (b) 9.67 per cent, (c) 93.00 per cent. 
 
 21. An engine and boiler test shows a heating surface efficiency 
 of 96 per cent. There are 11,400 units available for generating 
 steam, and 3100 units wasted, per pound of coal. The tempera- 
 
44 THE ELEMENTS OF STEAM ENGINEERING 
 
 ture of the steam was 342, and of the feed water 122. . Find: 
 (a) Efficiency of combustion. (&) Efficiency of boiler, (c) The 
 number of pounds of water evaporated per pound of coal. 
 Ans. (a) 78.62 per cent. (6) 75.48 per cent, (c) 9.98 per cent. 
 
 22. A triple-expansion engine with forced draft develops 20 
 I. H. P. per sq. ft. of grate surface, burns 35 Ibs. of coal per sq. ft. 
 of grate, per hour, and consumes 14 Ibs. of steam per I. H. P. per 
 hour. Absolute boiler pressure 160 Ibs., the temperature of which 
 is 363. Temperature of feed water 150, and 12,010 units are 
 available for generating steam per pound of coal. Find : (a) The 
 efficiency of the heating surface. (&) The evaporative power of the 
 boiler from and at 212. Ans. (a) 0.716. (6) 9.94 Ibs. 
 
 23. A boiler generates steam of temperature of 366 from feed 
 water at 152. There are 11,846 units available for generating 
 steam per pound of coal. The efficiency of the heating surface is 
 88 per cent-, and the consumption of coal is 4200 Ibs. a day. Find 
 the boiler horse-power. Ans. 54.739. 
 
CHAPTER V. 
 THE VALVE AND ITS MOTION. 
 
 70. In the steam engine, steam is admitted to, and exhausted 
 from,, each end of the cylinder by a valve actuated by a part of the 
 engine itself. 
 
 With the exception of the " drop " valves used with beam engines 
 of slow rotational speed, and the special cases of rotating valves, 
 such as those used with the Corliss engine, the ordinary locomotive 
 slide valve, and its modification known as the piston valve, are 
 exclusively used in stationary and marine practice. The inadapta- 
 bility of the drop and rotating valves to engines of high rotational 
 speeds, due to their trip gear, has limited their use. 
 
 The smooth and economical working of an engine depends, to a 
 large extent, upon the proper admission and release of steam to and 
 from the cylinder, and it is, therefore, very important that study 
 should be given to valves and their design. 
 
 In Figs. 2, 3, 4, 5, 6, and 7, the hatched sections represent the 
 valve seat and parts of the cylinder anrl piston, while the solid sec- 
 tion is that of an ordinary single-ported slide valve. 
 
 In the valve seat there are three passages, or ports, two leading 
 from the steam, or valve, chest to the cylinder, and the other lead- 
 ing either to the condenser, the atmosphere, or, in the case of 
 multiple-expansion engines, to the next cylinder in the order of the 
 expansion. The steam ports leading to each end of the cylinder are 
 marked 8 and 8' 9 and the exhaust port, through which the steam 
 leaves, is marked E. The flat face of the valve works steam-tight 
 on the flat seat on the side of the cylinder. 
 
46 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 The motion of the slide valve in distributing steam in the cylin- 
 der during the stroke of the piston P will be understood by reference 
 to the figures,, the arrows indicating the direction of motion of the 
 valve and piston. 
 
 In Fig. 2, the piston is just commencing its stroke to the right, 
 the valve showing a slight opening of the port at 8 for the admis- 
 sion of steam into the cylinder. This slight opening of the port 
 
 for steam when the piston commences its stroke, is known as the 
 " lead " of the valve, and is always provided for. The entering 
 steam starts the piston on its stroke, the valve moving in the same 
 direction. The port S' has been opened by the inside edge of the 
 valve, thus allowing the steam in the cylinder, which has been used 
 on the preceding stroke of the piston to the left, to escape through 
 
 the exhaust port E and exhaust pipe shown in circle. The valve 
 and piston continue their motions in the same direction until the 
 valve reaches the limit of its travel to the right, as shown in Fig. 3, 
 and comes for an instant to rest, the piston, however, continuing its 
 motion. The valve being now at its extreme point of travel to the 
 right, it will be seen that the port S' is wide open to the exhaust, 
 but that the port 8 is not wide open for the admission of steam. 
 
THE VALVE AND ITS MOTION 47 
 
 In the design of the valve it is always arranged to have a full open- 
 ing of the port for exhaust, but a full opening of the port for steam 
 is rarely provided for. 
 
 The eccentric, which is the actuating mechanism, now causes the 
 valve to commence its return stroke, the valve and piston moving in 
 opposite directions. A very important position of the valve is 
 shown in Fig. 4, when the steam edge of the valve has just closed 
 the port 8 and any further admission of steam to the cylinder is 
 " cut-off," the remainder of the stroke of the piston being accom- 
 plished by the expansive force of the steam already admitted. The 
 port 8', it will be noticed, is still open to exhaust. The valve and 
 piston continue their motions in opposite directions, and Fig. 5 
 shows the valve in mid-position, the piston having nearly completed 
 its stroke, and the exhaust, or inside, edges of the valve coinciding 
 with the inner edges of the steam ports 8 and 8'. This latter con- 
 
 dition results from there' being no inside, or exhaust, lap to the 
 valve, which, it will be seen later, is rarely the case. With the valve 
 in this position, two events of importance occur. The closure of 
 the port S' has entrapped within the cylinder some of the exhaust 
 steam of the preceding stroke, and as the piston proceeds this steam 
 is compressed, its pressure gradually increasing as the piston nears 
 the end of its stroke. This " compression " provides an elastic 
 cushion of steam which absorbs the momentum of the reciprocating 
 parts of the engine, and brings them to rest without shock. The 
 compressed steam assists the entering steam in starting the piston 
 on the return stroke ; and, as it fills the clearance space, it has an 
 appreciable effect on the amount expended, as a less quantity need 
 be drawn from the boiler at each stroke. 
 
 The left hand inner edge of the valve having arrived at the inner 
 edge of the port 8, the other event now occurs. The continued mo- 
 
48 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 tion of the valve in the direction of the arrow opens the port S, 
 and the steam that has been driving the piston is allowed to escape 
 through the exhaust port E, the pressure on the left of the piston 
 being suddenly reduced. This event is known as the " release " of 
 the steam. 
 
 The motions in opposite directions of the valve and piston con- 
 tinue, the port S opening wider to exhaust and the pressure of the 
 
 compressed steam increasing, until just before the arrival of the 
 piston at the end of the stroke, when the outside, or steam, edge of 
 the valve reaches the outside edge of the port S f , as shown in Fig. 6. 
 Now follows the event of " admission," and as the port 8' uncovers, 
 fresh steam is admitted, raising the compressed steam to full pres- 
 sure, which acts in opposition to the piston during the remainder of 
 the stroke, and brings it gradually to rest without shock at the end 
 of the stroke. 
 
 Fig. 7 shows the position of the valve and piston when the stroke 
 is completed, the port S' being open to the extent of the " lead." 
 All the events just described are repeated when the piston makes 
 the return stroke, at the completion of which the valve will be found 
 in the position shown in Fig. 2. 
 
 In this manner the piston is given a reciprocating motion in the 
 cylinder, which is communicated to the shaft by jneans of the con- 
 
THE VALVE AND ITS MOTION 
 
 49 
 
 necting-rod and crank mechanism, and the shaft revolves as long as 
 steam is supplied to the cylinder. 
 
 71. The Eccentric. Slide valves are usually actuated by an ec- 
 centric and rod. The eccentric consists of a disc keyed to the shaft 
 and revolving with it, but having its center eccentric to that of the 
 shaft. In Fig. 8, C is the center of the shaft, and E the center of 
 the eccentric. The eccentric disc is usually in two parts, D and D' ', 
 bolted together as indicated. As the shaft revolves, the eccentric 
 is carried with it, the latter turning within the strap, thus giving 
 to the end A of the rod a reciprocating motion which is communi- 
 cated to the valve. The end A, which is guided to move in a straight 
 line pointing to the center of the shaft, is directly connected to 
 the valve stem in engines designed to run only in one direction, 
 but in locomotives and marine engines, where the direction of mo- 
 
 8. 
 
 tion is often reversed, the reciprocating motion is communicated 
 through the agency of a link. 
 
 An examination of Fig. 8 will show that the motion of the end 
 A of the rod along the line CB will be twice the distance CE. CE 
 is the distance between the centers of the shaft and eccentric. It 
 is variously called the " throw of the eccentric," " eccentric arm," 
 and "eccentric radius," and it is equal to the half-travel of the 
 valve. 
 
 It will readily be seen that the motion given to the point A is 
 exactly equivalent to that which would be produced by a crank CE 
 and connecting rod EA, from which we conclude that the eccentric 
 and rod is a special case of the connecting-rod and crank mechanism, 
 in which the crank pin is made so large as to include the shaft 
 within its section. In order that this may be, the radius of the 
 4 
 
50 THE ELEMENTS OF STEAM ENGINEERING 
 
 eccentric must be greater than the sum of the radii of the crank and 
 of the shaft; that is, Er f > EC + Cr (Fig. 8). 
 
 72. Having considered the action and actuating mechanism of 
 the slide valve in the distribution of the steam in the cylinder, it is 
 next in order to fix the position of the eccentric on the shaft, rela- 
 tive to the crank, so that the events illustrated by Figs. 2, 3, 4, 5, 
 6, and 7 shall occur at the proper times. 
 
 For this purpose a slide valve with faces of just sufficient width 
 to cover, when in central position, the steam ports S and S', Fig. 9,. 
 will be used in illustration. 
 
 In order that alternate strokes of the piston in the cylinder shall 
 be in opposite directions, the valve must provide that, at each 
 stroke, steam shall be admitted to the driving side of the piston , 
 and the steam used in the preceding stroke shall be exhausted from 
 
 * 
 
 the opposite side. The valve in Fig. 9 is in its central position, 
 and if its movement were so timed that it would occupy that position 
 simultaneously with the arrival of the piston at the end of either 
 stroke, the provision required of the valve would be fulfilled ; pro- 
 vided the movement, or travel, of the valve, during one stroke of the 
 piston, were equal to twice the width of the steam port, and that 
 the first half of the valve movement were in the same direction as 
 that of the piston, and the last half in the contrary direction. Since 
 the throw of the eccentric is equal to the half-travel of the valve, 
 and as the crank moves with the piston, it is at once evident that 
 the desired motion to the valve of Fig. 9 would be obtained by plac- 
 ing the eccentric on the shaft 90 ahead of the crank. 
 
 73. We will now consider the distribution of the steam in the 
 cylinder by the valve of Fig. 9, the eccentric being placed on the 
 shaft 90 ahead of the crank. The piston being at the end of the 
 stroke, the eccentric will cause the valve to move to the right to 
 
THE VALVE AND ITS MOTION 
 
 51 
 
 admit steam to the cylinder through the port 8 to drive the piston 
 in the same direction. Simultaneously with the admission of steam 
 into 8, the right-hand inside edge of the valve will open the port 8' 
 to the exhaust, and there will not have been any " release " of the 
 steam before the piston arrived at the end of the preceding stroke, 
 nor was there any " lead " to the valve to give the full steam pres- 
 sure on the piston at the moment of commencing the stroke under 
 consideration. After the eccentric arm has revolved through 90, 
 the valve will have reached its extreme distance to the right, a dis- 
 tance just sufficient to open wide the port 8 to steam and the 
 port S' to exhaust. The crank and eccentric continue to revolve, 
 the piston proceeds on its stroke but the instant the eccentric 
 passes the 90 point of its rotation, the valve commences its travel 
 
 to the left, and after a further angular motion of 90 by the crank 
 and eccentric, the piston will have arrived at the end of its stroke 
 simultaneously with the arrival of the valve to its original position. 
 Thus, it is seen, there has been no " cut-off " of the steam before 
 the arrival of the piston at the end of its stroke, and, therefore, no 
 advantage taken of the steam's expansive force to complete the 
 stroke; nor has the exhaust closed before the completion of the 
 stroke to provide for that very necessary elastic cushion of com- 
 pressed steam to gradually bring the piston to rest without shock. 
 
 74. The relative movements of the crank and eccentric, as just 
 explained, will be better understood by referring to Fig. 10. CR 
 and CE are the original positions of the crank and eccentric arms, 
 the position of the valve being as shown in Fig. 9. CR' and CE' are 
 
52 THE ELEMENTS OF STEAM ENGINEERING 
 
 the crank and eccentric positions after the shaft has revolved 
 through 90 the valve having moved its extreme distance CE' to 
 the right, opening the ports wide to steam and exhaust. A further 
 rotation of 90 by the shaft brings the crank and eccentric to the 
 positions CE" and CE", respectively, the valve having returned 
 through a distance E'C to its original position, just as the piston 
 completed its stroke; for while the center of the crank pin has 
 rotated through the semicircle RR'R", its motion of translation has 
 been the diameter RE" of the crank pin circle, which, of course, is 
 equal to the stroke of the piston. If the shaft were to complete its 
 revolution, as indicated by the dotted semicircles, the crank and 
 eccentric would assume their original positions CR and CE, and 
 the valve would have moved its extreme distance CE'" to the left 
 and returned the same distance E'"C to its original position, the 
 piston, meanwhile, completing its return stroke, and the positions 
 of the valve and piston would be as represented in Fig. 9. 
 
 The action of this valve is clearly defective, and the difficulty 
 results from delaying until the end of the stroke the changes which 
 are necessary for its reversal. 
 
 In practice this difficulty is avoided by making the valve longer, 
 so that when in its central position, its faces will more than cover 
 the steam ports, as shown in Fig. 5 ; and in addition to this length- 
 ening of the valve, the eccentric is set on the shaft an angular dis- 
 tance greater than 90 ahead of the crank. 
 
 Definition. Lap of the valve called outside, or steam, lap is 
 the distance the outer, or steam, edge of the valve extends beyond, 
 or laps over, the steam edge of the port, when the valve is in the 
 mid-position of its stroke. 
 
 It has been pointed out that giving lead to the valve provides 
 that a full pressure of steam shall be exerted on the piston at the 
 commencement of its stroke, and that assistance be rendered the 
 compressed steam in bringing the piston momentarily to rest with- 
 out shock at the end of the stroke. 
 
 Definition. Lead of the valve is the distance the valve uncovers 
 the port for the admission of steam when the piston is at the end 
 of its stroke. 
 
 Eegarding the valve of Fig. 5 as that of Fig. 9 lengthened to 
 overlap the ports, it will be seen that in order to give it lead when 
 
THE VALVE AND ITS MOTION 53 
 
 the piston is at the end of its stroke, as shown in Fig. 2, page 45, 
 we must first advance the eccentric CE, Fig. 11, through an angle 
 EGA, called the angle of lap, to move the valve a distance CP equal 
 to the lap, and then advance it through the angle ACB, called the 
 angle of lead, to move the valve a distance PD, equal to the de- 
 sired lead. 
 
 Definition. Angular Advance of the eccentric is the number of 
 degrees in excess of a right angle that the eccentric is set on the 
 shaft in advance of the crank, and is equal to The Lap Angle -\- The 
 Lead Angle. 
 
 75. The primary object in giving the lap to a valve is to provide 
 means for cutting off the admission of steam into the cylinder be- 
 fore the end of the stroke, so that advantage may be taken of the 
 expansive power of the steam to continue the movement of the 
 piston to the end of the stroke. There is a limit,, however, to the 
 
 Fiq. II 
 
 amount of lap that should be given a valve. The addition of lap 
 necessitates an increase of the angular advance of the eccentric 
 in order to maintain the lead, and if the maximum port opening is 
 also to be maintained, the travel of the valve must be increased. It 
 is always desirable to have a small valve travel so that the work 
 expended in moving the valve shall be a minimum, and it is for 
 this reason that double-ported valves are resorted to, as by their use 
 only one-half the travel of the single-ported valve is necessary for 
 a given port opening. The increase in the angular advance of the 
 eccentric will have the effect of hastening all the operations of the 
 valve, and to whatever extent the opening of the exhaust has been 
 hastened its closure will likewise be hastened, and excessive com- 
 pression may result. With a fixed eccentric the limit of cut-off by 
 means of lap is about f of the stroke. 
 
 76. If, when the valve is in its mid-position, the inside edges 
 extend over the inside edges of the ports, the valve is said to have 
 inside, or exhaust, lap, and its effect is to delay the release of the 
 
54 THE ELEMENTS OP STEAM ENGINEERING 
 
 steam and hasten the compression. When in its mid-position, if 
 the inside edges of the valve do not overlap the inside edges of the 
 ports, but show, instead, an opening to the exhaust, the valve is 
 then said to have negative exhaust lap, and this is not infrequently 
 the case, particularly with vertical engines where a prompt exhaust 
 of the steam from the cylinder on the up stroke is desirable. An 
 examination of Fig. 5 will show that with negative exhaust lap, 
 there will be a period when both ends of the cylinder will be in 
 communication with the exhaust and with each other. This will 
 occur just before compression, and the released steam from the 
 driving side of the piston will flow into the exhausted side, causing 
 an increase in the back pressure, which will be shown by a rise in 
 the back pressure line of the indicator diagram just before compres- 
 sion. The period of this communication will be brief, as the mo- 
 tion of the valve is then at its quickest. 
 
 77. The necessity for giving lead to the valve has been shown, 
 and its amount is fixed arbitrarily, according to the type of engine 
 and the judgment of the designer. The rotational speed in 
 all engines arid the weight of the reciprocating parts compared to 
 piston area in vertical engines, are governing features in the deter- 
 mination of the amount of the lead. For slow-running stationary 
 engines the lead varies from -fa. to T *g- of an inch. For the high- 
 speed stationary engine it is seldom less than $ of an inch, and for 
 the locomotive it is commonly J inch when at full speed and linked 
 up. The weight of the reciprocating parts of a vertical engine act 
 with the steam on the top, or down, stroke, and against the steam 
 on the bottom, or up, stroke, and therefore the necessity for in- 
 creased lead at the bottom, or crank, end of the valve, to assist com- 
 pression in bringing the piston to rest without shock and to furnish 
 promptly a full pressure to start the piston on the up stroke. As 
 much as f of an inch for the bottom lead of a vertical marine 
 engine is not uncommon. 
 
 78. For reasons which will be explained later, the angularity of 
 the connecting-rod introduces an inequality in the movement of the 
 piston, which occasions a greater piston displacement on the for- 
 ward (toward the shaft) stroke than on the return stroke, for equal 
 angular positions of the crank, and this inequality is greater, the 
 greater the ratio of the length of the crank to the length of the 
 connecting-rod. Then, since the cut-off of both strokes is effected 
 
THE VALVE AND ITS MOTION 55 
 
 by the same eccentric, there will be an inequality in the two cut-offs, 
 later on the forward than on the return stroke. In stationary en- 
 gines, where the crank-connecting-rod ratio is comparatively small, 
 this inequality is not of much consequence, and a partial compen- 
 sation is usually provided by giving a trifle less lap on the crank 
 end of the valve than on the head end, without seriously disarrang- 
 ing the leads. To provide for an absolute equality in cut-off 
 would require so little lap on the crank end as to make the lead 
 excessive. In vertical marine engines, the crank-connecting-rod 
 ratio is comparatively large, and the inequality in cut-off more pro- 
 nounced; but since, as already stated, a large lead is desirable on 
 the bottom, or crank, end, the lap on that end may be sensibly less 
 than on the top end, which will afford a partial equalization of the 
 cut-off. In addition to this, it is common marine practice to make 
 the top cut-off a trifle later than that desired, which will make the 
 bottom cut-off also later, and therefore the mean cut-off more nearly 
 that desired. 
 
 79. Setting Slide Valves. The efficient working of the engine 
 depends largely on the proper adjustment of the valve on its stem, 
 and the placing of the eccentric in its proper place on the shaft. 
 The performance of these two operations is called " setting the 
 valve." It is first necessary to locate the dead points of the engine, 
 or put the engine " on its center," as it is called. When an engine 
 is on its center, a line joining the center of the crank pin and the 
 center of the shaft is in direct line with the axis of the cylinder, 
 and the piston is then absolutely at the end of its stroke. To ob- 
 tain this position the engine is jacked until the cross-head nearly 
 reaches the end of its stroke, when a mark is made on the slides to 
 correspond to one made, or existing, on the cross-head. At some 
 point on the engine framing near the crank disc, the pulley, or the 
 fly wheel, make a center punch mark, and with this mark as a center, 
 and a tram of convenient length as a radius, describe a small arc 
 on the revolving part selected. Now jack the engine past the end 
 of the stroke and until the mark on the cross-head returns to the 
 mark made on the slide. From the center-punch mark as a center, 
 describe another small arc with the tram on the selected revolving 
 part. The cross-head or the piston is now the same distance from 
 the end of the stroke as it was when the first arc was described on 
 the revolving part; so if the distance between these two arcs is 
 
56 THE ELEMENTS OF STEAM ENGINEERING 
 
 bisected, and the point of bisection marked with the center punch, 
 then when the engine is jacked so that the tram exactly spans the 
 distance between the two center-punch marks, the engine will be 
 on the center and the piston at the end of the stroke. A like pro- 
 cess will determine the other dead center. In putting an engine on 
 center, and during the whole operation of setting a valve, the engine 
 should be moved in the direction in which it is intended to run, in 
 order that all lost motion may be taken up in that direction. 
 
 Valves are usually set for equal leads, the exceptions being for 
 vertical engines, as already pointed out, and for cases where 
 attempts are made to equalize the cut-off. 
 
 To set a slide valve, the engine is placed on the center, and if the 
 eccentric be not fixed on the shaft in its proper position it should 
 be approximately so placed, making the angular advance, prefer- 
 ably, a trifle larger than that required. By means of the nuts on 
 the valve stem the valve is given the proper lead. Turn the engine 
 to the other center, and if the lead shown there is not the same, the 
 difference must be corrected half by altering the length of the 
 valve stem, and half by moving the eccentric. With the lead thus 
 equalized, the eccentric should be keyed fast on the shaft and the 
 nuts on the valve stem tightened; the operation will then be com- 
 plete. 
 
 Should the valve be designed for unequal leads, the method of 
 setting it is exactly the same as just described, the correction being 
 so made as to have the leads as desired. 
 
 80. The Link Motion. With marine engines and locomotives, 
 whose direction of running must often be reversed, the reciprocat- 
 ing motion furnished by the eccentric is communicated to the valve 
 through the agency of a link. The link motion was originally de- 
 signed as a means of reversing the motion of an engine, but it was 
 afterwards found to furnish a method of materially increasing the 
 range of expansion of the steam in the cylinder. 
 
 The action of a link motion in reversing an engine may be under- 
 stood by referring to Fig. 12, where CR represents the crank in 
 some one position. We have already seen that in order to have the 
 crank revolve in the direction indicated by the full line arrow, the 
 eccentric must be fixed on the shaft at some position CE ahead of 
 the crank. Now in order to have the crank turn in the direction 
 of the dotted arrow, the eccentric would have to be shifted on the 
 
THE VALVE AND ITS MOTION 
 
 57 
 
 shaft to the position CE f ahead of the crank in the opposite direc- 
 tion. As this is impracticable, the difficulty is overcome by having 
 two eccentrics keyed to the shaft, one having its center at E and 
 the other at E'. Each eccentric has its rod connected to one end 
 of a curved link which, as originally designed, is slotted to receive a 
 
 block directly connected to the valve stem, and by a suitable arrange- 
 ment of levers the link may be shifted so that the movement of the' 
 valve may be under the influence of either the go-ahead or the back- 
 ing eccentric as desired. 
 
 81. Figs. 13 and 14 are skeleton sketches of a link motion, in 
 which CR is the crank, E and E' the centers of the go-ahead and 
 backing eccentrics, respectively, and PQ the link, curved with a 
 radius equal to the length of the eccentric rod EP. 
 
 When the crank is pointing away from the link and the eccentric 
 rods are joined to the ends of the link nearest to them, as in Fig. 13, 
 the rods are said to be " open," and if, when the crank is in the- 
 same position, the rods are joined to the link as shown in Fig. 14,. 
 the rods are said to be " crossed." If, as is frequently the case with 
 piston valves, steam is taken at the inside edges, the rods as con- 
 nected in Fig. 13 would be called " crossed," and if as in Fig. 14 r 
 would be called " open." 
 
 When the link is shifted so that the valve block 5, Fig. 13, is 
 brought in line with the rod EP, it is in full forward gear, and 
 the motion of the valve is governed by the eccentric E. If the link 
 be shifted so that the block comes in line with the rod E'Q, it is in 
 
58 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 full backward gear, and the eccentric E' governs the motion of the 
 valve. When the block is midway between the two extreme posi- 
 tions the link is said to be in mid-gear, and the valve is influenced 
 equally by both eccentrics, with the result that it does not receive 
 sufficient motion to open the ports and the engine remains at rest. 
 
 VAL V ST M. 
 
 Should the link be shifted so that the valve block is brought to a 
 position intermediate between mid-gear and full forward gear, as 
 at 6r, the movement of the valve is influenced by both eccentrics, 
 but to such a large extent by eccentric E that the engine will con- 
 tinue to run ahead. The effect, however, of the slight influence of 
 the backing eccentric has been to decrease the travel of the valve, so 
 that all its operations will' be earlier than when working in full 
 gear and the cut-off will be, therefore, shorter. 
 
 The motion given to the valve-block when its position is inter- 
 mediate between mid and full gear is that due to a " virtual " eccen- 
 tric of less throw than that of the real eccentric. This question 
 has been investigated by designers, and it has been found that for 
 positions of the valve block intermediate between mid and full gear, 
 the centers of the virtual eccentrics lie in a parabolic curve. The 
 locus of these .centers will be represented with sufficient accuracy 
 by an arc of a circle passing through the centers E and E' of the 
 
THE VALVE AND ITS MOTION 59 
 
 eccentrics, and of a radius equal to the product of the length of 
 the eccentric rod and half the distance between the centers of 
 the eccentrics, divided by the distance between the eccentric rod 
 
 J7 1 P V" Tf 77" 
 
 pins in the link = - ~2PQ ^is ra -dius i g found to be equal 
 
 to OE, Figs. 13 and 14, and the arc EFE' has been drawn, concave 
 to the center of the shaft for open rods, and convex for crossed rods. 
 
 To find the virtual eccentric governing the motion of the valve 
 when the block is at G, divide the arc EFE' at F in the same ratio 
 that G divides PQ ; then CF is the throw, and DCF the angular 
 advance of the eccentric which is virtually actuating the valve. 
 The practical way of determining F is to produce PE and QE' 
 until they intersect at some point R in the center line; then EG 
 divides the arc EE' at F in the same ratio that G divides PQ. 
 
 The action of the link in providing a variable cut-off can now be 
 understood. 
 
 An investigation of the effects of the open and crossed rod con- 
 nections shows that with open rods the lead increases from full to 
 mid gear, and decreases with crossed rods; and that the longer the 
 rods are made and the shorter the link, the less change the lead will 
 undergo. It is the usual practice to make the open rod connection 
 for stationary engines. 
 
 82. The curvature of the Stephenson link occasions the variation 
 in the lead from full to mid gear, and this variation becomes greater 
 as the eccentric rods are shortened. Zeuner has shown analytically 
 in his treatise on Valve Gears that the length of the eccentric rod 
 should be the radius of the arc of the link in order that the leads 
 may be equal. It may be shown, too, that the motion of the valve 
 becomes more nearly harmonic as the eccentric rods and link are 
 lengthened. Practice, however, has about fixed the length of the 
 rods to about twelve times, and the length of the link to about four 
 times the throw of the eccentric. 
 
 83. The Stephenson link with open rod connection is well adapted 
 for locomotive practice. The adjustment should be such that 
 throwing the link into full gear, when starting the train, the cut-off 
 should be as long as from J to J of the stroke in order that, with a 
 partially open throttle, a uniform and moderate pressure be exerted 
 on the piston during a greater part of the stroke to overcome the 
 train friction without causing the driving wheels to slip. This, of 
 
60 THE ELEMENTS OF STEAM ENGINEERING 
 
 course, raises the terminal pressure and is the cause of the noisy 
 exhaust of a locomotive when starting a train. 
 
 The cranks of the locomotive being set at right angles the lead 
 at the start may be very small, but as the speed of the train increases 
 the throttle is gradually opened and then the cut-off is shortened 
 and the lead and compression increased by the process of raising the 
 link. The locomotive, being a high-speed engine, requires a con- 
 siderable amount of compression to overcome the inertia of the 
 reciprocating parts. 
 
 84. The crossed rod connection is commonly used for marine 
 engines, for then, with a decreasing lead from full to mid gear, the 
 engine will stop when the link is put at mid gear, which it would 
 not necessarily do with the rods open and the engine running with 
 light load. 
 
 85. The slotted Stephenson link is expensive to make and diffi- 
 cult to hang so that the centers of the eccentric rod pins and the 
 center of the valve block shall be in line and the block work easy 
 in the slot. These difficulties have led to the adoption for marine 
 purposes of the double-bar link, consisting of two solid bars of 
 rectangular section and of proper curvature. The bars are secured 
 together by distance pieces at the ends. The valve block is between 
 the bars and has flanges which bear upon the tops and bottoms of 
 the bars. The eccentric rods make a forked end connection to pins 
 projecting from the bars near the ends. 
 
 86. The Shifting Eccentric and Automatic Cut-Off. As stated 
 on page 36, quick running of an engine is one of the methods of de- 
 creasing the losses from condensation and re-evaporation, and it is 
 apparent that the higher the rotational speed consistent with safe 
 piston velocity the greater the efficiency of the steam action in the 
 cylinder; and since piston velocity is a factor of the power, any 
 increase in that direction admits of a decrease in the size of an 
 engine for a given power. Notwithstanding these advantages, the 
 introduction of the high-speed engine was long delayed. The belief 
 was prevalent that, as compared with one of lower speed, it was 
 more liable to frequent and serious accidents; and a practical ob- 
 jection to its introduction arose from the fact that the limit of 
 rotational speed was soon reached at which the " drop " cut-off 
 could be applied, and it became essentially necessary for the intro- 
 duction of the high-speed engine that some contrivance be devised 
 
THE VALVE AND ITS MOTION 61 
 
 by which the steam should have a positive and variable cut-off, auto- 
 matically controlled, in order that the speed should be constant 
 under the conditions of varying load and pressure. 
 
 The vast improvement in the character of the materials of con- 
 struction; the more perfect understanding of the stresses produced 
 in overcoming the inertia of the reciprocating parts; the skill of 
 the American workman all have made possible the production of 
 an engine of such perfect balance and adjustment, that there is no 
 longer a question of safety, even when running at the high rotational 
 speed that admits of a direct connection to the armature of a 
 dynamo-electric machine. 
 
 An inherent defect in the " fly-ball " governor precluded its 
 use in the solution of the cut-off problem, so effort was centered in 
 devising a mechanism in which the centrifugal action of revolving 
 weights, opposed by the varying tension of a spring, should control 
 the action of the valve to meet the exigency of a sudden variation 
 in the load on the engine or in the pressure of the steam. The 
 result was the production of what is known as the Shaft Governor, 
 now almost exclusively used with engines driving dynamo-electric 
 machines in producing electric light and power. 
 
 The control of the valve by the shaft governor is usually through 
 the medium of a shifting eccentric, and while the high-speed sta- 
 tionary engines of rival makers differ in detail, the underlying prin- 
 ciples in each are the same. 
 
 The governor is usually mounted within a pulley keyed on the 
 crank shaft, and its motion is, therefore, identical with that of the 
 shaft. The eccentric has two lugs diametrically opposite each 
 other, from one of which it is suspended in the center line of the 
 crank, and from the other the connection is made with the linkage 
 of the governor. The eccentric is slotted so that its center may 
 move across the shaft. 
 
 The linkage of the governor being properly proportioned, and 
 the weights adjusted for a definite speed, the action of the governor 
 is as follows : Any decrease in load, or increase in pressure, which 
 would tend to increase the speed of the engine and the centrifugal 
 force of the weights, is instantly met by the action of the springs, 
 which moves the center of the eccentric nearer the center of the 
 shaft, thus decreasing the throw and causing an earlier cut-off. An 
 increase of load, or decrease in pressure, causes an exactly reverse 
 action ; and thus the steam supply is adjusted to the load, and the 
 
62 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 engine kept at speed. When the engine is at rest the weights are 
 nearest the shaft and the eccentric is at its greatest throw, corres- 
 ponding to the full gear of the link motion : as the engine comes up 
 to speed the travel is reduced accordingly; and if a decreasing load 
 or an increasing pressure causes a still further reduction of travel, 
 the eccentric approaches its minimum throw corresponding to mid 
 gear of the link motion. As the eccentric moves from maximum to 
 minimum throw the effect on the lead may be investigated in a 
 manner similar to that employed for the link motion (Figs. 13 and 
 14). 
 
 87. Piston Valve. In order that the shaft governor shall be 
 sensitive there must be the minimum of work thrown on it, and this 
 
 can only be effected by having the least possible resistance offered 
 to the motion of the valve. This clearly points to the use of a bal- 
 anced valve, and the one largely used is a variety of the type known 
 as piston-valve. 
 
 Fig. 15 is a sectional view of the piston valve of the intermediate 
 cylinder of the triple-expansion engine in the mechanical laboratory 
 of the Baltimore Polytechnic Institute. The valve may be con- 
 ceived to be formed from the ordinary slide valve by curving .the 
 latter into cylindrical form. It is arranged to take steam at its 
 inside edges, though it could equally well be arranged to do so at 
 the outside edges. It will be seen that the valve is balanced. 
 
 When made for large marine engines the pistons of the valve are 
 fitted with springs to keep them steam tight, and the ports are cast 
 with diagonal bars to keep the rings from springing into the ports, 
 
THE VALVE AND ITS MOTION 63 
 
 and in order also to afford a continuous guide to the piston, so that 
 the rings may pass over the ports without catching on the edges. 
 
 The chief disadvantage of the piston valve is the difficulty in 
 keeping it steam-tight. 
 
 88. Radial Valve Gears. Efforts have been made from time to 
 time to supersede the use of the link motion and eccentrics in 
 actuating the valves of steam engines, and of these the most suc- 
 cessful have been the radial valve gears of Marshall and Joy. 
 
 In the Marshall gear the link is dispensed with and but one 
 eccentric is used. With the Joy gear the link and both eccentrics 
 are dispensed with, the valve motion being obtained from a point 
 in the connecting rod. 
 
 Eadial valve gears, when accurately proportioned, provide for any 
 degree of expansion with a uniform lead, but their parts are neces- 
 sarily heavy and difficult of correct adjustment, and the most recent 
 marine practice indicates a return to the old but reliable link motion 
 and eccentric. 
 
 PROBLEMS. 
 
 24. Diameter of shaft, 4"; diameter of eccentric, 10.5"; outside 
 diameter of strap, 12"; distance from outside to outside of lugs, 
 15.5"; distance from center line to center line of strap bolts, 13"; 
 width of lugs of each strap, 1.5"; depth of groove in strap to retain 
 eccentric, 0.25"; depth of enlargement of strap where rod is bolted, 
 6" at right angles to rod and 1.75" in line of rod; depth of rod at 
 strap end, 2.5", and at link end, 1.25"; diameter of boss at end of 
 rod, 2.5"; diameter of eye in boss, 1"; length of rod, 13 times .the 
 throw of the eccentric ; distance between stud bolts at strap end of 
 rod, 4.5"; travel of valve, 5"; angular advance, 38. Make a 
 sketch of the eccentric and rod, assuming the crank to be on the 
 head dead center. Scale, 1.5. 
 
 25. Throw of eccentric, 2"; length of eccentric rods, 24" ; length 
 of chord of link, 15"; angular advance of the eccentric, 38 ; stroke 
 of engine, 14". Make a skeleton sketch of the link motion, and find 
 the throw of the virtual eccentric when the link is half way between 
 full and mid gear, the connection of the rods being " open." Scale, 3. 
 
 Ans. Virtual eccentric, If". 
 
CHAPTER VI. 
 
 THE CONVERSION OF MOTION. ACTION OF THE CRANK 
 AND CONNECTING ROD. 
 
 89. If the point 'P 9 Fig. 16, moves in the plane of the paper about 
 C as a center, it will describe the circle RPB, and during the period 
 of describing the arc EP, the circular motion of P may be consid- 
 ered as compounded of two entirely independent movements at 
 right angles to each other that of R to Q in the line of the diam- 
 eter E8, and the other of Q to P. 
 
 Draw the diameter AB perpendicular to RS, and let P move with 
 .a uniform velocity. If we denote CP by r, and the angle PCR by 
 
 T 
 
 0, we may find the relation between the positions of P and Q as 
 follows : 
 
 RQ = RC QC = r r cos 6 r (1 cos 0) . 
 
 To find the position of Q when P has described f of the quad- 
 rant RA. 
 
 Here we have 9 = 60, and cos = $, consequently 
 
 That is, P describes f of the distance circumferentially from 
 R to A while Q moves \ the distance RC ; and since the motion of 
 P is uniform, it follows that the last half of RC will be passed over 
 by Q in one-half the time that it required to pass over the first 
 half. It is thus seen that while the motion of P is uniform, that 
 
THE CONVERSION or MOTION 65 
 
 of Q is variable; and if the relative positions of P and Q were 
 found for values of from to 180 it would be found that when 
 P is at R, the point Q is at rest, and as P continued its uniform 
 motion, the velocity of Q would gradually increase until the value 
 of reached 90, when the velocity of Q would be a maximum and 
 equal to that of P; as P moves from A to 8, Q would move ,from 
 C to 8 with diminishing velocity until its arrival at 8, when it 
 would again be at rest. During the movements just described, Q 
 is said to have a Simple Harmonic Motion. 
 
 When a point P moves uniformly in a circle, the perpendicular 
 PQ let fall at any instant to a fixed diameter R8, intersects the 
 diameter at a point Q, whose position changes by a Simple Har- 
 monic Motion. 
 
 To ascertain the relation existing between the velocities of Q and 
 P, draw the tangent PT, and from T let fall the perpendicular TD 
 upon RS. Suppose the velocity of P to be such that in a given 
 time it moves a distance PT. In the same time Q would move to 
 D, and we shall have : 
 
 Velocity of Q _ QD _ PE _ PQ _. . 
 
 velocity ^P~TT-"PT~TG~ 
 
 The sine of 6 varies from to 1, and we can therefore find at 
 any period of the motion how much the velocity of Q differs from 
 that of P. 
 
 90. It has been shown how the valve of a steam engine, actuated 
 by an eccentric, or its equivalent, admits steam alternately to the 
 two ends of the cylinder to drive the piston forward and backward 
 along a straight path of definite length called the stroke. This re- 
 ciprocating motion of the piston is converted into one of continuous 
 rotation of the shaft, through the agency of the mechanism of the 
 connecting rod and crank. In this mechanism the outer end of 
 the piston rod is connected to a cross-head, which is constrained by 
 guides to move in line with the axis of the cylinder. To a pin in 
 the cross-head the inner end of the connecting rod is so attached as 
 to permit an oscillating motion to that end of the rod. In a similar 
 manner the outer end of the connecting rod is connected to the 
 crank pin, and if the piston be driven forward or backward, the 
 crank will be caused to turn about the axis of the shaft by the push 
 or pull transmitted from the cross-head through the connecting rod 
 to the crank pin. 
 5 
 
66 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 In Fig. 17, let r denote the length CP of the crank, and c the 
 length PQ of the connecting-rod. If from R and 8 we lay off Ra 
 and Sb each equal to c, then ab will be the stroke of the cross-head, 
 and for purposes of discussion it is the stroke of the piston, the 
 latter being rigidly attached to the cross-head. Let CP be a posi- 
 tion of the crank, making an angle 6 with the center line Rb of the 
 engine. Then Q will be the corresponding position of the piston, 
 and is found by striking an arc from P as a center and with a 
 radius c. With Q as a center and the radius c, describe the arc Pi. 
 Then t8 is evidently equal to QB; and, since the diameter R8 of 
 the crank-pin circle is equal to the stroke of the piston, R8 may 
 equally well represent the piston stroke. Then t conveniently rep- 
 resents the position of the piston corresponding to the crank posi- 
 tion CP, and it is at once evident that the movement of the piston 
 
 is not the result of a simple harmonic motion. The angularity of 
 the connecting-rod introduces the inequality tv into the movement 
 of the piston, and this inequality gradually increases from its zero 
 value at the dead point 8 until the crank arrives at the position CT, 
 90 in advance of C8, when its value wC is a maximum. As the 
 crank continues its rotation the inequality gradually diminishes 
 until, at the dead point R, it is again zero. 
 
 An examination of Fig. 17 shows that when the crank is at the 
 mid-point of its revolution from 8 to R, the piston has traveled a 
 distance Cw beyond its mid-stroke; and it is seen that during the 
 first quarter of the revolution on the forward stroke the piston 
 travels a greater distance than during the second quarter. 
 
 This inequality can be avoided only by giving to the piston a 
 simple harmonic motion; but as that could be obtained only by 
 making the connecting rod infinitely long, the inequality must 
 always exist with the crank-connecting-rod mechanism. 
 
THE CONVERSION OF MOTION 67 
 
 The shorter the connecting-rod the greater, of course, will be the 
 inequality in the movement of the piston, and for this reason, if 
 for no other, the crank-connecting-rod ratio is made as small as 
 possible. In marine practice, where space limits the length of the 
 connecting-rod, this ratio is from 1 to 4 to -1 to 5, while for station- 
 ary engines it is not uncommonly as small as 1 to 7. 
 
 A further examination of Fig. 17 shows that during the return 
 stroke of the piston, conditions exactly the reverse of those of the 
 forward stroke obtain, as shown by the dotted lines. 
 
 It will be observed that the use of the connecting-rod of finite 
 length causes the cut-off to be considerably later on the forward 
 stroke than on the return stroke, and the reference to this question 
 on page 54 can now be understood. 
 
 The inequality in piston displacement for the same crank posi- 
 tions of the two strokes results in an unequal distribution of steam 
 in the cylinder and, therefore, an irregularity in the driving power 
 of the engine. The angularity of the connecting-rod is a disturb- 
 ing element in the consideration of every important dynamic ques- 
 tion of the steam engine, rendering more or less difficult the solution 
 of problems which would be otherwise simple. 
 
 91. With the aid of Fig. 17 we may find an expression for the 
 relative positions of the crank and piston at any instant. 
 
 Let a denote the angle made by the connecting-rod with the center 
 line of the engine, and denote by x' the distance the piston has 
 traveled on the forward stroke when the crank has the position CP. 
 
 x' = Cb (Gv + vQ) r -f- c (r cos + c cos a) 
 
 = r(l cos 0) + c c cos a (1) 
 
 We have h = r sin 0, and cos a = 
 
 therefore, c cos a = \/c 2 1? =. ^/c 2 r 2 sin 2 . 
 Substituting the value of c cos a in (1), we get : 
 
 x' = r(l cos 0) + c v' 6 ' 2 1* sin 
 
 = r(l-co80) +e ^i_ |y /i_^in'a 
 
68 THE ELEMENTS OF STEAM ENGINEERING 
 
 If x" denote the distance of the piston from the crank end of the 
 stroke for the same crank position, we shall have : 
 
 x" = abb = 2r-x' 
 
 - c [l - Jl - ^' 
 
 = 2r-r + rcostf 
 
 = r(l Hh eosfl) - .l-yl 
 In general terms we shall then have : 
 
 (A) 
 
 where the top sign must be taken if x is measured from the head 
 end of the stroke, and the lower sign if measured from the crank 
 end. 
 
 The position of the piston for any given crank position, or con- 
 versely, the crank angle for a given position of the piston, can be 
 found by equation (A). 
 
 The ratio r/c will become very small if raised to a power above 
 
 the square, so if in equation (A) the expression 1 s 
 
 be expanded by the binomial theorem, and the terms containing 
 higher powers of c than the square be rejected, no appreciable error 
 will result, and the formula will be more convenient for use. 
 Thus, 
 
 . , =,(!=,= COB fl). . : (B) 
 
 EXAMPLE I. Let c = 40", r 10", = 210 : find the position 
 of the piston, measured from either end of the stroke. 
 
 Here we have, ^ = | , cos 210 = cos 30 = O.SG6, and 
 
 sin 210 = sin 30 = . 
 
 Substituting in (B) we have: 
 
 x' = 10(1 + 0.866) + r V = 18 - 66 + 0.3125 == 18.9725 inches 
 when measured from the head end of the stroke; and 
 
THE CONVERSION OF MOTION 69 
 
 x" = 10(1 0.866) T V = 1.34 0.3125 = 1.0275 inches 
 when measured from the crank end of the stroke. 
 
 Equation (B) is of use in investigating the effect of the obliquity 
 of the connecting rod on the motion of the piston. For practical 
 purposes the relative positions of the crank and piston may be ob- 
 tained graphically. 
 
 It was shown on page 64 that r(l cos 0) would be the value of 
 x if the movement of the piston were the result of a simple har- 
 monic motion, but as that could be obtained only with a connecting 
 
 7* 2 sin 2 
 rod of infinite length, the term ~ i n equation (B) is the 
 
 measure of the error in piston displacement due to the angularity 
 of the connecting rod of finite length. 
 
 PEOBLEM. 
 
 26. Length of connecting-rod, 66"; stroke, 33". Find position 
 of the piston, measured from both ends of the stroke, when the 
 crank has passed through an angle of 135 on the head stroke. 
 
 Ans. x' = 29.1984". x" 3.8016". 
 
CHAPTEE VII. 
 STAGE EXPANSION ENGINES. 
 
 92. A stage expansion engine is one so designed that the steam, 
 after entrance into one cylinder and there partially expanded in the 
 performance of work,, is exhausted into a second, third, and not 
 infrequently into a fourth larger cylinder, for further expansion 
 and work before being finally exhausted into a condenser. 
 
 The term stage expansion is used to avoid the confusion which 
 sometimes arises from the use of compound. All stage expansion 
 engines are compound, but by common consent the term compound 
 is restricted to engines in which the expansion takes place in only 
 two cylinders, or in two stages, while the terms triple expansion and 
 quadruple expansion are applied to engines in which the expansion 
 takes place in three and four cylinders respectively. The first and 
 last cylinders in the order of the expansion are called, respectively, 
 the high-pressure and the low-pressure cylinders; any cylinder or 
 cylinders which intervene are known as intermediate cylinders. 
 
 93. There has been much discussion concerning the relative merits 
 of the simple and stage expansion types of engines, but the results 
 of comparative tests clearly indicate an economical advantage of 
 from 10 per cent to 20 per cent in favor of the compound system. 
 
 Perhaps the most convincing evidence of the advantage in the 
 use of the stage expansion engine is that it has almost entirely dis- 
 placed the simple engine for marine purposes, and has come largely 
 into use in mill engines, locomotives, and in stationary plants where 
 large powers are developed. 
 
 It has been shown that a given weight of steam in working from 
 a higher to a lower pressure is capable of doing a definite amount of 
 work, and while the number of cylinders through which the expan- 
 sion takes place can make no increase in this theoretical limit of 
 work, it does make a very material difference in the manner of its 
 performance. 
 
 A prerequisite to the use of the stage expansion system, and the 
 one to which its economy is solely due, is a high pressure of steam ; 
 
STAGE EXPANSION ENGINES 71 
 
 and since the terminal pressure desired is the same, whatever the 
 system, the higher the pressure the greater the rate of expansion. 
 
 It was pointed out on page 38 that the attainment of a high rate 
 of expansion in a single cylinder would be accompanied with such 
 a wide range of temperature as to occasion excessive initial conden- 
 sation and final re-evaporation. The consequent wide variation in 
 pressure would lead to very objectionable irregularity of rotational 
 effort on the crank-pin and excessive strains on the framing. 
 
 With a stage expansion engine this high rate of expansion would 
 be distributed through two or more cylinders, thus greatly reducing 
 the range of temperature and pressure in each, resulting in a reduc- 
 tion of condensation and the production of a more uniform turning 
 moment. 
 
 94. There are two classes of stage expansion engines, known as 
 the " continuous expansion " and the " receiver " types. The essen- 
 tial difference is that with the continuous expansion type the cranks 
 must be placed with each other or directly opposite each other on 
 the shaft, while with the receiver engine the cranks may be placed 
 at any angle with each other. 
 
 The continuous expansion engine is one of the two-stage type in 
 which the steam enters the low-pressure cylinder as fast as it is ex- 
 hausted from the high-pressure cylinder, a familiar example of 
 which is that of the tandem engine where the two pistons are on 
 one rod. 
 
 The receiver engine derives its name from the fact that originally 
 it was designed to have an intermediate reservoir to receive the 
 steam as it was exhausted from the high-pressure cylinder, and in 
 which it remained until the valve opened to admit it into the low- 
 pressure cylinder. The size of this receiver equalled that of the 
 high-pressure cylinder, and even larger. This separate reservoir 
 was found to be unnecessary, inasmuch as the low-pressure steam 
 chest, the exhaust pipe from one cylinder to the other, and the por- 
 tion of the high-pressure cylinder yet filled with steam when the 
 low-pressure valve opened, were found to furnish ample receiver 
 space. 
 
 95. The common, form of the compound, or two-stage expansion, 
 engine with cranks at right angles is shown in Fig. 18. When this 
 form of engine is designed for such large powers as require a low- 
 pressure cylinder greater than 100 inches in diameter, the three- 
 
72 THE ELEMENTS OF STEAM ENGINEERING 
 
 cylinder compound 'engine with two low-pressure cylinders, as shown 
 in Fig. 19, should be adopted. In this case the excessively large 
 low-pressure cylinder is replaced by two cylinders whose combined 
 volume equals that of the large cylinder they replace, and notwith- 
 
 standing the increased space occupied, and the greater number of 
 parts required with no increase in the power developed, this arrange- 
 ment is very advantageous, as the pieces are lighter and more easily 
 made, and the increase of the number of cranks from two to three 
 
 enables them to be placed at angles of 120 on the shaft, thus insur- 
 ing reduced straining and a steadier motion on the engine. With 
 this engine the work may be equally divided between the three cylin- 
 ders one-third being allotted to each by considerably increasing 
 
STAGE EXPANSION ENGINES 73 
 
 the receiver pressure, which has the effect of reducing the power 
 developed in the high-pressure cylinder and increasing that devel- 
 oped in each of the low-pressure cylinders. 
 
 The two-stage compound engine is advantageously employed 
 where the pressure varies from 80 to 120 Ibs. absolute. With pres- 
 sures varying from 120 Ibs. to 190 Ibs., the three-stage, or triple, 
 expansion type should be adopted, and it is the type which is very 
 largely used in the merchant marine and in vessels of war. For 
 pressures varying from 190 Ibs. to 275 Ibs., the four-stage, or quad- 
 ruple, expansion type is used. This type has not as yet come into 
 general use, and its employment is confined to small vessels where 
 lightness of the machinery admits of high rotational speeds. 
 
 The common arrangement of the three-cylinder triple-expansion 
 engine is shown in Fig. 20. The first receiver, R, leads from the 
 exhaust of the high-pressure cylinder to the valve chest of the inter- 
 mediate cylinder, and the second receiver, R', connects the inter- 
 mediate exhaust with the low-pressure valve chest. The cranks are 
 placed at angles of 120 with each other, and their sequence on the 
 shaft is a matter of dispute, but the order of high, low, intermediate, 
 seems to obtain the greatest favor from the fact that such arrange- 
 ment secures a more uniform receiver pressure. 
 
 96. The most serious inherent defect of the stage expansion sys- 
 tem is the loss occasioned by " drop." " Drop " means simply the 
 fall in pressure between the terminal pressure in one cylinder and 
 the initial pressure in the next succeeding cylinder in the expan- 
 
74 THE ELEMENTS OF STEAM ENGINEERING 
 
 sion. This is occasioned by friction in the exhaust passages and 
 pipes, and by the unrestricted expansion in the receiver. While it 
 is impossible to avoid entirely the loss from " drop/' measures should 
 be taken to prevent its being excessive. Earlier cut-offs and in- 
 creased compression in the cylinders succeeding the high-pressure 
 cylinder have a tendency to minimize " drop." 
 
 97. The mean pressure obtained in the stage expansion engine 
 with a given rate of expansion is less than would be obtained were 
 the expansion all to take place in one cylinder, and this difference is 
 due to " drop." The steam, when its expansion is completed, occu- 
 pies the low-pressure cylinder only and were there no loss from 
 " drop " the size of this cylinder for a given power would be the same 
 as that of a single-cylinder engine working with the same pressure 
 and the same .ratio of expansion. Owing, however, to the loss from 
 " drop " the size of the low-pressure cylinder must be made some- 
 what larger than would suffice for the single cylinder of a simple 
 engine of the same power. From this it is seen that the high and 
 intermediate cylinders add nothing to the power of the arrangement, 
 and that the low-pressure cylinder is the measure of the power. 
 
CHAPTEE VIII. 
 THE INDICATOR AND ITS DIAGRAM. 
 
 98. The steam engine indicator is an instrument devised to show 
 primarily the steam pressure within the cylinder at every point of 
 the stroke. 
 
 It consists essentially of a small steam cylinder containing a 
 piston whose vertical movement, when acted on by the pressure of 
 the steam beneath it, is opposed by a spiral spring of known tension. 
 
 The movement of the piston actuates a parallel motion consisting 
 of a system of light levers, a, point in which is constrained to move 
 in a straight line parallel to the motion of the piston. This parallel 
 point of the system carries a pencil which reproduces the vertical 
 movement of the indicator piston, magnified from 4 to 6 times by 
 means of the system of levers. In addition there is a cylinder, or 
 drum, to which a paper is attached, and which receives a forward 
 and backward motion of rotation on its own axis; the forward 
 motion by means of a string attached to the cross-head or other part 
 of the engine having a motion exactly coincident with that of the 
 engine piston, and the backward motion by means of the reaction 
 of a flat coiled spring which is attached to the base of the drum, 
 and which is drawn in tension by the forward motion of the drum. 
 
 It can readily be understood that when the instrument is in opera- 
 tion and the pencil pressed against the paper, the combination of 
 the vertical motion of the pencil and the horizontal motion of the 
 drum will cause a closed area to be described that will represent the 
 effective work done by the steam on the engine piston. 
 
 When communication is opened between the head end of the 
 engine cylinder and the lower end of the indicator cylinder, the 
 vertical motion of the indicator piston during the forward rotation 
 of the drum is that due to the pressure of the steam in the engine 
 cylinder during the forward stroke of the piston; and during the 
 backward rotation of the drum it is that due to the pressure on the 
 same side of the piston during the return stroke. The upper part 
 of the diagram is, therefore, a representation of the varying pres- 
 sure on the engine piston during its forward stroke, and the lower 
 part a representation of the back pressure opposed to the piston on 
 
76 THE ELEMENTS OF STEAM ENGINEERING 
 
 its return stroke. The length of the ordinate of the diagram at any 
 point is, therefore, when measured to the scale of the indicator 
 spring, the effective pressure on the piston at that point. 
 
 The closed area, or diagram, thus described represents the vary- 
 ing pressure on only one side of the engine piston during one revo- 
 lution. When communication is opened between the indicator and 
 the other end of the cylinder, a similar diagram will be obtained 
 which will represent the varying steam pressure on the other side 
 of the engine piston during a revolution. 
 
 99. Before giving an illustration of a diagram, several features 
 of the indicator will be noticed in order that the application of the 
 instrument to the steam engine may be understood. For a more 
 detailed description any one of the many publications on the sub- 
 ject may be consulted. 
 
 The pistons of the indicators in ordinary use are exactly one-half 
 square inch in area, and are fitted in the cylinders with great nicety, 
 requiring none but water packing. The parallel motion which 
 actuates the pencil receives its motion from the piston, a ball joint 
 connection being made with the piston rod. 
 
 The upper side of the piston has communication with the atmos- 
 phere by means of a small hole in the upper part of the cylinder. 
 
 The springs accompanying an indicator are numbered to cor- 
 respond to the number of pounds of pressure per square inch re- 
 quired to cause a vertical movement of exactly one inch to the 
 pencil. For example: a spring numbered 80 would require a 
 steam pressure of 80 pounds per square inch to cause a vertical 
 movement of one inch to the pencil. The tension of the spring to 
 be used in any particular case depends, of course, upon the height 
 of the diagram desired, which in turn depends upon the speed of 
 the engine. With a steam pressure of 160 Ibs. per square inch per 
 gauge, an 80 spring would give a diagram 2 inches in height above 
 the atmospheric line. In order to obtain satisfactory diagrams 
 from the modern high-speed engines, it has been found necessary 
 to use springs of high tension to limit the movement of the piston, 
 and thus obviate the loss from friction that would result from a 
 long and rapid movement. This provision for a small piston move- 
 ment is all the more necessary with the high pressures now in use, 
 in order that the pencil movement shall not be so great as to occa- 
 sion vibrations that will cause undulations in the line of the dia- 
 gram. 
 
THE INDICATOR AND ITS DIAGRAM 
 
 77 
 
 The height of the diagram must be sufficient for practical use; 
 and to provide for this, the parallel motion is designed to give to 
 the pencil a motion of from 4 to 6 times that of the piston, and it 
 is a matter of the greatest importance that the designed motion for 
 any instrument should not only insure the straight line motion of 
 
 K 
 
 the pencil, but should also maintain the fixed ratio of movement 
 between piston and pencil, for otherwise the vertical movement of 
 one inch to the pencil would not correspond in pounds pressure to 
 the number on the spring. 
 
 100. Figure 21 illustrates the construction of the Tabor indicator, 
 which consists of a steam cylinder A, containing a piston B and a 
 
78 THE ELEMENTS OF STEAM ENGINEERING 
 
 spring C. The spring is secured to the piston at one end and to 
 the cover D at the other end, and the pressure of the steam which 
 enters the indicator cylinder through the opening E compresses the 
 spring to a degree depending on the pressure. The movement of 
 the piston is transferred to the paper on the drum F, and multi- 
 plied five times by means of the arrangement of levers shown. The 
 most noticeable feature of this indicator is the means employed to 
 secure a straight-line movement of the pencil. A plate G contain- 
 ing a curved slot is fixed in an upright position, and a small roller 
 fixed to the pencil lever is fitted so as to roll freely in the slot. 
 The curve of the slot is so formed that it exactly neutralizes the 
 tendency which the pencil has of describing a circular arc in the 
 opposite direction, and the path of the pencil is a straight line when 
 the drum is not in motion. The pencil movement consists of three 
 pieces the pencil bar H, the back link K, and the piston rod link 
 L. The two links K and L are parallel to each other in all posi- 
 tions. The lower pivots of these links and the pencil point are 
 always in a straight line. The paper drum is attached by a cord 8 
 to a suitable reducing motion from the engine; the cord pulls the 
 drum around on its own axis with a motion corresponding to that 
 of the engine piston, and the return movement of the drum is ob- 
 tained by the internal coiled spring If. 
 
 The forward and backward motion of rotation on its axis of the 
 drum cylinder that carries the diagram must be exactly coincident 
 with the forward and backward motion of the engine piston, and 
 this is readily obtained by a suitable connection by means of cord 
 and pulley with the engine cross-head, the tension of the coiled 
 spring in the drum being sufficient to keep the cord stretched during 
 the return stroke of the engine piston. 
 
 The length of the diagram desirable, as well as the height, de- 
 pends largely upon the speed of the engine, slow speeds permitting 
 longer and higher cards. The ordinary length is from 3 to 4 inches, 
 so it is obviously necessary that the motion of the cross-head be 
 
 Length of stroke of engine 
 
 reduced in the ratio ^ -pr- ^ -5-^ : ^ . 
 
 Length of card desired 
 
 This arrangement may always be effected by a linkage suitable 
 to the construction of the engine, care being taken to avoid long 
 stretches of cord, and that the cord be led from its point of attach- 
 ment to the reducing motion in a direction parallel to the line of 
 motion of the cross-head, and then over a pulley in any direction 
 
THE INDICATOR AND ITS DIAGRAM 79 
 
 to the indicator; the object being to secure an equal motion of the 
 cord at the drum and at the point of its 'attachment to the reducing 
 motion. 
 
 The " lazy tongs " and other applications of the pantograph are 
 frequently used as reducing motions, and if carefully constructed 
 the results are accurate. 
 
 101. The proper interpretation of indicator diagrams reveals to 
 the engineer so many facts that are essentially necessary to secure 
 an economical and efficient performance of a steam engine, that a 
 careful study leading to a complete understanding of them is a mat- 
 ter of the greatest importance. 
 
 The principal value of the indicator diagram is that it furnishes 
 the means of ascertaining the mean effective pressure exerted on 
 the engine piston throughout the stroke, thus furnishing the factor 
 which permits the power of the engine to be determined; but an 
 inspection of the diagram reveals information concerning particu- 
 lars of the engine, of which the following named are the most im- 
 portant : 
 
 (1) Whether the valves are properly set; whether the admission 
 of steam is early or late; whether the initial pressure is below the 
 boiler pressure; and the degree to which the initial pressure is 
 maintained up to the point of cut-off. 
 
 (2) The point of the stroke at which the admission of steam to 
 the cylinder is cut off, and whether the cut-off is sharp or gradual. 
 
 (3) The point of the stroke at which release takes place, and 
 the pressure of the steam at that instant. 
 
 (4) The amount of back pressure opposed to the exhaust, the 
 point of the stroke at which the exhaust is closed, and the amount 
 of compression at the end of the stroke. 
 
 (5) Whether the steam ports are of adequate size, and whether 
 the valve or the piston leak. 
 
 (6) The approximate amount of steam consumed in a given 
 time, and a number of vital features concerning the balancing of 
 engines. 
 
 Figure 22 is a representation of the ideal indicator diagram show- 
 ing a proper distribution of steam in the cylinder, and it is the 
 duty of the engineer to set the valves of his engine to obtain, as 
 nearly as possible, such a diagram. 
 
80 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 102. The location of the indicator and the manner of its connec- 
 tion with the engine cylinder is governed largely by the construc- 
 tion of the engine to which it is to be applied ; but wherever its loca- 
 tion or the manner of its connection, the principles governing its 
 action are always the same. In the case of the ordinary horizontal 
 engine, the usual location is midway in a pipe connecting the two 
 ends of the cylinder. A three-way cock in the middle of the length 
 of this pipe has a projecting nipple to which the indicator is at- 
 tached. 
 
 103. When about to take a diagram the indicator should be 
 warmed, and the pipes blown through by alternately connecting the 
 indicator with the two ends of the cylinder by means of the three- 
 way cock. 
 
 Fig. 22 
 
 The length of the drum string being properly adjusted, it is 
 connected with the reducing motion, and if the pencil (HHHH) 
 be sharpened to a fine point, everything is in readiness to take the 
 diagram. 
 
 When the three-way cock is closed to both ends of the engine 
 cylinder, a small hole in the cock admits the pressure of the atmos- 
 phere under the indicator piston, and the piston being then in a 
 state of equilibrium the Atmospheric Line, A A', Fig. 22, may be 
 traced. 
 
 If now the cock be turned so as to admit steam from one end of 
 the engine cylinder into the indicator cylinder, the Admission Line 
 CD is instantly traced, and its height above the atmospheric line, 
 measured on the scale of the spring, represents the initial pressure 
 per gauge of the steam admitted to the cylinder. 
 
 The engine piston starting on its stroke, the Steam Line DEF 
 
THE INDICATOR AND ITS DIAGRAM 81 
 
 is traced during the time steam is being admitted into the engine 
 cylinder, or until cut-off takes place. 
 
 F is the point of cut-off, wnere the valve closes and prevents 
 any further admission of steam into the cylinder. The exact point 
 of cut-off is rather difficult to locate on the diagram, owing to the 
 fall of steam pressure due to the gradual closing of the port by the 
 valve, shown to a small extent in the fall in pressure from E to F. 
 
 FG is the expansion curve, representing the fall in the pressure 
 of the steam confined in the cylinder after cut-off, due to the ex- 
 pansion in volume in forcing the piston to the end of the stroke, 
 
 G is the Point of Release, or the point where the valve opens to 
 exhaust, thereby releasing the steam from the cylinder. 
 
 GH is the Exhaust Line,, which is traced in the interval between 
 release and the end of the stroke, the pressure falling rapidly to 
 that of the back pressure opposed to the exhaust. 
 
 HI is the Back Pressure Line, and indicates the pressure oppos- 
 ing the piston on the return stroke. In non-condensing engines 
 this line coincides with, or is slightly above, the atmospheric line, 
 and in condensing engines it is below the atmospheric line a dis- 
 tance corresponding to the vacuum obtained; but in either case it 
 is back pressure. Vacuum is expressed in inches of mercury, and 
 since one cubic inch of mercury weighs 0.491 pound, the inches of 
 vacuum multiplied by 0.491 will give the pressure equivalent to the 
 vacuum. 
 
 I is the point of exhaust closure, where compression begins. 
 This point, like those of cut-off and release, is difficult to locate 
 exactly. 
 
 1C is the Compression Curve, and represents the rise in pressure 
 due to the piston compressing the steam remaining in the cylinder 
 after the exhaust has closed. 
 
 For the study of the diagrams and for computations involving 
 pressures, it is necessary to locate the Vacuum Line 00', or line of 
 no pressure, from which all pressures should be measured in order 
 that they may be absolute. The location of this line is parallel to 
 the atmospheric line and at a distance from it equal to the pressure 
 of the atmosphere, the average value of which is 14.7 Ibs., measured 
 on the scale of the indicator spring. 
 
 Of equal importance is the establishment of the Clearance Line 
 OB, perpendicular to the atmospheric line, and at a distance from 
 the end of the diagram equal to the same percentage of the length 
 6 
 
82 THE ELEMENTS OF STEAM ENGINEERING 
 
 of the stroke that the volume of the clearance space of the cylinder 
 bears to the piston displacement. 
 
 104. Clearance, as we have already seen, is the volume of the 
 space between the valve face and the engine piston when the piston 
 is at the end of its stroke, and this clearance has to be filled with 
 steam at each end of the cylinder for every revolution of the engine. 
 This steam must come from the boiler, or from the steam left in 
 the cylinder by an early exhaust closure, or from both. The amount 
 of clearance varies in different styles of engines. In engines of 
 slow speed and long stroke, the variation is from 2 to 5 per cent; 
 for high-speed engines with short stroke, it may be as much as 10 
 per cent; while for marine engines, a clearance of 20 per cent is 
 not uncommon. 
 
 Engine builders usually measure the clearance volume very care- 
 fully, but in the absence of any data on the subject, an approxima- 
 tion of the clearance of an engine may be determined graphically 
 from the expansion or from the compression curves of a well-de- 
 fined indicator diagram. , 
 
 Thus in Fig. 22, select two points & and c in the compression 
 curve as far apart as possible, and through them draw an indefinite 
 line intersecting the vacuum line at a. From c lay off cd = ba, 
 and through d draw OB perpendicular to 00'. OB will be the 
 clearance line, and BD will represent the volume of the clearance 
 to the same scale that AA' represents the volume of the cylinder. 
 This construction, as well as the two that follow, assumes the curves 
 of expansion and compression to be hyperbolic, the asymptotes of 
 which are the vacuum and clearance lines. These assumptions are 
 sufficiently accurate for practical purposes, and as they greatly 
 simplify calculations, they are always made. 
 
 The method just given of locating the clearance line depends 
 upon the property of the hyperbola that, if any secant line be drawn, 
 the portions of the secant intercepted between the curve and its 
 asymptotes are equal. 
 
 Two other constructions are: (a) The line &c joining the two 
 selected points on the curve is a diagonal of a rectangle constructed 
 on the curve, the two sides of which are parallel to the vacuum line ; 
 if now a diagonal be drawn through the outer corners of the 
 rectangle, and produced until it intersects the vacuum line, the 
 point of intersection will be the center of the curve, or the inter- 
 
THE INDICATOR AND ITS DIAGRAM 83 
 
 section of the asymptotes; hence OB, perpendicular to 00', is the 
 clearance line. (&) If two points V and c' be taken on the expan- 
 sion curve, the construction is the same. 
 
 Though these constructions are mathematically correct when the 
 curves are hyperbolas, they give results by no means reliable when 
 applied to the curves of actual diagrams. 
 
 Clearance in an engine occasions a loss when the consumption 
 of steam per unit power is considered, but there are practical con- 
 siderations which render its existence highly desirable, if not neces- 
 sary. The clearance space between the piston and the cylinder 
 head when the piston is at the end of its stroke, gives space for the 
 variable amount of water which is always present in a cylinder and 
 doubtless prevents serious accidents which might otherwise occur. 
 
 Since the piston does not traverse the clearance space, the clear- 
 ance steam performs no initial work; that is, it does no work during 
 the period of admission, but after cut-off its effect is to raise the 
 pressure during expansion, and thus increase the area of the expan- 
 sion part of the diagram. If there were neither expansion nor 
 compression, the clearance steam would perform no work and 
 would be a total loss in the exhaust. On the other hand, if the ex- 
 pansion curve were carried down to the back pressure, and the com- 
 pression curve carried up to the initial pressure, there would be 
 absolutely no loss from clearance. 
 
 These latter conditions are rarely, if ever, realized in practice, 
 therefore there is always -a loss from clearance, and this loss is 
 greater as the clearance is proportionally large. 
 
 One effect of cushioning is to reduce the loss from waste of steam 
 in the clearance space ; but its most important effect is that it pro- 
 vides for smooth running of the engine by preventing shocks at the 
 end of the stroke. It is especially desirable that the diagram of a 
 high-speed single-cylinder engine should have its compression corner 
 well rounded. 
 
 105. The influence of clearance on the ratio of expansion is so 
 marked that it is essential that the amount of the clearance be 
 known and allowed for. 
 
 In any case the ratio of expansion is the volume of steam, includ- 
 ing clearance, at the end of the stroke, divided by the volume of 
 steam, including clearance, up to the point of cut-off. Denoting 
 the ratio of expansion by r, the initial volume, or volume up to 
 
84 THE ELEMENTS OF STEAM ENGINEERING 
 
 cut-off, including clearance, by v iy and the final volume, including 
 clearance, by v 2 , we shall have in all cases : 
 
 Keferring to Fig. 22, the ratio of expansion, if clearance were 
 
 PQI 
 
 neglected, would be ' ^ clearance were considered, the vol- 
 
 ume of steam that expanded after cut-off at F would be OP + PK, 
 and the final volume would be OP + PO' ; therefore the real ratio 
 
 OP + PO' 
 of expansion would be /i 4 PAT* Since ^ ne cross-sectional area 
 
 of the cylinder is uniform, the volume displaced by the piston at 
 any point is directly proportional to the fraction of the stroke 
 passed through up to that point, and volumes, therefore, may be 
 represented by the corresponding fractions of stroke. In like man- 
 ner the clearance volume, when divided by the cross-sectional area 
 of the cylinder, will be expressed in some fractional part of the 
 stroke. 
 
 If, therefore, we denote the full stroke of the piston by unity, it 
 may also represent the volume displaced by the piston in one 
 stroke, in which case the fraction of the stroke denoting the cut-off 
 will represent the volume displaced by the piston up to the point of 
 cut-off. If the fraction of the stroke performed by the piston up 
 to the point of cut-off be denoted by k, and if c be the fractional 
 part of the stroke representing the clearance, we shall have Tc -j- c 
 for our initial volume, and 1 + c for our final volume. The ratio 
 
 of expansion will then be r =. = --, 
 
 To illustrate the effect of clearance on the ratio of expansion, 
 and to show the application of the formula just given, we will con- 
 sider a numerical example. 
 
 EXAMPLE I. Diameter of cylinder, 42"; stroke, 4'; cut-off, 
 stroke ; clearance, 8 per cent. Find the ratio of expansion. 
 Here we have : 
 
 Stroke volume displaced by piston = - T^ X 144 - = 38.5 cu. ft. 
 
 Volume of clearance = 38.5 X 0.08 = 3.08 cu. ft. 
 V 2 = 38.5 + 3.08 = 41.58 cu. ft. 
 
THE INDICATOR AND ITS DIAGRAM 85 
 
 oo t 
 
 Volume displaced by piston up to cut-off = -^ = 9.625 cu. ft. 
 v 1 = 9.625 + 3.08 = 12.705 cu. ft. 
 
 Had the clearance been neglected, we should have had r = 
 
 38 5 
 g-g^g = 4 , a result which would affect the calculated mean pres- 
 
 sure. 
 
 The work would have been much simplified by the use of the 
 
 1 4- c 
 
 formula r = j - ; for, by substitution,, we would have 
 K + c 
 
 as before - 
 
 106. The question of expansion in the compound, triple, or any 
 other stage expansion system, is not different from that in the 
 single-cylinder engine, so far as the total rate of expansion is con- 
 cerned. To know the volumes of the high- and low-pressure cylin- 
 ders and their clearances, together with the point of cut-off in the 
 high-pressure cylinder is all that is necessary to know in order to 
 determine the total rate of expansion. The intervening cylinders, 
 receiver spaces, and point of cut-off in the low-pressure cylinder 
 have nothing to do with the question. 
 
 As in the simple engine, the total ratio of expansion in any stage 
 expansion engine is the ratio of the final volume occupied by the 
 
 steam to the initial volume, or r = , in which v 2 is the volume 
 
 of the low-pressure cylinder and its clearance. v is the initial vol- 
 ume, or the volume up to cut-off, including clearance, of the high- 
 pressure cylinder. 
 
 Neglecting, for the present, the question of compression and 
 cylinder condensation, the initial volume of steam must eventually 
 fill the whole of the low-pressure cylinder and its clearance space 
 at one end; therefore, to obtain the total ratio of expansion, we 
 have simply to divide this final volume v 2 by the initial volume v^. 
 This may be better understood if illustrated by a numerical example. 
 
 EXAMPLE II. A two-cylinder compound engine has a high-pres- 
 sure cylinder 26" in diameter, a low-pressure cylinder 50" in diam- 
 eter, and a stroke of 30". Initial absolute pressure of steam, 115 Ibs. 
 per sq. inch. Cut-off in high-pressure cylinder at 0.5 stroke, and 
 
86 THE ELEMENTS OF STEAM ENGINEERING 
 
 in low-pressure cylinder at 0.8 stroke. Clearance of each cylinder 
 5 per cent. It is required to find the total ratio of expansion. 
 
 The area of a 26" piston is 530.93 sq. inches, and of a 50" piston, 
 1963.5 sq. inches. The clearance of 5 per cent is equivalent to the 
 addition of 30 X 0.05 = 1.5 inches to the stroke. Neglecting com- 
 pression >and cylinder condensation, we shall have 
 
 530.93 X 16.5 _ - 07 
 1728 
 
 cubic feet of steam supplied the high-pressure cylinder per stroke, 
 which is the initial volume v x . 
 
 All of this steam must finally occupy the low-pressure cylinder 
 and its clearance at one end, hence its final volume v 2 will be 
 
 1963 5 X 31 5 
 
 '-iiyou = 35.793 cu. ft. The ratio of expansion is then 
 
 r = !!? = 35 ; 793 = 
 v l 5.07 
 
 No account has been taken of the volume of the high-pressure 
 cylinder filled with steam after cut-off, nor of that of the clearance 
 space between the cylinders, nor has the cut-off in the low-pressure 
 cylinder been considered, for the reason that they do not enter into 
 the question at all. 
 
 The effect of the receiver is to make the initial pressure lower in 
 the low-pressure cylinder than it would have been had the exhaust 
 been direct from the high-pressure to the low-pressure cylinder, 
 and this reduction is due to the unrestricted expansion of the steam 
 when it enters the receiver space. The receiver only plays the part 
 of a large clearance space. 
 
 A low-pressure cut-off will increase the receiver pressure and 
 therefore the initial pressure in the low-pressure cylinder, with a 
 consequent increase in power of the low-pressure engine. The in- 
 crease in receiver pressure is an increase in the back pressure on 
 the high-pressure piston, and occasions a decrease in the work of 
 the high-pressure engine. 
 
 So it is seen that the use of a low-pressure cut-off is to equalize 
 the power of the two engines, and has nothing to do with the total 
 rate of expansion. 
 
 This may be made more clear by considering the weight of steam 
 used per stroke in *bhe above example. 
 
 The specific volume of steam at 115 Ibs. pressure is 3.839 cu. feet, 
 
THE INDICATOR AND ITS DIAGRAM 87 
 
 5 07 
 consequently OOQ = 1.3207 Ibs. steam enters the high-pressure 
 
 cylinder per stroke. At the commencement of the return stroke of 
 the high-pressure piston the delivery of this steam into the receiver 
 will begin, and it will have been delivered when the return stroke is 
 completed. Its withdrawal from the receiver will begin with the 
 commencement of the stroke of the low-pressure piston, and must 
 all be drawn out by the time the low-pressure cut-off takes place. 
 If it were not all drawn out, the pressure in the receiver would in- 
 crease, as the engine ran, until no exhaust from the high-pressure 
 cylinder could take place. If more were drawn off, a point would 
 soon be reached at which a vacuum would exist in the receiver. As 
 both these assumptions are absurdly impossible, it follows that the 
 receiver has nothing to do with the total ratio of expansion. 
 
 The accuracy of our conclusions may be checked by a further 
 consideration of the numerical example. 
 
 We have found that 1.3&07 Ibs. steam are used per stroke, and 
 that the total ratio of expansion is 7.06. If these figures are cor- 
 rect, the calculated volume of steam found in the low-pressure cyl- 
 inder and clearance at the end of the stroke should be equal to the 
 combined volume of the low-pressure cylinder and clearance. 
 
 The relation between the pressure and volume of saturated steam 
 is pv& C:, therefore, p t v^ =p 2 v^*, in which p^ is the initial 
 pressure and v the initial volume in the high-pressure cylinder, 
 and p 2 the final pressure and v 2 the final volume in the low-pressure 
 cylinder. We have: 
 
 7.06 log 0.84880 Hog 9.92881 10 
 17 log 1.23045 
 16 colog 8.79588 10 
 
 (7.06)** log 0.90186 Hog 9.95514 10 
 115 log 2.06070 
 
 p 2 14.417 log 1.15884 
 
 The volume of one pound of steam at pressure of 14.417 Ibs. per 
 sq. inch is 26.82 cu. ft., and since there are 1.3207 Ibs., the total 
 volume it occupies is 1.3207 X 26.82 = 35.42 cu. ft. Allowing for 
 abbreviated decimals, this corresponds with the calculated volume 
 of the low-pressure cylinder and clearance, as it should. 
 
THE ELEMENTS OF STEAM ENGINEERING 
 
 To apply the formula, r = ^ * ^, in finding the total ratio of 
 
 expansion in a stage expansion engine, we must substitute for unity 
 in the numerator the volumetric ratio of the low-pressure cylinder to 
 the high-pressure cylinder, and for c the clearance percentage of this 
 ratio. Denoting the ratio of the low-pressure cylinder to the high- 
 pressure cylinder by ^, we would then have: 
 
 In the above example f == 3.698, hence r= 
 
 ., 
 
 (/CD) 0.5 + 0.05 
 
 = 7.06, as before. 
 
 In our calculations the effect of compression has been neglected, 
 but the only way this could affect the question would be to reduce 
 the quantity of steam withdrawn from the boiler at each stroke and 
 this may be regarded as virtually increasing slightly the ratio of 
 expansion, because a less weight of fresh steam would be used each 
 stroke. 
 
 The effect of cylinder condensation has also been neglected, but 
 this, too, would occasion a virtual augmentation of the ratio of ex- 
 pansion, because a smaller weight of steam than that delivered to 
 the high-pressure cylinder would be found in the low-pressure cylin- 
 der at the end of the stroke. 
 
 In either case, the steam must finally fill the low-pressure 
 cylinder and clearance whatever the weight. 
 
 The effect of cutting off in the low-pressure cylinder is to main- 
 tain the average pressure in the receiver higher than it would be if 
 there were no cutting off of the steam, -and therefore to reduce the 
 power developed in the high-pressure cylinder by increasing its back 
 pressure. Whether the steam is, or is not, cut off in the low-pres- 
 sure cylinder, the same weight of steam must find its way into that 
 cylinder at each stroke ; and if, by means of the cut-off, a less space 
 be allowed for the reception of the steam, the pressure will increase 
 accordingly. 
 
 107. Wire-drawing. Wire-drawing is a technical name for the 
 reduction in pressure of steam resulting from its passage through 
 contracted areas, and its effect is to reduce the efficiency of the 
 steam. 
 
 The initial pressure on the piston is always appreciably less than 
 
THE INDICATOR AND ITS DIAGRAM 89 
 
 the boiler pressure, and this difference is to be attributed to wire- 
 drawing in the steam pipe. The gradual falling of the steam line, 
 as the piston advances to the point of cut-off, is wire-drawing due 
 to the ports being of insufficient area to admit steam fast enough to 
 maintain a full pressure behind the piston, and also to the action 
 of the slide valve in closing the port gradually. This movement of 
 the slide valve prevents a sharp cut-off, and the exact point at which 
 the port closes is, therefore, difficult to locate on the diagram. To 
 make the ports, and therefore the valve, sufficiently large to avoid 
 wire-drawing would be a matter of more serious consequence than 
 the defect itself. 
 
 108. In order that the exhaust steam may flow from the cylinder 
 of a condensing engine to the condenser, or into the atmosphere 
 from the cylinder of a non-condensing engine, the actual back pres- 
 sure must be greater than the pressure in the condenser in the one 
 case, or greater than the atmospheric pressure in the other, and this 
 excess of pressure depends largely upon the freedom of passage for 
 the exhaust steam from the cylinder to the condenser or to the 
 atmosphere. 
 
 The release of the steam at from -^ to y 1 ^ of the stroke from the 
 end assists materially in the freedom of the exhaust; this is neces- 
 sary with a condensing engine to insure a nearly complete vacuum 
 when the piston starts on its return stroke, and with a non-con- 
 densing engine it enables the exhaust steam to begin its flow into 
 the atmosphere before the return stroke commences. 
 
 In practice the exhaust is closed at some point E, Fig. 23, before 
 the piston arrives at the end of its stroke, and the steam entrapped 
 in the cylinder is compressed by the advancing piston, its pressure 
 rising to some point F, when the valve opens to lead, the pressure 
 rising suddenly to A, and a new stroke commences. 
 
 109. Figures 23 and 24 represent diagrams from a non-con- 
 densing and from a condensing engine, respectively. 
 
 The diagrams taken from an engine of proper design and adjust- 
 ment do not differ materially from the theoretical diagram, but it 
 requires careful study and discriminating judgment to make proper 
 use of the information presented by them, a fact which may be ap- 
 preciated when it is considered that the only 'absolute information 
 a diagram gives is the varying pressure of the steam in the cylinder. 
 
90 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 The full line diagram of Fig. 25 would indicate a very satisfac- 
 tory performance of the engine from which it was taken. The 
 dotted lines illustrate some possible defects of an engine which 
 would readily be detected by the indicator. Thus, the fall of the 
 steam line to rib would indicate wire-drawing; the line cd would 
 
 Perfect 
 
 show that the " release " was too early, and the line ef that it was 
 too- late; the inclining of the admission line to the left, as at ga, 
 would show the lead to be too great, and its inclination to the right, 
 as at hi, would show insufficient lead. 
 
 24 
 
 110. Mean Effective Pressure from Diagram. There are two 
 methods of obtaining the mean effective pressure from a diagram : 
 (a) By the use of the planimeter. (&) By the measurement of 
 ordinates. 
 
 The planimeter 'is an ingenious instrument by means of which 
 the mean pressure may be obtained accurately and more quickly 
 
THE INDICATOR AND ITS DIAGRAM 
 
 91 
 
 than by the method of ordinates, though the latter method is the 
 more common. 
 
 To obtain the mean effective pressure by the method of ordinates, 
 erect perpendiculars to the atmospheric line touching the extreme 
 ends of the diagrams. Divide the space between these perpendicu- 
 lars into ten equal parts and erect perpendiculars to the atmospheric 
 
 line at the middle points of these divisions. The first and last of 
 these new divisions will be ^V the length of the diagram from the 
 ends, and the common interval between them -will be T V of the 
 length of the diagram. One-tenth the sum of the lengths of the 
 ordinates will be the length of the mean ordinate, or the mean effec- 
 tive pressure in pounds per square inch on the piston, measured to 
 the scale of the indicator spring. 
 
 Sccr/e /OO. 
 
 me.p. - 
 
 Jnch. 
 
 The diagrams of Fig. 26 were taken from the high-speed engine 
 in the mechanical laboratory of the Baltimore Polytechnic Insti- 
 tute. The sum of the lengths of the ordinates of the diagrams from 
 the head and crank ends of the cylinder are 3.2 and 3.25 inches 
 respectively, and the scale of the spring is 100 pounds to the inch. 
 The mean effective pressure is, therefore, for one revolution, 
 
92 THE ELEMENTS OF STEAM ENGINEERING 
 
 The diagrams from the two ends of the cylinder should be taken 
 simultaneously if two indicators are used, or one immediately after 
 the other if only one be used. 
 
 111. Having found from the diagram the mean pressure in 
 pounds per square inch acting on the piston throughout one revolu- 
 tion, the mean total pressure in pounds will be the product of this 
 mean pressure and the area of the piston in square inches ; and the 
 work per revolution in foot-pounds will be found by multiplying 
 this pressure by twice the length of the stroke in feet ; and the work 
 per minute will be found by multiplying the work per revolution by 
 the number of revolutions per minute. This last product divided 
 by 33,000 will express the work in horse-power. 
 
 The mean pressure having been found from the indicator dia- 
 gram, the horse-power is called Indicated Horse-Power, usually 
 written I. H. P., and expresses the gross work done on the piston 
 by the steam. It is usual to express the indicated horse-power by 
 an easily remembered formula, as follows : 
 
 y TT p pLaN 
 ~ 33,000' 
 
 in which p = mean effective pressure in pounds per sq. inch on the 
 
 piston. 
 
 L = length of stroke in feet. 
 a = area of piston in square inches. 
 N = number of strokes per minute = twice the number 
 
 of revolutions per minute. 
 
 It will be observed that the formula for I. H. P. is an algebraic 
 expression for the definition that work is " pressure into its path," 
 or " force into its distance." 
 
 For p X & is equal to the pressure, or force, in pounds ; L X N 
 is the path, or distance, in feet moved through in a minute; there- 
 fore, the product (p X a) (L X N) is given in foot-pounds per 
 minute; and as 33,000 foot-pounds of work per minute is the 
 
 measure of one horse-power it becomes evident that |^-QQQ is the 
 measure of the work of an engine in terms of I. H. P. 
 
 The I. H. P. formula also expresses that the external work of 
 steam is equal to PV, in which P is the pressure per square foot 
 
THE INDICATOR AND ITS DIAGRAM 93 
 
 and V the volume in cubic feet displaced by the piston in a minute. 
 
 _ PV 
 33,000 33,000 = 33,000 " 
 
 Only the net area of the piston must be used in calculating the 
 horse-power of an engine ; and as the steam on one side of the piston 
 acts only on an area equal to that of the piston diminished by the 
 cross-sectional area of the piston rod, this latter area must be de- 
 ducted from that side. 
 
 If a r denotes the cross-sectional area of the piston rod, then the 
 mean net area of piston to be used will be 
 
 In the stage expansion engine, the I. H. P. for each cylinder is 
 obtained, and their sum is the total power of the engine. 
 
 Should a diagram be looped as in Fig. 27, the area adc represents 
 negative work, and in obtaining the mean pressure from such a 
 diagram, the lengths of the ordinates included in the loop should 
 be subtracted from the total length of those in the area abe. 
 
 A loop like that in Fig. 27 would be occasioned by excessive 
 expansion. At the point a where the expansion curve crosses the 
 back pressure line, it is evident that the pressures on each side of the 
 piston are equal, and a cut-off which would occasion an expansion 
 so excessive as to reduce the steam pressure to a point below the 
 back pressure opposed to the piston would manifestly be too early. 
 The theoretical limit of expansion is such that the terminal pressure 
 should be just equal to the back pressure. In practice the terminal 
 pressure is in excess of this, varying from 24 to 28 pounds in non- 
 condensing engines, and from 9 to 12 pounds in condensing engines. 
 
94 THE ELEMENTS OF STEAM ENGINEEKING 
 
 In actual practice, a loop in a diagram would very likely indicate 
 that the engine was underloaded. 
 
 112. It has already been shown that mechanical work is produced 
 by a force working through a distance. In the case of any gas 
 working within a cylinder against a piston, the force will be the 
 mean value of the pressure of the gas multiplied by the area of the 
 piston, and the distance will be the stroke of the piston. In order 
 that the work may be expressed in foot-pounds, the force must be 
 expressed in pounds and the distance in feet. 
 
 It is seen, then, that the area of an indicator diagram is the 
 measure of the work performed in a cylinder; for its area is the 
 product of its mean ordinate and its length, factors representing the 
 mean effective pressure on the piston in pounds per square inch and 
 the distance moved through in feet respectively. 
 
 PROBLEMS, 
 
 27. Diameter of H. P. cylinder, 15"; diameter of L. P. cylinder, 
 30"; cut-off in H. P. cylinder, 0.25 stroke; clearance of each cylin- 
 der, 9 per cent: find total ratio of expansion. Ans. 12.82. 
 
 28. Diameter of H. P. cylinder, 42",- and of L. P. cylinder, 78"; 
 cut-off in H. P. cylinder, 0.3 stroke; clearance of H. P. cylinder, 
 10 per cent, and of L. P. cylinder 12 per cent: find total ratio of 
 expansion. Ans. 9.6575. 
 
 29. Eatio of expansion, 3.2727; cut-off, 0.25 stroke: find the 
 clearance. Ans. 8 per cent. 
 
 30. Eatio of expansion, 3.2727; clearance, 8 per cent: find the 
 point of cut-off. Ans. 0.25. 
 
 31. Diameter of cylinder, 9"; stroke, 10"; cut-off, 0.28 stroke; 
 clearance, 6.5 per cent. Initial pressure of steam, 95 Ibs. per square 
 inch absolute, the specific volume of which is 4.57 cubic feet. Find 
 ratio of expansion and terminal pressure, assuming the steam used 
 to be saturated. Prove the accuracy of the results by showing that 
 the calculated volume of the steam in the cylinder and clearance at 
 the end of the stroke is equal to the combined volume of the cylinder 
 and clearance. The specific volume of steam at the terminal pres- 
 sure is 14.04 cubic feet. Ans. r = 3.087. p 2 = 28.68. 
 
THE INDICATOR AND ITS DIAGRAM 95 
 
 32. A two-cylinder compound engine, the dimensions of which 
 are 15", 30", X 18", has <a clearance of 5 per cent in each cylinder. 
 The initial absolute pressure is 100 Ibs., and the cut-off is at f of 
 the stroke. Find the total ratio of expansion and the terminal pres- 
 sure in the low-pressure cylinder. Prove the accuracy of the work 
 by showing that the cubic feet of steam in the low-pressure cylinder 
 and clearance at the end of the stroke equals the volume of the low- 
 pressure cylinder and its clearance. The steam is assumed to be 
 saturated, and the specific volume may be found by the formula 
 = 482. Ans. r = 9.882. p 2 = 8.768. 
 
CHAPTER IX. 
 
 MEAN PRESSURE OF A GAS EXPANDING WITHIN A 
 CYLINDER, 
 
 113. With the understanding of work and of the indicator dia- 
 gram obtained from Chapter VIII, it will now be shown how the 
 mean pressure of any gas working within a cylinder may be calcu- 
 lated, the gas expanding according to any law. 
 
 Referring to Fig. 28, 
 
 L 
 
 . 28 
 
 Pi = initial pressure in pounds per square inch. 
 
 p 2 = terminal pressure. 
 
 p 3 = back pressure. 
 
 p m mean absolute pressure. 
 
 p e = mean effective pressure = p m p y 
 
 v^ = the initial volume, and v 2 = the final volume, of the gas. 
 
 r -^ * = ratio of expansion, and v 2 = rv^. 
 
 The area of the diagram represents the gross work performed by 
 the volume of gas v of initial pressure p in driving a piston in a 
 cylinder through a stroke AE against a back pressure p a , the vol- 
 ume v of pressure p^ expanding to the volume v 2 and pressure p 2 . 
 
MEAN PRESSURE OF GAS EXPANDING WITHIN CYLINDER 97 
 
 If p m denotes the mean absolute pressure of the gas, then 
 
 Gross Work = p m v 2 
 whence 
 
 _ Gross Work Area of Diagram 
 
 The area of the rectangle ABCF is p^v^, and if the equation of the 
 expansion curve be pv n = C, a constant, then the expansion area 
 
 FCDE is I pdv. 
 
 1) V n 
 
 But pv n p^v^ = p 2 v 2 n , whence p r 
 
 Substituting this value of p in the integral expression we get 
 
 Expansion area FCDE of diagram = / - n 
 
 Jvi v 
 
 Therefore, 
 
 Whole area of diagram p l v l -f pp* I 2 ~ 
 
 1 1 J ^ 
 
 /rUl 
 
 Pi v i + P\v* I v~"dv, since v. 2 rv l . 
 
 */Vl 
 
 By integration we get 
 
 Whole area = p^ 4- pp* | - X~v~ n + l 
 
 n + r ~ " 
 
 1 n 
 
 whence 
 
 Area of Diagram _ piVi(r l ~* w) _ p l (r l ~ n n~) 
 rv l "rt^CU^n)" r(l - n)~ 
 
 The value of the expansion area FCDE, or of the work done during 
 
 expansion, is often expressed as^ 1 - 1 _ ^ 2 2 . This value is imme- 
 
 diately deducible from equation (C), thus: 
 
 I" (TV V ~ " _ v l ~ " ~i 
 Expansion area = p^i\ n i _ l 
 
 n I n 1 n 1 
 
 p.v, 
 n l 
 
9s THE ELEMENTS OF STEAM ENGINEERING 
 
 114. Equation (D) is the general expression for the value of the 
 mean pressure, whatever the value of n. 
 
 In reality, n is the ratio of the specific heats of the gas at con- 
 stant pressure and constant volume, and has a yalue between unity 
 and 1.405. 
 
 If steam be the gas under consideration, there are three cases : 
 
 1. Adiabatic Expansion. When steam expands in a non-con- 
 ducting and non-radiating cylinder, so that it neither receives nor 
 parts with heat during the expansion, except that due to the work 
 performed, the expansion curve is adiabatic (without transference). 
 The pressure falls as a result of the increase in volume and in the 
 performance of the external work. 
 
 The law governing such expansion is that the pressure varies 
 nearly as the reciprocal of the ninth root of the tenth power of the 
 
 volume; or, pec -TO ; that is, pv } * = C . 
 
 This case is useful only in theoretical investigations, since non- 
 conducting and non-radiating cylinders do not exist in practice. 
 
 To find the expression for the mean pressure with adiabatic ex- 
 pansion, let n = ^- in equation (D), and we shall have: 
 
 ^(ri-V - W _ Pi(r~* - V) #(10 - 9r-*) 
 Pm ~ r(l - 1 /) _r r 
 
 9 
 
 2. Saturated Steam Expansion. When saturated steam expands 
 in a cylinder, doing work, but receiving from a jacket or other ex- 
 ternal source sufficient heat to prevent liquefaction, the pressure at 
 all points of the stroke is that due to the volume and temperature of 
 saturated steam. 
 
 According to Eankine's experiments the law governing the expan- 
 
 sion of saturated steam is that p <* ^- ; that is, pv & = C . 
 Letting n = ^J in equation (D), we have : 
 
 ) _ pl ( r - A -ffi _jP l (17- 16r-A) 
 
 3. Hyperbolic, or Isothermal, Expansion. When damp steam 
 expands in a cylinder, doing work meanwhile but receiving from 
 some external source an amount of heat equivalent to the external 
 work performed, the expansion curve is that of the rectangular 
 
MEAN PRESSURE OF GAS EXPANDING WITHIN CYLINDER 99 
 
 hyperbola, and the pressure varies inversely as the volume, or 
 P*\', that is, pv C. 
 
 This is a very common case in practice, and to facilitate calcula- 
 tions, the expansion curve of the indicator diagram is assumed to 
 be hyperbolic. 
 
 If, in this case, we substitute directly in equation (D) the value 
 n = 1, the result is $ , or indeterminate; but if we evaluate by 
 differentiation before substituting the value of n, we get the correct 
 result, thus : 
 
 _ ffi( rl ~ n ? _ d[p l (r' t - H n)'] __ ji( r l ~ " log e rdn dn ) 
 r(l - n) ~d\r(\~-ny\ 
 
 1 + r 1 log.r 
 
 ~ ~ 
 
 Substituting now the value of n, we have : 
 
 , 1 + log e r 
 P- = Pi X - -JJT*- 
 
 Or if, in the expression: 
 
 Whole area of diagram = p l v l + pj" I ^, we let n = 1, we have 
 
 9/v l v 
 
 f* rv ^dv r ~\ rc \ 
 
 Area of diagram = p^\ + pp" I ' = p l v l -f ppA log e ?; 
 
 t/t'! V JUl 
 
 = ^^(1 -f- log e r), whence 
 
 __ Area of diagram _p l v l (\ + log e r) _pi(l + log.,0 
 rvj rvj r 
 
 115. To construct the curves representing these three cases of ex- 
 pansion, let the vertical and horizontal lines OA and OF, Fig. 29, 
 be the lines of pressures and volumes respectively. 
 
 Suppose steam of 200 pounds absolute pressure per square inch 
 to be admitted into a cylinder having a volume of 10 cubic feet, and 
 suppose the piston to be driven a distance AB, equal to one-quarter 
 stroke, before the supply of steam is cut off. 
 
 There will be -^ = 2.5 cubic feet of steam confined in the cylin- 
 der, which will, by its expansive force, drive the piston to the end 
 of the stroke. 
 
 We will suppose the conditions of the expansion to be hyperbolic, 
 so that the pressures will be inversely as the volumes, according to 
 
100 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 Boyle's law. When the steam is expanded to 5 cubic feet, or twice 
 its initial volume, the pressure will have fallen to one-half the initial 
 
 pressure = -~- = 100 Ibs. = MN. When the expansion has been 
 
 carried to 7.5 cubic feet, or to three times the initial volume, the 
 
 20 
 pressure will be reduced to one-third the initial pressure = -=- 
 
 66| Ibs. = GK. When, finally, the piston reaches the end of the 
 
 cubic f&ei 
 
 FIG. 2g. 
 
 stroke, the volume of steam will be 10 cubic feet, or four times the 
 initial volume, and the pressure, therefore, will have fallen to 
 
 900 
 
 ~ = 50 Ibs. = FC. If the points B, N, K, and C be joined, the 
 curve will be that of the equilateral, or rectangular, hyperbola. 
 
 The rectangles OB, ON, OK, and OC, represent the conditions 
 of the steam as to volume and pressure when the piston is at the 
 positions JB, MN, GK, and FC respectively, and these rectangles 
 
MEAN PRESSURE OF GAS EXPANDING WITHIN CYLINDER 101 
 
 are all equal in area; for 200 X 2.5 = 100 X 5 = 66f X 7.5 
 = 50 X 10 = 500. 
 
 A sufficient number of points were not obtained to draw the curve 
 with any degree of accuracy, but we may easily obtain as many 
 points as we please. 
 
 116. The rectangular hyperbola may easily be described with- 
 out any calculation, thus: 
 
 Complete the parallelogram, OW, Fig. 29. Take any point, as 
 T, in the line of volumes, and draw T8 perpendicular to OF. Join 
 SO cutting JB, the vertical at the point of cut-off, at R. Through 
 R draw HR parallel to the line of volumes OF, and produce it until 
 it intersects the vertical TS at P; then P is a point of the curve. 
 
 117. To plot the Saturation Curve pv& = C. we may proceed as 
 follows: p -g, consequently log p log C ' *- = lg G 
 
 1.0625 log v. If v v the volume at cut-off, be denoted by 1, we shall 
 have C = p^v^ = p . Having now a value of (7, we shall have by 
 substitution log p = log p^ 1.0625 log v, by means of which we 
 obtain the saturation curve BD of Fig. 29, as follows : 
 
 The initial volume A B, or volume up to cut-off, being denoted by 
 
 / 
 
 unity, the volume OQ must be denoted by ~-F = 2.4, therefore, 
 
 log p = log 200 1.0625 log 2.4. 
 
 2.4 log 0.38021 llog 9.58002 10 
 1.0625 log 0.02633 
 
 (1.0625 log 2.4) log 0.40397 llog 9.60635 10 
 200 log 2.30103 
 
 p log 1.89706 
 Therefore, p = 78.9 = QL '. 
 
 In like manner the Adiabatic Curve, pv 1 * = (7, may be plotted. 
 
 TOT i, C , , n 10 log?; 
 
 We have p -^>, whence log p = log C <p- 
 
 = log C 1.111 logv. 
 
 If we let v = 1, we shall have log p = log p 1.111 log v, by 
 means of which we obtain the adiabatic curve BE of Fig. 29, as 
 follows : 
 
102 THE ELEMENTS OF STEAM ENGINEERING 
 
 log p log 200 1.111 log 2.4 
 
 2.4 log 0.38021 Hog 9.58002 10 
 1.111 log 0.04571 
 
 (1.111 log 2.4) log 0.42240 Hog 9.62573 10 
 200 log 2.30103 
 
 p log 1.87863 
 Therefore, p = 75.62 = QL" . 
 
 Figure 29 affords a ready means of comparing the three expansion 
 curves. They represent the expansions under the respective con- 
 ditions of an initial volume AB of steam of absolute pressure 
 OA = p 19 to a final volume OF, and a final, or terminal, pressure 
 p 2 -=2 FC, FD, and FE, respectively. The three curves will always 
 be relatively as shown. 
 
 PEOBLEMS. 
 
 33. Absolute initial pressure, 88 Ibs. per sq. inch; ratio of ex- 
 pansion, 4. Find the terminal pressures for the three cases of ex- 
 pansion, viz. : Adiabatic, Saturation, and Hyperbolic. 
 
 Ans. 18.660 Ibs.; 20.174 Ibs.; 22.000 Ibs. 
 
 34. Find the mean and terminal pressures of steam at 160 Ibs. 
 absolute pressure, after expanding 4 times, and when the conditions 
 of the expansion are: (a) Adiabatic; (&) Saturated; (c) Hyper- 
 bolic. Ans. (a) 91.40 and 34.33; (6) 93.12 and 36.70; 
 
 (c) 95.44 and 40.00. 
 
CHAPTER X. 
 BOILER AND ENGINE EFFICIENCY. 
 
 118. Boiler Efficiency. The question of boiler and engine effi- 
 ciency has already been adverted to in Chapter IV, and it is 
 here the purpose to consider it more fully. 
 
 The efficiency of a boiler is usually expressed in terms of the 
 weight of steam it produces per pound of coal it consumes, though 
 the efficiency in certain circumstances may very properly be ex- 
 pressed in terms of the steam production with respect to the 
 weight' or to the first cost of the boiler. 
 
 In estimating the efficiency of a boiler under the basis of coal 
 consumption, all heat due to the coal thrown into the furnace is 
 charged to the boiler, while it is credited only with the heat trans- 
 ferred through the heating surface and used in the generation of 
 steam. The ratio of these two quantities expresses the efficiency 
 of the boiler. If the two quantities were equal the efficiency would 
 be perfect and be expressed by unity, but owing to various causes 
 the actual efficiency varies from 0.55 to 0.75. 
 
 There are a number of sources to which this loss in efficiency 
 may be attributed. In the first place the losses in the furnace 
 under the most favorable conditions vary from 2.5 to 4 per cent. 
 The coal may be such that portions of it fall through the grate 
 unconsumed, and oftentimes the finer portions, when only partly 
 consumed, are carried from the furnace by the draft and lodged 
 in the tubes and uptakes. The supply of air at times may be in- 
 sufficient for complete combustion, resulting, as was stated on 
 page 24, in the production of carbonic oxide (CO) of a thermal 
 value of only 4500 units per pound, or less than one-third that due 
 perfect combustion. Imperfect combustion produces smoke con- 
 taining carbon only partly consumed and is, of course, lost through 
 the smoke pipe. 
 
 All the heat which is liberated in the furnace gases is not, by 
 any means, transmitted through the heating surface to the water 
 in the boiler. A small part is radiated to the surrounding atmos- 
 phere, and another and much larger part escapes through the 
 smoke pipe. The heat contained in the gases escaping through the 
 
104 THE ELEMENTS OF STEAM ENGINEERING 
 
 smoke pipe is the most serious loss of all and is inevitable, as its 
 prevention could only be accomplished by reducing the temperature 
 of the chimney gases to that of the outside air. This, of course, 
 cannot be done, since the temperature of the gases must be higher 
 than that of the water in the boiler to prevent a transference of 
 heat in the wrong direction. The temperature of the escaping 
 chimney gases will always be considerably higher .than that of the 
 water in the boiler, for in no instance would sufficient heating sur- 
 face be given to a boiler to reduce their temperature to that of the 
 steam. Aside from this, the natural draft of the boiler, occasioned 
 by the difference of weights between the columns of air within and 
 without the smoke pipe, depends directly upon the temperature of 
 the chimney gases, and this practical consideration fixes the tem- 
 perature between the limits of 500 and 600. 
 
 Under the most favorable conditions the losses above considered 
 aggregate little less than 25 per cent, and under the conditions of 
 insufficient air for combustion and incompetent firing the aggregate 
 may reach as much as 45 per cent. 
 
 In estimating the evaporative efficiency of boilers the condition 
 of the steam with respect to its dryness should be ascertained in 
 order to avoid erroneous results. In practice, the steam produced 
 by a boiler contains more or less moisture held in suspension, and 
 it is evident that the total heat required to produce such steam is 
 less than would be required to produce saturated steam, to the 
 extent of the latent heat which would be required to convert the 
 suspended moisture into steam. For example, suppose a boiler to 
 produce dry steam at a temperature of 366 from a feed water tem- 
 perature of 122. Then the total heat to produce one pound of 
 the steam would be H w = 1091.7 + 0.305(366 32) (122 32) 
 = 1103.57 units. 
 
 If the steam produced by the boiler contained 5 per cent of sus- 
 pended moisture, instead of being dry, then the heat required to 
 produce a pound of the moist steam would be : H w =0.95H L 
 + H 8 H f = 0.95[1091.7 0.7(366 -- 32)] + (366 32) 
 (122 32) = 1059 units. 
 
 If, in this instance, the 5 per cent of moisture were neglected 
 and the steam assumed to be dry, the evaporative efficiency would 
 
 be exaggerated in the ratio 1Q ^ 9 = 1.042, or to the extent of 
 4.2 per cent. 
 
BOILED AND ENGINE EFFICIENCY 105 
 
 In the above example the sensible heat, H s , or the heat in the 
 steam at the boiling point, is taken as t 32, in which t is the 
 temperature of the steam at the pressure considered. This value 
 is less than those given in the table of the properties of saturated 
 steam, and the difference arises from the fact that in the value, 
 t 32, the assumption is made that one degree rise in temperature 
 per pound of water is the equivalent of one thermal unit. This is 
 not exactly true, for at temperatures above the standard at which 
 the B. T. U. was measured (39, or more recently 62) the water 
 absorbs a small quantity of heat due to the work performed in 
 slightly expanding it. The sensible heats given in the table include 
 this small quantity, and are, therefore, equal to t 32 plus the 
 heat converted into work in expanding the water. The difference 
 is not great, however, and in the absence of tables the sensible heats 
 may be taken as t 32. For the same reason the table values of 
 the latent heat, or the heat of vaporization, of steam are slightly 
 smaller than those derived from the formula H = 1091.7 
 
 0.7 ( 32), since the table values are derived by subtracting the 
 sensible heat from the total heat (H L = H T H s ) . 
 
 In the examples which follow, the table values for the latent and 
 sensible heats will be taken, though in the absence of tables they 
 may be calculated without any material error in results. 
 
 119. We may modify the formula for finding the total heat re- 
 quired for the evaporation of one pound of water (H w ) so as to 
 meet the cases where the condition of the steam with respect to its 
 dryness is to be taken into account. Thus, if x denotes the "dry- 
 ness -fraction" of the steam we will then have H w = xH L -\-H 8 Hf. 
 If the steam is saturated, or dry, x=I 9 and the formula becomes 
 H w H L + H 8 H f = H, H f 1091.7 + 0.305( 32) 
 
 (tf 32), which is the formula previously given for H w . 
 
 120. The dryness fraction of steam may be determined by a 
 calorimeter test. There are three forms of calorimeters in use for 
 this purpose, known as the Barrel, the Separating, and the Throt- 
 tling calorimeters. The barrel calorimeter is the simplest, and if 
 great care is exercised in the tests the results obtained are nearly as 
 accurate as those derived from the other and more complicated 
 contrivances. 
 
106 THE ELEMENTS OF STEAM ENGINEERING 
 
 121. The Barrel Calorimeter consists of a barrel of hard wood, 
 about three-fourths filled with water and placed on accurate plat- 
 form scales. A half -inch pipe leads from the main steam pipe of 
 the boiler, the part of the pipe within the steam pipe being closed 
 at the end and perforated with holes about one-eighth inch in diam- 
 eter. This pipe should be covered with some non-conducting sub- 
 stance, and should be thoroughly heated before commencing the 
 test. The steam is admitted to the barrel through a hose attached 
 to the half-inch pipe. This hose reaches nearly to the bottom of 
 the barrel, is closed at the end and perforated for about 9 inches 
 of its length from the end. During the entrance of the steam the 
 hose should be moved about so as to promote the mixture of the 
 condensed steam and the condensing water. When the scales show 
 that the desired weight of steam (about one-fifteenth of the original 
 weight of the water in the barrel) has entered the barrel, the re- 
 sulting temperature of the water is taken by a thermometer deeply 
 immersed. The pressure of the steam is taken from the gauge, 
 and the corresponding temperature and latent heat is taken from 
 the table. The temperature of the water in the barrel, before and 
 after the mixture, should be taken with great care and with a ther- 
 mometer graduated to at least one-quarter of a degree, and it is 
 not advisable to have the temperature of the mixture much over 
 110 in order that radiation may be avoided. 
 
 Let W = original weight of water in barrel, 
 
 W^ = weight of steam and suspended moisture blown in, 
 w = weight of dry steam supplied, 
 t = original temperature of water in barrel, 
 tj_ = temperature of steam, 
 
 t 2 = temperature of the water in barrel after the addition 
 of the steam. 
 
 The heat lost by the steam will be the latent heat of the dry 
 steam supplied = wH L , plus the heat given up in cooling the lique- 
 fied dry steam and the water (moisture) carried in by the steam 
 from the temperature ^ to the temperature t 2 . This latter quan- 
 tity is equal to W^(t t 2 ). 
 
 The total heat lost by the steam will then be : 
 wH L +Wi(t, t 2 ). 
 
 The heat gained by the water will be W(t 2 t). 
 
BOILER AND ENGINE EFFICIENCY 107 
 
 If the heat due to cooling the liquefied steam and the water 
 (moisture) carried in it, be subtracted from the total heat lost, 
 the remainder will be the heat due to the latent heat of the dry 
 steam supplied, and we shall have wH L W(t 2 t) W^^ t 2 ), 
 
 i . , . , 
 
 whence w = - jf s - = weight of dry steam sup- 
 **i 
 
 plied. Calling y the fraction of moisture in the steam, we shall 
 
 (Wt-iti 
 
 have y - -, 
 
 And if x denotes the dryness fraction of the steam, we shall have 
 
 w 
 x==l-y= Wi . , 
 
 EXAMPLE I. The barrel of a calorimeter contains 175 pounds of 
 water at a temperature of 60. After 8 pounds of moist steam at 
 an absolute pressure of 165 pounds had been blown into the barrel 
 the temperature of the mixture, was found to be 108. Find the 
 dryness fraction of the steam. 
 
 Solution. From the tables the temperature of steam at 165 
 
 pounds pressure is found to be 366, and the latent heat 855.6 units. 
 
 W(t, /) W 1 (t l - t,) _ 175(108 60) - 8(366 - 108) 
 
 H~T 855.6 
 
 = 7.405 pounds of dry steam supplied. 
 
 Then, dryness fraction = = ?4^ = 0.9256 
 
 FT 1 O 
 
 122, Throttling Calorimeter. The operation of this calorimeter 
 depends upon the principle that the increase in volume of steam, 
 unaccompanied by the performance of work, occasions a liberation 
 of heat which may be utilized in the evaporation of any moisture the 
 steam may contain and in raising the temperature of the steam 
 above that due to its pressure; that is, superheat it. 
 
 The form of this calorimeter designed by Prof. E. C. Carpenter, 
 and manufactured by the Shaeffer and Budenberg Co., is shown in 
 Fig. 30. It consists of a collecting nipple screwed into the main 
 steam pipe; the calorimeter proper, a vessel to which steam is ad- 
 mitted through a converging orifice, and containing a deep cup filled 
 with oil, into which a thermometer is inserted to obtain the tem- 
 perature of the steam within the calorimeter; a manometer 
 attached to the calorimeter for determining the pressure within 
 the interior ; a globe valve in the pipe admitting steam to the calori- 
 meter, used merely to throttle the steam and thus regulate its pres- 
 
108 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 sure in the calorimeter ; and a pipe to carry off the steam from the 
 calorimeter. 
 
 Let PI be the boiler pressure taken from a gauge near the calori- 
 meter, H L the latent heat of vaporization, H s the sensible heat 
 corresponding to p , and x the weight of dry steam in one pound 
 of the mixture as taken from the boiler. Then 1 x denotes the 
 weight of moisture in a pound of the mixture, and xH L -\-H g is 
 the total heat in one pound of the mixture. 
 
 Let p z be the pressure in the calorimeter, and H* and t z the total 
 heat and temperature of saturated steam corresponding to p 2 . Let 
 
 THERMOMETER 
 
 Fig. 30. 
 
 t be the temperature of the superheated steam in the calorimeter, 
 as indicated by the thermometer. Then, the heat in one pound of 
 the steam in the calorimeter is : 
 
 in which the constant 0.48 is the specific heat of superheated steam. 
 Assuming that there is no heat lost from radiation, we shall have : 
 xH L + H a = H? + 0.48(7 t t ), whence 
 Hf + 0.48(1 tj - If, 
 
 X - TT > 
 
 * 
 
 and the percentage of moisture in the boiler steam will be 
 
 The values for H u H 8 , H* and i z are to be taken from the table 
 of the properties of saturated steam. 
 
OF THE 
 
 UNIVERSITY 
 
 BOILER AND ENGINE EFFICIENCY 
 
 EXAMPLE. The following data were taken from a test : 
 Absolute steam pressure, 155 pounds. Absolute pressure in calori- 
 meter, 14.8 pounds. Temperature of steam in calorimeter, 260.4. 
 Prom this we obtain : 
 
 - H46.7 + 0.48(260.8 212) -. 332.7 
 859.3 
 
 = 0.975 Ib. of dry steam in one pound of the boiler steam. 
 
 Percentage of moisture in steam = 100(1- 0.975) =2.5 per 
 cent. 
 
 If is obvious that if the heat liberated fails to superheat the steam 
 there can be no determination of its quality with the throttling 
 calorimeter. The limit beyond which this form of calorimeter is 
 inoperative is when t and t 2 are equal ; that is, when the heat liber- 
 ated is just sufficient to evaporate the moisture in the steam, leaving 
 it dry and saturated but with no degree of superheat. The practical 
 limit is when the steam contains little in excess of 2.5 per cent of 
 moisture, and the most convenient pressure to be maintained within 
 the calorimeter is that when the manometer registers at the zero 
 mark, or when the absolute pressure is that of the atmosphere, which 
 may be taken as 14.7 pounds in the absence of a barometer. If the 
 reading of the barometer is taken, the pressure is obtained by multi- 
 plying by 0.49. 
 
 123. Separating Calorimeter. For the determination of the 
 quality of steam containing a degree of moisture beyond the limit 
 of the throttling calorimeter, some form of steam separator is used. 
 The form largely in use is that devised by Prof. Carpenter. This 
 instrument provides for a mechanical separation of the moisture 
 from the steam, which permits the two quantities to be weighed 
 separately. 
 
 124 In making tests of boilers, allowance is made for the moisture 
 and ash in the coal, in order that the evaporative power of a pound 
 of the combustible may be obtained. 
 
 EXAMPLE. A boiler produces steam at 175 pounds pressure per 
 gauge from a feed water temperature of 142, evaporating 8.5 Ibs. 
 of water per pound of coal. The dryness fraction of the steam is 
 0.98. The coal contains 10 per cent of ash and 5 per cent of mois- 
 ture. Find the actual evaporation, and the equivalent evaporation 
 from and at 212, of a pound of the dry combustible. 
 
110 THE ELEMENTS or STEAM ENGINEERING 
 
 Here we have : 
 
 H w a= xH L + H s H f = 0.98 X 847 + 350 -- (142 -- 32) 
 = 1070 units. Equivalent evaporation from and at 212 is 
 
 8.5 X 1070 Q/I01 , 
 - m = 9.42 Ibs. 
 
 The evaporations of 8.5 pounds and 9.42 pounds were obtained 
 from 0.85 pound of dry combustible, therefore the actual and equiva- 
 
 o o 
 
 lent evaporations of a pound of dry combustible are Q-^T = 10 Ibs., 
 
 9 42 
 and ^- =i 11.08 Ibs., respectively. 
 
 125. Engine Efficiency. Mention has already been made of Car- 
 not' s Eeversible Cycle, or Perfect Heat Engine. In practice the 
 steam engine transforms only a portion of the mechanical energy 
 of the heat contained in the steam supplied to it. The greater the 
 portion of the heat transformed into work the greater the efficiency 
 of the engine, and it is the constant aim of the engineer to make 
 this portion as great as possible. 
 
 Unfortunately there are a number of causes which reduce the 
 heat converted into useful work to a small fraction of the heat sup- 
 plied. The principle of these causes is the Second Law of Thermo- 
 dynamics, which asserts that, " Heat cannot pass from a cold body 
 to a hot one by a purely self-acting process." It follows from this 
 law that no steam engine can convert into work all the heat of the 
 steam supplied to it, for as soon as the temperature of the steam 
 falls to that of the surrounding atmosphere, the heat remaining in 
 it is not available for doing useful work. Thus, if T^ be the highest 
 absolute temperature available, and T 2 the lowest, then the maxi- 
 
 T T 
 mum efficiency obtainable is "^m ? , or if tf be the temperature of 
 
 the steam of initial pressure, and t 2 the temperature of the steam of 
 final pressure, then the maximum efficiency of the most perfect 
 engine is 
 
 ft + 461) - (I, + 461) _ t, - t t 
 T, -~7T 
 
 As an illustration, suppose an engine working between the tem- 
 peratures of 366 and 142, corresponding to initial and final pres- 
 sures of 165 Ibs. and 3 Ibs. absolute, respectively. The ideal effi- 
 
 q ft ft - A f\ 
 
 ciency would then be nt^rrzar = 0.27, or little more than 25 per 
 cent. 
 
BOILER AND ENGINE EB^EICIENCY 111 
 
 126. The cycle of operations of this ideal engine, known as the 
 Carnot cycle,, consists of four stages, specified with simplicity by 
 Professor W. F. Durand, in his " Practical Marine Engineering/' as 
 follows : 
 
 (1) The first operation must consist of an expansion at constant 
 temperature, and all heat received from the source of supply must 
 be received during this operation (Isothermal expansion). 
 
 (2) The second operation must consist of an expansion with de- 
 crease of temperature, during which, however, no heat is allowed to 
 enter or leave the substance (Adiabatic expansion). 
 
 (3) The third operation must consist of a compression, during 
 which the temperature remains constant, and all heat removed from 
 the body must be removed during this operation (Isothermal com- 
 pression) . 
 
 (4) The fourth operation must consist of a compression with in- 
 crease of temperature, during which, however, no heat is allowed to 
 enter or leave the substance, and at the end the substance must find 
 itself in the same condition as at the beginning of number (1) 
 (Adiabatic compression). 
 
 It will be observed that work is done by the substance during 
 operations (1) and (2), and that work is done on the substance dur- 
 ing (3) and (4). The difference between work done by and on the 
 substance will be the net work obtained from the heat in the sub- 
 stance, and the ratio of this to the total heat supplied during (1) 
 is the efficiency which will be exactly measured by the difference of 
 the absolute temperatures of operations (1) and (3), divided by 
 the absolute temperature of ( 1 ) ; that is : 
 
 Efficiency of the Ideal Engine = 
 
 127. The conditions of the four operations of the ideal engine are 
 far from being obtained in practice, and under the most favorable 
 circumstances not more than from 60 per cent to 70 per cent of the 
 ideal efficiency is obtained in actual practice. The principal causes 
 which reduce the efficiency of the steam engine below that of the 
 ideal engine are enumerated on page 34. 
 
 Within the present limits of temperature used in the steam engine 
 the ideal efficiency varies from 25 to 30 per cent, while the actual 
 efficiency varies from 15 to 20 per cent. 
 
THE ELEMENTS OF STEAM ENGINEERING 
 
 128. Weight of Steam per I. H. P. per Hour. In the ideal engine 
 the efficiency is v-- -T^TO, and since one I. H. P. requires an ex- 
 
 ?1 ~T~ 401 
 
 penditure of * g - = 42.42 thermal units, it follows that the 
 minimum expenditure in the ideal engine to produce one I. H. P. 
 
 will be 42,42 +- = 
 
 46 ^ 
 
 thermal units per minute. 
 
 If H w denotes the units of heat required to produce one pound 
 of steam from the temperature t. 2 and at the temperature t lt and N 
 denotes the number of pounds of steam required per I. H. P. per 
 hour, we will have : 
 
 _ 60[42.42(^ + 461)] _ 
 ^~~~~ ~ 
 
 60[42.42 1 + 461)] 
 
 _ _ __ 
 
 [1091.7 + 0.305^ 32) 3 32)] [^ < a ] ' 
 
 The consumption of steam per hour of the ideal engine, both 
 condensing and non-condensing, given in the following table, were 
 calculated by the above formula. In the case of the condensing 
 engine the feed water is assumed to have been taken from the con- 
 denser at a temperature of 100, and in the non-condensing engine 
 it is assumed that the exhaust steam has been used to raise the tem- 
 perature of the feed to about 212. 
 
 The table shows a marked gain of efficiency due to the use of the 
 condenser, and this gain is greater for low than for high pressures. 
 
 POUNDS OF STEAM PER I. H. P. PER HOUR. 
 
 Initial pressure 
 in pounds. 
 
 Condensing 
 Engine. 
 
 Non-condensing 
 Engine. 
 
 tj Fahr. 
 
 70 
 
 8.62 Ibs. 
 
 19.32 Ibs. 
 
 303 
 
 90 
 
 8.13 Ibs. 
 
 16.55 Ibs. 
 
 320 
 
 110 
 
 7.73 Ibs. 
 
 14.76 Ibs. 
 
 335 
 
 1'30 
 
 7.43 Ibs. 
 
 
 347 
 
 150 
 
 7.19 Ibs. 
 
 
 358 
 
 180 
 
 6.89 Ibs. 
 
 
 373 
 
 Since the actual efficiencies of steam engines vary from 60 to 70 
 per cent of the ideal efficiencies, it will be seen from the table that 
 a consumption of from 11.5 to 12 Ibs. of steam per I. H. P. per 
 
BOILER AND ENGINE EFFICIENCY 113 
 
 hour may be expected at pressures commonly used with triple-ex- 
 pansion engines, and in the case of non-condensing engines a con- 
 sumption of 25 pounds per I. H. P. per hour may be expected under 
 favorable conditions. Assuming the boiler to evaporate 9.5 pounds 
 of water per pound of coal, the above figures would indicate a coal 
 consumption of 1.2 pounds per I. H. P. per hour in the case of the 
 triple-expansion engines and 2.7 pounds in the case of the non- 
 condensing engines. These figures conform fairly well with the 
 results obtained in actual practice. 
 
 129. Liquefaction in Cylinders On page 37 the object of jack- 
 eting cylinders was pointed out. It is not probable that the heat 
 from the j-acket steam ever entirely prevents initial condensation, 
 or supplies heat sufficient to replace all that transmuted into work 
 during expansion, and thus maintain the steam in a state of satura- 
 tion, but there can be no question as to the advantage derived from 
 jacketing the cylinders of engines of slow and moderately high 
 speeds. 
 
 Experiments have shown that the use of the jacket in such in- 
 stances increased the efficiency. of the steam from 6 to 25 per cent. 
 
 It is true that the transference of heat from the jacket to the 
 cylinder is accompanied by a corresponding liquefaction in the 
 jacket, and it would appear that the scene only of the liquefaction 
 has been changed, but the essential difference is that the steam 
 liquefied in the jacket passes off in the state of water at the tem- 
 perature of the jacket steam, each pound carrying with it a number 
 of units equal to the sensible heat of the steam, while that liquefied 
 in the cylinder escapes to the air or condenser in the state of steam, 
 carrying with it the latent heat of the jacket steam. The water 
 from the jacket is invariably drawn off and returned to the boiler 
 with the feed, and the heat supplied to the cylinder per pound lique- 
 fied in the jacket is the latent heat of the steam corresponding to the 
 pressure of the jacket steam. 
 
 The following example presents the theoretical consideration of 
 the action of steam in a jacketed cylinder. 
 
 A simple expansive engine has a jacketed cylinder 9" in diameter, 
 and a stroke of 10". Cut-off, 0.25 stroke; clearance, 6 per cent; 
 initial pressure of steam, 95 Ibs. absolute; back pressure, 18 Ibs.; 
 revolutions per minute, 375. The specific volume of steam at 95 Ibs. 
 pressure is 4.62 cubic feet, and it is assumed that only 0,9 of the 
 8 
 
114 THE ELEMENTS OF STEAM ENGINEERING 
 
 theoretical mean pressure is realized. Find: (1) The weight of 
 working steam per I. H. P, per hour; (2) the weight of jacket 
 steam per I. H. P. per hour, and its equivalent from the tempera- 
 ture of feed water; (3) the efficiency of the engine; (4) the total 
 weight of steam per I. H. P. per hour. 
 
 Solution. Stroke = |f = $ feet. 10 X 0.06 = 0.6 inch equiva- 
 lence of clearance in stroke. One-fourth stroke = 2.5 inches, and 
 
 ' i : - = -nr feet of stroke up to cut-off, including clearance. 
 
 9 
 Eadius = . . 1tl = f feet. 
 
 <i X 1/& 
 3>1416 *^ 3 J X = 0.368 cu. ft. volume swept by piston per stroke. 
 
 3 '(8)' 6 >Tl2 8 X *62 " = - 02473 lb ' steam used P er stroke ' 
 Eatio of expansion = 25 + +b06 = 3 ' 42 loge 
 
 =61921b 
 
 6Aa 
 
 Theoretical m. e. p. = 61.92 18 = 43.92 Ibs. 
 Mean effective pressure to be expected = 43.92 X 0.9 = 39.528 Ibs. 
 Work per stroke = Pv 144 X 39.528 X 0.368 = 2094.7 foot Ibs. 
 
 90Q4- 7 
 Work per pound of steam =002473= 84 > 704 foot lbs> 
 
 qq 
 
 Steam per I. H. P. per hour = g4 7 = 23 ' 37 lbs * 
 
 Or, the weight of steam per I. H. P. may be found thus : 
 Weight of steam used per hour = 0.02473 X 750 X 60 = 1112.85 
 Ibs. 
 
 2094.7X750 . 
 
 Steam per I. H. P. per hour = ' = 23.37 Ibs. 
 
 The steam in the jacket supplies the heat necessary to maintain 
 the working steam in a state of saturation throughout the stroke, 
 and the measure of this heat is the latent heat of the steam liquefied 
 in the jacket. 
 
 If x denotes the weight of steam liquefied in the jacket per 
 I. H. P. per hour, then the heat supplied by the jacket is x times 
 the latent heat of steam corresponding to the pressure, p lf of the 
 jacket steam. The total heat expended in the performance of 
 
BOILER AND ENGINE EFFICIENCY 115 
 
 effective work per I. H. P. per hour is then (H? H 2 ) 23.37 + xH[. 
 This expenditure of heat is balanced by the effective work done. 
 The gross work per stroke is 144/? w i; 2 , and the equivalent in work of 
 the heat energy contained in the p 2 steam is l4p. 2 v.,. The effective 
 
 , . ,, 1 44 v.( ") . va i Pi 
 
 work per stroke is then ~ > m wnlc " P* - " 
 
 34^ 
 
 27.78 pounds, and i> 2 is the displacement per stroke = 0.368 cubic 
 feet. Since the weight of steam used per stroke is 0.02473 pound, 
 
 n 3R8 
 there will be A (30473 = 14.8 cubic feet displaced by the piston for 
 
 each pound of steam used. Therefore, the effective work for the ex- 
 
 , 000fV , . , . 144X14.8(61.92-27.78)23.37 
 
 penditure of 23.37 pounds of steam is == 
 
 thermal units. Then we shall have : 
 
 - 37.78)23.37 wheuce 
 
 x = r 144 X 14.8 X 34.14 _ H T +H r~| 23.37 
 = (69.52 - 1181 + 1157) ^ 7 = 1.832 
 
 pounds. 
 
 T-, . , . Effective work 
 
 Efficiency of engine = TT T- 
 
 84,704 X 23.37 
 778 
 
 ~ [1181 (210 - 32)] 23.37 + 1.832 X 887 
 
 = 0.1015, the temperature of the feed 
 being taken at 210. 
 
 The steam liquefied in the jacket is returned to the boiler at the 
 temperature of the water in the boiler, and therefore the heat ex- 
 pended in the jacket is less than the total heat of formation of 
 the liquefied steam by the amount of heat required to raise the feed 
 water to the temperature of the boiler steam. Hence for each 
 pound of steam liquefied in the jacket there is an expenditure of 
 1181 (210 32) units. We shall have, therefore, 
 
 Weight of jacket steam per I. H.P. per hour ' , ^ 
 
 1.42 pounds. 
 
 The total weight of steam per I. H. P. per hour is then 
 23.37 + 1.42 = 24.79 pounds. 
 
116 THE ELEMENTS OF STEAM ENGINEERING 
 
 If the saturation curve, BD, of Fig. 29, is to be maintained dur- 
 ing expansion, heat must be obtained from a jacket or from some 
 other external source. Steam under ordinary conditions of practice 
 contains moisture, and if the steam jacket supplies sufficient heat 
 to evaporate this moisture and also to prevent liquefaction during 
 expansion, the curve approximates the hyperbola BC, which repre- 
 sents the expansion of a perfect gas at constant temperature. 
 
 It is found that considerable differences in the amount of heat 
 supplied by a jacket do not make very appreciable differences in the 
 form of the expansion curve, which fact detracts from the value of 
 an analysis of the performance of a steam engine from indicator 
 diagrams. 
 
 Experiments have shown that no particular advantage is derived 
 in having jacket steam of a pressure more than a few pounds higher 
 than that of the initial steam in the cylinder, so it is the practice 
 to make the jacket steam of the second and succeeding cylinders 
 of stage expansion engines pass through reducing valves. 
 
 The conditions of hyperbolic expansion are not infrequently real- 
 ized, and it is the universal practice to regard the expansion curve 
 as a hyperbola, which greatly facilitates calculations. 
 
 130. Condensation and Production of Vacuum. After using 
 steam in the cylinder of a condensing engine it is exhausted into a 
 condenser and brought into contact with a jet of water, as in the 
 case of a jet condenser, or else it comes in contact with a series of 
 tubes through which cold water is circulating, as in the case of a 
 surface condenser. The result in either case is the almost instant 
 condensation of the steam and the production of a vacuum. The 
 perfection of the vacuum depends almost entirely upon the existing 
 temperature in the condenser. If the temperature were 32, the 
 corresponding pressure would be 0.085 Ib. per square inch, and the 
 vacuum nearly perfect. No such vacuum as this is ever attained in 
 a condenser, nor would it be desirable, inasmuch as there must 
 always be an excess of pressure in the condenser over that in the 
 air-pump barrel in order that the valves should open. 
 
 With the surface condenser only the water resulting from the 
 condensation of the steam accumulates in the condenser, but more 
 or less air, due to leakage or liberated from the feed water, is also 
 present, and if it were allowed to accumulate, the vacuum would 
 soon be destroyed. For this reason the air-pump is necessary with 
 
BOILER AND ENGINE EFFICIENCY 117 
 
 both styles of condenser, and its function is to remove from the con- 
 denser the air and uncondensed vapor, as well as the water resulting 
 from the condensation. The ordinary temperature of the water 
 after condensation is about 126, corresponding to a pressure of 
 2 pounds per square inch. 
 
 -Heat Rejected into the Condenser. Steam is commonly exhausted 
 into the condenser at a pressure of from 4 to 5 pounds with a cor- 
 responding temperature of about 160. Assuming the temperature 
 of the water after condensation to be 126, each pound of exhaust 
 steam upon entering the condenser is reduced from 160 to 126, 
 which represents the liberation of 160 126 = 34 units. The lat- 
 ent heat of a pound of steam at 160 temperature is 1091.7 
 0.7(160 32) = 1002 units, so that each pound of steam gives up 
 1002 units by its condensation into water at 160. The total quan- 
 tity liberated by the condensation of one pound of the steam is 
 1002 + 34 = 1036 units, and this heat must all be destroyed, or 
 abstracted, by the injection water. This quantity of heat is known 
 as the heat rejected to the condenser per pound of steam used. The 
 total heat of formation of the steam will always be greater than that 
 rejected to the condenser, and the difference, neglecting the small 
 loss by radiation, is the amount converted into work. 
 
 Weight of Condensing Water Required Per Pound of Steam. 
 Suppose, in the case of a jet condenser, that the temperature of the 
 injection water is 60, then, using the figures given above, each 
 pound will absorb 126 60 = 66 units; therefore, to destroy 1036 
 
 i n^?fi 
 units given up by the pound of steam will require -^- = 15.7 Ibs. 
 
 of injection water. 
 
 In the case of a surface condenser a greater amount of water 
 will be required for condensation, due to the fact that the final tem- 
 perature of the injection water passing through the tubes must be 
 lower than that of the exhaust steam; while in the jet condenser 
 the injection water and the condensed steam form a mixture at the 
 same temperature. Suppose the injection water after passing 
 through the tubes was discharged at a temperature of 85. Then 
 each pound would absorb only 85 60 = 25 units, and there would 
 
 1 /"kQ/? 
 
 be required ~-^- = 41.44 Ibs. injection water per pound of steam 
 condensed. 
 
118 THE ELEMENTS OF STEAM ENGINEERING 
 
 PKOBLEMS. 
 
 35. The barrel of a calorimeter contains 175 Ibs. of water at a 
 temperature of 60. Ten pounds of steam at a pressure of 150 Ibs. 
 absolute are blown into the barrel, after which the temperature of 
 the mixture is found to be 120. The temperature of steam at 150 
 Ibs. is 358, and the latent heat 861.2 units. Find the dryness frac- 
 tion of the steam. Ans. 0.943. 
 
 36. The efficiency of a boiler is 0.7; steam pressure, 135 Ibs. per 
 gauge; temperature of the feed water, 120 ; thermal value of the 
 coal used, 14,200 units per pound. Find the evaporation of satu- 
 rated steam per pound of coal. Ans. 9.01 Ibs. 
 
 37. The efficiency of a boiler is 0.75; steam pressure, 150 Ibs. per 
 gauge; temperature of feed, 130; dryness fraction of the steam, 
 0.95 ; thermal value of the coal, 14,300 units per pound. Find the 
 number of pounds of steam of the given quality evaporated per 
 pound of coal. Ans. 10.19 Ibs. 
 
 38. A boiler under certain conditions evaporates 9.25 Ibs. water 
 per pound of coal into steam at 364 from feed water at 70, the 
 dryness fraction of the steam being 0.9. Under different circum- 
 stances 8.75 Ibs. of water are evaporated per pound of coal into 
 steam at 380 from feed water at 102, the fraction of moisture in 
 the steam being 0.05. Which of the two cases is the more efficient ? 
 
 Ans, First case is 4.13 per cent more efficient. 
 
 39. Steam containing 4 per cent of moisture is generated at 165 
 Ibs. gauge pressure from feed water at 152. The evaporation per 
 pound of coal is 8.5 Ibs. In another instance 8 Ibs. of water per 
 pound of coal are evaporated into steam at 125 Ibs. pressure from 
 feed water at 132, the dryness fraction being 0.97. Find the 
 equivalent evaporations from and at 212, and under the supposi- 
 tion that coal in the first instance cost $4.10 per ton and in the 
 second $3.40 a ton, show which case is the more economical. 
 
 Ans. Equivalent evaporation first case, 9.25 Ibs. 
 
 Equivalent evaporation second case, 8.81 Ibs. 
 First case 14.7 per cent more expensive. 
 
 40. A boiler generates steam at 185 Ibs. gauge pressure from feed 
 water of temperature of 152, evaporating 9 Ibs. of water per pound 
 of coal. The steam contains 3 per cent of moisture, and the coal 
 
BOILER AND ENGINE EFFICIENCY 119 
 
 contains 12 per cent of ash and 4 per cent of moisture. Find the 
 actual evaporation and the equivalent evaporation from and at 212 
 of a pound of the dry combustible. Ans. 10.7 Ibs. and 12 Ibs. 
 
 41. What would be the coal consumption per I. H. P. per hour in 
 the ideal engine working between the temperatures of 369 and 
 120, assuming the thermal value of the coal to be equal to that of 
 pure carbon, and the efficiency of the boiler to be perfect ? 
 
 Ans. 0.585 Ib. 
 
 42. How many pounds of water must be evaporated per hour per 
 I. H. P. with the ideal engine, the consumption of fuel being 0.585 
 Ib. of carbon per I. H. P. per hour, the temperature of the steam 
 369, and that of the feed water 120 ? Ans. 7f Ibs. 
 
 43. The efficiency of an engine is 14 per cent, and of the boiler 
 70 per cent. The coal used has a thermal value of 14,300 units 
 per pound. Find the number of pounds of coal required per 
 I. H. P. per hour. Ans. 1.816 Ibs. 
 
 44. A 9" by 10" engine uses steam at 80 Ibs. gauge pressure. Cut- 
 off, 0.25 stroke; revolutions, 375 per minute; clearance, 6 per cent; 
 I. H. P. developed, 46. The sensible heat and specific volume 
 of the initial steam are 294 units and 4.62 cubic feet, respectively. 
 The dryness fraction of the steam is 0.97, the thermal value of the 
 fuel 14,000 units, the temperature of the feed water 182, and the 
 efficiency of the boiler 70 per cent. Find (a) The pounds of steam 
 per I. H. P. per hour, (b) The pounds of water evaporated per 
 pound of coal, (c) The pounds of coal per I. H. P. per hour. 
 
 Ans. (a) 24.17 Ibs. (&) 9.77 Ibs. (c) 2. 47 Ibs. 
 
 45. Diameter of cylinder, 14"; stroke, 13"; cut-off, 0.'25 of the 
 stroke; clearance, 5 per cent; initial absolute steam pressure, 95 
 Ibs.; back pressure, 17 Ibs.; revolutions per minute, 300. The spe- 
 cific volume of steam at 95 Ibs. pressure is 4.62 cubic feet, and it is 
 expected that 0.9 of the theoretical mean effective pressure will be 
 realized. Find the weight of steam used per I. H. P. per hour. 
 
 Ans. 22.47 Ibs. 
 
 46. With a jet condenser the temperature of the exhaust steam is 
 193, of the injection water 60, and of the mixture -in the con- 
 denser 120. Find the weight of injection water per pound of ex- 
 haust steam. Ans. 17.53 Ibs. 
 
120 THE ELEMENTS OF STEAM ENGINEERING 
 
 47. With a surface condenser the temperature of the exhaust 
 steam is 170, and of the condenser 126. The injection water 
 enters at a temperature of 60 and is discharged at 85. Find the 
 weight of injection water per pound of exhaust steam. 
 
 Ans. 41.56 Ibs. 
 
CHAPTER XL 
 ENGINE DESIGN. 
 
 131. The Indicator Diagram in Preliminary Design. In the pre- 
 liminary design of the valve of an engine certain data are assumed, 
 from which other dimensions result as a consequence. It is then 
 possible to use the properties of the indicator diagram, regardless 
 of scale, to determine if the design of the valve will give satisfac- 
 tory results as to mean pressure and steam consumption, and thus 
 ascertain if the design is satisfactory. 
 
 In Fig. 31 let: 
 
 L = length of stroke. 
 
 Jc = length of stroke to cut-off. 
 
 c = clearance in fraction of stroke. 
 
 x =: distance from end of stroke at which exhaust closes. 
 p 1 = absolute initial pressure. 
 p 2 = absolute terminal pressure. 
 
 p 3 absolute back pressure =. initial pressure of compression. 
 p c = absolute terminal pressure of compression. 
 v = k -\- c = initial volume. 
 v 2 = L + c = terminal volume. 
 
 = r = ratio of expansion. 
 r c ratio of compression. 
 
122 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 The expansion and compression curves being hyperbolic, we shall 
 
 rn ry 
 
 have P^V! = p 2 v 2 , whence = = r\ 
 
 and in like manner r c . 
 
 Denote the area HBCDKR by A, the area WAHR by 5, the area 
 WRKGO by C, and the area GKDO by D. 
 
 The area A represents the effective work and is equal to p e X L, 
 from which we have : 
 
 A A + B+C+D (B + C+D) . 
 
 e 1S tne mean 
 
 effective pressure. 
 
 *r C -X 
 
 It has been shown on page 99 that the area of the whole diagram 
 A + B -\- C + D is p-iPid + lge r )? an d the area C is likewise 
 
 x). There- 
 
 - x)] 
 
 Area of B = c(p p c ), and area of D = p B (L 
 fore, 
 
 _ 
 
 The simplest way of finding p e is to calculate the areas A, B, C, 
 and D from the data given, and substitute their values in the 
 equation just found. 
 
 EXAMPLE I. Initial pressure of steam, 88 Ibs., and terminal 
 pressure, 27 Ibs. Stroke, 30"; clearance, 4 per cent of piston dis- 
 placement. A terminal pressure of compression of 60 Ibs. is de- 
 sired, the back pressure being 17 Ibs. All the pressures being 
 absolute, it is desired to find the point of cut-off, the point of 
 exhaust closure, and the mean effective pressure. 
 
ENGINE DESIGN 123 
 
 Draw Fig. 32 without regard to scale. We have r = J~f = 3.26, 
 the log e of which is 1.18. 
 c 30 X 0.04 = 1.2 inches. 
 
 p 1 v 1 = p 2 v 2) or SS(k + 1.2) = 27 X 31.2, whence k = 8.37 inches. 
 17 (a + 1.2)= 60 X l-2 5 whence x 3.04 inches. 
 
 r c ^ = 3.53, or r c = f- = 3.53. log e r c = 1.26. 
 
 Area (A + B + C + D) p^(l + log e r) = 88 X 9.57 X 2.18 
 
 = 1835.9. 
 
 Area B = c(p 1 p c )= 1.2(88 60)= 33.6. 
 Area C = p c v c (l + log c r e ) 60 X 1.2 X 2.26 = 162.72. 
 Area D p 3 (L x) = 17(30 3.04) =458.32. 
 Area A Area (A + B + C + D) Area (B + C + D) 
 = 1835.9 (33.6 + 162.72 + 458.32) = 1181.26. 
 
 p e = Are ? A = '= 39.375 pounds per square inch. 
 
 Jj 60 
 
 If it is determined that the engine shall make 170 revolutions per 
 minute and that the diameter of the cylinder be 26 inches, the 
 horse-power to be expected will be: 
 
 T TT T> pLaN 39.375 X 2 5 X r X 169 X 340 
 LILR = 3pOO = 33,000 
 
 132. To estimate the steam consumption per I. H.. P. per hour we 
 would proceed thus : 
 
 Referring to Fig. 32, the volume of steam in the cylinder at 
 
 cut-off is ^ ityoQ cubic feet, in which k and c are in inches, and 
 
 1 //co 
 
 a is the area of the piston in square inches ; and if s i be the specific 
 volume of the steam at pressure p i} the weight of the steam in the 
 
 cylinder at cut-off is T,y 9UN/ pounds. The volume of steam in 
 
 JL / /vo X i 
 
 i cr -\- c\ft 
 
 the cylinder at the moment of exhaust closure is v ^ ^J cubic 
 
 feet, and if s s be the specific volume of steam at pressure p s , the 
 weight of this steam is lo pounds. This last weight of 
 
 steam is compressed into the clearance space, and only the weight 
 (k + c)a (x 4- c)a . ,, . ,, , . 
 
 1728 X s 1728 X 1S a l^ owe ^ to esca P e in the exhaust, and is, 
 
 therefore, the weight of steam used per stroke. If R be the number 
 
124 THE ELEMENTS OF STEAM ENGINEERING 
 
 of revolutions per minute of the engine, we shall have : Pounds of 
 steam per I. H. P. per hour 
 
 ["(* + g > _ Qg + c >1g X 2 X 60 Ra 
 _ |_1728 X g| 1728 X 
 
 I. H. P. 14.4 I. H. P. 
 
 EXAMPLE. Find the number of pounds of steam used per hour 
 per I. H. P. by the engine of the example on page 122, the specific 
 volume of steam at pressures of 88 Ibs. and 17 Ibs. being 4.9 and 
 23.22 cubic feet, respectively. 
 
 Here we have : 
 Pounds of steam per I. H. P. per hour 
 
 170 X 3.1416 X 169 X 201.99 
 
 14.4 X 538.7 ' 14.4 X 538.7 X 4.y X 23.22 
 
 = 21.13 Ibs. Adding 25 per cent (see Sec. 149, p. 150), we get 
 26.4 Ibs. as an approximate result. 
 
 133. Size of Cylinder for a Given Power. The diameter of the 
 cylinder of an engine depends upon the piston speed and the mean 
 pressure permissible in obtaining the power desired. 
 
 The mean pressure depends upon the initial pressure and the 
 ratio of expansion, and may be calculated, under the assumption 
 that the expansion takes place according to Boyle's Law, by the 
 
 formula p m = ^ ' -^ - . Or, the mean pressure may be 
 
 obtained very approximately by drawing an ideal diagram from 
 the conditions of the case, from which the actual diagram to be 
 expected may be drawn and the mean pressure obtained. 
 
 The mean pressure derived theoretically is never realized in 
 practice, for there are always causes which make the mean pressure 
 from the actual indicator diagram less than that due the initial 
 pressure and ratio of expansion used in the design. 
 
 The principal causes which tend to make the actual mean pres- 
 sure less than that theoretically due the initial pressure and the 
 ratio of expansion are : Wire-drawing ; liquefaction in the cylinder ; 
 release of the steam before the piston arrives at the end of the 
 stroke ; clearance ; compression and back pressure ; friction in the 
 ports and passages ; and, in stage expansion engines, " drop." 
 
 Attempts are sometimes made to treat the effects of these causes 
 analytically, and though such treatment is interesting and to a 
 degree instructive, it involves so many assumptions which may or 
 may not be realized in practice that the results obtained are of little 
 value. 
 
ENGINE DESIGN 
 
 These causes have not, of course, the same effect in all cases, but 
 experience has shown that a factor, depending upon the type and 
 the working conditions of the engine, may be applied to the theo- 
 retical pressure obtained in any case, and the result be taken as the 
 mean pressure to be expected. 
 
 There is a substantial agreement among the authorities as to 
 these mean pressure factors, and they may be taken, under the con- 
 ditions named, as follows: 
 
 For high-speed stationary engines, from 0.87 to 0.94; for two- 
 stage expansion engines, 0.75 to 0.85; for triple-expansion engines 
 of the mercantile marine, 0.65 to 0.7; for triple-expansion engines 
 of war vessels, 0.6 to 0.65; for torpedo-boat engines, 0.45 to 0.55. 
 
 134. Piston Speed. The speed of the piston is determined from 
 considerations of convenience, the type of the engine, and the nature 
 of the work to be done. It depends, of course, upon the length of 
 the stroke and the number of revolutions, and is expressed in feet 
 per minute. The mean piston speed i k s 'equal to the product of 
 twice the length of stroke in feet and the number of revolutions per 
 minute. Experience has shown that a good piston speed for slow 
 pumping engines is from 125 to 175 feet per minute; for ordinary 
 horizontal engines, from 300 to 450 feet; for high-speed stationary 
 engines, from 500 to 650 feet; for marine engines, from 700 to 
 800 feet; for locomotives, from 900 to 1000 feet; and for torpedo- 
 boat engines a piston speed as high as 1200 feet has been attained. 
 
 135. Stroke. There is no standard rule for determining the 
 stroke of an engine, but the local condition as to space is often the 
 controlling influence in its determination. Strokes of stationary 
 engines vary from 10 inches to 60 inches, and those of marine en- 
 gines from 18 inches to 50 inches. 
 
 136. Cylinder Ratios. There is a lack of agreement among the 
 authorities as to the proper ratios between the cylinders of stage 
 expansion engines. It is the aim to equalize the power between the 
 cylinders, but this may be effected within practical limits by means 
 of the adjustment of the points of cut-off, and quite independently 
 of the cylinder ratios, but a too low cylinder ratio with equality of 
 powers may cause a pronounced inequality in the initial stresses 
 on the pistons. The important considerations of making the weight 
 of the machinery and the space it occupies as small as possible in 
 a vessel of war, together with the fact that engines of such vessels 
 
126 . THE ELEMENTS OF STEAM ENGINEERING 
 
 are seldom worked at their maximum power, make it desirable that 
 their cylinder ratios be made smaller than those of engines of the 
 mercantile marine, where the conditions are somewhat reversed. 
 
 The proposition has been advanced that, with given initial and 
 terminal pressures, percentages of clearance, point of cut-off in the 
 H. P. cylinder and the pressure at that point, the ratio of the net 
 areas of the high- and low-pressure pistons follows as a consequence. 
 The reasoning employed in the establishment of this proposition is 
 logical and interesting, but as the operation is long and tedious, 
 and as it involves several assumptions which can only be justified 
 by a reference to the results of successful practice, the question 
 very naturally arises, Why not at once select for any particular case 
 the cylinder ratio employed in successful practice? The construc- 
 tion of the theoretical diagram and the investigation of the initial 
 stresses on the pistons would then reveal the fact if the selected 
 ratio were improper. 
 
 Successful practice has shown that in the two-stage expansion 
 engine, with the initial pressure ranging from 80 Ibs. to 120 Ibs., 
 the ratio of the low-pressure cylinder to the high-pressure cylinder 
 should vary from 3 to 4.5 for engines of the mercantile marine, and 
 from 2.5 to 4 for engines of vessels of war. For triple-expansion 
 engines of the mercantile marine, with pressures ranging from 
 160 Ibs. to 180 Ibs., the ratio of the low-pressure cylinder to the 
 high-pressure cylinder varies from 6 to 7.5. For engines of war 
 vessels, where the pressures range from 160 Ibs. to 250 Ibs., the 
 ratio varies from 5 to 7. 
 
 Opinions are at variance concerning the proper size of the inter- 
 mediate cylinder of a triple-expansion engine. To make it a mean 
 between the H. P. and the L. P. cylinders, which is good practice, 
 its area of cross-section would be the square root of the product 
 of the cross-sectional areas of the H. P. and L. P. cylinders. Let 
 d and D denote the diameters of the H. P. and L. P. cylinders 
 respectively, and let x denote the diameter of the intermediate 
 cylinder. Tf <p denotes the ratio between the L. P. and H. P. 
 cylinders, then 
 
 , *D n - , M D- ,. , , D 1 D 
 
 But r- = IT > whence d 1 = . Therefore x 1 = /= , and x -77=-. 
 
 44' <p <* (> 
 
 That is, the diameter of the intermediate cylinder is equal to the 
 
ENGINE DESIGN 127 
 
 diameter of the low-pressure cylinder divided by the fourth root 
 of the ratio of the low-pressure cylinder to the high-pressure 
 cylinder. 
 
 EXAMPLE I. Eequired the dimensions of an engine to develop 
 46 horse-power; piston speed, 625 feet per minute; absolute pres- 
 sure, 95 Ibs. ; cut-off, 0.25 stroke; back pressure, 2 Ibs. above the 
 atmosphere; clearance, 5 per cent. 
 
 Here we have : 
 
 Eatio of expansion = --, = n 05 ^_ Q QK 3.5, the log e of which 
 
 is 1.2528. 
 Theoretical mean pressure = 95(1 + *- 2628 ) = 61.12 Ibs. 
 
 Theoretical mean effective pressure = 61.12 16.7 = 44.42 Ibs. 
 Using a mean pressure factor of 0.9, we shall have : 
 Expected mean effective pressure = 44.42 X 0.9 = 39.978, say 40 
 pounds. 
 
 40 X 625 X 3.1416 d* 
 
 ,, / 46 X 33,000 X 4 . . , 
 
 = V 40X625X3.1416 = 8 ' 8 > Say 9 mcheS ' 
 The piston speed is the product of twice the stroke in feet and the 
 number of revolutions per minute ; so if we assume the stroke to be 
 10 inches, we shall have : 
 
 -D T , . 625 X 12 
 Eevolutions per minute = ^ = 375. 
 
 EXAMPLE II. Eequired the dimensions of a compound engine to 
 develop 2000 I. H. P.; piston speed, 700 feet per minute; absolute 
 initial pressure, 112 Ibs.; back pressure in the condenser, 3 Ibs.; 
 cut-off in high-pressure cylinder, f stroke; clearance of high-pres- 
 sure cylinder, 14 per cent, and of low-pressure cylinder, 12 per cent. 
 
 As the engine is for the mercantile marine, a cylinder ratio of 
 4.25 and a mean pressure factor of 0.8 will be chosen. 
 
 Since the L. P. cylinder is the measure of the power of a stage 
 expansion engine, its diameter is first determined and made suffi- 
 ciently large to develop the whole of the power in that cylinder from 
 the mean pressure derived from the expansion. The whole power is 
 then apportioned equally between the cylinders. 
 
 Ratio of expansion = - = " + = 9.24, the log 
 
 of which is 2.224. 
 
128 THE ELEMENTS OF STEAM ENGINEERING 
 
 112(1 + 2.224) 
 Theoretical mean pressure = ^ ~ . - = 39.08 Ibs. 
 
 Theoretical mean effective pressure = 39.08 3 = 36.08. 
 
 Mean effective pressure to be expected = 36.08 X 0.8 = 28.864 Ibs. 
 
 pLaN ,, 33,000 X 2000 
 
 L H ' R = 3pOO> therefore > a = 28.864 X 700 r= 3266 ' 5 
 inches. 
 
 / QO O K 
 
 Diameter of L. P. cylinder = J '. = 64.49, say 64.5 inches. 
 
 K 
 
 Area of H. P. piston = ^ ' = 768.6 square inches. 
 
 Diameter of H. P. cylinder = $Tf$fo= 31.283, say 31.25 inches. 
 Assuming the stroke as 3 feet, the engine would then have to 
 
 iy(\(\ 
 
 make Q =117 revolutions per minute. 
 
 o X <6 
 
 EXAMPLE III. Find the dimensions of a triple-expansion engine 
 to be used in a ship of the mercantile marine, the horse-power to be 
 developed being 3600. Piston speed, 800 feet per minute; initial 
 absolute pressure, 155 Ibs. ; absolute back pressure, 3 Ibs. ; cut-off 
 in H. P. cylinder at 0.6 stroke. The estimated clearances are 18 
 per cent for the H. P. cylinder and 15 per cent for the L. P. 
 cylinder. 
 
 Assume a ratio of 7 between the volumes of the L. P. and H. P. 
 cylinders, and a mean pressure factor of 0.7, such assumptions 
 being in accordance with good practice. Then, 
 
 r = 7 (1 + O- 15 ) = 10.32, the log e of which is 2.337. 
 U.o +.v/lo 
 
 Theoretical mean pressure = -- 10 32* " ^'^ ^ s ' 
 Theoretical mean effective pressure = 50.12 3 = 47.12 Ibs. 
 Mean effective pressure to be expected = 47.12 X 0.7 = 32.984, 
 
 say 33 Ibs. 
 I. H. P.= S* Before, B = "ffOXMOO = 00 square 
 
 inches. 
 
 Diameter of L. P. cylinder = ,y/^f^ = ? 5 - 69 > sa J 75 - 75 inches - 
 Diameter of the I. P. cylinder = = 46 - 535 > sa J 46 - 5 inches. 
 
 Diameter of the H. P. cylinder = y 7 x Ut7854 = 28.61, say 28f 
 inches. 
 
ENGINE DESIGN 
 
 Assuming a stroke of 30 inches, the engine would have to make 
 
 800 
 2 fr x 2 ~ -^ revolutions per minute. 
 
 If triple-expansion engines of 3600 horse-power were required 
 for a vessel of war, a smaller cylinder ratio, a smaller mean pressure 
 factor, and a longer H. P. cut-off would have been assumed in 
 order to meet the changed conditions, thus:* 
 
 EXAMPLE IV. Find the dimensions of a triple-expansion engine 
 of 3600 horse-power to be used in a vessel of war. Piston speed, 
 800 feet per minute; initial absolute steam pressure, 155 Ibs.; 
 absolute back pressure, 3 Ibs.; cut-off in H. P. cylinder, 0.725 
 stroke. The estimated clearances are, 18 per cent for the H. P. 
 cylinder and 15 per cent for the L. P. cylinder. 
 
 Assume a ratio of 5.25 between the volumes of the L. P. and the 
 H. P. cylinders, and a mean pressure factor of 0.6. Then : 
 
 r = Q5 + o ffi = 6 ' 66 ' the loge f which is L898 ' 
 
 ,, - 155(1 + 1.898) 
 
 Theoretical mean pressure = v * = 67.43 Ibs. 
 
 Theoretical mean effective pressure = 67.43 3 = 64.43 Ibs. 
 Mean effective pressure to be expected = 64.43 X 0.6 = 38.658. 
 
 pLaN ,, 33,000 X 3600 
 
 L H " P '== 3boO' thereforeo= 38.658 X 800 = 3841-4 sq. 
 inches. 
 
 Diameter of L. P. cylinder = 
 
 69 93 
 
 Diameter of I. P. cylinder = 4 JLJ_ = 46.2, say 46.25 inches. 
 
 *y 5./c5 
 
 Diameter of H. P. cylinder = \/ 5 25^ 7854 = 30 - 875 > sa J 31 in * 
 
 137. Steam per I. H. P. per Hour. From the data of Example I, 
 Art. 136, the initial volume of steam per stroke, expressed in stroke. 
 is 10 X 0.25 + 10 X 0.05 = 3 inches. Then, 
 
 Steam used per hour = 8x8.1416X45X^x875x8X60 ^ ^ 
 
 1 / Ao 
 
 cubic feet. 
 
 The specific volume of steam at 95 pounds pressure is 4.62 cubic 
 feet, hence : 
 
 4-Q70 
 Steam per I. H. P. per hour = 4>6a ^ 46 = 23.39 Ibs. 
 
 9 
 
130 THE ELEMENTS OF STEAM ENGINEERING 
 
 It would probably be safe to increase this by 5 per cent, making 
 the steam per I. H. P. per hour 24.56 Ibs. 
 
 138. Coal per I. H. P. per Hour. Assuming, in the example just 
 considered, that the boiler evaporates 8 pounds cf water per pound 
 
 of coal, we would then have, ^ =3.07 pounds of coal expended 
 per I. H. P. per hour! 
 
 139. The steam space in a boiler depends upon the volume of 
 steam used by the engine and not upon the weight of steam used. 
 Two boilers having the same grate and heating surface will evapo- 
 rate the same weight of steam in a given time, but if the pressures 
 differ the volumes of the steam will differ. Slow-running engines 
 with comparatively low steam pressures will, therefore, require boil- 
 ers having larger steam space per unit of power than are necessary 
 for engines of higher speed and higher pressures using the same 
 weight of steam. There should be allowed 0.8 of a cubic foot of 
 steam space per I. H. P. for slow-running pumping engines. This 
 allowance is reduced as the character of the engines calls for higher 
 speeds and higher pressures, until a minimum, probably, of 0.35 
 cubic foot per I. H. P. is reached. 
 
 For the high-speed engine of Example I, Art 136, an allowance 
 of 0.5 cubic foot of steam space per I. H. P. would be sufficient. The 
 steam space of the boiler would then be 46 X 0.5 = 23 cubic feet. 
 
 The steam used per minute by the engine would be -g^- = 82.83 
 cubic feet. The boiler would, therefore, be emptied of steam 
 ?%P = 3.6 times a minute, giving the steam JE-gte 16.67 seconds 
 
 /CO 5.O 
 
 to dry. 
 
 140. Steam Port Dimensions. In determining the proper area of 
 steam ports the controlling influences are that the exhaust must be 
 free, thus minimizing the back pressure, and that excessive clearance 
 must be avoided. Almost invariably the exhaust steam from the 
 cylinder must pass through the same port through which it was 
 admitted, and for that reason the size of the port must be governed 
 by the velocity it is desirable the exhaust steam should have. A 
 mean velocity for the exhaust of 6000 feet per minute is in accord- 
 ance with good practice, so that the area of the ports through the 
 valve should be calculated from that basis. As will be seen later, 
 the port is never completely open for the admission of steam, 
 
ENGINE DESIGN 131 
 
 though the opening should not be so contracted as to give to the 
 admission steam a velocity exceeding 12,000 feet per minute. 
 
 It is evident that for the free escape of the exhaust steam the 
 relation, area of piston X velocity of piston = area of port X ve- 
 locity of steam, must obtain. From this we have : 
 
 Area of steam port = area of piston ^velocity of piston _ 
 
 We shall have, then, for the engine of Example I, Art. 136 : 
 
 A - , 3.1416 X 4.5 X 4.5 X 625 a co . . , 
 
 Area of steam port = - - = 6.624 sq. inches. 
 
 Taking the length of the port as 0.8 the diameter of the cylinder 
 we shall have : 
 
 Width of port = 6 ' 624 = 0.92, say jf inch. 
 
 141. Mechanical Advantage of the Stage Expansion Engine. 
 
 To illustrate the mechanical advantage of the stage expansion 
 engine in producing a reduction in the initial stress, and a decrease 
 in the ratio of initial to mean stress, over that produced with the 
 simple expansive engine, a comparison will be made between a sim- 
 ple expansive engine, with two cylinders, each having an area A/2, 
 and a compound engine of two cylinders, the ratio of whose piston 
 areas is 3. The area of the L. P. piston will then be A, and that of 
 the H. P. piston, A/3. In each case the initial pressure will be 80 
 Ibs. absolute, the ratio of expansion 6, and the back pressure in the 
 condenser 4 Ibs. The receiver pressure of the compound engine will 
 be taken as 22 Ibs. Each engine has the same number of working 
 parts, but a separate expansion valve would be required on each of 
 the cylinders of the simple engine to effect so early a cut-off as, at 
 least, J stroke. 
 
 ( 1 ) The simple expansive engine with two cylinders : 
 
 Mean pressure in each cylinder = - (T^ = 37.22 Ibs. 
 
 4 
 
 Effective initial load on each piston = (80 4)^- = 38 A Ibs. 
 
 Effective mean load on both pistons 33.22 X A Ibs. 
 
 Efficiency of the svstem Mean effec. load obtained = 33MA 
 
 Mean effec. load expected 33.22A 
 
 (2) The two-cylinder compound engine: 
 
 If the volume of the H. P. cylinder be denoted by unity, that of 
 the L. P. cylinder must be denoted by 3, and if v^ denotes the vol- 
 ume of the H. P. cylinder up to cut-off, we shall have : 
 
132 THE ELEMENTS OF STEAM ENGINEERING 
 
 3 3 
 
 - = r, or - = 6, whence v = 0.5 cut-off in H. P. cylinder. The 
 
 ratio of expansion in the H. P. cylinder is, therefore, TT-= = 2. 
 
 u.o 
 
 To insure a uniform pressure in the receiver,, the product of the 
 volume and pressure given to it by the H. P. cylinder must equal the 
 product of the volume and pressure taken from it, and we shall 
 have 80 X 0.5 = 22 X 3 X &, whence Tc = 0.606 cut-off in L. P. 
 cylinder to maintain a receiver pressure of 22 Ibs. 
 
 The ratio of expansion in L. P. cylinder = rA 1-65. 
 
 Mean pressure in H. P. cylinder = 80(1 +^' 6931 ) _ 67>7 2 lbs> 
 
 ,, 22(1 + 0.5008) 
 
 Mean pressure in L. P. cylinder = - .. -- - = 20.03 Ibs. 
 
 Mean effective pressure in H. P. cylinder = 67.72 22 =. 45.72 Ibs. 
 Mean effective pressure in L. P. cylinder =. 20.03 : 4 = 16.03 Ibs. 
 
 Effective initial load on H. P. piston = (80 22) ^ 19.331 Ibs. 
 
 Effective initial load on L. P. piston =(22 4) A 181 Ibs. 
 Mean effective load obtained on both pistons 
 
 = 45.72 X 4 + 16.031 = 31.271 Ibs. 
 o 
 
 Mean pressure due an initial pressure of 80 Ibs. and a ratio of 
 
 expansion of 6, is 80(1 + g 1 - 7918 ) = 37.32 Ibs. 
 Mean effective load expected = (37.22 4)1 = 33.221 Ibs. 
 Efficiency = =0.941. 
 
 In comparing the results it is found that the initial load on each 
 of the pistons of the simple engine is 381 Ibs., while that on the 
 H. P. piston of the compound engine is 19.381 Ibs., and on the 
 L. P. piston 181 Ibs., showing that the initial loads on the pistons 
 of the compound engine are practically equal, and their aggregate 
 less than that on either of the pistons of the simple engine. 
 
 The ratio of the mean load to the initial load on each piston of 
 
 33.22 X^ 
 
 the simple engine is ao . = ^ =-0.437, or 43.7 per cent. 
 
 OoA on 
 
 The same ratio for the pistons of the compound engine are as fol- 
 lows: 
 
ENGINE DESIGN 133 
 
 45.72 X 4 
 On H. P. piston = 19 33^ = - 788 > or 78 - 8 P er cent - 
 
 On L. P. piston = = 0.89, or 89 per cent. 
 
 In consequence of this reduction in the initial and mean loads, 
 the various parts of the compound engine whose size depends on 
 these factors may be made lighter than those of the simple engine, 
 and the friction on the guides and on the bearings will be much less 
 severe. The ratios of the mean to the initial loads being much 
 nearer unity in the case of the compound engine than with the 
 simple engine, the turning moment is much more uniform and the 
 engine will run more steadily. The question of uniform load and 
 steadiness in running is one of great importance in marine practice, 
 but with stationary engines where the use of the fly wheel is permis- 
 sible its importance is lessened. 
 
 The mean effective loads on the pistons of the compound engines 
 show the work to be very evenly divided between the cylinders, 
 but the total work falls short, of that of the simple engine by 
 1 0.941 = 0.059, or 5.9 per cent. This, loss may be attributed 
 to " drop," which, in this case, is - 8 / 22 = 18 Ibs. 
 
 A consideration of the range of temperatures in the cylinders will 
 show a very marked advantage in favor of the compound engine. 
 
 The temperature of steam at 80 Ibs. pressure is 312, and that of 
 steam at 4 Ibs. pressure is 153. The range of temperature is then 
 312 153 = 159 in each cylinder of the simple engine. 
 
 The temperature of steam at 22 Ibs. pressure is 233, therefore 
 312 233 = 79 is the range of temperature in the H. P. cylinder 
 of the compound engine, and 233 153 = 80 is the range of tem- 
 perature in the L. P. cylinder. It is thus seen that the range of 
 temperature in each cylinder of the compound engine is just about 
 one-half that in the cylinders of the simple engine. The advan- 
 tages which would accrue from this distribution of temperature are 
 set forth in Art. 58, p. 38. 
 
 142. Crank-pin -Piston Velocity Diagram. In some problems 
 connected with steam engine design, it is required to know the 
 velocity of the piston when the velocity of the crank-pin is known ; 
 this can very readily be done. 
 
134 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 Let (7P, Fig. 33, be the position of the crank when the connecting- 
 rod PQ makes an angle a with the center line of the engine. Pro- 
 duce the connecting-rod until it intersects the vertical through C 
 at R. The direction of P at any point is in the tangent at the point ; 
 and if we denote the velocity of the crank-pin by V, and that of the 
 
 piston by v, we shall have the resolved velocities of P and Q along 
 PQ equal, since PQ is a rigid rod. 
 Hence : 
 
 V sin <f> =. v cos a. 
 Consequently 
 
 s * n ^ _ sinp _ CR 
 V " 5ca . sin CRQ ~~ ~UP ' 
 
 Now if, to any scale, OP denotes the assumed constant velocity of 
 the crank-pin, then CR will, to the same scale, denote the velocity of 
 the piston. 
 
 We may construct a curve of piston velocities, as follows : 
 
 Let CP and PQ, Fig. 34, be the positions, respectively, of the 
 
 crank and connecting-rod, corresponding to the piston position Q. 
 
 Produce QP until it intersects the vertical through C at R. With 
 
 C as a center and CR as a radius, describe the arc Rr, intersecting 
 
ENGINE DESIGN 135 
 
 CP at r; then r is a point of the curve. The locus of r is the 
 polar curve of piston velocities, the radius vector in line of the 
 crank representing the piston velocity for that crank position, to 
 the same scale that the length of the crank represents the velocity 
 of the crank-pin. If the connecting-rod were of infinite length, the 
 locus of r would be two circles described on the vertical positions of 
 the crank as diameters. 
 
 The velocities of the crank-pin and piston are equal when the 
 crank is in its vertical positions, and again at crank positions Ct 
 and Ct', where the connecting-rod produced passes through 8 and 8'. 
 Between the points 8 and t, and 8' and t r , the velocity of the piston 
 is greater than that of the crank, and at all other points it is less. 
 
 The piston has its maximum velocity very nearly when the crank 
 and connecting-rod are at right angles, or when the connecting-rod 
 is tangent to the crank-pin circle. 
 
 PEOBLEMS. 
 
 48. Stroke of engine, 24 inches; initial absolute pressure of steam 
 at cylinder, 100 Ibs., but wire-drawing reduces the pressure to 95 Ibs. 
 at cut-off. Back pressure, 18 Ibs.; cut-off, 0.25 stroke; clearance, 
 10 per cent. Eelease takes place at 90 per cent of stroke, and ex- 
 haust closes after 75 per cent of the return stroke is completed. 
 The expansion and compression being hyperbolic, construct the ex- 
 pected indicator diagram, and find the terminal pressure, p 2 ; the 
 final pressure of compression, p c ; the mean effective pressure, p e , 
 by means of ordinates; and the estimated steam consumption per 
 I. H. P. per hour. 
 
 Ans. 2 = 31.8; p c = 63; p e 42.8;. steam 
 per I. H. P. per hour 22.47 Ibs. 
 
 49. Stroke of engine, 20"; diameter of cylinder, 18"; revolutions 
 per minute, 180; clearance, 5 per cent; terminal pressure, 28 Ibs.; 
 back pressure, 17 Ibs.; ratio of expansion, 3.5; ratio of compres- 
 sion, 4. The specific volumes of steam at the initial and back pres- 
 sures are 4.44 cubic feet and 23 cubic feet respectively. Using the 
 properties of the indicator diagram, it is required to find the mean 
 effective pressure, and assuming that 85 per cent of it will be real- 
 ized, find the I. H. P., and the number of pounds of steam used per 
 I. H. P. per hour. 
 
 Ans. p e 42.17; I. H. P. = 165.73; steam per I. H. P. 
 per hour, 22.6 Ibs. 
 
136 THE ELEMENTS OF STEAM ENGINEERING 
 
 50. An 18.5" x 30" engine makes 129 revolutions per minute, the 
 consumption of fuel being 5.5 long tons a day. The mean effective 
 pressure is 36 Ibs., and 26 pounds of steam are used per I. H. P. 
 per hour. Find the number of pounds of coal burned per I. H. P. 
 per hour, and the number of pounds of water the boiler must evapo- 
 rate per pound of coal. 
 
 Ans. 2.73 Ibs. of coal per I. H. P. per hour, and 9.524 Ibs. 
 of water evaporated per pound of coal. 
 
 51. An engine, 18.5" x 30", makes 129 revolutions per minute. 
 The mean effective pressure is 36.6 Ibs. per square inch, and the 
 consumption of coal per I. H. P. per hour is 3 Ibs. The weight of 
 steam used per stroke is 0.3478 Ib. Find the number of pounds of 
 steam used per I. H. P. per hour; the coal consumption (long tons) 
 per day; the number of pounds of water the boiler must evaporate 
 per pound of coal. 
 
 Ans. Pounds of steam per I. H. P. per hour, 28 ; tons of 
 coal per day, 6.18 ; pounds of water to be evapo- 
 rated per pound of coal, 9.33 Ibs. 
 
 52. Find the cylinder dimensions of an engine to develop 75 
 horse-power. Piston speed, 600 feet per minute ; absolute pressure 
 of steam, 95 Ibs.; cut-off, 0.3 stroke; back pressure, 2 Ibs. above 
 the atmosphere; clearance, 5 per cent. 
 
 53. Required the dimensions of a compound engine for use in the 
 mercantile marine to develop 1800 I. H. P. Piston speed, 725 feet 
 per minute; absolute initial pressure, 112 Ibs.; back pressure in the 
 condenser, 2 Ibs. ; cut-off in H. P. cylinder, 0.4 stroke. Estimated 
 clearances: in H. P. cylinder, 10 per cent; in L. P. cylinder, 12 
 per cent. 
 
 54. Find the dimensions of a triple-expansion engine to be used 
 in the mercantile marine, the horse-power to be developed being 
 4000. Piston speed, 800 feet per minute ; initial absolute pressure, 
 160 Ibs.; absolute back pressure, 3 Ibs.; cut-off in H. P. cylinder, 
 0.6 stroke. The estimated clearances are: 16 per cent for the 
 H. P. cylinder and 14 per cent for the L. P. cylinder. 
 
 55. What steam-port dimensions should be given to an engine 
 designed to have a piston speed of 800 feet a minute, the diameter 
 of the cylinder being 15 inches? 
 
CHAPTEE XII. 
 THE ZEUNER VALVE DIAGRAM. 
 
 143. The essential features and functions of the slide valve have 
 already been considered, and it is now the intention to consider, 
 with a degree of particularity sufficient for practical purposes, the 
 main features of its design. 
 
 No detail of the steam engine is of more importance than the 
 correct design of its valve gear, which includes the valve and the 
 mechanism that gives it motion. A defective design may occasion 
 a loss in fuel as great as 20 per cent, and cause an unevenness in 
 the motion of the engine, with a resulting wear and tear in the 
 working parts that may lead to serious casualty. 
 
 The angularities of the connecting and eccentric rods introduce 
 irregularities into the motion of the piston and of the valve. Mathe- 
 matical formula showing their relative positions during the dis- 
 tribution of the steam in the cylinder are too abstruse and com- 
 plicated for practical use, but graphic methods have been de- 
 vised to simplify the problem, of which that of Zeuner is 
 as simple and complete as any. It permits, without difficulty, 
 the angularity of the connecting-rod to be taken into account, 
 though not that of the eccentric rod; but since the length of the 
 eccentric rod is great compared with the throw of the eccentric, its 
 angularity introduces an inappreciable irregularity, and the motion 
 given to the valve is assumed to be harmonic. The rule and com- 
 passes are the only instruments needed in the construction of this 
 diagram, and its accuracy depends largely upon the skill of the 
 draftsman. 
 
 144. Take AB, Fig. 35, equal to the travel of the valve, and bisect 
 it at 0. Let represent the center of the shaft, OA the direction 
 of the crank when on the dead center, and OE the direction of the 
 center line of the eccentric, the angular advance being denoted by 0. 
 On AB as a diameter describe a circle. The intersection, E, of 
 this circle with OE will represent the center of the eccentric, since 
 the half-travel of the valve is equal to the distance between the 
 center of the shaft and the center of the eccentric. 
 
138 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 The motion of the valve being assumed harmonic, the perpen- 
 dicular, EC, let fall from E upon AB, gives 00 as the distance the 
 valve has moved from its mid-position (see Fig. 16). As the shaft 
 rotates the crank will assume the directions Oa, Oa', etc., and the 
 eccentric the positions Oe, Oe', etc.; and if from e and e' perpen- 
 diculars be let fall upon AB, the distances Of and Of intercepted 
 between the center of the circle and the feet of the perpendiculars 
 will represent the distances the valve has moved from its central 
 position when the crank has the directions Oa and Oa'. 
 
 Instead of the crank and eccentric moving in their circular paths, 
 suppose them to remain fixed and the radius OB to revolve in the 
 
 opposite direction and assume the positions 06 and 06', corres- 
 ponding to Oa and Oa'. From E let fall perpendiculars Es, ES* 
 on Ob and 06'. By similar triangles it is easily shown that 05, Os f 
 are equal to Of, Of', therefore Os, Os' represent equally well the 
 movements of the valve from its central position when the crank 
 assumes the directions Oa, Oa'. Since the angles OsE, Os'E are 
 right angles, the locus of the points s, s' is the circumference of a 
 circle described on OE as a diameter. 
 
 If, then, upon OE as a diameter a circle be described, the radii 
 vectors Os, Os' give the distances the valve has moved from its 
 central position for the angular positions 06, 06' of the crank, it 
 being remembered that for the eccentric positions Oe, Oe' the real 
 crank positions would be Oa, Oa'. 
 
THE ZEUNER VALVE DIAGRAM 
 
 139 
 
 This artifice is usual and convenient in the construction of the 
 Zeuner diagram and contributes to its simplicity. 
 
 A precisely similar construction and reasoning will show that a 
 circle described on a diameter OE', equal to and diametrically oppo- 
 site OW, so that EE' is a straight line, will give, by means of the 
 crank intercepts, the movement of the valve during the return 
 stroke. 
 
 We now have a diagram which shows the distance the valve has 
 moved from its central position for any position of the crank, and 
 therefore the corresponding position of the piston may be found. 
 
 145. The angular advance of the eccentric may be obtained as 
 follows : 
 
 Describe a circle, AEBE', Fig. 36, with the throw of the eccen- 
 tric as a radius. Let OA be the direction of the crank when on 
 
 36 
 
 the dead center. From set off in a direction away from the crank 
 the distance OC equal to the lap plus the lead. Draw the vertical 
 CE, cutting the circle at E. Then OE will be the position of the 
 eccentric radius when the crank is on the dead center, and the angle 
 DOE will be the angular advance of the eccentric. This follows 
 from the fact that the valve has moved from its central position a 
 distance equal to the lap plus the lead when the crank is on the 
 dead center. The above construction for the angular advance holds 
 when the valve takes steam at its outside edges; but if, as is not 
 infrequently the case, the valve takes steam at its inside edges, the 
 lap plus the lead must be laid off from towards the crank, and 
 OE' will then be the position of the eccentric, and DOE' the angle 
 of advance. 
 
 146. With a radius equal to the throw of the eccentric, or the half- 
 travel of the valve, describe the circle AEC', Fig. 37. Let OA be 
 
140 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 the direction of the crank when on the dead center, and OE the 
 direction of, and equal to, the throw of the eccentric. Draw the 
 diameter DD' perpendicular to OA. From E drop a perpendicular 
 to AB, intersecting it at N. ON is the distance the valve has 
 moved from its mid-position when the crank is on the dead center. 
 From set off ON' equal to ON, and draw N'C perpendicular to- 
 AB. Draw the diameter CO'. The angle COD angle DOE = 
 angular advance of the eccentric. If the crank move to Oa, the 
 eccentric will take the position Oe, such that the angles AOa and 
 EOe are equal. Drop ef perpendicular to AB. Then Of will be 
 the distance the valve has moved from its central position for the 
 
 crank position Oa. On Oa set off Of Of. From what has been 
 shown in Fig. 35, all points such as N' 9 f will lie in the circum- 
 ferences of two circles described on the diameters OC and OC' t 
 which are each equal to the half travel of the valve, and make an 
 angle with the perpendicular to the line of dead centers equal to 
 the angular advance of the eccentric, measured in a direction oppo- 
 site to that of the motion of the crank. 
 
 If, with as a center and a radius OL equal to the outside lap 
 of the valve, a circle be described, the distance the port is open for 
 the admission of steam from any crank position, as Oa, is given by 
 the intercepted part rf. This will be well understood from the 
 fact that the opening of the port is equal to the movement of the 
 valve from its central position, minus the lap. 
 
 Since ON' is the movement of the valve from its central position 
 when the crank is on the dead center, A, it must be equal to the 
 
THE ZEUNER VALVE. DIAGRAM 
 
 141 
 
 lap plus the lead, and since OL is the lap, N'L must, therefore, be 
 the lead. 
 
 We may now construct a complete diagram and, in order that 
 it be not too complicated, it will be drawn for the steam and 
 exhaust on only one side of the piston, the side remote from the 
 shaft, and the lap, the lead, etc., determined by the diagram will 
 be for the end of the valve remote from the shaft. 
 
 The stroke of the piston from the head end to the crank end of 
 the cylinder is known as the forward stroke (top stroke in vertical 
 engines), and from the crank to the head end, the return stroke 
 (bottom stroke in vertical engines). The diagram will, therefore, 
 give the necessary dimensions of the end of the valve to obtain the 
 
 38. 
 
 admission and cut-off of the steam for the forward stroke, and the 
 release and compression of the same steam on the return stroke. 
 
 With a center 0, Fig. 38, and radius OA equal to the half-travel 
 of the valve, describe a circle. A3 being the line of dead centers, 
 the diameter DD' is drawn perpendicular to it, and the angle DOE, 
 equal to the angular advance, is laid off from DO in a direction 
 contrary to that of the motion of the crank as indicated by the 
 arrow. On the radii OE and OE' of the diameter EE f as diam- 
 eters, describe the circles OyE and OXE'. For the stroke under 
 consideration the first of these circles is known as the primary valve 
 circle, or steam circle, and the second as the secondary valve circle, 
 or exhaust circle. 
 
14:2 THE ELEMENTS OF STEAM ENGINEERING 
 
 With a radius OL equal to the steam lap of the valve, describe 
 the arc sLsf. Now, since the admission and cut-off of the steam 
 must take place when the distance of the valve from its central 
 position is equal to the lap, it follows that s and s', the intersec- 
 tions of the lap circle with the steam circle, give the crank positions 
 OF and 00 for admission and cut-off respectively. In like man- 
 ner, release and compression occur when the distance of the valve 
 from its central position is equal to the inside lap. If, then, with 
 as a center and a radius OL' equal to the inside lap, an arc be 
 described, its intersections e and e' with the exhaust circle will give 
 the crank positions OQ and OQ' for release and compression respec- 
 tively. 
 
 Should the exhaust lap be negative, which is not infrequently 
 the case for the top stroke of vertical engines, the intersections of 
 the exhaust lap circle with the steam circle must be taken to get 
 the crank positions for release and compression. 
 
 This will become evident after a consideration of the fact that 
 the intercept on the crank position by the steam or the exhaust circle 
 gives, in every instance, the distance of the valve from its central 
 position. The valve is, therefore, centered when the intercept is 
 zero, that is, when the crank is in the position OP or OP', tangent 
 to both circles; PP' is, therefore, perpendicular to EE'. Now, if 
 the exhaust lap is positive, the crank will have to revolve beyond 
 OP before the port opens to release, and will be intersected by the 
 exhaust circle. If, however, the exhaust lap be negative, release 
 will have taken place before the valve is centered, or before the 
 crank reaches the position OP, and the intercept will, therefore, be 
 made by the steam circle. 
 
 For any position of the crank, as OF, Oy is the distance of the 
 valve from its central position, and the intercepted portion xy 
 between the lap and the steam circles is the amount the port is open 
 for the admission of steam. The intercepts in the hatched area 
 of the steam circle show port openings to steam, and therefore LG 
 is the measure of the lead. 
 
 From the relative positions of the crank and eccentric and the 
 harmonic motion of the valve, it is seen that from the crank posi- 
 tions OF to OE the opening of the port to steam continually in- 
 creases, at first quickly and finally slowly; from OE to 00 the 
 valve is closing the port, at first slowly and finally quickly. 
 
THE ZEUNER VALVE DIAGRAM 143 
 
 The intercepts in the hatched area of the exhaust circle show 
 openings of the port to exhaust. The port will, of course, be wide 
 open to exhaust when the valve has moved from its central position 
 a distance equal to the exhaust lap plus the width of the port; and 
 any further distance the throw of the eccentric may cause the valve 
 to move will result in overtravel. If with as a center and a 
 radius equal to the exhaust lap plus the width of the port, an arc 
 XZ be drawn, the port will be shown wide open to exhaust during 
 the period from X to Z, and E'E" will show the overtravel. 
 
 To draw the diagram for the return stroke, the exhaust circle 
 for the forward stroke becomes the steam circle for the return 
 stroke, and vice versa. 
 
 If the steam and exhaust laps are the same for both strokes, the 
 lap circles must be completed so as to intersect the opposite circles, 
 and thus give the corresponding events of the return stroke. Fre- 
 quently, however, they are different for the two strokes, in which 
 event the correct radii must be used. 
 
 This is essentially the Zeuner valve diagram, and by its applica- 
 tion the numerous problems of the slide valve may be solved. By 
 varying some of the points, the alterations in the others may easily 
 be found ; and by assuming certain elements the remainder may be 
 determined. 
 
 Some confusion exists at times as to what is meant by the differ- 
 ent names given to the strokes of the piston. It should be remem- 
 bered that the stroke known as -forward, outward, top, head, or 
 down, is that in which the piston moves toward the shaft; and the 
 stroke known as return, inward, bottom, crank, or up, is that in 
 which the piston moves away from the shaft. 
 
 147. There are a number of geometrical features of the diagram 
 which are very useful in the solution of problems. 
 
 1. The line joining the points of admission and cut-off is tan- 
 gent to the lap circle. 
 
 In Fig. 39 let OF and 00 be the crank positions, respectively, 
 for admission and cut-off, and let N'LN be an arc of the lap circle. 
 Join PC, cutting EE' at some point M. It is required to prove 
 that M is the point of tangency of PC and the arc N'LN. The 
 triangle FOG is isosceles, and, since ME is the greatest opening of 
 the port, it is midway between the admission and cut-off of the 
 steam; therefore EE f bisects the vertical angle FOC, and is per- 
 
144 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 pendicular to the base FC and bisects it at M.. Draw EN, and in 
 the right triangles EON and COM, we shall have the hypotenuse 
 CO equal to the hypotenuse EO. The acute angles at E and at C 
 are equal because their sides are respectively perpendicular to each 
 other ; therefore the triangles are equal, and ON = OM. But ON 
 is the lap ; therefore if, with as a center and ON as a radius, the 
 arc N'LN be described it will be tangent to FC at M . 
 
 A similar construction and reasoning will show that a line join- 
 ing the point of release, Q, with the point of exhaust closure, or of 
 compression, Q f , will be tangent to the exhaust lap circle. 
 
 2. If, from the dead point as a center, an arc be described that 
 will be tangent to the line joining the points of admission and 
 cut-off, the radius of this circle will be equal to the lead. 
 
 Fiy. 39 
 
 With dead point A, Fig. 39, as a center, describe an arc tangent 
 to FC', then its radius AT will be equal to the lead. Through A 
 draw A8 parallel to FC, and therefore perpendicular to EE'. The 
 right triangles ASO and EGO are equal, and GO 80. But 
 LO = MO, therefore GL = 8M = AT = the lead. 
 
 3. If the points of admission and cut-off, and the maximum 
 opening of the port to steam were given, or if the points of release 
 and compression and the width of port be given (taking the maxi- 
 mum opening of port to exhaust as the width of port), there would 
 not be sufficient data to construct the diagram. We can, however, 
 determine from the data given, in either case, the travel of the valve 
 and then construct the diagram. 
 
THE ZEUNER VALVE DIAGRAM 
 
 145 
 
 With center 0, Fig. 40, describe the indefinite circle EQQ'. Let 
 Q and Q' be the points, respectively, of release and compression. 
 These points would be given by the angles BOQ and AOQ f , or by 
 some definite part of the stroke of the piston, now represented in- 
 
 Fi'q. 
 
 definitely by AB. Join QQ', and from what has preceded we know 
 that QQ' will be tangent to the exhaust lap circle, and that EE', the 
 perpendicular to QQ' through 0, will give the position of the eccen- 
 tric, and, indefinitely, the diameters of the steam and exhaust cir- 
 cles. WE' will then represent to some scale the maximum opening 
 to exhaust, which is assumed to be "the width of the port, given in 
 this case as 1.5 inches. We would then have the proportion : 
 
 Travel of Valve _ EE' _ EE' Travel 2.5 
 
 Width of Port " WE' 
 
 The practical method of determining the travel would be as fol- 
 lows: 
 
 Draw ab and ac, Fig. 41, making any convenient angle with 
 each other. Lay off ad = WE' of Fig. 40 = || inch by actual 
 10 
 
146 THE ELEMENTS OF STEAM ENGINEERING 
 
 measurement. This represents the width of the port, in this case 
 1.5 inches. Lay off ae = 1.5 inches, and join de. Our scale is 
 now complete. Lay off on ac the distance of = EE r of Fig. 40 = 2.5 
 inches by measurement., and it will represent the travel of the valve. 
 
 Draw fg parallel to de ; then ag = 4 inches = travel of the valve. 
 The diagram can now be constructed. 
 
 If the point of admission and cut-off, F and (7, Fig. 40, and WE 
 the maximum opening of the port to steam were given, the travel of 
 the valve could be determined in a similar manner. 
 
 Travel of Valve EE' 
 
 have > HSTD^iS^^teSSTtot ~ WE ' 
 
 Should the point of admission, point of cut-off, and the lead be 
 given, the travel could be found from the proportion 
 Travel of Valve EE' 
 
 Given Lead 
 
 4. Since EGO, Fig. 39, is a semi-circle, and one extremity of 
 its diameter, if at the extremity, G, of the chord, OG, (equal to the 
 lap plus the lead) a perpendicular be erected, it will meet the cir- 
 cumferenc at E, the other extremity of the diameter. 
 
 148. An examination of Fig. 38 reveals 12 salient points of in- 
 formation, viz.: 
 
 1. Admission at F. 
 
 2. Cut-off at C. 
 
 3. Eeleaseat. 
 
 4. Compression at Q'. 
 
 5. Steam Lap, OL. 
 
 6. Exhaust Lap, OL'. 
 
 7. Steam Lead, LG = A T. 
 
 8. Exhaust Lead, L'G'. 
 
 9. Travel of Valve, EE'. 
 
 10. Maximum Opening of Port to Steam, EE. 
 
 11. Maximum Opening of Port to Exhaust, R'E'. 
 
 12. Angular Advance of the Eccentric, DOE. 
 
 Points 1, 2, >nd 4 are frequently given in angular units, and 
 these, with some other point in linear units, are sufficient to con- 
 struct the diagram. These points may be given as occurring at 
 certain points of the stroke, and in order to find the corresponding 
 crank positions the circle described on EE' (equal travel of the 
 valve) as a diameter may be taken to represent the travel of the 
 
Given 
 Connecting root, *4O 
 Crvnk , /O. 
 
 for 
 
 of the H P. Cy/. o-f o triple -X- 
 /fbsolute steom pressure. 195 1bs. 
 
 Steom Jap, -top H ; bottom 
 Exhaust tap, 1 'op ^ j 
 
 Travel,3%". 
 
 Mean cut-off, 7O %. 
 
 top. 
 
 Found 'from 
 
 91 " 
 
 opening, fop g^', bottom 
 Ex. open/'nq, Ful/ port. 
 "' 
 
 Release "top 0nd ' bottom, fa from end 
 
148 THE ELEMENTS OF STEAM ENGINEERING 
 
 valve on a new scale; or another circle described with as a 
 center, and to any convenient scale, may be taken as the crank 
 circle. 
 
 Points 1, 2, 5, 7, and 10 belong to the steam side of the diagram, 
 and points 3, 4, 6, 8, and 11 to the exhaust side, points 9 and 12 
 being common to both. 
 
 Three points are sufficient data to construct either the steam 
 side or the exhaust side alone, but for the complete diagram four 
 points must be given, and one or two of these must belong to a 
 different side than the others. One of the given points must be in 
 linear units, for if only angles were given the linear dimensions of 
 the valve might be made anything we please, so long as they bear 
 a certain ratio to each other. 
 
 If the width of the port in the face of the cylinder be given as a 
 part of the data, it may be taken as the maximum opening of the 
 port to exhaust, since the opening to exhaust will be greater than 
 the opening to steam, and the port is made only sufficiently wide 
 to allow a full opening. 
 
 If the cut-off be one of the given points it is understood to be 
 the mean cut-off. The obliquity of the connecting-rod occasions a 
 longer cut-off on the stroke toward the shaft than on the stroke 
 away from it, and a compromise must be made to secure the de- 
 sired mean cut-off. 
 
 Figure 42 is a complete diagram for the valve of the high-pressure 
 cylinder of a triple-expansion marine engine. The theoretical in- 
 dicator diagrams are shown dotted and the diagrams that may 
 reasonably be expected appear in full lines. 
 
 PEOBLEMS. 
 
 56. Travel of valve, 4.25 inches; lead angle, 6; angle at cut- 
 off, 105. Find the lap, lead, and the angular advance of the 
 eccentric. Scale, full size. 
 
 Ans. Lap, 1 A inches; lead, T \ inch; angular advance, 40 30'. 
 
 57. Stroke of engine, 3 feet; length of connecting-rod, 5 feet; 
 steam lap, 1 inch; exhaust lap, J inch; width of port, 2 inches; 
 overtravel of the exhaust port, -J- inch. Compression begins when 
 the piston is one-ninth of the stroke from the end. Find the travel 
 of the valve, angular advance, lead, and the point of cut-off ex- 
 
THE ZEUNER VALVE DIAGRAM 149 
 
 pressed as a decimal of the stroke. Scale of stroke, -J, and of dia- 
 gram, full size. 
 
 Ans. Travel, 4.75 inches; lead, ^ inch; angular advance, 
 8; cut-off, 84.5 per cent. 
 
 58. Crank angle at cut-off, 105; lead angle, 8; maximum 
 port opening to steam, inch. Find the travel of the valve, the 
 lap, the lead, and the angular advance. 
 
 Ans. Travel, 3.875 inches; lap, 1 ^ inches; lead, -g^-inch; 
 angular advance, 42. 
 
 59. Cut-off, 0.7 stroke; lap, 1 inch; lead, -^ inch. Show that 
 the angular advance is 36 25' 28", and that the travel of the valve 
 is 4 inches. 
 
 60. Travel of valve, 5 inches; width of port, 2 inches; opening 
 of port to steam, 1.25 inches; exhaust lap, f inch; lead, -f$ inch. 
 Find the angular advance, lap, point of cut-off, point of release, 
 point of compression, and overtravel. 
 
 Ans. Angular advance, 35 6'; lap, 1.25 inches; cut-off, 
 0.71; release, 94.5 per cent; compression, 86 per 
 cent; overtravel, -J inch. 
 
CHAPTEB XIII. 
 ECONOMY OF THE STEAM ENGINE AND BOILER. 
 
 149. Steam per I. H. P. per Hour from Actual Indicator Dia- 
 grams. One of the important uses of the indicator diagram is 
 to determine from it the consumption of steam per unit of power, 
 and thus get a measure of the economy of the engine. 
 
 The volume of steam in the cylinder at any point of the stroke 
 is equal to the displacement of the piston at that point, plus the 
 volume of the clearance space. If this volume be taken at cut-off, 
 or at any point after cut-off, and be divided by the specific volume 
 of steam at the absolute pressure at the point considered, the quo- 
 tient will be the weight in pounds ; deducting from this the weight 
 of the steam compressed at each stroke, we get the weight of steam 
 supplied to the cylinder per stroke measured at the point con- 
 sidered, and as shown by the indicator diagram. If the weight of 
 steam per stroke be multiplied by the number of strokes per hour, 
 and the product be divided by the horse-power of the engine, the 
 quotient will be the number of pounds of water used per I. H. P. 
 per hour. The pressure at the point considered is found from the 
 diagram, and the volume per pound may be taken from a table of 
 the properties of saturated steam, or it may be calculated by means 
 of the formula pv& = 482. 
 
 The consumption of steam per I. H. P., obtained as above ex- 
 plained, is based on the weight of steam shown by the indicator 
 diagram to have left the cylinder in the exhaust, but it does not 
 include the waste of steam due to liquefaction in the cylinder, 
 dampness of the steam, radiation, and other causes of loss. The 
 result is valuable as a means of comparing one engine with another, 
 or the performances of the same engine under different conditions. 
 
 If dry saturated steam were used, and if there were no lique- 
 faction during expansion, the quantity calculated from the diagram 
 would represent with approximate accuracy the weight of steam 
 used. The conditions of actual practice are such that from 20 
 to 30 per cent should be added to obtain results which are approxi- 
 mately correct. 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 151 
 
 Since there is always water in the cylinder at the point of cut-off, 
 and since more or less of this water is re-evaporated into steam as 
 the piston advances, the quantity of steam in the cylinder varies 
 during the stroke. During the exhaust, too, some re-evaporation 
 takes place. For these reasons, the consumption of steam shown 
 by the diagram is less than the quantity actually passing through 
 the cylinder. In stage expansion engines the steam used in the 
 low-pressure cylinder first passes through the high-pressure and 
 intermediate cylinders, and consequently the steam consumption of 
 the high-pressure cylinder will be the measure of consumption of 
 the whole engine. If the measure be taken also from the diagrams 
 of the intermediate and low-pressure cylinders, as it should be for 
 purposes of comparison, it will be found that the results from the 
 high-pressure diagram will be the greatest, and that the differences 
 between the cylinders may be regarded as fair measures of the loss 
 in transmission. 
 
 In computations concerning consumption of steam by an engine 
 in any given time, the volume of the cylinder and the number of 
 revolutions are, of course, factors; and as these same factors ap- 
 pear in the power developed by the engine, it follows that they 
 need not appear in computing the steam consumption per unit of 
 power. By reason of this fact a constant quantity, 13,750, has 
 been determined, the employment of which greatly simplifies cal- 
 culations, enabling us to ascertain the steam consumption per 
 I. H. P. of an engine from the indicator diagram alone, and quite 
 independently of the size and speed of the engine. 
 - The use of this constant is based on the consideration that, if a 
 piston one inch square in area move 12 inches, the displacement 
 
 will be 12 cubic inches, or ^^ of a cubic foot, and the work done 
 
 will be one foot-pound for each pound pressure per square inch. 
 
 Since there are 33,000 X 60 foot-pounds in a horse-power con- 
 tinuously developed for an hour, it follows that there must be a 
 
 piston displacement of ? 4 6 = 13,750 cubic feet per I. H. P. 
 
 per hour when the mean effective pressure is unity, regardless of the 
 size or speed of the engine; and since the volume of steam per 
 I. H. P. is inversely as the mean effective pressure, the quotient of 
 
 n displacement in cubic feet per I. H. P. 
 
152 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 when the mean effective pressure is p e pounds. Knowing the spe- 
 cific volume of steam at pressure p c , its reciprocal will be the weight 
 of a cubic foot, and denoting this weight by w, we shall have 
 
 13,750 X w 
 
 - - as the number of pounds of steam per I. H. P. per 
 
 hour, exclusive of losses from liquefaction,, leakage, etc., and mak- 
 ing no allowance for clearance, compression, and expansion. 
 
 The constant 13,750 is the number of cubic feet of steam to be 
 displaced by an engine in the development of one I. H. P. per hour, 
 the pressure being one pound per square inch, and there being 
 neither clearance, compression, release, nor expansion. 
 
 Engines, of course, are never worked under such conditions, but 
 the conditions of actual practice may be reduced to make the con- 
 
 +JD6* 
 
 stant 13,750 applicable to the determination of the steam consump- 
 tion per I. H. P. per hour of any engine, as shown by the indicator 
 diagram, if the diagram be given and its scale be known. 
 
 In computing the steam consumption from the indicator diagram 
 it will be well to make the calculations both at cut-off and at 
 release, choosing points on the expansion curve as close as possible 
 to the points of cut-off and of release, consistent with the conditions 
 that the port be closed to steam in the one case and not yet open to 
 exhaust in the other. Should there be no liquefaction during ex- 
 pansion the two quantities would be the same. 
 
 Figure 43 is an indicator diagram from a high-speed engine, the 
 scale of the diagram being 40, and the clearance of the engine 
 6 per cent. 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 153 
 
 To ascertain the steam consumption per I. H. P. per hour, we 
 would proceed as follows : 
 
 Determine the actual point of cut-off, x, as accurately as possible. 
 The scale of the diagram being 40, draw the perfect vacuum line, 
 00', 14.7 Ibs. below the atmospheric line. Erect perpendiculars to 
 the atmospheric line, touching the extremities of the diagram and 
 cutting the perfect vacuum line at A and '. By means of a plani- 
 meter, or by means of ordinates, the mean effective pressure is found 
 to be 42.4 pounds. 
 
 Then 40 4" = ^ e vo ^ ume ^ n cubic feet to be displaced per I. H. P. 
 per hour, and as this is perfectly independent of the volume of the 
 cylinder we may assume the stroke volume AO' as the unit volume 
 of a cubic foot. Lay off the clearance volume AO = 0.06 to the 
 same scale that AO' equals unity. The part of the stroke A'x, in- 
 cluding clearance, up to the point of cut-off is found to be 0.413 of 
 AO', and is the initial volume, v . The volume of steam remaining 
 in the cylinder and clearance at exhaust closure is NK = 0.3125 of 
 AO'. These volumes are proportionally the same as those in the 
 actual cylinder. 
 
 The pressure at cut-off, x, and at exhaust closure, K, are found 
 by measurement to be 84.5 Ibs. and 17 Ibs., respectively, and the 
 specific volumes of steam at these pressures are 5.153 cubic feet and 
 23.22 cubic feet. The weights per cubic foot are, therefore, 
 
 - = 0.194 Ib. and = 0.04307 Ib. respectively. 
 
 Then, 0.413 X 0.194 0.3125 X 0.04307 is the weight of steam 
 sent to the exhaust per cubic foot displaced, hence 
 
 10 7Kfk 
 
 %er;r (- 413 X 0.194 0.3125 X 0.04307)= 21.62 pounds 
 of steam consumed per I. H. P per hour, when measured at cut-off, 
 and the questions of clearance, expansion, and compression having 
 been considered. 
 
 This result might have been obtained by producing the compres- 
 sion curve by the eye until it intersects A'x at P". Then A'P" is 
 the volume of the cushion steam when compressed to the pressure of 
 the steam at cut-off, so that P"x = 0.342 of AO', is the volume of 
 steam sent to the exhaust. 
 
 m , 13,750 X 0.342 X 0.194 
 
 Then, - ^ ^ - = 21.62 Ibs. of steam consumed 
 
 per I. H. P. per hour, as before. 
 
154 THE ELEMENTS OF STEAM ENGINEERING 
 
 To make the measurement at release, we select a point Q in the 
 expansion curve just before the valve opened for release, and find 
 the pressure there to be 32 pounds, the specific volume at that pres- 
 sure being 12.78 cubic feet. The weight per cubic foot is then 
 
 .. a yg = 0.07821 Ib. QE is found by measurement to be just equal 
 
 to AO', hence ^|^(1 + 0.07821 0.3125 X 0.04307) = 21 Ibs. 
 
 steam consumed per I. H. P. per hour when measured at release, 
 indicating that 21.62 21 = 0.62 Ib. had liquefied during expan- 
 sion. 
 
 The measurement at release might have been made thus : 
 QE intersects the compression curve at P, and EP represents the 
 cushion steam when compressed to the pressure at release; hence 
 QP 9 which is found by measurement to be 0.83 of AO', is the vol- 
 ume of steam sent to the exhaust at the pressure of the release steam. 
 
 13,750 X 0.07821 X 0.83 
 
 Then, - = 21.05 Ibs. steam consumed per . 
 
 42.4 
 
 I. H. P. per hour when measured at release, as before. 
 
 The method frequently used, and the one that gives results more 
 nearly correct, is as follows : 
 
 Having determined the point of cut-off, x, we find, by a property 
 of the hyperbola already explained, the point c where the cut-off 
 would have taken place to produce the same expansion had there 
 been no wire-drawing. Construct the theoretical expansion and 
 compression curves, cQ' and KP'P". Draw Q'P* parallel to 00', 
 intersecting the theoretical compression curve at P'. Q'P' is found 
 by measurement to be 0.9 of A0 f . The theoretical terminal pres- 
 sure, 0'Q' f measures 33 Ibs., the specific volume at that pressure 
 
 being 12.41 cubic feet, and the weight per cu. ft., ^ ^ = 0.08051 
 
 13,750 X 0.9 X 0.08051 
 
 Ib. Then. - - = 23.5 Ibs. steam per I. H. P. 
 
 42.4 
 
 per hour. 
 
 To show that the estimate of the steam consumption per I. H. P. 
 per hour just made is independent of the size and speed of the 
 engine, we will assume that the engine from which the diagram, 
 Fig. 43, was taken has a cylinder 20" in diameter, a stroke of 2' and 
 was making 200 revolutions per minute when the diagram was taken. 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 155 
 Then, 3.1416 X 100 X 0.413 X *X 200 X 2 X 60 X 0.194 g ^ ^ 
 
 of steam used per hour when measured at cut-off, neglecting the 
 
 saving by compression. 
 
 3.1416 X 100 X 0.3125 X 2 X 200 X 2 X 60 X 0.04307 _ 1409 4 lbg 
 
 144 
 steam saved by compression. 
 
 Therefore, 8390 1409.4 = 6980.6 pounds of steam used per 
 hour when measured at cut-off. 
 
 T TT P - 3 - 1416 X 10 X 42 - 4 X 2 X 200 X 2 _ q99 Q0 
 I. H. P. - 33,000 m92 ' 
 
 *QO A / 
 
 Therefore, owo =21.617 pounds of steam per I. H. P. per 
 
 hour, which is the same result as that obtained by the use of the 
 constant 13,750. 
 
 150. Combining Diagrams of Stage Expansion Engines. Owing 
 to the difference in the initial pressures in the cylinders of stage ex- 
 p'ansion engines, springs of different tensions are used in the indi- 
 cators, and consequently the diagrams of the several cylinders are 
 to different scales. 
 
 A combination on the scale of the diagrams of a stage expan- 
 sion engine produces a single diagram which exhibits the changes 
 in pressure and volume the steam undergoes from the moment of 
 its admission into the H. P. cylinder until its final release from 
 the L. P. cylinder. Such a diagram enables a comparison to be 
 made between the work actually accomplished and that theoretically 
 due to the initial pressure and ratio of expansion, under the. suppo- 
 sition that the total expansion takes place in the L. P. cylinder. 
 Owing to fundamental defects in the steam engine itself, the results 
 obtained from the combined diagram cannot be more than approxi- 
 mate, but the process of the combination is instructive, and if care 
 be exercised in the interpretation of the results, valuable informa- 
 tion concerning the general performance of the engine may be ob- 
 tained and a factor determined which may be of value in subsequent 
 designs. 
 
 In the process of combining the diagrams, the pressure scale of 
 all the diagrams is the same, while the length of each must be 
 such as to represent the stroke-volume of the cylinder from which it 
 was taken. It is important that the clearance volume of each cylinj 
 
156 THE ELEMENTS OF STEAM ENGINEERING 
 
 der, expressed in percentages of stroke-volume, be added to the 
 length, of each diagram. 
 
 The diagrams, Figs. 44, 45, and 46, are from a triple-expansion 
 engine, the cylinder diameters of which are 26.5", 39", and 63", and 
 piston stroke 26". The recorded steam pressures were : 
 
 At H. P. cylinder 184.7 Ibs. absolute. 
 
 At 1st receiver 93.7 " " 
 
 At 2d receiver. . . 29.7 " 
 
 Diameter of piston rods, 4.75"; vacuum, 25"; revolutions per 
 minute, 174. Clearance: H. P. Blinder, 17.28 per cent; I. P. cylin- 
 der, 16.3 per cent; L. P. cylinder, 15 per cent. 
 
 ^From the data we deduce : 
 
 Net piston areas ____ 542.28 sq. in ____ 1216.56 sq. in. . . .3158.07 
 sq. in. Ratios of net piston areas referred to L. P. cylinder, 
 
 1 : 5.819 1 : 2.596, 1.000. 
 Volumetric cylinder ratios referred to L. P. cylinder, 
 
 . 3158.07 X 1.15 . - 
 : 542.68 X 1.1728' 5 ' 
 
 . 3158.07 X 1.15 . 9 
 ' 1216.56 X 1.163' 2 ' 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 
 
 157 
 
 The first step in the process of combination is to superpose the 
 diagrams from the two ends of each cylinder, and thus obtain the 
 mean full line diagrams shown in Eigs. 47, 48, and 49. Erect per- 
 pendiculars to the atmospheric lines, touching the diagrams at each 
 
 R 
 
 7** 
 
 ^- 
 
 T - 
 
 , 
 
 i^> 
 
 
 
 ^i 
 
 
 
 
 
 n 
 
 l\ 
 1 V 
 
 I V 
 
 \ 
 
 / 
 
 'i 
 i 
 
 ^ 
 
 i 
 
 
 i 
 
 ***. 
 
 
 
 
 \ 
 
 1 >. 
 
 1 / 
 
 ' V, 
 
 k 
 
 i 
 * 
 
 
 
 
 
 
 
 
 }, 
 
 z^_ 
 
 
 
 
 
 
 
 
 
 
 
 
 a 
 
 end. Lay off, to the pressure scale of each diagram, the perfect 
 vacuum lines, 14.7 Ibs. below the respective atmospheric lines. Lay 
 off ao, equal to the respective cylinder clearance percentage of ab, 
 and erect the clearance lines oc. Select the point x, as near after 
 
158 
 
 THE ELEMENTS or STEAM ENGINEERING 
 
 the actual point of cut-off as the eye can detect, and where, it is fair 
 to assume, the expansion is according to Boyle's law. Construct 
 the hyperbolic expansion curves for each diagram, from which a fair 
 idea may be obtained as to the character of the actual expansion in 
 each cylinder. Select the point y, as near after the L. P. exhaust 
 closure as the eye can detect, and construct the hyperbolic compres- 
 sion curve. Erect ordinates and obtain the mean pressure from 
 each diagram in the usual manner. 
 
 y 
 
 Draw the coordinate axes OX and OF, Fig. 50, representing the 
 lines of volume and pressure (from perfect vacuum), respectively. 
 Assume, arbitrarily, the pressure scale to be 20 Ibs. to the inch (that 
 of the L. P. diagram), and that OB, equal 10 inches, represents the 
 volume of the L. P. cylinder and its clearance. (Fig. 50 is drawn 
 to one-third scale.) 
 
 Then, in lineal measurement, 10 inches represents 1.15 times the 
 
 volume of the L. P. cylinder, therefore tr-y^ = 8.7", represents the 
 volume of the L. P. cylinder. Lay off OB = 10", and AB 8.7". 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 159 
 
 Then 10 8.7 1.3" = OA, represents the clearance of the L. P. 
 cylinder. 
 
 = 3.895" = PM, represents the volume of the I. P. cylin- 
 
 q OQK 
 
 der plus its clearance; therefore, ' ^ = 3.35" TM, represents 
 
 the volume of the I. P. cylinder, and 3.895 3.35 = 0.545" = FT, 
 represents the clearance of the I. P. cylinder. 
 
 5 *0 25 1.7524" = OK, represents the volume of the I. P. cyl- 
 
 inder plus its clearance; therefore ' = 1.4942" = RK, repre- 
 
 sents the volume of the H. P. cylinder, and 1.75241.4942 
 = 0.2582" = QR, represents the clearance of the H. P. cylinder. 
 
 Divide AB, TM, and RK each into ten equal parts, and erect 
 ordinates midway between the divisions. Transfer to these ordin- 
 ates the pressures obtained from the ordinates of Figs. 47, 48, and 
 49, and through the points thus obtained draw in the full line dia- 
 grams of Fig. 50. They will be the diagrams of Figs. 47, 48, and 
 49, drawn to the pressure scale of 20 Ibs. to the inch, and of lengths 
 corresponding to the relative volumes of the cylinders. 
 
 Transfer the point x of Fig. 47 to x' of Fig. 50, and determine 
 the theoretical point of cut-off, z, to produce the same terminal 
 pressure from an initial pressure of 184.7 Ibs. as that obtained from 
 the cut-off at x* from the initial pressure of 147.5 Ibs. Construct 
 the hyperbolic expansion curve zCS. The theoretical terminal pres- 
 sure, BS, is, by measurement, 17.6 Ibs. This may be checked by 
 
 applying to the curve the equation, xy = -, of the rectangular 
 hyperbola: thus, OB X BS = ^P-., whence BS = f 
 
 = 17.64 Ibs. to the scale of 20 Ibs. to the inch. The actual terminal 
 pressure is 12.8 Ibs., determined by drawing the hyperbolic expan- 
 sion curve of the L. P. cylinder. 
 
 A comparison of the area included between the hyperbola and the 
 axes with that of the full line diagrams, illustrates the losses due to 
 wire-drawing, resistances in the passages, " drop," and clearance^ 
 It will be noticed that the I. P. diagram overlaps the H. P. diagram. 
 This may very well occur in any instance, for any one point on the 
 forward pressure line of one diagram does not correspond with the 
 
COMB/MED 
 of a 
 
 TRIPLE EXP0/V3/0/V MARINE ENG/NE 
 
 **f 
 
 ** 
 
 /. Diameters of Cy/i n t/ers / 
 Z.D/amete.r of P/ston ffoc/s ' 
 
 * -r -T- ' 
 
 J/Ve/ P/'ston /?re0s ; S42. 68*, /2J6.56 ", 3158.07''. 
 
 4. Faf/os of/Ye/P/sfon flreas; /.'S.8/8 , /: 2.596, / i . 
 .C/ear&/?ces in percent, /78 , /6.3 t /^" . 
 6.Vo/vmefr/c Cy//nc/er fot/os; /.'S.706ZS, /: 2.5669, / 
 7. Stroke of Pistons > 2 6 " 
 
 8. Re. volutions per minute , / 7^ 
 
 d^Steam Pressure at Engine , I84-. 7 Ibs absolute 
 
 /O. in /st Receiver, 33. 7 " " 
 
 //. * " " 2nd f 29.7 " 
 
 JZ. Terminal Pressure by Hypenjolic Cun/e, / 7 64 * 
 
 l3./?cfao/ Termina/ Pressure , /% 8 " 
 
 /4. Theoretic*?/ n umber of Expansions, IO. 4- 7 
 of Expansions, //. 
 
 
 M. E. P. 
 
 1. H. P. 
 
 On Each 
 P/'aton 
 
 Reduced fo 
 i.PPis+on 
 
 Each 
 Cy/inc/er 
 
 ffe/ati've 
 per cenf: 
 
 /6. 
 
 H.PCylinder 
 
 73 
 
 JZ.545 
 
 90S 
 
 Z9.888 
 
 17, 
 
 I.P. 
 
 36.3 
 
 13.983 
 
 IO/9 
 
 33.652 
 
 18. 
 
 L.P > 
 
 15.3 
 
 I&.300 
 
 IIO4 
 
 36.460 
 
 19. 
 
 TOTAL 
 
 
 41.828 
 
 3028 
 
 /0O. OOO 
 
 . Tota/ /tf.F.f? ob fa/net/ from 
 
 . Mean Pressure Factor 
 
 - 0. 7O3 6 
 
 23. 
 
ECONOMY or THE STEAM ENGINE AND BOILER 161 
 
 point of the back pressure line of the diagram of the preceding 
 cylinder found on the same ordinate. The point of correspondence 
 is a matter of calculation, depending on the relative piston positions. 
 
 Theoretical number of expansions = jy^ 10.47. 
 
 147 5 
 Actual number of expansions -^ g = 11.52. 
 
 The equivalent of the mean effective pressures on the pistons of the 
 high and intermediate-pressure cylinders, when referred to the 
 
 73 36 3 
 
 L. P. piston, are, respectively, r-T = 12.545 Ibs. and 
 
 = 13.983 Ibs. The total mean effective pressure, referred to the 
 L. P. piston, is then 12.545 + 13.983 + 15.300 = 41.828 Ibs. The 
 theoretical mean effective pressure, due to an initial pressure of 
 184.7 Ibs. and a ratio of expansion of 10.47, is : 
 
 184.7(1 + log. 10.47) _ 184.7 X 3.37 _. 59 lg 
 10.47 10.47 
 
 41 828 
 Mean pressure factor = ' . = 0.7036. 
 
 The indicated horse-powers developed in the high, intermediate, 
 and low-pressure cylinders are 905, 1019, and 1104, respectively, 
 making a total for the engine of 3028 I. H. P. 
 
 The weight of a cubic foot of steam at pressure of 17.64 Ibs. is 
 found from the table to be 0.0446 Ib. Then, the approximate steam 
 consumption per I. H. P. per hour is : 
 
 13,750 X 0.0446 X 0.925 _ 
 41.8*8 
 
 151. The Dynamometer. In order to find the brake horse-power 
 of an engine, or the useful power delivered to the shaft, independent 
 of the power absorbed in friction in driving the engine itself, some 
 form of dynamometer is used. Fig. 51 is a representation of an 
 absorption dynamometer, known as the Prony Brake. A drum pul- 
 ley, A, keyed on the shaft, (7, has a strap, 88, partially encircling 
 it. Bolted to the strap, and bearing on the pulley, is a series of 
 blocks, &&, which have spaces intervening between them. A wooden 
 beam, B, has a wooden shoe, D, which also bears on the pulley, A. 
 The weight, w, is adjustable, and serves to balance the preponder- 
 ance of that part of the beam to the right of the center of the shaft. 
 11 
 
162 THE ELEMENTS OF STEAM ENGINEERING 
 
 If the strap, SS, be tightened about the pulley, A, by means of 
 the nuts, nn> one of two things must happen when the engine starts 
 and the shaft revolves in the direction of the arrow ; either the grip 
 of the strap must be sufficient to carry the blocks and beam round 
 with the pulley or the pulley must slip round over the blocks. In 
 practice the nuts are so adjusted that the pulley just slips round and 
 the beam kept horizontal. The resistance to the carrying of the 
 beam and blocks round with the pulley is the pressure, W, made 
 to act on a platform scales as shown, so that its amount may be 
 measured. The resistance to the pulley slipping round in the strap 
 is the friction, F. 
 
 At the moment of slipping we shall have, taking moments about 
 
 the center of the shaft, rF = aW, whence F = , in which r is 
 
 the radius of the pulley and a the distance from the center of the 
 shaft to the line of pressure on the scales. 
 
 For one revolution of the shaft, the work done is F X 2?rr 
 
 a, W 
 = - X Zirr = %iraW foot-pounds, when a is measured in feet 
 
 and W in pounds. For n revolutions the work is 27rnaW, and the 
 horse-power is 
 
 In practice the pulley may be kept cool by circulating water 
 within it. If both the rubbing surfaces are metallic, they should 
 be freely lubricated. An iron pulley rubbing over wooden blocks, 
 as in Fig. 51, requires little lubrication. 
 
 152. Engine and Boiler Tests. The primary object in testing a 
 steam engine is to determine the cost of the power. The mean 
 effective pressure as determined from the indicator diagram is the 
 exponent of the power, and the cost is expressed in thermal units, 
 in pounds of steam, or in pounds of coal per unit of power. 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 163 
 
 If the measure is to be in pounds of steam per I. H. P. per hour, 
 there are two methods one for the surface-condensing and one for 
 the non-condensing and for the jet-condensing types of engines. 
 With the surface-condensing engine the water resulting from the 
 condensation of the exhaust steam is carefully weighed, allowance 
 being made for all steam used for other purposes incidental to the 
 operation of the engine, and correction being made for moisture. 
 An hour's duration of such a test will give accurate results. In 
 testing non-condensing and jet-condensing engines the weight of 
 steam per I. H. P. per hour is determined by weighing the feed 
 water supplied to the boilers, always making allowance for the 
 steam used for purposes other than for driving the engine, such, for 
 example, as for pumping, for jackets, for heating, or for calori- 
 meter tests. 
 
 When the measure is made in pounds of coal per I. H. P. per hour 
 the test must be one combined of the engine and the boilers, in 
 which all the precautions stipulated for a boiler test are to be 
 observed, and the coal carefully weighed. The duration of such 
 trials is from 12 to 24 hours. 
 
 Tests may be made to determine the economy derived from the 
 use of special appurtenances, such as jackets and economizers. 
 The use of feed water heaters with non-condensing engines may 
 effect a saving in fuel as great as 12 per cent. 
 
 153. The installation of engine and boiler plants is usually done 
 by contract, and the only means of determining whether the stipu- 
 lations of the contract have been fulfilled is to test the engine and 
 boilers as a system. Preceding the commencement of such a test 
 the engine should be run for several hours under the test condi- 
 tions, and it is, of course, very necessary that the conditions remain 
 constant during the test. This is particularly true with respect to 
 the steam pressure. It is equally important that the measuring 
 instruments, such as thermometers, gauges, indicators, and scales 
 should be standardized before the commencement of a test. 
 
 Everything pertaining to the cost of the operation and mainten- 
 ance of a plant enters into its economy, and it is the province of the 
 engineer to obtain the best results at the minimum of cost. 
 
 154. The primary object in testing a boiler is to determine the 
 number of pounds of water evaporated per pound of coal, and the 
 conditions of the test should be such that the boiler be not driven 
 
164 THE ELEMENTS OF STEAM ENGINEERING 
 
 above what is supposed to be its rated capacity. To drive a boiler 
 to its utmost would be a test of its capacity, but since it is far 
 from economical to so work a boiler, such tests are not of common 
 occurrence. 
 
 At the meeting in 1899 of the American Society of Mechanical 
 Engineers, a revised code for conducting steam boiler trials was 
 adopted and termed the Code of 1899. 
 
 The Short Code itself, together with some extracts from the 
 report of the committee as published in the Journal of the Ameri- 
 can Society of Naval Engineers, will here be given. 
 
 EULES FOR CONDUCTING BOILER TRIALS CODE OF 1899. 
 
 1. Determine at the outset the specific object of the proposed 
 trial, whether it be to ascertain the capacity of the boiler, its effi- 
 ciency as a steam generator, its efficiency and its defects under 
 usual working conditions, the economy of some particular kind of 
 fuel, or the effect of changes of design, proportion, or operation; 
 and prepare for the trial accordingly. 
 
 2. Examine the boiler, both outside and inside; ascertain the 
 dimensions of grates, heating surfaces, and all important parts; 
 and make a full record, describing the same, and illustrating special 
 features by sketches. The area of heating surface is to be com- 
 puted from the surfaces of shells, tubes, furnaces, and fire boxes 
 in contact with the fire or hot gases. The outside diameter of 
 water tubes and the inside diameter of fire tubes are to be used in 
 the computation. All surfaces below the mean water level which 
 have water on one side and products of combustion on the other are 
 to be considered as water-heating surface, and all surfaces above 
 the mean water level which have steam on one side and products 
 of combustion on the other are to be considered as superheating 
 surface. 
 
 3. Notice the general condition of the boiler and its equipment, 
 and record such facts in relation thereto as bear upon the objects 
 in view. 
 
 If the object of the trial is to ascertain the maximum economy 
 or capacity of the boiler as a steam generator, the boiler and all its 
 appurtenances should be put in first-class condition. Clean the 
 heating surface inside and outside, remove clinkers from the grates 
 and from the sides of the furnace. Eemove all dust, soot, and 
 ashes from the chambers, smoke connections, and flues. Close air 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 165 
 
 leaks in the masonry and poorly fitted cleaning doors. See that the 
 damper will open wide and close tight. Test for air leaks by firing 
 a few shovelfuls of smoky fuel and immediately closing the damper, 
 observing the escape of smoke through the crevices, or by passing 
 the flame of a candle over cracks in the brickwork. 
 
 4. Determine the character of the coal to be used. For tests of 
 the efficiency or capacity of the boiler for comparison with other 
 boilers the coal should, if possible, be of some kind which is com- 
 mercially regarded as a standard. For New England and that 
 portion of the country east of the Allegheny Mountains, a good 
 anthracite egg coal, containing not over 10 per cent of ash, and 
 semi-bituminous Clearfield (Pa.), Cumberland (Md.), and Poca- 
 hontas (Va.) coals are thus regarded. West of the Allegheny 
 Mountains, Pocahontas (Va.) and New River (W. Va.) semi-bitu- 
 minous, and Youghiogheny or Pittsburg bituminous coals are 
 recognized as standards. (These coals are selected because they 
 are about the only coals which possess the essentials of excellence 
 of quality, adaptability to various kinds of furnaces, grates, boilers, 
 and methods of firing, and wide distribution and general accessi- 
 bility in the markets.) There is no special grade of coal mined in 
 the Western States which is widely recognized as of superior quality 
 or considered as a standard coal for boiler testing. Big Muddy 
 lump, an Illinois coal mined in Jackson County, 111., is suggested 
 as being of sufficiently high grade to answer these requirements in 
 districts where it is more conveniently obtainable than the other 
 coals mentioned above. 
 
 For tests made to determine the performance of a boiler with 
 a particular kind of coal, such as may be specified in a contract for 
 the sale of a boiler, the coal used should not be higher in ash and 
 in moisture than that specified, since increase in ash and moisture 
 above a stated amount is apt to cause a falling off of both capacity 
 and economy in greater proportion than the proportion of such 
 increase. 
 
 5. Establish the correctness of all apparatus used in the test for 
 weighing and measuring. These are : 
 
 (a) Scales for weighing coal, ashes, and water. 
 
 (&) Tanks or water meters for measuring water. Water meters, 
 as a rule, should be used only as a check on other measurements. 
 For accurate work the water should be weighed or measured in a 
 tank. 
 
166 THE ELEMENTS OF STEAM ENGINEERING 
 
 (c) Thermometers and pyrometers for taking temperatures of 
 air, steam, feed water, waste gases, etc. 
 
 (d) Pressure gauges, draft gauges, etc. 
 
 The kind and location of the various pieces of testing apparatus 
 must be left to the judgment of the person conducting the test; 
 always keeping in mind the main object, i.e., to obtain authentic 
 data. 
 
 6. See that the boiler is thoroughly heated before the trial, to 
 its usual working temperature. If the boiler is new and of a form 
 provided with a brick setting, it should be in regular use at least a 
 week before the trial, so as to dry and heat the walls. If it has 
 been laid off and become cold, it should be worked before the trial 
 until the walls are well heated. 
 
 7. The boiler and connections should be proved to be free from 
 leaks before beginning a test, and all water connections, including 
 blow and extra feed pipes, should be disconnected, stopped with 
 blank flanges, or bled through special openings beyond the valves, 
 except the particular pipe through which water is to be fed to the 
 boiler during the trial. During the test the blow-off and feed 
 pipes should remain exposed to view. 
 
 If an injector is used, it should receive steam directly through a 
 felted pipe from the boiler being tested. 
 
 If the water is metered after it passes the injector, its tempera- 
 ture should be taken at the point where it leaves the injector. If 
 the quantity is determined before it goes to the injector, the tem- 
 perature should be determined on the suction side of the injector, 
 and if no change of temperature occurs other than that due to the 
 injector, the temperature thus determined is properly that of the 
 feed water. When the temperature changes between the injector 
 and the boiler, as by the use of a heater or by radiation, the tem- 
 perature at which the water enters and leaves the injector and that 
 at which it enters the boiler should all be taken. In that case the 
 weight to be used is that of the water leaving the injector, com- 
 puted from the heat units if not directly measured, and the tem- 
 perature, that of the water entering the boiler. 
 
 Let w = weight of water entering the injector. 
 x =. weight of steam entering the injector. 
 jk = heat units per pound of water entering the injector. 
 7i 2 = heat units per pound of steam entering the injector. 
 h 3 = heat units per pound of water leaving the injector. 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 16 
 
 Then, w(h 3 h ) = units gained by the water. 
 x(h 2 h 3 ) units lost by the steam. 
 
 Therefore, ' x = 
 
 See that the steam main is so arranged that water of condensa- 
 tion cannot run back into the boiler. 
 
 8. Duration of the Test. For tests made to ascertain either the 
 maximum economy or the maximum capacity of a boiler, irrespec- 
 tive of the particular class of service for which it is regularly used, 
 the duration should be at least 10 hours of continuous running. 
 If the rate of combustion exceeds 25 pounds of coal per square foot 
 of grate surface per hour, it may be stopped when a total of 300 
 pounds of coal have been burned per square foot of grate. 
 
 In cases where the service requires continuous running for the 
 whole 24 hours of the day, with shifts of firemen a number of 
 times during that period, it is well to continue the test for at least 
 24 hours. 
 
 When it is desired to ascertain the performance under the work- 
 ing conditions of practical running, whether the boiler be regularly 
 in use 24 hours a day or only a certain number of hours out of 
 each 24, the fires being banked the balance of the time, the duration 
 should not be less than 24 hours. 
 
 9. Starting and Stopping a Test. The conditions of the boiler 
 and furnace in all respects should be, as nearly as possible, the 
 same at the end as at the beginning of the test. The steam pres- 
 sure should be the same; the water level the same; the fire upon 
 the grates should be the same in quantity and condition; and the 
 walls, flues, etc., should be of the same temperature. Two methods 
 of obtaining the desired equality of conditions of the fire may be 
 used, viz : those which were called in the Code of 1885 " the stand- 
 ard method " and " the alternate method," the latter being employed 
 where it is inconvenient to make use of the standard method. 
 
 10. Standard Method of Starting and Stopping a Test. Steam 
 being raised to the working pressure, remove rapidly all the fire 
 from the grate, close the damper, clean the ash pit, and as quickly 
 as possible start a new fire with weighed wood and coal, noting the 
 time and the water level while the water is in a quiescent state, just 
 before lighting the fire. 
 
 At the end of the test remove the whole fire, which has burned 
 low, clean the grates and ash pit, and note the water level when the 
 
168 THE ELEMENTS OF STEAM ENGINEERING 
 
 water is in a quiescent state, and record the time of hauling the fire. 
 The water level should be as nearly as possible the same as at the 
 beginning of the test. If it is not the same, a correction should be 
 made by computation, and not by operating the pump after the 
 test is completed. 
 
 11. Alternate Method of Starting and Stopping a Test. The 
 boiler being thoroughly heated by a preliminary run, the fires are 
 to be burned low and well cleaned. Note the amount of coal left 
 on the grate as nearly as it can be estimated ; note the pressure of 
 steam and the water level: Note the time, and record it as the 
 starting time. Fresh coal which has been weighed should now be 
 fired. The ash pits should be thoroughly cleaned at once after 
 starting. Before the end of a test the fires should be burned low, 
 just as before the start, and the fires cleaned in such manner as to 
 leave a bed of coal on the grates of the same depth, and in the same 
 condition as at the start. When this stage is reached, note the 
 time and record it as the stopping time. The water level and 
 steam pressures should previously be brought as nearly as possible 
 to the same point as at the start. If the water level is not the same 
 as at the start, a correction should be made by computation, and 
 not by operating the pump after the test is completed. 
 
 12. Uniformity of Conditions. In all trials made to ascertain 
 maximum economy or capacity, the conditions should be main- 
 tained uniformly constant. Arrangements should be made to dis- 
 pose of the steam so that the rate of evaporation may be kept the 
 same from beginning to end. This may be accomplished in a 
 single boiler by carrying the steam through a waste steam pipe, the 
 discharge from which can be regulated as desired. In a battery of 
 boilers, in which only one is tested, the draft may be regulated on 
 the remaining boilers, leaving the test boiler to work under a con- 
 stant rate of production. 
 
 Uniformity of conditions should prevail as to the pressure of 
 steam, the height of water, the rate of evaporation, the thickness 
 of fire, the times of firing and quantity of coal fired at one time, 
 and as to the intervals between the times of cleaning the fires. 
 
 The method of firing to be carried on in such tests should be 
 dictated by the expert or person in responsible charge of the test, 
 and the method adopted should be adhered to by the fireman 
 throughout the test. 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 169 
 
 13. Keeping the Records. Take note of every event connected 
 with the progress of the trial, however unimportant it may appear. 
 Eecord the time of every occurrence and the time of taking every 
 weight and every observation. 
 
 The coal should be weighed and delivered to the fireman in equal 
 proportions, each sufficient for not more than one hour's run, and 
 a fresh portion should not be delivered until the previous one has 
 all been fired. The time required to consume each portion should 
 be noted, the time being recorded at the instant of firing the last 
 of each portion. It is desirable that at the same time the amount 
 of water fed into the boiler should be accurately recorded, includ- 
 ing the height of the water in the boiler, and the average pressure 
 of steam and temperature of feed during the time. By thus re- 
 cording the amount of water evaporated by successive portions of 
 coal, the test may be divided into several periods if desired, and 
 the degree of uniformity of combustion, evaporation, and economy 
 analyzed for each period. In addition to these records of the coal 
 and the feed water, half-hourly observations should be made of 
 the temperature of the feed water, of the flue gases, of the external 
 air in the boiler room, of the temperature of the furnace when a 
 furnace pyrometer is used, also of the pressure of steam and of the 
 readings of the instruments for determining the moisture in the 
 steam. A log should be kept on properly prepared blanks containing 
 columns for record of the various observations. 
 
 When the " standard method " of starting and stopping the test 
 is used, the hourly rate of combustion and of evaporation and the 
 horse-power should be computed from the records taken during the 
 time when the fires are in active condition. This time is some- 
 what less than the actual time which elapses between the beginning 
 and end of the run. The loss of time due to kindling the fire at 
 the beginning and burning it out at the end makes this course 
 necessary. 
 
 14. Quality of Steam. The percentage of moisture in the steam 
 should be determined by the use of either a throttling or a separat- 
 ing steam calorimeter. The sampling nozzle should be placed in 
 the vertical steam pipe rising from the boiler. It should be made 
 of -J-inch pipe, and should extend across the diameter of the steam 
 pipe to within half an inch of the opposite side, being closed at the 
 end and perforated with not less than twenty -J-inch holes equally 
 distributed along and around its cylindrical surface, but none of 
 
170 THE ELEMENTS OF STEAM ENGINEERING 
 
 these holes should be nearer than J inch to the inner side of the 
 steam pipe. The calorimeter and the pipe leading to it should be 
 well covered with felting. Whenever the indications of the throt- 
 tling or separating calorimeter show that the percentage of moisture 
 is irregular, or occasionally in excess of 3 per cent, the results 
 should be checked by a steam separator placed in the steam pipe as 
 close to the boiler as convenient, with a calorimeter in the steam 
 pipe just beyond the outlet from the separator. The drip from the 
 separator should be caught and weighed, and the percentage of 
 moisture computed therefrom added to that shown by the calori- 
 meter. 
 
 Superheating should be determined by means of a thermometer 
 placed in a mercury-well inserted in the steam pipe. The degree of 
 superheating should be taken as the difference between the reading 
 of the thermometer for superheated steam and the readings of the 
 same thermometer for saturated steam at the same pressure as deter- 
 mined by a special experiment, and not by reference to steam tables. 
 
 15. Sampling the Coal and Determining its Moisture. As each 
 barrow load or fresh portion of coal is taken from the coal pile, a 
 representative shovelful is selected from it and placed in a barrel 
 or box in a cool place and kept until the end of the trial. The 
 samples are then mixed and broken into pieces not exceeding one 
 inch in diameter, and reduced by the process of repeated quarter- 
 ings and crushings until a final sample weighing about five pounds is 
 obtained, and the size of the larger pieces are such that they will 
 pass through a sieve with J-inch meshes. From this sample two 
 one-quart, airtight, glass preserving jars, or other airtight vessels 
 which will prevent the escape of moisture from the sample, are 
 to be promptly filled, and these samples are to be kept for subse- 
 quent determination of moisture and of heating value and for 
 chemical analyses. During the process of quartering, when the 
 sample has been reduced to about 100 pounds, a quarter to a half 
 of it may be taken for an approximate determination of moisture. 
 This may be made by placing it in a shallow iron pan, not over three 
 inches deep, carefully weighing it, and setting the pan in the 
 hottest place than can be found on the brickwork of the boiler 
 setting or flues, keeping it there for at least twelve hours, and then 
 weighing it. The determination of moisture thus made is believed 
 to be approximately accurate for anthracite and semi-bituminous 
 coals, and also for Pittsburg or Youghiogheny coal; but it cannot 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 171 
 
 be relied upon for coals mined west of Pittsburg, or for other coals 
 containing inherent moisture. For these latter coals it is im- 
 portant that a more accurate method be adopted. The method 
 recommended by the committee for all accurate tests, whatever the 
 character of the coal, is described as follows: 
 
 Take one of the samples contained in the glass jars, and subject 
 it to a thorough air-drying, by spreading it in a thin layer and 
 exposing it for several hours to the atmosphere of a warm room, 
 weighing it before and after, thereby determining the quantity of 
 surface moisture it contains. Then crush the whole of it by run- 
 ning it through an ordinary coffee mill adjusted so as to produce 
 somewhat coarse grains (less than ^ inch), thoroughly mix the 
 crushed sample, select from it a portion of from 10 to 50 grams, 
 weigh it in a balance which will easily show a variation as small 
 as 1 part in 1000, and dry it in an air or sand bath at a temperature 
 between 240 and 280 Fahrenheit for one hour. Weigh it and 
 record the loss, then heat and weigh it again repeatedly, at inter- 
 vals of an hour or less, until the minimum weight has been reached 
 and the weight begins to increase by oxidation of a portion of the 
 coal. The difference between the original and the minimum weight 
 is taken as the moisture in the air-dried coal. This moisture test 
 should preferably be made on duplicate samples, and the results 
 should agree within 0.3 to 0.4 of 1 per cent., the mean of the two 
 determinations being taken as the correct result. The sum of the 
 percentage of moisture thus found and the percentage of surface 
 moisture previously determined is the total moisture. 
 
 16. Treatment of Ashes and Refuse. The ashes and refuse are 
 to be weighed in a dry state. If it is found desirable to show the 
 principal characteristics of the ash, a sample should be subjected 
 to a proximate analysis and the actual amount of incombustible 
 material determined. For elaborate trials a complete analysis of 
 the ash and refuse should be made. 
 
 17. Calorific Tests and Analysis of Coal. The quality of the 
 fuel should be determined either by heat test or by analysis, or by 
 both. 
 
 The rational method of determining the total heat of combustion 
 is to burn the sample of coal in an atmosphere of oxygen gas, the 
 coal to be sampled as directed in Article 15 of this code. 
 
 The chemical analysis of the coal should be made only by an 
 expert chemist. The total heat of combustion computed from the 
 
172 THE ELEMENTS OF STEAM ENGINEERING 
 
 results of the ultimate analysis may be obtained by the use of Du- 
 long^s formula (with constants modified by recent determinations), 
 viz. : 14,600 C + 62,000 (H ) + 4000 S, in which (7, H, 0, 
 and 8 refer to the proportions of carbon, hydrogen, oxygen, and 
 sulphur, respectively, as determined by the ultimate analysis. 
 
 It is desirable that a proximate analysis should be made, thereby 
 determining the relative proportions of volatile matter and fixed 
 carbon. These proportions furnish an indication of the leading 
 characteristics of the fuel, and serve to fix the class to which it 
 belongs. As an additional indication of the characteristics of the 
 fuel, the specific gravity should be determined. 
 
 (Articles 18, 19, and 20 are omitted, as they relate to " Analysis 
 of Flue Gases," "Smoke Observations," and "Miscellaneous," 
 which are deemed unnecessary in the use of the short form of the 
 code.) 
 
 21. Calculations for Efficiency. Two methods of defining and 
 calculating the efficiency of a boiler are recommended. They are: 
 
 Heat absorbed per Ib. combustible 
 
 1. Efficiency of the boiler = -^-^ ^ 1 j-.ii t TvT' 
 
 Calorific value of 1 Ib. combustible 
 
 Heat absorbed per Ib. coal 
 
 2. Efficiency of boiler and grate =. Calorific value of 1 Ib. coal' 
 
 The first of these is sometimes called the efficiency based on com- 
 bustible, and the second the efficiency based on coal. The first is 
 recommended as a standard of comparison for all tests, and this is 
 the one which is understood to be referred to when the word " effi- 
 ciency " alone is used without qualification. The second, however, 
 should be included in a report of a test, together with the first, 
 whenever the object of the test is to determine the efficiency of the 
 boiler and furnace together with the grate (or mechanical stoker), 
 or to compare different furnaces, grates, fuels, or methods of firing. 
 
 The heat absorbed per pound of combustible (or per pound of 
 coal) is to be calculated by multiplying the equivalent evaporation 
 from and at 212 per pound of combustible (or coal) by 965.7. 
 
 DATA AND KESULTS OF EVAPORATIVE TEST. 
 
 Arranged in accordance with the Short Form advised by the 
 Boiler Test Committee of the American Society of Mechanical En- 
 gineers, Code of 1899. 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 173 
 
 Made by on boiler, at 
 
 to determine 
 
 Kind of fuel 
 
 Kind of furnace 
 
 Method of starting the test, " standard " or " alternate " 
 
 Grate surface sq. f t. 
 
 Water heating surface " 
 
 Superheating surface " 
 
 Total Quantities. 
 
 1. Date of trial 
 
 2. Duration of trial hours. 
 
 3. Weight of coal as fired, including equivalent of 
 
 wood used in lighting the fire, not including un- 
 burnt coal withdrawn from furnace at times of 
 cleaning and at end of test. One pound of wood 
 is taken to be equal to 0.4 pound of coal, or, in 
 case greater accuracy is desired, as having a 
 heat value equivalent to the evaporation of 6 
 pounds of water from and at 212 per pound. 
 (6 X 965.7 = 5794 B. T. U.) The term as 
 fired " means in its actual condition, including 
 moisture Ibs. 
 
 4. Percentage of moisture in coal per cent. 
 
 5. Total weight of dry coal consumed Ibs. 
 
 6. Total ash and refuse " 
 
 7. Percentage of ash and refuse in dry coal per cent. 
 
 8. Total weight of water fed to the boiler Ibs. 
 
 9. Water actually evaporated, corrected for moisture 
 
 or superheat in steam . . . . " 
 
 10. Equivalent water evaporated into dry steam from 
 
 and at 212 , . " 
 
 Hourly Quantities. 
 
 11. Dry coal consumed per hour Ibs. 
 
 12. Dry coal per sq. ft of grate surface per hour " 
 
 13. Water evap. per hr. corr. for quality of steam. . . . " 
 
 14. Equivalent evap. per hr. from and at 212 " 
 
 15. Equivalent evap. per hr. from and at 212 per sq. 
 
 ft. of water heating surface " 
 
174 THE ELEMENTS OF STEAM ENGINEERING 
 
 Average Pressures, Temperatures, etc. 
 
 16. Steam pressure per gauge Ibs. per sq. in. 
 
 17. Temperature of feed water entering boiler degrees. 
 
 18. Temperature of escaping gases from boiler " 
 
 19. Force of draft between damper and boiler ins. of water. 
 
 20. Percentage of moisture in steam or number of de- 
 
 grees of superheating per ct. or deg. 
 
 Horse-Power. 
 
 21. Horse-power developed (Item 14 -5- 34.5) .H. P. 
 
 22. Builder's rated horse-power " 
 
 23. Percentage of builder's rated horse-power devel- 
 
 oped per cent. 
 
 Economic Results. 
 
 24. Water apparently evaporated under actual condi- 
 
 tions per Ib. of coal as fired. (Item 8 -f- Item 
 3) Ibs. 
 
 25. Equivalent evaporation from and at 212 per 
 
 pound of coal as fired. (Item 9 -=- Item 3) . . . . " 
 
 26. Equivalent evaporation from and at 212 per 
 
 of dry coal. (Item 9 -f- Item 5) " 
 
 27. Equivalent evaporation from and at 212 per 
 
 pound of combustible. [Item 9 -f- (Item 5 
 
 -Item 6)] 
 
 (If Items 25, 26, and 27 are not corrected for 
 quality of steam, the fact should be stated.) 
 
 Efficiency. 
 
 28. Calorific value of the dry coal per pound B. T. U. 
 
 29. Calorific value of the combustible per pound 
 
 30. Efficiency of boiler (based on combustible) per cent. 
 
 31. Efficiency of boiler, including grate (based on dry 
 
 coal) 
 
 Cost of Evaporation. 
 
 32. Cost of coal per ton of pounds delivered in 
 
 boiler room $ 
 
 33. Cost of coal required for evaporating 1000 pounds 
 
 of water from and at 212 . . . . $ 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 175 
 
 155. Weighing the Feed Water. The apparently simple opera- 
 tion of weighing the feed water requires the utmost care, for the 
 omission of any one of the operations attending it may render the 
 results of the test valueless. 
 
 The arrangement of the necessary apparatus for a boiler test is 
 often governed by local conditions, but generally there are two 
 weighing tanks or barrels placed on scales, the water supply being 
 connected with the tanks. The capacity of these tanks may vary 
 rom 5 to 15 cubic feet say 33 to 112 gallons, or 312 to 937 
 pounds. These tanks should be labeled, tank No. 1 and tank No. 2, 
 and be placed above the level of a third, or feeding, tank having a 
 capacity somewhat greater than either of the two weighing tanks, 
 so that the contents of either of the weighing tanks may be dis- 
 charged into it before it is entirely emptied without danger of 
 overflowing. 
 
 Preparatory to a test both weighing tanks should be filled from 
 the source of supply and their gross weights recorded. When the 
 level of the water in the feeding tank is lowered to a well-defined 
 and easily observed mark, discharge tank Xo. 1 into it, and note the 
 time as the beginning of the test. The weight of tank No. 1 empty 
 should then be recorded opposite its recorded gross weight. As the 
 test proceeds tanks Nos. 1 and 2 are alternately discharged into the 
 feeding tank and filled from the source, great care being taken that 
 the times of each discharge and the weights of the tanks full and 
 the tanks empty be recorded. 
 
 To end the test the level of the water in the boiler is brought by 
 the feed pump or injector to where it was at the start, the discharge 
 from the weighing tank being so regulated that the level of the water 
 in the feeding tank shall be at the starting mark when the water in 
 the boiler is at its required level. The time is then noted as the end 
 of the test. It is obvious from this that all the water used during 
 the test has been weighed and recorded. 
 
 If the test is to determine the steam consumption by an engine 
 per unit of power the weight of the water supplied to the boiler 
 furnishing steam exclusively for the engine is measured as above 
 described. If the boiler is fed by a pump run by steam from the 
 boiler supplying the engine, such steam must be accounted for. 
 This can readily be done by having the exhaust from the pump dis- 
 
 CAl 
 
176 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 charge into a tank resting on scales and containing a known weight 
 of cold water. The weight of the tank and its contents can then be 
 taken at intervals, renewing the cold water when the temperature 
 of the mixture becomes high enough to vaporize. If the boiler is 
 fed by an injector the steam used for operating it is returned to the 
 boiler with but slight loss. 
 
 If the engine be of the surface condensing type the air pump 
 may be made to discharge the water resulting from condensation 
 into a filter, from which it may be drawn into weighing tanks, and 
 then fed to the boiler or otherwise disposed of. 
 
 156. Record of a Test. The following form of keeping a record 
 of the observations made during a test is the one used by the stu- 
 dents of the Baltimore Polytechnic Institute : 
 
 MECHANICAL LABORATORY, 
 BALTIMORE POLYTECHNIC INSTITUTE, 
 
 BALTIMORE, MD. 
 RECORD OF OBSERVATIONS MADE DURING THE TEST OF 
 
 Boiler, , 19, 
 
 
 Pressures. 
 
 Temperatures. 
 
 Fuel. 
 
 Feed- 
 water. 
 
 Time. 
 
 3 
 
 , 
 
 
 jj 
 
 1 
 
 
 
 
 
 
 
 6 
 
 
 1 
 
 I 1 
 
 Draugh 
 gauge 
 
 1 
 
 M 
 
 M 
 
 Boiler-r 
 
 i 
 
 & 
 
 i 
 
 I 
 
 J 
 
 1 
 
 5 
 
 Pounds 
 
 i 
 e 
 
 Pounds 
 c. ft. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 157. Report of a Test. The following is a form for the complete 
 record and report of a test : 
 
ECONOMY OF THE STEAM ENGINE AND BOILER 
 
 177 
 
 MECHANICAL LABORATORY, 
 
 BALTIMORE POLYTECHNIC INSTITUTE, 
 
 BALTIMORE, MD. 
 
 Report of Boiler Test made by 
 
 Class , 19 
 
 Kind of Boiler , 
 
 Manufactured by 
 
 
 Duration of trial Hours 
 
 
 ^ 
 
 Amount used Pounds 
 
 
 
 
 
 5 
 
 Evaporated dry steam Pounds 
 
 
 
 Grate surface, length ft., 
 width ft., Sq. ft. 
 
 
 p> w 
 
 Evap. from and at 212 Pounds 
 
 
 
 Water-heating surface .... SQ. ft. 
 
 
 
 PER POUND OF FUEL 
 
 
 K 
 
 Superheating surface Sq ft 
 
 
 
 * Actual . Pounds 
 
 
 r. 
 % 
 H 
 
 Area for draught (calorime- 
 ter) Sq. ft. 
 
 
 
 Equiv. from and at 212. . . .Pounds 
 PER POUND OF COMBUSTIBLE. 
 
 
 3 
 
 
 
 
 
 Actual Pounds 
 
 
 Q 
 
 Height chimney Ft. 
 
 
 H 
 
 Equiv. from and at 212. . . .Pounds 
 
 
 
 Ratio heating to grate surface. 
 Ratio air-space to grate surface. 
 
 
 i 
 
 PER SQ. FT. HEATING-SURFACE 
 PER HE. 
 
 
 ri 
 
 
 
 &> 
 
 Actual Pounds 
 
 
 3 
 
 D 
 
 Steam-^auge Pounds 
 
 
 H 
 
 Equiv. from and at 212. . . .Pounds 
 
 
 to 
 
 \ & 
 
 Draught-gauge Inches water 
 
 
 
 FROM 100 F. TO 70 POUND BY GAGE. 
 
 
 
 
 
 
 Per pound of fuel Pounds 
 
 
 PH 
 
 Absolute steam-pressure. . . .Pounds 
 
 
 
 Per pound of combustible Pounds 
 
 
 
 External air Degrees F 
 
 
 
 
 
 
 Boiler-room Degrees F. 
 
 
 
 PER SQUARE FOOT OF GRATE. 
 
 
 
 Flue Degrees F. 
 
 
 % 
 
 Actual, from feed-water tem- 
 
 
 ! M 
 
 
 
 Mg 
 
 perature Pounds 
 
 
 Pn 
 
 Feed-water Degrees F 
 
 
 $o 
 
 Equiv. from and at 212 Pounds 
 
 
 |l - a 
 
 Steam Degrees F 
 
 
 o 
 
 PER SQ. FT. OF WATER-HEATING 
 
 
 !' ^ 
 
 
 
 5|H 
 
 SURFACE 
 
 
 
 Total coal consumed Pounds 
 
 
 >b 
 
 
 
 
 Moisture in coal . Per cent 
 
 
 
 
 
 j 
 
 Dry coal consumed Pounds 
 
 
 
 Equiv. from and at 212. . . .Pounds 
 
 
 i, p 
 
 Total refuse dry Pounds 
 
 
 
 On basis 34y 2 Ibs. equiv. evap. 
 
 
 i' & 
 
 
 
 . 
 
 per hour HP 
 
 
 
 
 
 H 
 
 Builder's rating H.P. 
 
 
 
 
 
 w 
 
 
 
 
 Dry coal per hour Pounds 
 
 
 
 er's rating 
 
 
 
 p 
 
 
 
 Combustible per hour Pounds 
 Combustible per square foot 
 
 
 
 Heat generated per hour B.T.U. 
 Heat absorbed per hour B T U 
 
 
 a 
 
 of grate Pounds 
 
 
 
 Efficiency of boiler Per cent 
 
 
 1 S 
 
 Dry coal per square foot of 
 
 
 
 Efficiency of furnace Per cent 
 
 
 _! 
 9 
 
 Combustible per square foot 
 of heating-surface Per cent 
 
 
 
 
 
 fa 
 
 Dry coal per square foot of 
 heating-surface Per cent 
 
 
 
 
 
 
 
 
 
 
 
 
 Quality of steam Per cent 
 Superheat Degrees 
 
 
 
 
 
 
 
 
 
 
 
 
 Total water used Pounds 
 
 
 
 
 
 J ' 
 
 1 
 
 Total water used (by meter), cu. ft. 
 Total evap., dry steam Pounds 
 Factor of evaporation 
 
 
 
 
 
 
 Total from and at 212 Pounds 
 
 
 
 
 
 * Evaporation from temperature of feed-water to dry steam at gauge-pressure ; that is, it is the 
 pparent evaporation corrected from calorimeter test. 
 12 
 
178 THE ELEMENTS OF STEAM ENGINEERING 
 
 PKOBLEM. 
 
 61. The pulley of a dynamometer is 26 inches in diameter. The 
 revolutions of the engine are 300 per minute, and the length of the 
 beam from the center line of shaft to the point of application of 
 the pressure on the scales is 100 inches. What pressure will the 
 scales register when the engine is developing 70 brake horse- 
 power? Ans. 147 pounds, nearly. 
 
CHAPTEE XIV. 
 STEAM BOILERS. 
 
 158. Classification of Boilers. Steam boilers may be grouped 
 into three classes, viz. : Stationary, Locomotive, and Marine. 
 
 The stationary class includes a variety of forms, the most im- 
 portant being the horizontal and vertical fire tubular and the sec- 
 tional types. 
 
 The locomotive boiler in universal use consists of a rectangular 
 tire box attached to a cylindrical shell containing a number of tubes 
 through which the furnace gases pass directly to the smoke pipe. 
 This boiler possesses great steaming capacity, and is compact, 
 durable, and of moderate cost. It is not infrequently used in con- 
 nection with stationary and portable engines. 
 
 Marine boilers differ in form and setting from those in use on 
 land. The high steam pressures demanded by modern engineering 
 conditions have completely displaced the rectangular type of boiler 
 commonly in use for marine purposes as late as the year 1875. The 
 spherical form being impossible, the cylinder was naturally sug- 
 gested as the most practical form to give strength to boilers, and 
 the type known as the cylindrical return-fire-tubular boiler is used 
 largely on ships. Special types of the Babcock & Wilcox and of the 
 Mclausse sectional boilers have lately been used advantageously for 
 marine purposes, and it seems likely that some form of the sectional 
 water tubular boiler will eventually displace the boiler of the cylin- 
 drical or tank form. 
 
 159. Sectional Boilers. This type of water tube boiler is now 
 very largely used on land, and is passing through the experimental 
 stage prior to its eventual adoption for marine purposes. A sec- 
 tional boiler may be defined as one in which the contained water is 
 divided into numerous small masses connected with each other by 
 passages sufficiently large to permit free circulation, but not large 
 enough to permit so sudden a release of pressure, in case of rupture 
 in one of the sections, as to cause a disastrous explosion. 
 
 There are many varieties of sectional boilers, but generally speak- 
 
180 THE ELEMENTS OF STEAM ENGINEERING 
 
 ing they may be said to consist of one or more cylindrical drums 
 on top which are connected by a series of tubes with a drum or 
 drums at the bottom. The heating surface is made up almost 
 entirely of the tubes, which are of moderately small diameter and 
 therefore very strong, though made of thin metal. These boilers 
 may be made light and of great power, and, by reason of the small 
 volume of water they contain, steam may be raised very quickly, 
 but for the same reason variations in pressure quickly occur in cases 
 of unsteady feed or irregular firing. 
 
 160. Relative Advantages of Different Types of Boilers. As 
 
 already stated, the Scotch cylindrical form of boiler is largely in 
 use in ocean steamships and vessels of war, and it is regarded as a 
 type that fills the requirements of such service in a fairly satisfac- 
 tory manner. 
 
 In small craft, such as torpedo boats and yachts, where speed is 
 the desideratum, the advantage in weight possessed by the water 
 tube boiler has rendered the use of some one of its forms a necessity. 
 
 Professor Durand, in comparing the water tube boiler with the 
 boiler of the Scotch cylindrical fire tubular type, states that the 
 weight of the latter without water is usually from 25 to 30 pounds 
 per square foot of heating surface, while that for the lightest type 
 of the water tube boiler ranges from 12 to 20 pounds. The weight 
 of the contained water per square foot of heating surface is usually 
 from 12 to 15 pounds for Scotch boilers and from, say, 1.5 to 3 
 pounds for water tube boilers; so that Scotch boilers with water 
 will weigh from 35 to 50 pounds per square foot of heating surface, 
 while the weight for the water tube boiler will be from 13.5 to 23 
 pounds. It seems, however, that the heating surface in a Scotch 
 boiler is more efficient, square foot for square foot, than in a water 
 tube boiler, so that it is customary to give to the latter type 10 to 
 20 per cent more heating surface for equal powers. It must be 
 remembered, though, that the water tube boiler will stand forcing 
 to a much higher degree than the fire tube boiler. With the latter 
 supplying steam to triple-expansion engines the ratio of heating 
 surface to I. H. P. can hardly be reduced below 2, while with the 
 former this ratio has been reduced in many cases to 1.5. From the 
 nature of the construction of the water tube boiler it is better 
 adapted to stand the high pressures demanded by modern practice, 
 and less liable to disastrous explosion. From one-quarter to one- 
 
STEAM BOILERS 181 
 
 half hour is sufficient time in which to raise steam in a water tube 
 boiler, but from three to four hours should be taken with a fire tube 
 boiler. The water tube boiler has the additional advantage of being 
 portable in sections, the sections being readily assembled in place 
 upon arrival at destination. 
 
 It may be said to the disadvantage of the water tube boiler that 
 it requires fresh water feed; that it requires a uniform feed, due 
 to the small water space in the boiler ; that it cannot readily be made 
 in units larger than 1000 horse-power, while double that is not un- 
 common with the Scotch- boiler ; and that a rupture of a tube re- 
 quires the insertion of a new one, necessitating the drawing of the 
 fire and blowing down of the boiler. 
 
 161. To show the necessity for a steady feed for the water tube 
 boiler, we will assume a boiler having 50 square feet of grate 
 surface and burning 35 Ibs. of coal per hour per square foot of 
 grate. The consumption of coal would then be 35 X 50 = 1750 
 Ibs. per hour. With a consumption of 1.7 pounds of coal per 
 
 I. H. P. per hour would mean the development of = 1029.5 
 I. H. P. Assuming the consumption of steam per I. H. P. per 
 
 -| AOQ 5 y 1 4. * 
 
 hour to be 14 Ibs., an evaporation of - '? 6.435 tons of 
 
 water per hour would be necessary. If, when the water is at the 
 working level, the weight of water in the boiler is 0.6435 ton, the' 
 
 boiler would have to be filled ' = 10 times in an hour, so that 
 
 a cessation of the feed for only 6 minutes would empty the boiler. 
 
 162. Boiler Design. The problem of boiler design is too exten- 
 sive for present consideration, and only certain of its features, and 
 their inter-relation, will be presented. 
 
 There is no similarity in the process of designing " shell " and 
 " sectional " boilers. The sectional boiler seems to have reached 
 its present degree of perfection by a pure process of experiment, for 
 the question of its strength involves but few calculations. All its 
 tubes and their connections have much greater strength than that 
 necessary to withstand the steam pressure, so that the question of 
 design involves only those features leading to the production of a 
 form of boiler which will be simple in management, economical in 
 operation, and amply protected against injury. The volume of 
 
182 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 steam carried by a sectional boiler is so small compared with that 
 carried by the shell boiler that the question of circulation is of 
 great importance, and the form that secures the freest circulation, 
 other things being equal, is to be preferred. 
 
 163. The design of the cylindrical boiler involves a greater num- 
 ber of calculations than can be here attempted, but a statement of 
 the common proportions governing the problem will be of interest. 
 
 Let Fig. 52 represent a section perpendicular to the axis of a 
 cylindrical boiler shell of internal diameter D and thickness t, both 
 measured in inches. If p denotes the internal pressure in pounds 
 per square inch, then the radial pressures, each equal to p, above 
 the horizontal diameter are balanced by those below it. Consider 
 a very small arc db, of length x and one inch in width, as acted on 
 
 by the radial force px. The vertical component, px sin a, is alone 
 effective in producing stress in the sections of the metal at the ex- 
 
 
 tremities of the horizontal diameter. But sin a = , whence 
 
 flS 
 
 ac = eg = x sin a. Hence the component of the force acting on 
 the arc db which produces stress in the metal sections at the ex- 
 tremities of the horizontal diameter is p X eg. By summing up 
 the vertical components of all the elemental forces acting on the 
 upper semicircle, the resultant force tending to produce rupture at 
 the metal sections is pD. 
 
 If the length of the shell be L inches, this resultant force be- 
 comes pDL, and is the magnitude of the force tending to rupture 
 the shell in a longitudinal direction. The molecular forces which 
 
STEAM BOILERS 183 
 
 resist rupture are measured by the product of the tensile strength 
 of the metal and the areas of the sections made by the longitudinal 
 plane. If / denotes the tensile strength of the metal in pounds 
 per square inch, then the resisting force is 2ftL. At the point of 
 rupture we must have pDL 2ftL, whence pD = 2ft. 
 
 Considering the strength of the transverse, or circular, section of 
 the shell, the stress is occasioned by the pressure on the two ends. 
 Eegardless of the shape of the ends, the pressure tending to rupture 
 
 the shell transversely is ^-- . The resisting force is irDft, and 
 
 at the point of rupture we must have r*- = irDft, whence 
 
 pD = 4ft. It is thus seen that a cylindrical boiler shell is twice 
 as strong circumferentially as it is longitudinally. The tensile 
 strength of boiler steel may be taken as 65,000 Ibs. per sq. inch, 
 and the factor of safety should not exceed 5. 
 
 164. The first quantity usually determined in designing a shell 
 boiler is the grate surface, and the recent practice shows that for 
 each square foot of grate surface there should be from 8 to 12 
 I. H. P. developed by the engine, depending upon the nature of 
 the draft and type of boiler, the water tubular type requiring the 
 larger grate area. 
 
 The heating surface varies from 25 to 35 square feet for each 
 square foot of grate. 
 
 The coal burned per square foot of grate surface varies from 
 15 to 45 pounds per hour, according to the nature of the draft, 
 and the water evaporated per pound of coal is reckoned at from 
 6 to 10 pounds. 
 
 The sectional area of the tubes is taken as J to -f the area of the 
 grate surface, and the sectional area over the bridge wall, and of 
 the smoke pipe, from ^ to -J of the grate surface. 
 
 The volume of the combustion chamber is taken as from 3 to 4 
 cubic feet per square foot of grate surface, and the volume of the 
 steam space from 0.35 to 0.6 cubic foot per I. H. P. 
 
 165. The strength of a furnace tube to resist external pressure is 
 inversely proportional to its length and to its diameter, and before 
 the introduction of the corrugated furnace it was the practice to 
 make the furnace in sections, flanged at the ends and secured together 
 
184 THE ELEMENTS OF STEAM ENGINEERING 
 
 through the medium of the Adamson ring, Fig. 53 (a) or of t^ie 
 Bowling hoop, Fig. 53 (6). The ring or hoop gave sufficient stiff- 
 ness to the furnace and was universally used until the introduction 
 of the corrugated furnace. 
 
 Figure 54 illustrates the Fox corrugated furnace, which is very 
 largely used in boiler construction. It is an extension of the 
 principle of the Bowling hoop, and adds immensely to the strength 
 
 of the furnace. Its strength is calculated from the formula 
 
 14,000 . , 
 
 p = h , in which, p = pressure in pounds per square inch, 
 
 t = thickness in inches, and d = the least outside diameter in 
 inches. Corrugated furnace tubes have no longitudinal joint, being 
 rolled by special machinery to a true circular form. * 
 
 166. The ends of the Scotch boiler are secured by means of stay 
 rods running from end to end, screwing into the heads, and fur- 
 ther secured by nuts inside and outside, and a washer outside to 
 increase the area supported by the brace. The back end of the 
 combustion chamber is secured to the back end of the boiler by 
 means of screw stay bolts. 
 
 The tubes are secured into the tube sheets by expanding them 
 and beading over the back ends as a protection from the flame in 
 the combustion chamber. The Dudgeon tube expander is now 
 universally used. If no stay tubes are used, both the front and 
 back ends of the ordinary tubes must be beaded so as to brace the 
 tube sheets. Stay tubes are made of extra heavy material and are 
 threaded into the tube sheets. As an additional protection to the 
 stay tubes from flame they are usually fitted with cast iron ferrules 
 at the back ends. 
 
STEAM BOILERS 
 
 185 
 
 167. The flat top of the combustion chamber of a Scotch boiler 
 is braced with girder stays, such as shown in Fig. 55. It is a case 
 of a simple beam supported at the ends, with concentrated loads at 
 the stay bolts. The area supported by each girder is seen from the 
 figure to be 28.5 X 7.5 = 213.75 square inches, and as the pressure 
 is 200 Ibs. per square inch, the total load to be supported is 42,750 
 pounds. Let 'P denote this total load; -then the load P' at each of 
 
 P P 
 
 the three stay bolts is j. The reaction at each support is -^ . 
 
 The maximum bending moment will be at the middle stay bolt, 
 
 , r PL PL PL 
 
 and we shall have M max = - -=- = - 
 
 T) T O 7" 
 
 Hence -~- = . The safe working stress is taken as 8500 Ibs. 
 
 D C 
 
 per square inch. I = y~- , c = ~ , and L = 28.5 inches. 
 
 Then 42,750 x 28.5 = 8500 x 8* 
 
 18xf 
 
 X 28.5 = 85<MP. 
 
 The girders are arranged in pairs, and assuming the width of each 
 as J inch, we shall have & = 1.75 inches. Then for the depth of 
 
 742,750 X 28 5 
 
 the girder we shall have, d = J Q , An ^ -, ~ K - = 9.05 inches, 
 
 y oouu x J-. * o 
 
 which agrees with the depth of the girder as shown. 
 
 168. The shell being twice as strong circumferentially as it is 
 longitudinally, the circumferential joints are lapped and triple 
 riveted, while the longitudinal joints are double-butt-strapped and 
 triple riveted. 
 
 When flat surfaces are to be stayed, each stay must stand the 
 pressure on the area it supports, so the sectional area of the stay 
 must be such that its tensile strength will bear the strain safely. 
 If a denotes the pitch of the stays, d the diameter of the stay, / the 
 
186 THE ELEMENTS or STEAM ENGINEERING 
 
 safe stress of the metal, and p the pressure per square inch in 
 pounds, we shall have pa 2 = -^~- The pitch of the stays of flat 
 
 surfaces depends largely on the thickness of the plate to be stayed. 
 If the pitch be too great, or the plate too thin, there is danger of 
 the plate yielding by bulging. Stay rods such as those tying the 
 flat ends of a boiler are pitched from 13 to 19 inches, while the 
 screw stay bolts tying combustion chambers to <*ach other or to the 
 shell of the boiler are pitched from 7 to 8 inches. 
 
 Sometimes the ends of cylindrical boilers, and very generally 
 the heads of steam drums, are dished, or curved, to form a spherical 
 surface, in order that they may retain their form under internal 
 pressure ; in such cases no bracing is required for the heads. 
 
 Sometimes the flat ends are secured to the cylindrical shell by 
 gusset stays, which consist of flat plates riveted to angle irons 
 secured to the ends and shell. Gusset stays are also used at times 
 to secure the combustion chamber to the back end of the boiler. 
 
 169. The Eoberts boiler used in connection with the mechanical 
 laboratory of the Baltimore Polytechnic Institute is a good speci- 
 men of the water tube type of boiler. 
 
 The feed water from the pump is divided at the boiler into two 
 streams by a T, each stream entering a feed coil on one side or the 
 other of the drum and above the boiler proper. Passing through 
 the horizontal layers of these coils, it is then delivered into the 
 drum above the water line in order that any steam that may have 
 formed during the passage through the coils may rise to the top 
 of the drum. One of the feed coils delivers into one end of the 
 drum while the other delivers into the opposite end. 
 
 It is a feature of this boiler that the course of the water through 
 the feed coils is down, thus meeting the heated gases of the fire as 
 they rise. However much the gases may be cooled by contact with 
 the lower layers of the coils, they meet still cooler layers as they 
 rise, thus establishing the most favorable conditions for the trans- 
 ference of the heat of the gases to the water. 
 
 Under the action of gravity the water now flows from both ends 
 of the drum through T connections into the cross pipes and thence 
 through the large downflow pipes into the side pipes below, which 
 are on a level with the fire grate (see Fig. 56). These side pipes 
 are tapped for the upflow coils, through which the water now rises 
 again to the drum. It will be seen that the upflow coils rise ver- 
 
STEAM BOILERS 
 
 187 
 
 tically from the side pipes and then cross in layers under the drum 
 and directly over the fire. The holes in the side pipes for the up- 
 flow coils are "staggered" so that the spaces between the cross 
 layers of the coils directly over the fire are only one-half as wide 
 
 Fig. 56. 
 
 as those between the vertical pipes of the coils. It will be noted 
 too that the upflow coils leading from one side pipe deliver to the 
 drum on the opposite side. The feed, now arriving at the drum 
 for the second time, is in the condition of damp steam and bubbles 
 through the water in the drum and rises to enter the spray pipe. 
 
188 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 This spray pipe runs from one end of the drum to the other and is 
 connected to the heads at the highest points possible. It has a 
 series of small holes drilled in its center line along the top, their 
 purpose being to prevent water entering the pipe with the steam. 
 The steam now passes from both ends of the spray pipe into the 
 superheating coils which lie, one on each side, between the boiler 
 and its inclosing casing and above the fire brick of the* furnace. 
 
 
 Fig. 57. 
 
 The steam passes to the bottoms of the superheaters and thence rises 
 to a point nearly over the front end of the drum where a T con- 
 nection enables the superheated steam to be led to the safety valve, 
 engine, etc. 
 
 Figure 56 shows the drum, the large cross and downflow pipes at 
 front and rear, the side pipes, and the upflow coils leading from 
 them. Near the top of the front head of the drum will be noticed 
 
STEAM BOILERS 189 
 
 the L for connecting the spray pipe with the superheating coil on- 
 the left side of the boiler. 
 
 A perspective view of the complete boiler with casing removed 
 is shown in Fig. 57. The angular pipe on the right side leads 
 from the rear end of the spray pipe to the superheating coil on the 
 right side of the boiler, and the course of the stream through the 
 superheater and up to the large T over the head of the drum can 
 readily be traced. 
 
 PROBLEMS. 
 
 62. The steam pressure in a cylindrical boiler is to be maintained 
 at 210 pounds per square inch. Diameter of shell, 14 feet, and 
 the efficiency of the riveted joints, 85 per cent. Using a factor of 
 safety of 4, it is required to find the necessary thickness of shell. 
 
 Ans. 1.25 inches. 
 
 63. The Roberts boiler of the B. P. I. has 21 sq. ft. of grate sur- 
 race. What should be the diameter of the smoke pipe? 
 
 Ans. 22 inches. 
 
 64. A Scotch marine boiler is to supply steam for an engine of 
 820 I. H. P., the boiler to use forced draft; determine the follow- 
 ing: (a) Grate surface. (&) Heating surface, (c) Sectional area 
 of tubes, (d) Sectional area over bridge wall, (e) Sectional area 
 of smoke pipe. (/) Volume of combustion chamber, (g) Volume 
 of steam space. 
 
 Ans. (a) 68.33 sq. ft. (b) 2050 sq. ft. (c) '13.67 sq. ft. 
 (d) 11.39 sq. ft. (e) 11.39 sq, ft. (/) 273.32 cu. ft. 
 (g) 28.7 cu. ft. 
 
 65. The flat top of the combustion chamber of a Scotch boiler 
 is 3 feet wide and 2 feet deep, and is to be braced with girder stays, 
 the boiler pressure being 180 Ibs. per sq. inch. Assuming the 
 width of the stays to be J- inch and the pitch of the stay bolts 
 7 inches, find the number and depth of the girder stays and the 
 number and diameter of the stay bolts. 
 
 Ans. Five pairs of girders, 6f inches deep ; 10 1^-inch stay bolts. 
 
 66. The diameters of the corrugations of a furnace flue, from 
 outside to outside, are 39 inches and 36 inches, and the pressure 
 to be carried is 200 Ibs. per sq. inch. Find the thickness of the 
 flue. Ans. T 9 g inch. 
 
CHAPTER XV. 
 THE STEAM TURBINE. 
 
 170. The most important modern development in the utilization 
 of heat energy by mechanical means is that exemplified in the 
 steam turbine. 
 
 In the reciprocating steam engine the motion of the piston is 
 obtained by means of the pressure due to the static energy of the 
 steam., whereas in the steam turbine the aim is to convert the static 
 energy of the steam into kinetic energy and then to apply it to the 
 rotating parts of the turbine. 
 
 When turbines are classified according to the direction of flow 
 of the steam there are two classes : 
 
 (a) Radial-flow turbines, in which the steam flows from the 
 center to the circumference or from the circumference to the center 
 of the turbine wheel. 
 
 (.&) Parallel-flow turbines, in which the general direction of the 
 flow of the steam is parallel to the axis of the wheel. 
 
 171. The application of the turbine to marine propulsion has 
 attracted the attention of engineers, and, while it may yet be said 
 to be in the process of development, it cannot be regarded as ex- 
 perimental from the fact that the Allen Line has contracted to con- 
 struct a turbine-driven steamer for ocean service and the dinar d 
 Company is to add to its fleet two turbine-driven ships of 60,000 
 horse-power each and to attain a speed of 25 knots an hour. 
 
 The field of usefulness of the steam turbine in other directions 
 is extensive. It is highly successful in its application to belt trans- 
 mission in driving machine tools, to electric generators, to blowers, 
 and to the operation of centrifugal pumps. It is particularly suit- 
 able to the last named application, as the speeds required for the 
 best efficiency of these pumps are readily obtained. 
 
 The small turbines of one power-shaft of the De Laval type 
 when operating centrifugal pumps are quite efficient for lifts of 
 from 15 to 150 feet, the larger sizes, with two power-shafts, being 
 used for lifts of from 40 to 300 feet. For greater lifts the pump 
 is directly connected to the turbine shaft, the pump wheel then 
 revolving with a velocity of from 10,000 to 30,000 revolutions per 
 minute. The pump wheel will naturally be very small and will 
 
THE STEAM TURBINE 
 
 191 
 
 not produce any suction, but must be fed by another pump which 
 is connected to the gear shaft and running at a considerably re- 
 duced velocity. This auxiliar}^ pump sucks the water and presses 
 it into the high-speed pump wheel, which gives the high pressure 
 desired. Pumps of this type have been made for lifts up to a 
 normal head of 850 feet on a single wheel which, at a decreased 
 water quantity, can go up to 1000 feet, the small pump giving an 
 efficiency of about 64 per cent. 
 
 Fig. 58. 
 
 172. Figure 58 is a representation of the wheel and divergent noz- 
 zles of the De Laval turbine. The casing through which the nozzles 
 enter, and which encloses the wheel, is removed, as are also the 
 outer ends of some of the vanes, or buckets, in order that the general 
 idea of the course and action of the steam may be plainly illustrated. 
 The nozzles are set at an agle of 20 to the plane of the wheel j and 
 one of. them is shown transparent so that its divergent section may 
 be shown. The complete expansion of the steam takes place in the 
 nozzle. It enters the small end at initial pressure and is discharged 
 at the large end at about atmospheric pressure if the turbine is non- 
 condensing, and at a pressure corresponding to the vacuum if con- 
 
192 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 densing. The volume of the nozzle is, therefore, of the first im- 
 portance, as the entrance orifice must be of sufficient area to admit 
 the requisite amount of steam, and at the discharge end the area 
 must be such that enables the complete expansion of the steam to 
 take place during its passage through the nozzle. The proper length 
 of the nozzle to insure a regular flow of steam and to minimize 
 friction has been determined experimentally. 
 
 Pig. 59. 
 
 During the expansion of the steam within the nozzle, with the 
 consequent fall in pressure, its heat energy is converted from static 
 to kinetic, so that the issuing jet of steam, impinging on the 
 crescent-shaped vanes, causes the turbine wheel to revolve with a 
 velocity which, in practice, reaches 1350 feet per second for the 
 largest powers (300 H. P.) and the minimum of 600 feet per sec- 
 ond for the smallest powers. 
 
 173. Figure 59 shows the working parts of a De Laval turbine. 
 Mounted on the turbine shaft, A, is the turbine wheel, B, fitted 
 
THE STEAM TURBINE 193 
 
 with a series of vanes around its circumference. These vanes are 
 lettered F in Fig. 60. The pinion, C, is made solid with the shaft 
 and, when in position, engages the gear wheel, H, mounted on the 
 power shaft, 7. The pinion bearings, E and D, and the power-shaft 
 bearings, J and L, are in two parts, as shown. The ball bearing, 
 F (marked 8 in Fig. 61) bears most of the weight of the wheel and 
 is self-adjusting, being held in its seat by a cap and spring, as shown. 
 The flexible bearing, G (marked T in Fig. 61) is free to move in 
 response to any oscillations of the turbine shaft and is kept seated 
 by a cap and spring; its object is to prevent the escape of steam 
 when running non-condensing, or the admission of air when run- 
 ning in connection with a condenser. 
 
 Fig. 60. 
 
 174. Figure 60 is a section showing the divergent nozzle. The 
 initial steam from the chamber, D, enters the nozzle through the 
 valve, H, and after completing its expansion within the nozzle, im- 
 pinges with great velocity on the vanes, F. The number of these 
 nozzles varies from 1 to 15, according to the power of the machine, 
 and any of them may be put out of action by closing their inlet 
 valves, thus regulating the power developed. 
 
 175. Figure 61 is a plan section, and Fig. 62 an elevation section 
 of a 30 H. P. De Laval turbine dynamo. The steam entering 
 through the governor valve, (7, Fig. 62, passes into the chamber, D, 
 Fig. 61, and thence to the nozzles. After the transfer of its kinetic 
 energy to the vanes the steam passes into the chamber, 6r, and is 
 finally exhausted through H. The governor, 0, Fig. 62, is held in 
 the end of the power shaft, L, by means of a tapered shank, and 
 operates by means of the bell-crank, 'P, and the governor valve, C. 
 The power shaft is coupled to the dynamo at M. 
 
 13 
 
194 THE ELEMENTS OF STEAM ENGINEERING 
 
 176. In the construction of the wheel of the De Laval turbine the 
 problem of successfully resisting the stresses produced by the cen- 
 trifugal force arising from the high velocity presented itself. As 
 
 Fig. 61. 
 
THE STEAM TURBINE 
 
 195 
 
196 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
 these stresses vary with the square of the velocity it was evident 
 that a wheel in the form of a flat disc would burst at the normal 
 speed required for the operation of the turbine, so the form of the 
 wheels shown in Fig. 63 was devised to meet the requirements. As 
 
 Fig. 63. 
 
 shown by the sections, the wheel is heavy at the boss and tapers in 
 section toward the peripheral ring to which the vanes are attached. 
 The centrifugal force at the boss is relatively small and is a 
 maximum at the periphery. The gradual increase in section area 
 from the boss to the periphery provides an excess of material that 
 becomes available for the support of the highly stressed section at 
 
THE STEAM TURBINE 
 
 197 
 
 the periphery. The vanes are inserted into milled slots in the 
 peripheral ring, and the centrifugal force due to their velocity loads 
 the ring to an amount equal to the stresses produced in the body of 
 the wheel. The section of the wheel immediately under the ring 
 is purposely made small, and so designed that fracture will take 
 place at that section should the speed approximate double that of 
 the normal. A fracture of that description would break the ring 
 into such small pieces that no damage to the wheel case would result, 
 and, the vanes then gone, the action of the steam on the wheel 
 would immediately cease. 
 
 The wheels of the larger sizes are made solid in body in order 
 to give the maximum of supporting material at the boss, and, in 
 such cases, the shaft is made in two parts and secured to the wheel 
 as shown in Fig. 63. 
 
 177. Figure 64 is an approximate three-quarter size representa- 
 tion of two of the 200 vanes of a 300 H. P. De Laval wheel, one of 
 them shown in mid-section by a plane perpendicular to the axis of-' 
 rotation. The vanes are drop-forged from moulded steel and are 
 fitted into milled slots in the rim of the wheel. Fig. 65 is a front 
 view of a vane, and Fig. 66 a section on AB of Fig. 64. 
 
198 THE ELEMENTS OF STEAM ENGINEERING 
 
 The diameter of a 300 H. P. De Laval wheel is about 30 inches, 
 and the weight of one vane about 0.7 ounce. 
 
 178. Owing to the lack of uniform density of metals it is impos- 
 sible to produce a wheel which will balance about its axis of forma- 
 tion, and in consequence very serious vibrations would result at the 
 high speeds attained by the De Laval wheel should no provision be 
 made for their prevention. 
 
 The length and lightness of the shafts on which these wheels are 
 mounted admit of such flexibility that when the so-called critical 
 speed is reached the wheel is permitted to choose an axis of rotation 
 through its center of gravity, with the consequent absence of vibra- 
 tions and perfect smoothness of running. The high rotary speeds 
 of these shafts admit of slender dimensions, those of the smaller 
 powers being but f inch in diameter, while that of the maximum 
 of 300 H. P. is but If inches. There is always sufficient clearance 
 between the wheel and its casing to allow for the small displace- 
 ment of the wheel incident to its choosing its axis of rotation. 
 
 179. In the De Laval turbine the steam acts on a single set of 
 vanes, and since it can be shown that for maximum efficiency the 
 vane speed should be about half the velocity of the steam jet, it 
 follows that the use of even moderately high steam pressures pro- 
 duces such excessive velocities for the vanes, with the consequent 
 stresses due to centrifugal action, that a limit is put on the efficiency 
 by the strength of the material available for construction. It is 
 also a condition for efficiency with the De Laval wheel that the jet 
 shall leave the vanes at a minimum velocity, thus securing the 
 maximum transference of energy. 
 
 This latter condition may be dispensed with in a steam turbine 
 if it be so arranged that the steam acts on a series of sets of vanes, 
 their angles and velocities being such that a portion of the energy 
 of the steam is given up to each set. This is the basic principle of 
 the Westinghouse-Parsons, and of some other turbines, by means of 
 which high rotational speeds are avoided and the ability to develop 
 high powers secured. 
 
 180. Figure 67 illustrates the action of the steam on the vanes of 
 a Westinghouse-Parsons turbine, the dotted lines showing the course 
 of the steam and the horizontal arrows the direction of rotation of 
 the moving vanes. 
 
THE STEAM TURBINE 199 
 
 The fixed vanes project radially inward from the enveloping 
 casing. Mounted on the shaft, which is co-axial with the casing, 
 is a hollow cylindrical drum, from the outer surface of which the 
 moving vanes project radially outward. The entering steam first 
 encounters a ring of fixed vanes, from which it is deflected so as 
 to impinge in such a direction on the adjoining ring of moving 
 vanes as to give to them a rotary impulse. Leaving this ring of 
 moving vanes the steam next encounters another ring of fixed blades 
 and is by them deflected in the proper direction for impact on the 
 next ring of moving blades. The force rotating the moving vanes 
 is the double one of the impact of the entering steam and the 
 reaction of the effluent steam. The passages through the vanes are 
 of constantly increasing volume, corresponding to the increased 
 volume of the steam. This increase in volume of the passages is 
 
 ^ I I 
 
 F/XE0 
 MOy/HG 
 
 Ft 4. 67. 
 
 obtained by increasing the heights of the vanes until the mechanical 
 limit of such increase is reached, when the diameter of the turbine 
 is abruptly increased, thus allowing for another progression in the 
 heights of the vanes. 
 
 It is thus seen that the passage of the steam through the many 
 passages of gradually increasing volume permits a gradual expan- 
 sion of the steam and the conversion of its energy into velocity, 
 until, finally, the desired pressure and temperature is reached for 
 exhaustion into the atmosphere or into the condenser. 
 
 181. Figure 68 is a longitudinal section through a Westinghouse- 
 Parsons turbine, and is described in a paper by Mr. Francis Hodg- 
 kinson as follows : 
 
 " The steam enters at the governor valve and arrives at the cham- 
 ber, A, and passes out to the right through the turbine blades, even- 
 tually arriving at the exhaust chamber, B. On the left of the 
 steam inlet are shown revolving balanced pistons, C, C^, (7 2 , one 
 
200 
 
 THE ELEMENTS OF STEAM ENGINEERING 
 
THE STEAM TURBINE 201 
 
 corresponding to each of the cylinders in the turbine which, accord- 
 ing to size, may be 1, 2, 3, or 4 in number. The steam at A presses 
 against the turbine and goes through doing work. It also presses 
 in the reverse direction, but cannot pass the piston, (7, but at the 
 same time the pressure, so far as the steam at A is concerned, is 
 equal and opposite, so that the shaft is not subjected to any end 
 thrust. 
 
 " The pressure at D is equal to that at E by reason of the balance 
 port, F; similarly, so far as the steam pressure at E is concerned, 
 there is no end thrust. This same fact also applies to G. 
 
 " The areas of the balance pistons are so arranged that no matter 
 what the load may be, or what the steam pressure or exhaust pres- 
 sure may be, the correct balance is preserved and the shaft has no 
 end thrust whatever. 
 
 "At H is shown a thrust bearing which, however, has no thrust 
 to take care of, but serves to maintain the correct adjustment of the 
 balance pistons. 
 
 "At K is a pipe connecting the back of the balance pistons at L 
 with the exhaust chamber, to insure the pressure at this point being 
 exactly the same as that of the exhaust. 
 
 "At J are shown the bearings. They are unique in construction. 
 The bearing proper is a gun metal sleeve, which is prevented from 
 turning by a loose fitting dowel. Outside of this are three concen- 
 tric tubes having a small clearance between them. This clearance 
 fills up with oil and permits a vibration of the inner shell, at the 
 same time restraining same. The shaft therefore revolves about its 
 axis of gravity, instead of the geometric axis, as would be the case 
 were the bearings of every-day construction. The journal is thus 
 permitted to run slightly eccentric, according as the shaft may be 
 out of balance. This form of bearing in a very remarkable man- 
 ner performs the functions of De Laval's slender flexible shaft. 
 But in this case the shaft is built as rigidly as possible, so is not 
 liable to crystallization, which would result in eventual rupture. 
 
 " The bearings have ample surface, are continuously lubricated 
 under pressure, and it has been found in practice that they do not 
 wear. 
 
 "At R is shown a flexible coupling by, means of which the power 
 of the turbine is transmitted. In small sizes the two shafts have 
 a square cut on the ends, the coupling itself somewhat loosely fitting 
 
202 THE ELEMENTS OF STEAM ENGINEERING 
 
 over these. In larger sizes it is generally a modification of this 
 arrangement. 
 
 " The governor gear and oil pumps generally receive their motion 
 by means of a worm wheel gearing into a worm cut on the outside 
 of the coupling. 
 
 "At N is an oil reservoir into which drains all the oil from the 
 bearings. From there it runs into the pump, M, to be pumped 
 up to the chamber, 0, where it forms a static head which gives a 
 continuous pressure of oil to the bearings. The pump is single- 
 acting of the simplest possible construction, that will not become 
 deranged. The oil runs in by gravity, so that it is unlikely to fail 
 to continue pumping. 
 
 "A by-pass valve is provided, shown at iP, which admits high- 
 pressure steam by means of port Q to the steam space E. By open- 
 ing this valve as much as 60 per cent overload may be obtained, 
 and in the case of turbines operating condensing, full load may be 
 obtained should the condenser be at any time inoperative, due to 
 any cause, and the turbine allowed to exhaust into the atmosphere." 
 
 182. In comparison with the reciprocating engine it may be said 
 of the steam turbine : 
 
 (1) It is more simple in construction. 
 
 (2) It is more permanent, since there is the complete absence of 
 reciprocating parts, and the bearings so completely adjusted and 
 lubricated as to almost preclude the possibility of wear. 
 
 (3) The size and weight per horse-power is less than that for the 
 steam engine. 
 
 (4) The consumption of steam per horse-power is less than for 
 the simple reciprocating engine and about the same as that for the 
 triple-expansion engine. 
 
 (5) It permits the use of steam superheated to the highest de- 
 gree, since there is an absence of the necessity for internal lubri- 
 cation. 
 
PART II 
 
 TRANSMISSION OF POWER BY 
 BELTS AND GEAR-WHEELS 
 
CHAPTEE I. 
 BELTING. 
 
 1. The transmission of power and of motion by means of belts 
 running over pulleys is an important and familiar mechanical con- 
 trivance. It is practically noiseless, and may be used for trans- 
 mitting power through a distance as great as 30 feet without 
 intervening support. The principal disadvantage is that due to the 
 slip occasioned by the freedom of the belt to slip on the pulley, 
 rendering the transmission not so positive as that through the 
 medium of gear wheels; but this disadvantage, in cases where 
 mechanisms at rest are suddenly thrown into gear, becomes an 
 advantage in preventing shocks. Motion may be transmitted by 
 belting between shafts making various angles with each other, the 
 
 necessary provision being that the advancing side of the belt 
 be at right angles to the shaft it approaches, while the receding side 
 may deviate in any direction. . 
 
 2. Figure 1 is a side elevation and end view of a " quarter turn " 
 in power transmission by means of a belt and pulleys, the shafts 
 being at right angles. It will be noted that the belt in passing 
 from pulley A approaches pulley B in a direction at right angles to 
 the axis of 5, and in passing from B it approaches A in a direction 
 at right angles to the axis of A, as will be seen from the end view. 
 It will be noted, too, that a plane that is tangent to the face of one 
 pulley and perpendicular to the axis of the other cuts the other 
 pulley in the middle of its face. 
 
206 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 3. No definite rule can be given for the distance between shafts 
 that are to be connected with each other by belts. The rule that 
 ten times the diameter of the smaller pulley is an advantageous 
 distance is not of general application, for much depends on the 
 judgment of the engineer in applying general principles to meet the 
 conditions that confront him. For narrow belts running over small 
 pulleys, fifteen feet is a good average distance, and for larger belts 
 and larger pulleys, the maximum distance is about thirty feet. In 
 any event the distance should be such as to permit a gentle sag to 
 the belt when it is in motion. 
 
 4. Experiment has shown that belts tend to run on the highest 
 part of pulleys, and in order to keep them central on the rims a 
 convexity amounting to about one-twenty-eighth of the breadth is 
 given to the surface of the rims. 
 
 5. The belt connecting two pulleys may be open or crossed, the 
 resulting rotation of the pulleys being in the same or in opposite 
 directions respectively. 
 
 6. It is clear that, in the transmission of motion from one pulley 
 to another, the outer surface of the rim of each pulley will have the 
 same velocity as the belt; so if we denote by D t and Z> 2 the diame- 
 ters of the driver and follower respectively, and by N^ and iV 2 their 
 revolutions per minute, we shall have : 
 
 Speed of outer surface of driver rim = 7rP 1 A T 1 , and 
 Speed of outer surface of follower rim = irD 2 N 2 , 
 and since each of these is equal to. the speed of the band, we have 
 
 vD^ irD 2 N 2 , whence :& ss & . That is, the velocities of a 
 
 -Aj i/2 
 
 pair of pulleys are inversely as their diameters. 
 
BELTING 207 
 
 7. When motion is transmitted from one shaft to another by 
 belting, one or more shafts intervening, as in Fig. 2, we shall have 
 
 &a*-* 9 &*2* 9 and ~ 6 = 5s. Taking the product of these 
 JVi D 2 Jv s D 4 -A 5 D 6 
 
 equations, member by member, and remembering that N 2 = N 3 
 
 and 'N N 5) we have ^? = ^1 X -^ X ^ 5 . That is, the ratio of 
 iVj ) 2 .L> 4 .L/ 6 
 
 the speed of the last pulley to the speed of the first pulley is equal 
 to the continued product of the ratios of the diameters of each pair 
 of pulleys taken in order. 
 
 EXAMPLE I. The pulley on the shaft of a steam engine is 30 
 inches in diameter, from which a belt passes to a pulley 20 inches in 
 diameter on a shaft in a room above; a belt passes from another 
 20-inch pulley on this shaft to one of 8 inches in diameter on a 
 third shaft; an 18-inch pulley on the third shaft is belted to a 
 6-inch pulley on the spindle of a dynamo: find the speed of the 
 dynamo when the engine is making 100 revolutions per minute. 
 
 -a- Speed of Dynamo 30^20^.18 ,, , 
 
 Here, -f a yw -. = ^ X -^ X -*- > therefore. 
 
 Speed of Engine 20 86 
 
 30 20 1 8 
 Speed of Dynamo = X x -^X 100 = 11 25 revolutions per minute. 
 
 8. The thickness of a belt may have a very appreciable effect on 
 the velocity ratio of two pulleys. While the belt is in contact with 
 the pulley its inner surface is in compression and its outer surface 
 in tension, but the neutral surface midway between is of constant 
 length. It follows, then, that the velocity of the surface of the belt 
 in contact with the pulley is less than the velocity of the neutral 
 surface, and that the true velocity ratio of two pulleys connected 
 by a belt is obtained by increasing the diameters of the pulleys by 
 an amount equal to the thickness of the belt. 
 
 If the thickness of the belts in the above example were -fa inch, 
 and the thickness be taken into consideration, we shall have : 
 
 30 3 20 ^ 18i 
 Speed of dynamo = ^x^X i^X 100 
 
 *4 4 4 
 
 = 1083.76 
 
 m QQ 
 OU 
 
 revolutions per minute. The effect has been to decrease the calcu- 
 lated speed of the dynamo by 1125 ^^ 83 ' 76 = 0.0367 or by 3.67 f. 
 
208 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 9. Frictional Resistance between a Belt and a Pulley. Figure 3 
 represents two pulleys connected by a belt. With the pulleys sta- 
 tionary the belt is strained over them so that the initial tension 
 will insure against slipping. If, now, a force tends to turn the 
 driving pulley, D, in the direction of the arrow, the lower part of 
 the belt will be stretched so that its tension will be increased. The 
 
 tension in the upper part will be decreased an equal amount. Each 
 element of the belt in contact with the pulley assists the action of 
 the tension in the slack side of the belt in resisting the tension in 
 the tight side, and therefore the tension in that part of the belt in 
 contact with the pulley varies at every point. When the difference, 
 TI T 2 , becomes sufficient to overcome the resistance to motion 
 
 in the driven pulley, F, its rotation begins. The difference in ten- 
 sion in the two sides of the belt is the amount of friction between 
 the belt and the pulley, and is the measure of the driving force. If 
 the tensions were equal their moments about the center of the driven 
 pulley would be equal and there would be no rotation, since there 
 would be equal turning efforts in opposite directions. 
 
BELTING 209 
 
 Let TI and T 2 , Fig. 4, denote the tensions in the tight and slack 
 sides of the belt, respectively. Suppose that da be a very small 
 part of the central angle a subtended by the arc of contact of the 
 belt. 
 
 Considering the equilibrium of the very small arc subtending da, 
 we will assume the tensions at its extremities to be T and T + dT. 
 If the resulting reaction between the belt and pulley rim due to 
 
 these tensions be denoted by R, we shall have R = 2T sin ~ = Tda, 
 
 A 
 
 since sin and are very approximately the same. 
 
 <> a 
 
 Since slipping is about to take place, dT is the measure of the 
 friction over the small arc considered ; whence dT p,R =. fiTda , 
 in which p is the coefficient of friction. 
 
 j rp 
 
 We have then, for the small arc considered, -=- = pda, and 
 
 /TI ^T 7 /a 
 
 for the whole arc of contact I -,w- = I u.da, from which 
 
 JT Z * t/o 
 
 rp rp 
 
 log T l log T 2 = /xa, or log -=i //.a; whence -^ = e* a , where 
 
 *J *I 
 
 ^ 2.718, the base of the Naperian system of logarithms. 
 
 We shall then have, Iog 10 pa Iog 10 2.718 = 0.4343/>ta. 
 
 v *-/ 
 This equation shows that the ratio of the tensions depends only 
 
 upon the coefficient of friction and the angle at the center subtended 
 by the arc of contact of the belt, and is independent of the diameter 
 of the pulley. In the above equations a is expressed in circular 
 measure. 
 
 10. The results just obtained hold for a rope making several 
 turns about a post. Thus, in the equation log ( ~A 0.4343/m , if 
 n denotes the number of turns of the rope, then a = 2-rrn, and we 
 shall have, log(-^\ 0.4343 X 2^ ; whence, log TI log T 2 
 4-2.729n/x. 
 
 EXAMPLE II. A rope makes three turns about a post, one end 
 being pulled with a force of 25 Ibs. The coefficient of friction 
 being 0.4, what is the greatest force that can be resisted at the 
 other end ? 
 
 Here, T 2 = 25 , n = 3 , and p 0.4. 
 
 Then, log T^ = log 25 + 2.729 X 1.2 , whence 2\ = 47,000 Ibs. 
 14 
 
210 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 11. Transmission of Power by Belts. The friction between the 
 belt and pulley limits the amount of power that can be transmitted. 
 When overloaded, a belt will slip rather than break, therefore, at 
 the point of slipping, we shall have: 
 
 Horse-power transmitted = ^-^^ , in which F = T T 2 in 
 
 00,000 
 
 pounds, and V the velocity of the belt in feet per minute ; therefore, 
 
 (T-T)V 7\(l-p-U r,(l-->.-)r 
 Horse-power kz* _ * J \ 2 i / _ e ' 
 
 33,000 33,000 33,000 
 
 Substituting the value of e, and letting /* = 0.35 and a = 180 = TT 
 in circular measure, we shall have : 
 
 Horse-power - ~ - - 
 
 33,000 ' 33,000 * 
 
 TI may be taken as 80 Ibs. per inch width of single-ply belt, which 
 makes allowance for laced joints. If w = width of belt in inches, 
 then TI = Sow , and the equation for the horse-power becomes 
 
 Horse-power ' - for single-ply belting, and for 
 
 double-ply we shall have : Horse-power . 
 
 600 
 
 These two approximate equations are easily remembered, and are 
 useful for quickly estimating power. 
 
 The linear velocity of belts ranges from 3000 to 6000 feet per 
 minute. 
 
 12. Creeping of Belts. The driving side of a belt is necessarily 
 stretched more than the slack side, and in consequence the driving 
 pulley receives a greater length of belt than it gives off to the fol- 
 lowing pulley; therefore, the speed of the driving side is a trifle 
 greater than that of the slack side. But the speed of the rim of a 
 pulley is the same as that of the belt it receives; therefore, the 
 speed of the rim of the driver will be somewhat greater than that of 
 the rim of the follower, and, as a consequence, the speed of the 
 
 follower will be less than that given by the formula ^ 2 - = -^-. 
 
 j\ l DI 
 
 This difference between the speeds of the rims of the driver and 
 follower is known as creep, or slip, and amounts to about 2 per cent. 
 
 13. Strength of Belting. The ultimate strength of leather used 
 for belting varies from 3000 Ibs. to 5000 Ibs. per square inch of 
 
BELTING 
 
 section, and the strength of the laced joint may be taken as one- 
 third that of the solid leather. Taking a factor of safety of 5, the 
 safe working tension with a laced joint may be taken from 200 to 
 330 pounds per square inch of section. Single-ply belts range in 
 thickness from f$ inch to T 5 -g- inch, and the width must be made 
 sufficient to withstand the tension. The safe tension in pounds 
 per inch of width ranges from 50 to 100, according to the thickness 
 and the safe stress of section of the material. 
 
 14. Centrifugal Action in Belts. The stress in the rim of a fly- 
 wheel or of a pulley, due to centrifugal action, may be treated in 
 a manner similar to the stresses in a cylindrical boiler shell, if the 
 effects of the arms be neglected. 
 
 In this instance the radial force, instead of being the pressure of 
 
 the steam per square inch, is (see p. 385), in which r is the 
 
 mean radius of the rim in feet; and, as in the case of the boiler 
 shell, the resultant centrifugal force producing stresses in the 
 sections of the rim at the extremities of a diametral plane is 
 
 W*)P 9 Wit 2 
 
 ^LL. X2r= -:LZ_ (see p. 182). This force is balanced by the 
 tensions in the rim at the two sections, so that the whole tension in 
 
 the rim is - in pounds per square inch. 
 
 In precisely the same manner, that part of a belt embracing a 
 pulley is subjected to a stress due to centrifugal action, and at high 
 speeds part of the tension in the belt is expended in a tendency to 
 lift the belt as it passes over the surface of the pulley, thereby 
 decreasing the normal pressure on the pulley. In belt speeds in 
 excess of 50 feet per second the effect of this centrifugal action 
 should be determined, and the width of the belt increased accord- 
 ingly. 
 
 One linear foot of leather one square inch in section weighs 0.43 
 lb., so that one foot of belting a square inches in sectional area will 
 weigh 0.43a Ibs. 
 
 The effect of centrifugal action on the belt is then - , i n 
 
 which v is expressed in feet per second. The length of a foot of 
 belting is taken because the velocity is expressed in feet per second, 
 and the sectional area of the belt is taken in square inches because 
 the stress is expressed in pounds per square inch. 
 
212 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 In all cases where the effect of centrifugal action in belting is to 
 be considered, the real tensions are to be taken as T l -j : 
 
 t/ 
 
 andT 2 + QA3av * . It will be noted that the driving force, 
 
 }j 
 
 T T 2 , is not affected by the centrifugal action. 
 
 EXAMPLE III. Find the width of belt necessary to transmit 
 10 horse-power to a 12-inch pulley so that the greatest tension may 
 not exceed 40 pounds per inch of width when the pulley makes 1500 
 revolutions per minute, the weight of the belt per square foot being 
 1.5 Ibs., the coefficient of friction 0.25, and the arc of contact of 
 the belt 180. 
 
 0.43 
 
 Here, if t denotes the thickness of the belt, we have : 144 X 
 
 = 1.5 , whence t = 0.29 inch. 
 
 Then v = * X 12.29 X 1500 _ g(U3 ^ ^ gec(m(L 
 
 -L<y X OU 
 
 (ffi rp\ y 
 
 We have, Horse-power = ^ l 2; , whence 
 
 00,000 
 
 12 
 
 We have, S=: e**, whence T^ = T z e^= T 2 X 2.718- 257r =2.19T 2 . 
 
 Then, 2.19T 2 = 68.37 + T 2 , whence T 2 =. 57.46, and 
 
 TI 68.37 + 57.46 = 125.83 Ibs. 
 The effect of centrifugal action increases the tension by the 
 
 amount - 43gt; ' . If w denotes the width of the belt in inches, then 
 
 i 
 a = tw = 0.29w , and we shall have : 
 
 0.43 X 0.29w X 80.43 X 80.43 
 = #y.oo -t- - n 
 
 g 32.2 
 
 = 125.83 + 25.05w. 
 
 But from the conditions of the problem the tension must not exceed 
 40w ; hence 40z/; = 125.83 -(- 25.05w , whence w = 8.4 inches. 
 
 PEOBLEMS. 
 
 1. Two pulleys, 15 inches and 37.5 inches in diameter respec- 
 tively, are connected by a belt. If the former makes 500 revolu- 
 tions in 10 minutes, how many turns will be made by the other in 
 25 minutes? Ans. 500. 
 
BELTING 213 
 
 2. Sketch an arrangement of four pulleys with belts for driving 
 a fan at 1500 revolutions per minute from a shaft making 200 revo- 
 lutions per minute, giving the diameters of the pulleys to be used. 
 
 3. An engine shaft, making n revolutions per minute, carries a 
 56-inch pulley which drives, by means of a belt, a 36-inch pulley on 
 the line-shaft. The line-shaft carries another pulley, 42 inches in 
 diameter, which is belted to a 24-inch pulley on a counter-shaft. 
 Another pulley on the counter-shaft is 48 inches in diameter and 
 is belted to a 14-inch pulley on the spindle of a dynamo. Find the 
 number of revolutions made in a minute by the dynamo spindle. 
 
 Ans. 9.33n. 
 
 4. The flywheel of an engine is 28 inches in diameter and is 
 belted to a pulley 20 inches in diameter on another shaft. A 
 20-inch pulley on this second shaft is belted to a 10-inch pulley on 
 a third shaft, which carries an 18-inch pulley, which, in turn, is 
 belted to a 6-inch pulley on the spindle of a dynamo. Find the 
 speed of the dynamo when the engine is making 90 revolutions per 
 minute. If the belt thickness of three-sixteenths of an inch be 
 considered, find the per cent of loss in speed of the dynamo. 
 
 Ans. 756 revolutions ; 3.18 per cent. 
 
 5. The shaft of a high-speed engine, which is making 300 revo- 
 lutions per minute, carries a 14-inch pulley, over which passes a belt 
 to a 20-inch pulley on another shaft. A 10-inch pulley on this 
 shaft drives a 20-inch pulley on a third shaft carrying a 6-inch 
 pulley which is to be belted to the spindle of a machine so that the 
 revolutions of the spindle may be 45 per minute. Find the diameter 
 of the pulley on the spindle. Ans. 14 inches. 
 
 6. The pulley on an engine shaft is 5 feet in diameter and makes 
 100 revolutions per minute. The motion is transmitted from this 
 pulley to the main shaft by a belt running on a pulley, the difference 
 in tensions between the tight and slack sides of the belt being 115 
 pounds. What is the work done per minute in overcoming the re- 
 sistance to motion of the main shaft? Ans. 180,714 ft. Ibs. 
 
 7. A belt having a linear velocity of 350 feet per minute trans- 
 mits 5 H. P. to a pulley. Find the tension on the driving side, sup- 
 posing it to be double that on the slack side. Ans. 943 Ibs. 
 
214 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 8. A pulley 3 feet 6 inches in diameter, and making 150 revolu- 
 tions per minute, drives by means of a belt a machine which absorbs 
 7 H. P. What must be the width of the belt so that its greatest 
 tension shall be 70 pounds per inch of width, it being assumed that 
 the tension in the driving side is twice that in the slack side ? 
 
 Ans. 4 inches. 
 
 9. Find the H. P. that may be transmitted by a belt 8 inches wide 
 and passing over a 20-inch pulley on the shaft of an engine which 
 makes 350 revolutions per minute. The angle at the center sub- 
 tended by the arc of contact of the belt is 160, and the coefficient 
 of friction is 0.4. The tension on the driving side is not to exceed 
 80 Ibs. per inch of width. Ans. 23.94. 
 
 10. A belt laps 150 round a pulley of 3 feet diameter, making 
 130 revolutions per minute; the coefficient of friction is 0.35. 
 What is the maximum pull on the belt when 20 H. P. is being trans- 
 mitted and the belt is just on the point of slipping? 
 
 Ans. 900 Ibs. 
 
 11. A belt is to transmit 2 H. P. from a pulley 12 inches in diam- 
 eter on a shaft making 160 revolutions per minute. Find: (1) 
 The tensions on the driving and on the slack sides of the belt when 
 the belt embraces half the pulley rim and the coefficient of friction 
 is 0.3. (2) The width of the belt when the thickness is 0.25 inch, 
 the safe working tension being 320 Ibs. per square inch of section. 
 
 Ans. 215.15 Ibs.; 83.85 Ibs.; 2.69 inches. 
 
 12. What H. P. will a belt 8 inches wide transmit over an 18-inch 
 pulley making 300 revolutions per minute, the weight of 1 sq. ft. 
 of the belt being 1.29 Ibs., the coefficient of friction 0.3, the central 
 angle embraced by the belt 160, and the tension per inch of width of 
 belt not to exceed 80 Ibs? Take the weight of a foot of belting 
 1 sq. inch in section as 0.43 Ib. Ans. 15.8. 
 
 13. A leather belt is required to transmit 2 H. P. from a shaft 
 running, at 80 revolutions per minute to a shaft running at 160. 
 Find the stresses in the belt, assuming that the smaller pulley is 
 12 inches in diameter, and that the ratio of the tensions in the 
 tight and slack sides of the belt is 2.25 : 1. Then find the width 
 of belt, taking the working stress at 100 Ibs. per inch of width. 
 
 Ans. 236.3 Ibs.; 105 Ibs.; 2.36 inches. 
 
BELTING 215 
 
 14. Find the width of belt necessary to transmit 10 H. P. to a 
 pulley 12 inches in diameter so that the greatest tension may not 
 exceed 40 Ibs. per inch of width when the pulley makes 1500 revo- 
 lutions per minute, the weight of the belt per sq. ft. being 1.5 Ibs., 
 and the coefficient of friction 0.25. The weight of a cubic inch of 
 belting may be taken as 0.0358 lb., and the arc of contact of the belt 
 as 180. (The effect of the thickness of the belt and of the cen- 
 trifugal action of the belt must be taken into consideration.) 
 
 Ans. 8.4 inches. 
 
 15. An 8-inch belt, traveling over a 30-inch pulley making 174 
 revolutions per minute, transmits 18 H. P. The ratio of the ten- 
 sions in the tight and slack sides of the belt is 3.06 : 1, the arc of 
 contact 160, and the maximum tension allowed per inch width of 
 belt 80 Ibs. Taking the velocity of the neutral surface of the belt 
 as the true velocity, it is required to find: (1) The thickness of the 
 belt. (2) The coefficient of friction. Ans. 0.25 inch; 0.4. 
 
 16. A driving shaft, making 100 revolutions per minute, carries 
 a pulley 22 inches in diameter, from which a" belt communicates 
 motion to a 12-inch pulley on a counter-shaft. On the counter-shaft 
 is also a cone pulley having steps of 8, 6, and 4 inches in diameter, 
 which gives motion to another cone pulley, of equal steps, on a 
 lathe spindle. Sketch the arrangement in side and end elevations, 
 and find the greatest and least speeds at which the lathe spindle can 
 revolve. Ans. 366.66; 91.66. 
 
 17. Determine the H. P. that may be transmitted by a belt, 6 
 inches wide and 0.25 inch thick, running at a speed of 60 feet per 
 second, the tension in the slack side being 0.45 of that in the tight 
 side of the belt, the stress allowed being 280 pounds per sq. inch 
 of section. To what extent does the effect of centrifugal action 
 reduce the power transmitted? Ans. 20.87; 17.2 per cent. 
 
 18. A weight of 8000 pounds is suspended from one end of a 
 rope. How many turns of the rope must be taken around a circular 
 beam fixed horizontally, in order that a man, who can pull with a 
 force of 250 pounds, may keep the rope from slipping, supposing 
 the coefficient of friction to be 0.2? Ans. 2.75. 
 
 19. By taking 3 turns of a rope about a post, and holding back 
 with a force of 180 pounds, a man just keeps the rope from slipping. 
 
216 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 The coefficient of friction being 0.2, find the weight supported at 
 the other end of the rope. Ans. 7810 Ibs. 
 
 20. It is required to transmit 16 H. P. from a pulley 20 inches 
 in diameter by means of a belt which embraces only two-ninths of 
 the circumference of the pulley. Find the tensions of the two 
 parts of the belt when slipping is just prevented, and the width of 
 belt required, the thickness of the belt being three-eighths of an 
 Jnch, the safe working stress 300 Ibs. per sq. inch, the speed of the 
 pulley 120 revolutions per minute, and the coefficient of friction 
 0.35. Ans. 2174.2 Ibs.; 1333.8 Ibs.; 19.32 inches. 
 
CHAPTER II. 
 WHEELS IN TRAIN. 
 
 15. The circles in Fig. 5 represent the pitch circles of two toothed, 
 or spur, wheels in gear. The pitch circles of two wheels in gear 
 are circles which appear to roll upon each other and which pass, 
 approximately, through the middle of the elevation of the teeth. 
 They may be regarded as the outlines of two discs which roll to- 
 gether by the friction at their circumferences. 
 
 There can be no slipping of one pitch circle over the other, owing 
 to the teeth; therefore, the same length of circumference of each 
 must pass over the point of contact in a given time. If D and d 
 represent the diameters of the large and small wheels respectively, 
 and N and n the number of their revolutions per unit of time, then 
 -n-DN = irdn , or DN = dn . 
 
 The teeth on both wheels are of the same size, and their number 
 will be proportional to the diameters of the wheels, and we may 
 therefore write TN tn, in which T and t are the numbers of 
 teeth on the large and small wheels respectively. 
 
 16. The pitch of the teeth is the distance measured on the cir- 
 cumference of the pitch circle between the centers of two consecu- 
 tive teeth. The diameter of the pitch circle = 2 y in which p is 
 the pitch, and t the number of teeth. 
 
 17. The formulas for belt pulleys hold for toothed wheels, for the 
 belt performs the same office as the teeth it causes the circum- 
 ference of both pulleys to move over the same distance in the same 
 time. 
 
218 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 18. The usual arrangement of a train of toothed wheels in a 
 mechanism is shown in Fig. 6, where two wheels unequal in size are 
 placed on each axis, except the first and last, and where the larger 
 wheel of any pair gears with the next smaller wheel in the train. 
 On the first and last axes there is but one wheel. 
 
 19. Let T lt T 2 , T 3 , and N lf N 2 , N s denote the number of teeth 
 in the driving wheels of each pair in gear and their number of 
 revolutions in a period of time respectively; let t lf t 2 , t s , and n t , 
 n 2 , n B be like representations for the follower wheels of each pair. 
 
 From what has been shown, we shall have : 
 
 T^ = t^ , T 2 N 2 = t 2 n 2 , and T S N Z = * 8 n, . 
 Multiplying these equations, member by member, we have : 
 2\ X T 2 X T 3 X # X N 2 X N* = *! X t, X * 3 X n, X n 2 X fi a . 
 
 Since the first follower and the second driver are fixed on the 
 same spindle, we have, n i = N 2 , and likewise, n 2 = JV 3 . 
 Therefore, N X Z\ X T 2 X T s = n s X ^ X t X * 3 - 
 
 That is : 
 
 The number of revolutions of the first driver, multiplied by the 
 continued product of the numbers of teeth in all the drivers, is 
 equal to the number of revolutions of the last follower, multiplied 
 by the continued product of the numbers of teeth in all the 
 followers. 
 
 If, in Fig. 6, N = 16 per minute, TI = 72 , T 2 = 40 , T z = 36 , 
 24 , t z = 20 , t s = 18 , we shall have by substituting in the 
 general equation, 
 
 16 X 72 X 40 X 36 = ^ 3 X 24 X 20 X 18 , 
 
 whence, n z 192, or the last follower will make 192 revolutions 
 while the first driver is making 16. 
 
 20. The ratio of the number of revolutions of the last wheel of a 
 train in a given time to the number of revolutions of the first wheel 
 
WHEELS IN TRAIN 219 
 
 in the same time expresses the Velocity Katio, or the Value, of the 
 train, and will be denoted by e. 
 
 To find e in the above example, we have : 
 
 = **. == - 192 =- 12 = 72X40X36 
 A! 16 " 1" "" 24X20X18 
 
 Product of the number of teeth in all" the drivers. 
 ~ Product of the number of teeth in all the followers. 
 
 21. When any number of wheels are in gear, no two of them 
 being on the same axis, as in Fig. 7, the combination is the equiva- 
 lent only of a single pair of wheels, viz., the first wheel and the 
 last wheel; the intervening wheels simply transfer the motion and 
 determine the direction of rotation of the last wheel. If the num- 
 ber of idler wheels intervening between the first and last wheels be 
 
 odd, the direction of rotation of the first and last wheels will be the 
 same; if even, the rotation will be in opposite directions. 
 
 22. Generally speaking, the part of a machine to which the motive 
 force is applied is called the Driving End, and the part at which 
 the resistance is overcome, or at which the useful work is done, is 
 called the Follower End, or the Load End. 
 
 Movement of Driving End . ,, , , 
 
 The ratio, _ , _ . -^ , -, , is called the 
 
 Movement of Follower, or Load, End 
 
 Velocity Ratio of the machine. 
 
 Load 
 
 We have seen that Mechanical Advantage = . . 
 
 Driving Force 
 
 hence, 
 
 Mechanical Advantage _ Load X Movement of Load End 
 
 Velocity Ratio ~~ Driving Force X Movement of Driving End 
 
 Useful Work of Machine __ M ec i L(in : ca i 
 
 == ?j -= =-; y^ =r= j. aunfitwHVAM 
 
 Work Supplied to Machine 
 
220 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 23. Transmission of Power by Gear Wheels. In the different 
 mechanisms employing a train of toothed wheels for the transmis- 
 sion of power, either the movement of a small power, P, through a 
 comparatively great distance, is utilized in overcoming a much 
 greater resistance, W, through a much smaller distance; or, con- 
 versely, the movement of a large power, P, through a small dis- 
 tance is utilized in making a smaller resistance, W, move through 
 a much greater distance. The desiderata in the two cases are 
 power and velocity respectively, the underlying mechanical prin- 
 ciple being that, what is gained in power is lost in speed, and 
 conversely. 
 
 The power, P., may be applied by hand to the end of a lever or 
 to the crank pin of an engine, the lever or crank to be rigidly con- 
 nected to the first axis of the train. 
 
 24. Two toothed wheels of unequal size, fixed on the same axis, 
 is the mechanical equivalent of a lever with unequal arms, and 
 therefore modifies the power that may be transmitted. 
 
 The toothed wheels B and (7, Fig. 8, are fixed on the same axis, c. 
 The driving power, P, is applied through wheel A tangentially to 
 the pitch surface of the teeth in contact at 5, the tendency being 
 to turn wheels B and C in a contra-clockwise direction about axis c. 
 The action of P is resisted by the reaction P^ of the load, applied 
 tangentially to the pitch surface of the teeth of the wheels C and D 
 in contact at d, with a tendency to turn the wheels B and C in a 
 clockwise direction about axis c. With these opposite turning ten- 
 dencies, we shall have, by the principle of the lever : 
 
 , whence, - = 4 = 
 ' P cb 
 
 
 Radius of wheel B 
 Number of teeth in wheel C 
 Number of teeth in wheel B " 
 
WHEELS IN TRAIN 
 
 221 
 
 25. The double purchase wheelwork of Fig. 9 represents that 
 commonly applied to hoisting machinery, such as cranes. The 
 numbers attached to the wheels and pinions indicate the number of 
 teeth they contain. The length of the lever handles being 18 inches, 
 and the diameter of the drum 20 inches, the weight that can be 
 raised by the application of 40 pounds at each of the handles may 
 be calculated by the principle of the lever, as follows : 
 
 Since the radii of wheels are proportional to the number of teeth 
 
 in the wheels, we may denote by ISx and 120z the radii of the first 
 
 pinion and the last wheel of the train respectively. 
 
 Let R = Length of lever handles, 
 " r = Eadius of drum, 
 " P!, P 2 , P 3 = Tangential pressures at the points of contact of 
 
 teeth in gear. 
 
 Then, by the principle of the lever, we have : 
 
 P x R = P X 18ar (1) 
 
 P X 80 = P 2 X 40 (2) 
 
 P 2 X 120 = P 3 X 20 (3) 
 
 P 3 X 120z = W X r (4) 
 
222 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 Multiplying these equations, member by member, we have : 
 P X R X 80 X 120 X 120z = W X r X 18a? X 40 X 20. 
 w P _ r 1S# X 40 X 20 r _ r . 1 
 
 C6 ' ~W ~5'80X120Xl20i ~"%* -~E X W 
 
 w= 
 
 80(40 + 40)18 
 
 r 10 
 
 In the above calculations the effect of friction and the diameter 
 of the rope have been neglected. Friction would be likely to re- 
 duce the result as much as 30 per cent, and the effective drum 
 radius would be the radius of the drum plus the radius of the rope. 
 
 26. The tangential pressures, P 19 P 2 , P 3 may be found by means 
 of equations (1), (2), (3), and (4), if the pitch of the teeth be 
 known. 
 
 Thus, suppose the pitch of the teeth to be 1.25 inches. Then, 
 
 Radius of first pinion = A^X_1? 45 
 
 4rr 
 
 Radius of last wheel = !* X 120 = WO 
 
 2;: 2rr 
 
 From equation (1) we have: 
 
 80 X 18 = P! X |jp whence, P = 402.3 Ibs. 
 
 From equation (2) we have: 
 
 402.3 X 80 = 40 X P 2 , whence, P 2 = 804.6 Ibs. 
 From equation (3) we have: 
 
 804.6 X 120 = 20 X P 3 , whence P 3 = 4827.6 Ibs. 
 These results would indicate that the teeth of the wheels should 
 be made stronger as the drum shaft is approached. This could be 
 done by using a coarser pitch for the last pair of wheels in gear. 
 
 27. If P denotes the tangential pressure at the pitch surface of 
 any wheel in the train, and V the velocity in feet per minute of the 
 wheel at the pitch surface, then, 
 
 PV 
 
 Horse-power transmitted = , 
 
 oo^OOO 
 
 and this is constant for any stage of the transmission. 
 
 Thus, if the lever handles make 24 turns per minute, the horse- 
 power at the lever handle axle is : 
 
 PV _ 80 X STT X 18 X 24 _ n ,. 
 33^)00 ~ 33,000 X 12 
 
 Since the value of the train is - , the last wheel will make Q 
 
WHEELS IN TRAIN 223 
 
 of a revolution in a minute, hence, the horse-power at the drum 
 axle is : 
 
 4827.6 X 2* X 150 X 3 
 
 33,000 X 12 X 2- X 10 
 In like manner, the horse-power transmitted from axle A to axle 
 J9 may be shown to be the same, thus : 
 
 1 8 V 94. 
 
 Axle A makes - ~ - = 5.4 revolutions per minute, and the 
 oO 
 
 radius of the pinion of 40 teeth is - s - = -^ 
 
 2- 2x 
 
 Horse-power transmitted from axle A to axle B 
 
 804.6 X 27: X 50 X 5.4 
 
 33,000 X 2- X 
 
 28. The gearing of a lathe to cut a thread of a given pitch is a 
 familiar illustration of wheels in train. There are two systems of 
 lathe gearing simple and compound. The driving wheel of the 
 gearing is either fast on the lathe spindle, or derives motion by 
 means of intermediate gearing. The intermediate gearing af- 
 fords a ready means of throwing the gear wheels out of action when 
 the lathe is running at high speeds, as for polishing, and it also 
 serves the purpose of changing the direction of rotation of the 
 feed screw when desired. In either case the revolutions of the first 
 driver are the same as those of the lathe spindle. 
 
 Any wheels intervening between the driver and the wheel on the 
 feed screw, no two being on the same axis, have no influence other 
 than to convey the motion, and the only wheels to be considered are 
 the driver and the one on the feed screw. If the gearing be such 
 that the lathe spindle and the feed screw make the same number of 
 
224: TRANSMISSION or POWER BY BELTS AND GEAR-WHEELS 
 
 turns, the thread cut would have the same pitch as the thread on 
 the feed screw. In all other cases the pitch of the thread to be 
 cut will be finer or coarser than that on the feed screw in the 
 exact proportion that the revolutions of the lathe spindle in a unit 
 of time are greater or less than those of the feed screw. 
 
 Figure 10 is a representation of simple gearing for a lathe. The 
 wheel A is fast on the lathe spindle, or so connected with it as to 
 make the same number of revolutions as the spindle. The wheel 
 B is on the feed screw of the lathe, and the intermediate wheel I is 
 an idler, serving the purpose of making the wheels A and B rotate 
 in the same direction, and also of filling in the space between 
 A and B, the axis of I being adjustable in a slotted arm. 
 
 For each revolution of the wheel B the nut on the feed screw 
 advances a distance equal to the pitch of the screw, and it depends 
 entirely upon the ratio between wheels B and A as to the number 
 of turns the spindle will make while the tool moves through the 
 same distance. 
 
 Suppose the pitch of the feed screw is ^ inch, and it is desired to 
 cut a thread of ^ inch pitch. Then for two turns of the feed 
 screw the tool will advance one inch, and the lathe spindle must be 
 
 o o 1 o 
 
 made to turn 13 times. We have, then, e = ~ = -^r ^o 
 
 JLo O/c <o 
 
 Hence, giving 12 teeth to wheel A and 78 to wheel B is among the 
 possible solutions. 
 
WHEELS IN TRAIN 
 
 225 
 
 If it were desired to cut 10 J threads to the in^h, the feed screw 
 pitch being J, we would have e = T?r = = ^y = ~^-r. Hence, giving 
 
 10.0 A\- o* 
 
 16 teeth to the wheel A. and 84 teeth to the wheel B is a probable 
 solution. 
 
 29. Figure 11 is a representation of double gearing for a lathe. 
 The axis intervening between the lathe spindle and the feed screw 
 carries two wheels of different size and are, therefore, factors in 
 the value of the train. 
 
 Suppose the pitch of the feed screw to be -| inch, and it is de 
 sired to cut 20 threads to the inch. We would then have: 
 
 teeth, and 40, 24, and 120 to B, C, and D respectively, is a solution. 
 In order to cut a left-hand thread the lathe spindle and the feed 
 screw must turn in opposite directions, which may be effected by 
 interposing an idle wheel between C and D. 
 
 30. The back gear attachment to a lathe increases the power at 
 the expense of the speed. Fig. 12 is a sketch illustrating the action 
 of back gear. The speed pulleys, to which the pinion A is attached 
 are loose on the lathe spindle. The spur-wheel E is keyed to the 
 spindle. The spur-wheel C and pinion D, carried on the shaft F, 
 form the back gear and can, at will, be thrown into or out of gear 
 with A and E. 
 15 
 
226 TRANSMISSION OF POWER BY BELTS AND GEAR-WHEELS 
 
 The motion of. the speed pulleys may be conveyed to the lathe 
 spindle in two ways : (a) The back gear being disengaged, as shown 
 in the figure, the spur-wheel E is made to engage with the speed 
 cone by means of a bolt, thereby giving to the spindle the same 
 revolutions that are made by the cone. (6) Disengaging the wheel 
 E from the speed cone, and throwing the back gear into gear with 
 wheels A and E, the motion of the speed pulleys is then transmitted 
 to the lathe spindle by means of the train ACDE, making the entire 
 system consist of a train of which the pulley on the counter-shaft 
 is the first driver and the spur-wheel E the last follower. 
 
 PEOBLEMS. 
 
 21. Three spindles, A, B, and (7, are arranged parallel to one 
 another. On A there is a spur-wheel with 52 teeth which gears 
 with a wheel of 19 teeth on spindle B. On spindle B is another 
 wheel of 81 teeth, gearing with one of 21 teeth on spindle C. 
 While A is making 15 turns, how many will C make? How many 
 will B make? Ans. 158.33; 41.05. 
 
 22. If the change wheels of a lathe have 18, 30, 40, 50, and 88 
 teeth, show an arrangement for cutting a screw of 11 threads to the 
 inch, the lead screw having a pitch of one-third inch. 
 
 23. With a lead screw of half-inch pitch, and change wheels of 
 20, 24, 30, 40, 55, 60, 80, and 100, show how you would select 
 wheels to cut screws of 6, 11, and 16 threads to the inch. 
 
 24. The back gear of a lathe is in use. Diameter of pulley on 
 counter-shaft, 4.25 inches; diameter of pulley on lathe, 8.75 inches; 
 pinion on cone pulley of lathe, 18 teeth; spur-wheel on back shaft, 
 58 teeth; pinion on back shaft, 18 teeth; spur wheel on lathe 
 spindle, 58 teeth. How many revolutions per minute will the lathe 
 spindle make when the counter-shaft is making 150 revolutions per 
 minute? Ans. 7.017. 
 
 25. Two parallel shafts, whose axes are to be as nearly as pos- 
 sible 30 inches apart, are to be connected by a pair of spur wheels, 
 so that while the driver runs at 100 revolutions per minute the fol- 
 lower is required to run at only 25 revolutions per minute. Sketch 
 the arrangement, and mark on each wheel its ' pitch-diameter and 
 number of teeth, the pitch of the teeth being 1.25 inches. Deter- 
 mine, also, the exact distance apart of the two shafts. 
 
 Ans. 29.8295 inches. 
 
WHEELS IN TRAIN 227 
 
 26. Make a sketch of a back gear of a lathe. If the two wheels 
 have 63 and 63 teeth each, and each pinion 25 teeth, find the reduc- 
 tion in the velocity ratio of the lathe spindle due to the back gear. 
 
 Ans. I : 6.35. 
 
 27. The double purchase wheelwork of a crane consists of a pin- 
 ion of 16 teeth, on the handle axle; a wheel and pinion of 64 and 
 20 teeth respectively, on the first intermediate axle; a wheel and 
 pinion of 80 and 18 teeth, respectively, on the second intermediate 
 axle; and a wheel of 90 teeth on the drum axle. The power han- 
 dles are 20 inches long, the radius of the drum 11 inches, the 
 diameter of the rope 2 inches, and the pitch of the teeth 1.25 inches. 
 Neglecting friction, it is required to find: (1) the power that 
 must be applied at the handles to raise 9600 Ibs. at the drum; (2) 
 the tangential pressures, P 1? P 2 , P 3 , between the teeth of the wheels 
 in gear; (3) the H. P. transmitted, supposing the handles to make 
 20 turns per minute. 
 
 Ans. 72 Ibs; 452.4 Ibs.; 1447.7 Ibs.; 6434 Ibs.; 0.4569 H. P. 
 
 28. The wheelwork of a crane consists of a pinion of 11 teeth, 
 driven by a lever handle 18 inches long, gearing with a wheel of 92 
 teeth, and of a pinion of 12 teeth gearing with a wheel of 72 teeth 
 on the drum axle. The diameter of the drum being 18 inches, it is 
 required to find the ratio of the power to the weight raised, friction 
 being neglected. Ans. 1 : 100 nearly. 
 
 29. The motion of an engine shaft is communicated to a shaft, 
 A } - by means of an open belt passing over a fly-wheel 64 inches in 
 diameter to a 15-inch pulley. The motion of shaft A is trans- 
 mitted to a shaft, B, by means of spur-wheels of J-inch pitch and 
 of 72 and 24 teeth respectively, with an intermediate wheel. The 
 motion of shaft B is communicated to the spindle of a fan by 
 means of an open belt passing over pulleys of 24 and 12 inches 
 diameter respectively. The revolutions of the engine are 200 per 
 minute, the stroke 16 inches, and the mean pressure on the crank 
 pin 8000 pounds. Find the revolutions per minute of the fan, and 
 the driving tensions in the belts passing over the pulleys on the 
 shafts A and B. Will the fan and fly-wheel turn in the same or in 
 opposite directions? 
 
 Ans. 5120 -revolutions; 2000 Ibs.; 416.66 Ibs. 
 
PART III 
 MATERIALS OF ENGINEERING 
 
CHAPTEE I. 
 MATERIALS. 
 
 1. COPPER. In the pure state copper is red in color, soft, ductile, 
 and malleable, with a melting point at about 2000 F., and a tensile 
 strength of from 20,000 Ibs. to 30,000 Ibs. per square inch of sec- 
 tion. It is not readily welded, except electrically, but is easily 
 joined by the operation of brazing. Attempts have been made to 
 temper it, but without much practical success. It is readily forged 
 and cast, and, when cold, may be rolled into sheets or drawn into 
 wire. When in sheets or in small pieces it may be spun, flanged, 
 and worked under the hammer. 
 
 The tensile strength of copper rapidly falls off as the temperature 
 rises above about 400 F., so that from 800 to 900 the strength 
 is only about one-half what it is at ordinary temperatures. This 
 peculiarity of copper should be borne in mind when it is used in 
 places where the temperature is liable to rise to these figures. 
 Again, if copper is raised nearly to its melting point in contact 
 with air, it readily unites with oxygen and loses its strength in 
 large degree, becoming, when cool, crumbly and brittle. Copper 
 in this condition is said to have been burned. The possibility of 
 thus injuring the tenacity of copper is of the highest importance in 
 connection with the use of brazed joints in steam pipes. 
 
 2. In the 'operation of brazing a joint, the surfaces to be joined 
 are cleaned, bound together with wire or otherwise, then supplied 
 with brazing solder in small bits, mixed with borax as a flux, and 
 placed in a clear fire until the solder melts and forms the joint. 
 The brazing solder, or hard solder, as it is often called, is usually 
 a brass or alloy of copper and zinc. The melting point of all such 
 alloys is below that of copper, and when copper is joined to brass, or 
 two pieces of brass are joined together, the solder used must have 
 a melting point lower than either of these metals. In the operation 
 of brazing a copper joint, therefore, the greatest care must be taken 
 in the selection of a solder and in attention to the fire, so that there 
 may be no danger of burning the copper, and thus endangering the 
 quality of the metal in the joint. 
 
232 MATERIALS OF ENGINEERING 
 
 Copper unalloyed is used chiefly for pipes and fittings, especially 
 for junctions, elbows, bends, etc. For large sizes the material is 
 made in sheets, bent and formed to the desired shape and brazed at 
 the seams. Small sizes are either made by the same general process 
 or drawn from the solid, which then may be bent as desired after 
 drawing. Copper is also largely used as the chief ingredient of the 
 various brasses and bronzes. 
 
 3. CAST IRON. This material consists of a mixture and combina- 
 tion of iron and carbon, with other substances in varying propor- 
 tions. 
 
 Influence of Carbon. In the molten condition, the carbon is 
 dissolved by the iron and held in solution just as ordinary salt is 
 dissolved by water. The mixture or combination of the two ele- 
 ments is thus entirely uniform. The proportion of carbon which 
 pure melted iron can thus dissolve and hold in solution is about 
 3.5 per cent. If chromium or manganese is present also, the 
 capacity for carbon is much increased, while with silicon, on the 
 other hand, the capacity for carbon is decreased. In the various 
 grades of cast iron, the proportion of carbon is usually found to be 
 between 2 per cent and 4.5 per cent. 
 
 Now, when such a molten mixture cools and becomes solid, there 
 is a tendency for a part of the carbon to be separated and no longer 
 remain in intimate combination with the iron. The carbon thus 
 separated, or precipitated, from the iron takes that form known as 
 graphite, and collects in very small flakes or scales. The carbon 
 which remains in intimate combination with the iron is said to be 
 combined, while that which is separated is usually called graphitic. 
 
 The qualities of cast iron depend chiefly on the proportion of 
 total carbon and on the relative proportions of combined and 
 graphitic carbon. 
 
 With a high proportion of graphitic carbon the iron is soft and 
 tough, with a low tensile strength, and breaks with a dark and 
 coarse-grained fracture. In fact, the substance in this condition 
 may be considered as nearly pure iron with fine flakes of graphite 
 entangled and distributed through it, thus giving to the iron a 
 spongy structure. The iron thus forms a kind of continuous mesh 
 ' about the graphite, which decreases the strength by reason of the 
 decrease of cross-sectional area actually occupied by the iron itself. 
 Such irons are termed gray. 
 
MATERIALS 233 
 
 As the relative proportion of graphitic carbon decreases and that 
 of combined carbon increases, the iron takes on new properties, be- 
 coming harder and more brittle. Its tensile strength also increases 
 to a certain extent, and the fracture becomes fine-grained, or smooth, 
 and whiter in color. When these characteristics are pronounced, 
 the iron is said to be white. When about half the carbon is com- 
 bined and half separates as graphite, the effect is to produce a dis- 
 tribution of dark spots scattered over a whitish field. Such iron is 
 said to be mottled. 
 
 In a general way, with a large proportion of total carbon, there 
 is likely to be formed a considerable amount of graphitic carbon, 
 and hence such iron is usually gray and soft. With a large propor- 
 tion of carbon also, the iron melts more readily. and its fluidity is 
 more pronounced. As the proportion of total carbon decreases, the 
 cast iron approaches gradually the condition of steel, whose prop- 
 erties will be discussed in later paragraphs. 
 
 Of the special ingredients in cast iron the combined carbon is 
 one of greatest importance. It is that chiefly which, by uniting 
 with the iron, gives it new qualities, and the principal influence of 
 other substances lies in the effect which they may have on the pro- 
 portion of this ingredient. As between graphitic and combined 
 carbon, the former does not affect the quality of the iron itself, but 
 acts physically by affecting the structure of the casting, while the 
 latter by entering into combination with the iron, acts chemically, 
 and produces a new substance with different qualities. 
 
 The proportions of combined and graphitic carbon are influenced 
 by the rate of cooling, and by the presence or absence of various 
 other ingredients. Slow cooling allows time for the separation of 
 the carbon and thus tends to form graphitic carbon and soft gray 
 irons. Quick cooling, or chilling in the extreme case, prevents the 
 formation of graphitic carbon and thus tends to form hard, white 
 iron. 
 
 In addition to carbon, small particles of silicon, sulphur, phos- 
 phorus, manganese, and chromium may be found in cast iron. 
 
 4. Influence of Silicon. The fundamental influences of silicon 
 are two. (a) It tends to expel the carbon from the combined state 
 and thus to decrease the relative proportion of combined carbon and 
 increase that of graphitic carbon. (6) Of itself silicon tends to 
 harden cast iron and to make it brittle. 
 
234 MATERIALS OF ENGINEERING 
 
 These two influences are opposite in character, since an increase 
 in graphitic carbon softens the iron. In usual cases the net result 
 is a softening of the iron, an increase in fluidity, and a general 
 change toward those qualities possessed by iron with a high pro- 
 portion of graphitic carbon. This applies with a proportion of 
 silicon from 2 to 4 per cent. With more than this the influence on 
 the carbon is but slight and the result on the iron is to decrease the 
 strength and toughness, giving a hard but brittle and weak grade 
 of iron. 
 
 A chilled cast iron is an iron which if cooled slowly would be 
 gray and soft, but if cooled suddenly by contact with a metal mould 
 or other means, becomes white or hard, especially at and near the 
 surfaces. Certain grades of cast iron tend to chill when cast in 
 sand moulds. This property is usually undesirable. In such cases 
 the tendency can be prevented by the addition of silicon, which, by 
 forcing the carbon into the graphitic state on cooling, prevents the 
 formation of hard, chilled surfaces. In all cases the actual effect 
 of adding silicon will depend much on the character of the iron 
 used as a base, and only a statement of the general tendencies can 
 here be given. 
 
 To sum up, a white iron which would give hard, brittle, and por- 
 ous castings can be made softer, tougher, and more solid by the addi- 
 tion of silicon to the extent of perhaps 2 or 3 per cent. As the 
 silicon is increased, the iron will become softer and grayer and the 
 tensile strength will decrease. At the same time the shrinkage will 
 decrease, at least for a time, though it may increase again with 
 large excess of silicon. The softening and toughening influence, 
 however, will only continue so long as additional graphite is formed, 
 and when most of the carbon is brought into this state, the maxi- 
 mum effect has been produced and any further addition of silicon 
 will decrease both strength and toughness. 
 
 5. Influence of Sulphur. Authorities are not in entire agree- 
 ment as to the influence of sulphur on cast iron, some believing 
 that it tends to increase the proportion of combined carbon, while 
 others maintain that it tends to decrease both the combined carbon 
 and silicon. It is generally agreed, however, that in proportions 
 greater than about 0.15 to 0.20 of 1 per cent, it increases the shrink- 
 age and tendency to chill, and decreases the strength. Sulphur 
 does not, however, readily enter cast iron under ordinary condi- 
 tions, and its influence is not especially feared. An increase in the 
 
MATERIALS 235 
 
 proportion of sulphur in cast iron is most likely to result from an 
 absorption of sulphur in the coke during the operation of melting 
 in the cupola. 
 
 6. Influence of Manganese. This element by itself decreases 
 fluidity, increases shrinkage, and makes the iron harder and more 
 brittle. It combines with iron in all proportions. The combina- 
 tion containing less than 50 per cent of manganese is called spiegel- 
 eisen. With more than 50 per cent of manganese, it is called 
 ferromanganese. One of the most important properties of man- 
 ganese in combination with iron is that it increases the capacity of 
 the iron for carbon. Pure iron will only take about 3.5 per cent 
 of carbon, while with the addition of manganese the proportion 
 may rise to 6 or 7 per cent. Manganese is also believed to decrease 
 the capacity of iron for sulphur, and to this extent may be a de- 
 sirable ingredient in proportions not exceeding 1 to 1.5 per cent. 
 
 7. Influence of Chromium. This substance is rarely found in 
 cast iron, but it has the property, when present in large propor- 
 tion, of raising the capacity of the iron for carbon from 3.5 per 
 cent up to about 12 per cent. 
 
 8. Shrinkage of Cast Iron. At the moment of hardening, cast 
 iron expands and takes a good impression of the mould. In the 
 gradual cooling after setting, however, the metal contracts, so that 
 en the whole there is a shrinkage of about -J inch per foot in all 
 directions, though this amount varies somewhat with the quality 
 of the iron and with the form and dimensions of the pattern. In 
 a general way hardness and shrinkage increase and decrease to- 
 gether. 
 
 9. Strength and Hardness of Cast Iron. The hardness of cast 
 iron is chiefly dependent on the amount of combined carbon, as 
 already noted. 
 
 The strength is also chiefly dependent on the same ingredient. 
 The greatest crushing strength is obtained with sufficient combined 
 carbon to make a rather hard, white iron, while for the maximum 
 transverse or bending strength, the combined carbon is somewhat 
 less and the iron only moderately hard; and for the greatest ten- 
 sile strength, the combined carbon is still less and the iron rather 
 soft. Metal still softer than this works with the greatest facility, 
 but is deficient in strength. 
 
236 MATERIALS OF ENGINEERING 
 
 10. Use of Cast Iron .in Engineering. Cast iron is used for 
 cylinders, cylinder heads, liners, slide valves, valve chests and con- 
 nections, and generally for all parts having considerable complexity 
 of form. It is also used for columns, bed plates, bearing pedestals, 
 caps, etc., though cast and forged steel are to some extent displac- 
 ing cast iron for some of these items. It is also used for grate 
 bars, furnace door frames, and minor boiler fittings, and for a 
 great variety of special purposes usually connected with the sta- 
 tionary or supporting parts of machines. 
 
 11. Inspection of Castings. In the inspection of castings, care 
 must be had to note the texture of the surface, and to this end the 
 outer scale and burnt sand should be carefully removed by the use 
 of brushes or chipping hammer, or, if necessary, by pickling in 
 dilute muriatic acid. The flaws most liable to occur are blow 
 holes and shrinkage cracks. The latter, however, are not often 
 met with when the moulding and casting are properly carried out. 
 The parts of the casting most liable to be affected by blow holes 
 are those on the upper side or near the top. On this account a 
 sinking head or extra piece is often cast on top, into which the 
 gases and impurities may collect. This is afterward cut off, leav- 
 ing the sounder metal below. 
 
 The presence of blow holes, if large in size or in great number 
 and near the surface, may often be determined by tapping with a 
 hand hammer. The sound given out will serve to indicate to an 
 experienced ear the probable character of the metal underneath. 
 
 12. Cast iron may be brazed to itself or to most of the structural 
 metals by the use of a brazing solder of suitable melting point, and 
 with proper care in the operation. Cast iron may also be united 
 to itself or to wrought iron or steel by the operation of burning. 
 This consists in placing in position the two pieces to be united, and 
 then allowing a stream of melted cast iron to flow over the surfaces 
 to be joined, the adjacent parts being protected by fire clay or 
 other suitable material. The result is to soften or partially melt 
 the surfaces of the pieces, and by arresting the operation at the 
 right moment they may be securely joined together. 
 
 13. Malleable Cast Iron. If iron castings of not too great thick- 
 ness, and of such purity as to be low in sulphur, be embedded in 
 powdered red oxide of iron (red hematite) and maintained at a 
 red heat for two or three days, they become, in a measure, decar- 
 
MATERIALS 237 
 
 bonized as a result of the chemical action which ensues. The car- 
 bon first disappears from the outer layers, and as the process con- 
 tinues, decarbonization toward the innermost layers takes place. 
 If the process be carried to the extreme in the effort to withdraw 
 all the carbon from the interior, the outer layer is very liable to 
 become brittle, thus defeating the object to be attained. For this 
 reason there always remains a core of cast iron only partially de- 
 carbonized, while the outermost layers are left in the condition of 
 soft, or malleable, iron. The process has little effect upon the 
 sulphur, manganese, phosphorus, and other impurities of the cast- 
 ings, but the resulting product is a malleable casting much less 
 fusible than cast iron and possessing six times its ductility. The 
 best product can be twisted and bent to a considerable extent before 
 breaking, and its ability to withstand shocks is much greater than 
 that of cast iron. Pipe fittings, to some extent, are malleable 
 castings, and this material is largely used in appliances of car 
 construction where more strength and toughness are required than 
 cast iron affords. 
 
 14. WROUGHT IRON. Wrought iron is nearly pure iron mixed 
 with more or less slag. Nearly all the wrought iron used in mod- 
 ern times is made by the puddling process. For the details of this 
 process reference may be had to text-books on metallurgy. We can 
 only note here that, in a furnace somewhat similar to the open- 
 hearth, most of the carbon, silicon, and other special ingredients of 
 cast iron are removed by the combined action of the flame and of 
 a molten bath of slag or fluxing material, consisting chiefly of 
 black oxide of iron. As this process approaches completion small 
 bits of nearly pure iron separate from the bath of melted slag and 
 unite together. This operation is assisted by the puddling bar, 
 and after the iron has thus become separated from the liquid slag 
 it is taken out, hammered or squeezed, and rolled into bars or plates. 
 Some of the slag is necessarily retained in the iron and, by the 
 process of manufacture, is drawn out into fine threads, giving to the 
 iron a stringy or fibrous appearance when nicked and bent over 
 or when pulled apart. 
 
 The proportion of carbon in wrought iron is very small, ranging 
 from 0.02 to 0.20 of 1 per cent. In addition, small amounts of 
 sulphur, phosphorus, silicon, and manganese are usually present. 
 
 The proportion of sulphur should not exceed 0.01 of 1 per cent. 
 
238 MATERIALS OF ENGINEERING 
 
 Excess of sulphur makes the iron red-short, that is, brittle when 
 red hot. 
 
 The proportion of phosphorus may vary from 0.05 to 0.25 of 
 1 per cent. Excess of phosphorus makes the metal cold-short, 
 that is, brittle when cold. 
 
 The proportion of silicon may vary from 0.05 to 0.30 of 1 per 
 cent. 
 
 The proportion of manganese may vary from 0.005 to 0.050 of 
 1 per cent. The influence of the silicon and manganese is usually 
 slight and unimportant. 
 
 Special Properties. Wrought iron is malleable and ductile, and 
 may be rolled, flanged, forged, and welded. It cannot be hard- 
 ened as steel, though by the process of case-hardening a surface 
 layer of steel is formed and may be hardened. Wrought iron may 
 be welded, because for a considerable range of temperature below 
 melting (which takes place only at a very high temperature) the 
 iron becomes soft and plastic, and two pieces pressed together in 
 this condition unite and form, on cooling, a junction nearly as 
 strong as the solid metal. In order that this welding operation be 
 successful, the iron must be heated sufficiently to bring it to the 
 plastic condition, yet not overheated, and there must be employed 
 a flux (usually borax) which will unite with the iron oxide and 
 other impurities at the joint, and form a thin liquid slag, which 
 may readily be pressed out in the operation, thus allowing the 
 clean metal surfaces of the iron to effect a union as desired. 
 
 15. Uses in Marine Engineering. In modern practice, the place 
 of wrought iron in marine engineering has almost entirely been 
 taken by steel. Its former office was for all moving parts requir- 
 ing strength and toughness. It is still used to some extent for stay 
 bolts and braces of boilers, and for boiler tubes. 
 
 16. STEEL. The properties of steel depend partly on the propor- 
 tions of carbon and other ingredients which it may contain, and 
 partly on the process of manufacture. The proportion of carbon 
 is intermediate between that for wrought iron and for cast iron. 
 In the so-called mild or structural steel, the carbon is usually from 
 3^- to -J of 1 per cent. In spring steel the carbon proportion is 
 somewhat greater, and in high carbon grades, such as are used for 
 tool steel, etc., the carbon is from 0.6 to 1.2 per cent. In addition 
 
MATERIALS 239 
 
 to the carbon there may be sulphur, phosphorus, silicon, and man- 
 ganese in varying but very small amounts. 
 
 Steel may be made from wrought iron by increasing its propor- 
 tion of carbon, or from cast iron by decreasing the proportion of 
 carbon. The earlier processes followed the first method, and high- 
 grade steels are still made in this way by the crucible process. 
 
 17. Crucible Steel. In this process a pure grade of wrought 
 iron is rolled out into flat bars. These are then cut, and piled 
 and packed with intermediate layers of charcoal, and subjected to 
 a high temperature for several days. This recarbonizes or adds 
 carbon to the wrought iron, and thus makes what is then called 
 cement or blister steel. These bars are then broken into pieces of 
 convenient size, placed in small crucibles, melted, and cast into 
 bars or into such shapes as are desired. 
 
 18. Mild, or Structural Steel, is made wholly by the second gen- 
 eral method the reduction of the proportion of carbon in cast 
 iron. There are two general processes, known as the Bessemer, 
 and the Siemens-Martin or open-hearth. 
 
 19. Bessemer Process. In this process the carbon and silicon 
 are burned almost entirely out of the cast iron by forcing an air 
 blast through the molten iron in a vessel known as a converter. A 
 small amount of spiegeleisen, or iron rich in carbon and man- 
 ganese, is then added in such weight as to make the proportion df 
 carbon and manganese suitable for the charge as a whole. The 
 steel thus formed is then cast into ingots or into such forms as 
 may be desired. In this process no sulphur or phosphorus is re- 
 moved, so that it is necessary to use a cast iron nearly free from 
 these ingredients in order that the steel may have the properties 
 desired. 
 
 A modification by means of which the phosphorus is removed, 
 and known as the basic Bessemer process, is used to some extent. 
 In this, calcined or burnt lime is added to the charge just before 
 pouring. This unites with the phosphorus, removes it from the 
 steel, and brings it into the slag. In the basic process, the lining 
 of the converter is made of ganister or a calcined magnesia lime- 
 stone, in order that it may not also be attacked by the added lime- 
 stone and the resulting slag. 
 
240 MATERIALS OF ENGINEERING 
 
 In the Bessemer process first noted, and often known as the acid 
 process, in distinction from the basic process, the lining of the 
 converter is of ordinary fire clay or like material. 
 
 The removal of the phosphorus by the basic process makes pos- 
 sible the use of an inferior grade of cast iron. At the same time, 
 engineers are not altogether agreed as to the relative values of the 
 two products, and many prefer steel made by the acid process from 
 an iron nearly free from phosphorus at the start. 
 
 20. The Open-hearth Process. In this process a charge of ma- 
 terial consisting of wrought iron, cast iron, steel scrap, and some- 
 times certain ores, is melted on the hearth of a reverberatory fur- 
 nace heated by gas fuel on the Siemens-Martin, or regenerative 
 system. The carbon is thus partially burned out in much the 
 same manner as for wrought iron, and the proportion of carbon is 
 brought down to the desired point or slightly below. A charge of 
 spiegeleisen or of ferromanganese is then added in order that the 
 manganese may act on any oxide of iron slag which remains in the 
 bath, and which would make the steel red-short if allowed to form 
 a part of the charge. The manganese separates the iron from the 
 oxide, returns it to the bath, while the carbon joins with that already 
 present, and thus produces the desired proportions. 
 
 Here, as with the similar operation with the Bessemer con- 
 verter, there is no removal either of sulphur or of phosphorus, and 
 only materials nearly free from these ingredients can be used for 
 steel of satisfactory quality. With very low carbon, however, a 
 little phosphorus seems to be desirable to add strength to the metal. 
 This limitation of the available materials has led, as with the 
 Bessemer process, to the use of calcined limestone, which unites 
 with most of the phosphorus and holds it in the slag. Here, as 
 in the Bessemer process also, it is necessary to use a basic lining 
 for the furnace, and it is known as the basic open-hearth process. 
 The method without the use of the limestone has come to be known 
 as the acid open-hearth process. 
 
 As between the products of the two open-hearth processes, there 
 is much difference of opinion among engineers. Either will pro- 
 duce good steel with proper care, and neither will without it. It 
 is usually considered sufficient to specify the allowable limits for 
 the proportions of phosphorus and sulphur, and leave the choice of 
 the acid or basic process to the maker. 
 
MATERIALS 241 
 
 21. Open-hearth and Bessemer Steels Compared. Open-hearth 
 steel is usually preferred for structural material in marine engi- 
 neering. This is because: 
 
 (a) It seems to be more reliable and less subject to unexpected 
 or unexplained failure than the Bessemer product. 
 
 (&) Analysis shows that it is much more homogeneous in com- 
 position than Bessemer steel, and experience shows that it is much 
 more uniform in physical quality. This is due to the process of 
 manufacture, which is much more favorable to a thorough mixing 
 of the charge than in the Bessemer process. 
 
 (c) The open-hearth steel may be tested from time to time dur- 
 ing the operation, so that its composition may be determined and 
 adjusted to fulfil specified conditions. This is not possible with 
 the Bessemer process. 
 
 22. Influence of Sulphur on Steel. Sulphur makes steel red- 
 short or brittle when hot, and interferes with its forging and weld- 
 ing properties. Manganese tends to counteract the bad effects of 
 sulphur. Good crucible steel has rarely more than 0.01 of 1 per 
 cent of sulphur. In structural steel the proportion may vary from 
 0.02 to 0.08 or 0.10 of 1 per cent. When possible it should be 
 reduced to not more than 0.03 or 0.04 of 1 per cent. 
 
 23. Influence of Phosphorus on Steel. Phosphorus increases the 
 tensile strength and raises the elastic limit of low carbon or struc- 
 tural steel, but at the expense of its ductility and toughness, or 
 ability to withstand shocks and irregularly applied loads. It is thus 
 considered as a dangerous ingredient, and the amount allowable 
 should be carefully specified. This is usually placed from 0.02 to 
 0.10 of 1 per cent. 
 
 24. Influence of Silicon on Steel. Silicon tends to increase the 
 tensile strength and reduce the ductility of steel. It also increases 
 the soundness of castings and ingots, and by reducing the iron 
 oxide, tends to prevent red-shortness. The process of manufacture 
 usually removes nearly all of the silicon, so that it is not an element 
 likely to give trouble to the steel makers. The proportion allow- 
 able should not be more than from 0.10 to 0.20 of 1 per cent. 
 
 25. Influence of Manganese on Steel. This element is believed 
 to increase hardness and fluidity, and to raise the elastic limit and 
 increase the tensile strength. It also removes iron oxide and sul- * 
 
 16 
 
242 MATERIALS OF ENGINEERING 
 
 phur, and tends to counteract the influence of such amounts of 
 sulphur and phosphorus as may remain. It is thus an important 
 factor in preventing red-shortness. The proportion needed to ob- 
 tain these valuable effects is usually found between 0.20 and 0.50 
 of 1 per cent. 
 
 26. Semi-steel. A metal bearing this trade name has in recent 
 years attracted favorable attention among engineers and has come 
 into considerable use where somewhat greater strength and tough- 
 ness are required than can be provided by cast iron. It is made by 
 melting mild steel scrap, such as punchings and clippings of boiler 
 plate, with cast iron pig, in the proportion of 25 or 30 per cent 
 of the former to 75 or 70 per cent of the latter. The presence of 
 manganese and other special fluxes in small proportions is also 
 found to add essentially to the strength, toughness, and ability to 
 withstand shocks decidedly greater than for cast iron, and with 
 fairly good machine qualities. Semi-steel casts as readily as most 
 grades of cast iron, and its shrinkage and general manipulation are 
 about the same. The chief drawback seems to lie in the danger of 
 hardness under the lathe, planer, or boring tool, but with the 
 proper mixture this is avoided, and a material very satisfactory for 
 many purposes in marine engineering is thus produced. 
 
 27. Mechanical Properties of Steel. The tensile strength of 
 the lowest carbon steel, say about 0.10 of 1 per cent carbon, is 
 usually not above from 50,000 to 55,000 Ibs. per square inch of 
 section. The strength increases with the increase of carbon and, 
 with not above the usual proportions of sulphur and phosphorus, 
 quite uniformly. Experiment shows that under these circum- 
 stances the strength will increase up to 75,000 Ibs. per square inch, 
 or higher, at the rate of from 1200 to 1500 Ibs. per 0.01 of 1 per 
 cent of carbon added. At the same time, with the increase in 
 strength the ductility decreases, so that a proper choice must be 
 made according to the particular uses for which the steel is intended. 
 In the best grades of tool steel, with carbon ranging from 0.5 to 
 1.0 per cent or over, the strength ranges from 80,000 to 120,000 Ibs., 
 and even higher in exceptional cases. 
 
 28. Flange and Rivet Steel must be tough and ductile in the 
 highest degree. Such steel has usually a tensile strength between 
 50,000 and 60,000 Ibs. and an elastic limit of 30,000 to 40,000 Ibs. 
 Its elongation in 8 inches is from 30 to 35 per cent, and the reduc- 
 
MATERIALS 243 
 
 tion of area at the ruptured section from 50 to 60 per cent. It 
 will bend cold on itself and close down flat under hammer or press 
 without a sign of fracture, the thickness not exceeding 1 inch. 
 
 29. Shell Steel, used for boiler shells, etc., has usually a strength 
 between 55,000 and 65,000 Ibs.; elastic limit from 33,000 to 
 45,000 Ibs.; elongation in 8 inches of 25 to 30 per cent, and reduc- 
 tion of area at the fractured section of from 50 to 60 per cent. 
 
 30. For Shafting, the quality of the steel is about the same as 
 for shell plates. For piston and connecting-rods, the strength is 
 rather higher, but ductility somewhat lower. 
 
 31. For Steel Castings, the strength required is usually from 
 60,000 to 65,000 Ibs., with an elongation in 8 inches of from 10 
 to 15 per cent. 
 
 32. Special Properties of Steel. Mild or low carbon steel may 
 be welded, forged, flanged, rolled, and cast. It cannot be tem- 
 pered or hardened with a proportion of carbon lower than about 
 .f of 1 per cent. High carbon steel can be welded only imperfectly, 
 and, if very high in carbon, not at all. It can be forged with care, 
 and cast into forms as desired. It can be tempered or hardened 
 by heating to a full yellow and quenching in cold water or by other 
 means, and then drawing the temper to the point desired. 
 
 Mild steel should not be worked under the hammer or flanging 
 press at a low, or " blue," heat, as such working is found in many 
 cases to leave the metal brittle and unreliable. Steel, in order to 
 weld satisfactorily, should have a low proportion of sulphur, and 
 special care is required in the operation, because the range of tem- 
 perature through which the metal is plastic and fit for welding is 
 less than with wrought iron. 
 
 33. Tempering. In the operation of tempering, the steel after 
 quenching is very hard and brittle. In order to give the metal the 
 properties desired, the temper is drawn down by reheating it to a 
 certain temperature, and then quenching again; or better still, by 
 allowing it to cool gradually, provided the temper does not 
 rise above the limiting value suitable for the purpose desired. If 
 the reheating is done in a bath of oil, the conditions may be kept 
 under good control and the final cooling may be slow. If the re- 
 heating is in or over a fire, the control is lacking and the piece 
 must be quenched as soon as the proper temperature is reached. 
 
244 MATERIALS OF ENGINEERING 
 
 This is usually determined by the color of the oxide or scale which 
 
 forms on the brightened surface of the metal. The following table 
 
 shows the temperatures,, corresponding colors, and the uses for 
 
 which the various tempers are suited : 
 
 430 Faint yellow 1 
 
 450 Straw yellow L Hardest and keenest cutting tools. 
 
 470 Full yellow J 
 
 490 Brown yellow I Cutting tools requiring less hardness and 
 
 or orange J more toughness. 
 
 510 Purplish "1 Tools for working softer materials, or those 
 530 Purple J required to stand rough usage. 
 
 550 Light blue 1 Spring temper. Used for tools requiring 
 560 Full blue L great elasticity or toughness, or for work- 
 600 Dark blue J ing very soft materials. 
 
 34. Special Steels. In the common grades of steel, the valuable 
 properties are due to the presence of carbon modified in some degree 
 by other ingredients, as already described. There are other sub- 
 stances which, uniting with iron in small proportions, are able to 
 give to the combination increased strength, hardness, or other 
 valuable properties. We have thus various special steels in which 
 the properties may be due to the presence of both carbon and other 
 ingredients, or due chiefly to special ingredients other than carbon. 
 Of these special steels we may note the following : 
 
 35. Nickel Steel, containing somewhere about 3 per cent of nickel 
 and varying amounts of carbon, is found to have increased strength 
 and toughness as compared with ordinary steel. Nickel steel is 
 most extensively used for armor plate, though to some extent it has 
 been employed in government work for screw-shafts and for boiler 
 plates. For the former purposes it has given excellent satisfaction, 
 but for the latter use difficulty has been met with in obtaining 
 plates free from surface defects. 
 
 36. Chrome Steel, containing from 0.5 to 1.5 or 2 per cent of 
 chromium, may be made excessively hard, but it is not always 
 reliable, and is not regarded with general favor. 
 
 37. Tungsten Steel or Mushet Steel is a steel containing tungsten 
 and carbon, the former in proportions as high as 8 or 10 per cent. 
 This steel must be forged with care and is excessively hard. The 
 hardness is not increased by tempering, but is naturally acquired 
 as the metal cools. Hence it is said to be self-hardening. Some 
 
MATERIALS 245 
 
 specimens contain also small amounts of manganese and silver. Its 
 chief use is for lathe, planer, and other cutting and shearing tools 
 where excesssive hardness is required. 
 
 38. Uses of Steel in Marine Construction. In modern practice, 
 mild or structural steel is used entirely in the construction of ships. 
 The same general class of material is used for all parts of boilers, 
 though the tubes are still sometimes made of wrought iron. 
 
 Cast steel, as well as cast iron, is used for various parts of en- 
 gines, such as pistons, crosshead blocks, columns, bed-plates, bear- 
 ing pedestals and caps, propeller blades, and for many small pieces 
 and fittings. 
 
 Forged steel is used for columns, piston-rods, connecting-rods, 
 crank and line shafting, and for many other smaller and minor 
 parts. 
 
 39. LEAD. Lead is a very soft, dense metal, grayish in color after 
 exposure to the air, but of a bright silvery luster when freshly cut. 
 Commercial lead often contains small amounts of iron, copper, 
 silver, and antimony, making it harder than the pure metal. It is 
 very malleable and plastic. In engineering, lead is chiefly of value 
 as an ingredient of bearing metals and other special alloys. Lead 
 piping is also used to some extent as suction and delivery pipes for 
 water where the pressure is only moderate, and where the readiness 
 with which it may be bent and fitted adapts it for use in contracted 
 places. 
 
 40. TIN. Tin is a soft, white, lustrous metal with great mallea- 
 bility. Commercial tin usually contains small portions of many 
 other substances, such as lead, iron, copper, arsenic, antimony, and 
 bismuth. It is largely used as an alloy in the various bronzes and 
 other special metals. Tin resists corrosion well and in consequence 
 is often used as a coating for condenser tubes. It is also used for 
 coating iron plates, the product being the so-called " tin plate " of 
 commerce. It melts at about 450, which corresponds to a steam 
 pressure of about 400 Ibs. per square inch. Due to this low melting 
 point, tin is often used in the composition for safety plugs in boilers. 
 
 41. ZINC. Zinc, or "spelter," as it is often commercially called, 
 is a brittle and moderately hard metal with a very crystalline frac- 
 ture. The impurities most commonly found in zinc are iron, lead, 
 and arsenic. It is used chiefly as an alloy in the various brasses, 
 bronzes, etc., and as a coating for iron and steel plates, rods, etc. 
 
246 MATERIALS OF ENGINEERING 
 
 The process of applying zinc for such a coating is called " galvan- 
 izing/' and the product " galvanized " iron or steel. Electricity is 
 not used in the process, the articles, after being well cleaned, being 
 simply dipped in a tank of melted zinc and then withdrawn. Slabs 
 of zinc are also used in marine boilers to prevent corrosion. 
 
 42. ALLOYS. A mixture of two or more metals is called an alloy. 
 The properties of an alloy are often surprisingly different from 
 those of its ingredients. The melting point is sometimes lower 
 than that of any of the ingredients, while the strength, elastic 
 limit, and hardness are often higher than for any one of them. 
 
 Mixtures of copper and zinc are called brass. Mixtures of cop- 
 per and tin, or of copper, tin, and zinc, with sometimes other sub- 
 stances in small proportion, form gun metals, compositions, and 
 bronzes. These terms are, however, rather loosely employed. Vari- 
 ous mixtures of two or more of the metals copper, tin, zinc, lead, 
 antimony form the various bearing metals. 
 
 Brass and composition are used for piping and pipe-fitting; 
 globe, gate, check, and safety valves; condenser tubes and shells; 
 sleeves for tail shafts, and for a great number of small fittings and 
 attachments for which the metal may be suited. The bronzes are 
 employed for many of the uses of brass where more hardness, 
 strength, or rigidity are required. They are used with special 
 success as a material for propeller blades. 
 
 The white metals, supported or backed by some other metal, such 
 as brass, cast iron, or cast steel, to give the necessary strength, are 
 now very largely used for bearing surfaces. 
 
 43. TIMBER. The use of timber in modern engineering con- 
 struction is becoming limited. The advance made in the produc- 
 tion of steel, whereby its homogeneity is assured and its superior 
 strength unquestioned, has caused it to be substituted for wood 
 whenever it is possible and profitable to do so, 
 
 Generally speaking, the heaviest and darkest colored timber is 
 the strongest, and, in all cases, the strength of timber is greatest in 
 the direction of its grain. 
 
 The locality from which timber comes, the season of its cutting, 
 and the duration of its seasoning, are factors in its strength, and 
 the uncertainty arising therefrom makes it advisable to use a factor 
 of safety of not less than 10 in calculations relating to the dimen- 
 sions of timber to withstand given loads. 
 
MATERIALS 247 
 
 44. CONCRETE. Concrete is a mechanical mixture of cement, 
 sand, and broken stone or gravel, in such proportions as experiment 
 has shown to be proper for the conditions governing its use. It is 
 largely used for laying foundations for buildings and bridges in wet 
 ground, and for breakwaters and sea-walls. After laying, it soon 
 hardens to a strong mass which is little permeable to water. 
 
 Concrete cannot be purchased as a manufactured product, but 
 must be made as needed. Whatever the proportions used, there is 
 the utmost necessity for thorough mixing, such water being added 
 as may be necessary to secure coherency in the mixture. The use 
 of machines designed for the purpose secures a more perfect mix- 
 ture than that attained by the hand process. 
 
 There is not perfect agreement as to the most advantageous pro- 
 portions for the composition of concrete, but good results may be 
 expected from the proportions of 1 of cement and 4 of gravel and 
 sand, when used for columns; if used for beams, the proportions 
 should not differ much from 1 of cement, 2.5 of sand, and 4.5 of 
 broken stone. 
 
 45. STEEL-CONCRETE. The uncertainty concerning the protec- 
 tion from corrosion of the " skeleton " steel frame, so largely used in 
 modern building construction, is mainly responsible for the very 
 recent and extensive use of " steel-concrete " also known as " ferro- 
 concrete " and as " reinforced-concrete " for beams and columns 
 in the erection of buildings. 
 
 There can be no question as to the appropriate use of steel for 
 constructive purposes in such open structures as bridges and steam- 
 ships, but in cases where a steel skeleton is vested with the strength 
 of a structure, and is subsequently incased in terra-cotta, stone, or 
 brick, precluding visual inspection, there is serious question as to 
 the propriety of its use. 
 
 The use of concrete for constructive purposes was common among 
 the ancients, and the fact that in some ruins, as they stand to-day, 
 the concrete parts remain the stone having long since disappeared 
 is conclusive evidence of its durability. There is abundant evi- 
 dence also that iron embedded in concrete is protected from the 
 corrosive influence of moisture and from the ravages of fire. 
 
 It has been demonstrated experimentally that concrete is strong 
 in compression but quite weak in tension. In the case of a con- 
 crete beam supported at the ends and loaded at the middle, it has 
 been shown that the upper side, which is in compression, is capable 
 
248 MATERIALS OF ENGINEERING 
 
 of supporting ten times the load which would cause failure at the 
 lower side, which is in tension. 
 
 Having in steel a material of very high tensile strength, and in 
 concrete a material possessing high compressive strength as well as 
 the properties of durability and impermeability to moisture, the 
 problem arose of effecting their combination so as to produce a 
 composite material having the desirable qualities of both. The solu- 
 tion of this problem was the production of steel-concrete. 
 
 In the case of beams, steel bars are embedded in the area of the 
 concrete below the neutral axis, reinforcing the concrete subjected to 
 tensile stress, and thus enabling the full strength of the compression 
 area above the neutral axis to be utilized in supporting heavier loads 
 than would otherwise be possible. 
 
 In the case of columns, the well-known fact that the most 
 economical metal section is that of the hollow cylinder suggested at 
 once the conception of a column having three concentric parts : (a) 
 a central core of concrete; (b) an outer zone of concrete; (c) an 
 intermediate zone of steel. The steel is thus protected from fire 
 and moisture and is best disposed for the utilization of its maximum 
 strength. 
 
 Between the tubular reinforced concrete column just described 
 and the plain non-reinforced concrete column there are a variety 
 of possible combinations. In the form generally used, applicable 
 alike to columns and piles, the reinforcement consists of vertical 
 rods of steel tied together by a system of horizontal wires. These 
 horizontal ties not only keep the vertical rods in position, but ma- 
 terially assist in preventing flexure in them; they also prevent lat- 
 eral bulging of the concrete. 
 
 The reinforcement for both beams and columns is first placed in 
 position, after which they are enveloped by a wooden form, or 
 mould, into which the concrete is dumped, and rammed when 
 necessary. After a period of thirty hours the form may be removed 
 and the steel-concrete product allowed to season for several weeks 
 before being subjected to its load. 
 
 46. Adhesion of Concrete to Steel. It has been shown by experi- 
 ment that the concrete on the compression side of steel-concrete 
 beams may be subjected without rupture to a stress twenty times 
 that which would cause failure in a tension test in the concrete 
 alone. It has also been shown that reinforced concrete acquires 
 
MATERIALS 249 
 
 a power to resist crushing which is greater than the sum of the 
 resistances of the two materials taken separately. Such remarkable 
 results could not be obtained were it not for the adhesion between 
 the concrete and steel, such adhesion offering resistance to sliding 
 between the two surfaces and facilitating the transference of the 
 forces from one surface to the other. 
 
 The bond between the steel and concrete occasioned by adhesion 
 alone is liable to be destroyed by internal stresses due to shocks and 
 vibrations, and from unequal expansions resulting from thermal 
 changes, and it is with the idea of strengthening the adhesion bond 
 by mechanical means that the reinforcing bars are sometimes 
 twisted or corrugated. 
 
 47. Distinction between " Steel and Concrete Construction " and 
 " Steel-Concrete Construction." In steel and concrete construction, 
 beams and columns of large section form the basis of strength 
 of the structure, the concrete contributing no assistance, its weight 
 in fact forming a part of the load borne by the steel members. In 
 steel-concrete construction, the reinforcing steel takes care only of 
 the tensile stress in beams leaving the entire compressive stress 
 to be borne by the concrete while in columns it increases the com- 
 pressive strength and tends to prevent bulging of the concrete. To 
 secure a uniform distribution of the stress in the reinforced section 
 of concrete, the reinforcement should consist of steel bars of small 
 section distributed so that each shall bear its allotted part of the 
 stress. The proportion of steel to concrete depends directly upon 
 the ratio of the coefficients of elasticity of the two materials. The 
 value of E for steel is 30,000,000 Ibs. per sq. inch, implying that a 
 force of one pound would extend or compress a bar 1 sq. inch in 
 
 area by QQ QQQ QQQ = 0.000000033 of its original length. The 
 value for concrete may be taken as 3,000,000, so that for an equal 
 extension of the two materials the steel will bear ten times the 
 stress that can be borne by the concrete. 
 
 Suppose a bar of steel 1 sq. inch in area to be surrounded by a 
 ring of concrete 1 sq. inch in area; and suppose further that the 
 steel be subjected to a direct pull that would occasion in the con- 
 crete the safe allowable tension stress of 50 Ibs. per sq. inch. The 
 
 elongation in the concrete would then be 50 X 3 QQQ QQQ 
 = 0.0000167 inch. Considering the bond between the steel and 
 
250 
 
 MATERIALS OF ENGINEERING 
 
 concrete to be perfect, the steel would suffer an equal elongation, 
 occasioning a stress in the steel of 0.0000167 X 30,000,000 = 500 
 Ibs. per sq. inch, a result only one-thirtieth of the safe allowable 
 tension stress for steel; consequently, the sectional area of the 
 steel may be reduced to -^ sq. inch, thus raising its stress to 15,000 
 Ibs. per sq. inch without causing failure in the concrete. 
 
 48. Design of Steel-Concrete Beam, When a beam is deflected 
 the section above the neutral axis is in compression and that below 
 the neutral axis is in tension (see Art. 51, p. 312). The tension 
 is not the result of a direct pull, and in consequence the unit stress 
 is not the same throughout the section below the neutral axis but 
 varies directly as the distance from the neutral axis. 
 
 \ 
 
 In the case of a steel-concrete beam the tensile stress will be taken 
 both by the steel and concrete, but as the bond is assumed to be 
 perfect the concrete will be permitted to extend only to the same 
 amount as the steel. The amount of the total stress borne by the 
 concrete will depend upon the proportion of steel to concrete. 
 
 In a steel beam the resistance to tension and to compression are 
 equal and the neutral axis passes through the center of gravity of 
 the section, but in the case of a concrete beam where the compres- 
 sion strength of the material is much greater than its tension 
 strength, the neutral axis moves away from the axis through the 
 center of gravity as the stress increases. , 
 
 The formulae for the deflection of beams are based on the sup- 
 position that the neutral axis passes through the center of gravity 
 (see Part IV), so that some modification will be necessary to make 
 them applicable to steel-concrete beams. 
 
 Knowing that the ratio of the compressive and tensile strengths 
 of concrete is 10 : 1, we may inquire as to the probable location of 
 the neutral axis in a concrete beam. 
 
MATERIALS 
 
 251 
 
 Let oo, Fig. 11, denote the position of the neutral axis of the 
 section of a rectangular concrete beam; c and c v the distances 
 from the neutral axis of the outermost fibers of the compression 
 and tension areas of the section respectively; li and h t the dis- 
 tances from the neutral axis of the centers of pressure of the com- 
 pression and tension areas respectively; and & the breadth and d 
 the depth of the section. 
 
 The total stress in a section is the product of the mean stress act- 
 ing at the center of pressure and the whole area, and the moment 
 of this stress with respect to the neutral axis may be taken as the 
 measure of the resistance of the section to bending. 
 
 The compressive working stress of the concrete being 10 times 
 the tensile working stress, with equal deformations, the mean com- 
 
 Y 
 
 
 
 X 
 
 
 t 
 
 
 
 i 
 
 c/V 
 
 
 
 
 
 
 ** h * 
 
 Fi 9 . IZ. 
 
 pressive stress will be denoted by 10$ when S denotes the mean 
 tensile stress. The resistance to bending offered by the compres- 
 sion area will then be lOSbch, and that offered by the tension area 
 will be SbcJi-L. Since these resistances must be equal for equilib- 
 rium, we shall have : 
 
 WSbch = S1)cji 19 whence Wch = cj^. (1) 
 
 The center of pressure of a rectangle on which the pressure in- 
 creases uniformly from at one edge to p at the opposite edge may 
 be found as follows : 
 
 In the rectangle of breadth & and altitude h, Fig. 12, we have : 
 Elemental area = bdy, 
 
 Elemental pressure = bdy X ^ , 
 Whole pressure = -P- I ydy, 
 Moment of whole pressure ==-?/ y*dy . 
 
252 MATERIALS OF ENGINEERING 
 
 The distance y of the center of pressure from the edge where the 
 pressure is is then, 
 
 n y*~\ h 
 
 y* d y a 1 2h 
 
 3 ' 
 
 tf 5 I 
 
 fi 
 
 Then, in Fig. 11, we shall have li = ~, and h t = 2 |^. Substi- 
 
 tuting these values of c and c in (1), we have : 
 
 20? 2 2< 2 
 
 Q- = -*-, whence ^ = c\/10. 
 
 But c + c = c + cVIO d, whence c = 0.24d, and c = 0.76d, 
 which fixes the position of the neutral axis under our suppositions. 
 
 Conversely, if sufficient steel be added to the tension area of the 
 section to make the resistance of steel + the resistance of concrete 
 ten times greater than the resistance of the concrete in the compres- 
 sion area of the section, we shall derive, c = 0.76^ and c = 0.24d 
 
 From this we infer that the position of the neutral axis may be 
 made to depend upon the proportions of concrete and steel, and 
 upon the disposition of the steel relative to the neutral axis. We 
 may, therefore, determine the position of the neutral axis by calcu- 
 lation when these details have been settled, or we may determine 
 the proportions of steel and concrete for any chosen position of the 
 neutral axis. 
 
 EXAMPLE. A steel-concrete beam of rectangular section is to 
 have a clear span of 10 feet and support a uniformly distributed 
 load of 6000 Ibs. Assuming the depth of the beam to be 14 inches, 
 and that the neutral axis is 9 inches from the lower edge of the beam, 
 it is required to determine the breadth of the beam and the sectional 
 area of the reinforcement. 
 
 From Art. 30, p. 291, we have, for a section in which the neutral 
 
 WT 
 axis passes through the center of gravity, Bending Moment = ^ ; 
 
 * c 7" 
 
 and from Art. 53, p. 313, we have, Eesisting Moment = . For a 
 rectangular section, I = yo , and we shall have, for equilibrium, 
 
 since d = 2c. 
 Whence, 
 
 WL Sbd 3 
 
 "' 
 
MATERIALS 253 
 
 in which 8 is the extreme fiber stress. The mean unit stress be- 
 tween the neutral axis and the outermost fiber is - I = ^ , 
 
 and the total stress, F, is therefore, be X o > from which we 
 derive : 
 
 F- 3WL U\ 
 
 w 
 
 Considering the reinforcement to be placed in the axis passing 
 through the center of pressure, and the whole of the tensile stress 
 to be concentrated in the reinforcement, we have, by substitution 
 in (4) : 
 
 32 X 9 
 
 
 l 
 
 Since the safe allowable stress for steel is 15,000 Ibs. per sq. inch, 
 the requisite sectional area of the steel reinforcement is, 
 
 Taking the safe allowable compressive fiber stress in the concrete 
 as 500 Ibs. per sq. inch, we have, from (2) : 
 
 _ 3WL _ 3 X 6000 X 120 _ h 
 
 ~'~~ 16X500 X25 ~ 
 
 as the width of the concrete. 
 
 The dimensions of the beam will then be 14 inches in depth by 
 10.8 inches width, making 10.8 X 14 151.2 sq. inches area of 
 section. 
 
 A section of the beam is shown in Fig. 13, the reinforcement con- 
 sisting of two bars, each of 0.25 sq. inch section, placed in the axis 
 
254 MATERIALS OF ENGINEERING 
 
 through the center of pressure in the tension area, distant 3 inches 
 from the bottom of the beam. 
 
 This treatment of steel-concrete is neither satisfactory nor con- 
 clusive. Indeed, the use of this composite material has not yet 
 emerged from the experimental state, and the opinion is entertained 
 by some engineers that thirty years hence the now popular theories 
 concerning both steel and concrete will have undergone changes. 
 
CHAPTER II. 
 TESTING MATERIALS. 
 
 49. Stress. The application of external forces to a piece of ma- 
 terial tends to change its shape, and this tendency induces internal 
 forces, known as stresses, which offer resistance to the change. 
 These stresses may be of three kinds : 
 
 (1) If the external force be applied at right angles to the sec- 
 tion, and acts away from it, the stress is one of tension, or a tensile 
 stress. 
 
 (2) If the external force acts toward the section, the stress is 
 one of compression, or a compressive stress. 
 
 (3) If the external force acts parallel to the section, the stress is 
 one of shear, or a shearing stress. 
 
 It is a fundamental assumption that these direct stresses are uni- 
 formly distributed over the section, so that if F denotes the external 
 force, or load, A the area of section, and 8 the unit stress, we must 
 have, in the absence of rupture, F = AS. F is usually expressed in 
 pounds and A in square inches, so that we shall have for the unit 
 stress : 
 
 F 
 -J 
 
 in pounds per square inch. 
 
 50. Strain. A piece of material which is stressed by the appli- 
 cation of external force undergoes some change in its dimensions, 
 either lengthened or shortened, and the amount of this distortion is 
 known as the strain due to the external force, or load. 
 
 51. Different Kinds of Tests. Materials are tested for tension, 
 by pulling apart a test piece of specified dimensions ; for compres- 
 sion by crushing a piece of definite dimensions; in cross-breaking, 
 by supporting a piece at two points and breaking or bending it in a 
 testing machine by applying a load at an intermediate point; in 
 torsion, by twisting apart a piece in a machine designed for the 
 purpose; in direct shearing, by breaking a riveted or pin-joint con- 
 nection in a machine ; for impact, or shock, by letting a weight drop 
 through a definite height, and, by its blow, develop suddenly the 
 stress in the material. 
 
256 MATERIALS OF ENGINEERING 
 
 52. Ultimate Strength. The ultimate strength of a test piece is 
 the load required to produce fracture, reduced to a square inch of 
 original section; or, in other words, it is the ultimate or highest 
 load, divided by the original area. Thus, if the area of the cross- 
 section of a test piece is 0.42 sq. inch, and the load producing 
 
 fracture is 28,400 Ibs., the ultimate strength is ^'|| = 67,620 
 Ibs. per sq. inch. 
 
 53. Factor of Safety. The factor of safety is the quotient aris- 
 ing from dividing the ultimate strength by the unit stress. 
 
 54. Elastic Limit. The elastic limit is the load, per square inch 
 of area, which will just produce a permanent set in the material. 
 Thus, in a tension test, if the cross-sectional area of the test piece 
 be 0.7 square inch, and a permanent set just be produced by a load 
 of 28,000 Ibs., the elastic limit is 40,000 Ibs. per sq. in. 
 
 55. Elongation. The increase in length of a test piece, measured 
 just before rupture, divided by the original length of the piece, ex- 
 presses the elongation. 
 
 When a load is first applied to a test piece, the elongation is 
 nearly uniformly distributed throughout the whole length. This 
 continues until the piece begins to contract in area near the point 
 of final rupture, and nearly all the subsequent elongation is re- 
 stricted to the immediate vicinity of this point. 
 
 In expressing the elongation of any material, the length of the 
 test piece must be stated. The usual length of test piece is 8 
 inches, so that if an extension in length of 2 inches were noted just 
 before rupture, the elongation would be expressed as 25 per cent in 
 8 inches. 
 
 56. Reduction in Area. The reduction of area is found by divid- 
 ing the difference between the original and final sectional areas at 
 the point of rupture by the original area, expressing the fraction in 
 per cent. 
 
 57. Test Piece for Iron. The form of test piece for wrought iron 
 plate prescribed by the U. S. Board of Supervising Inspectors of 
 Steam Vessels is illustrated in Fig. 1. If the plate is y^mcn thick, 
 or less, the width at the reduced section must be 1 inch. If the 
 plate is over T 5 inch in thickness, the width of the piece must be 
 
TESTING MATERIALS 
 
 257 
 
 reduced so that the cross-sectional area at the reduced section shall 
 
 to* 
 
 ?. /. Test F<*ce for Iron, 
 
 be about 0.4 sq. inch, but it must not be greater than 0.45 sq. inch, 
 nor less than 0.35 sq. inch. 
 
 58. Test Pieces for Steel and Other Materials. Figure 2 shows 
 
 *0 
 
 i 
 
 ?. Trarf Piece /or Jteet. 
 
 the form of test piece for tension prescribed by the Navy Department 
 for tests of steel plates for naval uses. 
 
 Figure 3 shows the form prescribed by the Association of Ameri- 
 
 V- / RADIUS J 
 
 fc ^ 
 
 r 
 
 ^ 
 
 can Steel Manufacturers, and adopted by the U. S. Board of Super- 
 vising Inspectors. The test piece for plates is cut from a " coupon/' 
 
 fig. -4 Plate toit/i Coupon. 
 
 as it is called, left on one corner of the plate as shown at A, Fig. 4. 
 The U. S. law requires further that : 
 17 
 
258 
 
 MATERIALS OF ENGINEERING 
 
 " Every iron or steel plate intended for the construction of boil- 
 ers to be used on steam vessels shall be stamped by the manufac- 
 turer in the following manner: At the diagonal corners, at a dis- 
 
 8* 
 
 Round T<ut Pi**. 
 
 tance of about 4 inches from the edges and at or near the center 
 of the plate, with the name of the manufacturer, the place where 
 manufactured, the number of pounds tensile stress it will bear 
 to the sectional inch." 
 
 ij. 6. Bending Test 
 
 Figure 5 shows the usual round form of test piece for all materials 
 except plates. 
 
 7. Jh^ft 
 
 Figure 6 illustrates a cold bending test on a piece of steel plate. 
 A drift test is sometimes required. This is illustrated in Fig. 7, and 
 
 e. 
 
 uu 
 
 r Jntf* Test. 
 
 consists in driving tapered drifts of different sizes into a punched 
 or drilled hole until the diameter is increased to at least twice its 
 
TESTING MATERIALS 
 
 259 
 
 original size. The metal must stand this test without sign of frac- 
 
 ture about the edges of the hole. 
 T irons are illustrated in Fig. 8. 
 
 Bending tests for angle and 
 
 59. Ductility. Materials such as wrought iron, mild steel, cop- 
 per, and other metals which may be lengthened by the application 
 of an external force, with a corresponding decrease in thickness or 
 in diameter, possess the property of ductility. 
 
 60. Plasticity. If, in the process of stretching a ductile mate- 
 rial, the load be removed and none of the strain disappears, the ma- 
 
 Fig. 9. 
 
 
 
 terial is said to have passed beyond the ductile and to have entered 
 the plastic stage. The plasticity of the material is determined by 
 the final elongation and contraction in area of the test piece. In 
 structures subjected to live loads and shocks it is equally important 
 to know the power of the material used to resist deformation with- 
 out rupture as it is to know the ultimate strength, and for that 
 reason specifications for iron and steel usually require a certain 
 percentage of elongation and contraction of area in a stated length 
 of test piece. 
 
 61. Stress-Strain Diagram. If, in testing a material, the gradu- 
 ally applied loads be plotted as ordinates and the corresponding 
 
260 MATERIALS OF ENGINEERING 
 
 strains as abscissas, the resulting curve is known as a stress-strain 
 diagram. 
 
 For tension tests of wrought iron and mild steel such a diagram 
 will take the form shown in Fig. 9. 
 
 The load being gradually increased, it will be found that, within 
 a certain limit, the strains, or extensions, will be directly propor- 
 tional to the augmentations in the load, and that, if the stress 
 be relieved, the test piece will return to its original length. Dur- 
 ing this period the material is said to be perfectly elastic, and will 
 be so represented in the diagram by the straight line OA . By con- 
 tinuing the gradual increase in the load a point will be reached 
 where the proportionality between the strain and the augmenta- 
 tion in the load ceases, the strain increasing much more rapidly 
 than the load ; and if the stress be relieved the piece will not return 
 to its original length, but will acquire a permanent set. The load 
 at which this occurs is known as the elastic limit of the material. 
 
 A further increase in the load very soon develops a point where 
 the extension increases very rapidly as much as from 10 to 15 
 times its previous amount known as the yield point. This rapid 
 increase in the extension usually occasions an apparent reduction 
 in the stress, as shown by the fall in curvature beyond the point B. 
 For commercial purposes the yield point and the elastic limit are 
 taken as the same point, and the ordinate at B would represent, to 
 the scale of the diagram, this limit in pounds per square inch of 
 section of the material. 
 
 Passing the yield point the strains increase much faster than the 
 loads, but if the stress in ftie material be relieved a careful meas- 
 urement will show the disappearance of a small portion of the 
 extension, indicating the existence still of some elasticity and that 
 the specimen is passing through the ductile stage. 
 
 At about the time the maximum stress is reached at (7, the mate- 
 rial appears to have reached the plastic state, the extension in- 
 creasing, in time, without increase in load. Up to this point the 
 strain has been evenly distributed throughout the length of the 
 specimen, but here occurs an extension and reduction in section 
 purely local, immediately followed by rupture at D. 
 
 Within the elastic limit the extension of the specimen probably 
 
 would not exceed of its length, so it is quite impossible to 
 
TESTING MATERIALS 
 
 261 
 
 make direct measurements of the extensions corresponding to the 
 augmentations in load. Some form of extensometer is used to 
 make these measurements. 
 
 62. Tension Test of Mild Steel. For example, Fig. 10 is a stress- 
 strain diagram for a tensile test of mild steel. 
 
 So ooo 
 
 2.6000 
 
 * 
 
 I 
 
 $ 
 
 2.2.000 \-r~ 
 
 2oooo 
 
 /Qooo 
 
 /jooo 
 
 IOOOO 
 
 6*00 
 
 2.000 
 
 
 
 
 
 
 s* 
 
 *^- 
 
 
 
 
 
 \ 
 
 
 
 
 ^ 
 
 X 
 
 
 
 
 
 
 
 ^ 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 * 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 57/> 
 
 fee 
 
 / 
 
 Z7 
 
 "9 
 
 ra 
 
 m 
 
 
 
 
 
 
 
 
 r ^7 
 
 
 
 
 
 
 
 
 7i 
 
 /75 
 
 /o/- 
 
 r 7 
 
 ?5-7 
 
 c 01 
 
 
 >ee 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 /" 2' 
 
 xfenstons. Sca/e^ fc///$/ze. 
 
 ng. /o 
 
 The original dimensions of the test piece were : 
 
 Length, 8 inches; diameter, 0.75 inch; area of section, 0.4418 
 square inch. 
 
 The final dimensions were : 
 
 Length, 10.2 inches; diameter, 0.4843 inch; area of section, 
 0.1842 square inch. 
 
 The load was gradually applied with the uniform augmentation 
 of 2000 pounds, and the data tabulated as follows : 
 
262 
 
 MATERIALS OF ENGINEERING 
 
 Loads. 
 
 Extensions. 
 
 Remarks. 
 
 2,000 
 
 0.0012 
 
 
 4,000 
 
 . 0024 
 
 
 6,000 
 
 0.0034 
 
 
 8,000 
 
 0.0047 
 
 
 10,000 
 
 0.0060 
 
 
 12,000 
 
 0.0072 
 
 
 14,000 
 
 0.0083 
 
 
 16,000 
 
 0.0095 
 
 
 18,000 
 
 0.0106 
 
 
 20,000 
 
 0.0118 
 
 
 22,000 
 
 0.0220 
 
 Yield Point 
 
 26,000 
 
 . 5300 
 
 
 29,680 
 
 1 . 9200 
 
 Maximum Stress 
 
 25,820 
 
 2.2000 
 
 Breaking Stress 
 
 An inspection of the diagram shows the elastic limit to have 
 been reached at about 21,000 pounds. 
 
 t> I fifin 
 
 Elastic limit of specimen = Q 4418 = ^>^ ^s. P er sc l- i ncw - 
 
 maximum load 29,700 
 Ultimate stress = - = ^-^- a rs 67/220 Ibs. per sq. 
 
 inch. 
 
 original area " 0.4418 
 
 , . - Q u (10.2 - 8)100 _, K 
 
 Extension in 8. inches = g j - = 27.5 per cent. 
 
 (0.4418 - 0.1842)100 
 Contraction of area = 04418 ~ 58.31 per cent. 
 
 To find the modulus of elasticity we proceed as follows : 
 The sum of the extensions up to and including the 18;000 pound 
 load a point well within the limit of elasticity is 0.0533 inch, 
 and the sum of the loads is 90,000 pounds. The mean extension 
 
 0533 
 for 2000 pounds is, therefore, ' ^ 0.0012 inch. 
 
 T ^ 
 
 By Art. 50, page 311, we have, E = - , in which L is the orig- 
 
TESTING MATERIALS 
 
 263 
 
264 MATERIALS OF ENGINEERING 
 
 inal length and 8 the load producing the extension y. Here 
 L = 8 inches, S = 2000 pounds, and y = 0.0012. 
 
 Hence > E = = 30 > 180 > 000 lbs - p er s( i uare inch - 
 
 63. Compression Tests. In testing materials for compression the 
 specimens are not longer than from 1.5 to 3 times the diameter. If 
 the specimens are long the failure under compression will be by 
 buckling or bending, and for intermediate lengths partly by crush- 
 ing and partly by bending. 
 
PART IV 
 
 THE ELEMENTS OF THE MECHANICS 
 OF MATERIALS 
 
CHAPTER I. 
 MOMENTS. CENTER OF GRAVITY. 
 
 1. Moments. The moment of a force acting on a body may 
 be defined as the power of the force to turn the body about a point, 
 or about a fixed axis, and its measure is the product of the force by 
 the perpendicular distance from the point, or from the axis, to 
 the line of action of the force. The point, or axis, about which 
 the moments are taken is called the center of moments. 
 
 If there be a number of forces acting on the body, those tending 
 to turn it in one direction may be regarded as positive, and those 
 tending to turn it in the opposite direction as negative. By com- 
 mon consent forces with a turning tendency in a clockwise direction 
 are termed positive and those with a contra-clockwise tendency 
 
 negative. It is immaterial which kind of force is termed positive 
 and which negative, but having chosen one kind as positive in any 
 investigation the choice must be adhered to, and the opposite kind 
 must be regarded as negative. 
 
 2. In Fig. 1, let the forces, P, Q, and R act on a body in the 
 directions indicated. If the body remains in equilibrium the under- 
 lying principle of moments asserts that the algebraic sum of the 
 moments of the forces about any point as a center, or about any 
 line as an axis the point and the line being in the same plane 
 must be zero. In other words, the clockwise moments must be 
 equal to the contra-clockwise moments. 
 
 Let moments, be taken about the point 0. The force Q tends to 
 turn the body in a clockwise direction about and will be regarded 
 
268 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 as positive, and the tendency of the forces P and K is to turn it 
 in a contra-clockwise direction about and will he regarded as 
 negative. The equation of moments will then be, Qq Pp Rr 
 = 0. This follows directly from the meaning of the word equilib- 
 rium, which implies that the body is at rest, and this condition can 
 only result when there is no tendency to turn the body about the 
 point ; that is, when the algebraic sum of the moments about 
 is zero. Should the line of action of a force pass through the center 
 of moments, the moment of that force would vanish. 
 
 The principle of moments is very broad in its applications, and 
 if we denote a force by /, a mass by m, an area by a, a volume by v, 
 and the arm of each by r, then fr, mr, ar, vr will be the moment 
 of the force, mass, area, and volume, respectively. 
 
 In expressing the value of moments, the units of force, mass, 
 area, and volume are placed first, and the length units afterward. 
 For example, a moment may be expressed as so many pounds-feet, 
 and thus avoid confusion with work units. 
 
 3. Center of Gravity. The center of gravity of a body or of a 
 system of bodies is a point on which the body or system will balance 
 in all positions, supposing the point to be supported, the body or 
 system to be acted on only by gravity, and the parts of the body or 
 system to be rigidly connected to the point. 
 
 It follows from this definition that, all the particles of a body or 
 system of bodies are acted on by a system of parallel forces gravity 
 acting on each and that the algebraic sum of the moments of these 
 forces about a line must be zero when the line passes through the 
 center of gravity of the body or system ; otherwise the body or sys- 
 tem would not balance. 
 
 4. Let two heavy bodies of weights P and Q be situated as shown 
 in Fig. 2. Join them with a straight line and divide it at C, so 
 
 -Q = -. 
 
 x> r 
 
 weight, the system will, by the principle of the lever, balance when 
 
 that -Q = -. If the weights be joined by a rigid rod without 
 
 x> r 
 
MOMENTS. CENTER OF GRAVITY. . 269 
 
 supported at C. C is, therefore, the center of gravity of the system. 
 As the resultant of the weights is P -\- Q, the pressure on the sup- 
 port will be P + Q. The center of gravity of a uniform straight 
 rod is, evidently, at the middle point of its length. When consid- 
 ering a body at rest, we may assume its whole mass to be concen- 
 trated at the center of gravity. 
 
 5. Let the weights P and Q, Fig. 3, be attached as shown to a 
 balanced rod. Then, as has just been shown, the center of gravity 
 
 will be so situated that 7^ = , or Pm Qn\ that is, the algebraic 
 
 \J nt 
 
 sum of the moments is zero. In locating the position of the center 
 cf gravity of a system it will be convenient to take moments about 
 a point other than the center of gravity, or about a line other than 
 one passing through the center of gravity. Let x denote the dis- 
 tance of the center of gravity of the system from 0, and let p and q 
 be the distances, respectively, of the centers of gravity of P and Q 
 
 Fig. 3. 
 
 from 0. We have from the figure ra = x p, and n = q x. 
 Then P(x p) =Q(q x), from which we have x = p ,Q 
 
 T - ,, i , , - Pp + Qq + Rr + &c. 
 
 If the system be extended the result, x = p ^ 
 
 will be obtained. This formula is extensively used in the solution 
 of problems. 
 
 6. In considering a sheet of uniform thickness weighing M 
 pounds per unit of area, the weight of any given area will be Ma 
 pounds. We may substitute Ma^ for P and Ma 2 for Q in the for- 
 mula and obtain 
 
 - _ 
 
 + 2 a v + a* 
 
 in which A is the whole area. 
 This may be expressed as follows: 
 
270 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 'Distance, x, from of the center of gravity is 
 
 sum of the moments of all the elemental surfaces about 
 
 area of surface 
 the moment of the whole surface about 
 
 whole area of surface 
 Similarly, 
 
 the moment of the whole volume about 
 
 x = 
 
 whole volume 
 
 The moment of the whole surface and of the whole volume is ob- 
 tained by integration. 
 
 7. The center of gravity of a plane figure may be obtained 
 graphically as follows : 
 
 Let dbcde, Fig. 4, be any plane figure. Draw eb and ec. Let 
 A^ and g 19 A 2 and g 2 , and A s and g & denote the areas and centers 
 of gravity of the triangles eab, ebc, and ecd, respectively. Join g^ 
 and g 2 . The center of gravity of the figure abce must lie on this 
 line. From g^ lay off in any direction and to any scale, the dis- 
 tance g^t equal to A 29 and in the opposite direction, and to the same 
 scale, lay off from g 2 , parallel to g^t, the distance g. 2 s equal to A^. 
 Join st, and the intersection, g 19 2 , of st and g^g z is the center of 
 gravity of the figure abce. Join g lt 2 with g 3 , and the center of 
 gravity of A^ -(- A 2 + A 3 will lie in this line. From # 1? 2 lay off a 
 distance g ly 2 r equal to the area A M and from g^ lay off g a v parallel 
 to <7 1? z r and equal io A^-\- A 2 . Join vr, and its intersection with 
 ffi> 2^3 g ives ^u 2? 3 as tne center of gravity of the whole figure. 
 
MOMENTS. CENTER OF GRAVITY. 
 
 8. Should the surface contain a hole, as fmn in Fig. 5, we would 
 proceed as follows : 
 
 Denote by A t and g ly and A 2 and g 2 the areas and centers of 
 gravity of the whole figure abcde and of fmn, respectively. Join 
 g^g 2 . From g 2 lay off to scale, in any direction, g 2 s equal to A lf 
 and parallel to it, and on the same side of g^g 2 , lay off g^r. Join 
 
 sr and produce it to meet g^g 2 produced at g l9 2 . Then g lf 2 is the 
 
 required center of gravity. 
 
 9. The truth of these graphic methods may be shown as follows : 
 Let Oj and 2 , Fig. 6, be the positions of the centers of gravity 
 
 of two areas A and A Z9 respectively, and let their common center 
 
 of gravity be situated at g, distant r from and r 2 from 2 . 
 By the principle of moments we shall then have A^ = A 2 r. 2 . From 
 2 lay off 2 b, whose length represents the area A^ to some selected 
 scale, and from L lay off O^d parallel to 2 b and of such length as 
 
272 
 
 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 to represent the area A 2 to the same scale. Then the line joining b 
 and d will pass through g. The triangles O dg and 2 bg are similar, 
 
 and we have 2 b : O^d = 2 g : O^g, or 
 
 : A 2 = r 2 
 
 whence 
 
 A 1 r L = A 2 r 2 . It will be observed that the lines 2 b and O^d are 
 laid off on opposite sides of the line joining O t and 2 and at oppo- 
 
 site ends to their respective areas, and at any convenient angle. 
 Should one of the areas, as A 2 , be negative, i. e., is the area of a hole, 
 or of a part cut out of the surface, then 2 b and 0d must be laid 
 off on the same side of 2 , as shown in Fig. 7. 
 
 10. To find the center of gravity of a portion of a regular polygon 
 or of an arc of a circle, considered as a thin wire : In Fig. 8, let 
 
 riff. 8. 
 
 L = length of the sides of the polygon, or of the length of the 
 arc in the case of a circle; R = radius of the inscribed circle, or 
 the radius of the circle itself ; C = chord of the arc of .polygon or 
 circle; Y"= distance of the center of gravity from the center of 
 
 the circle. Then L : E C: Y, OT Y = -j- . For, let a denote 
 
 A 
 
 the length of a side of the polygon, Fig. 9. The center of gravity 
 of each side will be at its middle point, and distant y l9 y 2 , y s , &c., 
 from the diameter of the inscribed circle. Let x , x 2 , X B , &c., de- 
 
MOMENTS. CENTER OF GRAVITY. 273 
 
 note the projected lengths of the sides on the diameter. From the 
 similar triangles Ob c and edf we have de : Ob df : ~bc, or 
 
 a : R x l : y 1} whence y^ = --~; similarly, y 2 = , and so on. 
 
 Cl U, 
 
 If w denotes the weight of each of the n sides of the polygon, we 
 shal, have : T = %$ = ?*-** = ** n **. 
 
 Substituting the values of y ly y 2 , y 3 , etc., we have : 
 
 -p 
 y= (x 1 + x 2 -J- &c.) . But na = L, and x + x z + & c - C> 
 
 T>n 
 hence Y = j- . The polygon becomes* a circle when n is infinitely 
 
 great. 
 
 11. To find the center of gravity of a portion of a regular poly- 
 gon or of a sector of a circle when considered as a lamina or thin 
 sheet. 
 
 /o 
 
 Keferring to Fig. 10, and using the same notation as in Art. 10, 
 
 o r> 
 we shall have, -o- as the distance of the center of gravity of each 
 
 of the triangles from the center of the inscribed circle, and they 
 will be distant y lf y z , y z , &c., from the diameter. The base of each 
 of the triangles is a, and the projected lengths of these bases on 
 the horizontal diameter are x lf x 2 , x s , &c. From the similar tri- 
 angles 01 c and edf we have: 
 
 2R ZRx, "2Rx.i 
 
 -- : a = y 1 : x lf whence y - - ; similarly, y 2 - -, and 
 
 so on. 
 
 If w denotes the weight of each of the n triangles of the polygon, 
 we shall have: 
 
 w * + w * + &c - i + + &c -^ R 
 
 18 
 
274 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 12. To show the application of the foregoing principles to finding 
 centers of gravity, the solutions of a few problems will be given. 
 
 EXAMPLE I. Find the center of gravity of a triangle. 
 
 Conceive the triangle, Fig. 11, to be divided into a great number 
 of very narrow strips drawn parallel to one of the sides, as B. The 
 center of gravity of each strip will be at its middle point, therefore 
 the center of gravity of the triangle will lie in the locus of these 
 middle points; that is, in the median Oc. 
 The elemental area of the triangle = b-dh. 
 
 The moment of the elemental area h-b-dh. 
 
 From similar triangles we have, b : B = h : H ; whence, b = -- 
 
 Then, 
 
 B-tf-dh 
 Moment of elemental area = jj . 
 
 p 
 
 Moment of the whole area = T 
 
 Ji 
 
 ft FJ 
 
 The area of the triangle =. ~- . 
 
 The distance, x, of the center of gravity from the apex is, 
 
 BIT 
 
 r _ Moment of whole area _ 3 _ 2ff 
 Whole area = ~TT!l 3 
 
 2 
 
 That is, the center of gravity is at a perpendicular distance below 
 the apex equal to two-thirds of the altitude, and must, therefore, 
 be at #, the intersection of the median, Oc, and the parallel to the 
 
 n TT 
 
 base distant -g- from the apex. From similar triangles it is seen 
 
 that Og is f the length of the median, so that the center of gravity 
 is on the median at two-thirds its length from the apex. 
 
MOMENTS. CENTER OF GRAVITY. 
 
 275 
 
 EXAMPLE II. A square is divided into four equal triangles by 
 diagonals intersecting at 0; if one triangle be removed, find the 
 center of gravity of the figure formed by the three remaining tri- 
 angles. 
 
 Let w denote the weight of each of the remaining triangles, and 
 let a denote the side of the square, Fig. 12. The weight 2w of the 
 
 side triangles may be supposed concentrated at their common center 
 of gravity, 0. The distance of the center of gravity of the lower 
 
 triangle from is ^ X 5 = o , by Example I. Then if x denotes 
 the distance of the center of gravity of the three remaining triangles 
 
 Pp + Qq 2 to X -f w X " 
 
 from 0, we shall have, x = s ^ = s 
 
 P + Q 2w + w 
 
 That is, the required center of gravity is one-ninth the side of the 
 square from the center of the square. 
 
 a 
 
 9* 
 
 EXAMPLE III. A quarter of the area of a triangle is cut off by 
 a line drawn parallel to the base. Find the center of gravity of the 
 remaining quadrilateral. 
 
 Let EF be parallel to the base 5(7, Fig. 13, and let the triangle 
 AEF be the part cut off. The required center of gravity will lie 
 in the median AD. The triangles AEF and ABC are similar, and 
 
276 THE ELEMENTS OF THE MECHANICS OF MATEKIALS 
 
 are to each other in area as 1 : 4. Since the areas of similar tri- 
 angles are to each other as the squares of homologous sides, we have : 
 
 _ _ ^2) 
 
 1 : 4 = AG 2 : AD 2 ; whence, AG^^-. Let x denote the dis- 
 
 tance of the required center of gravity of the quadrilateral BEFC 
 from A. Then, 
 
 Pp-Qq 
 ~ 
 
 4 y % X AD - AD 
 ~~ ~" 
 
 4-1 9 
 
 That is, the required center of gravity is in the median AD and at 
 seven-ninths of its length from A. 
 
 EXAMPLE IV. Find the center of gravity of a cone. 
 Conceive the cone, Fig. 14, to be made up of a great number of 
 thin sections, each parallel to the base. The center of gravity of 
 
 each of these sections will be at its center, therefore the center 
 of gravity of the cone will lie in the locus of these sections; 
 that is, it will lie in the line joining the vertex with the center of 
 gravity of the base. The elemental volume =. 7rr~dh. The moment 
 of the elemental volume = irr^Tidli. From similar triangles we have : 
 
 r : R = h : H ; whence, r = -- . By substitution, we have : 
 
 Moment of elemental volume = 
 
 
 
 7rR* r H Ml *# 2 # 2 
 
 Moment of whole volume -gr / "<dti= - . 
 Volume of cone = . 
 
MOMENTS. CENTER OF GRAVITY. 277 
 
 If x denotes the distance of the center of gravity of the cone from 
 the vertex, we shall have : 
 
 Moment of whole volume 4 3H 
 
 hole volume 
 
 3 
 
 That is, the center of gravity is at a perpendicular distance below 
 the apex equal to three-fourths of the altitude, and must, therefore, 
 be at g, the intersection of the line Oc joining the apex with the 
 
 O TT 
 
 center of gravity of the base and the parallel to the base distant -j- 
 
 from the apex. From similar triangles it is seen that Og f Oc, so 
 that the center of gravity of the cone is in the line joining the 
 vertex with the center of the base and at three-fourths its length 
 from the vertex. 
 
 EXAMPLE V. Find the center of gravity of a semicircular arc, 
 or wire. 
 
 In Fig. 15 we have, from Art. 10, Y = , in which Y = dis- 
 tance of the center of gravity from the center, L = length of the 
 arc, and C = the chord of the arc. Then, Y = Rx R =^ R = D 
 
 ~K 7T 7T* 
 
 PKOBLEMS. 
 
 1. A rod 3 feet long and weighing 4 Ibs. has a weight of 2 
 Ibs. placed at one end ; find the center of gravity of the system. 
 
 Ans. One foot from the weighted end. 
 
 2. Find the center of gravity of a uniform circular disc out of 
 which another circular disc has been cut, the latter being described 
 on a radius of the former as a diameter. 
 
 Ans. One-sixth of radius of large circle from center. 
 
278 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 3. A heavy bar 14 feet long is bent into a right angle so that the 
 lengths of the portions which meet at the angle are 8 feet and 6 feet, 
 respectively; show that the distance of the center of gravity of the 
 bar so bent from the point of the bar which was the center of gravity 
 
 9 /2 
 
 when the bar was straight, is - feet. 
 
 4. The middle points of two adjacent sides of a square are joined 
 and the triangle formed by this straight line and the edges is cut 
 off; find the center of gravity of the remainder of the square. 
 
 Ans. -^Y of diagonal of square from center of square. 
 
 5. A piece of uniform wire is bent into the shape of an isosceles 
 triangle; each of the equal sides is 5 feet long, and the other side 
 is 8 feet long ; find the center of gravity. 
 
 Ans. -f% f the altitude from the base. 
 
 6. Find the center of gravity of a figure consisting of an equi- 
 lateral triangle and a square, the base of the triangle coinciding 
 with one of the sides of the square. 
 
 Ans. At a distance from the base of the triangle equal to 
 
 J? . 1 1 O + /C\A> 
 
 of the base. 
 
 7. A table whose top is in the form of a right-angled isosceles 
 triangle, the equal sides of which are three feet in length, is sup- 
 ported by three vertical legs placed at the corners; a weight of 20 
 Ibs. is placed on the table at a point distant fifteen inches from 
 oach of the equal sides ; find the resultant pressure on each leg. 
 
 Ans. 8J, 8J, 3J Ibs. 
 
 8. ABCD is a quadrilateral figure such that the sides AB, AD, 
 and the diagonal AC are equal, and also the sides CB and CD are 
 equal ; find its center of gravity. 
 
 Ans. ??!+A* units from C, in which a = AB and I = CB. 
 ba 
 
 9. ABC represents a triangular board weighing 10 Ibs. Suppose 
 weights of 5 Ibs., 5 Ibs., and 10 Ibs. are placed at A, B, and C, respec- 
 tively. Where is the center of gravity of the whole? 
 
 Ans. At five-ninths of the median drawn from C. 
 
 10. A rod of uniform thickness is made up of equal lengths of 
 three substances, the densities of which taken in order are in the 
 
MOMENTS. CENTER OF GRAVITY. 279, 
 
 proportion of 1, 2, and 3 ; find the position of the center of gravity 
 of the rod. 
 
 Ans. At T 7 of the whole length from the end of the densest part. 
 
 11. Find the position of the center of gravity of a piece of wire 
 bent to form three-fourths of the circumference of the circle of 
 radius E. 
 
 Ans. On a line drawn from the center of the circle to a point 
 bisecting the arc, and at a distance 0.3R from the 
 center. 
 
 12. A thin wire forms an arc of a circle, the radius of which is 
 10 inches, and subtends an angle of 60. Find the distance of the 
 center of gravity from the center. Ans. 9.549 inches. 
 
 13. Find the position of the center of gravity of a balance weight 
 having the form of a circular sector of radius E, subtending an 
 angle of 90. Ans. 0.6# from center of circle. 
 
 14. Find the center of gravity of a parallelogram. 
 
 Ans. At the intersection of the diagonals. 
 
 15. Find the center of gravity of a semicircular lamina, or sheet, 
 
 9 7) 
 
 of diameter D. Ans. ^ * 
 
 O7T 
 
 16. A square stands on a horizontal plane; if equal portions be 
 removed from two opposite corners by straight lines parallel to a 
 diagonal, find the least portion which can be left so as not to topple 
 over. 4ns. Three-quarters of the area of the square. 
 
 17. Find the center of gravity of a trapezoid. 
 
 Ans. On the line joining the middle points of the bases, and 
 at a perpendicular distance from the upper base equal 
 
 to -j , and from the lower base, 
 
 B and 6 being the lower and upper bases, respectively, 
 and H the altitude. If the distance be measured on 
 the line, 8, joining the middle points of the bases, 
 
 S H 
 
 a must be substituted for -^ 
 
 18. Find the center of gravity of a pyramid. 
 
 . Ans. In the line joining the vertex with the center of gravity 
 of the base and at three-fourths its length from the 
 vertex. 
 
,280 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 19. Find the center of gravity of a frustum of a cone. 
 
 Ans. In the line joining the centers of gravity of the upper 
 and lower bases, at a distance from the upper base 
 
 , , H 3^ + 2J5r + r 2 , . ,, . 
 equal to -^ pa I p \ z~ > an d from the lower 
 
 H 3r 2 + 2fir + #* _, 
 
 base, -- jp ' r ' ^ , R and r being the radii of 
 
 the lower and upper bases, respectively, and H being 
 the altitude. These results are true for the frustum 
 of any pyramid by substituting B for R and & for r, 
 B and & being homologous sides of the lower and up- 
 per bases, respectively. 
 
 20. A lever safety-valve is required to blow off at 70 Ibs. per 
 square inch. Diameter of valve, 3 inches ; weight of valve, 3 Ibs. ; 
 short arm of lever, 2.5 inches; weight of lever, 11 Ibs.; distance of 
 center of gravity of lever from fulcrum, 15 inches. Find the dis- 
 tance at which a cast-iron ball 6 inches in diameter must be placed 
 from the fulcrum. The weight of the ball-hook = 0.6 Ib. 
 
 Ans. 35.48 in. 
 
 21. A steel safety-valve lever 1 inch thick and 50 inches long 
 tapers from 3 inches in depth at the fulcrum to 1 inch at the end. 
 It overhangs the fulcrum 4 inches, the overhang having no taper. 
 Diameter of valve, 4 inches; weight of valve, 4.75 pounds; short 
 arm of the lever, 3.5 inches. Find the distance from the fulcrum 
 at which a cast-iron ball 9.5 inches in diameter must be placed in 
 order that steam shall blow off at 125 Ibs. pressure per square inch. 
 Weight of the ball-hook, 1.3 Ibs.; weight of a cubic inch of steel, 
 0.28 Ib. ; weight of a cubic inch of cast iron, 0.26 Ib. 
 
 Ans. 42.35 inches. 
 
 22. A trapezoidal wall has a vertical back and a sloping front 
 face; width of base, 10 ft. ; width of top, 7 ft. ; height, 30 ft. What 
 horizontal force must be applied at a point 20 ft. from the top in 
 order to overturn it, i. e., to make it pivot about the toe ? Width 
 of wall, 1 ft.; weight of masonry in wall, 130 Ibs. per cubic foot. 
 
 Ans. 18,900 Ibs. 
 
 23. Find the height of the center of gravity from the base of a 
 column 4 feet square and 40 feet high, resting on a tapered base 
 forming a frustum of a square-based pyramid 10 feet high and 
 8 feet square at the base. Ans. 20.4 feet from base. 
 
MOMENTS. CENTER OF GRAVITY. 281 
 
 Problems 24, 25, 26, and 27 are to be solved graphically. 
 
 24. Find the position of the center of gravity of an unequally 
 flanged beam-section; top flange, 3 inches wide, 1.5 inches thick; 
 bottom flange, 15 inches wide, 1.75 inches thick; web, 1.5 inches 
 thick; total height, 18 inches. 
 
 Ans. 5.72 inches from the bottom edge. 
 
 25. Find the height of the center of gravity of a T section from 
 the foot, the top cross-piece being 12 inches wide and 4 inches 
 deep ; the stem, 3 feet deep and 3 inches wide. 
 
 Ans. 24.16 inches. (Check the result by seeing if the mo- 
 ments are equal about a line passing through the sec- 
 tion at the height found. ) 
 
 26. A square board weighs 4 Ibs., and a weight of 2 Ibs. is placed 
 at one of its corners. Find the position of the center of gravity of 
 the board and weight. 
 
 Ans. On the diagonal drawn from the weighted corner and 
 at two-thirds its length from the opposite corner. 
 
 27. ABCD is a flat plate, right-angled at 5, and having the fol- 
 lowing dimensions : AB = 15 inches; AD = 12.5 inches, EG = 18 
 inches, and the perpendicular from D on EG measures 18.5 inches. 
 Construct the figure to a scale of 1.5 inches to the foot, and locate 
 the center of gravity of the plate, measuring its distance from the 
 side AB. Ans. 7.895 inches. 
 
CHAPTER II. 
 
 BENDING MOMENT. 
 DIAGRAM. 
 
 SHEAR. BENDING-MOMENT 
 SHEAR DIAGRAM. 
 
 13. Bending Moments. When forces act on a body in such a 
 manner as to tend to give it a spin or a rotation about an axis with- 
 out any tendency to shift its center of gravity, the body is said to 
 be acted on by a couple. A couple consists of two parallel forces 
 
 of equal magnitude acting in opposite directions, but not in the 
 same straight line. 
 
 A beam is subjected to a bending moment when it is so acted upon 
 at its ends by equal and opposite couples that there is a tendency 
 to turn it in opposite directions. 
 
 Thus the beam ran, Fig. 16, is acted on by the equal and oppo- 
 site couples, E and R, and TF and IF, the tendency being to turn 
 
 * 
 
 1 
 
 the beam in opposite directions about the point r; that is, to bend 
 it at r. In Fig. 17 the couples whose moments are R^ X mr and 
 R 2 X nr have the same effect on the beam as those of Fig. 16. The 
 beam of Fig. 16 is called a cantilever, from the nature of its sup- 
 port, while the beam of Fig. 17 is called a simple beam. 
 
 14. General Case of Bending Moments. The bending moment at 
 any section of a beam is the algebraic sum of all the moments of 
 
BENDING MOMENT. SHEAR. 283 
 
 the external forces acting to the left of the section. It is equal to 
 the moment of the reaction at the left support minus the sum of 
 the moments of the loads to the left of the section considered. It is 
 thus assumed that forces acting upward are positive, and those 
 acting downward are negative, so that bending moments may be posi- 
 tive or negative, depending upon whether the moment of the left 
 reaction is greater or less than the sum of the moments of the loads 
 to the left of the section. 
 
 15. In graphical constructions the signs of bending moments are 
 cf the first importance and are determined by the following rule : 
 
 Bending moments which tend to bend a beam or cantilever con- 
 cave upward, , are regarded as positive, and when they tend 
 
 to bend in the reverse way, , % , they are negative. 
 
 16. It should be observed that it is merely to avoid confusion in 
 the construction of diagrams that the external forces to the left of 
 the section were considered when defining the bending moment at 
 any section of a beam. Moments may equally well be taken to the 
 right of the section and the same value be obtained for the bending 
 moment at the section. When bending moments are obtained by 
 calculation, rather than by construction, the side involving the least 
 calculation in taking moments should be selected, though the cal- 
 culations for both sides afford a positive check as to the accuracy of 
 the work. 
 
 17. The beam, Fig. 18, is supposed to be without weight. The 
 bending moment at the section E is, taking moments to the left of E, 
 
 M^R^XaE WiXqE WzX pE, (1) 
 
 or, taking moments to the right of E, we have : 
 
 M = R 2 XbE W 3 XrE. (2) 
 
 Suppqse the distances and weights to be as shown in the figure. 
 Before finding the bending moments we must first find the reactions, 
 R! and R 2 . 
 
 Taking moments about a, we have : 
 
 R 2 X 10.5 = 80 X 8.5 + 50 X 5 + 60 X 2, whence 
 
 R 2 = 100 Ibs. 
 Taking moments about &, we have : 
 
 # X 10.5 = 60 X 8.5 + 50 X 5.5 + 80 X 2, whence 
 
 R = 90 Ibs. 
 which might have been expected, since R l -f- R should equal 
 
284 THE ELEMENTS OF THE MECHANICS or MATERIALS 
 
 By substitution in equation (1), we have for the bending mo- 
 ment at E, 
 
 M, = 90 X 6.5 60 X 4.5 50 X 1.5 =. 240 pounds-feet. 
 Substituting in equation (2), we have: 
 
 M 100 X 4 80 X 2 = 240 pounds-feet. 
 
 It is thus seen that the bending moment at a section is the same, 
 whether the moments be taken to the left or to the right. In this 
 instance the taking of moments to the right involved the least cal- 
 culation. 
 
 18. Bending-moment Diagrams are made to show graphically the 
 bending moment at any section of a beam. 
 
 For example, take the beam of Fig. 18. The bending moment, 
 Jf, at q is M R^ X 2 = 90 X 2 = 180 Ibs.-ft. 
 
 At p, M R^ X 5 F t X 3 = 90 X 5 60 X 3 = 450 180 
 = 270 Ibs.-ft. 
 
 At r, M = R^X 8.5 W , X 6.5 W 2 X 3.5 
 
 = 90 X 8.5 60 X 6.5 50 X 3.5 200 Ibs.-ft. 
 
 If, on a base line a'b', Fig. 19, and to a scale of 1 inch = 200 
 pounds-feet, we erect ordinates to represent these bending moments, 
 and then join their extremities with the broken line a'q'p'r'V, the 
 
BENDING MOMENT. SHEAR. 
 
 285 
 
 inclosed figure is a diagram whose ordinate beneath any section of 
 the beam will measure the bending moment at the section to the 
 scale adopted. Thus the ordinate for the bending moment at q 
 
 -i o/-v 2*70 
 
 measures <^ = 0.9 inch; that at p, ^^ = 1.35 inch; that at r, 
 
 200 
 200 
 
 = 1 inch. The ordinate beneath E measures 1.2 inch, which, 
 
 to scale, represents a bending moment of 1.2 X 200 = 240 pounds- 
 feet, as already found by calculation. 
 
 19. Shear. Shearing stresses exist when couples, acting like a 
 pair of shears, tend to cut a body between them. The application 
 of couples to beams in the manner already described subject the 
 beams to shearing actions as well as to bending moments. With 
 beams of lengths ordinarily encountered, the bending moment is far 
 
 more important than the shear, but with very short simple beams 
 and very short cantilevers, failure will invariably result from shear. 
 
 20. Shear Diagrams are made to show the amount of the shear at 
 any section of a beam, and in their construction attention must be 
 paid to the signs, so that in cases where the shear is partly positive 
 and partly negative the positive part may be placed above the shear 
 axis, or base line, and the negative part below. 
 
 21. General Case of Shear. The shear at any section of a beam 
 or of a cantilever is equal to the left reaction minus the sum of the 
 loads between the reaction and the section. In other words, it is 
 the algebraic sum of the forces to the left of the section, the upward 
 forces being regarded as positive and the downward negative. 
 
 22. If the short cantilever, Fig. 20, be loaded until it fails, there 
 will be a slight bending at the outer end, but the failure will be 
 due to the outer part shearing off bodily from the built-in part, as 
 
286 
 
 THE ELEMENTS OF THE MECHANICS OF MATEBIALS 
 
 indicated by the dotted lines. As there is no reaction at the left 
 end of the beam, the shear at all sections is constant and equal to 
 - W. The sign of the shear may be determined by a consideration 
 of the sliding tendency of the shear whether clockwise or contra- 
 clockwise. In Fig. 20, the sliding tendency is contra-clockwise, and 
 therefore negative. The ordinates of the shaded shear diagram are 
 of constant length and equal to W, and the whole of the dia- 
 gram, being negative, is below the shear axis mn. 
 
 23. In the case of the simple beam, Fig. 21, a failure from shear 
 will occasion the part between the supports to slide down, as shown 
 by the dotted lines. The shear at any section to the left of the 
 load, W, is R 1} and at any section between W and the right support 
 
 it is R 1 W = R 2 , since R 1 + R 2 W. The clockwise ten- 
 dency of the slide of the shear at the left support, and the contra- 
 clockwise tendency at the right support determine the signs of the 
 shears as shown by the arrows. 
 
 The shaded part of Fig. 21 represents the shear, and shows that it 
 changed sign under the load, the positive part being placed above 
 the shear axis, mn, and the negative part below. 
 
 24. As in the case when denning bending moments, only the 
 external forces to the left of a section were considered when defining 
 the shear at the section, but this, as in the case of bending moments, 
 was only for convenience. If W 8 denotes the sum of all the loads 
 between the left support and a given section, and W 8 ' the sum of 
 all the loads between the section and the right support, then evi- 
 dently R! + R 2 W 8 + W 9 ', whence 7^ W 8 (R 2 W s '). 
 The first member of this equation is the algebraic sum of the forces 
 
BENDING MOMENT. SHEAR. 28? 
 
 to the left of the given section and is the shear at the section, and 
 the second member is the negative of the algebraic sum of the forces 
 to the right of the section. The shear at a given section is then 
 the algebraic sum of the external forces to the right or to the left of 
 the section, but with contrary signs. 
 
 25. Relations between Bending-moment and Shear Diagrams. 
 The depth of the bending-moment diagram beneath any section 
 measures, as already shown, the bending moment at the section to a 
 given scale, and is equal to the algebraic sum of the moments of the 
 external forces to the left of the section. 
 
 The depth of the shear diagram beneath any section shows, to a 
 given scale, the shear at the section, and is equal to the algebraic 
 sum of the external forces to the left of the section. 
 
 Let the beam, Fig. 22, be loaded with W 19 W 2 , W 3 , at distances 
 d 19 d 2 , d 3 , respectively, from any given section, a. Denote the sup- 
 port reactions by R^ and R 2 , and their distances from a by r and r 2 , 
 respectively. 
 
 The shear at the extreme right end is, by our definition, 
 R i Wi W 2 W Q R 2 . On a base line, mn, Fig. 23, con- 
 struct separately the component parts of this shear, giving to .each 
 component part its algebraic sign. Thus the rectangle mq repre- 
 sents the shear throughout the beam due to R^ ; in like manner the 
 rectangles qs, tu, and uv, represent, respectively, the shears due to 
 TFj, W 2 , and W 3 , and the whole shaded figure is the shear diagram 
 of the beam, the algebraic sum of the shears beneath any section of 
 the beam measuring the shear at the section to the scale of the 
 figure. 
 
 The area of this shear diagram to the left of the section a is equal 
 to R^ W^. (1) 
 
 The area to the right of section a is equal to 
 Rir 2 -Ws 2 -W 2 (r 2 -d 2 )-W s (r 2 -d 3 ) 
 = R r 2 TP r F,r + W d 2 W s r 2 + W 3 d 3 
 = r 2 (R, W~ W 2 W 3 ) + W 2 d 2 + W 3 d 3 
 =R,rt + W& + W 9 d 9 
 = -(R 2 r 2 -W 2 d 2 -^W 3 d 5 ).' (2) 
 
 The second member of equation (1) is, by our definition, the 
 bending moment to the left of section a, and the second member of 
 equation (2) is the bending moment to the right of section a, and 
 since the bending moment at any section is the same, whether taken 
 
288 THE ELEMENTS OP THE MECHANICS OF MATEEIALS 
 
 to the right or to the .left of the section, it follows that the area of 
 the shear diagram to the right or to the left of a section is equal to 
 the bending moment at the section, having regard always to the 
 algebraic signs. 
 
 26. Figure 23 is an inconvenient form of the shear diagram. If 
 the algebraic sum of the shears be taken at every section of the beam, 
 and all those that are plus be plotted above the shear axis mn, and 
 those that are minus below, the ordinate of the resulting diagram 
 beneath any section will be the measure of the shear. The easiest 
 way of finding these algebraic sums is to revolve the negative part 
 of Fig. 23 about pq, the upper boundary of the positive part, so 
 that it assumes the position indicated by the dotted lines. By this 
 process of superposition the major part of the positive shear is neu- 
 tralized, and the resulting diagram is that shown in Fig. 24. Fig. 
 23 is a cumbersome way of obtaining the shear diagram, and was 
 constructed only to demonstrate the relations between the shear and 
 bending-moment diagrams. The proper procedure would be to 
 derive Fig. 24 at once from Fig. 22. Thus, R lf W 19 W 2 , and W 3 
 having been determined to scale, and the shear axis, or base line, 
 m'n', selected, we would proceed as follows : 
 
 The shear between the left support and TF is equal to R 19 which 
 gives the point &. Immediately the point of 'application of W^ is 
 passed the shear becomes R^ W 19 giving the point c, and this 
 shear continues up to the point of application of W 2 , but imme- 
 diately this point is passed the shear becomes R i W t W 2 , giv- 
 ing the point d. This shear continues until the point of application 
 of W s is passed when the shear becomes R^ TF W 2 W 3 
 = R 2 , giving the point e. 
 
 27. Figures 22, 23, and 24 were constructed to the following 
 scales : 
 
 Length, 1 inch = 4 feet; load, 1 inch = 150 IDS.; so that 1 sq. 
 inch = 150 X 4 = 600 Ibs.-ft. 
 
 The data of the beam and loads are as follows : 
 
 Length of beam = 12 ft. = - 1 / = 3 inches to scale; W^ = 60 
 
 Ibs. = = 0.4 inch to scale; W 2 = 45 Ibs. = = 0.3 inch; 
 
 QO 
 
 W = 90 Ibs. = ^ = 0.6 inch; ^ = 2 f t. = f = -J inch; d 2 = 3 
 ft. = } = 0.75 inch; d z = 6 f t. = = 1.5 inch; r = 5 ft = J 
 = 1.25 inch; r, = 7 ft. = J = 1.75 inch. 
 
r 
 
 cc, 
 
 m$ 
 
 
 i T 
 
 
 
 1 
 
 
 
 
 
 
 " w A* 
 
 -f 
 
 
 
 
 
 
 
 
 B^* 
 
 \ 
 
 
 
 
 
 z 
 
 
 ! I 
 
 1 
 
 H 
 
 f 
 
 Ml 
 
 
 
 
 
 ^ Fig-Z-^ 
 
 . 
 
 1 
 
 ill 
 
 
 
 c 
 
 -> 
 
 b, , 
 
 
 17? 
 
 
 
 
 
 
 Tjl- 1 ' 
 
 jf 
 
 ^ II 
 
 II 
 
 
 
 
 
 
 
 d, T 
 
 
 
 
 
 
 
 
 pi * 
 
 
 
 
 
 
 
 
 Fiq. Z5^ 6 
 
 ii in 
 
 
 1 
 
290 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 The support reactions were first found as follows: 
 Taking moments about the left support, we have : 
 
 *.(-i + ) = IF. ('i + *,) + ^,('1 +^) + ^(r, - dj, or 
 = 90 X 11 + 45 X 8 + 60 X 3, whence B 2 = 127.5 Ibs. 
 
 K 
 
 = 0.85 inch. Therefore, R = 60 + 45 + 90 127.5 
 = 67.5 Ibs. = ~ 0.45 inch. 
 
 . 10U 
 
 28. The shear diagram, Fig. 24, not only gives by its ordinate 
 under any section the shear at that section, but it also gives the bend- 
 ing moment at any section by taking the algebraic sum of the areas 
 to the right or to the left of the section. It follows from this that 
 the positive area of the diagram, or the area above the base line, is 
 numerically equal to the negative area, or the area below the base 
 line. 
 
 The algebraic sum of the areas to the right or to the left of sec- 
 tion a is found by measurement to be 0.3625 sq. inch; the bending 
 moment at that section is, therefore, 0.3625 X 600 = 217.5 Ibs.-ft. 
 
 29. The bending-moment diagram, Fig. 25, was constructed on 
 the base line m"n" to a scale such that 1 inch in depth = 300 Ibs.-ft. 
 The bending moments at W ly W 2 , W 3 were calculated as follows: 
 
 M Wi = RI(TI dj = 67.5 X 3 = 202.5 Ibs.-ft. = ^5 _ .675 
 
 inch to scale. 
 M 9 = R^fa + d 2 ) - - W 1 X d 2 = 67.5 X 8 60 X 5 = 240 
 
 240 
 pounds-feet = O?TQ = 0.8 inch. 
 
 *1 = ,(*! + d 3 ) - T^K + d t ) - W,(d s - d.) 
 
 1 27 5 
 
 = 67.5 X 11 60 X 8 45 X 3 = 127.5 Ibs.-ft. = 
 
 oUU 
 
 = 0.425 in. 
 
 Ordinates equal in height to the scale measurements representing 
 these bending moments were erected on m"n", as shown, and their 
 extremities joined by the broken line m'Jghlcn". The ordinate of 
 the resulting diagram beneath any section of the beam is a scale 
 measurement of the bending moment at the section. Thus the or- 
 dinate under the section a measures |$ of an inch, corresponding to 
 a bending moment of |- X 300 = 217.5 Ibs.-ft., as already found 
 from the shear diagram. 
 
BENDING MOMENT. SHEAR. 
 
 291 
 
 30. Figure 26 represents a simple beam supported at the ends and 
 uniformly loaded with w pounds per unit of length. The total load 
 is wL Ibs., and it is evident that the support reactions will each 
 
 ^um 
 
 o '' 
 
 wL 
 
 equal -^- Ibs. To find the bending moment at any section, a, dis- 
 tant x from the left support, we may assume the uniform load to be 
 made up of a number of parallel forces each equal to w. There are 
 wx of these forces between section a and the left support, and we 
 
292 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 may substitute for them their resultant, wx, acting at their center 
 of gravity, which is midway between the section and the support. 
 The bending moment, M a , at the section is then 
 
 M- T> .. x wLx wx* wx , T -. wxx' 
 
 M a = R,x wx X = -- - -- = - (L x) = 
 
 That is, the bending moment at the section is proportional to the 
 product of the segments into which the section divides the beam, 
 and the bending-moment diagram is, therefore, a parabola, as shown 
 in Fig. 27, having its axis vertical and under the middle of the beam. 
 For it is a property of the parabola that, if a diameter be drawn to 
 intersect a chord, then the product of the segments of the chord is 
 proportional to the length of that part of the diameter included 
 between its vertex and its point of intersection with the chord. 
 If, in the general expression for the bending moment, 
 
 a* wLx WX* i , L ,, ., 
 
 M a = -- , we let x = ^ , we shall have for the maximum 
 
 bending moment, M ma x, 
 
 _w!S_tj>IS = wL* = WL 
 
 Mmax - 4 y g y , 
 
 in which W = wL = the whole weight. 
 
 31. The shear diagram, Fig. 28, was constructed as follows: 
 The shear at the extreme right end is R 1 wL =. R 2 , or 
 
 ~2 -- wL = --. Constructing the component parts of this 
 shear we get the rectangle mg as the shear throughout the beam due 
 to . Conceiving the uniformly distributed load to be made up 
 
 of L units of w Ibs. each, the shears of these units will be repre- 
 sented by the horizontal lines of the triangular part of the diagram, 
 and are negative. Eevolving the negative part of the diagram about 
 rg, the upper boundary of the positive part, until it assumes the 
 position indicated by the dotted lines, we get the resulting shear 
 diagram of Fig. 29. 
 
 32. Figure 29 may readily be obtained at once from Fig. 26. 
 Thus the shear at any section distant from the left support is 
 
 Y = w ---- wx, Y denoting the shear. This is the equation of a 
 
 straight line, the origin being at the left support. It is seen that 
 the shear decreases as x increases, and has its maximum value. 
 
 ^5-, when x 0. When x = ~ the value of Y becomes 0, and 
 
BENDING MOMENT. SHEAR. 293 
 
 when x is greater than -~ the value of F becomes negative, and 
 when x = L the value of F becomes -^- . r's" is then the 
 straight line whose equation is F = -~ wx. 
 
 33. Bending-moment and Shear Diagrams of Cantilevers. 
 
 Figure 30 represents a cantilever with a concentrated load, W, at the 
 end, a concentrated load, TP , at an intermediate point between the 
 end and the support, and a uniformly distributed load of w pounds 
 per foot over a portion of its length. The construction of the bend- 
 ing-moment and shear diagrams of this beam will serve to illustrate 
 the three cases of a cantilever: (a) Loaded at the end. (&) A con- 
 centrated load at some point between the end and the support, (c) 
 Loaded uniformly with w pounds per lineal foot. 
 
 The cantilever of Fig. 30 has been drawn so that there is no reac- 
 tion at the left end, as can be done in all cases. From our defini- 
 tion of bending moment it will be seen then that the bending 
 moment is zero at the free end and is a maximum at the wall. The 
 tendency being to bend the beam concave downward all the bending 
 moments are negative. 
 
 The bending moment at any point, x, between W and W is 
 M = Wx. At any point, x lf between W^ and the point of com- 
 mencement of the uniform load of w Ibs. per foot is M = Wx 
 ^1(^1 a )> a being the distance between W and W . At any 
 point, ar 2 , the bending moment is 
 
 M= Wx 2 Wi(x t a) w \ Xt (a -j- 
 
 the factor X<L ~ ^? - being the arm of the center of gravity of 
 
 that portion of the beam that is uniformly loaded included by the 
 abscissa x z . The maximum bending moment at the wall is 
 
 M max = WL Wsi wr 2 X \ = WL W,r, ^ . 
 
 34. In the construction of the bending-moment diagram, Fig. 31, 
 the following data were used : 
 Lineal scale, 0.2 inch = 1 foot. 
 Load scale, 1 inch in depth = 800 Ibs. ft. 
 
294 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 L = 10 ft. = 0.2 X 10 2 inches to scale; r x = 7 ft. = 0.2 X 7 
 = 1.4 inches; r 2 = 4 ft. = 0.2 X 4 = 0.8 inch; F = 32 Ibs. ; 
 F z= 44 Ibs. ; w = 48 Ibs. per foot run. Then, the bending mo- 
 ment at the wall is : 
 
 M max = - - 32 X 10 + 44 X ? + -_ -= _ 1012 Ibs.-ft. 
 
 inches to scale, = ordinate of bending-mo- 
 ment diagram at the wall. 
 
 M Wi = W(L rj = 32 X 3 = 96 lbs.-ft. = ^ 
 
 = 0.12 inch. 
 
 M E = W(L r 2 ) - - Fifo -- r 2 ) = -- 32 x 6 44 X 3 
 = 324 lbs.-ft. = ??* = 0.405 inch. 
 
 oUU 
 
 Erecting these three ordinates the bending-moment diagram is com- 
 pleted, the curve under the part of the beam that is uniformly loaded 
 being parabolic. 
 
 The ordinates corresponding to the bending moments at W j. and 
 at E might have been found graphically as follows : 
 
 Divide the maximum bending moment (represented by the or- 
 dinate at the fixed end of the beam) into its component parts, WL, 
 
 F^, and ^p, equal respectively to 320, 308, and 384 lbs.-ft., and 
 
 equal, when reduced to scale, to 0.4 inch, 0.385 inch, and 0.48 inch, 
 respectively. The ordinates for the bending moments at F T and at 
 E are then found as shown by the dotted line of the figure. 
 
 35. In the construction of the shear diagram, Fig. 32, the lineal 
 scale of the bending-moment diagram, J inch = 1 foot, was adopted 
 but the load scale was taken as 1 inch = 160 Ibs., so that 1 sq. in. 
 of area of the shear diagram 160 X 4 = 640 lbs.-ft. 
 
 Since the sliding tendency is contra-clockwise all the shear is 
 negative, and is a maximum at the wall. Since there is no reaction 
 at the left end the maximum shear at the wall = W F wr 2 
 
 = 32 44 48 X 4 = 268 Ibs. = ^ = 1.675 inch 
 
 to scale. The component parts of this shear are the shears due to 
 F, W 19 and wr 2 , viz., 32 Ibs., 44 Ibs., and 192 Ibs., equal, respectively, 
 to 0.2 inch, 0.275 inch, and 1.2 inch, when reduced to scale. Con- 
 struct these component shears as shown in the figure, and the shear 
 
BENDING MOMENT. SHEAR. 
 
 295 
 
 diagram is complete. The are^ of this diagram to the left of any 
 section is equal to the bending moment at the section. Thus, the 
 area to the left of E is found by measurement to be 0.50625 sq. inch, 
 which, when reduced to scale, equals 640 X 0.50625 = 324 lbs.-ft, 
 the same as was found above by calculation. 
 
 36. The beam of Fig. 33 is supported at the ends and has a cen- 
 tral load W. 
 
 W 
 
 The reaction at each support is -^ . The bending moment at any 
 
 section, x, between the left support and the middle is M x = 9 
 
296 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 and increases directly as x increases. At the middle x = ^ , and 
 M = -. At any section, x , between the middle and the right 
 
 8 upport, M Xl = - W -=- 
 
 = 
 
 and decreases as x increases, becoming when x t =. L. The maxi- 
 mum bending moment is, therefore, at the middle and is equal to 
 
 . Since M x and M X1 = (L ar ) are straight line equa- 
 
 tions, the bending-moment diagram has the triangular form shown 
 in Fig. 34. 
 
 W 
 
 37. The shear equals - until directly after passing the middle 
 
 W W 
 
 section, when it becomes -- W = -- g- . The shear, therefore, 
 
 changes sign at the middle. The shear diagram is then as shown in 
 Fig. 35. 
 
BENDING MOMENT. SHEAR. 
 
 297 
 
 38. The beam of Fig. 36 is supported at the ends and has two 
 equal and symmetrically placed loads. The bending-moment dia- 
 gram, Fig. 37, and the shear diagram, Fig. 38, are left as studies 
 for the student. 
 
 1 
 
 
 
 
 1 
 
 
 
 tr 
 
 
 
 
 jShear 
 
 
 k 
 
 \ 
 
 
 
 
 tt<? 38 
 
 
 7T 
 
 39. The beam of Fig. 39 is uniformly loaded with w pounds per 
 unit of length, and overhangs the supports a distance a at each 
 end. Consider quite independently of each other the two over- 
 hangs as cantilevers and the part between the supports as a simple 
 beam. 
 
 Since the overhangs tend to bend the beam concave downward, 
 the bending moments due to them are negative. The bending mo- 
 ment of each overhang cantilever at the support will be wa X s 
 
 -c 
 
 wa 2 
 
 -<rj- , and, as we have seen, the parabolic curves joining the 
 
 extremities of the beam and the ordinates at the supports represent 
 the bending-moment diagrams of the overhangs. Disregarding the 
 
 load on the central span of the beam, the bending moment, ^ / - 
 
 <> 
 
 will be constant between the supports. Therefore, the negative part 
 
THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 below the base line of Fig. 40 represents the bending moments of 
 the whole beam due to the overhangs. 
 
 The uniformly loaded central span of length I occasions a reac- 
 tion at each of the supports of *|- , and as the tendency is to bend 
 the beam concave upward the bending moments are positive. The 
 
BENDING MOMENT. SHEAR. 299 
 
 bending moment at any section between the supports, distant x 
 
 -, f wbx wx z wx /7 N , . , 
 irom the leit support, is M - -~ - : -^- (o x) } which, 
 
 as we have seen, is the equation of a parabola having its axis ver- 
 tical under the middle of the beam. When x = ~ we have, for the 
 
 maximum bending moment, M max = -g- . The parabola above the 
 
 base line of Fig. 40 is the bending-moment diagram of the part of 
 the beam between the supports, and is positive. By superposition 
 the resultant bending-moment diagram of Fig. 41 is obtained. 
 
 The shear of the overhang at the left end is negative, because of 
 its contra-clockwise tendency, and at the support is equal to wa. 
 The shear of the overhang at the right end is positive, because of 
 its clockwise sliding tendency, and is equal to wa at the support. 
 The minus and plus triangles at the ends of Fig. 42 represent the 
 shears of the left and right overhangs, respectively. The shear at 
 
 the left support due to the load between the supports is -*-, and it 
 decreases an amount equal to w for each unit of length toward the 
 right support. At the middle it is, therefore,^- w *'= 0, and 
 
 at the right support it is -^ bw 5-. The shear diagram 
 
 of Fig. 42 is thus obtained. 
 
 EXAMPLE I. Figure 43 represents a beam overhanging both sup- 
 ports, and carrying a uniformly distributed load of 20 Ibs. per foot 
 and two concentrated loads as shown. Find the support reactions, 
 the maximum positive and negative bending moments, and con- 
 struct the bending-moment and shear diagrams. 
 
 Solution: Take moments about the supports to obtain the reac- 
 tions ; thus, 
 lSR t + 20 X 6 X 3 = 20 X 26 X 13 + 200 X 22 + 400 X 6. 
 
 # = 7331 Ibs. 
 18R 2 -[- 20 X 8 X 4 + 200 X 4 = 20 X 24 X 12 + 400 X 12. 
 
 R 2 506| Ibs. 
 
 In complicated problems like this it is the best practice to calculate 
 the bending moments at various sections of the beam, using the 
 results as ordinates to construct the diagram. 
 
 For the bending-moment diagram let the linear scale be 0.1 
 inch = 1 foot, and the load scale, 1 inch = 1000 Ibs.-ft. 
 
L y 
 
BENDING MOMENT. SHEAR. 301 
 
 The bending moments of the overhangs are negative because of 
 the tendency to bend the beam concave downward. 
 The bending moment at the section under W^ is : 
 
 M wi = -M>x4x2=-20X8=-160 Ibs.-f t. = - 0.16 inch to scale. 
 Jfjti =-wX8x4-WiX4=-640-800=- 1440 Ibs.-ft. = - 1.44 inch. 
 M 3 = R! X 3 - w X 11 X V - JPi X 7 = 2200 - 1210 - 1400 = - 4101bs.-ft. = - 0.41 inch. 
 Jf 4 = JZj x 4 - 10 X 12 X 6 - Wi X 8 = 2933^ - 1440 - 1600 = - 106% Ibs.-f t. = - 0.107 inch. 
 Jf 5 = ^ X 5 - w X 13 X - W l X 9 = 3666% - 1690 - 1800 = 176% Ibs.-ft. = 0.177 inch. 
 M 9 = Ri X 9 - w X 17 X - W l X 13 = 6600 - 2890 - 2600 = 1110 Ibs.-f t. = 1.11 inch. 
 Mwo = Si X 12 - w X 20 X 10 - W l X 16 = 8800 - 4000 - 3200 = 1600 Ibs.-f t. = 1.6 inch. 
 -Ku = -Bi X 15 - w X 23 X - Wi X 19 - TF 2 X 3 = 710 Ibs.-f t. = 0.71 inch. 
 MB* = RI X 18 - w X 26 X 13 - Wi X 22 - W s X 6 = - 360 Ibs.-f t. = - 0.36 inch. 
 
 Using these scale results as the lengths of ordinates, the bending- 
 moment diagram of Fig. 44 was constructed. The maximum posi- 
 tive bending moment, y, is under W 2 and is equal to 1600 Ibs.-ft. 
 The maximum negative bending moment, y', is at the left support, 
 and is equal to 1440 Ibs.-feet. The bending moment which is nu- 
 merically the greatest, whether positive or negative, is the one to be 
 considered in the design of the beam. It will be observed that the 
 bending moment changes sign at a section about 4^ feet to the right 
 of the left support, and again at a point about 1 foot to the left of 
 the right support. It will be shown later on, when treating of the 
 deflection of beams, how some bending moments between the sup- 
 ports are negative. 
 
 For the shear diagram the linear scale, 0.1 inch = 1 foot, of the 
 bending-moment diagram will be used, but the load scale will be 
 taken as 1 inch = 800 Ibs., so that 1 sq. inch of diagram area will 
 equal 800 X 10 = 8000 Ibs.-ft. 
 
 The uniform load of the left overhang = J^ = t ^1|? = 0.2 
 
 oUU 
 
 inch to scale. 
 
 The other forces affecting the shear diagram are : 
 
 _ oqi . , . TT 2 _400 
 
 800~800~ '' 800~~800- inch ' 800 ~ 800 = '~ 
 
 _6X20 5i nch .-gi-^i 
 
 800 ~ "~800~ '800"~800~ 800 
 
 = 0.916 - 0.25 0.2 = 0.467 inch ; A = = 0.64 inch. 
 
 oUU oUU 
 
 The shear of the left overhang is negative owing to its contra- 
 
 clockwise tendency, and that of the right overhang is positive from 
 
 its clockwise tendency. The shear of the left overhang is W^ 
 
 8w, and that of the right overhang is 6w. Construct these shears 
 
 as shown in Fig. 45. The shear of the beam between the supports 
 
 OF THE 
 
 UNIVERSITY 
 
302 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 is R! W 2 ISw. R! is made up of two component parts that 
 due to the left overhang and that due to the load between the sup- 
 ports. The resultant positive shear at the left support is, there- 
 fore, R! TF Sw 0.916 0.25 0.2 = 0.467 inch. The 
 shear at the right support, due to the loads between the supports, is 
 (#! Tf Sw) -- W 2 ISw = 0.467 inch 0.5 inch 0.45 
 inch. Construct the component parts of this shear as shown by 
 the shaded portion above the base line of Fig. 45. Revolve the 
 negative part of the shear about the upper boundary of the posi- 
 tive part so that it assumes the position indicated by the dotted 
 lines of Fig. 45. By this virtual superposition of the negative and 
 positive parts the final shear diagram, Fig. 46, is obtained. 
 
 40. Checks as to the accuracy of the shear diagram, Fig. 46, exist 
 in the facts that : ( 1 ) The sum of the positive areas must equal the 
 sum of the negative areas. (2) The algebraic sum of the areas to 
 the right of any section must equal the algebraic sum of the areas 
 to the left of the section. (3) The algebraic sum of the areas to 
 the right or to the left of any section must equal the bending 
 moment at the section to the scale of the diagram. Thus the alge- 
 braic sum of the areas of the shear diagram, Fig. 46, to the right 
 or to the left of the section under W 2 is found by measurement to 
 be 0.2 square inch. The bending moment at that section is, there- 
 fore, 8000 X 0.2 = 1600 lbs.-ft., as was found above by calcula- 
 tion. 
 
 PROBLEMS. 
 
 28. A uniform beam 25 ft. in length, whose weight is disre- 
 garded, is supported at the ends and has a concentrated load of 
 400 Ibs. at 9 ft. from the left support, and one of 500 Ibs. at 18 ft. 
 from the left support. Construct the bending-moment and shear 
 diagrams, and find the maximum bending moment in lbs.-ft. 
 
 Ans. 3564 lbs.-ft. 
 
 29. Construct the bending-moment and shear diagrams of the 
 beam of Example 1, the weight of the beam, 500 Ibs., being con- 
 sidered. Find also the maximum bending moment. 
 
 Ans. 5112.5 lbs.-ft. 
 
 30. A beam 12 ft. long is supported at the ends and loaded with 
 a weight of 3 tons at a point 2 ft. from one end. Find the bending 
 moment at the middle of the beam, and also the shearing force. 
 
 Ans. 3 tons-feet; 0.5 ton. 
 
BENDING MOMENT. SHEAR. 303 
 
 31. In a uniform beam of length L, supported at the ends, and 
 loaded at the center with a load W, show that the bending moment 
 
 WL, 
 
 is greatest at the middle of the beam and equal to ^-. Then 
 
 determine graphically the bending moment and shearing force at 
 the section 6 feet from either support of a beam of 25 feet span 
 and loaded with 5 tons at the middle. 
 
 A ns. 15 tons-ft. ; f tons. 
 
 32. A cantilever projects 10 feet from a wall and carries a uni- 
 form load of 60 Ibs. per foot; it also supports three concentrated 
 loads of 100, 300, and 500 pounds at distances from the wall of 2, 
 5, and 9 feet, respectively. Find the maximum bending moment, 
 and the maximum shear. Construct the bending-moment and 
 shear diagrams. Ans. 9200 Ibs. ft. ; 1500 Ibs. 
 
 33. A beam overhangs both supports equally, carries a uniform 
 load of 80 Ibs. per foot, and has a load of 1000 Ibs. in the middle, 
 the length of the beam being 15 feet, and the distance between the 
 supports 8 feet. Construct the bending-moment and shear dia- 
 grams, and find the maximum bending moment and the maximum 
 shear. Ans. 2150 lbs.-ft; 820 Ibs. 
 
CHAPTER III. 
 MOMENT OF INERTIA. RADIUS OF GYRATION. 
 
 41. Moment of Inertia. The moment of inertia of a surface is 
 the sum of the products of each elemental area of the surface by 
 the square of its distance from an axis about which the surface is 
 supposed to be revolving. If, instead of a surface, we have a body, 
 then we must substitute the elemental volume for the elemental 
 area in finding the moment of inertia. The moment of inertia 
 varies according to the position of the axis, being smallest when 
 the axis passes through the center of gravity. 
 
 It would be a crude and inaccurate method of finding the moment 
 of inertia by actually dividing an area or a volume into its ele- 
 ments, multiplying each by the square of its distance from the 
 axis, and then finally taking the sum of these products. We can, 
 however, find the moment of inertia with accuracy by means of 
 integration. 
 
 The strength of a beam or of a column depends upon the form 
 as well as the area of its section, and in all calculations respect- 
 ing the strength of beams and columns, the factor which gives ex- 
 pression to the effect of the form of the section is its moment of 
 inertia about an axis passing through the center of gravity. 
 
 The moment of inertia of a surface is universally denoted by I, 
 the axis being in the plane of the surface and passing through its 
 center of gravity. When the axis is perpendicular to the plane of 
 the surface, the moment of inertia is then termed the polar moment 
 of inertia, and is denoted by I p . 
 
 42. There are certain relations between the different moments of 
 inertia of the same section which are useful in the solution of prob- 
 lems. 
 
 43. Let I denote the moment of inertia of any surface about an 
 axis through the center of gravity of the surface, T the moment of 
 inertia of the surface about any other parallel axis, A. the area of 
 the surface, and H the perpendicular distance between the axes. 
 Then we shall have P = I + AH 2 . 
 
MOMENT OF INERTIA. EADIUS OF GYRATION 
 
 305 
 
 For, let yy, Fig. 47, be the axis through the center of gravity, 
 and yy the axis parallel to yy, both being in the plane of the 
 surface. 
 
 Denote the elemental areas of the surface by a 19 a 2 , a s , etc., and 
 their distances from the axis yy by r , r 2 , r 3 , etc. Then 
 
 r = *' # 2 a H r aH r etc - 
 
 etc. 
 
 + a B (H 2 + 2r z H 
 = asl + a 2 rl+ a s rl + H 2 (a, + a, + a 8 ) 
 
 The first term of the second member of this equation is, by our 
 definition, the moment of inertia, I, of the surface about the axis 
 yy through the center of gravity. The second term becomes AH 2 , 
 in which A denotes the whole area. The third term will reduce to 
 0, because the quantity within the parenthesis is the algebraic sum 
 of the moments of the elemental areas about a line passing through 
 their center of gravity. Hence, T = I + AH 2 . 
 
 44. Suppose an axis perpendicular to the plane of the surface 
 represented in Fig. 48 to pass through 0. Let a be an elemental 
 area of the surface, distant r from 0, and let X and Y be rectangu- 
 lar axes passing through and lying in the plane of the surface. 
 Then the polar moment of a is I p = ar 2 . The moment of inertia 
 of a about X is I x = ay z , and similarly I y = ax 2 . But we have 
 x 2 + f r 2 , therefore, ax 2 + ay 2 = ar 2 . That is, the polar 
 20 
 
306 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 moment of inertia of any surface is equal to the sum of the moments 
 of inertia of the surface about any two rectangular axes lying in 
 the plane of the surface and passing through the polar axis of 
 revolution ; that is, I p = I x -\- I y . 
 
 45. Radius of Gyration. We have seen that 1 = ^rj-f- a z r\ 
 + a 3 r l + e tc., in which ct 1 + & 2 + a s + etc. = A, the whole area. 
 Now if we can conceive the whole area to be condensed into a single 
 particle, distant K from the axis of rotation, we shall have I = AK 2 . 
 This imaginary point at which the particle is supposed to be situated 
 is known as the center of gyration, and its distance, K, from the axis 
 
 is the radius of gyration. We have then, K = A/-T, in which the 
 
 mass, M, the volume, V, or the weight, W, may be substituted for 
 the area, A. 
 
 46. If we denote the product of a force, area, volume, or weight 
 by its arm as the first moment, or simply the moment, of the force, 
 area, volume, or weight, then we may conveniently denote the pro- 
 duct of the force, area, volume, or weight by the square of its arm 
 as the second moment of the force," area, volume, or weight. Thus 
 the moment of inertia is sometimes known as the second moment. 
 
 EXAMPLE I. Find the least moment of inertia and the least ra- 
 dius of gyration of a parallelogram. 
 
 The least moment of inertia is that about an axis through the 
 center of gravity, g, Fig. 49. 
 The elemental area = 2Bdh. 
 Second moment of the elemental area =. 2B-h z -dh. 
 
MOMENT OF INERTIA. EADIUS OF GYRATION 307 
 
 Moment of Inertia 
 
 /*~ 7? ff 3 
 
 ia I 2B I tfdh = =^e- - 
 t/o M 
 
 IT- H 
 
 ^- 
 
 If the moment of inertia with respect to an axis coinciding* 
 with the base be desired, we have T = I + -41? 2 , in which JEf, in 
 
 rr 
 
 this instance, = -~ . 
 
 And^T 
 
 -/ 
 
 H 
 3 
 
 EXAMPLE II. Find the polar moment of inertia and radius of 
 gyration of a circle about an axis passing through its center. 
 Elemental area = %-rrrdr, Fig. 50. 
 Second moment of elemental area = 2-jrr 2 rdr. 
 
 R 
 
308 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 EXAMPLE III. Find the polar moment of inertia and radius of 
 gyration of a cone about its axis. 
 Elemental volume = 7rr 2 dh, Fig. 51. 
 
 r" 
 X 
 
 Second moment of elemental volume = VK 2 = 
 for the circle, K 2 = ^ . 
 
 7T i*R 
 
 Then / = r I r*dh. By similar triangles, we have 
 ^t/o 
 
 h r , 7 ^Tr j 77. ffdr 
 
 ~ff= -n> whence h = - - , and aA = - . 
 
 PEOBLEMS. 
 
 Find the moment of inertia and radius of gyration of the follow- 
 ing: 
 
 34 Triangle about an axis through vertex and parallel to base. 
 
 T , rr H - 
 
 Ans. 1 = 3 ; XL = -jr- ^/^ . 
 
 4 /i 
 
 35. Triangle about an axis through the center of gravity and 
 
 parallel to base. 
 
 BH*. K= H .VTT 
 36 ' " 18 
 
 Ans. / = 
 
MOMENT OF INERTIA. KADIUS OF GYRATION 309 
 
 36. Triangle about an axis coinciding with base. 
 
 A * -,>:; 
 
 37. Trapezoid about an axis coinciding with small base. 
 
 Ans. I' = ^ ; K j 
 
 38. Trapezoid about an axis coinciding with large base. 
 
 A Tf *-* \&U ~T~ *-*) 1? 
 
 JLns. A = ., - 5 J$. == 
 
 39. Trapezoid about an axis through center of gravity and par- 
 allel to base. 
 
 40. Square about its diagonal. 
 
 Am. 1 = j3>K = & 
 
 41. Circle about a diameter. 
 
 Ans. 1='^-- K=-. 
 
 42. Hollow circle about a diameter. 
 
 . /=*(!*-#); K = 
 
 + 
 
 64 v 4 
 
 Find the polar moment of inertia and radius of gyration of the 
 following surfaces : 
 
 43. Parallelogram about a pole passing through its center of 
 gravity. 
 
 44. Circle about a pole passing through its center. 
 
 Ans. I p = --g- ; K </$ . 
 
 45. Hollow circle about a pole passing through center. 
 
 Ans. I p = ~ (Z> 4 -d 4 )-K= J^^- 
 
 Find the polar moment of inertia and radius of gyration of the 
 following named solids : 
 
 46. Cylinder about a pole coinciding with its axis. 
 
 Ans. ^ = ^rx^=~V8. 
 
310 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 47. Hollow cylinder about a pole coinciding with its axis. 
 
 Ans. l r = ^(iy-^;K 
 
 48. Sphere about a diameter. 
 
 An, 4 . 
 
 49. A bar of rectangular section about a pole passing through the 
 center of figure. 
 
 
CHAPTER IV. 
 THE THEORY OF BEAMS. 
 
 47. Strain. The change of form which a load produces in a 
 body is called the strain due to the load. 
 
 48. Stress. The internal, or molecular, resistance which the 
 material of a body interposes to resist deformation is called a stress. 
 
 49. Coefficient of Elasticity. Within the elastic limit of all 
 materials the stresses are proportional to the strains, but since the 
 same intensity of stress does not produce the same strain in different 
 materials we must have some definite means of expressing the 
 amount of strain produced in a body by a given stress. The means 
 employed is to assume the body to be perfectly elastic and then 
 state the intensity of stress necessary to strain the body by an 
 amount equal to its own length. This stress is known as the 
 Modulus of Elasticity, or the Coefficient of Elasticity, and is 
 denoted by E. The values of E for different materials have been 
 determined and are practically the same for tension and compres- 
 sion. The mean values for E in pounds per square inch for the 
 materials most commonly used in engineering are as follows: 
 Timber, 1,500,000; cast iron, 15,000,000 ; wrought iron, 25,000,000; 
 steel, 30,000,000. 
 
 There are no materials of construction that are perfectly elastic, 
 and but few that will stretch 0.001 of their length and remain 
 elastic. 
 
 50. If y denotes the strain produced in a body by a stress S, 
 and Y the strain produced by the stress necessary to stretch the 
 body to double its length; then, since the stresses are proportional 
 to the strains, we shall have for a stress-strain diagram, Fig. 52, 
 
 = , whence E = = 
 
 E' y y 
 
 since the strain, or deformation, F, is equal to L. 
 
312 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 51. Neutral Axis. When a beam is deflected, as in Fig. 53, its 
 fibers on the convex side are in tension and those on the concave 
 side in compression. There must be a surface somewhere between 
 these conditions where the fibers are neither in tension nor in com- 
 
 pression. Such surface is known as the neutral surface, and the 
 line in which this surface intersects the plane of a section of the 
 beam is the neutral axis of the section. The neutral axis invariably 
 passes through the center of gravity of the section. 
 
 rig. S3 
 
 52. Suppose an elastic beam to be bent to the arc of a circle, 
 Fig. 54. Let r denote the radius of the neutral surface, and c the 
 distance from the neutral surface to the outermost surface. 
 
 The length of the neutral surface will not change in the bending, 
 hence %TTT =. original length of the outermost surface, and %ir(r-\-c) 
 = length of outermost surface after bending. 
 
THE THEORY OF BEAMS 
 
 313 
 
 The deformation = strain of outer surface = 2ir(r -\- c) 2irr 
 = 27rc. But we have seen that 
 
 Strain _Stress_ _ _S 
 
 E' 
 
 Original Length 
 
 Hence 
 
 Modulus of Elasticity 
 
 __-C __ S_ 
 
 r~r ~ E ' 
 
 (A) 
 
 -pi o 
 
 Then, from equation (A), we have = - unit stress at a 
 
 T C 
 
 distance unity from the neutral surface, 8 being the unit stress, 
 tension or compression, of the fibers of the outermost surface, and c 
 the distance from the neutral surface to the outermost surface. 
 
 . SS. 
 
 53. Resisting Moment. Suppose a beam section, Fig. 55, to be 
 made up of a great number of layers of areas a, a ly a 2 , &c., distant 
 y> y\> 2/2? &c., from the neutral axis. The unit stresses at distances 
 
 o o cr 
 
 y> Vi> y^ etc., are - X y, - X y,, - X y a , and the total stresses on 
 
 o rr nr 
 
 the elemental areas are - X ay, - X a^, - \ X 
 
 CO C 
 
 oo 
 
 moments of these stresses are - X ay\ - X c^y 
 
 c c 
 
 ^z, &c. The 
 
 o 
 
 - X a z yl, &c. 
 
 c 
 
 The sum of all these moments will be the resisting moment of the 
 
 o 
 
 section, hence Kesisting Moment = - (ay 2 + 0^1 + a$\ + &c.) 
 
 c * 
 
 RfJ" 
 
 = - , in which I is the moment of inertia of the section about 
 
 C/ 
 
 the neutral axis. The Bending Moment must be equal to the 
 Resisting Moment, since, for equilibrium, the internal stresses 
 must equal the external forces, therefore, 
 
 Bending Moment = M = 
 
 (B) 
 
314 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 54. Section Modulus. The factor - contains the dimensions of 
 
 c 
 
 the section and is the measure of its strength. It is called the modu- 
 lus of the section, and is denoted by Z. Hence, I = Zc, and 
 
 -o- = - = Z, whence S = -^ . Substituting this value of 8 in 
 equation (A), we have, - = -^., whence M = - == . (0) 
 
 The equation (C) just found is that of the elastic curve, showing 
 the relation between the bending moment at any section and the 
 radius of curvature of the beam. 
 
 55. In the theory of beams the important assumptions made are : 
 (a) That the strain increases directly as the distance from the 
 neutral axis. (&) That the stress is proportional to the strain. 
 (c) That the coefficient of elasticity is the same for tension as for 
 compression. 
 
 Experiments have shown that, within the elastic limit of any 
 material, the above assumptions may be regarded as perfectly true. 
 
 56. The equation, M = - - is the fundamental equation for 
 
 c 
 
 beam investigations, and since I is expressed in bi-quadratic inches, 
 M must be expressed in pounds-inches. The length of a required 
 beam, and the load to which it is to be subjected being known, we 
 can readily find the maximum bending moment, and then by assum- 
 ing the allowable working stress, 8, per unit of area at the outer- 
 
 most fiber, we shall have, -- = -. The numerical value thus ob- 
 
 tained for - must be satisfied by the dimensions assumed for the 
 
 c 
 cross-section of the beam. The smaller value of 8, whether for 
 
 tension or compression, should be taken. 
 
 57. Dangerous Section. The section of a beam at which the 
 bending moment is a maximum is known as the " dangerous sec- 
 tion." 
 
 In the case of a beam uniformly loaded it has been shown that 
 the bending-moment curve is parabolic and that, therefore, the 
 ordinate representing the bending moment is a continuous func- 
 tion of x. The determination of the dangerous section would then 
 be to find that value of x which would make the bending moment 
 a maximum. To do this we would place the first ^-derivative of M 
 
THE THEORY OF BEAMS 315 
 
 equal to zero, and the resulting value of x would be the abscissa of 
 the dangerous section. 
 
 For example, the beam of Fig. 56 is uniformly loaded with w 
 
 pounds per linear foot. The support reactions are each -^ . 
 
 The bending moment at any section distant x from the left sup- 
 port is, 
 
 wLx WX" 
 
 Then, -^ = ? ^ -- wx 0, whence %=-% 
 That this value of x makes M a maximum is evidenced by the 
 negative result of the second ^-derivative, thus : 
 
 = w. 
 
 Fif. 56. 
 
 The maximum bending moment, and therefore the dangerous 
 section, is then at the middle of the beam; see Fig. 27, p. 291. 
 It will be observed that the second member of the equation. 
 
 j- = -s wx, is the shear at the section distant x from the left 
 
 support. We infer from this that the first ^-derivative of the 
 moment is the shear, and that, in the case of uniform loading, the 
 maximum bending moment is at the section where the shear is 
 zero; see Fig. 29, p. 291. 
 
 If, in addition to the uniformly distributed load, the beam of 
 Fig. 56 were subjected to one or more concentrated loads, or if 
 there were no uniform load and the beam subjected simply to one 
 or more concentrated loads, the ordinate representing the bending 
 moment would no longer be a continuous function of x. The first 
 derivative of the moment at any section would, however, still be 
 the shear at that section, but the shear at the dangerous section 
 might nofc be zero because the shear is not necessarily zero at any 
 section. If the construction of the shear diagram shows that the 
 shear is not zero at any section, it will also show that at some one 
 
316 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 section the shear suddenly changes sign, or passes through zero, 
 and the maximum bending moment will be found at that section; 
 see Fig. 24, p. 289. 
 
 The construction of the shear diagram will at once locate the 
 dangerous section, or it may be found by taking the algebraic sum 
 of the forces to the left of various sections until the point is found 
 where the shear changes sign. 
 
 EXAMPLE I. A yellow pine beam of 18 feet span is to support 
 concentrated loads of 1200 Ibs. and 1500 Ibs. at points distant 4 
 feet and 10 feet from the left support. Determine the cross-sec- 
 tional dimensions of the beam. 
 
 From Fig. 57 we have: 18^ = 1200 X 14 + 1500 X 8, whence, 
 = 1600 Ibs. Also, 18R 2 = 1500 X 10 + 1200 X 4, whence, 
 R 2 = 1100. The construction of the shear diagram shows the dan- 
 gerous section to be under the load of 1500 Ibs. Hence, 
 
 10' 
 
 1500 
 
 M max 1600 X 10 1200 X 6 = 8800 Ibs.-ft. = 105,600 Ibs.- 
 inches. Using a factor of safety of 10, we shall have : 
 
 9000 
 S = -JQ- 900 Ibs. per square inch. Then, 
 
 /__ 105,600 _ 117 oo 
 
 - ~~- L17 - 33 - 
 
 Assuming a 5" x 12" beam, we shall have : 
 
 5 X 1728 
 
 _ 12ft 
 
 c ~ 12c ~ 12 X 6 
 
 This is so near the assumed value of the section modulus that it 
 would appear that the 5" x 12" beam would be secure. We shall, 
 however, investigate the influence of the weight of the beam itself. 
 Weight of beam = 5 X 12 X if X - 1 / = 300 Ibs. = 16| Ibs. per 
 foot, and # then becomes 1600 + 150 = 1750 Ibs. The consid- 
 eration of the weight of the beam will not change the position of 
 the dangerous section, as can be shown on the shear diagram. Then 
 
THE THEORY OF BEAMS 317 
 
 Mmax = 1750 X 10 1200 X 6 16.66 X 10 X 5 = 9466.66 
 Ibs.-ft. = 113,600 Ibs.-inches. 
 
 Mo 113.600 X 6 X 12 . , m , . 
 
 Then, 8 = -j- - 5 x 1728 = P er sq * m 
 
 is greater than the assumed safe stress of 900 Ibs., so a 5.5" x 12" 
 
 beam will be selected. Then, 
 
 Weight of beam = 5.5 X 12 X H X -^ = 330 Ibs. = 18.33 Ibs. 
 
 per foot, and R = 1765 Ibs. 
 
 Then, M max = 1765 X 10 - - 1200 X 6 - - 18.33 X 10 X 5 
 
 = 9534.33 Ibs.-ft. = 114,412 Ibs.-inches. 
 
 Then , fl== UX 
 
 result which shows the 5.5" x 12" beam to be safe. 
 
 58. Standard I Beams. The standard steel I beams so exten- 
 sively used in engineering work are rolled in light, intermediate, 
 
 . and heavy weights of thirteen different sizes. The different manu- 
 facturers issue hand-books containing tables of the properties of the. 
 beams they produce, and while there is an agreement in sizes as to 
 the depth, there are marked differences in the proportions of cross- 
 sections, and therefore also of the weights per foot and moments of 
 inertia. 
 
 59. The ultimate strength of structural steel may be taken as 
 65,000 Ibs. per square inch, and the elastic limit at about one-half 
 the ultimate strength. The working fiber stress per sq. inch may 
 be taken as 16,000 Ibs. for buildings and 12,500 Ibs. for bridges, 
 indicating factors of safety of 4 and 5.2, respectively. 
 
 60. The span and load of an I beam being given, the value of i 
 
 
 
 may be found by means of the formula, = -~-, and then the beam 
 in the manufacturer's tables which has an - equal to or next 
 
 
 
 greater than the one found is to be selected. 
 
 EXAMPLE II. What Cambria I beam should be selected for a 
 floor bearing a load of 180 Ibs. per square foot, the beams to have 
 a span of 25 feet, spaced 8 feet apart, and to have a maximum 
 unit stress of 16,000 Ibs. per sq. inch? 
 
 25 X 8 X 180 = 36,000 Ibs. load on each beam = ^%^ = 1440 
 
 /*o 
 
 Ibs. per foot. R^ 18,000 Ibs., and 
 
318 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 M max = 18,000 X 12-5 X 12 - - 1440 X 12.5 X 6.25 X 
 = 1,350,000 Ibs.-in. 
 
 A reference to the hand-book of the Cambria Steel Co. shows 
 that their 15-inch special I beam of 65 Ibs. per foot should be 
 selected. 
 
 EXAMPLE III. What is the proper size of I beam to carry a load 
 of 35,000 Ibs. concentrated at the middle of a span of 25 feet, the 
 fiber stress not to exceed 16,000 Ibs. per square inch? 
 M max = 17,500 X 12.5 X 12 = 2,625,000 Ibs.-inches. 
 
 L K 2,625,000 _ 161 56 
 c ~ IS " 16,000 
 
 This value of the section modulus would indicate that a 24-inch 
 standard Cambria I of 80 Ibs. per foot would do. We shall, how- 
 'ever, investigate the effect of the weight of the beam itself. 
 Weight of beam = 80 X 25 = 2000 Ibs., therefore = 18,500 Ibs. 
 Then M max = 18,500 X 12.5 X 12 80 X 12.5 X 6.25 X 12 
 
 = 2,700,000 Ibs.-inches. 8 = ^ = 2?7 ^g^ 12 - 15,500, 
 which is 500 Ibs. per sq. inch less than the safe allowable stress. 
 (The moment of inertia of the light 24-inch Cambria standard I is 
 2087.2.) The 24-inch standard I, weighing 80 Ibs. per foot, is, 
 therefore, the proper selection. 
 
 EXAMPLE IV. Heavy 12-inch Cambria I beams of 20 feet span 
 are to be used to support a floor bearing a uniform load, including 
 the weight of the beam, of 200 Ibs. per square foot. The fiber stress 
 is not to exceed 16,000 Ibs. per sq. inch. Find the spacing of the 
 beams. 
 
 From the table of properties of standard Cambria I Yearns, we 
 find for the beam selected, 
 
 Section Modulus = - = 41. 
 c 
 
 Let x denote the distance between center lines of consecutive beams. 
 Then 2Qx X 200 = 4000:r = load on one beam, and <^p = 200z 
 = load per foot. R^ 2000z. 
 M max 2000z X 10 X 12 200z X 10 X 5 X 12 = 120,000z. 
 
 Then i = f, or 41 = -?!!'nn!r > whence x = 5 - 47 feet - 
 C o lo,UUU 
 
THE THEORY OF BEAMS 319 
 
 The load on each beam is then 4000 X 5.47 = 21,880 Ibs. But 
 this includes the weight of the beam, which is 40 X 20 = 800 Ibs. ; 
 hence, 21,080 Ibs. may be placed on the beam. 
 
 PROBLEMS. 
 
 50. A rectangular beam, 9 inches deep, 3 inches wide, supports a 
 load of 0.5 ton, concentrated at the middle of an 8-foot span. Find 
 the maximum fiber stress. Ans. 0.3 ton per sq. inch. 
 
 51. A beam arranged with symmetrical overhanging ends is 
 loaded with three equal loads one at each end and one at the mid- 
 dle. What is the distance apart of the supports, in terms of the 
 total length, L, when the bending moment is equal over the supports 
 and in the middle, and at what section is the bending moment zero ? 
 
 Ans. -g-. At sections between the supports distant -^ from them. 
 
 52. A square timber beam of 12-inch side weighs 50 Ibs. per cubic 
 foot, is 20 ft. long, and supports a load of 2 long tons at the middle 
 of its span. Calculate the fiber stress at the middle section. 
 
 Ans. 1037 Ibs. per sq. inch. 
 
 53. Light 12-inch Cambria I beams of 20 feet span are to sup- 
 port a floor bearing a uniform load of 175 Ibs. per square foot. The 
 outermost fiber stress is not to exceed 16,000 Ibs. per sq. inch. At 
 what distance apart should the beams be placed ? 
 
 Ans. 5.5 ft. 
 
 54. How much stronger is a beam 4 inches wide, 8 inches deep, 
 and 8 ft. long, than one 3 inches wide, 5 inches deep, and 14 T ^- ft. 
 long? Ans. Six times as strong. 
 
 55. What safe uniform load can be placed on a wooden cantilever 
 5 feet long, 3 inches wide, and 5 inches deep, in order that the fiber 
 stress shall be 900 Ibs. per square inch? 
 
 Ans. 71 Ibs. per foot, nearly. 
 
 56'. What safe uniformly distributed load can be placed on a 
 standard Cambria channel beam weighing 20.5 Ibs. per foot, the 
 span being 16 feet, the web placed vertical, and the maximum fiber 
 stress not to exceed 12,500 pounds per square inch ? 
 
 Ans. 11,120 Ibs. 
 
 57. Find the factor of safety of a heavy 12-inch Cambria I beam 
 of 15 feet span when sustaining a uniformly distributed load of 
 30 tons. Ans. 2. 
 
CHAPTER V. 
 COLUMNS. SHAFTS. 
 
 61. Columns. A column, or strut, is a straight beam acted on 
 compressively at its extremities, and of such length compared with 
 its diameter or least sectional dimensions, that failure will result 
 from buckling, or lateral bending, instead of by crushing or by 
 splitting. In addition to the many familiar applications of col- 
 umns in structural work, the piston-rods and connecting-rods of 
 steam engines are also classed as columns. 
 
 62. If columns were initially absolutely straight, made of homo- 
 geneous material, and exactly centrally loaded, there would be no 
 difference in the character of their failure from that of short speci- 
 mens. These three conditions are never fulfilled in practice, and 
 in consequence columns are weaker than short blocks of the same 
 material. No satisfactory theoretical discussion of columns has 
 been made, and all the formula} used in their design contain em- 
 pirical constants determined by experiment. The formula having 
 the most rational basis is the one attributed to Rankine or to Gor- 
 don, and is, 
 
 AS 
 
 (D) 
 
 aK* 
 
 in which P is the load on the column expressed in pounds, A the 
 sectional area of the column in square inches, 8 the crushing 
 strength of a short block of the material, a a constant quantity 
 determined by experiment for different materials, L the length of 
 the column in inches, and K the radius of gyration of the section. 
 
 63. The strength of a column depends largely upon the manner 
 in which its ends are secured. If the ends are square, or flat, the 
 column is said to have fixed ends ; if one end be fixed and the other 
 end hinged, as in the case of a piston-rod, the end conditions are 
 known as " pin and square ; " if both ends be hinged or " rounded," 
 as in. the case of a connecting-rod, the end conditions are known as 
 "pin," or "round." The value of K 2 in formula (D) is found 
 
COLUMNS. SHAFTS 
 
 321 
 
 from the relation 7 = AK 2 , and the values for a, for 8, and for 
 suitable factors of safety for the three conditions of bearing, are 
 given in the following table. It will be observed from the table 
 that the ratios of the values of a for the fixed ends to the values of 
 pin and square and pin ends, are as 1.78 to 1 and 4 to 1, respec- 
 tively, and these ratios indicate the relative strengths of long col- 
 umns with these end conditions. 
 
 
 Timber. 
 
 Cast Iron. 
 
 Wrought 
 Iron. 
 
 Structural 
 Steel. 
 
 Hard 
 Steel. 
 
 s 
 
 a (Fixed Ends) .... 
 
 8000 
 3000 
 
 84000 
 5000 
 
 55000 
 6000 
 
 50000 
 36000 
 
 150000 
 25000 
 
 a (Pin and Square) 
 a (Pin) 
 
 1690 
 750 
 
 2810 
 1250 
 
 20250 
 9000 
 
 24000 
 18000 
 
 14060 
 6250 
 
 Factor of Safety for 
 Buildings 
 
 8 
 
 8 
 
 6 
 
 4 
 
 5 
 
 Factor of Safety for 
 Bridges 
 
 10 
 
 
 10 
 
 5 
 
 7 
 
 Factor of Safety for 
 Shocks 
 
 15 
 
 
 15 
 
 
 15 
 
 
 
 
 
 
 
 EXAMPLE I. A hollow cylindrical cast iron column, 16 ft. long, 
 with fixed ends, sustains a load of 200,000 Ibs. when used in a 
 building. Outside diameter, 10 inches; inside diameter, 8 inches. 
 Is it safe ? 
 
 A = 0.7854(100 64) = 28.27 sq. inches. 
 - 4096) _ 1ft 9 , 
 
 64 x 28.27 
 
 P = 
 
 28.27 
 
 incn ' 
 
 The factor of safety is g^- =10 nearly, which shows the column 
 to be perfectly safe. 
 
 EXAMPLE II. Find the safe load for a light 12-inch Cambria 
 I column, weighing 31.5 lbs. per foot, 16 feet long, and with fixed 
 ends, the column to be used in a building. 
 21 
 
322 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 Eeferring to the Cambria hand-book, we find that A = 9.26, and 
 K = 1.01. The factor of safety is 4. Hence, 
 
 p _ SA 12,500 x 9.26 _ 12,500 x 9.26 x 36,360 _ 5? ?55 
 
 1 Z 2 ~ j (16 x 12) 2 36,360 + 367864 
 
 aK' 1 36,000 x 1.02 
 
 per square inch. 
 
 64. In designing a column we have given the form of its section, 
 its length, the material of which it is made, the load it is to carry, 
 and the manner of securing its ends. The problem is then to deter- 
 mine the necessary area of cross-section. The general method of 
 procedure is to determine from the data given the necessary cross- 
 section area for a short block of the material; then, knowing that 
 the section area of the required column must be larger, assume 
 dimensions which give a larger area, and then, by means of formula 
 (D), ascertain the unit-stress resulting from such assumption. If 
 it is too great, the assumed area is too small ; if too little, a smaller 
 area must be chosen. Proceed thus by trial and error until a satis- 
 factory solution is obtained. 
 
 EXAMPLE III. A column of timber of square section, 14 ft. 
 long and with fixed ends, is required to support a steady load of 10 
 tons. Find the dimensions of the section. 
 
 8000 
 Using a factor of safety of 8, we have g := 1000 Ibs. per sq. 
 
 inch as the safe working unit stress. Hence, ^ 20 square 
 
 inches area of section needed for a very short column, or one less in 
 length than ten times the least dimension of the section. As the 
 required section must be larger, we will assume an area of 36 square 
 
 inches. Then A = (6) 2 , and K 2 = ~ = l2~x^6V = 3 ' Hence > 
 
 per square inch. Since this is much larger than the allowable unit 
 stress of 1000 Ibs., we must assume a larger area. Try an area of 
 
 64 square inches. Then A = (8) 2 , and I = 12 ^(SY = ^ 
 
 T-T a 20,000 / .. , (15 X 12) 2 X 3 1 
 
 Hence, S= ' 64 1 1 + 3000 x 16 J ~ P er S( l uare m - 
 
 This is a trifle small, so we will try a square area whose side is 7.9 
 inches. Then A = (7.9) 2 , and K 2 = 12 .v = 6 ' 2 ' 
 
COLUMNS. SHAFTS 323 
 
 20,000 f 1 , (15 X 12) 2 \ 
 Hence, 8 =- -TJfjfy j l ~r SQQQ x 5 2 J ~ ~ P er S( l uare mcn > 
 
 which is quite near the allowable unit stress. Therefore, a column 
 of square section, haying a side of 7.9 inches, will suffice. 
 
 65. Sometimes all the dimensions can be assumed except one, and 
 then, after expressing A and K 2 in terms of the unknown dimen- 
 sion, we can substitute them in formula (D) and solve the resulting 
 bi-quadratic equation for the unknown dimension. Thus, in Ex- 
 ample III, let x denote the side of the unknown square section. 
 
 x 4 x 2 
 Then A x 2 , and K 2 = ^- 2 = ^ . Substituting these values in 
 
 (D), we shall have: 
 
 i poo - 20,000 J 1 j_ (15 X12) 2 X 12 \_ 20,000 /, , 129.6\ 
 
 ~~^l 3000 x ^ / = ~^~l Tr 
 
 from which we get, x = 7.8 inches. 
 
 EXAMPLE IV. A hollow cylindrical cast iron column, 20 feet in 
 length, is required to sustain a steady load of 164,000 pounds. De- 
 termine its cross-sectional dimensions. 
 
 Using a factor of safety of 7, we have 12,000 Ibs. per 
 
 square inch as the safe unit working stress. Then 1 ' = 13.66 
 square inches of area needed for a very short column. Assuming an 
 area of 25 square inches, and assuming further that the outside 
 diameter of the column shall be 10 inches, we shall have, calling 
 d the inside diameter, 0.7854 X 100 0.7854 X d 2 25, whence 
 
 d = 8.126 inches. Then K 2 = = ^ 
 
 -A. 
 
 } = 13,8,0 Ib, per 
 
 Her.ce *= 
 
 inch, which is greater than the allowable unit stress of 12,000 Ibs. 
 
 Assuming an area of 29 square inches, we shall have : 
 0.7854 X 100 0.7854 X d 2 = 29, whence d = 7.94 inches, and 
 then K 2 = 10.19. 
 
 Hence S = ^^{ ! + JML } 12)048 Ibs. per . inc h, 
 a result sufficiently near the allowable stress to warrant making 
 the column 10 inches in outside diameter and the metal 1 inch 
 thick. 
 
 66. For the properties of the different forms of the built-up 
 beams and columns so largely used in structural work, reference 
 
324 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 should be made to the hand-books issued by the different steel 
 companies. 
 
 67. Shafts. When cylindrical shafts are used for the transmis- 
 sion of power they are subjected to torsion, or twist, which occa- 
 sions a shearing stress. The angle through which the elements are 
 twisted is, within the elastic limit of the material, proportional to 
 the applied force. The elements along the axis of the shaft remain 
 straight, and if P denotes the applied force, and r the perpendicular 
 distance of its point of application from the axis of the shaft, then 
 'Pr is the measure of the twisting moment. In the case of the 
 steam engine, P varies with the ratio of expansion of the steam in 
 the cylinder, and the shaft, in addition to its twisting moment, is 
 subjected to a slight bending moment due to its own weight. If 
 the twisting moment be calculated from the I. H. P., that is, from 
 the mean pressure in the cylinder, it is known as the mean twisting 
 moment. Since shafts must be strong enough to resist the maxi- 
 mum stress to which they may be subjected, the moment expressing 
 the maximum stress due to the combined twisting and bending 
 actions should form the basis of the calculation. This moment is 
 known as the maximum equivalent twisting moment. The ratio 
 between the maximum equivalent twisting moment and the mean 
 twisting moment varies somewhat with different kinds of engines, 
 but it is not far wrong to take it as 1.5. 
 
 68. It has been shown that the strength of a section to resist 
 bending is measured by its modulus, Z, and the relation -5- = - Z 
 
 has been shown. By reasoning analogous to that on pages 313 and 
 314, it may be shown that the strength of a section to resist torsion 
 is measured by its polar modulus, Z p , and we shall have : 
 
 ==*- (E > 
 
 69. In the case of a solid circular shaft for transmitting power, 
 we have I t = ?g, and c = |. Hence Z, = *j -s- f = yf = 0.1960. 
 
 For a hollow circular shaft, we shall have : 
 
 p ~ 32 ' 2 " 
 
 70. If, in the hollow shaft, we let - denote the fraction of the 
 solid section removed for the axial hole, we shall have : 
 
COLUMNS. SHAFTS 325 
 
 Solid Area 
 Eemoved Area = 
 
 n 
 
 That is, we shall have ~ = ^, whence d* = ~. 
 Then the modulus for the hollow shaft becomes : 
 
 (P) 
 
 It is the common practice for hollow shafts to make the internal 
 diameter equal to one-half the external diameter, in which case the 
 part removed is one-fourth the whole area of the section. We shall 
 
 then have n = 4, and Z p = ^- (1 iV) 9 g g = 0.1 84D 3 . 
 
 71. Horse-power Transmitted by Shafts. Let P be the average 
 force in pounds acting at a distance r inches from the axis of the 
 shaft. Then Pr = the mean twisting moment in pounds-inches. 
 
 If N be the number of revolutions per minute, we shall have: 
 
 , 2PmN 
 
 Work per minute in foot-pounds = ^ . 
 
 Then, horse-power transmitted = I. H. P. = 12 x 33 000 ' 
 whence Pr = 12 X ^'^jf L H " - . From equation (E) we have 
 
 Pr = M t * = ~SZ P 0.196&D 3 = Resisting moment for a 
 c 
 
 solid shaft. From equation (F), we have Pr = -r-l ~~-^ 
 
 = 0. 184:8 'D s , when the internal diameter is one-half the external, 
 and n 4. 8 may be taken as 7000 Ibs. per square inch for 
 wrought iron, and 10,000 for steel. Taking the maximum equiva- 
 lent twisting moment as 1.5 times the mean twisting moment, we 
 shall have : 
 
 0.196 x 70007> = 13X3S.OOOXLH.P. x 1.5 whMceB= 4- 
 
 for solid wrought iron shafts. 
 
 For hollow wrought iron shafts with inside diameter half the 
 outside, we shall have : 
 
 0.184 x 70002)3 = 1* >< 88,000 x I. H. P. x 1.5 
 
326 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 In like manner we shall have for steel : 
 
 D = '6. 64 A/ ' ,j for solid shafts. 
 
 ad 3 p 
 
 D _ 3.79 7 1 - ** r - for hollow shaft having external 
 V N diameter double the internal. 
 
 EXAMPLE V. A solid steel shaft is to transmit 4000 I. H. P. 
 while making 130 revolutions per minute, the maximum equivalent 
 twisting moment being 1.5 times the mean; find its diameter. 
 
 D = 3 .6 = 3.64=10.13 inches. 
 
 EXAMPLE VI. A steel shaft is to have 30 per cent of its section 
 removed in making an axial hole, and is then to transmit 9000 
 I. H. P. at 120 revolutions per minute. Find the outside and inside 
 diameters. 
 
 Let D and d denote the outside and inside diameters, respectively. 
 Then the area of the removed section = 0.7854Z) 2 X 0.3. 
 And d = -v/OaD' = 0.5477/X 
 
 We have, for hollow shafts, Z p = ^- (l -^\ , iu which - = frac- 
 tional area removed. Since the fractional part removed is 30 per 
 cent, we shall have - = ^, whence n=-^. Therefore, Z p 
 
 7~k3 
 
 = ^ (1 Tthr) = .1787D 3 , and the moment of resistance = SZ P 
 = twisting moment. 
 
 Hence 0.178 W X 10,000 = 12 X ^' 00 X ^ , whence 
 
 d 0.5477D = 0.5477 X 15.83 =. 8.67 inches. 
 
 EXAMPLE VII. Find the inside and outside diameters of a hol- 
 low steel shaft to transmit 8000 I. H. P. at 130 revolutions per min- 
 ute, the unit stress in the outermost fibers not to exceed 10,000 Ibs. 
 per square inch, and the interior diameter to be 98 per cent of one- 
 half the exterior diameter. Find also the diameter of a solid shaft 
 to transmit the same power under the same conditions. Show that 
 the two shafts are equal in strength, and that the hollow shaft is 
 20.87 per cent the lighter. 
 
COLUMNS. SHAFTS 327 
 
 Let D and d denote the outside and inside diameters, respectively. 
 
 .7854(.49Z)) 2 _ 6 
 Fractional section area removed by axial hole = 7354/^2 - 25 
 
 Hence - = _ -^ , whence n = ~ . 
 n 25 b 
 
 The resisting moment, 8Z P , is then, 10,000 X 0.185P 3 . 
 The maximum equivalent twisting moment 
 
 - - p _ 12 X 33,000 X8000 X 1.5 
 27TX130 
 
 Then, 10,000 X 0.186Z>- = 18X88,000X^00X1.6 , whence 
 
 n _ 7 12 X 33,000 X 8000 X 1.5 _ u 65 [ fa 
 
 -V 2* X 130 X 10,000 X 0.185 - 
 Then, d = 14.65 X 0.49 = 7.19 inches. 
 
 For the solid shaft, D = 3.64 y ^^' = 3.64 y / ^- 
 
 = 14.37 inches. 
 
 To be of equal strength the moduli of 'the sections must be equal ; 
 that is, we must have 0.185 (14.65) 3 == 0.196 (14.37) 3 . 
 
 The weights of the shafts are proportional to their sectional areas. 
 Sectional area of hollow shaft 0.7854 [( 14.65 ) 2 (7.19) 2 ] 
 = 128.36 square inches. 
 
 Sectional area of solid shaft = 0.7854(14.37) 2 =. 162.21 sq. in. 
 
 Hence the hollow shaft is C 16 ^ 1 128.36)100 _ 2Q g? cent 
 
 lo2.Icl 
 
 lighter than the solid shaft. 
 
 PKOBLEMS. 
 
 58. A hollow cylindrical cast iron column with fixed ends, whose 
 thickness of metal is 2 inches, length 24 feet, and outside diameter 
 24 inches, is to be used in a building. What safe load can it 
 sustain? Ans. 1,141,000 Ibs. 
 
 59. Find the diameter of a steel shaft to transmit 130 I. H. P. 
 at 300 revolutions per minute. (This being a small shaft it will 
 be lacking in stiffness, and in consequence 0.5 inch should be added 
 to the calculated diameter.) Ans. 3.25 inches. 
 
328 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 60. Find the outside diameter of a hollow steel shaft to transmit 
 10,000 I. H. P. at 125 revolutions per minute, the inside diameter 
 to be one-half the outside diameter. Ans. 16 inches. 
 
 61. Find the outside and inside diameters of a hollow steel shaft 
 to transmit 3000 I. H. P. at 200 revolutions per minute, the inside 
 diameter to be 54 per cent of the outside diameter, the maximum 
 equivalent twisting moment to be taken at 1.5 times the mean, and 
 the unit stress in the outermost fibers not to exceed 10,000 Ibs. per 
 sq. inch. Find also the diameter of a solid shaft to transmit the 
 same power under the same conditions. Prove the equality of 
 strength of the shafts, and show that the hollow shaft is 25 per cent 
 the lighter. 
 
 Ans. 9.25" and 5" for hollow shaft; 9" for solid shaft. 
 
CHAPTER VI. 
 THE DEFLECTION OF BEAMS. 
 
 72. Deflection of Beams. The value of the radius of curvature of 
 any plane curve of abscissa x, ordinate y, and length L } is given in 
 the differential calculus as 
 
 Wy d z y dx.d^y 
 
 ~dx* ~dx* 
 
 since the deflection is very small, and dx is dL in the limit. 
 
 Substituting this value of r in equation (C), Art. 54, we have, 
 M = - \ y , which is the form of the equation of the elastic curve 
 
 CL9& 
 
 for investigating the deflection of beams. 
 
 In any particular case the value of M must be expressed in terms 
 of x, and then, after two integrations, the deflection, y, will be 
 known for any value of x. 
 
 Thus for a simple beam, Fig. 58, loaded with W at the middle, we 
 
 Wx 
 
 have M = B x = ^- for any point between the left support and 
 
 the middle of the beam. 
 
 Then = . Integrating once, we have, .= 
 
 When x = -^ we shall have ^=0, whence C = -- -y^-. Then 
 
 _Wx* WI? 
 
 4 16 ' 
 
330 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 Wx* WL' 2 x 
 
 Integrating the second time, we get, Ely = -^ - jg f- & 
 
 = when x = 0, . . C' = 0. 
 WDx 
 
 Then Ely = 
 
 16 
 
 The deflection is a maximum when 
 
 = ' hence y- = - = 
 
 4817' 
 
 73. In the case of a cantilever, Fig. 59, loaded with W at the free 
 
 end, we shall have, M = Wx. Hence, - , ^ Wx . 
 
 Eldy _ Wx* n 
 ~dx~ '^-+ 6 - 
 
 || = when x = L, consequently C = ^L. Then, 
 J _ W, and */, = J^? _ J + 6". We shall 
 
 have y = when x - 0, .-. C' = 0. Then y = 
 The deflection is a maximum when x L, hence, 
 
 IF 
 
 -. 
 
 -*). 
 
 74. The expressions for the maximum deflections of the different 
 kinds of beams of uniform section, under different conditions of sup- 
 port and of loading, may be obtained by similar processes. 
 
 75. Points of Inflection. Cantilevers fixed, or built in, at the 
 end, bend concave downward, while simple beams fixed at the ends, 
 and those which overhang the supports, bend with a combined curva- 
 ture of concave downward and upward concave downward near 
 the supports and concave upward toward the middle. The 
 points at which these contrary curvatures separate, and at which 
 there is no bending whatever, are known as points of inflection. 
 They can be found by placing the general expression for the bending 
 moment equal to zero. Thus, in Example I, Art. 39, it is evident 
 
THE DEFLECTION OF BEAMS 
 
 331 
 
 from inspection that there will be a point of inflection between the 
 left support and W 2 , and one between the right support and W 2 . 
 The expression for the bending moment at any section distant x 
 from the left support, and between that support and W 2 is R x 
 
 aty 
 
 TT 1 (a; + 4) ^(x-^-8) 2 . Placing this expression equal to 
 
 zero, and substituting the values of R ly W 1? and w,. the resulting 
 quadratic gives a value of 4.368 for x, which shows that there is a 
 point of inflection at 4.368 feet to the right of the left support. The 
 expression for the bending moment at any section between W 2 and 
 the right support, and distant x^ from the left support, is R l x 1 
 
 -W,(x, + 4) - - W 2 (x, -- 12) - - Ifo + 8) 2 . Placing this 
 
 equal to zero, the resulting quadratic gives a value of 17.045 for x^ 
 showing that the other point of inflection is at the section distant 
 about 17 feet from the left support. 
 
 F/'p. 6O. 
 
 76. Beams with Fixed Ends. In the case of a beam built in, or 
 a fixed " at the ends, Fig. 60, and uniformly loaded with w pounds 
 per unit of length, we have a condition not heretofore encountered. 
 
 By " fixed " is meant that the parts of the beam built in are con- 
 strained to remain horizontal while the beam is being bent, but it is 
 understood that the beam is free endwise. 
 
 The reaction of the wall in keeping the built-in portion of the 
 beam horizontal is equivalent to the action of the couple, P, P, 
 causing an unknown bending moment, Pz, at the support, and nega- 
 tive because of its concave-downward tendency. 
 
 The bending moment, due to this reaction, at any section between 
 the left support and the middle, and distant x from the left sup- 
 port, is, 
 
 P(z-i-x) Px Pz. 
 
332 THE ELEMENTS OF THE MECHANICS OP MATERIALS 
 
 Hence, the fixing of the beam at the ends causes a constant bend- 
 ing moment over the entire length of the beam which is quite inde- 
 pendent of that caused by the load, but in the opposite direction. 
 
 Taking the origin at the support, the bending moment at any 
 section is, 
 
 Hence, 
 
 fS- 
 
 The first anti-z-derivative is, 
 
 !*!( 
 
 -| = when x and when x = , '. C 0. 
 
 Letting x = -^ 9 we have : 
 
 w(L* L 3 \ PzL wU ..*..!, A- 
 
 2 ( -g -- 24 ) --- 2~~ ~ ? whence Pz = -y~- , which is the bending 
 
 moment at the support. 
 
 Substituting the value of Pz in (1), and letting x = ~-, we have: 
 M wlU L>\ wL? wU n n w.L 2 
 
 -iVT^Tr-Tr* -ir^-^-^w 
 
 Therefore, the bending moment at the middle is but half that at 
 the support, and we have : 
 
 12 12 * 
 
 For the points of inflection we place the second derivative equal 
 to zero, thus : 
 
 (Lx - X2) _ "V = o, or X* - x 2 = ^, whence x = ^/1 
 
 ^ A- * 
 
 Integrating the second time, we get : 
 
 w/WB 8 a; 4N \ Pzx* , 
 
 ^-" T "" ' 
 
 i/ = when a; = . . C" = ; and i/ is a maximum when a; = ^- , 
 consequently, 
 
 F7 - _ w /i 4 _ Z* \ _ wi4 _ W 4 / 1 _ 1 _ 1 '\ wZ* _ PTi 
 
 ^M* - 2 V58 193/ "96" 2~ V48 192 48/ 384 ~ 384"' 
 
 and for the maximum deflection, we have: 
 
THE DEFLECTION OF BEAMS 333 
 
 77. Continuous Beams. The beams heretofore considered have 
 been either cantilevers or simple beams having two supports. A 
 beam resting on more than two supports is termed a continuous 
 beam. 
 
 The chief difficulties in the treatment of continuous beams are 
 the determination of the support reactions and the bending moments 
 
 or T 
 
 at the supports. These being determined, the formula M = - 
 
 is applicable to the question of the safe loading of the beam. 
 
 The treatment of continuous beams is somewhat simplified by the 
 use of Clapeyron's Theorem of Three Moments, and is fully set 
 forth in any complete treatise on the Mechanics of Engineering. 
 
 Only the special case of two spans uniformly loaded will here be 
 considered. 
 
 78. Beam Resting on Three Supports, having two equal spans 
 and loaded with w pounds per unit of length, Fig. 61. 
 
 i^/ 
 
 L- 
 
 The bending moment at any section distant x from the middle 
 support is, 
 
 Hence, 
 
 = when x = Q.:C = Q. 
 
 Lx* 
 
 aJ , , 
 
 4 
 
 y = when x = 0, .-. C" = 0. 
 The three supports being on the same level, it is known that y = 
 
 when x = -, consequently, 
 
334 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 R __-4- 
 1 \16 4:8) 2 \32 48 "*" 192 
 
 wL , A , -.x wL , a p 3wL 3W 
 ^ + 1) = =^, whence^ nr' S TT< 
 
 where TF is the whole distributed load. 
 
 q TJ7 
 
 From symmetry, R n = ^ = - . Then # 3 = V", since 
 
 ID o 
 
 B, + tf 2 + B 3 = W. 
 
 Placing the second derivative equal to zero to find the points of 
 inflection, we have: 
 
 L \ w/D T , \ 
 
 * iT ~ 
 
 =}-z, whence a; = . 
 
 Hence, the points of inflection are distant one-eighth of the 
 length of the beam from the middle support, and on either side of 
 that support. 
 
 To find the bending moment over the middle support, let x = 
 in (1), and we have: 
 
 wU _ wD WL 
 
 _ 
 * 3 32" " ~8~ 32 32 ' 
 
 The greatest bending moment between the supports will be at the 
 point where the shear passes through zero. We find this point by 
 placing the first ^-derivative of the moment equal to zero, thus: 
 
 dM 3L ,L A , 5Z ,,,.,, 
 
 ^- ~"Tg"+o" x= ' whence x = jg- ; that is, the maximum 
 
 Q J 
 
 bending moment between the supports is at the section distant -j^ 
 
 from the end support. To find its value, substitute the value of x 
 in (1), and we have: 
 
 M _ SwLfL _ 5ZA _ W( L* _ 51? . 
 
 W ^\ 4 " 16 
 
 wU ,> ftn , 9 ^ 
 " (64 - '- 
 
 16 
 
 9 WL 
 - 
 
 This result is numerically less than that found for the bending 
 
 (WT \ 
 -qK- j? hence the maximum bend- 
 
 ing moment is at the middle support, and we shall have for safe 
 loading : 
 
 81 _ WL 
 
 c " ~ 32 ' 
 
THE DEFLECTION OF BEAMS 
 
 335 
 
 The bending-moment and shear diagrams are now readily drawn, 
 as shown in Fig. 62. 
 
 79. The relative strength of beams may be obtained by a com- 
 parison of their maximum bending moments, and their relative 
 stiffness by a comparison of their maximum deflections. The 
 results of such comparisons for certain beams of uniform section are 
 given in the table which follows. 
 
 Kind of Beam and 
 Nature of Load. 
 
 Maximum 
 Bending 
 Moment. 
 
 Maximum 
 Deflection. 
 
 Relative 
 Strength. 
 
 Relative 
 Stiffness. 
 
 Cantilever, load at 
 
 WL 
 
 WL3 
 
 1 
 
 1 
 
 free end. 
 Cantilever, load 
 
 WL 
 
 3E1 
 WL3 
 
 2 
 
 2i 
 
 uniformly distributed. 
 Simple beam, load 
 
 2 
 WL 
 
 8EI 
 WL3 
 
 4 
 
 *3 
 
 16 
 
 at middle. 
 
 Simple beam, load 
 uniformly distributed. 
 
 Beam with fixed 
 
 4 
 
 WL 
 
 ~8~ 
 
 WL 
 
 48EI 
 
 5WL3 
 184EI 
 
 WL3 
 
 8 
 8 
 
 25| 
 64 
 
 ends, load at middle. 
 Beam with fixed 
 
 8 
 WL 
 
 192EI 
 WL 3 
 
 12 
 
 128 
 
 ends, uniform load. 
 
 12 
 
 384EI 
 
 
 
336 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 PEOBLEM. 
 
 62. A simple beam 30 ft. long weighs 20 Ibs. per foot, and over- 
 hangs each support 6 ft. It bears a superimposed load of 100 Ibs, 
 per ft. and a load of 1400 Ibs. concentrated at a point 3 ft. to the 
 right of the middle. Construct the shear diagram to a linear scale 
 of -J inch to the foot, and a load scale of 1 inch = 2000 Ibs. Find 
 the bending moment at the dangerous section, and the distances of 
 the points of inflection from the left support. 
 
 Ans. M m ax = 7760 lbs.-ft., and the points of inflection are 
 1.48 ft. and 16.9 ft. from the left support. 
 
CHAPTER VII. 
 RESILIENCE. 
 
 80. Resilience. If a bar be placed in a testing machine and sub- 
 jected to a gradually increasing load in a direction to produce a 
 tensile stress, the elongation produced will be proportional to the 
 load, provided the elastic limit of the material be not exceeded. 
 
 The load, or external force, will gradually increase from to P, 
 
 p 
 and its mean value will be -Q-. If y denotes the elongation, or dis- 
 
 placement, the expression for the work done is, -g . 
 
 -D *A on i_ Stress S & 8L 
 
 By Art. 50, p. 311, we have: ^^ = - = E.: y = ^ . 
 
 L 
 
 By Art. 49, p. 255, we have P = AS, so that the work done is 
 
 But AL is the volume of the bar, consequently the work done is 
 
 proportional to the volume of the bar, and therefore to its weight. 
 
 Within the elastic limit, the work done in stretching the bar is 
 
 S* 
 known as the Resilience of the bar, and the ratio, 7, is known as 
 
 the Modulus, or Coefficient, of Resilience, in which S is the stress 
 at the elastic limit of the material. 
 
 It is thus seen that the resilience of a bar is found by multiplying 
 its volume by the coefficient of resilience. 
 
 EXAMPLE I. A steel bar 10 feet long and 2 inches in diameter is 
 stretched to the elastic limit ; what is its resilience ? 
 
 Solution: The elastic limit of steel in tension is 50,000 pounds 
 per square inch, and the coefficient of elasticity 30,000,000 pounds 
 per square inch. 
 
 Then, Resilience = X 3.1416 X 10 = 1309 ft.-lbs. 
 
 81. Resilience from Sudden Loads or Shocks. If a bar be sub- 
 
 jected to a sudden load, P, that is, a load of the same intensity 
 
 throughout the period of the elongation, y, occasioned in the bar, 
 
 the external work is Py. The elongation, y, produces an internal 
 
 22 
 
338 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 stress, R, which increases gradually from. to R with a mean value 
 of -. The internal work is then -J^, and since this must equal the 
 
 external work, we find that R = 2P, and, therefore, the work done 
 on the bar is 2Py. Hence, the resilience due to a suddenly applied 
 load is just four times as much as that due to the same load applied 
 gradually. 
 
 82. Resilience of Beams. The work performed in bending a 
 beam to the maximum deflection within the elastic limit is the 
 resilience of the beam. 
 
 If the load be concentrated at one point, the resilience is the 
 product of half the load into the deflection. 
 
 Thus, if P denotes the load concentrated at the extremity of a 
 cantilever that will cause the maximum deflection within the elastic 
 
 cr T 
 
 limit, the maximum bending moment will be PL = , whence 
 
 P SI 
 
 = Lc' 
 
 pj'A &T i 
 
 The maximum deflection is 
 
 SI SU S* LI S 2 K 2 AJ 
 Consequently, Resilience = ^. 3^ = O ' 3? = 2#' W ' AL > 
 
 in which AK 2 is substituted for I; see Art. 45, p. 306. 
 
 For rectangular sections, c = ~ and K 2 . = -~ = y-g, so that 
 
 Then, for cantilevers of rectangular section and with concen- 
 trated load at the free end, we have : 
 
 Resilience =,., (B) 
 
 or, it is the product of the coefficient of resilience and one-ninth the 
 volume of the beam. 
 
 83. With a simple beam having a concentrated load at the middle, 
 
 ~P T Sf T 
 
 the maximum bending moment is -^- = -, whence P = 
 
 PL 3 SL S 
 The maximum deflection is 
 
 2SI SL 2 S 2 LI 
 Consequently, Resilience = -- . = . , which is the 
 
 same as that found for the cantilever with load at extremity. 
 
KESILIENCE 339 
 
 Hence, for simple beam with load concentrated at the middle, 
 we have : 
 
 Resilience = ~ . ^ . (C) 
 
 84. If the load be distributed, the half load on each elemental 
 length of the beam must be multiplied by the corresponding deflec- 
 tions and the resilience found from the summation of these products. 
 
 85. If a cantilever be loaded with w pounds per linear unit, the 
 elemental load is wdx 9 and if y be the corresponding deflection, the 
 
 elemental external work is ^ . 
 
 The bending moment for any section distant x from the free 
 
 n . wx 2 
 end is ;5 -. 
 
 A 
 
 Hence, El ^ <u ^j- , 
 
 Integrating once, we have, El -f- = -y -f- C. 
 
 ctx o 
 
 nr o wnen x ^^ LJ .*. C/ ^^ . 
 
 dx 6 
 
 y = when x = .-. C' = 0. 
 
 Then, y = oT^jC x * + 4ZAc), and we shall have: 
 
 7/ .2 f\Jj 
 
 Resilience = -J I ( a; 4 + 4L 3 ^)rfa; 
 
 w a f x* 
 481/L 5 
 
 M max = = , whence w = . 
 Consequently, Resilience = 
 
 7^ 
 
 For rectangular sections,^ = i, hence, 
 
 c 
 
 O2 
 
 Resilience =^. (D) 
 
 for cantilevers of rectangular section and uniform load. 
 
340 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 86. For a simple beam, uniformly loaded with w pounds per 
 linear unit, the elemental load is wdx and the elemental external 
 
 work for the deflection, y, is | 
 
 The bending moment at any section, distant x from the left sup- 
 
 , . wLx wx 1 w ,j 2 , 
 port, is -g- - 2 - = .g (Lx z 2 ). 
 
 Hence, EI = (Lx z 2 ). 
 ax A 
 
 Integrating once, we have, 
 
 y z= when x = . . C' = 0. 
 
 Then, y = n^Lj- (2Lx 5 x* L 3 x), and we have, 
 
 Eesilience = A ^ T I (%Ltf x* Dx^dx 
 
 __ 
 48^7 2 5 2 o 240^7 
 
 ,, -o .,. L S 2 8 K* AT 
 
 Consequently, Eesilience -= = . . - .AL. 
 
 For rectangular sections, y- = J, hence 
 ' Resilience =. 
 
 87. Comparing equations (B), (C), (D), and (E) with equation 
 (A) we observe that, for rectangular sections, the resilience of a bar 
 in tension is 9 times as great as that of a cantilever loaded at its free 
 end, 9 times as great as that of a simple beam loaded at the middle, 
 10 times that of a cantilever uniformly loaded, and 5f times that of 
 a simple beam uniformly loaded. 
 
 88. Resilience of Shafts under Torsion. If a shaft be twisted 
 by a force acting at right angles to the axis, each elemental area of 
 
EESILIENCE 341 
 
 its cross-section is subjected to a shearing stress which varies 
 directly as the distance of the elemental area from the axis. 
 
 The stress, R, on an elemental area, a, gradually increases from 
 
 p 
 to R, its mean value being -_ . 
 
 The angle through which the longitudinal elements are twisted 
 varies directly as their distance from the axis (Fig. 63). If tf 
 denotes the angle through .which the elements distant r from the 
 axis are twisted, then the displacement, y, of an elemental area, a, 
 for a length, dx, is dx tan & ; or since ff is very small, we shall have 
 y dxtf. 
 
 Within the elastic limit we have : 
 
 ? 5- 
 
 Then, Elemental Work = ^ay = ^a dxd' = . (1) 
 
 " -v &EJ 
 
 If 8 denotes the unit shearing stress at the remotest elemental 
 area, distant c from the axis, then is the unit shearing stress at 
 a unif s distance from the axis, and is the unit shearing stress 
 at the elemental area distant r from the axis. Hence, R = ; 
 
 and in like manner 7^ = ' , ^ = ^ , &c. 
 
 c c 
 
 Substituting in (1), we have: 
 
 Elemental Work = ^ ~ '~' dx ~ ' ~ ' Adx ' since 
 
 2JJJ ' ~<? ' 
 
 + a i r l + & c - = IP polar moment of inertia, and I p = AK 2 . 
 Integrating this over the whole length of the shaft, we have: 
 
 S* K* 
 Work Performed = Eesilience of Torsion = ~-. -^ . AL. 
 
 . , D , ^ I nD* ^Z) 2 Z> 3 
 
 For circular sections, c=^ and A 3 = = - -5 = , 
 
 so that ~ = \. 
 
342 THE ELEMENTS OF THE MECHANICS OF MATERIALS 
 
 Then, for shafts of circular sections, we have : 
 Resilience of Torsion = ^ . ^- 9 or, it is the product of the 
 coefficient of resilience and one-fourth the volume of the shaft. 
 
 89. The determination of the resilience of any given material is 
 regarded as the best measure of its capacity to withstand shocks. 
 
 PROBLEMS. 
 
 63. What work is required to subject a steel bar to a tensile 
 stress of 20,000 pounds per square inch 200 times a minute, the 
 length of the bar being 10 feet and its diameter 2 inches ? 
 
 Ans. 1.269 H. P. 
 
 64. For a rolled steel beam, symmetrical about the neutral axis, 
 the moment of inertia is 72 inch units. The beam is 8 inches deep, 
 is laid across an opening of 10 feet, and carries an evenly distrib- 
 uted load of 9 tons. Find the maximum fiber stress, also the central 
 deflection, taking E as 26,000,000. 
 
 Ans. 7.5 tons, and 0.216 inch. 
 
 65. A 10-inch Cambria I beam, weighing 25 Ibs. per ft. is laid 
 across an opening of 16 ft. In addition to a concentrated load of 
 1000 Ibs. at the middle it bears a uniformly distributed load of 
 14,000 Ibs. Find the maximum fiber stress and the maximum de- 
 flection, taking E as 29,000,000. Ans. 16,130 Ibs., 0.416 inch. 
 
 66. A beam of wood, 8 inches wide and 12 inches deep, and fixed 
 at the ends, covers a span of 14 ft., and bears, in addition to a con- 
 centrated load of 6000 Ibs. at the middle, a uniformly distributed 
 load of 5000 Ibs. Find the maximum fiber stress and the maximum 
 deflection, taking E as 1,500,000. Ans. 1020 Ibs., 0.128 inch. 
 
PART V 
 ELEMENTS OF GRAPHIC STATICS 
 
CHAPTEE I. 
 
 BOW'S SYSTEM OF LETTERING. FORCE DIAGRAM. 
 FUNICULAR POLYGON. 
 
 1. Graphic Statics. The methods by which static problems are 
 solved by means of scale drawings constitute Graphic Statics. 
 
 In very many cases, the determination by calculation of the 
 forces transmitted through the different parts of a structure involve 
 tedious and difficult processes,, with the consequent liability to error, 
 and, in the end the effort to check the accuracy of the results occa- 
 sions a procedure as tedious as the processes themselves. By the 
 graphic method, however, solutions are readily obtained, and the 
 process itself furnishes a check as to accuracy. 
 
 2. Bow's System of Lettering. To the method of lettering dia- 
 grams, devised by Mr. A. H. Bow, is due much of the facility in 
 making graphic solutions. 
 
 The two important features of the Bow system of lettering are: 
 (1) The placing of a letter in each of the spaces between the lines 
 of action of the external forces; (2) the naming in clockwise order 
 of each force by the two letters flanking it. 
 
 The forces of Fig. 1 (a) will serve to illustrate the Bow system. 
 The five forces are in equilibrium, and the letters A, B, C, D, and E 
 are placed in the spaces between them. Other letters, or even 
 numbers, would serve the purpose, and they might be placed in any 
 order, but it will be found convenient to commence at the left and 
 letter the spaces alphabetically. The force of 20 Ibs. is flanked 
 by the letters A and B and is known as the force AB, not as BA, 
 since the letters must be read clockwise. In like manner the re- 
 maining forces are known as BC, CD, DE, and EA. 
 
 3. Force Diagram. Taking the forces in clockwise order, and 
 denoting them by the corresponding small letters of the alphabet, 
 we may represent them in magnitude and direction in a diagram 
 by lines drawn to some chosen scale. Such a diagram is known as 
 a force diagram. 
 
346 ELEMENTS OF GRAPHIC STATICS 
 
 Thus, A being the first letter in clockwise order from the left in 
 Fig. 1 (a), the letter a will be the starting point of the force dia- 
 gram of Fig. 1 (6), and since the force AB acts downward, ab, to 
 the scale of 20 pounds to the inch, will represent it. The force BC, 
 next in clockwise order, and equal to 40 Ibs., acts upward and will 
 be denoted by be, two inches in length and measured upward. In 
 like manner, cd, measured downward and one-half inch in length, 
 denotes the force, CD, of 10 Ibs.; de, measured upward and one- 
 quarter inch in length, denotes the force, DE, of 5 Ibs. ; and, finally, 
 ea, which is found to measure three-quarters of an inch, properly 
 
 / 
 
 20/6s 
 
 ? \ B 
 
 Tc 
 
 tO/bs. 
 
 \ 
 
 \ & 
 
 
 c 
 e 
 d 
 
 L 
 
 /&s. 
 
 
 ^/ o / 
 
 
 \ 
 
 Slbs 
 
 a 
 
 
 
 i y 
 
 (a) 
 
 
 
 
 
 
 
 
 
 (d) 
 
 
 
 
 
 
 6 
 
 denotes the force, EA, of 15 Ibs. It will be noted that the force EA 
 is of just sufficient magnitude to fill, when drawn to scale, the space 
 between e and a and thus close the force diagram. This force dia- 
 gram is, in reality, a closed polygon which has resolved itself into 
 a straight line in consequence of all the forces being vertical. Had 
 it not closed, the system being in equilibrium, there would have 
 been an error in the construction. 
 
 With a correct construction, and the last point of a force diagram 
 not falling on the first point, would indicate a resultant force, equal 
 in magnitude to the scale distance between the last point and the 
 first point, and acting upward or downward according as the last 
 point had fallen above or below the first point. 
 
 4. If one of the forces of Fig. 1 (a) were unknown it could be 
 found by means of the force diagram. Suppose the force BO un- 
 
Bow's SYSTEM OF LETTERING. FORCE DIAGRAM 347 
 
 known. Then, commencing with the force CD, the force diagram 
 would be constructed by representing in clockwise order the forces 
 CD, DE, EA, and AB by the scale distances cd, de, ea, and ab, 
 respectively, of Fig. 1 (&), c being the first point of the diagram 
 and I the last thus determined. The missing force must then be 
 represented by ~bc, measured upward, and as it measures 2 inches, 
 it correctly does so for the force, BC, of 40 Ibs. 
 
 5. The system of lettering enables the resultant of any number 
 of the forces to be read at once from the force diagram. 
 
 Suppose the resultant of the forces EC, CD, and DE were re- 
 quired. The first and last letter in the naming of these forces are 
 B and E, respectively, so that be on the force diagram represents 
 the required resultant in magnitude and direction. The measure- 
 
 ment of be is found to be 1.75 inches, which, to the scale, repre- 
 sents 35 Ibs., the resultant of the forces 40 + 5 10, and it acts 
 upward since be is measured upward. 
 
 6. Should the system of forces act on an open framed structure, 
 such as the roof truss shown in Fig. 2, a letter must be placed 
 within each open space of the frame in addition to those placed in 
 the spaces between the external forces. The external forces, by 
 Bow's notation, are known as A B, BC, CD, DE, and EA. The 
 stress in the member connecting joints 1 and 2 is known as BG. 
 The stress in the member separating the spaces lettered F and G 
 may be known as GF or as FG according to the end of the member 
 under consideration. If the end considered is that at joint 1, then 
 the stress in the member is known as GF, because the forces about 
 that joint, taken in clockwise order, are AB, B G, GF, and FA. If 
 the other end of the member is under consideration, the stress is 
 known as FG, because the forces about the joint at that end, in 
 clockwise order, are known as FG, GH, HI, IE, and EF. 
 
348 ELEMENTS OF GRAPHIC STATICS 
 
 7. Funicular Polygon. Suppose a jointed frame, Fig. 3, to be 
 acted upon at its hinged joints by a system of forces in equilibrium, 
 the members, bars, or links of the frame being free to adjust them- 
 selves to the best position for withstanding the action of the forces. 
 
 Such a figure is called a funicular polygon, the word funicular 
 having no mechanical significance. Since the system is in equilib- 
 rium, the funicular polygon must, of course, be a closed polygon. 
 
 The equilibrium is occasioned by the balance between the internal 
 forces, or stresses, in the members and the external forces, and the 
 whole being in equilibrium, each joint in itself is in equilibrium. 
 
 Since the equilibrium at each joint is the result of the action of 
 three concurring forces, viz., the external force at the joint and the 
 stresses set up in the two members, it follows that a triangle may 
 
 be constructed for each joint which will represent these forces in 
 magnitude and direction. 
 
 If, for example, the force AB be known, the equilibrium of the 
 joint at which it is applied is maintained by the action of the force 
 AB and the stress forces in the members AO and BO. Draw ab 
 parallel to the force AB, and make its length, to some chosen scale, 
 represent the magnitude of AB. From a and b draw lines parallel 
 to AO and BO, respectively, intersecting at o. Then abo is the tri- 
 angle of forces for the joint ABO, and bo and oa represent not only 
 the directions of the stresses in the members BO and OA, but their 
 magnitudes as well, to the same scale that ab represents the force 
 AB. The stress bo in the end of the member BO at which AB is 
 applied occasions an equal and opposite stress ob at the other end 
 of BO, and, in fact, such is the case in all of the members of the 
 polygon, for in no other way could the equilibrium be maintained. 
 Taking the next joint in clockwise order, we have the external force 
 BC and the stress forces CO and OB in equilibrium. But ob 
 
Bow's SYSTEM OF LETTERING. FORCE DIAGRAM 349 
 
 has just been found to represent in magnitude and direction the 
 stress OB. Hence, by drawing from b and o lines parallel respec- 
 tively to BC and CO, intersecting at c, we shall have bco as the tri- 
 angle of forces for the joint BCO, and be and co will represent in 
 magnitude and direction the force BC and the stress in the member 
 CO, respectively. We now know the stress oc at the joint CDO, 
 and can construct the triangle of forces, cdo, for that joint. In 
 like manner we can proceed and determine all the external forces 
 and. stresses in the members, the last line, fa, closing the diagram, 
 thus proving that the system is in equilibrium. 
 
 8. It will be observed that the sides of the polygon just con- 
 structed represent the external forces of the funicular polygon, 
 and therefore abcdef is called the force diagram. Furthermore, 
 all the lines representing the stresses in the members of the funi- 
 cular polygon meet at a point called the pole. These stress lines 
 are known as vectors. 
 
 9. From what has preceded, it is seen that if all the external 
 forces that are applied to a funicular polygon are known in mag- 
 nitude and direction, and also the directions of two of its members, 
 the force polygon can be drawn. For, the force diagram can be 
 drawn from the known forces, and the intersection of the vectors 
 parallel to the known directions of the two members gives the pole. 
 The directions of the remaining members are then found by draw- 
 ing the rest of the vectors. 
 
 10. A force diagram of any system of forces in equilibrium can 
 be drawn, and by choosing any pole, a funicular polygon with respect 
 to .that pole can then be constructed to which the forces may be 
 applied. 
 
 11. The closed polygon of Fig. 4, having the forces AB, BC, CD, 
 DE, EF, and FA applied to its joints, has been drawn -to illustrate 
 some of the properties of the funicular polygon. Draw the force 
 diagram abcdef. The triangle of forces abo for the joint ABO de- 
 termines the pole, and the other vectors may then be drawn. 
 
 Let yy be any section of the polygon. The forces to the right 
 of the section are BC and CD and, by the triangle of forces, their 
 resultant is bd, acting from b to d. The forces on the left of the 
 section are DE, EF, FA, and AB, and, by the polygon of forces, 
 their resultant is db, acting from d to b. Hence, the resultant 6f 
 
 F THE 
 
350 
 
 ELEMENTS OF GKAPHIC STATICS 
 
 the forces on one side of a section has the same magnitude and line 
 of action as the resultant of the forces on the other side, but acting 
 in opposite directions to preserve equilibrium. 
 
 To find where this resultant acts, we replace the forces BC and 
 CD by their resultant Id, so that our force diagram now becomes 
 bdefa. The vectors od, oe, of, oa, and ob have o as their pole, 
 therefore a funicular polygon can be drawn having its sides par- 
 allel to these vectors. We have already OD, OE, OF, OA, and OB 
 parallel, respectively, to these vectors; but since a funicular poly- 
 gon must close, and since there must be a. force acting at each joint, 
 we produce OB and OD until they intersect, and at the joint thus 
 formed the resultant bd must act. 
 
 12. An inspection of the funicular polygon and the force dia- 
 gram shows: 
 
 (a) The resultant of the forces DE, EF, FA, and AB to the left 
 of the section is given in the force diagram by db, the first and last 
 letters of the forces when named in clockwise order; in like man- 
 ner the resultant of the forces BC and CD to the right of the sec- 
 tion is bd. 
 
 (b) The letters in the force diagram which give the resultant 
 also name the members of the funicular polygon which have to be 
 produced to intersection in order to get a point in the line of action 
 of the resultant. 
 
 Thus, the resultant of the forces DE, EF, and FA is da in mag- 
 nitude and direction, and producing the members D and A of the 
 funicular polygon to their point of intersection we get a point in 
 the line of action of this resultant. 
 
Bow's SYSTEM OF LETTERING. FORCE DIAGRAM 351 
 
 13. To make a practical application of the funicular polygon, 
 take the beam of Fig. 22, page 289,, reproduced in Fig. 5. The 
 linear scale is 1" = 4', and the load scale is 1" =150 Ibs., so that 
 Wi = 60 Ibs. = T 6 -gV 0.4 inch to scale, W 2 = 45 Ibs. = T 4 ^ = 0.3 
 inch, and W s = 90 Ibs. = -ff^ = 0.6 inch. Letter the beam ac- 
 
 cording to the Bow system. Then, W ly W 2 , W 3 , R 2 , and R will 
 be known as AB, BC, CD, DE, and EA, respectively. 
 
 To construct the force diagram, we set off vertically down- 
 ward, ab equal to 0.4 inch to represent W 19 or AB, to scale; ~bc 
 equal to 0.3 inch to represent W 2 , or BC ' ; and cd equal 0.6 inch to 
 represent W 3 , or CD. Then we know that da is the closing line 
 of the force diagram, all the forces being vertical, and that it rep- 
 resents the sum of the reactions R 2 and R 19 but we do not know the 
 proportion of the load borne by each support. 
 
 To find R! and R 2 , we must construct the funicular polygon. 
 
352 ELEMENTS OF GRAPHIC STATICS 
 
 Select at random some point o, distant oh from ad, and draw the 
 vectors oa, ob, oc, and od. From some point j in the line of action 
 of R 19 draw a line parallel to the vector ao, and from its point of 
 intersection, k, with the vertical from W draw a line parallel to 
 the vector bo, and from its point of intersection, f, with the line of 
 action of W 2 draw a line parallel to vector co, and from its point 
 of intersection, ra, with the line of action of W s draw a line par- 
 allel to the vector do. This last line intersects the line of action 
 of R 2 at n. Join n with /, and we have the funicular polygon 
 jkfmnj, with the external forces of the beam acting at its joints. 
 This funicular polygon has five sides, while there are but four vec- 
 tors in the force polygon, and since there must be a vector for each 
 side of the funicular polygon, the vector oe, parallel to nj, must 
 be drawn. The point e is thus determined, and de represents in 
 magnitude and direction the support reaction DE =-R 2 to the scale 
 adopted ; and in like manner ea represents in magnitude and direc- 
 tion the support reaction EA = R^. By measurement de is |-J- 
 inch, which, reduced to scale, equals X 150 = 127.5 Ibs. = R 2 ; 
 and ca measures f-J, which, to scale, equals fj X 150 =67.5 Ibs. 
 = RI. These are the same values found before for R and R 2 . 
 
 It will not be necessary to letter the funicular polygon in order 
 to know the names of the members, as an examination of the force 
 polygon at once discloses them. Thus, the member parallel to ao 
 is AO, the one parallel to bo is BO, and so on. The names of the 
 members may also be determined from the lettering of the beam, 
 since the length of each member is terminated. by the lines of action 
 of two forces of the beam. For example, the member Jef is ter- 
 minated by the lines of action of W and W 2 , and since the letter 
 B appears between W and W z , the member kf is named BO. 
 
 14. The Funicular Polygon a Bending-Moment Diagram. Con- 
 sider any section x of the beam of Fig. 5. Produce the members 
 jk and kf until they intersect the vertical through x at p and q, 
 respectively. Draw the horizontal lines rs, tu, and vw, and regard 
 them as the altitudes of the triangles jpz, kpq, and fqy, respectively. 
 All the triangles of the force polygon have the same altitude oh. 
 The triangles jpz and oea are similar, having their sides mutually 
 parallel. For the same reason the triangle kpq is similar to the 
 triangle oba, and triangle fqy is similar to the triangle ocb. Hence, 
 we shall have : 
 
Bow's SYSTEM OF LETTERING. FORCE DIAGRAM 353 
 
 ffi = * = 7^' whence RiXrs^pzXoh; 
 oJi = ^l = W' whence w iXtu = pqX oli ; and 
 ^= |^ = |r, whence F 2 X tw = ^ X oft. 
 The bending moment, M, at section x, is, 
 
 M = R^X rsW 1 XtuW 2 Xvw 
 
 pzX ohpq X oh qyXoh oh[pz (pq + gy)] 
 
 = o/t X 2/2- 
 
 Hence the bending moment at any section of the beam is equal 
 to the product of the polar distance, oh, and the ordinate of the 
 funicular polygon at the section. Thus, the ordinate g measures 
 |-{j- inch, and oh measures 1 inch = 150 Ibs. to scale; the bending 
 moment at the beam section above g is then, 150 Ibs. X IT X 4 ft. 
 = 217.5 lbs.-ft,, the same as was found on page 290. 
 
 15. The forces W 19 W 2 , and W 3 are, by the Bow system, known 
 as AB, BC, and CD, and their resultant, ad, acts through i (see 
 Art, 11). 
 
 16. The shear diagram can readily be constructed by projection 
 from the force diagram. 
 
 The shear at any section between the left support and W is 
 R^ = ea. This shear is plotted by projecting ea horizontally, as 
 shown. The shear at any section between W^ and W 2 is R^ W 
 = ea ab = eb, and this shear is plotted by projecting e~b. The 
 shear at any section between W 2 and W 3 is R W t W 2 = ea 
 - ab ~bc ec, and the shear at any section between W 3 and 
 the right support is R 1 W^ W 2 W 3 = ea ab be cd 
 = ed = R 2 . Projecting these two shears, the diagram is 
 completed. 
 
 17. It has been stated that the pole, o, may be selected at ran- 
 dom, but if it be so selected that its distance from the load line be 
 some decimal multiple of pounds or of tons, as the case may be, 
 then a bending moment scale may readily be obtained that will 
 enable the bending moment to be measured direct from the ordinate 
 of the diagram, thus saving the necessity of multiplying each meas- 
 urement by the polar distance. The pole, o, of Fig. 5 was taken at 
 a distance of 1. inch from the load line, ab, the polar distance, oh, 
 
354 
 
 ELEMENTS OF GRAPHIC STATICS 
 
 therefore representing 150 Ibs. Then, since the linear scale is 
 1" 4', the bending moment scale will be I" 4' X 150 Ibs. = 600 
 
 21 75 
 Ibs.-feet, or -fa inch = 10 Ibs.-feet. The ordinate g measures -- 
 
 f 8 
 \ A 
 
 inch, and the bending moment at that section of the beam is 217.5 
 Ibs.-feet. 
 
 EXAMPLE I. A beam, supported at the ends, is 20 feet long, 
 weighs 400 Ibs., and has concentrated loads of 360 Ibs. and 440 Ibs. 
 at 8 feet from the left end and 4 feet from the right end, respec- 
 
Bow's SYSTEM OF LETTERING. FORCE DIAGRAM 355 
 
 tively. Draw the bending-moment and shear diagrams, and meas- 
 ure the bending moments under the concentrated loads, and the 
 shear stress at the middle of the beam. 
 
 Select a linear scale of -J inch = 1 foot, and a load scale of 
 1 inch = 400 Ibs. 
 
 Set out the beam as shown in Fig. 6, and since the beam weighs 
 400 Ibs. it has, in addition to the concentrated loads, a uniformly 
 distributed load of 20 Ibs. per foot. 
 
 We shall first construct the funicular polygon for the concen- 
 trated loads, neglecting, for the present, the uniformly distributed 
 load. 
 
 Set off the load line ac by making ab = f $ = 0.9 inch, and 
 be = |-$ = l.l inches. Select the pole, o, at a distance of 1.25 
 inches from ac, so that the polar distance will represent 1.25 X 400 
 = 500 Ibs. We shall then have for the bending-moment scale, 
 J inch = 1 ft. X 500 Ibs. = 500 lbs.-ft, or T V inch = 100 Ibs.-ft. 
 
 Draw the vectors ao, ~bo, and co. From a point 1 in the line of 
 action of R^ draw a parallel to ao and produce it until it intersects 
 the line of action of W at the point 2. From 2 draw a parallel to 
 fro, producing it until it intersects the line of action of W 2 at the 
 point 3. From 3 draw a line parallel to co, intersecting the line 
 of action of R 2 at the point 4. Join 4 with 1 as the closing mem- 
 ber of the funicular polygon 1234. Draw od parallel to 4--1. 
 Then the polygon 1234 is the diagram of the bending moments, and 
 cd and da the support reactions due to the concentrated loads, W l 
 andF 2 . 
 
 The distributed load equals 20 X 20 = 400 Ibs. = - = * incn 
 to scale, and one-half of this is to be borne by each support. 
 
 The bending-moment diagram for the distributed load is of para- 
 bolic form, and will be placed above the diagram for the concen- 
 trated loads, so that the sum of the ordinates of the two diagrams 
 under any section of the beam will measure the bending moment at 
 the section. 
 
 Under the supposition that the whole of the distributed load of 
 400 Ibs. is concentrated at the middle point of the beam, the funicu- 
 lar polygon for this load must be constructed on the closing member 
 4--1 of the polygon 1234. To do so we will use the same pole, o, 
 that was used for the funicular polygon of the concentrated loads. 
 Lay off de and df, each one-half inch in length to represent the 
 halves of the distributed load. Draw the vectors oe and of. From 
 
356 ELEMENTS QF GRAPHIC STATICS 
 
 1 and 4 draw parallels to of and oe, respectively, intersecting at s. 
 Then s!4 is the funicular polygon for the distributed load, and the 
 ordinate st measures the bending moment at the middle point of 
 the beam due to the distributed load, supposing it all to be concen- 
 trated at the middle point. It can readily be shown that the bend- 
 ing moment at the middle of a simple beam, for a uniformly dis- 
 tributed load, is only half that due to the same load when concen- 
 trated at the middle. Hence, the ordinate 5t measures the bending 
 moment at the middle of the beam due to the distributed load, the 
 point 5 being the middle point of st. A parabola constructed on 1--4 
 as a chord, and passing through the points 1, 5, 4, is the bending 
 moment diagram due to the uniformly distributed load, 
 
 The complete bending-moment diagram for the beam is 12345. 
 
 The reaction, R,,, measured on the load line, is cd -(- df f-J- 
 inches; hence, R 2 = f-J X 400 = 696 Ibs. The reaction R^ = da 
 + de || inches; hence, R = } X 400 = 504 Ibs. 
 
 The ordinates y and y', of the bending-moment diagram, under 
 W and W 2 measure f inch and f -- inch, respectively. The bend- 
 ing moments under W and under W 2 are, therefore, 3400 Ibs.-ft. 
 and 2600 Ibs.-ft., respectively. These results may easily be checked 
 by calculation. 
 
 The shear at the left support due to the uniformly distributed 
 load is equal to half that load, and is positive because of its clockwise 
 iendency. This shear gradually decreases from left to right until, 
 .at the middle, it becomes zero, and at the right support it again 
 becomes equal to half the load, but negative owing to its contra- 
 'dockwise tendency. The diagram d'e'f'd" therefore represents the 
 :shear due to the distributed load. 
 
 The total shear at the left support is R^ and is equal to da + de. 
 'The parallel to e'f shows the gradual decrease of the shear from 
 the left support to W l due to the uniform load. Passing W^ the 
 shear suddenly drops to n and becomes negative, m'n being equal 
 to db. np parallel to e'f shows the gradual increase in the shear 
 from W to W 2 due to the distributed load. Passing W 2 the sheai 
 suddenly drops to q, pq being equal to be. qv parallel to e'f shows 
 the further increase in the shear due to the distributed load until, 
 at v, it becomes R 2 = (dc + df). The ordinate, uw, at the 
 middle of the beam measures -$ inch ; the shear at the middle sec- 
 tion is, therefore, ^ X 400 = 56 Ibs. 
 
Bow's SYSTEM OF LETTERING. FORCE DIAGRAM: 357 
 
 fe /r 
 
 18. The example of Fig. 43, page 300, of a beam with overhang- 
 ing ends, loaded uniformly with 20 Ibs. per foot and with two con- 
 centrated loads, may readily be solved by means of the funicular 
 polygon. 
 
 Set off the beam as shown in Fig. 7. Divide the beam into any 
 
358 ELEMENTS OF GRAPHIC STATICS 
 
 convenient number of parts, eight in this instance, so that each 
 part will be 4 feet in length, and will bear 80 Ibs. of the distributed 
 load. These eight parts constitute as many external forces, each 
 acting at its center of gravity. 
 
 Letter the beam according to the Bow system, and adopt the fol- 
 lowing scales : Linear, y 1 ^- inch = 1 foot ; load, 1 inch = 400 Ibs. 
 We shall then have: W^ = = 0.5 inch, W 2 = = 1 inch, 
 and each of the equal divisions will be -ffo = 0.2 inch. 
 
 Set off the load line, dk, accordingly. Select a pole, o, distant 
 1.5 inches from the load line, so that the polar distance will repre- 
 sent 1.5 X 400 = 600 Ibs. We shall then have : T V inch = 1' 
 X 600 Ibs. = 600 Ibs.-f t., or -fa inch = 100 Ibs.-ft. for the bending- 
 moment scale. 
 
 Draw the vectors of the force polygon. From some point m in 
 the line of action of E^ draw a parallel to ao, and produce it until 
 it intersects the line of action of AB at n. Through n draw a 
 parallel to &o, producing it to its intersection, p, with the line of 
 action of BC, or of W^. Draw parallels to the other vectors as 
 shown. The member qr, parallel to oj, intersects the line of action 
 of JK at r. The parallel to ok through r intersects the line of 
 action of R 2 at s. The closing member is then sm, and mnp- --qrs 
 is the funicular polygon, or the bending-moment diagram of the 
 beam. 
 
 Draw ol parallel to sm. Then, la and Tel represent to scale the 
 reactions R^ and R 2 , respectively. 
 
 It should be noted that ordinates within mnpt and sur give nega- 
 tive bending moments, and that t and u are the points of inflection. 
 
 The ordinates y and y' give the maximum negative and positive 
 
 144 . 
 bending moments, respectively. These ordinates measure ^ inch 
 
 and $ inch, so that the bending moments are 1440 and 1600 Ibs.-ft. 
 The shear of the left overhang is negative because of its contra- 
 clockwise tendency. Commencing at the left end of the beam, the 
 shear increases from zero to ab when TF is reached. Passing 
 TFj the shear becomes ac, and increases up to the left support, 
 where it becomes ad. Passing the left support it becomes 
 ft^ ad = Id and decreases up to TF 2 where it becomes Id dg 
 = Ig. Passing W 2 it becomes Ig gh = lh, and gradually in- 
 creases until the right support is reached, where it becomes 
 lh Jiz lz (there are but one and one-half divisions of the 
 
Bow's SYSTEM OF LETTERING. FORCE DIAGRAM 
 
 359 
 
 beam between W 2 and the right support). Passing the right sup- 
 port the shear again becomes positive and equals R 2 Iz = zk, 
 and then decreases to zero at the right end of the beam. 
 
 19. To draw the bending-moment and shear diagrams of the 
 cantilever with concentrated loads as shown in Fig. 8, we proceed as 
 follows : 
 
 Draw the load line ad, making ab, be, and cd equal, to some 
 selected scale, to the loads AB, EC, and CD, respectively. Select 
 
 some pole, o, and draw the vectors ao, bo, co, and do. From some 
 point m in the support line, draw a parallel to ao, producing it until 
 it intersects the line of action of AB at n. From n draw a parallel 
 to bo until it intersects the line of action of EG at p. From p draw 
 a parallel to co until it intersects the line of action of CD at q. 
 From q draw a parallel to do to meet the line of support at r. Then, 
 mnpqr is the bending-moment diagram of the cantilever, and the 
 ordinate under any section of the beam is the measure of the bend- 
 ing moment at the section. The maximum bending moment is, 
 of course, at the wall. 
 
 The shear diagram is drawn by projection from the load line and 
 presents no difficulties. 
 
360 
 
 ELEMENTS OF GRAPHIC STATICS 
 
 20. In the case of a cantilever it is found convenient to select the 
 pole at some chosen perpendicular distance from either extremity of 
 the load line. In Fig. 8 the pole, o, might just as well have been 
 chosen at o' and an equal bending-moment diagram, mrip'q'r', con- 
 structed, as shown by the dotted lines. 
 
 21. The example of Fig. 30, page 295, of a cantilever with a com- 
 bination of concentrated and uniformly distributed loads may 
 easily be solved by means of the funicular polygon. 
 
 Using the same linear scale of -f^ inch = I ft. as on page 294, 
 
 44lbs 
 
 . 9. 
 
 and a load scale of 400 Ibs. to the inch, we sliall first construct the 
 funicular polygon -for the concentrated loads. 
 
 Reduced to scale, AE = -fa = T W inch, and EG = ^ = T fo 
 inch. Set off the load line ac, Fig. 9, making ab and be equal to 
 yVg- inch and T ^ inch, respectively. 
 
 Choose a pole, o, at a perpendicular distance of one inch from c y 
 so that the polar distance, oc, represents 400 Ibs. We shall then 
 have for the bending-moment scale, -$ inch = 1 ft. X 400 Ibs. 
 = 400 lbs.-ft., whence -gV inch 40 Ibs.-ft. Draw the vectors ao, 
 bo, and co. 
 
 Commencing at some point, m, in the wall line, draw parallels 
 to ao, bo, and co in the usual manner, thus forming the funicular 
 
Bow's SYSTEM OF LETTERING. FORCE DIAGRAM 361 
 
 polygon, or bending-moment diagram, mnpq, for the concentrated 
 loads. 
 
 The cantilever is uniformly loaded with 48 Ibs. per foot for a 
 distance of 4 feet from the wall, making a total uniform load of 
 192 Ibs. Assuming this load concentrated at the outer extremity 
 of the 4 feet, set off cd equal to %$ f = f incn to represent it. 
 Draw the vector do, and from r, the intersection of pq with the 
 vertical at 4 feet from the wall, draw rt parallel to do. Since the 
 bending moment due to a concentrated load at the end of a canti- 
 lever is twice that due to the same load uniformly distributed, the 
 distance, qt, must be bisected at s, and the parabola having its apex 
 at r, and passing through s, gives rsq as the bending-moment dia- 
 gram due to the distributed load. Then, mnprs is the complete 
 bending-moment diagram for the cantilever. The ordinate, ms, 
 
 25 3 
 
 at the wall measures "-RTT- inch, and the bending moment is, there- 
 fore, 25.3 X 40 = 1012 lbs.-ft., the same as found on page 294. 
 
 The shear diagram is drawn by projection from the load line ; 
 and presents no difficulties. 
 
 PEOBLEMS. 
 
 1. A beam, 22 feet long, supports a load of 1000 Ibs. at a point 
 6 feet from the left end and one of 1200 Ibs. at 5 feet from the 
 right end. In addition, there is a uniformly distributed load of 
 100 Ibs. per foot. Construct the shear and bending-moment dia- 
 grams. What concentrated load must be placed at the middle of a 
 similar beam in order that the maximum B. M. shall be the same as 
 that of the given beam? Ans. 2190 Ibs. 
 
 2. A beam 30 feet long weighs 20 Ibs. per foot and overhangs 
 each support 6 feet. It bears a superimposed load of 100 Ibs. per 
 foot, and a load of 1400 Ibs. concentrated at a point 3 feet from 
 the right of the middle. Construct the bending-moment and shear 
 diagrams, and find graphically the bending moment at the danger- 
 ous section and the distances of the points of inflection from the 
 left support. Ans. 7760 lbs.-ft; 1.48 ft. and 16.9 ft. 
 
 3. A cantilever, 14 ft. long, supports three concentrated loads 
 500 Ibs. at 4 ft. from the wall, 600 Ibs. at 10 ft. from the wall, and 
 200 Ibs. at the extremity. In addition, it bears a uniformly dis- 
 tributed load of 80 Ibs. per foot run. Construct the bending-mo- 
 ment and shear diagrams. 
 
CHAPTEE II. 
 FRAMED STRUCTURES. RECIPROCAL DIAGRAM. 
 
 22. Framed Structures. A framed structure is an assemblage of 
 members for the transmission or modification of external forces, 
 the internal stresses occasioned thereby in the members being prin- 
 cipally those of tension and compression. A member in tension is 
 known as a tie ; if in compression, it is known as a strut. 
 
 23. A " frame " is a theoretical structure, the joints connecting 
 its members being supposed frictionless. There are, of course, no 
 such things as frames, since all joints offer some resistance to rota- 
 tion. In general, however, engineering structures approach so 
 nearly to frames that no sensible error results from treating them 
 as such. 
 
 24. The load on a structure consists of the weight of the struc- 
 ture itself and the external forces acting upon it. Wind pressure, 
 and the weight of the covering of the structure and of the snow it 
 may bear, are external forces. 
 
 The load on a roof truss must include, besides its own weight, the 
 weights of roof covering and of the lintels. In addition to this, an 
 allowance for wind and snow must be made. Wind pressure is as- 
 sumed to be uniformly distributed and is reckoned as so many 
 pounds to the square foot normal to the rafters. Knowing the 
 length of the rafters and the space between them, the load due to 
 the wind which each truss is to bear is known. 
 
 Authorities fairly well agree that no roof should be designed 
 to stand a total pressure of less than 40 pounds per square foot. 
 
 25. The reactions on a structure of its supporting foundations 
 are external forces, but they are distinguished from the external 
 forces constituting the load by calling them " supporting forces." 
 
 26. For the equilibrium of a structure, the external forces con- 
 stituting the load must equal the supporting forces, and the exter- 
 nal and internal forces must balance. 
 
 27. In the graphic method of determining the stresses in the 
 members of the framed structures here to be considered, the loads 
 
FRAMED STRUCTURES. RECIPROCAL DIAGRAM 363 
 
 are to be applied at the joints. If the true point of application of a 
 load on a member be at some point intermediate between its joints, 
 parallel forces equivalent to the load will be substituted at the 
 joints. If the point of application of a load be at the middle of a 
 member, the equivalent parallel forces at the joints will each be 
 equal to one-half the load; if the point of application be other 
 than at the middle, then the equivalent parallel forces at the joints 
 may be determined by moments. 
 
 28. Since the joints are to be considered frictionless, or perfectly 
 smooth, the pressures at the points of contact between the mem- 
 bers and a joint act normal to the surface, and their lines of action 
 must, therefore, pass through the center of the joint. The total 
 action of -the joint will then be a single force the resultant of 
 these pressures acting in line of one of the members. This result- 
 ant action in the member exerts an equal and opposite action at the 
 joint at the other end, thus maintaining the equilibrium of the 
 joint. It is evident, then, that a member in tension tends to pull 
 the joints at its ends toward each other, and one in compression 
 tends to separate them. 
 
 29. Reciprocal Diagram. Since there is equilibrium at each 
 joint of a framed structure, each joint center may be considered the 
 point of application of the load on the joint and of the stress forces 
 in its members, all acting in the plane of the frame. It follows, 
 therefore, that a closed polygon may be constructed whose sides 
 represent these forces in magnitude and direction. Such a polygon 
 is known as the Reciprocal Diagram of the joint, and is nothing 
 more nor less than the polygon of forces. Since the stresses in the 
 members of a joint are determined from its reciprocal diagram, 
 such diagram is also called a " stress diagram." 
 
 30. If a joint of a structure be selected at which but two mem- 
 bers meet, and the load on the joint be known, the reciprocal dia- 
 gram of the joint may be drawn by means of the triangle of forces, 
 and the magnitude and direction of the stresses in the members 
 can thus be determined. With this knowledge gained, the recip- 
 rocal of another joint may be drawn, provided that, if more than 
 two members meet at the joint there must be known, in addition to 
 the external forces acting at the joint, the stresses in all the mem- 
 bers except two. 
 
364 
 
 ELEMENTS OF GRAPHIC STATICS 
 
 31. As a preliminary to the construction of the reciprocal dia- 
 grams of the joints of a framed structure, a scale drawing of the 
 structure itself must be carefully made, which is known as the 
 Frame Diagram. 
 
 32. The complete preparation of the frame diagram comprises 
 the following: 
 
 (1) The determination of the magnitude and direction of the 
 total load at each joint, replacing any load that may be applied be- 
 tween two joints by equivalent parallel forces at the joints. These 
 equivalent parallel forces are determined in the same manner as 
 
 that employed in determining the support reactions of a loaded 
 simple beam. 
 
 (2) The determination of the supporting forces, or support re- 
 actions. These reactions can be, and often are, determined from 
 the reciprocal diagram, but 'their predetermination affords a check 
 as to accuracy, and not infrequently furnishes the known external 
 force at a joint where but two members meet, thus providing a 
 starting point for the construction of the reciprocal diagram. 
 
 (3) The correct lettering of the diagram according to the Bow 
 system. 
 
 (4) The marking with arrowheads of all the external forces, giv- 
 ing to each its value, and taking care that the arrowheads do not 
 cross the lines of the frame diagram. 
 
 33. It should be noted that when all the loads and support reac- 
 tions are vertical, loads coming over the supports of the structure 
 
FRAMED STRUCTURES. RECIPROCAL DIAGRAM 365 
 
 are omitted, as they have no influence on the stresses in the mem- 
 bers. Such loads must be deducted from the total support reac- 
 tions in order to obtain the proper values of R^ and R 2 to be used 
 in the determination of the stresses. V 
 
 34. Figure 10 represents a " King Post Truss " with a total uni- 
 form load of 8 tons on one side and 4 tons on the other. The loads 
 are applied to the joints as follows : 
 
 Of the 4 tons on the left hand rafter we may assume one-half of 
 it borne by the member AF and the other half by the member BG. 
 Distributing these two loads of two tons, giving a half of each to 
 the ends of its member, we get a total load of 2 tons at joint 1, 
 one ton at the apex end of the rafter, and one ton over the left 
 support. Proceeding in a precisely similar manner with the right 
 hand rafter, we get a total of 3 tons at the apex of the rafters, 4 
 tons at joint 3, and 2 tons over the right support. Rejecting the 
 loads over the supports, and denoting the span by a, we have : 
 
 E 1 X^ = ^X^ + 3x| + 4x|, whence R 4 tons, 
 
 and# 2 Xa = 4X^ + 3x| + 2x|, whence R 2 5 tons. 
 
 These are the values of R and R 2 to be used in determining the 
 stresses in the members, but the total wall reactions are 5 tons and 
 7 tons for R 1 and R 2 , respectively. 
 
 35. Drawing the Reciprocal Diagram. In drawing the recipro- 
 cal diagram we proceed as follows : 
 
 (1) Select a force scale as large as possible and draw the 
 force diagram of the external forces, marking the beginning and 
 ending of a line representing a force with the small letters of the 
 alphabet corresponding to the capital letters, taken in clockwise 
 order, found in the spaces flanking that force in the frame dia- 
 gram. This diagram represents the magnitudes and directions of 
 the external loads on the joints and of the support reactions, and, 
 if properly drawn, forms a closed polygon. In the most frequent 
 case, that of vertical loads, the polygon resolves itself into a straight 
 line, vertical in direction, as already explained. Determine the 
 magnitude of the support reactions, either by moments or by the 
 method of the funicular polygon. 
 
 (2) Choose a joint at which but two members meet, and resolve 
 the external force acting at the joint along the directions of the 
 
366 
 
 ELEMENTS OF GRAPHIC STATICS 
 
 members, lettering the components clockwise with the small letters 
 of the alphabet corresponding to the capital letters denoting the 
 members in the frame diagram. This will give a triangle, from 
 which the magnitude and direction of the stresses in the two mem- 
 bers may be determined. The joint at one of the supports usually, 
 but not always, furnishes a starting point, the known reaction at 
 the joint being the external force. 
 
 (3) One of the stresses thus found is now combined with the ex- 
 ternal or other known force at the next joint and resolved along the 
 directions of the other members of the joint, giving a polygon of 
 
 forces from which the stresses in two more members may be found. 
 This process is continued until the stresses in all the members are 
 determined. 
 
 36. It should be noted that in drawing the force polygon of a 
 joint of n forces, there are 2n conditions to be satisfied, viz. : 
 n magnitudes and n directions, and unless n 2 of these condi- 
 tions are known, the data is insufficient to construct the polygon. 
 
 EXAMPLE I. Suppose each rafter of the roof truss of Fig.' 11 to 
 support an evenly distributed load of 6 tons ; it is required to find 
 the magnitude and kinds of stresses in the members of the truss. 
 
 We commence by apportioning the loads to the joints of the frame 
 diagram. In doing so, we find a total load of 6 tons at the ridge 
 and a load of 3 tons directly over each supporting wall. As these 
 
FRAMED STRUCTURES. KECIPROCAL DIAGRAM 367 
 
 latter do not affect the stresses in the members, they will be omitted 
 from any further discussion, remembering that if the total reac- 
 tions at the walls are required, R l and R 2 must each be increased 
 by 3 tons. 
 
 The truss being symmetrical, the support reactions, R^ and R 2 , 
 are the same and each equal to half of the six tons at the ridge. 
 
 Letter the frame diagram, as shown, ignoring the vertical loads 
 over the supports, so that R 2 will be known as BC, and R^ as CA. 
 
 Adopt a load scale of J inch to the ton. 
 
 Since each of the three joints of the frame has but two members 
 and a known external force, either may be used as a starting point. 
 Selecting the joint at the left support, we draw ca, equal to f of 
 an inch, to represent the left reaction, CA, and we measure it up- 
 ward from c to a because R^ acts upward. From a and c draw 
 lines parallel to the members AD and DC, respectively. They in- 
 tersect at d, giving the triangle cad as the triangle of forces, or 
 the reciprocal diagram, for the joint at the left support. Hence, 
 ad and dc represent in magnitude and direction the stresses in the 
 members AD and DC. Knowing the direction of R lf as repre- 
 sented by ca, the direction of the stresses in the two members are 
 known by taking the sides of the triangle in order, as shown by 
 the arrows. The stress in AD acts in the direction ad, and that in 
 DC in the direction dc. Indicate these directions by placing arrow- 
 heads on the two members at points close to the joint. Knowing 
 that the stress in one end of a member is opposed by an equal and 
 opposite stress in the other end, we may now place arrowheads at 
 the other ends of the members AD and DC accordingly, as shown. 
 
 To draw the reciprocal of the joint at the right support, we draw 
 be upward, and make it f of an inch in length, to represent the 
 right reaction, BC, in magnitude and direction. From b and c 
 draw lines parallel to the members DB and CD, respectively. They 
 intersect at d, giving the triangle bed as the reciprocal of the joint 
 at the right support. Hence, cd and db represent in magnitude 
 and direction the stresses in the members CD and DB. The arrows 
 within the triangle bed show the direction of these stresses and they 
 are marked with arrowheads accordingly in the frame diagram. 
 
 By scale measurements of the reciprocals, the stresses in the 
 rafters are each found to be 4.4 tons, and in the tie, CD, the stress 
 is 3.2 tons. 
 
368 
 
 ELEMENTS OF GRAPHIC STATICS 
 
 The magnitudes and directions of the stresses in all the members 
 are now known and we have yet to determine only their kinds. 
 
 37. Kule for Determining the Kind of Stress in a Member. 
 If, after determining the directions of the stresses in all the mem- 
 bers by means of the reciprocals of the joints, and after marking 
 the ends of the members with arrowheads accordingly, it is found 
 that the arrowheads of a member point toward its joints, the 
 member is then in compression and is a strut; if they point away 
 from the joint, the member is in tension and is a tie. 
 
 Hence, it is found that the rafters, AD and BD, are in com- 
 pression,, and the member CD is in tension. 
 
 38. If, to the frame of Fig. 11, we add a vertical rod, called the 
 <( King Post," we obtain the frame of Fig. 12. 
 
 Proceeding, as before, we determine the stress in AD by means 
 of the triangle cad. 
 
 To draw the reciprocal of the joint ABEDA at the ridge, we 
 measure ab downward, and make it 1.5 inches long to represent, 
 to the chosen scale of 0.25 inch to the ton, the magnitude and direc- 
 tion of the external force of 6 tons. The stress in the rafter AD 
 has been found to be 4.4 tons, so we draw ad parallel to AD, and 
 make it 1.1 inches long to represent the 4.4 tons to scale. From 
 & we draw a line parallel to BE and we find that it intersects ad at d. 
 This shows that d and e are one and the same point, and that 
 there is no stress in the vertical post DE. Such a member is called 
 " redundant." 
 
FRAMED STRUCTURES. RECIPROCAL DIAGRAM 369 
 
 39. As a further illustration we will consider the King-post truss 
 of Fig. 10, p. 364, reproduced in Fig. 13. The loads and sup- 
 port reactions were found to be as shown. 
 
 Knowing the force EA, or R 19 we commence at the joint at the 
 left support and, with a scale of 0.25 inch to the ton, obtain eaf 
 as the triangle of forces for the joint. Marking with arrowheads 
 the kind of stresses in the members AF and FE, we proceed to the 
 next joint in clockwise order, that at which the load of 2 tons is 
 applied. 
 
 Of the eight conditions of this joint, the four directions and 
 two of the magnitudes are known, the magnitude of FA having just 
 been determined. We then readily obtain abgf as the force poly- 
 gon of the joint. Marking with arrowheads the kinds of stresses 
 in the members BG and BF, we proceed in clockwise order to the 
 other joints, obtaining for the joint at the ridge the force polygon 
 g~bch; for the joint HCDI the polygon hcdi', and for the joint at 
 the left support the triangle ide, thus completing the determination 
 of the stresses in all the members. 
 
 It is observed that each of the force polygons of the joints, 
 when taken in order, contains a side of the one immediately pre- 
 ceding it ; hence, each polygon can be built upon the one preceding 
 it, and thus produce one figure which will contain all the sides, 
 and be a graphic representation of the magnitudes and directions 
 of all the external forces and internal stresses of the structure. 
 Such a figure is known as the Reciprocal Diagram of the structure. 
 
 The reciprocal diagram of the truss of Fig. 13 is shown in Fig. 
 14, the load line ad having first been drawn and the lines repre- 
 senting the stresses then drawn in their regular order. The mag- 
 nitudes of the stresses were measured and found to be as shown in 
 the table. 
 
 40. It is convenient sometimes to determine the stress in a mem- 
 ber by the " Ritter Section " method. This method depends upon 
 the principle that, at any imaginary section of a frame there is 
 equilibrium between the external forces on one side of the section 
 and the stress forces on the same side that are in the members cut 
 by the section, and that, therefore, the algebraic sum of the mo- 
 ments of these forces about any point in the plane of the frame 
 is zero. 
 
 In Fig. 13, suppose an imaginary section xy to be taken. This 
 24 
 
Fiy. 
 
FRAMED STRUCTURES. KECIPROCAL DIAGRAM 
 
 371 
 
 section cuts the members EF, BG, and GF. Then, the left support 
 reaction and the force of 2 tons are balanced by the stresses in the 
 members EF, BG, and GF. Taking moments about the left sup- 
 port, we have : GF X r = 2 X s, whence Stress in GF = ^ . The 
 frame diagram of Fig. 13 was constructed to a scale of -^ inch to 
 the foot, and since s and r measure 6.25 and 7.5 feet, respectively, 
 
 X6.25 
 
 we get Stress in GF = 
 
 7.5 
 
 = 1.67 tons. 
 
 41. Any attempt to put arrowheads on the reciprocal diagram to 
 indicate the kinds of stresses in the members results in nothing but 
 confusion. If the kind of stress in a member cannot be discovered 
 by the eye, the polygon of forces for the joint in question must be 
 drawn as was done in connection with Fig. 13. 
 
 .15. 
 
 42. The stresses in the members of the braced cantilever of 
 Fig. 15, having a concentrated load of 2 tons at its outer extremity, 
 may be found as follows : 
 
 To a scale of -fa inch = -fa ton, lay off ab vertical and J inch 
 in length to represent the load of 2 tons. From b draw a parallel 
 to BC and from a a parallel to CA, intersecting at c. Then, abc 
 is the triangle of forces for the equilibrium of the joint ABC. The 
 force AB, acting downward, is denoted in direction and magnitude 
 by ab in the triangle, and the stresses in the other members, taken 
 in order, act in the directions be and ca, and their magnitudes are, 
 of course, denoted by the lengths of these lines to the chosen scale. 
 
372 ELEMENTS OF GRAPHIC STATICS 
 
 Mark these directions by arrowheads at the joint ABC. The di- 
 rections of the stress in the upper end of BC and in the outer end 
 of CA being now known, it follows that the stresses in the other 
 ends of these members are equal and opposite, and arrowheads must 
 be at once placed to indicate them. We now have sufficient data 
 to draw the force diagram for the joint CBD. The stress in the 
 lower end of CB has just been marked to act in the direction cb. 
 From b and c draw parallels to BD and DC, respectively, intersect- 
 ing at d; then the triangle cbd is the force diagram for the joint 
 CBD, and bd and dc are the directions of the stresses in the mem- 
 bers BD and DC, and have been marked accordingly in the frame 
 diagram. Arrowheads are at once placed at the other ends of BD 
 and DC to denote the directions of the equal and opposite stresses 
 at the wall and at the joint ACDE, respectively. The stresses in 
 AC and CD at the joint ACDE are now known, and their direc- 
 tions are ac and cd, respectively. From d draw a parallel to DE 
 and from a a parallel to EA. They intersect at e, and de and ea 
 are the directions and magnitudes of the stresses in the members 
 DE and EA, respectively. It should be noted that, in the recip- 
 rocal diagram, the stress line of one member may lie wholly or 
 partly on the stress line of another member, as was here instanced in 
 the stresses of AC and EA. The stresses in the different members 
 were measured to scale on the reciprocal diagram and marked on 
 the members of the frame. 
 
 The horizontal outward pull of 2.95 tons at the upper joint at 
 the wall occasions an equal and opposite reaction. At the lower 
 wall-joint there is a horizontal thrust of 2 tons and a diagonal thrust 
 of 2.25 tons. The horizontal component of this diagonal thrust is 
 hd = ac = 0.95 tons, making the two reactions equal but opposite 
 in direction. 
 
 43. the braced cantilever of Fig. 16 is 25 feet long, 10 feet deep, 
 and uniformly loaded on the top with 100 pounds per foot run. 
 
 In apportioning the total load of 2500 Ibs., it will be observed 
 that the member BJ, being but half the length of the members 
 CH and DF, sustains but J of the total load, CH and DF each sus- 
 taining f . 
 
 The apportionment of the load will therefore be : 500 pounds at 
 the joint at the outer end, 1000 pounds at the joint CDFGH, 750 
 pounds at the joint BCPIIJ, and 250 pounds at the joint ABJ at 
 
FRAMED STRUCTURES. EECIPROCAL DIAGRAM 
 
 373 
 
 the wall. This latter load, having no influence on the stresses in 
 the members, is rejected. 
 
 Commencing at the joint DEF, and with a scale -$ inch = 100 
 pounds, we get def as the force diagram for the equilibrium of the 
 joint, and at once mark arrowheads on the frame indicating the 
 directions of the stresses in EF, FD, FE, and DF. Having now 
 the magnitude of the stress FE at the joint FEG, we obtain feg as 
 the force diagram of the joint. Marking with arrowheads on the 
 frame the direction of the stresses thus obtained, we find that at the 
 
 I. 
 
 joint CDFGH the force CD and the stresses DF and FG are known 
 and, therefore, the force polygon cdfgh is readily obtained. The 
 reciprocal diagram may now be completed without difficulty. 
 
 44. The Warren Girder of Fig. 17 has a span of 27 feet, and 
 loaded with 1-J tons per foot, making a total load of 36 tons. The 
 members AF and DL each sustain but J of the load, and in the 
 apportionment, y 1 ^, or 3 tons, fall over the support at each end, and 
 the remainder of the load as shown. 
 
 At each end of the top flange there is equilibrium under the 
 action of the external force of 3 tons in line with the upright mem- 
 ber and the stresses in the two members at right angles to each 
 
374 
 
 ELEMENTS or GEAPHIC STATICS 
 
 other. Evidently the stress in each of the uprights is 3 tons, there- 
 fore there cannot be any stress in the members AF and DL if the 
 equilibrium is to be maintained. To indicate that there is no stress 
 in AF, the letters a and / must be placed at the same point in the 
 reciprocal diagram, and because of the absence of stress in DL, the 
 letters d and I are at the same point. It should not be forgotten 
 that the two external forces of 3 tons at the ends, coming directly 
 over the supports, are neglected, and that the reactions of 15 tons 
 
 i2.\+ons 
 
 9 \tons 3\+or# 
 
 \tons 3\ 
 
 f > 1 
 
 .2 tons. 
 Ci/- / 2.5 tons. 
 
 Fiy. /. 
 
 each are due solely to the loads of 9 tons, 12 tons, and 9 tons, the 
 forces which occasion the stresses in the members. In the con- 
 struction of the Warren Girder the end uprights and the members 
 AF and DL are omitted. 
 
 To the scale of -J inch = 3 tons, set off ef to represent in mag- 
 nitude and direction the left reaction, EF, of 15 tons. From / 
 and e draw parallels to FG and GE, respectively, intersecting at g. 
 efg is then the triangle of forces for the joint at the left support. 
 For the joint GFABH we get gfabJi as the polygon of forces, and 
 so on to the completion of the reciprocal diagram. 
 
 The triangles of the Warren Girder being equilateral, the stresses 
 found from the reciprocal diagram may easily be checked. Thus, 
 the stress in KL 15 sec 30 = 17.32 tons. 
 
FRAMED STRUCTURES. EECIPROCAL DIAGRAM 
 
 375 
 
 45. The Linville, or N, Girder, of Fig. 18 is irregularly loaded 
 on the bottom flange as shown. 
 
 To a scale of ^ inch = 1 ton, set off ab and be to represent in 
 magnitude and direction the loads of 20 tons and 30 tons, respec- 
 tively. 
 
 For convenience the frame is lettered, in this instance, in contra- 
 clockwise order, and the forces at the joints will be considered like- 
 wise. 
 
 OK *JB* 31.5 ton*; 
 
 Selecting a pole, o, we obtain, by means of the funicular polygon, 
 cd and da for the reactions R 2 and R lt respectively. 
 
 For the reason already given, there can be no stress in the 
 members CL and EA, and this is indicated by placing Z, in the 
 reciprocal diagram, at the same point with c, and e at the same 
 point with a. 
 
 At the joint at the middle of the top flange there is a state of 
 equilibrium under the action of the stresses in the members ID, 
 DH, and HI, HI being at right angles to each of the others. There 
 cannot, therefore, be any stress in HI, and h and i will fall at the 
 same point in the reciprocal diagram. 
 
376 
 
 ELEMENTS OF GRAPHIC STATICS 
 
 The members CL and EA add rigidity to the frame, and HI 
 resists the tendency of the top flange to bend. 
 
 For the equilibrium of the joint LDK we get the triangle Idle. 
 For the joint CLKJB we get the force polygon elk ft, and so on to 
 the completion of the reciprocal diagram. 
 
 46. The Fink Truss of four panels, Fig. .19, has a span of 64 
 feet, a depth of 12 feet, and is uniformly loaded with 1.5 tons per 
 foot, making 96 tons in all. The figure is constructed to a scale of 
 ^5- of an inch to the foot, and the load scale is taken as -$" = 1 ton. 
 
 tons 
 
 The apportionment of the load places 12 tons over each support, 
 and they are rejected. 
 
 There is insufficient data to begin at the joint at the left support, 
 but by means of the Eitter section, xy, and moments about the left 
 support, we obtain: 
 
 Stress in EH X 32 sin 6 = 24 X 16, whence, Stress in EH = 20 
 tons, and it is evident that the members EG, EM, and EN are sub- 
 jected to the same stress. 
 
 On the load line set off ab, be, and cd, each equal to || inch to 
 represent the external loads of 24 tons. Then, de = ea = the re- 
 actions to scale. 
 
 From e draw eg parallel to EG, and make it f $ inch long to rep- 
 resent the stress of 20 tons. Commencing at e, we now get eafge 
 
FRAMED STRUCTURES. KECIPROCAL DIAGRAM 377 
 
 as the stress polygon for the joint at the left support. Knowing af, 
 the stress polygon abif for the joint ABIF is readily drawn, and 
 so on to the completion of the reciprocal diagram. 
 
 By means of the Eitter section ut and moments about the upper 
 middle joint, we obtain: 
 
 Stress in JE X 32 sin a + 24 X 16 = 36 X 32, 
 
 TI , (36 X 32 24 X 16)^/73 
 whence, Stress in JE = - Q6 =68.35 tons, 
 
 which agrees with the scale length of je of the reciprocal diagram. 
 
 The stresses in AF, BI, CL, and DO are shown by the diagram to 
 be 80 tons each. This may be checked by the Ritter sections, ut 
 and vz. 
 
 Thus, for the section ut and moments about the joint GI1E, we 
 obtain: Stress in BI X 12 36 X 16 + Stress in JE X 6 cos a, 
 
 whence Stress in BI 48 + 68.35 X -4y = 80 tons. 
 
 From the section vz and moments about the joint FIHG, we 
 have: 
 Stress in EG X 2\/73 sin (6 a) + Stress in AF X 6 = 36 X 16, 
 
 whence, Stress in AF = - = 80 tons. 
 
 47. The reciprocal diagrams of all the frames that have been con- 
 sidered have closed, and the frames, therefore, complete. It will 
 be noticed that, in the cantilever frames, the number of members 
 equals twice the number of joints minus four, and that in the 
 frames supported at the ends the number of members equals twice 
 the number of joints minus three. 
 
 48. A frame having more than the requisite number of members 
 is redundant; that is, it contains more than the necessary number 
 of members to prevent distortion. 
 
 49. A deficient frame is one that has not sufficient members to 
 prevent distortion. 
 
 PROBLEMS. 
 
 4. Draw the reciprocal diagram of the Warren Girder, Fig. 20, 
 the triangles being equilateral. Span, 42 ft., and a uniformly dis- 
 tributed load of 1.5 tons per foot. Tabulate the stresses, and indi- 
 
378 
 
 ELEMENTS OF GRAPHIC STATICS 
 
 cate their kinds in the frame diagram. Scales : load, T V = 1 ton; 
 linear, T V = 1'. 
 
 5. Draw the reciprocal diagram of the roof truss of Fig. 21, 
 which is loaded as shown. Tabulate the stresses, and indicate their 
 kinds in the frame diagram. Scales: Load, \ inch = 1 ton; lin- 
 ear, Y 1 ^ inch = 1 foot. 
 
 6. The Fink truss of Fig. 22 has a span of 56 ft., a depth of 12 
 feet, and is uniformly loaded with 1.5 tons per foot. Draw the re- 
 ciprocal diagram to the scale of T y = 1 ton. Tabulate the 
 stresses, and indicate their kinds in the frame diagram. Linear 
 scale, T V inch = 1 foot. 
 
 I 
 
 7. The braced cantilever of Fig. 23 is 20 ft. long, 9 ft. deep, and 
 uniformly loaded on the top with 120 Ibs. per foot. Draw the 
 reciprocal diagram to a scale of -^" 100 Ibs. Tabulate the 
 stresses, and indicate their kinds in the frame diagram. Show that 
 
FRAMED STRUCTURES. RECIPROCAL DIAGRAM 
 
 379 
 
 the reactions at the wall are equal, but opposite in direction. Lin- 
 ear scale, T 3 -g- inch = 1 foot. 
 
 8. Draw the reciprocal diagram of the braced cantilever of Fig. 
 24, tabulating the stresses, and indicating their kinds in the frame 
 
 diagram. Give, in magnitude and direction, the resultant stress 
 on the pin of the upper joint at the wall. Scales : Load, T V inch 
 = 80 Ibs. ; linear, J inch = 1 foot. 
 
APPENDIX 
 
APPENDIX 
 
 MASS AND MEASUREMENT OF FORCE. 
 
 The Mass of a body is the invariable quantity of matter it contains. 
 
 The Weight of a body is the name given to the pressure which the 
 attraction of the earth causes a body to exert; or, in other words, it is 
 the force with which the earth attracts a body. 
 
 In the measurement of force there are two recognized units, viz., the 
 Absolute Unit and the Gravitation Unit. 
 
 Definition. The Absolute Unit of Force, called in England the Poundal, 
 is that force which, acting for one second on a mass of one pound, im- 
 parts to it a velocity of one foot per second. 
 
 Definition. The Gravitation Unit of Force, called the pound, is the 
 force required to support a mass of one pound against the attractive 
 force of gravity. 
 
 It has been found by experiment that if a body weighing one pound 
 be allowed to fall freely for one second it will acquire a velocity of 
 about 32.2 feet per second, this 32.2 feet per second being the accelera- 
 ting force of gravity, universally denoted by g. It follows from the 
 definitions that the gravitation unit of force is g times as great as the 
 absolute unit; that is, one pound is equal to g poundals. 
 
 The mass of a body being an invariable quantity, the absolute unit is 
 therefore invariable, and is the unit employed in theoretical mechanics, 
 in the solution of problems in which the results are to show a high 
 degree of accuracy and are to be independent of locality. 
 
 Since the accelerating force of gravity is a variable quantity, increas- 
 ing slightly as we pass from the equator toward the poles, the gravi- 
 tation unit of force is not constant. The variation is scarcely appre- 
 ciable, but it is none the less true that the gravitation unit does not 
 possess the quality of invariability. Notwithstanding this fact, the 
 gravitation unit of force is the one adopted by engineers in the science 
 of practical mechanics, in which forces are measured by the weights 
 they will support, the unit of measurement being a pound. 
 
 The pound weight has long been the standard of reference in the 
 measurement of force. The pressure of steam in a boiler is reckoned 
 in pounds per square inch; the tension in a string, the power of the blow 
 of a steam hammer, the force necessary to draw a railway train, are all 
 measured in pounds, and therefore the gravitation unit is the convenient 
 one for use, and it would be 1 very inconvenient to use the absolute unit 
 in calculations for practical purposes, though its use in theoretical in- 
 vestigations is desirable. 
 
384 APPENDIX 
 
 In gravitation measure, the unit of mass is the quantity of matter in 
 a body weighing g pounds, so that a body of mass 1 weighs g pounds, 
 and a body of mass M weighs Mg pounds. Then, if M denotes the mass 
 
 W 
 of a body weighing W pounds, we shall have, W = Mg, whence M = ; 
 
 9 
 
 so that the numerical estimate of the mass of a body is derived from its 
 weight, and although the weight of a body varies as it is carried from 
 
 W 
 place to place, the ratio remains constant. For, let W be the weight 
 
 9 
 
 of a body in a certain locality where the acceleration of gravity is g, 
 and let W be the weight of the same body at another locality where the 
 
 acceleration of gravity is g'. We would then have: - = ^ = ^ , 
 
 whence -2" ^-. Therefore, though W and g vary with the locality, 
 y y 
 
 TTT 
 
 the ratio _ remains constant, and the numerical value of M for the 
 9 
 
 same body is constant at all places. 
 
 A body of mass M moving with a velocity v is said to possess momen- 
 tum, or quantity of motion, which is measured by the product Mv. 
 
 By the First Law of Motion, the change of momentum of a moving 
 body is caused by the action of a force, and, by the Second Law, the 
 rate of change of momentum is proportional to the acting force. 
 
 Thus, if a force F act on a body of mass M for a period of t seconds, 
 changing its velocity from v to v 2 , then M(v 2 v t ) will express the 
 
 change in momentum, and "* 2 ~ V{ will be the measure of the rate 
 of change of momentum. Then, F = - ,~ cMf, in which f 
 v -i ~ v i __ t ne acceleration in feet per second, and c some constant 
 
 which it is desirable to make unity. If the absolute unit of force be 
 adopted, then M = l and f = l, so that c = l, and we have F = Mf; 
 that is, 
 
 Force Mass x Acceleration. 
 
 In the equation, F Mf, the force F is expressed in poundals, or in 
 absolute measure. It is, as has been shown, preferable to express 
 forces by their static measure, that is, by the weights they will sup- 
 port, so in order that the fundamental equation shall hold for gravita- 
 tion units it must be modified to suit them. 
 
 Let F and F' be the static measures of two forces that produce accel- 
 erations f and f on a mass M. Then, by the Second Law of Motion, 
 F : F' = f : f. If F' be the force due to gravity, viz., the weight, W, of 
 the body, then the acceleration is g, and we shall have F : W=f : g, 
 
 W 
 
 whence F = f, in which F is expressed in the same units as W, that 
 9 
 
 W 
 is, in the gravitation units of pounds, the invariable ratio denoting 
 
 9 
 the mass. 
 
APPENDIX 
 
 385 
 
 CENTRIPETAL AND CENTRIFUGAL FORCES. 
 
 If a body describes a circle of radius r, with a uniform velocity v, the 
 body is acted on by a force tending to the center of the circle the acceler- 
 ation of which is . 
 r 
 
 Let PQ, Pig. 1, be an arc of the circle such that P and Q are very near 
 together. Join P and Q with the center of the circle: then the angle 
 POQ is very small and will be denoted by a. Let t denote the small 
 interval of time during which the arc PQ was described. In the time 
 t a velocity parallel to PO, and equal to v sin , has been generated. 
 The acceleration, f, parallel to PO is, 
 
 _ v sin a _ va _ v arc PQ 
 
 ~ t ~T ~~ ~t' r 
 
 But arc PQ space described in time t = vt, hence f = - - . 
 
 . I. 
 
 Then, since F Mf, we have, as the force tending to the center, 
 
 r gr 
 
 That is, if a body of mass M describes a circle of radius r, with uni- 
 form velocity v, then, whatever the forces acting on the body, their 
 
 Mv* 
 
 resultant tends to the center of the circle, and is equal to 
 
 This 
 
 inward pull on the body toward the center of rotation is known as 
 centripetal force. The body being attached to the center, this inward 
 pull will be felt at the center, and will cause a reaction on the center 
 itself. This reaction is known as centrifugal force. Hence, the centrip- 
 etal and centrifugal forces acting on a revolving body are equal and 
 
 in which v is the 
 
 opposite, and are numerically equal to _^- _ 
 
 linear velocity of the body. 
 
 If the angular velocity of a revolving body, that is, the circular 
 measure of the angle described by any line of the body in a given time, 
 
 25 
 
386 APPENDIX 
 
 be given and be denoted by , then the arc described in the given time 
 by a point distant r from the center will be rw. But the length of the 
 arc is the linear velocity of the point in the given time; hence, the equa- 
 tion connecting linear and angular velocities is v = ru. 
 
 Then, Centripetal Force Centrifugal Force Wr * u * Wa) * r , in 
 
 gr g 
 
 which a) is the angular velocity of the body. 
 
 ACCUMULATED WORK. 
 
 A moving body is said to have work accumulated in it, and it can be 
 made to do work by parting with its velocity. 
 
 Let a body of mass M be moving with a velocity v; let a constant 
 force F, acting on the body through a space s, bring it to rest; then we 
 shall take Fs as the measure of the work accumulated in the body. 
 
 We have v 2 = 2fs, and f = ~ t whence v 2 = ^. Then, 
 
 M M. 
 
 Accumulated Work Fs = ^ . 
 
 & 
 
 The expression _JL is called the Vis Viva, or Kinetic Energy, of the 
 
 
 
 mass M, and is the exact equivalent of the work accumulated in the 
 body, and is also the measure of the work the force F will have to do 
 before bringing it to rest. 
 
 Let Ti be the height through which the body must fall to acquire the 
 velocity v. Then, since W = Mg and v z = 2gh, we have for the accu- 
 mulated work in the body, ^ 2 -_ _^ 2 _ _^ . 2gh = Wh. 
 
 & ag tig 
 
 Hence we say, that the work accumulated in a moving body is meas- 
 ured by the product of the weight of the body into the height through 
 which it must fall to acquire the velocity. 
 
 STEAM DISTRIBUTION BY THE SLIDE VALVE 
 
 AND 
 THE FORMATION OF THE INDICATOR DIAGRAM. 
 
 In figures 2 to 9 are shown the important positions of a plain slide 
 valve during a stroke of the piston, and a skeleton arrangement of the 
 piston-crank-eccentric-valve mechanism. The valve V moves to and fro, 
 admitting steam first to one side of the piston and then to the other 
 from the steam chest 8 through the steam ports s and s', and allowing 
 the used steam to escape through the same ports to the exhaust port e. 
 The varying pressures of the steam thus introduced move the piston P 
 alternately from one end of the cylinder to the other. This motion is 
 transferred to the outside of the cylinder by the piston rod PH. The 
 connecting-rod HC connects the outside end of the piston rod to the 
 crank pin G of the crank OG. As G is constrained to move about as 
 
APPENDIX 
 
 387 
 
 a center, the reciprocating motion of the piston is transmuted into 
 rotary motion of the crank. 
 
 To have the engine turn in the direction indicated by the arrow, the 
 eccentric must be placed ahead of the crank by 90 plus the angle of 
 advance, as at E. In the actual engine, the eccentric and crank are 
 each attached to the same shaft, but in order to show the mechanism 
 more clearly, the crank pin and eccentric circles have each been tufned 
 through 90 to bring them in the plane of the piston and valve rods. 
 The movement of the eccentric is transferred to the valve through the 
 eccentric rod EB and the valve rod BV. Thus the motion of the piston 
 
 Exhaust Lead 
 
 Fig. 2. 
 
 to make that motion 
 
 causes the valve to so distribute the steam 
 continuous. 
 
 The dimensions of the valve may easily be found when the laps are 
 given. Thus in Fig. 6, by laying off from the edge of the steam port 
 the steam lap on the steam edge of the valve, which in this case is the 
 outside edge, and the exhaust lap on the exhaust edge, the size of the 
 valve is determined. The throw of the eccentric and the angular ad- 
 vance will then determine how much lead and overtravel the valve will 
 have, and where the points of cut-off, release, compression, and admis- 
 sion will occur. 
 
 Below each of these figures are the horizontal lines ov and oV. The 
 distance of pb and p'b' above these lines represents to some scale the 
 boiler pressure. Also, the lines aa and a'a' show by their distances 
 
388 
 
 APPENDIX 
 
 above ov and o'v' respectively, the pressures against which the steam 
 is exhausted, which, in this case, is the atmospheric pressure. 
 
 Any point on this diagram will represent by its horizontal distance 
 from the end of the diagram the position of the piston, and from the 
 line op or o'p' the volume of steam in the cylinder. The vertical dis- 
 tances of any point above ov and aa, or above o'v' and a'a', will repre- 
 sent respectively the absolute and gauge pressures per square inch 
 acting on the piston. The upper diagram represents what is taking 
 place on the left-hand side of the piston, while the lower diagram is a 
 representation of what is taking place on the right-hand side. 
 
 Maximum Opening to Steam 
 
 b'- i/o' Full Opening to Exhousf 
 
 a' 
 
 Fig. 3. 
 
 In Fig. 2, the piston is at the head end of the stroke with the valve 
 open to steam lead at s, letting the steam in to start the piston on its 
 stroke; while at s' is shown the exhaust lead, letting the steam out so 
 as not to retard the piston's motion. The piston being in its initial posi- 
 tion and about to start on the forward stroke, the height of the point 
 1 above ov indicates the pressure which urges the piston forward, and 
 the height of the point 1' above o'v' shows the pressure tending to pre- 
 vent the forward movement. 
 
 The valve moves toward the right until the eccentric radius comes 
 in line with the eccentric rod, the piston reaching the position shown in 
 Fig. 3. The valve now has the maximum opening to steam at s, and 
 full opening to exhaust at s'. The ports are designed to let the steam 
 out, and therefore the exhaust should be fully open for at least part of 
 
APPENDIX 
 
 389 
 
 v'- 
 
 Exhaust Taking Place 
 
 Fig. 4. 
 
 Is I 
 
 Expansion Taking Place 
 
 P' Point of Compression 
 
 Fig. 5. 
 
390 
 
 APPENDIX 
 
 a 
 
 V - 
 
 Expansion Taking Place 
 
 ' p> Compression Taking Place 
 
 o' 
 
 Fig. 6. 
 
 6' 
 
 6' 
 
 L , ,..> 
 
 a' a' 
 
 v 
 
 Point of Release 
 
 Compression Taking Place 
 
 Fig. 7. 
 
APPENDIX 391 
 
 the stroke. This is accomplished by giving the valve overtravel, which 
 is the distance the exhaust edge of the valve goes beyond the steam 
 port when at the end of its travel. The positions of the points 2 and 2' 
 relative to ov and o'v', respectively, show the pressures on the piston. 
 
 As the crank continues to revolve, the valve moves toward the left. 
 When the steam edge comes in line with the steam port, Fig. 4, cut-off 
 occurs, and the position of the point 3 represents the pressure at that 
 point of the stroke. The drop in pressure from 1 to 3 was caused by 
 wire-drawing, due to small steam pipes and passages and to gradual 
 closure of the valve. The height of 3' above o'v' shows the back pres- 
 sure at cut-off, the exhaust still taking place through the small opening 
 at s'. 
 
 The next important position is shown in Fig. 5, where the port s' 
 closes and compression is just about to occur, the height of the point 4' 
 above o'v' indicating the pressure at the point of compression. Expan- 
 sion is taking place on the left-hand side, the pressure falling to the 
 point 4. At mid position of the valve, Fig. 6, expansion is taking place 
 on the left, the pressure falling, due to increase of volume of the steam. 
 On the right, compression is taking place, causing the pressure to rise 
 by decreasing the volume of the entrapped steam. In Fig. 7 is shown 
 the position of the piston when the point of release occurs at s; 
 the steam is just about to be released after having pushed the piston to 
 this point of the stroke, the pressure at release being shown by the 
 height of 6 above ov. Compression is still taking place on the other 
 side. 
 
 As the motion continues, exhaust takes place at s and the pressure 
 falls. The steam edge comes in line with the edge of the steam port at 
 s', Fig. 8; steam is just about to be admitted, the pressure having been 
 raised by compresion to 7'. 
 
 When the piston reaches the end of the stroke, Fig. 9, the valve is 
 open to the extent of the exhaust lead at s and to the extent of the 
 steam lead at s'. The pressure at 8 has fallen nearly to the back pres- 
 sure, and at 8' has rapidly risen to the boiler pressure from the terminal 
 pressure of compression at 7'. If the return stroke were now made, the 
 other part of the diagram shown in this figure would be formed, thus 
 completing the diagram for each side of the piston. 
 
 Any point on this diagram represents the pressure at a certain posi- 
 tion of the piston. The upper curve of the diagram shows the varying 
 pressure pushing the piston forward, and the lower curve shows the 
 varying pressure tending to hold the piston back. The diagram from 
 one end of the cylinder thus shows the pressure on one side of the piston 
 during two strokes. In balancing engines it is desirable to know the 
 effective pressure pushing the piston forward at each point of the stroke 
 and this is obtained by combining the top line of one diagram with the 
 bottom line of the other. 
 
392 
 
 APPENDIX 
 
 a' 
 
 Exhaust Taking Place 
 
 p 1 Point of Admission 
 
 o' 
 
 Fig. 6. 
 
 Steam Lead 
 
 fig. 9. 
 
APPENDIX 393 
 
 SOME USEFUL NOTES AND CONSTANTS. 
 
 Wrought iron weighs 480 pounds per cubic foot. Then, il|?. 36 
 
 cubic inches weigh 10 pounds. Therefore, 
 
 volume in cubic inches 1Q _ weight IQ pounds 
 
 36 
 Or, 
 
 sectional area in sq. inches x length in feet x 10 _ we i g ht. 
 
 3 
 
 Whence: 
 
 Sectional area in sq. inches weight per lineal foot x TO 
 Weight per lineal foot Sectional area x V 
 
 The weight of steel per cubic foot, 490 Ibs., is just 2% greater than 
 that for wrought iron. Hence, a lineal foot of steel, one square inch in 
 section, weighs -y- x 1-02 3.4 Ibs. We have, therefore, for steel : 
 
 Sectional area in square inches = wei g ht per j ineal foot . 
 
 3. * 
 
 Weight per lineal foot = sectional area x 3.4. 
 
 Cast iron, which weighs 450 Ibs. per cubic foot, is just ^ lighter than 
 wrought iron. 
 
 Wrought iron is 9.6 as heavy as white oak, 10.7 as heavy as yellow 
 pine, and 19.2 as heavy as white pine. 
 1 gallon = 231 cubic inches. 
 1 gallon of water weighs 8.355 pounds. 
 1 pound avoirdupois = 7,000 grains = 453.6 grammes. 
 1 cubic foot of water weighs 62.5 pounds, for practical purposes. 
 1 knot = 6,080 feet. 
 
 1 cubic foot of air, under atmospheric conditions, weighs 0.08 pound. 
 1 horse-power = 33,000 ft.-lbs. per minute = 746 watts = 0.746 K. W. 
 1 K. W. = 1.34 H. P. 
 Watts = volts x amperes. 
 TT = 3.1416 = 22/7, approximately. 
 
 The base of the Naperian, or hyperbolic, logarithms is 2.7183. 
 To convert common into Naperian logarithms, multiply by 2.3026. 
 V3 = 1.732 and V2~= 1.414. 
 A column of water 2.3 feet high corresponds to a pressure of one pound 
 
 per square inch. 
 A column of mercury 2.04 inches high corresponds to a pressure of one 
 
 pound per square inch. 
 
 Inches of vacuum x 0-49 = pounds pressure. 
 
 1 foot = 0.305 metre. 1 sq. inch = 6.541 sq. cms. 
 
 1 inch =25.4 millimetres. 1 pound = 0.4536 kilogrammes. 
 
 1 sq. ft. = 0.0929 sq. metre. 1 kilo. =2.2 pounds. 
 
 1 mile =1609.3 metres = 1.61 kilometres. 
 
 1 cu. inch = 16.387 cu. cms. 
 
 1 cu. foot =0.0283 cu. metre. 
 
INDEX 
 
 PAGE 
 
 Absolute temperature 4 
 
 unit of force 383 
 
 zero of temperature 4 
 
 Accumulated work 386 
 
 Adiabatic expansion 98 
 
 Admission line of indicator diagram 80 
 
 Air, quantity necessary for combustion 24 
 
 Alloys 246 
 
 American Society of Mechanical Engineers, boiler trial code of 164 
 
 Angular advance of the eccentric 53 
 
 Angularity of connecting-rod 66 
 
 Anthracite coal 22 
 
 Appendix 383 
 
 Area of test pieces, reduction in 256 
 
 Atmospheric line of indicator diagram 80 
 
 Automatic cut-off 60 
 
 Axis, neutral 312 
 
 Back pressure line of indicator diagram 81 
 
 Barrel calorimeter 106 
 
 Beam, cantilever 282 
 
 resting on three supports 333 
 
 simple 282 
 
 Beams, continuous 333 
 
 deflection of 329-335 
 
 resilience of 338-340 
 
 standard I 317 
 
 table of relative strength of 335 
 
 theory of 311 
 
 theory of, important assumptions made 314 
 
 with fixed ends 331 
 
 Belt and pulley, f rictional resistance between 208 
 
 Belt, thickness of 207 
 
 Belting 205 
 
 Belting, strength of 210 
 
 Belts, centrifugal action in 211 
 
 creeping of 210 
 
 transmission of power by 210 
 
 Bending moments 282 
 
 Bending moments, general case of 282 
 
 Bending-moment diagrams 284 
 
 Bending-moment diagram, the funicular polygon a 352 
 
 Bending-moment and shear diagrams, for different conditions of 
 loading . , 296-302 
 
396 INDEX 
 
 PAGE 
 
 Bending-moment and shear diagrams of cantilevers 293 
 
 Bending-moment and shear diagrams, relations between 288 
 
 Bituminous coal 22 
 
 Boiler, code for conducting trials of 30 
 
 design 181 
 
 efficiency 40 
 
 girder stays for 185 
 
 grate surface of 183 
 
 heating surface of 183 
 
 horse-power of 30 
 
 sectional area of tubes and of area over bridge wall 183 
 
 steam space in 130 
 
 strength of, longitudinally and circumferentially 183 
 
 test, record of 175 
 
 test, report of 176 
 
 thermal efficiency of engine and 39 
 
 The Roberts 186-189 
 
 trials, rules for conducting, code of 1899 164 
 
 volume of steam space and of combustion chamber 183 
 
 Boilers, classification of 179 
 
 efficiency of engine and 38 
 
 locomotive, marine, stationary, sectional 179 
 
 relative advantages of different types 180 
 
 Scotch 180 
 
 steam 179 
 
 Boiling point : 13 
 
 Bow's system of lettering 345 
 
 Boyle's law 3 
 
 Brake horse-power 10 
 
 Brake, Prony 161 
 
 Brazing 231 
 
 Brazing of cast-iron 236 
 
 Calorimeter, barrel 106 
 
 separating 109 
 
 tests 106-109 
 
 throttling 107 
 
 Calorimetry 8 
 
 Cantilever beam 282 
 
 Cantilevers, bend ing-moment and shear diagrams of 293 
 
 Carbon, graphitic 232 
 
 in fuel 23 
 
 on cast iron, influence of 233 
 
 Carnot, perfect heat engine of 33, 110 
 
 Castings, inspection of 236 
 
 Cast iron 232 
 
 Cast iron, brazing of 236 
 
INDEX 397 
 
 PAGE 
 
 Cast iron, influence of carbon on 233 
 
 influence of chromium on 235 
 
 influence of manganese on 235 
 
 influence of silicon on 233 
 
 influence of sulphur on 234 
 
 malleable 236 
 
 shrinkage of 235 
 
 strength and hardness of 235 
 
 uses of in engineering 236 
 
 Center, placing engine on 55 
 
 of gravity 268 
 
 of gyration 306 
 
 of pressure 251 
 
 Centrifugal action in belts 211 
 
 Centrifugal force 385 
 
 Centripetal force 385 
 
 Charles' law 3 
 
 Checks as to the accuracy of shear diagrams 302 
 
 Chromium, on cast iron, influence of 235 
 
 Circles, pitch 217 
 
 Clearance 82 
 
 Clearance line of indicator diagram 81 
 
 Clearance, on ratio of expansion, influence d 83 
 
 Coal 22 
 
 Coal, anthracite 22 
 
 bituminous 22 
 
 pounds per I. H. P. per hour 38 
 
 Code, short of 1899, for conducting boiler trials 164 
 
 Coefficient of elasticity 311 
 
 Coke 22 
 
 Columns 320 
 
 Columns, the design of 322 
 
 Combination of the laws of Boyle and Charles 5 
 
 Combining diagrams of stage expansion engines 155 
 
 Combustion 23 
 
 Combustion, efficiency of 40 
 
 quantity of air necessary for 24 
 
 Compounding 38 ' 
 
 Compression 47 
 
 Compression curve 81 
 
 Compression tests 264 
 
 Concrete 247 
 
 Concrete, adhesion of to steel 248 
 
 steel 247 
 
 Condensation and production of vacuum 116 
 
 Condensation and re-evaporation of steam in cylinders 35 
 
 Condenser, jet, surface 116 
 
 Condensing water, weight required per pound of steam 117 
 
398 INDEX 
 
 PAGE 
 
 Connecting-rod, action of crank and 64 
 
 angularity of 66 
 
 Conservation of energy 9 
 
 Constants, some useful notes and 393 
 
 Continuous beams 333 
 
 Continuous expansion type of stage expansion engines 71 
 
 Convection, transfer of heat by 11 
 
 Conversion of motion 64 
 
 Copper 231 
 
 Corrugated furnace 184 
 
 Crank and connecting-rod, action of 64 
 
 Crank and piston, relative positions of 67 
 
 Crank-pin-piston velocity diagram 133 
 
 Creeping of belts 210 
 
 Curve, compression 81 
 
 Curves of adiabatic, saturated, and isothermal expansion, construc- 
 tion of 99- 
 
 Cut-off 47 
 
 Cut-off, automatic 60 
 
 Cylinder ratios 125 
 
 Cylinder, size of for given power 124 
 
 Cylinders, liquefaction in 113 
 
 jt 
 
 Dangerous section 314 
 
 Deficient frame, a 377 
 
 Deflection of beams 329-335 
 
 De Laval turbine 191-198 
 
 Diagram, force 345 
 
 ideal indicator 79 
 
 indicator, horse-power from 92 
 
 indicator, in preliminary design 121 
 
 indicator, measure of work in cylinder 94 
 
 reciprocal 363 
 
 reciprocal, the drawing of 365 
 
 stress-strain 259 
 
 Diagrams, bending-moment 284 
 
 bending-moment and shear, of cantilevers 293 
 
 bending-moment and shear, relations between 287 
 
 indicator, interpretation of 79 
 
 indicator, steam per I. H. P. per hour from 150 
 
 of stage expansion engines, combining 155 
 
 Divergent nozzle of De Laval turbine 193 
 
 Dryness fraction of steam 105 
 
 Eccentric, angular advance of 53 
 
 Eccentric arm 49 
 
 Eccentric radius 49 
 
 Eccentric rods, crossed 57 
 
 Eccentric rods, open 57 
 
INDEX 399 
 
 PAGE 
 
 Eccentric, shifting 60 
 
 the 49 
 
 throw of ' 
 
 virtual 58 
 
 Economy of the steam engine and boiler 150 
 
 Efficiencies of engines and boilers 38, 103 
 
 Efficiency ' 
 
 Efficiency, boiler 40 > 103 
 
 engine 11G 
 
 losses affecting 34 
 
 maximum 34 
 
 Efficiency of combustion ' 
 
 Efficiency of engine, mechanical 40 
 
 Efficiency of heating surface 40 
 
 Efficiency, relative engine 40 
 
 thermal 39 
 
 total of system 4i 
 
 Elastic limit 256 
 
 Elasticity, coefficient of 3 
 
 modulus of Ml 
 
 Elongation in test pieces 256 
 
 Energy - 7 
 
 Energy, conservation of 9 
 
 kinetic 7 
 
 potential 7 
 
 Engine and boiler, thermal efficiency of 39 
 
 Engine, dead points of 55 
 
 Engine efficiency 110 
 
 Engine efficiency, relative 40 
 
 Engine, mechanical advantage of stage expansion 131 
 
 mechanical efficiency of 40 
 
 placing on center 55 
 
 reciprocating, compared with steam turbine 202 
 
 thermal efficiency of 39 
 
 Engines and boilers, efficiencies of 38 
 
 Evaporation, factor of 30 
 
 unit of 30 
 
 Evaporative effect of mineral oil 23 
 
 Evaporative power of fuel 26 
 
 Exhaust line of indicator diagram 81 
 
 Expansion, adiabatic, saturated steam, isothermal, hyperbolic 98 
 
 engines, relative merits of the simple and the stage. ... 70 
 
 engines, stage 70 
 
 influence of clearance on the ratio of 83 
 
 ratio of 83 
 
 ratio of in stage expansion engines 85 
 
400 INDEX 
 
 PAGE 
 
 Factor of evaporation 30 
 
 Factor of safety 256 
 
 Factors, mean pressure 125 
 
 Feed water, weighing the 175 
 
 Fink truss, the 376 
 
 Fixed ends, beams with 331 
 
 Foot pound 10 
 
 Force , 9 
 
 Force, absolute unit of 383 
 
 centrifugal 385 
 
 centripetal 385 
 
 diagram 345 
 
 gravitation unit of 383 
 
 measurement of 383 
 
 Frame, a 362 
 
 a deficient 377 
 
 a redundant 377 
 
 Framed structures 362 
 
 Fuel, evaporative power of 26 
 
 Fuels 22 
 
 Fuels, total heat of combustion of 26 
 
 Funicular polygon 348 
 
 Funicular polygon, a bending-moment diagram 352 
 
 Furnace, Adamson ring and Bowling hoop for 184 
 
 corrugated 184 
 
 Gas, a 3 
 
 Gas expanding within a cylinder, mean pressure of 96 
 
 Gases, kinetic theory of 3 
 
 Gay Lussac, law of 3 
 
 Gear, of lathe, back 225 
 
 wheels, horse-power transmitted by 222 
 
 wheels, transmission of power by 220 
 
 Gearing, lathe, compound 225 
 
 lathe, simple 223 
 
 Generation of steam under constant pressure 14 
 
 Generation of steam under constant volume 17 
 
 Girder, N, or Linville 375 
 
 Girder stays for boilers 185 
 
 Girder, Warren 373 
 
 Governor, shaft 61 
 
 Graphic statics 345 
 
 Graphitic carbon 232 
 
 Grate surface 183 
 
 Gravitation unit of force 383 
 
 Gravity, center of 268 
 
 Gyration, radius of 306 
 
INDEX 401 
 
 PAGE 
 
 Harmonic motion, simple 65 
 
 Heat, application of to water 11 
 
 kinetic theory of 6 
 
 latent 13 
 
 loss of by radiation 11 
 
 material theory of 6 
 
 mechanical equivalent of 
 
 necessary to evaporate a pound of water 29 
 
 of combustion of fuels, total 26 
 
 of steam, total 16 
 
 radiant 11 
 
 rejected into the condenser 117 
 
 sensible 12 
 
 specific 8 
 
 transfer by convection 11 
 
 Heat engine, Carnot's perfect 33 
 
 Heating surface 183 
 
 Heating surface, efficiency of 40 
 
 Horse-power 10 
 
 Horse-power, boiler 30 
 
 brake 10 
 
 from indicator diagrams 92 
 
 indicated 92 
 
 transmitted by gear wheels 222 
 
 transmitted by shafts 325 
 
 Hyperbolic expansion 98 
 
 I beams, standard 317 
 
 Ideal engine Ill 
 
 Ideal engine, pounds of steam per I. H. P. per hour, table of 112 
 
 Ideal indicator diagram 79 
 
 Indicator diagram 75 
 
 formation of 386 
 
 horse-power from 92 
 
 ideal 79 
 
 in preliminary design 121 
 
 measure of work performed in the cylinder 94 
 
 Indicator diagrams, interpretation of 79 
 
 Indicator diagrams, steam per I. H. P. per hour from 150 
 
 Indicator, location of 80 
 
 Indicator springs 76 
 
 Indicator, Tabor 77 
 
 the 75 
 
 Indicated horse-power 92 
 
 Indicated horse-power, pounds of coal per hour per 38 
 
 pounds of steam per hour per 39, 123 
 
 thermal units per minute per 39 
 
 Inequality in the motion of the piston 66 
 
 26 
 
402 INDEX 
 
 PAGE 
 
 Inertia, moment of 304 
 
 polar moment of 305 
 
 Inflection, points of 330 
 
 Iron, cast 232 
 
 wrought 237 
 
 test pieces for 256 
 
 Isothermal expansion 98 
 
 Jacketing, steam 37 
 
 Joy, valve gear of 63 
 
 Kinetic theory of gases 3 
 
 Kinetic theory of heat .- . . . 
 
 Lap, definition of 52 
 
 Latent heat 13 
 
 Latent heat of steam 16 
 
 Lathe, back gear of 225 
 
 gearing, compound 225 
 
 gearing, simple 223 
 
 Lead 245 
 
 Lead of valve 46 
 
 Lead of valve, definition of 52 
 
 Lettering, Bow's system of 345 
 
 Limit, elastic 256 
 
 Line, atmospheric, admission, steam, exhaust, back pressure, 
 
 vacuum, clearance 80, 81 
 
 Link motion 56 
 
 Link, Stephenson's 59 
 
 Linville, or N, girder 375 
 
 Liquefaction in cylinders 113 
 
 Machine, mechanical advantage of 219 
 
 mechanical efficiency of 219 
 
 Malleable cast iron 236 
 
 Manganese, influence of on cast iron 235 
 
 influence of on steel 241 
 
 Marshall, valve gear of 63 
 
 Mass 383 
 
 Materials 231 
 
 Materials, different kinds of tests of 255 
 
 testing 255 
 
 Mean effective pressure from indicator diagrams 90 
 
 Mean pressure factors 125 
 
 Mean pressure in stage expansion engines 74 
 
 Mean pressure of gas expanding within a cylinder 96 
 
 Mean pressure with adiabatic, saturated steam, and hyperbolic ex- 
 pansion 98 
 
 Measurement of force . . . . 383 
 
INDEX 403 
 
 PAGE 
 
 Mechanical advantage of a machine 219 
 
 Mechanical efficiency of an engine 40 
 
 Mechanical efficiency of a machine 219 
 
 Mechanical equivalent of heat 9 
 
 Mild steel, tension test of 261 
 
 Mild, or structural, steel 239 
 
 Mineral oil 23 
 
 Mineral oil, evaporative effect of 23 
 
 Modulus of elasticity 311 
 
 Modulus of resilience 337 
 
 Modulus, section for bending 314 
 
 section for torsion 324 
 
 Moment of inertia 304 
 
 Moment, maximum equivalent twisting 324 
 
 resisting 313 
 
 Moments 267 
 
 Moments, bending 282 
 
 bending, general case of 282 
 
 Motion, conversion of 64 
 
 link 56 
 
 simple harmonic 65 
 
 N, or Linville, girder 375 
 
 Neutral axis 312 
 
 Neutral surface 207, 312 
 
 Notes and constants, some useful 393 
 
 Nozzle of De Laval turbine, divergent 193 
 
 Oil, mineral 23 
 
 Oil, mineral, evaporative effect of 23 
 
 Phosphorus, influence on steel 241 
 
 Piston, inequality in motion of 66 
 
 relative position of the crank and 67 
 
 Piston speed 125 
 
 Piston valve 62 
 
 Pitch circles 217 
 
 Pitch of teeth 217 
 
 Plasticity 259 
 
 Point, boiling 13 
 
 Point of cut-off, release, exhaust closure 81 
 
 Points of inflection 330 
 
 Polar moment of inertia 305 
 
 Polygon, funicular 343 
 
 funicular, a bending-moment diagram 352 
 
 Port, steam, dimensions of 130 
 
 Power jo 
 
 Power, horse 10 
 
 horse, indicated 92 
 
 horse, transmitted by gear wheels . 222 
 
404 INDEX 
 
 PAGE 
 
 Power, size of cylinder for a given 124 
 
 transmission of by belts 210 
 
 transmission of by gear wheels 220 
 
 Pressure, formula connecting temperature with 18 
 
 formula connecting volume with 19 
 
 line, back 81 
 
 mean effective from indicator diagram 90 
 
 mean factors of 125 
 
 mean with adiabatic expansion, saturated steam expan- 
 sion, hyperbolic expansion 98 
 
 Prony brake 161 
 
 Pulley, frictional resistance between a belt and 208 
 
 Quick running : 38 
 
 Radial valve gears 63 
 
 Radiation, loss of heat by 11 
 
 Radius of eccentric 49 
 
 Radius of gyration 306 
 
 Ratio of expansion 83 
 
 Ratio of expansion, influence of clearance on 83 
 
 Ratio of expansion in stage expansion engines 85 
 
 Ratio, velocity 219 
 
 Ratios, cylinder 125 
 
 Receiver type of stage expansion engines 71 
 
 Reciprocal diagram 363 
 
 Reciprocal diagram, drawing the 365 
 
 Record of a boiler test 176 
 
 Reduction in area of test pieces 256 
 
 Redundant frame, a 377 
 
 Relative strength of beams, table of 335 
 
 Release 48 
 
 Report of a boiler test 176 
 
 Representative strengths and weights of materials, table of 263 
 
 Resilience 337 
 
 Resilience from sudden loads and shocks 337 
 
 Resilience, modulus of 337 
 
 Resilience of beams 338-340 
 
 Resilience of shafts under torsion 340 
 
 Resisting moment 313 
 
 Ritter section method of determining stresses 370 
 
 Roberts boiler, the 186-189 
 
 Running, quick 38 
 
 Rules for conducting boiler trials, code of 1899 164 
 
 Safety, factor of 256 
 
 Saturated steam ^ 6 
 
 Saturated steam expansion 98 
 
 Section, dangerous 314 
 
INDEX 405 
 
 PAGE 
 
 Section modulus for bending 314 
 
 Section modulus for torsion , 324 
 
 Separating calorimeter 109 
 
 Setting slide valves 55 
 
 Shaft governor 61 
 
 Shafts 324 
 
 Shafts, horse-power transmitted by 325 
 
 maximum equivalent twisting moment of 324 
 
 under torsion, resilience of 340 
 
 Shear 285 
 
 Shear and bending-moment diagrams for different conditions of 
 
 loading 296-302 
 
 Shear diagrams 285 
 
 Shear diagrams, checks to the accuracy of 302 
 
 relations between bending-moment and 288 
 
 Shear, first a>derivative of the moment is the 315 
 
 general case of 285 
 
 Shifting eccentric 60 
 
 Shrinkage of cast iron 235 
 
 Silicon, influence of on cast iron 233 
 
 influence of on steel 241 
 
 Slide valve, steam distribution by. 386 
 
 Smoke, cause and prevention of 24 
 
 Specific heat 8 
 
 Speed, piston 125 
 
 Stage expansion engines 70 
 
 Stage expansion engines, combining diagrams of 155 
 
 continuous expansion type 71 
 
 mean pressure in 74 
 
 mechanical advantage of 131 
 
 ratio of expansion in 85 
 
 receiver type of ; 71 
 
 relative merits of the simple and 70 
 
 Statics, graphic 345 
 
 Steam, distribution of by slide valve 386 
 
 dryness fraction of 105 
 
 generation, of under constant pressure 14 
 
 generation, of under constant volume 17 
 
 latent heat of 16 
 
 per I. H. P. per hour from actual indicator diagrams 150 
 
 per I. H. P. per hour for ideal engine, table of 112 
 
 pounds per I. H. P. per hour 39, 112 
 
 superheated 36 
 
 total heat of 16 
 
 weight of condensing water required per pound of 117 
 
 Steam boilers 179 
 
 Steam engine and boiler, economy of the 150 
 
 Steam jacketing . 37 
 
406 INDEX 
 
 PAGE 
 
 Steam line of indicator diagram 80 
 
 Steam port, dimensions of 130 
 
 Steam space in boiler 130 
 
 Steam turbine, the 190 
 
 Steel 238 
 
 Steel, Bessemer process 239 
 
 chrome 244 
 
 crucible 239 
 
 flange, or rivet 242 
 
 for castings 243 
 
 for shafting 243 
 
 influence of manganese on 241 
 
 influence of phosphorus on 241 
 
 influence of silicon on 241 
 
 mechanical properties of 242 
 
 mild, or structural 239 
 
 nickel 244 
 
 open-hearth process 239 
 
 proportions of carbon in 238 
 
 semi 242 
 
 shell 243 
 
 special properties of 243 
 
 table of temperatures and colors corresponding to tempers of. 244 
 
 tempering of 243 
 
 test pieces for 257, 258 
 
 tungsten, or mushet 244 
 
 in marine construction, uses of 245 
 
 Steel-concrete 247 
 
 Steel-concrete and steel and concrete construction, distinction be- 
 tween 249 
 
 Steel-concrete beam, design of 250 
 
 Steels, open-hearth and Bessemer compared 241 
 
 special 244 
 
 Stephenson link 59 
 
 Strain 255-311 
 
 Strength, ultimate 256 
 
 Stress J 11 
 
 Stress, rule for determining kind of 368 
 
 Stress-strain diagram 259 
 
 Stroke 125 
 
 Stroke, forward, outward, top, head, down, return, inward, bottom, 
 
 crank, up 143 
 
 Structure, a framed 362 
 
 equilibrium of a framed 362 
 
 load on a framed 362 
 
 reaction on a framed 362 
 
 Sulphur, influence of on cast iron , 234 
 
 influence of on steel 241 
 
INDEX 407 
 
 PAGE 
 
 Superheated steam 36 
 
 Surface; grate 183 
 
 heating 183 
 
 neutral 207, 312 
 
 System of lettering, Bow's 345 
 
 System, total efficiency of 41 
 
 Table of relative strengths of beams 335 
 
 Table of representative strengths and weights of materials 263 
 
 Table of temperatures and colors corresponding to tempers in steel . 244 
 
 Tabor indicator 77 
 
 Temper in steel, table of temperatures and colors corresponding to. . . 244 
 
 Temperature 7 
 
 Temperature, absolute 4 
 
 conversion of from one scale to another 8 
 
 formula connecting pressure with 18 
 
 Tempering 243 
 
 Test of mild steel, tension 261 
 
 Test piece for iron 256 
 
 Test pieces for steel and other materials 257, 258 
 
 Tests, calorimeter 106-109 
 
 compression 264 
 
 engine and boiler 162 
 
 Theory of beams, the 311 
 
 important assumptions made 314 
 
 Thermal efficiency 39 
 
 Thermal efficiency of engine 39 
 
 Thermal efficiency of engine and boiler 39 
 
 Thermal unit 8 
 
 Thermodynamics 8 
 
 Thermodynamics, first law of 8 
 
 second law of 9 
 
 Throttling calorimeter 107 
 
 Throw of eccentric 49 
 
 Timber 246 
 
 Travel of valve 49 
 
 Ultimate strength 256 
 
 Unit, British thermal 8 
 
 of evaporation 30 
 
 of work 10 
 
 Units per I. H. P. per minute, thermal 39 
 
 Vacuum, condensation and production of 116 
 
 Vacuum line of indicator diagram 81 
 
 Value of a train of wheels 219 
 
 Valve and its motion, the 45 
 
 Valve, lap of the 52 
 
 lead of the . 52 
 
408 INDEX 
 
 PAGE 
 
 Valve, piston 62 
 
 slide, distribution of steam by 386 
 
 slide, setting of 55 
 
 travel of the 49 
 
 Valve gear, Joy's 63 
 
 Marshall's 63 
 
 Valve gears, radial 63 
 
 Velocity ratio 219 
 
 Virtual eccentric 58 
 
 Vis viva 386 
 
 Volume, formula connecting pressure and 19 
 
 Warren girder 373 
 
 Water, heat necessary to evaporate a pound of 29 
 
 Water, weighing the feed 175 
 
 Weight of a body, the 383 
 
 Westinghouse-Parsons turbine, the 198-202 
 
 Wheels in train 217 
 
 Wheels, transmission of power by gear 220 
 
 Wire drawing 88 
 
 Wood 23 
 
 Work 9 
 
 Work, accumulated 386 
 
 percentage of applied to shaft 41 
 
 unit of 10 
 
 Wrought iron 237 
 
 Wrought iron, proportion of carbon in 237 
 
 special properties 238 
 
 uses in marine engineering 238 
 
 Zero of temperature, absolute 4 
 
 Zeuner valve diagram 137 
 
 Zinc . 245 
 

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