c!^ -pf o"^ A- .^ .0* V .n\ '-^^ ^^.S^ ^A^'^ \^ '/^, '''c^. .^<^, -? -fj .t-^^ ..^ \^ ^ THE STRENGTH OF MATERIALS BY THE SAME AUTHOR THE THEORY AND DESIGN OF STRUCTURES Third Edition. 618 pp. Demy 8vo. 95. net. FURTHER PROBLEMS IN THE THEORY AND DESIGN OF STRUCTURES 236 pp. Demy 8vo. 75. 6f/. net. ALIGNMENT CHARTS 32 pp. Crown 8vo. \s. 2,d. net. London: Chapman and Hall, Ltd., 11 Henrietta Street, W'.C. THE ELEMENTARY PRINCIPLES OF REINFORCED CONCRETE CONSTRUCTION 200 pp. Crown 8vo. 3^. net. CALCULUS FOR ENGINEERS (With Dr. H. BRYON HEYWOOD). 269 pp. Crown 8vo. ^s. net. [/// the Broadway Scries of Engineerhtg Handbooks.^ London : Scott, Greenwood & Son, 8 Broadway, Ludgate Hill, E.G. AN INTRODUCTION TO APPLIED MECHANICS 309 pp. Demy 8vo. 4J-. dd. net. \In the Cambridge Technical Series\ Cambridge : at the University Press. THE STRENGTH OF MATERIALS A TEXT-BOOK FOR ENGINEERS AND ARCHITECTS y BY .'i^' EWART S.^^'aNDREWS B.Sc, Eng. (Lond.) MEMBEk OF THE CONCRETE INSTITUTE; LECTUKEK IN THE ENGINEEUING DEPARTMENT OF THE GOLDSMITHS' COLLEtJK, NEW CROSS, AND AT THE WESTMINSTER TECHNICAL INSTITUTE ; FORMERLY DEMONSTRATOR AND LECTURER IN THE ENGINEERING DEPARTMENT OF UNIVERSITY COLLEGE, LONDON AUTHOR OF "theory AND DliSIGN OF STRUCTURES," "REINFORCED CONCRETE CONSTRUCTION," "CALCULUS FOR ENGINEERS," ETC. fVlTH NUMEROUS ILLUSTRATIONS, TABLES AND WORKED EXAMPLES NEW YORK D. VAN NOSTRAND COMPANY TWENTY-FIVE PARK PLACE 1916 \All rights reserved '\ Printed in Great Britain by Richard Clay & Sons, Limited, brunswick st., stamford st., s.e., and bungay suffolk. l^ It ^ PREFACE The advance in the application of scientific methods to architectural and engineering problems has made increasing demands upon the theoretical knowledge required by archi- tects and engineers; it is the aim of the present book to present in as simple a method as is consistent with accuracy, the principles which underlie the design of machines and structures from the standpoint of their strength. The subjects commonly called respectively the Strength of Materials and the Theory of Structures have much in common; much of the subject matter contained in the author's books upon the latter subject has, therefore, been Incorporated, the same general method involving the use and application of graphical methods in preference to purely mathematical methods having been adopted in the other branches of the subject. An attempt has been made to present more clearly than is general the various theories as to the cause of failure in materials and the effect of these theories upon design. Although the author hopes that the book will be especially useful for students reading for the Assoc. M. Inst. C. E., and University degree examinations in Engineering, he has attempted to present the subject in sufficiently practical form for it to be of greater assistance in practical design than is the case with an ordinary class book; with this in view many diagrams and tables have been incorporated for enabling the formulae to be applied with a minimum of time and trouble. A large number of numerical examples are worked out and further exercises are given ; the student is recommended to vi PREFACE work for himself all such examples and to pay particular attention to the assumptions which are made in deriving the various formulae. Nearly all engineering formulae are only approximately correct ; in the present branch of the subject this is chiefly because there is no material known which conforms exactly to the simple laws of elasticity upon which the subject is based. We cannot condemn too strongly the blind application to a particular practical problem of formulae which were never intended to be so applied ; the unfor- tunate distrust which practical engineers so often have to " theory " is to some extent brought about by the fact that the theories that they see employed are often inapplicable. It is essential for us to acknowledge the limits of theoretical methods and not to attempt to express our results to a greater degree of accuracy than the nature of the problem will allow. The author's thanks are due to Mr. J. H. Wardley, A.M.I. C.E., for much assistance and valuable criticism in the reading of the proofs; to ]\Ir. W. Mason, D.Sc, for the photograph from which Fig. 24 was made ; and to the various firms who have courteously assisted by supplying illustrations and descriptions of the various testing machines and apparatus with which their names are associated in the text. The author's indebtedness should also be recorded to the many text-books and periodicals that have been consulted and are referred to in the various portions of the book. The author will be grateful for the notification of clerical and other errors that may be found in the book. EwART S. Andrews. Qoldsmitha* College, New Cross, S.E. May 1915. CONTENTS CHAPTER I pa(;k Stress, Strain, and Elasticity 1 Kinds of stresses and strains — Hooke's Law — PoiSson's ratio — Stress-strain diagrams — Elastic moduli and the relations between them — Principal stresses — Ellipse of stress — Maximum strain due to combined strains — Resilience — Impact and temperature stresses — Stresses in heterogeneous bars. CHAPTER II The Behaviour of Various Materials under Test . 41 Properties other than elastic — The cause of failure of material — Cast iron — Steel, wrought iron and other ductile metals — Real and apparent tensile strength — Liider's Lines — Overstrain — Timber — Stone, concrete, cement and like brittle materials — EfiFect of relative breadth and height of compression specimens — Tensile and shear strength of concrete — Adhesion between concrete and steel. CHAPTER III Repetition of Stresses ; Working Stresses . . 84 Wohler's experiments — Natural elastic limit — Effect of speed upon results — P'atigue of metals — Sudden change of section — The conflict between theory and practice — Working stresses and factor of safety — Allowance for live loads. CHAPTER IV Riveted Joints ; Thin Pipes ..... 102 Forms of rivet heads and joints — Methods of failure — Efficiency of joint — Practical considerations — Thin pipes and cylinders — The collapse of thin pipes under external pressure. CHAPTER V Bending Moments and Shearing Forces on Beams . 121 Cantilevers and simply supported beams ; standard cases — Graphical construction — Relation between load, shear and B.M. diagrams — A template for B.M. diagrams — B.M. and shear diagrams for inclined loads. vii viii CONTENTS CHAPTEK VI PAGE Geometrical Properties of Sections .... 161 The determination of areas — Sum curves — Centroid- — Moment of inertia and radius of gyration — Momental ellipse or ellipse of inertia — Polar moment of inertia — Graphical constructions — Formulae for standard cases. CHAPTEE VII Stresses in Beams 192 Neutral axis — Assumptions in ordinary beam theory — Moment of resistance — Unit section modulus; beam factor — Discrepancies between theoretical and actual strengths of beams — Diagonal square sections — Influence of shearing force on stresses — Moment of resistance in general case. CHAPTER VIII Stresses in Beams {continued) ..... 215 Reinforced concrete beams — Straight line, no-tension method — General no-tension method — T beams — Combined bending and direct stresses — Beams with oblique loading. CHAPTEE IX Deflections of Beams ....... 248 General relation in bending — Curvature — Mohr's Theorem and graphical investigation of general and standard cases — Mathe- matical investigation of standard cases — Resilience of bending. CHAPTEE X Columns, Stanchions and Struts .... 279 Buckling factor and slenderness ratio — Methods of fixing ends, Euler's, Rankine, Straight line, Johnson, Gordon, Fidler and Lilly formulae — Reinforced concrete columns — Braced columns — Eccentric loading of columns. CHAPTEE XI Torsion and Twisting of Shafts 311 Stresses in a shaft coupling — General case of grouped bolts or rivets — Circular shafts ; horse-power transmitted ; solid and hollow shafts — Combined bending and torsion — Shafts with axial pull or thrust — Torsional resilience — Torsion of non- circular shafts — Effect of keyways in shafts. CONTENTS ix CHAPTEE XII PAGE Springs 336 Time of vibration — Close-coiled helical springs — Safe loads on springs — Open-coiled helical springs — Leaf or plate springs — PivSton rings — Plane spiral springs— Close-coiled helical springs under bending stress. CHAPTEE XIII The Testing of Materials .... . . 364 Testing machines of various types — Calibration of testing machines — Grips and forms of test piece — Extensometers — Autographic recorders — Torsion testing machines — Torsion meters — Repetition stress machines. CHAPTEE XIV The Testing of Materials {continued) . . . 396 Impact, ductility and hardness testing — Arnold ; repeated impact and Izod's impact machines — Brinell hardness test — Testing cement and concrete — Tension tests ; compression tests ; special gravity, fineness, soundness and setting tests for Portland cement — Thermal and optical methods of testing materials. CHAPTEE XV Fixed and Continuous Beams 416 Effect of fixing and continuity — Fixed beams with uniform, central and uniformly increasing loads — Fixed beams with asymmetrical loading — Continuous beams of two equal spans — Effect of lowering central support — Theorem of Three Moments — Table of results for equal spans and uniform loads — Graphical treatment of continuous beams — Beams fixed at one end and supported at the other. CHAPTEE XVI Distribution of Shearing Stresses in Beams . . 470 Horizontal shear^ — Distribution of shear in standard cases — Graphical treatment — Deflection due to shear-strain of cross section of a beam due to shear. CHAPTEE XVII Flat Plates and Slabs 488 Slab coefficients — Bach's theory for circular slabs — Grashof's results — Grashof-Rankine theory for square and rectangular slabs — Bach's theory for square and rectangular slabs with various modifications. X CONTENTS CHAPTEE XVIII PAGE Thick Pipes 510 Lame's Theory ; variation of hoop and radial stress ; maxiiauni shear stress ; maximum stress equivalent to strain ; diagram for designing thick tubes — Shrinkage stresses in a compound tube, and strengthening by initial compression. CHAPTEE XIX Curved Beams . 529 General analj'sis of stresses — Winkler's formula ; graphical treatment ; Resal construction, special cases of rectangular and circular section ; correction coefficients — Andrews -Pearson formula — Rings and chain links. CHAPTEE XX EoTATiNG Deums, Disks AND Shafts .... 552 Thin rotating drum or ring — Revolving disk ; disk of uniform strength — Whirling of shafts ; critical speeds ; centrall}'- loaded and unloaded shaft ; Dunkerley's approximate formulae. Exeecises ..... .... 571 Appendix — Tables of Properties of British Standard Sections Mathematical Tables Index THE STRENGTH OF MATERIALS Note. — Portions marked with an asterisk may be omitted on the first reading. CHAPTEE I STRAIN, STRESS, AND ELASTICITY Strain may be defined as the change in shape or form of a body caused by the application of external forces. Stress may be defined as the force between the molecules of a body brought into play by the strain. An elastic body is one in which for a given strain there is always induced a definite stress, the stress and strain being independent of the duration of the external force causing them, and disappearing when such force is removed. A body in which the strain does not disappear when the force is removed is said to have a permanent set, and such body is called a plastic body. When an elastic body is in equilibrium, the resultant of all the stresses over any given section of the body must neutralise all the external forces acting over that section. When the external forces are applied, the body becomes in a state of strain, and such strain increases until the stresses induced by it are sufficient to neutralise the external forces. For a substance to be useful as a material of construction, it must be elastic within the limits of the strain to which it will be subjected. Most solid materials are elastic to some extent, and after a certain strain is exceeded they become plastic. B 2 THE STRENGTH OF MATERIALS Hooke's Law — enunciated by Hooke in 1676 — states that in an elastic body the strain is proportional to the stress. Thus, according to this law, if it takes a certain weight to stretch a rod a given amount, it will take twice that weight to stretch the rod twice that amount ; if a certain weight is required to make a beam deflect to a given extent, it will take twice that weight to deflect the beam to twice that extent. is'i 7/7 1 / A / p ■^y Fo int A -^1 ,^ ^ \ y ^Y e/o' Fc int ,/ y "^ Ek fstn .1 ,fm it / < / V^ 1 6 / A - J c - -- ^ ^ / ■oc ■1 OS ■o 3/0 •oc •3 5/5 •oo Stra/n Jber Inch L^enoTh f'rocTure^d ForTi'on Fig. 2. — Stress-strain Diagram for Mild Steel. beyond the point rapidly gets smaller, so that the load may be decreased, and still keep the stress the same. In practice, it is very difficult to diminish the load so as to keep pace with the decrease in area, so that this last portion of the curve is very seldom accurate, and has, moreover, little practical importance in commercial testing because the 6 THE STRENGTH OF MATERIALS maximum stress is always taken as that given at d (see also p. 52). The specimen draws down at the point of fracture in the manner shown in the diagram. Before the test, it is cus- tomary to make centre-punch marks at equal distances apart along the length of the specimen. The distance apart of these points after fracture of the specimen indicates the distribution of the elongation at different points along the length. Four such marks, a, h, c, d, are shown on the figure. The greatest extension occurs at the point of fracture, so that on a specimen short length, the percentage total extension will be greater than on a longer specimen. We deal further with this point on p. 55. The stress-strain diagrams in compression and shear for mild steel are very similar to that for tension. In compres- sion it is difficult to get the whole diagram, because failure occurs by buckling, except on very short lengths, where it is very difficult to measure the strains, and in shear the test is best made by torsion, because it is almost impossible to eliminate the bending effect. Now, in torsion, the shear stress is not uniform, so that the metal at the exterior of the round bar reaches its yield point before the material in the centre, and this has the effect of raising the apparent yield point. We shall see later that the same occurs in testing for compression or tension by means of beams. The importance of the elastic limit has been overlooked to a great extent by designers of machines and structures ; but inasmuch as the theor}^ on which most of the formulae for obtaining the strength of beams are based, assumes that the stress is proportional to the strain, it must be remembered that our calculations are true only so long as Hooke's Law is true, so that the elastic limit of the material is a very impor- tant quantity. We shall deal further with this question in discussing working stresses (Chap. III.). Confusion between Elastic Limit and Yield Point. — In commercial testing, it is quite common to use no accurate STRAIN, STRESS, AND ELASTICITY 7 means for measuring the strains (instruments for such measurements are called extensometers , see pp. 371-379). The load on the steelyard of the machine is run out until the steelyard suddenly drops down on to its stops. The " steel- yard-drop " happens when the yield point is reached, but many people call this the elastic limit ; it is also sometimes called the " apparent elastic limit." As will be seen from the diagram, Fig. 2, there is no appreciable error made by this confusion in tension testing, but in cross bending the differ- ence is much more marked, and gives rise to confused ideas. We shall deal further with this point on pp. 207-209. The Elastic Constants or Moduli. — If a material is truly elastic, i. e. if the strain is proportional to the stress, then it follows that the intensity of stress is always a certain number of times the unital strain, or that the , . intensity of stress . . . tvt . i • . , • ratio 7—^1 — : — -. is constant. Now this stress-strain unitai strain ratio is called a modulus. That for tension and compres- sion is generally known as Young's modulus, and is given the letter E ; that for shear is called the shear, or rigidity modulus (G). There is an additional modulus called the hulk or volume modulus (K), which represents the ratio between the intensity of pressure or tension and the unital change in volume on a cube of material subjected to pressure or tension on all faces. Young's modulus is the one with which we shall be most concerned. Suppose that a tension member (a tie as it is called) or a compression member (a strut), of length I and cross-sectional area A is subjected to a pull or thrust P, and that the extension or compression is x, Fig. 1. Then the intensity of stress is -j, and the unital strain is y- P X PZ .* . Young's modulus = E = -r ^ -r = * — ^ A I Ax Numerical Example. — A mild-steel tie bar, 12 ins. loyig and of \\ ins. diameter, is subjected to a -pull of IS tons. If the extension is '0094 in., find Young's modulus. 8 THE STRENGTH OF MATERIALS Area of section of 1| ins. diam. = 1*767 sq. ins. 18 .-. Stress per sq. in. = ..^pr- = 10-19 tons per ' ' sq. in. 0004 Unital strain = .^ = 000783 .-. Young's modulus = ^-^^ = 13,000 tons per ^ -000783 sq. m. The value of Young's modulus can be found from the stress-strain diagram. Thus, in that for mild steel, Fig. 2, E = -i X -r stress Xow in the relation E = - ! , if the strain is equal to 1, strain ^ i. e. if the bar is pulled to twice its original length, we have that E = stress, and this accounts for the definition of Young's modulus that some writers have given, viz. : Young's modulus is the stress that is necessary to pull a bar to twice its original length. Some students find this definition more clear than the one previously given, but it must be remem- bered that no material of construction will pull out to twice its original length without fracture. Relation between Elastic Constants. — For an elastic material there will be certain relations between the elastic moduli E, G, K, and Poisson's ratio — . These relations can be found as follows — To first find a relation between E and K consider a cube of unit side subjected to a pull ff, Fig. 3 (a). Let the elongation along the axis be a, and let the trans- verse contraction be b. Then original volume of cube = 1 strained volume of cube = (1 + a) (1 — 6)- = 1-26 + 62 a-a-2o/;-a62 = I -T- a — 2b (nearly) because as the strains are ver}' small their product ma}' be neglected. .• . Increase in volume = (l + a — 2 6) — 1 = (a - 2 6) STRAIN, STRESS, AND ELASTICITY 9 Now apply a pull to each side of the cube. There will now be three pulls, each joroducing an increase of volume equal nearly to {a — 2 6). /I D >f) Fig. 3. Total increase in volume is nearly equal to 3 {a — 2 h) 2 b' 3a 1 - a AT ^ _ transverse strain _ a longitudinal strain 10 THE STRENGTH OF MATERIALS . • . Increase in volume = 3 a ( 1 — 2r}) .' . Since original volume = 1 increase in volume , . ^ . ■ « /i « ^ ^-. — J — -, = volume unital strain = Sail — 2'n) ongmal volume ,, intensity of pull /^ Now K = -T— I'S^ — • — = '5 — n — o~T unital strain S a [l — z-q) , ft tensile intensity of stress ^^ . , , and ' = ^— 1^ r,-^— 7 — -. = Young s modulus a unital tensile strain = E E •*• ^ = 3(1 -2-r,) ^^^ Now find the relation between E and G as follows — Suppose that two shearing forces of intensity s are applied to the faces of a unit cube a b c d, Fig. 3 (6). Now consider the equilibrium of the portion a d c, Fig. 3 (c). To balance the forces s there must be a force pulling /, along the diagonal A c, and the value of f, must be V2 x s. Now the area over which this force acts will be V2 since the cube is of unit Y^2 s side, so that there will be a tensile stress of - .— = s. V2 Similarly considering the portion BCD, Fig. 3 {d) there must be a compressing force /, along the diagonal b d, and ■\/2 s the compressive stress will be = , ^ s. Therefore we see that : Two shear stresses on planes at right angles to each other are equivalent to tensile and compressive stresses of in- tensity equal to that of the shear stress at right angles to each other, and at an angle of 45° to the shear stresses. Now the cube will be deformed to the shape AiBjCiDi, Fig. 3(e). The unital shear strain is measured by the angle of distor- tion 2 ?) (2) • • G Now we have already shown in (1) that E K 3(l-2.y) From (2) rj = ^ - 1 2G From (3) y = 1 E 2 6K A _ 1 _ 2G 1 E 2 6K 2VG^ 3k; 3 ' 2 1 1 3 G"^ 3K ~ E 9 3 1 or ^^ = ^. + , E G ^ Iv (3) (4) 12 THE STRENGTH OF MATERIALS This expresses the relation between the constants in its simplest form. It will be noted that if ry = |, as some authorities state, E then ^ = 2-5; this may be taken as true if the value of G for the material is not known. Strains in different Directions.^ — Suppose that the stresses in a material acting normally to, i. e. at right angles to, three planes at right angles to each other are /^, f,„ /.. Then the unital strain in the first direction is made up of the direct strain due to /j. and the transverse strains due to the other two. .*. unital strain in first direction = 5, = 1^ — ' ^^ " ^ E E / yj (f -L. f ) unital strain in second direction = s„ = J. — -^ ''' E E unital strain in third direction = 5, = (i — ^ ^1' 'y' E E Lateral Strain prevented in one Direction. — Now take the case of a piece of material which is free to dilate or expand in one direction, but is prevented from doing so in another at right angles, a compression stress /^ being applied in the third direction. Let the z direction be that in which strain is prevented. We then have and let the .t direction be the one with the stress /^. Then if /. is the stress caused in the z axis we have, since i,j = 0, because the material can expand freely in the y direction — E s, ^ j, - q j. * E ^, = - ^ (/. 4- /O 0=/.-r;/,. .-.E^, = -^(1 + .^)/,; -Es, = JA^-T) s. [l-v) STRAIN, STRESS, AND ELASTICITY 13 This result is interesting as showing that the ordinary definition of E only holds when lateral strain is not prevented. * Complex Stress. — Principal Stresses. — It can be shown that Avhen a body is under a complex system of stresses, such stresses will be the same as those due to the combination of three simple tensile or compressive stresses in planes at right angles to each other. Such simple stresses are called the principal stresses. Consider the case of a block of material subjected to pulls P and Q, Fig. 4, in two directions at right angles, and let Q ^ -B Fig. 4. — Principal Stresses. the pull ]jer square inch of the sectional area in each direction be p and q, respectively, these being principal stresses. Consider the stresses on a plane a b inclined at an angle to the force P. The stress p can be resolved perpendicularly and along A B, i.e. normally and tangentially to a b. Now consider 1 sq. in. of area perpendicular to p. The corresponding area along a b will be 1 sin 0' Now the component of p perpendicular to A b wiU be p 14 THE STRENGTH OF MATERIALS sin 0, and the component along a b will be p cos 0, but stress = component of force -^ area. .• . Normal or perpendicular component of stress p along a b = p sin - — : — 7 = p sin^ ^ sni ^ tangential or shear component of stress }j along a b = V cos -^ . ,, = p sin cos sm ^ Now considering stress q, its tangential component to a b will be opposite in direction to that of p, and since in this case the area is -. — j^^^^ — = - — - and the normal and sm (90 — 0) cos tangential comj)onents of q are respectively q cos and q sin 0, the normal comjoonent of stress will be q cos- 0, and the tangential component of stress will be — g sin cos 0, since in this case the tangential components are not in the same direction. .* . Total normal component = /„ = p sin"^ -{- q cos^ . . (1) Total tangential component ^ s = {p — q) sin cos . . (2) Now the resultant of the stresses /„ and s, which we will call /, will be given by a c. i.e./ = V"/;^ + 52 ^ V {p sin^~0^~Y^os^^~f'+'{p-qf sin^ 6* cos^ = V p^ (sin^ + sin2 cos^ 0) + q^ (cos^^m^T+cos^) + 2pq (cos^ sin^ — cos^ sin^ 0) — V p^ sin^ ^ (cos^ + sin^ ^) + q^ cos^ ^ (sin^ + cos2 ^) + - V /?2 sin2 + q^ cos'^ ^ (3) because cos'^ + sin^ ^ = 1. The inclination a of this stress is given b}' /„ p sin^ + Q cos^ d tan a = ^^ =- ; -X-- — zl - ^ 5 ip — q) sm 6^ cos f^ = ?L:^_^1^1^ + ^ (4, [p—q) tan ^ STRAIN, STRESS, AND ELASTICITY 15 If = tan {a - 0) = i ^ ^an a . tan ^ PJ^^^JAJ _ tan e _ {p — q) tan ^ ~ . , p tan^ 6 + q . ^ 1 + f r^r — I . tan ^ _ p tan^ ^ + g — (P — g) t^iJ-^ ^ ~ {p — q) tan ^ + tan {p tan^ ^ + g) _ g (1 + tan^J) _ q ~ p tan ^ (I + tan^ $)~ p tan ^ = ^cote (5) p Unlike Stresses.— If the stresses are unlike, i.e., one tension and one compression, we shall have by similar reasoning f„ = p sin^ ~ q sin^ s = [p -{- q) sin cos It will be noted that for p =' q and ^ = 45° we have /„ == and s == p. We have, therefore, a pure shear as equivalent to equal tensile and compressive stresses at right angles to each other, at 45° to the shear stress and equal in intensity to the shear stress (cf. p. 10). The Ellipse of Stress. — Draw circles of radius o x and o Y, Fig. 5, equal to q and p respectively, and let o R be drawn at angle ^ to o y. Draw a radius o f to the larger circle at right angles to o R and cutting the smaller circle in e. Draw F H at right angles to o y, and e g at right angles to F H, and join o G. Now o H = o F cos (90 — 0) = p sin 6 and G H = E K = o E sin (90 — ^) = q cos . • . o G = V o H^ + H G^ = V p^ sin^ + q^ cos^ 6 .• . by equation (3) o g = / 16 THE STRENGTH OF MATERIALS Now tan h O G = i^r = • . /, = cot ^ = tan /•5- *^ M Fig. 5. — Ellipse of Stress. centre o, at right angles to a given direction to the outer circle, and drawing f h horizontal to meet the ellipse of stress in g, then o g gives the resultant stress on a jDlane in the given direction, and the angle g o r = a gives the angle between such resultant stress and the plane. Numerical Example. — Suppose a square bar of 2 ins. side and 4 ins. long is subjected to pulls of 10 and 12 tons respectively in axial and transverse directions. Find the resultant stress on STRAIN, STRESS, AND ELASTICITY 17 a plane inclined at 60 degrees to the axis, and find the inclination of the stress to that plane. In this case V = ^ =" 2'5 tons per square inch, V and q 4 12 1*5 tons per 'square inch. Then Fig. 5 shows the elhpse of stress drawn to scale. Draw o L at 60° to o Y and draw o m at right angles to o l to cut the outer circle in m ; drawing m n horizontal to meet the ellipse of stress m n, then o n gives the resultant stress fS. Fig. 6. — Combined Normal and Shear Stress. and / L o N gives its inclination to the plane, o n will be found to be 2' 29 tons per square inch, and / l o n to be 79°. Now considering again the stresses p and q and the normal and tangential stresses f„ and s at an inclination 9 to p we see that p and q are the principal stresses corresponding to the stresses /„ and s. Now in practice we often require to find the magnitude and inclination of the principal stresses, because one of these stresses will be that of maximum intensity of stress. This is clear from the figure of the ellipse of stress, since o Y is obviously the maximum radius vector of the ellipse. We will now therefore find the principal stresses due to a normal stress /„ and a shear or tangential stress s at right angles to each other. 18 THE STRENGTH OF MATERIALS * Combined Normal and Shear Stress. — To investi- gate this problem Ave must first j^rove that a shear stress must always be accompanied by an equal shear stress at right angles to it. Take for example a unit cube, Fig. 6 (a), sub- jected to shearing forces S along opposite sides. These forces S form a couple, and the cube can be kept in equilibrium only by another couple of equal moment and opposite sense, which couple is given by shearing forces S^ at right angles to S. Now consider the case of a complex system of stress con- sisting of a normal stress / and a shear or tangential stress s. Let p N, Fig. 6 (6), represent a portion of the plane on which the stresses / and s act. Let one of the planes of principal stress be represented by p M, and let this principal stress be p. Then along m n there acts a shear stress also of intensity s. Then the resolved portions of the forces due to p and to the stresses / and s must be equal in the directions p n and m n. Therefore we have / . P N + 6- . M N = ;/J . P M COS (1) also 5 . p N = /> . p M sin (2) „ , , , P N M N . • . Jb rom ( 1 ) / ' + s =^ p cos ^ ' ' PM p M ^ i. e. / cos 6 -}- s sin 6 = p cos • ' • iV ~ f) cos 6 = s sin (3) P N From (2) s = p sin 6 ^ ' p M ^ .' . s cos = p sin (4) .*. Dividing (3) and (4) we have P - / _ « s p p {p — f) = s^ P" — p f — s^ =^ -^=i(i±Vi+f) (5) STRAIN, STRESS, AND •ELASTICITY 11) The minus sign corresponds to the second principal stress q, which will be in compression ; as we are concerned only with the maximum stress, we will take the positive value, viz. — • ^-K^+V^") (^) The direction of the plane at which this stress occurs is given by 6. This is found as follows — From (3) j) cos — f cos = s sin 6 From (4) p sin ^ s cos 6 s cos 6 .' . p =^ sin S COS" u ! p. • /I in\ .' . . ,, — / cos = s sni 6 ( / ) sm 6^ ' .' . s (cos^ — sin- 0) =^ f sin cos , sin 2 ^ .' . s cos 2 6 = f — ^— or tan 2 ^ - ^- (8) This will give two values of 0, 90° apart, and so gives the inclination of both planes of principal stress. Maximum Shear Stress. — Returning to the consideration of the principal stresses p and q, we saw that the tangential component on a plane at angle to p was given by {p — q) sin cos 6. (See p. 14, equation (2)). Now this will be a maximum when sin cos ^ is a maximum, i.e. when — ^ — is a maximum, or when = 45°. Therefore we see that the maximum shear stress occurs at 45° to the principal stresses, and is equal to ^^^— «— • In the problem that we are considering, we have proved that ^ = K^ + a/i + ^) ^""^ ^^^^ ^ = 2 (^ " a/i + ^^' 2 ~ 2 V ^ /2 20 THE STReJv^GTH OF MATERIALS . • . Maximum shear stress = ^ . i ^^^ (9) OJ- = \/ J+^' "(10) \4 The latter form is more convenient because in the case when / = 0, the former gives an indeterminate result. Numerical Example. — A steel bolt, 1 in. in diameter, is subjected to a direct pull of 3000 lbs. and to a shearing force of 1 ton. Find the maximum tensile and shearing stresses in lbs. per square inch, and the inclinations of the directions of the stresses to the longitudinal axis of the bolt. T ,1 . , 3000 3000 in this case / = . - . ~^.t = --^^ . area of 1 m. bolt '7854 = 3819 lb. per sq. in. s = .>y^w = 2852 lb. per sq. in. . • . Maximum tensile stress = p =^ ' i\ -\- ^J \ -\- --- 3819 /, . /,' . 4 X 28522\ {^-^^-^^) = 5342 lb. per square inch. Inclination of jmncipal plane to plane perpendicular to axis is given by , ^ . 2 5 2 X 2852 ^^^^^^=J- 3819- = 1-494 .-. 2^ = 56° 12' nearly .-.^ = 28° 6' .• . Inclination to longitudinal axis = 90 — 28° 6' = 61° 54' STRAIN, STRESS, AND ELASTICITY ? Maximum shear stress = + s' 38192 A/ "^ + 28522 - 2852 y 1 + 4 >,-2852^ = 2852 V 1 + -448 = 2852 X 1-203 = 3428 lb. per square inch. 21 ■ 'i, N s i A JoC' d Fig. 7. This stress will occur at 45° to the direction of principal stress, i. e. at 61° 54' — 45° == 16° 54' with longitudinal axis, or else at 90° to this, i. e. at 73° 6' with longitudinal axis. * Combined Shear Stress and two Normal Stresses at Rig-ht Angles to each other. Next consider the case of a shear stress s combined with normal stresses /^, f,, at right angles to each other (Fig. 7). Then resolving as before we have /:,?N + 5MN =2^.PM cos (11) /,/ N M + <5 P N = ;p . P M sin ^ (12) 22 THE STRENGTH OF MATERIALS From (11) /., cos -\- s sin --- p cos From (12) /, sin ~\- s cos = p sin i. e. {p — f^) cos ^ = 5 sin ^ (13) {p — f,) sinO =- s cos 6 (14) We therefore have by multiplying and cancelling sin 6 cos 6 [p - U) iP - h.) - ^' i.e. p^^-p (/. + /,) + /.,. /, = ^2 • (15) Solving this quadratic we get P = H(/' + /.") ± V1/.. + /,)— 4177/7^^)} ih + fy)^ /(/.-/.) 2 ±V''^T^+'' ^'^^ * In this case as in the previous one we usually take the positive sign. To get the direction of the principal stresses we have from (13) and (14) {p - /^.) = s tan^ (IV) ip-fy) =scote (18) .*. subtracting (18) from (17) (/. -fy)-^ (cot - tan 0). __ ^ _, 2tan^ 2 _ 2 Now tan 2 ^ = i-tan ^ ^ ZZT^a^O ~ ^"^ " *^^ ^ tan i.e. ^ot0-t^n6 = ^^^^ • • ^'^ '^'' tan -2 6* 29 z.e. tan2^= , _ , (19) \h I III [p — q) The maximum shear stress is as before equal to ^ — • . Maximum shear stress = ^J '" . — h 5^ Graphical Representation of Results.— The following STRAIN, STRESS, AND ELASTICITY 23 graphical construction, due we believe to Professor R. H. Smith, solves these equations. Fig. 8. — Graphical Construction for Combined Stresses. Set out o A, Fig. 8, to represent /;, to a convenient scale and o B to represent fy to a convenient scale ; if /^ and fy are opposite in sign they should be set out in opposite directions. 24 THE STRENGTH OF MATERIALS Bisect A B in c and at B set up b d at right angles to o a to represent the shear stress s ; then with centre c and c d as radius draw a semicircle, cutting o a in f and e. Then o f = ^ and o e = g. If E comes on the other side of o, the stress is negative. Join D F, then /d f b = 6, the angle of the principal stresses. Also maximum shear stress = c e. Proof,— b c = ^2^ = f^^' C D = Vb C2 + B d2 = j(t_^ + 5^ .'. OF = OC + CF = OC+CD (/. + /.) , Kh-h) 2 OE = OC — CE = OC — CD \2 + ^r-'^^r^ + s^ = p CE = C D = 2- - V 4 "^^ -^ if — f )^ " o ~ + 5^ = maximum shear stress. Now /b c D = angle at centre = 2 d f b , B D s 25 2 . • . From equation (19) b c d = 2 ^. . * . /d f b = ^. Application to a Single Normal Stress. — In this case, Fig. 9, B and o coincide so that oc = |oa = ^, and the construction comes as shown. This figure has been drawn for / = 3819 and 5 = 2852 as in the numerical example of p. 20. * Maximum Strain compared with Maximum Stress. — In questions involving complex stresses it is necessary to remember that the maximum strain does not occur on the same plane as the maximum stress. There is some con- STRAIN, STRESS, AND ELASTICITY 25 siderable divergence among elasticians (a term suggested by Professor Karl Pearson, F.R.S.) as to whether the ultimate criterion of strength of a material depends on the tensile or compressive stress exceeding a certain value, or the shear stress exceeding a certain value, or on the strain exceeding a certain value. This is dealt with on pp. 42-48. We have considered the cases of maximum tensile or compressive and shear stresses. We will now consider the question of maximum stress. Fig. 9. The given stresses are equivalent to simple stresses j), q at right angles to each other; we will assume p to be greater than q* . . Strain in direction of p P_vq E E [t] = Poisson's ratio). Simple stress in direction of p to cause same strain = Equivalent direct stress p, = p — -q q rjf = 2(i + V^ + 7? 2 \ 1-A 1 + 45' 7= 2 {{1 - V) +:(1 + v)^Jl + *f } (11) 26 THE STRENGTH OF MATERIALS In the direction at right angles there will be an equivalent stress equal to q — y p which comes equal to / / )(l_,,)_(l-f-^)yia-4f| (12) These formulae may also be derived from first princijiles as follows. Suppose a rectangular block a b c d receives two tensile strains at right angles and a slide strain in the same plane. Under the combined strain the block assumes the position ^,--^^- Ci i ' c ' ^ ^ ■ 1 y^^ y ' y^ :Xe 'y^ ac 1^, I / I I I ' I / I / I / I' B ■ " "B, "iz Fig. 10. — Combined Strains. A Do Co B^. Then, if ab = .r, b c = y, and a c = r, and Xi yi Ti are the strained lengths, and /Dj a d., = /? -i' J' Unital strain in direction .r = s^ = — X yx - y y _ 'I . • . We have x\ ^ x (1 - -5,) (1) (2) (3) yi = //(!- s.) r^ = r (1 -h s,) = rH\ -^s,: Since squares of strains may be neglected. (4) Now r^ = A Cg^ = A Bg^ + C2 B 2 STRAIN, STRESS, AND ELASTICITY 27 A B2 + C2 B2 = (A Bi + Bi B2)2 + A Di2 = (ABi + DiD2)2 + ADi^ Now A Bj = a^i = X (1 + Sx) A Di = 2/1 = y (1 + 5,) since ^ is small and therefore ^ x 5^ is of second order and therefore negligible. .-. ri2= {x(l+5.) +^2/}' + {2/(1 +^.)!' = x'{\ + 2s^) + 2xy/3 + y^l +2s,),...{5) neglecting all second powers of strains, but r^ = x^ + y^ .-. rj^^ = r^ + 2x^8, + 2y''-s, + 2xy /3 (6) . • . From (4) r2 (1 + 2 s,) = r^ + 2 x^ s, + 2 y^ s, + 2 X y f3 or..= g)%.+ (f)%,+ ^2^^ (7) Expressing this in terms of the angle we get S0 = Sr cos^ 6 + s,, sin^ 0^/3 sin cos ^ (8) Our next problem is to find the value of 6, for which the resultant unital strain Sq is a maximum. This occurs when ,-/ = a $ i.e. when s^ . 2 cos d ( — sin e) + Sy 2 sin a cos a + j8 (cos cos + sin [ — sin 6]) = i. e. when — Sr sin 2 ^ + 5^^ sin 2$ + /3 cos 2^=0 sin 2e{s, - Sy) = ^ cos 2 6* w or tan 2$ = — ^ — ..(9) This gives two values of at right angles, and so we see that the directions of maximum strain are at right angles. Now consider equation (8), reuniting and putting 1 = cos^ + sin^ 0, we get S0 (cos^ + sin^ 0) = s,. cos^ + Sy sin^ 6 + (3 sin $ cos $. 28 THE STRENGTH OF IMATERIALS Dividing by cos^ 6, we get Se (1 + tan^ 0) = s^ -\- s,, tan^ -{- /S tan or tan- 6 {s,j — Se) + /3 tan 6 + s^ ~ Se = I.e. tan Q - — ^{s,-s^ — For this to be real, /32 must be not /) ± (1 + ^) V(/.^ W^TIT^J . . (13) * Shear Strain equivalent to Two Direct Strains at Right Angles. — We will now consider from first principles STRAIN, STRESS, AND ELASTICITY 31 ill a similar manner the shear strain equivalent to two direct strains at right angles to each other. Let a rectangular block a b c d, Fig. 11, become strained to the form a b^ c^ d^, the extensions being shown to a greatly exaggerated scale, then we have AB, = y {I + s,) (14) A Bi = X (1 + S,r) (15) .*. Neglecting squares of strains A Cj^ = A Bi^ -f A Bj^ = X2 (1 + S,,)^ + 2/' (1 + 5,)2 = x^ -h y^ -h 2x^ s,,. -\-2ifs,, (16) = (a'2 + 2/2)(l + 2 5,) =. a;2 + 2/2 + 2 a;2 s. + ^y^-s, (17) .♦ . A Ci2 - A E22 = 2 X'2 {S,, - 5,) (18) Now A C^^ — A Eg^ = (a Ci + A Eg) (A C^ — A Eg) = 2 A Eg . Ci Eg approx. ^ X [S,. S,i) t -t r\\ .•. Ci Eo = ^ '-' 19) ^ ** 2AE2 Further Eg Eg = c^ Eg tan 6 (very nearly ; strictly X y {s,, — s„) tan^- 8$) = CjEg^ X . * , Eg Eg AE but SO = 2 — ( = — ^. — approx. ) A Eg \ radms ^ ^ / AEg^ {s, - Sy)xy (l + 25,)(x2 + 2/') 32 THE STRENGTH OF MATERIALS This will be a maximum when + " is a minimum, i. e y X tan $ + cot ^ is a minimum. sin , cos 6 sin^ d + cos^ S tan + cot cos 6^ ' sin sin 6^ cos 6 1 2 J sin 2 ^ sin 2 The minimum value of this = 2 when sin 2^ = 1, i.e., (9 = 45°. .• . Maximum shear strain occurs at 45° to direct strains. CJ ^x S,i • ■ • ' ~ Y{rv2s;) = ' ^ ' (1-2 5,) approx. = ^—^ — ^neglecting products of strains. There will be an equal angular distortion on the other diagonal. .' . Total angular distortion = shear strain = 28^ = {s^ — s,j). . • . Equivalent maximum shear stress = shear strain x ,G = ( tons per sq. in. 2240 „ = Tjge lb. per sq. m. . _ Stress _ 2240 •' • ^^^^"^ ~ "E ~ -196 X 30 X 10« STRAIN, STRESS, AND ELASTICITY 37 Now ^" = strain x original length r. - • ^^ 4.1. ^ '196 X 30 X 10« .*. Original lenefth = ?^~-.- = ^^.^ r. ^ ^ Strain 2240 x 8 .' . Volume = length x area of section. _ -196 x 196 X 30 X 10 « 8 X 2240 . = 64'33 cub. ins. Work done by 150 lbs. in falling 3 inches = 3 X 150 = 450 in. lbs. 64-33 X /2 = 450 /9( ^| 64-33 2E r _ /900lE 900 X 30 X 106 64-33 = 20,480 lbs. per sq. in. Ans.* Temperature Stresses. — Suppose a bar of length I is heated t° F. and a is the coefficient of expansion. Then, un- less prevented, the length of the bar will become I (1 x at), i. e. the increase in length will he at I. If the bar is rigidly fixed so that this expansion cannot take place, then there will be in the bar a strain equal to at I, and the unital strain will be -j— = at. This strain will produce a compressive stress of a ^ x E, where E is Young's modulus. Now for mild steel a = -00000657 per degree Fahrenheit, and E = 13,000 tons per square inch. .-. The stress per °F. = '00000657 x 13,000 = -0854 tons per square inch. Taking a range of temperature of 120° F., the stress due to temperature = 120 x -0854 = 10*25 tons per square inch. This is more than the safe stress for mild steel, so that the importance of designing so that the expansion may take place becomes quite evident. * This problem could be solved if E were not given; it would be found to cancel out. 38 THE STRENGTH OF MATERIALS "^ Heterogeneous Bars under Direct Stress. — If a bar, composed of two different materials — such as steel and concrete, or steel and coj)per — firmly connected to each other, be subjected to a pull or a thrust, the two materials must be strained by equal amounts, and since the values of Young's modulus for the two materials are different the stresses in the two materials will be different. Suppose one material has a cross-sectional area A and Young's modulus E, the resulting stress being / ; and let the corresponding quantities for the other material be A^, E^, Z^. Then, if iinder a pull or thrust P the unital strain is a*, we have I E h El and P = A / X = X = (1) Ai/r (2) (3) Fig 16. A / and Aj ii being the loads carried by each of the materials. E,/ Prom (1) and (2) /^ = Ej^x .-. P = /(A- E El A, ) E ' or / = Afl^^tV (4) (5) Now if a new bar is taken wholly of the first material of such area A., that the stress under a load P is the same as that in the compound bar, we have P A2 El A, ^ / or A2 = A ( 1 -T- EA (6) /A = — ^ (8) ^ + EA STRAIN, STRESS, AND ELASTICITY 39 This quantity Ag may be called the equivalent area of homo- geneous material, and the consideration of this problem has become in recent years much more important on account of the progress made in reinforced concrete construction. Returning to the general problem we see that p The load carried hj the first material then comes equal to P EA and that carried by the second comes equal to /iAi = --Va (9) Since these are not the same there will be an adhesive force tending to make one material slide relatively to the other. This adhesive stress may be computed as follows assuming that the load is applied uniformly. Load per sq. in. = — r — ■ — r— (A + Ai) . • . Load actually distributed to area A - A ^ (A + A,) p Load carried = t^ . (from 8) EA E Difference = load carried by adhesion calling ^ = _ P P. A '"'-, ,_A, A + Ai ^ + mA _ p J m A A "" ( m A + A'l ~ Ai + A _ P A A i (m - 1) ~ (A + Ai)XArrf wA) _ / iAAi( m -^1) (A + Aj) m 40 THE STRENGTH OF IMATERTALS Numerical Example. — A reinforced concrete column {Fig. 17) for which m = 15, is 20 inches square and has 4 1^" steel rods embedded in it. Find the load on the column when the stress in the concrete is 450 lbs. per sq. in. and the adhesive force. T Fig. 17, TT 4 X ^ X 1-252 = 4-91 in. 4 In this case A Ai = 20 X 20 - 4-91 = 395 nearly .-. P = 450 X 395 1 + 15 X 4-91 Adhesive force 395 210,870 lb. nearly 45 X 4-9 1 X 395 xU 400 (from 7] = 33,940 lbs. nearly CHAPTER II THE BEHAVIOUR OF VARIOUS MATERIALS UNDER TEST Properties other than Elastic. — In addition to the elastic properties of materials there are other strength properties which are of very great importance in the practical use of the materials. Ductility is the property of a material which allows it to be worked without cracking ; the strict use of the term refers to the capacity for being drawn out which a ductile metal possesses. Malleability is the property which allows a material to be hammered out and is very similar to ductility. Brittleness is lack of ductility or malleability. Hardness may be defined as the power of a material to resist denting by another material. (For tests for hardness see pp. 396-404.) The above properties are all relative ones and vary with the same material according to the treatment which it receives ; thus by "tempering " a metal we harden it and by " anneal- ing " it we soften it or render it more ductile, and some metals are hardened by plunging them into water when heated, whereas others are annealed by the same process. With reference to ductility it is important to remember that so long as a metal maintains its elasticity it has no ductility, i.e.au metal which possesses ductility cannot exhibit the fact until the yield point has been reached. In our calculations for the strength of various details we shall base 41 42 THE STRENGTH OF MATERIALS nearly all our formulae on the assumption that our material is elastic and so we must not expect the formulae to hold after the elastic Hmit has been reached. This is a point of very great importance. Apart from the convenience in the manufacture of articles which ductility gives, it has considerable value from the point of view of safety and strength, because a material does not lose its strength when it first starts drawing out, and the yield may either give us timely warning of excessive load- ing or, in the case of steam boilers and like fluid-retaining devices, the yielding may actually remove the excessive pressure. As we shall see later, however, the effect of taking a material beyond its yield point is to harden it. The usual test for ductility is the elongation in fracture by tension. * The Cause of Failure of Materials under Stress. — In recent years a very large amount of attention has been given to the question of the cause of failure of materials under test, and it is doubtful if the vital importance of this problem has been fully realised by practical engineers. As we shall see later, however, the choice of safe working stresses really depends in a large measure upon the view taken as to which of the various theories is correct. If we consider the question carefully we shall see that failure cannot occur by compression only ; if a material be prevented from escaping laterally, no amount of compression can rupture it. Even a fluid like water will resist a compression stress of very great magnitude if the vessel containing it is strong enough to resist failure by tension or shear. The late M. Armand Considere showed experimentally that concrete could not be crushed when given adequate lateral support, and he also proved that the very brittle material glass could be bent cold without fracture when placed in a liquid under great hydraulic pres- sure. Marble has been bent without fracture by Professor E. D. Adams of McGill University when placed in steel cylin- ders and compressed, and Professor Ira H. Woolson crushed BEHAVIOUR OF MATERIALS UNDER TEST 43 a cylinder of concrete encased in steel into the form shown in Fig. 18, and yet when the encasing cylinder of steel was removed the strength of the concrete was found to be not appreciably different from that of concrete which had not been similarly treated. The question therefore resolves itself whether tension or shear is the cause of failure, and we have reason to believe that in ductile materials such as mild steel failure occurs by shear and in brittle materials such as cement or concrete by tension. We will return to this after considering the various theories of failure ; there are four principal theories which we will consider. 1. Principal Stress or Rankine Theory. — ^According to Fig. 18. this theory, which was adopted by the great Glasgow professor, Rankine, the failure occurs when the maximum principal stress exceeds a certain value. We have seen (p. 19) that for a normal or direct stress / and shear stress s the principal stress is given by the relation / , 1 , or ?9 = -^ + 2 V/2 + 4^2 (16) and the inclination of this stress to the normal stress and to the shear stress is given by the relation tan 20 = —r. This stress p is the simple normal stress (tension or com- pression) equivalent in effect to the combined normal and shear stresses. 44 THE STRENGTH OF MATERIALS In the limiting case in Avhich the direct stress / is zero we get p = s and tan 2 $ = infinite, i. e. = 45°, i. e. a shear stress is equivalent to a normal stress of the same intensity and is at 45° to it, or the shear and tensile strengths of the material should be equal. 2. Principal Strain or St. Venant Theory. — According to this theory, which was favoured by the great French elas- tician after whom it is named, the failure occurs when the maximum principal strain exceeds a certain value. We have seen (p. 29) that for a normal stress / and a shear stress s the equivalent principal stress is given by the relation P>- L\a -yi) + a +-n)J} 4- ^^^\ 2 j^(l->y) + (l +'/?)yi + y, j- (2) and taking rj = I / f 3 , 5 / 4^1 ^• = 2i4 + 4V^+fJ ^^^^ orp = ^J + g V/^ + 4^2 (36) The inclination of this principal stress is the same as in the previous case. In the limiting case in which the direct stress / is zero we 5s . get p ^ -, i.e. a stress shear is equivalent to a normal stress of four-fifths of the shear stress, or the shear strength of a material is four -fifths of the tensile strength. 3. Equivalent Shear Stress or Guest or Tresca Theory. — According to this theory, which is associated with the name of Mr. J. J. Guest, who was one of the first to carry out careful experiments upon the subject, and is usually attributed to Tresca, failure occurs by sliding of the particles over each other, i. e. by shear. From p. 20 we get Equivalent shear stress = . ' +52 ^4^ and acts at an angle of 45° to the normal stress. BEHAVIOUR OF MATERIALS UNDER TEST 45 To compare a simple shear with a simple tension or compression by this formula we put s = o and we get Equivalent shear stress = ~ i. e. a shear stress is equivalent to a normal stress of twice its magnitude or the shear strength of a material is one-half of its tensile strength. Fig. 19. — Navier's Theory. It is interestmg to note that as shown on p. 32 the equivalent shear stress comes the same whether worked from the point of view of stress or of strain, and so there is logical support for this theory. 4. Sliding with Internal Friction or Navier Theory. — This theory deals with materials subjected to compressive stresses and attempts to explain the fact that short cylinders 46 THE STRENGTH OF MATERIALS of brittle material usually fail by sliding or shearing along a line e f, Fig. 19, inclined at an angle from 55° to 65°, on the ground that the particles are capable of exerting frictional resistances. Consider a short column a b c D of unit sectional area subjected to an ultimate compressive force u,, which causes failure, and consider the forces across a section E f, the ultimate or breaking shear stress in the material being u,. The force u,. acting along the section e f can be resolved into shear and normal components ac, cb respectively, equal to u, sin and u, cos 0. If /x is the angle of friction for the material, the normal component c b causes a frictional resistance equal to /x .c b, i.e. /x .u cos 6. Just before failure the shearing force acting upon E r = u^ X area of section. _ u, X normal section _ u, cos cos (because normal section is of unit area). When therefore failure is about to take place — Total force causing failure = ac = tc^ sin equals Total force resisting failure = au, cos -1 ^ * ^ cos i.e. u, (sin — ix cos 0) = ' ^ ' cos Q or n, = „ , . -.' r. ••••(!) cos Q (sm B — \j. cos b) Regarding w, and \}. as constant we now wish to find the value of which will make u, as small as possible ; this value of will be that at which failure will occur. ii will be as small as possible when cos B (sui — ix cos 0) is as large as possible. Let y — cos (sin — /x cos 0) sin 2 ^ , . = —IX cos- BEHAVIOUR OF MATERIALS UNDER TEST 47 T-i . d y ^ I or 2 ^ - 90° + <^ or == 45° + (3) In support of this theory experiments made by Bouton (Washington University, 1891) may be quoted as follows — Material. Number of Tests. Observed Value of (f) (degrees). Observed Value of e (degrees). 4r + ^ 55-3 53-4 61-7 58-6 58-5 Cast iron „ (different kind) Limestone Asphalte paving Milwaukee brick 24 24 4 3 4 20-6 16-9 33-4 27-3 27-0 54-8 55-0 62-2 59-7 58-2 Batio of shear to compressive strength on Navier theory. — From equation (1) we have Us = u, cos (sin — fx cos 0) Now put in this a = tan (h = — cot 2.^ = v~^ ^ ^ sm 2 rp, . . Us . / . . , COS 2 (9 .^ I his gives = cos ^ sm ^ + -^ — ?r^cos ^ Ur \ sm 2 (9 = cos ^ ( . ^ , (cos^ — sin^ 0) cos 0} 6' ] sm ^ + -^ 5-^ — n -^ } [ 2 sm cos J . r . , , (cos2 - sin2 0)] \ sm + ^ o-^—n } [ 2 sm J f 2 sin2 + cos2 - sm^ 0] ' ^ 1 2^nl / — cos cos ^ / . \ COS 2 sm ^ \ / 2 sm - I cot ^ (4) Taking 6^ = 60 for masonry, this would give shear strength = '289 compressive strength, 48 THE STRENGTH OF ]\L\TERIALS but the most recent experiments on shear strength of concrete (see p. 79) indicate that this is too low. Shear tests are very difficult to make without introducing bending, which tends to give the shear strength too low. An interesting theory which may be regarded as a modification of Navier's theory is outlined by i\Ir. H. Kempton Dj^son in a paper read before the Concrete Institute, December 1914. The first three theories have been very fully tested experi- mentally in recent years by Messrs. Guest, Hancock, Scoble, C. A. H. Smith, Mason and Turner,* and the result appears to be that the shear stress theory is most reliable for ductile materials while the strain theory is most reliable for brittle materials. Professor Ewing and ^Ir. Rosenhain have found by a microscopic examination of the crystals of a ductile metal under strain that beyond the yield point lines of slip are developed in the crystals, thus proving that the failure or yield is a slipping or shear one. Liider's lines (p. 54) are also indications that in ductile metals the failure is by shear. It is possible that ductility is a property of shear strength ; if a material is weaker in shear than in tension the shear causes the failure and slippage occurs before the tensile strength is reached, thus giving rise to ductility. If the material is relatively stronger in shear than in tension the material breaks before slippage occurs and thus cannot exhibit ductility. We will now consider the properties of various materials. CAST IRON We will deal with cast iron first because it is a brittle material and behaves differently under test from most other metals. The strength of cast iron varies considerably with its com- position, but like all brittle metals it is relatively weak in tension and strong in compression. * These experiments will be found fully described and discussed in various articles and letters in Engineering for 1909 and 1910. BEHAVIOUR OF MATERIALS UNDER TEST 49 Fig. 20 shows the stress-strain diagrams for cast iron in tension and compression, the resiiHs of the two tests being plotted on one diagram. It is clear from this diagram that the stress is never strictly proportioned to the strain in tension ; this has an important bearing upon the strength of cast-iron beams (see p. 209). The compression diagram is not continued to failure as the Fig. 20. — Stress-strain Curves for Cast Iron. failure would take place by buckling and injure the instru- ment for measuring the strain. When a cast-iron bar fails in tension, it breaks off " short," i. e. it does not produce a waist as indicated in Fig. 2 for mild steel. When compression tests are made on cylinders which are so short that buckUng effects are practically eliminated, the failure takes place by sliding diagonally as indicated in Fig. 21, and for shorter specimens still cracks sometimes develop which split off the outside portion leaving two inverted cones. Some observations on this kind of failure for concrete E 50 THE STRENGTH OF MATERIALS will be found on p. 69 and apply to cast iron. The diffi- culty in testing cast iron in compression in very short lengths is that it is so strong that difficulties arise as to the strength of the testing machine. The tensile strength of cast iron varies from about 7 to 15 tons per sq. in. in extreme cases, but more usually from 8 to 11 tons per sq. in. Figures for the compressire strength show more variation ; this is probably due to the fact that the size of the test piece, both as regards its length and breadth, affects the result. Fig. 21. — Compression Failure of Cast Iron, The following results of tests made upon J in. cubes by the American Foundrymen's Association show that specimens cut from bars of small cross section give much higher results than those from large. Cross Section Crushing strength in tons per sq. m. for cubes cut from of bar from which Cubes Middle First Second Third Fourth were cut. half inch. half inch. half inch. half inch. half inch. 4x J 69-0 ___ 1 X 1 44-5 49-8 — — — Hx li 37-0 39-4 37-0 — — 2x2 32*2 38-9 34-6 — — 2ix 2i 31-9 35-4 32-3 31-9 — ■ 3x3 28-6 32-5 30-1 28-7 — 3i X 3* 28-4 30-5 29-6 28-8 28-4 4x4 25-4 29-4 27-4 26-6 25-4 BEHAVIOUR OF MATERIALS UNDER TEST 51 Strength of Cast-Iron Beams. — Cast-iron when tested in bending shows an apparently greater strength than when tested in pure tension. This is due to the fact, as explained in greater length on p. 209, that the ordinary formula for beams is not strictly applicable for cast iron. The ratio of calculated breaking stress from bending . „ , ^ i r- p ^ — ., , ^ , . is usually about 1*5 tor tensile breakmg stress rectangular sections of depth twice the breadth ; it increases for round and square sections arranged diagonally and may rise to 3 ; the ratio becomes nearly 1 for I sections with a thin web. The following figures give the mean of a large number of tests made by Kirkaldy for the same kind of iron — Tensile breaking stress = 11 tons per sq. in. Compression ,, =54 ,, „ Calculated bending ,, =17 „ „ The early writers often called the breaking stress calculated from bending tests the " modulus of rupture," but the term is not to be recommended ; bending breaking stress is better. Effect of Temperature on Strength of Cast Iron. — The strength of cast iron increases slightly as the temperature is raised until about 900° F. is reached; it then diminishes rapidly, until at 1100° F. the strength is reduced by nearly 50 per cent, and at 1400° E. by about 75 per cent. Other properties of cast iron are tabulated on p. 83. Malleable Cast Iron. — Cast iron is rendered malleable by surrounding the casting with haematite or manganese dioxide and exposing it to red heat for many hours, depend- ing upon the size of the casting. The result of the process is to dicarbonise the iron and render it similar to mild steel. Some useful information on the subject, especially from the point of view of strength, is given by Mr. C. H. Day in the American Machinist of April 21, 1906. The following results of tests of Mr. Ashcroft are quoted from Vol. CXVII. Proc. I. C. E. 52 THE STRENGTH OF MATERIALS Tons per sq. in. Elongation, ^ on 10 in. Breaking Stress. Elastic Limit. Young's Modulus. 11,620 Tension . . Compression 20-6 21-6 8-94 2-8 10,240 — Bending . . 28-8 12,330 ! — Torsion . . 26-8 — Rigidity Modulus. 4,120 — Stanford, in Trans. Am. Soc. C. E., 1895, gives as the mean result of forty -two tests in tension an elongation of 6" 61 per cent, and an ultimate stress of 22 tons per sq. in. Similar results are given by Mr. Day. STEEL, WROUGHT IRON, AND OTHER DUCTILE METALS Real and Apparent Maximum Tensile Strength. — We have shown already, on p. 5, a stress-strain diagram for mild steel in tension and pointed out that the last portion D E of the diagram was usually inaccurate and of little com- mercial importance. The diagram sloi)es back because the load can be reduced as the area diminishes and the stress still be sufficient to cause fracture. Now this diagram can be corrected if its form is determined very carefully and the areas at the various points are measured. Taking any point a, Fig. 22, on the curve before the specimen began to draw down, we find 6 c by the relation ab X extended length original length tion that the volume keeps constant; so that original area x original length be This is based on the assump- reduced area x extended length ab X extended length original length ab X actual reduced area original area BEHAVIOUR OF MATERIALS UNDER TEST 53 The method cannot be used beyond the point at which the waist begins to form. Taking any point / beyond the point of drawing down, (U (!) sh rc^\n b G Fig. 22. d ^. csc^ r> 1 7 dfx actual area of bar , , , . h 12. 22, we find a e = —^ ^-. — ^ : and by doing ° origmai area "^ ° this for a number of points we get the corrected curve c e e', then f e' gives the real maximum stress as opposed to G D, which is the -apparent or commercial maximum stress. It is very difficult to get points / accurately. 54 THE STRENGTH OF MATERIALS It thus appears that the actual maximum stress at failure is considerably more than usually measured. Upon the Guest theory that failure occurs by shear, the tensile strength should be twice the shear strength, but ordinary commercial tests indicate that it is about 1 J times ; it is probable that if the true tensile strength were compared instead of the apparent or commercial strength, the agreement would be more in favour of the Guest theory.* The common form of fracture of a ductile metal is shown in Fig. 23 and consists of a kind of crater, the angle of the sides being approximately 45°. This strengthens the theory that failure is by sliding or shear. Liider's Lines. — When a highly polished specimen or one Fig. 23. — Tension Failure of Ductile Metal. provided with a very thin layer of scale is tested in tension or compression, lines at about 45° to the axis of the speci- men and of spiral form in the case of round sections are found to develop directly the yield point is reached. This was first noticed by Liider and supports the theory that the yield is really due to diagonal shearing. Fig. 24 shows these lines for thin mild-steel tubes in com- pression and were given in a paper by Mr. W. Mason, M.Sc, of Liverpool University. f Percentage Increase in Length and Decrease in Area. — The percentage elongation of the specimen and the decrease in area are usually regarded as reUable tests of ductility of the material, but it is clearly useless to specify the percentage elongation unless the diameter and the ♦ See a paper by Professor Carus Wilson, Proc. Roy. Soc. 1890. t Proc. Inst. M.E. 1909. Fig. 24. — Luder's Lines. [To face page 54. BEHAVIOUR OF MATERIALS UNDER TEST 55 original length are both specified, because most of the ex- tension occurs in the centre portion where the specimen draws down. This question has been treated very fully by Professor Unwin in Vol. CLV., Proc. Inst. C. E., and the following figures are taken from his paper as indicative of the general results — RoTHERHAM Steel Boiler Plate (area -5830 in.^) Extension in inches in each half inch. Number from Fracture 7 6 5 4 3 2 1 Frac- ture 1 2 •19 |'13 3 •12 4 •12 5 •11 6 7 8 9 •10 10 •09 Extension •07 •08 •10 •10 •11 •13 •18 •47 •11 •lO -09 Percentage elongation and different gauge lengths. Gauge Length (inches) % Elongation . . . 2 4 36^0 6 8 10 48^5 30^9 27-6 25^9 He suggests the formula — cVA % elongation = 100 {pi^ + h\ where A is the original area in sq. in. I is the gauge length in inches h and c are constants for a given material. The following values of c and h are given by Professor Unwin — Metal. c 6 18 10-6 9-7 •8 35 Mild steel Gun metal (cast) . . . Rolled brass .... Rolled copper .... Annealed copper . . . 70 8^3 101-6 84 125 56 THE STRENGTH OF IMATERIALS By the formula we can obtain the jorobable percentage extension on any length of any other metal if that on two specified lengths are known. The percentage contraction in area is not so commonly specified now as formerl}^ because the elongation is considered as giving sufficient indication of the ductility. Two Waists ix a Texsiox Specimen. — It happens very occasionalh^ that two waists form in a tension specimen, failure taking place at the one which draws down most rapidh'. This will affect the elongation and the test should be discarded. Fracture xear one Shoulder of Specimen. — If fracture occurs near one shoulder of the specimen (see Fig. 167) the elongation will be less than normal owing to the effect of the shoulder, and such a test should be discarded. Effect of Abrupt Change of Section upon the Tensile Strength. — The effect of an abrupt change of section such as in a screw thread or a sharp-edge groove does not have a ver}' marked effect upon the ultimate tensile stress of a material Avhen tested in the ordinar}^ way although it does affect the ductility. As we shall show later, however (p. 91), it has considerable effect in tests by repeated loading. The sharpness of the groove will have a weakening effect, but the presence of the larger area near it will have a strengthening effect . Fig. 25 shows the results of some tests by Sir Benjamin Baker * which are interesting ; the breaking stresses in tons per sq. in. are given below each figure. a is an ordinary tension specimen, h and c have saw cuts, at both and one side respectively, d has semicircular notches and e has a central hole with saw cuts at each side; after the saw cuts were made the bars were heated and the cuts closed, d is the strongest ; this would be expected, on the shear theory, as the diagonal line is relatively larger; * Proc. Inst. C. E., Vol. LXXXIV. BEHAVIOUR OF MATERIALS UNDER TEST 57 b is weaker than d on account of the abrupt change of section ; c is weaker still because the load is not central, and in e the hole probably gives an accentuation of the effect of the abrupt edges. Eor other results on the effects of holes see Chap. XIV. Effect of Overstrain. — If a ductile metal is loaded beyond the j'ield point and the load removed, and the speci- men is then loaded up again at once, it is found that the new yield point is higher, but the elastic limit is slightly lower. The overstrain also increases the ultimate or break- ing tensile stress. This is shown in Fig. 26 (a) in which b cc L d ^nV~7V~7^^ 32.-50 Lji\!_i -o 3l-^0 Fig. 25. 36 '30 L_J] is the initial or primitive elastic limit, and a is the initial yield point; at the point c the load is taken off and then the specimen is loaded up almost immediate^; the new elastic limit e is much lower than previously and the yield point d higher. If some hours had been allowed to elapse between taking off the load and reloading, the elastic limit would nearly return to its previous value b but the yield point would go higher still to the point g. In Fig. 26 (6) is shown the effect of keeping the load on for some time at the point c before increasing it further. The curve in full lines shows the effect of keeping the load jfixed for about ten minutes, and in dotted lines k I the effect of keeping it for ten days.* * For fuller information a paper by Professor Ewing, Proc. Roy. Soc. 1880, should be consulted. 58 THE STRENGTH OF MATERIALS This hardening effect of overstrain is well known in prac- tical work. Copper wire becomes very brittle by bending it backwards and forwards, and steel wire in the process of drawing becomes very hard indeed. K ?5 ( 1 ^/ k f'^' I ^ ^""^ ^ ,^ c '/' 20 c^y ^ ^l/ / b h 15 e 10 5 o 5 10 o 5 (h) /o Fig. 26. — Overstrain. Recovery of Elastic Limit from Overstrain. — As in- dicated above, the elastic limit slowly recovers its original value after it has been allowed to rest for a few hours ; it then will increase as the time of rest is extended and BEHAVIOUR OF MATERIALS UNDER TEST 59 ultimately gets above its final value and gets near to its new yield point. Mr. Muir * has shown that the temperature of boiling water gives an almost immediate recovery of the elastic limit to near the new yield point which will be as high as if the material had been allowed to rest for several days. Hardening by Quenching. — The hardening effect of over- strain is not the same as that effected by heating the metal to a high temperature and quickly cooling by quenching. This has the effect of making the metal very brittle, and if) Fig. 27. there is practically no yield point, the specimen breaking off short with practically no extension. Mechanical Hysteresis . — If a specimen of ductile material is loaded up beyond the elastic limit and the load is taken off slowly and the strains noted for descending loads, the stress-strain diagram for descending loads will be found not to coincide with that for the ascending load, the two curves forming a loop as indicated in Fig. 27 ; this, by analogy with magnetic hysteresis, is called a mechanical hysteresis loop. In experiments of this kind great care is necessary to eliminate errors of the instrument on the return, but many experimenters have found similar results well within the elastic limit. Very careful experiments by Mr. Bairstow, however, at the National Physical Labora- tory,! suggest that this phenomenon does not occur unless * Phil. Trans. Roy. Soc, 1889. t ^^id., vol. 210. 60 THE STRENGTH OF MATERIALS the " natural " elastic limit is passed. These " natural " elastic limits are dealt with on p. 87. Tensile Strength of Various Steels and Wrought Iron. — Fig. 28 shows typical stress-strain diagrams for various 60 50 40 30 ZO 10 /a k 1 /r» ^ ^ / ^ ^O^ ^ ^ ^y /i ^^\ \ Extension per cent. — A Tool steel (Unannealed). B Crucible steel. HO 30 C Medium steel. D Mild steel. E Wrought iron. 4o Fig. 28, — Stress-strain Diagrams for various Steels and Wrouffht Iron. kinds of steels and wrought iron; the strength properties depend to a large extent on the heat treatment and amount of " working " in manufacture. BEHAVIOUR OF MATERIALS UNDER TEST 61 Effect of Varying Amounts of Carbon on Strength. — The effect of increasing the carbon in the steel is to increase the strength at the expense of the ductility. The following figures are taken from Harbord and Hall's Metallurgy of Steel (Griffin) for normal steels. Stresses in Tons per sq. in. Percentage of Carbon. Breaking Stress. Elastic Limit. •09 21 9-4 •16 29 13-0 •15 33 13-1 •34 35 11-9 •44 41 16-2 •65 54 18-0 •79 57 20^0 •94 62 . 21-9 The proportion of carbon does not have an appreciative effect on the value of Young's modulus, nor does tempering or other hardening process. Alloyed Steels. — The following figures give mean values for some examples of various alloyed steels. Kind of Steel. Tons per sq. in. Breaking stress. Elastic Limit. Z Elongation. Nickel [•2 % C, 3-2 % M] 42 27 26 on 3 in. Tungsten [7-15 % C, -29 % Mn, -40 % W.] Annealed Unannealed [•46 % C, -28 % Mn, 8-33 % W.] Annealed Unannealed 25-5 310 42-5 64-0 30-6 18 24 25-5 45-0 39-6 on 2 in. 33 32-6 2-57 Vanadium [•20 % C, -27 % V, -48 % Mn] . 25-7 18 49 33-5 on 2 in. Chromium [•4 % C, 5 % Cr.] Annealed Hardened 55 32 24 on 2 in. 12 62 THE STRENGTH OF IVIATERIALS " Quality Factor." — This term has been used by some engineers for the result of adding the breaking stress to the percentage elongation. Stress-strain Diagrams for various Ductile Metals. — Fig. 28a shows tjrpical stress-strain diagrams for a number 40 35 3C c 25 o 20 15 10 O . ^___ — - / y /■ / / / // //b c^^ / / ^ / _D_ 1 / / / ^ //^ ^ lO Extension per cent. — A Aluminium Bronze. B Hard Brass. C Annealed Brass. 20 30 40 D Rolled Annealed Copper. E Rolled Aluminium. Fig. 28a. — Stress-strain Diagrams for various Metals. of ductile metals. These must be regarded as only average diagrams, because these metals vary in their elastic proper- ties to a considerable extent, depending on the method of working and upon their constitution in the case of alloys. In most cases the early portion of the stress-strain diagram is never quite straight, but there is usually a clearly defined yield point. BEHAVIOUR OF MATERIALS UNDER TEST 63 Effect of Temperature upon the Strength of Steel. — The effect of temperature upon the ultimate strength of steel is to first cause a slight diminution, then an increase up to about 500° F., and finally a progressive diminution in strength for temperatures beyond ; the elastic limit, however, falls progressively as the temperature increases. This is shown in Fig. 29, which represents the mean results of 60 50 40 ^0 20 10 _-^ _ _<^ _ ^N H ' ^oT \ \A ize of beam, e.g. l" x V x 12". Fig. 33. — Compression Faikire of Concrete Cube. STONE, CONCRETE, CEMENT AND LIKE BRITTLE MATERIALS Compressive Strength. — When stone, concrete, cement and like materials are tested in compression in the form of cubes or short cylinders, fracture nearly always occurs by splitting in diagonal planes in the manner indicated in Fig. 33. This is commonly referred to as a " shear failure,'" the failure being attributed to the shear stresses on the diagonal planes at 45" to the axis. We have seen already (p. 10) that on such planes there is a shear stress of equal intensity to the BEHAVIOUR OF MATERIALS UNDER TEST 69 compressive stress. There are, however, strong reasons for supposing that the fracture of brittle materials is due to tension and the most careful experiments on cement and concrete show that the shear strength is greater than the tensile strength (cf. p. 79). Tension Theory of Failure. — When a block of material is compressed longitudinally it swells laterally, as shown in dotted lines in Fig. 34a, the ratio between lateral swelling y and the longitudinal compression x being Poisson's ratio [r]), and one theory is that the [limit of compressive strength of ir-y Fig. 34. the material is reached when the tensile strain y reaches the limit of tensile strain for the material ; in the case of a block in a testing machine, there is nearly always a very large friction force F (Fig. 346) induced, which prevents the lateral expansion and causes the block to bulge as indicated in dotted lines, thus causing resultant stresses R in a diagonal direction, which cause the apparent shear fracture and make the specimen to appear stronger than it really is. It has been kno^vn for many years that the measured compressive strength of blocks depends upon the material placed between the press head and the block. The following figures quoted from Unwin's Testing of Materials are of interest in this connection — 70 THE STRENGTH OF IVL^TERIALS Crushing Load Material. in tons on 4 in. cubes. Material between cu1)e and preas head of i Testing Machine. Portland stone 1 57-7 52-6 450 33-5 Two millboards | One lead plate | One lead plate J in. smaller all round ! Three lead plates Yorkshire grit 79-7 800 56-2 35-9 1 Two millboards Cemented between two strong iron plates with plaster of Paris One lead plate Three lead plates The lead plates were '085 in. thick in each case, and the fracture was longitudinal, as in Fig. 34a, in each case with lead plates, and diagonal with the millboards. Professor Unwin, in commenting upon this, says that the "lead falsifies the result of the experiment," but we do not see why he should not consider the lead as giving the more correct result and the millboard figures as being false. Professor Perry apparently takes the latter view, for he says in Applied Mechanics (Cassell) : " There is much published information on the fracture by compression of blocks of stone, cement and bricks. In almost every case care is taken in loading the usually short specimens that friction at the ends shall prevent the material sweUing laterally. When sheet lead is inserted at the ends, it gives a small amount of lateral freedom, and in everj'- case the breaking load is lessened by its use, and therefore it is said to be wrong to use lead. I consider all this published information to be nearly valueless, except that there is some probability that half the usually published ultimate compressive strength for a cube is the true resistance to compression in the material.*' There is, of course, in the case of a pure compression, a shear stress across a diagonal plane, and for materials like mild steel, in which the shear strength is less than the com- pressive strength, tliis shear stress probably causes the ultimate failure, and thus determines the compressive strength. lateral strain = 2/ = 7] X = >7K e; tensile stress = Ut = ^ • u. • E, BEHAVIOUR OF MATERIALS UNDER TEST 71 Batio between Tensile and Compressive Strength for Con- crete. — The consideration of the transverse strain enables us , to calculate the ratio between the tensile and compressive strength of the material upon this theory. Let Uf = ultimate tensile strength of the material. u, = ultimate compressive strength of the material. E, = Young's modulus in tension at failure. E, = Young's modulus in compression at failure. r) = Poisson's ratio. Then, compressive strain = x = ^^- (1) Ratio of tensile to compressive strength -u:~~e: ^^^ Now this ratio, according to different authorities, varies from one-eighth to one-twelfth, according to the usual method of determining compressive stress, and depends on the age of the concrete, the higher value occurring usually at ages of three months and more, rj for concrete is not fully known, and in the absence of further information we will assign to it the value J, which is the theoretical value for a perfectly elastic solid. E< also has not been very fully determined, but Hatt, for a 1:2:4 mixture gives E< = 2*1 x 10" lb. per sq. in., which is approximately equal to the value usually accepted for E,. E, For a rough consideration, therefore, we will take -^p = 1. This would give -'-=■• Uc 4 If, as Professor Perry suggests, the actual compressive strength is about one-half that usually published, the above 72 THE STRENGTH OF MATERIALS u 1 figure compared with published figures gives ~ = q, which lie O bears comparison with the figures usually given, agreeing with the lower limit. All of these points are of very great importance, and it would be of great value if ver}^ careful experiments were made to determine ?;, E^ and E, for the same mixture at the same age, and to see how nearly true is the suggestion that the compressive strength is given in terms of the tensile strength by the above formula. Another consideration which enters into the problem is that tensile strengths as determined by the usual briquettes are somewhat less than the actual tensile strengths, the discrepancy being due to the variation in the distribution of the tensile stress across the specimen (see p. 78). According to various authorities the actual tensile strength is 1"5 to 1*75 the mean strength, and allowance for this would bring u 11 the value of — from ^ ^ to ^ . , which agrees very well with u, 12 14 ° "^ experimental values. We have already dealt with Navier's theory for this problem (p. 45). Effect of Relative Height and Breadth of Com- pression Specimens. — Very careful experiments in 1876 by Professor Bauschinger upon sandstone prisms have shown that the compressive strength of sandstone prisms decreases slightly when the relative height to breadth is greater than for a cube and increases when the relative height is less. Fig. 35 shows a curve expressing the results of Bauschinger's tests and is a modified form of a similar curve given by Professor Johnson,* and tends to support both the Navier theory and the transverse tension theor}'' that we have just given, because in one case we should have that the cube and more dumpy specimens prevent the rupture along the line given * The Materials of Construction (Wiley & Sons), BEHAVIOUR OF MATERIALS UNDER TEST 73 by the theory, and in the other case the effect of the friction in preventing the natural transverse swelling is greater with a dumpy section than with a comparatively tall one. I -5 O I Eatio 5eight Breadth Fig. .35. — Effect of Height upon Strength of Sandstone Blocks in Compression. Bauschinger recommended the following formula to repre- sent the results of these tests — u. 4A/ _^.V'k p \ h where u, = ultimate crushing stress A = area of cross-section p = perimeter of cross-section h = height of specimen a, b = constants. Strength of Cube with Chamfered Edges. — Fig. 36 shows the results of Bauschinger' s tests upon chamfered specimens, the part of the curves in full lines representing the range over which the actual experiments were carried. 74 THE STRENGTH OF MATERIALS The curve marked A is for comparison of the strength of the chamfered block with a cub^ of the same size as the large area, and the curve marked B is for comparison of the strength of the chamfered block with that of a cube of the same size as the small^area. L o •0 / \ / •8 A — ' 7- •f> V X z^ A < i ^- ^^_ •A- / Va / ^ y^ ^ / / \ yA .? •4 oocK - C3 "^^ 2000 \ CO .S 1000 I I f 2 3 ^ Fig. 39. shows on a diagram the results of experiments made at the Watertown Arsenal, U.S.A., in 1899. Curve A is for a mixture of one part of cement, two j)arts of sand, four parts of aggregate ; and curve B is for a mixtui-e of 1:3:6. The figures given are for the same brand of cement. Tensile Strength of Portland Cement and Concrete. •^The tensile strength of concrete is about -^^ of its com- pressive strength, but it is not usual to allow for any tension in the concrete in practice. The standard method of testing the strength of neat cement is, however, by tension, so that the tensile strength is of 78 THE STRENGTH OF MATERIALS considerable importance. (For method and apparatus for testing, see Chap. XIV.) The British standard specification requires the following strength of Portland cement. Briquettes 1 sq. in. in section (see Fig. 190) must develop at least the following strength — Neat cement. After 7 days (1 in moist air, 6 in water) . . 400 lbs. ,, 28 ,, ,, ,, 27 ,, . . oOO ,, Fig. 40. The increase from 7 to 28 days shall be at least — 25 % when 7 -day test gives between 400 lbs. and 450 lbs. 20 0/ 450 „ 500 „ 500 ,, 550 ,, 550 „ 600 „ 600 lbs. or upwards. 1^0/ J-" /O ?J J 5 5> 5 5 ^^ /O 5 5 5 5 5 5 5 3 K 0/ <-' /O 5 5 3 5 3 J 3 3 One 'part cement and three parts sand. After 7 days (1 in moist air, .6 in water) . . 150 lbs. „ 28 „ „ „ 27 „ . . 250 „ Increase between 7 and 28 days must be 20 % for stresses 200-250 and 5 % less for each 50 lbs. increase, 5 % being the minimum. Variation of Stress in Briquette. — It can be shown that the stress is not quite constant over the briquette, but varies somewhat as indicated in Fig. 40; this means that the test strengths are always a little less than their actual values.* Strength of Concrete in Shear. — Early experimenters found the shear strength of concrete to be from '12 to '2 of * See also Chap. XIV. BEHAVIOUR OF MATERIALS UNDER TEST 79 its compressive strength, but recent experiments suggest that the shear strength is considerably greater than this and depends upon the kind of aggregate (gravel, broken stone, broken brick, etc.) used. The most exhaustive investigation is that by Professor A. N. Talbot, who has also carried out many valuable experiments on the strength of reinforced concrete beams. The results of these experiments were pub- lished in a Bulletin of the Engineering Department of the University of Illinois. Two different methods of testing were used; in the first the shear strength was obtained by punching a* hole in a concrete plate, and in the second by means of concrete beam with the ends fixed. There was considerable variation in the results, as will be seen from the following summary taken from Engineering of June 6, 1907 — • SuMMAHY OF Shear Tests. (Professor Talbot.) rorm of Specimen. Plain plate Recessed block Reinforced i recessed ' block Restrained beam M o C o 1 o 1 1-3-6 1 1-3-6 : 1-3-6 1-3-6 ! 1-2-4 1-3-6 1-3-6 1-3-6 1-3-6 i 1-3-6 : 1-2-4 ! ! 1-3-6 1-3-6 1-3-6 j . 1-2-4 1 1-3-6 ' 1-3-6 1-2-4 Method of storing. Strength. Ratio of Shear to Compression. Air Water Damp sand Do. Do. Air Water Do. Damp sand Do. Do. Air Damp sand Do. Do. Do. Do. Do. g \ Shear. Compression. Cube. ^Jl^"- I der. lbs. per sq. m. 9 679 7 729 4 905 1 968 5 1193 17 796 6 692* 5 879 4 1141 1 910 5 1257 4 1051 4 1821 1 1555 5 2145 4 1313 1 1020 6 1418 lbs. per sq. in. 1230 1230 2428 1721 3210 1230 1230 1230 2428 1721 3210 1230 2428 1721 3210 2428 1721 3210 lbs. per sq. in. 1322 1160 2430 1322 I 1160 j 2430 1322 ': 1160 2430 : 1322 i 1160 I 2430 C-be. I CJ^- 0-55 0-59 0-37 0-56 0-37 0-65 0-56 0-71 0-47 0-53 0-39 0-86 0-75 0-90 0-67 0-54 0-59 0-44 0-68 0-83 0-49 0-86 0-78 0-52 1-38 1-39 0-88 1-00 0-88 0-58 * Specimens injured in removing the forms. 80 THE STRENGTH OF MATERIALS The conclusions to which these Illinois exj)eriments lead are that the resistance of concrete to shear is dej)endent on the strength of the stone used, as well as on the strength of the mortar ; and in the richer mixtures the stone appears to exercise the greater influence. With hard limestone and 1-3-6 concrete sixty days old the shearing strength may be expected to reach 1100 lbs. per sq. in.; and with 1-2-4- mixture 1300 lbs. per sq. in. There is reason to believe that if tests can be made with the load applied evenly over the shearmg section, so as to obtain the true resistance to simple shear, the results will be found to be higher than those already obtained. An important point brought out by Professor Talbot's investigations was the influence which variations in the constitution of the concrete have on the shearing strength. The compressive strength of concrete is largely affected by the strength of the cement, but the shearing strength is influenced more by the strength of the aggregate. Eor this reason it does not seem well to express the shearing strength in terms of the compressive strength. The method has the advantage, however, that an idea is gained of their relative action. If, as indicated by these experiments, the shear strength of concrete is greater than the adhesion between concrete and steel, then there is an advantage over plain bars for reinforce- ment in those bars such as the twisted or indented bars which cannot be withdrawn from the concrete without shearing it. Adhesion between Concrete and Steel. — It is absolutely necessary in a reinforced concrete structure that there shall be a good bond between the concrete and the steel, for the latter will bear its share of the stress only so long as there is no relative movement between the steel and the concrete. If a concrete beam were cast with holes throughout its length on the lower side and steel rods were inserted loosely into these holes, the strength of the beam would be practically BEHAVIOUR OF MATERIALS UNDER TEST 81 no greater with the rods than without, because relative move- ment between the steel and concrete would be possible. A concrete beam with the reinforcing bars loose is like a plate girder without an}^ rivets. The adhesive strength for plain bars can be found as follows : Let O be the perimeter of the bar and I its length ; then if / is the safe adhesive stress, the adhesive force T that can be carried is given by F = Olf f can be found experimentally by embedding a rod in concrete and finding the force necessary to pull it out and dividing the resulting stress by the factor of safety (usually taken as about 6). Most authorities take a safe adhesive stress of 60 lbs. per sq. in. Numerical Example. — Find the length in relation to the diameter of a round bar that must he embedded in concrete in order that the tensile stress of 16,000 lbs. per sq. in. will be reached as soon as the safe adhesion stress of 60 lbs. per sq. in. Let d = diameter of rod in inches. Let I = length of rod in inches. Then load to reach safe tensile stress == P = stress X area = 16000 X ^ 4 Load to reach safe adhesive stress = P = stress X length x perimeter ^ QO X I X 7rd Trd^ If these are equal, 601 x -n-d ^ 16000 x . 16000 wd^ ~ 60 X 4 ^7rd = 61 d nearly . • . the bar must be embedded for a length equal to 67 diameters. ^ 82 THE STRENGTH OF IVUTERIALS Strength of Tn^reER (Normal Values) {Stresses in thousarhds of 'pounds per sq. in.) • Ultimate Strength. ! 1 Kind of Timber. Young's Modulus (millions of lbs. per sq. in.). Weight in lbs. , per 1 cu.ft. Tensions along Grain. Com- pression along Grain. Shear Shear across along Grain. Grain. Bend- ing. Ash .... 8-14 6-8 2-4 •4--7 10-12 1-6 50 1 Beech . 9-18 4-6 — — 8-10 1-3 44 , Dantzig Fir 6-9 3-5 2-7 •4 4500 o 33 ! Elm . . . 6-12 5-8 3-5 •6--9 6-8 1-6 34 1 Oak. . . . 9-15 4-5-9 3-5-5 •6--9 7-10 1-7 58 Pitch Pine . . 8-12 4-6 4-9 •4 7-5 1-5 42 Red Pine . . 6-9 4-6 3 •4 4r-6 1-2 27 Teak . . . 12-14 12-14 4 1 12-16 2-4 49 Yellow Pine . 6-9 4-6 3-5 •4 4-8 1-7 32 1 Normal Crushing Strength of Cement, Stones, etc. Material. Brick (London stock) . ,, (Staffordshire blue) Brickwork in Cement . Cinder Concrete (1 : 2 : 4) Granite Gravel Concrete 1:2:4 1:3:6 Portland Cement . Portland Stone , Sandstone .... Slate Weight in lbs. per cu. ft. Ultimate Crushing Strength. (Thousands of lbs. per sq.in.) 115 140 100-150 97 170 120 130 90 145 135-145 175 2-5 7 1-25-2-5 1-8 after 28 days 12-20 2-4 after 28 days 1*8 ,f ,5 7 5 5-10 10 BEHAVIOUR OF MATERIALS UNDER TEST 83 « 2 < > 00 S M

H o S W ^ O 03 rn 2 -s ield. oint. O (M 1 1 •? 1 1 1 1 1 1 1 1 00 M 1 1 1 1 '-H'-il|(j!,IMMMI (Nilll 1^ ^H^ o o o o ooSSooo o2!5 oooo oo^ooo 1 1 1 o^ 1 oooo 1 I 1 o , O '=' • <>ioAi:^t-ol 1 |eo<-l|'^<='^<=><='S^<^^'^'^'^'^<^ 5iol'^'=>00000|00000000 ^CJ(=P001010 0AO>0 10 0000»0 o , w ^ o 1 1 CO o o CO lO ' Ui »0 O "* GO (N 00 03 c^ciJ, UN IS^^J. 1 1 12^ 1 1 I 1 1 A 1 1 m s CO C<) 1— 1 ^H 1— ( 1— 1 l-H . CO &i) o o 1 ^ e8 1 iSiiiSi i^is ill 2 1 a <5 ^ r— If— <(M ,-i r-((M '~',1( 1 .S;. . Sa.2 00 t^ ic>ioofocccc 1 lcocccocccocc5 ti CJ . . . fcfi . . ■? . g . . . E; . Tii . . . . . . .Ie .-^^Is .2||I .... . 03 -(J -o^ l-Sg^o^^. -i^ II §^ g :: ::2 ^ §^ - ^'^ ^ 3 .3 .S , s g §!:^0<3 pq M o P OHN OKI tn fi d 9s t^t^^Oi iCCi 1 I iCOl 1 1 1 1 1 1 ! I>" 1 C/3 -*^ V3 0C( J J5 Cast iron . 1 2 i 5J J> Oak . . . . Pine, yellow 16 3 13 6 r 600 J (bending) 500 '^ (direct) 35 5 (across grain) 3 (across grain) cwt. per sq. in. 5) ;? Cement concrete 1 1:2:4 . . j 60 60 lbs. per sq. in. Granite .... — — tons per sq. ft. Sandstone . . ) Yorkstone . . f — 20 — JJ J5 Limestone . — 15 — ; J 5j Brickwork in cement mortar '5 8 - (adhesion) in lime mortar , '4 (adliesion) * Many authorities allow J ton more for each of the stresses in mild steel. We have already seen that according to one theory the working shear stress for ductile metals should be half the tensile strength, viz. 3'75 tons per sq. in. for mild steel. This figure is not, however, in common use. 100 THE STRENGTH OF MATERIALS Allowance for "Live" Loads or Variable Loads. — There are two priiicijial methods of allowing for live loads which are in effect the same. (a) Equivalent Dead-load Method. — According to this method the static stresses are used and the loads are in- creased to give the equivalent dead load. The ways for allowing this, are — (1) equivalent dead load = dead + 2 live load. This may be called the dynamic formula. (2) equivalent dead load = w. n r + yJ n^ r'^ + 4: (w — -^J 2 where r is the variation of load, and w is maximum load, n being a constant which may be taken as I'S for steel. This formula is deduced from Unwin's formula for Wohler's experiments. For steel we get 1-5 r + J 2-25 r^ ^ 4.(w - 2) w,. = 2 When the variation is from zero to a maximum, we have r = w. Then w, = 21 w. (3) equivalent dead load = maximum load + variation. (6) Variable Working Stress Method. — According to this method the working stress is varied according to the relative amounts of live and dead loads. The common ways of allowing for this are — (1) Launhardt-Weyrauch method. ^^^ - . / /' minimum load \ Worknig stress = —^ 1 + o " • 1 ^ ° 1*5\ 2 X maximum load/ / being the static or dead-load working stress. (2) Dynamic method. Working stress = rV — . — ^ , / being as before. total load KEPETITION OF STRESSES 101 Take as a simple numerical example the case of a member of a roof truss in which the dead load is a tension of 5 tons, and the wind on one side causes a tension of 2 tons and on the other side a compression of 1 ton. The various methods give the following results : — (a) (1) Equivalent dead load = 5 + 2 x2 = 9 tons. 3\2 2 1-5 X 3 + J 2-25 X 9 +4(^7 y^) 5> 5J . 8-2 tons. 2 (3) J5 33 = 7 + 3 = 10 tons. (b) (1) Working stress = 6/ 7 (2) 5J 33 = / 7/ 9 Assuming the material to be mild steel. [h) (1) gives working stress = 6 tons per square inch. (2) ,5 5, = 5*4 ,, J, ,, Taking the material as mild steel, the requisite number of square inches in the sectional area of the tie are — 9 (a) (1) = 1-28 square inch. (2) 7 = 117 (3) '^^ = 1-43 ib) (1) I = 117 ^^' 5-4 = 1-30 If consideration of variation of stresses be neglected alto- gether, we should have — area = =1 square inch. CHAPTER IV RIVETED JOINTS ; THIN PIPES Forms of Rivet Heads. — The most common forms of rivet heads and their usual proportions are shown in Figs. 44, 45. For structural work the snap-headed rivets are most usual, but countersunk rivets are used where necessary to prevent CUP or SNAP HEAO CONiCAL HEAD PAN HEAD COUNTE R SU'SK HE A O Figs. 44, 45. — Forms of Rivet Heads. projections from the surface of the plate. Snap-heads take a length of rivet equal to about IJ times the diameter. It is usual in practice to adopt a diameter of rivet when cold equal to one-sixteenth of an inch less than the diameter of the hole, but in all calculations the diameter of the rivet is taken as being equal to that of the hole. Forms of Joints. — (a) Lap Joints and Butt Joints. — 102 RIVETED JOINTS; THIN PIPES 103 In the lap joint the plates overlap as shown in Fig. 46. This form of joint has the disadvantage that the line of pull is such as to cause bending stresses, tending to distort the joint as shown. In the hutt joint the edges of the plate come flush, and cover plates are placed on each side as shown, the thickness of each cover plate being usually five-eighths that of the main plates. In this form of joint the pull is central, so that there are no bending stresses. In the single cover joint, which is a cross between the lap I ' I ^^ I T 4- % I LAP JOINT. BUTT JOINT. Thickness of Cover SINGLE COVER JOINT. 5 t Thickness of Cover — - 4 Fifi. 40. — Forms of Riveted Joints. joint and the butt joint, there are bending stresses developed, tending to distort the joint as shown. It IS clear from the above that the butt joint should be adopted wherever possible. (6) Chain Riveting and Zig-zag or Staggered Rivet- ing. — The different rows of rivets in a joint may be arranged in chain form or zig-zag form, as shown in Figs. 47, 48. As we shall see later, the zig-zag form is more economical, and should be used whenever possible. The essential feature of zig-zag riveting is that the rivets in alternate rows are displaced laterally by half the distance between the rivets, i. e. by half the pitch of the rivets. In 104 THE STRENGTH OF ]\UTERIALS the form shown the joint is in a tie bar of a bridge and the rivets form a triangle ; it is common in boiler and like rivet- ing to make the pitch of the outermost of three rows twice that of the innermost row; the result is a special kind of zig-zag riveting which Ave may call triangular riveting. Methods by which a Riveted Joint may Fail. — A riveted joint may fail in any of the following ways : — o o o o o o o o o o o Fig. 47. — Chain Rivetine:. Fic 4S. — Zig-zag Riveting. (1) By tearing of the plate. (2) By shearing of the rivets. (3) By crushing of the rivets-. (4) B}' bursting through the edge of the plate. (5) By shearing of the plate. Fig. 49 shows these methods of failure. (4) and (5) are allowed for by the following rule : The minimum distance between the centre of a rivet and the edge of the plate is 1 J (h where d is the diameter of the rivet. If this rule is adhered to the joint will always fail first in one of the ways (1), (2), (3). The aim in designing a joint should be to make the force necessary to cause failure in the various ways equal. We will now consider the various ways of failure in detail, taking in each case a strip of plate equal to the pitch of the rivets. (1) Tearing of the Plate, — In this case the width along RIVETED JOINTS; THIN PIPES 105 which fracture will occur is {p — d), and as the thickness of the plate is t, the area of fracture ^ {p — d)t. VI /. / S P / z ^ '< ?■ ^^ DOUBLE Sfi£/^fi , P -A ,p 5 Fig. 49. ♦ P (2v- -A Therefore, if /, is the safe tensile stress in the material, the safe load which the joint can carry is equal to V =tt{V-d)t (1) U)() THE 8TRENGTH OF MATERIALS (2) Shearing of the Rivets. Til the case of single shear, the area sheared ^ double ,, 4 2jr_a2 - 4 - Therefore if /, is the safe shear stress on the rivet, the safe forces on the joint as regards shear are respectively P = * 4 J (3) Crushing or Bearing of Rivets. — In this case the crushing or bearing area is taken as the diameter of rivet multiplied by the thickness of the plate, i.e.d x t. There- fore, if /„ is the safe bearing stress on the rivet, the safe force on the joint as regards bearing is equal to V^j^^.d.t (3) The values of j, and s may be taken as given in Chap. III. For /,„ 10 tons per square inch may be taken for mild steel, and 8 tons per square inch for wrought iron. These figures are higher than for ordinary compression, and are obtained from the results of experiments. For structural work the strength of the joint as regards bearing will often be less than as regards shear, because the plates are often thin compared with the diameter of the rivet. Efficiency of Joint. — The efficiency of a joint is the percentage ratio of the least strength of a joint to that of a solid joint, i. e. -p,^ . _ _ Least strength of joint •^ ~ ^ "" Strength of solid plate Diameter of Rivets. — For the most economical joint the * A Board of Trade rule states that this should be taken as ^ , 4 and this rule is often though not universally adopted. This figure is based upon the results of tests. RIVETED JOINTS; THIN PIPES 107 diameter of the rivet should be such that the shear and bearing strength are equal. Unwin's Formula, which is in common use, gives d = l'2Vt (A) For single shear we have shear strength = a -^^ bearing strength =^ dt .f^. If these are equal c^ = '^^Z" = 2-54^ (if f,^2s) Thickness of Plate (inches). Fig. 50. — Diameter of Rivets. (D) y' y y y y ^ y ^^-^ y ^ ^ ^-^ ^,^ y y ^^-^ ^ ^ ^^ ,,.^-^ ^..^ ^ . / ^ :^^ ■ — ^ y^ ^^ ^:^ ^^^^-^ ^ ^^-^ <^ .^ -^p^ ^ ^ A P I "3 1 1 4- li 4 Z Ph Oq For double shear we shall have 2 7r6Z2 ^ -s = dt /,. d = TT S or on the Board of Trade rule dtf,, 1-75 TT 6^2 \21t 32 t TT . 1-75 5 7 TT (B) (C) 108 THE STRENGTH OF MATERIALS These values are plotted in Fig. 50. It is clear from this figure that for large thicknesses of plate for single shear the theoretical value gives diameters of rivet which are impossible in practice. Numerical Examples. — (1) A tie bar in a bridge consists of a flat bar of steel 9 inches wide by l^ inches thick. It is to be spliced by a double butt joint. Determine the diameter of the rivets and their number, and give sketches showing the proper pitch and arrangement of the rivets. {B.Sc. Lond.) According to Unwin's formula d = 1'2VT= 1'34 inches. This is, however, rather high for practice, and so we will adopt d =^ \ inch. Assuming that the rivets are arranged in zig-zag fashion, the strength of the joints against tearing through the outside rivet is equal to 7 (9 — 1) . IJ = 70 tons. Shear strength of each rivet = 5 . .^ • (1)^ = 7*85 tons. . • . Number of rivets required for shear = ^|g = 8-93 = say 9. Bearing strength of each rivet = 10 x 1 x IJ = 12*5 tons. 70 .-. Number of rivets required for bearing = ^^.r ^ ^^y ^• 9 rivets would thus be ample as regards bearing. The joint would then be arranged as shown in Fig. 51, the centre two rows being chain-riveted. We will now consider the strength of this joint under various ways of failure. If the plate tears along the line A a, the force necessary to reach the safe limit of stress is, as we have shown above, 70 tons. Now suppose that the plate tore along B b, shearing off the rivet in A A. 5 Then strength of line bb = 7(9-2)= 61'25 tons Strength of one rivet = 7*85 tons. RIVETED JOINTS; THIN PIPES 109 . • . Total strength against failure along b b = 61-25 + 7-85 = 69-1 tons. Now suppose plate tore along c c, shearing off the three rivets. 5 Then strength of line cc = 7(9 — 3). = 52-5 tons. Strength of three rivets = 23' 55, . • . Total strength against failure along c c = 52-5 + 23-55 = 76-05 tons. D C B /^ Rit/eis I Diam -1 1 — u J — u tT^ J L i"rf;d Cover P/oTo. L- 3 Fig. 51. Finally, suppose cover plates tore along d d, then strength 7 = 7(9 - 3). 2. 73-5 tons. From the above we see that the weakest section is along B B. mi nn ' p • • ^ Least strength of loint Inen emciency 01 lomt = c^, .i — r.^ — ,• v \ Ibtrength of sohd plate 691 69-1 9 X li X 7 78-8 87'8%. llu THE STRENGTH OF MATERIALS If instead of zig-zag riveting we had adopted chain riveting with three rows of three rivets (9 in aU) the least strength would be (9 - 3) IJ x 7 = o'2o tons. 52 ' 5 .'. efficiencv of joint = _ , = 667 ^o- If we had four rows of chain riveting with two rivets in each row (8 in all), the least strength would be (9 — 2)1 J X 7 = 6T2o tons. .•. efficiencv of joint = ^^^ = 77*7 ^o- too The above shows that the zig-zag riveting is considerably more efficient than the chain riveting, and is therefore more economical. (2) Design a douhle-rivefed lap joint to connect two steel plates J in. thick unth steel rivets, the tensile strength of the plates before drilling being 30 tons per sq. in.; the shearing strength of the rivets 24 tons per sq. in. : and the compressive strength of the steel 43 tons per sq. in. Find the efficiency of the joint. {A.M.I.C.E.) For J in. plates Unwin's formula would give d = 1"2 \ "5 = 'So in., say | in. The joint is a double-riveted lap, therefore there will be two rivets in single shear in a width of plate equal to the pitch. . • . Strength against tearing per pitch = /• {p-d)t = 30(p-(/).^ = \o{p-d)..{\) . ' . Strength against shearing per pitch _ 2 - r/2 ~ ' ■ 4 24 .2- 28-9 tons. II these are equal 15 ( p — „ ) = 289 28-9 _ 7 15 8 - 1-93 + -87 - 2-80,sav3ins. •■• ^^= 15 8 to RIVETED JOINTS; THIN PIPES 111 The bearing stress for a force of 28*9 tons would be equal 28-9 ■q" X 9: X ^ 33 tons per sq. in. 7 1 the bearing area of each rivet being q x -^ = *437 sq. in. . This is less than the allowable value of 43 tons per sq. in., showing that a larger diameter of rivet might be used with greater economy, but | in. diameter is in most cases more suitable in practice. The efficiency of joint in this case is equal to 28-9 _ 28-9 _ 30 X 3 X i ~ "45~ ~ ^* ^ /^ The joint then comes as shown in Fig. 52. ^ t) f :i Fig. 52. (3) A steel-plate tie bar in a bridge is subject to a tension due to dead load only 0/ 16 tons. The stress due to live load only varies from 36 tons tension to 10 tons compression. The tie bar is | in. thick and is to be joined to the side plate of a girder by means of a % in. gusset plate and double-cover butt joint- Select suitable working stresses and design the joint, arranging the rivets so that the tie bar is weakened by only one rivet section. {B.Sc. Lond.) The maximum load in this case is 36 + 16 = 52 tons, and the minimum load 16 — 10 ?= 6 tons. 112 THE STRENGTH OF MATERIALS Using the Launhardt-Weyraiich formula, we have „. , . ^, f /i , i^ii^- stress \ Working Stress = v r 1 + « ° I'o \ 2 max. stress/ = l4(l + 104) = -™^/ This gives a tensile stress of 4*93, say 5 tons per sq. in. ; a shear stress of 3"52^ say So tons per sq. in. ; and a bearing stress of 7 tons per sq. in. According to UnAvin's formula d = 1'2 V '15 = r04 in., but for practical reasons | in. would usually be adoi)ted. We now require to find the necessary width of the tie bar. Let this be w. Then Iw — ^ ) t ^^ ^^^ equivalent cross-sectional area. / 7\ 3 .' .(w — -q) • A- 5 must be equal to the maximum pull of 52 tons. V 8/ 3x0 .- . w =^ 13*89 + '875 = say 15 inches. The strength of each rivet in double shear is equal to 2 7r nv , . , ^ , . 3-5 = 4-22 tons. 4 \8/ 52 .' . Number of rivets required for shear = ,^^ = 12*3. We will use 14, as they give the best arrangement. The strength of each rivet in bearing is equal to 3 7 -7.^-7^ 4-58 tons. 4 8 .• . 14 rivets will be ample for bearing. The joint is then arranged as shown in Fig. 53. It is very important in such joints that the centre line of the rivets should coincide Avith the centre line of the tie bar, or else the pull in the bar would be eccentric. In such joints, therefore, the rivets should always be arranged symmetrically with regard to the centre line of the tie bar. RIVETED JOINTS; THIN PIPES 113 (4) Find the number of rivets necessary to the gusset plates, etc., at the base of a steel stanchion to the stanchion proper, the load carried being 150 tons. The diameter of the rivets is | in. and the thickness of the plate J i'^- The rivets usually have to be designed in such cases so that they will carry the whole load, so that if the stanchion itself Fig. 53. does not bear on the base plate the rivets will distribute the load satisfactorily. The strength of each _ tt rivet in single shear 4 The strength of each _ 7 rivet in bearing 8 v8 3-01 tons. 10 = 4-37 tons. 150 .'. Number of rivets necessary = „ , = 50 nearly. Some Practical Considerations in Riveted Joints. — Punching and Drilling of Rivet Holes. — It is quite common in this country for specifications to state that rivet 114 THE STRENGTH OF MATERIALS holes must be drilled out of the solid. Punching is known to injure to some extent the material in the neighbourhood of the hole, and is thus often objected to. The extent to which punched holes weaken a structure such as a plate girder compared with drilled holes does not appear to have been satisfactorily determined, although such determination from a practical point of view^ would seem to be absolutely neces- sary, since there is an increase in cost entailed in drilling the holes. In recent years punching machines and means for obtaining an accurate pitch of the holes have been improved considerably, and when we consider the increased cost of the drilling, punching is preferable in many cases. In recent years " gang " or multiple drilling machines have been introduced which lessen the cost of drilling; one great ad- vantage that drilling possesses is that the plates to be joined can be clamped together and drilled right through, thus ensuring accurate registering of the holes. A good com- promise is to punch the hole i to J inch less than required, and to reamer out to size, the damaged metal being thus removed ; but this is considerably more expensive than plain j)unching. A method of allowing for the damage of metal due to punching which has been suggested, and which we consider preferable, is to add J inch to the diameter of the hole in calculating the tearing or tensile strength. This adds very little to the size of the plate and saves in cost of pro- duction. The point that should be very carefully seen to is that the holes are accurately pitched, so that the holes will register well when the parts are assembled, and will not require excessive drifting as is the case when the spacing of the holes is inaccurate. It is probable that many more joints are unsatisfactory because the rivets do not fill the holes, owing to the latter not registering accurate^, than because the metal has been injured owing to punching the holes. There is considerable friction between the plates in a riveted joint, but this is not allowed for in calculations of the strength. RIVETED JOINTS; THIN PIPES 115 Pitch and Spacing of Rivets. — In order to prevent moisture getting between the plates and causing bulging due to rusting, or to prevent local buckling in the case of com- pression members, it is common to stipulate that the pitch of rivets shall not be greater than 6 ins., or sixteen times the thickness of the thinnest plate. The designer should re- member that pitches from 3 ins. upwards, increasing by half- inches, should be used, and odd fractional pitches avoided, except where absolutely necessary. As far as economically possible, the same pitch should be used throughout, and in many cases, for girder work, etc., 4 ins. is used unless special conditions require a different pitch. Working Strength of Steel Rivets Diam. Area in sq. ins. Strength in single shear at 5 tons per sq. in. Bearing Strength at 10 tons per sq. in. of Rivets Thickness in ins. of plate. in ins. tV 3 8 7 4 9 To 1 1 1 '^ f •1104 •55 M7 1-41 1^64 1^87 211 234 2^59 2^81 * •1963 •98 1^56 1-87 2^18 2^50 2-81 3^12 343 375 •3068 153 1^95 2^34 2^72 3^12 3-51 1 3^90 4^30 4^68 3 •i •4418 2-21 2^34 2^81 3^27 3^75 4-21 4^69 5^16 5^63 g •6013 301 2-72 3^27 3^82 4-37 4^91 5^46 6^02 6^56 1 •7854 3^93 312 375 4^37 5^00 5-62 6^25 6-87 7^50 Thin Pipes and Cylinders. — Suppose that a thin cylinder of diameter d, Fig. 54, and thickness t, is subjected to a pressure of intensity p. This pressure will tend to burst the pipe along a longitudinal section, and the pressure on the ends will tend to cause failure across the circumferential section x x. In thick pipes the stress will vary across the section and is dealt with in Chap. XVII. Longitudinal Section. — Consider a length I of the pipe. Then the radial pressure p at any point can be resolved into a component o b parallel to any diameter under consideration and a component a b normal to it. The resultant normal force will be the same as the pressure acting over the diameter. 116 THE STRENGTH OF IVIATERIALS If /, is the tensile stress across the section, we have . • . Force tending to burst pipe = p x area = p d I. Force resisting bursting = /, x area. = /, X 21 1 {It on each side). , p d I }) d ^ 2t This stress ft is often called the hoop stress. /^^ '>!}>»>;".:/•'! irrr. ii v..',;. .',"/A7///,/»/ ,'.'- '-ns ^ Tzz: t l^ / — >i -^^ (1) Fig. 54. — Stresses in Thin Pipes. Circumferential Section. — If // is the tensile stress across the section x x, we have Force tending to cause failure = p x area of end. Force resisting failure = // X area = // X TT d t (because the pipe is thin). RIVETED JOINTS; THIN PIPES 117 • • A =P -4- - Trdt = 1^ (2) 1 /< Therefore the stress across a longitudinal section is twice that across a circumferential section; for this reason longi- tudinal joints of boilers are made stronger than circum- ferential ones. * Equivalent Stresses on Strain Theory. — On any small cube of the material with sides parallel and normal to x x, there will be a hoop stress /„ a longitudinal stress \ ft, and a radial stress which is jj on the inside and o on the outside of the tube and may generally be neglected. . ' . Strain in longitudinal direction = -U — 4V, E V 2 7/, f 1 7 . • . Equivalent hoop stress = ^ ft o Similarly stress on circumferential section = '^ -\- q f^ 4 On the equivalent strain theory, therefore, the pressure on the ends strengthens the pipe. Numerical Exaiviple. — A holler 1 ft. Q ins. in diameter has to sustain a pressure of 80 lbs. per sq. in. If the efficiency of the joints is 70 per cent, and the safe stress is 4 tons per sq. in., find the thickness that the boiler should have and the necessary pitch of rivets on the longitudinal butt joint. . _pd ^ 80 X 90_ _ 90 2 / ~ 2 X 4 x" 2240 ~ 224 Efficiency of joint = 70 per cent. 118 THE STRENGTH OF MATERIALS rru- 1 ^ 1. ^ X 100 90 X 100 .'. Imckness must be — ^^ — — c^^. ^r. = "Svo in. 70 224 X 70 say f in. Diameter of rivets = 12 V t = 1 in. nearly. Efficiency = 70 per cent. T , . p — d „ .' . Intension — = "7 P i.e. p — I = '1 p .'. Sp = 1 p = 3" 3, say 3 J ins. TT Strength of each rivet in shear = . x l'15d^ x 5 = 6'9 tons. 5 Strength of plate per pitch = 7 x 2*25 x o = 9-85 tons. 9*85 .• . Number of rivets required per j)itch = /,.q =2 (as whole numbers only are possible). . • . A double row of rivets are required on each side of the joint with a pitch of 3 J ins. Collapse of Thin Pipes under External Pressure. — If thin pipes are subjected to external pressure there will be hoop compression stresses which may be cal- culated by the same formulae as we have obtained for the hoop tension due to internal pressure. If there is any inequality in the pipe bend- ing stresses will be induced which will cause failure by collapse before the crushing strength of the material is reached, this collapse being similar to the failure of columns by buckling. Fig. 55 shows some of the forms of collapse of such tubes. RIVETED JOINTS; THIN PIPES 119 Fairbairn's Formula. — The first well-known experiments on the subject were made about 1860 by Fairbairn. His formula is 806,300^219 p = collapsing pressure in lbs. per sq. in. d = diameter of tube in inches. t = thickness ,, ,, L = length of tube in feet. Board of Trade Rule. — Safe pressure = j . , for single-riveted lap-welded tubes. = for welded or double-riveted butt- (L + l)a jointed tubes. Stewart's and Illinois Experiments.* — These recent experiments were made with great care and proved that except for tubes of length less than about 5 diameters the collapsing pressure is practically independent of the length, so that Fairbairn's and the Board of Trade Rules are not applicable. Fig. 56 shows the results of Professor Stewart's experiments for 4 and 7 in. tubes ; it is clear from this that beyond a certain thickness the collapsing pressure is practically pro- portional to the thickness. The Illinois experiments confirmed this. The following formulae are given : — Very thin tubes ( 7 03 — d Brass : p = 93365 . - 2474 (Illinois) d 4" cooo / fiSOO .<(;/. 5C0O '-t^ SlOO 6200 6000 1800 4C00 «iOO / 47 ■^'k / 1 J3 / 4000 c- / 3600 / Q. / . ^ 34 00 3200 3000 2800 2600 -2 ►J I ?' /' I r ■< ^ L 1 -f^o \y 3 J V + / 2400 2200 2000 1800 a A y -^ '7; a. k A. ^ / o t 1 /^ i / Y 1400 / / \l y ^ 1200 / / 800 / + / / J H / ■¥ 600 400 / ^ * ,^ y ^,.- r- .-' '' Thick 1 nes£ of Wall', in 1 1 Dec niais of 1 1 an lnc^ 1 1 1 .(J 2 .C )4 .0 6 .i 3 .] .1 2 .1 i .1 .1 8 A .2 2 .2 4 .i C .S 8 .S .3 2 .J 14 .2 c .s 18 .40 Fig. 56. — Collapse of Tubes. Stewart's Experiments. Seamless cold-drawn steel: p = 95520^^ - 2090 (Illinois) Lap-welded Bessemer steel: f = 83270 ^ - 1025 (Illinois) j 1386 (Stewart) -f 86670 ^ d CHAPTER V BENDING MOMENTS AND SHEARING FORCES ON BEAMS Definitions. — The shearing force at any point along the span of a beam is the algebraic sum of all the perpendicular components of the forces acting on the portion of the beam to the right or to the left of that point. The bending moment at any point along the span of a beam is the algebraic sum of the moments about that point of all the forces acting on the portion of the beam to the right or to the left of that point. As the beam is in equilibrium under the forces acting on it, the algebraic sum of the forces at any point, and of the moments of the forces about the point, acting on both sides must be nothing; so that we shall get the same numerical values for the shearing force and bending moment from whichever side we consider them, but they will be opposite in sign. We will, wherever convenient, consider the shearing force and bending moment of the forces to the right of the section, and we will take an upward shearing force and an anti-clockwise bending moment as positive, the downward and clockwise being taken as negative. Bending Moment and Shearing Force Diagrams. — If the bending moment and shearing force at every point of the span be plotted against the span and the points thus obtained be joined up, we shall get two diagrams called the Bending Moment (B.M.) and Shear diagrams, and from these diagrams the values of these quantities can be read off at 121 122 THE STRENGTH OF MATERIALS any point of the span. We will consider the forms of these diagrams for various kinds of loading and for various ways of supporting the beam, but will only consider beams with fixed loads. We will use Mp and S,. to represent respectively the bending moment and shearing force at a point p. B.M. AND SHEAR DIAGRAMS WITH FIXED LOADS A. Cantilevers,* i.e. beams fixed at one end and free at the other, the loads being all at right angles to the length of the beam. Case 1. Cantilever with One Isolated Load. — Let a cantilever, fixed at the end b, Fig. 57, carry an isolated load W at the point a, at distance I from b. Consider any point p at distance x from a. Then we have S,. = W. This is constant throughout the span. .*-. Shear diagram is a rectangle of height W. Again M, = W x x This is proportional to x. .' . B.M. diagram is a triangle whose maximum ordinate is W I, this being the bending moment at the point b. Case 2. Cantilever with Two Isolated Loads. — Since the B.M. and shear at any point are defined as the sum of the moments and the forces to the left of that point, it follows that the B.M. and shear diagrams for a number of loads can be obtained by adding together the diagrams for the separate loads. In the present case, in which we have loads Wi and Wg at distances l^ and /g from the fixed end, the diagrams are obtained by adding together the separate diagrams as shown in Fig. 57 (2). Case 3. Cantilever with Uniform Load. — Let a uni- formly distributed load of w tons per foot run be carried by a * According to our convention the shears and B.Ms, for all the cases of cantilevers that we consider are negative. There is, however, no need to give the sign, unless both positive and negative values occur in the same beam. BENDING MOMENTS AND SHEARING FORCES 123 cantilever a b of span I. Consider a point p at distance x from the free end a. Then ?er Ft pnoc^hancp ^ffi a. »■-*- K.^^??^ Fig. 59o. — Simply Supported Beams. Uniform Load. Wa if Now consider the bending moment. X M^. = R„ X a; — t^;.T X Wlx WX^ _W , 2\ ^ "2 ~ 2 ^ 2^ ~ ^' This depends on x^, and therefore the B.M. diagram will be a parabola. The maximum B.M. will occur at the centre — i. e. when I X = 2' BENDING MOMENTS AND SHEARING FORCES 131 Then maximum B.M. = ^ y^J ~ ^ 2/ "" 2 ' 2 " 4/ w P- wP Wl 2^4=8 ^^T" Case 4. Uniform I^dad over Portion of Span. — Let a uniform load of w tona per foot run and of length e d equal to / be i^laced on a beam a B of span l, and let the centre c of the load be at distance a and b respectively from a and b. Then, if total load wl = W, K,, = ~ and K, = The shear between b and d will be constant, and will be Wa equal to ; between d and E the shear will decrease uniformly until at e the shear will be equal to R„-W = W«-W=^^* = -R„; between e and a the shear will be constant and equal to - Wb J , the shear diagram then coming as shown on the figure. The point k at which the shear is zero can be found as follows. Let it be at distance x from the centre c of the load. Then S^ = R„ - w (^ " ^) ^ ^ Wa wl , I.e. — ^ -[- 'W X ^ L 2 wl W a wl wla ">-^= 2 - ,, = 2 - T •'• ^~ 2 ~ T " ^ V2 ~ L The B.xM. diagram can be drawn by setting up a length W/ ^'2 ^'' '^ o > i- e. the bending moment at the centre of the o short span e d, then produce Cg f to o, making f o equal to 132 THE STRENGTH OF MATERIALS Cg F and join g to Ej and Dg, and produce to meet the reaction verticals in Ag and Bo. Join Ag Bg, and we then have the B.M. diagram as shoAvn. To prove that this gives the correct diagram, consider the bending moment at a point at distance x from the centre of the load. Then M, = R, (6 - .r) - | (| - xj Wa ,, , W/i Y n^ = -^- (6 - ^-l - 21 (.2 - •■^j W Now, °^ = Al« G C2 Eg Cg GCoXAoQ WZ a Wa Eg Cg 4 J^ 2 Similarly g r = -^ .-. QR = ^ (6 — a Q R X a W a ,, Wa , Wa,, .-. GH = GQ + QH = -^^~ + 2Y ^ ^^ W a / h — a Wa / L + 6-a \_Wa (6 + 6) _ Wa6 2 V L / 2l l . . J p h — X Agam = — -^ ^ GH b h — X W a (6 — x) .-. JP = — V — X GH = ■ ' h L , .JO o Do Agam = ^ G Cg Cg Dg ^ G Co X O D, W / 2 .-.JO = — ^ =-7— X -s — Cg Dg 4 Z W /7 T 2 - ^ BENDING MOMENTS AND SHEARING FORCES 133 Then since curve is a parabola — FC, — ON /0C2^^ F C2 \C2 D2 Wl 8 x^ Wl ~ l^ 8 4 W Z _ _ W Z 4.x^ _ Wjr2 • • "8 ^^ ~ S ^ l^ ~ 21 Wl Wx^ .•.NP=PJ — JO + ON Wa{b -x) W fl ^\ , WZ Wx^ L 2 V2 ; ' 8 2Z Wa ,, , W fx^ P P Ix = --ib-x) --^. ^^-g-+^^--- W a ,, , W /x^ Ix l^ Wa., . W fP ; , 2 (0 — a;) — ;^ J — I X + X'^ L ' ' 2Z \4 Wa ,, , W /Z V = -IT (^ - ^) - 21 12 - ^; Comparing this with (1) we see that N p gives the B.M. at the given point. We shall prove later (p. 149) that the B.M. is a maximum at that point of the span where the shear is zero, and so the vertical through k will give the maximum ordinate of the B.M. diagram. Alternative Construction. — The following alternative construction is usually more accurate in practice. On a hori- zontal base Ag B2, Fig. 60, set up Cg m equal to , ^. e. the Ij B.M. due to an isolated load W placed at c, the centre of the load. Join m Ag, M B2, cutting the verticals through E.and d in n and D respectively, and join n d. On Eg Do draw a parabola W I Eg Q D2 of height equal to -^ (the B.M. due to a uniform 134 THE STRENGTH OF MATERIALS load on a span e d), then the B.M. diagram comes as shown shaded. If it is desired to have the diagram on a straight base, the parabola may be drawn on the inclined base n d as indicated in dotted lines. The proof of this construction is left as an exercise to the student. Case 5. Irregular Load. — Graphical Construction. — Let a number of loads Wi, Wg, W3, and W^, be placed knj- where along a span ab. Number the spaces between the loads and set down, 0, 1 ; 1,2; 2, 3 ; 3, 4, as a vertical vector - I . ooocooorrxTrr) ^ 8 Fig. 60. — ^Alternative Construction for Uniform Load over Portion of Span. line to represent the loads to some convenient scale, and in any position take a point p at convenient polar distance p from the vector line, and join p 0, p 1, P 2, etc. • Across space then draw a h parallel to p ; across space 1 draw h c parallel to p 1 and so on until e / is reached, this being parallel to P 4. Join a f, then the figure a, b, c, d, e, /, a, will give the B.M. diagram for the given load system. Now draw p x parallel to a /,/the closing link of the link polygon then on the vector line, 4 :«: = R„ and xO = R^. To draw the shear diagram, draw a horizontal line through BENDING MOMENTS AND SHEARING FORCES 135 X right across the span : this gives the base line for shear. Now project the point horizontally across space ; project point 1 across space 1 and so on, the stepped diagram thus obtained being the shear diagram. Proof. — Produce the links cb, d c, e d, f e back to meet the vertical through a in b\ c\ d\ e\ and let the first link a b Fig. 61. — Graphical Construction for Shear and B.M. Diagrams. produced meet the last link e / in y. Then the point Y is the point through which the resultant of the loads acts. Now the triangles abb' and p 1 are similar. . a 6' _ 0, 1 h ~ V 0, 1 X Zi Wi X ^1 .-. aV = V P moment of first load about a - 136 THE STRENGTH OF IVIATERIALS q' *i 1 ^ 7 ' ' _ moment of second load about a imi ar ^ c - and so on _ sum of moments of loads about a _ ^_ but R„ X L = sum of moments of loads about a , R„ X L . • . a e = — P Now consider As a e' f and a; 4 p ; they are similar : a e' _ 4 X • • IT ^^ . p X a e' T^ .' . 4:X = ^ = R^ L Similarly a: = R^ Now consider any point r along the span. S„ = R3 - W4 but the ordinate s of the shear diagram is equal to 3 x, and therefore the stepped figure gives the correct shearing force at any point. Let the vertical through r cut the B.M. diagram in R^ Rg and / e produced in eg- Then by exactly similar reasoning as before T5 moment of R„ about r Ri e, = — P ^ moment of W 4 about R Kg 63 = P . • . Rj R2 = R^ ^2 — 1^2 ^2 _ moment of R^ — moment of W 4 about r ~ P P . • . M„ = ^ X Rj R2 . * . The ordinate of the B.M. diagram represents the B.M. at any point. BENDING MOMENTS AND SHEARING FORCES 137 Scales. — As in the case of the cantilever (p. 126), if r' = .T feet is the space scale and V ^ y tons is the force scale, and if the polar distance is p actual inches, then the vertical ordinates of the B.M. diagram represent the bending moment to a scale V^ = p x x x y it. tons. Note. — In this construction the bending moment R^ Rg is measured vertically and not at right angles to the closing line a f. Case 6. Irregular Load — Overhanging Ends. — The Fig. 62. — Beam with Overhanging Ends — Graphical Construction. construction just described is equally applicable to the case where the ends are overhanging. Fig. 62 shows such a case. Set out the loads down a vector line as before and take any pole p. Now draw a b parallel to p across space 0, i.e. between the support vertical and the first force line. Then draw h c parallel to p 1 across sj^ace 1 and so on, the last link e / being drawn between the last force line and the reaction vertical. Joining a f we get the B.M. diagram as shown. To get the shear diagram draw p 5 parallel to a /, then the horizontal through gives the base line for the shear between 138 THE STRENGTH OF MATERIALS A and B. The shears in the end spaces will be equal to the end forces 0, 1 and 3, 4 respectively, as shown on the figure. This graphical loading is applicable to all kinds of loading, and any of the previous standard cases can be worked by its means. In the case of a continuous load the latter should be divided up into a number of small portions, and the load in each portion treated as an isolated load acting down the centre of such portion. Parabola of 3''4 Order JB, Fig, 63. — Simply supported Beam with Uniformly Increasing Load. Case 7. Uniformly Increasing Load. — Suppose a beam A B carries a load which increases in intensity uniformly from the end b to the end a. Let the intensity of the load at unit distance from b he w tons per ft. run ; then the intensit}^ at any point p at distance x from b will be equal to w x (Fig. 63). The intensity of the load at a will be equal to w I, and the I w Z^ total load W will be equal to w I x ^ = -^ BENDING MOMENTS AND SHEARING FORCES 139 The resultant load W acts through the centroid of the load curve, i.e. at distance from a. W R. = -3 Then Sp = total load to right _ W w x'^ ~ 3 2~ This depends on x^ and therefore the shear curve is a parabola. The point c^ is obtained as follows — S/ = W w x^ ~ 3 2 ' x^ W wP 2~ 3 2x3 x^ P ~ 3 X = "L = -577 I V3 Mp = R3 X .T - 2 W .-r w x^ ' 3 3 6 This depends on x^, and so the B.M. curve is a parabola of the third order. The maximum B.M. occurs at the point of zero shear (see p. 149), i.e. when x = ,- V3 . . Maximum B.M. = >— — , 3V3 18V3 1 1 \ >,3V3 9V3/ _ 2W? _ 2WZa/3 ~9\/3^ 27 = 128 WZ. The B.M. and shear curves then come as shown in the figure. 140 THE STRENGTH OF MATERIALS Case 8. Uniformly Loaded Beam with both Ends Overhanging.^ — Let a beam of span l be loaded with a uniform load of w tons per foot run, and let it overhang a distance X at each end, the distance between the supports being I. The overhanging portions act as cantilevers, and the shear and B.M. diagrams for such portions will be as shown. The B.M. for the centre portion will be a parabola drawn on the base shown dotted, the resulting curve being as shown cross- hatched. urfcns per Ft nooooonoocYYTXTmoono. JC ^^z^. I L Sh ear B.M. X V Vol Fig. 63a. — ^Uniformly Loaded Beam with Overhanging Ends. If the load on the centre portion of the span were removed, the B.M. diagram would consist of the two end parabolas and the dotted line. This B.M. is opposite in direction to that due to the centre portion, and therefore on replacing the centre load and drawing the parabola, the resulting curve is the difference between the two as shown. To find the value of x to get the least resultant B.M. w^e proceed as follows. As X increases, the B.M. at the supports increases and the resulting B.M. at the centre decreases, so that the least B.M. will occur when the svipport B.M. is equal to the centre B.M. BENDING MOMENTS AND SHEARING FORCES 141 The support B.M. = ^-^ The centre B.M, = ^f^ - ^^- Tn , 1 , W X^ WP W X^ It these are equal —^r- = -^ ^- Z o Z „ wP- . • . w x^ — —^ o I ^ ~ 2 V"2 L I -\- 2x J I 1 2 . V2 2 + V2 ^+2 ^2J^-V2)^2-V2 = -586 z This gives the position at which the legs of a trestle table should be placed to give the maximum strength to the latter. Case 9. Uniformly Loaded Beam with One End Overhanging. — A beam a b, Fig. 64, of span l is supported at one end a and overhangs the support c at the other end ; we wish to find, with a uniformly distributed load, the position of the support c which will be most economical — i. e. give the least bending moment. If the length of the overhanging portion B c is /g ^^^ the distance a c is l^, the B.M. diagram will be as shown shaded in the figure ; the portion b^ d is a parabola tangential at b^ and is the familiar diagram for a cantilever with a uniformly dis- to I tributed load ; the portion a G c^ is a parabola of height — ^ , o the usual one for a freely supj)orted span a c ; and a^ d is a straight line. The maximum positive B.M. will be given by kj, which on (J Cl\ will be equal to — "^ q- since the B.M. between the point 142 THE STRENGTH OF MATERIALS Ai and the point e of contraflexure will be the same as for a freely supported beam of span a^ e, and the maximum negative value is given by CiD. Our problem resolves itself into find the position of c to make J k or c^ d the least j)ossible. Now, if 3"ou move the point c to the left, c^ d will increase IC joer unit lenqTh A L wiCj -a) ^ I Fig. 64. and K J will decrease, whereas if c moves to the right the converse happens. If, therefore, c^ d = k j, movement of c will increase one or the other, so that the least value of either occurs when they are equal. This gives 8 2 i.e. (/i - af = 4/2'- or [li — a) 21.-. (1) BENDING MOMENTS AND SHEARING FORCES 143 Again, by the property of the parabola ---("i^-f) (2) This can be found by taking the B.M. at e for the span Ai Ci ; also by similar As. E F _ Aj E Ci D Ai Ci I.e. E F = -^ X 11- (3) Combining (2) and (3) we get or, a = -? (4) h Putting this result in (1) we get n J~ ^^ ^ ^2 i.e. l^^ -21^1^-1^^ ^0 (5) The solution of this quadratic equation gives, taking the positive root — ... 1 + ^^ = ^^4-^2 _ L J ^ 2-^^^ _ 3.^^^ ^2 ^2 ^2 or, ?o = s— 7TT *• 6. ?o = '293 L ^ 3414 In this case the maximum B.M. will be equal to ivl^ _ w X (-293 L)^ wj.^ 2 " 2 ~ 23-3 It will be of some interest to compare this result with that which would occur if each end were overhung and the sup- ports were placed so as to give the least B.M. for this condition. In this case the best condition is given when the overhang is '207 L. This gives a maximum B.M. equal to w X (-207 L)2 w l2 2 = 46^6 ^PP^'°'^' which is half that for the previous case. 144 THE STRENGTH OF MATERIALS Steps in Shear Curves. — In practice it is impossible to get absolutely sharp steps in shear diagrams, because the load cannot be transmitted at a mathematical point, but must be distributed over a short length. This has the effect of slightly rounding off the corners of the shear diagram as shown exaggerated in dotted lines on Fig. 65a. Numerical Examples. — (1)^ freely supported beam of 20 ft. span carries a uniformly distributed load of 5 tons, and isolated loads of 3 and 2 tons, at distances respectively of 4 and 5 ft. from the ends {see Fig. 65). We have first to get the reactions R^ and R„. Take moments round b. R, x20-5xl0 + 3xl6 + 2x5 = 50 + 48 + 10 = 108 T> 108 . . ^ .-. R„ = 10 - 5-4 = 4-6 tons. The shear diagram then comes as shown in the figure, the amounts of the steps being equal to the isolated loads. The point at which the shear is nothing is found as follows — Let it be at distance x from b. Then S, = = R„ - 2 - i(; . X = 46-2- ^^ 20 = 2-6 - ^, 4 .-. ^ = 2-6 4 X = 10-4 feet. The B.M. at this point will be a maximum, and will be equal to 1 10-42 M X = R,. X 10-4 - 2 (10-4 - 5) - ^ . —^ = 47-84 - 10-8 - 13-52 - 23-52 ft. tons. The B.M. diagram will consist of a parabola for the uni- formly distributed load, the max. ordinate of which is equal BENDING MOMENTS AND SHEARING FORCES 145 5 X 20 to 8 = 12*5 ft. tons. The B.M. diagram for each of the isolated loads will be a triangle, the respective heights 3 X 4 X 16 (.a t4. 4. , 2 X 5 X 15 „ being ^^ = 9'6 ft. tons, and ^a = '^ *^- tons. Combining these three figures we get the B.M. diagram .sTons 2 Tons ^'Fb 4 -67 Fig. G5. ■shown on the figure, and on scaling off the maximum ordinate it will be found to be 23*5 tons. Note. — In all constructions where diagrams are going to be added together, such diagrams must of course be drawn to the same scale. (2) A girder oj 24 ft. span is supported at one end, and rests on a column at a point 6 jt. from the other end. The girder carries a uniformly distributed load of 6 tons and an isolated load of 2 tons at the free end. Draw the shear and B.M . curves. 146 THE STRENGTH OF MATERIALS /? A, 6 Tons distributed ^To ons P4 3Pf\^ . E> 3.M.j^or Isolated Load E> M for Unijorm Load C. Combined B.M. Fig. 65a. To jfind the reactions take moments round a (Fig. 65a), Then 18 R„ = 6 X 12 + 2 X 24 = 120 •"• "^" "" 18^ ^ ^^ ^^^^ .-. R^ = 8 - 6f = 11 tons. BENDING MOMENTS AND SHEARING FORCES 147 The shear at c will be = 2 tons. It then increases until the point B is reached, when its value becomes equal to 35 tons. It then suddenly changes sign to a value 3*17 tons and then decreases uniformly to the end A, where the value comes 1'33. The shear diagram then curves as shown in the figure, the dotted lines indicating what occurs in practice owing to the impossibility of getting the loads and reactions concentrated on a mathematical point. Considering first the B.M. for the isolated and uniform loads separately, the B.M. curve due to the isolated load will come as shown in the figure, the B.M. at b being equal to 6 x 2 = 12 ft. tons. Now, considering the uniform load, the diagram for the portion b c will be a parabola with vertex wP" 1 6^ at c, the ordinate b^ d at B^ being = = -^ x ^ = 4'5 ft. tons. Then between b and a the B.M. curve due to this overhanging load will be the straight line Aj d, as such over- hanging load requires an isolated balancing load at a. The B.M. curve for the portion a b will be a parabola of wV^ 1 18^ central height = —^ = v x „- = 1012 ft. tons, the shaded portion being the resulting curve for the central and over- hanging portions of the uniform load. Combining these diagrams we get the resulting B.M. curve as shown, the max. B.M. occurring at B, and being equal to 165 ft. tons. Relation between Load, Shear and B.M. Diagrams. — Let A c' T>' B, Fig. 66, represent the load curve on a span a b. Take any point p along the span, and consider a short piece c D of the load, the centre of which is at distance x from p. Then the shear at p due to this piece of the load will be equal to the area of the portion c d of the load curve. Therefore the total shear S,. at p will be equal to the area of the load curve up to that point. But a sum curve * is such that its ordinate at any point * See p. 162. 148 THE STRENGTH OF MATERIALS represents the area of the primitive curve up to that point. Therefore the shear curve is the sum curve of the load curve. Suppose b' F E G a' is the sum curve of the load curve. Now consider the B.M. at p. Fig. 66. — Relation between Load, Shear and B.M. Diagrams. The B.M. at p due to the portion c d of the load = given portion of load x x. Now if E and r are the corresponding points on the shear curve, the difference of the ordinates at E and f gives the load on the portion c d. . • . Load on portion c d = e f^. . • . B.M. at p due to portion c d = e f^ x a:. .' . Shaded portion e f Fg e^ represents the B.M. at p due to the portion c d of the load. BENDING MOMENTS AND SHEARING FORCES 149 .• . Total B.M. at p = M,. = area of shear diagram up to p. Thus the B.M. curve is the sum curve of the shear curve. So that by drawing the sum curve B J H of the shear curve we get the B.M. curve. Scales. — If V ^ x tons per foot is the scale of the load curve, and p-^ is the polar distance measured on the space scale for obtaining the shear curve, then the scale of the shear curve V "^ Pi ^ tons. If P2 is the polar distance from which the B.M. curve is obtained, measured on the space scale, the B.M. scale will be V^ = PiPi^ foot tons. Point of Maximum B.M. — If the B.M. is a maximum, the tangent to the curve at this maximum must be horizontal, and therefore the corresponding ordinate on the shear diagram must be zero in order for the line through the pole to be also horizontal. Thus we get the rule that the maximum B.M. occurs where the shear is zero. The base lines s s and m m of the shear and B.M. curves depend on the manner in which the ends are fixed. If one end is free, the shear and B.M. at this point are zero. If one end is freely supported the shear at this point will be equal to the reaction, and the B.M. will be zero. The above relations are expressed mathematically as follows : Let the load at any point at distance x from the origin be F {x) Then the shear at the point will be = / ¥ {x) d x -{- c^ and the B.M. will be =yyF {x) d x + c^x -{- c^. The integration constants c^ and Cg depend on the manner in which the ends are fixed, and correspond to the base lines above referred to. A Template for Bending Moment Diagrams. — For Various Cases of Uniformly Distributed Loads. — In designing beams carrying uniform loads it is necessary in order to draw the Bending Moment diagrams to draw para- bolas ; the usual procedure is to draw the parabolas for the 150 THE STRENGTH OF >L\TERIALS special arrangement of the loads and for the particular manner in which the beams are supported, this involving a good deal of geometrical construction. A temj^late can, however, be used to serve for a large number of cases in the following manner — On a base a b, Fig. 67 — for convenience say 5 ins. long — Fig. 67. draw by the usual construction a parabola a d c with vertex at A, the height b c being for convenience equal to a b. A template of the form a b c d can then be made, a 45° set-square being a convenient form to cut it from. A pro- jection is preferably j^rovided as shown to avoid a sharp point which is liable to break off. By means of this template and a suitable choice of scales, the Bending Moment (B.M.) diagrams for a large number of cases can be then drawn as follows — BENDING MOMENTS AND SHEARING FORCES 151 Case 1. Cantilever fully Loaded. — Draw the span e f, Fig. 68, to a suitable scale, so that e f is not larger than a b ; erect a vertical at the fixed end f and place the template on the paper with the point A coinciding with the free end e and draw in the curve to the point g where it meets the vertical through B. The B.M. diagram is then as shown shaded. Scales. — If the intensity of the load is w lbs. per foot run, then the B.M. scale will be the square of the space scale multiplied by Take for instance the case where the space scale is Fig. 68. 1 in. = 2 ft. and the load is 1000 lbs. per foot run ; then B.M 4 X 5 X 1000 scale is \" 10,000 ft. lbs. Case 2. Simply-supported Beam fully Loaded. — In this case, Fig. 69, we draw e^ f^ to represent the span and we draw as before k vertical f^ g^ at one end; the template is then placed on the paper with the point a coinciding with the point e^ at the other end of the span and the curve is drawn until the vertical is cut to the point G^. Now join Gi Ej, the B.M. diagram then coming as shown shaded, the Bending Moment at any point h being found by projecting vertically and measuring the height x as shown. The scales are obtained as previously described. 152 THE STRENGTH OF iLlTERIALS E. Fic. 69. Fig. 70. BENDING MOMENTS AND SHEARING FORCES 153 Case 3. Uniformly-loaded Beam Overhanging the Support at one End. — In this case, Fig. 70, we place the template on the paper with the vertex of the parabola at the overhanging end Fg and draw in the curve until we meet at L the vertical through the other end Eg; then join l to the other support point k, the shaded area giving the B.M. diagram, the Bending Moment at any point being read off by projecting vertically as explained in the previous case. Fia. 71. At points such as m where the B.M. diagram crosses itself, the Bending Moment changes sign ; this of course corresponds to a reversal of the tension and compression flanges of the beam. Case 4. Uniformly-loaded Beam Overhanging at EACH End. — To obtain the B.M. diagram in this case with the aid of the template, we place the template on the paper with the vertex of the parabola coinciding with one end E3 (Fig. 71) and draw the curve until we meet the vertical through the other end at q. 154 THE STRENGTH OF ^UTERIALS Join Q to the mid-point r of the span, the line Q r cutting the vertical through the support p at the point s. Einally join s to the other support point N", the B.M. diagram then coming as shown shaded in the figure. Case 5. Uniformly-loaded CoNTrs'Uors Beam of Two Equal Span's. — We can get the B.M. diagram in this case with the aid of the template by placing it on the paper with the vertex coinciding with the end support E4 and drawing in the curve mitil it intersects at the point G3 the vertical through the centre support G3 ; then by reversmg the template and commencing the curve at the other outside support E5 we shall get the reversed curve going from G3 to E5 as shown in Fig. 72. Fig. A length G3 t is then set down from G3 of length equal to J G3 F4 and, by joining t to E5 and E4, we get the B.M. diagram as shown shaded in the figure. The student should check the correctness of this method after reading Chap. XV. The scales are obtained as previously explained. Numerical Exa^iple. — Take a continuous beam of two equal spans each 16 ft. long, each span being covered by a load of 1500 lbs. per foot run. Taking a linear scale of T' = 4 ft.. E4 F4 will be 4 ins. Then the B.M. scale will be, as explained above, 42 X 5 w; 16 X 5 X 1500 1" = - \~ = = 60,000 ft. lbs. If the distance Gq t be measured, it will be found to come BENDING MOMENTS AND SHEARING FORCES 155 equal to '8 inch with a template of the dimensions suggested in Fig. 67. .-. Maximum B.M. = '8 x 60,000 = 48,000ft. lbs. The B.M. at any other point u can be obtained by reading off the vertical ordinate x to this scale. In the case of a beam supported freely at one end and securely fixed at the other, the B.M. diagram will come the same as one-half of that shown in Fig. 72, the point E4 being the freely-supported end and the point F4 the fixed end. A number of other, cases might be given, but we think that the above are sufficient to show that a template of this kind would be of considerable assistance to draughtsmen for obtaining the B.M. diagrams for a variety of cases. In some respects the template would be more easy to make if it were made of the shape a d c b. Fig. 67, because the con- vex curve can be somewhat more readily shaped. If, instead of the given dimensions for a b and B c, other values are taken, the rule for scales must be correspondingly amended, bearing A B^ in mind that B c should represent ^y— for the B.M. scale to be equal to the square of the space scale when w ^ 1. If b c has not this value, then the B.M. scale will vary in the inverse ratio. B.M. AND SHEAR DIAGRAMS FOR INCLINED LOADS In all the cases that we have considered up to the present all the loading has been at right angles to the length of the beam. We will now consider some cases in which this is not the case, and will take both horizontal beams with non- vertical loads and sloping beams. The principal difference in this case is that there will be thrust in the direction of the beam, and we shall have a curve of thrust in addition to the curves of shear and B.M. The general rule is to resolve all forces, including the reactions, along and perpendicular to the beam. From the forces along the beam a curve of thrusts can be dra^vn, and 156 THE STRENGTH OF MATERIALS from the forces perpendicular to the beam the curves of shear and bending moment are drawn in the ordinary manner. We will define the thrust at any point of a beam as the sum of the components in the direction of the beam of all the forces to the right of it, remembering that if the thrust is negative it becomes a pull. Thrust 0>iacjram Fig. 73. — Beam with Inclined Loads. Case 1. Horizontal Beam Freely Supported sub- jected TO Inclined Loads. — Let a beam a b have inclined forces Fi and F2 (Fig. 73) meeting the centre line in c and D. Let the end a rest on a free support and let the end b be freely supported, but prevented from longitudinal movement as shown. If the resultant of F^ and Fg acted towards the end A, then this end would have to be prevented from movement. BENDING MOMENTS AND SHEARING FORCES 157 Resolve the forces F^ and Fg into vertical and horizontal components W^ Wg and Q^ Q2 respectively. Then R„ will be inclined, the vertical component Wi, being that found by considering the forces W^ W2 in the ordinary way and the horizontal component Q^ being equal to Q^ + Qg. The reaction R^ will be vertical, and will be obtained by considering the forces W^ and Wg in the ordinary way. If the resultant of F^ and Fg were found it would pass through the intersection of R^ and R^, since three forces in equilibrium must pass through a point. The shear and B.M. diagrams are then found in the usual way for weights W^ and Wg, and are as shown. The thrust diagram is obtained by plotting up at each point the value of the thrust, and this comes as shown. The same method applies for any number of loads, two having been chosen to give simplicity of figure. Case 2. Inclined Beam with Vertical Loads. — Re- actions Parallel. — Let an inclined beam a b (Fig. 74) be supported freely at a and pin- jointed at b. Then if it be subjected to vertical forces F^ and Fg at c and d, the reaction at A, and therefore also that at b, must be vertical, their values being found in the ordinary manner. Now resolve the weights and reactions along and per- pendicular to the beam, obtaining weights W^, W^, W2, Wa? and thrusts Q3, Q^, Q2, Q.,- Then the B.M. diagram can be drawn either on a sloping base A B or the projected horizontal base A^ b^. M„ = W« X D b 1 D B R„ . • . Wb X D B = Rh Zi . ' . We see that for a sloping beam with vertical reactions the B.M. diagram is the same as for a horizontal beam of the same span as the horizontally projected length of the sloping beam. 158 THE STRENGTH OF MATERIALS The B.M. at a point p, for example, is obtained by drawing a vertical through it, a b representing the B.M. The shear and thrust diagrams are obtained as shown, and will be easily followed from the figure. Fig. 74. — Inclined Beam with Lower End freely supported. Case 3. Inclined Beam with Vertical Loads — Top Reaction Horizontal. — In this case the resultant load must first be found. Let this resultant act down the line x x (Fig. 75). The reaction R^, at B must be horizontal, so draw B X horizontal, then if this meets the line x x, R,v must also BENDING MOMENTS AND SHEARING FORCES 159 pass through x, so that by joining a x we get the direction of R,. The values of R, and R^, are then found by a triangle of forces a, b, c. Now resolve the weights and reactions as before along and perpendicular to a b. The perpendicular components will be Thrust Oiaaram Fig. 75. — Inclined Beam with Top End freely supported. the same as before, and so the B.M. and shear diagrams will be the same as in the previous case (Fig. 74). The thrusts will be different, and will be as shown on the figure, which will be clearly followed. Case 4. Sloping Cantilever. — This is worked in a similar manner. Consider, for example, a uniform load of intensity m; on a cantilever of length I at an inclination (Fig. 76). The 160 THE STRENGTH OF IMATERIALS B.M. curve will be a parabola. Its maximum ordinate will be iv I cos 6 rt J because the total weight will be w I, and it acts at a distance ^ from the abutment. The shear diagram CiT Fig. 76. will be a sloping straight line, the maximum shear being w I cos 6; the thrust diagram will also be a sloping straight line, the maximum thrust being w I sin 0. CHAPTER VI GEOMETRICAL PROPERTIES OF SECTIONS — AREA, CENTROID, MOMENT OF INERTIA, AND RADIUS OF GYRATION Before considering the relation between the Bending Moment and the stresses in a beam, we will consider some geometrical properties of sections which, as we shall find later, are involved in that relation. The Determination of Areas. — (a) Mathematical Method. — If F [x) represents a function of x and the graph of the function be drawn, then the area between graph and the axis of x is given by the expression A =y'F{x)dx In practice, in the determination of areas, this method may- become practically unworkable if the equation of the curve cannot be simply expressed or if the integration cannot be performed. When these conditions occur we have to rely on the planimeter or on the following. (b) Graphical Method. — If a curve be plotted on a hori- zontal base and a new curve be drawn, such that its ordinate at any point represents the area of the given curve up to that point, the new curve is called the Sum curve or Integral curve of the given curve, which is called the Primiitive curve. The sum curve can be obtained graphically as follows : Let A c D, Fig. 77, be any primitive curve on a straight base A B. Divide a b into any number of parts, not necessarily equal (but for convenience of working they are generally taken as equal). These so-called base elements should be taken so small that M 161 162 THE STRENGTH OF MATERIALS the portion of the curve above them may be taken as a straight line. About 1 cm. or '4 in. will usually be a suitable size and in most cases a smaller element 11 will come at the end. Find the mid-points, 1, 2, 3, etc., of each of the base elements and let the verticals through these mid-points meet the curve in 1 a, 2 a, 3 a, etc. Now project the points on to a vertical line A E, thus obtaining the points I b, 2 b, Z b, etc., and join such points to a pole p on a b produced and at some con- venient distance p from a. Across space 1 then draw a d Fig. 77. — Sum Curve Construction. parallel to p 1 6 ; d e across space 2 parallel to p 2 6, and so on, until the point n is reached. Then the curve a d e . . . n is the sum curve of the given curve, and to some scale b ?i represents the area of the whole curve. Proof. — Consider one of the elements, say 4, and draw / o horizontally. Now ilf,g,o is similar to the A p, 4 6, a ^o_46, A * / o ~ P A but FA = p and 4 6, A = 4, 4 a / o X 4, 4 a area of element 4 of curve go p p GEOMETRICAL PROPERTIES OF SECTIONS 163 ^. ., , , area of element 3 of curve , Similarlv j q = and so on . ' . Ordinate through g^go+fq-{-... area of first four elements of curve - ^ . • . The curve a d e . . . ?^ is the sum curve required. Then ii b n he measured on the vertical scale and p be measured on the horizontal scale, the area of the whole curve will be equal to ^^ x b n. It is obviously advisable to make p some convenient round number of units. The sum curve obtained by this method may have the same A /}• r c L___ < , J 1 J a ' 1 r-1 • ^ i * 1 I i i i / \ ' ' 1 ' •( ' ] ' \ 1 f } ' 1 B B' Fig. 78. operation performed on it, and thus the second sum curve of the primitive curve is obtained, and so on. If the operation be performed on a rectangle, the sum curve will obviously come a sloping straight line, and if the sum curve of a sloping straight line be drawn, it will be found to be a parabola. In the case in which it is required to apply this construction to a curve which is not on a straight base, the curve is first brought to a straight base as follows — Suppose A cB d, Fig. 78, is a closed curve. Draw verticals through A B to meet a horizontal base a' b'. Divide the curve into a number of segments by vertical lines at short distances 164 THE STRENGTH OF MATERIALS apart, and set up from the base a b lengths a^, h^^, etc., equal to the vertical portions a, h, etc., on the curve. Joining up the points thus obtained we get the corresponding curve a' Ci b', on a straight base. (c) Simpson's Rule. — Divide the base into an even number of equal parts (each equal to c) and measure all the corre- sponding ordinates. Then area of curve is equal to twice sum of four times sum of _ sum of first \ "^ odd ordinates and last ordinates J 3 (even ordinates Fig. 79. J AT -First Moment of an Area. X {d) Parmontier's Rule. — Divide up base and measure ordinates as above, then area of curve is equal to Op sum of odd _ ^ f/ second _ first ordinates 6(^\ordinate ordinate last _ preceding yi ordinate ordinate /J First Moment of an Area. — Let a small element of area a of any figure be situated at the point p. Fig. 79, and let x x be any straight line or axis. Then if p n is drawn j^erpen- dicular to x x, a x P N is the first moment of the element of area about the given line. Now, if the whole figure is divided up into elements of area such as a, and the moments of each element be taken about x x and the whole of these moments GEOMETRICAL PROPERTIES OF SECTIONS 165 be added together, the resulting sum is called the first moment of the area. . ' . The first moment of the whole area is the sum of quantities such as a x P N. This is expressed symbolically as follows — First moment of whole area = 'X {a x p n). Now the centroid or the first moment centre of an area is defined as the point at which the whole area can be con- sidered concentrated, in order that its moment about any given line will be equal to the first moment of the area about the same line. Thus if c is the centroid of the area, and c J is drawn perpendicular to x x, and the area of the whole figure is A, we have A + cj = :S(a.PN) . • . c J = — ^^ — T -' A This will not determine the exact position of c, but only its distance from the given line x x. If the exact position of the centroid is required we must also take moments about some other line, not parallel to x x, then the distance from the two lines will determine its position. In connection with the centroid it should be noted that the position of the centroid depends solely on the shape of the figure, and not on the position of the axes about which moments are taken. As in the case of forces, we have positive and negative moments in areas, the moment being positive when the given element of area is above or to the right of the given axis, and negative when it is below or to the left. First Moment about Line through Centroid. — Now consider the first moment of an area about a line c c, Fig. 80, through the centroid. The moments of elements of area above the line such as that at P will be positive, and the moments of elements of area below the line such as that at p' will be negative. Now in this case c J is zero, and therefore A x c j will also 166 THE STRENGTH OF :\IATERIALS be zero, and therefore we have the rule that the first moment of any area about a line tlirough its centroid is zero. PosiTio:?^ OF Centroid with Axes of Symmetry. — Suppose an area has an axis of s}Tnmetry Y Y, Fig. 81. Then this line y Fig. 81. di^^des the area into two exactly similar halves so that corresponding to each element of area at p having a positive GEOMETRICAL PROPERTIES OF SECTIONS 167 moment about Y y we have an equal element at p' having an equal negative moment about y y so that the total moment of the area about Y Y is zero, or Y Y passes through the centroid. If the figure has another axis of symmetry x x, the centroid also lies on this line, or we have the rule that the centroid of a figure is at the intersection of two axes of symmetry. For the determination of the position of the centroid for various cases see p. 175-189. It should be noted that the centroid of an area is the same as the centre of gravity of a template of the same shape as the area. Second Moments or Moments of Inertia. — The rforce(/) ^ product of a 4 "^^^^ (^) I by the square of its distance r from a [volume {v)j fforce ^ given point or axis is called the second moment of the - ^^^^ about the given line or axis. [ volume j Now, in considering rotating bodies the second moment of the mass has to be considered, and this quantity has been given the name of the moment of inertia. In the application of the second moment to the strength of materials we shall have nothing to do with inertia, but the term moment of inertia has been generally adopted, and so we shall use it; but we must remember that it is really a borrowed term and quite an unsuitable one. Application to Areas. — If an element of area a is situated at the point p, Fig. 82, and p n is drawn perpendicular to a line X X, then the second moment of this element of area about the line X X is equal to a x p n^. If, as in the case of the first moment, we divide the whole area up into elements and take the second moment of each, we see that the second moment of the whole area about x x is the sum of the second moments of the elements. The letter I is always used to denote the second moment, the line x x, about which the moments are taken, being indicated by writing it I^^. Thus we see I^^ = S (a x p n^). 168 THE STRENGTH OF ]\L\TERIALS In the same way, considering the line y y, we have Now suppose K is such a point that the whole area can be considered concentrated there so as to give the same second moments about x x and Y Y as the second moment of the area about these lines. Then A x K q2 = I,, and A X K R- = I,,. Then the x^oint k by analogy might be called the secondroid Fig. 82. — Second Moment or Moment of Inertia of an Area. of the area with regard to the axes x x and Y Y. The point of importance ^vith regard to the secondroid is that its position depends on the j)osition of the lines about which the moments are taken, whereas the position of the centroid does not. RADIUS OF GYRATION Now. the distances of the secondroid from the lines x x and GEOMETRICAL PROPERTIES OF SECTIONS 169 Y Y are called the second moment radii or radii of gyration about the given lines, and are written k^,. and k,j respectively. .'. We have A y^,2 = I,.^ = 2 (a X p n2) or k,. = ,, ^ A kj" = I,, - S (a X P m2) or k, = y "2- Now, in practice it is nearly always the second moment about a line through the centroid that is required, and this is obtained as follows — Given the Second Moment or Moment of Inertia of AN Area about a given Line, to find it about a Parallel Line through the Centroid. Suppose we know I^x- Now, I,^ = 2 (a X P n2) = :S [a X (p s + s N)2] = :s [a X (P S + d,,)^] = :S [a . (P S2 + 2 P S . 6^, + ^,2)] = :S (a . P S2) + ^ (a . 2 P S . 6?,) + S (a . ^/) Of the terms on the right-hand side S (a X P s)2 = I^.^ (which is required) 2 (a . 2 p s . d,) = 2 d^, 2 a . p s = 2 d,. (first moment of area about line c^ c^ through centroid) = 2 d_, X = ^ (a . d/) = d/ 5 a = d_,^ (area of whole figure) = d/.A . • . We have I^^ = I,^ + A d/ or I„ = I,^ - A ^,2 Similarly I,.,. = I,.,. — A d,/ * The Momental Ellipse or Ellipse of Inertia. — The principal axes of a section are defined as two axes at right 170 THE STRENGTH OF MATERIALS angles through the centroid, such that the sum of quantities such as a X p M, p N, or the 'product moment as it is called, is equal to zero. In the case of sections with an axis of symmetry, such axis determines one of the principal axes. Let X X and Y Y, Fig. 83, be the principal axes of a section and let k^ and k„ be the radii of gyration about the two axes. With o as centre draw an ellipse, o x being equal to ky and o Y Fig. 83. — Momental Ellipse of Ellipse of Inertia. being equal to k,.. Then this ellipse is called the momental ellipse or ellipse of inertia. To obtain the radius of gyration k._ about a line z z passing through o at an angle ^ to x x, draw 2; z a tangent to the ellipse parallel to z z, and draw o Q perpendicular to it. Then o q^ = k. for I,^ = 2 a . p r2 = 2 a (P S — S R)2 = 2 a (P S — N t)2 = 2 tt (1/ COS 6 — X sin O)"^ = -%a.x^ sin2 + 2ay^ cos^ ^ - 2 2 a; ?/ sin ^ cos ^ = sin2 e^a.x^ + cos2 O^ay^ -2sinecos01xy GEOMETRICAL PROPERTIES OF SECTIONS 171 Now, 2 X y is the product moment, and as x x and Y Y are the principal axes, this is zero. .• . I,, - sin2 e^{a.x^) + cos2 e^(a 2/2) = I^^ sin^ $ + I,^ cos^ .-. Ak,^ == A h^ sin2 + Ak/ cos^ k.} = A;/ sin^ $ 4- h,^ cos^ and therefore from the properties of the ellipse o Q is equal to k,. A rather more convenient construction (see Fig. 84) is to draw a circle with radius k^ and draw o d at right angles to z z to meet the circle in d. Draw d e horizontally to meet the ellipse in e, then o e = A;-. If many values are not required, the ellipse need not be drawn at all, but instead draw a second circle with radius k,, ; then draw F e vertically to meet d e in e, thus fixing the point e. To find the principal axes in the case where there is no axis of symmetry, the procedure is as follows — (a) By graphical methods or by calculation first find the value of the product moment and the radii of gyration about any two axes through the centroid at right angles. 172 THE STRENGTH OF MATERIALS Let the product moment be A p'^ and the radii of gyration k_r and k„. Then the angle of inclination B of the principal axes to k^ or k„ are given by the relation {b) By graphical methods or by calculation find the second moments of the given figure about lines x x and Y Y at right angles to each other and passing through the centroid and find it also about a third line z z at 45° to the other two. Then if is the inclination of the principal axes to x x and y y tan 2 ^ = L^+l^-^J^. CoxDiTiox THAT Peodijct Momext IS Zero. — It can be shown that the condition that the product moment about two lines is zero is that such lines form conjugate diameters of the moment al ellipse. A numerical example on the momental ellipse will be found on p. 242. Second Moments about any Two Lines through the Centroid at Right Angles. — A property of the second moments of a figure that is sometimes useful is that the sum of the second moments of an area, about two lines at right angles through the centroid, is equal to the sum of the second moments about any other pair of lines at right angles through the centroid. Second Moment or Moment of Inertia of Figure about an Axis perpendicular to its Plane. — The second moment or moment of inertia of an area about an axis o perpendicular to its plane is called the polar second moment or moment of inertia, and is equal to 2 a . p o^. Let any two axes x x and y y at right angles be drawn through o, and let perpendiculars p x, p m be drawn to these axes, Fig. 85. GEOMETRICAL PROPERTIES OF SECTIONS 173 Then p o'-^ = p n^ 4- n o^ = PN^ + P M^ . • . 2 a . P O^ --= :^ ft . P N^ + ^ a . P M^ = l\x + Ivv Fig. 85. — Polar Moment of Inertia. Therefore wc have the following rule — The polar second moment, or moment of inertia, about an axis perpendicular to the plane of any area, is equal to the Fig. 80. sums of the second niojiicnls about any two linos at right angles, drawn through the axis in the plane of the area. The Determination of Gentroids, Moments of Inertia, and Radii of Gyration. — (a) Mathematical. — Consider the curve of a function y = V (x). 174 THE STRENGTH OF ^UTERIALS Then considering a strip of width d x parallel to the axis of X, Fig. 86 Area of curve = / F {x) d x First moment of area about o y = / F (x) dx x x Second moment of area about o y = / F (x) dx x x^ Consider, for example, the parabola y- = 4:a x, and take the area between the curve and the axis of x, Fig. 87. Y 2 Fig. 87. Area of curve = fy dx = / 2 a^ x^ B = 2 a^J'x^dx = \ 2a^- . dx 2 .... „ x-^ a^ B- Now 2 «' b' = H Area of curve = ^ b H. Fig. 87. First moment about o Y = fx y dx ^ J 2ah x^-dx 2ah / x^dx ^ 2 «K ; x^ 4 2 ^ah B^ = ^ b^h o o -1 B H 6 . • . distance of centroid from o Y == ^V—— : = c B o' B H O GEOMETRICAL PROPERTIES OF SECTIONS 175 Second moment about o y = Cx^ y dx = f2 oh xr^ d x = 2 a^J x^ dx = 2aKl x^ 7 _ 4 , : 23 = ^ a^ B^ = ^, B^ H • ^2_tB^h |BH or k^ = Vf B. If the second moment is required about the base x z, we proceed as follows — T 2 3 lov - ;^ b3 H I,, =. I,, -k.d^ 232 9b2 - -^ H B 3 • B H . 25 I.\z ^ Ice + A. »]_ 2,6 „ , 8 3 = 7^^ -25^^ +75^^ ^ fl50 - 126 + 561 „ 80 \ 525 ^ ^^ 525 H B'* -I ,.7^-^ h = H B' 16 105''^^ A list of values of second moments, etc., for common figures will be found on p. 185. It often happens in practice that the mathematical method is unworkable, in which case the following graphical methods are necessary. (6) Gbaphical. — First and Second Moment Curve Method. — (1) Centroid. — Suppose we have any area p r q s, Fig. 88, and any two parallel lines x x and y y, at distance h apart. Draw a thin strip of the area parallel to x x and of thickness t and let its centre line be p Q. From one of the ends of this centre line, say Q, draw a perpendicular Q m to Y Y and from the other end draw p N perpendicular to x x. Join M N and let it cut p q in q^ and produce m q to cut x X in L. 176 THE STKENGTH OF MATERIALS Then the As p n q^, m n l are similar. p Qi _ NJ. _ P Q P N ~ M L ~ h PQi Multipljdng through by t we have p Qi X ^ = P Q X ^ X PN _ area of strip p Q x p n h h A^^^-P^ 4.- - £ ^ • First moment of strip about XX,,, Area oi portion p q^ of strip = , ^ (1) Fig. 88.— Graphical Determination of Centroid and Moment of Inertia, etc. Now divide the whole area up into strips and join up all the points corresponding to Q^, thus obtaining the First Moment Curve R Q^ s. Then the area to the left-hand side of the first moment curve Avill be the sum of the areas of portions of strips such as p Qj. Call this the First Moment Area (A^). Then we have Sum of first moments of strips about x x Ai = First moment of whole area h GEOMETRICAL PKOPERTIES OF SECTIONS 177 . * . First moment of whole area = A^ ^ , . , „ , . 1 p First moment about x x or distance oi eentroid irom x x = '^ area oi figure = -i^ (2) Draw any vertical line f b to cut x x in r and y y in b, and through F draw any inclined line, on which set out f a equal on some scale to A, and f a^ equal to A^. Join a B and draw a^ c parallel to it, then the eentroid lies on a line through c parallel to x x or Y Y. ^ C F F tti For — = — ± F B Fa C F _ Ai • ' • h ~ A A.h or c F = ~\~ A And this by relation (2) above gives the distance of the eentroid' from x X. (2) Second Moment. — ^If the second moment is required about the line x x draw q^ m^ perpendicular to Y Y and join Ml N, cutting p Q in Q2 and let m^ q^ produced cut x x in l^. Then the As p n q^, m^ n l^ are similar. • ^^2 ^ J^ Li ^ PQi * * P N Mj Li h P Qi X P N .-.PQ, =^^^ Multiply through by t, then we have P Ql X i X P N P Q2 X ^ = h T, , 1 ,, , , area of strip p q x p n ±>ut we have seen that p Qi x ^ = f p Q2 X i area of strip p q x p n^ second moment of strip p Q about X N .*. Area of portion p Q.2 of strip = —7-2 . .(o) Now repeat this construction for each of the strips and join up all the points corresponding to Q2, thus obtaining the second moment curve R Q2 s. N 178 THE STRENGTH OF MATERIALS Then the area to the left-hand side of the second moment curve will be the sum of the areas of portions of strips such as p Q2. Call this the second moment area (Ag). Then we have . _ Sum of second moments of strips about x x .-. Ix. = A2A2 (4) Some care is required in determining which area to read as A^ or Ag. It does not matter whether the verticals are drawn downward from p or from Q, but when the moments are re- quired about one of the lines, say x x, read, for the first moment area, the area on that side of the first moment curve from which the perpendiculars are drawn to x x, and in drawing the second moment curve draw from the first moment points, such as Qj, perpendiculars to the other line y y, again reading the area to the side from which the perpendiculars were drawn to x x. Now, on the line f a set out f a^^ equal to Ag on the same scale to which the other areas were drawn, and join a^ b, drawing ag d parallel to it. On D F describe a semicircle, and draw a line c e parallel to X X to meet it in e. Then c e will be equal to k, the radius of gyration about c c. Proof — F D _ F ^2 _ ^2 Now F B F «! Ax F D li X Ag Jl X Ixx A .CF k/ A X CF CF h F D F E F E FC • • F D . F C F E^ .* . F D =r FE2 CF .*. F E = kr GEOMETRICAL PROPERTIES OF SECTIONS 179 Now F E^ = F C^ + C E^ . • . k,^ = c E^ + d_r^ . • . c e2 - yfc,2 _ ^2 But we have already shown that k} = k/ - d/ . • . c E = A;,, Numerical Example. — Graphical Determination of Radius of Gyration of Rail Section about Centroid. x^./lrea of HalfSecfJoni. '■ = •ici-/8-Sti.ln I 7 Sum Curve o/-|-J.-\-L'?_r^ Halj Rail 4 T^i 1 V Fig. 89.— Rail Section. Fig. 89 shows the graphical determination of the radius of gyration about the centroid parallel to the base of a British Standard 85 lb. flat rail section. Since the section is symmetrical about a vertical centre line, the first and second moment curves need be drawn only for half the section, this simplifying the construction considerably. The lines x x and Y Y are taken as the horizontal lines, touching the section at top and bottom. 180 THE STRENGTH OF MATERIALS The areas A, Aj, Ag are next found b}^ planimeter or by sum- curve construction. (To avoid complication of figures, the sum curves for the first and second moment curves are omitted.) The first and second moment areas are to the left of the curves. When multij)lied by two, because only half the section was considered, we get A = 8*36 sq. ins. A^ = 4'02 sq. ins. Ag = 305 sq. ins. To the side of the figure a vertical r b is drawn between the X X and Y Y lines, and the points a, a^, a^ obtained as shown. X D C X Fig. 90. — ^IMoment of Inertia of Rectangle. Then by joining a b and drawing a^ c parallel, we get the point c, c F giving the distance of the centroid from x x ; and by joining a^ B and drawing a^ d parallel, we get the point d. On D F a semicircle is drawn, and c e is dra^\ii horizontal to meet the semicircle in e. Then c e = k, which on measuring will be found to be 1-91 ins. This construction should be gone through as an exercise. Application of above Method to Case of Rectangle. — Let A B c D, Fig. 90, be a rectangle of base h and height h, and take the lines x x and y y through c d and b a respectively. Then the first moment curve will be the diagonal bed, Avhile GEOMETRICAL PROPERTIES OF SECTIONS 181 the second moment curve will come a parabola b f d, so that A, ''^ A. 2 bh d.. = A 2 bh Fig. 91. — Mohr's Construction for Moment of Inertia. .-.I., = I,, - Ad/ _bh^ _ bh.h^ _bh^ ~ 3" 4 ~ T2 Alternative Graphical Construction — Mohr's Method. — The following graphical method for obtaining the second moment about the centroid is in some cases more convenient in use than the one previously given. Divide the area, Fig. 91 , 182 THE STRENGTH OF ^UTERIALS into a number of small strips of equal breadth, parallel to the direction about which moments are taken, and draw the centre line of each of said strips. Then if the strips are suffi- ciently small (we have only taken a few strips in the figure to avoid complication) the lengths of these centre lines represent the areas of the separate strips. Xow. on a vector line, to some scale, set out 0, 1, 1, 2, ... 6. 7 to represent the area of each strip, and take a pole p at distance = ^ total area 0, 7 from this vector line. Then anj'where across space draw and produce a line a k parallel to 0. p ; across space 1 draw a b parallel to p 1 ; across space 2, b c parallel to p 2, and so on until the point g is reached. Then draw the la?>t link g'h parallel to the last line p 7 to meet a 7i in h. Then the line c c through the centroid passes through h, and if a is the area of the shaded area, and A is the area of the figure, I- of figure = A >: a. Proof. — Consider one of the elemental areas, say 0, 1, and produce a b to meet the horizontal through k in b^. Then, by the law of the link and vector polygon construction, treating the areas of the elements as forces, 1 polar distance b'h = moment of first for 'ce about c c = 0, 1 X .r 1 ^ ^to1 :al area = 0. 1 < X Area of . la b' 0. h 1 _ 1 2 X X b' h X . ■2x2 r 2 A second moment of element about c c , , 11^ second moment of fisure about c c . • . Area of shaded figure = a = r-^ ?'. e. A a = second moment of ficfure about c c. • GEOMETRICAL PROPERTIES OF SECTIONS 183 The proof that Ji determines the centroid is based upon the fact that in the hnk and vector polygon construction the meet of the first and last links determines the resultant, and in this case the centroid is where the resultant of the separate areas considered as forces act. * Equivalent Centroid and Second Moment of Heterogeneous Sections. — Suppose that the cross section of a beam is composed of two materials for which Young's modulus is not the same, and let Young's modulus for one material B be m times Young's modulus for the second material C. Then in the case of direct stress we have seen that the material B behaves as if it were replaced by m times its area of the material C. In the case of a beam the same relation holds, so that we may replace the material B by an area m times as wide, the width being taken parallel to the line about which moments are taken. Then if A is the area of material B, and A^ that of material C, the equivalent area of homogeneous material C is given by A2 = Ai + m A To obtain the distance d of the equivalent centroid from a line X X, take first moments of the separate areas about x x and let them be M and M^ respectively. Then equivalent first moment of the second material is M2 = Ml + m M Ml + mM Aj + w A To obtain the equivalent second moment about a line x x, take the separate second moments about x x and let them be I and Ii respectively, then the equivalent second moment of the second material is given by I2 = Ii + m I We shaU give numerical examples and further explanation of this when dealing with flitched beams and reinforced beams. 184 THE STRENGTH OF MATERIALS The above reasoning ma}' be shown graphicall}" as follows — Let A B c D (Fig. 92) represent any area which has em- bedded in it two bars x and y of different material. For considering the moments about any line such as D B shown dotted, make a strip e r of the same depth as x, and of area equal to {m — I) area of x and also a strip G h of area equal to {m — 1) area of Y. Then the equivalent first and second moments of the heterogeneous section about the given line will be the same as a homogeneous section of form aefbghcd. We take e r = (?7z — 1) area of x because the bar already occupies an area equal to its area, so that equivalent area of second material = [{m — 1) -i- I] area of x = m x area of X. Calculation of Moment of Inertia and Radii o! Gyration of Sections used in Constructional Work. — The moments of inertia of sections composed of sections of known moment of inertia are found by adding up the moments of the separate parts, or subtracting when the area consists of the difference of known areas. GEOMETRICAL PROPERTIES OF SECTIONS 185 1 CO CO W 1 + N ^0 ^l« lO 1 ~* ^ r-H ' «t-l i o CO 4^ r« CO (3 ^ ^ a o CO 1 eo g M ' 3^ ' 1 ' ' 1 1— 1 + "* 1 ^ M m 1 '^ CO CO 1 »o 1 1 ^ ,-H 1 Si I-H hO i o 5 rOl-?^ 1 ^0 i(M 1 1 i 1 ^0 CO IQC ^ lTt< -ti M a ^ 6 «<-( ^^ "^^ o ><) I— 1 I-H .2 a >^ ica r< Cs| r^ [jO t+ 1 -^ |C<1 1 ^Se ^12 S o ^ 1 1 (r? iio Pm fH »< Ico "^ i Area 2!^ + r— 1 s> (M |C0 r-l |C0 S^ ID s ^ ^^ Figure. 1 O * 0) (D 02 Fh 00 186 THE STRENGTH OF MATERIALS Y 2 L— b ^ .© t X r X Fig. 93. — Properties of Common Figures. For the properties of British Standard Steel Sections, see Appendix. GEOMETRICAL PROPERTIES OF SECTIONS 187 The following examples should make the method of calcula- tion clear for any such case. See Figs. 93 and 94. f 1 h c -1 < 1 c V -at- O I — 2_ 10' ^^ 3-5 •© •475 •4 75' X -iz •375 5 /2i' Y 4i' ^' 10' X Y Fig. 94. — ^Moments of Inertia, etc. 188 THE STRENGTH OF MATERIALS (1) Box or I Section. — These are geometrically equi- valent as far as the line c c is concerned, because if the box section be cut in half vertically and the two halves be turned back to back, we get the I section. Then I„, = b^'-ib-lh){h-2tf (2) Hollow Circular Section. When the thickness of metal is small and equal to t, this approximates to I„ = ^^^ (3) Channel Section (neglecting inclination of sides and rounded corners). — Consider the section shown in Fig. 94 (3). Area = A = 35 x 475 + 505 x 375 + 35 x 475 = 5*219 sq. ins. To obtain distance d^ of centroid from x x take first moments about X X. Then A X d, An^ 3-5 , 505 x -375 x 375 , 3-5 x '475 x 35 = 3*5 X -475 X ^ H ^ ^ 2 = 2-910 + -363 + 2-910 = 6183 • ■ ^" = 5T2Y9 ^ 1185 ins. Second moment about x x = I^^ _ -4 75 X 3- 53 5-05 X -3753 -4 75 x 35 ^ - ^ + 3 +3 = 6-775 + -089 + 6-775 = 13-639 in. units. = 13-639 - (5-219 x 11852) = 13-639 - 7-323 = 6-316 in. units. / I .-. ^ = ^/ = 1010 nis. \ A (4) Cast-iron Beam Section. Area = A = 2 x IJ + 7 x 1 + 6 x li = 19 sq. ins. GEOMETRICAL PROPERTIES OF SECTIONS 189 Moments round base A d,. = 3 X 9-25 + 7 X 5 + 9 X -75 = 27-75 + 35 + 6-75 = 69-5 in. units. . • . d^. = ^^- = 3-658 ins. - 6 X 3 - 6583 _ 5 X 2-1583 2 x 6-342^ _ 1 x 4-892^ . • . Ice - 3 3 + 3 3 = 219-95 in. units. (5) Built-up Mild Steel Column Section. — Composed of two 10 X 3J X 28*21 channels and four 12 in. x \ in. plates. Required to find k^ and ICy. From the Table of Stand- ard Sections we obtain the following information concerning the Channel Sections — Area of each 8-296 sq. ins. I about centroid parallel to x x = 117-9 in. units 1 ,, ,, ,, y Y = O 194: ,, ,, Distance of centroid from web = -933 in. .-. Total area of section= (4 x 12 x i) -f- (2 x 8-296)= 40-592 sq. ins. Moment of Inertia about X X. 2 channels, 117-9 each = 235'8 2 pairs of 12 x | in. plates about centroid = ~ ^ — = 2-0 K X d^ for two pairs of plates = 2 x 12 x 5-5^ =726-1 Total . . . . . . = 963-9 in. units 7 / 963-9 . „. . •*'^'==V4ir592==^'^^''''- Moment of Inertia about Y Y. 4 X - X 12^ 4 plates 12 X J about centroid = ~~ := 288-0 2 channels about centroid = 2x81 94 =16-4 A X ^2 for each channel = 2 x 8-296 x 3-1832 = 168-5 Total = 472-9 4729 ^^ ^ V 40-592 ^ ^'^^ "^^* 190 THE STRENGTH OF ^UTERIALS (6) Built-up Beam Section. — Composed of two 14 in. x 6 in. 46 lb. I beams and four 14 in, x f in. plates (Fig. 95). Required I^^. From the Standard Section Tables we obtain the following information concerning the I beams — Area of each = 1353 ■•.XX 5 > 5 > = 440-5 Mean thickness of each flange = 698 in. a Fig. 95. I^^ OF WHOLE Section (not attow^no for Rivets). - Ivx of two I beams -= 2 x 440o = 881 ■' X 14 /5\ I of two pairs of plates about centroid ~ - — - — x ( - ) 12 \4/ 4-8 A d^ for two pairs of plates = 4x 14x — x 7 62o- = 2035 Total 2020 8 AixowANCE FOR RivETS (ucglcct I of each rivet -hole about its centroid). Area of each hole = (2 x -^- -f '698)4 = 1TU4 o o Dist. of centroid from xx = 7276 .-. I,, = 4 X 1-704 ■ 7-276- = 3608 . ■ . Xett I,, = 2920-8 - 3608 = 2560 GEOMETRICAL PROPERTIES OF SECTIONS 191 (7) Built-up Sections — Approximate Method. — The moment of inertia of built-up sections can be found approxim- ately by adding the moment of inertia of the I beams or channels to A d? for the plates, d being taken as the distance from the centre of one set of plates to x x and the nett area of the plates being taken for A. Taking the section of the previous example, we then get I^^ as follows — I„ of two I beams = 2 x 440-5 = 881 kd^ for plates = 4 x ^ (^14 - 2 x ^^ x 7-52 ^ 1875 Total approximate I^^ = 2756 in. units I CHAPTER VII STRESSES IN BEAMS We have seen in a previous chapter how the bending moment and shearing force at different points along a beam, loaded in various manners, can be found ; our next problem is to find the relations between these quantities and the stresses occurring in the beam. We shall get a good preliminary idea of the stresses occurring in beams b}^ considering a model devised by Professor Perr3\ Suppose that a beam fixed at one en(J carries a weight, W (Fig. 96), at the other end, and that it is cut through at a certain section. Then the right-hand portion can be kept in equilibrium by attaching a rope to the top and passing over a pullej^ a weight W being attached to the other end of the rope, and by placing a block B at the lower portion of the section and a chain a at the upper portion. Then the pull in the rope overcomes the shearing force ; and the block b carries a compressive force c, and the chain a carries a tensile force T. Since these are the onh' horizontal forces, they must be equal and opposite, and thus form a couple. Then the moment of this couple nuist be equal and opposite to the couple, due to the loading, which we have called the bending moment. In the actual beam, owing to the deflection which takes place, the material on one side of the beam will be stretched, and the material on the other side will be compressed, so that at some point between the two sides the material will not be strained at all, and the axis in the section of the beam at which 192 STRESSES IN BEAMS 193 no strain occurs is called the neutral axis (N.A.). We see, therefore, that : The neutral axis is the line in the section of a beam along which no strain^ and therefore no stress, occurs. In an elevation of a beam there is also a line of no strain or stress, which may also be termed a neutral axis. These two axes are really the traces of a neutral surface. If we kno^ the manner in which the strain varies from the neutral axis to the outer sides of the beam, from a knowledge of the relation between stress and strain we can find the stresses at different points across the beam, remembering that the total compressive stress must be equal to the total tensile Te n Fig. 96. — Stresses in Beams. stress, and the moment of their couple must be equal to the bending moment. The moment of the couple due to the stresses is often called the moment of resistance. Assumptions in Ordinary Beam Theory. — We will first make the following assumptions with regard to the bending of beams, and from such assumptions we will deduce a relation between the maximum stresses, due to bending at any cross section and the bending moment — o 194 THE STRENGTH OF IVIATERIALS (a) That for the material the stress is proportional to the strain, and that Young's modulus (E) is equal for tension and compression. (6) That a cross section of the beam which is plane before bending remains plane after bending, (c) That the original radius of curvature of the beam is very great compared with the cross-sectional dimensions of the beam. We will also for the present restrict our investigation to the case of simple bending, ^. e. that in which the following con- ditions hold — (1) There is no resultant thrust or pull across the cross section of the beam. (2) The section of the beam is symmetrical about an axis through the centroid of the cross section parallel to the plane in which bending occurs. To get a clear idea of the stresses in beams it is absolute^ necessary to have a clear idea of the assumptions involved in formulating any particular theory, and of the effect of such assumptions on the results. Let A B, Fig. 97, represent the cross section of a beam which has been bent (the amount of bending having been exagger- ated). Before bending, the line a b had the position of a^ b^ so that B Bi represents the maximum tensile strain, and a a^ the maximum compression strain. From our assumption (6), called Bernoulli's assumption, a^ b^ and a b are both straight lines. The neutral axis then passes through c, the point of no strain, and it follows from the above assumptions that the strains are proportional to the distances from the N.A. From assumption (a) it foUows that the diagram of intensity of stress is also a sloping straight line, Ag Bg, the portions Bg c and Cg A being continuous, because Young's modulus is equal in tension and compression. It is clear that the maximum stresses in compression and STRESSES m BEAMS 195 tension occur at the points A and b, and let these be /,. and ft respectively, d, and dt being the distance A c and B c . Position of Neutral Axis. — Now consider an element of area a at a point p at distance p n from the N.A. Then the stress at the points p is equal to Pj Pg Tension Cross -section Diagram of Intensity of Stress Fig. 97. — Stresses in Beams. But Pj^ C AC d, fc Pi P2 = J X Pi C = -^^ X P N d. . • . Stress carried by the element — a xH x P N .' . Total stress carried by section above N.A. = ]Sax^XPN fr = ■^ 2 a X p N dr = , X first moment of area above N.A. about N.A. dr Similarly if an element of area at a point p^ be considered, we see that I 196 THE STRENGTH OF IVIATERIALS Total stress carried by section below N.A. = 3^ X first moment of area below N.A. about N.A. But we have seen that the total tension T must be equal to the total compression C, and it follows from assumptions (a) (h) that d,. d, .'. we see that the first moment of the areas above and below the N.A. about the N.A. are equal and opposite in sign. Therefore, the total first moment of the whole area about the N.A. is zero. But we have seen that the first moment of an area is zero about a line through the centroid. Therefore, i7i simple bending with the given assumptions, the neutral axis passes through the centroid. The Moment of Resistance (M.R.) — We have proved that the stress carried on an element a of area about a point p is equal to a x j.' x p n The moment of this stress about the N.A. = stress X P N = a X ,' X pn2 d, . ' . Total moment of all the stresses over the cross section = 2 a . ^' X P n2 a, ' = ^' :S (a X P n2) ^ J (second moment of ^^hole area about the N.A.) a,. d. But the total moment of all the stresses is the moment of the couple w^hich we have called the moment of resistance. .*. we see that M.R. = , or ' d, a, I STRESSES IN BEAMS 197 The moment of resistance must, as has already been shown, be equal to the bending moment, which we will call M. .-. M ^y^or V (1) It will be seen that I, d, and dt depend merely on the shape II of the cross section, and -,- and -,- are called the compression U/Q (hi, modulus and tension modulus respectively of the section, and are written Z, and Z<. Thus our relation becomes M = /,Z, =/,Z, (2) In practice we usually want to know /<, and ft which give the maximum stresses across the section, and so we will write the result as /,. = ! (3) M f'-% (*^ In the case where the section is symmetrical about the N.A., d^ is equal to dt, so that Z^ and Z^ are equal. In this case, therefore, /, = ft, and we may write the relation as ' Z For values of section moduli for British Standard Beam Sections, see Appendix. Unit Section Modulus ; Beam Factor. — If instead of taking the section modulus as Z we took -v and called it the " unit section modulus," the quantity would be rather more useful and probably more easy to conceive to those who find difficulty in purely mathematical conceptions. We should then have Z I A;^ Unit compression section modulus = z, = -^ "= a ^ "^ ~d~ „ tension „ ,, = Zt = -j z, and Zi then become lengths 1*\^ rnK v^TKKNcrrn ov MVTK)n\i.s oi|iK-\tious (o^ ;\iul {4) ^ivi^ A /,.A 'A M Z, «, ivud ;< oau bo found giaphioally bv sotting ont c p. Fig. l>77i». i>f tho soot ion abovo tho >*.A.. joining u i\ \ v and drawing linos OK, DF ivsptvtivoly at riglit anglos to thoni. Thon i' k, c f are ;:, and ;. ivsjHvtivoly. Take, for oxaniplo, tho As r k i\ nc o: thoy aro similar. 6Vum factor. — Tho quantity "J = v r ^*^^^ '^ synunotrioal STR-KSSKS IN liKAMS ID!) ,. Z rniM. ,. , „ i • I i- 1 K(H!ti(tri, ()»• . , , lor- n, ])<",ih\ <»l ;i.Hyirimcl,ncal Hcotioii wIi(>h(5 A a Hafo bfMiding Hl/fOHHCiH an^ ilio HarrKJ in cotuprc-HHiou ;iikI (-criHioii, i.M Ji. (iicjiHiirfi of Mic vJiJiK! of I-Im! Hf5(!l,ioji n,H Ji f)ca,fri u.hd lun.y !»(«, cMlicd Mi(5 />(,(wi Idcior. 'i\il<(i, for inHJyarujf^ l/hc- If/' / (^ Hliuidard I Ixwirri fA()|M-ii(lix). Z 8:*>!) ;i,nd A I7'{. .'. boarn la(;l>or .^ •'J23 For IIk^ W / H^'Hiandard boairi Z ()()(), A 2()-2. .". beam factor ,., ' „,, ,, ".'{42 10 X 20-2 .'. tho 10'' X 8'' in a mom rusoMomicjal Hnoficm. Ar(5(5ian^ular HccMoii wonid ^dvc- u, beam facjtor ^ ^ \ "1^7. N(JMici{.K!Ai, i*]xAiviiMJO,s. - 'rii'5 following fi umc/fical ('/xam[)l(5H will ma,kc- il, cJcai- liow llio HlroHHOH in bcarriH loa,dc,d in ^dv in. unilH. 200 THE STRENGTH OF IVUTERIALS Example- I Examble 5 6 M. on each qirJer Fig. 98. — Examples of Beams. STRESSES IN BEAMS 201 Section b. This is composed of two triangles. •.1 = 2 X ^^,h in this case being the height of the triangle. _ 2 X 2-828 X 1-4143 ' 12 d = 1-414 7 _ 2 X 2-828 X 1-4JL43 _ 2;82^ ' 1-414x12 ~ 3 = -943 in. units. Section c. T _^ _^ x_2-26^ 64 ~ 64 " d = 113 64 X 113 = 1*13 in. units. Section d. 2 X 43 2 X ^ ~ 12 •8 X 2-53 12 = 10-67 - 208 = = 8-59 d = r . ^ 8-59 = 4-29 in. units. Section e. This is composed of three rectangles. T = •'75 X 23 2-5 X -43 -75 X 23 12 "^ 12 "^ 12 = -5 + -013 + -5 = 1013 d = V - . Z = r013 in. units. We see, therefore, that the order of the sections, from strongest to weakest, is d, a, c, e,b. t 202 THE STRENGTH OF MATERIALS We may take it, as a rule, that the strongest beam for a given area of cross section is that which has a depth as great as is practicalljr possible, and which has as much as possible of the metal at the outer portions of the beam. (2) A girder of 20 ft. span carries a uniformly distributed load of 10 tons, and a central load of 4 tons. Find a suitable British standard beam section for the girder if the maximum stress is to be 7 tons per sq. in. Its maximum B.M. due to the uniform load will be equal to WZ -g (see Eigs. 59, 59a, Cases 2 and 3) 10 X 20 X 12 . , = X m. tons = 300 in. tons. The maximum B.M. due to the central load = -— ^ 4 ^ 4 X 20 X 12 ~ 4 = 240 in. tons. These both occur at the same point, so that the maximum B.M. due to both loads = 540 in. tons. Now M - / Z 2. e. 540 = 7 Z .-. Z -= = 77"14 in. units. On referring to the table of standard sections (Appendix), we see that the section having the nearest modulus to this is a 14 X 6 X 57 lb. section for which Z = 76' 12, and we will adopt this section as being suihciently strong. (3) A tank which weighs J ton and measures 10' x 6' x 3' is filled with water, and carried on three girders placed length- wise, so that each girder takes an equal weight. If the girders are &' x 3'' x 12 lb. Standard Beams find the maximum stress in each. {A.M.I.C.E. Altered slightly.) STRESSES IN BEAMS 203 Weight of water in tank = 10 X 6 X 3 X 62-5 tons 2240 = 5-02 tons. . • . Total weight carried by girders = 5*02 + '5 = 5*52 tons i,r ' T^i^r u • ^ ^'52 10 X 12 .'. Maximum B.M. on each girder = — ^- x o o = 27-6 in. tons. Z for a 6" x 3" x 12 lb. beam is 6*736 in. units i 27-6 ... . • . / = 7r--o^ =41 tons per sq. m. ' 6-736 ^ ^ mi Compression Ih-nsion Cost Iron £>earn fliichea E>Qam . Fig. 99. (4) A cast-iron beam is the shape of an inverted T, 9 ins. deep over all, width of flange 6 ins., thickness of web and flange 1 in. If its length is 12 ft. find what weight at the centre will cause a tensile stress of 1 ton per sq. in. in the flange. What would the maximum compressive stress then be ? {A.M.I.C.E.) First find the centroid and second moment of the section. (See Fig. 99.) Area of section = A = 9 x 1 + 5 x 1 = 14 sq. in. 9 1 1st Moment about base = A fZ = (9 x 1) x ^ + 2 (2 J x 1) x ^ - 40-5 + 2-5 = 43 •'• ^ = 1^ = 3-07 ins. 14 204 THE STRENGTH OF MATERIALS 2nd Moment about base 1 X 93 2 X 21 X 13 ~ ' ~ 3 + 3 = 243 + 1-67 = 244-67 . • . 2nd Moment about parallel line through centroid = 1, = 1, - A (P = 244-67 - 14 X 3-072 = 244-67 - 13207 = 112-6 in. units. .-.z. 112-6 112-6 9 - 3-07 5-93 = 18-99 in. units. Z, == -^— r = 36-67 in. units. . • . Safe B.M. in tension = /, x Z^ = 36-67 in. tons. Neglecting weight of beam itself, if central load is W, the 4 maximum B.M. is . • . Maximum B.M. = ^ ^ = W x 1 22^1 2 _ gg ^ j^. tons. 4 4 ^^j 36-67 1-02 tons. ' • ^' 36 f X d 1 X 593 The compression stress —-^-^ — ' = — qTat — ~ -^"^^ ^^^^ P^^ ®^- ^^- (5) A pitched beam consists of two timbers, each 9 ins. thick and 16 ins. deep, and a steel plate placed symmetrically between them, the steel plate being 8 ins. deep and | in. thick. If E for timber is 1,500,000 lbs. per sq. in. and for steel 30,000,000 lbs. sq. per in., find the maximum tensile stress in the steel plate when the maximum tensile stress in the timber is 1000 lbs. per sq. in. Determine also for the same intensity of stress in the timber the percentage increase of load the flitched beam will carry as com- pared with the two timbers when not reinforced with the steel plate. {B.Sc. Lond.) Using the notation given on p. 183, we see that _ 30,000,000 _ ™- 1,500,000 -^^ (seei-ig. JJ). STRESSES IN BEAMS 205 .*. The steel plate is equivalent to a timber 20 times as wide, i. e. a timber 15 x 8 ins. . • . For the equivalent section of timber for the whole flitched beam Ig _ 2 X 9 X 163 (15 - I) 83 ~ 12 "^ " 12 = 6,144 + 608 = 6,752 in. units. . For the timber beam not reinforced I = 6,144. When the stress in the timber at the outside of the section is 1000 lbs. per sq. in,, that 4 ins. below the N.A., i. e. at the maximum depth of the equivalent timber plate, will be 4 ■^ X 1000 = 500 lbs. per sq. m. o But steel carries 20 times the stress in the timber for the strain. .'. Stress in steel = 20 x 500 = 10,000 lbs. per sq. in. For the flitch beam the equivalent modulus is 6752 8 844 in. units. S4.4. V 1000 .-. Safe B.M. in ft. lbs. = »^^ J^^ = 70^333 6 144 For the plain timber beam Z = ~~ — = 768 in. units o .-.Safe B.M. in ft. lbs. = ^^^ ^J^^^^ = 64,000 .'. Increased B.M. carried by flitched beam = 6,333 .-. % increase = ^^000 ^ ^^^ "" ^^ '^ "/^ We shall have further numerical examples on the stresses in beams at various points in the book. Approximate Value of Modulus of I Sections. — In practice girders are usually made of I section, because the most economical section is that in which as much as possible of the metal is placed in the edges or flanges. In this case an approximate formula for the modulus of the section can be found as follows : Let d (Fig. 100) be the distance between the I 206 THE STRENGTH OF MATERIALS centre of flanges of the section, the thickness of the flanges being t. Then if b is the breadth of the flanges, and t-^ the thickness of the web, we have I = B (D + tf (B - y (D - tf (1) .-. 12 I = B(D3 + 3D2^ + 3D^2_^^3)_(B_y (d3_3d2^_^3p^2_^3) = B (6 J)H + ^3) _!_ ^^ (j)3 _ 3 J) 2^ + 3 J) ^2 _ ^3) Now if t is small compared with d, f^ and t^ are negligible, Combression Fl^nac ^ t h — ^B .-^ Web D 3 FUnc?C , ension nanci FiQ. 100. Npw Z = I _ 21 2 21 (3) since i is small compared with d. 1 — nearly STRESSES IN BEAMS 207 ...z=.t{-'+'.-('-l')}('-3 = ^[6^t-^^''+hi>-3tt,-tt, + ^'^f} ..(4) = |6 B if + ^1 (d — ^) I to a first approximation, neglecting all remaining terms containing t^ oi tti. Now B X ^ = area of one flange = A and ^1 (d — ^) = area of the web = a =K^+S ('> Therefore we get the following rule : The modulus of an I section beam is approximately equal to the depth between the centres of the flanges multiplied by the area of one flange plus one-sixth of the area of the web. Discrepancies between Theoretical and Actual Strengths of Beams. — Many practical men have expressed considerable surprise that in testing beams the actual and theo- retical breaking strengths do not agree. A number of beams are tested, and a tension test is also made from the same material, and it is found that the load which, on the ordinary bending theory should cause the breaking stress in the beam, does not cause fracture, the amount of additional load depending on the shape of the cross section. This was the origin of the old " beam paradox," it being thought that the material must be stronger in bending than in tension. In fact, for cast- iron beams, an old erroneous theory which, for a rectangular beam, made M = — . — instead of - — ^ agrees con- siderably better with the breaking test than the modern theory. Now this discrepancy in the case of ductile metals is due to the fact that the ordinary bending theory is not applicable to breaking stresses, and no one who appreciated the value of I 208 THE STRENGTH OF MATERIALS the assumptions made in obtaining such theory would expect the theoretical and actual breaking strengths to agree. This is because the stress is not proportional to strain after the elastic limit is reached. Some experimenters who have measured the deflections of beams have stated that for mild steel the stresses at the elastic limit do not agree, but that is due to a confusion between the elastic limit and the yield point, and to the fact that the deflections were not measured with sufficient accuracy. In Chap. I we saw that for a tension test of mild steel the elastic limit and yield point were quite close to each other; but in bending this is not the case, the yield point occurring at a considerably later point than the elastic limit. Considerable error, therefore, arises if the yield point in bending be taken instead of the elastic limit. If the latter be carefully measured it will be found that the stresses in tension and bending at the elastic limit agree very closely. This point is proved, incidentally, in the Andrews-Pearson paper on Stresses in Crane Hooks, referred to in Chap. XIX. The reason for the yield point coming some distance after the elastic limit in bending is that only the material at the extreme edges has been stressed up to the yield point, and the whole section will not yield until the material nearer the centre has become stressed up to the yield point. We see, therefore, that there is no discrepancy between theory and tests so long as the conditions laid down in formulating the theory are fulfilled. If those conditions do not hold beyond a certain point, then, after that point, we must get a new theory if we wish to calculate the stresses. These so-called discrepancies between theoretical and actual strengths of beams point to the desirability of choosing the working stresses for ductile metals in terms of the stress at the elastic limit, and not of the breaking stress — as we pointed out in Chap. III. — because if the working stress in a beam is, say, one-half of the stress at the elastic limit in tension, then twice the load on the beam will cause the elastic limit in the STRESSES IN BEAMS 209 beam ; if, however, the workmg stress be taken as one-fourth of the breaking stress in tension, four times the load will not cause failure, the exact load to do this being more, and depending on the shape of the section. Cast-Iron Beams. — The discrepancy in the case of cast- iron beams is due to the fact that the stress -strain diagram is not a straight line except for very low stresses and that for given strain the stress is appreciably less in tension than in ConobressioK7. Tension. Fig. 101. compression. The result of this is that the stress diagram becomes more like that shown in Fig. 101 ; the neutral axis becomes raised above the centroid and the diagram is curved. In this figure the actual stress is shown about one-half of the calculated stress. This effect will be most marked in sections such as rounds or diagonal squares with a large amount of metal in the web; it is least in I sections with the com- pression flange smaller than the tensile, i. e. the discrepancy between theory and practice is least in a well-designed section. * Diagonal Square Sections. — It is an interesting fact that we can obtain as follows the apparently paradoxical I 210 THE STRENGTH OF MATERIALS result that by cutting away part of a beam of diagonal square section we increase it strength. Referring to Fig. 101a and the table on p. 185 1^ , of whole section = (V 2)4. 12 _ ^ ~48 .'. Z of whole section = .o -=~ o = o^ 48 2 24 A Fig. 101a. 2d^ 1^ ^ of two triangles removed = 2 x ^^ (about their own centroids) + 2 . cZ^ (^ ^ " V)' ^*° ^^^^^ ^^ ^'^^'^ .*. 1 01 remammg section — aq ~ q ~ '^d'^yn — o ) 2^2 48 t Z of remaining section I6d^ D2.48/, 4:dY] D^ f 48 (D -2(Z) [^ 3d4 16 (Z* 24 cZ2 D^ 4rf\2) 3d;/ Z of remaining section Z of original section D f _ 16^* _24cZ2 B-2d\ 3d* d2 1 - id\^] 3d) f STRESSES IN BEAMS 211 Now let - = X D ai..,{'-^*-"-('-r ,, -^^^ {1 - 24 a:2 + 64 a:^ - 48 x^} ( 1 — 2 cc) ' = 1 + 2 :r - 20 a:2 + 24 ^3 = (1 - 2a;)(l + 4 a; - 12 x'^) = (1 - 2 a;)2 (1 + 6 x) d X This is a maximum when -y^ = ax i. e. 2 - 40 a: + 72 ^2 = ^. e. 1 - 20 a; + 36 ^2 = i.e. (1 - 18 a;) (1 -2 x) = From this x = ^ or J and the maximum is for x = ^s because clearly x = ^ reduces the section to zero. From this we see that the strongest section is obtained by removing one-eighteenth of the depth from the top and bottom, i. e. one- ninth of the depth in all. When ^ = To 8\2 4 256 , .^„ , Therefore the section f of the depth of the original section appears to have a strength 1'053 times as much, i.e. about 5*3 % increase. The additional strength is, however, only apparent, because when failure starts at the edges we arrive at the stronger section. We do not know of any accurate tests that have been made to find to what extent this result holds in the actual beam. In a cast-iron beam of the original section under test a small crack will start on the tension side which will have the effect of cutting off one edge only. A complete study of this problem on a modified theory for cast-iron beams will be found in a paper by Mr. Clark in Proc. Inst. C.E. (1901-2). 212 THE STRENGTH OE MATERIALS * Influence of Shearing Force on Stresses in Beams. — It must be remembered, that up to the present we have considered only the tensile and compression stresses due to the bending moment. Besides these stresses there are tangential stresses due to the shearing force. The resultant stress at any internal point of the beam is the resultant or principal stress /^ ComJDression i ' \ / i ^ / 1 i / 7 i \ / : i „ ... ^^ / "^^-- Ihnsion ^ S.M. and Shear StrGSse'S 'RGSultanf' Fig. 102. — Principal Stresses in Beams. Oowbressior? Tension ^ Fig. 102 a. of the tangential and direct stresses, which resultant is fo.und as sho\\Ti in Chap. I. We shall deal in a subsequent chapter with the distribution of the shearing stresses across the section of the beam, but for the present we will assume that the shear stress is a maximum at the centroid and diminishes to zero at the extremities. Fig. 102 shows diagrammatically the shear and direct stresses across the cross section of the beam and also the resultant stresses which, as will be seen, STRESSES IN BEAMS 213 are parallel to the centre line of the beam at the extremities and are perpendicular to it at the centroid. If the principal stresses at various depths be found for a number of cross sections at various points along the span., and the directions of principal stress be joined up by a curve, we get a number of lines showing the manner in which the direc- tions of principal stresses vary from one point to another. Such curves will be found in Rankine's Applied Mechanics, and are of the form shown in Fig. 102a. In practice it will be found that, except for very short beams Strain Diaaram carrying heavy loads, the maximum tensile or compressive stress due to bending moment is usuaUy greater than the maximum shear stress, so that the consideration of stress due to bending moment is, as a rule, considerably more important than that of the shear stresses. * Moment of Resistance in General Case. — To follow the correct theory of beams it is not necessary to make any of the assumptions previously given, and we wiU now find the moment of resistance in the most general case. To investigate this, we must suppose that we know by experimental or other means the shape after distortion which is taken up by a cross I 214 THE STRENGTH OF MATERIALS section of the beam which was originally plane. We must also know the relation between stress and strain for the material of which the beam is composed. Let A B, Fig. 103, represent the elevation of a cross section of a beam which after bending is strained to the shape d c e. Then from the stress-strain curve and from the shape of the cross section draw a curve of stress d' c e'. This is obtained as foUows : let a 6 be any ordinate of the strain diagram ; then from the stress-strain curve find the stress corresponding to this strain, and multiply the stress by the breadth of the beam at the given point, and plot this equal to a' h' to some con- venient scale ; joining up points such as h' we get the stress diagram. Now let the area of the stress diagrams be Q and T and their centroids G^ and G2. Then, of course, in simple bending Q and T will be equal, and if q is the perpendicular distance between the centroids, the moment of resistance will be equal to T x g or Q X g. If the reader fully follows this general method with regard to the stresses in beams, he should not have the difficulty commonly experienced in following the more particular theories. CHAPTER VIII STRESSES IN BEAMS — (continued) * REINFORCED CONCRETE BEAMS There are many formulae for the strength of reinforced concrete beams, such formulae being deduced from certain assumptions with reference to the distribution of stress in the bent beam. We will consider three methods of calculating the stresses in reinforced concrete beams, working in each case the case of a rectangular section, this being most common, and in all three we will make the following assumptions — (1) That a section of the beam which is plane before bending remains plane after bending. (Bernoulli's assumption (see p. 194).) (2) That the beam is subjected to pure bending, i. e. that the total compressive stress is equal to the total tensile stress. Standard Notation. — Throughout the treatment we will adopt the following notation (see Fig. 104). Young's modulus for steel or other metal E, Young's modulus for concrete E^. t = Tensile stress per sq. in. in reinforcement. t, = ,, ,, ,, concrete, c = Compressive stress per sq. in. in concrete. At = Area of cross section of reinforcement. Ac = ,, ,, „ concrete. b = Breadth of beam. 215 210 THE STRENGTH OF MATERIALS df = Total depth of beam. d = Depth of beam to centre of reinforcement. n = Depth from compressive edge to neutral axis (N.A.). (^d — n) — ,, ,, centre of reinforcement to neutral axis. n-i = Ratio y. r = Proportional area of reinforcement to area above XXt, it bd 1, = Equivalent moment of Inertia of section. -* b >- ..1 >\ n V fu ■ ^ L_.0 ^__ 'f 79 Fig. 104. — Notation for Reinforced Concrete Beams. First Method — Ordinary Bending Theory. — The first method which we will consider is one which is not much used in practice because it gives safe loads which are lower than tests show to be necessary. It is, however, the general method applicable to beams formed of two elastic materials, and serves as a useful and instructive introduction to the subject. According to this method, we assume that the reinforced beam behaves exactly as an ordinary homogeneous beam with the reinforcement replaced by a narrow strip m times the area of the reinforcement, and at constant distance from the N.A. We showed how to find the centroid, moment of inertia, and radius of gyration of such an equivalent homogeneous section on p. 183. In the general case, let'7^^(^ig. 105) be the distance to the STRESSES IN BEAMS 217 neutral axis (equivalent centroid), and I^. the equivalent mo- ment of inertia about the centroid. Then*..=M(^;:=^) (1) c = t = Mn mM{d — n) (2) (3) where M is the bending moment. d N t n. d W Fig. 105. — Reinforced Concrete Beams. ■ Method 1. In the case of the rectangular beam we then get the following results — Equivalent area of section ^ h dt + {m — \) A., (4) As explained on p. 184, it is (m — 1) A^, because when we take away the reinforcement and replace it by m times its area of concrete, we have first to fill up the hole in which the reinforcement was, and this takes once A^, so that remaining additional area = (w — 1) A^. Take moments round the top, then we have n [h dt -^ {m — \) A, 1 = -f '^ + (m- l)A,d ''2'- +(m- 1)A,^ n = ,— , bd, + {m - I) A This fixes the position of the neutral axis. (5) 218 THE STRENGTH OF JVIATERIALS Taking second moments about the neutral axis, we have I = ^f + ^-i^^- + (m - 1) A, (d - «r- (6) In this formula we neglect the second moment of reinforce- ment about its own axis. Numerical Exa]viple. — Take the case of a beam 6 ins. wide and 12 ins. deep, the centre of the reinforcement being 2 ins. from the bottom and the area of reinforcement = 1*44 {see Fig. 106). Fig. 106. Taking m = 15, we get _ 6 X 144 + 2 X 14 X 1-44 X 10 ^ ~ 2(72 + 14 X 1-44) = 6-87 ins. ,'. {d-n) =^12 - 6-87 = 5-13 I = Lx (6:87)3 6_x (5-13)3 ^.^^ ^ 3.^3, 3 3 = 626 + 270 + 197 = 1093 nearly. . • . Taking a safe stress of 100 lbs. per sq. in. in tension for the concrete, Safe B.M. = ^^^^{^f^ ^ ^ '^^ ^*- ^^'• 100 X 6-87 Then t, = comp. stress in concrete — ^ ,„ 15 X 100 X 3-13 Then t = Tensile stress in steel = 513 = 134 lb./in.2 - = 915 lb./in.2 STRESSES IN BEAMS 219 It will be seen that we have taken t^ — 100, which is higher than usually allowed for concrete in tension ; but if the con- crete cracked the steel would still hold, and so we are justified in using a higher stress. The above example shows that on this method of calculation the beam is not very economical, as the steel is very little stressed and the concrete has only a small stress in compression. For this reason it is usual in practice to neglect the tensile stresses in the concrete, that is to say that it does not matter if the concrete does crack. Practice shows that such cracks, if present, do not matter so long as the adhesion between steel and concrete is good, and the tensile stress in the steel and the compressive stress in the concrete are within safe limits. We should like in this connection to point out that to neglect the tensile stresses in the concrete does not, as some writers state, increase the factor of safety. We shall see later that neglecting such stresses we get a much larger safe B.M. on the beam, and thus reduce the factor of safety. Strength of Same Beam not Reinforced. — To serve as a useful comparison we will find the strength of a 12''' x Q'^ concrete beam without reinforcement. M X 6 M If not reinforced, t = 6 X 123 144 12 In this case we must take safe t^ = 50 Ib./in.^ . • . Safe B.M. = ^"^ ^^^^^ = 600 ft. lbs. Therefore, calculating by our first method, the reinforced beam is roughly three times as strong. It would cost roughly twice as much, so that we see there is 50 % saved. Second Method — Straight-line, No-tension Method. — This method we name as above, because the additional as- sumptions are indicated by such name. We will now make the following additional assumptions — (a) All the tensile stress is carried by the reinforcement. 220 THE STRENGTH OF MATERIALS (6) For the concrete the stress is proportional to the strain (c) The area of reinforcement is so small that we may assume the stress constant over it. Fig. 107 shows the section, strain diagram, and stress diagram. We will first give the usual treatment which is based upon argument from first principles. In accordance with our first assumption a vertical plane section becomes an inclined plane section a.' b', the neutral axis (N.A.) being at the point c. Neutral /fxi's — ^ #- rt 1 Fig. 107. — Reinforced Concrete Beams. Method 2. What we first require to determine is the position of the N.A. Now A A.' and d d' represent the maximum strains in the concrete and the steel respectively, and since the line a' b' is assumed straight, these strains are proportional to their distances from the neutral axis. . We have max. strain in concrete n max. strain in steel but max. strain in concrete {d~n] c (7) and max. strain in steel = ^ n _ c E, • " • (^ -"ti) ~ t ' E, . • . nt = mc{d — 7i) m c (8) STRESSES IN BEAMS 221 , nt d-n = m c _t_ m c n=- f (9) d 1 + mc This determines the distance from the N.A. when both c and t are known ; but this will not always be the case. If the reinforcing bars are of given size, then t will depend on that size, and to determine the position of the neutral axis, we proceed as follows : The stress diagram shows the distribution of stress in the cross section. Since we have assumed the stress proportional to the strain, the stress diagram for the concrete will be a triangle. It will be seen therefore that the mean compressive stress is -^, and since the compression area is b n, we see that the total compressive stress is ^ cb n. As the stress in the steel is assumed uniform, we get that the total tensile stress in the steel is t A^, and if the beam is subjected to pure bending these must be equal. .-. tA, = ~ cbn (10) c 2 A, I.e. - = -,- . t on Comparing this with equation (8) we get 2 A, n b n m{d — n) .'.bn^ = 2mA,{d-n) (11) .-. bn^ = 2mAyd — 2'mA^n .' . b 71^ + 2 m A, . n — 2 m A., d = 0. The real solution of the quadratic equation gives »-6^{-^W(^ + ^a)} (-) 222 THE STRENGTH OF ]MATERIALS Since all the quantities in this expression are given, this fixes the position of the neutral axis. We may write this n _mA, r / ~2hd_ A d bd iV ^ mA, J i.e. n d ^ ''- = r m u 1+^ - r m = Wm^ r^ + 2 mr — in r For m = 15. This gives r = •007 •010 •015 •020 n d ^ ''^ •365 •417 •483 •530 1 (13) These values are plotted on Fig. 108. Moment of Resistance. — We can now find comparatively simply the moment of resistance. The resultant compression acts at the centre of gravity of its triangle. Therefore the distance between the resultant compressions and tensions is d — q- .•. If C and T represent these resultant compressions and tensions, we have that the moment of the couple due to the resisting stresses, which is called the moment of resistance, is given by MR ^c(d - ^^ = ^c69^((^-|) (14) or, MR = T(d - !? = ^A,(cZ-|) (15) And this moment of resistance must be equal to the maximum bending moment for the loading. STRESSES IN BEAMS 223 160 f II / 1 «7 c/ / '50 v W 1 !# ^ 5^ v^ 4 1 ^ - n I40 / /) 1 r ' 4 y // / ^ V^ 150 / 4 / X y P c\ 120 / // / C ■^ \J / J ii / ^ 110 / 1l 1 ^ / / / 1 i y 100 / / ■ X c -t / / / / \r ^ go / / 7/ ^ 1 ^, // y 80 j , / / / y 1 / / / ' ■ / 70 / 1 / / ^1 // / ^1 / bo // 1 // 1 / / // f 30 / / / , 7 1 ^^ "^ / k 1 .L •? cJ^ X' ^ ' III Cj! £> ^ • • __30 f II , ^ ^ ^ ^ ^ V ^ZO I / (0 0) y ¥ ^ ^ 10 ■50 ■40 ^0 20 rerceniaae neinjorceme.nr ?/o v£ /4i. /6 70 Fig. 108. — Rectangular Reinforced Concrete Beams. 224 THE STRENGTH OF IMATERIALS Numerical Example. — Take the same section as worked by the previous jormula {see Fig. 106) ; and take c = 500 lbs. per in. 2 Then equation (12) gives _ 1-44 X 15 f /, "72 X 6 X 10 ,) ""- 6 W^+ 1-44 X 15 ~^i = 5*61 inches. cZ - 71 = 10 - 5-61 = 4-39 inches. Then M.R. or safe B.M. considering the concrete is equal to 500 X 6 -^^1 f. on , 2^ „.l . „ — 2 X 0-61 4-39 + „5'61 m. lbs. = "^'^- X 5-61 X 8-13 = 5,700 ft. lbs. nearly. Comparing this with the safe B.M. by the first method we see that the present is more than three times as much. Stress in steel is then equal to y-^--- A, (^ - 3 5700 X 12 _^„ „ . , = 1-44 X 8-13 = ^'^^^^'-P"'^^'^ Assuming a span of 10 ft., the max. B.M. if the load is W X 10 uniformly distributed is „ ft. lbs. W X 10 ••• ^—"^-^ = 5700 .-. W - 4560 lbs. This includes the weight of the beam, which is roughly 10 X 12 X 6 ic,r. lU r.-A 11 TXi X 150 lbs. = 7oO lbs. 144 . • . Safe load uniformly distributed = 4,560 — 750 = 3,810 lbs. It will be seen from the stress in the steel that the area of reinforcement is more than it need have been. By combining equations (9) and (10) we could have found the value of A, to give the stress in the steel, say 16,000 lbs. per sq. in., when the compressive stress in the concrete is 500 lbs. per sq. in. STRESSES IN BEAMS 225 The above formula gives results which are in fairly good agreement with tests, and is the one most largely used in practice. Alternative Treatment for Straight-line, No-Ten- sion Method. — The following treatment follows more nearly the ordinary method of dealing with beams than the above, but it is not nearly so often used in this country. We shall, however, find it very useful in the case of beams other than rectangular ones with tension reinforcement only and so we give it here; its xalue has been scarcely sufficiently ap- FiG. 109. « predated. Fig. 109 shows the section of the beam and also the equivalent section. We find the position of the neutral axis for the equivalent section by the rule given on p. 196 that the total first moment of the cross section of the beam about the N.A. is zero. 71 .' . b n X j^ = 1st moment of compression area about N.A. — mAt{d — %) = 1st moment of equivalent tension area about N.A. b n^ •'• 2^ mA,{d — n) =0 (15) OT bn^ -i-2 7nA,n -2m A, d = [cf . p. 221] . . (12) This is the same relation as before. Q I 226 THE STRENGTH OF IVIATERIALS Equivalent moment of Inertia of the section = Ik = o + m A^ {d — n)^ Iv 71 = o + o {d — n) [from 15 above] h 71^ [ J 71 ^ 2 V^-3 Now M = ^ ^^ (15a) 71 ^ ' [compare ordinary bending formula p 197] =^cb7i^ld— „ j ' ~2n -lcnb(d-'^') (10) This agrees exactly with our previous result. Considering the stress on the reinforcement (156) M = m {d - 71) c t 71 7n {d - 71) 71 1 "'■ d ~ t ' 1 7n c as before. This is used if A^ is not given. Case in which Stresses are given and Area of Steel has to be found. In the case we have n^ ^ 1+ ' w c Suppose for instance t ^ 16,000 and c -^ 600 and ?n = 15 1 ' _ ^^ _ .Oft ^1 ~ 16,000 ~ 25 "^ 600 X 15 .-. n = -36 cZ. STRESSES IN BEAMS 227 Then from equation 10 we can calculate the necessary area of reinforcement by the relation K _ch n K en. 600 X -36 ^.^^^^ ' ' hd 2t 2 X 16,000 ' In this case the moments of resistance given by equations (14) and (15) will be equal and ^. -^ 600 6 X -36 ^(6^ - -12 d) M.R. = 2 .-. SafeB.M. = 95 6^^2 The coefficient 95 is called the resistance modulus and can be plotted in very convenient form as in Fig. 109. These curves may be likened to the tables of section moduli for steel sections. Numerical Example. — A reinforced concrete beam is required to carry a bending moment of 240,000 in. lbs. Design the section for stresses c = 600, t = 16,000, assuming that the breadth of the beam is 10 inches. By equation {U) d = ^ ^^-^ 240,000 M 95 X 10 = 15*9 inches. .-. A- ^ -00675 a A, = -00675 X 15-9 x 10 = 1*07 sq. inches. Adopt 2 — J"' bars [giving an area 1-2]. Third Method — General No-terision Method. — In this method we will, as before, assume that all the tensile stress is taken by the steel, but we will assume that the stress-strain curve for concrete is not straight but some other curve. 228 THE STRENGTH OF IVIATERIALS In this way we get the stress diagram, Fig. 110, from the strain diagram. Suppose that its area = k . c . ii and that its centroid is at distance y from the top. Then, as in equations (8), (9) we get {d — n) m .c n — . n d 1 + mc /. »~ A n d , d- r. cL, ------- -•-'- r Fig. 110. — Reinforced Concrete Beams. Methods. Now, since the total compressive stress must be equal to the total tensile stress, we have tAj = khnc (16) {d — n) m A^ k n . • . kh 11^ ^ m A^{d — n) .' . k b 71^ -{- m Aj 71 — m Aj d A^m ( ^ ^ 4:dkb '\ '' ^ 2X6 W ^ + m A, ~ ^/ or. d r TTi ( 2l\ r m J (H) (18) (19) Then moment of resistance = M.R. =tA,{d-y) for tension (20) = kb 71 c {d — y) for compression . . (21) STRESSES IN BEAMS 229 Numerical Example with Stress -strain Curve Para- bola. — Take the section that we have worked for the previous formulae. (See Fig. 106.) If the stress-strain curve is a parabola tangential at the compression edge we have 2 3ri » ■frko mTTon cr^/^'firk'n 'w — ^ j / 1 1 . 6 . tilt; glVCll OCOulUll fi — jj 1 A / X 1 1 ;? 1 ,AA 2.-3.6 ,3 = 5-12" .-. d -n = 10 - 5-12 = 4-88 Safe M.R. for concrete = 1 X 6 X 5-12 X 500 (4-88 -f 3-20) in. lbs. = 3 X j2 X 5'12 X 500 X 8-08 ft. lbs. ■} =: 6,900 ft. lbs. nearly Then stress in reinforcement 6,900 X 12 H TOAiu = ' = r44->^^ir8 = ^'^^^ ^^'- P^^ ^^- ^^- It will be seen that this method gives higher values still for the safe bending moments. The stress-strain curve for con- crete, although nearly parabolic, would not have the vertex of the parabola at a stress of 500 lbs. per sq. in. From the above we think that it should be clear that there is not much difficulty in finding the stress in reinforced concrete beams so long as we know accurately the properties of the concrete, and are clear as to what assumptions we are making. Reinforced Concrete T Beams. — Reinforced concrete floors usually consist of reinforced slabs with reinforced beams at definite intervals in a longitudinal direction, the whole being monolithic. Fig. Ill shows a section of such a floor, which may be regarded as a number of T beams. The 230 THE STRENGTH OF MATERIALS reinforcing bars a in a transverse direction in the slabs are arranged as shown to take the tension at the top where the bending moment reverses, due to the slabs being continuous. It is usual to take the effective breadth of the flanges of the T beams as less than B — J to | B — because the concrete be- tween the beams acts as a short beam in a direction at right angles, and so the centre portion is comparatively highly stressed for this reason. We will now consider the stress in the beam, adopting the no-tension, straight-line method. Case 1. If d, > n we get the same rules as given in method (2) for rectangular beams, 6, being substituted for h. Fig. 111. Case 2. If 6?^ <[ ?^ we proceed as follows — As before we have from a consideration of the strain diagram {d — n) m c • n = n = t d m c Now consider the total stress diagram, Fig. 112, i.e. hori- zontal lengths of compression figure = compressive stress per sq. in. X breadth of beam. Now total compressive stress on the section = C = area (k d h — h f g) c 6, n (6, — b,) XX - "2~~ 2 ^ w C 21 6s n {b. -b,)x^\ n J But C = T = ^ A, STRESSES IN BEAMS {bs -b,.)x^\ . ^ A^ = ^\b,n n 231 (22) 2 A. c _ {b,n — -~ — - I n J 2 A^m {d — n) n = \ b,n — {bs -b,)x^] n J n^ 0, n — ^-" '-—' ]■ = 2 A^m{d — n) nih, n + 2d, {b, - b,) + ^ {h, . - 5,)| = 2 A,m {d - n) V ft J i.e.b^n^ + 2n [A, m + d, {b, - b,\ =^2 A,md ^ [b,- b,) d,^ (23) A' 4 ^4— •*" I / - * AjC- TV /9 Fig. 112. — ^Reinforced T Beams. From this quadratic the value of n can be found. Then if the centroid of the compressive stress-strain curve area is at distance a from the centre of reinforcement Safe B.M. = C x « Let the centroid of the compressive stress-strain diagram be at distance y from the top. Now this centroid is the same as the centre of pressure on a similar body subjected to fluid pressure, the N.A. being the water line. In this case it is easily shown that 2nd Mt. of area above N.A. about top. y 1st Mt. of area above N.A. about top. b,n^ {b, — b,.) x^ ~3~ "^ 3 6., n^ {bs — b,) x^ * * ' '~Y "^ "~2 (24) This enables us to find a. 232 THE STRENGTH OF MATERIALS Many \vriters neglect the rib, i.e. neglect the portion f e h of the stress diagram, and others further assume y = '4: d, (the extreme limits between which y must lie are ' and ~\. This avoids the quadratic equation and makes the calculation much easier. We may put h,. = in equation (23) A A' d f jd TV /9 Fig. 112a. We then get at once n 2 m A, ^ + hs ds^ 2 {m A, + b, dS a then becomes equal to d — o ( n / ^ 3 \ 2 n — ds Considering compression we have Safe bending moment = C . a (25) = I j^ d, (2" - '^4 {d - i- (l^ -^f-)] (26) 2 n I 3\2n — ds /J ^ ' Alternative Treatment. — Applying the method of the STRESSES IN BEAMS 233 equivalent section we have (Fig. 113) a much simpler treatment. Moment of equivalent section about N.A. = n' i.e. 6.,^ — (6, — h,) {n - d,f m A^{d — n) This gives the same quadratic as equation (23) I ■ = ^|! _ (b^^MilLSzd^ + mAAd-n)K... (27) neglecting the rib we should have N u t ^ H d.-'TV Fig. 113. — ^Reinforced Concrete T Beams. This gives as before n = 2 m At ^ -\-h, d^ 2 {b, ds + m A,) Having found I^ we have as before cl Safe B.M. = — for concrete n m {d — n) for reinforcement 234 THE STRENGTH OF MATERIALS XuMERiCAL Example of T Beam.— Take the T beam of section shown in Fig. 114. In this case we Avill not assume the area of reinforcement (A,) to be given, but will calculate it so as to give c = 600 lbs. -per sq. in. t = 6000 lbs. per sq. in. m = 15 d m c 15 Then we have n = 1 + 1 + 6000 = 5-4 ins. 15 + 600 -^ ^ »^ 1 /' 3 — /o' — ' Fig. 114. . •. From equation (22) 16,000A, = ^00/5.4^^g_38xl:9:| 2, I 5*4 J .'. A^ = 4'38 sq. ins. .•. Adopt, say, 3 bars If'^' diameter. Then working by the equivalent moment of Inertia I, = 15 X 4-49 X 9-62 + 48 x \^ - ^^ /it - = load per unit length of the cable = say p. H -R = S (4' Now return to the case of the beam 1^_ M R ~ EI 1 M From equation (3) ^ = ^ j (5) The quantity E x I depends solely on the shape and material of the beam, and is called the " flexural rigidity." Then if this flexural rigidity is constant throughout the span, by 252 THE STRENGTH OF MATERIALS comparing statement A and equations (4) and (5) we see that : A loaded beam takes up the same shape as an imaginary cable of the same span which is loaded with the bending moment curve on the beam, and subjected to a horizontal pull equal to the flexural rigidity (EI). This is Mohr's Theorem, and the deflected form of the beam is called the elastic line of the beam. We see, therefore, that to obtain the elastic line of a beam our procedure is as follows — (1) Draw the bending moment curve for the beam. (2) Divide this curve up into narrow vertical strips, and set down mid-ordinates on a vector line, and take a polar distance equal to the flexural rigidity (EI). (3) Draw the link polygon for this vector polygon, and reduce it to a horizontal base, then this link polygon gives the elastic line to a scale which we shall determine later. For the present we will assume that the section of the beam is uniform along its length, or rather that the flexural rigidity is constant. We shall see later how to proceed when such is not the case. Standard Gases of Deflections. — In certain special cases we can calculate the maximum deflections by reasoning based on Mohr's Theorem, and we will deal with such cases now (Fig. 123). (1) Simply Supported Beam with Central Load W.— Let AB represent a simply supported beam of span I with a central load W. Then adb is the B.M. diagram, the maximum ordinate being equal to - . Let A^ c^ B^ be the elastic line of the beam ; then, according to Mohr's Theorem, the shape of this elastic line is the same as that of an imaginary cable of the same span loaded with the B.M. curve and subjected to a horizontal pull equal to the flexural rigidity. Now consider the stability of one half of this cable. It is kept in equilibrium by three forces : the horizontal pull H DEFLECTIONS OF BEAMS 253 at the point c^ ; the resultant load P on half the cable ; and the tension T at the point A^. Take moments about the point Ai, then we have H X 8 - P X ?/ ' ' H In this case P = area of one-half of B.M. diagram _l I Wl _ WJ^ ~ 2 ' 2 ^ "4 ' ~ 16 y = distance of centroid of shaded triangle from a ) ^P ^^ 1 ^^ T^ — j[ """"^ ITP '1 Fig. 123. — Deflections of simply supported Beams. I ~ 3 H = EI WP "I • ^ " 16 ^ 3. EI WP ~ 48 E 1 (2) Simply Supported Beam with Uniform Load. — Let AB represent a simply supported beam of span I, with a uniformly distributed load W. The B.M. diagram is a parabola, the height being equal to Wl '^-. Then considering the stability of half the imaginary cable, we have as before 8 =- P X «/ H 254 THE STRENGTH OF ]\LVTERIALS In this case P = area of one-half of B.M. diagram 12 W Z _ WJ2 ~ 2 ' 3 8 ~ 24 51 H = EI . WZ2 51 5WP • 24 ' 16 E I 384 E 1 •(3) Cantilever with ax Isolated Load not at Free End. — Let a cantilever of sj)an l (Fig. 124) carr3'ing a load W at a pomt at distance I from the fixed end a. Then the B.M. diagram is a triangle, a d being equal to W /, Ai B^ represents the elastic line of the beam and the imaginary cable. In this case we must imagine the load as acting upwards. The cable is horizontal at a^. Take moments round b^, then we have as before H X S = P X 2/ . . P xy . . 6 == jj-^ In this case P = = area of B.M. curve a c d Wl.l WP 2 2 I 2/ = L-3 H = EI • ^ ^^^' (t I \ In this case it should be noted that the portion of the beam beyond the load is straight. (4) Cantilever ^\^TH an Isolated Load at Free End. — This is the same as the previous case when Z = l. WP / L . ' . o = 2EIV 3 Wl3 3EI DEFLECTIONS OF BEAMS 255 (5) Cantilever with Uniform Load from Fixed End TO A Point before the Free End. — Let a b be a cantilever on span L, and let a load W be uniformly distributed from a to a point c, I being the length of a c. Then as before 8 = H In this case P = area of B.M. curve a c d 2 I = W?2 6 Fig. 124. — Deflections of Cantilevers. y L — I H = EI WZ2 I ^-4 6EI (6) Cantilever with Uniform Load over Whole Length. — This is the same as the previous case when ^ = l. 8=E^(L 6EI Wl2 3l * 4 Wl^ 6EI 4 8E I * (7) Simply supported Beam with Isolated Load ANYWHERE. — The reasoning in this case is somewhat long, but should not otherwise present any great difficulties. 256 THE STRENGTH OF MATERIALS The first important point to notice is that the maximum deflection will not occur under the load, so that, as it is the maximum deflection that we nearly always require, it is of very little use to find the deflection directly under the load as is commonly done. We have seen that the ordinate of the bending moment curve or link polygon of a beam is a maximum where the shear is zero, so that treating the B.M. curve as a load on the beam, the deflection will be a maximum where the shear due to this load is zero. Let a load W be placed at a point c on a beam A b of span I, Fig. 125, c being at distances a, b from A and B. Then a e B W ab is the B.M. diagram, c e being equal to — j — . The total load represented by this B.M. diagram treated as a load will x.t- . Wa6 I Wab be equal to the area of the A a e b = , x o ^ ~ 2 * It acts at the centroid G of the A. DEFLECTIONS OF BEAMS 257 2 The vertical through this point g is at distance :;^ r c o from c, F being the mid-point of the beam, so that the distance of this centroid from the end b is equal to , , 2 / / \ lb (l + b) ^ + 3 V2 - ^) = 3 + 3 = ^3— . • . The reaction at a due to this imaginary load is equal to Total load {I + b) _ W ab {I + b) 1 ^ 3 ~ Tf ' 3 Now let the deflection be a maximum at the point d at distance x from A. Then the shear at this point is zero. i.e. R, - I . K H = Wab fl + b\ X Wbx 2/ V 3 / 2 I a {I +■ b) 3 ^ or ^ ^ V 3 — ^^^ The maximum deflection S is then obtained by considering the stability of the portion a-^ d of the imaginary cable. Then we have as before 8 P-2/ H In this case P = area a k h y = H = 8 = Wbx X Wbx^ I 2 Wb.a{l + b) Ql 2 3^^ 21 Wab{l + b) 6Z 2 la {I + b) 3 V 3 EI Wab{l + b) 2 jail + b) 1 6/ 3V 3 EI W6 (a(l + b)Y 3EIZ I 3 J 258 THE STRENGTH OF MATERIALS This can be put into somewhat simpler form for use by putting al. Then h = {\ - a)L a Then^ _ W(l -a) ^al {2 - a) l^ ~ 3EI \ 3 j ^3~El(^-"M 3" / if' (cfnafi^ Fig. 126. — Deflections of Beams. * (8) Beam uniformly loaded from one End to the Centre. The B.M. diagram for this loading is given by the curve D T G E (Fig. 126), the curve D t g being a parabola tangential to the line e g. We have first to find where this maximum deflection will occur. We do this by the rule that the Maximum Bending Moment in a beam occurs at the point where the shear is zero. We will treat the diagram d g e therefore as the load on the beam. If E G be produced to h, the curve d t g will be a parabola DEFLECTIONS OF BEAMS 259 tangential at G, and it is most convenient in the present j)ro- blem to consider the B.M. diagram as made up of the difference between the triangle D h e and the parabolic segment d g h. The first step is to find the imaginary reaction R^, at e. To do this we consider the area of the triangle as a force Pg acting down its centre of gravity, which is at distance -^ from d. Then Pg = area ofADHE = Jdh.de = The area of the parabolic segment will be considered as an upwardly acting force p^ passing through its centre of gravity which will be at distance - from d. o Then P^ = area dtgh =Jhd.dk _ wl^ I _wP ~ 3^>r8 * 2 ^ 48 To get the imaginary reaction R^. at e, take moments about D. Then .-. R p, I "3 pi- = 1^. .1 'e ^ 3 Pi 8 ■ 48 w 8 X 48 IwP 384 (1) Suppose that the maximum defiection occurs at a point N at distance x from the centre. Then the imaginary shearing force S at this point = Shear at n = R^ — area n q e + area Q t G 384 8 2 "^2 '3 _1 wP wl fP J ^\ wx^ ~ 384 16 U "^ ' ^ "^ y ^ G~ 7 wP wP wP X wlx^ wx^ 384 64 16 16 ' 6 wP W X^ W P X wl x^ ^ 384 "^ "6 l6 16" or 260 THE STRENGTH OF MATERIALS If this = 0, we have on dividmg through by . and re- arranging the terms, 64: x^ - 24: xH - 24: xl' - P =0 '^(jf-^'(jf-^'&^' = '> (^> This is a cubic equation that cannot be solved by direct methods. We must proceed by trial as follows — If X = 0. left-hand side, which we will call y = — I If a- = l y ^ 064 - 24 - 24 - 1 = - 1-576 If .1- = -05 y = 008 - 06 - 12 - 1 = - -252 If X =04 y = 0041 - 038 - 96 - 1 = - 006 If the values of y are plotted against .v it will be found that y = for X = '0406 approximately, and for all practical purposes we may take x =04. Having determined the point of maximum deflection, we have next to calculate the value of the deflection S at this point . We first find the imaginary Bending Moment ^M^ at the point X M^ = Re (9 -r x) -}- moment of section Q t g — Moment of A q ^' e ""384 -^ ^*^ ' " ~ 2 ^ 3 ^ " 4 Wl .-AJs ^^^ ^'^^ -Q-- X 04 I X ^ X ^ = wl^ ;-00964 - -00001 - 00328; = -00637 ivl^ M, 00637 wl^ '.., •••^ = Ei=— EX- ^^^ As an interesting comparison, let us suppose that the ^^ hole load were spread right over the span. 5 W P Then 8 = ^„-, ^ ^ , according to the \\ell-known formula. 384 EI ° DEFLECTIONS OF BEAMS 26 J In this case W = ^ 5m;^4 _ -00651 wl"^ ''' ~ 768 El ~" EI '" ^^ We see therefore that we shall only make a slight error^- which is practically negligible considering the necessary devia- tions from theoretical conditions which occur in practice — if we treat for deflection purposes the present case as being the same as for the same load spread over the whole span, re- membering that the maximum deflection occurs at '54: I from the unloaded end. Graphical Construction for any Loading. — Let a c b be the B.M. curve for any given load system. Divide the base into a convenient number of equal parts and let e be the length of each base segment. The number is such that each piece of the B.M. diagram is approximately a rectangle. Now set down the mid ordinates of each section diminished in the ratio - on a vector line. These ordinates are diminished n in order to keep the vector diagram of a workable size. Now let the space scale be V = x feet, and let the B.M. scale be V = y foot tons. Then considering any section of the B.M. diagram, say 2, 3, the area of this section is e x mid ordinate. Therefore, on given scales, one inch in height of mid ordinate, since the area of each segment is proportional to the height of the mid ordinate, represents e x x x y square ft. tons. Since each portion of the vector line is — of the n ordinates, the portion 2, 3 of the vector line represents the area of its corresponding section of the B.M. diagram to a scale V = n X e X X X y square ft. tons. Now calculate the length of E I on this scale. This will be too large for EI practical use, so take a pole p at distance — , where r is some convenient whole number. With this pole p, draw the link polygon a' c' b\ then this is the elastic line of the beam for 262 THE STRENGTH OF MATERIALS the given loading, or, more strictly speaking, a' c' b', when reduced to a horizontal base, would give the elastic line. The scale to which the deflections are to be read is then obtained as follows — If the polar distance were taken equal to E I, the deflections would be to the space scale 1^' = x feet, but as the polar dis- . EI X tance is , the deflections will be to a scale V =— feet. The r r following numerical example should clear up the difficulty as to scale — Fig. 127. — Graphical Construction for Deflections. Numerical Example.—^ 16'' x 6'' x 62 lb. rolled steel joist of 24 ft. span carries a uniformly distributed load {in- cluding its own weight) of 8 tons, and also an isolated load of 5 tons, at a point 6 ft. from the left-hand support. Find the maximum deflection (Fig. 128). In this case E = 12,500 tons per sq. inch. I =- 725-7 inch units. 12,500 X 725-7 EI 144 = 62,980 sq. ft. tons. First draw the B.M. diagrams for each of the loads, taking as linear scale, say V = 4 ft., and for the B.M. scale, say V = 20 ft. tons. Now divide the B.M. diagram into a con- venient number of equal parts, say 12, and draw the mid DEFLECTIONS OF BEAMS 263 ordinate of each part, treating these as force lines, then set these ordinates down a vector line, 0, 1, 2, etc. . . 12 to a reduced scale, say one-fourth for convenience. Then 1 in. down the vector line represents 4 X 4 X 20 160 sq. ft. tons, because each base element is J in. 62,980 160 E I on this scale 373-9 ins. Majc.Defn.^ '^^ Fig. 128. — Example on Deflections. 393' 7 This is obviously not convenient, so take — ^x— 48 6'$6ins. Then 1 in. on the link polygon represents „^ in. deflection. The maximum ordinate of the link polygon will be found to be '58 in. .'. Maximum deflection = '58 x "8 = '46 in. Allowance for Deviation of Cross Section. — The cases up to the present have all been on the assumption that the section is constant, or rather that the Moment of Inertia, I, is the same all along the span. If such is not the case, the deflection can be found accurately by first altering the B.M. curve to make up for the variation in the section as follows — Suppose ABC (Fig. 129) is the B.M. curve on any beam 264 THE STRENGTH OF MATERIALS A D B, and suppose that I„ is the maximum moment of inertia or second moment of the section, this occurring at the point D. Then take any point along the beam at which B.M. is x Y X Y X I and moment of inertia I^ and find x y^ so that x y^ == — ^ "• Do this for a number of points along the span, and join up the points thus obtained, and we get the corrected B.M. curve from which the deflections can be found by the construc- tion given above. The value I,| is taken in obtaining the expression E I for this construction. Deflections of Girders of Uniform Strength and Constant Depth. — If the cross section of a beam varies so that the maximum stresses are constant along the span, r --^Correofed B.M- C6(rve. then the modulus of the section must vary in the same way M . as the B.M., and so the ratio y is constant. If the depth of M the girder is also constant, then the ratio -j- will also be constant. The corrected B.M. diagram will in this case be a rectangle, and the deflection can be found by Mohr's theorem as follows — As in the several previous cases we have F.y S = EI In this case P will be equal to — ^ and V =-7 since the curve is a rectangle. , Ml2 DEFLECTIONS OF BEAMS 265 W L In case of uniform loading M = -g— • 64 E I Wl In case of a central load M = — r- 4 • ?i - ^^' • 3 2 E I Another simple proof of this relation will be found on p. 272. Further numerical examples will be found at the conclusion of this chapter. DEFLECTIONS FROM MATHEMATICAL STANDPOINT M 1 From equation (3) -p, ^ = t^ Now when R is great, as it will be in this case, we have 1 _ d^y , d^y M • dx'-~^l 'M d X ET '^idx . -~- = slope of beam = / - r r y = deflection of beam = / / - EI Now consider the following standard cases (see Figs. 123, 124). (1) Simply Supported Beam with Central Load W. — Consider a point Q at distance x from the centre of the beam. ThenM = ^ (^-^ W/^ \ ^Ix Wa;2 , ^ ?r— X 2\2 J 4 4 / W I x'^ W X 8 12 — + Ci a: + c 266 THE STRENGTH OF MATERIALS The slope is zero when x = .-. Cy = 0, and the deflection is zero when X = I ±2 • WP ' ' 32 WP ^ 96 +^^ = ^ C2 = - W 48 P Then maximum deflection . occurs when x = Then S = C2 EI WP* ~ 48 EI (2) Simply Supported Beam with Uniform Load. — Taking a point as before at distance x from the centre, we have . M==f(J-.)-|(l-.> - ^f^^ - ~ 2 Vi ■' / ,^ , wP X wx^ , Mdx ^ — g 6~ ^^ + Cg as before c^ = wP x^ w x^ ^ "16 2^ = when x = • _ _u)P wP • " • ~ ^2 - g4 3g4 " ^ I 384 J 384 Then the maximum deflection occurs when x = c. 5wP 5 W P 8 = EI ~ 384 EI 384 EI * The minus sign indicates only that the deflection is downward, and need not be employed in calculations. DEFLECTIONS OF BEAMS 267 (3) Cantilever with an Isolated Load not at Free End. — Take a point q at distance x from load. M = - Wx .-. Slope X EI = f Mdx Wx^ 2 When x = — I, slope = -WP — WP . • . E I X slope under load = ^ — Mdx Wx^ WPx 6 "^ 2 +^V X ^j When X = — I, deflection = _ WJ3 _ WJ^ _ - WZ^ •'• ^2 - g 2 ~ 3 . • . Deflection under load, where a: = c, /-WZ3\ 1 r r deflection = / / _ ^2 _ X .EI V 3 /EI Deflection at free end = deflection under load + slope under load (l — I) _ /-WJ3 WZ2 ^ 1 ~1"^ 2"^''~^^J EI W /Z2^ _P\ Ell 2 6j W Z2 / _ I 2 E I V 3 or neglecting the minus sign, which indicates only that the deflection is downward, we get W /2 / I Maximum deflection = 8 = o-^ftt \'^ ~ ^ (4) Cantilever with Isolated Load at Free End. — This is obtained by putting Z = l in the above case. 268 THE STRENGTH OF MATERIALS (5) Cantilever with Uniform Load from Fixed End to a point before Free End. In this case M = — .-. Slope X EI =^yM(lx w x^ , = - g- + c, When X = — I, slope = c, = 6 w P E I X slope under load = — ^ b E I X deflection -= / /M d x _ wx'^ wP X ~ ~ 24 + '""6~ + ""2 When X == I, deflection = _wl^ wl^ _ ~ wl"^ •*• ^2- 24~~ 6 ~~8^ E I X deflection under load, when a: = - C2 - g- E I X deflection at free end — wl^ ^ — Q~ + slope under load x E I x (l — Z) o ~ -8 +^^~^^ "6 -_WP fL _ 11 2 Is 12/ wPf _l 6 V 4 - w / V / = -6"A^-4 Neglecting — sign Ave have WlV _ I 6 E I V 4 DEFLECTIONS OF BEAMS 269 (6) Cantilever with Uniform Load over Whole Length. — This is the same as the previous case when I = l. • • ^-6"Eir 4/ ~ 8 EI (7) Simply Supported Beam with Isolated Load anywhere. — Let a load W be placed at a point c on a beam A B, Fig. 130, of span I, and let it be at distance a I from the end A, the distance from the end b being (1 — a) I. Then R3 = — ^— = W a ^ R.=WJ-l-^I^^ = W(l-a) I /^^^ ^. ® a i C (/ -a.)L 'B R, Fig. 130. Consider a point at distance x from a between a and c. Then M^ = R, a; = W (1 - a) x .•.Elg = W(l-a)a; (1) .^j^^^Wd-a):.^^ ax 2 E 1 2/ = ^— g — ^ — + Ci a; + C2 (3) Now consider a point at distance x^ from a between c and B. Then M^^ = R, x^ - W {x^ - al) = W (1 - a)X-^ -W X^ + W al ^Wal -WaX^ E I -7—^, ^Wal -WaX^ a Xjf (4) 270 THE STRENGTH OF MATERIALS .-.EI ^^^^ Wa Lr, - Wa"^^' + C3 (o) EI?/- ^2 g + C3 .Ti + C4 . . . . (6) In equation (3) when .t = 0, ^ = .-. C2 = In equation (6) when x = I, y ^ WaP W a ^3 2 6 f Co / + c. = 0. 3 i' -r ^4 - W a ^3 ••• C4 = ^ C3Z (0 These two equations representing the elastic line on either side of the load have the same slope and the same ordinate when X = Xi — al . • . putting in these values in equations (2) and (0) and equating we have W (1 - a) a'-P ,^, , Wa3Z2 — 5^ — ^ — -i- Cl = ^^ ag /g 2~" + ^3 Ci = -2 + C3 (8) Putting in value x = x-^ = a Zin equations (3) and (6) and equating we have W(l-a)a3Z3 Wa3Z3 W a* Z3 Q +CiaZ = 2— ^ g— +C3aZ + C C^ a Z = g h C3 a Z ^ C3 I Wa3/2 Wa/^ , Ci a = ^ 3 r C3 (a — i) Wa3/, Wa/2 / WaoZ-X, 3 3 ' V^ 2 Wa3J2 ^ Wa/2 _ Wa3Z2 W a^ /^ 3 3 2 "^ 2 ^ V3 ' 6 2y' -WaZ2 g-- (2 - 3a + a^) -^^'"^'(1 -a) (2 -a) (9) 6 DEFLECTIONS OF BEAMS 271 . • . equation (3) becomes E I2, = Wg-") ^^ _ W aP X (1 _ „) (2 - a) (10) Assume a ^ ^j then the maximum deflection occurs between A and c. ?/ is a maximum when v^ = z. e. when ^^ ^ ^ — (1 — a) (2 — a) = , „ l^a{2 - a) I. e. when x"^ == ^-^ i.e. ^ = I J r ^^ ~ 3 (11) Putting this value in equation (10) we have °'^^ = 3Vl"|— 3— I (12) The deflection under the load is obtained by putting x = al in (10). rru i?T W (1 - a) a3 Z^ W a^ P ^, Then Ely = — ^ ^ -^ — (1 - a) (2 - a) ^^-^^7--)[a-2-a] _ Wa^(L:^) • -^ 3 EI ^^^^ Deflection of Girder of Uniform Strength with Parallel Flanges. — If the section of a girder varies along M its length so that the stress is constant, then -^ is constant, Li M so that if the depth is also constant -^ is also constant. Assuming also that E is also constant we have P = p ,- = constant. 272 THE STRENGTH OF MATERIALS E I . • . Wherever ^ is constant, the beam bends to an arc of a circle. • Let Fig. 131 represent a beam bent to an arc of a circle. (N.B. — The beam is shown vertical instead of horizontal for convenience of figure.) Then, if is the centre of the circle and c d the deflection of the beam, we have from the property of the circle c D (c o + o d) = a c2 Fig . 131. As c D is very small we may write 8 2R - 8R P 4 Now -p = M EI .-. 8 = MP 8 EI This result agrees with that obtained on p. 264 by reasoning from Mohr's Theorem. * Resilience of Bending-. — The work done in bending a beam to a given stress may be obtained as follows — The work done by a couple in moving through an angle is equal to the product of the moment of the couple into the DEFLECTIONS OF BEAMS 273 angle turned through. Therefore, if a short portion of a beam subjected to a bending movement M is bent to a slope 8 i, the work done in bending is ^ ? because M gradually increases from to M. . ■ . Total work done in bending over whole beam = P= • 2 JNow , a X 1 1 ~R' '•^- R^ .'. P : ^J 2R^^ but^: M "EI .-. P - = /2EI^^^ 1 rate of change of slope // the B.M. is constant, and the section is rectangular, then 2^1 J 2EI o But M2 = S^ d^ • 2 . E . ri2 " hh^ , h Now I = ^2 . ^ = 2 . p ~ E' 6E 4 / 2 / .V hh^ 12 X L Where V is the volume of the beam. P /2 .'. Resilience = ,, = /„ V 6E // the load is central and the section is rectangular. — Consider one-half of the beam, then x is the distance from the abutment 274 THE STRENGTH OF IVIATERIALS TIT W X p rw d X f 2 y 2 E I o L /w^ .T^ d X ~ J 8EI o ~ 192 E 1 96 E 1 2 A- ^r . . WL /I JNow M at centre = - . — = , 4 d • • 6. EI. ^2 _ f ][L ~ 6 E ' <^^ As before I = -r^, d = P = Resilience = 12' 2 /2 46 /i3 L 6E' 12/^2 ^^ V 18 E P _ /2 V ~ 18E Numerical Examples ox Deflections, etc. (1) A girder lias a span of 120 feet, and has to support a uniformly distributed load of 1^ tons per foot run. What depth must the girder have in the centre if the maximum deflection is not to exceed ytdu ^/ ^^^ span ? The maximum stress in the flanges is not to exceed 6J tons per sq. in. and E is 12,000 tons per sq. in. {B.Sc. Lond.) This question is not quite clear, because if the depth is not the same throughout, we cannot calculate the deflection until we kno\^' the Ava}^ it varies. DEFLECTIONS OF BEAMS 275 We will assume the depth constant — ^ ,^ W / 11 X 120 X 120 „^ ^ JNow at centre M = -^- == -^ 5 it. tons o o = 27,000 m. tons. M . • . If maximum stress = 6 J tons per sq. in. since / = ^ Li ry 27,000. .. Z — — -^. — ni. units. 5 W L^ ^^^^ ^ = 384EI 1,200 10 12 _ 5 X 150 X 120 X 120 x 120 •'• 10 ~ 384 X 12,0001 ^ '' .-. I = 405,000 in. units. Now jj = -^ where D = depth. 2j: _ 2 X 4 05,000 X 6-5 . •'• ^ ~ Z " 27,000 '''^• = 30x6;5 _ jg.^g ^^ This is a greater depth than would be usually adopted in practice for a solid web girder. (2) A cast-iron water pipe, 10 inches external diameter and J inch thick, rests on supports 40 feet apart. Calculate the maximum stress in the outer fibre of the material when empty and when full of water, also the corresponding deflections. {AJI.LC.E.) -r TT (D* - d^) IT (104 - 94) In this case 64 64 = 168-8 in. units. Z^\= ^— = 33-76 in. units. d 5 Volume of pipe = ^ (lOO — 81 j x -rj^ = 4-14 cub. feet. Volume of water = ^ • tt^ x 40 =- 17*67 cub. feet. 4 144 276 THE STRENGTH OF MATERIALS .*. Weight of pipe = w ^ ^ ^..^ ^ '832 ton. ° ^ ^ 2,240 Weight 01 water = w. = Tr^.r^ = "492 ton. ® ^ 2,240 .'. W = w; + 1^1 = 1324 tons (about) . • . Max. stress when empty M -832 X 40 X 12 , ,„ , ^Z= 8x33-76 = 1-48 tons per sq. in. Max. stress when full = — ^^r = 2' 35 tons per sq. in. Taking E as 8000 tons per sq. in. 5 WL^ 8 when empty = -334^^! _ 5 X -832 X 4 X 40 x 40 x 1 2 x 12 x 12 ~ 384 X 168-8 X 8000 = -89 inch. 8 when full = ^^^^r = 141 inches. (3) A pole 7nade of mild steel tube, 6 inches diameter and J inch thick, is firmly fixed in the ground, the top being 10 feet above the ground level. A horizontal pull of 2000 lbs. is applied at a point 6 feet from the ground. Find the deflection at the top. E = 13,500 tons per square inch. {B.Sc. Lond.) In this case I = — ^ — ^. = n. - - 64 64 = 32* 9 in. units. This is the same as Case (2). In this case ^ = 6 ft. L = 10 ft. W = 2000 lbs. = f ^i^.^ ton. 2,240 . _ 2000 6 X 6 X 12 X 12 x 8 x 12 • • ° ~ 2,240 ^ 13,500 X 32-9'x'2" = 5 inch, nearly. (4) What is the least internal radius to which a bar of steel DEFLECTIONS OF BEAMS 277 4 inches wide by f inch thick can he bent so that the maximum stress will not exceed 5 tons per square inch ? E = 13,000 tons per square inch. (A.M.I.C.E.) The general formula for bending is /ME d ~ I ~ R . i _ E ' ' d R ^ d^ or K = — 7- In this case d = distance from N.A. to extreme fibre, 3 . 3 13,000 = 488 inches. = 40-7 feet. It should be noted that the width of the bar is not necessary in this problem . The result is the radius of the centre line. (5) A cast-iron beam, has a rectangular cross section, the thick- ness being 1 inch and the depth of the section 2 inches. It is found that a load of 10 cwt. placed in the centre of a 3Q-inch span deflects this beam by '11 inch. Through what height would a weight of J cwt. have to fall on to the centre of the same span to produce a deflection of "30 inch ? {B.Sc. Lond.) It takes 10 cwt. to produce a deflection of "11 inch. 10 X "30 .'. It would take — yz. to produce a deflection of '30 inch. Now the work done in deflecting a bar when loaded in the centre = J W S, .*. Work done to produce '30 inch deflection 1 10 X -30 ^^ . = ^ . — 7Y — X 30 m. cwt. ■" * J. 1 = -341 ft. cwt. 78 THE STRENGTH OF MATERIALS If h is the height from which the J cwt. falls, work done by it = 9 (^ + T9 ) ^^' ^^^-^ because we shall take h as the height above the unstrained position of the beam. These two amounts of work must be the same, have K^^'S)-^^ h = -682 ^2 ^^0*' = 8-18 - -30 inches = 7*88 inches. CHAPTER X COLUMNS, STANCHIONS AND STRUTS The question of strength of columns of compression mem- bers is of very great importance, and has formed a field of discussion and investigation for many years. Interest in the subject has recently been aroused by the regrettable failure of the Quebec Bridge, and within the next few years many investigators will probably direct their energy towards giving us further information in this direction. Although the subject certainly presents difficulties, much of the confusion which is in the minds of many designers is undoubtedly due to insufficient grasp of the meaning of the various formulae in use. We will endeavour to make this subject quite clear by approaching it in the following manner, which was suggested by the author in 1908. In the design of a tie-bar we use a constant working stress, that is to say, the stress does not depend on the shape or the length of the tie ; but in struts or compression members the working stress depends on the shape and the length and the manner in which the ends are fixed. The quantity which determines the working stress, and thus the strength of a pin-jointed strut, column, or stanchion is equal to Length of column _ Least radius of gyration about centroid This quantity we will call the Buckling Factor of the strut. For struts with ends fixed in other ways the buckling factor is obtained by dividing the equivalent length of the strut by the least radius of gyration. We will show later how the equivalent length is obtained. 279 280 THE STRENGTH OF MATERIALS Slenderness ratio. — Some writers use this term in place of buckling factor. The slenderness ratio Length Least radius of gj^ration about centroid but does not take into account the method of fixing the ends, and so is not the same as the buckling factor. Two struts of the same material and having the same buckling factor may carry the same stress, no matter how their ends are fixed : this is not the case for two columns with the same slenderness ratio. The reason w^hy a variable working stress has to be used is that struts fail by buckling and not b}^ crushing, unless their length is extremely small. If for some reason the centre line of a strut is not quite straight or the load comes out of centre, there are bending stresses caused in the material, and the distortion due to these bending stresses tends to increase the eccentricity, and failure may ultimately occur due to this reason. Strut Formulae. — A large number of formulae, some theoretical and some empirical, have been proposed for ob- taining the w^orking stress in compression in terms of the buckling factor of the strut and of the crushing strength of the material. Before these formulae can logically be compared we must be careful to see that they are for the same crushing strength, and for the same manner of fixing the ends of the strut. We will consider the following : — (a) Euler's Formula. — This formula is intended for long struts in which the direct stress is negligible compared with the buckling stress. It is usually given in the following form — ^EI ^ " L'^ where P = the breaking load (not the working load) E = Young's modulus. I = least moment of inertia. L = length of pin- jointed strut. COLUMNS, STANCHIONS AND STRUTS 281 We will now put it into more convenient use for practice as follows — P , ,. ^ 7r2EAA;2 .*. -r = breaking stress = — * t^t— A ^ AL2 _ 7r2 E _ 7r2 E ~ /L\2 ~:f Adopting a factor of safety of 5, we get xxT 1 • i / breaking stress tt^E Working stress = /,, = J' = -— ^ ^ 5 5c^ For mild steel, E = 13,000 tons per sq. in. , TT^E 25,600^ • '. /j> = -^2, "^ 2 — tons per sq. m. For wrought iron, E = 12,500 24,600 • ' Ji' ~' Similarily for cast iron /^, = For timber /^, = 12,000 1,600 Proof of Euler's Formula. — The proof of Euler's formula is found by many students to be somewhat difficult to follow, as it involves the solution of a differential equation. Suppose that a column in some way or other becomes de- flected as shown in Fig. 132 (1) . Then there are bending stresses induced in it, and the strut will exert a force P on the supports tending to straighten itself. Now, if the load on the strut is less than P, the strut will straighten, and so is safe ; but if the load is greater than P, the strut will continue to deflect, and will ultimately break. When the load is equal to P, the strut is in unstable equilibrium, and so P is called the critical or buckling, or crippling load. Consider a point a on the strut. 282 THE STRENGTH OF MATERIALS The B.M. at A = M, = -Pa:.* .Now, if R is the radius of curvature, J^ _6Z2^_ M _ - Pa ; R ~ dy^~Wl~ EI • ' dj^^ ~ET • ^ = say - m2 . a: (1) assuming that I is constant, or that the strut is of uniform section. \ Z 5 4-5 -JC- 1 Fig. 132. — Methods of Fixing Ends of Columns. The general solution of this differential equation is X — A cos m ?/ + B sin my , (2) * The negative sign occurs because we take anti-clockwise moments to the right as positive, and x in the figure is, on the usual convention, negative. If the figure be turned round so that the column is horizontal and has a downward deflection it will be seen that according to the rule given on p. 121, M.^ is positive and x is negative. Fx therefore is negative, and to make the moment positive we must write M^ = — Pa;. If we had drawn the column buckled in the other direction, M,^ would have been negative and x positive, so that we still would have M., = - Pa;. COLUMNS, STANCHIONS AND STRUTS 283 where A and B are constants, which are obtained as follows — When y = -^ and -^^— , a: = ^2 2 „ . mL , -r, . mL .^. . • . = A cos -^ \- B sm -^r- (3) ^ . — mL , ^ . — mL ,., = A cos — ^ h B sm — ^— (4) . mL -r, . wL = A cos — ^ — 13 sm —^ . • . B must = . • . a; = A cos my (5) When 2/ = 0, a; is infinite, . * . A is not zero .p . mL ^ . • . II A cos — ^ = A mL , ^ cos — ^ must = A The general solution for this condition is that , mL _ wtt • • • w^ = -L2- • • EI~ L2 P = — L^ (6) The lowest value of P is given by n = 1, and as this is the most important for us, we write the result as ^--^ (7) It should be noted that P is independent of the quantity x, so that the force necessary to keep the strut deflected at large radius of curvature is the same as that to keep it at a small radius, and so if the load is the least amount greater than P the strut will go on deflecting, and so break. Use of Euler's Formula. — It must be remembered that in this formula we have not taken into account the direct 284 THE STRENGTH OF MATERIALS compression stress on the strut. If the safe stress given by Euler's formula is greater than the safe compressive stress for very short lengths of the material, then obviously we should not use Euler's result. Thus, if Euler for mild steel gives //, greater than 6 tons per sq. in. we should use 6 tons per sq. in. Another way of using it is as follows — Safe load = -- ,^- 5 i^ _p_2EI .-. I required = ^ -^ A required = P We have thus the least area and moment of inertia that the section must have and can so choose a suitable section from tables. Method of Fixing Ends — Equivalent Length of Strut. — In the above working we have considered the ends as pin-jointed. If the ends are fixed in any other way we must take as the length of the strut the length of the equi- valent pin-jointed strut; this we will call the equivalent length of the strut. Now consider the following methods of fixing the ends (see Fig. 132). (1) Pin Joints at Each End. — This is the standard case. (2) Both Ends Fixed in Position and Direction. — In this case the buckled form is as shown in the figure, and b c is the equivalent length, i.e. a pin- jointed strut of length B c is as strong as the fixed strut. . * . in this case equivalent length of strut = „- Buckling factor = c = ^ , (3) Both Ends Fixed in Direction only. — The buckled form in this case is as shown in the figure. On comparing COLUMNS, STANCHIONS AND STRUTS 285 with Case 1, it will be seen that the portion b c is equivalent to one-half the strut in Case 1, and so in this case, L since B c = A Li equivalent length of strut = L L I . Buckling factor == c (4) One End Fixed in Direction and Position, other End Pin- jointed. — It will be clear from the figure that in this case 2 L equivalent length of strut = — ^— 2L • . Buckling factor = c = Zk (5) One End Fixed in Direction and Position, other End Free. — In this case equivalent length of strut = 2 L 2L k . ' . Buckling factor c = Summary of Values of Buckling Factors Case 1. Case 2. Case 3. Case 4. Case 5. Buckling factor = c L k L 2k L k 2L dk 2L k These values should be used in Euler's and the other formulae involving the buckling factor. {b) Rankine's Formula. — This formula is sometimes called the Gordon-Rankine formula, and is of the form 1 + a . c^ /.= 286 THE STRENGTH OF MATERIALS Where /, = safe compressive stress for very short lengths of the material a = a constant depending on the material c == buckling factor of the strut ii, = working stress per sq. in. for the strut. The following values of a may be taken according to different authorities — Mild steel a = kftf^ to 7.7^K7^, /, = 6 tons per sq. in. Wrought iron a = ^^^ to -g^^^, /„ =4 „ „ „ Cast iron « = 2500 *%"io' ^'^ ^ " " " Timber a = ^qoo' ^' ^ *^ " " " In each case we prefer to use the higher value of the constant a. There is a very large amount of variation in the values of the constants as given by various authorities, and in com- paring the above with those given by others, the reader should be careful to compare the safe stresses given with the above figures with the safe stresses given b}^ others, because the value of /^ also varies in the various forms of the formula and thus, although the constants may be different, the re- sulting safe stress may be nearly the same. Care must also be taken to see whether pin-jointed or fixed ends are taken as the standard case. Construction of Rankine's Formula.— Rankine's formula may be looked upon as a corrected form of Euler's. If c is very small, i. e. if the strut is very short, the term a c^ is negligible, and so we get f^, = /,. This is, of course, the result which we ought to obtain. If c is great, ^. e. if the strut is very long, the term a c^ will be so great that 1 may be neglected in comparison with it, and so we get COLUMNS, STANCHIONS AND STRUTS 287 This will give the same result as Euler if a 5 . .„ 1 TT^E 25,600 , ^„„ ^. e. if - = -^-r = v, — = 4,267 a Djc o Although some writers state that constants obtained in this manner agree with experimental results, the constants are not usually calculated theoretically in this way, but are obtained from experiments. It is believed that the figures recommended above will agree well with the best practice. It is interesting to note that in one form of Rankine's formula, giving the breaking or crippling stress, viz. p / ^ + Hi) / is the stress at the elastic limit. In an earlier chapter we pointed out the desirability of obtaining the working stresses from elastic limit, i.e. basing the factor of safety on the elastic limit. An interesting and important paper by Mr. C. P. Buchanan, in Engineering News, December 26, 1907 — published after the Quebec Bridge disaster — gives the results of tests on full- size built-up columns such as are actually used in bridge practice. The tests extend over a period of fourteen years, and show that even for the short columns the buckling or crippling stress is not more than 90 per cent, of the tensile yield point (see p. 4). We thus see that in columns as actually used in practice, the buckling stress is certainly not more than the elastic limit stress, and so the only reasonable factor of safety is that based on the elastic limit. (c) Straight Line Formula. — These empirical formulae are used principally in America, and give very good approxi- mations for rough working. They are of the form = /, (1 - e . c) 288 THE STRENGTH OF IVIATERIALS Where f^, and /, are as before e = a constant depending on the material. The following values of e may be taken — For mild steel e = -0053 ,, wrought iron e = "OOoS ,, cast iron e = "008 „ timber e = -0083 As in Rankine's formula the values of constants vary con- siderably according to different authorities. {d) Johnson's Parabolic Formula. — This is also an empirical'formula devised to agree with Euler for long lengths, and to agree with the ordinar}^ compression strength for short lengths. It is of the form = /.(!-?• c^) (/ is a constant of such value as to make the curve of /,, plotted against c tangential to Euler, and the curve is used up to the point where it meets the Euler curve. The following values may be taken for g — For mild steel g = -000057 „ wrought iron g = -000039 ,, cast iron g = 00016 (e) Gordon's Formula. — This formula is often confused with Rankine's, and was used largely for some time, but it is now quickly going out of use in favour of the Rankine formula. This is probably due to the fact that designers are now more used to making calculations involving the radius of gyration, a quantity which practical men have usually looked upon with suspicion. Now that tables are published giving k for most sections, it is as easy to use as the diameter d. Gordon's formula is of the form , L _. COLUMNS, STANCHIONS AND STRUTS 289 Where /,., /^„ and L have their usual meaning. j is constant depending on the material and on the shape of the section, d is the least diameter or breadth of the section. The objection to this formula as compared with Rankine's lies in the fact that one has to use different constants for different shapes of section for the same material. Otherwise it is very similar to Rankine's. The following values for j may be taken, f, being the same as in Rankine — Shape of Section. j Mild Steel. , ^frwf,^* 1 ^^^* ^°^' Timber. Solid circle Hollow circle .... L, T, H, etc. .... Built-up sections . . . Rectangle (solid) . . . 1 370 1 600 1 300 1 400 1 500 1 500 1 800 1 400 1 550 1 700 1 110 1 180 1 90 1 120 1 125 1 200 1 100 1 160 (/) Fidler's Formula. — The reader is referred to Fidler's Bridge Construction for a very complete analysis of the strut problem. The formula which Mr. Fidler obtains gives the breaking stress, and is Minimum breaking stress = / + B + V(7^KF^2 m f B m Where / = ultimate pure compressive strength of material E R = Euler's breaking stress m — a constant of average value 1'2. u 290 THE STRENGTH OF MATERIALS The following values of f,,, the safe stress in tons per sq. in. for struts, are suggested by Fidler and are used by some authorities — L k Mild Steel. Wrought Iron. Cast Iron. i Pin ends. Fixed ends. Pin ends. ' Pixed ends. 1 Pin ends. Fixed ends. 1 20 5-20 5-29 3-92 3-99 8^07 8-65 40 4-76 5-09 3-64 3-89 5-68 7-56 ! 60 4-02 4-83 3-17 3-73 3-35 6-10 80 3-15 4-45 2-60 3-48 1-96 4-68 100 2-40 4-00 2-03 317 1^29 3-35 120 1-83 3-46 1-57 2-82 •93 2-37 140 1-42 2-96 1-24 2^48 •70 1-78 160 1-13 2-51 •98 2-14 •56 r40 180 •91 213 •80 1-84 •43 114 Lilly's Formula. — Professor Lilly of Dublin has devised formulae for columns which allow for secondary flexure or wrinkling in the column,* and may be regarded as a modifica- tion or correction of Rankine's formula. It is of the form 1 + w . - + ac^ V Where tn is a constant depending on the shape of section t is the thickness of the metal in the section a is -^-^. (Compare p. 287.) 7)1 5N/,. E '■ the values of N being given for some sections in Fig. 132a. Use of Strut Formulae.— Fig. 133 shows curves of f,, for mild steel for various values of the buckling factor according to the first four formulae. It is advisable to draw such a curve to a good scale, choosing one of the formulae — say * See Engineering, January 10, 1908, or a more complete paper and bibliography in Proc. Am. Soc. C.E. Vol. LXXVI (1913); also The Design of Plate Girders and Columns (Chapman & Hall, Ltd.). COLUMNS, STANCHIONS AND STRUTS 291 Rankine — with a = np,crr\\ such curve can then be used whenever the value of /^, is required. N. 50 iV. 60 iV^. 120 N.IQ Fig. 132a.— Lilly's Column Formula. It will be remembered that /^, gives the safe stress per sq. in. for struts with central loads. If the loads are eccentric we must proceed as described later. Then if A = area of section of strut, Safe load = P, = /^^ . A. If, as often occurs in practice, we are given the load but 292 THE STRENGTH OF MATERIALS have not designed the section, so that we do not know the buckling factor, we can often get a rough idea by taking a trial value of ^ equal to about f /,, i. e. 4 tons per sq. in. for steel, and finding the area requisite for this stress. This will give us an idea of the area required, and we can choose a section with roughly this area, and see by finding its buckling factor what is the safe load on it. ISO Buckling Factor = C- Fig. 133. — Curves for various Column or Strut Formulae. (Mild Steel) Many of the leading constructional steelwork firms publish tables of safe loads on various struts. Having previously checked one or two to see that these firms work with similar formulae, we can choose a suitable section for our case, and then apply our formula and see if such section is satisfactory. REINFORCED CONCRETE COLUMNS Short Columns Centrally Loaded. — We have shown on p. 38 that the safe load in a column in which buckling is COLUMNS, STANCHIONS AND STRUTS 293 negligible (the length being less than 15 times the least diameter, and the notation being modified) is given by P = c (A, + m A,) Gross-binding of Reinforcement. — In addition to the longitudinal reinforcementj some force of binding is necessary to keep the bars at the requisite distance apart. This is due to the following reason — Suppose that a reinforced column with bars a b, c d,. Fig. 133a, be compressed; then, quite apart from any buckling u R \ C B I D ■^"T Fig. 133a. of the whole column, the column will bulge out somewhat as shown, and the reinforcing bars will buckle because the value of ^ or the buckling factor for them will be large. If we bind the reinforcing bars together, as shown diagrammati- cally, so that they cannot buckle, the column will not bulge to anything like the same extent, and so will be considerably strengthened. From a large number of experiments M. Considere found that the best results are obtained when spiral coils are placed round the reinforcing bars at distances apart equal to i to ~o of the diameter of the coil. 294 THE STRENGTH OF MATERIALS M. Considere suggested the following allowance for the coils in the strength of the column — Let A/, be the equivalent area of longitudinal reinforce- volume of metal in coils \ length of column Then safe load = c (A, + m A, + 2 4 m A') Long Columns Centrally Loaded. — Some authorities use Euler's formula apj)lied to the homogeneous section, viz. — IT' . ,, . , ., / . A volume of metal m coils \ ment of the spiral coils [i.e. A,, = , ,, ,, , ^ \ length 01 column / Safe stress = fj, ^E 2 5 c c being the buckling factor. In obtaining c the radius of gyration of the equivalent homogeneous section (see p. 183) is used, "i: where A = A^ -f- m A,. I^. = equivalent second moment = L + (m — 1) A, r^ for section shown in Fig. 133a. Y being the moment of the section apart from the reinforce- ment, I^, = ^- (- [m — 1) A, r^ for circle b h^ = --— -f (m — 1) A, r^ for rectangle Then safe load = fj, x (A, + m A,) Rankine's formula can also be used in the form _ _ 5 00 ^ 8000 Braced Columns, Struts, and Stanchions. — Struts are often formed of roUed sections such as beams and channels braced together by diagonal bracing or plates. The strut that failed in the Quebec Bridge was a braced strut, and the report of the Commission states that there is not yet sufficient COLUMNS, STANCHIONS AND STRUTS 295 information for the design of such struts for very heavy loads.* For ordinary comparatively light work, however, . ^-N X y-^. Y — D ---^ ^ 5- Y ^^^^ -* P ^ I v^-^ ^^ i> o o a — o o o o o o s Fig. 134. — Columns with Open Webs. braced struts such as shown in Fig. 134 are commonly used but the diagonal bracing should preferably have one rivet * See Illinois University Bulletin, No. 44, G. Talbot and Moore, for an experimental investigation of the subject. 290 THE STRENGTH OF MATERIALS passing through the two diagonal bows as in the top of Fig. 135 instead of two rivets as shown. The unbraced length of one of the beams or channels must be such that the load per sq. in. on them is not more than the safe stress for them considered as struts. We can get an idea of the maximum unbraced length as follows — Let c = buckling factor of whole strut ,, ki = least radius of gyration of one channel or beam ,, P = total load carried b}^ strut ,, 2 A = total area of strut ,, S = maximum unbraced length of channel or beam. Then, using Euler's Formula, ^^-v- = '^ = ^ Each channel or beam carries ^ load stress 5S2 ~ S2 B _ B yfci^ • • C2 ~ S^ . • . S = k-^c r. ' Equivalent length of strut L yjT Since c = = ' Least radius of gryation of whole strut k S _h, L k L k .' . ^ = least number of panels = r- _ Least radius of gryation of whole strut Least radius of gyration of channel or beam As a rule a spacing of 2 to 3 times the breadth B or 30° to 45° inclination of the diagonals will be found to be satisfactory, and in practice would be adopted, unless the calculation required them to be less. The strength of the strut in this case is calculated as if the section consisted of the two channels or beams held at the requisite distance apart. See worked Example No. 4. Relative Values of Different Bracings. — Professor H. F. COLUMNS, STANCHIONS AND STRUTS 297 Moore, of Illinois University,* has given the results shown in Fig. 135 of the " flexual efficiency " of various forms of bracing. This flexual efficiency is the ratio of the calculated fibre stress to that obtained by measuring the strain, the calculation being made on the assumption that the braced section behaves as an integral one. The tests were made by ordinary cross bending and not as columns, but the comparative results may be taken as representing the relative values of the different ICO ^^ ^ — w /^ '/Tk yv' /?-■ /O N \ ■:mm>f^' \ ^ 60 '^t^^m """ , '^ ^^^ ^_ 1 ^ ^ 4U -— ^^ a ^^ _^ ^ ^- --^ . ■mam] "^ i > 4 \- « ? I i IC ) r > (4 Stress in thousands of pounds per sq. in. Fig. 135. — Efficiency of Bracing. kinds of bracing for column purposes. Particular attention is directed to the great advantage that single diagonal bracing with single rivets (the third from the top) has over that with the separate rivets (bottom) in which heavy secondary stresses occur. Least Radius of Gyration. — The least radius of gyration will be about or at right angles to an axis of symmetry if there be one, so that in this case we need only calculate k for * See a paper h»y Professor A. H. Basquin, Proc. Western Society of Engineers, 1914, on " The Design of Columns." This is one of the best papers which have been published on the subject. 298 THE STRENGTH OF MATERIALS the axis of symmetry and at right angles to it. If there is no such axis we should proceed as indicated on p. 172. Examples on Struts, etc., with Central Loads. — The following numerical examples should make the question of the design of struts, etc., clear. (1) A 10'' X C X 42 Standard I beam of mild steel is used as a stanchion, the length being 16 ft. and one end being fixed and one end pin- jointed. Find the safe load for it to carry. From the table of standard sections we see A = 12-35 sq. ins. Least k = 1"36 T, , ,. „ , equivalent length 2L .'. UucKilins; lactor = c = - ^ «^ = or ° l\5o 6 k 2 X 16 X 12 o^ o 1. + == Q ^ 1 .QA = 94-2 about 6 X 1 OO .•. Safe stress = f,, = 0^02" ^^sing Rankine's formula 1 + — ^ 6000 = 2"43 tons per sq. in. 1 + 1-47 .-. Safe load = 12-35 x 2-43 = 30 tons. (2) A solid cast-iron column, 6 inches in diameter and 15 feet long, is fixed at the lower end and carries a load at its free upper end. Calculate the load the column will safely carry, assuming a reasonable factor of safety. {B.Sc. Lond.) In this case k = . =1-5'' 4 Equivalent length = 2 L = 30 _ equivalent length _ 30 X^12 •'• ^ ^ k~ ~ V5 = 240 7 .-. Safe stress per sq. m. = /,, = 240">r240 ^ ^ 1,800 _ 7__ " 1 + 32 = -212 ton per sq. in. COLUMNS, STANCHIONS AND STRUTS 299 .-. Safe load = '212 x - 4" ^ = 6 tons. 4 According to Enler /,, = ^ ^ — o 7r2 E _ 12,000 5 c^ c^ 12,000_ _ 240 X 240 .-. Safe load = -'- = 5-88 tons. 4 (3) A steel rolled joist is used as a strut with built-in ends, the length of the strut being 15 feet. Find, from the data given below, the cross section of the joist, if it has to support a com- pressive load of 40 tons with a factor of safety o/ 4. (a) The total depth of the cross section of the joist is twice the width of the flanges, and the thickness of metal is to be f of the width of the flanges, {b) The crushing strength of a short strut of this quality of steel is 24 tons per square inch. (c) The constant in Bankine formula is ^^ ^^^. {B.Sc. Loud.) In this problem we must first find the breaking stress from the formula. In this case we do not use the equivalent length of the strut because the constant is given for fixed ends. ^ . . . 24 x^x^^cirvj-iig i 3UJ.COO - ^ 36,000 V A;/ • Sflfp ' stress ^ A breaking stress 6 Now let = area of section 1 /Ly 36,000 VA:/ and let B = breadth of flange then 2B = depth of beam B 8 = thickness of metal. Then A ^2B^xB^B(,^_ 2B\ 8 ) B2 7 B2 15 B2 ~ 4 "^ 32 ~ 32 •4687 B2 300 THE STRENGTH OF MATERIALS The least radius of gyration will be about an axis perpen- dicular to the flanges. -r B B3 7B /B\3 Then I=4-12+4^r2-U; ' B* 7 B* — 4- — 02111 B* - 48 + 12 X 2,048 " ^^^^^ ^ 3_I _ 02111 B^^ ' • " " A " -4687 B^ ^^^ ^ .-. Safe stress = ^?= ' A ~ 15 X 12 X 15 X 12 "^ 36,000 X 045 B2 40 6 •4687 B2 1 , 9 1 + 45 B2 1 + ^4TbO = 40--^^^^^^ = -15 X -4687 B2 .-. B*(-15 X -4687 X -45) - -45 B2 - 9 - 316B4 - 45B2 - 900 = The solution of this quadratic gives B2 = 18-2 nearly say B = 4J . • . Adopt a joist 10'' x 5'' with metal ^" thick. We could work this problem roughly by the given rule, as follows — Take /, = ^ X 6 = 4 .-. A = . = 10 sq. ms 15 B2 • 32 = 10 B2 10 X 32 64 ~ 15 "3 B = -^- = 4-62, say 5'' V3 (4) A steel column in a bridge-truss has pin-jointed ends and is 26 feet long. It consists of two standard W x S^' X 28-21 lb. COLUMNS, STANCHIONS AND STRUTS 301 channels 'placed 4 J inches apart. Find a safe load for the section. (See Fig. 134.) On looking up the tables, we see that for a 10'' x 3 J" x 28-21 lb. channel, A = 8-296 "'max. = O I I ^min. ^^ *Jy4: Dist. of C. G. from edge = P = -933 Then for whole strut k,,. = 3-77 = 3-1832 + -9942 . • . k,, = 3-33 Length _ 26 x 12 c = Least radius of gyration 3*33 93-6 • / ^ 6 • • '^ 93-6 X 93-6 "^ 6000 2-46 = 2-44 . Safe load = 2-44 x area = 2-44 X 2 X 8-296 = 40-4 say 40 tons. STRUTS WITH ECCENTRIC LOADING Simple Approximate Method. — If the thrust in the strut is out of the centre, i.e. where there is bending moment as well as direct thrust on the strut, we cannot use the same rules for design as in the ordinary case. In such case we may obtain approximate results by pro- ceeding as follows — Let the load P be at distance e from the centroid of the cross section, then M = P . e (Fig. 136). Case 1. Very Short Struts. — If the length is less than 302 THE STRENGTH OF MATERIALS 10 times the least diameter of the strut, the stresses are obtained as shown on p. 237. i. e. /, A ^Z. f, M P Z, A In this case /,. V Vx " K^ Z,. P P . e . d, ~ A "^ Ak'' A V^ ^ k^ ) P •'• A ed. ZPlufr.o IZy^a Y Ccniroia L D ■Cff^ — *■ -<+ B F Fig. 136. — Columns with Eccentric Loads, This gives the safe load P for a compressive stress /,. This ease is fully dealt with in Chap. VIII. Case 2, Struts Longer than 10 J)iameters. — In this case we must make some allowance for buckling tendencies, and ^\'e may proceed as follows — As in the previous case we have Combined compressive stress = . (l + , 2 j Now in this case this compressive stress should not be more COLUMNS, STANCHIONS AND STRUTS 303 than the safe stress x>er sq. in. obtained by considering the buckling formulae, A , , ed,. ' Safe central load on strut i.e. Safe eccentric load on strut = 7— — — ^ (1 + i where e = eccentricity of load dr = distance from centroid to edge of section nearest load k = radius of gyration about axis perpendicular to the plane containing the centroid and the load. This formula may be put into a form which is sometimes more useful as follows — Let P^ be the central load, which is equivalent to the eccentric load P. Then Px = P (1 + p' In this formula e should be taken as the effective eccentricity, /M i.e. ( p where M is the Bending Moment on the column), the value of e shown on the drawings is only true when the column is free at the top. For other cases see articles by the author in the Architect's and Builder's Journal, March 3 and 31, 1915. P Then p may be called the eccentricity factor for the strut. In using this formula it should be noted that it is Avorked on the assumption that the buckling will take place in the plane of the figure, and so the value of k for the strut in this direction should be used in finding the safe central load. If the safe eccentric load according to this formula comes more than the safe central load for the least value of k (this 304 THE STRENGTH OF MATERIALS can of course only occur when the least value of k is about the axis d b), the lower value should be used. Stanchions -with Web and Flange Connections. — The loads on stanchions are often communicated from girders connected by cleats, etc., to the web or flange of the stanchion. If such connections come on one side only, or if the loads communicated from the two sides are not equal, the load will not be central, and allowance for the eccentricity should be made. Numerical Example. — A mild steel stanchion 30 feet long and with ends fixed has the section shown in Fig. 136. Find the safe central load and also the safe loads communicated at the points b and c. In this case A = 40*59 sq. ins. ^xx ^^ 4'o7 ,, ,, fCyy = o'41 ,, ,, .*. Buckling factor = c = ^^ = „ ^—.^ = 52-8 * 2k 2 X 3*41 *** ~ 1 I ^^:3 X 52-8 " 1-464 ~ 6000 .-. Safe central load = 40-59 x 4'10 = 166 tons nearly. Load at c— e = 225 + 575 = 2-725 .-. 4 = 6'' ed,. 2-725 x 6 P 3-412 1-41 1 (KCK . • . Safe eccentric load at c = ,~ , , -. , == 69 tons nearly. 1 + 1 41 "^ Load at B. — We must now first calculate ^ as if k^^ were 30 X 12 minimum radius of gyration, i. e. c = „ v „_ = 36-9. &j ' 2 X 4-87 • f , ^ ^ — = 4-89 • • '" 36- X 36 -9 "^ 6000 X = 6" d,. = 6" X (Z, _ 6 X 6 , _^. •'* T2 ~ 4-872 COLUMNS, STANCHIONS AND STRUTS 305 c, n ^ • T J ^ 4-89 X 40-59 . • . bate eccentric load at b = 7 ^ ko ~ _ 4-89^ X 40-59 ~ 2-52 = 77-7 tons nearly. In this case the eccentricity factors for c and b are 2-41 and = 2-14 respectively. A rough rule sometimes adopted is to use 2 J and IJ as eccentricity factors for flange and web connections respec- tively, but such rule is not reliable. It is more nearly true for I beams used as stanchions than for built-up sections. Alternative Approximate Method. — Where the bending stress is large compared with the direct stress it seems reason- able to allow that instead of the previous treatment we shall subtract the bending stress from the value of /, used in the strut formula. The compressive bending stress for an effective eccentricity ' . using the Rankine formula we shall have , P e d, i =^ = ~ ~^^ '"A 1 + ac2 P / , , c, ^ ed,. AV ' ¥^ Cast-iron Struts Eccentrically Loaded. — In dealing with cast-iron struts with eccentric loads it must be remem- bered that they will probably fail by tension. The safe load P from the tension standpoint X dt , 306 THE STRENGTH OF :\L\TERIALS when /, is the safe tensile stress, and this should be compared with the safe load from the compression standpoint, and the lower value adopted. * Modified Euler Theory for Eccentric Loading. — In this case we have, Fig. 137, or putting Fig . 137. — Eccent ric Loading of Columns P (e + x) = M = - Elf" dy^ •*• d'-x dy'~ m = _ P ( EI^ /P \ E I X + e) d^-x dy'~ — m {x + e) ' '^j The general solution of this is (a- 4- e) = A cos my -\- 3 cos m y * Cf. p. 282, equation (1). (1) = COLUMNS, STANCHIONS AND STRUTS 307 Since x = for y ^ ± ^, B = as before . • . (x + e) = A cos my (2) . • ^ when .X' = 0, e = A cos ^^ . m L .'. A = e sec ^ m L . • . X = e sec — ^— . cos my — e = e (sec — ^ . cos my — 1 j (3) At point o where y = eccentricity = x„ + e = e^ / m L ,\ , m L = e ( sec — ^ I j + e = e sec -„ - ='^"^2^7© ••■•'^^ where f^ "^ ~\ . • . stress at o = . I 1 + -\-o~ A\ F = (putting ^ = c) A. (l + '^f sec ^ ^g (6) Now let = r^ ^ '— and call it the Eulerian angle. 2 Ve ^ Then, stress at o =/,.(! + ,2 ^^^ ^ ) C^) Values of sec 6 are given in the table on p. 308, taken from Professor Basquin's paper * previously referred to. c. c. J ■ 1 ^ Safe central load . • . bate eccentric load = — — ^ 1 + ^-^ sec This can only be used by trial if the load is not given. * Proc. Western Society of Engineers, 1914. 308 THE STRENGTH OF MATERIALS Buckling Factor c. 50 60 70 80 90 100 110 120 130 140 5 105 ] L-08 Ml 1-15 1-20 1-25 1-32 1-40 1-60 1-62 6 1-07 LIO 114 M8 1-24 1-31 1-40 1-51 1-63 1-82 7 1-08 111 1-16 1-22 1-29 1-38 1-50 1-64 1-83 2-08 8 109 ] L-14 M9 1-26 1-35 1-46 1-60 1-79 205 2-41 9 1-10 : L-15 1-22 1-30 1-41 1-54 1-72 1-97 2-32 2-85 10 1-12 1-17 1-25 1-34 1-47 1-63 1-86 2-18 2-65 3-46 ' 11 M3 L-19 1-28 1-39 1-54 1-74 2-02 2-44 3-12 4-38 12 M4 ] L'21 1-31 1-43 1-61 1-85 2-22 2-56 3-74 5-88 i 13 M5 1-23 1-34 1-49 1-69 1-98 2-42 3-16 4-63 8-69 14 1-17 : L-25 1-37 1-54 1-77 2-12 2-68 3-69 602 16-4 ^15 1-18 : 1-28 1-41 1-60 1-87 2-29 2-99 4-40 8-46 172, 1 §16 M9 ■ 1-30 1-45 1-66 1-97 2-47 3-38 5-43 14-4 o 17 1-21 1-32 1-49 1-72 2-09 2-69 3-87 704 39-0 U 18 1-22 1-35 1-53 1-79 2-21 2-95 4-51 9-86 g.19 1-24 L-37 1-57 1-87 2-36 3-25 5-39 16-4 1^0 1-25 1-40 1-62 1-95 2-51 3-62 6-65 47-4 0) f^21 1-27 1-43 1-66 2-04 2-69 407 8-63 -§22 1-28 ■ L-45 1-71 2-13 2-90 4-65 12-3 §23 1-30 1-48 1-77 2-24 3-13 5-40 20-9 (§24 1-31 1-51 1-82 2-35 3-40 6-41 65-9 Si 26 1-33 1-54 1-88 2-47 3-73 7-87 1-35 1-57 1-94 2-61 4-10 10-1 §27 1-37 1-61 2-00 2-76 4-57 13-8 ^28 ^29 1-38 1-64 2-08 2-93 5-15 22-9 1-40 1-68 2-15 311 5-84 57-3 30 1-42 1-72 2-23 3-32 6-79 31 1-44 1-75 2-32 3-56 8-04 05 32 1-46 1-79 2-41 3-83 9-86 2 33 1-48 1-84 2-51 4-14 12-7 3q34 1-50 L-88 2-61 4-50 17-3 ■■^35 1-52 L-92 2-73 4-93 28-7 ^ 36 1-54 1-97 2-83 5-43 68-7 Table of Eulerian Secants 1 §)37 g38 1-56 1-59 2-02 2-07 2-98 313 6-02 6-80 sec (^V4) ^39 ^40 1-61 2-13 3-29 7-73 E = 30,000,000. 1-63 2-18 3-46 9-07 41 1 1-66 2-24 3-66 10-8 7 42 ' 1-68 2-31 ■3-87 13-3 jVIultipliers of , „' 1 43 1-71 2-37 4-11 17-2 fC- 44 1-74 2-44 4-38 24-6 in Expression for 45 1-76 2-51 4-68 430 Maximum Stress. 46 1-79 2-59 503 172 47 1-82 2-67 5-42 48 1-85 2-76 5-88 49 1-88 2-85 6-42 50 1-91 2-95 7-06 COLUMNS, STANCHIONS AND STRUTS 309 Numerical Examples. — (1) Take the same case as dealt with on p. 304 for the load atB. We found load = 77' 7 tons = - — .^ ' lbs. per sq. in. = 4,300 lbs. per sq. in. nearly . • . sec = 1*03 about The effect of this upon the result is negligible. (2) Find the stress produced in the column of question (4), p. 300, if the load is 1'' out of centre, in the weak direction, ^ P 40x2,240 ip^cAnii. Here -r = — ^ ^ ' — = 10,800 lbs. per sq. m. A 8-296 ^ ^ c = 93*6 .*. sec = 1*6 approx. (from table) 1 X 1-6 X 5-75^ .-. stress = 10,800 (1 + 3.332 = 10,800 (1-83) = 19,800 lbs. per sq. in. nearly = 8'8 tons per sq. in. The approximate method would have given stress = 10,800 (l + ^^^3!^) = 10,800 x 1-52 = 16,400 lbs. per sq. in. nearly = 7'3 tons per sq. in. Johnson's Formula for Eccentric Loading. — This formula, due to Professor Johnson, is obtained by adding the additional eccentricity due to the deflection and is maximum stress in column = x "^ WT~2 (^) P P 1<^^__ A . ,,_PL2FA ^^^ 10 EPA Pfl+-- '^' I ^ V^ 10 E A k^ ■"• lOE A I kHl- ^ ^' (9) 310 THE STRENGTH OF MATERIALS A somewhat more correct but similar formula can be ob- tained by regarding the bending moment as uniform; this gives a deflection (see p. 264) = ^ = |^' = WE^A^B f, e c "SE" . effective eccentricity = 8 + e-=e(l+ '-^") V 8 E/ . I 1- ^ P(8 + e) d. . . bendmg stress = ^ .!. ' AF V" ' 8E - j^ (approx.) af(i-/^-; /, d. e ^ ^^ 8 E Total stress = direct stress + bending stress =: f t f,^d,e ^ ^^ 8E b{i /''^^J (1^) 8E, Professor Morley * obtained the same result by the expansion , . 0'\ 5e\ 610'' sec^ = l + 2,+^ + -gj-+ ... Taking first the two terms as an approximation sec ^ . / ' ' ^ I + ^ L 2 \E 8 E .*. Equation (7) becomes Total stress = /., |l + ^^' (l + g'-^)j * Theory of Structures (Longmans). CHAPTER XI TORSION AND TWISTING OF SHAFTS We have seen that in a beam the bending moment is resisted by a complex series of stresses in tension and compression which vary in intensity at different points in the depth of the beam. In the case of shafts we have twisting moment in place of the bending moment and the stresses are pure shear stresses which vary in intensity at different distances from the centre of the shaft. Stresses in a Shaft Coupling. — -As an introduction to Fig. 138. — Stresses in Coupling Bolts. the subject consider the case of the stresses in the bolts of the flange coupling shown in Fig. 138. Suppose that a twisting moment or torque T is being trans- mitted through the coupling from the shaft A to shaft B. The shaft b has a resistance to its motion which produces a reverse torque numerically equal to T and the effect of these opposite torques upon the coupling is a tendency to shear the bolts. Suppose that the bolts are at the same distance from 311 312 THE STRENGTH OF MATERIALS the centre and are so small that the shear stress over them may be regarded as constant and that they are equal in area and equally stressed. Then S = shearing force on each bolt .' . Taking moments about the axis of the shaft we have Resisting Torque = -^. x r .•.we have T = (1) or if n is the number of bolts T = "i-^"^'-- '■ (2) If the shaft is transmitting a horse-power, H.P., and is rotating at N revolutions per minute, the work done per revolution is 2 tt T so that the work done per minute is 2 tt N T , ^ H.P. X 33,000 „^ „ we have 1 = ^r ^-r ^^- l^^s. H.P. X 33,000 X 12 . ,, = 2^N ""■ "'^ '^) In calculations upon the strength of shafting it is always desirable to work in in. lbs., and taking the unit of 1000 lbs. as a " kip " we can work in in. kips to save writing a number of O's and thus running the risk of error in dealing with large numbers. In the case of the shaft coupling that we have considered it should be pointed out that we have made a great assumption in regarding the bolts as being equally stressed, because if, say, three of them are loose fits and the fourth is a good fit, the fourth one will carry all the load ; the same point holds with ordinary riveted joints. In practice, however, a number of bolts are always used and each is always regarded as carrying TORSION AND TWISTING OF SHAFTS 313 its proportion of the load, and the best way to meet the difficulty seems to be to have the workmanship as good as possible. General Case of Torque on Groups of Bolts or Rivets. — Suppose that we have any number of bolts or rivets of different areas and at different radii from the axis o about which a twisting action may be considered as taking place ; in Fig. 139 we have shown three such bolts. Then if we imagine a slight rotational movement of one part of the joint or coupling about the point o relatively to the other it will be Fig. 139. seen that the movements of the centres of the bolts will be proportional to their radii r^, r^, r^, etc., and therefore the strain and consequently the stress on any bolt is proportional to its distance from the a:»is. Let ' Couplings axd Joints. — (1) Find the diameter of holts necessary in a coupling which transmits 120 H.P. at 75 revolutions per minute. The diameter of the circle of the holt centres is lOJ inches (i. e. r = o"25 inches) and the coupling has 6 holts. The stress allowed is 2 J tons per sq. in. In this case from equation (3) 120 X 33,000 X 12 1 = — c^ 1-r^ iri • lbs. 2 IT X 75 = 100,000 in. lbs. nearly. Also from equation (2) T = '. X o'zi^ m. tons 4 6x2-5x7rX6Z2x2,240 ^ ^^ . „ = — — . x5-25m.lbs. 4 = 139,000 d~ in. lbs. nearly. , _ 100,000 •'• ~ 139,000 , /i6o,oo() ^^ . , ^ = Vr3970'00 = '^^^^^-^^"^^^^'' . • . Ado pt bolts Y ^^ diameter. (2) ExA3iPLE OF Cleat. — We ivill now take the case shown in Fig. 140 of the cleat given in the Handbook of Messrs. Dorman, Long d; Co., Ltd., for a 16 in. hy 6 in. standard I beam with a minimum span of 18 ft., the rivets being of f in. diameter. The safe uniformly distributed load given for this span and TORSION AND TWISTING OF SHAFTS 315 beam is 25 tons, so that the reaction at each end will be 25 ^ = 12-5 tons, and half of this will be carried by each angle. or the load P will be 6-25 tons. <^--4-> - 2i'-^Z^ McVi 625 Fig. 140. First find the position of the centre of gravity of the rivets. It is clearly on the horizontal line through the rivet 3, and its distance from the line 1, 3, 5 is obtained by moments thus — 5 ^ = 2 X 21 ^ 4-5 ^ . 1. e. a = -zr = 9 m. 5 316 THE STRENGTH OF MATERIALS Then we tabulate the dimensions as follows — No. of Rivet. r r2 1 2 3 4 5 4-58 2-62 -90 2-62 4-58 21-06 6-88 -81 6-88 21-06 2 7-2 = 56-69 Pa; _ 6^25 X 3^5 •*• * ~ 2>2 - 56^69 = -348 ton. The moment load will be a maximum on rivets 1 and 5 because they are farthest from x, and will be equal to Tg = -348 X 4-58 = 1-59 tons. The direct load W on these rivets 6-25 = 1-25 tons. Therefore resultant load = R5 = 2*20 tons. [See Fig. 140.] Now bearing area of a |-in. rivet in a |-in. plate 9 _3 3 ~4 ^ 8 32 sq. m. 2-20 X 32 Bearing stress on rivet = ^ = 7'82 tons per gq. in. Area of a |-in. rivet in section ^ -r x \~^] = '442 .4 2-20 . • . Shear stress on rivet = ^^rj^ = 4-98 tons per sq. in. The above calculation shows that the rivets are stressed just about up to what is commonly taken as a safe working stress for rivets in shear, viz. 5 tons per sq. in. The importance of allowing for the eccentricity of the stress wiU be clear from this example, because the resultant maximum stress on the rivets comes to nearly twice the value which would have been found if the eccentricity had not been taken into account. Torsion of a Circular Shaft. — Suppose that a circular shaft of length I and diameter d is subjected to a twisting TORSION AND TWISTING OF SHAFTS 317 moment or torque T (Fig. 141). To preserve the equilibrium of the shaft, equal and opposite torques must act at the two ends and each normal section of the shaft will be subjected to strains which will allow a slight twisting motion of such section without causing it to bend or warp out of its plane. We will therefore make the following assumptions in developing our theory of torsion — (1) That plane normal sections of the shaft remain plane after twisting. (2) That stress is proportional to strain, i. e. all the stresses are within the elastic limit. A line a b on the circumference of the shaft initially parallel to the axis becomes bent to the form A b' as a result of the Fig. 141. — Torsion. ajoplication of the torque, the end a being regarded for convenience as fixed. Then B b' may be called the arc of torsion and it subtends at the centre o an angle called the angle of torsion. This angle of torsion may be regarded as a torsional deflection. If we imagine the shaft divided up into a number of very small equal slices it follows that since each slice is exactly like every other slice the angle of twist in each sHce wiU be equal, and if we regard each slice for convenience as of unit length we have Angle of twist per unit length of shaft = I Now consider the slice contained between sections x x and Y Y, Fig. 142, the thickness x being regarded as very small, and consider a square abed of side x at distance r from the 318 THE STRENGTH OF :\L\TERIALS centre. In our figure 6 c is appreciably curved, but that is onlv because we cannot draw the fio^ure clearly without making X of appreciable size. The result of the twisting action is to make abed take up the form a h' c' d, this being the typical form (cf. Fig. 1) indi- cating pure shear strain, and the initial shear strain /? is given by the relation X Y -^ JU^ X -^ Y Fig. U2.— Torsion of Shafts. Shear stress — shear modulus x unital shear strain = y3G X G (1) Xow the angle hob' — angle of twist in length x = angle of twist per unit length x x _6x_ I and bb' = ar c = radius x angle = — ^ rOx .G xl rOG I . • . In (1) shear stress (2) TORSION AND TWISTING OF SHAFTS 319 This gives the important result that : The shear stress at any point in a shaft is proportional to the distance of that point from the centre. The shear stress is therefore the same at all points on circles concentric with the shaft and the variation of shear stress is indicated by the triangle o eg, the stress at the extreme fibre being s and that at any other radius r being equal to , _s r ^ ^ R Now consider a very small element of area a at a point p in the section at distance r from the centre, the area being so small that the stress over it is constant. r OG Then from (2) stress on element = — v- • . Force on element = S = stress x area rOG = -r- • " . * . Moment about o of force on element = S X r rOG = -^~.a.r G „ = -r ■"■' . ' . Total moment about the axis of all the forces on the section = Sum of separate moments = > — , - . ar^ 6 G sr\ = —-.— ^ ar^ because 0, G, and I are constant P = —j~ X Polar moment of Inertia of section ^GI,. I But the total moment about the axis of all the forces on the section must be equal to the twisting moment, so that we get T = -^4^" (3) 320 THE STRENGTH OF :NL\TERL.\LS From (2) we get ^^^^i = t^ and from (3) we get — = ^ ^ I, I By combining these results we obtain the following com- plete relation for torsion which should be compared with the corresponding relation on p. 249 for the bending of beams. stress T ^G ^^ = ip=i~ <■*> In practical calculations we are usually concerned with the maximum shear stress -s 5 T .-. we write t? — y T R ... J (o) By analogy with the method of dealing with the section modulus of a beam we call -^ the jjoJar modulus Z . T We thus get -s = ^ otT = f^Z (6) Gases in which the Formulae are Applicable. — These formulae are based upon the assumption that plane sections remain plane after twisting and this is true only for circular sections (soUd and hollow) ; for shafts of other section the approximate formulae given on p. 333 may be used. Solid Ciectxab Sectio>'. — This is of course by far the most common case of shafting which occiu^, and in this case . Z = = -196.5 03 (7 32 77 D^ _ D _ - D3 32 2 ~ 16 ■' 16 TORSION AND TWISTING OF SHAFTS 321 Hollow Circular Sectiox. — In this case, Fig. 143 {a), Z, = i. e.T = 16 Di 16 Di (8) When the metal is Yerj thin, Fig. 143 (6), we have Fig. 143. I^ = I D* _ (D _ ^j4 j and if t is so smaU that squares and higher powers of jt may be neglected we may -^Tite (D - ^)4 = D* - 4D3^ _ - BH I^, - D2 f *• e- Z,, = J) 2 5 7rD2i (9) Alternative Derivation of Formula for Solid Shaft. — The formulae (7) can also be derived as follows, and although we recommend the previous method as the more satisfactory 322 THE STRENGTH OF MATERIALS "we find by experience that some students find the alternative method more easy to follow. It is similar to the method wliich we have ahead}' used for beams in some cases (see pp. 220-222). Consider a very small sector a o b, Fig. 144, of the circle, so small that we may consider a o b as being practically a very narrow triangle. Set up a D, B c, to represent the maximum shear stress s and complete the pjTamid a b c d o as sho\^Tti. Then if this pyramid be considered as di\'ided up into a number of slices as indicated, the volume of each shce = area of piece of sector x stress on it = load on each Fig. 144. piece of sector; the volume of pyramid therefore re- jDresents the whole load acting on the sector, and to find the moment about the point o of the force on the sector we regard the volume of the pjTamid as acting at the centre of gravity of the p^Tamid which is at distance --. - from o. Now the volume of a pyramid = J area of base x height ab = g Ra A D X O A 1 R 5aR2 TORSION AND TWISTING OF SHAFTS 323 3R Moment about o = Volume x 4 3 4 5 a R^ 2 Now in the whole section there will be — of these sectors, a • because the whole circumference subtends an angle 2 tt at the centre. .*. Total moment about o of all the forces on the 5aR3 2 7r section = — -. — X — 4 a ~ 2 ~ 16 But this total moment must be equal to the torque T . rp _ SttW ' 16 This agrees with our previous result (equation (7)). Horse-Power Transmitted by Shafting. We have seen on p. 312 that ^ H.P. X 33,000 X 12 . „ T = ^^ m. lbs. H.P. X 33,000 X 12 . ^ = — c. — TVT — cti^at. m. tons. 2 TT N X 2,240 . • , putting this into equation (7) we have sjjrW _ H.P. X 33,000 X 12 16 ~ 2 TT N H.P. X 33,000 X 12 X 16 D3 = 27r2N 4 \ 4 / 4 7rD2 and of the solid shaft it will be — - — ; so that the area, and 4 therefore the weight per given length of the hollow shaft, is f of the corresponding value for the solid shaft. Summing up our results, therefore, we may say that " the hollow shaft has a weight of J that of the solid shaft and transmits yf of the horse-power; so that weight for weight 15 4 the hollow shaft will be .^ x ^ , ^. e. li times as efficient as lb o the solid shaft; the angle of torsion or torsional deflection will, however, be greater for the hollow shaft. This illustra- tion brings out the fact that we can easily see from a considera- tion of the stress diagram that the material at the centre of a shaft is not used so effectively as that at the outside. This result agrees with that which we found for beams (p. 201). Numerical Examples on Circular Shafting. — (1) Find diameter of wrought-iron shaft to transmit 90 H. P. at 130 revolu- tions per minute if the working stress is to be 5000 lbs. per sq. in. In this case H.P. = 90, N = 130, and s = 5000 .'. Working from the general formulae, which are much easier to remember than the particular one, we have T -- 5 X TT D3 5000 TT D3 16 16 also T = 90 X 33,000 X 12 2 TT X 130 • F)3 - 16 X 90 X 33,000 x 12 5000 TT X 2 TT x 130 = 44*45 ins.^ .-. D = = ^44-45 = 3-54 ins. Adopt 3J ins. diameter. 326 THE STRENGTH OF MATERIALS (2) A steel shaft 4 in. in diameter is running at 130 revolutions per minute and is found to have a twist or " spring " of 9 degrees measured upon a length of 30 feet. What horse-power is being transmitted, taking G = 12 x 10^ lbs. per sq. in., and what is the maximum stress in the shaft ? Our general formula is js _T_ _ GO R l,~~ I In our case we are given the following — K-Z, 1, _ g^- - -32- -^-^ ^ = ^° = w '^^^^^' = ^ * I = 30 ft. = 360 ins. ; G = 12 x 10^ lbs. per sq. in. ^ G ^ L 12 X 106 X TT ^ 2 7r2 X 105 . „ .'. I = — j—^ = — „„^ ^^ — X Stt = — ^z, in. lbs. / 360 X 20 15 , ^ ^ H.P. X 33,000 X 12 but T = H.P. = 27rN 2 TT X N X 2 TT^ X 105 33,000 X 12 X 15 4 TT^ X 130 X 102 33 X 12 X 15 = 271 To test whether the stress is within safe limits we write G<9R s = I _ 12 X 10« ^ 360 ^ 20 ^ = 10,500 lbs. per sq. in. nearly. (3) What diameter of hollow shaft would you use to transmit 5000 H.F. at 60 revolutions per minute if the maximum torque is IJ times the mean and the safe shear stress is 7,500 lbs. per sq. in. Take the internal diameter as half the external. TORSION AND TWISTING OF SHAFTS 327 J ... . H.P. X 33,000 X 12 . „ In this case mean torque = ^ — -^^ m. lbs. ^ 27rN 5000 X 33,000 X 12 , „ = iy ^7,^ m. lbs. 2 7r X 60 Tvr ^ + T, 1-5 X 5000 x 33,000 x 12 . ,^ Max. torque = T = ^ ^ m.lbs T>, . rp _ ^ '^ r' " V 2) J 7,500 TT 15 D3 ^''^^- 16D =T6~'~16~ r)3 _ 1;^ X 5000 X 33,000 x 12 x 256 2 TT X 60 X 7,500 TT X 15 This gives D = 17'9 inches. Adopt 18 inches external diameter . * Combined Bending and Torsion. — In a large number of cases in practice shafts are subjected to bending as well as torsional stresses; a common example occurs in the case of a crank shaft and also in the bending stresses caused by the weight of the shafts themselves or by pulleys carried between the points of support. Fig. 145 illustrates the action in an overhung crank shaft The driving force p is applied at the point a at the crank pin and causes a twisting moment equal to p . a b = p r about the axis B c. In the vertical plane we have the couple formed of p at A and p at b in an opposite direction ; to preserve equilibrium at B an equal and opposite force p must act which in the horizontal plane of B c combines with the reactionary force p at c. B c may therefore be regarded as a cantilever subjected to a bending moment equal to p x b c = p /. It is usual to assume the cantilever as extending to the centre of the bearing for the purpose of calculating the bending moment ; it is difficult to obtain a much more accurate result until we know the distribution of the pressures upon the bear- ing. Our procedure for the calculation of the combined stresses is as follows — First find the maximum bending moment M and the twisting moment T acting at any point along the length of 328 THE STRENGTH OF MATERIALS the shaft and calculate the corresponding maximum tensile, compressive and shear forces contributed by the bending moment and twisting moment respectively. If the shear stress contributed from the consideration of the shaft as a beam is at all appreciable we should add this stress to the maximum shear stress given by the torsion. Let / and s be the maximum bending and shear stresses, then, as explained on p. 44, we have three alternative formulae to apply, one of which gives the resultant shear stress and the other two the resultant tensile stress. Th( (1) Fig. 3y are Rankine's formula — Equivalent tensile stress St. Venant's formula — Equivalent tensile stress 145. 2^4 (2) 4s2\ + ,2 (3) Guest's formula — Equivalent shear stress 1 + = s ^l + 4^2 TORSION AND TWISTING OF SHAFTS 329 Equivalent Bending and Twisting Moments. — It is common to express these formulae in terms of equivalent bending or twisting moments. Now the polar modulus of a solid or hollow circular section is twice the ordinary or bending modulus of the same section. For the solid circle Z,, = -.,-— and Z = ^^ lb 3Z .'.we shall have _M _ T _ T ^ ~ Z' ^ ~Z^ ~2Z s T • • / ~ 2 M . • . taking Rankine's formula we have, if Mj, is the equivalent B.M. M, _ M /^ , /^ ^ T2^ IP. Equivalent tensile stress = ^^ = -^ (l +^/l + .■.M. = |(l + ^1+|;) (4a) '' 2 |(m + VM^ + T^) (46) St. Venant's formula would give or = g- + g VW^pT^ (56) • Guest's formula will give T M /~ T2 Equivalent shear stress = -^ = ^-^ /i j_ M / T2 or = VMM^T2 (66) For ductile materials, such as steel, formulae (6) are recom- mended for use in design, the safe shear stress being used for 330 THE STRENGTH OF MATERIALS determining the necessary value of the polar modulus Z^„ to carry the twisting moment. For brittle materials, such as cast iron, formulae (4) or (5) should be used, the St. Venant formula being recommended as the more reliable of the two ; the safe tensile stress should be taken in this case. There has been considerable confusion with these formulae because Rankine's formula is often given as an equivalent twisting moment and has been compared with Guest's formula for the same working stress; the point to keep in mind is that if Rankine's formula is used the safe tensile stress and bending modulus should be taken, but if Guest's formula is adopted the safe shear stress and polar modulus shoukl be used. Numerical Example. — What must be the diameter of a solid shaft to transmit a twisting moment of 160 ft. tons and a bending moment of 40 ft. tons, the tensile stress being limited to 4 tons per sq. in. ? What diameter would you use if the shear stress is limited to 3 tons per sq. in. Let D be the diameter of the shaft. Then s T ~7rD3 16 16 T / IVI 7rD3 32 M -U' 32 4 '^ 4 3 • • 4 4s^ + /2 ^ ^/l + 4 X (1()T)2 (32M)2 Vi + ^1 + \40/ = •75 + 4VI7 = 5-9 TORSION AND TWISTING OF SHAFTS 331 ,5-9/ 5-9 32 X 40 X 12 . • . Max. stress = 4 = —7,-^ = -^^ x tv^ 2 2 ttD'* -r.„ 5-9 X 32 X 40 X 12 ^ ^.„ , D^ = 5 = 3,600 nearly, O TT D — 15*3 inches about. /160\^2 Equivalent shear stress = <5 ( a/ 1+V40 _ 16 X 40 X 12 X 4-12 16 X 40 X 12 X 4-12 D3 = 37r D = 15 inches about. Shafts with Axial Pull or Thrust. — If in adition to the torsional stresses (and bending stresses if they occur) there P Q exists an axial pull (P) or thrust Q we add -. or . to the bend- ing stress to get the value of / to use in the formulae. In the case of end thrust acting, when the direct stress is taken as a criterion, the resultant direct stress should not exceed the safe stress upon the shaft considered as a column, and the design should be treated similarly to that of a column with an eccentric load (p. 301). In the case of steel shafts, which are most common, we suggest that a double test should be applied ; the shaft should be designed to carry the equivalent direct stress as a compressive stress on a column and also by the Guest formula, the diameter chosen being the greater of the two results. Torsional Resilience. — As we have already explained in connection with resilience in bending, the work done bj^ a couple is equal to the product of the couple into the angle turned through. If, therefore, the angle turned through is 0, the work done T6 by the couple which increases gradually from to T is -^ • R" I ~l, 332 THE STRENGTH OF MATERIALS . • . Work stored in shaft = ^ = c^' i^ • ?>. -r> = ~ '' ta 2 21, .sH (1 G D2 For a solid shaft -r^'l = , ^ D- lb the volume of the shaft = —j— .1 = Y 4 . 2jy_v • • D2 4 i.e. Work stored in solid shaft = -7 ^^ • V 4G 52 . • . Resilience = work stored in unit volume = -r-^ .... (2) 4G ^ ' 2 If we take G = E (cf . p. 12) this gives o 5^2 Resilience = -^^^- (3) For a hollow shaft of external diameter D and internal diameter D^ 2I,Z _ TT (D4 - D,4) . I _ ^ (D2 4- D^2) ( D2 _ 1) ^2) Z D2 ~ 16 D2 ~ 16 D2 and V = '^(5i^-»A^ 4 2 I, Z /D2 + Di2\ V • • D2 ^ p2 y • 4 .-. Resilience = ^^^ |l + (^J^^^ (4) In the limiting case of a very thin tube j} approaches 1 and we have ^2 Resilience = -^-p (5) Torsion of Non-Circular Shafts.— As we have akeady indicated on p. 320, the ordinary theory of torsion is true only for circular sections because in other sections the sections TORSION AND TWISTING OF SHAFTS 333 originally plane become bent out of the plane upon twisting. The following cases have been worked out fully by St. Venant, who has giVen the following approximate formulae — Section. Relation of Maximum Stress to Twisting Moment. T T = 5 . -208 S3 T~ H B Any symmetrical sec- tion not containing re- entrant angles. T = sBH2 3 + 1- H B T 5 A* 40 \, . y Angles of Torsion in Degrees. _292TZ((^2^D2) GD3# 410 TZ 205 TJ (B2+H2) G B3"H3 40I,,TZ A^G Where A = area of section ly = j)olar moment of Inertia y = distance of farthest edge from centre of shaft Numerical Example. — A square steel shaft is required for transmitting power to a SO-ton overhead travelling crane. The load is lifted at a rate of 40 ft. per minute. Taking the mechanical 334 THE STRENGTH OF IMATERIALS efficiency of the crane gearing as 35 %, calculate the necessary size of shaft to run at 160 revolutions per minute. The twist must not exceed 1° in a length equal to 30 times the side of the square. Take G = 13 x 10^ lbs. per sq. in. Work per minute to lift weight = 30 x 40 ft. tons. If rj of crane = 35 %. w w V. r 1 • . 30 X 40 X 100 Work to be supplied per minute = ^ .'.If revolutions per minute = 160 r^ 30 X 40 X 100 ,. , ^ Torque = ,^ ^ rrrAft. tons = T ^ 3t) X 2 TT X 160 I = 3-42 ft. tons. 410 X 3-42 X 2,240 x 12 x 30 S 13 X 10« . S* ^.3 _ 410 X 3-42 X 2,240 x 12 x 30 13 X 10« S = 4J inches (sa}^). Fig. 146. Effect of Keyways upon the Torsion of a Circular Shaft. — Professor H. F. ^Nloore of Illinois University has given the following results of experimental investigations upon shafts with kej'wa}^ grooves cut in them. Strength of cut shaft _ '26 lid Strength of uncut shaft D D Angle of torsion of cut shaft _ , * 4 6 , '7 d Angle of torsion of uncut shaft D D TORSION AND TWISTING OF SHAFTS 335 oooooooooooooooooooooooooooo ^ 'y^ r-. in '— 1 iri ■ — , »n ■ — j in < — i i_o <— j ifT> ri O C^ '^ 'O -* 'O :/s ^ •>< -t" cc oj cp '>4 -^ ,— (.— lF-(,— ((7s|(M(?-]CNC0C0CQC0'*'*'*-^OO>-HI>.(Ma0'^O»0OOCN ^i-H,— li— li— li— (1— li— It— lr-l(M(MC^CMCOCOCOCCrJoc5co'.i>cieoocicoO'*cO'— iCiooo-^co— IC503GO ■— l 10000»OOOOOOOOOOOOCOOCOOCOO(X)OCO»0000 C^lOl:^<:pC^Ol>-04(»0 I— I— if-(:5-^^CO'0 ■^lO-0;Ot--C»C;0-^'?^*Ot'COO'-HCO-*lOt- 1— (I— ii— — if-HGv|G<|! o X ^ — ^ 5^4 5^1 ■ri CO CO CO ^ -t". -r to O O O --C t- t^ X C; O O ^ Ol CO -^ -i< o > o eg © o o ><] 03 53 o .S3 o o p-t fH p-^o-iX'^p'poiX'+o'-p'rixriHppooooooooooo ooc;'^o-irt'00 CO -^ o o p r- X o CI o -H -H CI CO 't* '^< .-^ X o oi X o oi c; --S o) c; 'O «b O t» X C5 O f-i Ol M O "i l'- X Ct O --H -^ i C: O J '^ l'- O Ol O X r— I 1— I i-H I— I .— I ^H .^ r-H I— 1 VI 71 0< Ol Ol CO CO CO -T* 't -tl -t O O M'^'rib'ixo"— iM oooooooooooooooo»oooo»oo»oooooo »0Ot-XC3O'— l . • . the load W to cause a deflection 8 is given by W=7rf|3-.8 (5a) 64 R^ ?i ^ Grl^ = 8T5^-^ (3^) We might have obtained our result as follows from the con- sideration of the torsional resilience (p. 331) — Torsional resilience = -. ^ 4G . • . Work absorbed = ^-p x volume = Work done by weight _ W8 2 . _ 16 W R ^ _ Trd^l ^ ~ Tvd^ ' ^ ~ 4 W8 _ 162 w^ R2 . TT (^2 / •'•"2 " TT^d^ 4G.4 _ 16 W^ R^ I ~ TT # G , 32WR2? 6 = 7>ivm— as beiore. Time of Vibration. — Putting 8 = 1 in equation (5) we get the force to cause unit deflection ; this can then be put into equation (1) to find the time of swing, remembering that if all the other dimensions are in inch units, g (the gravity acceleration) must be also reduced to inch units. 340 THE STRENGTH OF MATERIALS Numerical Examples on Closely Coiled Springs. — (1) A closely wound helical spring is formed of 30 coils of |- in. round wire, the mean diameter of the coil being 4 ins. What axial load will produce a shear stress of 9 tons per sq. in., and if G is 4,700 tons per sq. in., wlmt will he the extension of the spring under the load ? s y ird^ T = Wx2= ,, 1(3 1\3 ^^^ ~ 16 9 TT W = Wo a A tons 32 X 64 lbs. _ 9 7r X 2,240 " 32 X 64 = 30-9 lbs. _ 64 \V W n ^ = ' Gd^ 64 X 30-9 X 30 X 8 X 256 , , _ . = 4,700-x 2,240 ^ \}21L2111 Before this extension occurred the spring would have ceased to be closely wound if the load were such as to stretch the spring, and the spring would have been fully closed if the load were such as to compress the spring. (2) // a closely wound helical spring made of wire J in. in diameter has 10 coils, each 4 ins. mean diameter, find the fre- quency of the free vibrations when it carries a load of 15 lbs. {taking G = 12 x 10® lbs. per sq. in.). \ ^ X Force to cause unit displacement Putting 8 = 1 in equation {oh) we have Force to cause unit displacement =jjpv3 (i)* X 12 X 106 Qi^^i, 8 X 43 X 10 Frequency = / — 9 - ^ /, seconds " \ 32 X 12 X 9-156 1 _ 1 / 32 x~12 X 9156 t 27r\ "15 = 2 43 per second. SPRINGS 341 It should be noted that in the above calculation we have neglected the weight of the spring itself, so that the result can only be regarded as approximate. Weight of Springs. — A cubic inch of steel weighs about •284 lb. = -284 X .'. Weight of spring = "284 x volume 7V_dH 4 TV d? irT> n 4 = -284 X = '^ d'^T> n very nearly. Safe Loads on Circular Springs. — The following are the highest safe torsion stresses upon steel spring wire found by experiments by Mr. Wilson Hartnell — Diameter of Wire. 1 2 Safe stress in lbs. per sq. inch. Safe Load on Spring (lbs.) 70,000 60,000 50,000 429 D 1,240 D 2,450 D D = mean diameter of spring in inches. we have W From the formula T 2 X '\^Qs(P WD 2 D from which the values in the third column of the table are obtained. Restriction on Use of FoRMUL.g]:. — When ine springs are used so that when stressed they are shorter than when unloaded {i. e. if the load were to act upwards in Fig. 147), it should be remembered that the spring may become shut before the safe load has been reached ; this does not diminish its strength, but impairs its value as a spring. 342 THE STRENGTH OF IVIATERIALS Alignment Charts for Close-coiled Circular Springs. DIAMETER OF SPRING \ «A ■■ ■ ll ■ I I I' M l I . I . I1 . .I I ■ ■ ^ - I ■ I I t ■ I ►. — -; ;C 00 -o / \ ALLOWABLE WORKING STRESS 5» > Sj5 / Jo / ?" / / r o p o c S 5 5 — i I i i i DIAMETER OF WIRE l -" l- ■■■ I ' ' ' / ' ' ' L ' — T / I I t I LOAD ON SPRING POUNDS § I MIII .1 I . Lhl 'l ■ I I || m/ ||[|| | I I . I h ll. M I . ■ |.|i|i I ' I ' |l|.'|ii . 'l ■ 5^0 0000000 o o 00 — CO <" J-O COOQOOOO O w Fig. 148. — Figs. 148, 149 show alignment charts * for the strength and * For an explanation of the principle of these charts see a booklet by the author on Alignment Charts, published by Messrs. Chapman and Hall, Ltd. [Price 1/3 net.] SPRINGS 343 deflection of round wire helical springs, taken from an article by Mr. F. Fitchett in Machinery for January 14, 1915. o ooooooo P ' 1 1 r I I I I I 1 1 1 1 I I I 1 1 1 1 1 1 1 1 1 I ' I X. §$gg s ss s g l " ' >' I LOAD ON SPRING. POUNDS DEFLECTION pF SPRIN9. INCHES .|... I ■ t ,\.\.i. V t ■ I ■ I.I .1.1 ^ V« 9> O QC tC w ♦ wo>^« ^ OB « — — — —~7i'^ o — >o ■■^■^i^en ai 00 P !^ ■'^ -* / r i'i'i ' " I """ "- '^ oc *J ^ tn ^ oo o o o o o NUMBER OF COILS iii|ii I I W iiiii I I 1 I I I 1 t I -r (C off ^J * (ji ^ I I FiG.^ 149. To use the charts we proceed as follows : Suppose, for instance, mean diameter of coil = 3 in., diameter wire = f in., 344 THE STRENGTH OF MATERIALS safe stress = 60,000 lbs. per sq. in. On Fig. 148 connect " diameter of spring " to " safe working stress '' and note the intersection on the dotted axis. Connect this point of axis through " diameter of wire " on to " safe load," which will give the result required, viz. 415 lbs. approximately. The chart can be used similarly to find anj^ one of the four quantities if the other three are given. To use the deflection chart of Fig. 149 take, for example, a load of 16 lbs., J in. diameter wire, 2i in. diameter of spring consisting of 20 coils, using G = 10,000,000 lbs. per sq. in. Connect " number of coils " to " coefficient of rigidity '' and note intersection on the right-hand intersection or " support line. Then join " diameter of wire " to " diameter of spring " and note intersection on centre support. Join these two intersections to meet the right-hand support and connect the point thus obtained to " load on spring '' and produce to the '' deflection " which gives 1 inch approx. Springs of Square Section Wire. — If our spring is like that shown in Fig. 147 with the exception that the wire is square instead of round in section, the length of each side being S, Ave can proceed as follows — 410 T / From p. 333 = r^(^^ (degrees) 711T/ , ,. - = G S* (^^^^1^^"^) andT = s -208 S3 WD Now T = W R = ^ „, s X -208 S3 -416 5 S3 .,. ••• ^^ = D" =-^- ^^^ 2 711R X WR; G S* GS^ 1-78 WD2/ "GS* •• ^^^ 8 = R^ = puttii ig I 8 SPRINGS r787rWD3 n GS* 5-6WD3 7i GS* 345 (3) Taking the same safe stresses as for round wire we get from formula (1) the following formulae for the safe loads on square wire springs — Side of Square (in.) Safe Load in lbs. 455 D 1316 D 2660 I) Comparison of Square Section and Circular Section Springs. — The volume of the square spring = S^ Z = V T B . Work absorbed = -^ 208 s S 3 7;^! IT I GS*"^^" 7-11 X -208 5 S3 Z X 2 •208_5_S3 2"~" G" 154 ^2 GS^ 154 Resilience = - —^ — G 154 <§2 G This is less than for a solid circular spring, the ratio of 25 circular to square resilience being 7 — ^ = 1 • 62 . Also for a spring of given diameter the load carried is greater for a circular section than for a square section of the same area. d T. 1128 • t; 4 X -208 346 THE STRENGTH OF ^L\TERIALS For the circular section we have rj. s X ird^ 8 d for square T = 208 s S^ = 208 5 S x area T _ d ■'• T. ~ 4 X -208 8 K — - = S-, /. e. if the areas are equal as suggested 4 1128 S = 1-36. Weight for iveight, therefore, a closely coiled circular section helical spring of given diameter is 136 times as strong and will absorb 162 tirnes as much energy as one of square section of the same diameter. Open-Coiled Helical Springs. — In the case of open- coiled helical springs the stress is principally a torsioned one, so that we wiU deal with it here although there is also bending stress. Referring to Fig. 150. let the centre line of the spring at any point be inclined at an angle a to the horizontal ; then the normal section plane x x of the wire will be at an angle a to the vertical load W. The load W has a moment W R about the line centre o of the section and this moment has a component o « = W R cos a which is a moment in the plane X X and causes a t^visting action, and a component a 6 = W R sin a which is a moment normal to the plane x x and causes a bending action. .-. T = WR cosal M = WRsina/ It is not altogether easy to follow this resolution into a twist- ing and a bending moment at first, parth* because it is not very easy to give a very clear diagram with the forces acting in different planes, but a Little consideration of the problem will probably remove the difficulty. SPRINGS 347 The angle of torsion in the plane x x will be given by .'. Work done against torsional stress = -^ = 901 W2 R2 C0S2 a I ' - ~^GX ^ ^ Fig. 150. — Open-coiled Springs. The bending moment is constant, so that (see p. 264) Work done against bending stress _ W2 R2 sin2 a I 2EI 2EI (3) 348 THE STRENGTH OF MATERIALS . • . Total work done against stress W2 R2 I /cos2 a sin2 a\ = 2 Ul, + ElJ (^) But total work done against stress = Work done by weight = 2 •■■^-^^^^Kov + ti") (^) Taking a solid circular section we have 2 I = I^, = . 32 W R2 Z /cos2 a , 2 sin2 a' _ 32 W R2 / (and taking ^ = 2*5 32WR2Z 2 ^ Q • 2 ^ .AN = — /T' -74 (COS^ a + '8 Sni^ a) (6) = ri ^4 (COS'^ a + '8 sni- a) = G.# <1- 2'™'") (') The movement due to tAvisting per unit length of the coil will be equal to R 6.,, where (9,, is the angle of torsion, and takes place in the -plsme x x. This is equivalent to a movement d e vertically and a horizontal movement e /. .-. S,, = R^,, cos a . ..(8) ^ / = R 0, sin a (9) This movement e f tends to increase the number of the turns of wire. The bending deflection upon a unit length will be equal to R ^B, where 6'„ is the angular change in the centre line of the beam. This movement takes place in the plane y y and will tend to unwind the coil ; it has a vertical component g h, which is the part of the deflection contributed by the bend- SPRINGS 349 ing and a horizontal component h j which is an unwinding tendency and opposes the winding-up strain ef. .' . 8„ = R, 0,^ . sin a h j = — R (9„ cos a T Now 0^ = ^ J- per unit length M ^„ = =^j per unit length because the bending moment is constant.* . • . per unit length T n X- o , cs R T cos a , RMsina denection = o^ + \ = ^ t H ^cTt — _ W^R2 cos2^ W R2 sin^ a ~ ~" GI,, ^ EI The deflection will be the same for each unit of length — This agrees with our equation (5) obtained in a different manner. Angular winding-up movement per unit length ■ _ e / — hj ~ R = 6^ sin a — ^,5 cos a ,^_ _, , sin a COS a sin a cos a W R GI„ EI Total winding-up movement = /3 = W R ? sin a cos «■ f p ^ ~ -^ -r 32 W R ? sin a cos a /-^ _ 2 G\ ,^^. 7r#G V E for a solid circular section _ 3 2 WR Z sin a cos a G _ 2 ~ 5 TT # G" ^^ E ~ 5 * See p. 249, and note in Fig. 121 that 9 = ^ = 4 ^f C C is unity .-. e= 1- ^. 350 THE .STRENGTH OF MATEKIALJS This is a maximum for a = 45"^, If }) is the i)itch of the coil and n is the number of turns / __ n^/ p'- - -MV^ Comparison of Close-coiled and Open-Coiled Springs. — Comparing result (7) with ecj^uation (36), p. 33*J, for the close-coiled spring, we have 8 for open-coiled spring , , ^ • o x . .. , ^ . ^ ^ (1 — 2 sm- a) = 8 for close-coiled spring in roi ^^^ 98 ^ ^. N N 96 \ \ \ "OJ \ II \ \ c JO "a B ' \- \ o 2 S : \ 2.0 30 "W O JO e (degrees) Fig. 151. — Correcting Coefificients for Open-coiled Springs. m may then be regarded as a correction coefficient, values of which are given in Fig. 151 for various values of a. Stresses in Wire. — The stresses in the wire can be found by calculating the separate bending and shear stresses and combining them in the manner described to find the equiva- lent simple direct or shear stress. The twisting and bending strains both cause a tendency SPRINGS 351 for the free end of the coil to turn about the vertical axis v v, thus altering the effective number of coils. BENDING SPRINGS Leaf or Plate Springs. — If we consider a leaf or plate spring of the type shown in Fig. 153 and note that the plates are bent to the same radius so that they contact only at their edges, we see that each plate may be regarded as supported at its point of contact with the one below it, the load trans- W mitted at its overhanging end being -^ . (See also Fig. 152.) The B.M. diagram for each plate comes therefore as shown. In order that the spring may close practically fiat, the curvature of each plate must remain constant after bending, i.e. the radius of each plate after bending must be the same. 1 M But from p. 249 ^ = ^^ ^ R EI M . . • . Since E is constant ^ is constant. Between b and b', the B.M. is constant so that the section is constant, but for A B and a' b' the B.M. varies in the triangular manner shown, so that the section must vary so as to keep ^ constant. This can be done by making the ends triangular in plan, the thickness being constant. Then at any point at distance x from a 12 M = ^\^ M _ 6 6 X rf3 •'• I ~ Wa: and ^ = 7, by similar As M 6c?3 . • . T =1X7 7/ = constant I WZ 352 THE STRENGTH OF MATERIALS Fig. 152. — Stresses in Plate Springs. Another way would be to keep the ends square and to vary the thickness as indicated. Then I ^ *jf , M = 'Y M For -p to be constant d/ SPRINGS d^ X X 353 and if the lap were I ^^ — "'•• ~ r contoured so that the relation held, the necessary conditions would be satisfied. Now suppose that there are n plates and that all but the top one are cut longitudinally through the centre and placed as shown in Fig. 152 they would make up the diamond-shaped Fig. 153. — Plate Springs. figure shown. The deflection, i. e. the upward vertical movement from its initial curved position, for such a single jDlate, which bends to a circular one will be very nearly equal MP to 8EI Now I ?i b d^ "12" 3MJ2 ~^2Ew6# A A 354 THE STRENGTH OF MATERIALS For a centrally loaded beam M = —^ ' ' SEnbd^ ^^ In a test of such a spring it will be found that the friction between the plates will cause the deflection to be less than this with an increasing load and to be more as the load is reduced. It is common to test such springs by loading them until the plate is flat; we then have 8 = 8,„ and we get from equation (1) the following value for the test or proof load W^ Stress in Plates. — The stress in the plates will be con- stant along their length because their depth as well as their moment of inertia is constant. •••/ = M d I ^ 2 Wl d . nb d^ ^' 12 3W? ' 2 ~ 2nbd'' ^^^ Derivation of Deflection from Resilience. — The formula for deflection may be derived from the resilience as follows — 42 Resilience (see p. 273) = J^ .*. Total work done in stressing = y!^ x vol. - ^-- Id — "" 6E •^'^- 2 JWS _ f^lndb ^ •*• 2 ~ 12 E dWU Hn'^db ~ In^b^ d^ xl2E ^ 3W ^ P ~ I6nbd^ . E .*. d = c,T->- 1. Ti as beiore. SPRINGS 355 Numerical Example. — A lamiyiated plate spring of 40 inches span has 12 plates, each -375 inch thick and 3*40 inches wide. Calculate the deflection when carrying a central load of 4 tons, taking E = 11,600 tons per sq. in. By formula (2) we have 3WP 8 = S^nbd^ 3 X 4 X 40 X 40 X 40 8 X 11,600 X 12 X 3-40 x -3753 = 3-84 inches. Fig. 154. — Piston Rings. Piston Rings. — Springs in the form of split-rings are placed around pistons in oil, gas and steam engines to prevent escape of the working fluid past the piston, and such rings should be designed so as to give as constant a pressure as possible all round the cylinder. The necessary variation in thickness has been investigated by Professor Robinson in the following manner. Let T>, Fig. 154, be the point of maximum thickness to at the centre of the ring which was initially circular on the outside and is sprung into position so that it is still circular on the outside. Consider a length a B of the ring, the thickness of the ring at the point b being te and the breadth throughout 356 THE STPxEXGTH OF MATERIALS being b. ^ is the angle which the arc b D subtends at the centre. Let R be the radius of the spring when bent and let R„ be the radius at b when in the unstrained condition indicated in dotted lines. If }) is the pressure per sq. in. we have B p = 7^ . A B (chord) . b = 2 pb R cos .^ (1) and the bending moment at p is equal to Fig. 155. — Piston Rings. By a modification of Fig. 121, p. 249, assuming a small initial radius of curvature R„ we shall get \R lij EI as a first approximation. I _ 1 R R, EI 2 pb Fx.^ cos^-^ E bte^ 12 24 p . R2 cos2 2 Ete' (3) SPRINGS 357 At the point d, ^ = and te = t^ ]. _ J. _ 24y R2 • • R R„ ~ E ^ 3 (*) Now R„ and R have to be the same in each position 24 25R2cos2|- .. _^. ^ 2 _ 24 2? R^ •'• El? ~ Ee If, therefore, the pressure 'p is to be constant te^ 2 ^ 1} = ^^^ 2 •••| == V'°''l ^^^ Fig. 156 shows vahies of te in terms of f^ for the various values of 6. Neglecting the additional stress due to the curvature of the bar (see ChajD. XIX.) we have at the point d ^ I ^VR R„ 2 E L / 1 1 •'• ^ 2 \R R, 112/ ^•^•r-r;=^e7: (') Putting this result in (4) we get I = l2pf-^ (7) .-. ^. = rVt^ <^' To find the necessary initial radius we have i- = i - ^ . (9) R„ R E^, ^^^ Numerical Example. — Taking E == 16 x 10^ lbs. per sq. in. and the working stress 4000 Ihs. per sq. in., find the necessary thickness and original external diameter for a cast-iron piston ring for a cylinder 20 inches in diameter, the necessary pressure being 3 lbs. per sq. in. 358 THE STRENGTH OF MATERIALS From equation (8) ^„ = 10 X 12 X 3 4000 = "95 in. nearl3^ This is less than is usually used in practice. /so /60 140 (20 IOC 3g GO "KP 20 . . . . V- 1 •2 o lo '8 O '^ Values of f^ — t^^. Fig. 156. — Thickness of Piston Rings for Uniforni Pressure. Unwin and Mellanby * give total depth of all packing ngs * Elements of Machine Design (Longmans), Part II (1912). SPRINGS 359 = , ^ + '6 in. for steam engines 15 = -f-- + 2*4 ins. for gas and oil engines = ^7^ for petrol engines. Taking, therefore, a steam engine with rings we should have = -64 in. From equation (9) 11 2 X 4000 • ■ R, R 16 X 10« X -95 •1900 = -09947 R„ = 10053 .*. original diameter = 20*11 nearly. Ring" of Uniform Thickness. — If the ring is of the same thickness t throughout, R will be constant, but R^ should vary in accordance with the following treatment — We have as before in equation (3) , , 24 ^ R2 cos^ ^ R ~ K "" El3 , , 24 2? R^ cos^ -^ •'•R, ^ R El3 ■ •■K^ ^''^ , (10) E ^3 _ 24 2? r2 1- For given values of t and f, R^ can be found by this formula for different angular positions, and it will be found that the curve for the initial shape of the ring differs considerably from a circle, so that rings made by cutting out parts from circular rings and springing into position will not give a uniform pressure. 360 THE STREXGTH OF MATERIALS * Plane Spiral Springs. — Consider a short length a b (Fig. 157) of a plane spiral spring, the free end of which is pulled with a force W. Then the bending moment acting on this short length Fig. 157. — Flat Spiral Springs. . • . If 8 ^ is the change in angle between the tangents at the two ends ^0 ==^ (seep. 250) Ss W.r X EI Total chano;e in ansle = ^-2 M 3 5 W EI EI if E and I are constant as is usual. .r 8 s SPRINGS 361 % X S s = 1st moment of spring constant about x x = length of spring x r (approx.) = Ir . Wlr EI (1) If the spring is wound up to produce a tensile force W at the end, the torque T which must be applied to the shaft will be equal to W r. Also 6 will be the angle turned through by the shaft, so that work stored up in spring 2 E I ^^ If the breadth of the spring is b and its thickness t, _ bj^ 6 6M 6M . • . Bending stress = / = bt' The maximum B.M. occurs at the point and is approxi- mately equal to 2 Wr. 12 Wr • 12 r Putting this result in (2) we have 144 r2 . ^ /2 bH"^ .1 /2 J2 li I Work stored = j-i .— «— fr^rr 144 r^ . 2 E I b t^ 288 E . ^- 24 E /2 Resilience = ^. „ 24 E Pbtl X volume 24 E 362 THE STRENGTH OF MATERIALS This is relatively small because the material is not used very economicalh^ parts of the spring being much more highly stressed than others. Close-coiled Helical Springs under Bending Stress. — If instead of subjecting a close-coiled helical spring to an axial load we subject it to a twisting action tending to unwind or wind up the spring, the whole spring will be subjected to a bending moment equal to the torque T applied. Therefore, if we neglect the effect of the curvature of the wire upon the stresses in it, we shall have, if 6 is the angle by which the spring winds up or unwinds — T f) T^ I Work stored = -^ = o^Fl" ^^^^ ^' ^^"^^ • • ^ - E"i Circular Section. — For a round wire of diameter d, d^ I = 64 and / • -oq" = 1 i.e./ 32 T Work stored = Resilience = 77 d^ 32 . 32 . 2 E Trd^ P .T-d^l _ P X volume 32 E ■ ~ 8E 8E Tl 64 T Z radians , " ~ E I 7T d^.E 1,168 TZ , — — I decrees. Rectangular Section.— If the depth of the section is b b t^ and the thickness is f , I = -, ^^ and T = Lipl SPRINGS 363 Work stored = 12 TZ E673 688T^Z E h t^^ radians degrees. Resilience = 2EI /2 62 ^4 .1 2 X 36. E.bt^ fnti f 6E ~ 6E P 12 X volume 6E SuMMAHY OE Resilience oe Vauious Types of Spring. Type of Spring. Pure tension Pure torsion on close-coiled helical spring (circular shaft) „ „ „ „ „ (square shaft) Plate spring Plane spiral spring Close-coiled helical springs with twisting action causing bending stresses — • Circular section . Rectangular section. Resilience. 2E 4G ■154 s^ ~G ■ P 6E _£_ 24 E 8E P 6E CHAPTER XIII THE TESTING OF MATERIALS Testing Machines. — In most types of testing machines the loads are applied through a system of levers and are so arranged that the levers are connected to one end of the specimen (or in the case of bending tests to the supports), and that a force is exerted by an hydraulic ram or screw gear to the other end, the lever system " floating " when the force exerted is equal to that applied to the levers. In this way additional weights can be put on to the levers without causing a shock in the specimen, because such additional weight does not come on to the specimen until the hydraulic ram or screw gear is operated further. We will describe some of the most common types of testing machines. WICKSTEED-BUCKTON SiNGLE LeVER VERTICAL TESTING Machine. — This type of testing machine, a photograph of which is shown in Fig. 158, was designed by Mr. J. H. Wick- steed, and is manufactured by Messrs. Joshua Buckton & Co., of Leeds. The form shown in the photograph is belt driven, the power being transmitted by toothed gearing to the screw at the base of the machine, but hydraulic rams are commonly employed to exert the necessary test force. This particular machine has a capacity of 30 tons, machines of this tj-pe being obtainable for capacities ranging from 5 to 100 tons, and can be employed for tests in tension, compression, bending, shear and torsion. Fig. 159 shows diagrammaticaUy the action of the machine, an hydraulic ram drive being shown. 364 fcJD > o ffl 00 THE TESTING OF MATERIALS 365 A horizontal lever A, Figs. 158, 159, is provided with a knife-edge b resting upon a strong vertical frame v; a jockey- weight w is movable along this lever and carries a vernier R by means of which the position of the weight can be read off upon a scale q. A second knife-edge c, carried by the lever, engages a link o connected to a cross-head. When operating for tension, one end of the specimen e is gripped in this cross-head, the s ZD r^ 7 irnr H K L m-. z^ , < G G r 1 Fig. 159. Sf^ecimen other end being gripped in a cross-head l connected by rods G to an hydraulic ram F. If the resultant of the jockey- weight w and the lever A is W and acts at a distance y from the knife-edge B, we have by moments V .X = W 2/ X The scale Q is graduated so as to read off values of P direct, because W and x are of fixed value. Stops s are provided for the lever a, which normally rests 366 THE STRENGTH OF IVIATERIALS on the lower one ; as the pressure in the ram is increased the force exerted upon the specimen gradually increases until it reaches the value P, Avhereupon the lever rises and '"' floats " between the two stops. A lower cross-head j is suspended from the cross-head H by rods K and is used for compression and bending tests. The diagram on the right-hand side of Fig. 159 shows how the force is applied in the case of a compression test. In a bending test the arrangement is similar, but the test-beam is placed on supports on the cross-head J and a load point or points is or are connected to the cross-head L. The jockey- weight w is adjusted along the lever A by a screw which runs through the latter and is driven from a shaft Fig. 160.— Werder Testing Machine. O operated by a hand-wheel u or by power from a counter- shaft X. Werder Horizontal Single Lever IVIachine. — This machine is used to a great extent on the Continent, and is shown in diagrammatic form in Fig. 160. The lever a is of bell- crank type and the two knife-edges B c are close together so that the leverage is great and comparatively small weights w can be employed. The knife-edge B is carried by the hydraulic ram F, and the force p is transmitted to the specimen through a cranked lever D. It is quite clear from this diagram that as the s^^ecimen stretches the load would go off it if the ram did not follow, i.e. if the pressure Avere not maintained in the cylinder. When, as is common, this pressure is generated by a small hand-pump, the operator goes on pumping until the lever floats between the stops s. Compound Lever Machines. — Riehle Type. — Fig. 161 shows a vertical type of compound lever testing machine, Fig. 161. — Riehle Testing Machine. [To jace paye 366. THE TESTING OF MATERIALS 367 made b}^ Riehle Bros., of Philadelphia, U.S.A., and used largely in America. The steelyard a is connected by a link with lever b, which is in turn connected with a lever c, which presses upwards upon the table or platen d. A cross-head E is operated by screws F, and according as the specimen is placed above or below this cross-head the test will be made in tension or compression. The machine shown is power driven by toothed gearing from an electric motor. These machines are controlled automatically by an electric contact device. At the outer end of the beam a are two contacts so arranged that when the beam reaches its highest position contact is made; this completes the circuit of an -S^ n ^^ G . ^rz D <^ 1 1 _ \ '^^ ) 1 9 Sf:>€cimen. I _c 1 1 — ^ — ' v^> //.'^y//'/^f^^ F Fig. 162. — Greenwood and Batley Compound Lever Testing Machine. electro-magnet which puts into gear with the driving mechanism the screw for moving the jockey -weight along the beam, but the movement of the jockey -weight can only follow up the extensions because contact is again broken as soon as the extension is more than is necessary to maintain the balance. Means are provided for varying the speed at which the weight is run out. An autographic recorder G is provided (see p. 379). Greenwood and Batley Horizontal Type. — This type of machine is made by Messrs. Greenwood and Batley, and was used by Professor Kennedy in the many researches which he carried out while at University College, London, this being one of the first testing machines installed in a college laboratory. The steelyard lever a, Fig. 162, has a knife-edge B and acts on a knife-edge c of a bell-crank lever d, which is pivoted upon a knife-edge e and is acted upon by a knife-edge F connected 368 THE STRENGTH OF JMATERIALS to a cross-head connected to the specimen. The other end of the specimen is carried by a cross-head operated by an hydraulic ram G. The usual leverage of the compound lever is 100 : 1 ; the jockey-weight w is generally moved along the steelyard, which carries a graduated scale, by means of a chain by a hand- wheel. WiCKSTEED-BucKTON HORIZONTAL Type. — This type is shown diagrammatically in Fig. 163 and by a photograph in Fig. 164. The steelyard lever a acts through a link c upon a bell-crank lever D, which connects by shafts shoT^n diagram- matically by G with the specimen. A massive carriage frame Fig. 163. — Wicksteed-Buckton Compound Lever Testing Machine. J is connected to the hydraulic ram r and carries a number of notches, into any of which can be fitted a cross- head K by which the other end of the specimen is carried. According to the position of the cross-head k, the specimen will be tested in tension or compression. This machine is very convenient for general testing on account of the ease with which it can be adjusted for different lengths of specimen and forms of test. Smaller Testing" Machines. — There are a large number of smaller testing machines in use, from which ver}' good results may be obtained in cases in which it is not essential for the specimens to be large ones. The student should remember that a great deal can be learnt with very simple apparatus. Fig. 165 shows a machine, designed by Professors Dixon and Hummel and manufactured by Messrs. W. and T. Avery, Ltd. ; &c o fa THE TESTING OF MATERIALS 369 it has an automatic load-indicating device in the form of two dished plates connected by a patented flexible metallic diaphragm; the space between the plates is filled with a non-elastic fluid and the pressure is recorded upon a sensitive gauge which is graduated to give the load on the specimen. The gauge can be tested by means of a small plunger which can be loaded with weights supplied with the machine to produce pressures corresponding to the total capacity of the machine. The machine shown has a capacity of 10,000 lbs. and the force is applied by a capstan acting through worm and wheel gearing to a central screw. This gear can be thrown out for quick return and the screw operated direct by the handle shown. Calibration of Testing Machines. — To ensure accurate results in the use of testing machines they should be calibrated periodically ; the vertical type of machine possesses advantage in this respect because a heavy weight can be hung on direct. The first test is for zero error. This is effected by moving the jockey- weight carefully to the zero mark and seeing if the lever floats ; if it does not we can correct for this by an adjustment of the vernier on the jockey- weight by moving the latter until the lever floats and then moving the vernier until it reads zero. The next point that we may test is the value of the jockey- weight. This can be effected without removing it from the machine in the machines shown in Figs. 158, 163, by finding the floating position and then moving the jockey-weight a carefully measured distance I along the lever ; then at a distance z from the fulcrum suspend weights w until the lever floats again. iv z Then weight of jockey-weight = W = -y-. V To test for the accuracy of the knife-edge distance x we may proceed as follows : Hang a heavy weight Wi from the shackles of the machine and note the distance u that the jockey- weight has to move to balance it; then X = — 1 - w BB 370 THE STRENGTH OF :\L\TERIALS Another imjoortant test is for sensitiveness, by whicli is meant the amount by which the load may vary without causing the lever to come against its stops. This may be tested at zero in vertical machines by placing the jockey -weight at zero and hanging small weights on to the shackles until the lever ceases to " float " ; this should be repeated for larger loads and should also be tried by taking weights off as well as by putting them on. Grips and Forms of Test-Piece in Tension. — ^'^lien tests are made on flat bars, as is very common for rolled sections, wedge grips are generally' emj)loyed. Fig. 166 (a) shows one form of wedge grip. \Yedges A, pro^-ided with serrations to grip into the specimen, are driven into a tapered («) Fig. 166. central passage through a block secured to the cross-head of the machine. In Fig. 166 (6) is sho^vn a grip suitable for a turned specimen provided with a collar. The collar bears against a washer c provided with a spherical end which bears against a tapered bush D, which engages in a similarly tapered central hole in the block B. This construction tends to keep the pull truty axial; a point of great importance. The ends of the speci- mens are very often screw-threaded, in which case they just screw into the blocks b. The British Engineering Standard Committee have specified the following rules, see Fig. 167, as to gauge length (cf. p. 55). (a) Flat Bars. — Gauge length = 8'" ; parallel for 9''. If the thickness is greater than J in., maximum width = IJ ins. Fig. 165. — Dixon and Hununel's Testing Machine. [To face page 370. THE TESTING OF MATERIALS 371 If the thickness is between f and J in., maximum width —■ 2 ins. If the thickness is less than f in., maximum width = 2 J ins. (6) Turned Sections. — Gauge length = 8 d; parallel for 9 d. (c) Turned Specimens from Forgings. — Area J in. ; gauge length = 2 ins. Area J in. ; gauge length = 3 ins. Area | in. ; gauge length = 3 J ins. Extensometers. — Extensometers are instruments for measuring the elastic strains of materials in tension or com- pression. In the types in most common use the strains are ^ P — at 8JL (b) (c) Fig. 167. ^ ~V magnified by an arrangement of levers and are measured by micrometer or by an indicator passing over a scale. We will describe a few of the most common types ; for other types a reference may be made to a paper by Mr. J. Morrow, in Proc. Inst. M. E. for 1904. An interesting report on the accuracy of various types of extensometers is given in the Report of the British Associa- tion for 1896. In these tests, different observers had bars of the same material sent for test. The results show very good agreement, some of the nearest results to the mean being obtained by instruments of very simple form. Goodman's Extensometer. — This extensometer is of very simple form and was designed by Professor Goodman, of 372 THE STRENGTH OE MATERIALS Leeds. It consists of two forked clips, a, b, Fig. 168, which carry pointed screws engaging in centre-punch marks in the specimen and are connected to rods c which join at their ends and carry a scale D on a projecting piece. Two light rods e, f form a fixed triangle, and the vertical rod E projects and has a small groove at its end which forms a bearing for a knife-edge carried by the pointer P. A second knife-edge on the latter rests upon a second vertical rod H Fig. 168. — Goodman's Extensometer. depending from the upper clip A. A small screw is pro- vided for bringing the pointer exactly to zero at the beginning of a test. The strain of the specimen causes the rod H to move slightly relatively to the rod e, this movement being magnified 100 times by the pointer lever. This and most other extensometers should be taken off the specimen as soon as the yield point is reached. Kennedy's Extensometer. — This extensometer was de- signed by Sir A. B. W Kennedy when professor at University College, London, and was one of the first lever extensometers. The instrument is for use in horizontal testing machines and THE TESTING OF MATERIALS 373 Fig. 169. — Kennedy's Extensometer. comprises two clips a^, Ag, Fig. 169, which carry triangular frames B^, Bg, which slide over and support each other. 374 THE STRENGTH OF MATERIALS The clips are as usual provided with pointed screws for engaging centre-punch marks in the specimen, lock-nuts being provided on the screws. As the specimen stretches, the frames slide relatively to each other, and a pointer-lever c which carries pins resting in depressions in each frame is thus caused to move over a scale d carried on an adjustable arm c Vj^m^ ©B i^m^^j H c. J. r. Co. £-w. Fig. 170. — Ewing's Extensometer. E. To give an adjustment for zero, the depression in the front frame is formed in a plate r which can be adjusted by a fine- pitch screw. Ewing's Extensometer. — This instrument was designed by Sir J. A. Ewing when professor at Cambridge. The principle involved is illustrated diagrammatically in Fig. 170. There are two clips B and c each attached to the test-piece A by the points of two set-screws. The slip b has THE TESTING OF MATERIALS 375 a projection b' ending in a round point P which engages with a conical hole in c ; when the bar extends this rounded point serves as a fulcrum for the clip c, and hence a point Q, equally- distant on the other side, moves, relatively to the clip B, through a distance equal to twice the extension. This dis- tance is measured by means of a microscope attached to the clip B . The microscope forms a prolongation of the clip B and the motion of the point Q is brought into the field of view by means of a hanging rod r. The rod R is free to slide on a guide in the chp B, and carries a mark on which the microscope is sighted. The displacement is read by means of a micrometer scale in the eye-piece of the microscope. The pieces B and b' are jointed to one another in such a way that the bar may twist a little, as it is sometimes liable to do during a test, without affecting the reading c. But the joint between B and b' forms a rigid connection so far as angular movement in the plane of the paper is concerned. This feature is essential to the action of the instrument : it is only then that P serves as a fixed fulcrum in the tilting of c by extension on the part of the specimen. Fig 171 is an illustration of the usual form of the complete instrument. The clips b and c in this standard pattern are set at 8 ins. apart. The object sighted is one side of a wire stretched horizontally across a hole in the rod b and illuminated by means of a small mirror behind. The distances cp and CQ are in this instance equal, with the effect that the movement of the sighted mark is double the extension of the test-piece. The length of the microscope is adjusted so as to give a constant magnification. This adjustment should be tested with the extensometer mounted on the specimen, and if necessary the length of the microscope tube can be altered by moving out or in the portion carrying the eye-piece. A complete revolution of the screw L, which has a pitch of -V of an inch, should cause a displacement of the mark through 50 divisions of the eye-piece scale, and when^this is the case the eye-piece is at the proper 376 THE STRENGTH OF MATERIALS distance from the objective. Readings are taken to tenths of a scale division, so that this displacement, which would also be given by ^i^ of an inch extension of the test-piece, corresponds to 500 units. Each unit then means ^,. ;.,,,. inch in the extension of the test-piece. A small extensometer based upon the same principle is used for measuring the compressive strain in short cylinders. Fig. 171. — Ewing's Extensometer. Dabwin's Extensometer. — This instrument has been designed by Mr. Horace Darwin, F.R.S., and is characterised by simplicity and solidity of construction, which make it suitable for heavy use. Another feature is that if the specimen should break unexpectedly when the extensometer is affixed little damage will result. The instrument is made in two separate pieces each of which is separately attached to the test-piece m, Fig. 172, by hard steel conical points P, P and p', p'. The steel rods carrying these points are mounted in] slides and after being driven __ THE TESTING OF MATERIALS 377 gently into the centre -punch mark in the test -piece are clamped in position by the milled heads r, r. Both parts of the instrument should be capable of rotating quite freely about the points, but there must be no backlash. The lower piece carries a micrometer screw fitted with a hardened steel point x and a divided head H. It also carries a vertical arm B at the top of which is a hardened steel knife- Fig. 172. — ^Darwin's Extensometer. edge. The upper and lower pieces work together about this knife-edge. A nickel-plated flexible steel tongue a forming a continuation of the upper piece is carried over the micrometer point X. This tongue acts as a lever magnifying the extension of the specimen, so that the movement of the steel tongue to or away from the steel point x is five times the actual extension of the specimen. To take a reading with the extensometer the thin steel tongue A is caused to vibrate and the divided head then turned till the point x just touches the hard steel knife-edge on the 378 THE STRENGTH OF MATERIALS tongue as it vibrates to and fro. This has proved to be a most delicate method of setting the micrometer screw, and the noise produced and the fact that the vibrations are quickly damped out indicate to yoVo mm. the instant when the screw is touching the tongue. After the load is applied a second reading is taken in a similar manner and the difference in the readings gives directly the extension of the test-piece. If the test-piece is of small diameter the spring does not vibrate in so satisfactory a manner; the cause of this is the flexibility of the test-piece, the instrument itself vibrating as well as the spring. Still, very delicate readings can be taken by simply deflecting the spring with the finger and noting the contact as it passes the point. No damage can be done by advancing the micrometer screw too far forward ; all that happens is that the point passes the knife-edge on one side or the other. In the usual form, the gauge length is 100 mm. ; it may be pointed out that over the elastic portion of the test for which extensometers are used, the gauge length is not a matter of importance. Unwin's Extensometer. — This extensometer, designed by Professor Unwin, is shown in Fig. 173, and makes its measure- ment by a micrometer acting in conjunction with two spirit- levels. Two clips a, b, are secured to the test-bar by pointed set- screws, c, d, and carry sensitive spirit-levels g. The lower clip is first set level by means of an adjusting screw e ; the upper clip is then levelled by the micrometer screw /, on the graduated head of which readings are taken. When placed midway between the two edges of the specimen the extensometer gives the mean strain, but if placed to one side or the other by adjustment of the screws in the clips, the strain at any point in the width can be found in the case of eccentric loading. Morrow's Mirror Extensometer. — A simple extenso- meter enabling great magnification of the strain to be obtained is that designed by Mr. J. Morrow. Clamping screws a, b, THE TESTING OF MATERIALS 379 Fig. 174, pass through rings c, d, to the latter of which a vertical strip is rigidly attached, a pointed rod e acting as a distance piece. Between the ring c and the strip r is placed a small diamond-shaped prism h which carries a mirror m, a light spring clip s maintaining the requisite pressure between the prism and the ring c and strip f. A second mirror isr is attached to Fig. 173. — ^Unwin's Extensometer. the strip f and any change in length of the specimen will cause the mirror m to rotate relatively to the mirror isr. By observing the images of a scale in both mirrors, we obtain the strain by the difference of the two readings. Autographic Recorders. — Many testing machines are provided with mechanism for drawing the stress-strain (or more accurately the load extension) diagram automatically as the test proceeds. One of the earliest mechanisms of this kind was one used by Professor Kennedy upon a horizontal 380 THE STRENGTH OF MATERIALS compound lever machine. The movements of a pointer upon a piece of smoked glass were obtained in one direction by the actual extensions of the bar, and the movement represent- ing the stress was obtained by multiplying up the strain in a ^ P; EL ^^-£^^ f £^- E ^ s tl Fig. 174. — Morrow's Extensometer. J^ longer bar coaxial with the test-bar, this longer bar being always stressed within the elastic limit and the load therefore being proportional to the extension. WiCKSTEED-BucKTON RECORDER. — This autographic re- corder is fitted to the Buckton machines and is shown on the Yia. 175. — Wicksteed-Buckton Autographic Recorder. [To face page 381. THE TESTING OF MATERIALS 381 extreme right-hand side of Fig. 158, and also to larger scale in Fig. 175. The record sheet is placed on a drum, around which passes a string which goes through a tube t and passes between pulleys upon the cross-heads between which the specimen is gripped. As the specimen becomes strained the distance between these cross-heads varies and this motion is communicated to the drum so that the rotation of the drum is proportional to the strain. The stress is measured by first putting the jockey-weight w near its extreme position and preventing the lever a from coming down upon its stop by means of a spring s connected up to the pencil carriage. As the specimen becomes stressed, spring s is proportionally re- lieved from load and thus shortens in length by an amount pro- portional to the load appHed to the specimen. The upward movement of the pencil is therefore proportional to the load, and the combined movement of drum and pencil traces out a load-extension curve which is generally — ^though not quite accurately — called the stress-strain diagram. Torsion Testing Machines. — Single lever testing machines are often provided with an attachment for enabling testing by torsion to be carried out. Torsion tests to failure can be made upon comparatively small machines. Fig. 176 shows diagrammatically the form of testing machine used by Professor Kennedy; most other machines are based upon the same principle. A graduated lever a is counter- weighted to balance about a knife-edge B coaxial with the specimen x, which usually has the form shown in Fig. 167 (6) with the exception that the ends are not screw-threaded. A jockey- weight \v runs along the lever and the specimen is clamped in a chuck which is connected to the lever at the pomt c. The other end of the specimen is secured by a chuck carried by a worm-wheel d which is operated through a Avorm from a handle e to apply the necessary torque. The jockey-weight is placed so as to exert a given torque, and the handle e is turned until the lever " floats " between the stops s. 382 THE STRENGTH OF MATERIALS Professor Thurston's Torsion Machine. — In this machine, Figs. 177, 178, the specimen is a short one with square ends, one of which is carried by a jaw rotated by worm gear, the other being carried by a jaw connected to a weighted pendulum, the angular movement of which determines the torque applied. X~ ^ Fig. 17G. — Torsion Testing Machine. An autographic diagram is obtained by securing a pencil to the pendulum in such a manner that the pencil moves parallel to the axis of the specimen as the pendulum swings outwards. A cylinder carrying a paper strip is secured to the jaw carried by the worm-wheel. The paper thus rotates by an amount equal to the angle of torsion, and the pencil moves at right angles by an amount which is a measure of the torque applied. THE TESTING OF MATERIALS 383 A templet of the form shown in Fig. 178 is employed to obtain a standard size of specimen. Avery's Reverse Torsion IMachine. — Fig. 179 shows two views of a torsion machine, patented by Messrs. Avery, by means of which a torque can be applied in either direction. Fig. 177. — Thurston's Torsion Testing Machine. The specimen x is gripped between special three- jaw chucks g, g', and the torque is applied from a hand-wheel i through a worm-wheel h mounted upon an adjustable standard a*. The torque is thus communicated to a lever / and thence through a supplementary lever k and rod I to the steelyard b, upon which the usual jockey d is mounted. 384 THE STRENGTH OF MATERIALS The main torsion lever / is fulcrummed on ball-bearings /^. Passing through the centre of this bearing is a spindle e of cruci- form section to which the chuck g is attached. This spindle is connected with the lever / by means of rollers e^ whereby it has limited longitudinal movement through the lever, but cannot revolve therein. This longitudinal distance is to allow Iflb Fig. 178. of adjustment due to the shortening of specimens undergoing tests, the collar e^ on the spindle e preventing the withdrawal or extended movement of the spindle. The main lever / is provided with knife-edges p and /^ through Avhich connection is made to the supplementary lever k ; this lever k is w ithin the main lever / and has a ball-bearing fulcrum k^ on a bracket a^. It is suspended by means of the link m from the knife- edge /2 of the main lever, and at its opposite end it is connected with the knife-edge of the main lever P by a link m^. The links m and m^ connect with the lever k through Imife-edges k^ and P. Between the knife-edge P and the ball-bearing k^ is another knife-edge n which forms the connection from the levers / and k to the steelyard b by means of the rod Z. The THE TESTING OF MATERIALS 385 lever / is counterbalanced by the adjustable weight /*, and the lever ^ by a similarly arranged weight k^. Assuming the torque to be applied as indicated by the arrow y, it lowers the end of the lever / which is in contact with the ^>ci ^1 --------- Fig. 179.— Avery's Reverse Torsion Machine. lever k ; the point of greatest depression in contact with the lever k will be the knife-edge /^ which through the link m^ and knife-edge k^ lowers this end of the lever k, about its fulcrum k^, so that this movement of the lever k lowers the knife-edge n, thereby exerting a downward pull on the connecting rod I and raising the free end of the steelyard b. cc 386 THE STRENGTH OF MATERIALS If the torque is a^oplied in the direction indicated by the arrow z the end of the lever / which is in contact with the lever Ic is raised and the knife-edge /^ also raised ; this upward move- ment of ths knife-edge /^ raises the link m dependent there- from and also the knife-edge k^ of the supplementary lever A:, causing the lever Ic to move about its fulcrum 1^ as before. The knife-edge n is again depressed by this movement of the lever and exerts as before a downward pull on the connecting rod. By this arrangement, in whichever direction the tor- sional stress is applied to the main lever /, the resultant direction of force on the connecting rod I is the same. Professor Lilly's Reverse Torsion LIachine. — This reverse torsion machine, patented by Professor Lilly of Dublin, is a very simple machine for obtaining autographic diagrams in torsion and is particularly of value when working within the elastic limit. A circular table a. Fig. 180, is fixed to any convenient bench or stool, and has through its centre a hollow steel cylinder B. In the central part of the cylinder is placed the specimen c to be tested, one end being secured to it by the key at d and the other end passing through the adjustable bearing E ; it is secured to the lever g H i by the key at F. The lever consists of a solid shank h, which is rigidly connected to the spring i ; the weight G with its connecting arm forms part of the solid shank, and is for the purpose of balancing the lever. Fixed to the spring i at J is a light arm k, at the end of which is an adjustable spring Q carrying the recording pencil l. This pencil is adjusted to slide along the straight edge N which forms part of the frame s. A circular drum m revolves on its outer edge o on the table A, and is connected by adjustable pivot bearings to the frame s which is connected by adjustable pivot bearings R to the arms of the solid shank h. The manner of carrying out a torsion test with the machine is as follows : The specimen c to be tested is placed in position in the cylinder B, and secured to it by driving up 'the key at d ; the lever G h i is now placed in position on the top THE TESTING OF MATERIALS 387 end of the specimen c and secured to it by driving up the key at F. A sheet of squared paper is fixed on the drum m, which o H > P5 o M (D O o 00 is then put in position on the table a, and the pivot bearings adjusted; the pencil l is now placed in the central position 388 THE STRENGTH OF MATERIALS on the drum and in contact with the straight edge n by adjust- ing the arm k and the spring q. The torsion test on the specimen is carried out by applying a pull or push to the handle p; the pencil L then automatically graphs the stress-strain diagrams on the squared paper. The movement of the pencil L along the straight edge n is proportional to the push or pull on the handle p and gives to scale the magnitude of the torsional or twisting moment ; this may be shown as follows — Regarding the spring i as a cantilever with a load at the free (l ij end, the value of y and -—- at the point J is proportional to the force applied at p. Calling the length of the arm z, the deflec- d u tion of the pencil end is y ~ z ^^ which is proportional to the applied force. The roll of the drum m under the pencil is proportional to the angle of torsion of the specimen. The pencil graphs the combination of these two movements at right angles to one another, and the resulting stress-strain diagrams are thus obtained to rectangular co-ordinates. To calibrate the machine a known. pull is applied to the handle p by means of a spring or otherwise, and the dis- tance traversed by the pencil along the straight edge gives to scale on the squared paper the magnitude of the apphed twisting moment. The scale of the angle of torsion is obtained by observing the number of turns of the drum during one complete revolution on the circular table. For examples of diagrams taken with this machme the reader is referred to a paper in Froc. Inst. C. E. Ireland, Vol. 41. Professor Coker's Combined Bending and Torsion Machine.* — This machine has been devised by Professor Coker for experiments upon combined bending and torsion. The various parts are supported in a built-up frame con- sisting of two planished steel shafts, a, Fig. 181, secured in * Phil. Mag., April 1909. THE TESTING OF MATERIALS 389 cast-iron cross frames b, mounted on four standards, one of which latter is adjustable in height to secure steadiness on an uneven floor. U23on the steel shafts are two castings c, D, each of which has a cylindrical bearing E encircling one of the shafts and resting with a flat face F in line contact with the other shaft, and secured in j)osition by a cross-bar G threaded on studs. This connection is perfect!}^ rigid, since it removes all degrees of freedom, and it is readily released by simply turning back one of the cross-bar nuts, leaving the casting free to slide into a new position. It also has the advantage Fig. 181. — Coker's Combined Bending- and Torsion Machine. that no accurate fitting is required for the supporting frame. The casting c carrying the worm-wheel gear w has trunnion bearings h at right angles to and intersecting the axis of the specimen. The bearings are fitted with friction rollers, and when the machine is used simply for torsion the worm-wheel is kept in a vertical position by an arm i keyed to the bearing H and locked in position by a thumb -screw. A weight J attached by an arm to the second bearing balances the pivoted casting in all positions. The weigh-levers are supported from a vertical standard K of the frame d by a wire l, terminating in a thin plate M 390 THE STRENGTH OF MATERIALS with a keyhole slot encircling the spindle N. Formerly a roller bearing was used for this spindle, but this is an un- necessary rejBnement, as the friction is extremely small and can be easily taken into account. The casting supported in this way has three levers, p, q, and r, the first two of which are for the application of twisting moments s, and the third R, in the line of the specimen, is for applying a bending moment. All the loading levers are provided with knife-edges of circular form, made by turning an ordinary Whitworth nut down to form a disk with a V-shaped edge. These disks carry rings t with wdde-angled V-shaped recesses on the inner sides, and light rods v screwed into these rings carry the weights. This arrangement of knife-edge is very easy to adjust accurately, and when bending and twisting stresses are applied simultaneously the rolling line contact adjusts itself to the bending and twisting of the specimen. The bending of the specimen causes a change in the effective arm of the bending levers, which is generally negligible, but a correction may be necessary wdth a very long specimen. For if a is the length of the lever-arm, and h is the radius of the circular knife-edge, an angular deviation of amount will cause a change of a — {a cos ^ + 6 sin 0) in the lever- arm, and this is zero when ^ = and also when a = a cos ^ -f 6 sin ^. In the machine described a is 10 inches and h is 0*5 inch, and the angles ^ = and 6 = 515° both correspond to an effective length of 10 inches. The maximum correction be- tween these values is easily shown to be at an angle given by the equation b cos = a sin 6, in the present case 29° approximately, for which value the correction is 0*12 per cent. In the majority of tests the angular change at the ends rarely exceeds 5°, and the correction is therefore so very small as to be practical!}^ negligible. The worm-wheel w and the casting v for the weigh-levers are bored out to receive the ends of the specimen, and are THE TESTING OF MATERIALS 391 provided with fixed keys which slide in corresponding key- ways cut in the specimen. When tubes are subjected to stress they are provided with solid ends secured by transverse pins, thereby avoiding brazed joints, since these latter are trouble- some owing to the state of the metal being altered by the brazing. The end of the specimen projecting through the worm-wheel is fitted with a lever x for applying bending moment, and both levers for bending may be loaded in- dependently or by a cross-bar suspended from stirrups as shown. CaPiefomelgr Fig. 182. — Simple Torsion Meter. Torsion Meters. — The elastic angular strain in torsion requires less magnification than the elastic longitudinal strain in tension, and so comparatively simple apparatus can be used. Fig. 182 shows a simple apparatus made by Mr. A. Macklow- Smith. Two arms a, b, connected together by an extensible sleeve, are secured by pointed screws to the specimen. The arm A carries a cathetometer or telescope and the arm B carries an ivory scale upon which the angle of torsion is read. Professor Coker's Torsion Meter.* — This apparatus, * Phil. Mag., April 1909. 392 THE STRENGTH OF MATERIALS which can be used also for measuring the strain in combined bending and torsion, consists of a graduated circle A, Fig. 183, mounted on the specimen b by three screws c in the chuck- plate D. A sleeve e provided with three screws grips the specimen at a fixed distance away from the first set. The spacing of these two main pieces on the specimen is effected by a clamp, not sho^Mi in the figure, which grips the double ones P, G, and maintains them at the correct distance apart until the set -screws are adjusted. Fig. 183.— Coker's Torsion Meter. The clamp is afterwards removed, leaving the plane of the graduated circle perpendicular to the axis of the specimen and the sleeve correctly set and ready to receive the reading microscope H. The vernier plate carries a sliding tube i, on which a wire j is mounted, and the movement of the latter due to bending or twist is measured by a scale in the eye-piece K, the divisions of which are calibrated by reference to the graduated circle. It is found convenient to have the microscope-tube pivoted about an axis perpendicular to its central line at L, so that any slight difference due to imperfect centring can be adjusted THE TESTING OF MATERIALS 393 by the screw m to make the calibration value agree for a series of specimens. Torsion Dynamometers (or torsiometers as they are sometimes called) are instruments for indicating the horse- power being transmitted by a shaft rotating at a known speed by measuring the angle of torsion over a given length. They are often provided with autographic record devices. The horse-power is derived by a combination of formula (3), p. 312, and formula (11), p. 324. Repetition Stress Machines. — We described on p. 85 one of the forms of machine used for rotating beams by Wohler in his experiments in repetition of stress. Similar machines are in use in many engineering laboratories ; in most cases the spring is replaced by a weight which has a spherical socket resting on a spherical bearing fixed to the end of the specimen. By this arrangement the weight remains free from oscillation as the specimen sags under load. Professor J. H. Smith's Machine. — In this machine, which is shown in Fig. 184, the variation of stress is caused by the variation of the longitudinal component of the centrifugal force of the rotating weights e. The specimen which is to be tested connects the two pieces c and B. The upper piece is of circular cross section, and has a locking arrangement consisting of a cap and set screws. The lower piece b is of circular section in the upper bearing l, and of square section in the lower bearing m. The two pieces c and b are supported by the frame- work F of the machine; and the specimen is inserted with- out straining it by first locking it to the piece b, then by locking c to the specimen and afterwards locking c to the frame. The piece b has a bearing n at right angles to its length in which a spindle revolves; there are two plates k, k, and weights E, e, rigidly attached to this spindle. The rotating spindle is driven by means of a pair of pieces in contact, one, a crank pin diametrically opposite to e, on 394 THE STRENGTH OF MATERIALS one of the rotating plates K, and one a radial slot on a plate connected with either another unit or to a shaft rotating in fixed bearings. By this arrangement the driving force is not transmitted to the specimen. The component of the centrifugal force exerted by the Fig. 18-i. — Smith's Stress Repetition Machine. rotating weights e, e, produces an alternating stress in tlie specimen. The spring h, and tightening device j, enable the operator to put any desired amount of tension or compression in the specimen. This spring may be replaced by weights and levers or by an hydraulic cylinder. The lead rings or springs at G, G, act as buffers and receive the blow when the specimen breaks. THE TESTING OF MATERIALS 395 The complete machine consists of one or more units together with the necessary balancing rotating masses which may be either parts of other units or parts connected to a revolving shaft mounted on the framework. See also Professor Arnold's machine, p. 399. CHAPTER XIV THE TESTIIS'G OF MATERIALS {cOTltd.) Impact, Ductility, and Hardness Testing. — As we have indicated on p. 54, the elongation under a tensile test is commonly taken as a measure of the ductilit}-, but experience shows that this is not sufficient in all cases, and in recent years a number of simple machines have been devised for carrying out tests upon small specimens. Cold and hot bending tests are commonly specified by the various authorities and purchasers of steel and iron, giving the angle through which the specimen must bend without cracking. In the specification for structural steel issued by the British Standards Committee, for instance, there is a clause that test- pieces must without fracture withstand being doubled over until the sides are parallel and the internal radius is not greater than IJ times the thickness of the test -piece, the latter being not less than IJ inches wide. Tests of this kind have the advantage that they can be made in the workshop without special apparatus, but the disadvantage that the results are rather negative. Captain Sankeys Hand Bendi>'g Machixe. — In this machine, patented by Captain Sankey, a j)iece of metal is bent backwards and forwards through a fixed angle until it is broken ; the bending moment being measured by the deflection of a spring and recorded upon a paper drum. The standard angle is 9H°, i.e. 16 radians, so that the work done to make a complete bend is obtained b}' multiplying the bending moment b}' 16. 396 THE TESTING OF MATERIALS 397 At one corner of the base of the machine there is a grip A for securing one end of a flat steel spring b . The other end of the spring is fitted with a special grip c for holding one end of the test-piece d. The other end of the test-piece is fixed into a handle E, about 3 feet long, by means of which it is Fig. 185. — Sankey's Hand Bending Testing Machine. bent backwards and forwards through the standard angle. A graduated arc F is provided to show this standard angle. Alongside of the spring, and fixed to the bed-plate, there is a horizontal drum g, to carry the recording paper, and the pencil H has a horizontal motion actuated by the motion of the grip c and conveyed by the steel Avires l and m and the multiplying pulley n, the wires being kept taut by a weight. The zero line is in the middle of the paper, and the pencil H 398 THE STRENGTH OF MATERIALS moves in one direction when the bending is from right to left and in the opposite direction when it is from left to right. The drum has a ratchet wheel K, with a detent (not shown) worked by the motion of the pencil carrier. The result of the combined motion of the pencil and of the drum is to produce an autographic diagram such as shown. Obviously, the greater the stiffness of the test-piece the more the flat spring B will have to be bent before its resistance is equal to the resistance to bending of the test-piece. Hence the motion of the pencil is proportional to the bending moment required to bend the test-piece. In operation, the test -piece is properly secured in the handle E by means of the set screw ; it is then inserted into the grip c, and the free length (1| inches) is adjusted by means of a gauge provided for the purpose, after which the grip c is tightened. The handle is slowly pulled towards the left until the specimen is felt to be " yielding " — this action can be distinctly felt, and this bend is known as the " yield bend." Without altering the pressure on the handle, the record cyHnder is now rotated two teeth by working the detent by hand, and the first bend is completed by making the mark on the handle coincide with the pointer indicating the " standard " angle. The bending is then reversed, and the test-piece is bent until the mark on the handle coincides with the second pointer. The bending is again reversed, and so on until the specimen breaks. The point at which the test-piece breaks should be noted in decimals of one bend, which are marked on the graduated arc. The machine is cahbrated by fixing the handle end of the lever in the jaws and applying a known force by means of a spring-balance and comparing the record made on the strip with the actual moment appHed. If there is any discrepancy between the two results, the spring is adjusted until such discrepancy disappears. The number of bends which a give^ material can endure before fracture is a measure of the ductihty, and experiment shows that this is approximately THE TESTING OF MATERIALS 399 proportional to the percentage elongation multiplied by the percentage reduction in area in a standard tensile test. The following empirical results have been found from experiment to be approximately true — Yield-point stress in tons per sq. in. _ First bending moment in lb. ft. ~ r55 Ultimate tensile stress in tons -per sq. in. _ Longest line in lb. ft. ^ r55 Fig. 186. — Arnold's Testing Machine. The energy required to cause rupture is equal to 1'6 multi- plied by the number of bends, multiplied by the mean range of bending moment in lb. ft. Professor Arnold's Reverse Bending Machine. — In this machine, which may also be regarded as a repetition of stress machine, a bar a, Fig. 186, f in. in diameter, is firmly held in a clamp b and passes through a slot in a slide c which is reciprocated by a shaft d running at a standard speed of 650 revolutions per minute. The distance between line of contact of the slot with the specimen and the point where the 400 THE STRENGTH OF rilATERIALS latter enters the clamp is 3 inches and the slot is adjusted to cause a deflection of | in. in the specimen on each side. The number of bends which the sj^ecimen endures before fracture is taken as a measure of the capacity of the material to resist failure by shock. Repeated Impact Testing Machine. — The machine shown in Fig. 187 is made by the Cambridge Scientific Instru- ment Co., Ltd., and is a modification of a machine described by Messrs. Seaton and Jude * and used by Dr. Stanton j of the National Physical Laboratory. .-1 --1 ,z:L ■«i -r-- J s -^ Mr M Fig. 187, — Repeated Impact Testing Machine. The machine is fitted with a cone-pulley a, so that it can be driven by a belt from a hne shaft or small electric motor. One end of the spindle driven by this cone-pulley carries a crank b which is connected to the lifting rod c. This lifting rod is supported on a roller d, at some point in its length, so that the circular motion imparted to the rod at the crank end causes it to rock and shde on the roller. Thus an oval path, shown in dotted lines, is traced by the free end of the lifting rod. At this end the rod is bent at right angles so that on the upstroke it engages with and hfts up the hammer head e. This hammer head is fixed to the rod f, which is hinged at the end g. Having reached the top of its path, the lifting rod c moves * Froc. I. Mech. E., 1904. t Proc. I. Mech. E., 1908. Fig. 188. — Impact Testin.o- Machine. [To face page 401. THE TESTING OF MATERIALS 401 forward and disengages the hammer, which then falls freely on to the specimen h under test. This cycle is repeated from 70 to 100 times a minute. The height through which the hammer falls can be varied by moving ing the roller d along a scale m which is calibrated to read directly the vertical height through which the hammer falls. Adjustment can be made by this means up to a maximum of 3 J inches (90 mm.). The specimen h is usually about J^' (12 mm.) in diameter, with a groove turned in it at its centre to ensure its fracture at this point in its length. It is supported on knife-edges 4 J'' (114 mm.) apart, the hammer striking it midway betw^een these knife-edges. The knife-edges are cut slightly hollow, and a finger spring holds one end of the specimen in place. The other end is held in a chuck which is hinged in such a manner that it does not take any portion of the hammer blow, all of which comes on the knife-edges. The specimen remains stationary whilst the blow is struck, but between the blows it is turned through an angle of 180°. A revolution counter to register the number of blows struck is fixed to the bed plate of the instrument. When fracture occurs, the specimen falls away, and the hammer head continues to fall, first tripping an electric switch, and finally coming to rest on a steel stop-pin h. Izod's Impact Testing Machine. — This machine, which is made by Messrs. Avery, tests the impact-resisting qualities of a material by measuring the energy absorbed from a pendu- lum which breaks a projecting notched specimen as it swings past it. The specimen is 2 inches long, -f^ inch thick and f inch broad, and has a notch cut in it by means of a templet ; it is held in the \ace shown at the base of the machine. Fig. 188, and the pendulum is then released from a fixed height by means of a trigger. The energy required to fracture the specimen takes some of the swing out of the pendulum, and the height to which DD 402 THE STRENGTH OP MATERIALS the latter swings on the other side is indicated by the pointer passing over the scale at the top of the specimen. The scale is graduated to give the energy directly in foot pounds. A brittle material will not absorb much of the energy, whereas a tough material will absorb a good deal. Inasmuch as the base is not absolutely rigid, the results of tests in this machine are relative rather than absolute, but it gives very useful results in practice and has the advantage that the tests can be made in a very short time. The same machine can be adapted for testing hardness by impact. A strong cast-iron anvil is provided in which a specimen 1 inch in diameter and 1 inch long is placed. The pendulum strikes a loose plunger which carries a hardened steel ball; this ball is placed in contact with the specimen prior to the impact, which causes an indentation in the specimen. The diameter of this indentation is taken as a measure of the hardness as in the Brinell machine next to be described. Brinell's Hardness Testing Machine. — In this machine a hardened steel ball is pressed with a predetermined force against the plate whose hardness is required. The diameter of the resulting curved depression is then found and from this the " hardness number " is obtained in the manner described below. Fig. 189 shows the Brinell machine made by Messrs. J. W. Jackman «fe Co., Ltd. The specimen is placed upon the top • of the stand, which is then adjusted by the hand-wheel to bring the specimen into contact with the hardened steel ball (10 mm. diameter) which projects from the conical end of the plunger of the machine. The upper portion of the machine comprises a small fluid-operated testing-machine, oil being the working fluid. By means of a small projecting pump-handle the pressure of the fluid is increased until the cross-piece " floats," the pressure being indicated on the dial. Weights are provided with the machine, which make the floating occur at a force of 500 to 3000 kg. (increasing 500 kg. at a time). '~1 Fig. 189. — Brinell Hardness Testing Machine. [To face page 402. THE TESTING OF MATERIALS 403 The pressure depends only on the weight applied and not upon the accuracy of the gauge. If P is the load, D the diameter of the ball and d that of the impression, the quantity H = D _ V D2 - d^ D is called the Brinell Hardness Number. The following table gives values. BRINELL'S HARDNESS NUMBERS (for load 3000 kg.) Diameter of Steel Ball =10 mm. Diameter Diameter Diameter Diameter of Hard- of Hard- of Hard- of Hard- Ball Im- ness Ball Im- j ness Ball Im- ness Ball Im- ness pression Number pression Number pression Number pression Number mm. mm. mm. mm. 2-0 946 3-25 351 4-50 179 5-75 105 2-05 898 3-30 340 4-55 174 5-80 103 2-10 857 3-35 332 4-60 170 5-85 101 215 817 3-40 321 4-65 166 5-90 99 2-20 782 3-45 311 4-70 163 5-95 97 2-25 744 3-50 302 4-75 159 6-0 95 2-30 713 3-55 293 4-80 156 6-05 94 2-35 683 3-60 286 4-85 153 6-10 92 2-40 652 3-65 277 4-90 149 6-15 90 2-45 627 3-70 269 4-95 146 6-20 89 2-50 600 3-75 262 5-0 143 6-25 87 2-55 578 3-80 255 5-05 140 6-30 86 2-60 555 3-85 248 5-10 137 6-35 84 2-65 532 3-90 241 5-15 134 6-40 82 2-70 512 3-95 235 5-20 131 6-45 81 2-75 495 4-0 228 5-25 128 6-50 80 2-80 477 4-05 223 5-30 126 6-55 79 2-85 460 4-10 217 5-35 124 6-60 77 2-90 444 4-15 212 5-40 121 6-65 76 2-95 430 4-20 207 5-45 118 6-70 74 30 418 4-25 202 5-50 116 6-75 73 3-05 402 4-30 196 5-55 114 6-80 71-5 3- 10 387 4-35 192 5-60 112 6-85 70 315 375 4-40 187 5-65 , 109 6-90 69 3-20 364 4-45 183 5-70 1 107 6-95 68 For other test loads, the hardness numbers are proportional to those in the table. Within certain limits the Brinell Hardness Number of a 404 THE STRENGTH OF MATERIALS material gives a very fair indication of its tensile strength. Thus for steels with a hardness number less than 175, the ultimate tensile stress in tons per sq. in. is obtained approxi- mately by multiplying the hardness number by -23. TESTING CEMENT AND CONCRETE Tension Tests. — The form of briquette in accordance with the Specification of the British Engineering Standards Committee is shown in Fig. 190, the cross-section being 1 sq. in. at the weakest point. We have given on p. 78 the require- FiG. 190. — Cement Briquette. ments as to tensile strength in accordance with this specifica- tion. As the strength obtained under test is found to depend upon the rate of loading, being higher for quick loading, the above specification stipulates that the loading shall be at a rate of 500 pounds per minute. A simple form of lever machine, made by W. H. Bailey & Co., is illustrated in Fig. 191. The specimen is gripped in the shackles and the load is applied by allowing shot to fall into the bucket, the leverage being such that the tension applied is fifty times the weight of the shot. The shot-hopper is provided with a valve, the operating arm of which passes over the lever, so that when the specimen breaks the supply of shot is automatically cut off. The shot is then weighed, a THE TESTING OF MATERIALS 405 spring-balance being often used which gives readings equal to fifty times the weight of the shot, thus giving the breaking stress direct. A rather more accurate form of machine, made by the same firm, is shown in Fig. 192. In this machine water is allowed to run slowly into a long graduated can placed at the end of the lever. The supply of water is cut off when fracture occurs Fig. 191. — Cement Testing Machine (Tension). and a gauge glass placed outside the can is provided with a scale graduated to enable the breaking stress to be read off direct. In another common form of testing machine a jockey-weight is moved automatically along a lever arm by means of a weight controlled by an adjustable dashpot which enables the rate of loading to be varied. On the fracture of the specimen the weight becomes stationary and the breaking stress is read off on a scale attached to the lever. Compression Tests. — Compression tests are not usually 406 THE STRENGTH OF MATERIALS specified for pure cement, although they are becoming more common. For concrete in reinforced concrete works, however, compression tests are nearly always required. A common specification is that cubes, the area of each side being 50 sq. cm., of 3 parts sand to 1 part cement by weight Fig. 192. — Cement Testing Machine (Tension). shall develop at 28 days at least 10 times the standard tensile strength (^. e. 2000 lbs. per sq. in.). The following test results for the concrete cubes (with area THE TESTING OF MATERIALS 407 of each side 50 sq. cm.) are recommended by the Concrete Institute. Proportion by Volume. Crushing Strength in lbs. per sq. in. Cement Sand Coarse Material 28 days after mixing 120 days after mixing 1 1-2 1-5 2 2 2 2 2 4 4 4 4 1600 1800 2000 2200 2400 2600 2800 3000 Fig. 193. — Compression Tension Press for Cement, etc. A common form of special machine for crushing tests of cement, concrete, etc., is shown in Fig. 193. The cube is first fixed by means of the upper hand- wheel and the side -press 408 THE STRENGTH OF MATERIALS screw is then operated to compress the operating fluid (oil or glycerine). The crushing pressure is recorded by the pressure gauge which is constructed so as to maintain its reading after fracture has occurred. Specific Gravity, Fineness, Soundness, and Setting Tests for Portland Cement. Specific Gravity. — According to the British Standard Specification, the specific gravity of Portland cement shall not be less than 3* 15 when fresh or 3* 10 after 28 days from grinding. • There is considerable doubt as to the value of this test; some very useful information on this and other matters in cement testing will be found in a paper on " Common Fallacies in Cement Testing," read by Mr. W. L. Gadd, F.I.C., before the Concrete Institute, December 11, 1913. A chemical analysis appears to be a much more reliable test. Fineness. — The fineness test is applied by means of standard sieves. In the British Standard Specification not more than 18 % residue is allowed for a sieve of square mesh with 180 wires, each '002 in. in diameter, per inch, and not more than 3 % for a 76 mesh with wire '0044 in. in diameter. Le Chatelier Soundness Test. — This test is conducted with a piece of split brass tube a, Fig. 194, 30 mm. internal diameter and 30 mm. long, the thickness of metal being J mm. Pointers b b are attached to the tube, the length from their points to the centre of the tube being 165 mm. This tube is used as a mould, and is filled with wet cement, one end being placed previously on a piece of glass ; the other end is then covered with a weighted piece of glass and the whole is placed in water at a temperature from 58° to 60° F. and left for 24 hours. The distance between the pointers is then measured and the mould is then placed in water which is heated to boihng point and maintained in that condition for 6 hours. The British Standard Specification stipulates that after boiling the increase in the distance apart of the pointers shall not be greater than 6 mm. Setting Tests. — Setting tests are often made by finding THE TESTING OF MATERIALS 409 the time before a standard weighted needle fails to make an impression in the cement. The Standard Specification requires that before any sample is submitted to setting test it shall be spread out for a depth of 3 inches for 24 hours in a temperature from 58 to 64° F., and that the setting time shall not be less than 2 hours nor more than 7 hours. Mr. Gadd in B' ■B Fig. 194.— Chatelier Cement Test. the paper referred to above has shown that this spreading for the purpose of aeration leads to very variable results, depending largely on the locality and humidity. The Thermal Method of Testing Materials. — It was apparently first noticed by Magnus that changes of stress are accompanied by a change of temperature ; when a body is stretched its temperature lowers very slightly, and when it is compressed or squeezed its temperature rises. By 410 THE STRENGTH OF MATERIALS means of a thermopile and a very delicate galvanometer, therefore, the changes of temperature at various points of a structure, and therefore the stresses, can be determined. Care should be taken to distinguish the phenomenon under con- sideration from the well-known heating that occurs at the yield point in tension experiments ; this is very much greater in magnitude and is reverse in sign, the elastic tension being accompanied by a fall in temperature. Lord Kelvin has deduced the following formula to deal with the problem — AT. Ta ^ ^T=-js^-^^ In this formula A T = change in temperature. T = mean absolute temperature. J = the mechanical equivalent of heat. a — coefficient of expansion of material. d = density of material. S = specific heat of material. A 2? = change of stress. In centigrade units for a temperature of about 20° C. this formula gives for steel A T = - -000012 A p ■ Corresponding to a stress change of 20,000 lbs. per sq. in., therefore, we have a temperature change of only '24° C. The extreme delicacy of the method makes it suitable for use only under circumstances in which great care is taken to exclude draughts. This subject is dealt with in detail by Professor Coker in his Cantor Lectures before the Society of Arts 1913. In this paper the results of thermal tests upon a channel section were given, and there was a marked departure from the straight- line relation in the stress diagram, this being accounted for by the asymmetry of the section. Other experimental resuhs were quoted which showed that the tensile yield point as determined by the thermal method agree very closely with that THE TESTING OF MATERIALS 411 found by extensometer. In the curve of temperature plotted against load, there is a sharp cusp at the yield point because the temperature then rapidly rises instead of falling. In these experiments, the observed results have to be corrected for cooling eSects . Although of very great interest, the method seems rather too delicate for very extended application. The Optical Method of Testing Materials. — Sir David Brewster discovered a hundred years ago that when plane polarised light is sent through a piece of glass under stress, an effect is produced upon the light which is detected by the appearance of colour bands when viewed through an analyser. The optical aspects of the subject are beyond our present scope, but the reader will be able to study these from the bibliography given below. The mathematical problems in- volved were dealt with by Clerk Maxwell and others, and in recent years particular attention has been given to the applica- tion of the method to the. determination of stresses in various machine and structural details by Professors Alexander, Filon and Coker. If a beam of plane-polarised light is passed through a speci- men of transparent material, such as glass or xylonite, and Nicol's prisms in the polariser and analyser are set with their principal planes at right angles so as to cut off the light, no effect is produced if there is no stress in the material, but if there is any stress, a colour effect is produced, and regions of zero stress or equal and opposite principal stresses, such as the neutral axis of a beam, can be detected by a dark patch or line. If the material is elastic, the colour produced will be a measure of the difference of the principal stresses at any point. The stress corresponding to a given colour is determined numerically by experiment by uniformly loading a small specimen until the colour produced is the same as at any particular point of the model under consideration where the stress is required. In cases where one of the principal stresses is negligibly small, the stress thus obtained can be taken as equal to the greater principal stress required. 412 THE kSTRENGTH OF MATERIALS In Professor Coker's experiments very successful results have been obtained with models of various structional and machine details cut out of sheet xylonite. Fig. 195 * shows the results of his experiments upon a tie-bar eccentrically loaded ; the resulting curves of stress agree very well with the theoretrical straight-line variation, with the so Fig. 195. — Stresses in Eccentrically Loaded Tie-bar. exception of that at the highest load, in which the material begins to yield at the edges and the straight line bends over as shown. In this case the test-piece showed residual stress at this edge after the load had been removed. In experiments upon models of standard cement briquettes Professor Coker found t that the maximum stress was about 1*75 times the mean stress (cf. p. 78). * Engineering, January 6, 1911. t Ibid., December 13, 1912. THE TESTING OF MATERIALS 413 Fig. 196 "^ shows the results of the experiments by the same investigator upon the distribution of stresses in Fig. 196. — Effect of Holes on Stress in Beams, rectangular beams with holes cut through them. In the upper specimen a hole is in the centre and in the lower one holes are formed half way between each edge and the neutral axis. In this case it wiU be noted that the maximum stress * Engineering, March 3, 1912. 414 THE STRENGTH OF MATERIALS does not occur at the outermost fibre, but at the outer edge of each hole. Professor Coker and Mr. W. A. Scoble, B.Sc.,* have also experimented upon the stresses in tie-bars with holes in them , ...,.., T-p width of plate such as occur m riveted lomts. it c = -,• -^ y ^ — r- "^ diameter of hole and f is the mean stress over the whole width of plate, they find that their results for one central hole may be expressed by the formula maximum stress _ 3 c mean stress c + 1 Also if a is the radius of the hole, the longitudinal stress at a distance r from the centre of the hole on a normal section of the plate is expressed by the formula /, 2+^:+ 3 a* where p is the stress at a long distance from the hole ; while there is also a radial stress given by It will be noted that at the edge of the hole /,. = 3p, i.e. three times the stress some distance from the hole. The following results were obtained wdth a strij) 1 in. wide and '186 in. thick, the load being 100 lbs. Diameter of Central Hole (inches) 1 i Stresses in lbs. per sq. in. P mean , maximum 549 547 568 570 613 584 620 724 868 1035 1470 1560 1770 1850 2040 * Engineering, March 28, 1913; Society of Arts Journal, January 16, 1914. THE TESTING OF MATERIALS 415 Bibliography Brewster, Phil. Trans., 1816. Clerk Maxwell, Collected Pampers, Vol. I. P. Alexander, Phil. Soc, Glasgow, 1887.* KeTT,Phil.3Iag., 1888. Filon, Proc. Camb. Phil. Soc, Vols. 11, 12. Phil. Mag., 1912. Coker, Engineering, January 6, 1911; March 3, 1912; December 13, 1912; March 28, 1913. Journal Society of Arts, Cantor Lecture, 1913. * Reprinted in Alexander and Thomson's Elementary Applied Mechanics (Macmillan). CHAPTER XV FIXED AND CONTINUOUS BEAMS If the ends of a beam are fixed in a given direction so that they are not able to take up the inclination due to free bending, or if a beam rests on more than two supports, the B.M. and shear diagrams will be different from the cases of simply supported beams that we have considered up to the present. In the first case the beam is said to be iixed, built-in, or encastre, and in the second it is said to be continuous. We will consider how the shear and B.M. diagrams can be found for such beams, and will point out their advantages and disadvantages compared with simply supported beams. FIXED BEAMS If the ends of a beam are fixed in a horizontal direction, then the beam when bent takes up some form such as a b c (Fig. 197). If the ends were free it would assume the dotted Fig. 197. — Fixed Beams. form a' b &, and to get it back to the form a b c, negative bending moments, shown diagrammatically as due to forces Fj, Fg, have to be imposed upon it. The ends of the beams will therefore be subjected to bending moments which will be negative because they cause curvature in an opposite direction 416 PIXED AND CONTINUOUS BEAMS 417 to that due to the load. This change in sign of the bending moment means that the tension and compression sides of the beam are reversed. We will consider the cases of fixed beams both from the graphical and the mathematical standpoint, as we did in the case of the deflections of beams. INVESTIGATION FROM GRAPHICAL STANDPOINT According to Mohr's Theorem, the deflected form of a beam is the same as that of an imaginary cable of the same span loaded with the bending moment curve of the beam, and subjected to a horizontal pull equal to the flexural rigidity (E I) . If the ends of a beam are fixed in a horizontal direction, the first and last links of the hnk polygon determining the elastic line will be parallel ; this means to say that the first and last points on the vector line on which the elemental areas of the bending movement curve are set down must coincide. But this is equivalent to saying that the total area of the bending moment curve for the fixed beam must be zero. This enables us to enunciate the following rule — // the ends of a beam are fixed in a horizontal direction at the same level, and the section of the beam is constant along its length, there will be negative bending moments induced, and the area of the negative bending moment diagram will be equal to that due to the load for the beam if considered freely supported. We will speak of the negative bending moment diagram as the " end B.M. diagram," and that for the beam freely supported as the " free B.M. diagram." The problem now divides itself into two cases : (a) That in which the loading is symmetrical. (6) That in which the loading is irregular or asymmetrical. Symmetrical Loading. — If the loading is symmetrical, then the beam looks the same from whichever side it is viewed, and so the end bending moments will be equal, and their value can be found by dividing the area of the free B.M. diagram by the span. This will be made more clear by considering the following cases — EE 418 THE STRENGTH OF :\L\T£RIALS (1) Uniform Load on Fixed Beam. — Let a uniform load of intensit}- w cover a span a b (Fig. 198) of length /. The free B.M. curve is in this case a parabola a c b, with maximum IV l~ ordinate -^• o Therefore, since the area of a parabola is two-thirds of the Fig. 198. — Fixed Beam with Uniform Load. area of the circumscribing rectangle, area of free B.M. curve 2 , w P w P = 3' ''-J- --12 End B.M. = wP 12 J IV P IV P 12 Then setting u^^ a e and b f equal to , ^ and joining e f we get the end B.M. diagram, and the effective B.M. curve is the difference as sho^Ti shaded. At the points G and h the B.M. is zero, and these points are called the points of contra- flexure, the curvature of the elastic line changing sign at these points. FIXED AND CONTINUOUS BEAMS 419 Suppose the point g is at distance x from the centre of the beam, then the ordinate of the parabola must be equal to -,^ XjLI 10 /P o\. w P 1-e. 2 1^4 ^'j - 12 w X- X' IV P 24 12 I X = 2V3 Distance of g from e = ^ — x 2 ^ 2 2V3 = -211 Z Shear Diagram. — With symmetrical loading the shear diagram will be the same as for the simply supported beam. This is because the shear at any point of a beam is equal to the slope of the B.M. curve at that point, and the slope of the B.M. is not altered in the case of symmetrical loading because the base line of the diagram is merely shifted vertically. Deflection. — The deflection at the centre can be found as before by considering the stability of the imaginary cable a^ Bj^. Considering the stability of the left-hand half of the cable, then taking moments about a^, we have E I X 8 = P ^1 - P ^, = P (2/i - ^2) In this case P = area of one-half of the free B.M. curve, 2 I wP wP ~ 3 ■ 2 ■ 8 " ~ ^24^ Vi 51 ~ 16 2/2 I ~4 EI X 8 wl^ /5l l\ wP 24" Ue "iJ ~ 384 .-. 8 wP WP ~ 384EI "" 384 EI 420 THE STRENGTH OE MATERIALS It will be noted that this is one-J&fth of the deflection for a freely supported beam with the same loading. (2) Isolated Central Load on a Eixed Beam. — In this case the area of the free B.M. curve 1, WJ 2^ ^4 ~ 8 .-. End B.M. = 8 Z = Wl Fig. 199. — Fixed Beam with Central Load. The B.M. and shear diagrams are as shown in Fig. 199, the points of contraflexure being at J and J span. Deflection. — As in the previous case we have E I X 8 - P (2/1 - 2/2) Wl I Wl^ In this case P = 8 16 Ui EI.8 = 8 = I 3 / WP 16 Wl 192 EI 4 WP 192 FIXED AND CONTINUOUS BEAMS 421 This is one-fourth of the deflection for a freely supported beam with the same loading. * Asymmetrical Loading. — In this case the end B.M.s will not be equal, and in this case, in addition to the condition that the areas of the end B.M. diagram and free B.M. diagram must be equal, we have the further condition that their centres of gravity must fall on the same vertical line. This can be proved as follows : Considering the imaginary cable and taking moments about one end, the tension at the Fig. 200. — General Case of Fixed Beams. other end passes through the point so that its moment is zero. Therefore the moment of the B.M. diagrams about this point must be zero. Since the areas of these diagrams are equal, their centres of gravity must be at the same distance from the given point. Let a span a b, Fig. 200, of length /, be subjected to any irregular load system which produces a free B.M. curve AcdB, and let the centre of gravity of that diagram lie upon the vertical line G G. Suppose the end B.M.s are M, and M,„ and A a and b b are set up equal to these end B.M.s, then the trapezium a a 6 b is the end B.M. diagram, and the conditions that have to be satisfied are that the area of the trapezium 422 THE STRENGTH OF MATERIALS shall be equal to the area of the curve a c d b, and that its centre of gravit}^ shall lie upon the line g g. Join a b, thus dividing the trapezium into two triangles, and draw verticals X X and y y at distances equal to from a and b. The centres of gravity of the triangles a a b, b « 6 lie on the lines x x and Y Y respective^, and our problem resolves itself into dividing the total area of the curve a c fZ b (which area we will denote by a) into two areas acting down the lines x x and Y y. This is effected bj^ treating the areas as vertical forces, and setting down a vector line 0, 1, to represent the area a. Taking any convenient pole p, we then join p and 1 p and draw x g, g y across the verticals x x, g G, Y y parallel to p, 1 p respectivel}', and joinxy; then drawing p 2 parallel to xy, 1, 2 gives us the area which must act down the vertical y y, and 2, that down X X. Then M, x ^ = area of triangle a a b = 2, • M =2,0x2 Similarly M„ = ' , This enables the B.M. diagram to be drawn. Shear Diagram. — In this case as the end B.M.s are not equal the shear diagram will not be the same as for a freely supported beam, but the base line will be shifted. Since the shear at any point is the slope of the B.M. curve, the base line of the shear curve will be shifted downwards by an amount " ' " " because this is the change in slope of the base line t of the B.M. diagram between the freely supported and the fixed beam. If in the figure ac e d b represents the shear diagram for a freely supported beam with the given loading, then the effect of building-in the ends of this diagram is to ]M — M lower its base line by an amount a a' = b b' = ^ ' , ^ ", thus giving the diagram a' c e' d b'. FIXED AND CONTINUOUS BEAMS 423 Special Gases. — 1. Fixed Beam with Uniformly In- creasing' Load. — Let a beam ab of span I be subjected to a load of uniformly increasing intensity, the intensity at unit distance from b being w tons per ft. run, the total load being W. Then, as shown on p. 139 for a freely supported W 2 W beam, R„ = ,^ , R^ = — and the free B.M. diagram is a para bola of the 3rd order, the maximum B.M. being equal to •128 WZ and occurring at a distance '577 I from b. Then the area of this free B.M. diagram is equal to -r^- and its centre of gravity occurs at a distance ^ from b. This can be proved mathematically as follows — Area of B.M. curve = / M d x / 'w P X 10 x^\ , "6 6 ) '^ ^ w P x'^ w x* ;~I2 "2T + ^_ The area = when x =^ 0. . ' .- c = _wl^ _wl^ _wl^ _^P •■• ^^^^ - 12 24 " 24 ~ 12 First mt. of B.M. curve about vertical through b = jM.xdx I fw P X^ IV X^\ -. V 6 ~ Q^ '^"^ I / wP X^ w x^ f8 30 + ^\ Moment = when x = 0. .* . c^ = -r^. . , wl^ wP wl^ '. 1^ irst moment = ,„ — „^ = .^ 18 30 4o 424 THE STRENGTH OF IVIATERIALS 1st moment . ■ . Distance of centroid from vertical throngrh b area wl^ wl^ 45 ■ 24 24 Z 8Z 45 15 This fixes the Hne g g, and the areas that must be considered W P W l^ as actmg up x x and y y respectively are thus ^ and -^ \YP 31 since the total area -^^ acts at distance ^ „ from y y. . • . Taking moments about y Y we have- Area actmg down x x x 7^ = -tt. x , ^ 6 iZ ID . ^. , WP I WP . • . Area actmg down x x = ^„ -^ v. = >./^ bi) 6 ZO „ WZ2 2 WZ M — V — = 20 Z 10 ,, WZ2 2 Wl M — - V = 30 Z 15 The resulting B.M. diagram then comes as shown shaded in Fig. 201. The amount of shifting of the base line for shear will be wz wn w 7W ^TT. — , K -^ Z = ^^ SO that the shears at the ends are - ^ „ _ 10 15 J SO 10 3 W and ^ respectively, the shear curve for the fixed beam then coming as shown. 2. Non-central Isolated Load. — The following construc- tiori can be used for this case : Let a b, Fig. 202, represent a fixed beam of span Z carrying an isolated load W at a point p, the distances of which from a and b are a and b respectively. First draw the "free bending moment" diagram adb, i. e. set up p D = — , — , and join d to a and B. Project d horizontally to meet the vertical through the support b at E and join e a. FIXED AND CONTINUOUS BEAMS 425 Then pf = Reverse or end bending moment at b = M^; and F D = reverse or end bending moment at A = M,. Fig. 201. — Fixed Beam with Uniformly Increasing Load. Therefore set up b h = r p and a G^ = f d and join g^ h, the complete bending moment diagram then coming out as shown shaded. This may be done by projecting f horizontally and drawing d g^ parallel to F a. 420 THE STRENGTH OF MATERIALS Proof. — To prove this construction ^\c must obtain values for the reverse bending moments for this case. Referring to Fig. 203, it will be remembered that in fixed beams the con- Jl ./J Fig. 202. — ^Fixed Beam with Isolated Load. ditions to be satisfied are that the " free '' and " end " bending moment diagrams a b d and a G^ H b respectively must be equal and op^iosite in area, and must have their centroids upon the same vertical line. Fig. 203. — ^Fixed Beam with Isolated Load. The first condition gives us —^^^ — — - = area of A a d b = i a b . r ]) I .Wab Wab 1 2 • (1) FIXED AND CONTINUOUS BEAMS 427 The end bending moment diagram a Gj. h b may be con- sidered as divided up into two triangles a g^ B, b g^ h, whose centroids act in the " third Unes " x x and y y respectively. We have next to calculate the position of the centroid g of the free bending moment diagram. According to the ordinary rule, the centroid g will be one-third of the way up the median line c D. 1 1 fl, •*• ^ " 6 ^^ ~6 3V2 ""y ~3 Regarding the areas of the triangles ag^b, bg^h, adb as concentrated in the lines x x, y y, and g g respectively, we have by taking moments about the line x x Area ofAADB xa; = area of A b g^ h x ^ o i. e. from (1), — ^ — . .t = | M,, ? x ^ M,J2 ^Wab _Wab a ■ • 6 ~ 2 '^ ~ '2' 3 _ w an ~ 6 • M = ^^^ • • -^'-Ljj 72 Similarly, M^ = „ — As a check (M, + M„) = '^f ^ + ^ |^ Wa6. ,, Wab , Wab = — ^ (a 4- 6) = ^2 '^= I (M, -f MJ I Wab (as in (I)). Wab Now, smce — -. — = p d „ _PD X a. PD X 6 428 THE STRENGTH OF MATERIALS but in Fig. 202, by similar As P F AP a B E AB ~ r P F B E I .a p D . a Similarly, e h = d f = — , " = M, Position of Load for !MAxiMr^r Reverse Bending Moment.- — The position of the load for a maximum value of the reverse or the end bending moment M, is obtained by putting -^' = and noting that a = {I — b), dh " - ^ W d{ah^) _ i^ dh i.e. ^fP = do 2 6 Z - 3 62 = I. e, = o • or Taking the first value, which is clearly the maximum, then M _^3V3y _4WJ _ W| P ~ 27 ~ 6-75 Therefore the maximum reverse bending moment for an Wl isolated load is equal to w;^^, and occurs when the load is at one-third of the span. !MAXoir>E Positive or Intermediate Bending Moment. — Referring to Fig. 202, the maximum intermediate bendmg moment occurs at the load point p and is equal to D J. . • . Maximum intermediate bending moment = !Mp = D J = P D — J p Wab FIXED AND CONTINUOUS BEAMS 429 Now, remembering that {a + b) = I JP=PF + FJ = M3 + (M.-MjJ = M„.'; + ^^* _ Wtt^6 a W.ab'b = Waft ^^, _^ j,^ ~ ^ \ Z2 = — y^ — .2 ah _ 2 . W g^ 6^ To get the maximum vahie of this for any position of the load put ", - = '• '• ~l^' \ d^ / " ^ d.a^a^ -2al -\- a") I.e. J = da i.e. 2a?-^ - 6ft^Z + 4a3 _ i. c. Z^ ~ 3 a ^ + 2 a'-^ - (^ - a) (Z - 2a) = I . I.e. a — ^, or I. Taking the former value, which is the only one possible, wc have Therefore the maximum intermediate bending moment W I occurs when the load is at the centre and is equal to -^— . o 430 THE STRENGTH OF MATERIALS Graphical Method of finding g g. — If the nature of the loading is such that the position of the hne g g cannot be calculated without difficulty we may proceed as follows : Divide the free B.M. diagram acb up into a number of vertical strips, not necessarily equal, and draw vertical force lines through the centres of these strips and set down the ordinates on a vector line, and wdth any pole draw a link polygon. The point Avhere the first and last links meet will be a point on the line G G. This is the same method as adopted in finding the centroid of a figure by Mohr's method (Chap. LX). The area of the B.M. diagram can be found by sum-curve construction, and the problem completed as indicated with reference to Fig. 200. INVESTIGATION FROM MATHEMATICAL STANDPOINT As we have previously seen r M Slope of beam = / -^ ^ ^ ^ If the end of the beam is built-in this slope must come zero at the two ends. Consider the following special cases — (1) Uniform Load on Fixed Beam. — Taking the in- tensity of load as w and the centre of the beam as origin, theh considering a point at distance x from the centre, for the freely supported beam we have M,. = |(^' - x^) (See p. 266.) Let the effect of the building-in be to cause an end B.M. = M^ Then for the fixed beam M., = ^ (4 " ^') " ^^ Slope -= / -p, -J- d X 1 (w P" X w x"^ EIV 8 6 Slope is when x = 0. . • . c = 0. Also slope must be when ^ = 9 - M,, X -f c j FIXED AND CONTINUOUS BEAMS 431 .-. ' 48 i'" I '. c. M, X 2 wP 16 w P ~ 24" wP ~~ 12 wP 48 • To obtain the deflection we integrate again, and we get deflection =/ /EI^^^ 1 ~EI /i^; P x^ \ 16 w x^ 24 M, a;2 2 + Ci^ 1 "~EI Z!/; P x^ V 16 w x^ w P x^ 24 + Ci^ 1 ~ EI /i/; P x^ V 48 w x^ \ 24 +^V This is when x I ~ 2 •*• 192 wP 384 + ^1 ~ 384"" = ^ .'. When .T = 0, deflection = =ri\ E I "" 384 E I .-. Maximum deflection ^ 3^-^-^ ^ ^ 3^^^ ^ The B.M. and shear diagrams are then as shown on Fig. 198. (2) Isolated Central Load. — Taking as before a span I and the centre as the origin, if the load is W, for a freely supported beam we have ^■'" "" 2 \2— ^. .'.If the end B.M. due to fixing the ends is M,, we have for the fixed beam M. = ^ (2^ - a;W M. 432 THE STRENGTH OF IMATERIALS /M When X ^ O,«slope = 0. . • . Cg = 0. When X = x, slope also =- in this case. He have : = -^ -~ M, ^ V 8 16 ■' 2/ "" E I X I W/2 M . - =: - ' 2 16 M ^^^ 8 To get the deflection we integrate again, then deflection = / / t^i j d x 1 /Wlx^ Wx^ M. a;2 This is when x = ^ EIV 8 12 2 _ J^/Wl .t2 _ W^ ~EI\ 16 12 I + c, + C3 WJ3 _ WP - 64 96 + ^3 - ^ ' ^3 ^- 192 c. When X = 0, deflection -^ --t E I W /■■ .". Maximum deflection = 192 E I * (3) Fixed Beam with Uniformly Increasing Load. — Let a span a b of length I have a uniformly increasing load, of zero intensity at the point B, and let the intensity of load at iHiit distance from B be i:; units per ft. run. Then taking the end B as origin, we have in the case of the freely supported beam ,^ wl'^x w x^ M'= 6—6 FIXED AND CONTINUOUS BEAMS 433 Now let M, and M„ be the end B.M.s, then the negative B.M. at distance x from b is equal to . * . for the fixed beam /M =Frj • d X _ 1 I'l^Z^a;^ wx^ /M^j-MAa;2 ^ ~ EI 1 12 ~ 24 ~ ^-^"^ "" V I J 2 + ^^/ • • ^^' When X = 0, slope = 0. . • . c^ = 0. Also when x = l, slope = • • 12 24 ^^^^'^~ 2Z ~^ 2 ■ 2' ~ 24 .-. M. +M3 = '^j2' (^) To get another relation between M, and M^, consider the deflection ; then deflection = / / =nrj d x _ 1 \wl^^^ wx^ M,x^ ( M, - M, \ ^ , 1 .ON ~EI\ 36 120" 2 V I y-e"^^^/ "^^^ Deflection = when x = 0. . '. C5 = 0. Also deflection = when a; = Z 36 120 2 V Z / 6 - M., Z^ _ M, j2 M„J2 ^ ^^^Z5 _ w; Z5 2 6 ^ 6 ~ 120 36 M, Z2 M, j2 ^ ^wl^ •'• 6 "^ 3 ""360 .-. M. +2M,= ^^^^ (4) FF 434 THE STRENGTH OF MATERIALS . • . Combining (3) and (4) Ave get ^ _ 1 W P IV P " "~ "60^ 12 ~ 30 ~ 15 • ' ^ ^- 12 30 ~ 10 The B.M. diagram then comes as shown in Fig. 201. In all the above cases we have assumed that the beam is of constant cross section along its length. If such is not the case, the end B.M.s can be found by taking the corrected B.M. diagram as explained in the chapter on the deflections of beams. Advantages and Disadvantages of Fixed Beams. — We have seen that, in the examples that have been considered, a fixed beam is stronger than the corresponding freely sup- ported beam, and that the fixed beam has smaller deflections and is thus more rigid. In most cases, moreover, the maximum B.M. occurs at the abutments, Avhere the beam can be strength- ened -wdthout adding materially to the bending moments and thus increasing the stresses. In the freely supported beam, on the other hand, the maximum B.M. occurs at the centre. Avhere an addition of weight to strengthen the section would add materially to the B.M. The reason Avhy such beams are not more commonly adopted is because, in fixing in the ends securely, the tangents at each end to the beam must be absolutely horizontal, and any deviation from this will alter the stresses, and any difference of level at the two ends due to unequal settlement would cause considerable stresses in the beam. There is also considerable stress due to change in temperature if the beam is securely built-in to the masonry, and all these points make the actual stresses in any practical case somewhat uncertain, so that many designers do not use this type of beam. All the above objections can be obviated by cutting the beam through at the points of contraflexure and resting the centre portion on the two end portions. This is the principle of the cantilever girder construction and for FIXED AND CONTINUOUS BEAMS 435 large spans is very economical. This is shown diagrammatic - ally in Fig. 204, in which a fixed beam a b is shown divided at the points of inflexion c and d and the centre portion is represented as hanging from the end portions. The B.M. in the centre portion will be the same as for a freely supported beam of span I loaded in the given manner. The B.M. for the cantilever portions will be the same as for cantilevers of span Zj loaded with the given loading and also with loads at the ends equal to the reactions at the ends of the centre portions. In the figure, uniform loading is shown, and in such case these reactions are each equal to -^ . It will be Fig. 204. found that the resulting B.M. and shear curves obtained in this way will be the same as shown in Fig. 198. The deflec- tions can also be found by adding together the deflections at the centre of the centre portion and at the end of one of the cantilever portions. Fixed Beam with Ends not at same Level. — Suppose that a fixed beam a b. Fig. 205, has its ends at a different level, then apart from the loading on the beam, the deflected form of the beam will be as shown in the figure, the point of contraflexure being at the centre point c. The deflection 6 e of the portion a c, assuming the beam divided at c, will be equivalent to that due to a weight P hanging downwards at c, but for a cantilever with load at end WZ3 8 = 3 EI 436 THE STRENGTH OF MATERIALS In this case we have eh = . P - Z\3 3EI 24 EI X eb P 12^1 X cf 12EI X /9 7x3 The upward deflection due to R^, = S^ = ~vo ^i^-~ Then 8 — 8^ = difference in level between final posijtions of A, B, and c. Now let R3 = if Z -f 2 /, 2 / being the additional reaction due to the beam being continuous, then R. = R.- = ^' - / . • . Sink of central column = Sink of end columns e w I '2 - ' e Difference = S — S^ = ( ^^ -f 3 / 1 /3 R3 ; f-ivl 1 /S R« ~ — tv e ; ? ) = S - 8^ R. _ 5w]^ _ R„ P ~ 24 E I 6 EI P 3 \ DWl^ . wl 6EI'2e/ 24 EI' e 5 IV l^ . IV I .' . R« = 24 EI e J3 _^ 3 6EI ' 2e = IV I 24 EI e I P I 6 E 5 . 6EI = IV I \ I ' 2eJ [5 , 6En J 4 ' eP I I - ^ M FIXED AND CONTINUOUS BEAMS 445 Reasoning as before, we then get M„= 2 5 6E^ 4~^ eP 1 + 9EI - 1 It will of course be noted that if the piers had been of the same material and of areas proportional to the reactions, the amount of sinking due to their elasticity would have been equal, and the B.M. diagram therefore would remain as shown in Fig. 208. * The Theorem of Three Moments. — We will now find the relation which must exist between the support bending moments and the loading for a continuous beam of any number of spans, the supports all being on the same level. Let A B and b c be any two consecutive spans of length l^ and ^2 of a continuous beam of any number of spans, and let A e B, B / c (Fig. 210) be the free B.M. diagrams for the loading on these spans. Let G^ and G2 be the centroids of these free B.M. diagrams, and let them be at distances y-^, y<^ respectively from A and c, the areas of the diagrams being respectively Si and Sg. Then, if M^, M3, M^ are the support moments at A, B, and c respectively, Cla'peyroni' s Theorem of Three Moments states that M, ?i + 2 M3 {h + h) + MJ2 = 6 1 ^y^ + ^f^) We can prove this with the aid of Mohr's Theorem* as follows : Let a! b' c' be the deflected form or elastic line of the beam, then if the beam is of the same material throughout, and of constant cross section, the elastic line is of the same shape as that of an imaginary cable loaded with the B.M. diagrams and subjected to a horizontal pull equal to E X I. Now the * See p. 252. 446 THE STREXGTH OF >L\TERIALS tangent to the imaginary cable is common at the point b'. Let such tangent be at angle 6 to the line a' b', and let the perpendicular from a' on to it be of length ii\, the tension in such cable at b' being T„ ; then considering the stability of the imaginary cable we have by taking the span a b and taking moments round a' Tb' ^ Vi = moment of B.M. diagram about a' = Sji/i — ■ moment of support B.M. diagram about a' Mc e .^ h < K Ma ^^ f^ ^p:'^=-^-^ Pn /I \ B c '-^'^ / 7 *-/2^ h 6^ ^ ^.B' % ^--*^ ■- »^ £-/o, ^^^"^ r//c Line c Fig. 210. — Theorem of Three Moments. = Si2/i-M,^'x J Ar / - / - ^*^«2 3 (1) because the support B.M. diagram can be di\-ided with two triangles of area ^-^ and ^— ^^, the distances of their centroids from a' being respectively -^ and ^ \ Xow p^ = I^ sin 0, and T - EI cos , E I being the horizontal pull in the cable. FIXED AND CONTINUOUS BEAMS 44: m E I Zi sm • , T„, X ;Pi = ^-. = E I Zi tan ^ cos .•.EItan^=%i^-^A_^i (2) Now by considering the second span, as is the same for both spans and E I is constant, we get EItan^ = -(S^f -^^'^-2M^) (3) The — sign is used because the moments are taken in opposite directions. Then, combining equations (2) and (3) we get ?! 6 6 V ^2 ^ ^ or M, Zi + 2 M, {I, + Z2) + M, l, = 6 (^"^-^^^^ + ^f ^^) • • • • (4) TA^5 z5 jj 2, 3 = ,, ,, abB = S„ Then with pole p at polar distance (/;) = E I if A^d is drawn parallel toOT,dh to If, hg to 2 f and ^ b^ to 3 p, cZ h and h g are called the mid links, and a^ d and g b^ give the support tangents. Fig. 213. — Continuous Beams — Graphical Treatment. Now in our problem Ave do not know the position of the points and 3, and we see that these would be known if the mid links were found, so that our j^roblem now reduces to that of finding these mid links. On both sides of the centroid vertical g g draw lines at distance p, and set down lengths l^, 2^ equal to 1, 2 and join them across, intersecting on the centroid vertical. These lines are called the cross lines. Now draw any vertical u u, then clearly the intercepts made by the vertical on the mid links and cross lines are equal. FIXED AND CONTINUOUS BEAMS * 453 From this it follows that if a point on one mid link is known, a point vertically below it on the other mid link can be found. Again, let the right-hand mid link of this span meet the left-hand mid link of the next span in a point j, Fig. 214, on a vertical line q q. Fig. 214. — Continuous Beams — ^Fixed Points. They are similar Then consider the triangles g j k, f 2 3. j k _ x-^ p X j k = 2, 3 X X;^ = Xj^ X area 6 as 454 THE STRENGTH OF MATERIALS Similarly considering the triangle on the other side of Q Q we should have Where Zg is the length of the next span. Further, x-^^ -\~ X2 = k (li + h) h • • ^1 ~ 3 x<, - 3 . • . Q Q is at a distance = -^ from y y, and is thus called an inverted third line. Determination of "Fixed Points." — Let ab c (Fig. 214) represent two consecutive spans of a continuous beam, and let the third lines be drawn as shown. Suppose that we know that the right-hand mid link of the span A B passes through a fixed point r. Let this mid link cut the inverted third line q q in J and the third line y Y in l, then L b' must be a support tangent. Produce l b' to meet the first third line of the span b c in l', then j l' is the left- hand mid link; and then join fb' and produce it to meet J l' in f', then f' will be a fixed point on the mid links of the second span. This is shown as follows — Let the vertical through f' be at distances z-^, 23 from the third lines. Then the triangles f' j n, f' k' l' are similar. • • k'l' 22 ^^ and triangles b' k' l', b' k l are similar. (2) k' 1/ I2 further, the triangles f l k, f j n are similar. - , (3) JN /2 FIXED AND CONTINUOUS BEAMS 455 Multiplying together ( 1), (2) and (3), we get 1 X r ^2 12 h . ^1 ~ ih _ h /2 ^2 hh also Z^ -T ^2 h + h 3 .-. Zi - 1Z2 = ~Z2 + ^h . -^2 + ih _^l/2 ^2 ~hfl = 1 + ^^^2 •' ' ^2 — 1~T~^JIjT ~ constant. h / 2 + ^2 /I . • . r' is a fixed point. In this way a number of fixed points right along the various spans can be found as hereinafter further explained. A fixed point is found at the terminal spans, as follows — Case 1. Freely Supported End. — The end B.M. here must be zero, therefore support tangent and mid link must be collinear, so that a' is the first fixed point. Case 2. Built-in or Fixed End. — Support tangent is horizontal, so that first fixed point is where horizontal through a' cuts the first third line. Graphical Construction for any Given Case. — We are now in a position to set out the construction for obtaining the B.M. diagram, which is as follows — Draw the free B.M. diagrams and the third lines, the inverted third lines and the centroid verticals. Fig. 215 shows a continuous beam of three spans, one end being freely supported and the other fixed, x x representing the left-hand third lines, Y Y the right-hand third lines, q q the inverted third lines, G G the centroid verticals. Now draw the cross lines at the bottom of the paper, such lines being obtained by setting down the areas S^, Sg, etc., of the free B.M. curves on vertical lines at each side of the centroid verticals at distances representing the value of E I 456 THE STRENGTH OF MATERIALS CO C a, eS « CO o •S '43 o O eg to e3 o a, O 6 ^^^ FIXED AND CONTINUOUS BEAMS 457 reduced in some convenient ratio, the scale of E I being the same as that of the areas. If the support moments only are required and not the deflections, and E I is the same for each span, E I need not be calculated, any convenient polar distance being taken. P, P^, and Pg are the intersections of the cross lines. Now find the fixed points. The end a is fixed, so that f is the first fixed point ; now set down r r' equal to the intercept ■ / /i on the cross lines and draw any line r' J^ to the inverted third line, cutting y Y in l ; join L b' and produce to meet the third line x^ x^ in l^ ; then the intersection of l^ j^ and e' b' gives the fixed point Fj^ on the second span. This is repeated as shown, and the points f^', Fg, Fg' found. ^ow start the other end d. This is freely supported, therefore, as we have seen before, d' is the first fixed point Hg. By means of the cross lines, we then get the corresponding fixed point Hg', and by repeating the same construction as for the points F, we get a number of other fixed points, Hj', H^, h', h. The mid links and support tangents are now drawn in, and there will be two checks on the accuracy of the construction, viz. — (a) Mid links must meet on centroid verticals. [b) When adjacent mid links are joined, they must pass through points of support. Now, from the points 1, 2, etc., on the cross lines, draw parallels to the mid links, and obtain the poles r, r^, Rg and then draw parallels to the support tangents, thus obtaining the points 0, 3, etc. Then and so on, the support moments then being set up and the true B.M. curve for the continuous beam thus being found. Another interesting graphical method of finding the support moments in a continuous beam has been devised by Professor Claxton Eidler, and will be found in his book on Bridge Construction. 458 THE STRENGTH OF MATERIALS Advantages and Disadvantages of Continuous Beams. — It will be seen by considering the B.M. diagrams for continuous beams that the maximum B.M. is less than that which would occur if a number of separate simply supported beams were placed across the same supports (ex- cept in the case of two uniformly loaded equal spans, when it is the same), and that such maximum B.M. occurs at the abutments. The principal disadvantages are — (a) It is not easy to ensure all the supports remaining at exactly the same level. (6) The method of calculation of the stresses assumes that the beam is of uniform cross section throughout, this condition not being an economical one. Experimental investigations in Germany have shown that if the beam is not of uniform cross section, the method described may still be employed without great error, (c) In the case of rolling loads, which occur frequently in bridge design, the • calculations are much more difficult than in the case of separate spans. Beams Fixed at one End and Freely Supported at the Other. — If a beam is fixed at one end and freely sup- ported at the other, the B.M. and shear diagrams will be the same as for the half of a continuous beam of two equal spans of the same span as the given beam, and loaded in the same manner. This is because fixing the end of a beam makes such end horizontal, and this is what happens at the central support of a continuous beam with two equal spans loaded in the same manner. The consideration of the following two standard cases should make this clear. (a) Beam Fixed at one End and Freely Supported at THE Other, Subjected to a Uniform Load. — The B.M. and shear diagrams in this case are the same as for one span of the first case of continuous beams that we have considered, and will therefore be as shown in Fig. 216. FIXED AND CONTINUOUS BEAMS 459 (6) Beam Fixed at one End and Freely Supported at THE Other, Subjected to a Central Load. — Let the central load be W and the span I. Then, if b is the fixed end, a the freely supported end, and a' the imaginary freely supported end existing beyond the fixed end, we have, by the Theorem of Three Moments, MJ + 2M,(? + Z) + M,Z = 6|^ix|x2^^ + ^x|x^^ Now M, = M,, = .-. 2M,..2? = 6 W Z3 W ^3^ . M. IQl 16 16 Z J } Fi€t. 216. Fig. 217. Beams Fixed at one End and Supported at the Other, The B.M. diagram then comes as shown in Fig. 217. To get the shear diagram we first work out the reactions. j^ _ W , M, - M„ . ' . R,. ~ 2^ I W 2 3WZ IQl 5W 16 11 W 16 The shear diagram then comes as shown in the figure. 460 THE STRENGTH OF :\L\TERIALS (c) to find the maximum deflection for a uniformly Loaded Beam fixed at one End and Supported AT the Other. The bending moment diagram for this case is as shown shaded in Fig. 218. The curve b d c is a parabola of height -^ where iv is the load per unit length of the beam, this being the B.M. diagram for the downward miiform load on the canti- FiG. 218. — Deflections of Beams. lever, and a j is equal to 3 ic P ! J B being a straight line ; this 3 ic I being the B.M. diagram for the reaction at b, which is ^— 8 Our first problem is to find the point n at which the deflec- tion has its maximum value. Consider the position Aj^n of the imaginar}^ cable. The forces acting on it are a horizontal tension equal to e i at n and an equal horizontal tension at the point a^, since the beam must be horizontal at the fixed end a; also an upward vertical force equal to the negative FIXED AND CONTINUOUS BEAMS 461 area cjb, and a downward vertical force equal to the positive area d f g. If these forces are in equilibrium, since the horizontal forces are equal and opposite, the vertical forces are also equal and opposite, so that we get the following rule — The maximum deflection will occur at the point where the area D F G ^5 equal to the area d J c. This is the same as saying that the area a h g J is equal to the area a h f c. Now, if H B = X and a b = Z HG_AJ ^ _a;AJ — y— . * . H G y- X I I Area a h g J = -^ (a J + g h) V '- X I -I , X A J 1 + .2 ^" V Z {I - x) {I + x ) SwP '2~~ I ' 8 3wl 16 (P -x^) (1) Also Area a h f c == Area a b d c — Area f h b 1 1 = ;^AC.AB — „ FH.HB Iwl^ 1 W X^ ^ 3~2 "" 3 ~T~ If (1) = (2) (2) ^^ (l^ - ^') =^ I (^' - ^') • Factorising, we get ^^Q^ {I + x)(l- X) =^~{l- ^) (l' + l^ + ^^') w . ■ . dividing through by ^ (^ — ^) ^^^ multiplying across we get 9?2 + 9?a: = 8Z2 + 8Za: + 8x2 i.e. Sx^ -Ix -l^ = (3) 462 THE STRENGTH OF MATERIALS The general solution of this quadratic equation is X ^ _ I (1 ± V33) 16 The negative value is inadmissible .•.. = L(L+/33)^. ^2 nearly. . • . The maximum deflection occurs at a distance = '422 1 from the simply supported end. We now proceed to find the maximum deflection S by con- sidering the stability of the portion n Bj of the imaginary cable. The forces acting on it are a tension at B, the horizontal tension E I at n, and the area of the bending moment diagram B F G. By taking moments about the point b^, we eliminate the tension at this point and get E I x 8 = moment about B^ of area b F G. Now, this area is made up of the difference between the A B H G and the parabola b h f. X T __ I x^ S wl^ _ 3w x^l The area of the A = jGH.BH = J.^.Aj.a; M • 8 16 2 X The centroid of the A is at distance -^ from b , p ^ , , Zwx^l2x w x^ I .' . moment oi A about b, — — ^tt^ — . -^ = --;: — lb 3 8 The area of the parabola = J f h . b H _ I w x^ _ w x^ The centroid of the jDarabola is at distance — ~ from b. zv X 3 X . • . moment of parabola about B, = —7, . -^ o 4 w x^ FIXED AND CONTINUOUS BEAMS 463 .* . moment about b^ of area b f g w x^l w x^ E I X 8 = -j^ (Z - a;) putting X = '422 I ^ T ^ w X -4223 P (-578 I) . • . EI X S = ^ — ^^ ^ o = -00543 w Z* , -00543 w l^ •*• ^^ — Eir~ putting wl = total load = W . -00543 W P ^ = EI _ WP 184 EI For a uniformly loaded beam, simply supported at each 5 W Z^ end, we should get 8 = „- . t^ j , while for one similarly loaded, WP but fixed at each end- we should get 8 = ttttttTtj so that ^ 384 E I we see that, in the case under consideration the deflection is between these two values. This is, of course, what one would expect. The same method may be applied to the case of an isolated central load W on a beam similar fixed. In this case the maximum deflection = >_ and occurs at ^7^ from the 48 VS E I a/S simply supported end. We will conclude this chapter with a further number of worked examples of fixed and continuous beams. Worked Examples. — (1) A beam of 20 ft. span is built-in at one end and is supported at a point 5 feet from the other end. Draw the B.M. and shear diagrams for a uniform load of J ton per foot run. 464 THE STRENGTH OF MATERIALS Let AB (Fig. 219) be the beam, fixed at the end a and supported at the point c. The portion b c of the beam acts as a cantilever, and there- fore the B.M. at c = M, 1 5x5 ^ X ^ = 625 ft. tons. To find the B.M. at a, we imagine a span a c' exactly similar to A c to exist within the wall. Fig. 219. Then, by the Theorem of Three Moments, we have M, X 15 4- 2 M, (15 + 15) + M, . 15 = ^ (15^ + 15^) o But M,. = M, - 6-25 .-. 60 M, + 30 X 6-25 - ^(2 x 15^) o .-. 4M, + 12-5 = ^f 4 1 ^12 ... 4M, - , - 12 5 4 = 56-25 - 12-5 =43-75 .■ . M, = 10-94 ft. tons nearly. The B.M. diagram is then as shown in the figure. To get the reaction at c we proceed exactly as in the case of con- tinuous beams FIXED AND CONTINUOUS BEAMS 465 . ^ 1 15 , M, - M^ 1 5 M, -M, I.e. J:^, - 2 . 2 "^ '""'15 '^2'2 "^ 5 = 3-75 - -31 + 1-25 + 1-25 = 3-44 + 2-5 = 5*94 tons. The shear diagram then comes as shown in the figure. (2) A roiled joist is firmly built-in at one end, and the other end rests freely on the top of a cast-iron column. The span of the joist is 16 feet, and it carries a single load of 10 tons, 12 feet Fig. 220. — Example of Beam Fixed at one End and Supported at Other. from the column ends. Determine the reaction on the column, and draw the B.M. and shear diagrams. {B.Sc. Lond.) Let A B represent the beam, fixed at the end a, the load being at the point c (Fig. 220). Then the free B.M. diagram is a triangle a d b, c d being equal to 30 ft. tons. Wa& 10 X 12 X 4 I 16 Then area of B.M. diagram = J X 30 X 16 - 240 sq. ft. tons. The centroid g of the B.M. diagram occurs at a distance HH 466 . THE STRENGTH OF MATERLILS J E c from E the centre of the beam, i. e. at a distance 9J ft. from A. Then, imagming a span exactly similar to ab to exist beyond the fixed end, we have, by the Theorem of Three Moments 16 M3 + 2 M. (16 + 16) + 16 M, = 6 {^^\l^'' + ^^'} M3, = M, = .. T^ 6 X 2 X 240 X 28 ^ ^,^ 64 M, = — ^ = 7 X 240 16 X 3 M = 7 X 240 _ 210 64 " 8 = 26-25 ft. tons. The reaction on the end b for a freely supported beam = ^B = — T^ — = 2*5 tons. Id . '. In this case R^ = r^ H -~~ = 2-5 + I - 26-25 16 = 2-5 - 1-64 = *86 tons. (3) A continuous girder consists of two unequal spans of 100 ft. and 120 ft. respectively. The girder is 300 ft. long and overhangs the end supports at each end, and is loaded as shown {Fig. 221). Draw the B.M. and shear diagrams and show the points of inflexion and magnitude of the supporting forces. {B.Sc. Lond.) In this case the end pieces A B, d e act as cantilevers. ,, 40 X IJ X 40 , ^^^ J., . .• . M„ = ^ = 1,200 ft. tons. 40x2x40 ^1^600 ft. tons. . FIXED AND CONTINUOUS BEAMS 467 The free B.M. curve for span b c is a parabola with maximun' ,. ^ li X 100 X 100 , ^_^ „^ ^ ordinate = -^ ~ = 1,875 ft. tons. The free B.M. curve for span c d is a parabola with maxi- ,. , 2 X 120 X 120 o^AA^^ ^ mum ordmate = = 3,600 ft. tons. lit^ns l^cr Ft. run ^*"^ M ^'trun nnmoononnoOnnOOOOOOO ^ 40-»- lOC B.M. Dioqram 12.0 Shear l^loaram Fig. 221. Da — >■ ■<- 40 -> Then applying the Theorem of Three Moments we have 100 M„ + 2 M, (100 + 120) + 120 M, = i (1 J X 100^ + 2 x 120^) .-. 120,000 + 440 Mo + 192,000 = 375,000 + 864,000 440 M, = 927,000 M, = 2,107 ft. tons nearly. 468 THE STRENGTH OF MATERIALS We now proceed to the determination of the reactions, o - 1 V X X = X X b. . ' . Mean shearing stress along d d D = F b X = S X a.y . . x bx .1 Sj, = S .a.y 1.6 • (4) We can express this in terms of the mean stress m = -^ over the whole section as follows — _S . a .y ^" ~ AVkH a.y ,_, We may call j~ the shear coefficient. It will be noted that a x y increases up to the neutral axis and then decreases, because the first moment of the area below the N.A. is negative. We thus see that the shear stress is a maximum at the neutral axis. It must be remembered that 5^ gives only the mean shear stress along d d. This stress is not uniform along d d, but for sections which are narrow at the neutral axis, the sections used in practice generally falling under this head, the maximum shear along the neutral axis will be not much greater than the value of s^ at the neutral axis as given by the above result. For sections like the square and the circle the maxi- mum shear along d d will be from 5-10 % greater than the mean shear, while for sections such as an oblate ellipse or a broad rectangle the difference may amount to as much as 25 %. It is beyond our present scope to go further into the question as to the variation of shear stress along d d, but we should remember that such stress is not uniform; the maximum stress for various cases has been worked out by St. Venant. 474 THE STRENGTH OF MATERIALS Consider the following special cases (Figs. 225, 226). (1) Rectangular Section. — Mean shear along a line at distance x from N.A. of a rectangle of height h and breadth b _ _ ^ -y - -5. - w . -p-^- In this case a -x]b y = X + l/h k^ = 12 2V2 X = + x fxeclarkjle Parabola Cinclt Fig. 225. s. m .[^ — xjh . ^[^^^ X 12 6 m ( . — x^ = 6 m ( J - ^, _ 3m /, _4:X^ ~ 2 \ 'W. This depends on x^, so that the curve showing the mean shear stress at various depths will be a parabola. The maximum value of s,. occurs when a: = 0, ^. e. at the neutral DISTRIBUTION OF SHEAR STRESSES 475 axis. This gives 5<, = —^ — 1*5 m. Thus we see that in a rectangular beam the maximum shear stress occurs at the centre, and is equal to 1*5 times the shearing force divided by the area of the section. (2) Circular Section. — This case is not quite so simple as the previous case, but we can find the shear stress at the N.A. simply as follows — In this case we have 7rD2 2D 6 =D ttD^ 2D 8 • Stt . • . 5„ . = m D2 16 4 w .D = 1-33 m So that the mean shear stress along the N.A. is 1 J times the mean shear stress over the whole section. In this case it is interesting to note that the maximum shear stress along the N.A. is 1*45 m. (3) Pipe Section. — Let a thin pipe be of mean diameter D and thickness t (Fig. 226). Then a = —^ D y ^^ IT D2 ^ ~ 8 6=2^ ttD^ D . • . 5v A = 'W* X -:^ = 2 m ^ X 2* 476 THE STRENGTH OF MATERIALS So that the mean stress shear along the N.A. is twice the mean shear stress over the whole section. (4) I Section. — To calculate the proportion of the shearing force carried by the flanges and web, respectively. Take a beam of I section of breadth h and height h, and let the thickness of the flanges and the web be t and w, respectively. First consider a horizontal line p p in the flange at distance X from the top edge (Fig. 226). i^ Pi P-h7--f — ?4^ uy T -?, B \K Ml Fig. 226. Then mean shear along p p = m . ^^g ; s,. = m .b X {h — x) " 6 F 2 — 9 y^ 2 V ^ ^ ) (1) This depends on x^, so that the curve showing the variation of stress is a parabola. When X = t, i. e. at the junction of web and flange, St = 2^2 (^ ^ - ^') (2) Now consider a horizontal line Pj^ p^ in the web at distance x^ from the top. Then mean shear along pj p^ — -' '— DISTRIBUTION OF SHEAR STRESSES 477 In this case ay = first moment of area above p^ p^ about N.A _b t {h — t) ^ ^ ,_ ^^ \h /^ ^ , ^1 — ^ + w{x,-t)\^^-[t+^-^) _b t {h — t) w {Xi — t) (h — x^ — t) - 2 + 2~' also b = w in general expression for shear stress. 'n 9 1,2 m fbt{h — t) (% — t) (h — x^ — t) 2 k^ I w w _ m , mt (h — t) {b — w) .ox - 2 p (^ ^1 - ^-1 ) + -^2k^w « ^^^ The second term of this expression is constant for all values of x-^^ and the first term is the shear stress which would occur if the fianges extended down to p^ p^. We thus see that the diagram of distribution stress is obtained as follows — First draw a parabola a k d, the centre ordinate J K of h which is obtained by putting a; = ^ in equation (1). I.e. JK = 2PV2 47 8 F At the points b and c corresponding to the inside edges of the flanges set out g e and h f equal to the expression o r.2 ^^^ re -draw the portion g k h of the parabola between the points e and f, then the curve A G E L F H D givcs the shear stress at the various depths of the cross section. Then total shear carried by web is equal to area of piece B E L F c of curve multiplied by width of web. Now take the case in which ^ = v^ and w — ^ and b =^ ^, 478 THE STRENGTH OF MATERIALS this being about the proportions for a rolled steel joist, then BG =s,= ^^^^ {h t - f") _ m /h^ h^ \ ^ 2k^ VlO ~ 100/ _ m 9 ^^ "" 2 A:2 • 100 _ m /h^ 9 h^\ __ m 16 h^ _ m 4 A^ •" • ^ ^ "" 2 F [J - 100 j ~ 2k^ • loo ~ 2l2 • "25" also mt{h-t){h -w) also GE - ^^^~ _ m h 9h 9h 20 ~ 2 F • rd • 10 * 2¥ ^ 7j _ m Slh^ ~ 2T2 • 100' _ m 9 h^ m 81 h^ ''' ^^^ 2k^' 100 '^ 2 F • Too m 9h^ 2 k^ ' 10 . • . Area of curve belfc =BcfBE +^mk _ 4^ m /9h^ , §^^ (A) " 5 • 2 P VTO ^ 75 y • • • • ^ ^ Now m this case i = -j^ yo = ^ _ ?i* (thy L ~ 24 20 • V 5 j • 12 = -0417 h* - -0192 h^ = 0225 h^ The area of the section = b h — {b — w) {h — 2 t) __ 7^2 _ 9^ 4^ ~ 2 20 • 5 = •14^2 r. k^= ^=-?l^f = '1608^' A T4/j2 Returning to equation (4) we get area of curve belfc 4 m A r9 y^2 8 ^2 + 10 F I 10 ' 75 DISTRIBUTION OF SHEAR STRESSES 479 " 10 X -leosl^ ' , 4 X 1007 ^^^•^^-1-608 = 2-505 m^ (5) . • . Shear carried by web = 2*505 mil x width of web = 2-505 mh X ^^ = -1262 mh^ (6) Now area of whole section = • 14 ^^ .'. Total shear S on section = "W/i^ x m Shear carried by web _ • 1252 _ r.^ ^ o/ •*• Tot^rshear O^ ~ /°* It is commonly assumed in practice that in plate and box girders the whole of the shear is carried by the web, and the above calculation shows that in an I beam, in which the flanges are larger in proportion to the depth than in most plate and box girders, this is true within 10 %. It must, however, be remembered that in girders built up of joists and plates, such as the comparatively shallow and heavy girders used in buildings, this assumption will not be so nearly true. Shear Stresses in Reinforced Concrete Beams. — ^The usual treatment is as follows — Consider two vertical sections of a reinforced concrete beam made at points ab a short distance x apart (Fig. 227), the section not changing appreciably from a to b. At the point A, the total stresses due to the bending moment are C and T, and at b they are C and T' ; then if the corresponding bending moments are B and B', we see that T = C = ? a a ...T-T'=^^ ~ (1) Adhesion Stress due to Shear. — But T — T' is the 480 THE STRENGTH OF J^IATERIALS difference in the pulls in the reinforcement at the two points ; that is, it is the force which tends to pull the reinforcement out of the concrete. T -^r X adhesive or shear force per unit length X . a B T a> X. T' ^JS a. Fig. 22: but B -B' X the rate of change of the bending moment shearing force = S the adhesive stress per sq. in. and the total perimeter of the reinforcement, and if /„ O we have /„ X O = adhesive stress per unit length = T - T' X ■■ /„ = s O .a (2) DISTRIBUTION OF SHEAR STRESSES 481 In the case of rectangular beams with tension reinforce- ment only or with double reinforcement where the top re- inforcement is placed at J depth of N.A. a = (d — ^ . our formula becomes /„ = — ; -^ (3) ou - This deals Avith what is known as the horizontal shear as regards the adhesion between steel and tension. Shear Stress in Concrete.- — In addition to this we have / > ^=!*S^ / ^ V \ 1 / / \ n \ \ \ 1 \ d. \ \ \ * / / CL < ^ > / \ ccb / / / / \ \ i/y//////} v//f/ //////// / Fig. 228. to consider the shear stress in the concrete itself, w^hich will be constant from the reinforcement to the N.A. and will then vary towards the top as indicated in Fig. 228. The difference of pulls T — T' is distributed over a hori- zontal rectangular area one side of which is x and the other side of which is b, the breadth of the beam. . • . If 5 is the shear stress we shall have s xb X X = T ~-T but T - T' == ^'^ (from (1)) (B - B') s = X . a .b (4) II 482 THE STRENGTH OF IVIATERIALS But as before we may put — - — = shearing force S X S a and for reactangular beams as above S ,9 = b d n (5) (6) GRAPHICAL TREATMENT FOR FINDING DISTRIBUTION OF SHEAR STRESS ON A CROSS SECTION Consider the section, composed of joists and plates, shown in Fig. 229. The first step is to " mass the section up " about p \Q ft ,- — ; 1^ '^L^ .-' ir i _j' /I' '■1 1 A IS B \ -^ c- 1 1 1 JC «- ^ 1 1 1 ^ ^ r Fig. 229. a vertical centre line : this is done by drawing horizontal lines across the joists, and adding on each side of the centre joist the corresponding horizontal ordinate of the outside joists. This gives the section shown in the figure {i.e.a d = ah -]-hc -{- c d). Consider any line p p. We have shown that the mean shear stress along p p = t = -^-p . — ^ (i: DISTRIBUTION OF SHEAR STRESSES 485 As we have previously shown, the deflection 8 in this case due to B.M. is equal to .^ ^-i x ^ 48 E I •*• a ~ AG • 4 • 48 EI _ 12E ^I ~ G • A Z2 E 5 Taking ^ = ^ and noting that I = A A;^ 1=30^^1-' (2) (2) Continuous Loading. In this case //, = ^'-^^ • o 5 W Z3 384 E I /x _48j8 E I ''' S~ 5 • G • A> Taking ^ = o as before, ^ = 24/3.^' (3) For rectangular section jS = 1'5 and k^ = y^, h being the depth of the beam. .-. (2) becomes^ = 3-75 (y)' (3) becomes^ ^ ^ ( I It follows from this that if y = ^^, the deflection due to shear is 3* 75 per cent, and 3 per cent, respectively of that due to B.M. in the two cases. We see, therefore, that for solid rectangular beams in which the span is more than 10 times the depth, the deflection due to shear is negligible. It must, however, be remembered that for rolled joists, 486 THE STRENGTH OF IVIATERTALS plate girders, and the like, the deflection due to shear will be quite appreciable for sections which are deep compared with their span. Bridge engineers often state that the deflection of a bridge is more than the calculated deflection. Part of this difference may be due to the giving in the riveted con- nections, but certainly the measured deflection would agree better with the calculated deflection if the latter included the shear deflection. It has been suggested that this could be remedied by taking E about 10,000 tons per sq. in. instead of 12,500 in the ordinary deflection formula. It should also be noted that we have taken only the strain due to the maximum shear stress, neglecting the fact that it is variable. This gives results a little too high, but is better than taking the mean shear stress. Distortion of Gross Section of Beam due to Shear, etc. — In finding an expression for the relation between the stresses and the B.M. on a beam, we made use of Bernoulli's assumption that the cross section remains plane after bending. The two causes tending to distort the cross section are (1) shear stress, (2) differences in lateral compression due to extension in fibres. Consider two cross sections of a beam at distance x (Fig. 231) apart, and let the B.M. at the sections be M and M^ respec- tively, and consider points p and p^ at distance y from the centre line, the section being the same at the two points. Then stress at p = -y" , at p^ = — -i — T ^ 1 . ^ . ^ M?/ ^ Ml?/ .' . Lrateral compression strani at p = 77 -nrj^ ^'^ ^i =^ "^ -c« j stress because longitudinal strain, = ' — :^ — and lateral or transverse strain = 7; x longitudinal strain. . • . Difference in lateral compression strain = ^j . (M^ — M) y .'.On a short length d y of the section, the difference in lateral compression = p' p^" = J/j . (M^ — M.).y.dy DISTRIBUTION OF SHEAR STRESSES 487 .• . a = slope of p Pi = ^^ = -g'j • --'— ^ • 2/ • ^ 2/ but we have shown that when x is ven^ small ^—^ = the shearing force S X j7 J .^ .y..dy To find the total change in angle between any section and the line originally parallel to the centre line, we must add all the elementary changes in angle. S Total change = : = Eiy^ .dy 2EI m . Tjy^ '' 2EF because m • S A Now we have previously shown that due to the shear there is a change of angle equal to -p— -7 Vf . • . Total change due to both causes _m ( ay , rjy^ k^\hG ' 2E _ m / ay r] y^ G ^ GkA b'^ 2E ^ 0" 1 771 / a tj ij \ putting E = and v = a ^-^^^ comes to ^-r^ ( -, - "^ 90 ) From this relation the slope at any portion of the section can be found, and the distorted form of the cross section can be obtained. Our present scope prevents us from dealing with this interesting problem further, but what we have given should serve as an indication of the method in which the problem may be attacked. CHAPTER XVII * FLAT PLATES AND SLABS In the beams that have been considered up to the present there is a support along two edges only, that is the support is at most along two parallel lines; when a plate or slab is supported upon more than two straight edges we have to consider the strengthening effect of the side supports, and for this purpose slab formulae are required. CIRCULAR PLATES AND SLABS Slab Coefficients. — In many problems it is convenient to use slab coefficients to compare the bending moments in a slab supported on its edges with the corresponding cases in which the slab is supported on two edges only as in the ordinary beam. We then have /? = slab coefficient B.M. on slab ~ B.M. on corresponding beam resting upon two edges Bach's Theory. — Bach obtains formulae in a very simple manner by assuming the supporting pressure imiformly dis- tributed along the edge of the plate or slab and calculating the bending moment over various sections. In the use of these results it should be remembered that they give the mean bending stress across what corresponds to the breadth of the beam, but that they do not give the absolute maximum. A similar point occurs in considering shear stresses in beams (see p. 473). We will take the following standard cases. A. Circular Slabs supported on the Outside Edge. — 488 FLAT PLATES AND SLABS 489 (1) Uniformly Distributed Load W. — The reaction pressure W W per unit length will be P = ^ -n ^ o — t>- Then considering TT U Z TT Jtv w the forces in one half of the slab we have a load ^ acting downwards at G^ (Fig. 232), the " load-centre " or centroid of W the semi-circular area and a resultant reaction = -^ acting at the " reaction-centre " G„ or centroid of the semi-circular arc. W onocxiooooooooo Fig. 232. — Circular Slab supported at the Edge with Uniform Load. . • . Taking moments about x x we have W Bending moment on x x = M^ = -^ . o G^, — Wg o G^ 2 4R (OG„- OG,) ^^. , 2 R o C}^ = ^ — and o Gj, = O TT TT . O G, - O G, WR w; TT R2 . R R / 4\ _ 2 R ^ ~ SJ ~ 3 tt" or, if w is the load intensity of load — 3 (1) 490 THE STRENGTH OF MATERIALS The mean stress on x x = / = -^^ = .g^ = ^-^^ . ••/ = ",?' (2) If the slab were freely supported at y and y we should have the reaction acting at y and the load at g, . - -288 W R .• . Slab coefficient = /3 = ^^ -^ 288 W R = -368 (2) Central Load on Radius r. — In" this case as before we have ^ W B.M. on X X (Fig. 233) = M, = -^ (o g,. - o gJ _ W/2 R _ 4r 2 \ TT 3 TT -'=s-t;r('-ia i« In the limiting case where r = we have the point load for which '='j i») In this case if the supports were at y y we should have _WR/ ir ~ 2 V 37rR 2(1-^^''- ^=T'-4?^^ ^^^ \ SttR. = "636 for the point load when r = FLAT PLATES AND SLABS 491 B. Circular Slab supported on a Circular Pillar. (1) Uniformly Distributed Load W. — In this case the tension and compression edges will be on opposite sides of those in the previous case, the present one being the equivalent of the uniformly loaded cantilever. OOOf)CY:CYYTrT^ w Supported at edge ; central load. Supported at centre ; uniform load. Fig. 233. Fig. 234. Circular Slabs. By moments as before about x x (Fig. 234) we have W W _ W /4R _ 2 r 2 ^Stt _ WR/2 _ r If the load is w per unit area, W ^ wttVJ^ '2 r .'. M. =wW (7) (8) In the limiting case of r = 0, which corresponds to a point support, this gives 492 THE STRENGTH OF i\L\TERIALS Taking the corresponding beam we should have G^ acting on the edge of the supporting circle /3 = WR (5- 77 r TT 2R 2 r 3 R 2 irr 3 ~ 2R There is no slab coefficient corresponding to ;• = 0, or rather it would be more correct to saj' that it will be 1 in this case . /_ 6M _22rR-^ • • ^ ~2R.r^ ~ /2 i^) (2) Load Distributed uniformly along the Edge. — This case is the same as case A (2) with the loads and reactions reversed. C. Oval Plate supported on Edge (Fig. 235). — Load Uniform. — In this case we can obtain an approximate solution by assuming that the jioints G^ and g^ will be the same for Mj as for a circle of radius ^ and for M,, as for a circle of radius 2 FLAT PLATES AND SLABS 493 6 TT 6 M, W 6 Wl This gives M^ = O TT ^' It^ Tvlt^ O TT Similarly /, = -^ = ^^^, If we put W = — ^ — we have M -^fill" •" ~ 24 M,= wb P 24 Corresponding to these we have /. w b^ h It will be noted that the stress is greatest across the short axis. This should not be used for ovals with Z ]> 2 6 which should be treated as ordinary beams of span b, giving M = ^ 8 Another way of dealing with this problem is as follows : If I is so great that the effect of the edges is negHgible, we have ID b t for the short span of a unit width B.M. = —^ and Z = -„ * • ^" ~ 8 ■ 6 ~ i^ For the circle, we have by Grashof 's theory (table on p. 495) _3dw'R^ _39wb^ _-3wb^ '' - 32 i^ - 1:287^ ~ ~^^ approx. . • . We may make up an empirical formula which is correct for the extreme values b — I and r = 494 THE STRENGTH OF I\L4TERIALS Grashof's Theory. — Grashof investigated the strains in flat plates by an investigation of the deflected form and adopted the stress equivalent to the maximum strain as the criterion of the resultant stress (see p. 44), the result being different from that obtained by means of the calculation of the principal stresses. The derivation of the formulae is too complicated for our present scope, so that we will give the results only and refer the reader to Professor Morlej-'s Strength of Materials (Long- mans) for the mathematical deduction of the formulae. SQUARE AND RECTANGULAR SLABS Square and rectangular slabs are of importance in several cases in practical design, particular^ in reinforced concrete construction and in tanks and valve-chest covers. Rigorous methematical methods cannot be applied to these cases, so that we have to fall back on approximate methods of which the two following are the most common. These hold for uniformly distributed loads onh\ Grashof-Rankine Theory. — These formulae for slabs supported on their edges are obtained by the following reason- ing and should not be confused with the rigorous Grashof treatment for circular slabfe. Consider two narrow central strips (Fig. 236) of width x parallel to the axes x x and y y, thus forming one strip passing over the other so that the two strips must have the same deflection. The whole slab is considered as divided up into strips at right angles to each other, the strips passing one over the other. The load w per unit area may then be considered as divided up into two portions Wi, w,, carried respectively on the long and short spans. For the long span we have 8 = ^q-j-U t (1) ,, short span we have S = '' ^ (2) oo4 hi i FLAT PLATES AND SLABS 495 < o w < Ph <1 1^ u M O o CO O w cc o "SCO ^1 ^fl 22 Til p^ § w o i;D 00 lO i-H (N q3 '^ ^ r-H i"* o a c^ o p. a a2 o Oh o &, P ^1 o o ■ O CO ^§ 4J 1— I (D ,— , '+-' S-l -(J biD© ^ O S -1-3 © £3 bc.aJ §'W Oi CO j3 ^^ ^ ^ Stra (N M ^|P^ S CO o t'pH K 1 ;3 -* Tt< 1— 1 Valu Maxii P5 - Ph « P^l^ + P3!^ be 11 1—1 'rH o + be '" O 4^ — 1 M ^- — ^ s ot Ti4 !lO '^ Ico o^ ■ess equi ^ .CO IN Ph ^ ^ »o ^^ 02 ^ ; « !^ ^i^ 5* IN ^,a P^ ^ 1 02 T+< . I-H OQ 1 1—1 incipal Stre IM Ph «,. CO CO IN , p^ «, CO \-^ P^|5^ «) bO + be 4^ O M CO ,Ti< + p^l^ bJO Ph Th !io f^ (M S I-H 00 ^ ^ lO ^ j f^l-. —1 loo S loo 1 CO ' C =H-I o o GO T3S II c3 o o THE STRENGTH OF MATERIALS w,, l^ 496 These must be equal .*. also iVi, -^ Wi = w i.e. Will + , 4 j = w Wt = w 1 + Y vC- "^ Y Fig. 236. — Grashof-Rankine Theory of Rectangular Slabs. .'. M^ = B.M. on short section, i.e. B.M. on long span. 'w, P- w P 1 + (3) Similarly w,, w 1 + M, = B.M. on short section, i.e. B.M. on long span. _ w,, b^ _ IV b^ 1 + 6* '. For long span, i. e. short axis — B.M. on slab _ ^ b^ ^' ~ B.M. on corresponding beam ^* + 6* For short span, i. e. long section I* + h*' (4) (6) FLAT PLATES AND SLABS 497 Some confusion is likely to arise if we do not clearly keep in mind the fact that if we are considering the stresses across the long section we take the B.M. on the short span, /• I M I n n 1 1 1 1 m 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n H I 1 1 1 1 n 1 1 1 1 1 1 1 1 n 1 1 m • AC .. M =^^^^ '' 12 AC W/6 12 ^/l^ + 62 (3) Neglecting the support on the short side we should have • ^ M,^^^ (4) To get a reasonable comparison between M^^ and M^. we ought to compare the B.M.s on the same length because, of course, a c is greater than Y Y. M M . • . B.M. per unit length along a c for slab = — ^ = --=^^ WZ6 i2 (Z2 + 62) (5) W6 wb"^ B.M. per unit length along Y Y for beam = -^,- = — g- Diagonal slab coefficient - f^^^ - 12 (^2 _|_ 62) • 8^ 3 (^2 + 62) 2 3(1 +(f^ (6 FLAT PLATES AND SLABS p has the following values — 501 I Diagonal b Slab Coefficient. 1 •333 1-25 •407 1-5 •461 1-75 •502 2 •533 To use these figures we find 8 and treat that when multiplied by /? as the mean B.M. per unit length along the diagonal. Numerical Example. — Take the same case as worked out on p. 498, and adopting a stress of 16,000 lbs. per sq. in. find the necessary thickness of a rectangular metal slab, comparing the results by the two formulce. On Bach's Theory — — ^ = Q = 3,600 ft. lbs. per foot width o o = 3,600 in. lbs. per inch width. .'. B.M. per inch length on diagonal = '461 x 3,600 = 1,660 in. lbs. nearly. M / ^ = 16,000 X 1 X ^^ ' ' 6 ~ 6 2 _ 6 X 1,660 = -62 16,000 t = •\/-62 = -79 in. nearly, say y^ (2) Sections Parallel to Sides. — The following modifica- tion of Bach's treatment is more suitable for reinforced concrete slabs where the reinforcement is parallel to the edges. We can consider in a similar manner the strength of the section x x (Fig. 239). The reactions on the sides will have reactions at the mid- points equal to ^ at d and E and pb &tY. The load acts at the point r. 502 THE STRENGTH OF 1VL4TERIALS Therefore, taking moments about the hne x x we have pi I ^ I \N I 2 4 ' ^ 2 2 4 WZ = |'(26+/) _ W Z (2 6-fJ) _ W I ~ 8 (r+ 6) 8 Wbl S{1 ^b) Y ± B (7) Fig. 239. — Rectangular Slab : Modified Bach Treatment. Neglecting side support on long side, we should have WJ 8 b 1 M. .*. Slab coefficient for x x = ^/ == ,— , 7 = 7 l+b ^^l (8) Similarly, if we consider the strength of the section y Y we should get Slab coefficient for y y = yS^ = 7 7 = z (^) ^ +7 FLAT PLATES AND SLABS These results can be tabulated as follows — 503 I b Slab Coefficients. Short Section and Long Span pi Long Section and Short Span 1 •500 •500 1-25 •444 •556 1-5 •400 •600 1-75 •364 •636 2 •333 •667 It follows from this that the B.M. comes the same on the two spans, so that the long span is the weaker as the breadth is less. Experiments do not bear this out. Numerical Example. — Taking the previous case we shall have /3i = -400, ft = •600 .-. B.M. on long span = -4 x 1,166,400 = 466,600 in. lbs. B.M. on short span = "6 x 777,600 = 466,600 in. lbs. iMf = 466,600 6 466,600 on long span 16,000 x P = 1-22 384,000 t = V'l'22 = 1| ins, nearly. This agrees fairly well with the Rankine value, although it is deduced from a different span. It is interesting to note that although Bach worked upon the diagonal as being the weakest section the above treatment requires a greater thickness. Variation of Bending" Moment. — To obtain an idea of the variation of the Bending moment in this case consider a portion a b u u (Fig. 240) of the slab. Then the load on I W 2{l + h) the shaded area =^ As before, we have p 504 THE STRENGTH OF MATERIALS . We have at x a force p h and at d and e a force ^^, and W:r at F a downward force I . • . Taking moments about u u Ave have M,=phx +2'P^ .^ = Wx 21 b X x^ 2 {I + b) "^ 2 {I + 6)"~ 2l[ Wr {bl + xl -x{l + b)} 2l{l + b) ^^ fhJ -h X _^bx{l-x) 27(^ + 5)^ ^^ ~ 2l{l +b) Y d U D A 2 Y Fig. 240. putting X = ^ this gives (10) M^ = oi^j—TTx = o /7 , rx ^s beiore, thus checkmer 8 Z (/; + 6) 8 (? + 6) ' ^ W6 ^" = 21(1+6) i^^-^'> • (11) A diagram showing the variation of M^ with y would come a parabola with vertex at the centre, similar to the ordinary B.M. diagram for a simply supported beam. For a corre- sponding consideration for the other span we should get Wl Mj, = 6/h(f I ;.\ 1^ ^ "" ^^\ which would also be a parabola. FLAT PLATES AND SLABS 505 We may therefore assume that the stress will vary across a section approximately in the form of a parabola, so that the maximum stress will be 1"5 times the mean stress ; this would require the thickness at the centre to be \/l'5 = 1"22, say 1"25 times the thickness given by the ordinary treatment of Bach's Theory. In our example this would make t for diagonal consideration = "96, which agrees quite well with the Grashof-R-ankine theory. Variations of Bach's Theory. — There is reason to believe that the pressure in rectangular slabs is greater at the centres of the supporting edges than at the corners; we will there- fore consider two variations in pressure. Variation I. — Pressure varies according to a Parabola. — We will now therefore assume the pressure to vary in the form of a parabola as shown in Fig. 24L We will take, as before, the total pressure on each side proportional to its length, so that the total pressure on each long side = V ^ WZ ' 2 [l + b) and that on each short side ^P = W6 ' 2 (I + b) The pressure at the centre of each side is therefore 1*5 ^3, p being the value given in equation (1). The resultant pres- sure along A B will act at the point y, while that on the half sides A X, B X will be at the centroids of the parabolas, i. e. 31 , ,- irora X. Id Therefore, taking moments about x x we have M =P ^+?J^ ^-"^ ^ * ■ 2 ^ 2 ' 16 2*4 Wbl 3WZ2 Wl 4 (/ + 6) ' 32 (? + b) Wl [2b + ^^-{l + b)} S{1 + b) 8ir+T)r 4j ^^^^ 506 THE STRENGTH OF MATERIALS Neglecting side support, we have as before M =WJ . • . Slab coefficient for x x = ^/ [l + b) I 1 - 4 6 1 + / (13) Fig. 241. — ^Rectangular Slab with Parabolic Distribution of Edge Pressure. Similarly, we get for y Y by reversing I and b b I Slab coefficient for y Y = (5i, = I + b I _ 1 b 4 1 + .^ (14) FLAT PLATES AND SLABS These results can be tabulated as follows — 507 I b Slab Coefficients. Short Section and Long Span |3/ Long Section and Short Span ^6 I—" •375 •375 1-25 ■306 •444 1-5 •250 •500 1-75 •205 •545 2 •167 •583 4- Fig. 242. — Rectangular Slab with Triangular Distribution of Edge Pressure. Variation II. — Pressure varies according to a Triangle. — In this case we will assume the pressures to be even more concentrated at the centres than in the previous case, and assume the pressure distribution shown in Fig. 242. As before, we take total pressure on each long side = P,. = ^^m , 7v and that on each short side = P, = 0/7 , i.\ 2{l + h) 2 {I + b) 508 THE STRENGTH OF MATERIALS Taking the side pressures as acting as the centroids of the triangles, we get W I 2 4 M — P - -]- — '^ - ^ • 2 2 '6 'Whl Wl^ 4 (r+ b) "^ 12 {I + b) Wl U. 2 1 S{l + b)\^^ 3 8 (l + b) Wl S{1 + b)\ 3 . • . Slab coefficient for x x = F/ (15) b - I I + b lJ ■ Similarly, by reversing I and b, slab coefficient for y y I _ 1 b 3 (16) = r, = / (17) + 1 These results can be tabulated as follows- I b Slab Coefficients. Short Section and Long Span Pi Long Section and Short Span P" 1 •333 •333 1-25 •259 •407 1-5 •200 •467 1-75 •151 •515 2 •111 •555 Numerical Example. — J'aH?i^ the parabolic variation, calculate the thickness of the slab previously considered. We have, neglecting slab action — B.M. on long span = 1,166,400 in. lbs. B.M. on short span = 777,600 in. lbs. FLAT PLATES AND SLABS 509 For I = 1-5, 13, = -250, B, - 500 .-. M, = 388,800 ill. lbs. M.^ = 291,600 in. lbs. .-. Short span 16,000 x ^^ ^ = 388,800 16,000x1^2x12^^^33^^^^ 388,800 _ 16,000 X 24 t = Vl'Ol = 1 in. nearly. Rectangular Slabs Clamped on their Edges. — An approximate treatment commonly adopted in practice for W I W6 this case is to regard the B.M.s at the edges to be - and -y^- respectively, multipHed by the slab coefficient derived for supported ends ; those at the centre to be — and reduced 24 24 in the same ratio. In cases where the load may be on only a part of the slab the B.M.s at the centre are usually taken as the same as for the edges. CHAPTER XVIII * THICK PIPES We have considered already the strength of a thm pipe and obtained simple formulae by assuming that the stress was constant throughout the length and thickness of the pipe. WHien a pipe is not very thin compared -^ith its diameter we have to allow for the variation of stress across the section. Lame's Theory.— Let a pipe be of internal radius r (Fig. 243) and external radius R and let it be under pressure either from the inside or from the outside. Xow consider an imaginary thin ring of thickness 8 x and internal radius x. This rmg will be subjected to a radial pressure p on the inside which hj considerations of sj'mmetry must be the same all round, and on the outside it will be subjected to a radial pressure which will differ sUghtly from y and which we ma}' call ;p -^ hp. This assumes that the tube is subjected to pressure on the outside ; if it is on the inside the same formulae hold with appropriate change of sign as explained later. We may therefore apply to this imaginary hoop the same treatment as for a thin pipe, the circumferential stress, or hoop stress, being /. Considering a unit length of pipe we have Force tending to cause collapse of ring = {p -{- Sp) X 2 {x + S X) Force resisting collapse of ring =^ 2 f 8 x ^ p x .2 x These must be equal 510 THICK PIPES 511 . • . dividing by 2 and neglecting the product, S p .Sx,oi two very small quantities we have p.x-{-x.Sp-\-pSx ^ f Sx + p . X .' . {f — p) h X = X h p (/ - ^) = T^ Fig. 243. — Stresses in Thick Pipes. In the limit when the increments are infinitely small this gives if-p) = ''ff w This is one relation between / and p. Now let us assume that the strains along the length of the pipe will be such that a plane section before subjection to pressure remains plane after subjection to pressure, i.e. that Jongitudinal strain is constant. i. e. '^J^ + "^-J = constant (2) 512 THE STRENGTH OF MATERIALS because both / and p will cause transverse strains in the direction of the length of the pij)e, and they will have the same sign. . ■ . Since -q and E are constant, if our stresses are within the elastic limit we may write f + p — constant = 2 a (say) ••• f = {2a-p) (3) Put this value in (1) and we get 2o X d p a — 2 p =^ , - ax 2a = 2p + ='^J (4) Cu tX) (aj JO = x( 2 2> + 1 \ = 2ax . • . d {px^) = 2 ax . dx (5) Integrating we get p x^ = a x^ +6 where 6 is a constant -'• P -CL + ^2 ••• (6) but / = 2 a - 2> (by 3) i.e. f =^a --^ (7) by calculating a and h in any particular case we can find formula for p and /. Special Gases. — (1) Pressure Inside = pi\ Pressure Outside = i. e. p = Pi for X = r p = for a; = R ••.^. = « + J W b THICK PIPES 513 Put this value in (8), then ■■■^-^ (9) . • . Hoop stress at inside — /,• is obtained b}^ putting a; = r in (7) I.e. /. =a---2 _ fi r^ Pi R2 - - R2^ir^2 - R2 _ ^2 ~ R2-r2 ^^^^ The negative sign indicates that the stress is a tension. Hoop stress at outside = fo = a — p^ = - (R2 _ r^) ~ (R2~_r72) This is also a tensile stress and is clearly less than /,, so that with internal pressure the maximum stress occurs on the inside. At any intermediate radius x , h — Pj ^^ _L P' "^^ ^^ ~ ~ (R2""^2j + (R2 _ ^2) ^,2 - ^'-^^ r5!_i\ n.N (R2 _^2) \^2 -^j U'^; It should be noted from equation (11) that no matter how LL 514 THE STRENGTH OF MATERIALS great the thickness of the tube may be the hoop stress is always greater than the internal pressure, so that for any given material there is a certain maximum pressure which must not be exceeded. It should also be noted that the assumption of the constancy of longitudinal strain holds only while the stress is within the elastic limit. Numerical Examples. — (1) A cast-steel cylinder 2 ft. in external diameter and 3 inches thick is subjected to an internal pressure of 2 tons per s(l- in. Where and of what magnitude is the m^aximum stress ? The maximum stress is on the inside and is given by the formula p (R2 + r^) ti (R2 - 2(242 r2) f- 182) (242 _ 182) '7 2 X 62 (16 +^) 62(16 -9) 7*14 tons per sq. in. (2) Plot a curve showing the maximum stress in terms of the internal pressure in a tube whose ratio of external to internal radius varies from TIO to 4. _ p, (R2 + r2) U (R2_r2) /, _ R2 + ^2 ' Pi R2 - r ?)• R^2 r + 1 This gives the following values- R r MO 1-20 1-30 1-5 2 00 2-50 300 3-50 400 f' Pi 10-52 5-55 3-90 2-60 1-67 1-38 1-25 118 113 THICK PIPES 515 If- = 110,^^-^ = 110, r r 10. T) V f . ' . The thin pipe formula / = ^~~- would srive — = 10, so t p that the thin pipe formula would be about 5 % in error. The above figures give the curve shown in Fig. 244. These results should be compared with Example 2, p. 521. 10 9 8 7 6 S 4 3 Z 1 I \ "^ \^ 1 Valuea of R -^ r z-5 3'0 3-5 4o Fig. 244. — Variation of Hoop Stress for various ratios of External to Internal Radii of Pipes with Internal Pressure. Curves of Variation of Radial and Hoop Stress for Internal Pressure for R == 2 r. T3 ^. ,^, b RV2 4r4 4r2 By equation (9) ^- = ^,— ^ = 37^ = ^ By equation (10) ^ = - ^2 Pi Pi PiX^ 516 THE STRENGTH OF MATERIALS 1 4 ^2 ~ 3 "^ 3 .1-2 -sKx) 3~3rv.i-7 / / h = ^-7^ L (1 4/rYl li.fr These results are plotted in Fig. 245 and show clearly the variation of the stresses across the section. 10 •47 1 t \ \ \ \ \ \ ... . N \ \ \, \ k, \ "^ - '^ ^ ^ 1-17 ro7 •67 K b7 > 1 + 1-8 10 Values of 2 4- r Fig. 245. — Variation of Stress in a Thick Pipe with Internal Pressure. Maximum Shear Stress.— The stresses / and p are the principal stresses in the material and, as we showed on p. ID the maximum shear stress will be equal to s = ^-^ = ^,, (from (6) and (7)) ^ X" p, R^r2_ ~ (R2 - r2).T2 The maximum shear stress will occur Allien x has its least value, i. e. at the inside edge where x — r (15) THICK PIPES 517 Max. shear stress = p^B,^ (R2 - r2) Maximum Stress equivalent to Strain. a strain in the circumferential direction equal to — ^ . * . Total circumferential strain = 4^ — %^ E E . • . Equivalent stress = Total strain x E = fe = f-VP b (1 + 7}) (16) There will be IP = a{l -rj) x" (17) This gives a maximum equivalent stress at the inside of p, (R2 + ^2) fe (R2 - r2) /RM-^2 '^ 1112 _ ^ y'i -P2 _ ^2 + vj (18) putting y} = \ this gives /. = - Vi R2 + r^ 1) R2 _ ^2 + 4j NuMERiCAL Example. — Take the case of a tube with internal radius 4 in. and external radius 12 in. and take p = 10,000. This case was given by Professor C. A. M. Smith,* and we have added the equivalent stresses. This gives 6 = 180,000, a = - 1,250. The results may then be tabulated as follows — Radius. Maximum Stresses. I 1 Hoop Tension Compression. Hoop Tension. Shear. equivalent to P / s Strain. 4 10,000 12,500 11,250 ... 15,000 5 5,950 8,450 7,200 10,950 6 3,750 6,250 5,000 8,750 7 2,420 4,920 3,670 7,420 8 1,560 4,060 2,810 6,560 9 970 3,470 2,220 5,970 10 550 3,050 1,800 5,550 11 240 2,740 1,490 5,240 12 1 1 2,500 1,250 5,000 * Engineering, September 2, 1910. 518 THE STRENGTH OF 1VL4TERIALS Formula for Thickness of Pipe in Terms of Internal Diameter and Pressure. — On the maximum stress theory we have, if ft is the safe tensile stress, _ p, (R^ + r^) '' R"2-r2 .ft R2 + r2 1. e. U + Vi ft - Vi R r r + t R!» - 2R2 2r2 U±Vi U Vi t ft + Vi r r y ft — Vi t Ift+Vi 1 --'{^'ikr') tt+Vi il ft - Vi On the Maximum Strain Theory we have , /R2 + r2 \ h Vi V A + (1 - - -n) Vi A - (1 + >?) p, //. + (1- - v) Vi A- (1 v) Vi .'. t ^ r t =r -\ If q = -3 t = r v t ^ R^ + r^ ~ R2 - 7-2 _ R2 (V!:^ r + (1 (19) If -. = i -(!+>?) V, 4/7+3y. \ A - 1-3 p, - 1 - 1 - ' KV 3 /! 3 /, + 2 p, 4p, - 1 ..(19a) ..(196) ..(19c) THICK PIPES 519 Fig. 246 shows a chart reproduced from Machinery, January 14, 1915, for determining the necessary thickness of a cyHnder or pipe in accordance with Formula 19. In using the chart to determine the constant, the horizontal line through the proper pressure value is located, and the curve starting from the desired value of the stress is next found; this curve is then followed to the point where it intersects the horizontal pressure line, after which the vertical line is followed to the bottom of the chart to determine the constant. This constant multiplied by the inside radius r of the cylinder gives the required thickness of the cylinder wall. Case 2. — Outside Pressure = %, Inside Pressure = In this case p = p^ when x = R and p ^ when x — r b I.e. Po = a + ^^ = , b a == b r2 Po = P» - P"^'''' (20) J^ _ 1 \ (R2 - r^) R2 rV a = ^Jt^:^ (21) (R2 - r2) Hoop Stress at Inside. At inside, where x = r, fi = a b (R2 _ ^2) + (J^2 _ ^2) 2 Po R /Qnv (R2 -r2) • ^^"^^ 520 THE STRENGTH OF :\L\TERIALS 1 STRESS IN CYLINDER WALL IN POUNDS PER SQUARE INCH. PRESSURE IN CYLINDER IN POUNDS PER SQUARE INCH. — 1 — \a / J A < / ^ ^ 1 _4 / 1 A y V\ .A X ^ ^^ ^ > . ^ / A ^^ ^ ^ .^ ^ - /^ '>^ -=i li^ 1! / y^^ i^ ^-J ^ 1 >--r-' 1 \ / / X 1 /I ■^ J 1 j-^ >^^' // V A\ a-:a. A i J X^ ll\U'l±^ X'-X^^\yA, \ \ \ 1 / ^^ \>^ "! i 1 1 // /// Y X y^ ^ -^ li- -f^ \ 1 1 //// /y^y\^^ X ^ M 1 ! ^i,^ ^Ihf/l/y/ '/ / A / ^ ^ y 'illll//'// . / y y / i i/////7// /' / / ^ ^ X^ \ ///////;/ / /\ / / ^ X ^ r" / // // \/ ■ / ^/i \ /////(/ /fi A .^ -^ \ ///// /\/ Y \x ^ ^ ^ ^ '/////// / ^ ^ ^ III//////X y V _^ .^ H^ fT 1 11 //////y / y^ ^ - "^ i _ llli/// r- V ^ ^i^^'^n^ ^ — — —- ' r _^ — — i \ S ^ ^ <>t,<>i ^. ^. ^. ^. '^. "^ '-■'. '-''. ® ^. R ^. *"•. =°. °°. ^. ^. '^. '- 0* 0* s' © 0" 1- chinery Fig. 246.— Diagram for Thick Tubes. THICK PIPES 521 Hoop Stress at Outside. At the outside, where x = R, hoop stress = /„ _ _ b - ct ^2 -^^^' + ''^ (23) In this case the maximum hoop stress, which is compressive throughout, also occurs on the inside and is greater than the radial pressure ; irrespective of the thickness of the pipe, the external pressure may never be more than half the safe compressive stress on the material. Putting in the values of a and b in the general formula for this case we have at a radius x _ p„ R2 p„ R2 r^ ^ ^ (R2 - r2) - '(lR'2'372y ^2 P"^^ A-fYj (24) (R2 - r2) I \x ' = iBp^, !■ - ©■} « Numerical Examples. — (1) A cast-steel cylinder 2 ft. internal diameter and 3 in. thick is subjected to an external pressure of 2 tons per sq. in. Where and of what magnitude is the maximum stress? {Compare Example 1, p. 614.) The maximum stress occurs on the inside and is given by the formula - 2 po R' _ 4 X 24^ _ 4 X 42 X 6^ I' - (1^2 „ ^2) - 242 - 182 ~ 62(16 - 9) 4 X 16 = — ^ — = 9' 14 tons per sq. in. (2) Plot a curve showing the maximum stress in terms of the external pressure in a tube whose ratio of external to internal radius varies from 110 to 4. 2^^ p, ~ R2 - 7-2 ~ f-Ry' r 522 THE STRENGTH OF IVIATERIALS This gives the following values — R r 110 1 1-20 1-30 loO 3-60 2 00 2-50 3-CO 3-50 218 4 00 1 11-52 6-55 4-93 2-67 2-38 2-25 213 R R For = MO we have ^ = MO r R — ^ •. t = R ir IZ — 10 — 8 — 1 . ■ 6 - 1 \ a \ •1-4 \ 1 \ o 'x,^ «Q ^^^^ 3 . _^ ^^ n \'5 Values of R 4-r zo rs 30 35 40 Fig. 247. — Variation of Hoop Stress for various ratios of External to Internal Radii of Pipes with External Pressure. The thin pipe formula would give = 11 SO that the thin pij)e formula would be about 5 % in error. The above results gives the curve shown in Fig. 247, which should be compared with that shown in Fig. 244 for internal pressure. THICK PIPES 523 Curves of Variation of Radial and Hoop Stress for External Pressure for R = 2 r. h _ _ R^r ^ _ _ iL^ r2 - 3 By equation (20) Vo R2 . £ * ' Vo R2 4 R2 - r^ 3 '^1 Vo vo ■^ "8 ^ ^ ,/ ^ •6 •1- / / / / / / '3 P^ -2- o ^ / / > / 1-4 1-8 Vbl •S7 Z'oy Z-27 S5. 02 ft 2-47 I VO rfo7 Values olx-^r Fig. 248. — ^Variation of Stress in a Thick Pipe with External Pressure. 4 4r2 ~ 3 3^2 _4/ /ryi 1 Vo = («-|.)-2^- 4 4r2 ~ 3 ^ 3 a;2 (26) =J{'+(iT) ™ These give the curves shown in Fig. 248, which should be compared with Fig. 245, 524 THE STRENGTH OF MATERIALS Strengthening Thick Pipes for Internal Pressure by Initial Compression. We have seen that the maximum hoop stress is always greater than the internal pressure, so that if we are to be able to make pipes sustain very high pressures we must devise some method of reducing the hoop stresses. This may be effected by bringing the metal of the tube into a state of initial compression. In the early days guns were cast around chills to cause the metal to solidify immediately on the inside and so come into compression when the remainder of the metal contracted upon cooling. Another method, now commonly adopted, is to wind strong steel wire under heavy tension on the outside of a tube, thus bringing it into compression which will balance to some extent the pressure caused inside the gun by the explosion. A further method is to shrink an outer tube on to an inner one ; this puts initial tension stresses into the outer tube and initial compression stresses into the inner tube. The effect of the shrinking is shown in Fig. 249. The top diagram indicates the distribution of the hoop stresses across the section for a solid tube ; the shrinkage stresses are shown in the central diagram ; these being obtained by applying the condition that the radial pressures of the junction are equal and opposite. The combined stresses are shownin the bottom diagram, from which it is seen that the maximum tensile stress is very much reduced and that the tensile stresses are more nearly constant. Necessary Difference in Radius for Shrinkage. — For the outer and inner tubes, the same general formulae will hold, but the constants will be different. i.e. For the outer tube we have p = a, + ^ (1) f =a.-k (2) THICK PIPES 525 For the inner tube V ^(^i+ ~2 (3) ens/on Shr/'nnaqe Stresses I M I I I i I I I I M I Combress/on I Stresses y/Uulhide Tuhe'^nsidc TuLe\^ R -^ -H Fig. 249. — Shrinkage. Stresses in Compound Tube. f ^ Cli - x" (4) 526 THE STRENGTH OF MATERIALS We must have the same value of p for the junction where • n A. ^" — 1 ^' i. e. {a,. — a^) r^ = (6, — 6,.) (5) Next consider the circumferential strains at the junction. For the outer tube we have Unital circumferential strain = != -\-^^ (6) Similarly for inner tube Unital circumferential strain = — (^ + ~f j . . .(7) The value of y is the same in each case, and in (6) / is as in (2) and in (7) as in (4). .*. Increase in circumference of outer tube 2 7rri 4(-'j + >7P E Decrease in circumference of inner tube .". Difference in circumference of two tubes before heating and shrinking on . • . Corresponding difference in radius _ difference in circumference 2 r = (from (5)) -^ (a, - aj 2 i.e. Proportional difference in radius = -p, (a^ — a^). . . . (8) This will probably be made more clear by the following numerical example. THICK PIPES 527 Numerical Example. — A compound mild-steel cylinder con- sists of a tube 6 ins. in external radius and 4*5 internal radius shrunk on to another tube which has an internal radius of 3 ins. If the radial compression at the common surface is 4000 lbs. per sq. in. after shrinking find the circumferential stress at the inner and outer surfaces and at the common surface, and find also the original difference in radius necessary to effect the given radial pressure at the junction. Outer tube. p = for X ^ 6 - 4000 for X - 4-5 . n — _L -^ 4h / 4 1 \ 4000=..+ 3j^=^(,-^--3,)6. 324 X 4000 bo = - a„ = — 7 36,000 X Inner tube. 7 Interior hoop stress {x = 4*5) _ _ 36,000 _ 324 X 4000 ~ 7" 7 '^ 81 = 14,286 lbs, per sq. in. (tension) . Exterior hoop stress {x = 6) - _ 36,000 _ 324 X 4000 JL ~ ~^7 " 7 ^36 = 10,286 lbs, per sq. in. (tension) . p = for X =^ 3 = 4000 for X = 4-5 .-. = a, + g .-. a, = — 4:b 4000 = a, + ^ _ 4 6. _ 6, _ _ 5 b^ 81 9^ 81 ,,= _400^2iAl._ 64,800 o a, = 7,200 528 THE STRENGTH OF ^UTERIALS .'. Interior hoop stress {x = 3) = 14.400 lbs, per sq. in. (compression). Exterior hoop stress (.r = 45) = 7.2W - : - ^-iii) = 10,400 lbs, per sq. in. (compression). Taking E = 30 x 10^ lbs. per sq. in. and v; = J. the cir- cumferential strain at 4 J ins. radius in the outer cylinder _ 14,286 , 4000 ~ 30 X 10« "^4 X 30 X W ^^,^,.,^ 1000 Circumferential compressive strain at 4 J ins. radius in the iiuier cvlinder _ 10,400 _ 4000 ~ 30 10^ 30 X 10« = ^103467 - 3,-^^-^e .• . original difference in diameter = 9 (0004762 -^ •0 + R Then from equation (6) '" A VRi R/ V?/ + R I/ + R R.i/2 Now let kh'^ = %, ^ .y + R where h is defined by the above relation, and may be called the link radius. It corresponds to the radius of gyration in the ordinary case. „, Ah'' ^f y"- .-.We see ^ = :S — ^-^ R \2/ + R ^ 2/R 2/ + R _y \ _ _Ah^ y ^-r)-"-- R2 E /I l\Ah^ Th-/ =1.(^-1) A • VRi R/ R 534 THE STRENGTH OF MATERIALS .•.Wehave/.,=^jJ':(^^-^) (9) From equation (7) M = :s ^^^ ^ 1 1 1 -^y R 2/-tt ER(i--i^2 y'^ Ri r; R + 2/ Ri R/ " R + 2/ = E-^Ki-K) ■• ■•<^*^) returning to equation (8) we see R VRi R/ ' \R Ri; 2/ + R ^ M M^ / R_y R. A^ A^2- VR +2// ^^ General Graphical Solution. — Let Fig. 251 represent the section a d b e of a beam, and o the centre of curvature of the centre line d e, the beam being, of course, curved in the plane of bending. Now consider a very narrow strip P Q of the half section at distance y from c D. Join p o, cutting c d in s and draw s R parallel to Q c to cut p q in r. ^, p R R s y Then ^ p Q Q o R + 2/ Repeating this construction for a number of strips such as p Q, and joining up the points obtained, we get a curve A R D Ri B which is one-half of a curve called the link rigidity curve. In a symmetrical section, which is the most common, the two halves will be identical. * This is the stress due to bending only; in the case of hooks we have to add the direct stress over the whole section. CURVED BEAMS Then the area of this link rigidity curve = - A,* = -^ y R+2/ 535 Fig. 261. — Curved Beams, etc. ■•■A„ = Now let -*- = R2 area of half link rigidity curve area of half section = L i. e. A /^2 _ A X L X R2 * The area is negative because the area of the portion d b^ b is greater than that of a p d . a,, represents the excess of the former area over the latter, i. e. area a r d Rj b — area a p d Pj^ d. 536 THE STRENGTH OF MATERIALS Now put these values in the equation (11) for stress. Then we have '" ~ R. A + A.L;R2(R + y) ' ^MM y _ A \R "^ R L (R + y) ^ M f y I AR\' "^L(R+7)J ' Then if d, and dt are the distances from the hne D E to the extreme compression and tension fibres respectively, assum- ing the inside to be in tension and the outside to be in compression, we have Maximum compressive stress _ / _ M f d, 1 Maximum tensile stress = ft = ir^^l r~-/^~ — yr ~ 1 < • • (1^) A R \^Li (R — dt) I Position of Neutral Axis. — The value of y to make /, = gives the distance y,, of the neutral axis from d e. '■ '• L (R + 2/.) " Vo = — L R — L t/„ LR R 1 + L , R2 This enables us to find the position of the neutral axis. Alternative Formula. — The stress formula on Winkler's assumption that yi = y can be put in a number of alternative forms. Suppose, for instance, that the neutral axis is at distance y^ below the centroid line c c-^ (Fig. 250). Then total strain at p Q = Pi Qi — p Q will be proportional to {y + y„) the distance from the N.A. . • . We write P^ Q^ — p q = m (?/ + y„). Moreover, p q = (?/ + R) x /e o d CURVED BEAMS 537 h _ JPiQi - P„9 _ (y + y.) m E P Q (?/ + R) * /e O D /. ^ • V T^{ where ti is a constant (2/ + ^) (2/ + I^) (13) Fig. 252. Since 2 /„ . a = Eii y + y ^ + R = 1. e. y. y «■ ^"+R a .V + 1/ + R (14) 538 THE STRENGTH OF MATERIALS Again, taking moments about the neutral axis we have Zj y y -^ Bo J . / = M (y + y,.) ' ■ i»+«{2'i^"}- Resal's Construction. — ^According to this construction we proceed as before and find the link rigidity curve. The line n n passing through the centroid g of the curve A R Ri B r/ r' is then found by graphical or other methods, and the moment of inertia I^ of the link rigidity curve is then found about the line n n. In practice it is sufficient to apply the construction to one half only. Then N N is the neutral axis line and _MR {y + y„) ^'!~ I, '(y + R) ^''^ The rule for the line n n passing through the centroid of the link rigidity curve is 2/o %a^y Now aj a X R Q a R .R + y yo ^ <^y _Z/R + y ja^ +y This is the relation required by equation (14) Inn = S "1 (2/ + yo? /inN i MR (1/ + yo) .• . In equation (16) /, = i^Tj^I^y CURVED BEAMS 539 This formula looks simpler, but it involves the determination of i/o and I^, which are rather more troublesome than the calculations in the previous method. SPECIAL CASES (1) Rectangular Section. — If the section is a rectangular T -;-y dv -jt Fig. 253. one of breadth h and depth d, we may proceed by mathematical analysis as follows — A, = - b y R"T^ dy = -b - b / 1 - R n + y dy y-n log. {y + R) 2n + d + 2 J d 2 -bid -n log. 7 / 7 T-> 1 2 R -(- cZ b{ - d -\-R log, 2^ _ 2n-d d d . L = K bd - ... 2R + d R, R d 1 + A 2R 1 - R - 1 540 THE STRENGTH OF MATERIALS = |{log,.ll _Rr /^ ~ d\ V2R _d 2R )-log„(l J VI _i. 2R/J ^ d^ , rf5 (^6 d-' A. 2R (^ IR ' 1 /^ 12 VR 2(2R)2 ' 3(2R)'' 4(2R)4 ' 5(2R)5 ' 6(2R)^ 7(2R)' _ d-^ _ ^3 d^_ _ rfs _ rfs _ d-' "'2(2R)2~3(2R)3 4(2R)^~o(2R)5 6(2R)^ 7(2R)- 12 VR; ^ 80 VR/ "^ 448 VR/ / 2 l.fd\^ . 1 /^\^ r> + 80 VR + 448 (if- (2) Circular Section. Fig. 254. A, = - h y dy t-{^y + r -r -r -r + r "Rhdy "^kbdy -r + r R + 2/ R-fi/ h dy .. -f r The second term / h dy = area of circle = ttt- }- (1) * See, for instance, C. Smith's Treatise on Algebra (Macmillan), p. 383. CURVED BEAMS 541 R + 9 , r h dy _ -p /"2 r cos d {r sin 0) 1/ R+^~ y R+TsiiT^^ + ■ = 2r2R / C0S2 ^ ^ ^ R + r sin 61 (2) _ cos2 e _ (l-si n^^) _ 1 __^( • 0_ I^sin ^ R4-rsin6' R+ r sin ^ ~ R + rsin(9 r\ R + r sin ^ 1 sin ^ R / sin R + r sin 1 r sin ^ R + r sin r R2\ 1 + r VR + ^ sin R/, R R + r sin r^/ R + r sin y • cos^ OdO R + r sin ^ + 2 sin^ R2 R2\ _d^ r^ / R» + r sin (3) + sin OdO r + + 'B.dO (4) + Now - f4^^ ^^ + f RdO ^2 cosO 2 + R^- = 0+^-f..(4a) A • f d^ r Again / ^ , . ^ = / ^ y R + r sm (9 / __ dO^ K + 2 r sm . cos ^^ do sec^ n (^ ^ R + 2 r tan ^ . cos^ ^ R sec^ ^ + 2 r tan ^- seC / dO ( put tan ^ = it, then R(l + tan2 j +2rtan2 du = Jsec^ s 542 THE STRENGTH OF ^lATERIALS du 2 R 2 R du du J R (1 + «2) +%ru ^ J R + 2 r M + R m2 J { ' d u u + ^) +(l R2 2^ R ■-(.'■-£)■ (5) This is of the form /d X ■/ dO R ^ rsin^ R /: + 13' P tan tan X — a U -r R \ ^ R2 — B WTien ^ = '1. tan = 1 = w ^ = — ^ tan;^ = — 1=2^ z, z 1 - R2' cZ^ R + r sin ^ x/R2-r tan-^ 1 R r . \ I' r'- ■M ^ ~ R2- tan R V^ ~ R2' /..n-_^L VR2 - r2 I VR2 - 4- tan 1 _AilJL.\ \/R2 - 7-2] (6) , _a /R +r , ^ , /R - rl The portion inside the bracket is of the form tan'^ y + tan"^ * See Lamb's Infinitesimal Oulculus (Camb. Uiiiv. Press), p. 172. CURVED BEAMS 543 If tan a = -, cot a = ?/ = tan ( ^ — y ^ V2 y .' . tan~^ ?/ = — — a = ^— tan"^ ^2 2 y .' . tan~^ y + tan"^ = t^ (6) becomes / dO + R + r sin ^ VR^ — ^2 w 2 _ 5^^^ ^"^ n^ -r r^jR+rsine r^ VR^ — TT VR2 - 7-2 r^ + 2 /.x J /^ X r Gos^ede Rtt TrVR^-r^ .'. from (4) and (4a) / --^-- ^ — ^ = -- „ --o ^ ' ^ ^^ R + r sm (9 r^ r^ + r m "^' ^ (2) R f^^ =27rR2-27rR VR^ - ^' ^ J^ + y In (1) A, = 2 TT R2 - 2 TT R VR^ - r^ - ir r^ 2 R2 2 R /7RV = .r^i-^---7(7T-l-l}----(^) I r^ r Correction Coefficients for Ordinary Beam Form- ulae.^ — -If the bending moment on an ordinary beam is M, we have for a rectangular section the bending stress , _ 6M ' ~ hd'' and for a circular section of diameter d _ 32 M 544 THE STRENGTH OF MATERIALS Let /,, /o be the correct stresses at inside and outside respectively of the curve, then we may write /. = " / 2-0 1-8 re V . \ X ^ C'l re le \\ inside. \ \ \ \\ V \ •*3 \^ s <» 1 I-a o / <^ ^ / f ?ec iro' pgi e ^^ :::;::;;- o — Con Values of It ^ cZ 2*5 5*5 4*5 Fig. 255. — Correcting Coefficients at Inside. where a and ^ are correcting coefficients by which the stresses calculated in the ordinary manner should be multiplied to give the corrected values. CURVED BEAMS 546 The following values of a and ^ have been calculated — - •75 1 2 3 4 5 Circular Section. Rectangular Section. 1 a •62 •70 •84 •89 •91 •93 a /3 211 1-62 123 114 110 1^08 1-92 1-52 1-20 112 1-09 107 •65 •73 •85 •90 •92 •95 j These figures are plotted in Figs. 255, 256. 10 '90 o '80 00. 1 70 p a •60 _ - -;::: — ^^ -— - ^^ ■ — _^ -^ --^ "^ Re cYa na\ 3, ,^ --^ -^ t ^ -^ V A y } '/ // / / // -^{ "ir ok i. y /^ / // / 1 1 5 2. Z 5 % J 3 •5 ^ 1. 4 •5 5 Values of E. -^ tZ. Fig. 256. — Correcting Coefficients at Outside. The nearness of the curves for the two sections shows that the same coefficients may be used as a first approximation" for other sections. ^ Andrews-Pearson Formula. — In this theory,* published * A Theory of the Stresses in Crane and Coupling Hooks, Draper's Company Research Memoirs, Technical Series 1 (Dulan & Co., Lon- don). For other experimental investigations see " Maximum Stresses in Crane Hooks," by Professor Goodman, Proc. Inst. C. E., Vol. CLXVII (1906-7); "An Investigation of Strength of Crane Hooks," American Machinist, Vol. 32, October 30, 1909. 546 THE STRENGTH OF :\L\TERIALS in Um)4 by the author and Professor Karl Pearson, F.R.S., a correction is made for the fact that owing to lateral strain it is not quite correct to say that y = y^ The resulting formula is ^r\ where A 71 = Ay, ' R 1 + 1 R 1 -^ ^ R i/\i-'j For simplification of resuJts we write 73 = 7i - 72 For a rectangular section we have _ R _ R )7 2R + t^y-^ /2R -d.^-^\ ^^^ ~ (1 ->7)c/\^ 2R y '' 2R / ' Then A y, = ^ R A somewhat similar graphical construction can be employed to that in the Winkler formula, but still greater care has to be taken to ensure accuracy. These formulae are extremely troublesome to use and require the utmost care to avoid arithmetical error, and the additional accuracy over the Winkler formula is so small that it is doubtful if the revised formula is worth the additional trouble and risk of error. This point is dealt with very fully by Professor Morley in Engineering, September 11 and 25, 1914. The Strength of Rings and Chain Links. — In deter- mining the stresses in a ring or link we have, in addition to CURVED BEAMS 547 the problem of finding the stresses in a curved member, that of finding the bending moment at any point. Approximate Theory of Circular Ring. — We will calculate the manner in which the bending moment varies in a ring in which the effect of the curvature on such variation is negligible. By designing the ring for the bending moment thus obtained, allowance for the curvature effect upon the stress by the correction coefficients given in Figs. 255, 256, we shall obtain a very fair approximation to the stresses. Consider a radial section intersecting the centre line at c ^^\ / ^ x^'"'^ r-\ \ \ / ' /^ WX \ \ I ' /^ \ ^ \ {'•■ I \ ] • 1 \ 1 \ \ ' \ V \ \ ) '] \ ^ ^'"~" X N ,^ / -^ Fig. 257. — Circular Rings. (Fig. 257). The stresses in each quadrant will be similar by symmetry. The bending of the link must be such that the sections o E, o B remain parallel to their original positions. But the angular change due to bending is given by the /M ^^d s (see p. 360) /m d s J EI ^^ we have (1) Now the B.M. on the section o c = M, = B.M. at B - '' • ^ c w 2 Rsin^ (2) 548 THE STRENGTH OF MATERIALS M„ - Y R sin ^ ] . R d! (9 _.yM..^y-^M^ EI EI M„ . R ^ + ^ R2 cos ^ /MT.Y-_^^^WR^^_^^| ^ 1 = -[M,R(;i-0)+ "o (o-i)y X M„R TT W R2\ 1 2 2 ; EI If this - 0, M3 - ^^ - -318 W R (3) TT W R .• . M, = M, - -\p - W R (-318 - -5) CI = - -182 WR ..(4) The point of zero bending moment is given by putting M, =0 .• . in equation (2) -318 W R - '5 W R sin ^ =. sin B = — p- = '636 •5 i.e,. 6 = 39" 5 degrees approximately. W sin ^ W The direct stress at b = ; at c = — . ; and at E = ^. The mean shear stress at B W ^ Wcos^ . ^^ „ = -^ ; at c = ^ . - and at L = 2 ' 2A Numerical Example. — Find the safe load upon a mild steel ring of 4 inches mean diameter formed of round rod 1 inch in diameter. We will take the safe tensile stress in the material as 7 tons per sq. in. Let the safe load be W Then M, = 318 W R = 636 W . • . By ordinary bending formula "^~32^' = -636 W CURVED BEAMS 549 In our case , = 2 a .' . from the table on p. 545 a = 1-23 .-. -636 W X 1-23 = ^^ •'• ^ = 32 X -636" X 1-23 = '^L^^^^^^J- Considering the section at E and taking this value of W we have Direct stress = ^^ ^ rrxn = '^^ *on per sq. in. approx. M, = -182 WR = -364 W , .. , 7 X M, 7 X -364 . ^^ bending stress = — j^— - = — ^636~~ ^ approx. .* . total stress = 4*00 + '56 = 4'56 tons per sq. in. Chain Links with Straight Sides. — We can apply as follows the same approximate theory to the determination of the stresses in a chain link composed of semi-circular ends and straight sides. Considering one quarter of the link as before, we have that the angular change due to bending between the points B and e (Fig. 258) must be zero. Between f and B we have as before Angular change = / ^j^^ '^{m ^_wr^\ 1 im„. . - 'EI The bending moment will be constant over the straight part of the link. . * . Angular change between E and f = ^^ . ^ WR and M^ = M, = M« - ^- as before , /.. R^ WR2 M,L\ /. wehave^M..-^ 2^" + ^ ) Rtt WR2 . M.. L\ J^ EI ,,^ Rtt WR2 M„L WRL\ 1 2 2 ' 2 4 y • EI 550 THE STREXC4TH OF MATERIALS If this is zero, WR(R + M. = ttR -h L W R /2 R -f L 2 Vtt R + L Fig. 258. — Oval Chain Links. _ WR f 2R ^ L _ 1 ~ 2 I TT R -^ L ^' WR/2R--R, WR^, 2-- ""2 WR+L (5) WR^ 11416 \ 2 VttR+L;" 2 VttR+L/ 2 WR+L/"^^ If L = these formulae reduce to the same result as in the previous case. Experiments on Chain Links. — The strength of chain links has been investigated ver}' thoroughh^ by Professors G. A. Goodenough and L. E. Moore,* who give a very complete theoretical treatment of the subject. * University of Illinois Bulletin, Xo. 18. CURVED BEAMS 551 Their summary and conclusions are as follows — 1. The experiments on steel rings confirm the theoretical analysis employed in the calculation of stresses. 2. The experiments on various chain links confirm the analysis and show that the pressure may be taken as uniform over the arc of contact. 3. A load on the link produces an average intensity of stress K -. in the cross section of the link containing the minor 2 A ^ axis, and with an open link of usual proportions the maximum tensile stress is four times this value. 4. The introduction of a stud in the link equalises the stresses throughout the link, reduces the maximum tensile stresses about 20 per cent, and reduces the excessive com- pression stress at the end of the link about 50 per cent. 5. The stud-link chain of equal dimensions will, within the elastic limit, bear from 20 to 25 per cent, more load than the open-link chain, but the ultimate strength of the stud link is probably less than that of the open link. 6. In the formulae for the safe loading of chains given by the leading authorities on machine design, the maximum stress to which the link is subjected seems to be under- estimated and the constants are such as to give stresses from 30,000 to 40,000 lbs. per sq. in. for full load. 7. The following formulae are applicable to chains of the usual form P = 0'4:d^ s for open links 'P = 5 d^ s ,, stud „ where P = safe load, d = diameter of stock and s the maxi- mum permissible tensile stress. CHAPTER XX * ROTATING DRUMS, DISKS AND SHAFTS Thin Rotating Drum or Ring. — If a thin drum or ring of radius r (Fig. 259) rotates T^ith a velocity v, there will be acting on each unit length of the rhig a centripetal pressme p equal to , where iv is weight of unit length of the ring. 1/ Thus pressure p will cause a hoop stress / and on any dia- FiG. 259.— Thin Rotating Cylinder. metral section the resulting force tending to cause bursting of the ring will be equal to p d, as in the case of thin pipes dealt with on p. 115. The force resisting bursting will be equal to / X 2 A w^here A is the cross-sectional area of the ring. We have, therefore, / x 2 A = p d = '— = , _ 2. w V- _ ic V ''' f ^ YglK ~ gA ' ROTATING DRUMS, DISKS, AND SHAFTS 553 Since w is the weight per unit length of the ring, we have, if p is the weight per unit vohime of the material, w ^ p K, since the volume of a unit length of the ring is 1 x A • f = P^ 9 If p is the weight in lbs. per cu. in. of the material, v is in feet per sec. and g = 32'2 feet per second per second, we have, bringing everything to inch units , p ?;2 X 12 X 12 12 p v2 ^ =^ ~3F2'^12" ^ "32^ ^^'- P^" '^- ^^• = f^approx. for cast iron. = qTk approx. for mild steel. This stress is often called the centrifugal stress. Numerical Example. — At what peripheral speed may a thin mild-steel ring be rotated if the centrifugal stress is not to exceed 16,000 lbs. per sq. in. ? We then have 16,000 = ^ v^ = 9-5 X 16,000 V = x/9*5 X 16,000 = 390 feet per sec. approx. Revolving Disk. — The consideration of the stresses in a revolving disk bears considerable resemblance to that of the stresses in a thick pipe, but it presents greater difficulty. If the disk is uniform in breadth and such breadth is compara- tively small we may proceed as follows — Considering, as in the case of Lame's theory (p. 510), an elemental ring at radius x of thickness 8 x (Fig. 260) and of unit breadth, we have a resultant centrifugal tension on the section given by -ij, w v^ J 2wv^ ±,= ~ — . a = 9/ 9 554 THE STRENGTH OF MATERIALS If the angular velocity = o), i; = o) x 2 9 Jb,. = . 6 X since iv = p h x 9 For convenience we will write -- ^ a 9 .-. F. = 2qo>^x-Sx (1) Then by the same reasoning as in Lame's theor}^ we shall have Force tending to cause bursting of ring = F, + {p + Sp).2{x + S x) Force resisting bursting of ring ^2fSx + p.2x Fig. 260. These must be equal • ' • -^ + iv + ^P) (^ + S x') ^ j Sx + px neglecting products of small quantities F (V + p X + p S X + X ^ p = f S X + p X F . • . if — p)Sx=xSp+ 2 = ^ + g -2 x^^ (by (1)) ROTATING DRUMS, DISKS, AND SHAFTS 555 . • . In the limit f = p + ^/- + ^ co^ .t^ (3) Next consider the strains. If the radius x increases to {x + u), the circumference increases from 2 tt a; to 2 tt {x + u) . ' . increase in circumference = 2 tt u . • . Unital circumferential strain = ^ = - Z TT X X Also the thickness of the ring increases from 8a:;toSa: + Su o IJ u U .' . Unital radial strain = ^— = -, — in the limit. ox dx Now the principal stresses acting in an element of this ring are / and p . • . (as shown on p. 25) Unital circumferential strain ^ ^ [f — r] p) Unital radial strain = ^ {p — v f) i-e. -^ == ^ (/ -'V2?) (4) §-:=e(p-''/) (5) . • . solving these two simultaneous equations we have E fu . 7] du ^-,-r^-)a" + fS <^) Putting these results in (3) we have E /u 7] du (1 - rj^) \X "^ ~d^ _ E /rj U du\ x'E / 7] U rj d u d^ u {I — r)^)\ X dx) \ — y]'^\ x^ X dx d x^ + g w^ x"^ E i. e. multiplying through by t, ^ U , ridu 7] U , du 7]U , 7]d u xd^u {\ — rr-) „„ X dx X dx X dx dx^ E 556 THE STRENGTH OF MATERIALS To solve this differential equation we write it x^ d^u , xdu (1—772) Now assume u = C x^ = X^.6CX+X.3CX^-CX^+ ^^ ^^ . g 0)2 x3 Equation (9) will be solved if C = - ^^~~f2 ^ "^^ .... (10) This equation is of the"! kind dealt with in Forsyth's Differential Equations (Macmillan), §§ 38, 39. The '' complementary function " is x^ d^ u X du _ d x^ d X d^u 1 du '^ _ ci ' ' d x^ X d X x^ d^u d /u\ _ ' ' d x"^ d X \x/ Integrating, we have -J 1- - = constant = Cj (11) ax X To integrate again, write it X a u ^-j _ + w =r (Ji a: dx ^ 1. e. --^ — - = Ci .T dx C a;2 . • . integrating u x = ■— \- C2 ■ , •••I = ^' + & (12) . • . putting this in (11) 4j* _ Ci _ C2 ,,0 dx~ 2 x^ ^ KOTATING DRUMS, DISKS, AND SHAFTS 557 We also have from " particular integral '' u = C x^ - = C X2 X t^ = 3 C a;2 a X . ' . adding the complementary and particular integral we have ^ = C a;2 + ^^ + % X 2 x^ (1 - v^) q 0) 2 ^^ , Ci C2 ,..s = 8E + T + ^ ^^^^ du _ 3(1— 7)^) q 0)2 x^ , Ci _ C2 /, K\ .dx~ 8E "^2 0:2 ^ ^ Special Cases. — (1) External radius R, internal radius r. We must have 2? = at the inside and outside . • . p = for a; — R and a; = r . * . at inside in equation (7) o — -^ /rju du ~ 1 — rj^\ r dr E r (1 -r}^)qo,^r^ , C^rj 0^7} r '•^' ^ - (1 - ^2) \ ^g-E -^ + 2 + _ 3(l-ry2)go,2^2 Ci _ C2\ 8E "^2 r2j i.e.5(l+>;)-jMl-.)=^^-^^^^^3+,) ..(16) Similarly at outside where a; = R we shall have ^ (1 - 77) (r2 - R2) . 8 E (1 + r)){3 +r;)ga)2R2r2 8E Putting this in (16) and simplifying (17) ^^_ (1 - r,) {S + r,) q o>^ (R^ + r^) ^jgj 4 E 558 THE STRENGTH OF MATERIALS Now put these values into the equations (6), (14) and (15) for / - (r=:^2) 1^ - 8 E + 2 (^ + ^^ + x^ ^^ - ^^ 3 77 (1 — rf) q la^ X^\ 8E / + (3 + ';) -/- - 3 , x^} Similarly to obtain p we use equations (7), (14) and (15) and take the given values of the constants. E ^riu dn\ ^ {I -yj^)\x ^ dxf -{l-r,')\- 8 E + 2 (1 + ^) - :.2 (1 - V) _ 3^1 - 7y2) qj^^ " 8E f R2r (3+^)- 2 -3 a: X p w2 (3 + 7?) f-^ R^r^ ^ It is clear from equation (19) that the greatest hoop tension occurs at the inside where x — r 9 and if r is very small ••• /-.v. = ^ " {(3 +v)^' + a-l)r'} (21) _ p w2 R2 (3 + ^) /max. - -4^ " — l^^J ROTATING DRUMS, DISKS, AND SHAFTS 559 (2) Disk without central hole. — If the disk is a solid one of radius R, ^ = for a; = R and u the circumferential strain must be for x = .'. from equation (14) Cg = .'. Equations (14) and (15) become u _ (1 - rf)qio^x''- Ci x~ ~ 8E +2 ^^"^^ du 3 (1 — 7)^) q iJi^ x^ , C, ,^.- di=-- 8E- + 2 (24) . • . Since p = for x = H, equation (7) becomes E f7?Ci 7y(l -7?^)go>2R^ Ci 3{l -yj^)qu>^U^ '\ (1_^2)\ 2 "SE" ' "^2 8E j t. e. J (1 + ^) = ^^^^^ g co^ R2 (3 + ,7) (1 -r;)(3+>?)go>^R^ ,^.. ^1 - ~ 4E ^ ^ . • .^Equation (6) becomes ^ f _ (1 - rjf q oi^X^ (1 -^) (3 + r]) g (o^ R2 '''{i-7]^)\ 8E "^ 8E _ 3 ry (1 - 7/2) g (o^ a; ^ (1 - ry) (3 + 7?) g (o^ R^l 8E ' "'"'^ "8E ^ j = ^f 1(3 + .7) R2 - (1 + 3 r;) a;-^} = ^ 1(3 + >?) R^ - (1 + 3 r;) a:2j (26) Similarly equation (7) becomes P = ''^g {B + v) Cii' - ^') (26) Each of these is a maximum at the centre where x ^ 0, where we have Lax. = Z^Huax. = Sj- (3 + ^) (27) This is exactly one half the value given by equation (22) for a disk with a very small hole, so that on this theory a disk has its strength reduced by one half by having a hole made 560 THE STRENGTH OF ]MATERL\LS through it. For this reason the De Laval turbine drums are made without a central hole. If V is the peripheral speed at the outside we have r = w R .-. Equation (27) gives / = ^- (3 -f v) ^ti^h is ^^-—^ of the stress in a thin drum of the same external radius (of. p. 553). Taking , = 1, i^+2) = g NuMEEiCAL Example. — At wlmt peripheral speed may a narrow mild-steel disk he rotated so th-^ bj^e' ^^/ R-i / . • . At any radius ^ =b^e £7-2/ Whirling of Rotating Shafts; Critical Speeds.— If a shaft rotates at a high speed, the lack of mathematically exact balancing results in an eccentricity of load which causes Fig. 262.— Whirling of Centrally Loaded Shaft. centrifugal forces to be induced and these centrifugal forces will cause deflections which increase the eccentricity to be increased ; this increased eccentricity causes further deflection and so on, the deflection increasing indefinitely and giving rise to whirling at certain speeds called critical speeds. In certain cases the whirling speed is the same as the natural frequency of transverse vibration of the shaft. The centrifugal forces may be regarded as having a neutralis- ing effect upon the elastic forces tending to return the shaft to its natural shape, so that when whirhng occurs the effective stiffness of the shaft is reduced to zero. Flexible Shaft loaded at Centre. — Referring to Fig. 262 let a shaft a b of length I be loaded at the centre with a disk of weight W and let the shaft be provided with flexible bearings ROTATING DKUMS, DISKS, AND SHAFTS 563 which do not interfere with the natural deflection of the shaft. Then if 8 is the deflection caused by the centrifugal force and e is the eccentricity of the load, i.e. the distance from the centre of the shaft to the centre of gravity of the load, the effective eccentricity is (S + e). If the angular velocity is w . • . Centrifugal force = F = — ^ 9 Ai . F^' ■ T? 48EI8 ^^^^=48EI '•'■^- ' p ' 48EI8 Wco^, , , /48EI W(o2^ Wco^e l^ 9 y 9 W 0)2 e 48 E I .8 = 9 ' l^ 9 4:SElg -w^WP ^^^ From this equation it is clear that 8 will become indefinitely great if 48 E I (/ - w^ W Z^ = ^.6. if.= Z^^LX (2) This value of w gives the critical speed. For mild steel E = 30 x 10^ lbs. per sq. in. and g = 32*2 X 12 ins. per sec. per sec. ; if therefore W is in lbs. and I and I in inch units 48 X 30 X 106 X 32-2 x 12 . 1 WP = 746,000. / radians per sec (3) 74,600 X 60 /nr 2 TT MWP = •711 X W^L revolutions per minute (4) For a round shaft of diameter d inches we have I ird^ 64 •158 X 10^ d-^ n = j-^=^^ — 564 THE STRENGTH OF :VIATERIALS Working from the transverse vibration we have, as on p. 337, ^ = 2 77 / ' Weight ^ g X force to cause unit displacement 48 E I . • . Force = W .-. t = 2- V48EI^ for unit deflection Frequency 1 1 't = o-a/—^-^ per second 2-V WZ3 ^ -^n J^^^V^T minute. 77 X 60 \ WZ^ Fig. 263. This,, for the given value of E, is exactly the same result as is obtained in equation (4) above. If the critical speed is exceeded either by providing guides which prevent the excessive deflection or by speedmg up so quickly that the inertia of the shaft prevents the dangerous deflections from developing, the shaft will '" settle down " and nui smoothly in a deflected form (Fig. 263), the weight rotating about an axis which gradually approaches its centre of gravity as the speed increases. This fact is made use of in the flexible shaft of the De Laval turbine. If t-j is the critical velocity we may put in equation (1) ROTATING DRUMS, DISKS, AND SHAFTS 565 i. e. 8 = W P 0..2 -WP 0.2 o.;2 (u2 — O) 2 •e (5) This gets numerically less as o) increases, so that as the speed increases more and more the shaft tends to straighten out. If the shaft is horizontal and the weight is perfectly balanced, W P and there is an initial deflection S„ = j^ ^ y , this will give rise to a centrifugal force which causes an additional deflection 8^ ...r = Wo>2 8.) g FP ^""^ ^1^48 EI Fig. 264.— Whirling of Unloaded Shaft. 48EI8i Wo)2 . , W(o2 . P 9 g '48 EI Wo)^ _ Wco^ 9 ^ ~ 9 , /48EI Wo)^ W(o2 , I.e. Oj — ^ = . do Woj2 ^ 48 EI Wo.2 9 P 9 W 0)2 Z3 8, ~ 48 E I ^ - 0.2 w ^3 (6) This, as one would expect, gives the same result as before "with e = 8„. Unloaded Shaft. — In this case we obtain at certain critical speeds a condition of insta?jility which is very similar to that which occurs in a loaded column. Suppose that the shaft of length I is initially straight and that due to some cause it becomes deflected so that at some point p (Fig. 264) at distance x from a convenient origin, say the centre point c, the deflection is y. 566 THE STRENGTH OF MATERIALS If w is the weight per unit length of the shaft this deflection will cause at the point p a load equal to per unit length. But since the B.M. diagram is the second integral of the load diagram (p. 149) we shall have dyM. _ ww^y dx^ g Moreover, .pr^ = ^-^ EI d x^ w ui^ y d^ y ''' EI> ^ dx^ 2 putting -^ J- == m*j we get the differential equation »'2'=rf.^ w The general solution of this is 2/ = A cosh m X + B sinh mx -\- C cos m x -{- D sin m x . . (2) Ends freely supported. — If the ends are freely supported, the deflection and B.M. are each zero at each end. . • . 2/ = when x = — ^ and x = + ^ -,%, = when .r = — ^ and x = ^ dx^ Z ^ d v also the slope -^ = when .r = Cv X 7nl ^ . , ml ^ ml ^ . ml .-. = A cosh -^ + Bsmh- 2" +C cos - -2- + D sm — ^ ^ ml ^ . ^ rnl , ri ml -^^ . ml = A cosh -o^ + B smh ^^ + C cos ^ + D sm ^ ^ ml 1 ml ^ ^ ^^c "^ ^ Now cosh Y = cosh — ^ ; cos ^ = cos — ^ ml . , ml . ml ^\r.'^^ smh - 2 = - ^1^^ 2^ ' ^^^ "" 2 ^ ~ sm 2 D a nd B each = . • . A cosh '2 + C cos 2 =0 (^) ROTATING DRUMS, DISKS, AND SHAFTS 567 . • . our equation becomes y = A cosh m X -\- C cos m x ,-r d cosh X . , d cos x Now — ^ = smh X ; —^ = — sni x Cb X Cv X d sinh X , d sin x — -^ = cosn X ; — ^ = cos x Cb X Cv X CJ tJ . ' . -7-^ = Am^ cosh m X — Cm^ cos m x dx^ .-.putting X = + ^OT - - .' .0 = Am^ cosh ;-- — C m^ cos — 2 2 i. e. A cosh -^ — C cos -— = (4) 2 2 .• . Comparing this with (3) we see that A must = .' .y = C cos m x Further, C cos -~- = ml TV , — = 2 rtc. Taking the lowest vahie l^"" m (w 0)2 YJ EIi7 o) = ^ A / ^ radians per second l^ \ w . • . If ?i is the number of revohitions per minute Itt n IT n • 30 (D 30 TT /EI a 1 • • ' • ^ 72 " A/ ^^ revolutions per mmute (5) If the shaft is of diameter d inches and I is in inches Taking E =- 30 x 10^ lbs. per sq. in. g = 322 X 12 ins. per sec. per sec. 568 THE STRENGTH OF IVL.\TERIALS 64 tc = -f_ X -28 lb. 4 , 4-8 X 106 f/ ,^, we have n = (6) 9 — "^ — 4. — The higher critical speeds -v^ill occur for m = *" " , " , " , ^ JU ^ etc., giving values of n multiplied by 4, 9, 16, etc. Both ends fixed-. — In this case ,^ = for x — + in dx ~ 2 addition to the condition that v = for x = 2 .• . B and D are each equal to as before and A cosh ^^ -f- C cos -^ = 2 2 when .r = - , — " = m A suih ~-^ — m C sin - = 2' dx 2 2 I d y . . ^ ml ^ . 7?i Z ^ — ni A sum — — ^n n oi^-. 2 «. e. A sum ^ — C sm -- = . m I ' ' C . , m I suih — m Z A ""^ 2 Also from (3) 7^ = — — , cosh , ml , -, ml .• . — tan = tann iu ^ 3 TT The solution of this gives w Z = 474 = " aj^prox. Taking the approximate value, this will give the first critical speed about nine times that in the case of the freely supported shaft. Dunkerley's Empirical Formulae. — Professor Dunker- ley,* who was one of the first investigators of the theoretical * Phil. Trans. Roy. Soc. 1895, Liverpool Engineering Society, 1894-5. ROTATING DRUMS, DISKS, AND SHAFTS 569 and actual whirling speeds of shafts, has given the following empirical formulae which agreed very well with his experiments. Let oji be the critical angular velocity for a given unloaded shaft; let o)^ be the critical angular velocity of the same shaft carrying a wheel at any position, neglecting the mass of the shaft. Then the critical velocity w^ of the loaded shaft will be given by or if Uo, n^, n^ are the corresponding number of revolutions per minute Tio^ n-^^ n^ If a second load be keyed at another position, the critical angular velocity of which is wg with the first load removed neglecting the weight of the shaft, then (Oi (Oo 0)r or m general — = ?, — For further information on this subject, the reader may refer to Professor Dunkerley's paper and to Stodola's Steam Turbines (Constable) . EXERCISES CHAPTER I 1. A tie rod in a roof structure has to stand a total pull of 40 tons. If the stress in the material is to be not greater than 5 tons per sq. in., find a suitable diameter. Ans. 3| ins. diam. 2. Taking the shearing strength of mild steel to be 20 tons per sq. in., calculate the force necessary to punch a f in. hole in a f in. plate. Find also the stress in the punch. Ans. 29*4 tons ; 66' 7 tons per sq. in. 3. A bar of mild steel fin. diam. and 10 ins. long stretches '00816 in. when carrying a load of 5 tons. Calculate Young's modulus (E) in lbs. per sq. in. Ans. 30 X 10^ lbs. per sq. in. 4. If E is 29,000,000 lbs. per sq. in. for wrought iron, what decrease in length of a column 20 ft. high and 12 sq. ins. sectional area takes place when carrying a load of 36 tons ? Ans. '0556 in. 5. What load in lbs. is hung on an iron wire 50 ft. long and *! in. diameter to make it stretch -^o- in. ? Ans. "076 lb. 6. Plot a stress-strain diagram for the following test of a specimen from a mild-steel boiler plate — Load lbs 4,000 •0009 8,000 •0020 " 12,000 •0033 16,000 20,000 24,000 28,000 Extension ins. . . •0044 •OOSe ^0070 1 •0082 Load lbs 30,000 34,000 36,000 1 40,000 44,000 48,000 52,000 Extension ins. . . •0103 •016 •7 ■ •19 ^30 -47 •75 Load lbs 56,000 1-3 59,780 54,900 g^ , /Luads-1" = 10,000 lbs. ocaies (^Extensions— up to yield Extension ins. . . 25 2^9 point 500 times full size. Beyond = 4 times do. Orig. dimens. Length = 10 ins., width = 1*753 ins., thickness = '64 in. Final „ „ = 12-9 ins. „ = 1*472 ins. „ = '482 in. Find stress at elastic limit, maximum stress. Young's modulus, and per- centage extension and reduction of area. 57] 572 EXERCISES 7. In a plate girder the maximum intensity of stress at right angles to the vertical cross section of the web is 5 tons per sq. in., and the in- tensity of shearing stress is 2 tons per sq. in. Find the position of the planes of principal stress at that point and their intensities. (A.M.I.C.E.) Ans. 19° 20' aTid 70° 40' to vertical ; 5" 7 and 0*7 tons per sq. in. 8. The limit of elasticity of a W.I. bar was found to be 20,000 lbs. per sq. in., the strain at that point being 0*0006; what was the resilience of the material? (A.M.I.C.E.) Ans. 6 in. Ihs. 9. Two rods, one of copper and the other of steel, are fixed at their top ends, 24 ins. from one another, and hang vertically downwards. They are connected at their bottom ends by a horizontal cross-bar, and on this bar is to be placed a weight of 2000 lbs. If each rod is 18 ins. long, and if the diameter of the copper rod is 1 in. and of the steel rod f in., find where the weight must be placed so that the cross-bar may remain hori- zontal. E for copper = 16 X 10^ lbs. per sq. in. ; for steel = 29 X 10^ lbs. per sq. in. (B.Sc. Lond.) Ans. 11*9 ins. from the steel rod. 10. A load of 560 lbs. falls through ^ in. on to a stop at the lower end of a vertical bar 10 ft. long and 1 sq. in. in section. If E = 13,000 tons per sq. in., find the stresses produced in the bar. Ans. 5*45 tons per sq. in. 11. A bar of iron is at the same time under a direct pull of 5000 lbs. per sq. in., and a shearing stress of 3,500 lbs. per sq. in. What will be the resultant tensile stress in the material ? Ans. 6,800 lbs. per sq. in. 12. In Question 11, find the resultant tensile stress from the strain consideration. Ans. 7,250 lbs. per sq. in. 13. Find whether, in the problem of Questions 11 and 12, on the as- sumption that the shear strength of the material is 4 of the tensile strength, the resultant shear stress is more serious than the resultant tensile stress or strain. Ans. Res. shear stress = 4,300 lbs. per sq. in. Not so serious. 14. Steel rails are welded together and are unstressed at a temperature of 60° F. They are prevented from buckling and cannot expand or con- tract. Find the stresses when the temperature is : (1) 20° F., (2) 120° F., taking steel as expanding '0012 of its length for a temperature change of 180° F. E = 30 X 106 lbs. per sq. in. If the elastic limit is 40,000° F., at what temperature ould it be reached ? (A.M.I.C.E.) Ans. 8000, 12,000 lbs. per sq. in. ; 260° F. 15. If the stress p at a point on one plane is inclined at an angle of 60° to that plane and on a plane at right angles to the former the stress is a simple shear, find the principal stresses at the point and their direction. -4n5. 1 (l ± \/|) ; tan2e= y EXERCISES 573 CHAPTER III 1. In a roof truss a certain tie lias in it a pull of 3*05 tons due to the dead weight alone. When the wind is on the left of the truss it alone causes a pull of 5*5 tons in the same tie, and when it is on the right side it causes a compression of 1*2 tons. Work out what you would consider a satisfactory section for the tie if it is made of mild steel. Ans. 3 ins. x f in. flat. 2. Estimate the dead load equivalent to a tensile dead load of 15 tons and a live load of 20 tons ; if the strain is not to exceed '001, find the area of section required, E being 13,500 tons per sq. in. Ans. 55 tons ; 4" 07 sq. ins. 3. A 3-girder bridge to carry a double line of rails has an effective span of 38 ft. 6 ins. Find a suitable working stress assuming that the weight of the girders is 5^^ of the weight to be carried ; that the flooring weighs 7 cwt. per ft. run of the whole width of the bridge ; that the permanent way, etc., weighs 160 lbs. per foot run for each line of rails; and the live load is 40 cwt. per foot run per line of rail. Ans. 5 tons per sq. in. 4. What load, suddenly applied, will produce in a mild steel bar an extension of -^ of an inch ? The bar is 5 ft. long and 1| sq. ins. in section. Take E = 13,000 tons per sq. in. Ans. 4*06 tons. CHAPTER IV 1. Two lengths of a flat steel tie bar, which has to carry a load of 50 tons, are connected together by a double butt joint. The thickness of the plate is I in. Find the diameter and the number of rivets required, and the necessary width of the bar for both chain and zigzag riveting. What is the efficiency of each and the working bearing pressure ? Make a dimen- sioned sketch of the joint. Ans. I in. rivets, 12 and 10| in^. wide ; 75 per cent, and 87 per cent. ; 9*2 tons per sq. in. 2. A diagonal tie in a lattice girder has to carry a load of 15|^ tons and is I in. thick. Using f in. rivets, find the necessary width of tie and calculate the number of rivets required (in single shear) and sketch the arrangement. Aiis. 5 J ins. wide, 7 rivets. 3. Plates 1 in. thick are connected by a treble riveted butt joint, the pitch in outside rows being twice that in the others, and d = lin. Taking shear resistance in double shear = 1*75 times that in single shear, determine p for equal shear and tearing resistance. Find also the efficiency. Ans. 6 1 i7is. ; 85 per cent. 4. For equal strengths in tension and shear calculate the pitch for a 574 EXERCISES butt joint, given the following data: Plates 1 in. thick; rivets 1| ins. diam. ; two rows of rivets on each side of joint ; f, = 54,000 ; /< = 65,000 lbs. per sq. in. Ans. 5g- ins. 5. A steel boiler 4 ft. in diameter, and subject to a pressure of 200 lbs. per sq. in., is \ in. thick. Find the intensity of tlie circumferential and longitudinal stresses, the efficiency of the joints being 75 per cent. Ans. 12,800; 6,400 Ihs. per sq. in. 6. Find a suitable thickness of plate and design a double riveted lap joint (longitudinal) for a cylindrical drum 5 ft. in diameter, subjected to an internal gauge pressure of 250 lbs. per sq. in. Take a working stress of 5 tons per sq. in. (A.M.I.C.E.) Ans. Plates 1 in. thick ; rivets 1^- ins. diameter ; 3 J ins. pitch. 7. Calculate the thickness of shell of a boiler 4 ft. 6 ins. in diameter to resist a pressure of 150 lbs. per sq. in. Assume an efficiency of riveted joints of 70 per cent, and take the working stress as 6 tons per sq. in. (A.M.I.C.E.) Ans. -^^ in. 8. Determine the stresses across the longitudinal and transverse sections of the plates of a boiler drum 3 ft. in diameter and J in. thick, subject to a steam pressure of 200 lbs. per sq. in., assuming that the drum is long and that it has no longitudinal seam. Ans. 7,200 ; 3,600 lbs. per sq. in. CHAPTER V 1. A cantilever whose weight may be neglected carries isolated loads of 2 tons and ^ ton at distances of 5 ft. and 8 ft. respectively from its built- in end, the cantilever being 10 ft. long. Sketch shear and B.M. diagrams. Ans. Max. B.M. = 14 ft. tons ; shear = 2|- tons. 2. A certain joist used as a cantilever weighs 18 lbs. per foot, and the max. B.M. which it can carry is 63 "56 in. tons. Find how long the span may be for the cantilever to be able to safely sustain its own weight. Ans. 36-3 ft. 3. A beam of 12 ft. span carries loads of 3 and 4 tons at distances of 5 and 8 ft. from the left-hand support. Draw the shear and B.M. curves. Ans. Max. B.M. = 15*66 ft. tons ; reaction, 3"91 and 3*09 tons. 4. A beam of 25 ft. span carries a load of ^ ton per foot run, and an isolated load of 6 tons at a distance of 4 ft. from the left-hand support. Find the maxihium bending moment, and sketch the shear and B.M. curves. Ans. Max B.M. = 25*2 ft. tons. 5. A beam of 40 ft. span carries a uniformly distributed load of 20 tons ; at points 11 ft. 3 ins. from each end isolated loads of 11 tons are carried, and between these points and each end additional loads of 4*5 tons are uniforml}" distributed. Draw the B.M. diagram. Ans. Max. B.M. = 250 ft. tons nearly. EXERCISES 575 6. A beam 25 ft. long is anchored down at one end and rests over a support 6 ft. from the other end. It carries a load of 15 tons at the free end, and a uniform load of 5 cwt. per foot run. Sketch the shear and bending moment curves. Ans. Max. B.M.— 94*5 ft. tons. 7. A beam 34 ft. long overhangs one support by 6 ft. and carries a load of 10 tons uniformly distributed. In addition it carries a load of 3 tons at the overhanging end and a load of 12 tons uniformly distributed along a length of 12 ft. commencing from the other end. Find the maximum bending moment on the beam. Ans. 62*2 ft. tons. 8. A beam is laid horizontally upon two supports which are 12 ft. apart, and projects at each end 6 ft. beyond the support. A load of 2 tons is carried upon each of the projecting ends, and 1 ton at the centre of the span. What is the B.M. at the centre and at each support ? Sketch the B.M. diagram. (A.M.I.C.E.) Ans. 9 ft. tons ; 12 ft. tons. 9. A plate girder is built of depth = J^ span. The maximum per- missible B.M. in ft. tons in such girder is roughly given by formula : B.M. = 7 X area of flange in inches X depth in feet. Find the maximum span for such a girder to carry its own weight : {a) neglecting its web altogether; (6) taking its web as half the sectional area of one flange. Neglect all angles, rivets, and stiffeners. Take steel as weighing 490 lbs. per cub. ft. Ans. {a) 1,536 ft. ; (b) 1,229 ft. 10. A beam of 20 ft. span carries a uniform load of 5 cwt. per ft. run and an additional load of 9 tons spread over 12 ft. starting from the right-hand end. Draw the B.M. and shear diagrams. Ans. Max. B.M. 38*7 ; Reactions 8*8 and 5-2 tons. CHAPTER VI 1. Find the moment of inertia about the centroid of an I beam 8 ins. deep, the width of flanges being 5 ins. The flanges are '575 in. and the w^eb '35 in. thick. Ans. 89' 1 in. units. 2. A stanchion section consists of two standard channels 11 ins. X S^ ins. placed back to back at 6^ ins. apart and two 14 ins. X ^ in. plates riveted to each flange. Find the least radius of gyration. Ans. 4*12 ins. 3. Find the radius of gyration of a hollow cylindrical column with an external diameter of 12 ins. and a thickness of 1 in. ; also of a solid square column 4 ins. hj 4 ins. Ans. 3*90 ins. ; 1*15 ins. 4. A cast-iron girder has an upper flange 4 ins. by 1 in. ; a lower flange 8 ins. by 1|- ins. and a web 6 ins. by 1 in. Find its moment of inertia and radius of gyration about an axis through the centroid parallel to the flanges. Ans. 195 ins.'^ ; 2*98 ins. 5. A channel section has a base of 10 ins. ; sides 3 ins. ; the thickness of 576 EXERCISES metal being f in. Find the position of the centroid and the moment of inertia about a line through the centroid parallel to the base. Ans. '726 in. from hase ; 6*62 ins.^. 6. A column is built up of two I beams 10 ins. deep and with flanges 5 ins. wide, the centres of the beams being 10 ins. apart. The area of each is 8*82 sq. ins., and the greatest and least moments of inertia are 145"7 and 9" 78 in. units respectively. Riveted at the top of each pair is a plate 12 ins. wide. Xeglecting the rivets, find the thickness of the plate if the greatest and least moments of inertia are the same. Ans. yV in. 7. A column is built up of two channel sections 12 x 82^ X i^ in., with a plate ^ in. thick riveted to the flanges at top and bottom. Find the distance x apart that the channels must be for the moments of inertia to be equal about the two axes of symmetry, the width of the plates being ic + 7J ins. Ans. 9f iiis. CHAPTER VII 1. A 20 in. X 7J in. joist is supported at both ends. The weight per foot of this section is 89 lbs., and the moment of inertia = 1,646 ins.*. Find the distributed load in a 25 ft. span which vnll cause a max. flange stress of 7 tons per sq. in. Ans. 29*7 tons net. 2. The moment of inertia of a 12 in. X 5 in. X 32 lb. joist is 221 ins.*. Two such joists are placed side by side, and support a water-tank which weighs 1 ton when empty. Eflective span = 15 ft. What is the weight of the water in the tank when the stress in the extreme fibres of the joist is 6*5 tons per sq. in. ? Ans. 19*8 tons. 3. Two 6 X 3 X ^ in. "fs are used back to back as a girder on which a light crane runs. Compare the safe load which such a beam would carry with that of a joist of same span, depth, width, and thickness of metal. Ans. Joist 5*36 times as good. 4. Find the bending moment which may be resisted by a cast-iron pipe 6 ins. external and 4^ ins. internal diameter when the greatest intensity of stress due to bending is 1,500 lbs. per sq. in. Ans. 21,750 in. lbs. 5. A rolled-steel joist 16 ins. deep, with flanges 6 ins. wide and 1 in. thick (the web being | in. thick), is used to support a uniformly distributed load of 2 tons per ft. run. If the span is 12 ft. 6 ins., what is the maximum stress in the lower flange ? (A.M.I.C.E.) Ans. 4^'^ tons jjer sq. in. 6. Find what diameter of axle should be employed if the wheels are 4 feet 9 ins. apart and the loads on them are 7 and 3 tons respectively, the axle boxes projecting 9 ins. beyond the wheels. Draw the B.M. diagram. Ans. Max. B.M. = 58' 68 in. tons ; diameter = 5 ins. 7. A gallery is carried by two 9 in. x 3 in. timber cantilevers, each 5 ft. long. What distributed load may the gallery carry if the safe stress is 10 cwt. per sq. in, Ans. 27 cwt. EXERCISES 577 8. Either of the following sections is available for a beam which is required to be as strong as possible : (a) Circular, 2 ins. diam. ; (6) rect- angular, 2 ins. deep, 1"178 ins. wide. Which would you use ? (A.M.I.C.E.) Ans. Circular. 9. A cast-iron beam section is 20 ins. deep ; top flange 4 ins. x 1 in. ; bottom flange, 16 ins. X 1| ins. ; web, 1 in. Find the safe distributed load which a cast-iron girder of the above section, and of 20 ft. span, could safely carry. Take the safe stresses as 1 ton/in.^ in tension, and 4 tons/in.^ in compression. Ans. 10*6 tons net ; 11*9 tons gross. 10. Find the bending stress in a locomotive coupling-rod 8 ft. long, 2 ins. broad and 4| ins. deep. It runs at 200 revolutions per minute, the crank radius being 11 ins. • Ans. 2*6 tons per sq. in. CHAPTER VIII 1. A tie bar 9 ins. wide and 1|^ ins. thick is curved in the plane of its width. If there is a total tensile load on the bar of 30 tons, and if the mean line of pull passes 3 ins. to one side of the geometrical axis at the middle of the bar, find the maximum and minimum stresses at the centre section of the bar. (A.M.I.C.E.) Ans. 6f tons per sq. in. tension ; 2f tons per sq. in. compression. 2. An upright timber post 12 ins. in diameter supports a vertical load of 18 tons, 3 ins. from the vertical axis of the post. Determine the maxi- mum and minimum stresses on a normal cross section and show by a diagram how the intensity of stress varies across the section. Ans. '477 and, '159 ton per sq. in. 3. A cast-iron post 12 ins. in external diameter and 10 ins. internal diameter carries an axial load of 40 tons and also an eccentric load of 5 tons, parallel to the axis at an eccentricity of 12 ins. Find the maximum stress. ■Ans. 1*98 tons per sq. in. 4. A short wooden pillar is 20 ins. high, and rectangular in cross section, the thickness of the section is 6 ins., and the width 12 ins. Two vertical loads act on the top of the pillar, both loads act in the middle of the thick- ness, one of them, W^, acts at a point 1|- ins. on one side of the centre, and the other, Wg, acts at a point 2| ins. on the other side of the centre. If the stress over the base of the pillar is everywhere compressive and varies uniformly, its intensity being twice as great at the 6 in. edge near the line of action of W2 as it is at the 6 in. edge near the line of action of Wj, what is the ratio of W^ to Wj ? (B.Sc. Lond.) Ans. 13:11. 5. A reinforced concrete beam, 8 ins. X 11 ins. deep, has four ^ in. bars, with centres at 1 in. from the bottom. Calculate for a span of 12 ft. the safe load (a) on the modified beam formulae; (6) on the no-tension, straight-line formulae. Take t = 15,000, c = 100, t, = 500, m = 15. Ans. (a) 1,205 lbs. ; (6) 3,920 lbs., including weight of beam. PP 578 EXERCISES 6. A reinforced concrete T beam has a flange 4 ft. X 3| in., the width of web being 10 ins. If the centre of reinforcement is 15 ins. below the top, calculate its necessary area, using the above figures. Ans. 7" 94 sq. ins. 7. Find the relation between the depth of slab and effective depth of a T beam in terms of the stresses and reinforcement for the neutral axis to curve at the bottom of the slab. . d,\ 2 r t d / c 8. A T beam is required to carry a B.M. of 320,000 in. lbs. The deptli to centre of reinforcement is 16 ins., and the depth of slab is 4 ins. If c = 600 and t — 16,000, what area of reinforcement and effective breadth of slab would you use ? Ans. 1'39 sq. ins. ; 12| ins. 9. A reinforced concrete floor is 9 ins. thick, the centre of the reinforce- ment being 2 ins. from the bottom edge. If c = 600, t = 15,000, and m = 15, calculate the reinforcement necessary, and the load can that be safely carried. Ans. "63 sq. in. per ft. width ; 386 lbs. per sq. ft. 10. A beam of rectangular section of breadth one half the depth is bent by a couple in a plane at 45° to the axes of the section. Find the safe B.M. in terms of those about the principal axes. ^ 9a. /^ 7 \ / ^ CHAPTER IX 1. If two precisely similar beams of rectangular section, one of cast iron and the other of wrought iron, w^ere laid across the same span and loaded with the same load (within the elastic limit), what would be the relative deflections of the two beams? (A.M.I.C.E.) Ans. As E,. : E„ = about 8 : 13. 2. A beam is of 20 ft. span and tlie movement of inertia of its section is 300 in. units ; what will be the central deflection for a uniformly dis- tributed load of 16 tons ? (A.M.I.C.E.) Ans. -72 i7i. 3. A beam of cast iron, 1 in. broad and 2 ins. deep, is tested upon supports ' 3 ft. apart, and shows a deflection of ^ in. under a central load of 1 ton. Calculate the modulus E. (A.M.I.C.E.) Aiis. 5,832 tons per sq. in. 4. Suppose that three beams or planks, A, B, and C, of the same material are laid side by side across a span L = 100 ins., and a load W = 600 lbs. is laid across them at the centre of the span so that they all bend together. The beams are all 6 ins. wide, but two are 3 ins. and one 6 ins. deep. What will be the load carried by each beam, and what will be the extreme fibre stress in each ? (A.M.I.C.E.) Ans. 480 lbs., 60 lbs. ; 1,333 lbs. per sq. in., 667 lbs. per sq. in. 5. Calculate the least radius to which a 1 in. round bar of wrought iron [E = 28 X 10^ lbs. per sq. in.l may be bent, in order that the skin stress EXERCISES • 579 may not exceed 15 tons per sq. in. What is then the moment of resistance of the section? (A.M.I.C.E.) Ans. 34-7 ft. ; 1-47 in. tons. 6. A beam of uniformly rectangular section is supported freely at the ends and carries a uniformly distributed load. Find the ratio of depth to span so that when the maximum stress at the centre section due to bending is 4 tons per sq. in,, the deflection at the centre is ^^^ of the span. E = 12,000 tons per sq. in. (B.Sc. Lond.) Ans. Span = 24 X depth. 7. Find the greatest deflection in inches of a rectangular wooden beam carrying a load of 2 tons at the centre of a span of 20 ft., with a limiting intensity of stress of 1000 lbs. per sq. in. The depth of the beam is 14 ins. Calculate the breadth. E = 6000 tons per sq. in. (A.M.I.C.E.) Ans. ^ in. nearly ; 8' 2 ins. wide. 8. A 16 in. X 6 in. x 62 lb. R.S.J, carries a load of 12 tons at quarter span, the span being 24 ft. Find graphically the maximum deflection and compare that calculated for the same beam with the load at the centre. (I for this section = 725*7 in. units, E = 12,500 tons/in.^) Ans. '46 in. ; "66 in. at centre. 9. A simply-supported beam of uniform section and 30 ft. span is found to deflect 6 ins. under its own weight. Find the slope of the beam at the supports and also the slope which would arise if the same deflection were caused by a central load instead of a uniform one. Ans. '0533 ; '05. 10. A vertical post, 24 ft. in height, supports at its upper end a horizontal arm projecting 6 ft. from the post. Find the horizontal and vertical displacements of the free end of the horizontal arm when a load of 6000 lbs. is suspended from it. E for post and arm = 28 X 10^ lbs. per sq. in. ; I for post = 412, for arm = 360 (inch units). Neglect direct compression of the post. (B.Sc. Lond.) Ans. Horizontal 1*55; vertical '85 in. 11. A cantilever of circular section is of constant diameter from the fixed end to the middle, and of half that diameter from the middle to the free end. Estimate the deflection at the free end due to a weight W there. OQ WJ 73 Ans. " — , where I is that at fixed end. 24 E I 12. A timber beam 30 ft. long and 12 ins. square in cross section rests on a support at each end. If a load of 1 ton is placed in the centre of the beam, find the work done in deflecting it. Ans. 705*6 in. lbs. CHAPTER X 1. A cast-iron column has its ends securely built in. It is 12 ins. in external diameter, and 18 ft. long. What total load could you place on it if the factor of safety is 10, and the thickness of metal If ins. ? The constant for the Gordon formula is ^u- (B.Sc. Lond.) Ans. 163 tons. 2. A mild steel strut, rectangular in cross section, the breadth being 680 EXERCISES four times its thickness, is 9 ft. long, and has pin ends. Determine the cross section for 24 tons, and a factor of safety of 5. Use Rankine's formula, and take /,, = 67,000 lbs. per sq. in., and the constant ^^j^-^. (B.Sc. Lond.) 3. Which would carry the heavier load for jBxed ends : {a) a solid mild- steel column 9 ins. diam. ; (6) a built-up mild-steel stanchion consisting of two 14x61 beams, at 8| ins. centres, with two 16 x |- in. plates each side ? Length in each case 14 ft. A7is. The built-up one. 4. Discuss the formula of Gordon and Rankine in connection with the buckling of struts of moderate lengths, and state its limiting conditions. Four wrought-iron struts, rigidly held at the ends, all of section 1 in. X 1 in., and of lengths 15'0, 30'0, 60*0, and 90'0 ins. respectively, are found to buckle under loads of 15'9, 11*3, 7*7, and 4'35 tons. Test whether these satisfy the formula quoted, and, if so, find average values of the two empirical constants involved. (B.Sc. Lond.) 5. A stanchion for a workshop has to carry a small stanchion 10 ft. long from the roof which carries 5 tons, and also the girder for a 15-ton crane. If the centre line of the roof load and crane girder are 13 ins. apart, design a suitable section for the stanchion. Ans. Two 10 ins. X 5 ins. X 30 I beams 13 ins. apart. 6. A hollow cylindrical steel strut has to be designed for the following conditions. Length 6 ft., axial load 12 tons, ratio of internal to external diameter == '8, factor of safety, 10. Determine the necessary external diameter of the strut and thickness of the metal if the ends are securely fixed in. Use Rankine's formula, taking/ = 21 tons per sq. in., constant for rounded ends = y-V^ Ans. 4^ ins. diam. ; | in. thick. 7. A steel column is built up of two 10 X 3| ins. X 28'21 lbs. channel sections placed 4| ins. apart, and two 12 X I in. plates at each end. If the ends are pin-jointed, what would you consider a safe load on a length of 22 ft. ? Ans. 122 tons. 8. What would be the safe load on the column of Question 7 if the load were 3 ins. out of centre ? Ans. 48*8 tons. 9. Find what thickness a hollow circular cast-iron column should have for an axial load of 60 tons, the factor of safety being 8. The column is 20 ft. long and is securely fixed at each end. Ans. 2 ins. CHAPTER XI 1. A shaft 3 ins. in diameter, running at 250 revolutions per minute, transmits 50 H.P. Find the maximum stress and the twist of the shaft in degrees in a length of 100 ft. Take G = 12 X 10« lbs. per sq. in. Ans. 2,380 lbs. per sq. in. ; 28*5^ EXERCISES 581 2. A turbine is connected to a dynamo placed vertically above it by a shaft, 2 ft. in diameter, made of steel plate | in. thick. Calculate the diameter of solid shaft required to transmit the same power at the same speed with the same maximum stress due to twist. Find the relative weights. (A.M.I.C.E.) Ans. ISSQ i7is. diameter ; -304:1. 3. If a rod J in. in diameter and 20 ins. long is fixed at one end and the other end is twisted through an angle of 15° relatively to it, what is the unital strain in the outer fibres of the rod ? Ans. '00164. 4. A bar of iron, ^ in. diameter, is twisted to destruction. Calculate what twisting movement is required for this purpose assuming that the shearing stress becomes uniform over the whole section and equals in the limit 19 tons per sq. in. (A.M.I.C.E.) Ans. '622 in. tons. 5. Through what angle will a 2| inch steel shaft be twisted if it is 80 ft. long and the twisting movement is 19,000 in. lbs. ? Ans. 23° nearly, 6. If the end of a rod 1 in. in diameter is twisted by the turning effort of a force of 80 lbs. acting at the end of a 12 in. lever, find the force which, when applied to the end of the same lever, would twist equally a rod of the same material, but of 1| ins. diameter and half the length. Ans. 810 lbs. 7. A bar of mild steel 1 in. in diameter twists through an angle of 2*2 degrees in a 20 in. length when subjected to a torque of 2,200 in. lbs. An exactly similar bar of the same material deflects "03 in. when loaded at the centre of a 20 in. simply-supported span with a load of 264 lbs. Calculate the value of Young's Modulus, Rigidity Modulus, Bulk Modulus and Poisson's Ratio. (B.Sc. Lond.) Ans. E = 29*88, G = 11*67, K = 32*62 million 'pounds per sq. in. n = *28. 8. A shaft which runs at 135 revolutions per minute transmits 50 H.P. and is subjected to a bending moment equal to *75 of the twisting moment. What diameter of shaft is necessary on the principal stress theory ? Ans. 2|- ins. CHAPTER XII 1. A steel wire |- in. in diameter is coiled into a spiral spring 5 ins. mean diameter. What weight could such a spring carry to produce a maximum stress in the wire of 5 tons per sq. in. ? (A.M.I.C.E.) Ans. 5*45 tons. 2. Find the pull required to cause a deflection of 1 in. in a closely- wound helical spring of 2*5 in. mean diameter made of 120 turns of J inch round wire, taking G = 12 X 10^ lbs. per sq. in. Ans. 3-| lbs. 3. A helical' spring of 3 ins. diameter is composed of 20 turns of steel wire '258 in. diameter. If a load of 25 lbs. is hung on it what will the deflection and maximum stress ? Ans. 3*18 ins. ; 5,560 lbs. per sq. in. 582 EXERCISES 4. A laminated spring composed of 20 plates each f in. thick and 2*95 ins. wide has a span of 3 ft. Find the deflection under a load of 5 tons if E is 12,000 tons per sq. in. Ans. 2' 37 ins. 5. How many plates f inch thick and 3 ins. broad, the largest being 30 ins. long, are required in a leaf spring whose maximum stress is to be 30,000 lbs. per sq. in. with a load of 1 ton ? Ans. 8. 6. A steel clock spring J in. wide and -^ in. thick is wound on a spindle fff in. in diameter. If the safe stress is 48,000 lbs. per sq. in. what is the maximum moment available for driving the clock ? Ans. 2 i/i. lbs. approx. CHAPTER XV 1. A girder 100 ft. long is supported at each end and in the middle, and carries a uniform load of 2 tons per ft. run. Draw the B.M. and shear diagrams, and find the pressure on each support. (A.M.I.C.E.) Alls. Max. B.M. 625 ft. tons.; reaction 37*5; 125; 37'5 tons. 2. A continuous girder consists of four spans, the two outer spans are each 20 ft. long, and the two inner spans are each 4(> ft. long; the girder carries a uniformly distributed load of H tons per ft. run. Find (a) The reactions at each of the piers; (b) The bending moment and shear at each of the piers ; (c) The position of the points of zero bending moment. Sketch complete bending moment and shear diagrams for the girder. (B.Sc. Lond.) 3. A balk of timber, 30 ft. long, rests on two end supports, and is supported also by a prop which acts at a point 12 ft. from the left-hand end. If the balk of timber carries (including its own weight) a load of 2 cwt. per ft. run, and if the tops of the three supports are level, determine the reactions at the three supports, and the bending moment at the point at which the prop is applied. Draw complete l^ending moment and shear diagram. (B.Sc. Lond.) 4. A horizontal girder of uniform section 25 ft. lonL' is firmly fixed at one end, and supported by a column at 18 ft, from the fixed end. The girder carries a uniform load of 2 tons per ft. run of its length, and, in addition, a concentrated load of 30 tons at 14 ft. from the fixed end. When unloaded, the girder just touches, but does not exert any pressure on the supporting colimin. Find the pressure on the column, and draw bending moment and shearing force diagrams for the girder. (B.Sc. Lond.) 5. A beam of 20 ft, span is built in at one end a, and is freely supported at other end b. It carries a imif orm load of h ton per ft. run, and a central isolated load of 10 tons. Draw the bending moment diagram, first finding the bending at end a, and show where the maximum intermediate bending moment occurs. Draw also the shear diagram. EXERCISES 583 6. A continued girder of 2 spans, 20 ft. and 10 ft., has an overhang of 5 ft. from the smaller span. It carries a uniformly distributed load of ^ ton per ft. run, and an isolated load of 1| tons at the free end (d). Find the support moments, and draw the shear and B.M. diagrams. Determine whether this arrangement is stronger than that in which the support c comes below the point d. Arts. Max. B.M. 16*77 ft. tons ; not so strong. 7. A beam of span I is fixed horizontally at both ends. Two equal loads W are placed at equal distance h from the ends of the beam. Prove that the greatest deflection of the beam is equal to — -— - (3 Z — 4 A), and ^ ^ 24 E I ^ ^ that the bending moment at the centre of the beam is equal to — - — V (B.Sc. Lond.) 8. A beam of 20 ft. span is fixed at the end and carries a uniformly- distributed load of 1 ton per ft. run from one abutment to the centre. Find the end B.M.s. Ans. 10-4, 22*9 ft. tons. 9. In a continuous beam of three spans, the centre span is 72 ft. and the end spans 36 ft. each. A dead load of ^ ton per ft. run covers the whole span. Determine the support moments when a live load of 1 ton per ft. run covers (a) the first span ; (6) the first two spans; (c) the whole beam. Ans. (a) 243, 162; (6) 567, 486; (c) 547 ft. tons. CHAPTER XVI 1. A C.I. beam has the following section: top flange, 4 x 1-| ins.; web, 12 X If ins. ; bottom flange, 12x2 ins. The centroid of the section is 5*5 ins. from the base of the bottom flange, and the moment of inertia of the section about a line through its centroid, at right angles to the depth, is 1200 ins.*. Draw a curve showing the intensity of shear at all points of the section, and find the ratio of maximum to mean shear stress. What proportion of shearing force is carried by the web ? (B.Sc. Lond. ) 2. Find the greatest intensity of shear stress at a section of an I beam at which the total shear is 15 tons; the overall depth is 8 ins. ; flanges, 6 ins. X -61 in.; web, -44 in. thick; I = 111-6 in.^. (B.Sc. Lond.) 3. Find the ratio of the maximum to the mean shear stress on the section of a cast-iron beam of the following dimensions : Top flange, 2 X ^ ins.; bottom flange, 6x1^ ins.; web, 7x1 ins. Aiis. 2'46. 4. A beam of uniform rectangular section, 6 ins. broad by 12 ins. deep, is supported at the ends, and has a span of 12 feet. It carries a uniformly distributed load of 20 tons. At a point in the cross section, 4 feet from 584 EXERCISES one end and 3 ins. vertically above the neutral axis, calculate the maximum intensity of compressive stress. Ans. I'll tons per sq. in. 5. An I beam, 20 ins. deep, has flanges l^ ins. wide and 1 in. thick, and a web If in. thick. Tlie greatest moment of inertia is 1,650 in. units and the total shear over the section is 80 tons. Show by a diagram the intensity of shear stress at all points of the section. (A.M.I.C.E.) Ans. Stress at N.A. — 8'44 tons per sq. in. 6. Allowing a bending stress of 1,500 lbs. per sq. in. and a shearing stress with the grain of 120 lbs. per sq. in., what uniform load can be carried by a timber beam 12 ins. deep, 3 ins. wide and of 12 ft. span ? Ans. 5,760 lbs. 7. A plate girder 4 ft. deep over the 3|ins. X 3|ins. X |in. angles has at each flange one plate 16 ins. x ^ in., the web being | in. thick. Find the distribution of shear stress on a section at which the shearing force is 44 tons. Ans. Stress at N.A. = I'Ql ; at junction of angle and web = 1*25; just above = "481; at bottom of flange of angle = "348; just above = '014; junction of angle and plate = '049 ton per sq. in. CHAPTER XVII 1. Show on the Bach Theory that the maximum central concentrated load that can be carried by a circular plate of given thickness is independent of the radius of the plate. 2. What must be the thickness of a mild steel plate covering an opening 4 ft. square if the load is 200 lbs. per sq. ft. and the safe stress is 16,000 lbs. per sq. in. ? Ans. '70 on the Bach Theory. 3. Prove that on the Bach Theory the uniform load that a square plate of given thickness can carry is independent of the size of the plate. 4. A rectangular reinforced concrete slab 10 ft. by 15 ft. has to carry a load of 300 lbs. per sq. ft. including its own weight. For what bending moment would you design the reinforcement in each direction ? Ans. 560,000 and 162,000 in. lbs. [Ranhine). 5. A cylinder end is 12 ins. in diameter and | in. thick. Compare the maximum stresses on the Bach and Grashof theories if the steam pressure is 100 lbs. per sq. in., the end being taken as freely supported. Ans. 6,400 lbs. per sq. in. {Bach), 7,800 lbs. per sq. in. {Grashof). CHAPTER XVIII 1. A solid steel gun has an inside diameter of 7'5 ins. and a thickness of r75 ins. What is the greatest tensile stress carried by an explosion pressure of 10,000 lbs. per sq. in. ? Ans. 27,300 lbs. per sq. in. EXERCISES 585 2. What should be the external diameter of a gun whose internal diameter is 3|- ins., if the explosion causes a pressure of 15,000 lbs. per sq. in. and the allowable stress is 30,000 lbs. per sq. in. ? Ans. 5'63 ins. 3. A solid gun has an external diameter of 12 ins. and an internal diameter of 6 ins. What inside pressure ^^dll cause a hoop tension of 30,000 lbs. per sq. in. ? Ans. 18,000 lbs. per sq. in. 4. Find the safe internal pressure for an hydraulic press cylinder of external diameter 7 ins. and internal diameter 5 ins., the maximum safe stress being 3,000 lbs. per sq. in. Ans. 973 Ihs. per sq. in. 5. An hydraulic press has an external diameter of 16 ins. and an internal diameter of 8 ins. If the pressure is 3 tons per sq. in., find the principal stresses at the external and internal circumference. Ans. External 0, 2 {tens.); internal 3 {comp.) 5 {tens.) tons per sq. in. 6. Find the equivalent tensile hoop stresses in the problem of the above question if Poisson's ratio is — Ans. 5*86; 2 tons per sq. in. 7. Find the necessary thickness of a pipe of 8 ins. internal diameter subjected to an internal pressure of 520 lbs. per sq. in. Adopt the maximum strain theory taking tj = ^ and maximum tensile stress 10,000 lbs. per sq. in. Ans. '22 in. 8. A tube whose internal and external radii are 2 and 3 ins. is hooped so as to cause an initial hoop compression on the inside of 18,000 lbs. per sq. in. What will be the tensile stress at the inside if the explosion causes a pressure of 25,000 lbs. per sq. in. ? Ans. 29,000 lbs. per sq. in. CHAPTERS XIX AND XX 1. Show that for a curved beam of rectangular section of depth d the deviation of the neutral axis from the centre is given approximately by , R being the radius of curvature of the centre line. 12 sx 2. A rectangular bar 1 in. wide and 2 ins. deep is bent to a radius of 4 ins. and is used as a hook. Find the maximum stress caused by a load of 1,000 lbs. acting through the centre of curvature of the bar. Ans. 7,700 lbs. per sq. in. 3. Find the maximum bending moments upon a chain link made of f in. circular stock, the link being 5 ins. deep and 3 ins. broad. Treat the ends as circular and the sides as straight. The load carried is 3,000 lbs. Ans. 1,.300 in. lbs. ; 387 in. lbs. 4. Find the stress due to centrifugal force in the rim of a cast-iron flywheel 8 ft. in diameter running at 160 revolutions per minute. Ans. 437 lbs. per sq. in. 586 EXERCISES 5. A steel shaft is of 13*4 in. diameter and is 98 ft. long. Find its critical speed when unloaded. Ans. 46 revolutions per minute. 6. Find the angular velocity at which whirling will start in an unloaded steel shaft 3 ins. in diameter and 11 ft. long. Ans. 75 radians per second. 7. Find the whirling speed of a shaft carrying a central load of 1,170 lbs. between swivelled bearings 7 ft. apart. The shaft is 3-J- ins. in diameter. Ans. 725 revolutions per minute. APPENDIX. The following Tables give the properties of the British Standard Sections which are usually listed by makers. -t Pi k \ Y r v t \ f\ Y^ V I T \ W:^ J X — 5-1 \ I r4-- -X ::i c J -1 ^ . Fig. A. — Properties of British Standard Sections, 5s: AppeJidix. BRITISH STANDARD SECTIONS.* (See Fig. A.) Properties of British Standard I Beams. Size \Vt. per 1 foot r Thickness 1 1 1 II i Moments of Inertia i Section Moduli Radii of (jyration H. B. 1 I 1 1 I / ins. T X XX vv j XX YY 1 XX i 1 "1 ins. YY inches. lb. ins. 1 I sq. ins. ^ ins. ins. ins. ins. ins. 3.x i| ■ 4 •16 •248 i-i8 1-66 •124 I'll •165 I-I9 •325 3x3, 8| •20 •332 2-50 379 1-26 2-53 •841 1-23 710 4 X i| 5 •17 •240 1-47 3-67 •194 1-84 •222 1-58 •363 4x3; 1 9* •22 •336 2 -80 7-53 1-28 376 •854 1*64 •677 4|x if| 6| •18 ..23 1-91 ' 677 •263 2-85 •300 1-88 •371 5 X 3 1 1 •22 •376 3-24 1 13-6 I "46 5*45 •974 2-05 •672 5 X 4i 18 •29 •448 5-29 227 5-66 9-08 2-51 2-07 1-03 6x3 12 •26 •348 3-53 20*2 I '34 674 •892 2*40 •616 6 X 4| 20 'Zl •431 5*88 347 5*41 11-6 2*40 2-43 •959 6x5 25 •41 •520 7-35 43*6 9-II 14-5 3*64 2*44 rii 7x4 16 '^-':> •387 471 39-2 3*41 I I -2 171 2-89 •851 8x4 18 •28 •402 5-30 557 3'57 13-9 179 3*24 •821 8x5 28 •35 •575 8-24 89-4 10-3 22'3 4-IO 3-29 1*12 8x6 35 •44 •597 10-3 II I 17-9 27-6 5-98 3-28 1-32 9x4 21 •30 •460 6-i8 8i-i 4'20 i8-o 2*IO 3-62 •824 9x7 58 •55 •924 17T 230 46-3 51-1 13-2 3-67 1-65 10 X 5 30 •36 •552 8-82 146 978 29'I 3-91 4 "06 1-05 10 X 6 42 •40 •736 12-4 212 22*9 42-3 7-64 4-14 1-36 10 X 8 70 •60 •970 20'6 345 71-6 69*0 17-9 4-09 1-87 12x5 32 •35 •550 9-41 220 974 367 3-90 4-84 I -02 12x6 44 •40 •717 12*9 315 22 '3 52-6 7 "42 4*94 1-31 12x6 54 •50 •883 15-9 376 28-3 62-6 9 -43* 4-86 1-33 14x6 46 •40 •698 13*5 441 21 "6 62-9 7 -20 571 1-26 14x6 57 •50 •873 i6-8 553 " 27-9 76-2 9"3i 5-64 1-29 15 X 5 42 •42 •647 12-4 428 II -o 57-1 478 5-89 •983 15x6 59 •50 •880 17-3 629 28-2 83-9 9-40 6 '02 1-28 16 X 6 62 *55 •847 1 8-2 726 27-1 907 9*02 6-31 1-22 18 X 7 75 1 '55 •928 22*1 1 1 50 46-6 128 ^ys 7-22 I '45 20 X 7I 89 ' -Go i-oi 26*2 1671 62-6 167 167 7 '99 1-55 24 X r\ 100 1 -60 1 lro7 29-4 2655 66-9 221 17-8 9-50 1-51 • Published by permission of the Engineeniig Standards Committee. The Tables of British Standard I Beams, Channels, and Zed Bars are reprinted from Report No. 6 as issued by the Committee. Additional calculations have been inserted in the Tables of British Standard Unequal Angles, Equal Angles, and Tee Bars for thicknesses other than those calculated by the Coir'»iittee, such calculations having been taken by permission from the Pocket Companion issued by Messrs. Dorman, Long & Co., Ltd. 588 Appendix, Properties of British Standard Channels. Size Standard Thicknesses Weight per foot Area c H 5 Moments of Inertia Section Moduli Radii of Gyration A X B t T 1 About XX About YY About XX About YY About XX ins. About YY ins. ins. ins. lbs. sq. ins. ins. ins. ins. ins. ins. ins. 15x4 •525 •630 41-94 12-334 •935 377-0 14-55 50-27 4-748 5-53 1-09 12x4 •525 •625 36-47 10727 I -03 1 218-2 13-65 36-36 4*599 4-51 I-13 I2X3i •500 •600 32-88 9-671 -867 190-7 8-922 31-79 3-389 4*44 -960 12x3^ •375 •500 26'IO 7-675 -860 158-6 7-572 26-44 2-868 4*55 ■993 11x3-2 •475 •575 29-82 8-771 •896 148-6 8-421 27-02 3-234 4-12 -980 10x4 •475 •575 30-16 8-871 I-I02 1307 12-02 26-14 4*147 3*84 I-16 10 X il •475 •575 28-21 8-296 ■933 117-9 8-194 23-59 3-192 3'77 *994 10x3^ •375 •500 23-55 6-925 •933 102-6 7-187 20-52 2 -800 3*85 I -02 9x3! •450 •550 25*39 7-469 -971 88-07 7-660 19-57 3-029 3*43 I -01 9x35 •375 •500 22-27 6-550 •976 79-90 6-963 17-76 2-759 3*49 1-03 9x3 •375 •437 19-37 5-696 •754 65-18 4-021 14-48 1-790 3*38 -840 8x3* •425 •525 22-72 6-682 I -Oil 63-76 7-067 15-94 2-839 3-09 1-03 8x3 •375 •500 I9-3C 5 '67 5 •844 53*43 4-329 13-36 2-008 3*07 'S73 7x3! •400 •500 20-23 5 "950 I -061 44-55 6-498 12-73 2-664 274 1*04 IX-:, •375 •475 17-56 5-166 -874 37-63 4-017 10-75 1-889 2-70 -882 6x3i •375 •475 17-9 5-266 1-119 29-66 5-907 9-885 2-481 2-36 1-06 6x3 •312 •437 14-49 1 4-261 -938 24-01 3-503 8-003 1-699 2-37 -907 Properties of British Standard Zed Bars. Size Standard Thicknesses Area Weight per foot Moments of Inertia Section Moduli c ■~ in c -a < 1 "2 -5.2 4-1 >^ A X B / T About XX About YY About XX About YY ins. ins. ins. sq. ins. lbs. ins. ins. ins. ins. ins. 10 X 3^ *475 •575 8-283 28-16 117-865 12-876 23-573 3-947 14 -839 9x3! -450 -550 7-449 25-33 87-889 12-418 19-531 3-792 i6i *843 8x31 -425 -525 6-670 22-68 63729 12*024 15-932 3-657 i9i -845 7x31 •400 -500 5-948 20-22 44-609 ii-6i8 12-745 3-521 23 -840 6X3^ •375 ;475 5-258 17-88 29-660 11-134 9-887 3-361 28i •821 5x3 •350 ^-450 4-169 14-17 16-145 6-578 6-458 2-328 29i -698 589 Appendix. Properties of British Standard Unequal Angles. Size and Weight Dimensions Moments of Inertia Section Moduli - Thickness Area per 01 u. >> foot J P About About About About 3-72 25 •96 6| X 3i X 3 8 3-610 12-27 2-22 -741 15-7 3-27 y(^i I-18 16^ '75 ii 11 J. 2 4-750 16-15 2-28 •792 20-4 4-20 4-83 1-55 16^ •75 j» 'J 5 s 5-860 19-92 2*33 -841 24-83 5-06 5-95 1-90 16 •74 6 X 4 X 3 s 3-610 12-27 I-9I •923 13-2 4-73 3-23 1-54 23i -87 11 j» 1 2 4-750 16-15 1-96 •974 17-1 6-10 4-23 2-02 23i •86 11 ii 5 5-860 19-92 2 -02 I -02 20-8 7-36 5-23 2-47 23I •86 6 X 3| X 3, 3-424 11-64 2-01 ll-':> 12-6 3-22 3-16 1-18 19 •76 " n 1 4-502 15-31 2-06 •823 16-4 4-14 4-16 1-55 19 •75 1' ?j 5 8 5-549 18-87 2-II •872 19-88 4-97 5-II 1-89 18^ •75 5! >' 3i X 3 S 3-236 1 I -00 1-80 -807 9-93 3-15 2-68 1-17 22 •76 M -5 \ 4-252 14-46 1-85 -857 12-80 4-05 3-51 i'53 22 •75 ;5 11 5 8 5-236 17-80 1-90 -905 15-6 4-86 4 jj 1-87 21^ •75 5i X 3 X 3. 3-050 10-37 1-90 -662 9-45 2-02 2-62 •86 17 -64 J? ?> 1 2 4-003 13-61 1-95 -711 12-2 2-58 3 "44 1-13 16^ -64 ?? V 5 8 4-925 16-74 2-00 -759 14-7 3-08 4-20 XV \b\ •63 5 X 4 X 3 8 3-236 II -oo 1-51 i-oi 7-96 4-53 2-28 1-52 32 -85 » 5? 4 4-252 14-46 1-56 1-06 10-3 5-82 2-99 1-98 32 •84 « 11 5 8 5-236 17-80 I -60 i-ii 12-4 7-OI 3-66 2-43 32 -83 5 X 3^ X 3 8 3-050 10-37 1-59 -848 7-64 3-09 2-24 I-I7 25^ •75 » 5> 1 2 4-003 13-61 1-64 •897 9-86 3-96 2-93 1-52 25i •75 )5 M 5 8 4-925 16-74 1-69 -944 II-9 4-75 3-60 1-86 25 •74 5 X 3 X Vb 2-402 8-17 1-66 •667 6-14 1-68 1-84 •72 20 •65 5) 5> 3 8 2-859 ! 9-72 1-68 -693 7-24 1-97 2-i8 -85 i9i •65 )) V 1 2 3"749 12-75 ■742 9*33 2-51 2-85 i-ii 19^ •64 J: 35 5 8 4-609 15-67 1 1-78 •789 11-25 3-00 3-49 1-36 19 -64 590 Appendix, Unequal Angles {continued). Size and Thickness 4^ X ins. 1>\ X 4 X J 2 4 X X -\ 13 32 X Z\ X 2; 8 \ 2 5. 8 X 16 X 2.^ X X 2 Area I I {Weight i per foot sq. in. 2 "402 2-S59 3749 4*609 2'246 2'67i 3 '499 4-295 2*091 2-485 3-251 3-985 I '934 2*298 3-001 3*673 1-799 2*111 2-752 1-312 1*921 2^ X 2 X )» 2 2-499 X I 1*187 ?) 3 6 ^'733 55 1 2 2*249 2 X 1 4 1*063 55 I'e 1*309 V s 1*547 li X 1% -622 51 1 4 •814 55 I'fi •997 lb. 8-17 9-72 12*75 15*67 7*64 9*o8 11*90 14*61 7*11 8*45 11*05 13-55 6-58 7-8i 10-20 12*49 6-05 7-18 9-36 4-46 O DO 8*50 4*04 5-89 7-65 3-6i 4-45 5*26 2*11 2*77 3-39 Dimensions ins. -36 -39 -44 -48 *i6 ins. *866 *89i -940 •987 -915 Moments of Inertia About X X 1*19 •941 1*24 •990 1-28 1*04 1*24 •746 1*27 *77i 1*31 *8i9 1*36 *865 1*04 *792 1*07 ■*8i9 I*II -867 1*16 -912 1*12 *627 1*15 *652 1-20 -699 •895 *648 -945 -6,7 •992 -744 -976 -482 1*03, •532 1*07 *578 '774 -527 •799 •552 •823 •575 •627 •381 -653 •407 *678 -431 ins. 4*22 5-69 7*31 8*8r 3-46 4*08 5-23 6-28 3*31 3-89 4-98 5-96 2*27 2*67 3-40 4*05 2*15 2*52 3-20 1*14 I -62 2-05 I -06 1*50 1-89 ■636 -770 •895 -240 •308 •369 About Y Y ms. 2*55 3*00 3-84 4*61 2*47 2*90 371 4-44 1*59 1-87 2*37 2*83 1-53 1*80 2*28 2*71 •910 I -06 1-34 *7i6 1*02 1-28 '373 *525 *656 •359 •433 •502 *ii5 '1 46 -174 Section Moduli About X X ins. 1-54 1-83 2-39 2-92 1*22 1-45 1-89 2*31 1*20 1-42 1-85 2-26 *92 1*10 1*42 i-73 -90 I -07 ■-39 -54 -79 1*02 •52 •76 -98 '37 *45 -53 •17 •23 About Y Y *97 1-15 1*5 1*83 -96 1*13 1*48 1-80 •71 •84 I -09 1*33 •69 -83 1*07 1-30 -49 •57 •74 *39 -57 '73 -25 -36 -46 •24 •30 •35 •10 *i3 •16 c •— {/I o f 302 3oh 30 30 37 37 37 3H 28^ 28^ 28^ 28 35i 3Sh 35i 35 26| 26 26 34 34 33h 23* 23 22.^ 32 3ii 31I 28^ 28 28 '-5.2 ft: 2 »^ QJ <^ J ° ins. *74 -74 *74 •74 *72 *72 •71 •71 -64 -64 *63 ■63 *62 *62 •61 *6i •54 *53 *53 •52 -52 •52 *43 *42 -42 •42 •42 •42 •32 591 Appendix. British Standard Equal Angles. Sizes Area Weight per foot T J 1 ! 1 Section Modulus about X X Least Radius of Gyration ins. sq. ins. lb. ins. ins. 1 ins. I ins. 8 X 8 X i 775 26-35 2-15 47'4 8-IO 1-58 8 X 8 X 5. 9-61 32-67 2-20 1 58-2 10-03 ^'11 8 X 8 X 3 4 11-44 38-89 2-25 68-5 11-91 1-56 6 X 6 X I'S" 5-06 17-21 1-64 17*3 3*97 I-18 6 X 6 X 5 S 7-II 24-18 1-71 23-8 5-55 I-18 6 X 6 X 3 4 8-44 28-70 1-76 27-8 6-56 I-I7 5 X 5 X 3 S 3-6i 12*27 1-37 8-51 2-24 '98 5 X 5 X 1 475 16-15 1-42 I i-o 3-07 •98 5 X 5 X 5 S 5-86 19-92 1-47 13-4 3-80 -98 4ix 4ix 3 S' 3'24 I I -oo 1 -22 6-14 1-87 -88 4|x 45 X 1 2 4-25 14-46 1-29 7-92 2-47 -87 45 X 45 X 5. S 5-24 17-80 I '34 9-56 3-03 '^1 4 X 4 X 3 S 2-86 972 1-12 4-26 1-48 78 4 X 4 X 1 375 12-75 I-I7 5-46 1-93 : '17 4 X 4 X 5 S 4-6i 15-67 I -22 6-56 2-36 •77 35 X 3^x "t'« 2*09 7-II •97 2-39 '95 -68 3ix 3ix 3 & 2-48 8-45 I "OO 2-80 1*12 •68 35 X 3^x 1 2 3-25 11-05 1-05 3'57 1-46 : -68 35 X 3l X 5 S 3-98 i3'55 1-09 4-27 177 -68 3 X 3 X 1 4 1-44 4-90 •827 I-2I •56 •59 3 X 3 X 3 8 2'II 7-18 •877 1-72 -81 •58 3 X 3 X 1 2 275 9-36 -924 2-19 1-05 •58 3 X 3 X 5 3-36 11-43 •970 2-59 1-28 -58 l\ X 2^ ><: 1 4- I-I9 4-04 703 ■677 •38 , •48 2^ X l\ X t''« 1-46 4-98 •728 -822 •46 , •48 2^X 2^X 3 8 173 5-89 752 -962 '55 •48 2| X 2^ X 1 2-25 ■7-65 799 I-2I -71 •48 2^X 2\ X 16 •809 275 •616 •378 •23 •44 2f X 2} X 1 4 I '06 3-6i •643 •489 •30 •44 2^X 2|X IB i'3i 4-45 -668 •592 •^:>7 •43 2r X 2|X 3 S i'55 5-26 •692 •686 •44 •43 592 Appendix. British Standard Equal Angles (continued). Sizes Area WViifht per foot J I.XX Section Modulus about X X Least Radius of Gyration ins. sq. ins. lb. in. in. in. in. 2 X 2 X i« 715 2-43 •554 •260 -18 •39 2 X 2 X 1 4 -938 3*19 •581 •336 •24 •39 2 X 2 X tV I-I5 3-92 •605 •401 •29 •38 2 X 2 X 3 s 1-36 4-62 •629 •467 •34 •38 ifx i|x .'5 .1« •622 2-II •495 •172 •U •34 ijx i^x 1 4 -814 2-77 •520 •220 •]8 •34 i|x if X 5 16 •997 3*39 •544 •264 •22 •34 I^X I^ X IG •526 179 •434 •105 •10 •29 [*X I^X 1 4 •686 2-33 -458 •134 •13 •29 i|x 4x "16 •839 2-85 -482 •159 •16 -29 I4-X IjX 16 •433 1-47 •371 •058 •07 •24 I^X \\ X 1 4 •561 1-91 •396 •073 -C9 •23 British Standard Tees. SizfS Area Weight per foot J Moments of Inertia Section Moduli Radii of Gyration XX YY ins. XX YY XX YY ins. sq ins. lb. ins. ins. ins. ins. ins. ins. 6 X 4 X 3 s 3-634 12-36 -915 4-70 6*34 1-52 2-II I-I4 1-32 6 X 4 X 1 2 4-771 l6-22 -968 6-07 8-62 2-00 2-87 I-13 1-34 6 X 4 X 5 s 5-878 19-99 I -02 7*35 10-91 2-47 3-64 I - 1 2 1-36 6 X 3 X 8 3-260 II -08 •633 2 -06 6*39 -87 2-13 -795 1-40 6 X 3 X \ 4-272 14-53 •684 2-63 8-65 I-14 2-88 -785 1-42 6 X 3 X 5 S 5-256 17-87 •732 3*14 10-94 1-39 3-65 'ir:> 1-44 5 X 4 X 8 3-257 11-07 -998 4'47 3-69 1-49 1-48 I-I7 I -06 5 X 4 X 1 4-268 14-51 1-05 5*77 5-02 1-96 2-01 I-I6 1-08 5 X 3 X 3 8 2-875 9-78 •691 1-97 3*71 -85 1-49 -828 1-14 5 X 3 X 1 2 3762 12-79 -741 2-52 5-03 i-ii 2-01 -818 1-16 4 X 4 X 3 8 2-872 9-77 l-II 4-19 1-90 1-45 •95 I-2I -814 4 X 4 X 1 2 3758 12-78 i-i6 5-40 2-59 1-90 1-29 I -20 •830 4 X 3 X 5. 2-498 8-49 •767 1-86 1-91 •83 -96 -863 -875 4 X 3 X h 3-262 11-08 -816 2-30 2-60 1-08 1-30 -851 -893 QQ 593 Appendix. British Standard Tees ( confiniied). Sizes 1 Area j Weight per 1 foot ] J Moments of Inertia Section Moduli Radii of Gyratioa 1 XX YY XX YY XX YY :ti3. sq. ins. lbs. ins. I'lS. 1 05. ins. ins. ins sns. 3i X 3| X f 2-496 8*49 -98 2*79 1*28 no 'IZ 1*05 •717 3? X 3^ X ^ 3-259 11*08 1*04 3-54 1-75 1-44 I'OO 1*04 •733 3 X3 xf 2*121 7*21 •868 1*70 *8i6 *8o -54 •897 •620 3 X3 x^ 2*760 9-38 *9i8 2*16 i*ii 1*04 •74 *886 *636 3 X2|xf 1*929 6*56 •695 i-oi •814 -56 •54 •725 ■650 3 X2^X^ 2*506 8*52 •742 1-28 1*12 'l?i •74 •713 •665 2^ X 2^ X ^ I-I97 4-07 •697 •677 *302 •38 •24 -752 *502 2^ X 2| X -T% 1-474 5*01 •724 •832 •387 •46 •31 •747 *5I2 2^ X 2|x f 1*742 5*92 •750 -959 •473 -55 •38 •742 •521 2ix 2^X^ 1*071 3-64 •638 *488 *224 -30 •20 ■675 •457 2^X2^X1 1-554 5-28 *689 *685 •349 •44 ■31 *664 •474 2 X 2 X J •947 3*22 •579 'lyi •157 •24 •16 -597 -407 2 X2 xf 1*367 4*64 •628 •469 •246 •34 ■25- •5S6 •424 1^X2 X J •820 2*79 *648 •307 •068 -23 *o9 •612 *288 I4X2 X-,% 1*003 3-41 •674 •369 *oS8 *28 •12 ■607 •296 ifx i|x^ *820 2*79 -519 *22I •107 •18 •12 •520 •361 i|xi|Xi% -999 3-40 •544 •265 •137 •22 *i6 ■515 •370 i^x iix^% -531 1*81 -435 *io6 *048 •ID *o6 •447 •301 I^X T^X^ *692 1 2-35 1 •460 •135 *o67 -.3 *o9 •442 •312 .^94 MATHEMATICAL TABLES 595 Angle Chord. Sine. 1 Tangent. Co-tangent. 1 Cosine. i De- grees Radians. 0^ X 1 1-414 1^5708 90' 1 •> 3 ■0175 •0349 •0524 •0698 •017 •035 •052 •070 •0175 •0349 •0523 •0698 •0175 •0349 ■0524 •0699 57-2900 28-6363 19-0811 14-3007 -9998 -9994 j -9986 -9976 1-402 1-389 1-377 1-364 1"5533 1-5359 1-5184 1-5010 89 88 87 86 5 •0873 •087 •0872 •0875 11-4301 i -9962 1-351 1-4835 85 6 7 8 9 •1047 •1222 •1396 •1571 •105 •122 •140 •157 •1045 •1219 •1392 •1564 •1051 •1228 •1405 •1584 9-5144 8-1443 7-1154 6-3138 -9945 , -9925 •9903 -9877 1-338 1-325 1-312 1-299 1-4661 1-4486 1-4312 1-4137 84 83 82 81 10 •1745 •174 •1736 •1763 5-6713 •9848 1-286 1-3963 80 11 12 13 li •1920 •2094 •2269 •2443 •192 •209 •226 •244 •1908 •2079 •2250 •2419 •1944 •2126 •2309 •2493 5-1446 4-7046 4-3315 4-0108 -9816 -9781 -9744 -9703 1-272 1-259 1-245 1-231 1-3788 1-3614 1-3439 1-3265 79 78 77 76 15 •2618 •261 •2588 •2679 3-7321 •9659 1-218 1-3090 75 16 17 18 19 •2793 •2967 •3142 •3316 •278 •296 •313 •330 •2756 •2924 •3090 •3256 •2867 •3057 •3249 •3443 3-4874 3-2709 3-0777 2^9042 -9613 -9563 -9511 •9455 1-204 1-190 1-176 1-161 1-2915 1-2741 1-2566 1-2392 74 73 72 71- 20 •3491 •347 •3420 •3640 2^7475 •9397 1-147 1-2217 70 21 22 23 24 •3665 •3840 •4014 •4189 •364 •382 •399 •416 •3584 •3746 •3907 •4067 •3839 •4040 •4245 •4452 2-6051 2-4751 2-3559 2-2460 1 •9336 -9272 -9205 •9135 1-133 1-118 1-104 1-089 1-2043 1-1868 1-1694 1-1519 69 68 67 66 25 •4363 •433 •4226 •4663 2-1445 •9063 1-075 1-1345 65 26 27 28 29 •4538 •4n2 •4887 •5061 •450 •467 •484 •501 •4384 •4540 •4695 •4848 •4877 •5095 •5317 •5543 2-0503 1-9626 1-8807 1-8040 •8988 •8910 •8829 •8746 1-060 1-045 1-030 1-015 1-1170 1-0996 1-0821 1-0647 64 63 62 61 30 •5236 •518 •5000 •5774 1-7321 •8660 1-000 1-0472 60 31 32 33 34 •5411 •5585 •5760 •5934 •534 •551 •568 •585 •5150 •5299 •5446 •5592 •6009 •6249 •6494 •6745 1-6643 1-6003 1-5399 1-4826 •8572 •8480 •8387 •8290 •985 •970 •954 •939 ' 1-0297 1-0123 -9948 -9774 59 58 57 56 35 •6109 •601 •5736 •7002 1-4281 •8192 •923 -9599 55 36 37 38 39 ■6283 •6458 •6632 •6807 •618 •635 •651 •668 •5878 •6018 •6157 •6293 •7265 •7536 •7813 •8098 1-3764 1-3270 1-2799 1-2349 •8090 •7986 •7880 •7771 •908 •892 •877 •861 -9425 •9250 •9076 -8901 54 53 52 51 40 •6981 •684 •6428 •8391 1-1918 •7660 •845 -8727 50 41 42 43 44 •7156 •7330 •7505 •7679 •700 •717 •733 •749 •6561 •6691 •6820 •6947 •8693 •9004 •9325 •9657 1-1504 1-1106 1-0724 1-0355 •7547 •7431 •7314 •7193 •829 •813 •797 •781 •8552 •8378 •8203 •8029 49 48 47 46 45- •7854 •765 •7071 1-0000 1-0000 •7071 •765 •7854 45 Cosine Co-tangent Tangent Sine Chord Radians Degrees Ang 'le 596 MATHEMATICAL TABLES Logarithms 1 2 3 4 5 6 7 8 9 12 3 4 5 6 7 8 9 10 11 12 13 14 UCKXJ 0043 0086 I 0128 0170 0212 0253 0294 0334 0374 4 9 13 17 4 8 12 16 21 20 26 30 34 38 24 28 32 37 0411 0453 0492 0531 0569 0607 0645 1 0682 0719 4 8 12 15 0755 4 7 11 15 19 19 23 27 31 35 22 26 30 33 0792 1139 0828 0864 0899 0934 0969 1 1004 1038 1072 1106 Is 7 11 14 1 3 7 10 14 18 17 21 25 28 32 20 24 27 31 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 7 10 13 3 7 10 12 16 16 20 23 26 30 19 22 25 29 14G1 1492 1523 1553 1584 1 1614 1644 1673 1703 1732 3 6 9 12 3 6 9 12 15 15 18 21 24 28 17 20 23 26 15 16 1761 1790 1818 1847 1875 1 1903 1931 1959 1987 2014 3 6 9 11 3 5 8 11 14 14 . . 17 20 23 26 16 19 22 25 •2011 2068 2095 2122 2148 1 2175 1 2201 2227 2253 2279 3 5 8 11 3 5 8 10 14 13 16 19 22 24 15 18 21 23 17 18 2301 2553 2330 2355 2380 2405 1 2430 1 2455 2480 2504 2529 3 5 8 10 2 5 7 10 13 12 15 18 20 23 15 17 19 22 2577 I 2601 2625 2648 f 1 2672 2695 2718 2742 2765 2 5 7 9 2 5 7 9 12 1 14 16 19 21 11 14 16 18 21 19 20 2788 2810 2833 2856 2878 1 1 2900 2923 2945 2967 2989 2 4 7 9 11 2 4 6 8 11 13 16 18 20 13 15 17 19 3010 3032 3054 3075 3096 1 3118 3139 3160 3181 3201 2 4 6 8 11 1 13 15 17 19 21 22 23 2i 3222 3243 3263 3424 3444 3464 3617 3636 ' 3655 3802 3820 3838 3284 3483 3674 3856 3304 J 3324 3502 3522 3692 I 3711 3874 1 3892 3345 3541 3729 3909 3365 3560 3747 392-7 3385 3579 3766 3945 3404 3598 3784 3962 2 4 6 8 , 10 12 14 16 18 2 4 6 8 ; 10 12 14 15 17 2 4 6 7 ! 9 ! 11 13 15 17 2 4 5 7 ' 9 i 11 12 14 16 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2 3 5 7 9 ! 10 12 14 15 26 27 28 29 30 1 31 S2 33 34 35 4150 4314 4472 4624 4166 4330 4487 4639 4183 4346 4502 4654 4200 4362 4518 4669 1 4216 4378 4533 4683 4232 4393 4548 4698 4249 4409 4564 4713 4265 4425 4579 4728 4281 4440 4594 4742 4298 4456 4609 4757 2 3 5 7 8 ! 10 1113 15 2 3 5 6 8 1 9 11 13 14 2 3 5 6 8 9 11 12 14 13 4 6 7 9 10 12 13 4771 1 4786 4800 4814 1 ^ 4829 1 4843 4857 4871 4886 4900 13 4 6 7 9 10 11 13 4914 5051 5185 5315 4928 j 4942 5065 5079 5198 1 5211 5328 5340 1 1 4955 5092 5224 5353 4969 4983 5105 5119 5237 5250 5366 5378 4997 5132 5263 5391 5011 5145 5276 5403 5024 5159 5289 5416 5038 5172 5302 5428 13 4 6 17 13 4 5 7 13 4 5 6 13 4 5 6 8 10 11 12 8 9 11 12 8 9 10 12 8 9 10 11 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 1 1 2 4 5 ' 6 7 9 10 11 36 55(33 37 5U8:3 38 5798 39 5911 5575 5587 5694 5705 5809 ; 5821 5922 j 5933 5599 5717 5832 5944 5611 5729 5843 5955 5623 5740 5855 5966 5635 5752 5866 5977 5647 5658 5763 5775 5877 5888 5988 : 5999 5670 5786 5899 6010 1 2 4 5; 6 12 3 5 6 12 3 5 6 12 3 4 5 7 8 10 11 7 8 910 7 8 910 7 8 9 10 40 1 41 42 43 44 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 12 3 4 5 6 8 9 10 6128 6232 6335 6435 1 6138 6149 6243 6253 6345 ' 6355 6444 6454 j 6160 6263 6365 6464 6170 6274 6375 6474 6180 6284 6385 6484 6191 6294 6395 6493 6201 6304 6405 ' 6503 6212 6222 6314 6325 6415 i 6425 6513 j 6522 i 12 3 4 12 3 4 12 3 4 12 3 4 5 5 5 5 6 7 8 9 G 7 8 9 6 7 8 9 6 7 8 9 45 6532 6542 1 6551 €561 6571 6580 6590 6599 1 6609 6618 12 3 4 5 I 1 6 7 8 9 46 47 48 49 6628 6721 6812 6902 6637 6730 6821 6911 6646 6656 6739 6749 6830 6839 6920 1 -6928 6665 6758 6848 6937 6675 6767 6857 6946 6684 6776 6866 6955 1 6693 6702 6785 6794 6875 1 6884 6964 6972 6712 6803 6893 6981 12 3 4 12 3 4 12 3 4 12 3 4 5 5 4 4 6 7 7 8 5 6 7 8 5 6 7 8 5 6 7 8 50 6990 6998 j 7007 7016 1 7024 7033 7042 7050 7059 7067 12 3 3 4 5 6 7 8 MATHEMATICAL TABLES 597 Logarithms 1 2 3 4 3 6 7 8 i 9 12 3 4 5 6 7 8 9 51 52 63 54 7076 7160 7243 7324 7084 7168 7251 7332 7093 7177 7259 7340 7101 7185 7267 7348 7110 7193 7275 7356 7118 7202 7284 7364 7126 7210 7292 7372 7135 7218 7300 7380 7143 7226 7308 7388 i 7152 7235 7316 7396 12 3 3 12 2 3 12 2 3 12 2 3 4 4 4 4 5 6 7 8 56 7 7 5 C 6 7 5 6 6 7 55 7-404 7412 7419 7427 7435 1 7443 7451 7459 7466 i 7474 12 2 3 4 5 5 6 7 56 57 58 59 7482 7559 7G34 7709 7490 7566 7642 7716 7497 7574 7649 7723 7505 7582 7657 7731 7513 7520 7589 7597 7664 7672 7738 7745 7528 7604 7679 7752 7536 7612 7686 7760 7543 7619 7694 7767 7551 7627 7701 7774 12 2 3 12 2 3 112 3 112 3 4 4 4 4 5 5 7 5 5 6 7 4 5 6 7 4 5 6 7 60 7782 7789 7796 7803 7810 1 7818 7825 7832 7839 7846 112 3 i 4 4 5 6 6 61 62 63 64 7853 7924 7993 8062 7860 7931 8000 8069 7868 7938 8007 8075 7875 7945 8014 8082 7882 j 7889 7952 j 7959 8021 1 8028 8089 j 8096 7896 7966 8035 8102 ' 7903 ' 7973 8041 i 8109 7910 7980 8048 ■ 8116 7917 7987 8055 8122 112 3 112 3 112 3 112 3 4 3 3 3 4 5 6 6 4 5 6 6 4 5 5 6 4 5 5 6 65 8129 8136 8142 8149 8156 j 8162 8169 8176 8182 8189 112 3 3 4 5 5 6 63 67 68 69 70 8195 8261 8325 8388 8202 8267 8331 8395 8209 8274 8338 8401 8215 8280 8344 8407 8222 8228 8287 8293 8351 8357 8414 8420 8235 8299 8363 8426 8241 8306 8370 8432 8248 8312 8376 8439 8254 8319 8382 8445 112 3 112 3 112 3 112 2 3 3 3 3 4 5 5 6 4 5 5 6 4 4 5 6 4 4 5 6 8451 8457 8463 8470 8476 1 8482 8488 1 8494 8500 J 8506 1 1 1 2 2 3 4 4 5 6 71 72 73 7i 8513 8573 8G33 8692 8519 8579 8639 8698 8525 8585 8645 8704 8531 8591 8651 8710 8537 8597 8657 8716 8543 8603 8663 8722 8549 8609 8669 8727 8555 8615 8675 8733 8561 8621 8681 8739 8567 8627 8686 8745 112 2 112 2 112 2 112 2 3 3 3 3 4 4 5 5 4 4 5 5 4 4 5 5 4 4 5 5 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 1 1 2 2 3 3 4 5 5 76 77 78 79 8808 8865 8921 8976 8814 8871 8927 8982 8820 8876 8932 8987 8825 8882 8938 8993 8831 8887 8943 8998 8837 8893 8949 9004 8842 8899 8954 9009 8848 8854 8904 8910 8960 8965 9015 9020 8859 8915 8971 9025 112 2 112 2 112 2 112 2 3 3 3 3 3 4 5 5 3 4 4 5 3 4 4 5 3 4 4 5 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 112 2 3 3 4 4 5 81 82 83 84 9085 9138 9191 9243 9090 9143 9196 9248 j 9096 9149 9201 9253 9101 9154 9206 9258 9106 9159 9212 9263 9112 9165 9217 9269 9117 9170 9222 9274 9122 9128 9175. 9180 9227 9232 9279 9284 9133 9186 9238 9289 112 2 112 2 112 2 112 2 3 3 3 3 4 4 5 3 4 4 5 3 4 4 5 3 4 4 5 85 9294 9299 i 9304 9309 9315 9320 9325 9330 9335 9340 112 2 3 1 3 4 4 5 1 86 87 88 89 9345 9395 9445 9494 9350 9400 9450 9499 9355 9405 9455 9504 9360 9410 9460 9509 9365 9415 9465 9513 9370 9420 9469 9518 9375 9425 9474 9523 9380 9430 9479 9528 9385 9435 9484 9533 9390 9440 9489 9538 112 2 112 112 112 3 2 2 2 3 4 4 5 3 3 4 4 3 S 4 4 3 3 4 4 90 9542 9547 9552 9557 '• 9562 9566 9571 9576 9581 9586 1 1 1 2 ; 2 3 3 4 4 91 92 93 91 9590 9638 9685 9731 9595 9643 9689 9736 9600 9647 9694 9741 9605 9652 9699 9745 9609 9657 9703 9750 9014 9661 9708 9754 9619 9666 9713 9759 9624 9671 9717 9763 9628 9675 9722 9768 9633 9680 9727 9773 112 112 112 112 1 2 3 3 4 4 2 3 3 4 4 2 3 3 4 4 2 3 3 4 4 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 112 2 3 3 4 4 96 97 98 99 9823 9868 9912 9956 9827 9872 9917 9961 9832 9877 9921 9965 9836 , 9881 9926 9969 9841 9886 9930 9974 9845 9890 9934 9978 9850 i 9894 9939 9983 9854 9899 9943 9987 9859 9903 9948 9991 9863 9908 9952 9996 112 112 112 112 2 2 2 2 3 3 4 4 3 3 4 4 3 3 4 4 3 3 3 4 598 MATHEMATICAL TABLES A>fTIL0GARITH3IS 12 3 4 •5 6 7 8 9 12 3 4 5 6 7 8 9 •00 vy»-> l i':'i2 1"14 1016 1019 1021 M n 1 1 1 1 ■. ■ ■- •01 1V23 •02 l':»47 •03 1072 •04 lt»'. 10-26 1(K8 1 1030 1033 1050 105-2 lO-M 1057 1074 1076 1079 lOSl 1099 ll'>2 ll'H 111' 7 1035 1059 1084 iit:'9. 1 1 1038 1 1(M0 1042 1045 1062 1064 1067 1069 lt)86 1089 1091 1094 1112 1114 1117 1119 11 11 11 111 112 2 2 112 2 2 112 2 2 1 2 2 2 2 ■05 1122 1125 1 U27 ll:5r. 1132 11:55 ll:>^ 1140 1143 1146 "111 1 2 2 2 2 •06 11 tS •07 1175 •08 la^yi •09 12:J») 1151 U-53 1156 1159 117S 11S«:» ll'^S 11S6 1205 12«>? 1211 1213 1233 1238 1239 1242 1161 1189 1216 1245 1164 1167 1169 1172 1191 IIW 1197 1199 1219 12-22 12-25 1227 1247 1250 1253 1256 111 111 111 111 12 2 2 2 12 2 2 2 12 2 2 3 1 2 2 2 3 •10 1251 1262 j 12>>5 12>'^^ 1271 1274 127t' li79 1282 12--'> l' 1 1 1 12 2 2 3 •11 12^8 •12 131S •13 1:549 ■14 i:>-'-^ 1291 12W 1297 WX) 1321 1324 1327 1330 1352 1355 135S 1361 13S4 13S7 1390 1393 1.3i)3 1334 136v5 1396 lZty> 1309 1312 1315 1:537 1340 1343 1346 i:56^ 13n 1374 1377 14'» 1403 1406 I4i>9 111 f) 1 1 1 111 '"' 1 1 1 2 2 2 2 3 2 2 2 2 3 2 2 2 3 3 ■15 1 1413 1416 1419 li22 142^' 1429 1 1432 1435 1439 1U2 '^ 1 1 1 2 2 2 3 3 •16 1445 ■17 1479 •18 1-514 •19 1-549 1449 1452 ' 1455 1459 14S3 1486 1489 1493 1517 1521 1524 152S 155J 1556 1560 1563 1462 1496 1-531 1-567 1466 1469 1472 1476 1500 1503 1507 1510 1-535 1538 1542 1545 1570 1574 1578 15^1 111 111 111 111 2 2 2 3 3 2 2 2 3 3 2 2 2 3 3 2 2 3 3 3 ■20 1">5 15S9 1592 1596 1600 i^^.s 1>V'7 l.Ul 1614 l.;l> i. 1 1 1 2 2 3 3 3 21 1'22 •22 1»'.'V' •23 1»;'.'> •24 173> 16-26 16»53 17t>2 1742 1629 1633 1637 1667 1671 1675 1706 inO 1714 1746 17-50 I7.54 i>:v4i 1679 1718 17-5S 1>U4 1648 1652 lt'.56 1683 16'<7 1690 1694 1722 1726 1730 1734 17H2 1766 I77t> 1774 » 1 1 2 112 112 112 2 2 3 3 3 2 2 3 3 3 2 2 3 3 4 2 2 3 3 4 ■25 177- 17-1 17-'> 1791 1795 17'j9 1.^:13 lNi7 I'-ll l>lr; i< 1 1 2 2 2 3 3 4 •26 1-2'. •27 1-2 •23 r.v»5 29 1950 1S24 1S2S 1S»56 i 1871 1910 ' 1914 1954 1959 1832 1S37 1875 IS 79 1919 19-23 1963 196.^ 1>4] 1S.<4 19"2^ l-<45 1,S49 1JS54 18-58 1888 1892 1S97 1923 ' 2<>28 2032 2ii37 1:' 112 2 3 3 4 4 ■31 21)42 ■32 2089 •33 213-S ■34 21SS 2046 2(^1 2«)94 2099 2143 ' 2148 2193 2198 2056 1 2061 2104 2109 21>3 2158 2-203 2208 2«>55 2113 216« 2213 2070 2075 2060 20W 2118 -2123 2128 2133 2168 2173 2178 2183 2218 2-2-23 2-2-2S 2234 112 112 112 112 2 2 3 3 4 4 2 3 3 4 4 2 3 3 4 4 3 3 4 4 5 •35 2239 •36 -2291 37 2Ui •38 2399 ■39 24-55 2244 1 2249 2254 2259 ■1-^,:', 2270 -2275 _ 22.^.' 22>^; 112 2 3 3 14 5 2296 2350 2404 2460 2o<:>l 2307 2312 2355 2360 2366 2410 -2415 2421 2466 2472 2477 2317 2371 2427 24>3 2323 2328 ! 2333 "2339 2377 2382 2388 2393 2432 24.38 -2443 -2449 24*^9 2495 -Z'/*^ 2-V>6 112 2 112 2 112 2 112 2 3 3 4 4 5 3,3445 3 3 4 4 5 3 3 4 5 5 • ■40 2512 2.n- 2-523 2529 2535 2->4l 2-547 2-5-53 2 -55 'J 2-'>'-4 1 1 2 2 344 5 5 41 2570 42 -2630 43 2>?92 44 27->4 •45 2S1> 2576 2582 2588 2594 2636 2642 2649 2655 2698 2704 2710 2716 2761 2767 2773 27SO 26ij"» 2661 2723 2606 2*512 2618 2624 2667 -2673 2679 2685 2729 2735 2742 2748 2793 2799 2805 -2812 112 2 112 2 112 3 112 3 3 4455 3 4 4 5 6 3 4 4 5 6 3 4 4 5 6 2>25 2S31 2838 2844 2>51 28-58 2864 2>71 2877 112 3 3 4 5 5 6 •46 2SS4 ■47 -yyc'i ■48 :;.'LV •49 :.■■>: •^91 2897 2904 2911 2958 2965 2972 2979 3»>27 3034 3041 3048 3^97 3105 3112 3119 2917 2985 3055 31-26 2924 2931 2938 29+4 2999 2992 3006 3013 30»)2 3069 3076 3083 31-^3 3141 3148 31-55 112 3 112 3 112 3 112 3 3 4 5 5 6 3 4 5 5 6 4 4 5 6 6 4 4 5 6 8 MATHEMATICAL TABLES Ajsttilogarithms 599 1 ; 2 3 i i 4 5 1 6 7 8 9 1 2 34 5 6 7 8 9 •50 3162 1 1 3170 1 3177 3184 3192 3199 i 3206 j 3-314 3221 3228 1 1 2 3 4 4 5 6 7 •51 •52 •53 •54 3236 3311 3388 3467 3243 ' 3251 3258 3319 i 3327 3334 3396 ' 3404 3412 3475 , 3483 3491 3266 3342 34-20 3499 3273 3350 3428 3508 3281 3357 3436 3516 3289 3365 3443 3524 3296 3373 3451 3532 3304 3381 3459 3540 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 4 4 ■ 4 4 5 5 6 7 5 5 6 7 5 6 6 7 5 6 6 7 •55 3548 3556 1 3565 3573 3581 3589 3597 3606 3614 3622 1 2 2 3 4 1 5 6 7 7 •56 •57 •58 59 3631 3715 3802 3890 3639 3648 3724 3733 3811 3819 3899 3908 3656 3741 3828 3917 3664 3750 3837 3926 3673 3758 3846 3936 3681 3767 3855 3945 3690 3776 3864 3954 3698 3784 3873 3963 3707 3793 3882 3972 1 1 1 1 2 2 2 2 3 3 3 3 3 3 4 4 4 4 . 4 5 5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8 •60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1 2 3 4 5 6 6 7 8 •61 •62 •63 •64 4074 4169 4266 4365 4083 4178 4276 4375 4093 4188 4285 4385 4102 4198 4295 4395 4111 4207 4305 4406 4121 4217 4315 4416 4130 4227 4325 4426 4140 4236 4335 4436 4150 4246 4345 4446 4159 4256 4355 4457 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 7 8 9 6 7 8 9 6 7 8 9 6 7 8 9 •65 4467 4477 4487 4498 4508 4519 4529 4539 4550 4560 1 1 2 3 4 5 6 7 8 9 •66 •67 •68 •69 4571 4677 4786 4898 4581 4688 4797 4909 4592 4699 4808 4920 4603 4710 4819 4932 4613 4721 4831 4943 4624 4732 4842 4955 4634 4742 4853 4966 4645 4753 4864 4977 4656 4764 4875 4989 4667 4775 4887 5000 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 5 6 6 6 7 910 7 8 9 10 7 8 910 7 8 910 •70 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 1 2 4 5 6 7 8 911 •71 •72 •73 •74 5129 5248 5370 5495 5140 5152 5260 5272 5383 5395 5508 5521 5164 5284 5408 5534 5176 5297 5420 5546 5188 5309 5433 5559 5200 5321 5445 5572 5212 5333 5458 5585 t 5224 5346 5470 .5598 5236 5358 5483 5610 1 1 1 1 2 2 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 8 10 11 7 9 10 11 8. 9 10 11 8 9 10 12 •75 5623 5636 5649 5662 5675 5689 5702 j 5715 5728 5741 1 3 4 5 7 8 9 1012 •76 •77 ■78 •79 5754 5888 6026 6166 5768 5902 6039 6180 5 781 5916 6053 6194 5794 5929 6067 6209 5808 5943 6081 6223 5821 5957 6095 6237 5834 5970 6109 6252 5848 5984 6124 i 6266 5861 5998 6138 6281 5875 6012 6152 6295 1 1 1 1 3 3 3 3 4 4 4 4 5 5 6 6 7 7 7 7 8 9 11 12 8 10 11 12 8 10 11 13 9 10 11 13 •80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 1 3 4 6 7 9 10 12 13 •81 •82 •83 •84 6457 6607 6761 6918 6471 6622 6776 6934 6486 6637 6792 6950 6501 6653 6808 ' 6966 6516 6668 6823 6982 6531 6683 6839 6998 6546 6699 6855 7015 ! 6561 i 6714 ; 6871 7031 6577 6730 6887 7047 6592 6745 6902 7063 2 2 2 2 3 3 3 3 5 5 5 5 6 6 6 6 8 8 8 8 9 11 12 14 9 11 12 14 9 11 13 14 10 11 13 15 85 7079 7096 7112 7129 , 7145 7161 7178 7194 7211 7228 2 3 5 7 8 10 12 13 15 -88 •87 •83 •89 7244 7413 7586 7762 7261 7430 7603 7780 7278 7295 7447 1 7464 7621 7638 7798 ; 7816 7311 7482 7656 7834 7328 7499 7674 7852 7345 7516 7691 7870 : 7362 I 7534 7709 i 7889 7379 7551 7727 7907 1 7396 . 7568 7745 7925 2 2 2 2 3 3 4 4 5 5 5 5 7 7 7 7 8 9 9 9 10 12 13 15 10 12 14 16 11 12 14 16 11 13 14 16 •90 7943 7962 7980 i 7998 , 8017 8035 8054 1 8072 8091 8110 2 4 6 7 9 11 13 15 17 •91 •92 •93 •94 8128 8318 8511 8710 8147 8337 8531 8730 8166 8356 8551 8750 8185 8375 8570 8770 8204 8395 8590 8790 8222 8414 8610 8810 8241 8433 8630 8831 1 8260 8453 8650 8851 8279 8472 8670 8872 1 8299 ' 8492 8690 1 8892 2 2 2 2 4 4 4 4 6 6 6 6 8 8 8 8 9 10 10 10 11 13 15 17 12 14 15 17 12 14 16 18 12 14 16 18 •95 8913 8933 8954 8974 .8995 9016 9036 9057 9078 9099 1 2 4 6 8 10 12 15 17 19 •96 •97 •98 9120 9333 9550 9141 9354 9572 9162 9376 9594 9183 9397 9616 ! 9204 9419 9638 9226 9441 9661 9247 9462 9683 1 i 9268 1 9484 ' 9705 9290 9506 9727 9311 9528 9750 2 2 2 4 4 4 6 7 7 8 9 9 11 11 11 13 15 17 19 13 15 17 20 13 16 18 20 INDEX Abrupt change of section, 56, 90 Adams's experiments on marble, 42 Adhesion between concrete and steel, 80 Alignment charts for springs, 342 Aluminium, 62 Andrews -Pearson formula for curved beams, 545 Areas — mathematical determination, 161 Parmontier's rule, 164 Simpson's rule, 164 sum curves, 162 table of various sections, 185 Arnold's testing machine, 399 Autographic recorders, 379 Avery's reverse torsion machine, 385 Bach on plates and slabs, 488-509 Bairstow, 59, 88 Baker, B. — abrupt change of section ex- periments, 56 repetitions of stress, 86 Basquin, 307 Bauschinger on strength of stone, 72 Beam factor, 197 Beams. See Bending moments; deflections ; inclined beams ; shear; stresses, bending. Bending moments — cantilevers, 122-127 continuous beams, 430-458 fixed beams, 414-438, 458 simply supported beams, 127- 160 Bernoulli's assumption, 194 Bolts, coupling, 311 Brinell hardness test, 402 Buckton testing machines, 364, 368, 381 Built-in beams, 416-438, 458 Bouton experiments on compres- sion, 47 Bracing of columns, 294 Brass, 62, 83 Brittleness, 41 Bronze, 62, 83 Buchanan on columns, 287 Buckling factor, 279 Bulk modulus, 7 Cantilevers, 122-127 Cast iron, 49, 209, 305 Centroid, 165 Chain links, 546-551 Cleat, stresses in rivets, 315 Coker — thermal and optical testing, 409, 411 torsion testing, 388 Collapsing pressure for pipes, 118 Columns — centrally loaded, 279-301 eccentrically loaded, 301-310 Compressive strength, see various materials. Concrete — compressive strength, 69-77, 82 reinforced, 215-235, 292, 479 shear strength, 78 tensile strength, 77 Considere, 42 Continuous beams, 438-458 Critical speed of shafts, 562-569 Curvature of beams, 250 Curved beams — Andrews- Pearson formula, 545 correction coefficients, 543 general conditions of strain, 529 Resal's construction, 538 601 602 INDEX Ctirved beams {continued) — rings and chain links, 546 Winkler's formula, 533 Cylinders, see Pipes. Darwin's extensometer, 376 Deflections — beams, simply supported and cantilever, 248-278 fixed beams, 419, 420, 460 shear, due to, 483 Diagonal square beam section, 210 Disks, rotating, 552-562 Distribution of shear stress in beams, 470-487 Dixon and Hummel testing machine, 368 Drums, rotating, 552 Ductility, 41 Dunkerley on whirling of shafts, 568 Dynamic stress and strain, 33 Dyson on internal friction, 47 Eden on stress repetition, 89 Elastic bodies, 1 Elastic moduli, 7 relation between, 8 Ellipse — of inertia, 169 of stress, 15 Elongation and gauge length, 54 Encastre beams, 416^38, 458 Euler's column formula, 280, 306 Ewing's extensometer, 374 Failure, cause of, 42 Fairbaim, 84, 119 Fatigue of metals, 89 Fidler column formula, 289 Filon on optical testing, 415 Fitchett, F., on springs, 343 Fixed beams, 416-438, 458 Flitched beams, 203 Foster on stress rej)etition, 89 Gadd on cement tests, 408 Glass, 83 Goodenough on chain links, 551 Goodman extensometer, 371 Gordon column formula, 288 Grashof on plates and slabs, 494 Greenwood and Batley testing machine, 367 Grips for test-pieces, 370 Guest theory of stress, 44, 328 Hardness, 41, 402 Hartnell, W., on springs, 341 Heterogeneous sections — direct stress, 38 geometrical properties, 183 Hooke's Law, 2 Horse- power of shafting, 323, 335 Hysteresis, mechanical, 59, 89 Illinois experiments, 119, 297, 551 Impact, strain and stress due to, 35 Impact testing, 400 Inclined beams, 155-160 Inertia, moment of, 169-191 Internal friction in materials, 45 Izod — exjjeriments on shear, 64, 67 impact testing machine, 401 Johnson — column formula, 288 eccentric loading, 309 Kennedy's extensometer, 372 Keyways in shafting, 334 Lame's theory for thick pipes, 510 Lilly- column formula, 290 torsion testing machine, 386 Live loads, 100 Liider's Lines, 54 Macklow-Smith torsion meter, 392 Malleability, 41 Malleable cast iron, 51 Modulus — elastic, 7 section, 197 Mohr, 181, 252 Moment of inertia, 167-191 Moment of resistance, 196, 213, 222 Momental ellipse, 167 Moore, 297, 334, 551 Morley, 3 on columns, 310 on curved beams, 545 on slabs, 494 Muir on overstrain, !59 INDEX 603 Navier internal friction theory, 45 Non-circular shafts, 333 Oblique loading on beams, 241 Overstrain, 57 Permanent set, 1 Perry, 70, 192 Pipes and cylinders — collapse of, 118 initial pressure in, 524 Lame's theory, 510 shear stresses in, 51 6 thick, 510-538 thin, 115 Piston rings, 355 Plastic bodies, 1 Plates — Bach theory, 488^93, 499-509 circular, 488 Grashof and Rankine, 494 oval, 492 square and rectangular, 494 Poisson's ratio, 3 Polar moment of inertia, 173 Portland cement, strength of, 68- 79, 404 Principal stresses, 13 Quality factor, 62 Radius of gyration,' 169-191 Rankine — column formula, 285 combined stress theory, 43, 320 slab formula, 494 stress lines, 213 Reinforced concrete — beams, 215-235 columns, 292 shear in beams, 479 Repetition of stress, 84 Resal's construction, 538 Resilience — definition, 33 in bending, 272 in torsion, 331 summary for various springs, 363 Rigidity modulus, 7 Rings, 546 Riveted joints, 102-115 Rotating drums and disks, 550- 562 Rotating shafts, 562-569 St. Venant, 44, 328, 333 Sankey bending machine, 396 Scoble on optical testing, 404 Secondroid, 168 Section modulus, 197 Shaft coupling, stresses in, 311 Shafting, stresses in, 316-335 Shear — diagrams for cantilevers, 122- 127 diagrams for simply supported beams, 127-160 diagrams, steps in, 144 strain and stress, 3 stress and strain equivalent to complex stresses, 19, 30 stress in beams, 470 Shrinkage stresses in pipes, 525 Slabs, see Plates. Slate, 83 Slenderness ratio, 279 Smith, C. A. M., 48, 517 Smith, J. H., repetition testing machine, 392 Smith, R. H., construction for combined stresses, 23 Springs — closed-coiled helical, 337, 362 leaf on plate, 351 open-coiled helical, 346 piston rings, 355 plane spiral, 360 summary, 363 time of vibration, 337 Stanchions, see Columns. Stewart on collapse of pipes, 119 Stone, compressive strength, 68, 82 Straight line column formula, 287 Strain — definition and kinds, 1-3 in different directions, 12 maximum strain equivalent to combined strains, 25 transverse, 3 Stress — bending, 192-247 cause of failure under, 42 604 INDEX Stress (continued) — ■ combined bending and direct, 235-240, 328 complex, 13 definitions and kinds, 1-3 dynamic, 33 ellipse of, 15 impact, 35 principal, 13, 19 repetition of, 84 shear, 3, 212 shear in beams, 470 temperature, 37 working, 93-100 Stress-strain diagrams, 5, 49, 53, 60, 62, 65 Struts, see Columns. Sum curve, 162 Superposition, principle of, 244 T beams, reinforced concrete, 229 Temperature stresses, 37 effect on strength, 63 Tensile strength, real and appar- ent, 52 of materials, see various ma- terials. Testing — calibration of machines, 369 cement and concrete, 404 extensometers, 371-380 grips and forms of test- pieces, 370 impact, hardness and ductility, 396 machines (general), 364-369 Testing (contimied)— optical, 411 thermal, 409 torsion machines, 382-393 Theorem of three moments, 445 Thick pipes, see Pipes. Thurston testing machine, 383- 386 Timber, 65, 82 Torsion, 311-335 Turbine shaft, settling down, 564 Twisting, 311-335 Unit section modulus, 197 Unital strain, 3 Unwin — elongation formula, 55 extensometer, 379 on piston rings, 359 on stone cubes, 70 Waist in tensile fracture, 54, 56 Werder testing machine, 366 Whirling of shafts, 562-569 Wicksteed-Buckton testing- ma- chines, 364, 368, 381 Winkler's formula for curved beams, 533 Wohjer's experiments, 84 Woo J son experiments on concrete, 43 Working stresses, 93-99 Wrought iron, 82 Yield point, 4 Young's Modulus, 7 Printed ix Gkeat Bkitain- bv Richaud Clav & Soss, Limited. IIRLXSWICK. ST., STAMFOKD ST., S.E., AND hlNGAV, Sl'FFOLE. ^^ -^A 0'