STRENGTH OF MATERIAL AN ELEMENTARY STUDY PREPARED FOR THE USE OF MIDSHIPMEN AT THE U. S. NAVAL ACADEMY BY H. E. SMITH * PROFESSOR OF MATHEMATICS, U. S. NAVY SECOND EDITION, REVISED NEW YORK JOHN WILEY & SONS, INC. LONDON : CHAPMAN & HALL, LIMITED 1914 COPYRIGHT, 1908, 1909, BY H. E SMITH, U. S. N. Stanhope Ipresft F. H. G1LSON COMPANY BOSTON. USA PREFACE. THIS book has been prepared for the use of the Midshipmen at the U. S. Naval Academy, and is designed to cover a short course in the subject taken up in the Department of Mathe- matics and Mechanics preliminary to the work in the Depart- ments of Ordnance and Gunnery and of Steam Engineering at the Academy. In arranging the subject matter many of the methods in- troduced by officers previously on duty in the Department of Mathematics have been employed, and the endeavor has been to lead the student to the opening point for the professional work carried on by the other Departments. iii 360375 INTRODUCTION. BEFORE beginning the study of Strength of Material, let us see what has been discovered by experimenting with test pieces of material and note some of the conclusions arrived at from the results obtained in this way. Experiment shows us that whenever a force acts on a body formed of any substance the dimensions of that body are changed. In mechanics all bodies were assumed rigid and the results obtained under this assumption were true, for mechanics taught us to find the action of one body on another or the force transmitted by one body to another, while strength of material will teach us the effect in the body itself of a force acting upon it. Mechanics showed us that by means of a piece of material force could be moved from one point to another, and strength of material will show us that in trans- mitting the force the substance forming the conveyance suffers some slight temporary deformation if the force is with- in certain limits; that beyond these same limits the substance suffers permanent deformation and if the force be great enough will be completely ruptured. The study of strength of material will include finding the safe limits of a force to be transmitted by any particular piece of material, finding the deformation caused by transmitting any force, and finding the dimensions of a piece of material in order that it may safely transmit any particular force. With regard to deformation materials differ greatly, for example, the force which will double the length of a piece of rubber will not apparently change the length of a piece of steel of the same size, though both of these substances are elastic, and each if stretched within limits will return to its VI INTRODUCTION. original length when the stretching force is removed. The force which makes an indentation in a piece of putty will scarcely affect a piece of lead, but in this case the indentation made will remain in both these substances as they are plastic. Obviously, then, we must experiment with the different ma- terials and find out some of their physical properties before proceeding with a mathematical investigation. The materials used in building are elastic and we determine their physical properties by experimenting with small pieces of them in machines made for the purpose. Take steel, for example; small test pieces of different shapes are tried under different forces. A pull is applied and we find the test pieces stretch; if we apply pressure the test pieces are compressed. Having applied all sorts of forces we compare our results and find that the stretching or compres- sion is always of the same amount if the same value of force on unit cube of the steel is used. We also find that up to a certain limiting value of the force the material will always return to its original shape when the force is removed, but if we go beyond this value we find the piece will not return to its original shape; in other words it is not perfectly elastic for forces beyond this value. If we go through the whole list of materials used in build- ing and test each kind in the same way we will find that they will all behave in a similar manner, the difference will lie in the amount of deformation for any force and the value of the limiting force beyond which they are not per- fectly elastic. Experimenting further we find the value of this limiting force for the different materials and having it we can compare the various substances as to their usefulness under different circumstances. If we continue to experiment, again and again with a single substance we find that we may apply as often as we like a force less than the limiting one we found, and that the piece INTRODUCTION. vii will always return to its original shape when the force is removed; but when the force used exceeds the limiting one found, if by ever so little, and the force is applied and re- moved often enough the piece will break, though the deforma- tion caused by the first application of the force was too small to be measured or even noticed. From this fact we see that we must never use a piece of material which will have to sustain a force which is in the least greater than the limiting value found by experimenting. Materials differ in other ways. A piece of glass is easily broken by a light blow and is therefore called brittle or fragile; wrought iron can be twisted and bent into almost any shape without rupture and therefore is called malleable or ductile. Material which has been melted and cast into desired shapes cools quicker at some parts thaa it does at others, thereby setting up within it internal stresses which are irregularly distributed. Such castings can be broken by a comparatively light blow though they can usually withstand a large pressure. These internal stresses can be removed by a process called annealing, which consists in heating the body to a red heat and allowing it to cool slowly, thus allowing the particles an opportunity to rearrange themselves. Metals have the peculiarity that if overstrained they harden in the vicinity of the overstrain and this hardening goes on with time. Thus a bar which is sheared off while cold will finally become extremely hard and brittle near the sheared end, as will a plate in the neighborhood of cold-punched rivet holes. To avoid this, bore the rivet holes and saw the bar, or, if feasible, anneal the cold-sheared bar and the plate near the cold-punched rivet holes. Now in all practical cases we must of course use material that will not break, but in addition we must have material which will not change its dimensions to any considerable extent under the applied load. Vlil INTRODUCTION. The experiments by which the constants used in this study have been determined were very carefully conducted and are f the results of many independent efforts on the parts of many different scientists. In this work the mathematical investi- gation only will be touched upon, the experimental part being beyond its scope. CONTENTS. PAGE INTRODUCTION v CHAPTER I. STRESS AND STRAIN, TENSION AND COMPRESSION 3 II. SHEARING FORCE 15 III. TORSION 26 IV. STRESS DUE TO BENDING 34 V. COMBINED STRESSES 43 VI. SHEARING STRESS IN BEAMS 52 VII. BENDING MOMENTS, CURVES OF SHEARING STRESS AND BENDING MOMENTS 61 VIII. SLOPE AND DEFLECTION OF BEAMS 73 IX. SLOPE AND DEFLECTION (Continued) 83 X. CONTINUOUS BEAMS 94 XI. COLUMNS AND STRUTS 105 XII. STRESS ON MEMBERS OF FRAMES 112 XIII. FRAMED STRUCTURES 123 XIV. FRAMED STRUCTURES (Continued) 133 MISCELLANEOUS PROBLEMS 141 Reinforced Concrete Beams; Poisson's Ratio; Stress in Thick Cylinders and Guns; Built-up Guns; Stress due to Centrifugal Force; Bending due to Centrifugal Force; Flat Plates. IX STRENGTH OF MATERIAL. STRENGTH OF MATERIAL. CHAPTER I. TENSION AND COMPRESSION STRESS AND STRAIN. 1. In the study of strength of material we must consider two ways of arranging the different pieces used: first, when there is to be motion between the parts, and second, when the parts are to be relatively at rest. In the first case, force is transmitted from one piece to another and the combination of pieces is called a machine, the study of which involves the principles of dynamics; the second arrangement is called a framed structure, or simply structure, and we must employ the principles of statics in its investigation. In either arrangement, any two parts in contact have a mutual action between the touching surfaces, and the effects produced by this action depend in great measure upon the way in which it is applied. In any case it tends to change the shape or dimen- sions of the parts involved and, if the force is great enough, to crush or break them. So for permanence the machine or structure must be strong enough in each part to withstand any force to which it may be subjected. If then we can find the greatest force that any particular piece of material can endure without breaking or suffering a permanent change in shape we can be sure of its remaining intact for all force within that limit. In the first four chapters we will apply, separately, all the different forces to which a piece of material can be subjected and as any piece in our machine or struc- ture may have more than one of these forces to sustain at any 3 4 .. .STRENGTH OF MATERIAL. given instant we will, in the fifth chapter, show the effect of the combined action of two or more of them. 2. Let us first investigate the effect on a piece of material of an external force applied to it, and we will choose for our investigation a straight rod of uniform cross-section and will not consider the force of gravity as acting. We will apply to the end of our rod a pull, F (not sufficient to break or per- manently change its shape), which, in order that the rod remain stationary, will require an equal pull in the opposite direction at the other end. These forces tend to tear apart the particles of the material, and as the bar remains intact the particles must be in a different condition, relative to each other, from that in which they were before we applied the pull. If instead of a pull we exert a pressure on one end, an equal and opposite pressure must be applied to the other end to keep the bar in equilibrium, and the particles of the material will now tend to crowd together and crush each other. In both of these cases we have arranged our forces so that the bar does not move and they have been taken small enough so as not permanently to change its dimensions. Now as the length of the bar separates the points of application of our forces, there must be some action set up among the par- ticles of the bar itself which transmits the force from one end Pull H H In tension Pressure In compression Fig. 1. Fig. 2. to the other, or to a common point of action. Let us now imagine a plane passed through the bar perpendicular to its axis: in the first case (the pull, Fig. 1) our forces would tend TENSION AND COMPRESSION. 5 to pull apart the two pieces of the bar; in the second (not shown) they would press them together, each piece would tend to move in the direction of the external force acting on it and the amount of this tendency would be equal to that force so that the total action in the bar between the particles on either side of any imaginary section would be equal but opposite to the external forces applied. The action of all the particles on the right side of any section would be equal to but opposite to the force on the right end, and of those on the left side equal to but opposite to the force on the left end. This must be true in order that the bar remain intact, i. e. in equilibrium. 3. Stress. When any such action as the above is set up among the particles of a piece of material that piece is said to be under stress, and the external force which causes the stress is called the load. A rod under the action of a pull is said to be in tension, and the pull is the tensile load. A rod to which pressure is applied is said to be under compression, and the pressure is called the compressive load. Hereafter in using the word stress we will mean the amount of action between the particles in unit cross-sectional area, and will use " total stress" for the action over the area of the whole cross-section. The stress per unit cross-sectional area is sometimes called " intensity of stress" and is equal to the external force, F, divided by the cross-sectional area, A. The forces on one side of the section only are used, so, F p (stress) = A. Of course we would get the same result by using the forces on the other side of the section, as the bar must be in equi- librium; they, however, would act in the opposite direction. 4. Now let us see what will happen if we gradually increase the load, F. We know that all materials are more or less elastic, so as F is increased the bar will stretch or shorten 6 STRENGTH OF MATERIAL. according as F puts it in tension or compression; and up to a certain value of F, the bar will spring back to its original length when F is removed. If we experiment carefully with any material we will find that there is for it a certain value of F, after reaching which, and then having removed F, the bar will be found a little longer or a little shorter than it was originally; in other words, it will have a "permanent set." This value of F for unit sectional area is called the " limiting stress" or "elastic limit" of that material. It is obvious that no part of our machine or structure must be subjected to a force equal to this, for if it be the piece so used is afterward unfit to do its work. If we go on increasing F the bar will finally break or crush, but at present we are interested only in the elastic limit, the stress for which we will call /, and for several materials its value will be found in the table at the end of this chapter. F We know that any stress, p, is equal to , so to find the A. tensile or compressive load any piece of material can sustain F unhurt, we put/ = or F (the limiting load) = fA. This A is true for all ordinary lengths of material under tension, and for short pieces under compression. We will later (Chapter XI) consider long pieces under compression, in which the stress due to bending 'must be taken into account. 5. In the preceding article we saw that, within the elastic limit, a piece of material would, when the load was removed, return to its original length. After voluminous experiments, it has been found that within the elastic limit the extension or compression of a rod under stress varies directly as the stress, or stress is equal to a constant multiplied by the ex- tension or compression. This is known as Hooke's Law, from its discoverer. If we let x denote the total extension of a rod of length I, the extension per unit length will be , and TENSION AND COMPRESSION. 7 this extension per unit length is called strain, and we will denote it by e; then if p denote the stress per unit area of cross-section, E being a constant found by experiment and called the modulus of elasticity. Its value for several materials will be found in the table at the end of this chapter, E = e 6. Work Done on a Piece of Material by a Load. In all cases of overcoming resistance work is done, and we will now find how much work is done on a piece of material on which a load is acting. By Hooke's Law p = Ee so that, as we have equilibrium within the elastic limit, the resistance of the material to the load must equal p per unit area. Before the load is applied the resistance is nothing. Now let us apply the load very slowly. The re- sistance will increase with the load from zero to the equal of the final load. It follows that the mean resistance will be equal to one-half of the final load. Work is equal to force multiplied Fig. 3. by the space through which it acts. Due to the load, our rod has stretched through the distance x. Hence the work done is equal to one-half the final resistance multiplied by x, or if L be the final load L the work is equal to X x. & *- The following graphic method may be clearer: Let the abscissa represent the extensions, and the ordinates the loads 8 STRENGTH OF MATERIAL. causing them. When x = 0, L = 0, and as the load in- creases so will the extension in accord with the formula p == Ee, giving us a straight line for the load curve. The work done will now be represented by the area between the load curve, OA, the axis of x, and the ordinate, Ab, represent- ing the final load, L. extension Work = X L = area of triangle OAb. The following is the method by calculus: L (load) = pA (Art. 4), p = E~ = Ee (Art. 5). The load curve then is L = pA = AE j=AEe. The work done by any load L acting through a space dx is Ldx or dW = Ldx, but which integrated between the limits for x gives EA X 2 ~] total extension ' U which as p =- and L = pA gives W for the work as before. TENSION AND COMPRESSION. 9 If our bar be stretched to just within the elastic limit we will have also /= orL =fA; hence, from the above, f 2 Al f 2 W = - = - XF. (V equals volume of bar). 2h 2h/ As we have stretched our bar to just within the elastic limit, it will return to its original length when we remove the load. This load then does the greatest amount of work that can be done on a piece of material without injuring it. While stretched, the rod has stored in it an amount of energy equal to the work done in stretching it to its elastic limit. This stored energy is called the resilience of the bar, / 2 and the part is the modulus of resilience. 2E 7. The work done by forces quickly applied is much greater than if the force is slowly increased, for, suppose a vertical rod having a collar round its lower end is stretched by a weight W falling from a height h upon the collar. The work done by the falling weight is W (h + x) and this Lx must equal the work done on the rod or We have then or the slowly applied load which would stretch the rod to the same extent as the falling weight would have to be con- siderably greater than the weight. 10 STRENGTH OF MATERIAL. Again, if we imagine a load L to be instantaneously applied and to cause a strain equal to x and a load L, slowly applied which would cause the same strain, the work done in the first case or Lx (force times space) must equal T ~ the work done in the second case or - Equating we have L t = 2L or a suddenly applied load has twice the effect of a slowly applied one. Definitions: STRESS. When due to the load on a body, there is mutual action between the particles on either side of a sec- tion through the body, so that the particles on one side exert a force on those on the other side, stress is said to exist in that body, and the intensity of the stress is the force per unit area of cross-section. Briefly, then, stress is force per unit area of cross-section. STRAIN. The ratio of change of length in a body due to the load on it, to the original length; briefly, change of length per unit length. MODULUS OF ELASTICITY. The stress which would double the length of a rod provided Hooke's Law held good /load\ for that extension: or, it is the ratio of unit stress I - - to \area/ /total extension\ unit strain I - ; or ratio of stress to strain \original length / within Hooke's Law. ELASTIC LIMIT. That stress, which if exceeded will pro- duce a permanent change of length; or the maximum stress a material can suffer without being permanently deformed. RESILIENCE. When a bar is loaded to its elastic limit, the work done in stretching it is called the resilience of the bar, and the modulus of resilience is this work divided by the TENSION AND COMPRESSION. 11 volume of the bar; or, resilience is the capacity of a body to resist external work. ULTIMATE STRENGTH is the stress which produces rupture. WORKING LOAD is the maximum stress to which a piece of material will be subjected in actual practice. STRENGTH OF MATERIAL Average values in pounds per square inch. Modulus Material. Weight per cubic foot. Elastic limit. / Modulus of elasticity. of elasticity for shear and Ultimate fiber stress. Limiting shearing stress. Pounds. E. torsion. q C. Steel 490 I 35,000 to ) | 50,000 j 29,000,000 10,500,000 110,000 50,000 Iron, cast .... 450 i 6,000 T) \ 20,000 C } 15,000,000 5,000,000 35,000 7,850 Iron, wrought 480 25,000 25,000,000 10,000,000 54,000 20,160 Brass, cast . . 9,500,000 20,000 4,030 Brass, drawn 520 14,500,000 5,600,000 70,000 Copper, cast . Copper, drawn 540 12,000,000 15,000,000 '6,000,000 22,000 65,000 2,890 Stone, granite. 'l60 2,000 ' 6,000,000 1,800,000 2,000 Timber 40 3,000 1,500,000 140,000 10,000 ' 1,200 Examples: 1. A steel bar, 5 ins. long, sectional area \ sq. in., stretches .007 in. under a load of 20,000 Ibs., and shows no permanent elongation when the load is removed. What is the modulus of elasticity of the metal? Solution: Ft 20,000 X 5 X 2 E = - = 28,571,428.571 in.-lbs. A.X .1)07 2. A vertical wrought-iron rod 200 ft. long has to lift suddenly a weight of 2 tons. What is the area of its cross- section if the greatest strain to which wrought iron may be 12 STRENGTH OF MATERIAL. subjected is .0005 for unit length. E = 30,000,000. Neglect the weight of the rod. (See Art. 7.) F Solution: Stress = 8960 Ibs. = p = Ee. A. F 8960 = E~e = 30,000,000 X. 0005 r A = ^ Sq ' in " 3. Find the area of cross-section in example 2, taking the weight of the rod into account. Ans. .611 sq. in., or .625 if rod also is suddenly raised. 4. A steel rod J sq. in. in sectional area and 5 ft. long is found to have stretched T fo in- under a load of 1 ton. What is the modulus of elasticity of steel? Ans. 35,840,000 in.-lbs. 5. A chain 30 ft. long and sectional area } sq. in. sus- tains a load of 3900 Ibs.; an additional load of 900 Ibs. is suddenly applied. Find the resilience at the instant the 900 Ibs. is applied. E = 25,000,000. Ans. 25.992 ft.-lbs. 6. A steel rod 3| ft. long, 2 sq. ins. sectional area, reaches the elastic limit at 125,000 Ibs., with an elongation of .065 in. Find the stress and strain at the elastic limit, the modulus of elasticity, and the modulus of resilience of steel, and express each in its proper units. Ans. / = 50,000 Ibs. per sq. in.; e = .00166; ^E = 30,000,000 Ibs. per sq. in.; and modulus resilience = 41 Ibs. per sq. in. 7. A piston rod is 10 ft. long and 7 ins. in diameter. The diameter of the cylinder is 5 ft. 10 ins., and the effective steam pressure is 100 Ibs. per sq. in. Find the stress pro- duced and the total alteration in length of the rod for a complete revolution. E = 30,000,000. Ans. Change in length .0796 in. TENSION AND COMPRESSION. 13 ( 8. How much work can be done in stretching a com- position rod 5 ft. long and 2 in. in diameter, without injury, if the proof stress of the metal is 2.8 tons per sq. in.? E = 4928 ton-ins. 22 Ans. .15 ton-ins., using x = 9. The proof strain of iron being TT nrc, what is the shortest length of rod 1^ sq. ins. sectional area, which will not take a permanent set if subjected to the shock caused by checking the weight of 36 Ibs. dropped through 10 ft., before beginning to strain the rod? E = 30,000,000. Ans. 192.31 ins. 10. Find the shortest length of steel rod, 2 sq. ins. sec- tional area, which will just bear, without injury, the shock caused by checking a weight of 60 Ibs. which falls through 12 ft. before beginning to strain the rod. E = 30,000,000. Modulus of resilience = 15 Ib.-ft. Ans. 24.05 ft. 11. A brass pump rod is 5 ft. long and 4 ins. in diameter and lifts a bucket 28 ins. in diameter, on which is a pressure of 6 Ibs. per sq. in., in addition to the atmosphere, against a vacuum below the bucket which reduces the atmospheric pressure to 2 Ibs. What is the stress in the rod and the total extension per stroke? E = 9,000,000. Ans. Stress 925 Ibs. per sq. in., and extension .0061 in. 12. Assuming a chain twice as strong as the round bar of which the links are made, what size chain must be used on a 20-ton crane with three sheaved blocks if / = 6000? Ans. Diameter of section of metal .8899 in. 13. A piston rod is 9 ft. long and 8 ins. in diameter. The diameter of the cylinder is 88 ins. and the effective pressure is 40 Ibs. per sq. in. What is the stress produced 14 STRENGTH OF MATERIAL. and the total alteration in length of the rod per revolution? E = 29,000,000. Ans. x = .0359 in. 14. Find the work done in Example 13, and find the resil- ience of the rod if/ = 12 tons. Ans. W = 3.88 in.-tons. R = 52.5 in.-tons. 15. The stays of a boiler in which the pressure is 245 Ibs. per sq. in. are spaced 16 ins. apart. What must be their diameter if the stress allowed is 18,000 Ibs. per sq. in.? Ans. 2J ins. 16. What is the area of the section of a stone pillar carry- ing 5 tons if the stress allowed is 150 Ibs. per sq. in.? Ans. 75 sq. ins. 17. What is the length of an iron rod (vertical) which will just carry its own weight? / = 9000 Ibs. per sq. in. Ans. 2700 ft. 18. The coefficient of expansion of iron is .0000068 in. per degree F. An iron bar 18 ft. long, 1 ins. in diameter, is secured at 400 F., between two walls. What is the pull on the walls when the bar has cooled to 300 F.? Ans. 34,862 Ibs. 19. Coefficient of expansion of copper is .0000095 in. E = 17,000,000. A bar of iron is secured between two bars of copper of the same length and section at 60 F. What are the stresses in the bars at 200 F.? Ans. In copper 2958, iron 5916 Ibs. per sq. in. 20. A bar of iron is 3 ft. long, 2 ins. in diameter. The middle foot is turned down to one inch diameter. Compare the resilience with the original bar and with a uniform bar of the same weight. Ans. (1) 1 to 8; (2) 1 to 6. CHAPTER II. SHEARING. 8. We will next apply force to produce shearing. In study- ing shearing we will use the same sort of rod we used in the preceding chapter but will apply our force in a different way. To get the proper effect we will slip our rod through holes bored in two extra pieces of material as in Fig. 4, and then apply opposite and equal pulls to the two extra pieces so that the effect on the rod will be to pull one part up and the other down. If we pull hard enough our rod will be sliced off smoothly as though a plane had been passed through it per- pendicular to its axis. It will be sheared off. Practically this is as near as we can approximate to pure shearing. In theory the two forces, F, should act on either side of a section of the rod in parallel planes indefinitely close to each other so that their pull would induce no tendency to turning or bending; then whether we actually sheared the bar through or not, the stress in the bar on one side of the section would be the force F divided by the F area A, of the section of the bar, or, q = - A. It will be noticed that the direction of the shearing stress is parallel to the section, while in tensile or compressive stress the direction is normal to the section ; for this reason shearing is sometimes called tangential stress. 9. Two plates bearing a longitudinal load and held together by a riveted joint present a good illustration of shearing. 15 16 STRENGTH OF MATERIAL. The rivet is under almost pure shear if it closely fits the rivet holes, and the bearing surfaces of the plates are plane. Fig. 5 shows a section and plan of I [I a single-riveted lap joint. The j_> * distance between the centers of the rivet holes is called the pitch p, and it is obvious that each rivet supports the load on a strip of plate equal in width to the pitch of the rivets. So that if F is the force acting Fig. 5. along this strip of plate, the F shearing stress, q, on the rivet supporting it is equal to A. as before, or as A is , where d is the diameter of the rivet, ^ -o F*- ?; F 1 ^ O we have for the shearing stress q = . If we put the limiting shearing stress for the material of which our rivet is made for q, we can find the diameter of the rivet we must use under the conditions, d = 2 V nq In working joints, such as pin joints, the action is not pure shear, but, due to the clearance necessary for a working fit, there is some bending. Ex- periment has proved that the stress at the center of the pin of such a joint is $ of that found by the above formula. To find the dia- Fig 6 meter of the pin necessary for a joint like that in Fig. 6, we first notice that there are two sections of the pin to be sheared, SHEARING. 17 then, , _ I shearing stress ^ I at center of pin 4 F 3 " 2~A SF or, d = The limiting shearing stress for several materials is given in the table at the end of Chapter I. 10. Usually the value of F for the above is readily found, as the thrust on a piston rod, etc.; but for the force acting on riveted joints the ordinary steam boiler will give us a good example. Fig. 7 is a section through a boiler designed to carry a steam pressure of s pounds per square inch, its radius is r, and we wish to find the tangential force at any point A . Draw the per- pendicular diameters AB and CD, then on a ring 1 inch wide there will be a pressure of s pounds on each inch of the circumference. Now if we should resolve Fig. 7. horizontally and vertically the pressure on each one of these square inches, the sum of all the horizontal components would be zero. If, however, we take the sum of all the horizon- tal components on the right half of the ring we will get all the forces which act toward the right; of course those to the left of the diameter A B will be equal but opposite in direction. It will be further noticed that to support these forces we have the boiler material at A and also at B and both these points support equal "amounts, so we can divide the force acting on the semicircle by 2 to get that part which acts at A or 5; or, 18 STRENGTH OF MATERIAL. we can take the sum of the horizontal components of the pressure on that part of the ring from C to A which will give the same result. Finding this sum is most readily done by calculus. Take any point on the circumference and let its angular distance from C be 0, then rdd (remembering our strip is 1 inch wide) will be an element of area on which the pressure is s X rdd acting radially and of which the horizontal component is srdO cos 6, and if we integrate this expression between the limits and - we will get the sum* of all these components, or the force H. H = sr I z cos Odd = sr sin p = = sr. Having H, the force on a strip 1 inch wide, the force on a strip supported by one of pur rivets is found by multiplying this by the pitch, or F = psr, so that the shearing stress on the rivet is psr The vertical components have no tensile effect at A, but form all the tensile effect at C. 11. In this connection we can find the required thickness of a cylindrical shell, such as a steam pipe, remembering that it is under tensile stress, not shearing, and changing our con- stants accordingly. If I is the length considered and t the thickness of the plate, the sectional area of the metal will be It, and if / is the tensile strength allowed, the resistance of the shell will befit; this must be balanced by the force IF, which from Art. 10 is Isr: sr so fit = Isr; or t = SHEARING. 19 If we wish the thickness of boiler plates, we must, since boilers are built of plates with riveted joints, divide our result by the efficiency of the joint, strength of joint being equal to the strength of the solid metal multiplied by the efficiency of the joint. 12. We have seen how to find the shearing stress on a rivet of a single-riveted lap joint, but we must remember we have taken out of our plate a piece of metal equal to the diameter of the rivet hole, so that we have left in the strip of plate, to bear the whole force F, a section whose area is (p d) t, and the tensile stress on this section must not exceed the tensile F strength of the material, or to balance; / = ; t(p d) transposing, the pitch for a single-riveted la~p joint is There is another kind of joint called the butt joint, where a narrow strip of material, or strap, covers the edges of the 1 J 1 Fig. 8. \ i i I u i plates and is riveted to both. This joint is called a single butt joint, single-riveted (Fig. 8), and is treated exactly as a single-riveted lap joint. If there are two straps used, one on either side of the plates as in Fig. 9, the joint is called a double butt strap joint, 20 STRENGTH OF MATERIAL. single-riveted, and here it will be noticed we have two sec- F tions of rivet to shear, so our formula becomes q = 2 A Now suppose we have two rows of rivets in either a lap joint or single butt joint (where more than one row is used the rivets are generally staggered as in Fig. 10, shown in the o o o o o o o 1 o 1 Fig. 10. plan) we will have two sections of rivets to shear, and our formula will be the same as for the single-riveted double butt strap, F and if we have for these joints n rows of rivets, the formula becomes F If we use double butt straps we will have double the number of sections to shear, so have to divide the above value of F by 2; or, for double butt straps, F SHEARING. 21 Single-riveted . Double-riveted . Lap joint or single butt strap. F qxd* Double L q ,., or I< - -. Tra 4 4 7T Cl 2 ^T F _ v nqxd 2 ^ - 4> rf Q or 7' 1 or 2 nqxd* 13. If we have more than one row of rivets the plate will, if it be ruptured, carry away along the outer row of rivets in every case; for (see Fig. 11) if it did not, there would be Fig. 11. one or more rows of rivets to be sheared in addition to the carrying away of the plate, and clearly the rupture would occur where it would require the least force. Consequently, as far as the stress on the plate is concerned, it may always be computed as we have it in the final part of Art. 12, using the limiting stress allowed for the material; or, f(p-d)t =F. At the end of Art. 12 we have formula giving F for the shear- ing stress of the material of the rivets, and if we equate these two values of F we can find the pitch of our rivets for equal strength (that is, rupture will be as likely to occur by shearing the rivets as by the plate carrying away), if we know their diameter and the thickness of the plates. 22 STRENGTH OF MATERIAL. Equating we have which gives the pitch where particular plates and rivets are to be used. In these formulae p = pitch of rivets, d diameter of rivets, n = number of rows, q = limiting shearing stress for rivets, / = limiting tensile stress for plate, t = thickness of plate. 14. When a bar is subjected either to tension or compres- sion there is shearing stress along any oblique section. Let the angle BDC = 0, then the intensity of the force F (acting F in either direction) along the section DB is - - , but the area DB area DB equals the area of BC X cosec 0, or if A is the D c Fig. 12 area of a right section, the intensity of the force on DB is F . The component of this force resolved along DB is A cosec F F F -cos d = sin cos =- - sin (2 0). This being A cosec A 2 A F a tangential force is shearing stress. resolved per- A cosec SHEARING. 23 F F pendicular to DB is- sin = -sin 2 #, the normal A cosec A stress. When = or - the shearing stress is 0, but for any i other value of 6 it is a finite quantity, which proves our propo- sition. When = the normal stress is 0, but increases with d to a maximum when = - , which is as it should be. 2 Examples : 1. Two wrought-iron plates, 3 ins. wide by J in. thick are lap-jointed by a single rivet, 1 in. in diameter. What will be the pull required to break the joint, if the tensile strength is 18 tons per sq. in.? What is the efficiency of the joint? Solution: Area (plate) =3Xi--lXi = lsq. in. 4 .'. Strength = 18 tons. Area of section without rivet hole = | sq. in. /. Strength = | X 18 = 27 tons. Efficiency of the joint is || or = 66%. Area of rivet section = - 4 If it is of the same material as the plate the force necessary and 2 18 X 22 to shear it is F = = - - 14.13 -f tons and the 4 4X7 joint would break by shearing the rivet. 2. A cylindrical vessel with hemispherical ends, diameter 6 ft., is exposed to internal pressure 200 Ibs. above the atmos- phere. It is constructed of solid steel rings riveted together. If / = 7 tons per sq. in., 'how thick must the metal be, and what is the longitudinal stress in the metal of the ring joint whose section is ^ that of the solid plate? sr 200 X 6 X 12 Solution: e== = = .459 in. 24 STRENGTH OF MATERIAL. ., ,. , 10 nr 2 p 5rp Longitudinal stress = 7rr 2 p = ^ . 2 xrtq, or, q = 7.2 irrt 1 1 5 X 3 X 12 X 200 q= 7 X. 459X2240 3. What must be the thickness of a copper pipe, j in. in diameter, to sustain a pressure of 1350 Ibs. per sq. in.? / = 950 Ibs. per sq. in. Ans. .533 in. 4. A single-riveted lap joint, plate J in. thick, is under a load of 3 tons per sq. in. of plate section. Rivets f in. in diameter, pitch 1J ins. What is the shearing stress on the rivets and the efficiency of the joint? Ans. Shearing stress 3.82 tons. Efficiency 60%. 5. Two plates, f in. thick, are double-riveted with double butt straps, rivets 1 in. in diameter. The shearing strength of the rivets is f the tensile strength of the plates. Find the pitch and the efficiency of the joint. Ans. Pitch 4 ins. Efficiency .7778. 6. What is the pitch of the rivets in a treble-riveted, double butt strap joint between plates J in. thick, rivets f in. in diameter, if the tensile stress of the plates is limited to 10,000 Ibs. per sq. in., and the shearing stress of the rivets to 8000 Ibs. per sq. in.? What is the efficiency of the joint? Ans. Pitch f f $ in. Efficiency 83 + %. 7. A cylindrical boiler 8 ft. 4 ins. in diameter is under 100 Ibs. per sq. in. pressure. What must be the thickness of the plates that the stress may not exceed 4000 Ibs. per sq. in.? Ans. 1J ins. 8. What is the pitch of 1-in. rivets in a double-riveted lap joint between |-in. steel plates? Tensile stress 13.2 tons and shearing stress 10.5 tons per sq. in. What is the efficiency of the joint? Ans. Pitch 3 ins. Efficiency 71+ %. SHEARING. 25 9. The steel plates of a girder are J in. thick, treble-riveted with double butt strap, with 1-in. rivets. Shearing stress of the rivets is f the tensile strength of the plate. What is the pitch? Ans. 6.24 ins. 10. In a pin joint (Fig. 6) the shearing stress of the pin is | the tensile stress of the rod. Compare the diameters. Ans. Nearly equal. 11. A square bar of steel is under a tensile stress of 4 tons and a compressive stress of 2 tons at right angles to its axis. What are the shearing and normal stresses on a plane making 1 the angle tan" 1 =with the axis? V2 Ans. S. = 2\/2 tons. N = 0. 12. What should be the pitch of 1.25-in. rivets in a treble- riveted, double butt strap joint between 1-in. plates, if the resistance to shearing is f the resistance to tearing? Ans. 6.77 ins. 13. What pressure in a copper pipe, T 4 5 in. in diameter and .02 in. thick, will stress the copper to 8000 Ibs. per sq. in.? Ans. 800 Ibs. CHAPTER III. TORSION. 15. In Chapters I and II we have seen the effects of tension or compression, and of shearing. In this chapter we will fix firmly one end of our rod and twist it by applying a turning moment to the free end. This will subject it to tor- sion, with the effect that any right section will be turned about its center by an amount depending upon the distance of the section from the fixed end.* The stress between the particles on either side of any section will be parallel to the section, or it is tangential stress. Clearly, then, torsion is a kind of shear. If before ap- plying our turning moments we draw a line, AB (Fig. 13), on our test piece, parallel to its Fio . 13 axis, it will, after the moment is applied, be found to have deformed into helix or screw thread, and the point B will now be at 6. The angle BAb, or <, is the angle of torsion, and is proportional to the radius of the rod. The angle at the center, 6, is called the angle of twist, and is proportional to the length of the rod. The distance Bb is equal to rd, but it is also equal to l(f>; as (j> is a very small angle, so that AB and Ab are practically equal, rO = ty; or < = ~ . (1) * We might have applied a turning moment to both ends in oppo- site directions, in which case the section at the middle of the rod (the fixed end as taken above) would have remained stationary, though it would have been stressed as the others. 26 TORSION. 27 We have, by experiment, the equation q = Ccf) for torsion, where q is the stress at the surface and C is the modulus of distortion or torsion, just as we have p = Ee to hold in ten- sion or compression, <, the angle of torsion, being a meas- ure of the strain. We have then two values of , M'^< -I'/' 7 -'- (2) It is obvious that for any material, d, on any section, will be constant for any stress. This being true, our equation for 6 proves that q varies with r, or if q, q l} q 2 be the stress on the surface rods of the same length of radii r, r l7 r 2 , then ?=*=?>. (3) r r, r 2 This is also true by experiment, for radii drawn on any section remain straight lines after torsion. 16, Let us find the stress in a tube so thin that the intensity of the stress due to torsion will be practically the same over the whole sectional area. If we call this inten- sity of stress q, the mean radius of the tube r, and t the thickness, then the area of any section will be 2 nrt, and the stress on this area will be q.2 nrt, and the moment of this stress about the center of the section will be q. 2 nrt.r. This must balance the moment of the force applied to twist the tube, and, calling this moment T, (D Now putting this volume of q in equation (2) of Art. 15, which is true for either hollow tubes or solid rods, we have IT for the angle of twist of a thin tube. (Values of C at end of Chapter I.) 28 STRENGTH OF MATERIAL. 17, Let us now take a tube whose thickness must be considered. Let r, be its external and r 2 its internal radius (Fig. 14), and let q equal the stress at a distance r from the center of the tube. Call q l the maximum stress which is at the surface, then from equation (3), Art. 15, Fig. 14. r i 7 i The element of area at a distance r from the center is 2 xrdr, the stress on it is q. 2 nrdr, but from above q = > r i so in terms of the maximum stress the stress on the element is .2 xrdr and the moment of this stress is .2 7ir 2 dr or r, r, -^-r 3 dr, which if integrated between the limits of r, which are r 2 and r,, will give us the moment of the resistance offered by the tube to the twisting moment applied, or 4 r t Knowing T this gives us q^ = - r 2 <) If now r 2 = 0; or in other words, if the tube is solid, 2 T 18. It is clear that if we give to q { the limiting value for the material, we can find the diameter of the solid rod that will carry a given twisting moment, This formula is used to TORSION. . 29 design shafts to transmit a given horsepower. Let a be the length of the crank arm, and F the mean force acting on it during a revolution; then the mean twisting moment is equal to aF, and the work done per revolution is equal to aF.2 TT (care must be taken to use the same units through- out). The energy of the machine per revolution is H. P. X 33,000 foot-pounds, where H. P. is the indicated N horsepower and N the number of revolutions per minute. This is also the work done per revolution, and if divided by 2 n will give the mean twisting moment. The mean twist- ing moment is less than the maximum twisting moment, but the shaft must be designed to carry the greatest, which is equal to a constant times the mean, or T = KT m ' We put the greatest moment equal to - substitute the limit- ing stress allowed for q and solve for r. We can get the greatest twisting moment if we take the force acting on the piston, through the piston and connecting rods to the end of the crank, when the crank is perpendicular to the connect- ing rod, and multiply this by the length of the crank, remem- bering that the force acting along the connecting rod is the force acting on the piston multiplied by the secant of the angle the piston rod makes with the connecting rod. 19. To compare solid and hollow shafts, let the radii of the hollow shaft be r^ and r 2 , and that of the solid shaft be r; then if T is the resistance offered by the solid shaft and T l that by the hollow shaft, * q (r < r <) T 2 r ' 2 r 4 r * * 1 _ ^ r \ _ r l ~ "2 xqr 3 r/ 2 which gives us the ratio for shafts of the same material. 30 STRENGTH OF MATERIAL. Now if the sectional areas are the same nr 2 = n (r 2 r 2 2 ) or r = Vrf r 2 2 , substituting Il T which shows that the nearer r, approaches r n or the thinner the metal of the hollow shaft becomes, always retaining the same sectional area as the solid, the nearer the ratio of strength of hollow to strength of solid would approach GO. There is, however, a limit to the thinness of our shaft, as it must be able to support its own weight with- out buckling. If we assume that the ratio of the radii of the hollow shaft is as 2 to 1, and that the sectional areas T are equal, we get = 1.44, or the hollow shaft is nearly half again as strong as the solid one under these conditions. 20. Substitute in equation (2) of Art. 15 the value of q found in Art. 17 for the solid shaft, and we get the angle of twist 2 Tl = for a solid shaft. If we use q found in that article, for the hollow shaft we get 2TI for the hollow shaft. Now if T is known and 6 can be meas- ured, we can find the value of C; or, if C is known and measured, we can find T, and thence the horsepower that a rotating shaft is transmitting. TORSION. 31 Examples : 1. What must be the diameter of a shaft to transmit a twisting moment of 352 ton-ins., the stress allowed being 3 tons per sq. in.? What H. P. would this shaft trans- mit at 108 revolutions per minute, the maximum twisting moment being H of the mean? Solution: 2T 2T 2X352X7X2 q = or r 3 = = - = 64. Trr 3 nq 22 X 7 .'.r = 4 ins.; d = 8 ins. _55 H. P. X 33,000 X 12 ~36 ' N X In X 2240 352 X 2240 X 108 X 2 X 22 X 36 33,000 X 12 X 7 X 55 2. Find the size of a hollow shaft to replace the preced- ing, exterior diameter to be I the interior. What is the weight of metal saved in a steel shaft 60 ft. long? (Cu. ft. steel = 480 Ibs.) Ans. Exterior diameter 8.454 ins. Interior diameter 5.284. Weight saved 3213 Ibs. 3. Find the diameter of a solid shaft to transmit 9000 H. P. at 140 revolutions per minute, the stress allowed being 10,000 Ibs. per sq. in., and the maximum twisting moment I the mean. Ans. d = 14.5695 ins. 4. Find the size of a hollow steel shaft to replace the above, internal diameter being T 9 s of the external. What is the saving in weight for 60 ft. of shafting? Ans. External diameter = 15.091 ins. Internal diameter = 8.4886 ins. Weight saved 8893.5 Ibs, 32 STRENGTH OF MATERIAL. 5. Compare the strength of a solid wrought-iron shaft with a hollow steel shaft of the same external diameter, the in- ternal diameter of the steel shaft being \ the external and the elastic strength of steel that of iron. Ans. As 32 is to 45. 6. A solid shaft fits exactly inside a hollow shaft of equal length. They contain the same amount of material. Com- pare their strengths when used separately. Ans. As 3 is to \/2. 7. The resistance of a twin-screw vessel at 18 knots is 44,000 Ibs. At 95 revolutions per minute what will be the twisting moment on each shaft? What is the H. P.? Ans. T = 360 in.-tons. H. P. = 2432. 8. The pitch of a screw propeller is 14 ft.; the twist- ing moment on the shaft is 120 ton-ins.; the mean dia- meter of the thrust bearing rings is 15 ins.; coefficient of friction is .05; find the thrust and the efficiency of the thrust bearing. Ans. Thrust = 4^ tons. Efficiency of bearing 98t%. 9. The angle of torsion of a shaft is not to exceed 1 for each 10 ft. of length; what must be its diameter for a twisting moment of 16,940 ft.-lbs.? f 221 (7=10,976,000 in.-lb. units. * = I. Ans. Diameter = 6 ins. 10. A solid steel shaft, 10.63 ins. in diameter, is transmit- ting 12,000 H. P. at 200 revolutions per minute. If the maximum twisting moment is I the mean, what is the maxi- mum stress (torsion) in the shaft ? Ans. 20,000 Ibs. per sq. in. TORSION. 33 11. If the modulus of rigidity be 4800 in in.-ton units, what is the greatest stress to which a shaft should be sub- jected that the angle of torsion may not exceed 1 for each 10 diameters of length? Ans. 4.2 tons per sq. in. 12. In changing engines in a ship, the number of revolu- tions is increased J; H. P. is doubled; the ratio of maximum to mean twisting moment is changed from f to ; and the strength of material of the shaft is 25% greater. What is the relative size of the new shaft? Ans. The same size. 13. Find the angle of torsion of a steel tube, 6 ft. long, J in. thick; mean diameter 12 ins., shearing stress allowed 4 tons per sq. in. Ans. 0.0478, CHAPTER IV. BENDING. 21. There is one more way in which we can introduce stress into our rod and that is by bending it. If we rest our rod on supports at its ends and load it at its middle, it will bend into a curve. The plane of this curve is called the plane of bend- ing (OABO, Fig. 15). The particles on the concave side of Fig. 15. the rod will tend to crowd together and will be in compres- sion, while those on the convex side will tend to pull apart and will be in tension; obviously, there will be some intermediate plane, perpendicular to the plane of bending, where there will be neither tension nor compression, a plane of no stress. This plane is called the neutral surface. The neutral line is the intersection of the neutral surface with the plane of bending, and it gives us the curve into which the rod bends, and is therefore called the elastic curve. The intersection of the neutral surface with any section of our rod perpendicular to its axis is called the neutral axis of that section. In Fig. 15, the lines AO and BO meet, being their point of intersec- 34 BENDING. 35 tion. AOB is the plane of bending. CEFD is the neutral surface. A B is the neutral line or elastic curve; and if HH is any section of the rod perpendicular to AB, then GK is its neutral axis. 22. Now before bending, all sections of our rod are parallel, but after bending these same sections (assuming that they remain plane) will be inclined to each other, being nearer together on the concave side, the same distance apart as before bending at the neutral surface, and farther apart at the convex side. If R is the radius of curvature of the neutral line AB (Fig. 16), 6 the angle between the planes of any two sections, and y the distance from the neutral plane to any point in the rod, then the length of the elastic curve between Fig. 16. these two sections is R6, and the length between these two sections, along a line all points of which are at a distance y from the neutral surface is (R y) 0, or Rd yd, so that yd is the total change of length between the two sections at a distance y from the neutral surface. The change of length per unit length is the total change divided by the original length, or the strain is = From Chapter I, Art. 5, Rd R P we have the strain due to tension or compression equal to . E p y Hence =7-, or the stress due to bending at a distance y hi R from the neutral plane will be * E 36 STRENGTH OF MATERIAL. The stress varies then as the distance from the neutral surface, and must not at the outer surface in the plane of bending exceed the elastic limit of the material, or the surface fiber stress must be less than /. 23. We have above our formula for the fiber stress, but do not know what value to use for y, as where the' neutral axis lies is not known. Neither do we know the length of R. We will first prove that the neutral axis of any section of a beam passes through the center of gravity of the section. We know that the stress is maximum compressive at the surface on the concave side; decreases with the distance from the neutral surface, where it is zero; there changes to tension; and in- creases to a maximum at the surface on the convex side. The rod is in equilibrium, a law of which is that the sum of the horizontal components of all the forces acting must equal zero. Now the loads which cause the bending are all vertical, so can have no horizontal components; the only other forces are the horizontal stresses at the section, and their sum must be equal to 17 zero, the stresses on the opposite sides being equal but opposite in direction. Fig. 17 shows the stress on one side only, and the section may be of any shape whatever. E Let AB be the neutral axis. The formula p = y gives us R the stress per unit area on any element of area, dA, which is throughout at a distance y from the neutral axis. The stress E on the element is then pdA, or yd A. If we integrate this R between the limits for our section, we will get the total stress on the section, which we know to be equal to zero, or E r 11 "* 1 ' H = -JydA=0. (1) ** limit BENDING. 37 Now the value of the integral, f yd A, divided by the area of the section will give us the distance of the center of gravity of the section from the line AB, and if this distance were zero the line would pass through the center of gravity of the section. Now equation (1) equals 0; we know E and R have /limit ydA must be equal to limit zero. Hence the neutral axis always passes through the center of gravity of the section. 24. Determination of R. Our rod is in equilibrium, one of the laws of which is that the sum of the moments of all the forces about any axis must equal zero. Being true for any axis we will take the neutral ,w axis for the axis of moments. We will first find the sum of the moments of the external forces. In Fig. 18 let AB be the rod of length /, loaded with W pounds in the middle, so that the supporting forces are W each ; then to the left of the section HH the moment of W W the supporting force is x; and the moment of the forces to the right is Wx T' the same value as before. The moment to the left tends to turn the part of the beam on the left of the section in the direction of motion of the hands of a watch, while that to the right tends to turn the right part of the beam in the opposite direction; therefore, if we call the first direction positive, the other is negative. As in the other cases of equilibrium we will use the values found on one side only of the section. In 38 STRENGTH OF MATERIAL. this case, the moment of the external forces about the neutral axis is the bending moment for that section, and we will Wx designate it by M, so M = -- Considering then the part to the left of this section, the moment of the stress in the sec- tion must balance the moment of the external forces about the neutral axis. Fig. 19 shows enlarged the part of the rod to the left of the section. We see that the moment of the external forces tends to turn this part of the rod in the direction indicated by the arrow marked (1), while the moment of the stress in Fig. 19. the section tends to turn it in the direction of the arrow (2), and for equilibrium these moments must be equal. El As in Art. 23, the stress on the area dA is yd A, and the it Tjl moment is ydA.y, which integrated between the limits for H ft /*imit the sections gives I y 2 dA. The integral f\fd A between R /limit limits is the moment of inertia of the area of the section about the- neutral axis, and we will designate it by /, so that E the moment of the stress in the section is equal to /. This R W V must equal M , so M -= /, or R = I. In Art. 22 we found E p = _, so we now have a general formula for bending, R y PME BENDING. 39 In which p stress at any point at a distance y from the neutral surface; M = bending moment at any section due to exter- nal loads; / = moment of inertia of the area of the section about the neutral axis; R = radius of the arc into which the rod is bent; E = modulus of elasticity of the material of the rod. 25. We have now seen the effect of applying forces to our rod in all the different ways. If the stresses caused have the same line of action, that is, if they act on any section in the same or 'in a diametrically opposite direction, their algebraic sum will give us the total stress on the section. For example, the stress due to several longitudinal loads is the algebraic sum of the loads divided by the area of the section; or, if a rod is under a tensile stress and there is also a fiber stress due to bending, the maximum stress would be at the convex sur- face and equal to the sum of the two stresses, while at the con- cave surface, which would be in compression due to the bending, the stress would be the difference between the two, and would act in the direction of the greater. Another fact must be considered in actual practice : When the temperature changes, a free rod expands or contracts without stress, but if the rod be prevented from expanding or contracting, stress is produced. The method of taking the algebraic sum when the lines of action are the same is called the principle of superposition, which may be stated as follows: The effect due to a combination of forces is equal to the sum of the effects due to each force taken separately. When the stresses caused by our forces have not the same line of action, such as combinations of shearing or torsion with bending, we must arrive at their maximum effects in some other way. In the next chapter we will endeavor to show how to find the maximum stress together with its direc- 40 STRENGTH OF MATERIAL. tion, which is due to the combined effects of two or more stresses which act at right angles. We will then be able to combine the effects of any of the forces which we have applied separately to our rod in these first four chapters. We will find that the maximum stresses due to any of these combina- tions are greater than any of the stresses acting singly, and that they act on planes which are inclined to those on which any of the single stresses act. This will account for the apparently erratic manner in which material sometimes car- ries away; so in deciding if a single part of a machine or structure is strong enough, we must first find to what forces it is subjected, and then if it be able to sustain the total stresses they induce. I Examples : 1. A beam 2 ins. wide by 3 ins. deep is subjected to a bending moment of 72 ton-ins. What is the maximum fiber stress? Solution: p M My 3 y --j P-- y=2- 1 9 72 X 3 X 2 I =i2 Ah =2- ' p - ~2irr =24tons - 2. An iron I-beam (without weight) of 12-ft. span has flanges 4 ins. by 1 in., and web 8 ins. by \ in. * What is the greatest central load it can carry if the stress is limited to 4 tons per sq. in.? Solution : y. M max = = W X 3 ft.-lbs. =W X 36 in.-lbs., _ 552 4_TFX36X3 l=.y- 5. - = = = 5X36X3 BENDING. 41 3. A cast-iron I-beam has a top flange 3 ins. by 1 in. ; bot- tom flange 8 ins. by 2 ins.; web, trapezoidal, J in. thick at top and 1 in. thick at bottom; total depth of beam 16 ins. Find the position of the neutral axis and the ratio of maxi- mum tensile to compressive stresses. Ans. Neutral axis 4.81 ins. from bottom. Ratio T to C is 3 to 7. 4. A wooden beam of rectangular cross-section is 15 ft. long and 10 ins. wide. If the maximum bending moment is 16.5 ton-ft., and the allowed stress is \ ton per sq. in., what is its depth? Ans. 15.4 ins. 5. An I-beam is 25 ft. long, top flange 3 ins. by 2 ins.; bottom flange 10 ins. by 3 ins.; web 12 ins. by 1 in.; total depth 17 ins. If the stress is limited to 4J tons per sq. in., find the greatest central load it can support in addition to its own weight (take weight of beam as 2000 Ibs. acting at its center). Ans. 6.48 tons. 6. What is the radius of the smallest circle into which a rod of iron 2 ins. in diameter may be bent without injury, the stress being limited to 4 tons per sq. in. E = 13,000 ton-ins. Ans. R = 270 ft. 10 ins. 7. A spar, 20 ft. long, is supported at the ends and sus- tains a maximum bending moment of 3147.5 Ib.-ft. If the stress be limited to J ton per sq. in., what is the diameter of the spar? Ans. 7.0025 ins. 8. What is the diameter of the smallest circle into which a ^-in. steel wire may be coiled, keeping the stress within 6 tons per sq. in.? (E for steel wire being 35,840,000 in in.-lb. units.) Ans. 111J ft. 42 STRENGTH OF MATERIAL. 9. A rectangular beam 12 ft. long, 3 ins. wide, 9 ins. deep, is supported at the ends. Stress is limited to 3 tons per sq. in. Find the load which can be carried at the center; also find the load if the beam lies the flat way, i.e., 3 ins. deep and 9 ins. wide. Ans. 1st case 3f tons; 2d case 1J tons. 10. Find the breadth and depth of the rectangular beam of maximum strength which can be sawed from a log 2 ft. in diameter, and compare its resistance to bending with that of the largest square beam that can be sawed from the same log. 2 2\/2 Ans. Top is -ft., depth is =- ft. Resist bending in V3 A/3 ratio of 1.089 to 1. 11. A steel I-beam 30 ft. long; flanges 7 ins. by .8 in.; web 24 ins. by .5 in.; carries 31,900 Ibs. at its center. What is the maximum fiber stress? Ans. 16,000 Ibs. per sq. in. 12. Compare the resistance to bending of a wrought-iron I-beam; flanges 6 ins. by 1 in.; web 8 ins. by J in., when up- right, and when laid on its side. Ans. 4.6 to 1. 13. A round steel rod, 2 ins. in diameter, can only with- stand a bending moment of 6 ton-ins. What is the greatest length of such a rod which will just carry its own weight when supported at the ends? Ans. 28J ft. CHAPTER V. COMBINATION OF STRESSES. 26. In a rod suffering tension and shear let HH (Fig. 20) be any section which has normal stress due to the tensile load, F, and tangential stress due to shear. Let the square shown represent the base of an elementary cube of volume, whose height dz is perpendicular to the plane of the paper, and on whose faces dydz are the stresses P and S as shown. It is H \ >F P J[ dot fr t Fig. H 20. clear that the stresses P balance each other, but it will be noticed that the tangential stresses S form a moment whose effort is to turn the cube to the right. Now the forces are assumed within the elastic limit and we know the cube to be in equilibrium, so there must be an equal and opposite moment tending to turn it to the left. In other words, we must have tangential stresses equal to S on the faces dxdz, acting in a direction such that their moment will turn the cube to the left. These latter stresses are called longitudinal shear, and because of the resistance offered to it a solid beam will bend less, when supported at the ends, than a pile of thin boards of the same volume. We may then concede that the shearing stresses at any point within a body in a state of stress are equal and act in planes at right angles with each other. 43 44 STRENGTH OF MATERIAL. 27. We can now find the plane on which the resultant stress, due to a number of stresses acting in planes at right angles, is a maximum. Fig, 21 is, enlarged, the elementary cube of Fig. 20, with the additional stress P lf acting so as to put it in vertical tension. -& Let AB be any plane making the angle a with the direction *-P of the stress P, and let the sum of the components of all the stresses on one side of this plane, when resolved ____> c w | along and perpendicular to p i it, be N and T as shown. The stresses on the other side of A B would give equal and opposite components to N and T (equilibrium). So that our cube is now in equilibrium under the stresses shown. Let us resolve these stresses horizontally and vertically, then the sum of both the hori- zontal and the vertical components will be zero. For example, the intensity of the stress P on AB is P sin a, remembering that the area of the inclined plane is the area of face dydz X cosec a. (See Art. 14.) Resolving then we have N sin a + T cos a P sin a S cos a. = (1) horizontally. N cos a T sin a P l cos a S sin a = (2) vertically. Eliminating T between (1) and (2) we get N = P sin 2 a + P l cos 2 a + 2 S sin a cos a. (3) Reducing (sin 2 a = 1 cos 2 a a 1 + cos 2 a and 2 sin a cos a = sin 2 a j cos 2 a + S sin 2 a. (4) COMBINATION OF STRESSES. 45 By the same method we get P - Pi T = - - sin 2 a + S cos 2 a. (5) (4) and (5) give us the stresses along and perpendicular to any plane due to the combined stresses P, P l and S. Putting equal to zero the first derivative with regard to N and a or T and a in (4) and (5) will give us the value of a for which N or T is a maximum or minimum. = o = -- - -- 2 sin 2 a + 2 S cos 2 a da 2 gives tan 2 a = JL or a = J tan" 1 A-+ . (6) 7/T7 P p = = - - 2 cos 2 a - 2 S sin 2 a da 2 gives p _ p D _ E) i- . _ tan 2 a = -^-^ or a = i tan' 1 -^-i^+ _ . (7) 28. Equation (6) gives us the angle with P of a plane on which the normal stress due to the combined load is a maxi- mum (the plane 90 from it giving the minimum), and equation (7) the angle with P of a plane on which the tangential stress due to the combined load is a maximum (minimum 90 from it). Obviously, from these equations, the two planes are at 45 from each other (tan 2 a being in one case the negative reciprocal of what it is in the other), or the planes of maximum normal and maximum tangential stresses lie at angles' of 45 with each other. Equation (6) shows the maximum and minimum normal stresses to be at right angles, and if we substitute the value of 2 a found from it, in equation (5) we find that on this plane of maxi- 46 STRENGTH OF MATERIAL. mum normal stress the tangential stress is zero, or the shear is zero when the normal stress is a maximum. The normal stresses found in equation (6) and acting on planes at right angles are called principal stresses and their directions prin- cipal directions. Principal stresses are always at right angles. If we substitute in equations (4) and (5) the values of sin 2 a and cos 2 a, obtained from equations (6) and (7) respectively, we will get the maximum value of the normal and tangential stresses. andT 7 - 2 j v < o T \^r r i) {J (9) -f sigii max. - sign min. i V4S 2 + (P - , Pi) 2 The + sign of (8) gives the maximum tension (or compres- sion), and the sign gives the maximum compression (or tension) on a plane at right angles, i.e., principal stresses. Using the formula of this article we can find the maximum value and its direction of any combination of stresses due to the loads used in the preceding chapters. 29. The Stress Ellipse. Let us assume that at any point within a body there are two normal stresses of intensity, P and P l (Fig. 22), acting at right angles. Then by Art. 14, x, the intensity of the stress P on any plane making an angle a. with the direction of P, is P sin a, and ?/, the intensity of the stress P l on the same plane, is P l cos a, x y - = sin a, and = cos a. Squaring and adding we have 2 H -- - = sin 2 a + cos 2 a = 1. The above is the equation of an ellipse of which the semi- axes are P and P 1; and of which x and y, the coordinates of COMBINATION OF STRESSES. 47 any point (C, Fig. 22), represent respectively the intensity of the stresses P and P l on a plane making the angle a with P. The radius vector, OC, represents on the same scale the amount and direction of the resultant intensity of stress ,' Fig. 22. on the plane AB. The equation of this ellipse in terms of the eccentric angle found from the ratio = , and the equation of the a x a wx load curve (w acting downward) would be L = For a a uniform load of w Ibs. per ft.-run, L = w. The equations for load curves are easily found and are very useful, for let us notice the relation between the loads and the shearing force at any section. As a general formula for concentrated loads, we found the S. F. at any section to be equal to P %W. With a beam loaded with any continuous load, if L be the load per unit length, then Ldx is the load on any elementary length, dx, of the beam, and the total load from the origin to a section at any distance, x, from the origin is (*x I Ldx, the value of L being given by the equation to the load curve. Now the value of this integral is the algebraic 58 STRENGTH OF MATERIAL. sum of all the loads to the left of the section, and that is also our definition of shearing force. As a general rule, then, for any continuous load the shearing force is S. F. = C *Ldx. In using the formula we must remember that the constant of integration is the value at the origin of the quantity, repre- sented by the integral in this case, that is, the value of the supporting force P. For concentrated loads we must still use S. F. = P 2 W, and if a continuously loaded beam supports also a number of concentrated loads we must find the shearing forces sepa- rately, and get the total shearing force by algebraically adding the results (principle of superposition). In this latter case also the supporting forces for both the continuous and con- centrated loads must be found and used separately. Examples: 1. A beam 15 ft. long is supported at the ends and carries 4 tons 5 ft. from the left end, 1 ton 8 ft. from the left end, and J ton 10 ft. from the left end. Find the supporting forces and draw the shearing curve. Ans. P = 3.3 tons. Q = 2.2 tons. 2. A spar 20 ft. long is supported at the ends and loaded with 500 Ibs., 4 ft.; 250 Ibs., 9 ft.; and 900 Ibs., 18 ft. from the left end. Find the supporting forces and draw the shear- ing curve. Ans. P = 627.5 Ibs. Q - 1022.5 Ibs. 3. A plank 16 ft. long is laid across a ditch and a man weighing 192 Ibs. walks across it. Find the shearing curve when he is 3 and when he is 8 ft. from the left end. SHEARING STRESS IN BEAMS. 59 4. A weight of 384 Ibs. is placed 5 ft. from the left end of the above plank. What is the shearing force at a point 6 ft. from the left end when the man is 8 ft. from there? Ans. - 24 Ibs. 5. A beam 18 ft. long has a uniform load of 50 Ibs. per ft.-run. Draw curve of shearing force and give value at points 7 ft. and 13 ft. from the left end. Ans. At 7 ft. 100 Ibs.; at 13 ft. - 200 Ibs. 6. On the beam of example 5, weights of 300 Ibs. and 500 Ibs. are placed 6 ft. and 12 ft. respectively from the left end, in addition to its uniform load. Draw S. F. curve and give value of S. F. 8 ft. from left end. Ans. 116 Ibs. 7. A beam 5 ft. long has its left end fixed in a wall and supports a load of 1000 Ibs. at its free end. What is the S. F.? and draw curve. 8. The beam of example 7 has a distributed load of 100 Ibs. per ft.-run. What is the maximum S. F.? and draw the curve. 9. An oak beam 15 ft. long and 1 ft. square floats in sea water. It is loaded at the center with a weight which will just immerse it wholly. Draw curve of S. F., and give maxi- mum value. (35 cu. ft. of sea water weighs 1 ton; 1 cu. ft. of oak weighs 48 Ibs.) Ans. Maximum S. F. = 120, Ibs. 10. A pine beam 20 ft. long and 1 ft. square floats in sea water, and is loaded at the middle with a weight which will just immerse it. Draw a curve of S. F. and give value 5 ft. from left end. (Cu. ft. pine weighs 39 Ibs.) Ans. S. F. 5 ft. from left end = 125 Ibs. 60 STRENGTH OF MATERIAL. 11. A beam 20 ft. long, supported at the ends, carries a load which uniformly increases from at the left end to 50 Ibs. per ft.-run at the right. Draw the curve of S. F., and find its maximum value, also find the point of the beam where its value is zero. Ans. Maximum value = 333J Ibs. Value zero where x = 11.5+ ft. 12. The buoyancy of an object floating in the water is at the ends and increases uniformly to the center, while its weight is at the center and increases uniformly to the end. Draw curve of S. F., and give maximum value. Ans. Maximum S. F. = 4 CHAPTER VII. CURVES OF BENDING MOMENTS AND SHEARING FORCE. 36. In Chapter II it has been shown that when a beam is loaded there are horizontal stresses set up on any trans- verse section, these stresses being compressive on one side and tensile on the other side of the neutral plane. Art. 14 of the same chapter shows that the moment about the neu- tral axis of the external forces on one side of any section must be equal to the moment about the neutral axis of the stresses on the same side of the section. Definition: The bending moment at any section is the moment about the neutral axis of that section of all the external forces on one. side of that section. By definition then the bending moment at the section HH of a beam loaded as in Fig. 29 would be M = Px - W l (x - x,) - W 2 (x - x 2 ), or finding the line of action of the resultant of all the forces W^ W 2 , etc., and calling its distance from the section d M =Px - 2 which is true for any kind of load. 61 62 STRENGTH OF MATERIAL. If we now consider the increment AM which the bending moment receives if we take our section a distance Ax to the right, the bending moment about the neutral axis of the new section will be M + AM = P (x + Ax) - STP (d + Ax) = Px - ZWd + Ax (P - STT), but Px 2Wd is the original bending moment equal to M, and (P 2 FT) is our formula for the shearing force (F) at any section, so M + AM = M + FAx, or AM = FAx, and passing to the limit dM = F dx; integrating, we have for a general formula for bending moment M =/F dx. This equation is true for any kind of load, but care must be taken to use the correct constant of integration, which is the value of M at the origin. This for beams free at the origin is zero, but if a beam is fixed (prevented from moving in any way) at the origin, there is a bending moment there. We will devote the rest of this chapter to some examples of shearing force and bending moments in beams loaded and supported ia different ways. 37. A beam supported at the. ends is loaded as shown in Fig. 30. Find the curves of shearing force and bending moment. By definition the S. F. = P SPF, by which we get the curve of S. F. to be ABCDEF (Fig. 30, a). Consid- ering each load separately, the supporting forces for the 25-lb. load are P = 17.5, Q = 7.5, and the S. F. curve is A BCD (Fig. 30, 6). The 35-lb. load gives supporting forces p = y ; Q = 28, and the S. F. curve is ABCD (Fig. 30, c). If we add algebraically the ordinates given by these curves at any point distant x from the origin we will get the ordi- nates for the S. F. curve for that point given in Fig. 30, a. 251bs 351bs (63) 64 STRENGTH OF MATERIAL. Obtaining the ordinates by definition, the curve of B. M. for the beam with these loads is given by OGHX in Fig. 30, a. Let us get separately the curves of B. M. due to each load. For the 25-lb. load we get by definition the curve in Fig. 30, d, the maximum B. M. being directly under the load and the curve being positive at all points. For the 35-lb. load we get the curve in Fig. '30, e, this curve also being positive at all points. If we add together the ordinates given by these curves at any point distant x from the origin we will get the ordinate of the curve of B. M. for that point as given in Fig. 30, a. Now by definition, the bending moment at the middle of the beam, for example, is 24.5 X 5 - 25 X 2 = 72.5 Ib.-ft. and by the formula fF dx it is - .5 X 5 = - 2.5, which is evidently incorrect, but is due to the fact that having taken the two concentrated loads together, the shearing force of one being positive and the other negative, we have not multiplied the total shearing force by x. The moments of the two shearing forces cause bending in the same direc- tion, therefore, if we neglect the negative sign (which we have assumed to indicate direction only) the shearing force at this point will be 7 + 7.5 = 14.5, which multiplied by 5 gives 72.5 as before. With many loads to get the shearing force separately would be a tedious operation, so it is better to make it a rule to get the S. F. and B. M. for concentrated loads from the definition. We will find no trouble with beams having continuous loads unless there are concentrated loads in addition, in which case we get the curves due to the continuous loads by formula, and those due to the concen- trated loads by definition, and apply the principle of super- position. 38. A beam 10 ft. long supported at the ends is loaded with a uniform load of 25 Ibs. per ft.-run. Find the curves CURVES OF BENDING MOMENTS. 65 of S. F. and B. M. In this case, a continuous load, our for- mula is very convenient. L = - w = - 25 Ibs. F = /Ldx = - 25 x + C. C being the S. F. at the origin is equal to P = 125 Ibs. .-. F = 125 - 25 x. (1) Fig. 31. (1) is the equation to the curve of S. F. This being an equation to the first degree is a straight line and plots as the line AB of Fig. 31. The bending moment is M = /Fdx =/(125 - 25 x) dx, or, M -= 125 x 25 x 2 [(7=0, (2) C is zero because B. M. is zero at the origin. (2) is the equation of the curve of B. M., and being of the second degree is a conic (parabola) and plots as OCX of Fig. 31. The maximum B. M. is at the middle of the beam where the S. F. is zero. To get values for either S. F, or 66 STRENGTH OF MATERIAL. B. M. at any section of the beam, substitute for x the dis- tance of the section from the origin in equation (1) or (2) respectively. 39. A beam 5 ft. long, fixed at the left end, with the right end unsupported, carries a weight of 50 Ibs. on the right end. Find the curves of S. F. and B. M. In this case the left end supports the whole load, so P = 50, and the shearing force by definition (concentrated loading) is F = 50 Ibs., the curve being the straight line A B of Fig. 32. The bend- ing moment by definition is M = Px, which being an equa- tion of the first degree is a straight line, the ordinates varying from zero to 250 Ib.-ft. Q 2 This would give us a line in- clined upward from 0, but we know the greatest B. M. is at the origin. Now notice that the curvature of this beam is just the opposite of that of a beam supported at the ends, the center of curvature being below the beam in this case, while it is above a loaded beam supported at the ends. In fact if we turn Fig. 32 "upside down" we will have just one-half of the beam supported at the ends and loaded with 2 P in the middle, W being one of the supporting forces; so this kind of bending is called neg- ative, and instead of the curve of B. M. inclining upward from 0, it inclines downward from X as in the figure. The curve plots directly if we take the origin at X and move the section to the left, for this arrangement is the same as a beam twice as long supported in the middle and loaded with 50 Ibs. at both ends. 40. A beam 5 ft. long fixed at the left end, with the right end unsupported, carries a uniform load of 25 Ibs. per ft.-run. Find the curves of S. F. and B. M, CURVES OF BENDING MOMENTS. 67 A continuous load; so we will use the formula. L = - w = -25 Ibs. F = /Ldx = - 25 x + C. C being the S. F. at the origin is equal to P, =125 Ibs. .-. F = 125 - 25 x. (1) Fig. 33. (1) being an equation of the first degree is a straight line and plots as AX in Fig. 33. The bending moment is M = /Fdx =/(125 -25 x) dx = 125 x - 25 x- C r Here C t is not zero. But we know there is no bending moment at the right end of the beam, so we substitute M = and z=5, and solve for C 1; = 125 X 5 - + C v from which C l = 312.5; substituting this value of C l we get ^jti *U (2) 25 x 2 M = l25x 312.5 for the equation of the curve of B. M., which being of the second degree is a conic (parabola) and plots as XC in Fig. 33. This equation gives a maximum negative value of B. M. at the origin, where we know it should be. 68 STRENGTH OF MATERIAL. 41. A beam 10 ft. long, supported at the ends, is loaded with a uniform load of 25 Ibs. per ft. -run and a concentrated load of 100 Ibs. 4 ft. from the left end. Find the curves of S. F. and B. M. Fig. 34. For the uniform load (Art. 38) the curve of S. F. is AB (Fig. 34). For B. M. the curve is 01 X. For the concentrated load the S. F. curve is CDEF, and the curve of B. M. is OJX. Adding algebraically the respective ordinates, we get the S. F. curve for both loads to be GHMN, and for B. M. the curve of both loads is OLX. 42. A beam 10 ft. long is supported at the ends and has a load uniformly increasing from zero at the left end to 100 Ibs. per ft.-run at the right end. Find curves of S. F. and B. M. Here (see Art. 35) L = - 100 x 10 - 10 x, 10 S. F. =/Ldx =/-lOxdx = + C, CURVES OF BENDING MOMENTS. 69 where C being the S. F. at the origin is equal to P. We can find P by the method used in the example at the end of Art. 32, as follows: In Fig. 35 OL is the load curve. The Fig. 35. center of gravity of the load then acts at the length of the beam from 0, and the total load is 100 X - = 500 Ibs.; fit therefore, taking moments about 0, 500 X . 10 - Q X 10 = 0; and Q = 333J, the other supporting force being 500 - 333 J; or P = 166. But we can get P in another way, for S. F. = P - 5 x 2 and B.M.= (Fdx= ( (P - 5x 2 )dx=Px J J 5x 5 C l is equal to zero, for the bending moment at the origin is zero, and M is also zero at the right end of the beam, so that if we substitute x = 10 and solve the equation 5 X 1000 P X 10 -- - - = 70 STRENGTH OF MATERIAL. we get P = 166 as before. Putting this value of P in the equation for S. F. and B. M. we get S. F. = 166 - 5 x 2 (1) and B. M. = 166 x - , (2) 3 which equations give the curves ACB for shearing (Fig. 35) and ODX for bending. If we put (1) equal to zero and solve for x we get the point on the beam where the shearing force is zero; and if we put this value of x in (2) we will get the maximum B. M., for the maximum B. M. occurs where the S. F. is zero (see Art. 28), for by the principles of maxima and minima, the first derivatives of the B. M. with respect to x, put equal to zero, will give us the shearing equation (1) equal to zero, d (BM) I 500 10 _b ' = 166 - 5z 2 =0; from which x=\/- = , dx V 3 x 5 V3 the value of x where B. M. is a maximum. Examples : 1. A beam 10 ft. long, supported at the ends, carries a load of 1000 Ibs. 4 ft. from the left end. Find curves of S. F. and B. M. 2. A beam 5 ft. long, fixed at the left end and unsupported at the right end, carries 1000 Ibs. at the right end. Find curves of S. F. and B. M. 3. A beam 10 ft. long, supported at the ends, carries a uni- form load of 100 Ibs. per ft.-run. Find curves of S. F. and B. M.; give values at ends and center. Ans. S. F., ends 500; center 0. B. M., ends 0; center 1250 Ib.-ft. CURVES OF BENDING MOMENTS. 71 4. A beam 5 ft. long, fixed at the left end, unsupported at the right end, carries a uniform load of 100 Ibs. per ft.-run. Find curves of S. F. and B. M. and give values at ends. Ans. S. F., left end 500 Ibs.; right end 0. B. M., left end - 1250 lb.-ft.; right end 0. 5. A beam 10 ft. long, supported at the ends, carries 400 Ibs. 4 ft. from the left end, and 600 Ibs. 6 ft. from the left end. Find curves of S. F. and B. M. and give values at the ends and center. Ans. S. F., left end 480 Ibs.; right end 520; center 80. B. M., left end Ibs.; right end 0; center 2000 lb.-ft. 6. What is the longest steel bar of cross-section of 1 sq. in. that can be supported at its center without being perma- nently bent, the greatest allowable bending moment for the bar being 2000 Ib.-ins.? Ans. 19 ft. 7. A beam 20 ft. long, supported at the ends, carries 2000 Ibs. 5 ft. from the left end, and 5000 Ibs. 4 ft. from the right end. Find curves of S. F. and B. M. and give values at the ends and at the loads. Ans. S. F., left end 2500 Ibs.; right end 4500 Ibs.; be- tween loads 500 Ibs. B. M., end 0; first load 12,500 lb.-ft.; second load 18,000 lb.-ft. 8. If the beam of example 7 were loaded with 200 Ibs. per ft.-run, find the curves of S. F. and B. M. and give values at ends and center. Ans. S. F., ends 2000 Ibs.; center 0. B. M., ends 0; center 10,000 lb.-ft. 9. A round steel rod of 2 ins. diameter can only withstand a bending moment of 6 ton-ins. What is the greatest length of such a rod which will just carry its own weight when sup- ported at the ends? Ans. 29.2 ft. 72 STRENGTH OF MATERIAL. 10. A beam 20 ft. long, supported at the ends, carries a uniformly distributed load of 5 tons, and a concentrated load of 5 tons, 4 ft. from the left end. Find curves of S. F. and B. M. and give values at the center. Where is the greatest bending moment? and give its value. Ans. S. F., center - 1 ton. B. M., center 22^ ton-ft. B. M., maximum 24^ ton-ft., 6 ft. from left end. 11. A pine beam 20 ft. long and 1 ft. square floats in sea water. It is loaded at the center with a weight just sufficient to immerse it wholly. Find curves of S. F. and B. M. and give maximum values. A cu. ft. of pine weighs 39 Ibs., of sea water 64 Ibs. Ans. Maximum S. F., 250 Ibs. Maximum B. M., 1250 Ib.-ft. 12. A beam 54 ft. long, supported at the ends, is loaded with 15 cwt. per ft.-run, for a distance of 36 ft. from the left end. Find the curves of S. F. and B. M. What is the maxi- mum B. M. and the B. M. at 6, 12, and 36 ft. from the left end? Ans. B. M. maximum = 216 ton-ft.; B. M. 6 = 94.5 ton- ft.; B. M. 12 = 162 ton-ft.; B. M. 36 = 162 ton-ft. 13. A steel beam 5 ft. long is fixed at one end, unsup- ported at the other end, and is of rectangular section 2 ins. wide and 3 ins. deep. What weight at the free end will de- stroy the beam if the limiting stress is 24 tons per sq. in.? Ans. 1.2 tons. 14. The buoyancy of a floating object is at the ends, and increases uniformly to the center, while the weight is at the center and increases uniformly to the ends. Find the curves of S. F. and B. M. and give maximum values in terms of the displacement, Z>, and the length, /, of the object. D ID Ans. S. F. maximum = ; B. M. maximum = - CHAPTER VIII. SLOPE AND DEFLECTION. 43. The slope at any section of a loaded beam is the angle between the tangent to the neutral line at that section and the straight line with which the neutral line would coincide if the beam were not bent; or, slope is the angle between our axis of X and the tangent at any section to the curve into which the beam is bent. The deflection at any section of a loaded beam is the dis- tance from the axis of X to the point where the neutral line pierces that section; or, it is the ordinate at that section of the neutral line (see Fig. 36). From Chapter IV we have the general formula for bending, M E ^ El = , from which R = > E being a constant and / also, in this case, as we are consid- ering beams of uniform cross-section; M, of course, varies at different sections of the beam. By calculus the formula for the radius of curvature at any point of a curve given by its rectangular equation is dx 2 Equating these two values of R we have TM n M d?y dx* 73 74 STRENGTH OF MATERIAL. which by integration will give us the equation to the curve into which the beam is bent. As this integration would be somewhat complicated, and as in properly built structures the dimensions of the different pieces of material are such that the bending is very slight, we can without appreciable error dy simplify the operation considerably as follows: being the ax tangent of the angle that the bent beam at any point makes with the axis of X, is a very small fraction, and being squared in equation (1) becomes so small that it may be neg- lected, in comparison with unity, so that equation (1) for all practical purposes becomes El I . d*y ---- ; or, El = M; M d 2 y dx z dx 2 and integrating this equation we get EI=/Mdx, (2) dx which will give us the tangent of the slope at any point when we substitute for M its value in terms of x as found in Chap- ter VII, and integrate. The integral f Mdx is called the slope function, and we will designate it by S. When S is divided by El we have the tangent of the angle of slope. If we write equation (2) and integrate again we get Ely =/Sdx. (3) This is the equation of the curve into which the beam is bent, and the value of y given by this equation is the ordinate of the neutral line at any section distant x from the origin. y is usually negative. SLOPE AND DEFLECTION. 75 44. In integrating to get values for equations (2) and (3) we must not forget the constants of integration. If a beam is fixed at the origin, the slope there is zero, but a beam supported at the ends and loaded will bend into a curve like that of Fig. 36. Clearly the slope is greatest at the ends and there will be a constant of integration for equa- tion (2). We will usually know from the way in which the beam is loaded a value of x at which the slope is zero. o H #- X * ^ -^ X -4 ' ' ~ "-^l^LL. ~ ~^^~-^~^ f Q Fig. 36. For example, with a uniform load the beam of Fig. 36 would bend so that at its middle the slope would be zero; therefore, if we substitute (having integrated the right mem- ber) for x, in equation (2), half the length of the beam and put the equation equal to zero, we can solve for C, the con- stant. If we do not know a value of x where the slope is zero, we will know a point where the deflection is zero (one of the supports for example), and substituting in equa- tion (3) the value of x for this point, solve for the constant of integration of equation (2). An example will be solved illustrating the method. There is usually no difficulty in finding the constant of integration for equation (3), because the deflection of a beam at the origin, whether the beam be fixed or supported there and no matter how loaded, is zero. Of course a beam could be propped up somewhere along its length, so that the left end would not rest on its support. 76 STRENGTH OF MATERIAL. 45. A beam 10 ft. long, supported at the ends, is loaded with a uniform load of 25 Ibs. per ft. -run. Find the slope and deflection. Here the bending is perfectly symmetrical. Let Fig. 36 represent the bent beam, being the origin and OX the position of the neutral line before the load was applied, the dotted line representing its position after the application of the load. p = Q = 125. L = - 25, F = - 25x + C = -25.T + 125. [C = P], oe ~2 M = - - + 125 x+ [C = 0, dx 6 2 Knowing the bending is symmetrical, - 1 - = 0, where x = 5 dx (the middle), so In case we do not know the bending is symmetrical, we carry C l down through the next integration for deflection, which gives Elv = ~ 2 ^f + nr + CtX + [C ' = - (2) C 2 is zero, for the deflection is zero at the origin. The deflec- tion is also zero where x = 10 (the other support), and substituting this value we get 25 X 10000 125 X 1000 o = ---- r ~ + - - - +ioc,, from which C l = - 1041 as before, and equation (1) becomes *?! + !?. -10411 (3) SLOPE AND DEFLECTION. 77 from which by substituting the abscissa of any section of the beam for x and dividing the result by El we get the tangent of the angle of slope for that section. Substitut- ing the value of C l in equation (2) gives (4) from which the ordinate of the neutral line (curve of bend- ing) at any section may be found. In finding the value of this ordinate, or of the slope, care must be taken to use the same linits throughout; for example, if we have a steel beam and E is given as 30,000,000, it is in pounds per square inch; we must therefore use x in inches, find / for the section in inch units, and reduce the bending moment to inch-pounds; this will give us the deflection in inches. It is convenient to solve this problem by using letters to represent the numerical data and to make the substitutions later in the equation which gives the desired result; for example, in the above problem let the load per foot-run equal "w" and the length of the beam "a," then L = w, wa F = -wx + C = -wx + P = -wx-l -- , 2i wx 2 wax M = - ++[C l =0, dy wx 3 wax 2 /a\ 3 ( ) \2J a wa . ^ . wa { wx wax wax 78 STRENGTH OF MATERIAL. By this latter method, the particular result desired may be obtained without doing the numerical work required for the other equations. In the above beam the maximum 7/Y7 slope is at the ends, and the tangent of the angle is -- The maximum deflection occurs at the middle of the beam, 5 iva 4 and there j, = - 46. A beam, supported at the ends, carries a single con-.. centrated load W at a distance a from the left end and b from the right end. Find equation for slope and deflection. a+b a+b Fig. 37. In this problem the beam will be bent as shown in Fig. 37, and the elastic curve will consist of two branches, the part from to the weight and that from the weight to the other end of the beam. The shearing force on each part will be constant, but the values will be different and they will have different signs. W T e must consider the two parts sepa- rately. With the origin at 0, the bending moment for any section HH to the left of the load W is Wbx M = - -, a + b Wbx 2 dy _ dx 2 (a + 6) Ely Wbx s 6 (a + 6) [C 2 = 0. SLOPE AND DEFLECTION. 79 For the part to the right of the load W we will take our origin at the right end, then, letting ;r, be the distance from the right end to any section H'H' ', the bending moment will be M- Qx ~ X ~ the sign being negative because the moment tends to turn the right end of the beam counter clockwise, or in the oppo- site direction to that of the other end; we have then for the right end * *&, a + b d,i Wax* dx 2 (a + b) If we move the origin for these equations back to the left end of the beam, by substituting {x (a + b)} for x l we get Wa (a + 6 - x) M = a + b dy Wa dx 2 (a + b} Now for the section under W the slope and deflection are the same, using either the equations for the part of the curve to the left of W, or those for the part to the right of W, so we 80 STRENGTH OF MATERIAL. can put the corresponding values equal to each other when x = a Wba- 2 (a + 6) Wba 3 6 (a + 6) Wab 2 2 (a + 6) (1) 6 (a + 6) or, C> + C/6 = (6 - a) (2) From the simultaneous equations (1) and (2) we get Wab (a + 2 6) , _ Tf afc (2 a + 6) 1- 6 (a + 6) / 8 ' " 6 (a + 6) and substituting these values of C, and (?/ we g e * for the right end, ;/ <% == _TFa(a+6-a;) 3 for the left end, Wbx* Wab (a + 2 6) 2 (a + 6) 6 (a + 6) EIy = 6 (a + 6) 6 (a + 6) dx Ely 2 (a + 6) 6 (a + 6) Wa (a + b - xY 6 (a + 6) TFa6(2a + b)(a + b - x] 6 (a + 6) and from either set of these equations we get the deflection at the load to be 3 El (a + 6) Since the load is not at the middle of the beam, the maxi- mum deflection will occur in the longer segment, and at the section of maximum deflection the slope will be zero; there- SLOPE AND DEFLECTION. 81 fore, supposing a to be greater than b if we put the equation of the slope for the part of the curve to the left of W equal to zero and solve for x we will get the abscissa of the section of maximum deflection; and substituting this value of x in the equation for the deflection will give us the maximum deflection, as follows: Wbx 2 Wab (a + 2 b) a (a + 2 6) and this value of x substituted in Wbx 3 Wab (a + 2 6) x Ely = --- 6 (a + b) 6 (a + b) shows the maximum deflection to be Wb /a (a + 2 ymax ~ ~ ' 3 El (a + 6) or, if a = 6, Wa 3 /a (a + 2 b)\ ( 3 / Umax 6 El Examples: 1. A steel beam, 10 ft. long, supports a concentrated load of 25 tons at its middle. What is the deflection under the load if / in inch-units is 84.9 and E= 30,000,000? Find equation of elastic curve. Ans. .077 ins. 2. A steel beam of / section, 20 ft. long, 8 ins. deep, 3 ins. wide, with flanges and web each J in. thick, is used to support a floor weighing 200 Ibs. per sq. ft. The beams are supported at the ends and spaced 3 ft. apart. What is the maximum deflection? E = 30,000,000. Find equation of elastic curve. Ans. 1.3 ins. 82 STRENGTH OF MATERIAL. 3. A beam, supported at the ends, carries a uniform load of 150 Ibs. per ft. -run. It is 10 ft. long, 6 ins. deep, and 4 ins. wide. E = 1,200,000. Find equation of elastic curve and the maximum deflection. Ans. .39 ins. 4. A beam, 12 ft. long, 8 in. deep, and 12 in. wide, is sup- ported at the ends and carries a load of 1200 Ibs., 3 ft. from the left end. E = 1,200,000. Find the elastic curve and the maximum deflection. Ans. .102 + ins. 5. A steel beam, 60 ft. long and weighing 100 Ibs. per ft., carries a concentrated load of 14,400 Ibs. at the middle. If / in inch-units is 1160, find the maximum deflection. E = 30,000,000. Ans. 4.055 ins. 6. Beams, 20 ft. long, support a floor weighing 100 Ibs. per sq. ft. The beams are spaced 4 ft. apart. Supposing the section of the beam to be square, find the side of the square in order that the deflection may not exceed \ in. E = 700 in.-tons. Ans. Side is 12.18 ins. CHAPTER IX. SLOPE AND DEFLECTION Continued. 47, Example: A beam of length a is fixed at the left end and unsupported at the right. It is loaded with a uni- form load of w Ibs. per ft.-run. Find the equation of the elastic curve. L = w, F = wx + C = wx + wa (P = whole load), M = + wax + [C l = (C l is found in Art. 40), dii wx 3 wax 2 wa' 2 x El dx=--& + ^r -2- + [ c > = (beam fixed at origin), wx 4 wax 3 wa 2 x 2 E/J/= ~2? + -6" ^+^=0 (beam fixed at origin). The maximum deflection obviously occurs where x = a. wa 4 I/max = 8 El Let us suppose the free end of the above beam is propped up to the same level as the fixed end. The prop now sup- ports some of the load, so we do not know P. L = w, F = - wx + C = - wx + P, M = - ^f + Px + C t . a 83 84 STRENGTH OF MATERIAL. We know the bending moment at the end supported by the prop is zero, so putting M = and substituting x = a, from which , >--- Pa. Substituting this value of C l we get M = wx* f Px H wa* h - Pa, 2 2 m *a wx 3 D,y,2 w;a 2 a: Pax - J dx 6 2 2 wx* Px 3 wa 2 2 i Pax 2 , . . . . (beam fixed at origin), (beam fixed at origin). We can now find P, for the deflection is zero where x = a, and all the other constants are determined. wa* Pa : load Fig. 38. from which P = 5 wa and substituting this value in the equations will give us the curves desired and the slope and deflection. The curves are shown plotted in Fig. 38, AB being the curve of S. F., and CDX that of bending moment. The B. M. at the origin is wa negative and equal to . The greatest positive B. M. SLOPE AND DEFLECTION. 85 occurs where the shearing force is zero at x = , and substi- tuting this value in the equation for B. M. gives the greatest positive B. M. equal to . The maximum B. M. is there- 128 fore the negative one, and the beam will break, if overloaded, at the fixed end. It will be noticed that the B. M. is zero at the free end, and also at a point between there and the fixed end. Putting the equation for B. M. equal to zero and solving for x gives a a x = a and x = - . The point where x = - is called a virtual 4 4 joint, because, if a beam loaded and supported in this way had a hinge at this point the bending moment would not cause it to turn. Putting = and solving for x gives CttC x = .58 a and a value greater than the length -of the beam. Where x = .58 a the slope is zero, and this is the point of maximum deflection. Substituting this value of x in the equation for deflection gives .006 wa 4 Compare the curve of B. M. for this beam with that in Fig. 33. As soon as the prop under the free end of this beam begins to bear a part of the load there exists a positive B. M., and the virtual joint, which is then near the right end of the beam, moves toward the fixed end, as the right end is propped up and the load on the prop increases; but the results due to change in conditions are readily followed. 48. A beam a ft. long is fixed at both ends and carries W a concentrated load W at its middle. Here P= Q= 86 STRENGTH OF MATERIAL. W .*. S. F. = and is constant from the ends to the load, but has opposite directions on opposite sides of the load; there- fore, as in Art. 44, we must consider the two parts of the beam separately. To the left of the load, then, M (C is not known), (beam fixed at end). dx 4 The loading being symmetrical, the slope at the middle is a zero ; substituting x = i 2 The equation of the curve of bending to the left of the load, then ' is Wx Wa /\ B\ / y. H ; K\ p / \ Q ' \ "fe Fig. 39. This is an equation of the first degree, and therefore repre- sents a straight line (EF in the figure). We know the B. M. is symmetrical, and that the curve for the right end will have equal ordinates, but of opposite sign (the curve being FG in SLOPE AND DEFLECTION. 87 the figure). Now it is clear we cannot embrace both of these straight lines in one equation of the first degree, and for this reason we must consider the two parts separately. Before proceeding, we will notice that the bending moment has the same value with different signs at the ends and in the middle. This shows that the beam is as likely to break at the ends as in the middle if overloaded, and also that the virtual joints are at quarter span. Wx Wa The maximum deflection is at the middle and equal to Wa 3 ' /max ~ IQ2YI' 49. The Cantilever Bridge. In the cantilever bridge the joints are placed at the points where the bending moment would be zero if the bridge were continuous over the whole span. We will consider a beam fixed at both ends, having two fixed joints and carrying a uniform load of w Ibs. per ft.-run. There being joints at F and G (Fig. 40) we must consider the part FG as a separate beam supported at the ends, the load on it being equally divided and supported at F and G by the beams OF and GX. From the conditions we could assume P and Q with this loading to be equal to w (a + 6 + c) - j but we can get their values in another way: L = w, F = - wx + C = - wx + P, 88 STRENGTH OF MATERIAL. The bending moment at the joint is zero, so 0= w (a + 6) 2 and from equations (1) and (2) we get +C 19 = -(2a + 6),andC 1 -- -- ^ substituting, w = - wx + -(2 a + 6), ,4 (2) The joints being where B. M. is zero, the curves of S. F. and B. M. will be continuous as shown in Fig. 40, but for slope and deflection we must consider each part as a separate beam, Fig. 40. the parts at ends having in addition to their uniform load a concentrated load at their ends, F and G, equal to half the total load on the middle part. For equal strength, the bend- SLOPE AND DEFLECTION. 89 ing moments at the ends and middle of the above beam should have equal values, and this requires a to be equal to c and wa wb 2 (0 + 6) = whence = ?r; or, 4 a = 2 6 2 ; or, (2 a or the square of the whole span must equal 2 6 2 , and the joint span must be distant from the middle of the beam =. - 2V2 . 50. Traveling Loads. If a load W moves from left to right across a beam of length a the positive shearing force = (-*> Fig. 41. for any position of W is equal to P (Fig. 41). P is ob- viously greatest when W is just over it, and as W moves across the beam P's value drops uniformly until when W is just over Q, P's value is zero. We can then represent the change of the positive S. F. by the line AX, and by the same reason- ing the change in the negative S. F. would be represented by OB. Considering the B. M. we know that for any posi- tion of W the B. M. is greatest for the section under W and W for that section is M = Px (a x) x. By the principles of maxima and minima we can find the value of x for which 90 . STRENGTH OF MATERIAL. M is a maximum by putting the first derivative of M with respect to x equal to zero and solving for x, as dM W = (a - 2 z) = 0, dx a which gives the value of x, or the position of W which gives the maximum B. M. If the traveling load is continuous, such as a train of cars crossing a bridge, the maximum value occurs when, if the train is long enough, it extends completely over the bridge; if it is not long enough for this the maximum positive S. F. occurs when the whole train has just got on to the bridge and the maximum B. M. when the middle of the train is just over the middle of the bridge. 51. Oblique Loading. Loading is said to be oblique when a principal axis of any cross-section of a beam does not lie in the plane of the external bending moment. Let Fig. 42 represent the section of a beam and let the arrow-head, marked M, represent the direction of the B. M. The axis of X is perpen- dicular to the plane of the paper through 0, and the axes of Y and Z Fi 42 are shown. Resolve M into compon- ents parallel to the axes of Y and Z. The direction of the fiber stress due to bending is perpen- dicular to the plane of the paper, and that part of it, due to M sin a, which acts normal to the side AD of the section is M sin a : z p.-- -- -> SLOPE AND DEFLECTION. 91 which is obtained by substituting M sin a, the component moment, in - = , the general equation for bending, I y y l being the moment of inertia of the section about the axis of Y, and z the distance of the line AD from the axis of Y. The normal fiber stress perpendicular to the plane of the paper on AB is M cog a Applying the principle of superposition the total fiber stress at A would be M sin a . z M cos a . y p = PV+ PZ= j -j- 52. The work done in bending a beam is evidently equal to M6, where M is a uniform bending moment and 6 is the angle of slope at the ends. If I is the length of the beam and R the radius of the curve into which it is bent, then 6 = , 2 R ME El and from = we have R = - Hence the work done I R M MH by a uniform bending moment is equal to But bend- 2 El ing moments are seldom uniform, so for a variable bending moment i Work= /Af'fa. Examples: 1. A dam is supported by a row of uprights fixed at their base and their upper ends held vertical by struts sloping at 45. Water pressure varies as the depth, or L = wx. Find the equation for deflection. Length of upright a. What is the thrust on the struts? Ans, Thrust = f horizontal pressure of .water, 92 STRENGTH OF MATERIAL. 2. A railroad is inclined at 30' to the horizontal. The stringers are 10.5 ft. apart and the rails are 1 ft. inside the stringers. The ties are 8 in. deep and 6 in. wide. The load transmitted by each rail to one tie is 10 tons. What is the maximum normal stress in each tie? Ans. 6428.8 Ib.-ins. 3. A beam, 2. a ft. long, fixed at the ends, is uniformly loaded with w Ibs. per ft.-run. Find the maximum deflec- tion and the virtual joints. 1/Jfl ^ Ans. Maximum deflection = - - Virtual joints /- 24 El where x = a (1 v J). 4. A beam, a ft. long and fixed at the ends, carries a uniformly increasing load from zero at the left end to w Ibs. per ft.-run at the right. Find the maximum deflection and the virtual joints. A -if 1/7 ^ Ans. Maximum deflection = Virtual joints a 2 a 3125 El where x = - and . 5. A beam, a ft. long, fixed at one end and the other end propped up to the same level, carries a uniformly decreasing load from w Ibs. per ft.-run at the fixed end to zero at the propped end. Find the equation for deflection, the position of the maximum positive B. M., and the position of the virtual joints. Ans. Maximum positive B. M. where x = a \/$. Vir- tual joints where x a >/|. 6. A beam, a ft. long, fixed at both ends, carries a single concentrated load W at a distance d from the left end. Find the deflection under the load. Wd 3 (a - d) 3 Ans. 3 = 3 Ela* SLOPE AND DEFLECTION. 93 7. If the load of the beam in example 6 had been at the middle of the beam, what would have been the maximum deflection? Wa 3 Ans. d = 192 El 8. A single load of 50 tons crosses a bridge of 100-ft. span. Draw the curves of maximum S. F. and B. M., and give the values of those quantities at half and quarter span. 9. A train, weighing 1 ton per ft. -run and 112 ft. long, crosses a bridge of 100-ft. span. Draw curves of maximum S. F. and B. M., and give values at half and quarter span. 10. A steel shaft carries a load equal to k times its own weight, first uniformly distributed, second concentrated at its middle; considering it as a beam fixed at the ends, find the distance apart of the bearings that the ratio of deflec- tion to span may be Ans. (1) Span in feet= 10.5 3 / y * /C ~T~ 1 3 /~J2 (2) Span = 8.3 \/ (d in inches). K + T? CHAPTER X. CONTINUOUS BEAMS. 53. A continuous beam is one which extends over several supports. In this chapter we will consider as heretofore only such beams as have a uniform cross-section and, as in all practical constructions the supports are adjusted so as to allow as little strain as possible in the material, we will assume all the supports to be at the same level. It is obvi- ous that the bending moments at the intermediate supports will be negative, as at these points the conditions are similar to those of a beam balanced over a single support. If the end spans are short we may have the supporting forces at the ends equal to zero, and if the ends are " anchored " (fastened down to the support) we may have a negative sup- porting force, that is, one acting in the same direction as the Fig. 43. loads. In Fig. 43 we see approximately the curves of bending moments and shearing force for a continuous beam carrying a uniform load and having five equally spaced sup- ports. There are two virtual joints between supports ex- cept for the end spans, and we will find that the value of the deflection in each span is somewhere between that for a 94 CONTINUOUS BEAMS. 95 uniformly loaded beam of span length supported at the ends and one uniformly loaded of span length fixed at the ends. The difficulty in working with continuous beams lies in find- ing the supporting forces, for the usual method of taking moments will not do, there being two unknown quantities in each equation, and the equations reducing to identities when we attempt to solve them. For beams having only three supporting forces, the solutions are not difficult, as will be shown; but as the number of supports increases the cal- culations become more and more tedious. 54. A beam of length 2a carries a uniform load and is supported by three equidistant supports. A beam of length 2a carrying a uniform load of w per unit length and supported at the ends, has a maximum deflection equal to --- (see Art. 45). A like beam supported at the ends and carrying a concentrated load R at its middle has R (2a) 3 a maximum deflection equal to - - (Art. 46). Now 48 El the above continuous beam may be considered as a uni- formly loaded beam which has a prop at its middle by which the middle point is propped up to the same level as the end supports. Obviously then the deflection at any point will be that due to the uniform load minus that due to the thrust of the prop, and as the prop deflects upward we have ~~ ~48W ~ 384 El or R, the thrust of the prop, or the middle supporting force The supporting forces at the ends are obviously equal, so w (2a) - 1 wa P = Q =- = ft wa. 96 STRENGTH OF MATERIAL. Having the supporting forces we have now no difficulty in getting the equations for slope and deflection, for, taking the origin at the middle, the B. M. for any section distant x from L_I Fig. 44. the origin is, by definition (remembering that the part of R R\ which is due to the loading on one side of the origin is J, dy Rx 2 WX 3 f from symmetry El =- - + MQ X + [C = . . . . < the slope is zero dx 46 r at the origin. 77, T "& WX X \ deflection is zero J= ~~ ~ + M [ l= ^ the Substituting the value of R we have for the B. M. at any section M = f wax - - + M Q . And as the B. M. is zero at the ends, we have, substituting wa 2 x = a, the value of M equal to - ; or o wx 2 wa 2 M = ft wax - ; CONTINUOUS BEAMS. 97 putting this value of M equal to zero and solving for x a gives us x = as the position of the virtual joints. In- tegrating we get the equations for slope and deflection, dy 5 wx 3 wa 2 x El f- = wax 2 - - + [C - 0, dx 16 68 and 5 3 wx 4 wa 2 x 2 48 " "24"" 16 ^ l ~ To show how carefully the level of the supports must be adjusted, suppose there is left a deflection over the middle support, equal, let us say, to of the total deflection which would occur were there no middle support; then 1 5w (2' a) 4 _5w (2 a) 4 R (2 a) 3 5 384 El 384 El 48 El ' w (2 a) from which R = or one-half the total load. If this deflection were in the other direction, or , we would have R = 4 the total load. Remembering that the deflection in any case is very small when we see that a variation. of it up or down causes R to change from j to J the total load, we can understand how carefully the supports must be ad- justed to the same level. 55. With concentrated loads we must proceed in a differ- ent way. Suppose a beam which is supported at the middle and ends carries a concentrated load W midway between the supports, . as represented in Fig. 45. Taking the origin at the middle as before and considering the right half of the beam, the bending moment at any section HH between the origin and the load is (Art. 46) Q (a - x). (I) 98 STRENGTH OF MATERIAL. Integrating, dy Wax and For a section between W and the end of the beam dx 2 Integrating, - ^T + (C = ?. (5a) W ^3 -i Q=Xw Fig. 45. The slope under the load is the same in both branches of the curve; therefore, substituting x = - in equations (2) and (5a) and equating them we have Wa 2 Wa 2 Qa 2 Qa 2 _Qa 2 Qa 2 Wa 2 ~ ~^~ ~2~~ ~g~ = ~2~ ~8~ + = ~8~' Substituting, dy Qx 2 Wa 2 CONTINUOUS BEAMS. 99 Integrating, E-M Cl = , (6a) The deflection under the load is the same in both branches, so as before, using equations (3) and (6a) we have Wa^ TFo 3 Qa 3 Qa 3 = Qa 3 Qa 3 Wa 3 Q '~16 48" 8 ' 48 " 8 " 48 " 16 lf or, Wa 3 Ll= 48 ' Substituting, Qax 2 Qx 3 Wa 2 x Wa 3 Now the deflection at the right support is equal to zero, and putting equation (6) equal to zero, substituting x = a, solv- ing for Q, gives Q = T 6 F W. From symmetry, P equals Q, and R must support the rest of the load, hence R=2W-(P + Q) = V W. The methods of this and the preceding article will cover most of the cases of continuous beams found in actual prac- tice, but there are many cases where beams have more than three supports, and for these a method known as the " Method of Three Moments " must be employed. This method came into use about 1857, and is a general method for obtaining the supporting forces. 56, Theorem of Three Moments. Let Fig. 46 represent the part of a beam over any three consecutive supports, and let the loading be uniform between adjacent sup- ports. Take the origin at P and let the bending moments 100 STRENGTH OF MATERIAL. at P, R and Q be M ly M 2 and M 3 respectively, the distance between supports to be a and a lt the loading between P and R be w Ibs. per unit length, and between R and Q be w l Ibs. J L T ---- 4 |p ^ [R |Q Fig. 46. per unit length; then the bending moment at any section HH between P and R is (Art. 45) wx z M = - + Px + C. But C in this case is equal to M lf so Integrating, El = M.x + P - + [C = ?, (2) dx 26 and B/y-ar^+P^-^+Cx + tc.-o 6 24 (level supports). (3) Now 2/ = when x = a, hence ATa 2 Pa 3 im 4 M.a Pa 2 i^a 3 = h + Ca. .'. C = + 2 6 24 2 6 24 CONTINUOUS '-SSAMS. 101 Substituting, dy Px 2 wx 3 M^a Pa 2 wa 3 dx~ lX ~2~ 6 2 6 24 and M/Y2 D/y*3 'jpy 1 ^ /I//" /Tf O* P/Tf^'Y* 1/1/7 "7* . ,*/ X^ // UUJU rJ. 4 Lt Ju JL C6 Ju Cc/C* *(/ From (1), if we let x = a, we get M 2 in terms of M lt or, M 2 = M, + Pa - ^f (6) jfc Now if we imagine the beam reversed and use Q for the origin, we can in the same way find the equation for bend- ing moment, slope and deflection; for the part of the beam between Q and R and over R the values of the bending moment, slope and deflection will be the same for both branches of the curve. We can therefore equate these values if we substitute x = a in those of the left branch and x = a t in those of the right. Making these substitutions and equating we get for the left branch, for the right branch, M l + P a -^ = M 2; =M 3 + Qa 1 -^, (7) M.a Pa 2 wa 3 , .. = s ,opeover/ J i-L + -Li.. (8) If we substitute the values of Q and P from equation (7) in equation (8) and reduce we get 7/V7 ^ I 7/5 // ^ M,a + 2 M 2 (a + a,) + M 3 a, = - "^ 1 1 (9) This is known as the equation of the three moments, and by using different sets of three consecutive supports we can get as many equations, less two, as there are supports. 102 STRENGTH -Q& MATERIAL. From our knowledge of cortditkms at the ends of the continu- ous beam we can get two more equations involving the moments over the supports, and thus having as many equa- tions as there are unknown quantities we may find all the supporting forces. For example, let the beam of Fig. 43 be supported at equal intervals and uniformly loaded with w Ibs. per unit length. From equation (9) or, (1) (2) (3) The bending moment is zero at the ends, hence M, M 5 =0. From (1) and (3) Substituting, we get from (2) and (1) From (6), using the first three supports, wa 2 3 wo? Now Q is equal to the change in S. F. at the second support. The negative part which is due to the load to the left of the support is P wa, and the positive part is given by the equa- CONTINUOUS BEAMS. 103 tion M 3 = M 2 + Q R X a - , and the sum of the two parts ft gives Q = 3 wa , In the same way R = f | wa, or from symmetry we know S = Q and T = P, so the total load less (P + Q + T + S) = R, from which R = f f wa. Examples : 1. A continuous beam of five equal spans is uniformly loaded with w Ibs. per ft.-run. If the beam is 38 ft. long, what are the supporting forces and the bending moments at the supports? 43 37 Ans. P = 3 w. Q= w. R = w. 5 5 2. Find the side of a steel beam of square section to span four openings of 8 ft. each, the total load per span being 44,000 Ibs. and the greatest horizontal fiber stress not to exceed 15,000 Ibs. per sq. in. Ans. Side = 5.66 in. 3. Find the depth of a steel beam of rectangular section twice as deep as broad, to span three openings each 12 ft. wide, the total load on each span being 6000 Ibs. and the greatest fiber stress allowed being 12,000 Ibs. per sq. in. Ans. 4.4 in. 4. A continuous beam of three spans is loaded only on the middle span with a uniform load. What are the sup- porting forces? wl Ans. At ends 20 104 STRENGTH OF MATERIAL. 5. A square steel beam, / = 268.9 in inch-units, is 36 ft. long and spans four openings, the end spans being each 8 ft. and the middle ones each 10 ft. Find the maximum B. M. of the beam. What will be the uniform load per foot to make a maximum fiber stress of 15,000 Ibs. per sq. in.? 261 w Ans. 31 6. A continuous beam carries a uniform load. If the end spans are each 80 ft. what will be the length of the middle span in order that the B. M. at its middle may equal zero? Ans. 1= 67. 14 ft. 7. How much work is done on a beam 10 ft. long, 10 ins. deep and 8 ins. wide, which carries a uniform load of 250 Ibs. per ft.-run? 5 Ans. Work = CHAPTER XL COLUMNS AND STRUTS. 57. When a prismatic piece of material, several times longer than its greatest breadth, is under compression, it is called a column or strut; a column being such a piece placed vertically and carrying a static load, and all others being struts. A column or .strut of material which is not homogeneous, or one on which the load does not act exactly inthe geometric axis, will bend or buckle. For mathematical investigation we must assume our material homogeneous, and instead of assuming a slight deviation of the line of action of our load from the geometric axis, we will assume that the column has first been bent by a horizontal force, then such a load applied to its end as will just keep it in the bent form after the removal of the horizontal force. The assump- tions we make will be three: (1) the column perfectly straight originally; (2) material per- fectly homogeneous; and (3) load applied ex- actly over the center of the ends. These conditions are never exactly fulfilled in practice, Fig. 47. so a rather large factor of safety must be applied. 58. Eider's Formula for Long Columns. We will choose a column with rounded or pivoted ends, so that they will be free to move slightly as the column is bent. Let Fig. 47 represent such a column, held in the bent position by the load W. Let the origin be at the center of the column, the axis of X vertical and that of Y horizontal ; d is the maxi- 105 106 STRENGTH OF MATERIAL. mum deflection, and ?/ the ordinate of the neutral line at any section HH distant x from the origin. From Art. 43 we have the general equation of bending, The bending moment for the section HH is, from the figure, M == W(d - y), so d ^ = Ti (8 - y) - Multiplying both members of this equation by 2 and inte- grating, we get now y is zero at the origin and is also zero there, the tangent to the curve of bending being parallel to the axis of X at that point, therefore the constant of integration, being this tangent, is zero. Extracting the square root of both members of equation (1) and transposing, we get _Jw -i-r; a which integrated gives (2) C 2 is zero, for where x = 0, y = and cos~ l 1=0. COLUMNS AND STRUTS. 107 Letting I equal the length of the column, when x = - , we have y = d; substituting these values in equation (2) we get The angle whose cosine is zero is - from which we get . El'%' 'El "F which is Euler's formula for long columns with round ends. Transposing equation (2) we get for the equation of the elastic curve y = d ] 1 cos, 59. A column loaded with W as per the above formula will just retain any deflection which may be given it, therefore this value of W is called the critical load, for the column will straighten out if there be any decrease in this load, and for any increase it will keep on bend- ing until it breaks. This is the load then which puts the column in neutral equilibrium. The bending of columns is perfectly uniform, so we can derive from the formula of the preced- ing article, which is for a column with rounded or pivoted ends, the formula for columns with one or both ends fixed. Let Fig. 48 represent the elastic curve of a column with both ends fixed. There will be two points of inflection, B and D, and as the bending 108 STRENGTH OF MATERIAL. is perfectly uniform these points will be at quarter the length of the column from the ends. The part BCD will represent the elastic curve of a column such as we considered in the preceding article, so the critical load of a column with fixed ends will be the same as for a column of half its length with pivoted ends, or, The part BCDE of Fig. 48 would approximately represent the elastic curve for a column with one end (E) fixed ; so the critical load for it would be the same as for a column of two- thirds its length with pivoted ends, or This latter formula is not quite so accurate as the others, for the ends are not in the same vertical line. It has been shown by experiment that when the length of a pillar is greater than 100 diameters the theoretical values are closely approached, while with shorter lengths these values are much too large. Columns having flat ends, if the ends are prevented from lateral movement, are considered as having fixed ends. 60. As Euler's formula is applicable only in the case of very long columns, another formula has been obtained in different ways by several different writers, which gives more accurate results for short columns. Obviously a column may be so short as to fail by crushing alone, in which case the crushing load would be W = fA, where / is crushing strength of the material and A the area of the cross-section. For ordinary columns, then, W must lie between fA and the value given by Eider's formula. The following formula is most used for short columns. If p l be the compressive stress due to W, and p 2 the fiber COLUMNS AND STRUTS. 109 stress due to bending, the maximum stress will be Pi + p 2 ', this must not exceed the strength of the material, so for the limiting stress, /= p 1 + p 2 . W If A is the cross-sectional area of the column, p l = -. A. p M My The equation of bending, = , gives us p 2 = - , remem- = ft = |, sin 6 = cos <> = - = . 116 STRENGTH OF MATERIAL. For the equilibrium of the joint at the peak the sum of the vertical components of the stresses Aa and Ba must be equal to the load 1750, but must act in the opposite direc- tion; therefore, Aa cos cj) + Ba cos = 1750, or, f Aa + f Ba = 1750. (1) Considering the joint at the right support, the sum of the horizontal and of the vertical components of the forces acting there must be zero, or as before the vertical compo- nent of Ba must equal Q, and the horizontal component must equal the stress Ca. Resolving, horizontally, Ca Ba cos (f>, or, Ca = \ Ba. (2) Vertically, Ba sin = 1120, or, f Ba = 1120. .'. Ba = 1400 Ibs. Substituting in (1), | Aa + f . 1400 = 1750. .'. Aa = 1050 Ibs. Substituting in (2), Ca = | . 1400 = 840. /. Ca = 840 Ibs. Now it is obvious that the stresses Aa and Ba will be com- pressive, and that Ca will be tensile, but in many cases it is not obvious, and with the graphic method will be shown a way of ^accurately distinguishing. Taking the joint at the peak, the external load acts down, so the resultant of the stresses Aa and Ba must act up; to do this they must push on the joint; a stress then which pushes on a joint is com- pressive and one which pulls is tensile. This method of finding stresses will hereafter be called the method by calculation, to distinguish it from Hitter's method. 65. Hitter's Method. This may be called the method of sections, as it involves our fourth rule for the equilibrium STRESS ON MEMBERS OF FRAMES. 117 of a structure. The moments may be taken about any axis, and if possible we choose an axis about which the moments of all the unknown stresses cut by the section except one disappear. We will solve the problem of the preceding arti- cle by this method and to do so will first consider a section Q=1120 Fig. 52. HH as shown in Fig. 52. If now we take moments about the joint at the peak, the moments of the stresses Aa and Ba will be zero and we will have to the left of the section only the external force P and the stress Ca to deal with; or to the right the external force Q and the stress Ca, there being no moment of the 1750-lb. load about the joint at the peak. Taking moments, then, the perpendicular distance from the peak to the stress Ca being 12 ft., and that to the line of action of the force P being 16 ft., we have 630 X 16 - Ca X 12, or, Ca = 840, as before. Taking moments about the joint at the right support will eliminate the stress Ca, and still working with the forces to the left of the section we have 630 X 25 = Aa X 15, or, Aa = 1050, as before. Care must be taken to use the perpendicular distance to the line of action of the forces. We have used 15 in this case 118 STRENGTH OF MATERIAL. because, it will be noticed, this frame is right-angled at the peak. To get the stress in Ba we must use another section, for it is clear that the only internal stresses involved are those of the members through which the section passes. We will use the section KK and take the moments about an axis through, let us say, the point L. (We may use any point.) Here the perpendicular distance to the stress Ba is 7.2 ft., and to the supporting force Q it is 9 ft.; so 1120 X 9 = Ba X 7.2, or, Ba = 1400, as before. Instead of choosing L for the axis of moments, we could just as well have taken the joint at the left supporting force from which the moments are 1120 X 25 = Ba X 20, or, Ba = 1400. This method will be found convenient where the stress in particular members is desired, and is frequently necessary in the graphic method. 66. The graphic method, invented by Clerk Maxwell, is based upon the principle of mechanics that a number of forces acting at a point are in equilibrium only when they can be represented in amount and direction by the consecutive sides of a closed polygon. In all frames the forces acting at the joints must be in equilibrium, and the line of action of the internal forces must be in the direction of the axis of the member in which they are found. All the external forces taken in order around the frame will form a closed polygon, because by our first rule for the equilibrium of structures they must balance. Draw carefully a diagram of the frame and at each joint indicate by an arrow-head the load, if there be one; indicate also the supporting forces. This figure is called the frame diagram, and will be denoted by F. D. The diagram representing the polygons of forces acting at STRESS ON MEMBERS OF FRAMES. 119 the joints is called the reciprocal diagram, and it will be de- noted by R. D. Having determined the amount and direc- tion of the external forces, plot them to a convenient scale, forming the external force polygon. In case the loads are all vertical, as in the example of Art. 63, the external force polygon is a vertical straight line, as in Fig. 53, the load R.D. c F.D. Q=1120 Fig. 53. i_LB 1750 Ibs. being represented by AB, the supporting force Q, acting up, by BC, and P by CA, all of which forces have the same distinguishing letters in the R. D. as in the F. D. Now the stress in the left-hand rafter is in the direction of the rafter itself, so from A of the external force polygon draw a line parallel to this rafter to represent the line of action of the stress Aa, and from B draw a line parallel to the right- hand rafter to represent the line of action of the stress Ba. The intersection of these two lines will be the point a, and if we connect C and a, the line will be found to be parallel to the tie rod of the frame, and Ca will represent the stress in the tie rod to the same scale as that used to plot the ex- ternal force polygon, just as Aa and Ba will represent the stress in the rafters. This figure is called the Reciprocal 120 STRENGTH OF MATERIAL. Diagram, and that the lines do represent the stresses may be proved, for the angles BAa and ABa are equal respectively to and 6 of the F. D. by construction, and as AC is to scale equal to 630, Aa will equal 630 sec < = 630 X f = 1050; Ba will equal 1120 sec - 1120 X I =1400; and Ca will equal 630 tan , or 1120 tan 6, either of which will give 840. The results are then the same as those obtained by other methods. Of course, if the diagram is drawn accurately, the stresses may be measured off to scale. To find the kind of stress in any member, notice that the forces acting on the joint at the peak are the load A B and the stresses Aa and Ba. If we pick these lines out on the R. D., they will form the polygon of forces (in this case a triangle) for the joint at the peak. Knowing the direction of one of these forces, we will indicate it by an arrowhead; for example, we know AB acts down, so we put an arrowhead pointing down on the line AB of the R. D. For the equi- librium of the joint at the peak the direction of the forces acting on it will be indicated if we suppose the arrowhead of AB to move in order around the sides of the polygon for this joint, starting in the direction in which it points. When on each side it will indicate the direction in which the stress for the corresponding member acts on the joint, and remem- bering that a push indicates compression and a pull tension, we can at once state what kind of stress exists in any member acting on the joint. The arrowheads are marked in the R. D. of Fig. 53 for the joint at the peak. They have been indi- cated for a single joint only, because if we consider any other joint involving the stress of any of the members acting on this one we would have two arrowheads on the same line pointing in opposite directions. This at first glance appears to be in- correct, but when we remember that the stress in any member acts in opposite directions on the joints at its two ends, and that we would n and 6 re- spectively, and the span S, and taking moments about the joint at the left support (Fig. 56, a), Sf QScos # =- 4 S> Sf Sf -X4-f-f X4 + 2S--X 3 - -X 1.5. 2 42 .-. Q cos 6= 6.5, and, by the same method, P cos $= 9.5, which shows the sum of the vertical components of the sup- porting forces to be equal to the vertical loads as it should be, also P sin Now the cylinder has an internal pressure of S Ibs. per unit area and an external pressure of S l Ibs. per unit area, so if r r 2 we have R at the inside surface equal to S, and if r = r l we have R at the outside surface equal to S iy and, sub- stituting, we get and SI=- + -T; (6) 2 TV*' solving (5) and (6) we have Ci _r.VCS.-fl) (7) 7*o 7*i MISCELLANEOUS PROBLEMS. 149 and C= 2 ^f ' "" ^ S} , (8) 1 2 substituting the values from (7) and (8) in (3) and (4) we get R = r*S l -rSS rW (,-) and r.'S, - r,'S , r.V,' (S, - S) .'-' r-r If there is no external pressure, as in the cases of cast guns and hydraulic pipes, S l will be equal to zero and our for- mulas become and From equations (11) and (12) we see that the greatest stress is at the inner surface of the cylinder, and here the total extension found by the method of Art. 79 must not exceed the elastic limit of the material. It will be noticed that the values of H and R found above do not take into considera- tion the lateral contraction of the material mentioned in the preceding article. If we consider this lateral contraction, and call H l the stress which would produce an extension equal to the total elongation found by the methods of Art. 79, we would have H, = H - kR - kT. For the material used in the construction of the modern naval gun, the value of Poisson's ratio is J. Putting in this 150 STRENGTH OF MATERIAL. value for k and also the values of H and R from equations (9) and (10), the value of T being found directly from the ex- ternal and internal end pressures and the sectional area of the cylinder, to be T = we have, after reducing, for the true hoop stress r,'S t -r,'S , 4r,V(S,-S) which is the formula used for the hoop stress for naval guns. 81. Built-up Guns. The guns at present used in the navy are known as " built-up " guns; that is, they are built of several pieces which are made separately. The inner piece is called the tube, outside of it is the jacket, and outside of the jacket are the hoops. The exterior diameter of the tube is made a little greater than the interior diameter of the jacket, the exterior diameter of the jacket a little greater than the interior diameter of the first hoop, etc. The pro- cess known as assembling consists in heating the jacket until it expands sufficiently just to slip over the tube, after which it is allowed to cool. In cooling the jacket contracts and grips the tube firmly, putting considerable external pressure on it. The hoops are then put over the jacket in the same way. This way of securing the several parts together is called the method of hoop shrinkage, and obvi- ously if two parts are assembled by this method, the inner one will be under hoop compression and the outer one under hoop tension. Considering a cylinder formed of two parts, let the radii after assembling be r z , r 2 and r lt the least radius being r 3 . Let R 3 , R 2 and R l be the radial stresses at the inner, intermediate and outer surfaces. We will call the difference between the inside radius of the jacket and the outside radius of the tube before assembling, e:then if MISCELLANEOUS PROBLEMS. 151 after assembling e 2 is the decrease in the length of the radius of the tube and e l the increase in the length of the radius of the jacket, we have e = e 2 + e r (1) If H 2 is the hoop compression on the outside surface of the tube and H l the hoop tension on the inside surface of the jacket, e 2 , the change of length of the radius of the tube due to the stress H 2 , will be (Chapter I), and e^ the change E of length of the radius of the jacket due to the stress H lt will be ; hence E E or, #2 + ^=-' (2) There being no internal pressure in this cylinder, equation (13) of the preceding article will give us a formula for the hoop compression at the outer surface of the tube, if we let S = 0, Si = R 2 and r = r 2 , at the same time changing r 2 of equation (13) to r 3 and r x to r 2 . This gives us the equation 2 o r 2 ,- 2 r 2 -- r 3 6 r 2 (r 2 - r 3 ) There being no external pressure on the jacket, equation (13) of the preceding article will also give us a formula for the hoop tension at the inner surface of the jacket if we put S l = 0, S = R 2 and r = r 2 . This gives us the equation 152 STRENGTH OF MATERIAL. Taking the factor R 2 from the right side of equations (3) and (4) all the quantities that remain are known (they being the different radii). Calling the value of these remaining quantities in equation (3) C i and those for equation (4) C 2 , we have #2=^1 (5) and #i = -ttA; (6) dividing (5) by (6) to eliminate R 2 we have !--! Solving the simultaneous equations (2) and (7) for H l and #2 gives CEe and Equations (8) and (9) give the hoop stresses at the joint, and having these we can from either of the equations (3) or (4) find the value of R 2 . Having R 2 , equation (13) of the preceding article will give us the value of the hoop com- pression at the inner surface of the tube, at which place the hoop compressive stress is obviously greatest when the gun is at rest. At the instant of firing a gun the internal pres- sure becomes very great and both the tube and jacket will then be under hoop tensile stress, but obviously the first effect of the internal pressure must be to overcome the initial compressive stress in the tube, though it increases the tensile stress in the outer hoops where the effect of the internal pressure is least; thus the built-up gun can sustain greater internal pressures than the solid gun. MISCELLANEOUS PROBLEMS. 153 82. Stress Due to a Centrifugal Force. If a body of mass m, at the end of a cord, is whirled round in a circle the cord will be put in tension, the centrifugal force, as we have learned from mechanics, being equal to , where v is the linear velocity and r the distance of the center of gravity of the mass from the axis of rotation. If n is the number of revolutions per second, the angular velocity aj will be equal to 2 nn. As the linear velocity is raj, the tension of the cord in terms of the angular velocity will be equal to 4 WxWr I W / \ [m and a> = 2nn 1 V g I Consider a fly-wheel, the spokes of which represent a com- paratively small part of the total weight, so that we may consider all the weight as being at the rim of the wheel. Let the thickness of the rim be t and its distance from the axis of rotation be r. Let the weight of the material be w Ibs. per unit volume, and the breadth of the rim a units. Then the total weight of the rim will be w . 2 nrta. The total radial force from the above formula is w . 2 nrta X 4 xWr and the radial force per unit area will be _ w . 2 xrta X 4 x 2 n 2 r w . 4 n 2 n 2 rt 9(2xra) ~ This will give us the radial stress for a rim so thin that the stress may be considered uniform throughout the section. If, however, we consider a thick rim whose inside radius is r 2 and whose outside radius is r 1? the increment of radial stress on a circular element whose thickness is dr will be 154 STRENGTH OF MATERIAL. given by the above formula. We will have then for the increment of radial stress on this elemental volume w . 4 xWrdr dR = - -, (1) the value of this increment having the negative sign be- cause the radial stress decreases as the distance from the axis of rotation increases. To prove this latter statement, suppose a bar of uniform section and length / rotates about an axis through its end. Let a be the sectional area, w the weight per unit volume, and x the distance of any point P from the axis of rotation. The stress at any point P is due to the centrifugal force of the part of the rod beyond P; therefore, as the center of gravity of this part is at a dis- l + x tance from the axis and its weight is a (I x}w, the l centrifugal force, by the formula, at the beginning of this article is c.r.- TO( * 2 ~* 2)a ' 2 . It is clear from this equation that when x = Z the C. F. is zero, and when x = it is a maximum. Proceeding then the integration of (1) gives 4 7i 2 n 2 wr 2 C t being the constant of integration. We know the value of the radial stress at the outer edge of the rim is zero; therefore when r = r MISCELLANEOUS PROBLEMS. 155 from which hence, R = ^ W - H). (2) ^ This equation gives the radial stress at any distance r from the axis of rotation. Now considering the equilibrium of any particle, we have, by the same reasoning employed in Art. 80, the same two equations, H + R = C (3) arid rdR + Rdr = Hdr. (4) Substituting in (4) the values of R and dR from (1) and (2) we have 4 xWwrdr 4 n*n 2 w r - + (r* - r 2 ) dr = Hdr. 9 20 Integrating, the constant of integration being zero when r = 0, we have H - ^ W + r>). (5) ^ Equations (2) and (5) give the radial and hoop tensile stress respectively for any point at a distance r from the axis of rotation, and both are independent of r 2 . Therefore these equations can be applied to solid wheels, such as a grindstone, as well as to fly-wheels. As in the preceding article these formulae have been deduced without considering the lateral deformation of Art. 79. The true hoop tension, taking this lateral deformation into consideration, would be H l =H kR 156 STRENGTH OF MATERIAL. and the true radial stress R t = R kH. The value of the radial stress is greatest where r = r 2 (equation (2)), and for this value of r we have R,= -(V - krS - r 2 2 - A;r 2 2 ). (6) 5/ The hoop tension is greatest where r = r l ; hence From equations (6) and (7) it is clear that the hoop tension, being the greater stress, will be the cause of rupture and also that the rupture will begin at the outside surface of the rim. 83. Bending Due to Centrifugal Force. We will have a bending moment, due to centrifugal force, if both ends of a straight- rod are constrained to move in circles, for example the, horizontal rod between the two driving wheels of a loco- motive. In this case each point of the rod at any instant is moving in a circle, and the centrifugal force due to this motion acts in the direction of the center of that circle for that instant. The stress caused, acts in a diametrically oppo- site direction. Obviously the effect will be greatest when the horizontal rod is at its lowest point, for here the weight of the rod itself acts down, as does also the stress due to centrifugal force. Calling the radius of the circle described by the ends of the rod r, and w the weight per unit length of the rod (considered of uniform cross-section) , the radial stress due to centrifugal force will be gr The value of the linear velocity v is obtained from our knowledge of the speed of the engine, radius of the drive wheels, etc. For example, if the speed of the engine is S, MISCELLANEOUS PROBLEMS. 157 radius of the drive wheels r 1; and radius of motion of the ends of the rod r, the linear velocity of any point in the horizontal rod will be S R , and this gives wS 2 r This value of R is uniform throughout the length of the rod; therefore the rod may be considered as a beam loaded uni- formly with R per unit length, and will have in addition Fig. 71. when at its lowest position a uniform load due to its own weight. The stress at any point can then be readily found p M from the formula for bending, = y l A connecting rod offers an important example of bending stress due to centrifugal force, but in this case one end of the rod is constrained to move in a straight line, while the other end moves in a circle. At the instant the connecting rod is at right angles with the crank arm (obviously the greatest effect is produced when the load is perpendicular to the con- necting rod), if the end of the connecting rod is above the center of motion of the crank arm, the weight of the con- necting rod will act down, while the line of action of the 158 STRENGTH OF MATERIAL. stress due to the centrifugal force, will act upward, in a direc- tion parallel to the crank arm. Therefore, as before, the greater stress will be produced when the end of the connect- ing rod is below the center of motion of the crank arm. In this position, however, the force acting on the piston will put the connecting rod in tension, so that it can readily sustain this bending stress. When the end of the connecting rod is above the center of motion of the crank arm, the stress, produced by the pressure on the piston, is compressive and the connecting rod is in the condition of a strut; any bend- ing due to centrifugal force will now become an important matter (Chapter XI). Taking the position then as shown in Fig. 71, the centrifugal load at any point of the con- necting rod will vary directly as the distance of the point from A; for as the radial stress at any point due to the wv 2 motion of the connecting rod is equal to where r is the gr radius of the circle being described by the point at the in- stant and the value of v is the linear velocity of the point B, because the whole connecting rod, at the instant, is mov- ing in a direction tangential to the circle described by B. Obviously r varies from a at B to an infinite length at A; therefore, we must have the load on the connecting rod due to the centrifugal force vary uniformly from zero at A, to a wv 2 maximum at B, where it is equal to - - The bending ga moment due to this load (Art. 42), taking A for the origin, is wv 2 x\ < = (Ix ) 6 ga\ I I being the length of the connecting rod and x the distance along it from A to any point. This bending moment tends to bend the rod so that the middle of it will move upward. The weight of the rod itself acts vertically downward, and the MISCELLANEOUS PROBLEMS. 159 component of the weight which acts perpendicular to the rod will be w cos per unit length. This load causes a bending moment equal to (Art. 38) the effect of which is to cause the middle of the rod to move downward. The total bending moment at any point distant x from A is then Having the bending moment at any point due to the motion and weight of the rod, the stress due to these causes is readily found from = But it may be repeated that a con- necting rod, in addition to the above stress, suffers a com- pressive stress, when in this position, due to the pressure on the piston, and also a bending stress due to this compressive load, which puts the rod in the condition of the column or strut discussed in Chapter XI. Referring to that chapter, it will be seen that the comparatively slight bending, due to centrifugal force, assumes important dimensions when we consider that the least variation of the axis of the connecting rod, from the line of action of this large compressive load, makes the opportunity for the load to cause bending. 84. Flat Plates. Experiment has proved that a circular flat plate when subjected to too great a pressure on one side always breaks along a diameter. Any diameter, then, of a circular plate, is perpendicular to the greatest tensile stress due to bending, caused by the pressure on one side of the plate. Let us consider a circular plate simply supported at the cir- cumference, and subjected to a uniformly distributed load on the side opposite to the support. If the plate were fixed at 160 STRENGTH OF MATERIAL. the circumference, it would be stronger than the one we are considering, just as a beam fixed at the ends is capable of supporting a greater load than if it were free at the ends. We will consider our support as a ring on which the plate rests, the pressure being uniformly distributed on the upper side. Let Fig. 72 represent such a plate, the support being at the circumference, and the load w Ibs. per unit area. We must Fig. 72. first find the bending moment at the diameter AB. The T area of the triangular element shown is . rdd, and the load on it is w dd. The supporting force under the arc ft rdd must be equal to this load, and its distance from the diameter AB is r cos 6. The center of gravity of the uni- form load on the triangular element is at a distance r from the center and r cos from the diameter AB. The bend- ing moment at the diameter AB due to these two forces is then WT dM= -- MISCELLANEOUS PROBLEMS. 161 Integrating between the limits - - and for 6, 2 2 wr 3 M = which is the bending moment due to the loads on one side of the diameter AB. Letting t be the thickness of the plate the moment of inertia of the section through AB about the rt 3 neutral axis is / = - We have then from the equation of bending, = , ly in this equation being equal to j y I \ 2 / _wr 3 t_ 6 r> P ' : ~3~"~2'7? ~ W T^ which gives the stress on any section through a diameter. In designing a plate of this kind we would know the pressure to which it would be subjected and the dimensions of the open- ing it would cover, so if we put the limiting tensile strength of the material we are to use for p, we can solve the above equation for t, the thickness necessary. If the load instead of being uniform is a concentrated one, calling it W and supposing it is to be at the most effective position, the center, and to bear uniformly on a small circle of radius ^ concentric with the supporting ring, the part of W the load on one side of any diameter will be ,and its center 4 r of gravity will be at a distance ! from the diameter. The 3 7i supporting force will be uniform throughout the length of the semicircle of the ring on the same side of the diameter as this load, and the center of gravity of this semicircle is at a 162 STRENGTH OF MATERIAL. 2r distance - - from the diameter, the total supporting force W on this semicircle being . The bending moment at the diameter will be t being the thickness of the plate, and the moment of inertia rt s of the section being 7 = as before, we have for the stress in the section My W / 2 rA t 6 3 W / 2 r,\ p = - - = [ r _I|. _ . _ 1 1 M . 7 TT V 3 / 2 rt* xt 2 \ 3 r ) If r, = r, or, which is the same thing, if the plate bears a uniform load, we have, remembering W will now be equal to nr 2 w , r 2 /2 and if r t = 0, or the load is concentrated at a point, 3 W The stress given by these formulae is the maximum stress, which occurs, as one would expect, at the center of the plate. Experiment proves this to be the case, for such plates begin to crack at the center and the crack extends from there, along a diameter, to the edge of the plate. Experiment also proves the above formula to give very approximate results, so, though this may be a tentative method of deducing them (the assumptions not being strictly true), they will serve for all practical purposes. MISCELLANEOUS PROBLEMS. 163 ---a Fig. 73. Rectangular Plates. Experiment shows that rectan- gular plates, when the length is not more than about twice the width, will crack along a diagonal. Therefore, if the sides are a and b (Fig. 73), the diagonal section will sup- port the greatest bending stress. Suppose the plate of thickness t to support a uni- form load of w per unit area. wab The load on one side of the diagonal will be and its JL * or center of gravity will be at a distance - from the diagonal. o Assuming the supporting force to be uniform along the edges, the part acting along the edge a will act at its center of gravity, which is at a distance - from the diagonal, as is also fj the center of gravity of the part acting along the edge b. These two edges will support the whole load on this side of the diagonal, or the whole supporting force here will equal The bending moment about the diagonal then is wab x wab x wabx TM . . I 23 2 2 " 12 Calling g the length of the diagonal, the moment of inertia at 3 of the section through it about its neutral axis is /= 1 2, and y = - . The equation of bending gives us My wabx t 12 wabx 12 *2 'at* ~ 2qt 2 ' 164 STRENGTH OF MATERIAL. Referring to the figure, g = \/a 2 + b 2 , and from similar Da triangles x = - ; substituting these values, Va 2 + b 2 wa 2 b 2 p = 2 (a 2 + b 2 ) t 2 If the plate is square, a = b, and we have wa 2 P = 4?' These formulae should for practical purposes have a factor on the right member of the equation of 1.5 if the support is & fixed one (riveted joint), and of 5 if there is only a simple support. The formula for elliptical plates, which experiment shows break along the major axis, is, for wrought iron and steel plates simply supported around the edges, wa 2 b 2 P ~*' t 2 (a 2 + b 2 } (if cast iron is used the coefficient is 3 instead of f ) , where w is the load per unit area, t the thickness, and a and b the semi-major and semi-minor axes respectively. The theoret- ical solution for elliptical plates is very difficult because it involves elliptical integrals and because the supporting force is not uniform around the edge, as indeed is also the case for rectangular plates, but the variation for the latter is not excessive unless the plate is more than two or three times as long as it is wide. In fact all the above deductions are approximations, for the assumptions made are not strictly true. The formula?, however, agree closely with the results obtained experimentally and may be assumed correct for all practical purposes. The subject of flat plates is probably the most unsatis- factory one in the study of strength of material, and practi- cal engineers have different methods for each form of plate. MISCELLANEOUS PROBLEMS. 165 Fault may be found with the assumptions of most of these methods, though they all give approximate results. 85. The principle of least work may be stated as fol- lows: For stable equilibrium the stresses in any structure must have such values that the potential energy of the sys- tem is a minimum. The stresses of course must be within the elastic limit of the material. When forces act upon bodies which conform to Hook's law, the principle of least work may be applied to determine some unknown reaction. Perhaps the best-known example is that of a four-legged table which supports an unsymmetrically placed load. We can get from our knowledge of statics three equations to find the part of the load supported by each leg, and the solu- tion can be arrived at by means of the fourth condition fur- nished by the principle of least work. Distribution of Stress We have all along assumed that the stress is uniform throughout a section. As a matter of fact this is not the case. It can be mathematically proved that the shearing stress in a rod of square section varies as the ordinates of a parabola, being zero where the normal stress due to bending is a maximum, and a maximum where the normal stress due to bending is zero (at the neutral surface). Again, the shear parallel to the neutral axis in a rod of square section is zero, but in a rod of circular section it has a finite value. Velocity of Stress. When a force is applied to a piece of material the stress is not instantaneously produced, but moves with a wave motion through the mass. The velocity of this motion can be found, and it is shown to depend upon the stiffness and density of the material. The velocity of stress should be taken into account in problems involving impact and suddenly applied loads. Internal Friction. When material is subjected to force and deformation occurs the molecules of the material move and this motion is resisted by internal friction. Heat is 166 STRENGTH OF MATERIAL. produced and for the time between the application of the load and the complete rearrangement of the molecules the stresses at planes through the material are changing; for this time then our formulae for planes of maximum stress are not correct. Fatigue of Materials. Experiment proves that material will break if subjected to repeated stress, even if the stress be somewhat below the ultimate strength of the material. Experiment also shows that the greater the num- ber of applications the less becomes the value of the stress necessary to cause rupture. For example, about a hun- dred thousand applications of a stress of 49,000 Ibs. to wrought iron will cause rupture, but if 500,000 applications were made the stress need be only 39,000 Ibs. The loss of strength due to repeated stress is known as the fatigue of materials. This fatigue is more marked if the stress alter- nates from tension to compression and back again. The preceding facts have been mentioned to give the student the knowledge of their existence so that if inclined he may look them up in more complete works on the sub- ject. The time limit of the course for which this book has been written precludes the possibility of entering into dis- cussion of many subjects of importance, such as the stress in hooks, links, springs, rollers, foundations, arches and many others. Miscellaneous Examples: 1. A reinforced concrete beam is 48 ins. wide, 54 ins. long and 5 ins. deep. It has a sectional area of 3.6 sq. in. of steel, the center of gravity of which is I ins. below the top of the beam which is uniformly loaded with 2400 Ibs. per CTjl \ Use =10.) Find the position of the neu- E c ) tral surface, and the stress in the steel. Ans. Neutral surface 3.45 ins k below top. MISCELLANEOUS PROBLEMS. 167 2. A reinforced concrete beam is 60 ins. long, 48 ins. wide and 5 ins. deep. The sectional area of the steel is 2.4 sq. in. and its center of gravity is f ins. below the top of the beam. The load is uniform and equals 3600 Ibs. per in. length. Find the position of the neutral surface and the 77 1 stress in the concrete and steel. - = 10. ^c Ans. Neutral surface 1.68 ins. from top. 3. If the allowable unit stress in concrete be 500 Ibs. per sq. in., and that for steel be 10,000 Ibs. per sq. in., what per- centage of steel must be put in a beam 20 ins. wide by 10 ins. deep if the steel is at a distance of 9 ins. from the top? Ans. 0.75%. 4. If the allowable unit stresses for concrete and steel are 500 and 25,000 Ibs. per sq. in. respectively, what is the resisting moment of a beam 7 ins. wide and 10 ins. deep if the reinforcement of steel is 1% of the sectional area and placed at the lowest point of the beam? Ans. 92,800 in.-lbs. 5. A beam is 8 ft. long and 1 ft. wide. The concrete is 5 ins. thick and the steel reinforcement is 1 in. from the bot- tom of the beam. What area of steel section will be neces- sary? Using J in. square bars, what should be the distance between them if a floor is made in this way? 6. The cross-section of a steel bar is 16 sq. in. The bar is put under a stress of 27,000 Ibs. per sq. in. If k J, what is the sectional area while under stress? Ans. 15.99 sq. in. 7. A steel bar is 2.5 in. in diameter and 18.5 ft. long. What are its length and sectional area under a pull of 64,000 Ibs.? 168 STRENGTH OF MATERIAL. 8. The inside radius of a cylinder is 6 ins., its outside radius is 1 ft. The internal pressure is 600 Ibs. per sq. in., and the external pressure is that due to the atmosphere. What are the hoop and radial stresses at the outside and inside surfaces, also midway between these surfaces? Ans. Hoop stress inside, 960; outside, 375 Ibs. per sq. in. 9. A solid cylinder is under a uniform external pressure of 14,000 Ibs. per sq. in. Show that the hoop and radial stresses are uniform and give values. 10. A gun is built of a tube, inside radius 3 ins., outside radius 5 ins., and a jacket 2 ins. thick. Before assembling, the difference between the outside radius of the tube and the inside radius of the jacket was .004 in. What are the stresses at the outside and inside surfaces? Ans. Hoop compression, 14,400 Ibs. per sq. in. 11. In example 10 what powder pressure when firing the gun will just reverse the stress at the inner surface of the tube? 12. The radii of a gun composed of tube and jacket are r 3 = 3.04 ins., r 2 = 5.8 ins., and ^ = 9.75 ins. The allow- able unit stress is 50,000 Ibs. per sq. in. for both tension arid compression. What are the radii before assembling? Ans. Radius of bore, 3.0451 ins.; outside radius of tube, 5.805 ins.; inner radius of jacket, 5.7915 ins.; outer radius, 9.7436 ins. 13. What internal powder pressure will stress the gun of example 12 to just 50,000 Ibs. per sq. in. tension? Ans. 51,100 Ibs. per sq. in. 14. What is the greatest tangential stress in a cast-iron fly-wheel 30 ft. in diameter, rim 1 in. thick and 4 ins. wide, when it is making 60 revolutions per minute? Ans. Very roughly, 800 Ibs. per sq. in. MISCELLANEOUS PROBLEMS. 169 15. What is the stress in a cast-iron fly-wheel rim having a linear velocity of 1 mile a minute? Ans. Roughly, 750 Ibs. per sq. in. 16. What diameter should a fly-wheel have if it is to make 100 revolutions per minute, and the maximum allow- able linear velocity is 6000 ft. per min.? Ans. Roughly, 19 ft. 17. A cast-iron bar 9 ft. long, 3 ins. wide and 2 ins. thick revolves about an axis } in. in diameter passed through it at a distance of 4| ft. from one end. How many revolutions per second will produce rupture? 18. A solid steel circular saw is 4 ft. in diameter, and makes 2700 revolutions per minute. What is the stress at the circumference? How many revolutions per minute will cause a stress of 35,000 Ibs.? 19. An engine is making 750 revolutions per minute; the connecting rod is 2 ft. long, and the crank arm 6 ins., ma- terial steel. What is the bending stress, due to centrifugal force, on the connecting rod, if the area of its section be 1.5 sq. ins.? 20. A cast-iron fly-wheel, mean diameter of rim 20 ft., makes 90 revolutions per minute. The cross-section of the rim is 10 sq. ins. What is the stress? 21. What must be the thickness of a cast-iron cylinder head 36 ins. in diameter, allowable stress 3600 Ibs. per sq. in., to sustain a load of 250 Ibs. per sq. in.? Ans. 5 ins. 22. The allowable stress for steel being 12,000 Ibs. per sq. in., how thick would a steel head for the cylinder of example 21 have to be? Ans. 2.6 ins. 170 STRENGTH OF MATERIAL. 23. A circular steel plate is 24 ins. in diameter and 1.5 ins. thick. It carries a load of 4000 Ibs. at the center rest- ing on a circle of 1 in. diameter. What is the maximum stress? Ans. 1650 Ibs. per sq. in. 24. Suppose the load of example 23 were distributed on a surface of 3 ins. diameter. What would be the stress? Ans. 1555 Ibs. per sq. in. 25. What must be the thickness of a steel plate 4 ft. square to carry 200 Ibs. per sq. ft. Ans. 0.2 in. 26. What uniform load can be carried by a wrought-iron plate f in. thick, 5 ft. long and 3 ft. wide? Ans. About 12 Ibs. per sq. in. 27. A floor 18 ft. long, 15 ft. wide is made of concrete 4 ins. thick, with 1 in. square wrought-iron rods, spaced 1 ft. apart and at .75 in. from the bottom of the concrete. The floor carries a load of 150 Ibs. per sq. ft. What is the maxi- E s mum stress? = 15. E c Ans. Stress is. about 450 Ibs. per sq. in. 28. An elliptical plate (cast iron) has a major axis 24 ins. long, a minor axis 16 ins., and is under a uniform pressure of 22 Ibs. per sq. in. The allowable stress is 3000 Ibs. per sq. in., and the plate is simply supported at the edges. What must be the thickness? Ans. 1 in. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW Books not returned on time are subject to a fine of 50c t>er volume after the third day overdue, increasing to $1.00 per volume after the sixth day. Books not in demand may be renewed if application is 'made before expiration of loan period. DEC 13 t EC ' IEC 8 H FER 20 1926 OCT 28 1! OCT 20 50m-7,'16 3C0375 UNIVERSITY OF CALIFORNIA LIBRARY