UNIVERSITY OF CALIFORNIA AT LOS ANGELES Gift Of Prof .Harry Thomas Cory -rA PART III. STRENGTH OF MATERIALS. [OB MECHANICS OF MATERIALS]. CHAPTER I. ELEMENTARY STRESSES AND STRAINS. 178. Deformation of Solid Bodies. In the preceding por- tions of this work, what was called technically a " rigid body," was supposed incapable of changing its form, i.e., the positions of its particles relatively to each other, under the action of any forces to be brought upon it. This sup- position was made because the change of form which must actually occur does not appreciably alter the distances, angles, etc., measured in any one body, among most of the pieces of a properly designed structure or machine. To show how the individual pieces of such constructions should be designed to avoid undesirable deformation or injury is the object of this division of Mechanics of En- gineering, viz., the Strength of Materials. ^Q ^L g^ X^X 5 ^^/ Fm. 198. 17& As perhaps tne simplest instance of the deformation or distortion of a solid, let us consider the case of a prismatic rod in a state of tension, Fig. 192 (link of a surveyor's 1 i I i^a 196 MECHANICS OF ENGINEERING. chain, e.g.). The pull at each end is P, and the body is said to be under a tension of P (Ibs., tons, or other unit), not 2P. Let ABGD be the end view of an elementary parallelepiped, originally of square section and with faces at 45 with the axis of the prism. It is now deformed, the four faces perpendicular to the paper being longer* than before, while the angles BAD and BCD, originally right angles, are now smaller by a certain amount d, ABG and ADC larger by an equal amount 3. The element is said to be in a state of strain, viz.: the elongation of its edges (parallel to paper) is called a tensile strain, while the alter- ation in the angles between its faces is called a shearing strain, or angular distortion (sometimes also called a slid- ing, or tangential, strain, since BC has been made to slide, relatively to AD, and thereby caused the change of angle). [This use of the word strain, to signify change of form and not the force producing it, is of recent adoption among many, though not all, technical writers.] 179. Strains. Two Kinds Only. Just as a curved line may be considered to be made up of small straight-line ele- ments, so the substance of any solid body may be consid- ered to be made up of small contiguous parallelepipeds, whose angles are each 90 before the body is subjected to the action of forces, but which are not necessarily cubes. A line of such elements forming an elementary prism is sometimes called a, fibre, but this does not necessarily imply a fibrous nature in the material in question. The system of imaginary cutting surfaces by which the body is thus subdivided need not consist entirely of planes ; in the sub- ject of Torsion, for instance, the parallelopipedical ele- ments considered lie in concentric cylindrical shells, cut both by transverse and radial planes. Since these elements are taken so small that the only possible changes of form in any one of them, as induced by a system of external forces acting on the body, are * When a is nearly (or 90") BG and AD (or AB and DC) are shorter than before, on account of lateral contraction; see 193. ELEMENTARY STRESSES, ETC. 197 elongations or contractions of its edges, and alteration of its angles, there are but two kinds of strain, elongation (contraction, if negative) and shearing. 180. Distrihuted Forces or Stresses. In the matter preced- ing this chapter it has sufficed for practical purposes to consider a force as applied at a point of a body, but in reality it must be distributed over a definite area ; for otherwise the material would be subjected to an infinite force per unit of area. (Forces like gravity, magnetic at- traction, etc., we have already treated as distributed over the mass of a body, but reference is now had particularly to the pressure of one body against another, or the action of one portion of the body on the remainder.) For in- stance, sufficient surface must be provided between the end of a loaded beam and the pier on which it rests to avoid injury to either. Again, too small a wjre must not be used to sustain a given load, or the tension per unit of area of its cross section becomes sufficient to rupture it. Stress is distributed force, and its intensity at any point of the area is where dF is a small area containing the point and dP the force coming upon that area. If equal dP's (all parallel) act on equal dfsof a plane surface, the stress is said to be of uniform intensity, which is then where P= total force and F the total area over which it acts. The steam pressure on a piston is an example of stress of uniform intensity. 198 MECHANICS OF ENGINEERING. For example, if a force P= 28800 Ibs, is uniformly dis- tributed over a plane area of F= 72 sq. inches, or ^ of a sq. foot, the intensity of the stress is =400 Ibs. per sq. inch, 72 (or p = 28800-^=57600 Ibs. per sq. foot, or _p=14400-5 ^=28.8 tons per sq. ft., etc.). 181. Stresses on an Element ; of Two Kinds Only. When a solid body of any material is in equilibrium under a sys- tem of forces which do not rupture it, not only is its shape altered (i.e. its elements are strained], and stresses pro- duced on those planes on which the forces act, but other stresses also are induced on some or all internal surfaces which separate element from element, (over and above the forces with which the elements may have acted on each other before the application of the external stresses or " applied forces "). So long as the whole solid is the "free body " under consideration, these internal stresses, being the forces with which the portion on one side of an imag- inary cutting plane acts on the portion on the other side, do not appear in any equation of equilibrium (for if intro- duced they would cancel out); but if we consider free a portion only, some or all of whose bounding surfaces are cutting planes of the original body, the stresses existing on these planes are brought into the equations of equilib- rium. Similarly, if a single element of the body is treated by itself, the stresses on all six of its faces, together with its weight, form a balanced system of forces, the body being supposed at rest. Fie. 193. ELEMENTARY STRESSES, ETC. 199 As an example of internal stress, consider again the case of a rod in tension ; Fig. 193 shows the whole rod (or eye- bar) free, the forces P being the pressures of the pins in the eyes, and causing external stress (compression here) on the surfaces of contact. Conceive a right section made through US, far enough from the eye, G, that we may con- sider the internal stress to be uniform in this section, and consider the portion BSG as a free body, in Fig. 194. The stresses on US, now one of the bounding surfaces of the free body, must be parallel to P, i.e., normal to US', (otherwise they would have components perpendicular to P, which is precluded by the necessity of 2Y being = 0, and the supposition of uniformity.) Let F = the sec- Fio. 195. tional area RS, and p = the stress per unit of area ; then -T.Y= gives P Fp, i.e., p= F (2) The state of internal stress, then, is such that on planes perpendicular to the axis of the bar the stress is tensile and normal (to those planes). Since if a section were made oblique to the axis of the bar, the stress would still be parallel to the axis for reasons as above, it is evident that on an oblique section, the stress has components both nor- mal and tangential to the section, the normal component being a tension. 200 MECHANICS OF ENGINEERING. The presence of the tangential or shearing stress in ob- lique sections is rendered evident by considering that if an oblique dove-tail joint were cut in the rod, Fig. 195, the shearing stress on its surfaces may be sufficient to over- come friction and cause sliding along the oblique plane. If a short prismatic block is under the compressive ac- tion of two forces, each =P and applied centrally in one base, we may show that the state of internal stress is the same as that of the rod under tension, except that the nor- mal stresses are of contrary sign, i.e., compressive instead of tensile, and that the shearing stresses (or tendency to slide) on oblique planes are opposite in direction to those in the rod. Since the resultant stress on a given internal plane of a body is fully represented by its normal and tangential components, we are therefore justified in considering but two kinds of internal stress, normal or direct, and tangen- tial or shearing. 182. Stress on Oblique Section of Rod in Tension. Consider free a small cubic element whose edge =a in length; it has two faces parallel to the paper, being taken near the middle of the rod in Fig. 192. Let the angle which the face AS, Fig. 196, makes with the axis of the rod be = a. This angle, for our present purpose, is considered to remain the same while the two forces P are acting, as before their action. The re- sultant stress on the face AB hav- ing an intensity p^P-^-F, (see eq. 2) per unit of transverse section of rod, is = p (a sin a) a. Hence FlG - 197 - its component normal to AB is pa 2 sin 2 a ; and the tangential or shearing component along ELEMENTARY STRESSES, ETC. 201 A B =-jt>a 2 sin a cos a. Dividing by the area, a 2 , we have the following : For a rod in simple tension we have, on a plane making an angle, a, with the axis : a Normal Stress =p sin 2 a per unit of area . . (1) and a Shearing Stress =p sin a cos a per unit of area . (2) " Unit of area " here refers to the oblique plane in ques- tion, while p denotes the normal stress per unit of area of a transverse section, i.e., when a=90, Fig. 194. The stresses on CD are the same in value, as on AB, while for BG and AD we substitute 90 a for a. Fig. 197 shows these normal and shearing stresses, and also, much exaggerated, the strains or change of form of the element (see Fig. 192). 182a. Relation between Stress and Strain. Experiment shows that so long as the stresses are of such moderate value that the piece recovers its original form completely when the external forces which induce the stresses are re- moved, the following is true and is known as Hookas Law (stress proportional to strain). As the forces P in Fig. 193 (rod in tension) are gradually increased, the elonga- tion, or additional length, of RK increases in the same ratio as the normal stress, p, on the sections US and KN> per unit of area [ 191]. As for the distorting effect of shearing stresses, consider in Fig. 197 that since p sin a cos a p cos (90 a) sin (90 a) the shearing stress per unit of area is of equal value en alt four of the faces (perpendicular to paper) in the elementary block, and is evidently accountable for the shearing strain, i.e., for the angular distortion, or difference, d, between 90 and the present value of each of the four angles. Ac- cording to Hooke's Law then, as P increases within tta limit mentioned above, 3 varies proportionally to p sin a cos a, i.e. to the stress. 202 MECHANICS OF ENGINEERING. 182b. Example. Supposing the rod in question were of a kind of wood in which a shearing stress of 200 Ibs. per sq. inch along the grain, or a normal stress of 400 Ibs. per sq. inch, perpendicular to a fibre-plane will produce rup- ture, required the value of the angle which the grain must make with the axis that, as P increases, the danger of rupture from each source may be the same. This re- quires that 200:400::^) sin a cos a :p sin 2 a, i.e. tan. a must = 2.000.'.=63^. If the cross section of the rod is 2 sq. inches, the force P at each end necessary to produce rup- ture of either kind, when a=63^, is found by putting p sin a cos a= 20Q.'.p =500.0 Ibs. per sq. inch. Whence, since p=P+F, P=1000 Ibs. (Units, inch and pound.) 183. Elasticity is the name given to the property which most materials have, to a certain extent, of regaining their original form when the external forces are removed. If the state of stress exceeds a certain stage, called the Elastic Limit, the recovery of original form on the part of the ele- ments is only partial, the permanent deformation being called the Set. Although theoretically the elastic limit is a perfectly defi- nite stage of stress, experimentally it is somewhat indefi- nite, and is generally considered to be reached when the permanent set becomes well marked as the stresses are in- creased and the test piece is given ample time for recovery in the intervals of rest. The Safe Limit of stress, taken well within the elastic limit, determines the working strength or safe load of the piece under consideration. E.g., the tables of safe loads of the rolled wrought iron beams, for floors, of the New Jersey Steel and Iron Co., at Trenton, are computed on the theory that the greatest normal stress (tension or com- pression) occurring on any internal plane shall not exceed 12,000 Ibs. per sq. inch ; nor the greatest shearing stress 4,000 Ibs. per sq. inch. ELEMENTARY STRESSES, ETC. 203 The intimate Limit is reached when rupture occurs. 184. The Modulus of Elasticity (sometimes called co-efficient of elasticity) is the number obtained by dividing the stress per unit of area by the corresponding relative strain. Thus, a rod of wrought iron y 2 sq. inch sectional area being subjected to a tension of 2^ tons =5,000 Ibs., it is found that a length which was six feet before tension is -= 6.002 ft. during tension. The relative longitudinal strain or elongation is then= (0.002)-=- 6=1 : 3,000 and the corres- ponding stress (being the normal stress on a transverse plane) has an intensity of pt=P+F= 5,000-r- > =10,000 Ibs., per sq. inch. Hence by definition the modulus of elasticity is (for ten- sion) # t =jVf-=10,000-r 3 -^=30,000,000 Ibs. per sq. inch, (the sub-script " t " refers to tension). It will be noticed that since E is an abstract number, E t is of the same quality as p t , i.e., Ibs. per sq. inch, or one di- mension of force divided by two dimensions of length. (In the subject of strength of materials the inch is the most convenient English linear unit, when the pound is the unit of force ; sometimes the foot and ton are used to- gether.) The foregoing would be called the modulus of elasticity of wrought iron in tension in the direction of the fibre, as given by the experiment quoted. But by Hooke's Law^> and e vary together, for a given direction in a given ma- terial, hence within the elastic limit E is constant for a given direction in a given material. Experiment confirms this appr oxim ately . Similarly, the modulus of elasticity for compression E* 204 MECHANICS OF ENGINEERING. in a given direction in a given material may be determined by experiments on short blocks, or on rods confined lat- erally to prevent flexure. As to the modulus of elasticity for shearing, E*, we divide the shearing stress per unit of area in the given direction by d (in it measure) the corresponding angular strain or distortion ; e.g., for an angular distortion of 1 or d=.0174, and a shearing stress of 1,566 Ibs. per sq. inch, we have E A = gg= 9,000,000 Ibs. per sq. inch. Unless otherwise specified, by modulus of elasticity will be meant a value derived from experiments conducted within the elastic limit, and this, whether for normal stress or for shearing, is approximately constant for a given di- rection in a given substance.* 185. Isotropes. This name is given to materials which are homogenous as regards their elastic properties. In such a material the moduli of elasticity are individually the same for all directions. E.g., a rod of rubber cut out of a large mass will exhibit the same elastic behavior when subjected to tension, whatever its original position in the mass. Fibrous materials like wood and wrought iron are not isotropic ; the direction of grain in the former must always be considered. The " piling " and welding of nu- merous small pieces of iron prevent the resultant forging from being isotropic. 186. Resilience refers to the potential energy stored in a body held under external forces in a state of stress which does not pass the elastic limit. On its release from con- straint, by virtue of its elasticity it can perform a certain amount of work called the resilience, depending in amount upon the circumstances of each case and the nature of the material. See 148. 187. General Properties of Materials. In view of some defi- nitions already made we may say that a material is ductile * The moduli, or " co-efficients," of elasticity as used by physicists are well explained in Stewart and Gee's Practical Physics, Vol. I., pp. 164, etc. Their "co-efficient of rigidity" is our E t . ELEMENTARY STRESSES, ETC. 205 when the ultimate limit is far removed from the elastic limit ; that it is brittle like glass and cast iron, when those limits are near together. A small modulus of elasticity means that a comparatively small force is necessary to produce a given change of form, and vice versa, but implies little or nothing concerning the stress or strain at the elastic limit ; thus Weisbach gives E VJ Ibs. per sq. inch for wrought iron == 28,000,000= double the E,. for cast iron while the compressive stresses at the elastic limit are the same for both materials (nearly). 188. General Problem of Internal Stress. This, as treated in the mathematical Theory of Elasticity, developed by Lame, Clapeyron and Poisson, may be stated as follows : Given the original form of a body when free from stress, and certain co-efficients depending on its elastic proper- ties ; required the new position, the altered shape, and the in- tensity of the stress on each of the six faces, of every element of the body, when a given balanced system of forces is applied to the body. Solutions, by this theory, of certain problems of the na- ture just given involve elaborate, intricate, and bulky analysis ; but for practical purposes Navier's theories (1838) and others of more recent date, are sufficiently exact, when their moduli are properly determined by experiments covering a wide range of cases and materials. These will be given in the present work, and are comparatively sim- ple. In some cases graphic will be preferred to analytic methods as more simple and direct, and indeed for some problems they are the only methods yet discovered for ob- taining solutions. Again, experiment is relied on almost exclusively in dealing with bodies of certain forms under peculiar systems of forces, empirical formulse being based on the experiments made ; e.g., the collapsing of boilei flues, and in some degree the flexure of long columns. 206 MECHANICS OF ENGINEERING. 189. Classification of Cases. Although, in almost any case whatever of the deformation of a solid body by a balanced system of forces acting on it, normal and shearing stresses are both developed in every element which is affected at all (according to the plane section considered,) still, cases where the body is prismatic, and the external system con- sists of two equal and opposite forces, one at each end of the piece and directed away from each other, are commonly called cases of Tension; (Fig. 192); if the piece is a short prism with the same two terminal forces directed toward each other, the case is said to be one of Compression ; a case similar to the last, but where the prism is quite long (" long column "), is a case of Flexure or bending, as are also most cases where the "applied forces" (i.e., the external forces), are not directed along the axis of the piece. Rivet- ed joints and " pin-connections " present cases of Shearing; a twisted shaft one of Torsion. When the gravity forces due to the weights of the elements are also considered, a combination of two or more of the foregoing general cases may occur. In each case, as treated, the principal objects aimed at are, so to design the piece or its loading that the greatest stress, in whatever element it may occur, shall not exceed a safe value ; and sometimes, furthermore, to prevent too great deformation on the part of the piece. The first ob- ject is to provide sufficient strength ; the second sufficient stiffness. / 190. Temperature Stresses. If a piece is under such con- straint that it is not free to change its form with changes of temperature, external forces are induced, the stresses produced by which are called temperature stresses. TENSION. 207 TENSION. 191. Hooke's Law by Experiment. As a typical experiment in the tension of a long rod of ductile metal such as wrought iron and the mild steels, the following table is quot- ed from Prof. Cotterill's " Applied Mechanics." The experi- ment is old, made by Hodgkinson for an English Eailway Commission, but well adapted to the purpose. From the great length of the rod, which was of wrought iron and 0.517 in. in diameter, the portion whose elongation was observed being 49 ft. 2 in. long, the small increase in length below the elastic limit was readily measured. The succes- sive loads were of such a value that the tensile stress p=P- : rF, or normal stress per sq. in. in the transverse section, was made to increase by equal increments of 2657.5 Ibs. per sq. in., its initial value. After each application of load the elongation was measured, and after the removal of the load, the permanent set, if any. TaDle of elongations of a wrought iron rod, of a leugth=49 ft. 2 in. p A Jl =l+l X Load, (Ibs. per square inch.) Elongation, (inches.) Increment of Elongation. e, the relative elongation, (ab- stract number.) Permanent Set, (inchea) 1X3667.5 .0485 .0485 0.000082 2X " .1095 .061 .000186 3X " .1675 .058 .000283 0.0019 4X " .224 .0565 .000379 .002 5X " .2805 .0565 .000475 .002? ex " .337 .0565 .000570 .003 TX " .393 .056 .004 8X " .452 .059 .000766 .007S 9X " .5155 .0635 .0195 10X " .598 .0825 .049 MX " .760 .162 .1545 12X " 1.810 .550 .667 etc. 208 MECHANICS OP ENGINEERING. Eeferring now to Fig. 198, the notation is evident. P is the total load in any experiment, F the cross section of the rod ; hence the normal stress on the transverse section is p=P-r.F. When the loads are increased by equal in- crements, the corresponding increments of the elongation A should also be equal if Hooke's law is true. It will be noticed in the table that this is very nearly true up to the 8th loading, i.e., that J^, the difference between two con- secutive values of A, is nearly constant. In other words the proposition holds good : if P and Pi are any two loads below the 8th, and /I and i, the corresponding elongations. The permanent set is just perceptible at the 3d load, and increases rapidly after the 8th, as also the increment of elongation. Hence at the 8th load, which produces a ten- sile stress on the cross section of p= 8x2667.5= 21340.0 Ibs. per sq. inch, the elastic limit is reached. As to the state of stress of the individual elements, if we conceive such sub -division of the rod that four edges of each element are parallel to the axis of the rod, we find that it is in equilibrium between two normal stresses on its end faces (Fig. 199) of a value =pdF~ (P^F}dF where dF is the hor- izontal section of the element. If dx was the original length, and dh the elongation produced by pdF, we shall have, since all the dx's of the length are equally elongated at the same time, dl X TENSION. 209 where 1= total (original) length. But dX-^dx is the rela- tive elongation e, and by definition ( 184) the Modulus of Elasticity for Tension, E l} =p-r-e dx Eq. (1) enables us to solve problems involving the elonga- tion of a prism under tension, so long as the elastic limit is not surpassed. The values of E^ computed from experiments like those just cited should be the same for any load under the elas- tic limit, if Hooke's law were accurately obeyed, but in reality they differ somewhat, especially if the material lacks homogeneity. In the present instance (see Table) we have from the 2d Exper. E=p+ 6=28,680,000 Ibs. per sq. in. 5th " E,= " =28,009,000 8th " E t = " =27,848,000 " " If similar computations were made beyond the elastic limit, i.e., beyond the 8th Exper., the result would be much smaller, showing the material to be yielding much more readily. 192. Strain Diagrams. If we plot the stresses per sq. inch (p) as ordinates of a curve, and the corresponding relative elongations (e) as abscissas, we obtain a useful graphic re- presentation of the results of experiment. Thus, the table of experiments just cited being utilized in this way, we obtain on paper a series of points through which a smooth curve may be drawn, viz. : OBC Fig. 200, for wrought iron. Any convenient scales may be used for p and e ; and experiments having been made on other metals in tension and the results plotted to the same scales 210 MECHANICS OF ENGINEERING. as before for p and e, we have the means of comparing their tensile properties. Fig. 200 shows two other curves, rep- resenting (roughly) the average behavior of steel and cast iron. At the respective elastic limits J?, B , and B ', it will be noticed that the curve for wrought iron makes a sudden turn from the vertical, while those of the others curve away more gradually ; that the curve for steel lies nearer the vertical axis than the others, which indicates a higher value for E t ; and that the ordinates BA', B'A', and B 'A " (respectively 21,000, 9,000, and 30,000 Ibs. per sq. inch) in- dicate the tensile stress at the elastic limit. These latter quantities will be called the moduli of tenacity at elastic limit for the respective materials. [On a true scale the point C would be much further to the right than here shown. Only one half of the curve for steel is given, for want of space.] Within the elastic limit the curves are nearly straight (proving Hooke's law) and if a, a, and a" are the angles made by these straight portions with the axis of X (i.e., of e), we shall have (E t for w. iron) : (E t c. iron) : (E t steel) : : tan a : tan a' : tan a" TENSION. 211 as a graphic relation between their moduli of elasticity (since E u =^-). Beyond the elastic limit tins wrought iron rod shows large increments of elongation for small increments of stress, i.e., the curve becomes nearly parallel to the horizontal axis, until rupture occurs at a stress of 53,000 Ibs. per sq. inch of original sectional area (at rupture this area is some- what reduced, especially in the immediate neighborhood of the section of rupture ; see next article) and after a rel- ative elongation e= about 0.30, or 30%. (The preceding table shows only a portion of the results.) The curve for steel shows a much higher breaking stress (100,000 Ibs. per sq. in.) than the wrought iron, but the total elongation is smaller, e= about 10%. This is an average curve ; tool steels give an elongation at rupture of about 4 to 5%, while soft steels resemble wrought iron in their ductility, giving an extreme elongation of from 10 to 20%. Their breaking stresses range from 70,000 to 150,000 Ibs. or more per sq. inch. Oast iron, being comparatively brit- tle, reaches at rupture an elongation of only 3 or 4 tenths of one per cent., the rupturing stress being about 18,000 Ibs. per sq. inch. The elastic limit is rather ill denned in the case of this metal ; and the proportion of carbon and the mode of manufacture have much influence on its be- havior under test. 193. Lateral Contraction. In the stretching of prisms of nearly all kinds of material, accompanying the elongation of length is found also a diminution of width whose rela- tive amount in the case of the three metals just treated ia about y^ or ^ of the relative elongation (within elastic limit). Thus, in the third experiment in the table of 191, this relative lateral contraction or decrease of diameter = y z to y of s, i.e., about 0.00008. In the case of cast iron and hard steels contraction is not noticeable ex- 212 MECHANICS OF ENGINEERING cept by very delicate measurements, both within and with- out the elastic limit ; but the more ductile metals, as wrought iron and the soft steels, when stretched beyond the elastic limit show this feature of their deformation in a very marked degree. Fig. 201 shows by dotted lines the original contour of a wrought iron rod, while the con- tinuous lines indicate that at rupture. At the cross section of rupture, whose position is determined by some local weakness, the drawing out is peculiarly pronounced. The contraction of area thus produced is some- times as great as 50 or 60% at the fracture. 194. "Flow of Solids." When the change in re- lative position of the elements of a solid is ex- treme, as occurs in the making of lead pipe, drawing of wire, the stretching of a rod of duc- tile metal as in the preceding article, we have FIG. 201. instances of what is called the Flow of Solids, in- teresting experiments on which have been made by Tresca. 195. Moduli of Tenacity. The tensile stress per square inch (of original sectional area) required to rupture a prism of a given material will be denoted by T and called the modulus of ultimate tenacity ; similarly, the modulus of safe tenacity, or greatest safe tensile stress on an element, by T'; while the tensile stress at elastic limit may be called T". The ratio of T to T" is not fixed in practice but depends upon circumstances (from % to ^). Hence, if a prism of any material sustains a total pull or load P, and has a sectional area =F, we have P= FT for the ultimate or breaking load. \ p=FT' " " safe load. V . . (2) p=FT"" " load at elastic limit. ) Of course T' should always be less than T". TENSION. 213 196. Resilience of a Stretched Prism. Fig. 202. In the gradual stretching of a prism, fixed at one extremity, the value of the tensile force P at the other necessarily de- pends on the elongation A at each stage of the lengthening, according to the relation [eq. (1) of 191.] within the elastic limit. (If we place a weight G on the flanges of the unstretched prism and then leave it to the action of gravity and the elastic action of the prism, the weight begins to sink, meeting an increasing pressure P, proportional to X, from the flanges). Suppose the stretching to continue until P reaches some value P" (at elastic limit say), and X a value A". Then the work done so + far is FI I 202. U = mean force x 8 P ace - # P r ' ' ( 4 > But from (2) F = FT', and (see 184 and 191) .-. (4) becomes U=y 2 T e". Fl=% T" e" V . . (5) where V is the volume of the prism. The quantity ^ T"s", or work done in stretching to the elastic limit a cubic inch (or other unit of volume) of the given material, Weis- bach calls the Modulus of Resilience for tension. From (5) it appears that the amounts of work done in stretching to the elastic limit prisms of the same material but of difl'er- ent dimensions are proportional to their volumes simply. The quantity %T"t" is graphically represented by the area of one of the triangles such as OA'B, OA'B" in Fig. 200 ; for (in the curve for wrought iron for instance) the modulus of tenacity at elastic limit is represented by A' B, and e" (i.e., e for elastic limit) by OA'. The remainder of 214 MECHANICS OF ENGINEERING. the area OBC included between the curve and the hori- zontal axis, i.e., from B to C, represents the work done in stretching a cubic unit from the elastic limit to the point of rupture, for each vertical strip having an altitude =p and a width =ds, has an area =pde, i.e., the work done by the stress p on one face of a cubic unit through the dis- tance ds, or increment of elongation. If a weight or load = O be " suddenly "applied to stretch the prism, i.e., placed on the flanges, barely touching them, and then allowed to fall^ when it comes to rest again it has fallen through a height ^, and experiences at this instant some pressure Pi from the flanges; PI=? The work Gh has been entirely expended in stretching the prism, none in changing the kinetic energy of G, which =0 at both beginning and end of the distance ^. Since Pi=%Gr, i.e., is > 6?, the weight does not remain in this position but is pulled upward by the elasticity of the prism. In fact, the motion is harmonic (see 59 and 138). Theoretically, the elastic limit not being passed, the oscillations should continue indefinitely. Hence a load G " suddenly applied " occasions double the tension it would if compelled to sink gradually by a sup- port underneath, which is not removed until the tension is just = G, oscillation being thus prevented. If the weight G sinks through a height =h before strik- ing the flanges, Fig. 202, we shall have similarly, within elastic limit, if ^= greatest elongation, (the mass of rod being small compared with that of G). G(h+^^y 2 P^ '. . . . (6) If the elastic limit is to be just reached we have from eqs. (5) and (6), neglecting ^ compared with h, F . ', . , . (7) TENSION. 215 an equation of condition that the prism shall not be in- jured. Example. If a steel prism have a sectional area of ^ 6q. inch and a length 1=10 ft. =120 inches, what is the greatest allowable height of fall of a weight of 200 Ibs., that the final tensile stress induced may not exceed T"= 30,000 Ibs. per sq. inch, if e" =.002 ? From (7), using the inch and pound, we have 197. Stretching of a Prism by Its Own Weight. In the case of a very long prism such as a mining- pump rod, its weight must be taken into account as well as that of the terminal load P , see Fig. 203. At (a.) the prism is shown in its unstrained condition ; at (6) strained by the load P l and its own weight. Let the cross section be =F, the heaviness of the prism =;-. Then the rela- tive extension of any element at a distance A () P FIG. 208. j from o is (1) (See eq. (1) 191) ; since P^-\-F?x is the load hanging upon the cross section at that locality. Equal dx's, therefore, are unequally elongated, x varying from to I. The total elongation is _ P.* I 72 ur ~ FE t 1FE, Le., A= the amount due to P,, plus an extension which half the weight of the prism would produce, hung at the lower extremity. 216 MECHANICS OF ENGINEERING. The foregoing relates to the deformation of the piece, and is therefore a problem of stiffness. As to the strength of the prism, the relative elongation =ctt.--dx [see eq. (1)], which is variable, must nowhere exceed a safe value e'= I ' +E L (from eq. (1) 191, putting P=FT , and /=/). Now the greatest value of the ratio dX : dx, by inspecting eq. (1), is seen to be at the upper end where x=l. The proper cross section F, for a given load P lt is thus found. Putting j we have p =L _ m (2) 198. Solid of Uniform Strength in Tension, or Twinging body of minimum material supporting its own weight and a terminal load P,. Let it be a solid of revolution. If every cross-section F at a distance x from the lower extrem- ity, bears its safe load FT', every element of the body is doing full duty, and its form is the most economical of material. The lowest section must have an area FIG. 204. F =P^ T', since P, is its safe load. Fig. 204. Consider any horizontal lamina ; its weight is fFdx, (f= heaviness of the material, supposed homogenous), and its lower base F must have Pj+ G for its safe load, i.e. G+P^FT' . . : , (1) in which G denotes the weight of the portion of the solid below F. Similarly for the upper base F+dF, we have . . (2) By subtraction we obtain r Fdx=T'dF', i.e. lcte= ** 1 F TENSION. 217 in which the two variables x and F are separated. By in- tegration we now have (3) i.e., F=F Q er =j er' (4) from which J^may be computed for any value of x. The weight of the portion below any F is found from (1) and (4) ; i.e. while the total extension ^ will be rrri l=*"~l (6) the relative elongation dX-^-dx being the same for every dx and bearing the same ratio to e" (at elastic limit), as T' does to T". 199. Tensile Stresses Induced by Temperature. If the two ends of a prism are immovably fixed, when under no strain and at a temperature t, and the temperature is then low- ered to a value t' t the body suffers a tension proportional to the fall in temperature (within elastic limit). If for a rise or fall of 1 Fahr. (or Cent.) a unit of length of the material would change in length by an amount 7 (called the co-efficient of expansion) a length =1 would be con- tracted an amount ),=f t l(t-t'} during the given fall of tem- perature if one end were free. Hence, if this contraction is prevented by fixing both ends, the rod must be under a tension P, equal in value to the force which would be 218 MECHANICS OF ENGINEERING. necessary to produce the elongation /, just stated, under ordinary circumstances at the lower temperature. From eq. (1) 191, therefore, we have for this tension due to fall of temperature P= For 1 Cent, we may write For Cast iron y = .0000111 ; " Wrought iron .0000120 ; " Steel = .0000108 to .0000114 ; " Copper rj = .0000172 ; " Zinc = .0000300. COMPRESSION OF SHORT BLOCKS. 200. Short and Long Columns. In a prism in tension, its own weight being neglected, all the elements between the localities of application of the pair of external forces pro- ducing the stretching are in the same state of stress, if the external forces act axially (excepting the few elements in the immediate neighborhood of the forces ; these suffering local stresses dependent on the manner of application of the external forces), and the prism may be of any length without vitiating this statement. But if the two external forces are directed toward each other the intervening ele- ments will not all be in the same state of compressive stress unless the prism is comparatively short (or unless numerous points of lateral support are provided). A long prism will buckle out sideways, thus even inducing tensile stress, in some cases, in the elements on the convex side. Hence the distinction between short blocks and long columns. Under compression the former yield by crush- ing or splitting, while the latter give way by flexurp fi- e ' bending). Long columns, then will be treated separately COMPRESSION OF SHORT BLOCKS. in a subsequent chapter. In the present section the blocks treated being about three or four times as long as wide, all the elements will be considered as being under equal compressive stresses at the same time. 201. Notation for Compression. By using a subscript c, we may write JE C = Modulus of Elasticity;* i.e. the quotient of the compressive stress per unit of area divided by the relative " shortening ; also C= Modulus of crushing ; i.e. the force per unit of sec- tional area necessary to rupture the block by crushing ; C'= Modulus of safe compression, a safe compressive stress per unit of area ; and C"= Modulus of compression at elastic limit. For the absolute and relative shortening in length we may still use A and e, respectively, and within the elastic limit may write equations similar to those for tension, F being the sectional area of the block and P one of the ter- minal forces, while p = compressive stress per unit of area of F, viz.: within the elastic limit. Also for a short block Crushing force =FC Compressive force at elastic limit FG" \ . (2) Safe compressive force =FC' 202. Remarks on Crushing. As in 182 for a tensile stress, so for a compressive stress we may prove th-at a * [NOTE. It must be remembered that the modulus of elasticity, whether for normal or shearing stresses, is a number indicative of stiff- ness, not of strength, and has no relation to the elastic limit (except that experiments to determine it must not pass that limit).] 220 MECHANICS OF ENGINEERING. shearing stress =p sin a cos a is produced on planes at an angle a with the axis of the short block, p being the com- pression per unit of area of transverse section. Accord- ingly it is found that short blocks of many comparatively brittle materials yield by shearing on planes making an angle of about 45 with the axis, the expression p sin a cos a reaching a maximum, for a=45 ; that is, wedge- shaped pieces are forced out from the sides. Hence the necessity of making the block three or four times as long as wide, since otherwise the friction on the ends would cause the piece to show a greater resistance by hindering this lateral motion. Crushing by splitting into pieces parallel to the axis sometimes occurs. Blocks of ductile material, however, yield by swelling out, or bulging, laterally, resembling plastic bodies some- what in this respect. The elastic limit is more difficult to locate than in ten- sion, but seems to have a position corresponding to that in tension, in the case of wrought iron and steel. With cast iron, however, the relative compression at elastic limit is about double the relative extension (at elastic limit in tension), but the force producing it is also double. For all three metals it is found that E c =E t quite nearly, so that the single symbol E may be used for both. EXAMPLES IN TENSION AND COMPRESSION. 203. Tables for Tension and Compression. The round num- bers of the following tables are to be taken as rude averages only, for use in the numerical examples following. (The scope and design of the present work admit of nothing more. For abundant detail of the results of the more im- portant experiments of late years, the student is referred to the recent works 'of Profs. Thurston, Burr, Lanza, and Wood). Another column might have been added giving the Modulus of ^Resilience in each case, viz.: y 2 e"T" /77"2\ (which also = ^ ) ; see 196. e is an abstract num- 2 At / EXAMPLES IN TENSION AND COMPRESSION. 221 her, and =X-l, while E tt T", and T are given in pounds per square inch: TABLE OF THE MODULI, ETC., OF MATERIALS IN TENSION. e" e % T" T Material. (Elastic limit.) At Rupture. Mod. of Blast. Elastic limit. Rupture. abst. number. abst. number. bs. per sq. in. Ibs. per sq. in. bs. per sq. in. Soft Steel, .00200 .2500 26,000,000 50,000 80,000 Hard Steel, .00200 .0500 40,000,000 99,000 130,000 Cast Iron, Wro't Iron, Brass, .00066 .00080 .00100 .0020 .2500 14,000,000 28,000,000 10,000,000 9,000 22,000 ( 7,000 \ to j 19,000 18,000 (45,000 \ to (60,000 16.000 to 50,000 Glass, 9,000,000 3,500 Wood, with the fibres, 1 .00200 < to .01100 .0070 to .0150 200,000 to 2,000,000 3,000 to 19,000 6,000 to 28.000 Hemp rope, 7,000 [N.B. Expressed in kilograms per square centim., E t , T and T" would be nu merically about V ]4 as large as above, while e and e" would be unchanged.] TABLE OF MODULI, ETC.; COMPRESSION OF SHORT BLOCKS. e" e E c C" C Material. Elastic limit. At lupture. Mod. of Blast. Elastic limit. Rupture. abst. number abst. number. Ibs. per sq. in. Ibs. per sq. in. Ibs. per sq. in. Soft Steel, 0.00100 30,000,000 30,000 Hard Steel. 0.00120 0.3000 40,000,000 50,000 200,000 Cast Iron, 0.00150 14,000.000 20,000 90,000 Wro't Iron, 0.00080 0.3000 28,000,000 24,000 40,000 Glass, 20,000 Granite, Sandstone, | See 10,000 5,000 Brick, J 3,000 Wood, with the fibres, ( 0.0100 < to ( 0.0400 350,000 2,000,000 2,000 to 10,000 Portland / Cement, f (I 813a) 4,000 222 MECHANICS OF ENGINEERING. 204. Examples. No. 1. A bar of tool steel, of sectional area =0.097 sq. inches, is ruptured by a tensile force of 14,000 Ibs. A portion of its length, originally ^ a foot, is now found to have a length of 0.532 ft. Required T, and e at rupture. Using the inch and pound as units (as in the foregoing tables) we have jT=y^?=144326 Ibs. per sq. in.; (eq. (2) 195) ; while e=(0.532 0.5)xl2-=-(0.50xl2)=0.064. EXAMPLE 2. Tensile test of a bar of " Hay Steel " for the Glasgow Bridge, Missouri. The portion measured was originally 3.21 ft. long and 2.09 in. X 1.10 in. in section. At the elastic limit P was 124,200 Ibs., and the elongation was 0.064 ins. Required E lt T", and e" (for elastic limit). " =.00165 at elastic limit. _ 2"'=124,200H-(2.09x 1.10)= 54,000 Ibs. per sq. in. lbs - er in Nearly the same result for E t would probably have been obtained for values of p and e below the elastic limit. The Modulus of Resilience of the above steel (see 196) would be % s" T "=44.82 inch-pounds of work per cubic inch of metal, so that the whole work expended in stretch- ing to the elastic limit the portion above cited is #= ^ e" T" F=3968. inch -lbs. An equal amount of work will be done by the rod in re- covering its original length. EXAMPLE 3. A hard steel rod of ^ sq. in. section and 20 ft. long is under no stress at a temperature of 130 EXAMPLES IN TENSION AND COMPRESSION. 223 Cent., and is provided with flanges so that the slightest contraction of length will tend to bring two walls nearer together. If the resistance to this motion is 10 tons how low must the temperature fall to cause any motion ? i) be- ing = .0000120 (Cent, scale). From 199 we have, ex- pressing P in Ibs. and F in sq. inches, since E t = 40,000,000 (6s. per sq. inch, 10x2,000=40,000,000 x # X (130-*') X 0.000012; whence r=46.6 Centigrade. EXAMPLE 4. If the ends of an iron beam bearing 5 tons at its middle rest upon stone piers, required the necessary bearing surface at each pier, putting C' for stone =200 Ibs. per sq. inch. 25 sq. in., Ans. EXAMPLE 5. How long must a wrought iron rod be, supported vertically at its upper end, to break with its own weight ? 216,000 inches, Ans. EXAMPLE 6. One voussoir (or block) of an arch -ring presses its neighbor with a force of 50 tons, the joint hav- ing a surface of 5 sq. feet ; required the compression per sq. inch. 138.8 Ibs. per sq. in., Ans. 205. Factor of Safety. When, as in the case of stone, the value of the stress at the elastic limit is of very uncertain determination by experiment, it is customary to refer the value of the safe stress to that of the ultimate by making it the w'th portion of the latter, n is called a factor of safety, and should be taken large enough to make the safe stress come within the elastic limit. For stone, n should not be less than 10, i.e. (7 / =(7-i-w; (see Ex. 6, just given). 206. Practical Notes. It was discovered independently by Commander Beardslee and Prof. Thurston, in 1873, that if wrought iron rods were strained considerably beyond the elastic limit and allowed to remain free from stress 224 MECHANICS OF ENGINEERING. for at least one day thereafter, a second test would show higher limits both elastic and ultimate. When articles of cast iron are imbedded in oxide of iron and subjected to a red heat for some days, the metal loses most of its carbon, and is thus nearly converted into wrought iron, lacking, however, the property of welding. JBeing malleable, it is called malleable cast iron. Chrome steel (iron and chromium) and tungsten steel pos- sess peculiar hardness, fitting them for cutting tools, rock drills, picks, etc. By fatigue of metals we understand the fact, recently dis- covered by Wohler in experiments made for the Prussian Government, that rupture may be produced by causing the stress on the elements to vary repeatedly between two limiting values, the highest of which may be considerably below T (or (7), the number of repetitions necessary to produce rupture being dependent both on the range of variation and the higher value. For example, in the case of Phoenix iron in tension, rupture was produced by causing the stress to vary from ) to 52,800 Ibs. per sq. inch, 800 times ; also, from to <4,000 Ibs. per sq. inch 240,853 times ; while 4,000,000 va- riations between 26,400 and 48,400 per sq. inch did not cause rupture. Many other experiments were made and the following conclusions drawn (among others): Unlimited repetitions of variations of stress (Ibs. per yjq. in.) between the limits given below will not injure the metal (Prof. Burr's Materials of Engineering). Wrought iron I From 17 ' 600 Com P- to 17,600 % Tension. ' ( " to 33,000 ( From 30,800 Comp. to 30,800 Tension. Axle Cast Steel.-] " to 52,800 ( " 38500 Tens, to 88,000 (See p. 232 for an addendum to this paragraph.) SHEARING. SHEARING. 225 207. Rivets. The angular distortion called shearing strain in the elements of a body, is specially to be provided for in the case of rivets joining two or more plates. This distortion is shown, in Figs. 205 and 206, in the elements near ';he plane of contact of the plates, much exaggerated. FIG. 206. In Fig. 205 (a lap-joint) the rivet is said to be in single shear ; in Fig. 206 in double shear. If P is just great enough to shear off the rivet, the modulus of ultimate shear- ing, which may be called S, (being the shearing force per unit of section when rupture occurs) is (1) in which F= the cross section of the rivet, its diameter being =d. For safety a value >S"= ]/^ to y> of S should be taken for metal, in order to be within the elastic limit. As the width of the plate is diminished by the rivet hole the remaining sectional area of the plate should be ample to sustain the tension P, or 2P, (according to the plate considered, see Fig. 206), P being the safe shearing force for the rivet. Also the thickness t of the plate should be such that the side of the hole shall be secure against crushing ; P must not be > C'td, Fig. 205. Again, the distance a, Fig. 205, should be such as to prevent the tearing or shearing out of the part of the plate between the rivet and edge of the plate. 226 MECHANICS OF ENGINEERING. For economy of material the seam or joint should be no more liable to rupture by one than by another, of the four modes just mentioned. The relations which must then subsist will be illustrated in the case of the " butt- joint " with two cover-plates, Fig. 207. Let the dimen- sions be denoted as in the figure and the total tensile force on the joint be = Q. Each rivet (see also Fig. 206) is ex- posed in each of two of its sections to a shear of ^ Q, hence for safety against shearing of rivets we put (1) Along one row of rivets in the main plate the sectional area for resisting tension is reduced to (b 3d)t lt hence for safety against rupture of that plate by the tension Q, we put ' ...... (2) Equations (1) and (2) suffice to determine d for the rivets and t { for the main plates, Q and b being given; but the values thus obtained should also be examined with refer- ence to the compression in the side of the rivet hole, i.e., yh Q must not be > C't t d. [The distance a, Fig. 205, to the edge of the plate is recommended by different authorities to be from d to 3d] Similarly, for the cover-plate we must have . . . (3) < and / 2 Qnot > C'td~ SHEARING. 227 If the rivets do not fit their holes closely, a large margin should be allowed in practice. Again, in boiler work, the pitch, or distance between centers of two consecutive rivets may need to be smaller, to make the joint steam-tight, than would be required for strength alone. 208. Shearing Distortion. The change of form in an ele- ment due to shearing is an angular deformation and will be measured in /r-measure. This angular change or dif- ference between the value of the corner angle during strain and *-TT, its value before strain, will be called d, and is proportional (within elastic limit) to the shearing stress per unit of area, p s , existing on all the four faces whose angles with each other have been changed. Fig. 208. (See 181). By 184 the Modulus of Shearing Elasticity is the quotient obtained by dividing p K by d ; i.e. (elastic limit not passed), ..... (1) or inversely, S=p s +E s (1)' The value of E & for different substances is most easily determined by experiments on torsion in which shearing is the most promi- nent stress. (This prominence depends on the position of the bounding planes of the element considered ; e.g., in Fig. ] /_ i_ 208, if another element were considered within the one there shown and with FIG. 908. its planes at 45 with those of the first, we should find tension alone on one pair of opposite faces, compression alone on the other pair.) It will be noticed that shearing stress cannot be present on two opposite faces only, but exists also on another pair of faces (those perpendicular to the stress on the first), forming a couple of equal and opposite moment to the first, this being necessary for the equilibrium of the element, even when 228 MECHANICS OF ENGINEERING. tensile or compressive stresses are also present on the faces considered. 209. Shearing Stress is Always of the Same Intensity on the Four Faces of an Element. (By intensity is meant per unit of area ; and the four faces referred to are those perpen- dicular to the paper in Fig. 208, the shearing stress being parallel to the paper.) Let dx and dz be the width and height of the element in Fig. 208, while dy is its thickness perpendicular to the paper. Let the intensity of the shear on the right hand face be =y s , that on the top face =p s . Then for the ele- ment aw a free body, taking moments about the axis per- pendicular to paper, we have q s dz dy X dx p s dx dy x dz =0 .*. g s =p s (dx and dz being the respective lever arms of the forces q s dz dy and p s dx dy.) Even if there were also tensions (or compressions) on one or both pairs of faces their moments about would balance (or fail to do so by a differential of a higher order) independently of the shears, and the above result would still hold. 210. Table of Moduli for Shearing. if E a 8* 8 Material. i.e. 6 at elastic limit. Mod. of Elasticity for Shearing. (Elastic limit.) (Rupture.) Ibs. per sq. in. arc in ^-measure. Ibs. per sq. in. Ibs. per sq. in Soft Steel, 9,000,000 70,000 Hard Steel, 0.0032 14,000,000 45,000 90,000 Cast Iron, 0.0021 7,000,000 15,000 30,000 Wrought Iron, 0.0022 9,000,000 20,000 50,000 Brass, 5,000,000 Glass, Wood, across! fibre, 1 1,500 to 8,000 Wood, along ( fibre, 1 500 to 1,200 SHEARING. 229 As in the tables for tension and compression, the above values are averages. The true values may differ from these as much as 30 per cent, in particular cases, accord" ing to the quality of the. specimen. 211. Punching rivet holes in plates of metal requires the overcoming of the shearing resistance along the convex surface of the cylinder punched out. Hence if d = diam- eter of hole, and t= the thickness of the plate, the neces- sary force for the punching, the surface sheared being F= fad, is P=Sfad . . . . (2) Another example of shearing action is the " stripping " of the threads of a screw, when the nut is forced off lon- gitudinally without turning, and resembles punching in its nature. 212. EandE s ; Theoretical Relation. In case a rod is in tension within the elastic limit, the relative (linear) lateral contraction (let this =m) is so connected with E t and E m that if two of the three are known the third can be de- duced theoretically. This relation is proved as follows, by Prof. Burr. Taking an elemental cube with four of its faces at 45 with the axis of the piece, Fig. 209, the axial half-diagonal AD becomes of a length AD I =AD+.AD under stress, while the transverse half diagonal contracts to a length B'D'=AD m.AD. The angular distortion d E A PIG. 209. 212. FIG. 210. 230 MECHANICS OF ENGINEERING. is supposed very small compared with 90 and is due to the shear p s per unit of area on the face BG (or BA\ From the figure we have tan(45_) = ==l_^e, appro*. [But, Fig. 210, tan(45 x)=l 2# nearly, where a; is a small angle, for, taking CA= unity= AE, tan AD= AF= AEEF. Now approximately EF= 'EG. ^/2 and EG= ~BI)^/2=x^/2 .'. AF= l2x nearly.] Hence 1 <5= 1 m e ; or S= m+e . . (2) Eq. (2) holds good whatever the stresses producing the deformation, but in the present case of a rod in tension, if it is an isotrope, and if p = tension per unit of area on its transverse section, (see 182, putting =45), we have E t =p+s and fi*=(p s on BC)+3=y^p+d. Putting also (m : )=r, whence ra=r, eq. (2) may finally be written Prof. Bauschinger, experimenting with cast iron rods, found that in tension the ratio m : e was = fib* as an average, which in eq. (3) gives His experiments on the torsion of cast iron rods gave # s = 6,000,000 to 7,000,000 Ibs. per sq. inch. By (4), then, E t should be 15,000,000 to 17,500,000 which is approxi- mately true ( 203). Corresponding results may be obtained for short blocks in compression, the lateral change being a dilatation in- stead of a contraction. SHEARING. 231 13. Examples in Shearing. EXAMPLE 1. Required the proper length, a, Fig. 211, to guard against the shearing off, along the grain, of the portion ab, of a wooden tie-rod, the force P being = 2 tons, and the width of the tie = 4 inches. Using a value of S' = 100 Ibs. per sq. in., we put 5a#'= 4,000 cos 45; i.e. FIG. an. a= (4,000 xO. 707) -f- (4x100)= 7.07 inches. EXAMPLE 2. A fy in. rivet of wrought iron, in single shear (see Fig. 205) has an ultimate shearing strength P= FS = i^THftS^ i^X }& ) 2 x 50,000= 30,050 Ibs. F or safety, putting #'=8,000 instead of S,P'= 4,800 Ibs. is its safe shearing strength in single shear. The wrought iron plate, to be secure against the side- crushing in the hole, should have a thickness t, computed thus : P'=tdC'; or 4,800= *.^xX2,000 .% *=0.46 in. If the plate were only 0.23 in. thick the safe value of P would be only ^ of 4,800. EXAMPLE 3. Conversely, given a lap-joint, Fig. 205, in which the plates are * in. thick and the tensile force on the joint = 600 Ibs. per linear inch of seam, how closely must y^ inch rivets be spaced in one row, putting #'=8,000 and C"=12,000 Ibs. per sq. in. ? Let the distance between centres of rivets be =x (in inches), then the force upon each rivet =600o;, while its section 1^=0.44 sq. in. Having regard to the shearing strength of the rivet we put 600x= 0.44x8,000 and obtain #=5.86 in.; but considering that the safe crushing resistance of the hole is = i^^. 12,000= 2,250 Ibs., 600x=2,250 gives z=3.75 inches, which is the pitch to be adopted. What is the tensile strength of the reduced sectional area of the plate, with this pitch ? 17SS87 232 MECHANICS OF ENGINEERING. EXAMPLE 4. Double butt-joint ; (see Fig. 207) ; % inch plate; ^ in. rivets; T= (7=12,000 ; '=8,333; width of plates =14 inches. Will one row of rivets be sufficient at each side of joint, if #=30,000 Ibs.? The number of rivets = ? Here each rivet is in double shear and has therefore a double strength as regards shear. In double shear the safe strength of each rivet =ZFS'= 7,333 Ibs. Now 30,000 -r- 7,333=4.0 (say). With the four rivets in one row the re- duced sectional area of the main plate is =[14 4x ^] X 3 /a =4.12 sq. in., whose safe tensile strength is =^ T 2 7 '=4.12x 12,000=49,440 Ibs.; which is > 30,000 Ibs. .-. main plate is safe in this respect. But as to side-crushing in holes in main plate we find that C't.d (i.e. 12,000 X 3 / 8 X % =3,375 Ibs.) is <^Q i.e. <7,500 Ibs., the actual force on side of hole. Hence four rivets in one row are too few unless thickness of main plate be doubled. Will eight in one row be safe ? 213a. (Addendum to 206.) Elasticity of Stone and Cements. Experiments by Gen. Gillmore with the large Watertown testing-machine in 1883 resulted as follows (see p. 221 for notation) : With cubes of Haverstraw Freestone (a homogeneous brown- stone) from 1 in. to 12 in. on the edge, E was found to be from 900,000 to 1,000,000 Ibs. per sq. in. approximately ; and about 4,000 or 5,000 Ibs. per sq. in. Cubes of the same range of sizes of Dyckerman's Portland cement gave E from 1,350,000 to 1,630,000, and C from 4,000 to 7,000, Ibs. per sq. in. Cubes of concrete of the above sizes, made with the Newark Cc.'s Rosendale cement, gave E about 538,000, while cubes of cement-mortar, and some of concrete, both made with National Portland cement, showed E from 800,000 to 2,000,- 000 Ibs. per sq. in. The compressibility of brick piers 12 in. square in section and 16 in. high was also tested. They were made of common North River brick with mortar joints f in. thick, and showed a value for E of about 300,000 or 400,000, while at elastic limit C" was on the average 1,000, Ibs. per sq. in. TORSION. 233 CHAPTER II. TORSION. 214. Angle of Torsion and of Helix. When a cylindrical beam or shaft is subjected to a twisting or torsional action, i. e. when it is the means of holding in equilibrium two couples in parallel planes and of equal and opposite mo- ments, the longitudinal axis of symmetry remains straight (and the elements along it exper- ience no stress (whence it may be called the "line of no twist"), while the lines originally parallel to it assume the form of helices, each element of which is distorted in its angles (originally right angles), the amount of distortion being assumed pro- portional to the radius of the helix. The directions of the faces of any element were originally as follows : two radial, two in consecutive transverse sections, and the other two tangent to two consecutive circular cylinders whose com- mon axis is that of the shaft. E.g. in Fig. 212 we have an unstrained shaft, while in Fig. 213 it holds the two 234 MECHANICS OF ENGINEEKIXG. couples (of equal moment P a = Q b) in equilibrium. These couples act in parallel planes perpendicular to the axis of the prism and a distance, l t apart. Assuming that the transverse sections remain plane and parallel during tor- sion, any surface element, m, which in Fig. 212 was entire- ly right-angled, is now distorted. Two of its angles have been increased, two diminished, by an amount 8, the angle between the helix and a line parallel to the axis. Suppos- ing m to be the most distant of any element from the axis, this distance being e, any other element at a distance z from the axis experiences an angular distortion = - d. 6 If now we draw B' parallel to O'A the angle B B', =a, is called the Angle of Torsion, while d may be called the helix angle; the former lies in a transverse plane, the latter in a plane tangent to the cylinder. Now tan d = (linear arc B B')--l ; but lin. arc B B'= ea\ hence, putting 8 for tan 8, (8 being small) -? d) (8 and a both in ic measure). 215. Shearing Stress on the Elements. The angular distor- tion, or shearing strain, 8, of any element (bounded as al- ready described) is due to the shearing stresses exerted on it by its neighbors on the four faces perpendicular to the tangent plane of the cylindri- cal shell in which the element is situated. Consider these neighboring elements of an outside element removed, and the stresses put in ; the latter are accountable for the dis- tortion of the element and so^ TORSION. 235 hold it in equilibrium. Fig. 214 shows this element "free." "Within the elastic limit d is known to be propor- tional to j9,, the shearing stress per unit of area on the faces whose relative angular positions have been changed. That is, from eq. (1) 208, d=p^E s ', whence, see (1) of 214, (2) In (2) p a and e both refer to a surface element, e being the radius of the cylinder, and p,. the greatest intensity of shearing stress existing in the shaft. Elements lying nearer the axis suffer shearing stresses of less intensity in pro- portion to their radial distances, i.e., to their helix-angles. That is, the shearing stress on that face of the element which forms a part of a transverse section and whose dis- tance from the axis is z, is p } = p s , per unit of area, and the total shear on the face is pdF, dF being the area of the face. 216. Torsional Strength. We are now ready to expose the full transverse section of a shaft under torsion, to deduce formulae of practical utility. Making a right section of the shaft of Fig. 213 anywhere between the two couples and considering the left hand portion as a free body, the forces holding it in equilibrium are the two forces P of the left-hand couple and an infinite number of shearing forces, each tangent to its circle of radius z, on the cross section exposed by the removal of the right-hand portion. The cross section is assumed to remain plane during tor- sion, and is composed -of an infinite number of dF's, each being the area of an exposed face of an element ; see Fig. 215. MECHANICS OF ENGINEEKING. Each elementary shearing force = p*dF, and z is its lever arm about the axis Oo . For equilibrium, 2* (mom.) about the axis Oo must =0 ; i.e. in detail or, reducing, P* fz*dF=Pa; or, M. eJ e (3) Eq. (3) relates to torsional strength, since it contains p s , the greatest shearing stress induced by the torsional couple, whose moment Pa is called the Moment of Torsion, the stresses in the cross section forming a couple of equal and opposite moment. Jp is recognized as the Polar Moment of Inertia of the cross section, discussed in 94 ; e is the radial distance of the outermost element, and = the radius for a circular shaft. 217. Torsional Stiffness. In problems involving the angle of torsion, or deformation of the shaft, we need an equa- tion connecting Pa and a, which is obtained by substitut- ing in eq. (3) the value of p B in eq. (2), whence Pa . TORSION. 237 From this it appears that the angle of torsion, a, is propor- tional to the moment of torsion, Pa, within the elastic limit ; a must be expressed in ^-measure. Trautwine cites 1 (i.e. a= 0.0174) as a maximum allowable value for shafts. 218. Torsional Resilience is the work done in twisting a shaft from an unstrained state until the elastic limit is reached in the outermost elements. If in Fig. 213 we imagine the right-hand extremity to be fixed, while the other end is gradually twisted through an angle a\ each force J^* of the couple must be made to increase gradually from a zero value up to the value P l} corresponding to a t . In this motion each end of the arm a describes a space = Ki, and the mean value of the force = ^4P l (compare 196). Hence the work done in twisting is . . (5) By the aid of preceding equations, (5) can be written or = &L (6) 1 If for p s we write S' (Modulus of safe shearing) we have for the safe resilience of the shaft If the torsional elasticity of an originally unstrained shaft is to be the means of arresting the motion of a moving mass whose weight is G, (large compared with the parts intervening) and velocity =v, we write ( 133) V=.ti <7 2' as the condition that the shaft shall not be injured. 238 MECHANICS OF ENGINEERING. 219. Polar Moment of Inertia. For a shaft of circular cross section (see 94) 7 p =i^^r 4 ; for a hollow cylinder I^y^rtrf) ; while for a square shaft 7 p =i^6 4 , b being the side of the square ; for a rectangular cross-section sides b and h, I p =^bh(b 2 -\-h 2 ). For a cylinder e=r; if hol- low, e=r , the greater radius. For a square, 6 220. Non-Circular Shafts. If the cross-section is not cir- cular it becomes warped, in torsion, instead of remaining plane. Hence the foregoing theory does not strictly ap- ply. The celebrated investigations of St. Venant, how- ever, cover many of these cases. (See 708 of Thompson and Tait's Natural Philosophy ; also, Prof. Burr's Elas- ticity and Strength of the Materials of Engineering). His -results give for a square shaft (instead of the . . *? . (1) and Pa^/jPp., instead of eq. (3) of 216, p n being the greatest shearing stress. The elements under greatest shearing strain are found at the middles of the sides, instead of at the corners, when the prism is of square or rectangular cross-section. The warping of the cross-section in such a case is easily veri- fied by the student by twisting a bar of india-rubber in his fingers. 221. Transmission of Power. Fig. 216. Suppose the cog- wheel B to cause A, on the same shaft, to revolve uni- formly and overcome a resis- tance Q, the pressure of the teeth of another cog-wheel, B being driven by still another FIG. 216. wheel. The shaft AB is un- TORSION. 239 der torsion, the moment of torsion being =Pa= Qb. (P l and $1 the bearing reactions have no moment about the axis of the shaft). If the shaft makes u revolutions per unit-time, the work transmitted (transmitted ; not expend- ed in twisting the shaft whose angle of torsion remains constant, corresponding to Pa) per unit-time, i.e. the Power, is L=-P.2xa.u=27tuPa . . . (8) To reduce L to Horse Power ( 132), we divide by N t the number of units of work per unit-time constituting one H. P. in the system of units employed, i.e., Horse Power =H. P.= For example ^=33,000 ft.-lbs. per minute, or =396,000 inch -Ibs. per minute ; or = 550 ft.-lbs. per second. Usually the rate of rotation of a shaft is given in revolutions per minute, But ei. (8) happens to contain Pa the moment of torsion acting to maintain the constant value of the angle of tor- sion, and since for safety (see eq. (3) 216) Par=S'I p +e, with J p = y^Ttr^ and e=r for a solid circular shaft, we have for such a shaft (9) which is the safe H. P., which the given shaft can trans- mit at the given speed. /S" may be made 7,000 Ibs. per sq. inch for wrought iron ; 10,000 for steel, and 5,000 for cast- iron. If the value of Pa fluctuates periodically, as when a shaft is driven by a connecting rod and crank, for (H. P.) we put wX(H. P.), m being the ratio of the maximum to the mean torsional moment ; m = about 1V 2 under ordi- nary circumstances (Cotterill). 240 MECHANICS OF ENGINEERING. 222. Autographic Testing Machine. The principle of Pro Thurston's invention bearing this name is shown in Fig FIG. 217. 217. The test-piece is of a standard shape and size, its central cylinder being subjected to torsion. A jaw, carry- ing a handle (or gear-wheel turned by a worm) and a drum on which paper is wrapped, takes a firm hold of one end of the test-piece, whose further end lies in another jaw rigidly connected with a heavy pendulum carrying a pen- cil free to move axially. By a continuous slow motion of the handle the pendulum is gradually deviated more and more from the vertical, through the intervention of the test-piece, which is thus subjected to an increasing tor- sional moment. The axis of the test-piece lies in the axis of motion. This motion of the pendulum by means of a properly curved guide, WR, causes an axial (i.e., parallel to axis of test-piece) motion of the pencil A, as well as an angular deviation /9 equal to that of the pendulum, and this axial distance CF,=sT, of the pencil from its initial position measures the moment of torsion=f > a=jPc sin )tf. As the piece twists, the drum and paper move relatively to the pencil through an angle sUo equal to the angle TORSION. 241 of torsion so far attained. The abscissa so and ordinate sT of the curve thus marked on the paper, measure, when the paper is unrolled, the values of a and Pa through all the stages of the torsion. Fig. 218 shows typical ANG. TORSION 60 iurves thus obtained. Many valuable indications are given by these strain diagrams as to homogeneousness of composition, ductility, etc., etc. On relaxing the strain at any stage within the elastic limit, the pencil retraces its path ; but if beyond that limit, a new path is taken called an "elasticity -line," in general parallel to the first part of the line, and showing the amount of angular re- covery, BC, and the permanent angular set, OB. 223. Examples in Torsion. The modulus of safe shearing strengtn, S', as given in 221, is expressed in pounds per square inch ; hence these two units should be adopted throughout in any numerical examples where one of the above values for S' is used. The same statement applies to the modulus of shearing elasticity, .Z7 S , in the table of 210. EXAMPLE 1. Fig. 216. With P = 1 ton, a = 3 ft., I = 10 ft., and the radius of the cylindrical shaft r=2.5 inches, required the max. shearing stress per sq. inch, p a the shaft being of wrought iron. From eq. (3) 216 which is a safe value for any ferrous metal. 242 MECHANICS OF ENGINEERING. EXAMPLE 2. What H. P. is the shaft in Ex. 1 transmit- ting, if it makes 50 revolutions per minute ? Let u = number of revolutions per unit of time, and N= the num- ber of units of work per unit of time constituting one horse-power. Then H. I*.=Pu%7ra-r-N, which for the foot- pound-minute system of units gives H. P.=2,000x50x2^x3--33,000=57^: H. P. EXAMPLE 3. What different radius should be given to the shaft in Ex. 1, if two radii at its extremities, originally parallel, are to make an angle of 2 when the given moment of torsion is acting, the strains in the shaft remaining con- stant. From eq. (4) 217, and the table 210, with a=^= 0.035 radians (i.e. TT -measure), and .Z^y^rr 4 , we have 2,000x36x120 _ ^0.035x9,000,000 .. r=2.04 inches. (This would bring about a different p., but still safe.) The foregoing is an example in stiffness. EXAMPLE 4. A working shaft of steel (solid) is to trans- mit 4,000 H. P. and make 60 rev. per minute, the maximum twisting moment being 1^ times the average; required its diameter. d=14.74 inches. Ans. EXAMPLE 5. In example 1, p A = 2,930 Ibs. per square inch ; what tensile stress does this imply on a plane at 45 with the pair of planes on which p s acts ? Fig. 219 shows p.dx* p.dx 1 FIG. 220. TORSION. 243 a small cube, of edge =dx, (taken from the outer helix of Fig. 215,) free and in equilibrium, the plane of the paper being tangent to the cylinder ; while 220 shows the portion BDC, also free, with the unknown total tensile stresspcfcc 2 ^ acting on the newly exposed rectangle of area =dxXdx^"2, p being the unknown stress per unit of area. From sym- metry the stress on this diagonal plane has no shearing component. Putting 2 [components normal to BD]=Q, we have pdx z */2=2dx 2 p s cos4:5 =dx t p s ^2~.'.p=p s . (1) That is, a normal tensile stress exists in the diagonal plane BD of the cubical element equal in intensity to the shearing stress on one of the faces, i.e., =2,930 Ibs. per sq. in. in this case. Similarly in the plane AC will be found a compressive stress of 2,930 Ibs. per sq. in. If a plane surface had been exposed making any other angle than 45 with the face of the cube in Fig. 219, we should have found shearing and normal stresses each less than p a per sq. inch. Hence the interior dotted cube in 219, if shown " free " is in tension in one direction, in compression in the other, and with no shear, these normal stresses having equal intensities. Since S' is usually less than T' or C', it p s is made = >S" the tensile and compressive actions are not injurious. It follows therefore that when a cylinder is in torsion any helix at an angle of 45 with the axis is a line of tensile, or of compressive stress, according as it is a right or left handed helix, or vice versa. EXAMPLE 6. A solid and a hollow cylindrical shaft, of equal length, contain the same amount of the same kind of metal, the solid one fitting the hollow of the other. Compare their torsional strengths, used separately. The solid shaft has only ^ the strength of the hollow one. Ans. 244 MECHANICS OF CHAPTER III. FLEXURE OF HOMOGENEOUS PRISMS UNDER PERPENDICULAR FORCES IN ONE PLANE. 224. Assumptions of the Common Theory of Flexure. When a prism is bent, under the action of external forces per- pendicular to it and in the same plane with each other, it may be assumed that the longitudinal fibres are in tension on the convex side, in compression on the concave side, and that the relative stretching or contraction of the ele- ments is proportional to their distances from a plane in- termediate between, with the understanding that the flex- ure is slight and that the elastic limit is not passed in any element. This " common theory " is sufficiently exact for ordinary engineering purposes if the constants employed are prop- erly determined by a wide range of experiments, and in- volves certain assumptions of as simple a nature as possi- ble, consistently with practical facts. These assumptions are as follows, (for prisms, and for solids with variable cross sections, when the cross sections are similarly situated as regards a central straight axis) and are approximately borne out by experiment : (1.) The external or " applied " forces are all perpendicu- lar to the axis of the piece and lie in one plane, which may be called the force-plane; the force-plane contains the axis of the piece and cuts each cross-section symmetri- cally; (2.) The cross-sections remain plane surfaces during flexure ; (3.) There is a surface (or, rather, sheet of elements) which is parallel to the axis and perpendicular to the force-plane, and along which the elements of the solid ex- FLEXURE. 245 perieuce no tension nor compression in an axial direction, this being called the Neutral Surface ; (4.) The projection of the neutral surface upon the force plane (or a || plane) being called the Neutral Line or Elastic Curve, the bending or flexure of the piece is so slight that an elementary division, ds, of the neutral line may be put =dx, its projection on a line parallel to the direction of the axis before flexure ; (5.) The elements of the body contained between any two consecutive cross-sections, whose intersections with the neutral surface are the respective Neutral Axes of the sections, experience elongations (or contractions, accord- ing as they are situated on one side or the other of the neutral surface), in an axial direction, whose amounts are proportional to their distances from the neutral axis, and indicate corresponding tensile or compressive stresses ; (6.) E t =E c -, (7.) The dimensions of the cross-section are small com- pared with the length of the piece ; (8.) There is no shear perpendicular to the force plane on internal surfaces perpendicular to that plane. In the locality where any one of the external forces is Applied, local stresses are of course induced which demand separate treatment. These are not considered at present. 225. Illustration. Consider the case of flexure shown in Fig. 221. The external forces are three (neglecting the 246 MECHANICS OF ENGINEERING. weight of the beam), viz.: P lt P 2 , and P 3 . P l and P 3 are loads, P 2 the reaction of the support. The force plane is vertical. N^L is the neutral line or elastic curve. NA is the neutral axis of the cross -section at m ; this cross-section, originally perpendicular to the sides of the prism, is during flexure ~| to their tangent planes drawn at the intersection lines ; in other words, the side view QNB, of any cross-section is perpendicular to the neutral line. In considering the whole prism free we have the system P lt P 2 , and P 3 in equilibrium, whence from 2T=0 we have P 2 =Pi+P 3 , and from I* (mom. about O) = 0, P 3 Z 3 = PI?!- Hence given P l we may determine the other two external forces. A reaction such as P 2 is some- times called a supporting force. The elements above the neutral surface N^LS&re in tension ; those below in com- pression (in an axial direction). 226. The Elastic Forces. Conceive the beam in Fig. 221 separated into two parts by any transverse section such as QA, and the portion NiON, considered as a free body in Fig. 222. Of this free body the surface QAB is one of PLEXURfc. 247 the bounding surfaces, but was originally an internal sur- face of the beam m Fig. 221. Hence in Fig. 222 we must put in the stresses acting on all the dF's or elements of area of QAB. These stresses represent the actions of the body taken away upon the body which is left, and according to assumptions (5), (6) and (8) consist of normal stresses (ten- sion or compression) proportional per unit of area, to the distance, 2, of the cLF's from the neutral axis, and of shear- ing stresses parallel to the force-plane (which in most cases will be vertical). The intensity of this shearing stress on any dF varies with the position of the dF with respect to the neutral axis, but the law of its variation will be investigated later ( 253 and 254). These stresses, called the Elastic Forces of the cross-section exposed, and the external forces P l and P 2 , form a system in equilibrium. We may therefore ap- ply any of the conditions of equilibrium proved in 38. 227. The Neutral Axis Contains the Centre of Gravity of the Cross-Section. Fig. 222. Let e= the distance of the outer- most element of the cross-section from the neutral axis, and the normal stress per unit of area upon it be =p, whether tension or compression. Then by assumptions (5) and (6), 224, the intensity of normal stress on any dF is = -1 p and the actual normal stress on any c^is^ - pdF . (1) This equation is true for dF'a having negative 's, i.e. on the other side of the neutral axis, the negative value of the force indicating normal stress of the opposite char- acter ; for if the relative elongation (or contraction) of two axial fibres is the same for equal z's, one above, the other below, the neutral surface, the stresses producing the changes in length are also the same, provided E^E C \ see 184 and 201. 248 MECHANICS OF ENGINEERING. For this free body in equilibrium put IX=Q (Xisa horizontal axis). Put the normal stresses equal to their X components, the flexure being so slight, and the X com- ponent of the shears = for the same reason. This gives (see eq. (1) ) C- pdF= ; i.e. IL CdFz= ; or, Fl=0 (2) / e e J e IH which 2= distance of the centre of gravity of the cross- section from the neutral axis, from which, though un- known in position, the a's have been measured (see eq. (4) 23). In eq. (2) neither p^-e nor F can be zero .-. z 'must = ; i.e. the neutral axis contains the centre of gravity. Q. E. D. [If the external forces were not all perpendicular to the beam this result would not be obtained, necessarily.] 228. The Shear. The " total shear," or simply the " shear," in the cross-section is the sum of th.e vertical shearing stresses on the respective dF's. Call this sum J, and we shall have from the free body in Fig. 222, by putting 2T=0 (Y being vertical) P 2 P l J=Q.'.J=P 2 P l . . (3) That is, the shear equals the algebraic sum of the ex- ternal forces acting on one side (only) of the section con- sidered. This result implies nothing concerning its mode of distribution over the section. 229. The Moment. By the "Moment of Flexure" or simply the Moment, at any cross- section is meant the sum of the moments of the elastic forces of the section, taking the neutral axis as an axis of moments. In this summa- tion the normal stresses appear alone, the shear taking no part, having no lever arm about the neutral axis. Hence, Fig. 222, the moment of flexure 1028 FLEXURE. 249 This function, CdFz 2 , of the cross-section or plane figure is the quantity called Moment of Inertia of a plane figure, 85. For the free body in Fig. 222, by putting ^(mom.s about the neutral axis NA)=0, we have then =0, or in general, ^=M . (5) in which M signifies the sum of moments, about the neutral axis of the section, of all the forces acting on the free body considered, exclusive of the elastic forces of the exposed section itself. 230. Strength in Flexure. Eq. (5) is available for solving problems involving the Strength of beams and girders, since it contains p, the greatest normal stress per unit of area to be found in the section. In the cases of the present chapter, where all the exter- nal forces are perpendicular to the prism or beam, and have therefore no components parallel to the beam, i.e. to the axis X, it is evident that the normal stresses in any section, as QB Fig. 222, are equivalent to a couple ; for the condition ^Y=0 falls entirely upon them and cannot be true unless the resultant of the tensions is equal, parallel, and opposite to that of the compressions. These two equal and parallel resultants, not being in the same line, form a couple ( 28), which we may call the stress-couple. The moment of this couple is the " moment of flexure " p , and it is further evident that the remaining forces in Fig. 222, viz.: the shear J and the external forces P l and P 2 , are equivalent to a couple of equal and opposite moment to the one formed by the normal stresses. M.T. 250 MECHANICS OF ENGINEERING. 231, Flexural Stiffness. The neutral line, or elastic curve* containing the centres of gravity of all the sections, was originally straight ; its radius of curvature at any point, as N, Fig. 222, during flexure may be introduced as fol- lows. QB and U'V are two consecutive cross-sections, originally parallel, but now inclined so that the intersec- tion (7, found by prolonging them sufficiently, is the centre of curvature of the ds (put =dx) which separates them at N, and CG=p= the radius of curvature of the elastic curve at N. From the similar triangles U'UGf&nd GNCwe have dX : dx : : e :. p, in which dX is the elongation, U' U, of a portion, originally =dx, of the outer fibre. But the rela- tive elongation e= -= of the latter is, by 184, within the elastic limit, =-^L.*. = and eq. (5) becomes E E p (6) From (6) the radius of curvature can be computed. E= the value of E t =E c , as ascertained from experiments in bending. To obtain a differential equation of the elastic curve, (6) may be transformed thus, Fig. 223. The curve being very AXI8X flat, consider two consecutive ds's with equal dx's ; they may be put = their dx's. Produce the first to intersect the dy of the second, thus cutting off the d*y, i.e. the difference between two consecutive cfa/'s. Drawing a per- pendicular to each ds at its left extremity, the centre of curva- ture G is determined by their in- tersection, and thus the radius of curvature p. The two shaded triangles have their small angles FLEXURE. 251 equal, and d?y is nearly perpendicular to the prolonged ds ; hence, considering them similar, we have and hence from eq. (6) we have (approx.) ElQl=M . . (7) as a differential equation of the elastic curve. From this the equation of the elastic curve may be found, the de- flections at different points computed, and an idea thus formed of the stiffness. All beams in the present chap- ter being prismatic and homogeneous both E and / are the same (i.e. constant) at all points of the elastic curve. In using (7) the axis JTmust be taken parallel to the length of the beam before flexure, which must be slight ; the minus sign in (7) provides for the case when d-y+dy? is es- sentially negative. 232. Resilience of Flexure. If the external forces are made to increase gradually from zero up to certain maximum values, some of them may do work, by reason of their points of application moving through certain distances due to the yielding, or flexure, of the body. If at the be- ginning and also at the end of this operation the body is at rest, this work has been expended on the elastic resis- tance of the body, and an equal amount, called the work of resilience (or springing-back), will be restored by the elasticity of the body, if released from the external forces, provided the elastic limit has not been passed. The energy thus temporarily stored is of the potential kind ; see 148, 180, 196 and 218. 232a. Distinction Between Simple, and Continuous, Beams (or "Girders"). The external forces acting on a beam consist 252 MECHANICS OF ENGINE EIIING. generally of the loads and the " reactions " of the sup- ports. If the beam is horizontal and rests on two supports only, the reactions of those supports are easily found by elementary statics [ 36] alone, without calling into ac- count the theory of flexure, and the beam is said to be a Simple Beam, or girder ; whereas if it is in contact with more than two supports, being " continuous," therefore, over some of them, it is a Continuous Girder ( 271). The remainder of this chapter will deal only with simple beams. ELASTIC CURVES. 233. Case I. Horizontal Prismatic Beam, [Supported at Both Ends, With a Central Load, Weight of Beam Neglected. Fig. 224. First considering the whole beam free, we find each (. m IB FIG. 224. 233. reaction to be = >P. AOB is the neutral line ; required the equation of the portion OB referred to as an origin, and to the tangent line through as the axis of X. To do this consider as free the portion mB between any sec- tion m on the right of and the near support, in Fig. 225. The forces holding this free body in equilibrium FIG. 225. ELASTIC CURVES. 253 are the one external force ]^P, and the elastic forces act- ing on the exposed surface. The latter consist of 7, the shear, and the tensions and compressions represented in the figure by their equivalent " stress-couple." Selecting N, the neutral axis of TO, as an axis of moments (that J may not appear in the moment equation) and putting Jf (mom) =0 we have Fig. 226 shows the elastic curve OB in its purely geomet- rical aspect, much exaggerated. For axes and origin as in figure d^y-^-dx 2 is positive. Eq. (1) gives the second x-derivative of y equal to a function of x. Hence the first x-derivative of y will be equal to the oj-anti-derivative of that function, plus a con- stant, C. (By anti-derivative is meant the converse of de- rivative, sometimes called integral though not in the sense of summation). Hence from (1) we have (El being a con- stant factor remaining undisturbed) (2)' is an equation between two variables dy-- dx and x, and holds good for any point between and B\ dy-r-dx de- noting the tang, of a, the slope, or angle between the tan- gent line and X. At the slope is zero, and x also zero ; hence at (2)' becomes #7x0=0 0+ C which enables us to determine the constant C, whose value must be the same at as for all points of the curve. Hence (7=0 and (2)' becomes 254 MECHANICS OF ENGINEERING. dx from which the slope, tan. a, (or simply a, in /r-measure ; since the angle is small) may be found at any point. Thus at B we have x=y 2 l and dy-s-dx=a l , and _ ''"'"IB ' El Again, taking the aj-anti-derivative of both members of eq. (2) we have and since at both x and y are zero, 0' is zero. Hence the equation of the elastic curve OB is To compute the deflection of from the right line join- ing A and B in Fig. 224, i.e. BK, =d, we put x=tfl in (3), y being then =rf, and obtain Eq. (3) does not admit of negative values for x ; for if the free body of Fig. 225 extended to the left of 0, the ex. ternal forces acting would be P, downward, at ; and }4P, upward, at B, instead of the latter alone ; thus altering the form of eq. (1). From symmetry, however, we know that the curve AO, Fig. 224, is symmetrical with OB about the vertical through Q. ELASTIC CURVES. 255 233a. Load Suddenly Applied. Eq. (4) gives the deflection d corresponding to the force or pressure P applied at the middle of the beam, and is seen to be proportional to it. If a load G hangs at rest from the middle of the beam, P= G ; but if the load G, being initially placed at rest upon the unbent beam, is suddenly released 'from the ex- ternal constraint necessary to hold it there, it sinks and deflects the beam, the pressure P actually felt by the.beam varying with the deflection as the load sinks. What is the ultimate deflection d m ? Let P m = the pressure be- tween the load and the beam at the instant of maximum deflection. The work so far done in bending the beam = i^P m c? m . The potential energy given up by the load = Gd m , while the initial and final kinetic energies are both nothing. .'. Gd m =*/ 2m Pd m . . (5) That is, P m =26r. Since at this instant the load is sub- jected to an upward force of 2 G and to a downward force of only G (gravity) it immediately begins an upward mo- tion, reaching the point whence the motion began, and thus the oscillation continues. We here suppose the elas- ticity of the beam unimpaired. This is called the "sud- den " application of a load, and produces, as shown above, double the pressure on the beam which it does when grad- ually applied, and a double deflection. The work done by the beam in raising the weight again is called its re- silience. Similarly, if the weight G is allowed to fall on the mid- dle of the beam from a height h, we shall have Gx(h+d m ), or approx., Gh,= and hence, since (4) gives d m in terms of P 96 ' - - OT h = 256 MECHANICS OF ENGINEERING. This theory supposes the mass of the beam small com- pared with the falling weight. 234. Case II. Horizontal Prismatic Beam, Supported at Both Ends, Bearing a Single Eccentric Load. Weight of Beam Neg- lected. Fig. 227. The reactions of the points of support, P and PU are easily found by consider- ing the whole beam free, and put- ting first JT(mom.) u 0, whence P l =Pl+l l ,a,nd then J(mom.) B =0, F ' 23 ~- whence P =P(Z l Z)- 5 -*i- P and P! will now be treated as known quantities. The elastic curves OC and CB, though having a common tangent line at C (and hence the same slope c ), and a com- mon ordinate at C, have separate equations and are both referred to the same origin and axes, as shown in the figure. The slope at 0, o, and that at B,a lt are unknown constants, to be determined in the progress of the work. Equation of OC. Considering as free a portion of the beam extending from B to a section made anywhere on OC, x and y being the co-ordinates of the neutral axis of that section, we conceive the elastic forces put in on the exposed surface, as in the preceding problem, and put 2'(mom. about neutral axis of the section) =0 which gives (remembering that here d-y-^-dx 2 is negative.) whence, by taking the x anti-derivatives of both members El *L ^Iv- To find C, write out this equation for the point 0, where dy-t-dx=OQ and x=0, and we have C=EIao', hence the equation for slope is FLEXURE ELASTIC CURVES. 257 , (2) Again taking the x anti-derivatives, we have from (2) Ely =P _-P_ +EIa0+(C'=0) (3) (at Oboth x and y are .-. C"=0). In equations (1), (2), and (3) no value of x is to be used <0 or >7, since for points in CB different relations apply, thus Equation of CB. Fig. 227. Let the free body extend from B to a section made anywhere on CB.2\moms.), as before, =0, gives (N.B. In (4), as in (1), ElcPy-r-dx 2 is written equal to a neg- ative quantity because itself essentially negative ; for the curve is concave to the axis X in the first quadrant of the co-ordinate axes.) From (4) we have in the ordinary way (aj-anti-deriv.) (5)-' To determine C", consider that the curves CB and OC have the same slope (dy-^dx) at C where x=l; hence put a- I in the right-hand members of (2) and of (5)' and equate the results. This gives C" = ^PP-\-EIfj^ and .'. (5) (6) 258 MECHANICS OF ENGINEERING. At C, where x=l, both curves have the same ordinate ; hence, by putting x=l in the right members of (3) and (6)' and equating results, we obtain C'"= ^PZ 3 . .-. (6)' be (6) as the Equation of CB, Fig. 227. But a^ is still an unknown constant, to find which write out (6) for the point B where x=l, and y=0, whence we obtain != a similar form, putting P for P,, and (^ Z) for /. 235. Maximum Deflection in Case IL Fig. 227. The or- dinate y m of the lowest point is thus found. Assuming > /^i it will occur in the curve 0(7. Hence put the dy-^dx of that curve, as expressed in equation (2), =0. Also for write its value from (7), having putPj=P?-=-Z,, and we have whence [x for max. y] = Now substitute this value of x in (3), also OQ from (7), and ^ =Pl-r-li, whence Max. Deflec.=i/max=V 9 . [P 3^+2^] 236. Case III. Horizontal Prismatic Beam Supported at Both Ends and Bearing a Uniformly Distributed Load along its Whole Length. (The weight of the beam itself, if considered, FLEXUKB. ELASTIC CUKVES. 259 constitutes a load of this nature.) Let 1= the length of the beam and w= the weight, per unit of length, of the loading ; then the load coming upon any length x will be =wx, and the whole load =wl. By hypothesis w is constant. Fig. 228. From symmetry we know that the reactions at A and D are each =j4wl, that the middle of the neutral line is its lowest point, and the tangent line at is horizontal. Conceiving a section made at any point m of the neutral line at a distance x from 0, consider as free the portion of beam on the right of m. The forces holding this portion in equilibrium are ywl,t1o.Q reaction at B ', the elastic forces of the exposed surface at m, viz.: the tensions and compressions, forming a couple, and J the total she?r ; and a portion of the load, iv( l /J< x). The sum of the me tnents of these latter forces about the neu- tral axis of m, is the same as that of their resultant ; (i.e., their sum, since they are parallel), and this resultant acts in the middle of the length y 2 l x. Hence the sum of these moments =w(yl x) ]/2(%l x). Now putting JT (mom. about neutral axis of ?ra)=0 for this free body, we have (i) 260 MECHANICS OF ENGINEERING. Taking the a>anti-derivative of both sides of (1), EI Tx = I /2( I /4 l2x - l /^+( c =V ( 2 ) as the equation of slope. (The .constant is =0 since at both dy-^-dx and x are =0.) From (2), ^=^(^-Yi2* 4 )+[C'=0] . . (3) A which is the equation of the elastic curve ; throughout, i.e., it admits any value of x from x=-\-^l to x= y 2 l. This is an equation of the fourth degree, one degree high- er than those for the Curves of Cases I and II, where there were no distributed loads. If w were not constant, but proportional to the ordinates of an inclined right line, eq. (3) would be of the fifth degree ; if w were propor- tional to the vertical ordinates of a parabola with axis vertical, (3) would be of the sixth degree ; and so on. By putting x= l /^l in (3) we have the deflection of be- low the horizontal through A and B, viz.: (with W= total load =wl) , 5 # " 384 " El 384 ' El 237. Case IV. Cantilevers.^-A horizontal beam whose only support consists in one end being built in a wall, as in Fig. 229(a), or supported as in Fig. 229(6) is sometimes called a canti- lever. Let the student prove that in Fig. 229(a) with a single end load P, the deflection of B below the tangent at Ois d=^jFV s -r-.7;the same state- ment applies to Fig. 229(6), but the tangent at is not horizontal if the beam was originally so. It can also FIG. 229. "be proved that the slope at B, Fig. 229(a) (from the tangent at 0) is FLEXURE ELASTIC CURVES. 261 PP The greatest deflection of the elastic curve from the right line joining AB, in Fig. 229(6), is evidently given by the equation for y max. in 235, by writing, instead of P of that equation, the reaction at in Fig. 229(6). This assumes that the max. deflection occurs between A and 0. If it occurs between and B put (l\l) for I. If in Fig. 229(a) the loading is uniformly distributed along the beam at the rate of w pounds per linear unit, the student may also prove that the deflection of B below the tangent at is = 238. Case V. Horizontal Prismatic Beam Bearing Equal Ter- minal Loads and Supported Symmetrically at Two Points. Fig. 231. Weight of beam neglected. In the preceding cases we have made use of the approximate form ElcPy+dx 1 in determining the forms of elastic curves. In the present case the elastic curve from to G is more directly dealt with by employing the more exact expression EI-^p (see 231) for the moment of the stress-couple in any section. The reactions at and C are each = P, from symmetry. Considering free a portion of the beam extending from A to any section m between and C (Fig. 232) we have, by putting - (mom. about neutral axis of w)=0, ^62 MECHANICS OP ENGINEERING. That is, the radius of curvature is the same at all points of 0(7; in other words 0(7 is the arc of a circle with the above radius. The upward deflection of F from the right line joining and C can easily be computed from a knowl- edge of this fact. This is left to the student as also the value of the slope of the tangent line at (and C). The deflection of D from the tangent at C= l / z P^-7-EI, as ID Fig. 229(a). SAFE LOADS IN FLEXURE. 239. Maximum Moment. As we examine the different sec- tions of a given beam undar a given loading we find differ- ent values of p, the normal stress per unit of area in the outer element, as obtained from eq. (5) 229, viz.: . (1) in which / is the " Moment of Inertia " ( 85) of the plane figure formed by the section, about its neutral axis, e the distance of the most distant (or outer) fibre from the neu- tral axis, and M the sum of the moments, about this neu- tral axis, of all the forces acting on the free body of which the section in question is one end, exclusive of the stresses on the exposed surface of that section. In other words M is the sum of the moments of the forces which balance the stresses of the section, these moments being taken about the neutral axis of the section under examination. For the prismatic beams of this chapter e and / are the same at all sections, hence p varies with M and becomes a maximum when M is a maximum. In any given case the location of the " dangerous section," or section of maximum M, and the amount of that maximum value may be deter- mined by inspection and trial, this being the only method (except by graphics) if the external forces are detached. FLEXURE SAFE LOADS. 263 If, however, the loading is continuous according to a de- finite algebraic law the calculus may often be applied, taking care to treat separately each portion of the beam between two consecutive reactions of supports, or detached loads. As a graphical representation of the values of M along the beam in any given case, these values may be conceived laid off as vertical ordinates (according to some definite scale, e.g. so many inch-lbs. of moment to the linear inch of paper) from a horizontal axis just below the beam. If the upper fibres are in compression in any portion of the beam, so that that portion is convex downwards, these or- dinates will be laid off below the axis, and vice versa ; for it is evident that at a section where M= 0, p also =0, i.e., the character of the normal stress in the outermost fibre changes (from tension to compression, or vice versa) when M changes sign. It is also evident from eq. (6) 231 that the radius of curvature changes sign, and consequently the curvature is reversed, when ./If changes sign. These mo- ment ordinates form a Moment Diagram, and the extremities a Moment Curve. The maximum moment, M m , being found, in terms of the loads and reactions, we must make the p of the " dan- gerous section," where M= M mt equal to a safe value B', and thus may write (2) Eq. (2) is available for finding any one unknown quanti- ty, whether it be a load, span, or some one dimension of the beam, and is concerned only with the Strength, and not with the stiffness of the beam. If it is satisfied in any given case, the normal stress on all elements in all sections is known to be = or <7?', and the design is therefore safe in that one respect. As to danger arising from the shearing stresses in any 264 MECHANICS OF ENGINEERING. section, the consideration of the latter will be taken up in n subsequent chapter and will be found to be necessary only in beams composed of a thin web uniting two flanges. The total shear, however, denoted by J, bears to the mo- ment M, an important relation of great service in deter- mining M m . This relation, therefore, is presented in the next article. 240, The Shear is the First x-Derivative of the Moment. Fig. 233. (x is the distance of any section, measured parallel to the beam from an arbitrary origin). Consider as free a ver- tical slice of the beam included between any two consecutive vertical sections whose distance apart is dx. The forces acting are the elastic forces of the two internal surfaces now laid bare, and, possibly, a portion, ivdx, of the loading, which at this part of the beam has some intensity =10 Ibs. per running linear unit. Putting ^(mom. about axis .AT')=0 we have (noting that since the tensions and compressions of section N form a couple, the sum of their moments about N' is just the same as about N t ) pdF But P- = M, the Moment of the left hand section,^ = M> , that of the right ; whence we may write, after dividing through by dx and transposing, 2 dM -, dx T =.7; (3) for 10 If vanishes when added to the finite J, and M' M= dM= increment of the moment corresponding to the incre- ment, dx, of x. This proves the theorem. FLEXURE. SAFE LOADS. 205 Now the value of x which renders M a maximum or minimum would be obtained by putting the derivative dM--rdx= zero ; hence we may state as a Corollary. At sections where the moment is a maximum or minimum the shear is zero. The shear J at any section is easily determined by con- sidering free the portion of beam from the section to either end of the beam and putting ^'(vertical components) =0. In this article the words maximum and minimum are used in the same sense as in calculus ; i.e., graphically, they are the ordinates of the moment curve at points where the tangent line is horizontal. If the moment curve be reduced to a straight line, or a series of straight lines, it has no maximum or minimum in the strict sense just stated ; nevertheless the relation is still practically borne out by the fact that at the sections of greatest and least ordinates in the moment diagram the shear changes sign suddenly. This is best shown by drawing a shear diagram, whose ordinates are laid off vertically from a horizontal axis and under the respective sections of the beam. They will be laid off upward or downward according as 7is found to be upward or downward, when the free body con- sidered extends from the section toward the right. In these diagrams the moment ordinates are set off on an arbitrary scale of so many inch-pounds, or foot-pounds, to the linear inch of paper ; the shears being simply pounds, or some other unit of force, on a scale of so many pounds to the inch of paper. The scale on which the beam is drawn is so many feet, or inches, to the inch of paper. 241. Safe Load at the Middle of a Prismatic Beam Support- ed at the Ends. Fig. 234. The reaction at each support is *4P. Make a section n at any distance x< from B. Consider the portion nB free, putting in the proper elas- tic and external forces. The weight of beam is neglected. T?rorn ^(mom. about w)=0 we have 26G MECHANICS OF ENGINEERING. P = x;Le. t M=%Px Q A Evidently M is proportional to x, and the ordinates repre- senting it will therefore be limited by the straight line B'R, forming a triangle B'EA. From symmetry, another triangle RA' forms the other half of the moment dia- gram. From inspection, the maximum M is seen to be in the middle where x= y 2 l, and hence (1) Again by putting 2Yvert. compons.)=0, for the free body nB we have J= and must point downward since ~ points upward. Hence the shear is constant and = y 2 P at any section in the right hand half. If n be taken in the left half we would have, nB being free, from ^'(vert. com.)=0, FLEXURE. SAFE LOADS. 267 the same numerical value as before ; but e/must point up- ward, since ~ at B and J at n must balance the downward P at A. At A, then, the shear changes sign suddenly, that is, passes through the value zero; also a,i A, M is a maximum, thus illustrating the statement in 240. Notice the shear diagram in Fig. 234. To find the safe load in this case we write the maximum value of the normal stress, p,= R', a safe value, (see table in a subsequent article) and solve the equation for P. But the maximum value of p is in the outer fibre at A, since Jf for that section is a maximum. Hence (2) is the equation for safe loading in this case, so far as the normal stresses in any section are concerned. EXAMPLE. If the beam is of wood and has a rectangu- lar section with width b= 2 in., height h-*= 4 in., while its length Z= 10 ft., required the safe load, if the greatest nor- mal stress is limited to 1,000 Ibs. per sq. in. Use the pound and inch. From 90 I=> l / a ^ 3 =Yi2X2x64=10.66 biquad. inches, while e=l=2 in. . p== 4x1,000x10.66 _ le 120x2 242. Safe Load Uniformly Distributed along a Prismatic Beam Supported at the Ends. Let the load per lineal unit of the length of beam be =w> (this can be made to include the weight of the beam itself). Fig. 235. From symmetry, each reaction = }4wl. For the free body nO we have, put- ting l T For safety M ra must = , in which R = ^ ton per sq. c inch, e=i^h = ^ of unknown depth of beam, and /, 90, = I b h 3 , with 6=10 inches ,:^. y 2 .jLxlOA 3 =198; or A 2 =237.6/. A=15.4 inches. 245. Comparative Strength of Rectangular Beams. For such a beam, under a given loading, the equation for safe load- ^ ^ T = .; i. e. % R bW=M n (1) FLEXURE. SAFE LOADS. 273 whence the following is evident, (since for the same length, mode of support, and distribution of load, M m is propor- tional to the safe loading.) For rectangular prismatic beams of the same length, same material, same mode of support and same arrange- ment of load : (1) The safe load is proportional to the width of beams 'having the same depth (A). (2) The safe load is proportional to the square of the depth of beams having the same width (6). (3) The safe load is proportional to the depth of beams having the same volume (i. e. the same bh). (It is understood that the sides of the section are hori- zontal and vertical respectively and that the material is homogeneous.) 246. Comparative Stiffness of Rectangular Beams. Taking the deflection under the same loading as an inverse measure of the stiffness, and noting that in 233, 235, and 236, this deflection is inversely 'proportional to I=^bh s = the " moment of inertia " of the section about its neutral axis, we may state that : For rectangular prismatic beams of the same length, same material, same mode of support, and same loading : (1) The stiffness is proportional to the width for beams of the same depth. (2) The stiffness is proportional to the cube of the height for beams of the same width (6). (3) The stiffness is proportional to the square of the depth for beams of equal volume (bJil). (4) If the length alone vary, the stiffness is inversely proportional to the cube of the length. 247. Table of Moments of Inertia. These are here recapitu- lated for the simpler cases, and also the values of e. the distance of the outermost fibre from the axis. Since the stiffness varies as /(other things being equal), 274 MECHANICS OP ENGINEERING. while the strength varies as I-t-e, it is evident that a square beam has the same stiffness in any position (89), while its strength is greatest with one side horizontal, for then e is smallest, being =y^b. Since for any cross-section 7= CdF z\ in which 2=the distance of any element, dF, of area from the neutral axis, a beam is made both stiffer and stronger by throwing most of its material into two flanges united by a vertical web, thus forming a so-called " I-beam " of an I shape. But not without limit, for the web must be thick enough to cause the flanges to act together as a solid of continuous substance, and, if too high, is liable to buckle sideways, thus requiring lateral stiffening. These points will ba treated later. SECTION. I 6 Bectangle, width = b, depth = h (vertical) Vn M %h Hollow Rectangle, symmet. about neutral axis. See 1 Fig. 238 (a) ) Via [i Ai*-6 8 AJ %h t Triangle, width =&, height = h, neutral axis parallel 1 to base (horizontal). 1 Vie M %h Circle of radius r K*r r Ring of concentric circles. Pig. 238 (b) X*(r* , r t ) ?i Rhombus; Fig. 238 (c) h = diagonal which is vertical. /48 6A Xh. Square with side b vertical. Vu !>* y*b " " 6at45withhoriz. Vis * 4 WV2 248. Moment of Inertia of I-beams, Box-girders, Etc. In common with other large companies, the N. J. Steel and FLEXURE. SATE LOADS. 275 Iron Co. of Trenton, N. J. (Cooper, Hewitt & Co.) manu- facture prismatic rolled beams of wrought-iron variously called /-beams, deck -beams, rails, and " shape iron," (in- cluding channels, angles, tees, etc., according to the form of section.) See fig. 239 for these forms. The company ilL T CHANNEL. DECK-BEAM-. FIG. 239. publishes a pocket-book giving tables of quantities rela- ting to the strength and stiffness of beams, such as the safe loads for various spans, moments of inertia of their sections in various positions, etc., etc. The moments of inertia of /-beams and deck-beams are computed accord- ing to 92 and 93, with the inch as linear unit. The /-beams range from 4 in. to 20 inches deep, the deck- beams being about 7 and 8 in. deep. For beams of still greater stiffness and strength com- binations of plates, channels, angles, etc., are riveted to- gether, forming " built-beams, :: or " plate girders." The proper design for the riveting of such beams will be ex- amined later. For the present the parts are assumed to act together as a continuous mass. For example, Fig. 240 shows a " box-girder," formed of two " channels " and two plates riveted together. If the axis of symmetry, N, is to be horizontal it becomes the neu- tra l ax ^ s - -ket @= the moment of iner- tia of one channel (as given in the pocket-book mentioned) about the axis L_^ N perpendicular to the web of the chan- HT ' \P ne l- Then the total moment of inertia oj FIG. 340. the combination is (nearly) (1) 27G MECHANICS OF ENGINEERING. In (1), b, t, and d are the distances given in Fig. 240 (d ex- tends to the middle of plate) while d' and t' are the length and width of a rivet, the former from head to head (i.e., d' and t' are the dimensions of a rivet-hole). For example, a box-girder of wrought-iron is formed of two 15-inch channels and two plates 10 inches wide and 1 inch thick, the rivet holes ^ in. wide and 1^ in. long. That is, 6=10; t=l; d=8; t' = %-, and d'=l% inches. Also from the pocket-book we find that for the channel in question, (7=376 biquadratic inches. Hence, eq. (1) I N = 752+2xlOxlx64 4x^X^(8 X) 2 =1737biquadr.in. Also, since in this instance e = 8^ inches, and 12000 Ibs. per sq. inch (or 6 tons per sq. in.) is the value for R' (=greatest safe normal stress en the outer element of any cross-section) used by the Trenton Co. (for wrought iron), we have ^1^X1737=^,00 inch-lbs. ' That is, the box -girder can safely bear a maximum mo- ment, M m , = 2451700 inch-lbs. = 1225.8 inch-tons, as far as the normal stresses in any section are concerned. (Proper provision for the shearing stresses in the section, and in the rivets, will be considered later). 249. Strength of Cantilevers. In Fig. 241 with a single w-w? concentrated load P at the projecting extremity, we J easily find the moment at , to be M =Px, and the / max. moment to occur at the section next the wall, {f its value being M m =PL The shear, J, is constant, and = P at all sections. The moment and shear diagrams are drawn in accordance with these results. FLEXURE. SAFE LOADS. If the load W = id is uniformly distributed on the can- tilever, as in fig. 242, by making nO free we have, putting . about n) = 0, =tfwP= */ 2 Wl. P --=wx . ~ .-. Hence the moment curve is a parabola, whose vertex is at O 7 and axis vertical. Putting I (vert, compons.) = we obtain J = wx. Hence the shear diagram is a triangle, and the max. J= wl = W. 250. Resume" of the Four Simple Cases. The following table shows the values of the deflections under an arbitrary load P, or W, (within elastic limit), and of the safe load ; Cantilevers . Beams with two end supports. With one end loadP Fi- ail With unif. load W= wl Fig. 242 Load Pin middle Fig. 234 Unif. load W=wl Fig. 235 Deflection H.g /8 'lli 1 PI* 5 WP 384' El ( Safe load (from ?J[ 1 = Xm) ifi le R'l *ar 8 ^T Relative strength I 2 4 8 j Relative stiffness '( under same load ( Relative stiffness '( under safe load 1 1 A 16 4 16 5 (Max. shear = Jm, (and 1 location, P, (at wall) W, (atwaU) HP, (at supp). X W, (at supp). also the relative strength, the relative stiffness (under the same load), and the relative stiffness under the safe load, for the same beam. The max. shear will be used to determine the proper web-thickness for /-beams and " built-girders." The stu- dent should carefully study the foregoing table, noting especially the relative strength, stiffness, and stiffness under safe load, of the same beam. Thus, a beam with two end supports will bear a double 278 MECHANICS OF ENGINEERING. load, if uniformly distributed instead of concentrated in the middle, but will deflect ^ more ; whereas with a given load uniformly distributed the deflection would be only ^ of that caused by the same load in the middle, provided the elastic limit is not surpassed ii? either case. 251. R', etc. For Various Materials. The iormula^2/= M mt e from which in any given case of flexure we can compute the value of p m , the greatest normal stress in any outer element, provided all the other quantities are known, holds good theoretically within the elastic limit only. Still, some experimenters have used this formula for the rupture of beams by flexure, calling the value of p m thus obtained the Modulus of Rupture, R. R may be found to differ considerably from both the T or C of 203 with some materials and forms, being frequently much larger. This might be expected, since even supposing the relative extension or compression (i.e., strain) of the fibres to be proportional to their distances from the neutral axis as the load increases toward rupture, the corresponding stresses, not being proportional to these strains beyond the elastic limit, no longer vary directly as the distances from the neutral axis ; and the neutral axis does not pass through the centre of gravity of the section, necessarily. The following table gives average values for R, R, R", and E for the ordinary materials of construction. E, the modulus of elasticity for use in the formulae for deflection, is given as computed from experiments in flexure, and is nearly the same as E t and E c . In any example involving R', e is usually written equal to the distance of the outer fibre from the neutral axis, whether that fibre is to be in tension or compression ; since in most materials not only is the tensile equal to the compressive stress for a given strain (relative extension or contraction) but the elastic limit is reached at about the same strain both in tension and compression. FLEXURE. SAFE LOADS. 279 TABLE FOB USE IN EXAMPLES IN FLEXURE. Timber. Cast Iron. Wro't Iron. Steel. Max. safe stress in outer fi- ) V 1,000 bre -.R'dbs. per sq. inch). J 6,000 in tens. 12,000 in comp. 12,000 15,000 to 40,000 Stress in outer fibre at Has. | limit -J^/Ubs. per sq. In.) f 17.000* to 35,000 30,000 and upward. " Modul. of Rupture " =-.ff-=lbs. per sq. inch. #=Mod. of Elasticity, =lbs. per sq. inch. ) 4,000 V to \ 0,000 ) 1,000,000 V to j 3,000,000 40,000 17,000,000 50,000 25,000,000 120,000 Hard Steel. 30,000,000 In the case of cast iron, however, (see 203) the elastic limit is reached in tension with a stress =9,000 Ibs. per sq. inch and a relative extension of ^ of one per cent., while in compression the stress must be about double to reach the elastic limit, the relative change of form (strain) being also double. Hence with cast iron beams, once largely used but now almost entirely displaced by rolled wrought iron beams, an economy of material was effected by making the outer fibre on the compressed side twice as far from the neutral axis as that on the stretched side. Thus, Fig. 243, cross-sections with unequal flanges were used, so proportioned that the centre of gravity was twice as near to the outer fibre in tension as to that in compression, i.e., e?=2e i ; in other words more material is placed in tension than in compression. FIG. m The fibre A being in tengion ( w ithin elas- tic limit), that at B, since it is twice as far from the neu- tral axis and on the other side, is contracted twice as much as A is extended ; i.e., is under a compressive strain double the tensile strain at A, but in accordance with the above figures its state of stress is proportionally as much within the elastic limit as that of A. Steel beams are gradually coming into use, and may ul- timately replace those of wrought iron. * In the tests by U. S. Gov. in 1879 with I-beams. R" ranged from 25,000 to 38,000, and tlie elastic limit was reached with less stress in the Innre than in the smaller beams. Also, for the same beam, R' decreased with larger spans. 280 MECHANICS OF ENGINEERING. The great range of values of E for timber is due noi only to the fact that the various kinds of wood differ widely in strength, while the behavior of specimens of any one kind depends somewhat on age, seasoning, etc., but also to the circumstance that the size of the beam un- der experiment Las much to do with the result. The ex- periments of Prof. Lanza at the Mass. Institute of Tech- nology in 1881 were made on full size lumber (spruce), of dimensions such as are usually tak^n for floor beams in buildings, and gave much smaller values of R (from 3,200 to 8,700 Ibs. per sq. inch) than had previously been ob- tained. The loading employed was in most cases a con- centrated load midway between the two supports. These low values are probably due to the fact that in large specimens of ordinary lumber the continuity of it& substance is more or less broken by cracks, knots, etc., the higher values of most other experimenters having been obtained with small, straight-grained, selected pieces, from one foot to six feet in length. The value R' =12,000 rbs. per sq. inch* is employed by the N. J. Iron and Steel Co. in computing the safe loads for their rolled wrought iron beams, with the stipulation that the beams (which are high and of narrow width) must be secure against yielding sideways. If such is not the case the ratio of the actual safe load to that computed with 72' =12,000 is taken less and less as the span increases. The lateral security referred to may be furnished by the brick arch-filling of a fire-proof floor, or by light lateral bracing with the other beams. 252. Numerical Examples. EXAMPLE 1. A square bar of wrought iron, 1*^ in. in thickness is bent into a circular arc whose radius is 200 ft., the plane of bending being par- allel to the side of the square. Required the greatest nor- mal stress p m in any outer fibre. Solution. From 230 and 231 we may write t =P .. p=eE-r-n, i.e., is constant. P e * For '.heir soft-steel bc;ims this com|>:my uses 16,000 Ibs. persq. inch. FLEXURE. SAFE LOADS. '281 For the units inch and pound (viz. those of the table in 251) we have e=% in., p =2,400 in., and ^=25,000,000 Ibs. per sq. inch, anl .-. p=p m = ^X 25,000,000-^ 2,400 =7,812 Ibs. per sq. in., which is quite safe. At a distance of }4 inch from the neutral axis, the normal stress is =^-^-^p m = 2 /^p m = 5,208 Ibs. per sq. in. (If the force-plane (i.e., plane of bending) were parallel to the diagonal of the square, e would = l / 2 X 1.5^2 inches, giving jt> m = [ 7,812x^2 ] Ibs. per sq. in.) 238 shows an instance where a portion, OC, Fig. 231, is bent in a circular arc. EXAMPLE 2. A hollow cylindrical cast-iron pipe of radii 3 '/2 and 4 inches* is supported at its ends and loaded in middle (see Fig. 234). Required the safe load, neglecting the weight of the pipe. From the table in 250 we have for safety . *** From 251 we put #=6,000 Ibs. per sq. in.; and from 247/=*-(r, 4 r 2 4 ); and with these values, r 2 being =4 r \ 4 } e n=4, ~=-y- and Z=144 inches (the inch must "be the unit of length since .#' = 6,000 Ibs. per sq. inch) we have P=4x6,000x^- -5- (256- 150)-r- [144x4] .-. P= 3,470 Ibs. The weight of the beam itself is G= 7?, ( 7), i.e., G=-(r?-r*}l r = f (16-12^)144x^=443 Ibs. (Notice that f, here, must be Ibs., per cubic inch}. This weight being a uniformly distributed load is equivalent to half as much, 221 Ibs., applied in the middle, as far as the strength of the beam is concerned (see 250), .'. P must be taken =3,249 Ibs. when the weight of the beam is consid- ered. * And 12 feet iu length between supports. 282 MECHANICS OF ENGINEEBLNG. EXAMPLE 3. A wrought-iron rolled I-beam supported at tlie ends is to be loaded uniformly Fig. 235, the span being equal to 20 feet. Its cross-section, Fig. 244, has a depth parallel to the web of 15 inches, a flange width of 5 inches. In the pocket book of the Trenton Co. it is called a 15-inch light I- beam, weighing 150 Ibs. per yard, with a moment of inertia = 523. bi- quad. inches about a gravity axis perpendicular to the web (i.e., when the web is vertical, the strongest position) and = 15 biq. in. about a gravity axis parallel to the web (i.e., when the web is placed horizontally). First placing the web vertically, we have from 250, pi T W^= Safe load, distributed, =8'. With #=12,000, fe, /!=523, Z=240 inches, e { = 7}4 inches, this gives JF 1 =[8xl2,OOOx523][240x^]=27,893 Ibs. But this includes the weight of the beam, =20 ft. x'-fUbs. =1,000 Ibs.; hence a distributed load of 26,902 Ibs., or 13.45 tons may be placed on the beam (secured against lateral yielding). (The pocket-book referred to gives 13.27 tons as the safe load.) Secondly, placing the web horizontal, of W l or only about 1 / 12 of W^ EXAMPLE 4. Kequired the deflection in the first case of Ex. 3. From 250 the deflection at middle is __ =: _L=A R i L 384 ' El, 384 ' fe, ' El^ 48 ' E ' e l FLEXUKE. SAFE LOADS. 283 .-. ^=0.384 in. EXAMPLE 5. A rectangular beam of yellow pine, of width b=4: inches, is 20 ft. long, rests on two end supports, and is to carry a load of 1,200 Ibs. at the middle ; required the proper depth h. From 250 P-lRI-lRW 1 ~U ~ T 12 * p" .-. h 2 =6Pl-^-4:Rb. For variety, use the inch and ton. For this system of units P=0.60 tons, .#'=0.50 tons per sq. in., Z=240 inches and b= 4 inches. .-. /r=(6x0.6x240)-K4x0.5x4)=108sq. in. .-. =10. 4 in. EXAMPLE 6. Suppose the depth in Ex. 5 to be deter- mined by the condition that the deflection shall be = Ysoo of the span or length. We should then have from 250 , 1 , 1 PP 500 48 Ei Using the inch and ton, with E= 1,200,000 Ibs. per sq. in., which = 600 tons per sq. inch, and /= 1 / 12 fcA 3 , we have 500X0.60X240X240X12 As this is > 10.4 the load would be safe, as well. EXAMPLE 7. Eequired the length of a wro't iron pipe supported at its extremities, its internal radius being 2^ in., the external 2.50 in. } that the deflection under its own weight may equal l / m of the length. 579.6 in. Ans. EXAMPLE 8. Fig. 245. The wall is 6 feet high and one foot thick, of common brick work | i | i | i | i | i | i | [ (see 7) and is to be borne by an ] i ] i | i ] i ] i jjzj /-beam in whose outer fibres no .1 , =r ~l greater normal stress than 8,000 ^ ~"E Ibs. per sq. inch is allowable. If FIG. 245. a number of I-beams is available, 284 MECHANICS OF ENGLNEEKING. ranging in height from 6 in. to 15 in. (by whole inches), which one shall be chosen in the present instance, if their cross-seetions are Similar Figures, the moment of inertia of the 15-inch beam being 800 biquad. inches ? The 12-inch beam. Ans. SHEARING STRESSES IN FLEXURE. 253. Shearing Stresses in Surfaces Parallel to the Neutral Surface. If a pile of boards (see Fig. 246) is used to sup- port a load, the boards being free to slip on each other, it ;.s noticeable that the ends overlap, although the boards are of equal length (now see Fig. 247) ; i.e., slipping has occurred along the surfaces of contact, the combina- tion being no stronger than the same boards side by side. If, however, they are glued together, piled as in the former figure, the slipping is prevented and the deflection is much less under the same load P. That is, the com- pound beam is both stronger and stiffer than the pile of loose boards, but the tendency to slip still exists and is known as the " shearing stress in surfaces parallel to the neutral surface." Its intensity per unit of area will now be determined by the usual " free-body " method. In Fig. 248 let AN' be a portion, considered free, on the left of any Pio. 848. SHEAR IX FLEXURE. 285 section N f , of a prismatic beam slightly bent under forces in one plane and perpendicular to the beam. The moment equation, about the neutral axis at N', gives 2 =M' ; whence jt)'= M'e (1) Similarly, with AN as a free body, NN 1 being =dx, = w p and p' are the respective normal stresses in the outer fibre in the transverse sections N and N' respectively. Now separate the block NN', lying between these two consecutive sections, as a free body (in Fig. 249). And P a furthermore remove a portion of the top of the latter block, the portion lying above a plane passed parallel to the neu- tral surface and at any distance z" from that surface. This latter free body is shown in Fig. 250, with the system of forces representing the actions upon it of the portions taken away. The under surface, just laid bare, is a portion of a sur- face (parallel to the neutral surface) in which the above men- tioned slipping, or shearing, tendency exists. The lower por- tion (of the block NN'} which is now removed exerted this 280 MECHANICS OF ENGINEERING. rubbing, or sliding, force on the remainder along the under surface of the latter. Let the unknown intensity of this shearing force be JT(per unit of area) ; then the shearing force on this under surface is =Xy"dx, (y",= oa in figure, being the horizontal width of the beam at this distance z" from the neutral axis of N') and takes its place with the other forces of the system, which are the normal stresses between ' , and portions of J and J', the respective \_z= z" total vertical shears. (The manner of distribution of J over the vertical section is as yet unknown ; see next arti- cle.) Putting S (horiz. compons.) = in Fig. 250, we have f e -p'dF C e Z -pdFXy"dx=Q J z" e J z" e .-.Xy"dx=P fldF e z" But from eqs. (1) and (2), / p = (M'M).=^ r dM, while from 240 dM = Jdx ; .... (3) as the required intensity per unit of area of the shearing force in a surface parallel to the neutral surface and at a distance z" from it. It is seen to depend on the " shear " J and the moment of inertia I of the whole vertical section; upon the horizontal thickness y" of the beam at the sur- face in question ; and upon the integral J zdF, which (from 23) is the product of the area of that part of the vertical section extending from the surface in question to the outer fibre, by the distance of the centre of gravity of that part from the neutral surface. SHEAR IX FLEXURE. 287 It now follows, from 209, that the intensity (per unit area) of the shear on an elementary area of the vertical cross section of a bent beam, and this intensity we may call Z, is equal to that X, just found, in the horizontal section which is at the same distance (z") from the neutral axis. 254. Mode of Distribution of J, the Total Shear, over the Verti- cal Cross Section. The intensity of this shear, Z (Ibs. per sq. inch, for instance) has just been proved to be =X=~^, C l J ' zdF (4) To illustrate this, required the value of Z two inches above the neu- tral axis, in a cross section close to the abutment, in Ex. 5, 252. Fig. 251 shows this section. From it we AS have for the shaded portion, lying above the locality in question, y" = 4 inches, and =5.2 r j 2 // = zdF = (area of shaded portion) X (distance of its centre of gravity from 2JA) = (12.8 sq. in.) x (3.6 in.) = 46.08 cubic inches. The total shear J = the abutment reaction = 600 Ibs., while / = L W = . x 4 x (10.4) 3 = 375 biquad. inches. Both t/and /refer to the whole section. 7 600x46.08 -.Q.o!, .'.Z= ._ - =18.42 Ibs. per sq. in., d75 x4 qui+e insignificant. In the neighborhood of the neutral axis, where z" = 0, we have y" = 4 and C e zdF= r e zdF = 20.8 x 2.6=54.8, J z"=0 J wh__e J and / of course are the same as before. Hence for z" =0 ^ 2SS MECHANICS OF ENGINEERING. Z=Z = 21.62 Ibs. per sq. in. At the outer fibre since /* *dF=0, z" being = e, Z is = / e for a beam of any shape. For a solid rectangular section like the above, Z and z ' bear the same relation to each other as the co-ordinates of the para- bola in Fig. 252 (axis horizontal). Since in equation (4) the horizontal thickness, y", from side to side ef the sec- tion of the locality where Z is desired, occurs in the denominator, and since / zdF t/,// increases as z' 1 grows numerically smaller, the following may be stated, as to the distribution of 7, the shear, in any vertical section, viz.: The intensity (Ibs. per sq. in.) of the shear is zero at the outer elements of the section, and for beams of ordi- nary shapes is greatest where the section crosses the neu- tral surface. For forms of cross section having thin webs its value may be so great as to require special investiga- tion for safe design. Denoting by Z Q the value of Z at the neutral axis, (which = X in the neutral surface where it crosses the vertica section in question) and putting the thickness of the sub- stance of the beam = 1 at the neutral axis, we have, T ( area above ) ( ,, -,. , , ., , ) Z =X =-J- X 1 neutralaxis t x J thed l st - fi*soent. ( (6) H> ( (or below) f ( grav.from that axis p 255. Values of Zo for Special Forms of Cross Section. From the last equation it is plain that for a prismatic beam the value of Z is proportional to J, the total shear, and hence to the ordinate of the shear diagram for any particular case of loading. The utility of such a diagram, as obtain SHEAB IK FLEXURE. 289 ed in Figs. 234-237 inclusive, is therefore evident, for by locating the greatest shearing stress in the beam it enables us to provide proper relations between the load- ing and the form and material of the beam to secure safety against rupture by shearing. The table in 210 gives safe values which the maximum Z in any case should not exceed. It is only in the case of beams with thin webs (see Figs. 238 and 240) however, that Z Q is likely to need at- tention. For a Rectangle we have, Fig. 253, (see eq. 5, 254) b =b, I=Y, 2 6A S , and. .:Z =X =-^- 2. i.e., = |- (total shear)-;- (whole area) Hence the greatest intensity of shear in the cross-section is A as great per unit of area as if the total shear were uniformly distributed over the section. FIG. 254. FIG. 355. FIG. 256. "For a Solid Circular section Fig. 254 7= J r e ^F= Ib J . 2r 2 3,T 3 [See 26 Prob. 3]. For a Hollow Circular section (concentric circles) Fig. 255, we have similarly, 290 MECHAiaCS OF ENGINEERING. J VTIT frxr 1 -r 2 ) 2 ' 2 3 Applying this formula to Example 2 252, we first have as the max. shear 7 m = */ 2 P =1,735 Ibs., this being the abut- ment reaction, and hence (putting x = (22 -r- 7)) which cast iron is abundantly able to withstand in shear- ing. For a Hollow Rectangular Beam, symmetrical about its neutral surface, Fig. 256 (box girder) The same equation holds good for Fig. 257 (I-beam with square corners) but then 6 2 denotes the sum of the widths of the hollow spaces. 256. Shearing Stress in the Web of an I-Beam. It is usual to consider that, with I-beams (and box- beams) with the web vertical the shear J, in any vertical section, is borne exclusively by the web and is uniformly distributed over its section. That this is nearly true may be proved as follows, the flange area being comparatively large. Fig. 258. Let FI be the area of one flange, and F that of the half web. Then since FIG. 258. SHEAR IX FLEXURE. 291 (the last term approximate, ^ J^ being taken for the radi- us of gyration of F } ,} while F=^V +^o-j-, (the first term approx.) we have f f rr ** /4- "VV**-*- 1 I -* U/ ^o== 7f = i M ?. /a y .1.0 *M which = if we write (2^+^,) -4- (6^+2^) =K But Mo is the area of the whole web, .'. the shear per unit area at the neutral axis is nearly the same as if J were uniformly dis- tributed over the web. E. g., with FI = 2 sq. in., and F Q = 1 sq. in. we obtain Z = 1.07 (J-t-baho). Similarly, the shearing stress per unit area at n, the upper edge of the web, is also nearly equal to J -f- b^ (see eq., 4 (254) for then \T (zdF)l= F^/h, nearly, while / remains as before. The shear per unit area, then, in an ordinary I-beam is obtained by dividing the total shear J by the area of the web section. EXAMPLE. It is required to determine the proper thickness to be given to the web of the 15-inch wrought- iron rolled beam of Example 3 of 252, the height of web being 13 inches, with a safe shearing stress as low as 4000 Ibs. per sq. in. (the practice of the N. J. Steel and Iron Co., for webs), the web being vertical. The greatest total shear, J mt occurring at either support and being equal to half the load (see table 250) we have with 6 = width of web, Z max.= -; i.e. 4000 = ... ^ = .26 inches. 292 MECHANICS OF ENGINEERING. (Units, inch and pound). The 15-inch light beam of the N. J. Co. has a web ]/% inch thick, so as to provide for a shear double the value of that in the foregoing example. In the middle of the span Z^ = 0, since J = 0. 257. Designing of Riveting for Built Beams. The latter are generally of the I-beam and box forms, made by riveting together a number of continuous shapes, most of the ma- terial being thrown into the flange members. E. g., in fig. 259, an I-beam is formed by riveting together, in the manner shown in the figure, a "vertical stem plate" or web, four " angle-irons," and two " flange-plates," each of these seven pieces being continuous through the whole length of the beam. Fig 260 shows a box-girder. If the riveting is well done, the combination forms a single rigid beam whose safe load for a given span may be found by foregoing rules ; in computing the moment of inertia, how- ever, the portion of cross section cut out by the rivet holes must not be included. (This will be illustrated in a subsequent paragraph.) The safe load having been com- puted from a consideration of normal stresses only, and the web being made thick enough to take up the max, total shear, J m , with safety, it still remains to design the riveting, through whose agency the web and flanges are caused to act together as a single continuous rigid mass. It will be on the side of safety to consider that at a given SHEAR IN FLEXURE. 293 locality in the beam the shear carried by the rivets con- necting the angles and flanges, per unit of length of beam, is the same as that carried by those connecting the angles and the web ("vertical stem-plate"). The amount of this shear may be computed from the fact that it is equal to that occurring in the surface (parallel to the neutral sur- face) in which the web joins the flange, in case the web and flange were of continuous substance, as in a solid I- beam. But this shear must be of the same amount per horizontal unit of length as it is per vertical linear unit in the web itself, where it joins the flange ; (for from 254 Z =X.) But the shear in the vertical section of the web, being uniformly distributed, is the same per vertical linear unit at the junction with the flange as at any other part of the web. section ( 256,) and the whole shear on the ver- tical section of web = J, the " total shear " of that section of the beam. Hence we may state the following : The riveting connecting the angles with the flanges, (or the web with the angles) in any locality of a built beam, must safely sustain a shear equal to Jon a horizontal length equal to the height of iveb. The strength of the riveting may be limited by the re- sistance of the rivet to being sheared (and this brings into account its cross section) or upon the crushing resist- ance of the side of the rivet hole in the plate (and this in- volves both the diameter of the rivet and the thickness of the metal in the web, flange, or angle.) In its practice the N. J. Steel and Iron Co. allows 7500 Ibs. per sq. inch shear- ing stress in the rivet (wrought iron), and 12500 Ibs. per sq. inch compressive resistance in the side of the rivet- hole, the axial plane section of the hole being the area of reference. In fig. 259 the rivets connecting the web with the angles are in double shear, which should be taken into account in considering their shearing strength, which is then double ; those connecting the angles and the flange plates are in 294 MECHANICS OF ENGINEERING. In fig. 260 (box -beam) where the beam is built of two webs, four angles, and two flange plates, all the rivets are in single shear. If the web plate is very high compared with its thickness, vertical stiffeners in the form of T irons may need to be riveted upon them lat- erally [see 314]. EXAMPLE. A built I-beam of wrought iron (see fig. 259) is to support a uniformly distributed load of 40 tons, its extremities resting on supports 20 feet apart, and the height and thickness of web being 20 ins. and y 2 in. re- spectively. How shall the rivets, which are y in. in di- ameter, be spaced, between the web and the angles which are also y 2 in. in thickness ? Keferring to fig. 235 we find that J = y W = 20 tons at each support and diminishes regularly to zero at the middle, where no riveting will there- fore be required. (Units inch and pound). Near a sup- port the riveting must sustain for each inch of length of beam a shearing force of (J -r- height of web) = 40000 -r- 20 in. = 2000 Ibs. Each rivet, having a sectional area of y^ 7i (7/%f = 0.60 sq. inches, can bear a safe shear of 0.60 X 7500 = 4500 Ibs. in single shear, and .-. of 9000 Ibs. in double shear, which is the present case. But the safe compressive resistance of the side of the rivet hole in either the web or the angle is only ^ in. x l /i in. x 12500 = 5470 Ibs., and thus determines the spacing of the rivets as follows : 2000 Ibs. -r- 5470 gives 0.36 as the number of rivets per inch of length of beam, i.e., they must be 1 4- 0.36 = 2.7 inches apart, centre to centre, near the supports; 5.4 inches apart at ^ the span from a support ; none at all in the middle. However, " the rivets should not ba spaced closer than 2^ times their diameter, nor farther apart than 16 times the thickness of the plate they connect," is the rule of the N. J. Co. As for the rivets connecting the angles and flange plates, being in two rows and opposite (in pairs) the safe shear- FLEXUKE. BUILT BEAM. 295 ing resistance of a pair (each in single shear) is 9,000 Ibs., while the safe compressive resistance of the sides of the two rivet holes in the angle irons (the flange plate being much thicker) is 10,940 Ibs. Hence the former figure (9,000) divided into 2,000 Ibs., gives 0.22 as the number of pairs of rivets per inch of length of the beam ; i.e., the rivets in one row should be spaced 4.5 inches apart, centre to centre, near a support ; the interval to be increased in inverse ratio to the distance from the middle of span, ^bearing in mind the practical limitation just given). If the load is concentrated in the middle of the span, instead of uniformly distributed, 7is constant along each half-span, (see fig. 234) and the rivet spacing must accord- ingly be made the same at all localities of the beam. SPECIAL, PROBLEMS IN FLEXURE. 258. Designing Cross Sections of Built Beams. The last par- agraph dealt with the riveting of the various plates ; we now consider the design of the plates themselves. Take for instance a built I-beam, fig. 261 ; one vertical stem- FIQ. 261. 296 MECHANICS OF ENGINEERING, plate, four angle irons, (each of sectional area = A, re- maining after the holes are punched, with a gravity axis parallel to, and at a distance = a from its base), and two flange plates of width = b, and thickness = t. Let the whole depth of girder = h, and the diameter of a rivet hole =1f. To safely resist the tensile and compressive forces induced in this section by M m inch-lbs. (M m being the greatest moment in the beam which is prismatic) we have from 239, *.= ? a) E 1 for wrought iron = 12,000 Ibs. per sq. inch, e is = */ 2 h while 1, the moment of inertia of the compound section, is obtained as follows, taking into account the fact that the rivet holes cut out part of the material. In dealing with the sections of the angles and flanges, we consider them concentrated at their centres of gravity (an approx- imation, of course,) and treat their moments of inertia about N as single terms in the series CdF z* (see 85). The subtractive moments of inertia for the rivet holes in the web are similarly expressed ; let b = thickness of web. 7 N for web = -Jyb (h 2) 3 26/ [} t a'] 2 7 N for four angles = 4 A [~ t a] 2 7 N for two flanges = 2(6 2f) t (1) 2 the sum of which makes the 1^ of the girder. Eq. (1) may now be written -^= 7 * (2) which is available for computing any one unknown quan- tity. The quantities concerned in / N are so numerous and they are combined in so complex a manner that in any numerical example it is best to adjust the dimensions of the section to each other by successive assumptions and FLEXURE BUILT BEAM. 297 trials. The size of rivets need not vary much in different cases, nor the thickness of the web-plate, which as used by the N. J. Co. is " rarely less than ^ or more than ^ inch thick." The same Co. recommends the use of a single size of angle irons, viz., 3" X 3'' X ^", for built girders of heights ranging from 12 to 36 inches, and also y in. rivets, and gives tables computed from eq. (2) for the proportionate strength of each portion of the com- pound section. EXAMPLE. (Units, inch and pound). A built I-beam with end supports, of span = 20 ft. = 240 inches, is to support a uniformly distributed load of 36 tons = 72,000 Ibs. If y inch rivets are used, angle irons 3'' x 3" X y 2 ", ver- tical web ^4" in thickness, and plates 1 inch thick for flanges, how wide (b = ?) must these flange-plates be ? taking h = 22 inches = total height of girder. Solution. From the table in 250 we find that the max. M for this case is * Wl, where W = the total distributed load (including the weight of the girder) and I = span. Hence the left hand member of eq. (2) reduces to Wl h_ 72000 x 240 x 22 16 ' X" ~ 16 x 12000 That is, the total moment of inertia of the section must be = 1,980 biquad. inches, of which the web and angles supply a known amount, since b = *", t =1", <*= ^", a' = 1%", A= 2.0 sq. in., a = 0.9', and h = 22", are known, while the remainder must be furnished by the flanges, thus determining their width b, the unknown quantity. The effective area, A, of an angle iron is found thns : The full sectional area for the size given, = 3 x */4 + 2^ X YZ =2.75 sq. inches, from which deducting for two rivet holes we have A= 2.75 2 x Y X # = 2.0 sq. in. The value a = 0.90" is found by cutting out the shape 298 MECHANICS OF ENGINEERING. of two angles from sheet iron, thus : and balancing it on a knife edge. (The gaps left by the rivet holps maybe ignored, without great error, in finding a). Hence, substituting we have I v for web =i I N for four angles = 4x2x [9.10] 2 =662.5 I N for two flanges=2(^7)xlX(10X) 2 =220.4(& 1.5) .-. 1980= 282.3+662.5+(fr 1.5)220.4 whence b = 4.6 + 1.5 = 6.1 inches the required total width of each of the 1 in. flange plates. This might be increased to 6.5 in. so as to equal the united width of the two angles and web. The rivet spacing can now be designed by 257, and the assamed thickness of web, ^4 in., tested for the max. total shear by 256. The latter test results as follows : The max. shear J m occurs near either support and = y 2 W= 36,000 Ibs. .-., calling Z/o the least allowable thickness of web in order to keep the shearing stress as low as 4,000 Ibs. per sq. inch, b' x 20" x 4000 =36000 .-. 6' =0.45" showing that the assumed width of ^ in. is safe. This girder will need vertical stiffeners near the ends, as explained subsequently, and is understood to be sup- ported laterally. Built beams of double web, or box- form, (see Fig. 260) do not need this lateral support. 259. Set of Moving Loads. "When a locomotive passes over a number of parallel prismatic girders, each one of which experiences certain detached pressures corresponding to the different wheels, by selecting any definite position of the wheels on the span, we may easily compute the reac- tions of the supports, then form the shear diagram, and finally as in 243 obtain the max. moment, M m) and the FLEXUKE. MOVING LOADS. 1>99 max. shear J m , for this particular position of the wheels. But the values of M m and J m for some other position may be greater than those just found. We therefore inquire which will be the greatest moment among the infinite number of (Jf m )'s (one for each possible position of the wheels on the span). It is evident from Fig. 236 from the nature of the moment diagram, that when the pressures or loads are detached, the M m for any position of the loads, which of course are in this case at fixed distances apart, must occur under one of the loads (i.e. under a wheel). We begin /. by asking : What is the position of the set of moving loads when the moment under a given wheel is greater than will occur under that wheel in any other po- sition ? For example, in Fig. 262, in what position of the a* .-i t=L_MJ.,fr_ O ffi 1 00 r =r~i p j loads PI, P 2 , etc. on the span will the moment M 2 , i.e., under P 2 , be a maximum as compared with its value under P 2 in any other position on the span. Let R be the resultant of the loads which are now on the span, its variable distance from be = x, and its fixed distance from P 2 = a'\ while a, b, c, etc., are the fixed distances between the loads (wheels). For any values of x , as the loading moves through the range of motion within which no wheel of the set under consideration goes off the span, and no new- wheel comes on it, we have S 1 =--^- R, and the moment under P 2 (1) 300 MECHANICS OF ENGINEERING. In (1) we have M. 2 as a function of x, all the other quan- tities in the right hand member remaining constant as the loading moves; x may vary from x= a+a' to xl (c+& a'). For a max. M 2 , we put dM 2 ^-dx=Q, i. e. ~(l-2x~+a')=0 .-. x (for Max M,}=y 2 l+y 2 a' (For this, or any other value of x, cPM.^dx 2 is negative, hence a maximum is indicated). For a max. M 2 , then, R must be as as far (y 2 a f ) on one side of the middle of the span as P 2 is on the other ; i.e., as the loading moves, the moment under a given wheel becomes a max. when that wheel and the centre of gravity of all the loads (then on the span) are eg ui -distant from the middle of the span. In this way in any particular case we may find the- respective max. moments occurring under each of the wheels during the passage, and the greatest of these is the M m to be used in the equation M^^R'I-^e for safe loading.* As to the shear J, for a given position of the wheels this will be the greatest at one or the other support, and equals the reaction at that support. When the load moves toward either support the shear at that end of the beam evidently increases so long as no wheel rolls completely over and beyond it. To find /max., then, dealing with each support in turn, we compute the successive reactions at the support when the loading is successively so placed that consecutive wheels, in turn, are on the point of roll- ing off the girder at that end ; the greatest of these is the max. shear, J m . As the max. moment is apt to come under the heaviest load it may not be necessary to deal with more than one or two wheels in finding M m . EXAMPLE. Given the following wheel pressures, A< . . & . . >B< . . 5' . . >C< . . 4 . . 2. " " " " A, B, C, and D ; 3. " " " " B, (7, and D. 4. In what position of the wheels on the span will the moment under B be a max. ? Ditto for wheel C? Eequired the value of these moments and which is M m ? 5. Eequired the value of J m , (max. shear), its location and the position of loads. Results. (1.) T.ff to right of A. (2.) 10* to right of A. (3.) 4.4' to right of B. (4.) Max. M B = 1,273,000 inch Ibs. with all the wheels on ; Max. M c = 1,440,000 inch-lbs. with wheels B, C, and D on. (5.) J m = 13.6 tons at right sup- port with wheel D close to this support. 260. Single Eccentric Load. In the following special cases of prismatic beams, peculiar in the distribution of the loads, or mode of support, or both, the main objects sought are the values of the max. moment M m) for use in the equation J4=^(see 239); and of the max. shear J m , from which to design the web riveting in the case of an / or box-girder. The modes of support will be such that the reactions are independent of the form and material of the beam (the weight of beam being neglected). As before, the flexure is to be slight, and the forces are all perpendicular to the beam. The present problem is that in fig. 263, the beam being prismatic, supported at the ends, with a single eccentric load, P. We shall first disregard the weight of the beam itself. Let the span = ?!+,. First considering the whole beam free we have the reactions 72, = P^ -f- I and R 2 PI, - Z. Making a section at m and having Om free, x being < 1 2 , -T (vert, compons.) = gives FIG. sea. 302 MECHANICS OF ENGINEERING. R, e/=0, Le., J=R 2 ; while from 2 (mom.) m =0 we have P- Z -R,x= .\M= fi&=S& e I These values of J and M hold good between and (7, / being constant, while M is proportional to x. Hence for G the shear diagram is a rectangle and the moment dia- gram a triangle. By inspection the greatest M for 00 is for x Z 2 , and = Pl^ -r- 1. This is the max. M for the beam, since between G and B, M is proportional to the dis- tance of the section from B. is the equation for safe loading. J R l in any section along CB, and is opposite in sign to what it is on 0(7; i.e., practically, if a dove-tail joint existed anywhere on OC the portion of the beam on the right of such section would slide downward relatively to the left hand portion ; but vice versa on CB. Evidently the max. shear J m = fy or R 2 , as 1 2 or Z : is the greater segment. It is also evident that for a given span and given beam the safe load P', as computed from eq. (1) above, becomes very large as its point of application approaches a sup- port ; this would naturally be expected but not without limit, as the shear for sections between the load and the support is equal to the reaction at the near support and may thus soon reach a limiting value, when the safety of the web or the spacing of the rivets, if any, is considered. Secondly, considering the weight of the beam, or any uniformly distributed loading, weighing w Ibs. per unit of length of beam, in addition to P, Fig. 264, we have the reactions P P1 2 . W , PI, , W ^ = T + T ;and7?2= T+lT Let 1 2 be >l l ; then for a portion Om of length x =- ( min. ) (3) 4 & It remains to be seen whether a value of M may not exist in some section between and C, (i.e., for a value of x l& we are not concerned with R ] the corresponding value of M, and the greatest M on OC would be M c . For the short portion BC, which has moment and shear diagrams of its own not con- tinuous with those for 0(7, it may easily be shown that M G is the greatest moment of PIG. 864. an J section. Hence the M 304 MECHANICS OF ENGINEERING. max., or M m) of the whole beam is either M c or M n> according as x n is > or < l z . This latter criterion may be expressed thus, [with 1 2 y 2 I denoted by 1 3 , the distance of P from the middle of the span] : (e q . and since from (4) and (2) The equation for safe loading is and .... (6) I j Se ^eqs (3)and(5) ' for J/ c and M n If either P, W, l s , or ^ is the unknown quantity sought, the criterion of (6) cannot be applied, and we .. use both equa- tions in (6) and then discriminate between the two results. The greatest shear is J m ^H lt in Fig. 264, where 1 2 is > Zi. 261. Two Equal Terminal Loads, Two Symmetrical Supports Fig. 265. [Same case as in Fig. 231, 238]. Neglect weight of beam. The reaction at each support being =P, (from symmetry), we have for a free body Om with x < ^ Px . . . (I) while where x > ^ and < Ii Px-P (xl,} ^Lo .-. Jf=PZ, . (2) That is, see (1), M varies directly with x between and (7, while between C and D it is constant. Hence for safe loading FLEXURE. SPECIAL PROBLEMS. 305 The construction of the B moment diagram is evident p from equations (1) and (2). As for J, the shear, the same free bodies give, from 2 (vert, forces )=0. On OG . J=P ... (4) On CD . J=PP=zero(5) (4) and (5) might also be ob- FIG. 265 tained from (1) and (2) by writing J=d M-^-dx, but the former method is to be preferred in most cases, since the latter requires M to be expressed as a function of x while the former is applicable for examining separate sections without making use of a variable. If the beam is an I-beam, the fact that 7is zero any- where on G D would indicate that we may dispense with a web along C D to unite the two flanges ; but the lower flange being in compression and forming a " long column " would tend to buckle out of a straight line if not stayed by a web connection with the other, or some equivalent brac- ing. 262. Uniform Load over Part of the Span. Two End Supports. Fig. 266. Let the load= W, extending from one support over a portion = c, of the span, (on the left, say,) so that W=> we, w being the load per unit of length. Neglect weight of beam. For a free body Om of any length x < B (i.e. < c), I moms m =0 gives wx* 2 (1) which holds good for any section on B. As for sections on B C it is more simple to deal with the free body m'C. of length x' < C B from which we have M=R t x' . . (2) 306 MECHANICS OF ENGINEERING. which shows the moment curve for B C to be a straight line DC, tangent at D to the -T parabola (7 D representing eq. (1.) (If there were a con- centrated load at B, C D would meet the tangent at D at an angle instead of co- inciding with it ; let the stu- dent show why, from the shear diagram). The shear for any value of x on B is : On OB while on B G E 1 wx .. ^2= constant (3) (4) The shear diagram is constructed accordingly. To find the position of the max. ordinate of the para- bola, (and this from previous statements concerning the tangent at the point D must occur on B, as will be seen and will .-. be the M m for the whole beam) we put J=0 in eq (3) whence (5) and is less than c, as expected. [The value of B ls= ~(l^) t =(wc -7-Z) (I ), (the whole beam free) has been substi- tuted]. This value of x substituted in eq. (1) gives Jk=(i # j?. y 2 . we.-. ^I=y 2 \\y 2 '^-YWc....(Q} & L is the equation for safe loading. The max. shear J m is found at and is = R u which is evidently >/2 2 , at C. FLEXURE. SPECIAL PROBLEMS. 307 263. Uniform Load Over Whole Length With Two Symmetrica Supports. Fig. 267. With the notation expressed in the fig- ure, the following results may be obtained, after having divided the length of the beam into three parts for sepa- rate treatment as necessitated by the external forces, which are the distributed load W, and and the two reactions, each = Y?, W. The moment curve is made up of parts of three dis- tinct parabolas, each with its axis vertical. The central par- abola maj sink below the hori- zontal axis of reference if the supports are far enough apart, in which case (see Fig.) the elas- tic curve of the beam itself becomes concave upward be- tween the points E and F of " contrary flexure." At each of these points the moment must be zero, since the radius of curvature is GO and M = El -f- p (see 231) at any sec- tion ; that is, at these points the moment curve crosses its horizontal axis. As to the location and amount of the max. moment ?J m , inspecting the diagram we see that it will be either at H t the middle, or at both of the supports B and C (which from symmetry have equal moments), i.e., (with I = total length,) , R'f] ( and,.-J=j either ^\y 4 l*-l*] ..... at H JZ at S and G according to which is the greater in any given case ; i.e. according as ^ is > or < ^ ^/g. The shear close on the left of S = wl l} while close to the right of B it = # W - wl,. (It will be noticed that in this case since the beam overhangs, beyond the support, the shear near the support is not equal to the reaction there, as it was in some preceding cases.) 308 MECHANICS OF ENGINEERING. Hence J m = j ^ ^_^ J according as 264, Hydrostatic Pressure Against a Vertical Plank. From elementary hydrostatics we know that the pressure, per unit area, of quiescent water against the vertical side of a tank, varies directly with the depth, x, below the surface, and equals the weight of a prism of water whose altitude = x, and whose sectional area is unity. See Fig. 268. Pis. 268. Ttie plank is of rectangular cross section, its constant breadth, b, being r~ to the paper, and receives no sup- port except at its two extremities, and B, being level with the water surface. The loading, or pressure, per unit of length of the beam, is here variable and, by above defini- nition, is = w= ?xb, where f = weight of a cubic unit (i.e. the heaviness, see 7) of water, and x = Om depth of any section m below the surface. The hydrostatic pres- sure on dx = ivdx. These pressures for equal dx's, vary as the ordinates of a triangle OR^B. Consider Om free. Besides the elastic forces of the ex- posed section m, the forces acting are the reaction R Q , and the triangle of pressure OEm. The total of the latter is /*. wdx fbf xdx =' ' Jo (1) and the sum of the moments of these pressures about m is equal to that of their resultant ( = their sum, since they are parallel) about m, and .*. fb . . - 2 o FLEXURE, SPECIAL PROBLEMS. 309 [From (1) when x = Z, we have for the total water pres- Z 2 sure on the beam W { = fb ^ and since one-third of this z will be borne at we have R =^ f&Z 2 .] Now putting I( moms, about the neutral axis of w)=0, for Om free, we have (2) (which holds good from x = to x = 1). From I (horiz. forces) = we have also the shear J=S W K =y 6 r b (Z 2 3s 2 ) .... (3) as might also have been obtained by differentiating (2), since J = dM -f- dx. By putting J = ( 240, corollary) we have for a max. M, x = I -* -^3, which is less than I and hence is applicable to the problem. Substitute this in eq. 2, and reduce, and we have R'l R'l I 1 v =J* in ,i.e. = g - 7r -y .... (4) as the equation for safe loading. 265. Example. If the thickness of the plank is h, re- quired h = ?, if R' is taken = 1,000 Ibs. per sq. in. for timber ( 251), and I = 6 feet. For the inch-pound-second system of units, we must substitute R' = 1,000 ; I = 72 inches ; r = 0.036 Ibs. per cubic inch (heaviness of water in this system of units); while I =bh? -* 12, ( 247), and e = y 2 h. Hence from (4) we have 1000 W 0.0366 x72 3 , _ Q Q7 . 12^>P = -97T" ' ' ~ It will be noticed that since x for JIf m = Z -5- Vs, and not ^ Z, M m does not occur in the section opposite the resul- tant of the water pressure ; see Fig. 268. The shear curve is a parabola here ; eq. (3). 310 MECHANICS OF ENGINEERING. 266. The Four x-Derivatives of the Ordinate of the Elastic Curve If y = func. (a;) is the equation of the elastic curve for any portion of a loaded beam, on which portion the load per unit of length of the beam is w = either zero, (Fig. 234) or = constant, (Fig. 235), or = a continuous func. (x) (as in the last ), we may prove, as fol- lows, that w = the ie-derivative of the ^ shear. Fig. 269. Let N and N' be two * consecutive cross-sections of a loaded - beam, and let the block between them, bearing its portion, wdx, of a distributed ~ load, be considered free. The elastic forces consist of the two stress-couples FIG. 269. (tensions and compressions) and the two shears, e/and J -j- dJ t ck7being the shear-increment conse- quent upon x receiving its increment dx. By putting ^(vert. components) = we have J+dJwdxJ=Q .-. w= dx Q. E. D. But J itself = dM -4- dx, ( 240) and M = [d?y -r- dx 2 ] EL By substitution, then, we have the following relations : y=func.(:r)= ordinate at any point of the elastic curve (1) _i= a = slope at any point of the elastic curve . . (2) El -T^- = M = ordinate (to scale) of the moment curve (3) /73 W i 7- f the ordinate (to scale) ) / EI = ' of the shear ' ' ( ( the load per unit of length } EI ^ = w = -s of beam = ordinate (to scale) > . (5) dx* | of a curve of loading. ) If, then, the equation of the elastic curve (the neutral line of the beam itself ; a reality, and not artificial like the FLEXURE. SPECIAL PROBLEMS. 311 other curves spoken of) is given ; we may by successive differentiation, for a prismatic and homogeneous beam so that both E and / are constant, find the other four quan- tities mentioned. As to the converse process, (i.e. having given w as a function of x, to find expressions for J t M and y as func- tions of x) this is more difficult, since in taking the x-anti-derivative, an unknown constant must be added and determined. The problem just treated in 264, however, offers a very simple case since w is the same function of x, along the whole beam, and there is therefore but one elas- tic curve to be determined. We .-. begin, numbering backward, with T?T _ rfrr W = X ' 8e6 (*.a\ ~dtf~ ~ r ^ ( last and Fig. 268 ( ' ' [N. B. This derivative (dJ-r-dx) is negative since dJ and dx have contrary signs.] .-. (shear=)JE /&=_ ? b ~+ Const. But writing out this equation for x=Q, i.e. for the point 0, where the shear= J? , we have HQ= + Const. .-. Const. = R^ and hence write . (Shear) . (4a) Again taking the x-anti-derivative of both sides (Moment =) E I ^V= T b+X&+(Con8t.=0) . (3o) ajr 6 [At 0, x=0 alsoM,.-. Const. =0]. Again, At 0, where x=0 dy -z-dx=a <) =t}ie unknown slope of the elastic line at 0, and hence C'=EI a^ (2o) 312 MECHANICS OF ENGINEERING. Passing now to y itself, and remembering that at 0, both y and x are zero, so that the constant, if added, would= zero, we obtain (inserting the value of E^ from last . (la) the equation of the elastic curve. This, however, contains the unknown constant a, = the slope at 0. To determine write out eq. (la) for the point B, Fig. 268, where x is known to be equal to I, and y to be = zero, solve for a p , and insert its value both in (la) and (2a). To find the point of max. y (i.e., of greatest deflection) in the elastic curve, write the slope, i.e. dy -f- dx, = zero [see eq. 2o] and solve for x ; four values will be obtained, of which the one lying between and I is obviously the one to be taken. This value of x substituted in (la) will give the maximum deflection. The location of this maximum deflection is neither at the centre of action of the load nor at the section of max. moment (# = The qualities of the left hand members of equations (1) to (5) should be carefully noted. E. g., in the inch -pound- second system of units we should have : 1. y (a, linear quantity) = (so many) inches. 2. dy-r-dx (an abstract number) = (so many) abstract units. 3. M (a moment) == (so many) inch-pounds. 4. J (a shear, i.e., force) = (so many) pounds. 5. w (force per linear unit) = (so many) pounds per run- ning inch of beam's length. As to the quantities E, and 7, individually, E is pounds per sq. in., and /has four linear dimensions, i.e. (so many) bi-quadratic inches. FLEXUKE SPECIAL PROBLEMS. 313 267. ResiUence of Beam With End Supports. Fig. 270. If a mass whose -weight is G (G large com- pared with that of beam) be allowed to fall freely through a height = h upon PG t< . j j 3 T"1" S tne centre of a beam supported at its $_J ' p m extremities, the pressure P felt by the FIG. 270. beam increases from zero at the first instant of contact up to a maximum P m , as already stated in 233a, in which the equation was derived, d m being small compared with h, The elastic limit is supposed not passed. In order that the maximum normal stress in any outer fibre shall at most be=./?', a safe value, (see table 251) we must put Tf'T ~p I = -j [according to eq. (2) 241,] i.e. in equation (a) above, substitute P m = [4 R'l] -t-le, which gives ' ' R" W V /M ' ~E'^ V having put I=FJc? (k being the radius of gyration 85) and Fl= V the volume of the (prismatic) beam. From equation (b) we have the energy, Gh, (in ft.-lbs., or inch- Ibs.) of the vertical blow at the middle which the beam of Fig. 270 will safely bear, and any one unknown quantity can be computed from it, (but the mass of G should not be small compared with that of the beam.) The energy of this safe impact, for two beams of the same material and similar cross-sections (similarly placed), is seen to be proportional to their volumes; while if further- more their cross-sections are the same and similarly placed, the safe Gh is proportional to their lengths. (These same relations hold good, approximately, beyond the elas tic limit.) It will be noticed that the last statement is just the con- 314 MECHANICS OF ENGINEERING. verse of what was found in 245 for static loads, (the pressure at the centre of the beam being then equal to the weight of the safe load) ; for there the longer the beam (and /. the span) the less the safe load, in inverse ratio. As appropriate in this connection, a quotation will be given from p. 186 of " The Strength of Materials and Structures," by Sir John Anderson, London, 1884, viz.: " It appears from the published experiments and state- ments of the Railway Commissioners, that a beam 12 feet long will only support ^ of the load that a beam 6 feet long of the same breadth and depth will support, but that it will bear double the weight suddenly applied, as in the case of a weight falling upon it," (from the same height, should be added) ; " or if the same weights are used, the longer beam will not break by the weight falling upon it unless it falls through twice the distance required to frac- ture the shorter beam." 268. Combined Flexure and Torsion. Crank Shafts. Fig. 271. Let OiB be the crank, and N0 l the portion projecting beyond the nearest bearing N. P is the pressure of the connecting-rod against the crank-pin at a definite in- stant, the rotary motion be- ing uniform. Let a= the perpendicular dropped from the axis 00 l of the shaft upon P, and 1= the distance of P, along the axis 00! from the cross-section NmN' of the shaft, close to the bearing. Let NN' be a diameter of this section, and parallel to a. In considering the portion NO^ free, and thus exposing the circular section NmN', we may assume that the stresses to be put in on the ele- ments of this surface are the tensions (above NN'} and the compressions (below NN'} and shears "| to NN', due to the bending action of P ; and the shearing stress tan^ FLEXURE. SPECIAL PROBLEMS. 315 gent to the circles which have as a common centre, and pass through the respective dF's or elementary areas, these latter stresses being due to the twisting action of P. In the former set of elastic forces let p = the tensile stress per unit of area in the small parallelopipedical ele- ment m of the helix which is furthest from NN' (the neu- tral axis) and /= the moment of inertia of the circle about NN'; then taking moments about NN' for the free body, (disregarding the motion) we have as in cases of flexure (see 239) [None of the shears has a moment about NN'.~\ Next taking moments about 00 X , (the flexure elastic forces, both normal and shearing, having no moments about OOj) we have as in torsion (216) in which p f is the shearing stress per unit of area, in the torsional elastic forces, on any outermost dF, as at m; and /p the polar moment of inertia of the circle about its centre 0. Next consider free, in Fig. 272, a small parallelepiped taken from the helix at m (of Fig. 271.) The stresses [see 209] acting on the four faces f" to the paper in Fig. 272 are there represented, the dimensions (infinitesimal) being n I to NN*, b || to 00,, and d ~\ to the paper in Fig. 272. ./ ^ Fig. 278. 316 MECHANICS OF ENGINEERING. By altering the ratio of b to n we may make the angle what we please. It is now proposed to consider free the triangular prism, GHT, to find the intensity of normal stress q, per unit of area, on the diagonal plane GH, (of length =c,) which is a bounding face of that triangular prism. See Fig, 273. By writing 2' (compons. in direc- tion of normal to GH)=0, we shall have, transposing, qcd=pnd sin 6+p a bd sin 6-\-p & nd cos 6 ; and solving for q q=p ~ sin d+p & [| sin 0+^ . cos 0] ; . (1) but n : c= sin 6 and b : c= cos 6 .'. q=p sin 2 0+p s 2 sin cos . (2) This may be written (see eqs. 63 and 60, O. W. J. Trigo- nometry) n20 . . (3) As the diagonal plane GH is taken in different positions (i.e., as 6 varies), this tensile stress q (Ibs. per sq. in. for instance) also varies, being a function of 6, and its max. value may be >^>. To find 6 for q max. we put -^ , =p sin 20+2p a cos 20, . . (4) = 0, and obtain : tan [2(0 for q max)] --- ^ . . . (5) Call this value of 0, 0'. Since tan 20' is negative, 20' lies either in the second or fourth quadrant, and hence sin 20'= , ^ and cos 20' = T . P (6) vy+4$ f vy+4pj [See equations 28 and 29 Trigonometry, O. W. J.] The FLEXURE. CRANK SHAFT. 337 upper signs refer to the second quadrant, the lower to the fourth. If we now differentiate (4), obtaining . (7) we note that if the sine and cosine of the [20'] of the 2nd quadrant [upper signs in (6)] are substituted in (7) the re- sult is negative, indicating a maximum ; that is, q is a max- imum for 0= the 6' of eq. (6) when the upper signs are taken (2nd quadrant). To find q max., then, put ff for 6 in (3) substituting from (6) (upper signs). "We thus find q max = X[>+Vp 2 +4p 8 2 .] . . (8) A similar process, taking components parallel to GH, Fig. 273, will yield q s max., i.e., the max. shear per unit of area, which for a given p and p s exists on the diagonal plane GH in any of its possible positions, as 6 varies. This max. shearing stress is v. (9) In the element diametrically opposite to ra in Fig. 271, p is compression instead of tension ; q maximum will also be compression but is numerically the same as the q max. of eq. 8. 269. Example. In Fig. 271 suppose P=2 tons = 4,000 Ibs., a=6 in., 1=5 in., and that the shaft is of wrought iron. Required its radius that the max. tension or com- pression may not exceed 7?' =12,000 Ibs. per sq. in.; nor the max. shear exceed S'= 7,000 Ibs. per sq. in. That is, we put y=12,000 in eq. (8) and solve for r : also y s =7,000 in (9) and solve for r. The greater value of r should be taken. From equations (a) and (6) we have (see 219 and 247 for 7 P and 2) 318 MECHANICS OF ENGINEERING. which in (8) and (9) give max. q=y 2 *L [4Z+4/(4^) a +4(2a) a ] . . (8a) KIT n and max. q t =% s\/(4Z) 2 +4(2a) 2 W With max. g=12,000, and the values of P, a, and Z, already given, (units, inch and pound) we have from (8a), ^=2.72 cubic inches .-. r=1.39 inches. Next, with max. g s =7,000 ; P, a, and I as before ; from (9a), r*=%.84: cubic inches .*. r=1.41 inches. The latter value of r, 1.41 inches, should be adopted. It is here supposed that the crank-pin is in such a position (when P=4,000 Ibs., and a=6 in.) that q max. (and q s max.) are greater than for any other position ; a number of trials may be necessary to decide this, since P and a are different with each new position of the connecting rod. If the shaft and its connections are exposed to shocks, R and JS' should be taken much smaller. 270. Another Example of combined torsion and flexure is shown in Fig. 274. The P 1 ' r /7^\ ^~~"^k~~ p wor k ^ *ke working force oj-^// i_J\ /T\~ 1 '' 1 p i( vertical cog-pressure) is y^\ tv^T^Tri^ij - 2 ?4. That is, the rigid body consisting of the two wheels and shaft is employed to transmit power, at a uniform angular velocity, and since it is symmetrical about its axis of rotation the forces act- ing on it, considered free, form a balanced system. (See 114). Hence given PI and the various geometrical quan- FLEXURE. CRANK SHAFT. 319 titles !, &i, etc., we may obtain $ and the reactions P and P B , in terms of P t . The greatest moment of flexure in the shaft will be either P Z,, at (7; or P B ^, at D. The portion CD is under torsion, of a moment of torsion =P 1 a 1 = Qfa. Hence we proceed as in the example of 269, simply put- ting P li (or P B ?3, whichever is the greater) in place of PZ, and P&i in place of Pa. We have here neglected the weight of the shaft and wheels. If Q l were an upward ver- tical force and hence on the same side of the shaft as P lt the reactions P and P B would be less than before, and one or both of them might be reversed in direction. 320 MECHANICS OF ENGINEERING. CHAPTEK IV. FLEXURE, CONTINUED, CONTINUOUS GIRDERS. 271. Definition. A continuous girder, for present pur- poses, may be denned to be a loaded straight beam sup- ported in more than two points, in which case we can no longer, as heretofore, determine the reactions at the sup- ports from simple Statics alone, but must have recourse io the equations of the several elastic curves formed by its neutral line, which equations involve directly or indirect- ly the reactions sought ; the latter may then be found as if they were constants of integration. Practically this amounts to saying that the reactions depend on the man- ner in which the beam bends ; whereas in previous cases, with only two supports, the reactions were independent of the forms of the elastic curves (the flexure being slight, however). As an Illustration, if the straight beam of Fig. 275 is placed on three supports 6>, B, and (?, at the same level, the reactions of these supports seem at first sight indeterm- inate ; for on considering the p ./ H whole beam free, we have three rH3^^H~~^ ^ unknown quantities and only B /^ f-~ A ^"""""f~""'''''2A two equations, viz : S (vert. ' Fl0 275 compons.) and S (moms, about some point") = 0. If now be gradually lowered, it receives less and less pres- FLEXURE. CONTIGUOUS GIRDERS. 321 sure, until it finally reaches a position where the beam barely touches it ; and then O's reaction is zero, and B and C support the beam as if were not there. As to how low must sink to obtain this position, depends on the , stiffness and load of the beam. Again, if be raised above the level of B and C it receives greater and greater pressure, until the beam fails to touch one of the other supports. Still another consideration is that if the beam were tapering in form, being stiffest at 0, and pointed at B and 6 y , the three reactions would be different from their values for a prismatic beam. It is therefore evident that for more than two supports the values of the reactions de- pend on the relative heights of the supports and upon the form and elasticity of the beam, as well as upon the load. The circumstance that the beam is made continuous over the support 0, instead of being cut apart at into two independent beams, each covering its own span and hav- ing its own two supports, shows the significance of the term " continuous girder." All the cases here considered will be comparatively simple, from the symmetry of their conditions. The beams will all be prismatic, and all external forces (i.e. loads and reactions) perpendicular to the beam and in the same plane. All supports at the same level 272. Two Equal Spans ; Two Concentrated Loads, One in the Mid- dle of Each Span. Prismatic Beam. Fig. 275. Let each half- span = y 2 I. Neglect the weight of the beam. Required the reactions of the three supports. Call them P B , P and P c . From symmetry P B = P c , and the tangent to the elastic curve at is horizontal ; and since the supports are on a level the deflection of C (and B} below O's tangent is zero. The separate elastic curves OD and DC have a common slope and a common ordinate at D. For the equation of OD, make a section n anywhere be- tween and D, considering nC a free body. Fig. 276 (a) 322 MECHANICS OF ENGINEERING. PC | Y Fie. 876. with origin and axis as there indicated. From 2 (moms about neutral axis of ri) = we have (see 231) (1 =V) . (2) The constant = 0, for at both x, and dy -*- dx, = 0. Taking the x-anti-derivative of (2) we have (B) Here again the constant is zero since at O,x and y both =0. (3) is the equation of OD, and allows no value of x <0 or> -A. It contains the unknown force P c . For the equation of DC, let the variable section n be made anywhere between D and C, and we have (Fig. 276 (V) ; x may now range between y 2 l and I) (5)' To determine C", put x = y 2 l both in (5)' and (2), and equate the results (for the two curves have a common tangent line at D) whence C' % PI? (5) FLEXURE. CONTINUOUS GIRDERS. 31*3 Hence Ely = # PPx-P c !_ + (7" . . (6)' At D the curves have the same y, hence put x = -L in the right hand member both of (3) and (6)', equating results, and we derive C"'= 1 PP (6) which is the equation of DC, but contains the unknown reaction P c . To determine P c we employ the fact that O's tangent passes through C, (supports on same level) and hence when x = I in (6), y is known to be zero. Making these substitutions in (6) we have From symmetry P B also = -JP, while P must = ~P, since P B + P + P c = 2 P (whole beam free). [NoTE. If the supports were not on a level, but if, (for instance) the middle support were a small distance = ^ below the level line joining the others, we should put x = I and y = ho in eq. (6), and thus obtain P B = P c = -? 6 P + 3EI^, which depends on the material and form of the C prismatic beam and upon the length of one span, (whereas with supports all on a levd, P B = P c = A P is independent of the material and form of the beam so long as it is ho- mogeneous and prismatic.) If P , which would then = g P 6 J577(Ao-hP), is found to be negative, it shows that requires a support from above, instead of below, to cause it to occupy a position Jio below the other supports, i.e. the beam must be " latched down " at 0.] The moment diagram of this case can now be easily con- structed ; Fig. 277. For any free body nG, n lying in DC, we have 324 MECHANICS OF ENGINEERING. i.e., varies directly as x, un- c til x passes D when, for any point on DO, M=*/ l6 Px-P(x- 1 -) which is =0, (point of in- flection of elastic curve) for x= 8 / n I (note that x is PIS. 277. measured from C in this figure) and at 0, where x=l, becomes |,-PZ Hence, since M max. =^Pl, the equation for safe loading is 6 ^ p/ ~e~ 32 (7) The shear at C and anywhere on CD=^P, while on DO it = l ^P in the opposite direction : which is less than M , although M B is the calculus max. (negative) for M, as may be shown by writ- ing the expression for the shear (J=% W- wx) equal to zero, etc. FLEXURE. CONTINUOUS GIUDERS. 331 Hence M m =^ Wl, and the equation for safe loading is ^=r^ ^1 (5) Since (with this form of loading) if the beam were not built in but simply rested on two end supports, the equa- tion for safe loading would be [R'I-t-e~] = * Wl, (see 242), it is evident that with the present mode of support it is 50 per cent, stronger as compared with the other ; i.e., as re- gards normal stresses in the outer elements. As regards shearing stresses in the web if it has one, it is no stronger* since t7 m =y Win both cases. As to stiffness under the uniform load, the max. deflec- tion in the present case may be shown to be only -i- of that in the case of the simple end supports. u~~ It is noteworthy that the shear diagram in Fig. 283 is identical with that for simple end supports 242, under uniform load ; while the moment diagrams differ as fol- lows : The parabola KB' A, Fig. 283, is identical with that in Fig. 235, but the horizontal axis from which the ordi- nates of the former are measured, instead of joining the extremities of the curve, cuts it in such a way as to have equal areas between it and the curve, on opposite sides i.e., areas \_KC'H>+AG'0']=vxQ& H'G'B' In other words, the effect of fixing the ends horizontally is to shift the moment parabola upward a distance = M (to scale), = ~ Wl, with regard to the axis of reference, O'B', in Fig. 235. 276. Remarks. The foregoing very simple cases of con- tinuous girders illustrate the means employed for deter- mining the reactions of supports and eventually the max. moment and the equations for safe loading and for deflec- tions. When there are more than three supports, with spans of unequal length, and loading of any description, the analysis leading to the above results is much more complicated and tedious, but is considerably simplified 332 MECHANICS OF ENGINEERING. and systematized by the use of the remarkable theorem of three moments, the discovery of Clapeyron, in 1857. By this theorem, given the spans, the loading, and the vertical heights of the supports, we are enabled to write out a rela- tion between the moments of each three consecutive sup- ports, and thus obtain a sufficient number of equations to determine the moments at all the supports [p. 641 Rankine's Applied Mechanics.] From these moments the shears close to each side of each support are found, then the reactions, and from these and the given loads the moment at any section can be determined ; and hence finally the max. moment M m , and the max. shear J m . The treatment of the general case of continuous girders by graphic methods, however, is comparatively simple, and its presentation is therefore deferred, 391. THE DANGEROUS SECTION OF NON-PRIS- MATIC BEAMS. 277. Remarks. By " dangerous section " is meant that sec- tion (in a given beam under given loading with given mode of support) where p, the normal stress in the outer fibre, at distance e from its neutral axis, is greater than in the outer fibre of any other section. Hence the elasticity of the material will be first impaired in the outer fibre of this section, if the load is gradually increased in amount (but not altered in distribution). In all preceding problems, the beam being prismatic, /, the moment of inertia, and e were the same in all sections, hence when the equation -=M [239] was solved for p, > giving we found that p was a max., = p m , for that section whose M was a maximum, since p varied as M, for the moment FLEXURE XON-PRISMATIC BEAMS. 333 of the stress-couple, as successive sections along the beam were examined. But for a non-prismatic beam /and e change, from sec- tion to section, as well as M, and the ordinate of the moment diagram no longer shows the variation of p, nor is p a max. where M is a max. To find the dangerous section, then, for a non-prismatic beam, we express the J/, the /, and the e of any section in terms of x, thus obtain- ing j3=func. (x), then writing dp-t-dx=Q, and solving for x. 278. Dangerous Section in a Double Truncated Wedge. Two End Supports. Single Load in Middle. The form is shown in Fig. 284. Neglect weight of beam ; measure x from one sup- port 0. The t reaction at r_~^r4 7~j[w~jDl eac h support is y 2 P. The width of the Fio.284. beam - b at all sections, while its height, v, varies, being = h at 0. To express the e = y 2 v, and the / = 1 btf (247) of any section on OC, in terms of x, conceive the sloping faces of the truncated wedge to be prolonged to their intersec- tion A, at a known distance = c from the face at 0. We then have from similar triangles v : x + c : : h : c, .-. v = ~ (x + c) . . (1) C and .-. e = y 2 - (x+c) and 1= 1 6 -O+c] s . (2) c 12 c' For the free body nO, I (moms. n ) = gives [That is, the M = y 2 Px.] But from (2), (3) becomes By putting dp -*- dx == we obtain both x = c, and 334 MECHANICS OF ENGINEERING. x = -f c, of which the latter, x = + c, corresponds to a maximum for p (since it will be found to give a negative result on substitution in d*p -4- dx*). Hence the dangerous section is as far from the support 0, as the imaginary edge, A, of the completed wedge, but of course on the opposite side. This supposes that the half -span, )4l, is > c\ if not, the dangerous section will be at the middle of the beam, as if the beam were prismatic. He- with \***^**^*H>^ ) at middle) ( while with ) f he Aquation for safe ( R ^ [2A]2 /7 V loading is : (put x=c < i i-= y 2 PC (6) jand|>=;.B'ia[3]) ( (see 239.) 279. Double Truncated Pyramid and Cone. Fig. 285. For FIG. 285. the truncated pyramid both width u, and height = v, are variable, and if b and h are the dimensions at 0, and c = OA distance from to the imaginary vertex A, we shall have from similar triangles u=~ (a?+c)and v= -(cc+c). in the moment Hence, substituting e=}4v and I=L equation . . (7) ,~ ' ' ' FLEXURE. NON-PKISMATIC BEAMS. 335 Putting this = 0, we have x = c, x = c, and x = + y 2 c, hence the dangerous section is at a distance x = y 2 c from 0, and the equation for safe loading is either *?*L-> ft Fl . . . . if # Z is < */ 2 c .... (9) b (in which b' and A' are the dimensions at mid-span) . . For the truncated cone (see Fig. 285 also, on right) where e = the variable radius r, and / = % it r*, we also have / =[Const.] . ^- f .... (11) and hence p is a max. for x = }4 c, and the equation for safe loading either = % Pl,tor %l < % c ..... (12) (where r' = radius of mid -span section) ; or x = % p Cf for I/4 i > (where r = radius of extremity.) NON-PRISMATIC BEAMS OF "UNIFORM STRENGTH." 280. Remarks. A beam is said to be of " uniform strength " when its form, its mode of support, and the dis- tribution of loading, are such that the normal stress p has the same value in all the outer fibres, and thus one ele- ment of economy is secured, viz. : that all the outer fibres may be made to do full duty, since under the safe loading, p will be = to R' in all of them. [Of course, in all cases of flexure, the elements between the neutral surface and 336 MECHANICS OF ENGINEERING. the outer fibres being under tensions and compressions less than R' per sq. inch, are not doing full duty, as regards economy of material, unless perhaps with respect to shearing stresses.] In Fig. 265, 261, we have already had an instance of a body of uniform strength in flexure, viz. : the middle segment, CD, of that figure ; for the moment is the same for all sections of CD [eq. (2) of that ], and hence the normal stress p in the outer fibres (the beam being prismatic in that instance). In the following problems the weight of the beam itself is neglected. The general method pursued will be to find an expression for the outer -fibre-stress p, at a definite sec- tion of the beam, where the dimensions of the section are known or assumed, then an expression f or p in the varia- ble section, and equate the two. For clearness the figures are exaggerated, vertically. 281. Parabolic Working Beam. UnsymmetricaL Fig. FIG. 886. CBO is a working beam or lever, B being the fixed fulcrum or bearing. The force P being given we may compute P c from the mom. equation P 1 PJn while the fulcrum reaction is P B =P -\-P c . All the forces are ~[ to the beam. The beam is to have the same width b at all points, and is to be rectangular in section. Eequired first, the proper height ^, at B, for safety. From the free body BO, of length = 1 , we have I (moms B ) = 0; i.e., (1) FLEXURE. NOX-PRISMATIC BEAMS. 337 Hence, putting p B = R', h t becomes known from (1). Required, secondly, the relation between the variable height v (at any section n) and the distance x of n from 0. For the free body nO, we have (2' moms a = 0) pJ =PcX . or PM> =Pj , ^a ... ft But for " uniform strength " p a must = p a \ hence equate their values from (1) and (2) and we have 5 = A, which may be written (% vf = MW x (3) so as to make the relation between the abscissa x and the ordinate }4 v more marked ; it is the equation of a para- bola, whose vertex is at 0. The parabolic outline for the portion BC is found simi- larly. The local stresses at C, B, and must be proper- ly provided for by evident means. The shear J = P , at 0, also requires special attention. This shape of beam is often adopted in practice for the working beams of engines, etc. The parabolic outlines just found may be replaced by trapezoidal forms, Fig. 287, without using much more ma- terial, and by making the slop- ing plane faces tangent to the p^T'"" ~f= parabolic outline at points T gK^ZZj 1 '"'' ~~ and T 1} half-way between and "^^T^ "~" B, and C and B, respectively. It PIS. 237. can be proved that they contain minimum volumes, among all trapezoidal forms capable of circumscribing the given parabolic bodies. The dangerous sections of these trape- zoidal bodies are at the tangent points TO and TV This is as it should be, (see 278), remembering that the subtan- gent of a parabola is bisected by the vertex. 282. I-Beam of Uniform Strength. Support and load same as in the preceding . Fig. 288. Let the area of the 338 MECHANICS OF ENGINEERING. Fio. 288. flange-sections be = F and let it be the same for all values of x. Considering all points of F at any one section as at the same distance z from the neutral axis, we may write 7 = z 2 F, and assuming that the flanges take all the tension and compression while the (thin) web carries the shear, the free body of length x in Fig. 288 gives (moms, about n) P-=P c x ; i.e. - = P e x : or, since p is to be constant, e z z = [Const.], x . . ., . ' (1) i. e. z must be made proportional to x. Hence the flanges should be made straight. Practically, if they unite at G, the web takes but little shear. 283. Rectang. Section. Height Constant. Two Supports (at Ex- tremities). Single Eccentric Load. Fig. 289. b and h are the dimensions of the section at B. With BO free we have e B PIG. 289. At any other section on BO, as n, where the width is u, the variable whose relation to x is required, we have for wOfree Equating p K . and p n we have u : b :: x : 1 Q (2) (3) FLEXURE. BEAMS OF UNIFORM STRENGTH. 339 That is, BO must be wedge-shaped with its edge at 0, ver- tical. 284. Similar Eectang. Sections. Otherwise as Before. Fig. 289 a, b and h are the dimen- sions at B; at any other section n, on BO, the height v and width u, are the variables whose relation to x is desired and by hypoth- esis are connected by the relation u : v :: b : h (1) (since the section at n is a rectangle similar to that at B). For the free body BO p B = For the free body nO p n = --^ Writing p lt = p B we obtain IQ -r- bh 2 = x put u = bv -s- h, from (1) ; whence (2) .. . ' (3) uv 2 , in which ' (4) which is the equation of the curve (a cubic parabola) whose abscissa is x and ordinate l /^ v ; i.e., of the upper curve of the outline of the central longitudinal vertical plane section of the body (dotted line BO) which is sup- posed symmetrical about such a plane. Similarly the central horizontal plane section will cut out a curve a quarter of which (dotted line B 0) has an equation (5) That is, the height and width must vary as the cube root 340 MECHANICS OF ENGINEERING. of the distance from the support. The portion CB will give corresponding results, referred to the support G. [If the beam in this problem is to have circular cross- sections, let the student treat it in the same manner.] 286. Uniform Load. Two End Supports. Rectangular Cr. Sections. Width Constant. "Weight of beam neglected. How should the height vary, the height and width at the middle being h and 5 ? Fig. 289 &. From symmetry each reaction = y z W = ^ wl. At any cross section n, the width is = b, (same as that at the middle) and the height = v, variable. S (moms. n ) =0, for the free body nO, gives t W But for a beam of uniform strength, p n is to be p B as computed from 2 (moms. B ) = for the free body . . BO, i.e from _ 2 2 w Hence solve (1) for p n and (2) for p B and equate the results, whence t^=*L[Z*-rf] ; or (^) 2 = ( [Z*-* 2 ] . (3) This relation between the abscissa a? and the ordinate yv, of the curve CBO, shows it to be an ellipse since eq. (3) is that of an ellipse referred to its principal diameter and the tangent at its vertex as co-ordinate axes. In this case eq. (3) covers the whole extent of both upper and lower curves, i.e. the complete outline, of the curve CBOB', whereas in Figs. 286, 289, and 289 a, such is not the case. FLEXURE. BODIES OF UNIFORM STRENGTH. 341 287. Cantilevers of Uniform Strength, Beams built in at one end, horizontally, and projecting from the wall with- out support at the other, should have the forms given be- low, for the given cases of loading, if all cross-sections are to be Rectangular and the weight of beam neglected. Sides of sections horizontal and vertical. Also, the sections are symmetrical about the axis of the piece, b and h are the dimensions at the wall. 1= length. No proofs given. Width constant. -, Vertical outline parabolic. Single end load. Height constant. , Single end load. Horizontal outline triangular. Constant ratio of height v to width u. Both outlines cu- bic parabolas. Fig. 290, (a). Fig. 290, (b). i 29(X (1) (2) (3) T ' 342 MECHANICS OF ENGINEERING. Uniform Load. angular. Uniform Load." Height constant. Horiz. outline is two parabolas meet- ing at (vertex) with geomet. axes II to wall. Fig. 291, (6). Uniform Load.*. Both outlines semi- cubic parabolas. I Fig. 291, (c). Sections similar rectangles. (#)'=(#*)' . .(5) T (6) ? (67 289. Beams and cantilevers of circular cross-sections may be dealt with similarly, and the proper longitudinal outline given, to constitute them " bodies of uniform strength." As a consequence of the possession of this property, with loading and mode of support of specified character, the following may be stated ; that to find the equation of safe loading any cross-section loJiatever may be employed. This refers to tension and compression. As regards the shearing stresses in different parts of the beam the condition of " uniform strength " is not necessarily ob- tained at the same time with that for normal stress in the outer fibres. DEFLECTION OF BEAMS OF UNIFORM STRENGTH. 290. Case of 283, the double wedge, but symmetrical, i.e., ?,=$,= jZ, Fig. 292. Here we shall find the use of the BEAMS OF UNIFOKM STRENGTH. 343 ^ p t EI form : (of the three forms for the moment of the stress P couple, see eqs. (5), (6) and (7),- 229 and 231) of the most direct service in determining the form of the elastic curve OB, which is symmetrical, and has a common tangent at B y with the curve BC. First to find the radius of curva- ture, p, at any section n, we have for the free body nO, J?(moms. n 0), whence we have >/ 12 | uk*= %P $ and .'. p- % % . ^ (1) from which all variables have disappeared in the right hand member ; i.e., p is constant, the same at all points of the elastic curve, hence the latter is the arc of a circle, having a horizontal tangent at B. To find the deflection, d, at B, consider Fig. 292, (6j where d=KB, and the full circle of radius BH*-*p is drawn. The triangle^O.Bjs similar to YOB, and .-. KB : OB : : OB : TB But OB=y 2 l, KB=d and Y=2/> ' d= ^ and ' from e( i- From eq. (4) 233 we note that for a beam of the same material but prismatic (parallelopipedical in this case,) having the same dimensions, b and h, at all sections as at 1 PT? PP the middle, deflects an amount =JQ- ^^^i under a 344 MECHANICS OF ENGINEERING. load P in the middle of the span. Hence the tapering beam or the present has only ^ the stiffness of the pris- matic beam, for the same 6, A, ?, E, and P. 291. Case of 281 (Parabolic Body), WithZ l =7 () , i.e., Symmet- rical. Fig. 293,(a). Bequired the equation of the neutral ^T~ ,,1 B 4- ^ T> ** i " () 8 t^-H* line 05. For the free body nO, J(moms. n )=0 gives us 27/_ != y 2 Px . , . (1)' Fig. 293, (6), shows simply the geometrical relations of the problem, position of origin, axes, etc. OnB is the neutral line or elastic curv<^ whose equation, and greatest ordinate d, are required. (The right hand member of eq. (1)" is made negative because d?y-+dy? is negative, the curve being con- cave to the axis X in this, the first quadrant.) Now if the beam were prismatic, /, the " moment of in- ertia " of the cross-section would be constant, i.e., the same for all values of x, and we might proceed by taking the x- anti-derivative of each member of (1)" and add a constant ; but it is variable and is a;T , (from eq. 3, 281, putting ^=^0 hence (1)" becomes To put this into the form Const, x ^=func. of (x\ we need BEAMS OF UNIFORM STRENGTH. 345 3 only divide through by x 2 ,,(and for brevity denote -L Ebh\+ (y 2 l)T by A) and obtain We can now take the a>anti-derivative of each member, and have A^=y 2 P(^x + ^)+C .... (2)' To determine the constant C, we utilize the fact that at B, where x=}l, the slope dy-r-dx is zero, since the tangent line is there horizontal, whence from (2)' ,-. (2)' becomes A - =P[^l-x ] ..... (2) Ay=P WJi.x-y 3 x*]+[C'=0] ... (3) (C"=0 since for x=0, y=Q). We may now find the deflec- tion d (Fig. 293(6)) by writing x= y 2 l and y =d, whence, after restoring the value of the constant A, D73 and is twice as great [being=2. _ r ]* as if the girder 4:Ebhi i 3 * See $ 233, putting / = bh in eq. (4). were parallelopipedical. In other words, the present girder is only half as stiff as the prismatic one. 292. Special Problem. (I.) The symmetrical beamjn Fig. 294 is of rectangular cross-section and constant width = ft, 346 MECHANICS OF ENGINEERING. but the height is constant only over the extreme quarter spans, being ^=^4 h, i. 6' half the height li at mid-span. The convergence of the two truncated wedges forming the middle quarters of the beam is such that the prolongations of the upper and lower surfaces ivould meet over the supports (as should be the case to make A=2A,). Neglecting the weight of the beam, and placing a single load in middle, it is required to find the equation for safe loading ; also the equations of the four elastic curves ; and finally the deflec- tion. The solutions of this and the following problem are left to the student, as exercises. Of course the beam here given is not one of uniform strength. 293. Special Problem. (II). Fig. 295. Eequired the man- ner in which the width of the beam must vary, the height being constant, cross-sections rectangular, weight of beam FIG. 295. neglected, to be a beam of uniform strength, if the load is uniformly distributed ? FLEXURE. OBLIQUE PORCBS- 347 CHAPTER V. FLEXURE OF PRISMATIC BEAMS UNDER OBLIQUE FORCES. 294, Remarks, By " oblique forces " will be understood external forces not perpendicular to tho beam, but these external forces will be confined to one plane, called the force-plane, which contains the axis of the beam and also cuts the beam symmetrically. The curvature induced by these external forces will as before be considered very slight, so that distances measured along the beam will be treated as unchanged by the flexure. It will be remembered that in previous problems the proof that the neutral axis of each cross section passes through its centre of gravity, rested on the fact that when a portion of the beam having a given section as one of its bounding surfaces is considered free, the condition of equilibrium ^' (compons. || to beam)=0 does not introduce any of the external forces, since these in the problems re- ferred to, were "| to the beam ; but in the problems of the present chapter such is not the case, and hence the neutral axis does not necessarily pass through the centre of gravity of any section, and in fact may have only an ideal, geomet- rical existence, being sometimes entirely outside of the section ; in other words, the fibres whose ends are exposed in a given section may all be in tension, (or all in compres- sion,) of intensities varying with the distance of each from the neutral axis. It is much more convenient, however, to take for an axis of moments the gravity axis parallel to the 348 MECHANICS OF ENGINEERING. neutral axis instead of the neutral axis itself, since this gravity axis has always a known position. 295. Classification of the Elastic Forces. Shear, Thrust, and Stress-Couple. Fig. 296. Let AKM\>& one extremity of a portion, considered free, of a prismatic beam, under oblique forces. C is the centre of gravity of the section ex- posed, and GC the gravity axis ~| to the force plane GAK. The stresses acting on the elements of area (each =dF ) of the section consist of shears (whose sum=7, the "total shear") in the plane of the section and parallel to the force plane, and of normal stress parallel to AK and proportion- al per unit of area to the distances of the dF's on which they act from the neutral axis NC", real or ideal (ideal in this figure). Imagine the outermost fibre KA, whose dis- tance from the gravity axis is=e and from the neutral axis -Pi-- S96. ==e +a, to be prolonged an amount AA', whose length by- some arbitrary scale represents the normal stress (tension 01 compression) to which the dF at A is subjected. Then, ii a plane be passed through A' and the neutral axis NC", tne lengths, such as mr, parallel to AA', intercepted between this plane and the section itself, represent the stress-inten- FLEXURE. OBLIQUE FORCES. 349 sities (i. e., per unit area) on the respective dF's. (In thia particular figure these stresses are all of one kind, all ten- sion or all compression ; but if the neutral axis occurs within the limits of the section, they will be of opposite kinds on the two sides of NC."} Through C", the point determined in A'NC" by the intercept CC' of the centre of gravity, pass a plane A"M"T" parallel to the section it- self ; it will divide the stress-intensity AA' into two parts Pi and p 2 , and will enable us to express the stress-intensity mr, on any dF at a distance a from the gravity -axis GC, in two parts ; one part the same for all dF's, the other depen- dent on 2, thus : [Stress-intensity on any dF] = PI+ p 2 . . (1) and hence the [actual normal stress on any dF] = p { d F + p 2 dF (2) For example, the stress-intensity on the fibre at T, where 2 = e,, will be pi ^- /%, and it is now seen Low we may find the stress at any dF when p l and p^ have been found. If the distance a, between the neutral and gravity axes is desired, we have, by similar triangles p 2 : e :: C'O'.a whence a =& . e . . . . (3) It is now readily seen, graphically, that the stresses or elas- tic forces represented by the equal intercepts between the parallel planes AMT and A" M" T", constitute a uniform- ly distributed normal stress, which will be called the " uni- form thrust," or simply the thrust (or pull, as the case may be) of an intensity = p lt and .. of an amount= Cp\dF = Pi fdF = p l F. It is also evident that the positive intercepts forming the 350 MECHANICS OF ENGINEERING. wedge A" A' G' G' and the negative intercepts forming the wedge M"M.'G'G' form a system of "graded stresses" whose combination (algebraic) with those of the " thrust " shows the two sets of normal stresses to be equivalent to the actual system of normal stresses represented by the small prisms forming the imaginary solid AMT . . A'M' T'. It will be shown that these graded stresses constitute a " stress-couple." Analytically, the object of this classification of the nor- mal stresses into a thrust and a stress-couple, may be made apparent as follows : In dealing with the free body KAM Fig. 296, we shall have occasion to sum the components, parallel to the beam, of all forces acting (external and elastic), also those ~| to the beam; and also sum their moments about some axis ~| to the force plane. Let this axis of moments be GC the gravity -axis of the section (and not the neutral axis) ; also take the axis X || to the beam and Y~\ to it (and in force-plane). Let us see what part the elastic forces Flo 2 97. will play in these three summations. See Fig. 297, which gives merely a side view. Referring to eq. (2) we see that [see eq. (4) 23]. But as the z's are measured from G gravity axis, z must be zero. Hence [The IX of the Elastic forces] = Pl F= (4) FLEXURE. OBLIQUE FORCES. 351 Also, [The I T of the Elastic forces] = e7== the shear; . (6) while for moments about G [see eq. (1)] [The 2 (moms. G ) of the elastic forces] = and hence finally where 7 G , C z 2 dF, is the "moment of inertia " of i the section about the gravity axis G, (not the neutral axis). The expression in (6) may be called the moment of the stress-couple, understanding by stress-couple a couple to which the graded stresses. of Fig. 297 are equivalent. That these graded stresses are equivalent to a couple is shown by the fact that although they are X forces they do not appear in eq. 4, for IX; hence the sum of the tensions ^-f zdF ~| equals that of the compressions f^ C^zdF "I in that set of normal stresses. We have therefore gained these advantages, that, of the three quantities J"(lbs.), p v (Ibs. per sq. inch), and p 2 (Ibs. per sq. inch) a knowledge of which, with the form of the section, completely determines the stresses in the section, equations (4), (5), and (6) contain only one each, and hence algebraic elimination is unnecessary for finding any one of them ; and that the axis of reference of the moment of inertia /is the same axis of the section as was used in former problems in flexure. Another mode of stating eqs. (4), (5) and (6) is this : The sum of the components, parallel to the beam, of the exter- nal forces is balanced by the thrust or pull; those perpen- 352 MECHANICS OF ENGINEERING. dicular to the beam are balanced by the shear; while the sum of the moments of the external forces about the gravity axis of the section is balanced by that of the stress- couple. Notice that the thrust can have no^moment about the gravity axis referred to. The Equation for Safe Loading, then, will be this : (a) . (pip>) max. "1 whichever "] is r-jr , . ( 7) greater< J For E', see table in 251. The double sign provides for the cases where p { and p 2 are of opposite kinds, one tension the other compression. Of course (pi+pz) max is not the same thing as [p Y max. +p z max.]. Inmost cases in prac- tice e l =e, and then the part (b) of eq. (7) is unnecessary. 295a. Elastic Curve with Oblique Forces. (By elastic curve is now meant the locus of the centres of gravity of the sec- tions.) Since the normal stresses in a section differ from those occuring under perpendicular forces only in the ad- dition of a uniform thrust (or pull), whose effect on the short lengths (=dx) of fibres between two consecutive sec- tions V V and U V , Fig. 297, is felt equally by all, the loca- tion of the centre of curvature R, is not appreciably differ- ent from what it would be as determined by the stress- couple alone. Thus (within the elastic limit), strains being proportional to the stresses producing them, if the forces of the stress- couple acted alone, the length dx=G G' of a small portion of a fibre at the gravity axis would remain unchanged, and the lengthening and shortening of the other fibre-lengths between the two sections U V and IF V , originally parallel, would occasion the turning of TJ' V through a small angle (relatively to U V^) about G', into the position which it oc- cupies in the figure (297), and GR l would be the radius of curvature. But the effect of the uniform pull (added to that of the couple) is to shift U' V parallel to itself into the position UV, and hence the radius of curvature of the FLEXURE. OBLIQUE FORCES. 353 elastic curve, of which G G is an element, is G^R instead of GJt'. But the difference between G R and G^R' is very small, being the same, relatively, as the difference between G Q G and G G' ; for instance, with wrought-iron, even if p lt the intensity of the uniform pull, were as high as 22,000 Ibs. per sq. in. [see 203] G G would exceed G Q G' by only l / u of one per cent. (=0.0008) ; hence by using GB' instead of GR as the radius of curvature p, an error is introduced of so small an amount as to be neglected. But from 231, eqs. (6) and (7), = EI^ = M, the ft CLXr the sum of the moments of the external forces ; hence for prismatic beams under oblique forces we may still use (i) as one form for the J(moms.) of the elastic forces of the section about the gravity-axis ; remembering that the axis X must be taken parallel to beam. 296. Oblique Cantilever with Terminal Load. Fig. 298. Let 1= length. The " fixing" of the lower end of the beam is its only support. Measure x along the beam from 0. Let Fio. 299. n be the gravity axis of any section and nT, =x sin a, the length of the perpendicular let fall from n on the line of action of the force P (load). The flexure is so slight that n T is considered to be the same as before the load is al- ,354 MECHANICS OP ENGINEERING. lowed to act. [If a were very small, however, it is evident that this assumption would be inadmissible, since then a large proportion of nT would be due to the flexure caused by the load.] Consider nO free, Fig. 299. In accordance with the pre- ceding paragraph (see eqs. (4), (5), and (6)) the elastic forces of the section consist of a shear J, whose value may be obtained by writing ^'7=0 whence J=P sin ; . -. .+ * , (1) of a uniform thrust =piF, obtained from 2X=Q, viz : P cos p l F=Q .'. piF=P cos a. ; . (2) and of a stress-couple whose moment [which we may write either t=-, or El ^ ] is determined from ,T(moms. n )=0 or P^Px sin a=0, or P^=Px sin a . . (3) As to the strength of the beam, we note that the stress-in- tensity, pi, of the thrust is the same in all sections, from to L (Fig. 298), and that p,, the stress-intensity in the outer fibre, (and this is compression if e=no' of Fig. 299) due to the stress-couple is proportional to x ; hence the max. of [/>!+#>] will be in the lower outer fibre at L, Fig. 298, where x is as great as possible, =1 ; and will be a compres- sion, viz. : . = P [ cos + foin a) e~\ ^ ^ (4) L F I J .. the equation for Safe Loading is R =P r cosg j_^(sin ) #>] max. FLEXURE. OBLIQUE FORCES. 355 can not exceed, numerically, [pi+p-H max. The stress- intensity in the outer fibres along the upper edge of the beam, being p\p^ (supposing e l =e} will be compressive at the upper end near 0, since there p 2 is small, x being small ; but lower down as x grows larger, p 2 increasing, a section may be found (before reaching the point L) where Pz=pi and where consequently the stress in the outer fibre is zero, or in other words the neutral axis of that section passes through the outer fibre. In any section above that section the neutral axis is imaginary, i.e., is altogether out- side the section, while below it, it is within the section, but cannot pass beyond the gravity axis. Thus in Fig. 300, O'L' FIG. 300. FIG. 301. is the locus of the positions of the neutral axis for successive sections, while OL the axis of the beam is the locus of the gravity axes (or rather of the centres of gravity) of the sections, this latter line forming the " elastic curve " un- der flexure. As already stated, however, the flexure is to be but slight, and a must not be very small. For in- stance, if the deflection of from its position before flex- ure is of such an amount as to cause the lever-arm OR of P about L to be greater by 10 per cent, than its value (=1 sin a) before flexure, the value of p 2 as computed from eq. (3) (with x=T) will be less than its true value in the same proportion. The deflection of from the tangent at L, by 237, Fig. 229(a) is d=(P sina)P-r-3EI, approximately, putting P sin a 356 MECHANICS OF ENGINEERING. for the P of Fig. 229 ; but this very deflection gives to the other component, P cos #, || to the tangent at L, a lever arm, and consequent moment, about the gravity axes of all the sections, whence for -' (moms. L )=0 we have, (more ex- actly than from eq. (3) when x=l) (6) ("We have supposed P replaced by its components || and "I to the fixed tangent at L, see Fig. 301). But even (6) will not give an exact value for p. 2 at L ; for the lever arm of P cos a, viz. d, is >(P sin a)P-3EI, on account of the presence and leverage of P cos a itself. The true value of d in this case may be obtained by a method similar to that indicated in the next paragraph. 297. Elastic Curve of Oblique Cantilever with Terminal Load. More Exact Solution. For variety place the cantilever as in Fig. 302, so that the deflection OY= d tends to decrease the moment of P about the gravity axis of any section, n. AVe may replace P by its X and Y components, Fig. 303, || and ~| respectively to the fixed tangent line at L. The origin, D Jp */ WX ^ o, is taken at the free end of Fl - 808- Fm.303. the beam. Let a= angle bet- ween P and X. For a free body On, n being any section, we have 2' (moms. n )=0 whence ; P(cos a)y P(sin a) x (1) [See eq. (1) 295a]. In this equation the right hand member is evidently (see fig. 303) a negative quantity; this is as it should be, for Eld^y-^dy? is negative, the curve being concave to the axis X in the first quadrant. (It must be noted that the axis X is always to be taken || to the beam, for EId 2 y -t-dx? to represent the moment of the stress-couple.) FLEXURE. OBLIQUE FORCES. 357 Eq. (1) is not in proper form for taking the a>anti-deri- vative of both members, since one term contains the vari- able y, an unknown function of x. Its integration is in- cluded in a more general case given in some works on cal- culus, but a special solution by Prof. Kobinson, of Ohio, is here subjoined for present needs.* ~\Ve thus obtain as the equation of the elastic curve in Pig. 303, * ^ + e^ir(sin)x-(cos%l=smarer e^l (2) In which e denotes the NaperianBase=2.71828, an abstract number, and q for brevity stands for To find the deflection d, we make xl in (2), and solve for y; the result is d. The uniform thrust at L is piF=Pcosa .... (3) while the stress intensity p^ in the outer fibre at L, is ob- * Denoting Pcoa+EIlyy q* and P&ma+EIby p*, eq. (l)becomea-~^=g t yp t x . . (6) Differentiate (6) then ^=? 9 ^- p". Differentiate again : whence- ~=g-*-^|p . (7) Letting^- See (7) < j?L=q* u , which ' mult - b y 2 <*' ^es -^ 2 du d*u=2g*vdu ' ( dx constant J, ^ /2dudM=2? / vdit+C .:'j^=q*v*+C. whe . qx qx r n Square each side of (9); then C'e We w+=u*+-V, .'. u=K ? _5_ n n 9 n ZC's* -. (see eqs. 6 and 8) ff V-J*-X ^t* * ^^lldatlng the 358 MECHANICS OF ENGINEERING. tained from the moment equation for the free body L viz: &L=P(<$ma}l P(cosa)d ..... (4) in which e= distance of outer fibre from the gravity axis. The equation for safe loading is written out by placing the values of p lt p^> an d d, as derived from equations (2)> (3), and (4) in the expression To solve the resulting equation for P, in case that is the unknown quantity, can only be accomplished by successive assumptions and approximations, since it occurs trans- cendentally. In case a horizontal tension-member of a bridge-truss is subjected to a longitudinal tension P' (due to its position in the truss and the load upon the latter) and at the same time receives a vertical pressure P" at the middle, each half will be bent in the same manner as the cantilever in Fig. 302 ; ^P" corresponding to P sin a, and P'to Pcoso. ~fl qx yx constant factors we now write y = -^x-^me ne . . the equation required . (10) To determine the constants m and n(m= C'-t-2q*; n= C-t-2C'q*) we first find dy-t-dx, i.e. by differentiating (10) -^=~^+Q me n +V n .... (11) z=0 for y=0 .'. (10) gives . . . 0=0+m n n n =0 i.e. m =0 .-. m= . (12) Also for x=l dv n 1 Q l y l ~\ ^-=0.'. (11) gives 0=^+0 wi n -fn n . . (13); .'.with n=m we have m=n= -(14). The equation of the curve, then, substituting (14) in (10) isy=x- - - __ . (15) .-., Substituting for p and q we 4+n nave, (as in 897) /&jf V*"n aisina-ycosa =sin **-*"** FLEXURE. OBLIQUE FORCES. 359 298. Inclined Beam with Hinge at One End. Fig. 304. Let t i. Bequired the equation for safe loading ; also the ru3ximum shear, there being but one load, P, and that in tirj middle. The vertical wall being smooth, its reaction, B, Ai is horizontal, while that of the hinge-pin being un- known, both in amount and direction, is best replaced by its horizontal and vertical components H and F , unknown in amount only. Supposing the flexure slight, we find these external forces in the same manner as in Prob. 1 37, by considering the whole beam free, and obtain =? cota ; H also = cota ; F = P (1) For any section n between and B, we have, from the free body nO, Fig. 305, uniform thrust = j^F = H cos a and from 2* (moms. n ) = 0, (2) =Hx sin a (3) and the shear = J = 5" sin a = % P cos a (4) The max. (/>i+ft) to be found on OB is .-. close above B, where x = ^ I, and is 360 MECHANICS OF ENGINEERING. = p co In examining sections on CB let the free body be Cn' t Fig. 306. Then from 2 (longitud. comps.) = (the thrust=) p^F = F sin a + H cos a (6)' i.e. p l F=P[ain a + }4 cos a cot a] (6) while, from 2'(moms. n ') = 0, sin a (7)' (7) Hence (p } + P*) for sections on CB is greatest when x' is greatest, which is when x' = ]/ 2 I, x' being limited be- tween x' = and x' = y 2 I, and is max. on <7=P cos tan g + Cot a +^ (8) which is evidently greater than the max. (pi see eq. (5). Hence the equation for safe loading is B' = P cos aL+X .... (9) in which R' is the safe normal stress, per square unit, for the material. The shear, J, anywhere on CB t txom2 (transverse comp.) =0 in Fig. 306, is J= PO cos a HO sin a = tf P cos a . . (10) .301 FLEXURE. OBLIQUE FORCES. As showing graphically all the results found, moment, thrust, and shear diagrams are drawn in Fig. 307, and also a diagram whose ordinates represent the variation of (p^p^ along the beam. Each ordi- nate is placed vertically under the gravity axis of the section to which it refers. 299. Numerical Example of the Forego- ing. Fig. 308. Let the beam be of wrought iron, the load P = 1,800 Ibs., hanging from the middle. Cross sec- tion rectangular 2 in. by 1 in., the 2 in. being parallel to the force-plane. Required the max. normal stress in -any outer fibre ; also the max. total shear. This max. stress -intensity will be in FlG . 307. the outer fibres in the section just below B and on the upper side, according to 298, and is given by eq. (8) of that article ; in which, see Fig. 308, we must substitute {inch-pound-second-system) P = 1,800 Ibs.; F 2 sq. in.; I = ^120* + 12 2 = 120.6 in.; e = 1 in., / = | 2 W = fi = 2 A biquad. inches ; cot a = 1 ; cos a = .0996 ; and tan a = 10. ,. ma,. FIG. 808. Fio. 309. FIG. 310. 9000 Ibs. per sq. inch, very nearly, compression. This is w 3G2 MECHANICS OP ENGINEERING. the upper outer fibre close under B. In the lower outer fibre just under B we have a tension = p. 2 PI = 7,200 Ibs. per sq. in. (It is here supposed that the beam is secure against yielding sideways.) 300. Strength of Hooks. An ordinary hook, see Fig. 309, may be treated as follows : The load being = P, if we make a horizontal section at AB, whose gravity axis g is the one, of all sections, furthest removed from the line of action of P, and consider the portion C free, we have the shear = J = zero . . . . . . (1) the uniform pull = p,F=P . . . (2) while the moment of the stress-couple, from 2 (moms. g ) = 0,is &*=Pa (3) For safe loading p -f p 2 must = R', i.e. Tt is here assumed that e = e l} and that the maximum occurs at AB. 301 Crane. As an exercise let the student investigate the strength of a crane, such as is shown in Fig. 310. FLEXURE. LONG COLUMNS. CHAPTER VL FLEXURE OF " LONG COLUMNS." 302. Definitions. By " long column " is meant a straight beam, usually prismatic, which is acted on by two com- pressive forces, one at each extremity, and whose length is so great compared with its diameter that it gives way (or " fails ") by buckling sideways, i.e. by flexure, instead of by crushing or splitting like a short block (see 200). The pillars or columns used in buildings, the compression members of bridge-trusses and roofs, the " bents " of a trestle work, and the piston-rods and connecting-rods 'of steam-engines, are the principal practical examples of long columns. That they should be weaker than short blocks of the same material and cross-section is quite evident, but their theoretical treatment is much less satisfactory than in other cases of flexure, experiment being very largely relied on not only to determine the physical constants which theory introduces in the formulae referring to them, but even to modify the algebraic form of those formulae, thus rendering them to a certain extent empirical. 303. End Conditions. The strength of a column is largely dependent on whether the ends are free to turn, or are fixed and thus incapable of turning. The former condi- tion is attained by rounding the ends, or providing them with hinges or ball-and-socket-joints ; the latter by facing off each end to an accurate plane surface, the bearing on which it rests being plane also, and incapable of turning. 364 MECHANICS OF ENGINEERING. In the former condition the column is spoken of as having round ends ; Fig. 311, (a) ; in the latter as having fixed ends, {or flat bases ; or square ends), Tig. 311, (Z>). FIG. 312. Sometimes a column is fixed at one end while the other end is not only round but incapable of lateral deviation from the tangent line of the other extremity ; this state of end conditions is often spoken of as "Pin and Square," Fig. 311, (c). If the rounding of the ends is produced by a hinge or *' pin joint," Fig. 312, both pins lying in the same plane and having immovable bearings at their extremities, the column is to be considered as round-ended as regards flex- ure in the plane ~| to the pins, but as square-ended as re^ gards flexure in the plane containing the axes of the pins. The " moment of inertia " of the section of a column will be understood to be referred to a gravity axis of the sec- tion which is ~| to the plane of flexure (and this corres- ponds to the " force-plane " spoken of in previous chap- ters), or plane of the axis of column when bent. 303. Euler's Formula. Taking the case of a round-ended column, Fig. 313 (a), assume the middle of the length as an origin, with the axis X tangent to the elastic curve at that point. The flexure being slight, we may use the form -s-dx 1 for the moment of the stress-couple in any FLEXURE. LONG COLUMNS. 365 (a FIG. 314. section n, remembering that with this notation the axis X must be || to the beam, as in the figure (313). Considering the free body nC, Fig. 313 (6), we note that the shear is zero, that the uniform thrust =P, and that ,T(nioms. n )=0 gives (a being the deflection at 0) (1) Multiplying each side by dy we have El - T - i dycPy=Pady Py dy . . (2) dx Since this equation is true for the y, dx, dy, and (Py of any element of arc of the elastic curve, we may suppose it written out for each element from where y=0, and<%=0, up to any element, (where dy=dy and y=y) (see Fig. 314) and then write the sum of the left hand members equal to that of the right hand members, remembering that, since dx is assumed constant, 1-t-dx 2 is a common factor on the left. In other words, integrate between and any point of the curve, n. That is, f12 J e/o e/o ajr / dy-0 The product dy d*y has been written (dy}d(dy\ (for d?y is 3G() MECHANICS OF ENGINEERING. the differential or increment of dy] and is of a form like xdx, or ydy. Performing the integration we have which is in a form applicable to any point of the curve, and contains the variables x and y and their increments dx and dy. In order to separate the variables, solve for dx, and we have dx = IEI VI .,a?=- v /-- (vers. sin x . -. . (6) .e (6) is the equation of the elastic curve DO C, Fig. 313 (a), and contains the deflection a. If P and a are both given, y can be computed for a given x, and vice versa, and thus the curve traced out, but we would naturally suppose a to depend on P, for in eq. (6) when x= y 2 l, y should =a. Mak- ing these substitutions we obtain y 2 l= JM (vers. sin 1.00) ; i.e., */ 2 l= ^ \ (7) Since a has vanished from eq. (7) the value for P ob- tained from this equation, viz.: P O =EI^ . . . ./ .; ... (8) is independent of a, and is ,'. to be regarded as that force (at each end of the round- ended column in Fig. 313) which will hold the column at any small deflection at which it may previously have been set. FLEXURE. LONG COLUMNS. 367 In other words, if the force is less than P no flexure at all will be produced, and hence P is sometimes called the force producing " incipient flexure." [This is roughly ver- ified by exerting a downward pressure with the hand on the upper end of the flexible rod (a T-square-blade for in- stance) placed vertically on the floor of a room ; the pres- sure must reach a definite value before a decided buckling takes place, and then a very slight increase of pressure oc- casions a large increase of deflection.] It is also evident that a force slightly greater than P would very largely increase the deflection, thus gaining for itself so great a lever arm about the middle section as to cause rupture. For this reason eq. (8) may be looked upon as giving the Breaking Load of a column with round ends, and is called Euler's formula. Referring now to Fig. 311, it will be seen that if the three parts into which the flat-ended column is di- , vided by its two points of inflection A and B are considered free, individually, in Fig. 315, the forces acting will be as there shown, viz.: At the points of inflection there is no stress- couple, and no shear, but only a thrust, =P> and hence the portion AB is in the condition of a round-ended column. Also, the tangents to the elastic curves at and C being pre- served vertical by the f rictionless guide-blocks and guides (which are introduced here simply as a theoretical method of preventing the ends from turning, but do not interfere with verti- cal freedom) OA is in the same state of flex- ure as half of AB and under the same forces. Hence the length AB must = one half the total length I of the flat-ended column. In other words, the breaking load of a round- ended column of length =^l, is the same as that of a flat-ended column of length =1. Hence for the I of eq. (8) write l / 2 l and we have as the breaking load of a column with flat-ends and of length =1. PIG. 315. 368 MECHANICS OF ENGINEERING. (9) Similar reasoning, applied to the " pin-and-square " mode of support (in Fig. 311) where the points of inflec- tion are at B, approximately ^ / from (7, and at the extremity itself, calls for the substitution of 2 /^ I for I in eq. (8), and hence the breaking load of a "pin-and-square " column, of length = I, is P,=| EI* ., ' . . (10) Comparing eqs. (8), (9), and (10), and calling the value of PI (flat-ends) unity, we derive the following statement : The breaking loads of a given column are as the numbers f 1 I 9/16 I ^ i according to the I flat-ends \ pin-and-square \ round-ends \ mode of support. These ratios are approximately verified in practice. Euler's Formula [i.e., eq. (8) and those derived from it, (9) and (10)] when considered as giving the breaking load is peculiar in this respect, that it contains no reference to the stress per unit of area necessary to rupture the material of the column, but merely assumes that the load producing " incipient flexure ", i.e., which produces any bending at all, will eventually break the beam because of the greater and greater lever arm thus gained for itself. In the canti- lever of Fig. 241 the bending of the beam does not sensibly affect the lever-arm of the load about the wall-section, but with a column, the lever-arm of the load about the mid- section is almost entirely due to the deflection produced. 304. Example. Euler's formula is only approximately verified by experiment. As an example of its use when considered as giving the force producing " incipient flex- ure " it will now be applied in the case of a steel T-square- blade whose ends are free to turn. Hence we use the round-end formula eq. (8) of 303, with the modulus of elasticity .7=30,000,000 Ibs. per sq. inch. The dimensions FLEXURE. LONG COLUMNS. 369 are as follows : the length I = 30 in., thickness = i of an inch, and width = 2 inches. The moment of inertia, I r about a gravity axis of the section || to the width (the plane of bending being || to the thickness) is (247) /. , with TT = 22 -5- 7, -- 30,000,000 22 2 1 _ 72- go0-2.03 Ibs. Experiment showed that the force, a very small addition to which caused a large increase of deflection or side-buck- ling, was about 2 Ibs. 305. Hodgkinson's Formulae for Columns. The principal practical use of Euler's formula was to furnish a general form of expression for breaking load, to Eaton Hodgkin- son, who experimented in England in 1840 upon columns of iron and timber. According to Euler's formula we have for cylindrical columns, /being =% KT* = ~ nd* (247), CVx for flat-ends .. , P l =1 ET? . f 16 t i.e., proportional to the fourth power of the diameter, and inversely as the square of the length. But Hodgkinson's experiments gave for wrought-iron cylinders J3.55 /T 3 -* PI = (const.) x ; and for cast iron P!=(const.)x j^j Again, for a square column, whose side = b, Euler's for- mula would give whilo Hodgkinson found for square pillars of wood 370 MECHANICS OF ENGINEERING. Hence in the case of wood these experiments indicated the same powers for b and I as Euler's formula, but with a dif- ferent constant factor ; while for cast and wrought iron the powers differ slightly from those of Euler. Hodgkinson's formulae are as follows, and evidently not homogeneous ; the prescribed units should .-. be care- fully followed, d denotes the diameter of the cylindrical columns, b the side of square columns, 1= length. ( For solid cylindrical cast iron columns, flat-ends ; -< Breaking load in tons ) A A n a , 7 . . , N i55 n . ., \ 7 ( of 2,240 Ibs. each | =44.16 X (^ in inches) + (Zmft.) . ( For solid cylindrical wrought iron columns, flat-ends ; -< Breaking load in tons ) -, 04 ,-, . -, \ 3 - M n . ,, N 2 ( of 2,240 Ibs. each | -134 X (dm inches) + (I in ft.) {For solid square columns of dry oak, flat-ends ; ["For solid square columns of dry fir, flat-ends ; Hodgkinson found that when the mode of support was " pin-and-square," the breaking load was about ^ as great ; and when the ends were rounded, about ^ as great as with flat ends. These ratios differ somewhat from the theoretical ones mentioned in 303, just after eq. (10.) Experiment shows that, strictly speaking, pin ends are not equivalent to round ends, but furnish additional strength ; for the friction of the pins in their bearings hinders the turning of the ends somewhat. As the lengths become smaller the value of the breaking load in Hodg- kinson's formulae increases rapidly, until it becomes larger than would be obtained by using the formula for the crushing resistance of a short block (201) viz., FC, i.e., the sectional area X the crushing resistance per unit of area. In such a case the pillar is called a short column, or " short block," and the value FC is to be taken as the breaking FLEXURE. LONG COLUMNS. 371 load. This distinction is necessary in using Hodgkinson's formulae ; i.e., the breaking load is the smaller of the two values, FC and that obtained by Hodgkinson's rule. In present practice Hodgkinson's formulae are not often used except for hollow cylindrical iron columns, for which with d 2 and d { as the external and internal diameters, we have for flat-ends Breaking load in tons ) _ _, (c? 2 in in.) 355 (d, in in.) 355 of 2,240 Ibs. each \~ ~ ( i fr feet) - in which the const. = 44.16 for cast iron, and 134 for wrought, while n = 1.7 for cast-iron and = 2 for wrought. 306. Examples of Hodgkinson's Formulae. Example 1. Ee- quired the breaking weight of a wrought-iron pipe used as a long column, having a length of 12 feet, an internal diameter of 3. in., and an external diameter of 3^ inches, the ends having well fitted flat bases. If we had regard simply to the sectional area of metal, which is F = 1.22 sq. inches, and treated the column as a short block (or short column) we should have for its com- pressive load at the elastic limit (see table 203) P"=FC" = 1.22 x 24,000=29,280 Ibs. and the safe load P 1 may be taken at 16,000 Ibs. But by the last formula of the preceding article we have Breaking load in ) _ 1% n x (3.25) 355 - 3 s - 55 _ lg 07 tons of 2,240 Ibs. each f iff i.e.= 15.07 x 2240= 33,768 Ibs. Detail, [log. 3.25] x 3.55= 0.511883x3.55= 1.817184 ; [log. 3.00] x 3.55=0.477,121x3.55=1.693,779 ; and the corresponding numbers are 65.6 and 49.4 ; their difference = 16.2, hence Br. load in long tons = 134 *| 6 ' 2 = 15.072 long tons. Irxrx. =33,768 Ibs. 372 MECHANICS OF ENGINEERING. With a " factor of safety " (see 205) of four, we have, as the safe load, p' = 8,442 Ibs. This being less than the 16000 Ibs. obtained from the " short block " formula,should be adopted. If the ends were rounded the safe load would be one- third of this i.e., would be 2,814 Ibs ; while with pin-and- square end-conditions, we should use one-half, or 4,221 Ibs. EXAMPLE 2. Kequired the necessary diameter to be given a solid cylindrical cast-iron pillar with flat ends, that its safe load may be 13,440 Ibs. taking 6 as a factor of safety. Let d = the unknown diameter. Using the proper formula in 305, and hence expressing the breaking load, which is to be six times the given safe load, in long tons we have (the length of column being 16 ft.) 13440 x 6^ 44.16 (d in inches) 3 - 55 (1) (2) or log.rf= 3 -ij[log. 36+1.7xlog. 16-log. 44.16] . . (3) .-. log.d= 3 -y 1.958278] =0.551627 .-. d = 3.56 ins. This result is for flat ends. If the ends were rounded, we should obtain d =4.85 inches. 307. Rankine's Formula for Columns. The formula of this name (some times called Gordon's, in some of its forms) has a somewhat more rational basis than Euler's, in that it in- troduces the maximum normal stress in the outer fibre and is applicable to a column or block of any length, but still contains assumptions not strictly borne out in theory, thus introducing some co-efficients requiring experimental de- termination. It may be developed as follows : Since in the flat-ended column in Fig. 315 the middle portion AB, between the inflection points A and B, is acted on at each end by a thrust = P, not accompanied by any shear or stress-couple, it will be simpler to treat tha* FLEXUKE. LONG COLUMNS. 373 portion alone Fig. 316, (a), since the thrust and stress- couple induced in the section at R, the middle of AB, will be equal to those at the flat ends, and G, in Fig. 315. Let a denote the de- flection of R from the straight line AB. Now consider the portion AR as a free body in Fig. 316, (&), putting in the elastic forces of the section at R, which may be clas- sified into a uniform thrust = , and a stress couple of moment FIG. 316. (see 294). (The shear is evidently zero, from S (hor coinps.) = 0). Here p l denotes the uniform pres- sure (per unit of area), due to the uniform thrust, and p 3 the pressure or tension (per unit of area), in the elastic forces constituting the stress-couple, on the outermost element of area, at a distance e from the gravity axis (~| to plane of flexure) of the section. F is the total area of the section. / is the moment of inertia about the said gravity axis, g I (vert, comps.) = gives P = p,F . . (1) J(moms. g ) = Ogives Pa =&L .... (2) For any section, n, between A and R, we would evidently have the same p t as at R, but a smaller p 2 , since Py < Pa while e, /, and F, do not change, the column being pris- matic. Hence the max. (Pi+p jfc2= b-h 2 Fig. 318 (6). #=-!* A Fig. 318 (c). ^=^-3^ ribs. Cross of equal arms. I-Beam as a pillar. Let area of web =. " " " both flanges =A. Channel Iron. Let area of web =B ; of flanges =A (both), h extends from edge of flange to middle of web. Fig. 318 (d). tf=h 2 L-_ FIG. 319. 310. Built Columns. The " compression members " of wrought-iron bridge trusses are generally composed of several pieces riveted together, the most common forms being the Phoenix column (ring-shaped, in segments,) and combinations of channels, plates, and lattice, some of which are shown in Figs. 319 and 320. Experiments on full size columns of these kinds were made by the U. S. Testing Board at the Watertown Arse- nal about 1880. The Phoenix columns ranged from 8 in. to 28 feet in length, and from 1 to 42 in the value of the ratio of length to diameter. The breaking loads were found to be some- what in excess of the values computed from Rankine's formula ; from 10 to 40 per cent, excess. In the pocket- book issued by the Phoenix company they give the follow- ing formula for their columns, (wrought-iron.) FLEXUKE. LONG COLUMNS. 379 Breaking load in Ibs. ) _ 50,000 F for flat-ended columns ) 1 _j_ j 8 3,OOOA 2 where F = area in sq. in., I = length, and A = external diameter, both in the same unit. Many different formulae have been proposed by different engineers to satisfy these and other recent experiments on columns, but all are of the general form of Bankine's. For instance Mr. Bouscaren, of the Keystone Bridge Co., claims that the strength of Phoenix columns is best given by the formula Breaking load in ) _ 38,000 F fe) ^ j ~ Ibs. for flat-ends, 1 + 100,000/fc* (F must be in square inches.) The moments of inertia, /, and thence the value of k* = / -f- F, for such sections as those given in Figs. 319 and 320 may be found by the rules of 85-93, (see also 258.) 311. Moment of Inertia of Built Column. Example. It is pro- posed to- form a column by joining two I-beams by lattice- work, Fig. 321, (a). (While the lattice-work is relied upon to cause the beams to act together as one piece, it is not regarded in estimating the area F, or the moment of iner- tia, of the cross section). It is also required to find the proper distance apart = x, Fig. 321, at which these beams must be placed, from centre to centre of webs, that the liability to flexure shall be equal in all axial planes, i.e. that the 1 of the compound section shall be the same about all gravity axes. This condition will be ful- filled if 7 Y can be made = 1^* (89), being the centre of gravity of the compound section, and X perpendicular to the parallel webs of the two equal I-beams. Let F' = the sectional area of one of the I-beams, P v (see Fig. 321(a) its moment of inertia about its web-axis, J x ' that about an axis "| to web. (These quantities can be * That is, with flat ends or ball ends ; but with pin ends, Fig. 312, if the pin is || to X, put 4/ Y = / x ; if II to T, put 47 X = /Y . 380 MECHANICS OF ENGINEERING. found in the hand-book of the iron company, for each size of rolled beam). Then the total J x = 2/' x ; and total J Y = 2[> v + (see 88 eq. 4.) If these are to be equal, we write them so and solve for x, obtaining x = I'- (i) 312. Numerically; suppose each girder to be a 10^ inch light I-beam, 105 Ibs. per yard, of the N. J. Steel and Iron Co., in whose hand-book we find that for this beam J' x = 185.6 biquad. inches, and I' v 9.43 biquad. inches, while F' = 10.44 sq. inches. "With these values in eq. (1) we have (185.6-9.43) 10.44 _ -v/67.5 = 8.21 inches. The square of the radius of gyration will be & 2 =2/' x -^2 J F'= 371.2 -5-20.88 =17.7 sq. in. . (2) and is the same for any gravity axis (see 89). As an additional example, suppose the two I-beams united by plates instead of lattice. Let the thickness of the plate = t, Fig. 321, (b). Neglect the rivet-holes. The distance a is known from the hand-book. The student may derive a formula for x, imposing the condition that (total 7 X )= 7 T - FLEXURE. LONG COLUMNS. 381 313. Trussed Girders. When a horizontal beam is trussed FIG. 323. in the manner indicated in Fig. 322, with a single post or strut under the middle and two tie-rods, it is subjected to a longitudinal compression due to the tension of the tie- rods, and hence to a certain extent resists as a column, the plane of whose flexure is vertical, (since we shall here sup- pose the beam supported later ally. )Taking the case of uni- form loading, (total load = JP)and supposing the tie-rods screwed up (by sleeve nuts) until the top of the post is on a level with the piers, we know that the pressure between the post and the beam is P' '= % W (see 273). Hence by the parallelogram of forces (see Fig. 322) the tension in each tie-rod is Q- _6 W 16 * cos a (1) 2 cos a At each pier the horizontal component of Q is P= Q sin a= JFtan a 16 Hence we are to consider the half -beam BO as a "pin-and- square " column under a compressive force P= 5 /is W tan a } as well as a portion of a continuous girder over three equidistant supports at the same level and bearing a uni- form load W. In the outer fibre of the dangerous section, 0, (see also 273 and Fig. 278) the compression per sq. inch due to both these straining actions must not exceed a safe limit, E\ (see 251). In eq. (6) 307, where P 2 is tne breaking force for a pin-and-square column, the great- 382 MECHANICS OF ENGINEERING. est stress in any outer fibre = C ( = the Modulus of Crush- ing) per unit of area. If then we write p wl . instead of G in that equation, and 6 / 16 JFtan a instead of P 2 we have j max. stress due ) _ _ 5 JFtan an , 16 Q Z 2 ( to column action j "" P CO] ~ Jg ^ " ~' ** " while from eq. (3), p. 326, we have (remembering that our present W represents double the W of 273). ( max. stress due ) = = l_^e = ^ { to girder action j ^" [ 16 7 16 By writing p col .+_p Ki = R'= a safe value of compression per unit-area, we have the equation for safe loading tan !+. ^+-16JW . . (2) Here I = the half -span OS, Fig. 322, e = the distance of outer fibre from the horizontal gravity axis of the cross section, k* the radius of gyration of the section referred to the same axis, while F = area of section. fj should be taken from the end of 307. EXAMPLE. If the span is 30 ft. 360 in., the girder a 15 inch heavy I-beam of wrought iron, 200 Ibs. to the yard, in which e = }4 of 15 = 7 l / 2 inches, ,F=20 sq. in., and F = 35.3 sq. inches (taken from the Trenton Co.'s hand-book), required the safe load W> the strut being 5 ft. long. From 307, ft = 1 : 36,000; tan = 15-J-5 = 3.00. Hence, using the units pound and inch throughout, and putting R' = 12,000 Ibs. per sq. in. = max. allowable compression stress, we have from eq. (2) 16x20x12,000 " 16 1 JClflO "" O)h .ISO .3 J" 1 " 3 9 36,000 * 35.3 "" 35.3 i. e., 69,111 Ibs. besides the weight of the beam. If the middle support had been a solid pier, the safe load would have been 48 tons ; while if there had been no middle support of any kind, the beam would bear safely FLEXURE. LONG COLUMNS. 383 only 11.5 tons, the strut)]. [Let the student design the tie-rods (and 314. Buckling of Web-Plates in Built Girders. In 257 men- tion was made of the fact that very high web plates in built beams, such as /beams and box-girders, might need to be stiffened by riveting T-irons on the sides of the web. (The girders here spoken of are horizontal ones, such as might be used for carrying a railroad over a short span of 20 to 30 feet. An approximate method of determining whether such stiffening is needed to prevent lateral buckling of the web, may be based upon Rankine's formula for a long column and will now be given. In Fig. 323 we have, free, a portion of a bent I-beam, between two vertical sections at a distance apart= hi -= the height of the web. In such a beam under forces L to its axis it has been proved (256) that we may consider the web to sustain all the shear, J, at any section, and the flanges to take all the tension and compression, which form the "stress-couple" of the section. These couples and the two shears are shown in Fig. 323, for the two exposed sections. There is supposed to be no load on this portion of the beam, hence the shears at the two ends are Ntl 1 J I, 1 j h- > + equal. Now the shear acting between each flange and the horizontal edge of the web is equal in intensity per square inch to that in the vertical edge of the web ; hence if the web alone, of Fig. 323, is shown as a free body in Fig. 324, we must insert two horizontal forces = 7, in opposite 384 MECHANICS OF ENGINEERING. directions, on its upper and lower edges. Each of these = J since we have taken a horizontal length h^ = height of web. In this figure, 324, we notice that the effect of the acting forces is to lengthen the diagonal BD and shorten the diagonal AC, both of those diagonals making an angle of 45 with the horizontal. Let us now consider this buckling tendency along AC t by treating as free the strip A C, of small width = b t . This is shown in Fig. 325. The only forces acting in the direc- tion of its length AC are the components along AC of the four forces J' at the extremities. AVe may therefore treat the strip as a long column of a length I = hi V% of a sec- tional area F = bb l} (where b is the thickness of the web plate), with a value of Ar 2 = y ]2 6 2 (see 309), and with fixed (or flat) ends. Now the sum of the longitudinal components of the two J'.'s at A is Q = 2 J' y 2 \/2 = J' V2; but J' itself = ^. b */ 2 h V2, since the small rectangle on which J' acts has an area = b }4 \ V2, and the shearing stress on it has an intensity of (J -f- bh^ per unit of area. Hence the longitudinal force at each end of this long column is (1) According to eq. (4) and the table in 307, the safe load (factor of safety = 4) for a wought-iron column of this form, with flat ends, would be (pound and inch) P_ #Mi36,000 _ 9,00066, , 9 . 1 - 1 - 9JT -- \pj 11 + *fli 11 ' If, then, in any particular locality of the girder (of wrought-iron) we find that Q is > P lt i.e. if L is > -i' 0005 .. (pound and inch) . , (3) FLEXURE. LONG COLUMNS. 385 then vertical stiffeners will be required laterally. When these are required, they are generally placed at intervals equal to A 1} (the depth of web), along that part of the girder where Q is > P lt EXAMPLE Fig. 326. Will stiffening pieces be required in a built girder of 20 feet span, bearing a uniform load of 40 tons, and having a web 24 in. deep and ^ in. thick ? From 242 we know that the _ w -40 TONS greatest shear, ./max., is close to either pier, and hence we investi- | [*- 10-- gate that part of the girder first. J max. = y* W = 20 tons =40,000 Ibs. .-. (inch and lb.), see (3), Pro 826. , ...... (4) while, see (3), (inch and pound), 1 . ^1,500 ' which is less than 1666.66. Hence stiffening pieces will be needed near the extremi- ties of the girder. Also, since the shear for this case of loading diminishes uniformly toward zero at the middle they will be needed from each end up to a distance of - of 10 ft. from the middle. 586 MECHANICS OF ENGINEERING. CHAPTER VII. LINEAR ARCHES (OF BLOCKWOKK). 315. A Blockwork Arch, is a structure, spanning an opening or gap, depending, for stability, upon the resistance to compresssion of its blocks, or voussoirs, the material of which, such as stone or brick, is not suitable for sustain- ing a tensile strain. Above the voussoirs is usually placed a load of some character, (e.q. a roadway,) whose pressure upon the voussoirs will be considered as vertical, only. This condition is not fully realized in practice, unless the load is of cut stone, with vertical and horizontal joints resting upon voussoirs of corresponding shape (see Fig. 327), but sufficiently so to warrant its assumption in theory. Symmetry of form about a vertical axis will also be assumed in the following treatment. 316. Linear Arches. For purposes of theoretical discussion the voussoirs of Fig. 327 may be considered to become PIQ. 327. infinitely small and infinite in number, thus forming a " linear arch," while retaining the same shapes, their depth ~| to the face being assumed constant that it may not appear in the formulae. The joints between them are "1 to the curve of the arch, i.e., adjacent voussoirs can exert pressure on each other only in the direction of the tangent -line to that curve. LINEAlt ARCHES. 387 317. Inverted Catenary, or Linear Arch Sustaining its Own Weight Alone. Suppose the infinitely small voussoirs to have weight, uniformly distributed along the curve, weigh- ing q Ibs. per running linear unit. The equilibrium of such a structure, Fig. 328, is of course unstable but theo- retically possible. Required the form of the curve when equilibrium exists. The conditions of equilibrium are, obviously : 1st. The thrust or mutual pressure T between any two adjacent voussoirs at any point, A, of the curve must be tangent to the curve ; and 2ndly, considering a portion BA as a free body, the resultant of H the pres- Fio. 328. Fio. 329. Fig. 330. sure at B the crown, and T at A, must balance E the re- sultant of the II vertical forces (i.e.,weights of the elementary voussoirs) acting between B and A. But the conditions of equilibrium of a flexible, inexten- sible and uniformly loaded cord or chain are the very same (weights uniform along the curve) the forces being reversed in. direction. Fig. 329. Instead of compression we have tension, while the II vertical forces act toward in- stead of away from, the axis X. Hence the curve of equi- librium of Fig. 328 is an inverted catenary (see 48) whose equation is (1) See Fig. 330. e = 2.71828 the Naperian Base. The "par- ameter " c may be determined by putting x = a, the half span, and y= Y, the rise, then solving for c by successive 388 MECHANICS OF ENGINEERING. approximations. The " horizontal thrust" or H Q , is = gc, while if s = length of arch OA, along the curve, the thrust T at any point A is From the foregoing it may be inferred that a series of vous- soirs of finite dimensions, arranged so as to contain the catenary curve, with joints ~| to that curve and of equal weights for equal lengths of arc will be in equilibrium, and moreover in stable equilibrium on account of friction, and the finite FIG. 331. width of the joints ; see Fig. 331. 318. Linear Arches under Given Loading. The linear arches to be considered further will be treated as without weight themselves but as bearing vertically pressing loads (each voussoir its own). Problem. Given the form of the linear arch, itself, it is required to find the law of vertical depth of loading under which the given linear arch will be in equilibrium. Fig. 332, given the curve ABC, i.e., the linear arch itself, re- quired the form of the curve MON, or upper limit of load- ing, such that the linear arch ABC shall be in equilibrium under the loads lying between the two curves. The load- ing is supposed homogeneous and of constant depth ~| to paper ; so that the ordinates z between the two curves are proportional to the load per horizontal linear unit. Assume a height of load z at the crown, at pleasure ; then required the z of any point m as a function of z and the curve ABC. LINEAR ARCHES. <389 Practical Solution. Since a linear arch under vertical pressures is nothing more than the inversion of the curve assumed by a cord loaded in the saine way, this problem might be solved mechanically by experimenting with a light cord, Fig. 333, to which are hung other tieavy cords, or bars of uniform weight per unit length, and at equal horizontal distances apart when in equilibrium. By varying the lengths of the bars, and their points of attachment, we may finally find the curve sought, MON. (See also 343.) Analytical Solution. Consider the structure in Fig. 334. A number of rods of finite length, in the same plane, are in equilibrium, bearing the weights P, P lt etc., at the con- FIG. 334. Fio. 335. necting joints, each piece exerting a thrust T against the adjacent joint. The joint A, (the " pin " of the hinge), im- agined separated from the contiguous rods and hence free, is held in equilibrium by the vertical force P (a load) and the two thrusts T and T', making angles = and 6' with the vertical ; Fig. 335 shows the joint A free. From ^hor- izontal comps.)=0, we have. ^sin dT' sin 0'. That is, the horizontal component of the thrust in any rod is the same for all ; call it H . .'. sin (1) 390 MECHANICS OF ENGINEERING. Now draw a line As T to T' and write S ( compons. I to As)=0 ; whence P sin 6'= T sin , and [see (1)] (2) Let the rods of Fig. 334 become infinitely small and infi- nite in number and the load continuous. The length of each rod becomes =ds an element of the linear arch, ft is the angle between two consecutive ds's, d is the angle be- tween the tangent line and the vertical, while P becomes the load resting on a single dx, or horizontal distance be- tween the middles of the two cfe's. That is, Fig. 336, if 7-= weight of a cubic unit of the loading, P=fzdx. (The lamina of arch and load considered is unity, "| to paper, in thickness.) H =a constant = thrust at crown ; 6=6', and sin fi=ds+f>, (since the' angle between two consecutive tan- gents is that between two con- secutive radii of curvature). Hence eq. (2) becomes but dx=ds sin Fio. 836. (3) Call the radius of curvature at the crown p Qi and since there Z=ZQ and 0=90, (3) gives rVo ^foJ hence (3) may be written z = p sin 3 (4) This is the law of vertical depth of loading required. For a point of the linear arch where the tangent line is verti- cal, sin tf =0 and z would = oo ; i.e., the load would be in- LINEAR ARCHES. 391 finitely high. Hence, in practice, a full semi-circle, for in- stance, could not be used as a linear arch. 319. Circular Arc as Linear Arch. As an example of the preceding problem let us ap- ply eq. (4) to a circular arc, Fig. 337, as a linear arch. Since for a circle /> is con- stant = r, eq. (4) reduces to (5) 837. Hence the depth of loading must vary inversely as the cube of the sine of the angle d made by the tangent line (of the linear arch) with the ver- tical. To find the depth z by construction. Having z given, G being the centre of the arch, prolong Ca and mate ab = z ; at b draw a ~| to Cb, intersecting the vertical through a at some point d ; draw the horizontal dc to meet Ca at some point c. Again, draw ce ~| to Cc, meeting ad in e ; then ae = z required ; a being any point of the linear arch. For, from the similar right triangles involved, we have z a =ab=ad, sin 6=ac sin d. sin 6=ae sin 6 sin d sin Q i.e., ae=z. Q.RD. [see (5.)] 320. Parabola as Linear Arch. To apply eq. 4 318 to a parabola (axis vertical) as linear arch, we must find values of p and p the radii of curvature at any point and the crown respectively. That is, in the general formula, we must substitute the forms for the first and second dif- ferential co-efficients, derived from the equation of the 392 MECHANICS OF ENGINEERING. PIG. 338. FIG. 339. curve (parabola) in Fig. 338, i.e. from x 2 = 2 py ; whence we obtain -,or cot ft, and ^L _ o p dx* p Hence p = p cosec.0, i.e. p= I+p * f (6) At the vertex 6 = 90 .. p = p. Hence by substituting for p and p in eq. (4) of 318 we obtain z =z = constant [Fig. 339] (7) for a parabolic linear arch. Therefore the depth of homo- geneous loading must be the same at all points as at the crown ; i.e., the load is uniformly distributed with respect to the horizontal. This result might have been antici- pated from the fact that a cord assumes the parabolic form when its load (as approximately true for suspension bridges) is uniformly distributed horizontally. See 46 in Statics and Dynamics. 321. Linear Arch for a Given Upper Contour of Loading, the arch itself being the unknown lower contour. Given the upper curve or limit of load and the depth z at crown, re- quired the form of linear arch which will be in equili- brium under the homogenous load between itself and that upper curve. In Fig. 340 let MON be the given upper contour of load, z is given or assumed,^' and z" are the respective ordinates of the two curves BA C and MON. Required the eqation of BAG. LINEAR ARCHES. 393 As before, the loading is homogenous, so that the weights of any portions of it are proportional to the corresponding areas between the curves. (Unity thick- ness ~| to paper.) Now, Fig. 341, regard two consecutive ds's of the linear arch as two links or consecutive blocks bearing at their junction m the load dP = 7- (z r + z") dx in which f denotes the heaviness of weight of a cubic unit of the loading. If T and T are the thrusts exerted on these two blocks by their neighbors (here supposed removed) we have the three forces dP, T and T', forming a system in equilibrium. Hence from 2X =0. Tcos

' T sin

)=dP (4) But tan 9' =$^ and tan w' = dz '+^ z> , (dx constant) dx dx while dP = f (z r -f- ") dx. Hence, putting for convenience H = ya 2 , (where a = side of an imaginary square of the 394 MECHANICS OF ENGINEERING. loading, whose thickness = unity and whose weight = H) we have. '=>'+*"> ....... < 5 > as a relation holding good for any point of the linear arch which is to be in equilibrium under the load included between itself and the given curve whose ordinates are %", Fig. 340. 322. Example of Preceding. Upper Contour a Straight Line. Fig. 342. Let the upper contour be a right line and hor- izontal ; then the z" of eq. 5 becomes zero at all points of ON. Hence drop the accent of z' in eq. (5) and we have y? a 2 Multiplying which by dz we obtain This being true of the z, dz, d?z and dx of each element of the curve O'B whose equation is desired, conceive it writ- ten out for each element between 0' and any point m, and put the sum of the left-hand members of these equations = to that of the right-hand members, remembering that a 2 and dx 2 are the same for each element. This gives _ /*_! / x 1 I a' c/ close the triangle in Fig. 345 must be = and || to P 3 , since that tri- angle is nothing more than the left-hand half-parallelogram of Fm.344. FIG. 345. Fig. 344. Also, in 345, *to dose the triangle properly the directions of the arrows must be continuous Point to Butt, round the periphery. Fig. 345 is called a force polygor ; of three sides only in this case. By means of it, given any two of the three forces which hold the point in equilibrium, the third can be found, being equal and || to the side necessary to " close " the force polygon. Similarly, if a number of forces in a plane hold a mate- rial point in equilibrium, Fig. 346, their force polygon, FIG. 346. FIG. 347. Fig. 347, must close, whatever be the order in which its sides are drawn. For, if we combine P t and P 2 into a re- sultant Oa, Fig. 346, then this resultant with P 3 to form a resultant Ob, and so on ; we find the resultant of P lt P 2 , P 3 , and P 4 to be Oc, and if a fifth force is to produce equilib- rium it must be equal and opposite to Oc, and would close the polygon Odabc.O, in which the sides are equal and par- GRAPHICAL STATICS. 399 allel respectively to the forces mentioned. To utilize this fact we can dispense with all parts of the parallelograms in Fig. 346 except the sides mentioned, and then proceed as follows in Fig. 347 : If P 5 is the unknown force which is to balance the other four (i.e, is their anti-resultant], we draw the sides of the force polygon from A round to B, making each line paral- lel and equal to the proper force and pointing the same way ; then the line BA represents the required P 5 in amount and direction, since the arrow BA must follow the continuity of the others (point to butt). If the arrow BA were pointed at the extremity B, then it gives, obviously, the amount and direction of the result- ant of the four forces P l . . . P 4 . The foregoing shows that if a system of Concurrent Forces in a Plane is in equi- librium, its force polygon must close. 326. Non-Concurrent Forces in a Plane. Given a system of non-concurrent forces in a plane, acting on a rigid body, required graphic means of finding their resultant and anti- resultant ; also of expressing conditions of equilibrium. The resultant must be found in amount and direction ; and also in position (i.e., its line of action must be determined). E. g., Fig. 348 shows a curved rigid beam fixed in a vise at T, and also under the action of forces Pj P 2 P 3 and P 4 (besides the action of the vise); required the resultant of By the ordinary parallelogram of forces we com- bine P l and P, at a, the intersection of their lines of FIG. 348. action, into a re- sultant .Z? a ; then 7? a with P 3 at b, to form H b ; and finally 7? b with P 4 at c to form R c which is .. the resultant required, i.e., of P l . . . . P t ; and c . . . F is its line of action. MECHANICS OF ENGINEERING. The separate force triangles (half-parallelograms) by which the successive partial resultants R & , etc., were found, are again drawn in Fig. 349. Now since E c , acting in the line c..F, Fig. 348, is the resultant of PI . . PI, it is plain p < that a force R c ' equal to R c and act- ing along c . . Jf^but FIG - 349 - in the opposite di- rection, would balance the system P l . . . P 4 , (is their anti- resultant). That is, the forces P t P 2 P 3 P 4 and M e ' would form a system in equilibrium. The force fi c ' then, repre- sents the action of the vise T upon the beam. Hence re- place the vise by the force R c ' acting in the line . . . F . . . c to do which requires us to imagine a rigid prolongation of that end of the beam, to intersect F . . . c. This is shown in Fig. 350 where the whole beam is free, in equilibrium, under the forces shown, and in precisely the same state of stress, part for part, as in Fig. 348. Also, by combining in one force diagram, in Fig. 351, all the force triangles of Fig. 349 (by making their common sides coincide, and putting R c f instead of 7? c , and dotting all forces other than those of Fig. 350), we have a figure to be interpreted in connection with Fig. 350. o' P, A|P 2 SPACE DIAGRAM FIG. 350. FORCE DIAGRAMS! FIG. 351. Here we note, first, that in the figure called a force-dia- gram, P! P 2 P 3 JP 4 and R,.' form a closed polygon and that GRAPHICAL STATICS. 40 1 their arrows follow a continuous order, point to butt, around the perimeter ; which proves that one condition of equilibrium of a system of non-concurrent forces in a plane is that its force polygon must close. Secondly, note that ab is || to Oa', and be to Ob' ; hence if the force-diagram has been drawn (including the rays, dotted) in order to deter- mine the amount and direction of R c ', or any other one force, we may then find its line of action in the space-diagram, as follows: (N. B. By space diagram is meant the figure show- ing to a true scale the form of the rigid body and the lines of action of the forces concerned). Through a, the intersec- tion of P l and P 2 , draw a line || to Oa' to cut P 3 in some point b ; then through b a line || to Ob' to cut P 4 at some point c; cF drawn il to Oc' is the required line of action of 7? c ', the anti- resultant of P,, P,, P 3 , and P 4 . obc is called an equilibrium polygon; this one having but two segments, ab and be (sometimes the lines of action of P l and R c ' may conveniently be considered as segments.) The segments of the equilibrium polygon are parallel to the respect- ive rays of the force diagram. Hence for the equilibrium of a system of non-concurrent forces in a plane not only must its force polygon close, but also the first and last segments of the corre- sponding equilibrium polygon must coincide with the resultants of the first two forces, and of the last two forces, respectively, of the system. E.g., ab coin- cides with the line of action of the resultant of P l and P 2 ; be with that of P 4 and R' c . Evidently the equil. polygon will be different with each different order of forces in the force polygon or different choice of a pole, 0. But if the order of forces be taken as above, as they occur along the beam, or structure, and the pole taken at the " butt " of the first force in the force polygon, there will be only one ; (and this one will be called the special equilibrium polygon in the chapter on arch-ribs, and the " true linear arch " in dealing with the stone arch.) After the rays (dotted in Fig. 351) have been added, by joining the pole to each 402 MECHANICS OF ENGINEERING. vertex with which it is not already connected, the final figure may be called the force diagram. It may sometimes be convenient to give the name of rays to the two forces of the force polygon which meet at the pole, in which case the first and last segments of the corresponding equil. polygon will coincide with the lines of action of those forces in the space-diagram (as we may call the representation of the body or structure on which the forces act). This " space diagram " shows the real field of action of the forces, while the force diagram, which may be placed in any convenient position on the paper, shows the magnitudes and directions of the forces acting in the former diagram, its lines being interpreted on a scale of so many Ihs. or tons to the inch of paper ; in the space-diagram we deal with a scale of so many/ee to the inch of paper. We have found, then, that if any vertex or corner of the closed force polygon be taken as a pole, and rays drawn from it to all the other corners of the polygon, and a cor- responding equil. polygon drawn in the space diagram, the first and last segments of the latter polygon must co-incide with the first and last forces according to the order adopted (or with the resultants of the first two and last two, if more convenient to classify them thus). It remains to utilize this principle. 327. To Find the Resultant of Several Forces in a Plane. This might be done as in 326, but since frequently a given set of forces are parallel, or nearly so, a special method will now be given, of great convenience in such cases. Fig. 352. Let P! P 2 and forces whose resultant is re- qed. Let us first find their an^i - resultant, or force which will balance GRAPHICAL STATICS. 403 them. This anti-resultant may be conceived as decom- posed into two components P and P r one of which, say P, is arbitrary in amount and position. Assuming P, then, at convenience, in the space diagram, it is required to find P'. The five forces must form a balanced system ; hence if beginning at O lt Fig. 353, we lay off a line O^A = P by scale, then A\ - and II to P,, and so on (point to butt), the line BOi necessary to close the force polygon is = P' re- quired. Now form the corresponding equil. polygon in the space diagram in the usual way, viz.: through a the intersection of P and PJ draw db || to the ray Oi . . . 1 (which connects the pole O l with the point of the last force mentioned). From b, where ab intersects the line of P^ draw be, II to the ray { . . 2, till it intersects the line of P 3 . A line me drawn through c and || to the P' of the force diagram is the line of action of P'. Now the resultant of P and P' is the anti-resultant of P 1} P 2 and P 3 ; .'. d, the intersection of the lines of P and P', is a point in the line of action of the anti-resultant re- quired, while its direction and magnitude are given by the line BA in the force diagram ; for B A forms a closed poly- gon both with P! P 2 P 3 , and with PP'. Hence a line through d \\ to BA, viz., de, is the line of action of the anti- resultant (and hence of the resultant) of P,, P 2 , P 3 . Since, in this construction, P is arbitrary, we may first choose Oi, arbitrarily, in a convenient position, i.e., in such a position that by inspection the segments of the result- ing equil. polygon shall give fair intersections and not pass off the paper. If the given forces are parallel the device of introducing the oblique P and P' is quite neces- 328. The result of this construction may be stated as follows, (regarding Oa and cm as segments of the equil. polygon as well as ab and be): If any two segments of an equil polygon be prolonged, their intersection is a point in the line of action of the resultant of those forces acting at 404 MECHANICS OF ENGINEERING. the vertices intervening between the given segments. Here, the resultant of PI P 2 P 3 acts through d. 329. Vertical Reaction of Piers, etc. Fig. 354. Given the vertical forces or loads PI P 2 and P 3 acting on a rigid body (beam, or truss) which is supported by two piers having smooth horizontal surfaces (so that the reactions must be vertical), required the reactions V and F n of the piers. For an instant suppose V and V n known ; they are in FIG. 354. equil. with P l P 2 and P 3 . The introduction of the equal and opposite forces P and P' in the same line will not dis- turb the equilibrium. Taking the seven forces in the order P V P l P 2 P 3 V n and P', a force polygon formed with them will close (see (b) in Fig. where the forces which really lie on the same line are slightly separated). With 0, the butt of P, as a pole, draw the rays of the force dia- gram OA, OB, etc. The corresponding equil." polygon begins at a, the intersection of P and V Q in (a) (the space diagram), and ends at n the intersection of P' and F n . Join an. Now since P and P' act in the same line, an must be that line and must be || to P and P' of the force diagram. Since the amount and direction of P and P' are arbitrary, the position of the pole is arbitrary, while PI, P 2 , and P 3 are the only forces known in advance in the force diagram. Hence F and'F" n may be determined as follows: Lay off the given loads P 15 P 2 , etc., in the order of their occur- rence in the space diagram, to form a " load-line " AD GRAPHICAL STATICS 405 (see (6.) Fig. 354) as a beginning for a force-diagram ; take any convenient pole 0, draw the rays OA, OB, OC and OD. Then beginning at any convenient point a in the vertical line containing the unknown V , draw ab || to OA, be || to OB, and so on, until the last segment (dn in this case) cuts the vertical containing the unknown F n in some point n. Join an (this is sometimes called a closing line) and draw a || to it through 0, in the force-diagram. This last line will cut the " load-line " in some point n', and divide it in two parts n' A and Dn', which are respectively V and V u required. Corollary. Evidently, for a given system of loads, in given vertical lines of action, and for two given piers, or abut- ments, having smooth horizontal surfaces, the location of the point n' on the load line is independent of the choice of a pole. Of course, in treating the stresses and deflection of the rigid body concerned, P and P' are left out of account, as being imaginary and serving only a temporary purpose. 330. Application of Foregoing Principles to a Roof Truss.- Eig. 355. W l and JF 2 are wind pressures, P l and P 2 aro loads, while the remaining external forces, viz., the re- FIG. 355. 40G MECHANICS OF ENGINEERING. actions, or supporting forces, V ot V a and H n , may be found by preceding . (We here suppose that the right abut- ment furnishes all the horizontal resistance ; none at the left). Lay off the forces (known) W lt W 2 , P 1} and P. 2 in the usual way, to form a portion of the closed force polygon. To close the polygon it is evident we need only draw a horizontal through 5 and limit it by a vertical through 1. This determines H IL but it remains to determine n' the point of division between V and V a . Select a convenient pole 0,, and draw rays from it to 1, 2, etc. Assume a con- venient point a in the line of V n in the space diagram, and through it draw a line || to Ojl to meet the line of W l in some point b ; then a line II to t 2 to meet the line of W 2 in some point c ; then through c \\ to 0$ to meet the line of PI in some point d ; then through d \\ to 0^ to meet the line of P 2 in some point e, (e is identical with d, since P { and P 2 are in the same line) ; then ef \\ to Ofi to meet H D in some point/; then/gr || to 0,6 to meet V n in some point g. abcdefg is an equilibrium polygon corresponding to the pole Oi. Now join ag, the " closing-line," and draw a II to it through O l to determine %', the required point of division between V and V n on the vertical 1 6. Hence V and V n are now determined as well as H n . [The use of the arbitrary pole 0^ implies the temporary employment of a pair of opposite and equal forces in the line ag, the amount of either being = Ojw']. Having now all the external forces acting *on the truss, and assuming that it contains no " redundant parts," i.e., parts unnecessary for rigidity of the frame-work, we proceed tc find the pulls and thrusts in the individual pieces, on the following plan. The truss being pin-connected, no piece extending beyond a joint, and all loads being con- sidered to act at joints, the action, pull or thrust, of each piece on the joint at either extremity will be in the direction of the piece, i.e., in a known direction, and the pin of each GRAPHICAL STATICS. 407 joint is in equilibrium under a system of concurrent forces consisting of the loads (if any) at the joint and the pulls or thrusts exerted upon it by the pieces meeting there. Hence we may apply the principles of 325 to each joint in turn. See Fig. 356. In constructing and interpreting the various force polygons, Mr. R. H Bow's convenient notation will be used ; this is as follows : In the space diagram a capital letter [ABC, etc.] is placed in each tri- angular cell of the truss, and also in each angular space in the outside outline of the truss between the external forces and the adjacent truss-pieces. In this way we can speak of the force W l as the force BC, of W 2 as the force CE, the stress in the piece a/3 as the force CD, and so on. That is, the stress in any one piece can be named from the letters in the spaces bordering its two sides. Corresponding to these capital letters in the spaces of the space-dia- gram, small letters will be used at the vertices of the closed force-polygons (one polygon for each joint) in such a way that the stress in the piece CD, for example, shall be the force cd of the force polygon belonging to any joint in which that piece terminates ; the stress in the piece FO by the force fg in the proper force polygon, and so on. In Fig. 356 the whole truss is shown free, in equili- brium under the external forces. To find the pulls or thrusts (i.e., tensions or compressions) in the pieces, con- sider that if all but two of the forces of a closed force polygon are known in magnitude and direction, while the directions, only, of those two are known, the ichde force polygon may be drawn, thus determining the amounts of those two forces by the lengths of the corresponding sides. We must .-. begin with a joint where no more than two pieces meet, as at a ; [call the joints a, /9, f, d, and the cor- correspouding force polygons a', $' etc. Fig. 356.] Hence at ' (anywhere on the paper) make ab \\ and = (by scale) to the known force AB (i.e., F ) pointing it at the upper end, and from this end draw be = and || to the known force BC (i.e., W,) pointing this at the lower end. 408 MECHANICS OP ENGINEERING. FIG 356. To close the polygon draw through c a' II to the piece CD, and through a a || to AD ; their intersection deter- mines d, and the polygon is closed. Since the arrows must be point to butt round the periphery, the force with which the piece CD acts on the pin of the joint a is a force of an amount = cd and in a direction from c toward d ; hence the piece CD is in compression ; whereas the action of the piece DA upon the pin at a is from d toward a (direction of arrow) and hence DA is in tension. Notice that in constructing the force polygon a' a right-handed (or clock-wise) rotation has been observed in considering in turn the spaces AEG and D, round the joint a. A similar order will be found convenient in each of the other joints. Knowing now the stress in the piece CD, (as well as in DA) all but two of the forces acting on the pin at the joint ft are known, and accordingly we begin a force polygon, /j', for that joint by drawing dc,= and || to the dc of polygon a', but pointed in the opposite direction, since the action of CD on the joint ft is equal and opposite to its action on the joint (this disregards the weight of the piece). Through c draw ce and || to the force CE (i.e., W 2 ) and GRAPHICAL STATICS. 409 pointing the same way ; then ef, = and || to the load EF (i.e. PI) and pointing downward. Through f draw a || to the piece FG and through d, a || to the piece GD, and the polygon is closed, thus determining the stresses in the pieces FG and GD. Noting the pointing of the arrows, we readily see that FG is in compression while GD is in tension. Next pass to the joint d, and construct the polygon d', thus determining the stress gh in GH and that ad in AD ; this last force ad should check with its equal and oppo- site ad already determined in polygon a'. Another check consists in the proper closing of the polygon f, all of whose sides are now known. [A compound stress-diagram may be formed by super- posing the polygons already found in such a way as to make equal sides co-incide ; but the character of each stress is not so readily perceived then as when they are kept separate]. In a similar manner we may find the stresses in any pin- connected frame-work (in one plane and having no redun- dant pieces) under given loads, provided all the support- ing forces or reactions can be found. In the case of a braced-arch (truss) as shown in Fig. 357, hinged to the abutments at both ends and not free to slide laterally upon them, the reactions at and B de- pend, in amount and direc- tion, not only upon the- equations of Statics, but on the form and elasticity of the arch-truss. Such cases will be treated later under arch-ribs, or curved beams. 332. The Special Equil. Polygon. Its Relation to the Stresses in the Rigid Body. Eeproducing Figs. 350 and 351 in Figs. 358 and 359, (where a rigid curved beam is in equilibrium under the forces P,, P 2 , P s , P 4 and R',.} we call a . . b . . c 410 MECHANICS OF ENGINEERING. the special equil. polygon because it corresponds to a force- diagram in which the same order of forces has been ob- served as that in which they occur along the beam (from left to right here). From the relations between the force SPACE DIAGRAM FIG. 358. FORCE DIAGRAM FIG. 3o9. diagram and equil. polygon, this special equil. polygon in the space diagram has the following properties in connec- tion with the corresponding rays (dotted lines) in the force diagram. The stresses in any cross-section of the portion O'A of the beam, are due to P alone ; those of any cross-section on AB to P! and P 2 , i.e., to their resultant R a , whose mag- ^itude is given by the line Oa' in the force diagram, while its liue of action is ab the first segment of the equil. poly- gon. Similarly, the stresses in BG are due to P I} P 2 and P s , i.e., to their resultant 7? b acting along the segment be, its magnitude being = Ob' in the force diagram. E.g., if the section at ra be exposed, considering O'ABm as a free body, we have (see Fig. 360) the elastic stresses for inter- nal forces) at m balancing the exterior or " applied forces " PI, P 2 and P 3 . Obviously, then, the stresses at m are just GRAPHICAL STATICS. 411 the same as if JR b the resultant of P lt P 2 and P 3 , acted upon an imaginary rigid prolongation of the beam intersecting be (see Fig. 361).7? b might be called the "anti-stress-result- ant " for the portion BC of the beam. We may .. state the following : If a rigid body is in equilibrium under a sys- tem of Non-Concurrent Forces in a plane, and the special equi- librium polygon has been drawn,, then each ray of the force diagram is the anti-stress-resultant of that portion of the beam ichich corresponds to the segment of the equilibrium polygon to i. cinch the ray is parallel ; and its line of action is the seg- ment just mentioned. Evidently if the body is not one rigid piece, but com- posed of a ring of uncemented blocks (or voussoirs), it may be considered rigid only so long as no slipping takes place or disarrangement of the blocks ; and this requires that the " anti-stress-resultant " for a given joint between two blocks shall not lie outside the bearing surface of the joint, nor make too small an angle with it, lest tipping or slipping occur. For an example of this see Fig. 362, show- ing a line of three blocks in equilibrium under five forces. The pressure borne at the joint MN, is = R & in the force-diagram and acts in the line ab. The con- struction supposes all the forces given except FIG. 362. one, in amount and posi- tion, and that this one could easily be found in amount, as being the side remaining to close the force polygon, while its position would depend on the equil. polygon. But in practice the two forces P l and R',. are generally unknown, hence the point 0, or pole of the force diagram, can not be fixed, nor the special equil. polygon located, until other considerations, outside of those so far presented, are brought into play. In the progress of such a problem, as will be seen, it will be necessary to use arbitrary trial po- sitions for the pole 0, and corresponding trial equilibrium polygons. MECHANICS OF ENGINEElilifG. CHAPTER IX. GRAPHICAL STATICS OF VERTICAL FORCES. 333. Remarks. (With the exception of 378 a) in prob- lems to be treated subsequently (either the stiff arch -rib,, or the block-work of an arch-ring, of masonry) when the body is considered free all the forces holding it m equil. will be vertical (loads, due to gravity) except the reactions at the two extremities, as in Fig. 363 ; but for convenience each reaction will be replaced by its horizontal and verti- cal components (see Fig. 364). The two TTs are of course equal, since they are the only horizontal forces in the system. Henceforth, all equil. polygons under discussion will be understood to imply this kind of system of forces. P l} P.>, \ etc., will represent the " loads " ; V and V n the vertical components of the abutment reactions ; // the value of either horizontal component of the same. (We here sup- pose the pressures T and T D resolved along the horizon- tal and vertical.) GRAPHICAL STATICS. 413 334. Concrete Conception of an Equilibrium Polygon. Any equilibrium polygon has this property, due to its mode of construction, viz.: If the ab and be of Fig. 358 were im- ponderable straight rods, jointed at b without friction, they would be in equilibrium under the system of forces there given. (See Fig. 364a). The rod ab suffers a compression equal to the R A of the force diagram, Fig. 359, and be a compression = R b . In some cases these rods might be in tension, and would then form a set of links playing the part of a suspension-bridge cable. (See 44). 335. Example of Equilibrium Polygon Drawn to Vertical Loads Fig. 365. [The structure bearing the given loads is not shown, but simply the imaginary rods, or segments of an equilibrium polygon, which would support the given loads in equilibrium if the abutment points A and JB, to which the terminal rods are hinged, were firm. In the present case this equilibrium is unstable since the rods form a standing structure ; but if they were hanging, the equilibri- um would be stable. Still, in the present case, a very light bracing, or a little friction at all joints would make the equilibrium stable. p, .1 ; Given three loads P lt P 2 , and P 3 , and two " abutment verticals " A' and B', in which we desire the equil. poly- gon to terminate, lay off as a "load-line," to scale, P,, P 2 , and P 3 end to end in their order. Then selecting any pole, 414 MECHANICS OF ENGINEERING. O, draw the rays 01, 02, etc., of a force diagram (the F's and P's, though really on the same vertical, are separated slightly for distinctness ; also the ZTs, which both pass through and divide the load-line into F and F n ). We determine a corresponding equilibrium polygon by draw- ing through A (any point in A'} a line || to . . 1, to inter- sect Pi in some point b ; through 6 a || to . . 2, and so on* until B' the other abutment-vertical is struck in some point B. AB is the " abutment-line " or " closing-line" By choosing another point for 0, another equilibrium polygon would result. As to which of the infinite number (which could thus be drawn, for the given loads and the A and B' verticals) is the special equilibrium poly- gon for the arch-rib or stone-arch, or other structure, on which the loads rest, is to be considered hereafter. In any of the above equilibrium polygons the imaginary series of jointed rods would be in equilibrium. 336. Useful Property of an Equilibrium Polygon for Vertical Loads. (Particular case of 328). See Fig. 366. In any equil. polygon, supporting vertical loads, consider as free any number of consecutive segments, or rods, with the loads at their joints, e. g., the 5th and 6th and portions of the 4th and 7th which, we sup- pose cut and the compressive forces in them put in, T\ and in order to consider 4567 a free body. For equil., according to Statics, the lines of action of r l\ and T 7 (the com- pression in those rods) must in- tersect in a point, G, in the line of action of the resultant of P 4 , P 5 , and P 6 ; i.e., of the loads occurring at the -inter- vening vertices. That is, the point C must lie in the ver- tical containing the centre of gravity of those loads. Since the position of this vertical must be independent of the particular equilibrium polygon used, any other (dotted lines in Fig. 366) for the same loads will give the same re- GRAPHICAL STATICS. 415 suits. Hence the vertical CD, containing the centre of gravity of any number of consecutive loads, is easily found by drawing the equilibrium polygon corresponding to any convenient force diagram having the proper load-line. This principle can be advantageously applied to finding a gravity -line of any plane figure, by dividing the latter into parallel strips, whose areas may be treated as loads applied in their respective centres of gravity. If the strips are quite numerous, the centre of gravity of each may be considered to be at the centre of the line joining the mid- dles of the two long sides, while their areas may be taken as proportional to the lengths of the lines drawn through these centres of gravity parallel to the long sides and lim- ited by the end-curves of the strips. Hence the " load- line " of the force diagram may consist of these lines, or of their halves, or quarters, etc., if more convenient ( 376). USEFUL RELATIONS BETWEEN FORCE DIA- GRAMS AND EQUILIBRIUM POLYGONS, (for vertical loads.) 337. R6sum6 of Construction. Fig. 367. Given the loads PI, etc., their verticals, and the two abutment verticals A' and B', in which the abutments are to lie ; we lay off a load-line 1 ... 4, take any convenient pole, 0, for a force- diagram and complete the latter. For a corresponding equilibrium polygon, assume any point A in the vertical A', for an abutment, and draw the successive segments Al, 2, etc., respectively parallel to the inclined lines of the force diagram (rays), thus determining finally the abut- ment JS, in B", which (B) will not in general lie in the hor- izontal through A. Now join AB, calling AB the abutment-line, and draw a parallel to it through 0, thus fixing the point n' on the 416 MECHANICS OF ENGINEERING. g i FIG. load-line. This point n', as above determined, is indepen- dent of the location of the pole, 0, (proved in 329) and divides the load-line into two portions ( V = 1 . . . n', and V' n = n' ... 4) which are the vertical pressures which two supports in the verticals A' and B' ivould sustain if the given loads rested on a horizontal rigid bar, as in Fig. 368. See 329. Hence to find the point n' we may use any convenient pole 0. [N. B. The forces V tt and F D of Fig. 367 are not identi- cal with V and F' n , but may be obtained by dropping a "] from to the load-line, thus dividing the load-line into two portions which are V ' (upper portion) and V n . However, if A and B be connected by a tie-rod, in Fig. 367, the abutments in that figure will bear vertical press- ures only and they will be the same as in Fig. 368, while the tension in the tie -rod will be = On'.] 338. Theorem. The vertical dimensions of any two 'equili- brium polygons, drawn to the same loads, load-verticals, and abutment-verticals, are inversely proportional to their H's (or " pole distances "). We here regard an equil. polygon and its a,butment-line as a closed figure. Thus, in Fig. 369, we have two force-diagrams (with a common load-line, for convenience) and their corresponding equil. polygons, for the same loads and verticals. From 337 we know that On' is || to AB and n' is || to A S . Let CD be any ver- tical cutting the first segments of the two equil. polygons. GKAPHICAL STATICS. 417 Denote the intercepts thus determined by 2' and ' , respect- ively. From the parallelisms just mentioned, and others more famil- iar, we have the triangle \n' sim- ilar to the triangle Az' (shaded), and the triangle ln' similar to the tri- angle A z. Hence FIG. 369. ^ ) ~h f the proportions between ( In' _z f -, lw' bases and altitudes ( ~ff ^ ~Jf o .-. z f : z' : : H : H. The same kind of proof may easily be applied to the vertical intercepts in any other segments, e. g., z" and z" Q. E. D. 339. Corollaries to the foregoing. It is evident that : (1.) If the pole of the force-diagram be moved along a vertical line, the equilibrium polygon changing its form in a corresponding manner, the vertical dimensions of the equilibrium polygon remain unchanged ; and (2.) If the pole move along a straight line which con- tains the point n', the direction of the abutment-line remains constantly parallel to the former line, while the vertical dimensions of the equilibrium polygon change in inverse proportion to the pole distance, or H, of the force- diagram. [His the ~| distance of the pole from the load- line, and is called the pole-distance]. 340. Linear Arch as Equilibrium Polygon. ( See 316.) If the given loads are infinitely small with infinitely small horizontal spaces between them, any equilibrium polygon becomes a linear arch. Graphically we can not deal with these infinitely small loads and spaces, but from 336 it is evident that if we replace them, in successive groups, 418 MECHANICS OF ENGINEERING. by finite forces, each of which = the sum of those com- posing one group and is applied through the cen- of gravity of that tre of gravity of group, we can draw an equilibrium polygon whose segments will be tangent to the curve of the corresponding linear arch, and indicate its posi- tion with sufficient exactness for practical purposes. (See Fig. 370). The successive points of tangency A, m, n, etc., lie vertically under the points of division between the groups. This relation forms the basis of the graphical treatment of voussoir, or blockwork, arches. 341. To Pass an Equilibrium Polygon Through Three Arbitrary Points. (In the present case the forces are vertical. For a construction dealing with any plane system of forces see construction in 3780.) Given a system of loads, it is re- quired to draw r/1 -.2 an equilibrium polygon for them through any three points, two of which may be consid- ered as abut- ments, outside of the load- verticals, the third point being between the verticals of the first two. See Fig. 371. The loads P lt etc., are given, with their verticals, while A, p, and B are the three points. Lay off the load-line, and with any convenient pole, 1? construct a force-diagram, then a corresponding preliminary equilibrium polygon beginning at A. Its right abutment B lt in the vertical through B, is thus found. O v n' can now be drawn || to AB V to determine n'. Draw n'O II to BA. The pole of the required equilibrium polygon must lie on n'O ( 337)- FIG. 371. GRAPHICAL STATICS. 419 Draw, a vertical through p. The H of the required equili- brium polygon must satisfy the proportion H : H^ : : ~rs : pin. (See 338). Hence construct or compute H from the proportion and draw a vertical at distance H from the load-line (on the left of the load-line here) ; its inter- section with n' gives the desired pole, for which a force diagram may now be drawn. The corresponding equilibrium polygon beginning at the first point A will also pass through p and B ; it is not drawn in the figure. 342. Symmetrical Case of the Foregoing Problem. If two points A and B are on a level, the third, p, on the middle vertical between them ; and the loads (an even number) symmetrically disposed both in position and magnitude, about ), we may proceed more simply, as follows : (Fig. 372). From symmetry n' must occur in the mid- dle of the load-line, of which we need lay off only the upper half. Take a convenient pole 0i, in the horizontal through n', and draw a half force diagram and a corres- ponding half equilibrium polygon (both dotted). The up- per segment be of the latter must be horizontal and being prolonged, cuts the prolongation of the first segment in a point d, which determines the vertical CD containing the centre of gravity of the loads occurring over the half -span on the left. (See 336). In the required equilibrium poly- gon the segment containing the point p must be horizon- tal, and its intersection with the first segment must lie in CD. Hence determine this intersection, C, by drawing the vertical CD and a horizontal through p ; then join CA, which is the first segment of the required equil. polygon. A parallel to CA through 1 is the first ray of the corres- ponding force diagram, and determines the pole on the horizontal through n'. Completing the force diagram for 420 MECHANICS OF ENGINEERING. this pole (half of it only here), the required equil. poly- gon is easily finished afterwards. 343. To Find a System of Loads Under Which a Given Equi- librium Polygon Would be in Equilibrium. Fig. 373. Let AB be the given equilibrium polygon. Through any point , , , as a pole draw a parallel to each segment of the equilibrium polygon. Any vertical, as V, cutting these lines will have, intercepted upon it, a load-line 1, 2, 3, whose parts 1 . . 2, 2 . . 3, etc., are proportional to the successive loads which, placed on the corresponding joints of the equilibrium polygon would be supported by it in equilibrium (unstable). One load may be assumed and the others constructed. A hanging, as well as a standing, equilibrium polygon may be dealt with in like manner, but will be in stable equi- librium. The problem in 44 may be solved in this way. ARCHES OF MASONRY. 421 CHAPTER X. RIGHT ARCHES OF MASONRY. 344. In an ordinary "right" stone-arch (i.e., one in which the faces are "I to the axis of the cylindrical soffit, or under surface), the successive blocks forming the arch- ring are called voussoirs, the joints between them being planes which, prolonged, meet generally in one or more horizontal lines; e.g., those of a three -centred arch in three ii horizontal lines ; those of a circular arch in one, the axis of the cylinder, etc. Elliptic arches are sometimes used. The inner concave surface is called the soffit, to which the radiat- ing joints between the voussoirs are made perpendicular. The curved line in which the soffit is intersected by a plane "1 to the axis of the arch is the Intrados. The curve in the same plane as the intrados, and bounding the outer ex- tremities of the joints between the voussoirs, is called the Extrados. Fig. 374 gives other terms in use in connection with a 422 MECHANICS OF ENGINEERING. stone arch, and explains those already given. " springing-line." AS is the 345. Mortar and Friction. As common mortar hardens very slowly, no reliance should be placed on its tenacity as an element of stability in arches of any considerable size ; though hydraulic mortar and thin joints of ordinary mortar can sometimes be depended on. Friction, however, between the surfaces of contiguous voussoirs, plays an essential part in the stability of an arch, and will there- fore be considered. The stability of voussoir-arches must .*. be made to depend on the resistance of the voussoirs to compresssion and to sliding upon each other ; as also of the blocks composing the piers, the foundations of the latter being firm. 346. Point of Application of the Resultant Pressure between two consecutive voussoirs ; (or pier blocks). Applying Navier's principle (as in flexure of beams) that the press- ure per unit area on a joint varies uniformly from the extremity under greatest compression to the point of least compression (or of no compression) ; and remembering that negative pressures (i.e., tension) can not exist, as they might in a curved beam, we may represent the pressure per unit area at successive points of a joint (from the intra- dos toward the extrados, or vice versa) by the ordinates of a straight line, forming the surface of a trapezoid or tri- angle, in which figure the foot of the ordinate of the cen- tre of gravity is the point of application of the resultant pressure. Thus, where the least compression is supposed FIG. 376. MASONRY ARCHES. 423 to occur at the intrados A, Fig. 375, the pressures vary as the ordinates of a trapezoid, increasing to a maximum value at B, in the extrados. In Fig. 376, where the pressure is zero at B, and varies as the ordinates of a triangle, the result- ant pressure acts through a point one-third the joint- length from A. Similarly in Fig. 377, it acts one-third the joint-length from B. Hence, when the pressure is not zero at either edge the resultant pressure acts within the middle third of the joint. Whereas, if the resultant press- ure falls without the middle third, it shows that a portion Am of the joint, see Fig. 378, receives no pressure, i.e., the joint tends to open along Am. Therefore that no joint tend to open, the resultant press- ure must fall within the middle third. It must be understood that the joint surfaces here dealt with are rectangles, seen edgewise in the figures. 347. Friction. By experiment it has been found the angle of friction (see 156) for two contiguous voussoirs of stone or brick is about 30 ; i.e., the coefficient of fric- tion is / = tan. 30. Hence if the direction of the press- ure exerted upon a voussoir by its neighbor makes an angle a less than 30 with the normal to the joint surface, there is no danger of rupture of the arch by the sliding of one on the other. (See Fig. 379). 348. Resistance to Crushing. When the resultant pressure falls at its extreme allowable limit, viz. : the edge of the middle third, the pressure per unit of area at n t Fig. 380, is double the mean pressure per unit of area. Hence, in de- i * signing an arch of masonry, N. / we must be assured that at 7lf 379 every joint (taking 10 as a factor of safety) | Double the mean press- ) b legg than y \ ure per unit of area I 424 MECHANICS OF ENGINEERING. C being the ultimate resistance to crushing, of the material employed ( 201) (Modulus of Crushing). Since a lamina one foot thick will always be considered in what follows, careful attention must be paid to the units employed in applying the above tests. EXAMPLE. If a joint is 3 ft. by 1 foot, and the resultant pressure is 22.5 tons the mean pressure per sq. foot is p=22.5-^3=7.5 tons per sq. foot .*. its double=15 tons per sq. foot=208.3 Ibs. sq. inch, which is much less than J / 10 of C for most building stones ; see 203, and below. At joints where the resultant pressure falls at the middle, the max. pressure per square inch would be equal to the mean pressure per square inch ; but for safety it is best to assume that, at times, (from moving loads, or vibrations) it may move to the edge of the middle third, causing the max. pressure to be double the mean (per square inch). Gen* Gillmore's experiments in 1876 gave the following results, among many others : NAME OF BUILDING STONE. C IN LBS. PER SQ. INCH. Berea sand-stone, 2-inch cube, - - - - 8955 4 " " - 11720 Limestone, Sebastopol, 2-inch cube (clialti), - - 1075 Limestone from Caen, France, - - - .- 3650 Limestone from Kingston, K Y., - - - - 13900 Marble, Vermont, 2-inch cube, - 8000 to 13000 Granite, New Hampshire, 2-inch cube, 15700 to 24000 349. The Three Conditions of Safe Equilibrium for an arch of uncemented voussoirs. Recapitulating the results of the foregoing paragraphs, we may state, as follows, the three conditions which must be satisfied at every joint of arch-ring and pier, for each of any possible combination of loads upon the structure : (1). The resultant pressure must pass within the middle- third. (2). The resultant pressure must not make an angle > 30 with the normal to the joint. (3). The m^an pressure per unit of area on the surface AKCH OF MASONRY. 425 of the joint must not exceed l / w of the Modulus of crush- ing of the material. 350. The True Linear-Arch, or Special Equilibrium Polygon; and the resultant pressure at any joint. Let the weight of each voussoir and its load be represented by a vertical force passing through the centre of gravity of the two, as in Fig. 381. Taking any two points A and B, A being in the first joint and B in the last ; also a third point, p t in the crown joint (supposing such to be there, although gener- ally a key-stone occupies the crown), through these three points can be drawn [ 341] an equilibrium polygon for the loads given ; suppose this equil. polygon nowhere passes outside of the arch-ring (the arch-ring is the por- tion between the intrados, mn, and the (dotted) extrados m'n') intersecting the joints at b, c, etc. Evidently if such be the case, and small metal rods (not round) were insert- ed at A, b, c, etc., so as to separate the arch -stones slight- ly, the arch would stand, though in unstable equilibrium, the piers being firm ; and by a different choice of A, p, and B, it might be possible to draw other equilibrium poly- gons with segments cutting the joints within the arch- ring, and if the metal rods were shifted to these new inter- sections the arch would again stand (in unstable equilib- 'rium). In other words, if an arch stands^ it may be possible to draw a great number of linear arches within the limits of the arch-ring, since three points determine an equilibrium polygon (or linear arch) for given loads. The question arises then : which linear arch is the locus of the actual re- sultant pressures at the successive joints ? [Considering the arch-ring as an elastic curved beam inserted in firm piers (i.e., the blocks at the springing-line 426 MECHANICS OF ENGINEERING. are incapable of turning) and having secured a close fit at all joints before the centering is lowered, the most satisfac- tory answer to this question is given in Prof. Greene's " Arches," p. 131 ; viz., to consider the arch-ring as an arch rib of fixed ends and no hinges ; see 380 of next chapter; but the lengthy computations there employed (and the method demands a simple algebraic curve for the arch) may be most advantageously replaced by Prof. Eddy's graphic method (" New Constructions in Graphical Statics," published in Van Nostrand's Magazine for 1877), which applies to arch curves of any form. This method will be given in a subsequent chapter, on Arch Bibs, or Curved Beams ; but for arches of masonry a much simpler procedure is sufficiently exact for practical purposes and will now be presented]. I y If two elastic blocks I / \ of an arch-ring touch at ~\/^ one ed g e > Fi g- 382 > tneir Jm adjacent sides making a ^ VSS V N / small angle with each FIG- 382. FIG. 383. other, and are then grad- ually pressed more and more forcibly together at the edge m, as the arch-ring settles, the centering being gradually lowered, the surface of contact becomes larger and larger, from the compression which ensues (see Fig. 383); while the resultant pressure between the blocks, first applied at the extreme edge m, has now probably advanced nearer the middle of the joint in the mutual adjustment of the arch- stones. With this in view we may reasonably deduce the following theory of the location of the true linear arch (sometimes called the " line of pressures " and " curve of pressure") in an arch under given loading and with^rw piers. (Whether the piers are really unyielding, under the oblique thrusts at the springing-line, is a matter for sub- sequent investigation. 351. Location of the True Linear Arch. Granted that the- voussoirs have been closely fitted to each other over the ARCH OF MASONRY. 427 centering (sheets of lead are sometimes used in the joints to make a better distribution of pressure); and that the piers are firm ; and that the arch can stand at all without the centering ; then we assume that in the mutual accom- modation between the voussoirs, as the centering is low- ered, the resultant of the pressures distributed over any joint, if at first near the extreme edge of the joint, advances nearer to the middle as the arch settles to its final posi- tion of equilibrium under its load ; and hence the follow- ing 352. Practical Conclusions. I. If for a given arch and loading, with firm piers, an equilibrium polygon can be drawn (by proper selection of the points A, p, and B, Fig. 381) entirely within the mid- dle third of the arch ring, not only will the arch stand, but the resultant pressure at every joint will be within the middle third (Condition 1, 349) ; and among all possible equilibrium polygons which can be drawn within the mid- dle third, that is the " true " one which most nearly coin- cides with the middle line of the arch-ring. II. If (with firm piers, as before) no equilibrium poly- gon can be drawn within the middle third, and only one within the arch-ring at all, the arch may stand, but chip- ping and spawling are likely to occur at the edges of the joints. The design should .-. be altered. III. If no equilibrium polygon can be drawn within the arch-ring, the design of either the arch or the loading must be changed ; since, although the arch may stand, from the resistance of the spandrel walls, such a stability must be looked upon as precarious and not countenanced in any large important structure. (Very frequently, in small arches of brick and stone, as they occur in buildings, the cement is so tenacious that the whole structure is vir- tually a single continuous mass). When the " true " linear arch has once been determined,, the amount of the resultant pressure on any joint is given, by the length of the proper ray in the force diagram. 428 MECHANICS OF ENGINEERING. ARRANGEMENT OF DATA FOR GRAPHIC TREATMENT. 353. Character of Load. In most large stone arch bridges the load (permanent load) does not consist exclusively of masonry up to the road-way but partially of earth filling above the masonry, except at the faces of the arch where the spandrel walls serve as retaining walls to hold the earth. (Fig. 384). If the intrados is a half circle or half- FIG. 384. ellipse, a compactly -built masonry backing is carried up beyond the springing-line to AB about 60 to 45 from the crown, Fig. 385 ; so that the portion of arch ring below AB may be considered as part of the abutment, and thus AB is the virtual springing-line, for graphic treatment. Sometimes, to save filling, small arches are built over the haunches of the main arch, with earth placed over them, as shown in Fig. 386. In any of the preceding cases it is customary to consider that, on account of the bond- ing of the stones in the arch shell, the loading at a given distance from the crown is uniformly distributed over the width of the roadway. ARCHES OF MASONRY. 429 354. Reduced Load-Contour. In the graphical discussion of a proposed arch we consider a lamina one foot thick, this lam iiia being vertical and ~] to the axis of the arch ; i.e., the lamina is || to the spandrel walls. For graphical treatment, equal areas of the elevation (see Fig. 387) of this lamina must represent equal weights. Taking the material of the arch-ring as a standard, we must find for each point p of the extrados an imaginary height z of the arch-ring material, which would give the same pressure (per running horizontal foot) at that point as that due to the actual load above that point. A number of such or- dinates, each measured vertically upward from the extra- dos determine points in the "Reduced Load-Contour," i.e., the imaginary line, AM, the area between which and the extrados of the arch -ring represents a homogeneous load of the same density as the arch-ring, and equivalent to the actual load (above extrados), vertical by vertical. 355. Example of Reduced Load-Contour. Fig. 388. Given an arch-ring of granite (heaviness = 170 Ibs. per cubic foot) with a dead load of rubble (heav. = 140) and earth (heav. = 100), distributed as in figure. At the point p, of the extrados, the depth 5 feet of rubble is equivalent to a depth of [|* x5]=4.1 ft. of granite, while the 6 feet of earth is equivalent to [j^x6]=3.5 feet of granite. Hence the Reduced Load-Contour has an ordinate, above p, of 7.6 feet. That is, for each of several points of the arch -ring extrados reduce the rubble ordinate in the ratio of 170 : 140, and the earth ordinate in the ratio 170 : 100 and add the re- sults, setting off the sum vertically from the points in the extrados*. In this way Fig. 389 is obtained and the area *This Is most conveniently done by graphics, thus : On a right-line set off 17 equal parts (of any convenient magnitude.) Call this distance OA. Through draw another right line at any convenient angle (30 to 60) with OA, and on it from O set off OB equal to 14 (for the rubble ; or 10 for the earth) of the same equal parts. Join AB. From O toward A set off* all the rubble ordinates to be reduced, (each being set off from 0) and through the other extremity of each draw a line par- allel to AB. The reduced ordinates will be the respective lengths, from 0, along OB, to the intersections of these parallels with OB. * With the dividers. 430 MECHANICS OF ENGINEERING. there given is to be treated as representing homogeneous granite one foot thick. This, of course, now includes the arch-ring also. AB is the " reduced load-contour." 356. Live Loads. In discussing a railroad arch bridge the " live load " (a train of locomotives, e.g., to take an ex- treme case) can not be disregarded, and for each of its po- sitions we have a separate Reduced Load-Contour. EXAMPLE. Suppose the arch of Fig. 388 to be 12 feet wide (not including spandrel walls) and that a train of lo- comotives weighing 3,000 Ibs. per running foot of the track covers one half of the span. Uniformly * distributed later- ally over the width, 12 ft., this rate of loading is equiva- lent to a masonry load of one foot high and a heaviness of 250 Ibs. per cubic ft., i.e., is equivalent to a height of 1.4 ft. of granite masonry [since ^ x 1.0 1.4] over the half span considered. Hence from Fig. 390 we obtain Fig. 391 in an obvious manner. Fig. 391 is now ready for graphic treatment. FIG. 391. 357. Piers and Abutments. In a series of equal arches the pier between two consecutive arches bears simply the weight of the two adjacent semi-arches, plus the load im- * If the earth-filling is sLallcw, the Ir.minae directly under the track prob- ably receive a greater pressure than the others. ARCHES OF MASONRY. 431 mediately above the pier, and .. does not need to be as large as the abutment of the first and last arches, since these latter must be prepared to resist the oblique thrusts of their arches without help from the thrust of another on the other side. ' In a very long series of arches it is sometimes customary to make a few of the intermediate piers large enough to act as abutments. These are called " abutment piers," and in case one arch should fall, no others would be lost except those occurring between the same two abutment piers as the first. See Fig. 392. A is an abutment-pier. nn-nnn GRAPHICAL, TREATMENT OF ARCH. 358. Having found the " reduced load-contour," as in preceding paragraphs, for a given arch and load, we are ready to proceed with the graphic treatment, i.e., the first given, or assumed, form and thickness of arch-ring is to be investigated with regard to stability. It may be necessary to treat, separately, a lamina under the spandrel wall, and one under the interior loading. The constructions are equally well adapted to arches of all shapes, to Gothic as well as circular and elliptical. 359. Case I Symmetrical Arch and Symmetrical Loading. (The " steady " (permanent) or " dead " load on an arch is usually symmetrical). Fig. 393. From symmetry we need 432 MECHANICS OF ENGINEERING. deal with only one half (say the left) of the arch and load. Divide this semi-arch and load into six or ten divisions by vertical lines ; these divisions are considered as trape- zoids and should have the same horizontal width = b (a convenient whole number of feet) except the last one, LKN, next the abutment, and this is a pentagon of a different width ft,, (the remnant of the horizontal distance LC). The weight of masonry in each division is equal to (the area of division) x (unity thickness of lamina) x (weight of a cu- bic unit of arch-ring). For example for a division having an area of 20 sq. feet, and composed of masonry weighing 160 Ibs. per cubic foot, we have 20x1x1603,200 Ibs., applied through the centre of gravity of the division. The area of a trapezoid, Fig. 394, is ^&(Ai+^ 2 )> and its cen- tre of gravity may be found, Fig. 395, by the construction of Prob. 6, in 26 ; or by 27. The weight of the pen- tagon LN, Fig. 393, and its line of application (through centre of gravity) may be found by combining results for the two trapezoids into which it is divided by a vertical through K. See 21. Since the weights of the respective trapezoids (except, ing LN) are proportional to their middle vertical in- tercepts [such as ^(^1+^2) Fig. 394] these intercepts (trans- ferred with the dividers) may be used directly to form the load-line, Fig. 396, or proportional parts of them if more convenient. The force scale, which this implies, is easily computed, and a proper length calculated to represent the weight of the odd division LN ; i.e., 1 ... 2 on the load- line. Now consider A, the middle point of the abutment joint, Fig. 396, as the starting point of an equilibrium polygon (or abutment of a linear arch) for a given loading, and re- quire that this equilibrium polygon shall pass through p, the middle of the crown joint, and through the middle of the abutment joint on the right (not shown in figure). Proceed as in 342, thus determining the polygon Ap for the half-arch. Draw joints in the arch-ring through those points where the extrados is intersected by the ver- AKCIIES OF MASONBY. 433 FIG. 390. FIG. 397. tical separating the divisions (not the gravity verticals). The points in which these joints are cut by the segments of the equilibrium polygon, Fig. 397, are (very nearly, if the joint is not more than 60 from p, the crown) the points of application in these joints, respectively, of the resultant pressures on them, (if this is the " true linear arch " for this arch and load) while the amount and direction of each such pressure is given by the proper ray in the force -dia- gram. If at any joint so drawn the linear arch (or equilibrium polygon) passes outside the middle third of the arch-ring, the point A, or p, (or both) should be judiciously moved (within the middle third) to find if possible a linear arch which keeps within limits at all joints. If this is found impossible, the thickness of the arch -ring may be increased at the abutment (giving a smaller increase toward the crown) and the desired result obtained ; or a change in the distribution or amount of the loading, if allowable, may gain this object. If but one linear arch can be drawn within the middle third, it may be considered the " true " one ; if several, the one most nearly co-inciding with the middles of the joints (see 351,and 352) is so considered. 360. Case II. Unsymmetrical Loading on a Symmetrical Arch; (e.g., arch with live load covering one half -span as in Figs. 390 and 391). Here we must evidently use a full force diagram, and the full elevation of the arch -ring and load. 434 MECHANICS OF ENGINEERING. See Fig. 398. Select three points A, p, and B, as follows, to determine a trial equilibrium polygon : Select A at the loiccr limit of the middle third of the abutment-joint at the end of the span which is the more heavily -loaded ; in the other abutment-joint take B at the upper limit of the middle third ; and take p in the middle of the crown-joint. Then by 341 draw an equilibrium polygon (i.e., a linear arch) through these three points for the given set of loads, and if it does not remain within the middle third, try other positions for A, p, and J5, within the middle third. As to the " true linear arch " alterations of the design, etc., the same remarks apply as already given in Case I. Very frequently it is not necessary to draw more than one linear arch, for a given loading, for even if one could be drawn nearer the middle of the arch- ring than the first, that fact is most always apparent on mere inspection, and the one already drawn (if within middle third) will furnish values sufficiently accurate for the pressures on the respective joints, and their direction angles. 360a. The design for the arch-ring and loading is not to be considered satisfactory until it is ascertained that for the dead load and any possible combination of live-load 'in addition) the pressure at any joint is ARCHES OF MASONRY. 435 (1.) Within the middle third of that joint ; (2.) At an angle of < 30 with the normal to joint- surface. (3.) Of a mean pressure per square inch not > than l / x of the ultimate crushing resistance. (See 348.) 361. Abutments. The abutment should be compactly and solidly built, and is then treated as a single rigid mass. The pressure of the lowest voussoir upon it (considering a lamina one foot thick) is given by the proper ray of the force diagram (0 .. l,e. g., in Fig. 396) in amount and direc- tion. The stability of the abutment will depend on the amount and direction of the resultant obtained by com- bining that pressure P a with the weight G of the abutment and its load, see Fig. 399. Assume a probable width RS for the abutment and compute the weight G of the corresponding abutment OBRS and MNBO, and find the centre of gravity of the whole mass C. Apply G in the vertical through (7, and combine it with P a at their in- tersection D. The resultant P should not cut the base It Sin a point beyond the middle third (or, if this rule gives too massive a pier, take such a width that the pressure per square inch at S shall not exceed a safe value as FIG. 399. computed from 362.) After one or two trials a satisfactory width can be obtained. We should also be assured that the angle PDG is less than 30. The horizontal joints above ES should also be tested as if each were, in turn, the lowest base, and if necessary may be inclined (like mn) to prevent slipping. On no joint should the maximum pressure per square inch be > than l / m the crushing strength of the cement. Abut- ments of firm natural rock are of course to be preferred where they can be had. If water penetrates under an abutment its buoyant effort lessens the weight of the lat- ter to a considerable extent. 436 MECHANICS OF ENGINEEEING. 362. Maximum Pressure Per Unit of Area When the Resultant Pressure Falls at Any Given Distance from the Middle ; according to Navier's theory of the distribution of the pressure ; see 346. Case I. Let the resultant pressure P, Fig. 400, (a), (a) FIG. 401. fall within the middle third, a distance = nd ( < % d) from the middle of joint (d depth of joint.) Then we have the following relations : p (the mean press, per. sq. in.),/?,,, (max. press, persq. in.), and p a (least press, per sq. in.) are proportional to the lines h (mid. width), a (max. base), and c (min. base) respectively, of a trapezoid, Fig. 400, (b), through whose centre of gravity P acts. But ( 26) 6 a+c ' Pm=P (6+l). Hence the following table : = *6 d press. p m = 2 V.d 7m d then the max. times the mean pressure. Case II. Let P fall outside the mid. third, a distance = nd (> l /6 d) from the middle of joint. Here, since the joint is not considered capable of withstanding tension, we have a triangle, instead of a trapezoid. Fig. 401. First compute the mean press, per sq. in. or from this table: (lamina one (l-2n) 18 d inches foot thick). ARCHES OF MASONRY. 437 For nd = A<* &d ^d ft* &<* A^ P = i p 10 ' d 1 P 8 ' d 1 P 6 ' d 1 P 4 ' d 1 P 2 ' d infinity. (d in inches and Pin Ibs. ; with arch lamina 1 ft. in thickness.) Then the maximum pressure (at A, Fig. 401) p m , = %p, becomes known, in Ibs. per sq. in. 362a. Arch-ring under Non-vertical Forces. An example of this occurs when a vertical arch-ring is to support the pressure of a liquid on its extrados. Since water-pressures are always at right angles to the surface pressed on, these pressures on the extradosal surface of the arch- ring form a system of non-paral- lel forces which are normal to the curve of the extrados at their respective points of application and lie in parallel vertical planes, parallel to the faces of the lamina. We here assume that the extradosal surface is a cylinder (in the most general sense) whose rectilinear elements are ~| to the faces of the lamina. If, then, we divide the length of the extrados, from crown to each abutment, into from six to ten parts, the respective pressures on the corresponding surfaces are obtained by multiplying the area of each by the depth of its centre of gravity from the upper free surface of the liquid, and this product by the weight of a unit of volume of the liquid ; and each such pressure may be considered as acting through the centre of the area. Finally, if we find the resultant of each of these pressures and the weight of the corresponding portion of the arch-ring, these resultants form a series of non-vertical forces in a plane, for which an equilibrium polygon can then be passed through three assumed points by 3Y8a, these three points being taken in the crown-joint and the two abutment- joints. As to the " true linear arch" see 359. As an extreme theoretic limit it is worth noting that if the extrados and intrados of the arch-ring are concentric circles; if the weights of the voussoirs are neglected ; and if the rise of the arch is very small compared with the depth of the crown below the water surface, then the circularcentre-line of the arch-ring is the " true linear arch" 438 MECHANICS OF ENGINEERING. CHAPTEE XL ARCH-RIBS. 364. Definitions and Assumptions. An arch-rib (or elastic- arch, as distinguished from a block-work arch) is a rigid curved beam, either solid, or built up of pieces like a truss (and then called a braced arch) the stresses in which, under a given loading and with prescribed mode of sup- port it is here proposed to determine. The rib is sup- posed symmetrical about a vertical plane containing its axis or middle line, and the Moment of Inertia of any cross section is understood to be referred to a gravity axis of the section, which (the axis) is perpendicular to the said vertical plane. It is assumed that in its strained condi- tion under a load, the shape of the rib differs so little from its form when unstrained that the change in the ab- scissa or ordinate of any point in the rib axis (a curve) may be neglected when added (algebraically) to the co- ordinate itself ; also that the dimensions of a cross-section are small compared with the radius of curvature at any part of the curved axis, and with the span. 365. Mode of Support. Either extremity of the rib may be hinged to its pier (which gives freedom to the end-tangent- line to turn in the vertical plane of the rib when a load is applied); or may \yejixed, i.e., so built-in, or bolted rigid- ly to the pier, that the end-tangent-line is incapable of changing its direction when a load is applied. A hinge may be inserted anywhere along the rib, and of course ARCH RIBS. 439 destroys the rigidity, or resistance to bending at that point. (A. hinge having its pin horizontal "| to the axis of the rib is meant). Evidently no more than three such hinges could be introduced along an arch- rib between two piers ; unless it is to be a hanging structure, acting as a suspension-cable. 366. Arch Rib as a Free Body. In considering the whole rib free it is convenient, for graphical treatment, that no section be conceived made at its extremities, if fixed ; hence in dealing with that mode of support the end of the rib will be considered as having a rigid prolongation reach- ing to a point vertically above or below the pier junction, an unknown distance from it, and there acted on by a force of such unknown amount and direction as to preserve the actual extremity of the rib and its tangent line in the same position and direction as they really are. As an illustra- tion of this Fig. 402 shows free an arch rib. ONB, with its extremi- ties and B fixed in the piers, with no hinges, Q\ and bearing two loads P. and P 2 . The other ces of the sys- tem holding it in equi- librium are the horizontal and vertical components, of the pier reactions (H t V, H UJ and V n ), and in this case of fixed ends each of these two reactions is a single force not in- tersecting the end of the rib, but cutting the vertical through the end in some point F (on the left ; and in G on the right) at some vertical distance c, (or d\ from the end. Hence the utility of these imaginary prolongations OQF, and BRG, the pier being supposed removed. Compare Figs. 348 and 350. The imaginary points, or hinges, F and G, will be called abutments being such for the special equilibrium polygon FIG. 402. 440 MECHANICS OF ENGINEERING. (dotted line), while and B are the real ends of the curved beam, or rib. In this system of forces there are five unknowns, viz.: V t V M B H , and the distances c and d. Their determina- tion by analysis, even if the rib is a circular arc, is ex- tremely intricate and tedious ; but by graphical statics (Prof. Eddy's method ; see 350 for reference), it is com- paratively simple and direct and applies to any shape of rib, and is sufficiently accurate for practical purposes. This method consists of constructions leading to the loca- tion of the " special equilibrium polygon " and its force diagram. In case the rib is hinged to the piers, the re- actions of the latter act through these hinges, Fig. 403, i.e., the abutments of the special equilibrium polygon coincide with the ends of the rib and B, and for a given rib and load the unknown quantities are only three V, V llt and H\ (strictly there are four ; but IX = gives H a = H). The solution FlG . 403 . by analytics is possible only for ribs of simple algebraic curves and is long and cumbrous ; whereas Prof. Eddy's graphic method is comparatively brief and simple and is applicable to any shape of rib whatever. 367. Utility of the Special Equilibrium Polygon and its force diagram. The use of locating these will now be illustrated [See 332]. As proved in 332 and 334 the compres- sion in each " rod " or segment of the " special equilibrium polygon" is the anti-stress resultant of the cross sections in the corresponding portion of the beam, rib, or other struc- ture, the value of this compression (in Ibs. or tons) being measured by the length of the parallel ray in the force diagram. Suppose that in some way (to be explained sub- sequently) the special equilibrium polygon and its force diagram have been drawn for the arch-rib in Fig. 404 hav- ing fixed ends, and B, and no hinges ; required the elastic stresses in any cross-section of the rib as at ra. Let the ARCH KIBS. 441 H-" FiG. 404. scale of the force-diagram on the right be 200 Ibs. to the inch, say, and that of the space -diagram (on the left) 30 ft. to the inch. The cross section m lies in a portion TK, of the rib, cor- responding to the rod or segment be of the equilibrium polygon; hence its anti-stress-resultant is a force R 2 acting in the line 6c, and of an amount given in the force-diagram. Now R 2 is the resultant of V, H, and P lf which with the elastic forces at m form a system in equilibrium, shown in Fig. 405 ; the portion FO Tm being considered free. Hence Pis. 405. taking the tangent line and the normal at m as axes we should have Z (tang, comps.) = ; I (norm, comps.) = ; and I (moms, about gravity axis of the section at ra) = 0, and could thus find the unknowns p lt p. 2 , and J", which ap- pear in the expressions piF the thrust, ^-- the moment of 442 MECHANICS OF ENGINEERING. the stress-couple, and J"the shear. These elastic stresses are classified as in 295, which see. p y and p 2 are Ibs. per square inch, e/is Ibs., e is the distance from the horizontal gravity axis of the section to the outermost element of area, (where the compression or tension is p 2 Ibs. per sq. in., as due to the stress-couple alone) while 1 is the " mo- ment of inertia " of the section about that gravity axis. [See 247 and 295 ; also 85]. Graphics, however, gives us a more direct method, as follows : Since R 2 , in the line be, is the equivalent of V, H, and P lt the stresses at m will be just the same as if R 2 acted directly upon a lateral pro- longation of the rib at T (to intersect ftcFig. 405) as shown in Fig. 406, this prolongation Tb taking the place of TOF in Fig. 405. The force diagram is also reproduced here. Let a denote the length of the "| from ra's gravity axia upon be, and z the vertical intercept between m and be. For this imaginary free body, we have, from - (tang. compons.^O, R, cos a= and from 2 (norm. compons.)=0, J R 2 sin a=J while from -T( moms, about) ) -, p 2 I the gravity axis of m)=0, \ We have ** = ~ e ' But from the two similar triangles (shaded ; one of them is in force diagram) a :z :: H:R, .'. R. 2 a=Hz, whence we may rewrite these relations as follows (with a general state- ment), viz.: If the Special Equilibrium Polygon and Its Force Diagram Have Been Drawn for a given arch-rib, of given mode of support, and under a given loading, then in any cross-section of the rib, we have (F = area of section): The projection of the proper /i \ mu rm, of tlie force (1.) The Thrust, f tho drawn at the given section. ARCH RIBS. 443 (2.) The Shear, i.e., J, = (upon which dependsthe The projection of the proper ray (of the force diagram) up- '3.) The Moment of the stress couple, i.e., ?- , = on the normal to the rib curve shearing stress in the web). (See 253 and ' v at the given section. 256). The product (Hz) of the H (or pole-distance) of the force- diagram by the vertical dis- tance of the gravity axis of the section from the spec, equilib- rium polygon. By the " proper ray " is meant that ray which is parallel to the segment (of the equiL polygon) immediately under or above which the given section is situated. Thus in Fig. 404, the proper ray for any section on TK is B 2 ; on KB, RS ; on TO, B v . The projection of a ray upon any given tangent or normal, is easily found by drawing through each end of the ray a line ~| to the tangent (or normal) ; the length between these ~] 's on the tangent (or normal) is the force required (by the scale of the force diagram). We may thus construct a shear diagram, and a thrust diagram for a given case, while the successive vertical intercepts between the rib and special equilibrium polygon form a moment diagram. For example of the z of a point m is ^ inch in a space diagram drawn to a scale of 20 feet to the inch, while H measures 2.1 inches in a force diagram con- structed on a scale of ten tons to the inch, we have, for the moment of the stress-couple at ra, M =Hz= [2.1 X 10] tons X[^x20]ft.=210ft. tons. 368. It is thus seen how a location of the special equili- brium polygon, and the lines of the corresponding force- diagram, lead directly to a knowledge of the stresses in all the cross-sections of the curved beam under consideration, bearing a given load ; or, vice versa, leads to a statement of conditions to be satisfied by the dimensions of the rib, for proper security. It is here supposed that the rib has sufficient la-ceral 444 MECHANICS OF ENGINEERING. bracing (with others which lie parallel with it) to prevent buckling sideways in any part like a long column. Before proceeding to the complete graphical analysis of the differ- ent cases of arch-ribs, it will be necessary to devote the next few paragraphs to developing a few analytical rela- tions in the theory of flexure of a curved beam, and to giving some processes in " graphical arithmetic." 369. Change in the Angle Between Two Consecutive Rib Tan- gents when the rib is loaded, as compared with its value before loading. Consider any small portion (of an arch rib) included between two consecutive cross- sections; Fig. 407. KHG W is its unstrained form. Let EA, = ds, be the original length of this portion of the rib axis. The length of all the fibres (j| to rib-axis) was originally =ds (nearly) and the two consecutive tangent-lines, at E and A, made an angle = dti originally, with each other. While under strain, however, all the fibres are shortened equally an amount dA lt by the uniformly distributed tangential thrust, but are unequally shortened (or lengthened, accord- ing as they ai e on one side or the other of the gravity axis E, or A, of the section) by the system of forces making what we call the " stress couple," among which the stress at the distance e from the gravity axis A of the section is called pi per square inch ; so that the tangent line at A' now takes the direction A'D "\ to H'A'G' instead of A'Q (we suppose the section at E to remain fixed, for convani- ARCH RIBS. 445 ence, since the change of angle between the two tangents depends on the stresses acting, and not on the new posi- tion in space, of this part of the rib), and hence the angle between the tangent-lines at E and A (originally = dd] is now increased by an amount CA'D = dy (or G'A'R = d(p}\ G'H' is the new position of GH. We obtain the value of d as follows: That part (rf/ 2 ) of the shortening of the fibre at G, at distance e from A due to the force pjlF, is 201 eq. (1), dL = ?. But, geometrically, d^ also =ed. Next let the section D, now at D', turn through it proper angle d, and from similar right-triangles, 8 x : dv : : y : u and dy : dv : : x : u .'. dx = ydy and dy=xd

etc., select any Pole, 0,, and join 0j ... 1, 0i ... 2, etc. ; also, beginning at FIG - 414 - any point FI in the vertical F', if we draw F l . . . a \\ to O l . . 1 to intersect the line of P l ; then ab \\ to 0, . . 2, and so on until finally a point GI, in G', is determined; then the figure FI .abc Cnis an equilibrium polygon for the given loads and load verti- cals, and 0j ... 1234 is its " force diagram." The former is so called because the short segments F& ab, etc., if considered to be rigid and imponderable rods, in a vertical plane, hinged to each other and the terminal ones to abut- ments F { and # would be in equilibrium under the given loads hung at the joints. An infinite number of equilib- AKCH-RIBS. 451 riura polygons may be drawn for the given loads and abutment-verticals, by choosing different poles in the force diagram. [One other is shown in the figure ; 2 is its pole. (J^ G! and F 2 G 2 are abutment lines.)] For all of these the following statements are true : (1.) A line through the pole, || to the abutment line cuts the load-line in the same point n' t whichever equilibrium polygon be used ( /. any one will serve to determine n'). (2.) If a vertical CD be drawn, giving an intercept z' in each of the equilibrium polygons, the product Hz' is the same for all the equilibrium polygons. That is, (see Fig. 414) for any two of the polygons we have H { :# 2 ::s/: V; or H 2 z 2 ' =H l z l f . (3.) The compression in each rod is given by that " ray " (in the force diagram) to which it is parallel. (4.) The " pole distance " H, or ~| let fall from the pole upon the load-line, divides it into two parts which are the vertical components of the compressions in the abutment- rods respectively ( the other component being horizontal) ; // is the horizontal component of each (and, in fact, of each of the compressions in all the other rods). The compressions in the extreme rods may also be called the abutment reactions (oblique) and are given by the extreme rays. (5.) Three Points [not all in the same segment (or rod)] determine an equilibrium polygon for given loads. Hav- ing given, then, three points, we may draw the equilibrium polygon by 341. 375. Summation of Products. Before proceeding to treat graphically any case of arch-ribs, a few processes in graphical arithmetic, as it may be called, must be pre- sented, and thus established for future use. To make a summation of products of two factors in each by means of an equilibrium polygon. 452 MECHANICS OF ENGINEERING. Construction. Suppose it required to make the summa- tion 2 1 (x z) i. e. y to sum the series x l Zj-f- x. z z 2 + a? 3 % + . . . by graphics. Having first arranged the terms in the order of magni- tude of the a?'s, we proceed as follows : Supposing, for illustration, that two of the z'a (% and z 4 ) are negative (dotted in figure) see Fig. 415. These quantities x and z may be of any nature whatever, anything capable of being represented by a length, laid off to scale. First, in Fig. 416, lay off the z's in their order, end to end, on a ver- tical load-line taking care to J off z 3 and i J_ 2 4 upicard in _their turn. Take any con- FIG. 416. venient pole 1, ... 2, etc.; then, having pre- viously drawn vertical lines whose horizontal distances from an extreme left-hand vertical F' are made = x {) x, r a? 3 , etc., respectively, we begin at any point F, in the verti- cal F' t and draw a line || to ... 1 to intersect the x } ver- tical in some point ; then 1' 2' II to ... 2, and so on, fol- lowing carefully the proper order. Produce the last seg- ment (6' ... G in this case) to intersect the vertical F' in some point K. Let KF =k (measured on the same scale as the x's), then the summation required is S (xz) = Hk. H is measured on the scale of the z's, which need not be the same as that of the a?'s ; in fact the z's may not be the same kind of quantity as the o?'s. [PROOF. From similar triangles H: 2, :: x l : & .'. x^^Hlc, \ and " " " H:z*::x 2 : k. 2 , .-. xjs. 2 =Hk, , ; draw the rays . -CH-RIBS. 453 and so on. But H(k l -\-k,+etG.}=HxFK=Hk]. 376. Gravity Vertical From the same construction in Fig. 415 we can determine the line of action (or gravity vertical) of the resultant of the parallel vertical forces z lt %, etc.- (or loads); by prolonging the first and last segments to their intersection at (j. The resultant of the system of forces or loads acts through C and is vertical in this case ; its value .being = JT (2), that is, it = the length 1 ... 7 in the force dia- gram, interpreted by the proper scale. It is now supposed that the z's represent forces, the a?'s being their respective lever arms about F. If the z's represent the areas of small finite por- tions of a large plane figure, we may find a gravity -line (through (7) of that figure by the above construction; each 2 being-applied through the centre of gravity of its own portion. Calling the distance a? between the verticals through C jmd F, we t, have also x . 2 (2) = I (xz) because I (z) is the resultant of the II z's. This is also evident from the proportion (similar triangles) H : (I . . 7) :: x : L 454 MECHANICS OF ENGINEERING. 376 a. Moment of Inertia (of Plane Figure) by Graphics. Fig. 416 a. I y = ? First, for the portion on right. Divide OR jnto equal parts each = Ax. Let z } , z. 2 , etc., be the middle ordinates of the strips thus obtained, and x l} etc. their abscissas (of middle points). Then we have approximately / N for OH = AX.Z&I+ Ax.z. 2 x.?+ ...... ].. (1) But by 375 we may construct the products z^x^z^x^ etc., taking a convenient H', (see Fig. 416, (6)), and obtain Jc l} Jc 2 > etc., such that ZjXi = H'k lt z?x 2 = H'k 2 , etc. Hence eq. (1) becomes : 7, for OB approx.=7Pzte[& 1 ;r 1 +& 2 a?2+ ...]... (2) By a second use of 375 (see Fig. 416 c) we construct I, such that lew + k. 2 x 2 + ____ = H"l \H" taken at con- venience]. /. from eq. (2) we have finally, (approx.), 7 N for OR=H'H"lAx ____ (3) For example if OR ~ 4 in., with four strips, Ax would = 1 in.; and if H' = 2 in., H" = 2 in., and I = 5.2 in., then 7 N for OR = 2x2x5.2x1.0=20.8 biquad. inches. The 7 N for OL, on the left of N, is found in a similar manner and added to 7 X for OR to obtain the total 7 V The position of a gravity axis is easily found by cutting the shape out of sheet metal and balancing on a knife edge ; or may be obtained graphically by 336 ; or 376. 377. Construction for locating a line vm(Fig. 417) at (a), in the polygon FG in such a position as to satisfy the two following conditions with reference to the vertical inter- cepts at 1, 2, 3, 4, and 5, between it and the given points 1, 2, 3, etc., of the perimeter of the polygon. ARCH-RIBS. 455 Condition I (Calling these intercepts w,, u^, etc., and their horizontal distances from a given vertical F, x lt x.,, etc.) 2 (u) is to = 0; i.e., the sum of the positive w's must be numerically = that of the negative (which here are at 1 and 5). An infinite number of positions of vm will satisfy condition I. Condition H S (ux) is to = ; i.e., the sum of the moments of the positive w's about F must = that of the negative w's. i.e., the moment of the resultant of the posi- tive w's must = that of the resultant of the negative ; and .'. (Condit. I being already satisfied) these two resultants must be directly opposed and equal. But the ordinates u in (a) are indi- vidually equal to the differ- ence of the full and dotted ordinates in (b) with the same x's .'. the conditions may be rewritten : I. 2' (full ords. in(Z>))= I (dotted ords. in (b)) II. 2 [each full ord. in (6) X its x] = - [each dotted ord. in (b) x its x] i.e., the centres of gravity of the full and of the dotted in (b) must lie in the same vertical Again, by joining vG, we may divide the dotted ordi- nates of (b) into two sets which are dotted, and broken, re- spectively, in (c) Then, finally, drawing in (cZ), R, the resultant of full ords. of (c) T, " " " broken " " " T', " " " dotted " " " ^^j | ! T ! R f, i 1 T ('0 FIG. 417. 456 MECHANICS OF ENGINEERING. we are prepared to state in still another and final form the conditions which vm must fulfil, viz. : (I.) T+T' must = R; and (II.) The resultant of T and T' must act in the same vertical as R. In short, the quantities T, T', and R must form a bal- anced system, considered as forces. All of which amounts practically to this : that if the verticals in which T and T' act are known and R be conceived as a load supported by a horizontal beam (see foot of Fig. 417, last figure) resting on piers in those verticals, then T and T' are the respec- tive reactions of those piers. It will now be shown that the verticals of T and T' are easily found, being independent of the position of vm; and that both the vertical and the mag- nitude of R, being likewise independent of vm, are deter- mined with facility in advance. For, if v be shifted up or down, all the broken ordinates in (c) or (d) will change in the same proportion (viz. as vF changes), while the dotted ordinates, though shifted along their verticals, do not change in value ; hence the shifting of v affects neither the vertical nor the value of T', nor the vertical of T. The value of T, however, is proportional to vF. Similar- ly, if in be shifted, up or down, T' will vary proportionally to mG, but its vertical, or line of action, remains the same. T is unaffected in any way by the shifting of m. R, de- pending for its value and position on the full ordinates of (c) Fig. 417, is independent of the location of vm. We may /. proceed as follows : 1st. Determine R graphically, in amount and position, by means of 376. 2ndly. Determine the verticals of T and T' by any trial position of vm (call it v 2 w 2 ), and the corresponding trial values of T and T (call them T 2 and T' 2 ). 3rdly. By the fiction of the horizontal beam, construct ( 329) or compute the true values of T and T', and then determine the true distances vF and mG by the propor- tions vF : v,F : : T : T, and mG : mG : : T' : T' z . ARCH-RIBS. 457 Example of this. Fig. 418. (See Fig. 417 for s and t.) From A toward B in (e) Fig. 418, lay off the lengths (or lines proportional \* to them) of the full /f\ ordinates 1, 2, etc., of (/). Take any pole Oi, and draw the equilibrium p o 1 y - g o n (/")' and pro- long its extreme seg- ments to find C and thus determine 7?'s vertical. R is repre- sented by AB. In (g) [same as (/) but shifted to avoid complexity of lines] draw a trial v.,m., and join v 2 G.,. Deter- mine the sum T 2 of the broken ordi- 'FIG. 4is. nates (between v. 2 G 2 ana F 2 G 2 ) and its vertical line of ap- plication, precisely as in dealing with R ; also T' 2 that of the dotted ordiuates (five) and its vertical. Now the true T=Rt+(8+t) and the true T'=Rs+(s+t). Hence com- pute vF=(T+T 2 } v,F 2 and mG=(T'+T't} m^G 2 , and by laying them off vertically upward from F and G respec- tively we determine v and m, i.e., the line vm to fulfil the conditions imposed at the beginning of this article, rela- ting to the vertical ordinates intercepted between vm and given points on the perimeter of a polygon or curve. Note (aX If the verticals in which the intercepts lie are equidistant and quite numerous, then the lines of action of T 2 and T' 2 will divide the horizontal distance between F and G into three equal parts. This will be exactly true in the application of this construction to 390. Note (b). Also, if the verticals are symmetrically placed about a vertical line, (as will usually be the case) tww 2 is 458 MECHANICS OF ENGINEERING. best drawn parallel to FG, for then T^ and T' 3 will be equal and equi-distant from said vertical line. 378. Classification of Arch-Bibs, or Elastic Arches, according to continuity and modes of support. In the accompany- ing figures ihefull curves show the unstrained form of the rib (before any load, even its own weight, is permitted to come upon it) ;the dotted curve shows its shape (much ex- aggerated) when bearing a load. For a given loading Three Conditions must be given to determine the special equilibrium polygon ( 366 and 367). Class A. Continuous rib, free to slip laterally on the piers, which have smooth horizontal surfaces, Fig. 420. This is chiefly of theoretic interest, its consideration being therefore omitted. The pier reactions are neces- sarily vertical, just as if it were a straight horizontal beam. Class B. Rib of Three Hinges, two at the piers and one intermediate (usually at the crown) Fig. 421. Fig. 36 also is an example of this. That is, the rib is discontinuous and of two segments. Since at each hinge the moment of the stress couple must be be zero, the special equilibrium polygon must pass through the hinges. Hence as three points fully determine an equilibrium polygon for given load, the special equilibrium is drawn by 341. FIG. 420. FIG. 421. [ 378a will contain a construction for arch-ribs of three hinges, when the forces are not all vertical.] Class C. Rib of Two Hinges, these being at the piers, the rib continuous between. The piers are considered im- movable, i.e., the span cannot change as a consequence of loading. It is also considered that the rib is fitted to its AllCil KIBS. 459 hinges at a definite temperature, and is then under no con- straint from the piers (as if it lay flat on the ground), not even its own weight being permitted to act when it is fi- nally put into position. AVhen the "false works" or temporary supports are removed, stresses are in- duced in the rib both by its loading, including its own weight, and by a change of temperature. Stresses due to temperature may be ascertained separately and then combined with those duo to the loading. [Classes A and B are not subject to temperature stresses.] Fig. 422 shows a rib of two hinges, at ends. Conceive the dotted curve (form and position un- der strain) to be superposed on the continuous curve (form before strain) in such a way that B and its tangent FIG - 422 - line (which has been dis- placed from its original position) may occupy their pre- vious position. This gives us the broken curve O n B. 00 a is .'. O's displacement relatively to B and .Z?'s tangent. Now the piers being immovable O n B (right line)=O.B ; i.e., the X projection (or Ax) of 00,, upon OB (taken as an axis of X} is zero compared with its Ay. Hence as one condi- tion to fix the special equilibrium polygon for a given load- ing we have (from 373) r (1) The other two are that the ) must pass through . (2) special equilibrium polygon ) " " " B . (3) Class D. Rib with Fixed Ends and no hinges, i.e., continu- ous. Piers immovable. The ends may be fixed by being inserted, or built, in the masonry, or by being fastened to large plates which are bolted to the piers. [The St. Louis Bridge and that at Coblenz over the Rhine are of this class.] Fig. 423. In this class there being no hinges we 400 MECHANICS OF ENGLNEE11IXG. have no point given in advance through which the special equilibrium polygon must pass. However, since O's dis- placement relatively (and absolutely) to B and J?'s tangent is zero, both Ax and Ay [see 373] zero. Also the tan- gent-lines both at and B being fixed in direction, the angle be- tween them is the same under loading, or change of temperature, as when the rib was first placed in position under no strain and at a definite temperature. Hence the conditions for locating the special equilibrium polygon are FIG. 423. Mds _ El ls = o r M y as o c r " c/o #/ ' Jo Mxds = 0. 7 e/o El In the figure the imaginary rigid prolongations at the ends are shown [see 366]. Other designs than those mentioned are practicable (such as : one end fixed, the other hinged ; both ends fixed and one hinge between, etc.), but are of unusual occur- rence. 378a. Rib of Three Hinges. Forces not all Vertical. If the given rib of three hinges upholds a roof, the wind-press- ure on which is to be considered as well as the weights of the materials composing the roof-covering, the forces will not all be vertical. To draw the special equil. polygon in such a case the following construction holds : Re- quired to draw an equilib- rium polygon, for any plane system of forces, through three arbitrary points, A, p and B ; Fig. B 423a. Find the line of action of 7?,, the resultant of all the forces occurring FIG. 423a. ^ between A and p ; also, AKCH-RLBS. 401 that of R 2 , the resultant of all forces between ,p and B ; also the line of action of R, the resultant of BI and R 2 , [see 328.] Join any point M in R with A and also with B, and join the intersections N&nd 0. Then A N will be the direction of the first segment, B that of the last, and NO itself is the segment corresponding to p (in the de- sired polygon) of an equilibrium polygon for the given forces. See 328. If A N' p 0' B are the corresponding segments (as yet unknown) of the desired equil. polygon, we note that the two triangles MNO and M'N'C/, having their vertices on three lines which meet in a point [i.e., R meets R l and R 2 in C"], are homological [see Prop. VII. of Introduc. to Modern Geometry, in Chauvenet's Geometry,] and that . . the three intersections of their corresponding sides must lie on the same straight line. Of those inter ^ sections we already have A and B, while the third must be at (7, found at the intersection of AB and NO. Hence by connecting C and p, we determine .A 7 ' and 0'. Joining WA and O'B, the first ray of the required force diagram will be || to NA, while the last ray will be II to O'B, and thus the pole of that diagram can easily be found and the cor- responding equilibrium polygon, beginning at A, will pass through p and B. (This general case includes those of 341 and 342.) 379. Arch-Rib of two Hinges; by Prof. Eddy's Method.* [It is understood that the hinges are at the ends.] Re- quired the location of the special equilibrium polygon. We here suppose the rib homogeneous (i.e., the modulus of Elasticity E is the same throughout), that it is a " curved prism " (i.e., that the moment of inertia I of the cross- section is constant), that the piers are on a level, and that the rib-curve is symmetrical about a vertical line. Fig. 424. For each point m of the rib curve we have an x and y (both known, being the co-ordinates of the point), and also a z (intercept ;-f between rib and special equilib- rium polygon) and a z' (intercept * P. 25 of Prof. Eddy's book ; see reference in preface of this work. 4G2 MECHANICS OF ENGINEERING. between the spec. eq. pol. and the axis X (which is OS}. The first condition given in 378 for Class C may be transformed as follows, remembering [ 367 eq. (3)] that M = Hz at any point m of the rib (and that El is con- stant). CJJL Myds = 0, i.e., * fzyds = . - . fzyds = JLJ. e/o e/o but z = y z' ' C (y z')yds=0; i.e., f yyds = C yz'ds . (1) e/o e/o e/o In practical graphics we can not deal with infinitesimals ; hence we must substitute As a small finite portion of the rib-curve for ds; eq. (1) now reads I* yy As = J? B yz' As. But if we take all the As's equal, As is a common factor and cancels out, leaving as a final form for eq. (1) S(yy} = Z*(yz'} . . . (1)' The other two conditions are that the special equilibrium polygon begins at and ends at B. (The subdivision of the rib-curve into an even number of equal As's will be ob- served in all problems henceforth.) 379a. Detail of the Construction. Given the arch-rib B, Fig. 425, with specified loading. Divide the curve into FIG. 425. AKCH RIBS. 463 eight equal /te's and draw a vertical through the middle of each. Let the loads borne by the respective Js's be PI, P 2 , etc., and with them form a vertical load-line A C to some convenient scale. With any convenient pole 0" draw a trial force diagram 0" AC, and a corresponding trial equilibrium polygon F G, beginning at any point in the vertical F. Its ordinates z/', z 2 ", etc., are propor- tional to those of the special equil. pol. sought (whose abutment line is OS) [374a (2)]. We next use it to de- termine n' [see 374a]. We know that OB is the " abut- ment-line " of the required special polygon, and that . . its pole must lie on a horizontal through n'. It remains to determine its H, or pole distance, by equation (1)' just given, viz. : S* yy = -\yz'. First by 375 find the value of the summation -*(yy\ which, from symmetry, we may write = 2^V y l Hence, Fig. 426, we obtain Next, also by 375, see Fig. 427, using the same pole dis- tance H as in Fig. 426, we find " +y* Z( " Again, since 2| (yz")= y^" + 2/7*7" + y<&" + 2/5*5" which " from symmetry (of rib) we obtain, Fig. 428, ,; I!(yO = #a*/',(Bame J); and .-. 2{ (yz"}=H (,"+&/') If now we find that kS'+k T "=Zk, 464 MECHANICS OF ENGINEERING. the condition 21 (yy) = If (yz") is satisfied, and the pole distance of our trial polygon in Fig. 425, is also that of the special polygon sought; i.e., the z" 's.are identical in value with the s"s of Fig. 424. In general, of course, we do not find that &,"+V 2&. Hence the z" 's must all be increased in the ratio 2&: (V+V) to become equal to the z"s. That is, the pole distance H of the spec, equil- polygon must be TT_ &/'+&/' rrtr (in which H" = the pole distance of the %k trial polygon) since from 339 the ordi- nates of two equilibrium polygons (for the same loads) are inversely as their pole distances. Having thus found the Hoi the special polygon, knowing that the pole must lie on the horizontal through n', Fig. 425, it is easily drawn, beginning at 0. As a check, it should pass through B. For its utility see 367, but it is to be remembered that the stresses as thus found in the different parts of the rib under a given loading, must afterwards be combined with those resulting from change of temperature and the shortening of the rib axis due to the tangential thrusts, before the actual stress can be declared in any part. [NoTE. If the " moment of inertia," /, of the rib-sec- tion is different at different sections, i.e., if /is variable, foreq. (1)' we may write 2*((y.^}= -*(y. *'} . . . (1)' \ "/ \ -H/ (where n = -, 7 being the moment of inertia of a particu- *0 lar section taken as a standard and /that at any section of rib) and in Fig. 426, use the ^ of each Js instead of y in the vertical " load-line," and for z" in Figs. 427 and n 428]. ' ARCH-IUBS. 465 330. Arch Rib of Fixed Ends and no Hinges, Example of Class D. Prof. Eddy's Method.* As before, E and I are constant along the rib Piers immovable. Bib curve symmetrical about a vertical line. Fig. 429 shows such a rib under any loading. Its span is OB, which is taken as an axis X. The co-ordinates of any point m' of the rib curve are x and y, and z is the vertical intercept between m' and the special equilibrium polygon (as yet unknown, but to be constructed). Prof. Eddy's method will now be given for finding tha spe- v cial equil. polygon. The three conditions it must satisfy (see 378, Class D, remembering that E and I are constant and that M Ih fr jin FIG. 429. 367) are flds= ; Cxzds= ; and fyzds=0 . . (1) e/o e/o e/o Now suppose the auxiliary reference line (straight) vm to have been drawn satisfying the requirements, with respect to the rib curve that / B fP /*'ds=0;and xz'ds=0 .... (2) e/o e/o in which z' is the vertical distance of any point m' from vm and x the abscissa of m' from 0. From Fig. 429, letting z" denote the vertical intercept (corresponding to any m') between the spec, polygon and the auxiliary line vm, we have z=zfz", hence the three conditions in (1.) become -z"}ds=0', i.e., see eqs. (2) z"ds=Q . (3) * P. 14 of Prof. Eddv's book ; see reference in preface of this work. 466 MECHANICS OF ENGINEERING. Cx (z'-z")ds=0 ; i.e., see eqs. (2) C*xz"ds=o e/o e/ o provided vm has been located as prescribed. For graphical purposes, having subdivided the rib curve into an even number of small equal J,s's, and drawn a verti- cal through the middle of each, we first, by 377, locate vm to satisfy the conditions 1?(0=0 and I?(a*')=0 . . (6) (see eq. (2) ; the As cancels out) ; and then locate the special equilibrium polygon, with vm as a reference-line, by making it satisfy the conditions. 2ft")=0 (7); 2Ka*")=0 . (8); S* (yz")=S* (yz'} . (9) (obtained from eqs. (3), (4), (5) by putting ds=4s, and can- celling). Conditions (7) and (8) may be satisfied by an infinite number of polygons drawn to the given loading. Any one of these being drawn, as a trial polygon, we determine for it the value of the sum 2*(yz") by 375, and compare it with the value of the sum 2*(yz') which is independent of the special polygon and is obtained by 375. [N.B. It must be understood that the quantities (lengths) x, y, z, z', and z", here dealt with are those pertaining to the verticals drawn through the middles of the respective ^s's, which must be sufficiently numerous to obtain a close result, and not to the verticals in which the loads act, necessarily, since these latter may be few or many according to circumstances, see Fig. 429]. If these sums are not equal, the pole distance of the trial equil. polygon must be altered in the proper ratio (and thus change the '"s in the inverse ratio) neces- sary to make these sums equal and thus satisfy condition (9). The alteration of the z'"s, all in the same ratio, will ARCH-RIBS. 4G7 not interfere with conditions (7) and (8) which are already satisfied. 381. Detail of Construction of Last Problem. Symmetrical Arch- Rib of Fixed Ends. As an example take a span of the St. Louis Bridge (assuming /constant) with " live load" cov- ering the half span on the left, Fig. 430, where the vertical FIG. 430. scale is much exaggerated for the sake of distinctness*. Divide into eight equal Js's. (In an actual example sixteen or twenty should be taken.) Draw a vertical through the * Each arch-rib of the St. Louis bridge is a built up or trussed rib of steel about 52o ft. span and 52 ft rise, in the form of a segment of a circle . Its moment of inertia, however,' t ia not strictly constant, the portions near each pier, of a length equal to one twelfth of the span, having a value of /one-half greater than that of the remainder of the arc. 468 MECHANICS OF ENGINEERING. middle of each Js, P 1} etc., are the loads coming upon the respective Js's. First, to locate vm, by eq. (6) ; from symmetry it must be horizontal. Draw a trial vm (not shown in the figure) and if the (+z')' s exceed the ( z')'s by an amount z ', the true vm will lie a height - ' above the trial vm (or be- low, if vice versa) ; n = the number of Js's. Now lay off the load-line on the right, (to scale), take any convenient trial pole 0'" and draw a correspond- ing trial equil. polygon F'"G m . In F'"G'" t by 377, locate a straight line v'"m" so as to make 2*(z'")=0 and (xz'")=Q (see Note (b) of 377). [We might now redraw F"'G"' in such a way as to bring v'"m'" into a horizontal position, thus : first determine a point n"' on the load-line by drawing 0"'ri" \\ to v'"m'" take a new pole on a horizontal through ri", with the same H' 1 ', and draw a corresponding equil. polygon ; in the lat- ter v"'m'" would be horizontal. We might also shift this new trial polygon upward so as to make v"'m''' and vm coincide. It would satisfy conditions (7) and (8), having the same a""s as the first trial polygon ; but to satisfy con- dition (9) it must have its z""s altered in a certain ratio, which we must now find. But we can deal with the indi- vidual z"''s just as well in their present positions in Fig. 430.] The points E and L in vm, vertically over E'" and /,"' in v'"m'", are now fixed ; they are the intersections of the special polygon required, with vm. The ordinates between v'"m'" and the trial equilibrium polygon have been called %'" instead of z"\ they are pro- portional to the respective z"'s of the required special polygon. The next step is to find in what ratio the (z"')'s need to be altered (or H'" altered in inverse ratio) in order to be- come the (s")' s J i- e ' * n or( ier to fulfil condition (9), viz.: AKCH-R1BS. 4G9 i 9 > -- -^ s 4 -**. x N r ^- ^----" ^ ^ () 1 i \^ ^^ ^ ^> 11.- H| y - tJ ~>j- S((yz")=Z\(yz') . (9) This may be done pre- cisely as for the rib with two hinges, but the nega- tive (O's must be prop- erly considered ( 375) See Fig. 431 for the de- tail. Negative z"s ors""s point upward. From Fig. 431a .-. from symmetry From Fig. 4316 we have Fio. 431. and from Fig. 431c [The same pole distance H is taken in all these construc- tions] .'. IV) = #,(> + &r). If, then, H (fc,+Jfe r ) = ZHJc condition (9) is satisfied by the z""8. If not, the true pole distance for the special equil. polygon of Fig. 430 will be =. 2k With this pole distance and a pole in the horizontal through '" (Fig. 430) the force diagram may be completed for the required special polygon ; and this latter may be con- structed as follows : Beginning at the point E, in vm, through it draw a segment I! to the proper ray of the force diagram. In our present figure (430) this " proper ray " would be the ray joining the pole with the point of meet- ing of P 2 and P 3 on the load-line. Having this one seg- 470 MECHANICS OF ENGINEERING. ment of the special polygon the others are added in an obvious manner, and thus the whole polygon completed. It should pass through L, but not and B. For another loading a different special equil. polygon would result, and in each case we may obtain the thrust, shear, and moment of stress couple for any cross-section of the rib, by 367. To the stresses computed from these, should be added (algebraically) those occasioned by change of temperature and by shortening of the rib as occasioned by the thrusts along the rib. These " temperature stresses," and stresses due to rib-shortening, will be con- sidered in a subsequent paragraph. They have no exist- ence for an arch-rib of three hinges. [NOTE. If the moment of inertia of the rib is variable / z" we put for z' and for z" in equation (6), (7), (8), and (9), n having the meaning given in the Note in 379 a, which see ; and proceed accordingly]. 381a. Exaggeration of Vertical Dimensions of Both Space and Force Diagrams. In case, as often happens, the axis of the given rib is quite a flat curve, it is more accurate (for find- ing M) to proceed as follows : After drawing the curve in its true proportions and pass- ng a vertical through the middle of each of the equal z/s's, compute the ordinate (y) of each of these middle points from the equation of the curve, and multiply each y by four (say). These quadruple ordinates are then laid off from the span upward, each in its proper vertical. Also multiply each load, of the given loading, by four, and then with these quadruple loads and quadruple ordinates, and the upper extremities of the latter as points in an exagge- rated rib-curve, proceed to construct a special equilibrium polygon, and the corresponding force diagram by the proper method ( for Class B, C, or D, as the case may be) for this exaggerated rib -curve. The moment, Hz, thus found for any section of the ex- ARCH-RIBS. 471 aggerated rib-curve, is to be divided by four to obtain the moment in the real rib, in the same vertical line. To find the thrust and shear, however, for sections of the real rib, besides employing tangents and normals of the real rib we must draw, and use, another force diagram, obtained from the one already drawn (for the exaggerated rib) by re- ducing its vertical dimensions (only), in the ratio of four to one. [Of course, any other convenient number besides four, may be adopted throughout.] 382. Stress Diagrams. Take an arch-rib of Class D t 378, i.e., of fixed ends, and suppose that for a given loading (in- cluding its own weight) the special equil. polygon and its force diagram have been drawn [ 381]. It is re- quired to indicate graphically the variation of the three stress-elements for any section of the rib, viz., the thrust, shear, and mom. of stress- couple. I is constant. If at any point m of the rib a section is made, then the stresses in that section are classified into three sets (Fig. 432). (See 295 and 367) and from 367 eq. (3) we see that the ver- tical intercepts between the rib and the special equil. polygon being proportional to the products Hz or moments of the stress-couples in the corresponding sec- tions form a moment diagram, on inspection of which we can trace the change in this moment, Hz = ?- , and hence the variation of the stress per square inch, p^, (as due to stress couple alone) in the outermost fibre of any section (tension or compression) at distance e from the gravity axis of the section), from section to section along the rib. By drawing through lines On' and Of parallel re- spectively to the tangent and normal at any point m of the rib axis [see Fig. 433] and projecting upon them, in turn, the proper ray (R, in Fig. 433) (see eqs. 1 and 2 of 367) 472 MECHANICS OF ENGINEERING. we obtain the values of the thrust and shear for the sec- tion at m. When found in this way for a number of points along the rib their values may be laid off as vertical lines from a horizontal axis, in the verticals containing the re- spective points, and thus a thrust diagram and a shear dia- gram may be formed, as constructed in Fig. 433. Notice that where the moment is a maximum or minimum the shear changes sign (compare 240), either gradually or suddenly, according as the max. or min. occurs between two loads or in passing a load ; see m', e. g. Also it is evident, from the geometrical relations involv- e 1, that at those points of the rib where the tangent-line is parallel to the " proper ray " of the force diagram, the thrust is a maximum (a local maximum) the moment (of ARCH-RIBS 473 stress couple) is either a maximum or a minimum and the shear is zero. .From the moment, Hz = &?, & ~ may be computed. From the thrust = Fp lt ^ 1 = th ^ st > (F = area of cross-section) may be computed. Hence the greatest compression per sq. inch (^i+/> 2 ) may be found in each section. A separate stress-diagram might be con- structed for this quantity (PI+PI). Its max. value (after adding the stress due to change of temperature, or to rib- shortening, for ribs of less than three hinges), wherever it occurs in the rib, must be made safe by proper designing of the rib. The maximum shear J m can be used as in 256 to determine thickness of web, if the section is I-shaped, or box-shaped. See 295. 383. Temperature Stresses. In an ordinary bridge truss and straight horizontal girders, free to expand or contract longitudinally, and in Classes A and B of 378 of arch- ribs, there are no stresses induced by change of tempera- ture ; for the form of the beam or truss is under no constraint from the manner of support ; but with the arch- rib of two hinges (hinged ends, Class C) and of fixed ends (Class D) having immovable piers which constrain the dis- tance between the two ends to remain the same at all tem- peratures, stresses called " temperature stresses " are in- duced in the rib whenever the temperature, t, is not the same as that, t tl , when the rib was put in place. These may be determined, as follows, as if they were the only ones, and then combined, algebraically, with those due to the loading. 384. Temperature Stresses in the Arch-Rib of Hinged Ends. (Class C, 378.) Fig. 434. Let E and /be constant, with 474 MECHANICS OF ENGINEERING. ,rt o, H n , other postulates as in 379. S^~"~~'\^\ ^ e * ^ = temperature of /f y _ z \ \\ erection, and I = any other O//H H \B temperature; also let I = go "S length of span = OB (in- Fio"~434~~* variable) and r^ co-efficient of linear expansion of the material of the curved beam or rib (see 199). At tempera- ture t there must be a horizontal reaction H at each hinge to prevent expansion into the form O'B (dotted curve), which is the form natural to the rib for temperature t and without constraint. We may /. consider the actual form OB as having resulted from the unstrained form O'B by displacing 0' to 0, ie., producing a horizontal displace- ment O'O =1 (t-tjft. But O'O = Ax (see 373 and 374) ; (KB. 'a tangent has moved, but this does not affect Ax, if the axis X is horizontal, as here, coinciding with the span ;) and the ordinate y of any point ra of the rib is identical with its 8 or intercept between it and the spec, equil. polygon, which here consists of one segment only, viz. : OB. Its force diagram consists of a single ray 0\ n' ; see Fig. 434. Now ( 373) Ax = CMyds ; and M=Hz = in this case, Hy , ,. . x H / B 2 T ( hence for graphics, and .-. I (t-Q y= ffds ; \ , X * , ' e . . (1) From eq. (1) we determine //, having divided the rib-curve into from twelve to twenty equal parts each called Js . For instance, for wrought iron, t and , being expressed in Fahrenheit degrees, y = 0.0000066. If E is expressed in Ibs. per square inch, all linear quantities should be in inches and H will be obtained in pounds. 2ly* may be obtained by 375, or may be computed. H being known, we find the moment of stress-couple = Hy t ARCH-RIBS. 475 at any section, while the thrust and shear at that section are the projections of H, i.e., of O^n' upon the tangent and normal. The stresses due to these may then be determined in any section, as already so frequently explained, and then combined with those due to loading. 385. Temperature Stresses in the Arch-Ribs with Fixed Ends. See Fig. 435. (Same postulates as to symmetry, E and 1 constant, etc., as in 380.) t and t have the same meaning as in 384. Here, as before, we consider the rib to have reached its ac- tual form under tem- perature t by having had its span forcibly , shortened from the _AJ length natural to "^ * FiiTtts temp, t, viz.: O'B', to the actual length OB, which the immovable piers compel it to assume. But here, since the tangents at and B are to be the same in direction under constraint as before, the two forces H, representing the action of the piers on the rib, must be considered as acting on imaginary rigid prolonga- tions at an unknown'distance d above the span. To find H and d we need two equations. From 373 we have, since M=Hz=H(y-d), . (2) or, graphically, with equal ^/s's ' (3) Also, since there has been no change in the angle between end-tangents, we must have, from 374, 476 XECHAKICS or ENGINEERING. or for graphics, with equal Av's, -ly = nd . . . (4) in which n denotes the number of Js's. From (4) we determine d, and then from (3) can compute H. Drawing the horizontal F G, it is the special equilibrium polygon (of but one segment) and the moment of the stress-couple at any section = Hz, while the thrust and shear are the projections of H= 0\n' on the tangent and normal respect- ively of any point m of rib. For example, in one span, of 550 feet, of the St. Louis Bridge, having a rise of 55 feet and fixed at the ends, the force H of Fig. 435 is = 108 tons, when the temperature is 80 Fahr. higher than the temp, of erection, and the en- forced span is 3j^ inches shorter than the span natural to that higher temperature. Evidently, :.f the actual temp- erature t is lower than that t ia of erection, Hmnst act in a direction opposite to that of Figs. 435 and 434, and the *' thrust " in any section will be negative, i.e., a pull. 386. Stresses Due to Rib-Shortening In 369, Fig. 407, the shortening of the element AE to a length A'E, due to the uniformly distributed thrust, p\F, was neglected as pro- ducing indirectly a change of curvature and form in the rib axis ; but such will be the case if the rib has less than three hinges. This change in the length of the different portions of the rib. curve, may be treated as if it were due to a change of temperature. For example, from 199 we ee that a thrust of 50 tons coming upon a sectional area of F = 10 sq. inches in an iron rib, whose material has a modulus of elasticity = E = 30,000,000 Ibs. per sq. inch, and a coefficient of expansion # = .0000066 per degree Fahrenheit, produces a shortening equal to that due to a fall of temperature (t t) derived as follows : (See 199) (units, inch and pound) ft -t}= P - 100 ' 000 -50 10x30,000,000x.0000066~' Fahrenheit. Practically, then, since most metal arch bridges of cla-5S3s C and D are rather flat in curvature, and the thrusts AKCH-KIBS. 477 due to ordinary modes of loading do not vary more than 20 or 30 per cent, from each other along the rib, an imagin- ary fall of temperature corresponding to an average thrust in any case of loading may be made the basis of a con- struction similar to that in 384 or 385 (according as the ends are hinged, or fixed) from which new thrusts, shears, and stress-couple moments, may be derived to be combin- ed with those previously obtained for loading and for change of temperature. 387 Resume It is now seen how the stresses per square inch, both shearing and compression (or tension) may be obtained in all parts of any section of a solid arch-rib or curved beam of the kinds described, by combining the re- sults due to the three separate causes, viz.: the load, change of temperature, and rib-shortening caused by the thrusts due to the load (the latter agencies, however, com- ing into consideration only in classes C and D, see 378). That is, in any cross-section, the stress in the outer fibre is, [letting T } ,(, T^ f , T h '", denote the thrusts due to the three causes, respectively, above mentioned ; (Hz)', (Hz)", (Hz)'", the moments] (B*)" (Ifr)'" . . . (1) Le., Ibs. per sq. inch compression (if those units are used). The double signs provide for the cases where the stresses in the outer fibre, due c to a single agency, may be tensile. Fig. 436 shows the meaning of e (the same used heretofore) /is the moment of in- ertia of the section about the gravity axis (horizontal) C. F = area of cross- section. [0! = e ; cross section symmet- rical about C]. For a given loading we may find the maximum stress in a given rib, or design the rib so that this maximum stress shall be safe for the ma- terial employed. Similarly, the resultant shear (total, not 478 MECHANICS OF ENGINEERING. per sq. inch) = J' J" J'" is obtained for any section to compute a proper thickness of web, spacing of rivets, etc. 388 The Arch-Truss, or braced arch.. An open-work truss, if of homogeneous design from end to end, may be treated as a beam of constant section and constant moment of inertia", and if curved, like the St. Louis Bridge and the Coblenz Bridge (see 378, Class D), may be treated as an arch-rib.* The moment of inertia may be taken as where F Y is the sectional area of one of the pieces (I to the curved axis midway between them, Fig. 437, and h = dis- tance between them. FIG. 438. FIO. 437. Treating this curved axis as an arch-rib, in the usual way (see preceding articles), we obtain the spec, equil. pol. and its force diagram for given loading. Any plane ~] to the rib-axis, where it crosses the middle m of a " web- member," cuts three pieces, A, B and C, the total com- *The St Louis Bridge u not strictly of constant moment of Inertia, being somewhat strengthened near each pier ARCH-RIBS. 479 pressions (or tensions) in which are thus found : For the point ra, of rib-axis, there is a certain moment = Hz, a thrust r h , and a shear = J, obtained as previously ex- plained. We may then write Psin/3 = J (1) and thus determine whether P is a tension or compres- sion ; then putting P'+P" P cos ft = T h 2 (in which P is taken with a plus sign if a compression, and minus if tension); and (3) we compute P 1 and P", which are assumed to be both com- pressions here. /? is the angle between the web member and the tangent to rib-axis at m, the middle of the piece. See Fig. 406, as an explanation of the method just adopted. HORIZONTAL STRAIGHT GIRDERS. 389, Ends Tree to Turn. This corresponds to an arch- rib with hinged ends, but it must be understood that there is no hindrance to horizontal motion. (Fig. 439.) In Fio 439. treating a straight beam, slightly bent under vertical forces only (as in this case with no horizontal constraint), as a 480 MECHANICS OF ENGINEERING. particular case of an arch-rib, it is evident that since the pole distance must be zero, the special equil. polygon will have all its segments vertical, and the corresponding force diagram reduces to a single vertical line (the load line). The first and last segments must pass through A and B (points of no moment) respectively, but being vertical will not intersect P t and P 2 ; i.e., the remainder of the special equilibrium polygon lies at an infinite distance above the span AE. Hence the actual spec, equil. pol. is useless. However, knowing that the shear, J, and the moment M (of stress couple) are the only quantities pertaining to any section ra (Fig. 439) which we wish to determine (since there is no thrust along the beam), and knowing that an imaginary force H' y applied horizontally at each end of the beam, would have no influence in determining the shear and moment at m as due to the new system of forces, we may therefore obtain the shears and moments graphically from this new system (viz.: the loads P lt etc., the vertical reactions V and V M and the two equal and opposite /T's). [Evidently, since H 1 ' has no moment about the neutral axis (or gravity axis here), of m, the moment at m will be unaffected by it ; and since H" has no component ""] to the beam at m, the shear at m is the same in the new system of forces, as in the old, before the introduction of the tf"s.] Hence, lay off the load-line 1 . . 2 . . 3, Fig. 439, and con- struct an equil. polyg. which shall pass through A and B and have any convenient arbitrary H" (force) as a pole distance. This is dona by first determining n on the load- line, using the auxiliary polygon A'a'F, to a pole (7 (arbi- trary), and drawing O'n 1 \\ to A'B'. Taking O 1 ' on a hori- zontal through n, making 0''n'=H", we complete the force diagram, and equil. pol. AaB. Then, z being the ver- tical intercept between m and the equil. polygon, we have: Moment at m~M=H"z (or=ITz' also), and shear at m, or /, 2 . . n' t i.e., projection of the proper ray J? 2 > or 0" . . 2, upon the vertical through m. Similarly we ob- tain M and J at any other section for the given load. (See ARCH-KI13S ; SPECIAL CASE ; STRAIGHT. 481 329, 337 and 367). The moment of inertia need not be constant in this case. 390. Straight Horizontal Prismatic Girder of Fixed Ends at Same Level. No horizontal constraint, hence no thrust. / con- stant. Ends at same level, with end-tangents horizontal. We may consider the whole beam free (cutting close to the walls) putting in the unknown upward shears J and J w i 2 and the two stress couples of unknown moments JH and M,, at these end sections. Also, as in 388, an arbitrary H" horizontal and in line of beam at each extremity. Now (See Fig. 33) the couple at and the force H" are equiv- alent to a single horizontal H" at an unknown vertical dis- tance c below ; similarly at the right hand end. The special polygon FG is to be determined for this new sys- tem, since the moment and shear will be the same at any section under this new system as under the real system. The conditions for determining it are as follows : Since the end-tangents are fixed, -Mds=Q .*. JTJ//s=0 and since 482 MECHANICS OF ENGINEERING. O's displacement relatively to J5's tangent is zero we nave 2Mxds=0 .-. ZH"zxAs=0 .'. 2'xzJs=0. See 374. Hence for Equal 4s\ l'(z)=Q and I(xz)=0. Now for any pole 0'" draw an equil. pol. F'"G'" and in it (by 377; see Note) locate v'"m,'" so as to make 2'(3"')=0 and 2"(2*'")=0. Draw verticals through the intersections E'" and L"' t to determine E and Z on the beam, these are the points of inflection (i.e., of zero moment), and are points in the re- quired special polygon FG. Draw 0"'n" II to v'"m'" to fix n". Take a pole 0" on the horizontal through n", making / 'y,"=H" (arbitrary), draw the force diagram 0" 1234 and a corresponding equilibrium polygon beginning at E. It should cut L, and will fulfil the two requirrnents 2*(z)=Q and ~*(xz)=Q, with reference to the axis of the beam O'B'. The moment of the stress-couple at any section m will be M=H"z, and the shear J = the projection of the " proper ray " of the force diagram 0" . . 1, 2, etc., upon the vertical (not in the trial diagram 0'". . 1, 2, etc.). As far as the moment is concern- ed the trial polygon F'" G'" will serve as well as the special polygon FG ; i.Q.,M=H'"z'" as well as H"z t H'" being the pole-distance of 0'" ; but for the shear we must use the rays of the final and not the trial diagram. The peculiarity of this treatment of straight beams, considered as a particular case of curved beams, consists in the substitution of an imaginary system of forces in- volving the two equal and opposite, and arbitrary H's, for the real system in which there is no horizontal force and consequently no " special equilibrium polygon," and thus determining all that is desired, i.e., the moment and shear at any section. In the polygon FG the student will recognize the " mo- ment-diagram " of the problems in Chaps. Ill and IV. He will also see why the shear is proportional to the slope - of the moment curve in those chapters. For ax example, the " slope " of the second segment of the poly- gon FG, that segment being II to 0" 2, is ARCH-KIBS ; SPECIAL CASE ; STRAIGHT. 483 tang, of angle 20'V'=2n"-f-0'V=shear -r- H" and similarly for any other segment ; i.e., the tangent of the inclination of the " moment curve," or line, is propor- tional to the shear. It is also interesting to notice with the present problem of a straight beam, that in the conditions Jf(sJs)=0 and 2(zAs)x=Q, for locating the polygon FG, each 4s is 1 to its z, and that consequently each zAs is the area of a small vertical strip of area between the beam and the polygon, and (zAs)x is the " moment" of this strip of area, about 0' the origin of x. Hence these conditions imply ; first, that the area EWL between the polygon and the axis of the beam on one side is equal to that (CfFE+LB'G) on the other side, and, secondly, that the centre of gravity of EWL lies in the same vertical as that of O'FE and LB' G combined. Another way of stating the same thing is that, if we join FG, the area of the trapezoid FO'B 1 G is equal to that of the figure FEWLG, and their centres of gravity lie in the same vertical. A corresponding statement may be made (if we join F"'G"'^ for the trapezoid F"'v'"m"'G"\ and figure 484: MECHANICS OF EKGLNEEKIXG. CHAPTER XII. GRAPHICS OF CONTINUOUS GIRDERS. [MAINLY DUE TO PROF. MOHR. OF AIX-LA-CHAPELLE ] 391. The Elastic Curve of a Horizontal Loaded Beam, Homoge neous and Originally Straight and Prismatic, is an Equilibrium Polygon, whose "load-line" ixl20x40x80=192,000 (sq. in.) (Ibs.), the scale of our "moment-area-diagram," as we may call it, must be 192,000^6.4=30,000 (sq. in.) (Ibs.) per linear inch of paper. Now JET=21,333,333 (sq. in.) (Ibs.), which on the above scale would be represented by 711 linear inches of paper. With tt=100, however, we lay off ST=EI+n = 7.1l inches of paper as a pole distance, and with a trial pole 0" in 490 MECHANICS OF ENGINEERING. the vertical TW draw the trial equilibrium polygon or elastic curve A"B", and with it determine n', then the final polygon A'B' as already prescribed. In A'B' we find the greatest ordinate, NK, to measure 0.88 inches of paper, which represents an actual distance of 0.88 x 20 = 17.6 inches But the vertical dimensions of the exaggerated elastic curve A'B' are rc=100 times as great as those of the actual, hence the actual max. deflection is d=17.6-^w=0.176 in. [This maximum deflection could also be obtained from the oblique polygon A"B" whose vertical dimensions are equal to those of A'B'. By the formula of 235 we ob- tain d=0.174 inches.] 395. Direction of End-Tangents of Elastic Curve in the Foregoing Problem. As an illustration bearing on subsequent work let us suppose that the only result required in 394 is tan , i.e., the tangent of the angle B'A'T', which the tangent-line B'T' to the elastic curve at the extremity A\ Fig. 343, makes with the horizontal line A'B', (tan. is called the "slope," at A.} Let 7?'$' be the tangent-line at B'. These two " end-tangents " are parallel respectively to EO' and FO', and intersect at some point R. Now since A' KB' is an equilibrium polygon sustaining an imaginary set of loads represented by the successive vertical strips of the moment-area ACB, the intersection It must lie in the vertical containing the centre of gravity, U, of that mo- ment-area [ 336 j . Hence, if the vertical containing U is known in advance, or, as in the present case, is easily constructed without making the strip-subdivision of 394, we may determine the end-tangents very briefly by considering the whole moment-area, M.A., (considered as a load) applied in the vertical through U t as follows : Since ACB is a triangle, we find U by bisecting AB in X, joining CX, and making XU ^ XC, and then draw a vertical through U. Laying off EF=6A inches [so as to represent a moment-area of 192,000 (sq. in.) (Ibs.) on a scale of 30,000 (sq. in.) (Ibs.) per linear inch of paper], CONTINUOUS GIRDER BY GRAPHICS. 491 and making ST=7.ll inches as before, we assume a trial pole 0" on TW, draw the two rays 0"E and 0"F, construct the corresponding trial polygon of two segments A"R"B", for the purpose of finding n'. With a pole 0' on TW&nd on a level with n' we draw the two rays O'E and O'F, and the corresponding segments A'R, and RB'. (B' should be on a level with A', as a check.) These two segments are the end-tangents required. We have, therefore, tan = 'B T T r -=- AtW^ In the present numerical problem we find B'T' to mea- sure 3 in. of paper, i.e., 60 in. of actual distance for the exagg. elastic curve, and therefore 0.60 in. in the real elas- tic curve (with n = 100) It is now evident that the position and direction of the end-tangents of the elastic curve lying beticeen any two sup- ports are independent of the mode of distribution of the moment-area so long as the amount of that moment-area and the position of its centre of gravity remain unchanged. This relation is to be of great service. 396. Re- Arrangement of the Moment-Area. As another illus- tration conducing to clearness in later constructions, let us determine by still another method the end-tangents of the beam of 394 and 395. See Fig. 444. As already seen, their location is independent of the arrangement of the moment-area between. Let us re-arrange this moment- area, viz., the triangle ACB, in the following manner : By drawing AX parallel to BC, and prolonging BCio P in the vertical through A, we may consider the original moment-area ACB to be compounded of the positive mom.- area VBXA, a parallelogram, with its gravity-vertical passing through D, the middle of the span ; of the negative mom.-area VCA, a triangle whose gravity-vertical passes 492 MECHANICS OF ENGINEERING. through A making ADi=% AD; and of another negative mom. -area, the triangle ABX, whose gravity-vertical passes through D 3 at one-third the span from B. That is, the (ideal) positive load A CB is the resultant of the positive load (M.A.) 2 and the two negative loads (or upward pulls) (M.A.\ and (M.A.) 3 , and may therefore be replaced by them without affecting the location of the end-tangents, at A and B, of the elastic curve AB. These three moment- areas are represented by arrows, properly directed, in the figure, but must not be confused with the actual loads on the beam (of which, here, there is but one, viz., P). From the given shapes and dimensions, since ACB = 192,000 (sq. in.) (Ibs.), we easily derive by geometrical principles : (M.A.\= +576,000 (sq. in.) (Ibs.) (M.A.\= - 96,000 = -288,000 Hence, with a pole distance El -4- n = 7.11 in. as before, and a " moment-load-line " formed of 1'2' = (M.A.\, (on scale of 30,000 (sq. in. Ibs.) to one inch) 2'3' = (M.A.\, and 3'4' (M.A.) 3 , first with a trial pole 0", construct the trial CONTINUOUS GIRDER BY GRAPHICS. 493 polygon A"B", and find n' in usual way ( 337) ; then take a pole 0' on the horizontal through n' and the vertical TW, and draw the new polygon A' 123 B'. It should pass through B' on a level with A', and A'\ and B'3 are the required end-tangents (of the exagg. elastic curve). [NOTE If B' were not at the same level as A', but (say) 0.40 in. below it, B' should be placed at ra, a distance = 2 inches (on the paper) below its present posi- tion, (since the distance scale is 1:20 and n = 100, in this case) and the " abutment -line " of final polygon would be A'm~\. Of course, this special re-arrangement of the moment- area is quite superfluous in the present problem of a dis- continuous girder (not built in), but considerations of this kind will be found indispensable with the successive spans of a continuous girder. 397. Positive and Negative Moment- Areas in Each Span of a Continuous Girder (Prismatic). In the foregoing problem of a discontinuous girder (covering one span only) not built in at the ends (otherwise it would be classed among continuous girders), the moment-curve, or equilibrium polygon of arbitrary H, is easily found by 389 without the aid of the elastic curve, and the end moments are both zero ; (i.e., the moment-curve meets the beam in the end- verticals) but in each span of a continuous girder the end-moments are not zero (necessarily), and the points in the end-verticals where the moment-curve must terminate (for an assumed H) can not be found without the use of the elastic curve (or of some of its tangents) of the whole beam, dependent, as it is, upon the loading on all the spans, and the heights of the supports. Let Fig. 445 show, in general, any one span of a pris- matic continuous girder (prismatic ; hence /is constant), between two consecutive supports A and B . P lt P 2 , etc., are the loads on the span. [If the displacement of A n relatively to the end-tangent at J? , and the angle between the end-tangents (of elastic 494 MECHANICS OF ENGINEERING. curve) were known, the. moment-curve or equilibrium polygon FWG, (AS being the axis of beam) might be found by a process similar to that in 390, but the elastic curves in successive spans are so inter-dependent that the above elements can not be found directly.] I" fli ' L_LJ-K We now suppose, for the sake of discussion, that the whole girder has been investigated (by a process to be presented) for the given loads, spans, positions of supports, etc., and then the moment-curve FEWLG found, with the corresponding force-diagram, for the span in the figure and some arbitrary H. The horizontal line AB represents the axis of the beam (for this purpose considered straight and horizontal) as an axis from which to measure the moment ordinates. _Thus, the moment (of the stress-couple) at A is = H X AF\ at B, Hx BG ; at E and L, zero (points of inflection). Now, according to the usual conceptions of analytical geometry, we may consider the portion EWL, of the mo- ment-area, above AB as positive, and those below, AEF and LB G, as negative ; but since not one of these three CONTINUOUS GIRDER BY GRAPHICS. 495 areas, nor the position of its gravity-vertical, is known in advance, since they are not independent of the other spans, a more advantageous re -arrangement of the moment- area may be made thus : Join FG and FB, and we may consider the original moment-area replaced by the following three component areas: the positive moment-area FE WLGF (shaded by ver- tical lines); the negative triangular moment-area AFB\ and the negative triangular moment-area BFG (the nega- tive moment-areas being shaded by horizontal lines). (In subsequent paragraphs, by positive and negative moment- areas will be implied those just mentioned.) These three moment-areas, treated as loads, each applied in its own gravity -vertical, and considered in any order, may be used instead of the real distributed moment-area, as far as determining, or dealing with, the end-tangents of the elastic-curve at A and B is concerned ( 396), and the fol- lowing advantages will have been gained : (1.) The amount of the positive moment-area, (M.A.\ in Fig. 445, (depending on the area lying between the polygon FEWG and the abutment-line FG of the latter), and the position of its gravity -vertical) are independent of other spans, and can be easily found in advance, since this moment-area and gravity -vertical are the same as if the part of the beam covering this span were discontinuous and simply rested on the supports A and B (> , as in 389. (2.) The gravity-vertical of the left-hand negative mo- ment-area, (M.A.\, is always one-third the span from the left end-vertical, A A'; that of the other, (M.A.\, an equal distance from the right end-vertical, BB'. (3.) The two (right and left) negative moment-areas are triangular, each having the whole span I for its alti- tude, and for its base the intercept AF(or BG) on which the end-moment depends. Hence, if the amounts of thesa negative moment-areas have been found in any span, we may compute the values which AF and BG must have for a given H, and thus determine the terminal points F and G of the moment-curve of that span (for that value of H\ 496 MECHANICS OF EXGINEEKIXG. For example, if (M.A.\ has been found (by a process not yet given) to be 160,000 (sq. in.) (Ibs.) while AB I - 160 in., then the moment M x which AF represents, is computed from the relation (M.A.\ = y 2 ~AB x M^ If IT has been chosen = 100 Ibs. we put HxAF = 2000 and obtain AF = 20 inches of actual distance, so that with a scale of 1:20 for distances AF would be one linear inch of paper. (Of course, in computing BG the same value of H must be used.) With H= 100 Ibs., then, and F and G as known points of the equilibrium polygon FEW G, it is easily drawn by the principles of 341. We thus notice that the amounts of the two negative moment-areas are the only elements affected by the con- tinuity of the girder, in this re-arrangement of the actual moment-areas. In the lower part of Fig. 445 A' and B' represent the extremities of the (exagg.) elastic curve, the vertical dis- tance B'B"', of B' from the horizontal through A' (in case the two supports A and B n are not at same level, as we here suppose for illustration) being laid off in accordance with the principles of the note in 396. NOTE. It is now evident that if the "false polygon " (as it will be called) A'I23B f has been obtained (and means for doing this will be given later) in which the first and last segments are the end tangents of the (exagg.) elastic curve, and which bears the same relation to the three moment areas just mentioned, as that illustrated in Fig. 444, we may proceed further to determine the amounts of (M.A.) l and (M.A.\ as follows, by completing the moment-area dia- gram : Having laid off the known (M.A.\ (or positive moment- area) =2'3', and ST=EI-n, a line parallel to 12 drawn through 2', determines the pole 0', through which paral- CONTINUOUS GIKDEB BY GRAPHICS. 497 leis to A'l and B'3 will fix 1' and 4' on the vertical SV, and thus determine I'2=(M.A.\ and 3''=(M.A.} 3 . Their numerical values are then computed in accordance with the scale of the moment-area diagram. The polygon A'Vl^B' will be called the "false polygon " of the span in question, its end-segments being the end- tangents of the elastic curve. 398. Values of the Positive Moment-Area in Special Cases. For several special cases these are easily computed, and as an illustration, Fig. 446 shows a continuous girder, AF of five spans, all six supports on a level, and the weight of the beam neglected. At the extremities A and F, as at the other supports, the beam is not built in, but simply touches each support in one point ; hence the moments at A and F are zero, i.e., the moment curve must pass through A and F t so that in the first span the left negative mo- ment-area, and in the fifth span the right negative mo- ment-area, are zero. The positive moment-areas are shaded. On the first span is placed a uniformly distributed load W over the whole span I. .-. the positive moment-area for that span is the same as in the case of Fig. 235 [see (1) 397] and being represented by a parabolic segment whose area is two-thirds that of the circumscribing rec- tangle, its value is 498 MECHANICS OF ENGINEERING. (M.A.}' l = %.rtWl'xl'= l /uWT* . . (1) [see eq. (2) 242], while its gravity- vertical bisects the span. The only load on the second span is a concentrated one, P", at distances h" and l^ from the extremities of the span; hence the positive moment-area is triangular and has a value (M.*X'=#P"W . . . (2) as in 393. Its gravity vertical may easily be constructed as in Fig. 443 [see (1) 397]. The third span carries no load ; hence its positive mo- ment-area is zero, and the actual moment-area is composed solely of the two triangular negative moment-areas CDH and DHI t the moment-curve consisting of the single straight line HI. The fourth span carries a uniform load W IV =tu lv l lv t and .-. has a positive moment-area (M.4.J?=yuW"(F? ... (3) as in eq. (1), acting through the middle of the span (grav- ity-vertical). Since the fifth and last span carries no load, its positive moment-area is zero, the moment curve being the straight line JF, so that the actual moment-area is composed of the left-hand negative moment-area. At F it is noticeable that the reaction or pressure of the support must be from above downward to prevent the beam from leaving the point F\ i.e., the beam must be " latched down," and the reaction is negative. If the beam were built in at A (or F) the moment at that section would not be zero, hence the left (or right) neg. moment-area would not be zero in that span, as in our present figure. But in such a case the tangent of the elastic curve would have a knotcn direction at A (or F) and the problem would still be determinate as will be seen. CONTINUOUS GIRDER BY GRAPHICS. 499 A' , . . F' gives an approximate idea (exaggerated) of the form of the elastic curve of the entire girder. A change in the loading on any span would affect the form of this curve throughout its whole length as well as of all the moment curves. NOTE. It is important to remark that any two of the triangular negative moment-areas which have a common base (hence lying in adjacent spans) are proportional to their altitude i.e., to the lengths of the spans in which they occur ; thus the neg. mom.-areas (Fig. 446) GCH&udDCH have a common base CH, = r_ } r (The notation explains itself ; see figure.) It also follows, that the resultant of these two neg. moment-areas (if re- quired in any construction ; see 400) acts in a vertical which divides the horizontal distance between their gravity ver- ticals in the inverse ratio of the spans to which they belong [ 21 and eq. (4) above]. Hence, since this horizontal distance is /^"~J~K^ their resultant must act in a vertical Y'", whose distance from the gravity-vertical of GCH is %l'", and from that of CHD, ftl. 399. Amount and Gravity-Vertical of the Positive Moment Area of One Span as Due to Any Loading. Since we can not deal directly with a continuous load by graphics, but must subdivide it into a number of detached loads sufficiently numerous to give a close approximation, let us suppose that this has already been done if necessary, and that P 1} P 2 , etc., are the detached loads resting in the span AB in question ; see Fig. 447. Since [by (1), 397] the positive moment-area is the same as the total moment-area would be if this portion of the beam simply rested on the extremities of the span, not extending beyond them, we may use the construction in 389 for finding it, remembering that in that paragraph the 500 MECHANICS OF ENGIXEEKING. oblique polygon in the lower part of Fig. 439 will serve as well as the (upper) one whose abutment -line is the beam itself, as far as moments are concerned. Hence, Fig. 447, lay off the load-line LL', take any pole 0, with any convenient pole-distance H, and draw the equilibrium polygon FWG. After joining FG, FWGF will be the positive moment-area required. To find its gravity vertical, divide the span AB, or FG', into from ten to twenty equal parts (each As] and draw a FIG. 447. vertical through the middle of each. The lengths z,, 2 etc., on these verticals, intercepted in the moment-area, are proportional to the corresponding strips of moment- area, each of width = Js, and of an amount = Hzds. Form a load line, SK, of the successive a's, and with any pole 0', draw the equilibrium polygon A'E' (for the z-verticals). The intersection, R, of the extreme segments, is a point in the required gravity -vertical ( 336). The amount of the moment-area is (M,A.\= 2[Hzds] CONTINUOUS GIRDER BY GRAPHICS. 501 For example, with the span =Z=120 in., subdivided into twelve equal ^/s's, we have Js= 10 inches (of actual distance). If H=4 inches of paper and SK= J()=10.2 inches of paper, the force-scale being 80 Ibs. to the inch, and the distance- scale 15 inches to the inch (1:15), we have (M.A. ) 2 = [4 x 80] x 10 x [10.2x15] =489600. (sq. in.)(lbs.) 400. Construction of the "False-Polygons" For All the Spans of a Given (Prismatic) Continuous Girder, Under Given Loading, and With Given Heights of Supports. [See note in 397 for meaning of " false-polygon ". j Let us suppose that the given girder covers three unequal spans, Fig. 448, with supports at unequal heights, and that both extremities A and D are built-in, or " fixed," horizontally. To clear the ground for the present construction, we suppose that, from the given loading in each span, the positive moment-area of each span has been obtained in numerical form [so many (sq. in.) (Ibs.) or (sq. in.) (tons)] and its gravity-ver- tical determined by 398 or 399 ; that the horizontal distances (i.e., the spans I', I", and I" and the distance be- tween the above gravity-verticals and the supports) have been laid off on some convenient scale ; that El has been computed from the material and shape of section of the girder and expressed in the same units as the above mo- ment-areas ; that a convenient value for n has been se- lected (since El-r-n is to be the pole distance of all the moment-area-diagrams), and that the vertical distances of B, C, and D, from the horizontal through A, have been laid off accordingly (see note in 396). In the figure (448) verticals are drawn through the points of support ; also verticals dividing each span into thirds, since the unknown negative moment-areas (sub- scripts 1 and 3) act in the latter ( 397) ; and the gravity- verticals of the known positive moment-areas. The ver- ticals Y', Y", V", and V"', are to be constructed later. The problem may now be stated as follows : Given the positions of the supports, the value of El and n, 502 MECHANICS OF ENGINEERING. the fact that the girder is fixed horizontally at A and D, tht heights of supports, the location of the gravity-verticals of all the positive and negative moment-areas, and the amounts of the positive moment-areas ; it is required to Jind graphically the " false-polygon" in each span. The " false-polygons," viz.: A 123 B for the first span (on the left), B 123 C for the second, etc., are drawn in the figure Pis. 448. for the purpose of discussing their properties at the out- set. Since SB and BI are both tangent to the elastic curve at B, they form a single straight line ; similarly 01 is but the prolongation of 3(7. Also Al and 3Z> must be hori- zontal since the beam is built in horizontally at its ex- tremities A and D. That is, the three false polygons form a continuous equilibrium polygon A . . . D, in equilibrium under the "loads" (M.A.y, (KA.) 2 ' t (M.AX, etc., so that we might use a single mom. -area-diagram in con- nection with it, but for convenience the latter will be CONTINT'OUS G1KDER BY GRAPHICS. 503 drawn in portions, one under each span, with a pole dis- tance = il. n Of this polygon A ... D, we have the two segments A\ and 3Z> already drawn, and know that it passes through the points B and C ; we shall next determine by construc- tion other points (called " fixed points ") p ', p", p ", p'", and PQ'" in (the prolongations of) certain other segments. To find the "fixed point " p^', where the segment 23 in the first span cuts the vertical V, the gravity -vertical of (M.A.\ f . The vertex p', or 1, is already known, being the intersection of Al with V. Lay off 2' 3' = (M.A.\ r which is known, and take a trial pole O t 'with a pole-distance El-^n ; join t ' 2' and O t ' 3'. Draw 12 || to 2'0 t to determine 2o on the vertical (M. A.\' t then through 2 a line || to 0/3' to cut V in PQ. The unknown segment 23 must cut V in the same point ; since all positions of O/ on the vertical T'U' will result in placing^/ in this same point, and one of these positions must be the real pole 0' (unknown). [This is easily proved in detail by two pairs of similar triangles]. To determine the "fixed-point " p", in the prolongation of segment 12 of second span. The prolongations of the segments 12 (of first span) and 12 (of second span) must meet in a point k in the (vertical) line of action of the resultant of (M. A.)J and (M. A.\" ( 336). Although the amounts of (M. A.) 3 ' and (M. A.\" are unknown, still the vertical line of action of their resultant (by 398, Note) is I" known to be Y', a horizontal distance to the right of o .y ; hence Y f is easily drawn. Therefore, the unknown triangle ^31 has its three vertices on three known verti- cals, the side '3 passes through the known point p ', and the side 13 through the known point B. Now by pro- longing PQ 2o (or any line through p ') to cut (M. A.\' and T in 3 and &/> respectively, joining 3 i? and prolonging this line to cut (M. A.\" in some point 1 , and then joining ko' 1 , we have a triangle & '3 1 of which we can make a 504 MECHANICS OF ENGINEERING. statement precisely the same as that just made for k' 3 1 But if two triangles [as F31 and V 3 1 ] have their vertices on three parallel lines (or on three lines which meet in a point) the three intersections of their correspond- ing sides must lie on the same straight line [see reference to Chauvenet, 378 a]. Of these intersections we have two,^9 ' and B ; hence the third must lie at the intersection of the line p f B (prolonged) with k ' 1 , and in this way the "fixed point" p", a point in &'l and .'. in the segment 12 (of second span; prolonged, is found. Draw a vertical through it and call it V". The fixed point p " (in prolongation of segment 23 of sec- ond span) lies in the vertical V" and is found from p" and tne known value of (M.A.) 2 " precisely &sp f was found from p'. That is, we lay off vertically 2" 3" = (M.A.) a ", and join 2" and 3" to O t ", which is any point at distance El n to the right of 2"3". Through /' draw a line || to 2" O t " to cut (M. A.\" in 2 , then 2^" II to O t " 3" to determine p" on the vertical V". The fixed points p'" andp/" in the third span lie in the prolongations of the segments 12 and 23, respectively, of that span, p'" being found from the points p " and C and the verticals (M.A.) 3 ", Y", and (M.A.\'" y in the same manner asp" was determined with similar data, while p '", in the same vertical V" as p'", depends on (M. A.) 2 '" and its gravity vertical as already illustrated ; hence the d,etail need not be given ; see figure. In this way for any number of spans we proceed from span to span toward the right and determine the succes- sive fixed points, until the points p and p Q of the last span have been constructed, which are p'" and p" in our present problem. Since p '" is a point in the segment 23 (prolonged) of the last span, we have only to join it with 3 in that span, a point already known, and the segment 23 is determined. Joining the intersection 2 with p'" we determine the next segment 21 and of coarse the vertex 1, which is then joined with C and prolonged to intersect M. A.\" to fix the segment 1(73 and the point 3. Join CONTINUOUS GIKDEll BY GRAPHICS. 505 3 p ", and proceed in a similar manner toward the left, until the whole equilibrium polygon (or series of " false polygons ") is finally constructed ; the last step being the irrhig of 2 with p'. 401. Treatment of Special Features of the Last Problem. (1.) If the beam is simply supported at A, Fig. 448, instead of built-in, (M. A.)' becomes zero, and the two segments A\ d 12 of that span form a single segment of unknown direction. Hence, the point A will take the place of p 1 , and the vertical F A that of V. (2.) Similarly, if D, in the last span, is a simple support (beam not built in) (M. A.) z r " becomes zero, and the seg- ments D3 and 32 form a single segment of unknown direction, so that after p "' has been found, we join p" and D to determine the segment Z)2 ; i.e., in this last span, D takes the place of 3 of the previous article. (3.) If the first span carries no load (M. A.) a ' is zero, and the segments 12 and 23 will form a single segment 23. Hence if the beam is built in at A, pj will coincide with the known point p' (i.e., 1), while if A is a simple support p and p' coincide with A, since, then, (M. A.) L ' is zero and A 123 is a single segment. (4.) If the last span is unloaded (third span in Fig. 448), (M. A.). 2 '" is zero, 123 becomes a single segment, and hence PQ'" will coincide with p'" ; so that after p"' has been con- structed it is to be directly joined to 3, if the beam is built in at D, and will thus determine the segment 13 ; or to D, if D is a simple support, (for then (M. A.) z "' is zero and the three segments 12, 23, and 3 D form a single seg- ment.) (5.) If an intermediate span is unloaded (say the second span, Fig. 448) the positive mom. -area, (M. A.) 2 ", is zero, 123 becomes a straight line, i.e. a single segment, and therefore p " coincides with p" ; hence, when p" has been found we proceed as if it were po". 402. To Find the Negative Mom. -Areas, the Mom. -Curves, Shears, and Reactions of the Supports. (1.) Having constructed the 506 MECHANICS OF ENGINEERING. false polygons according to the last two articles, the nega- tive moment-areas of each span are then to be found by the note in 397, Fig. 445, and expressed in numerical form. [If the positive inom.-area of the span is zero the points 2' and 3' will coincide, Fig. 445, and in the case mentioned in (3), (or (4)), of 401, if A (or D) were a simple support, in Fig. 448, the mom. -area-diagram of Fig. 445 would have but two rays.] (2.) The moments at the supports (or "end-moments " of the respecthe spans) depending, as they do, directly on the negative mom. -areas, can now be computed as illustrated in (3) 397. The fact that each " end-moment " may be obtained from two negative mom.-areas, separately, one in each adjacent span (except, of course at the extremities of the girder) forms a check on the accuracy of the w.ork. - The two values should agree within one or two per cent. (3.) The " moment-curve " of each span or equilibrium polygon formed from a force-diagram whose load-line con- sists of the actual loads on the span laid off in proper order, can now be drawn, a convenient value for H having been selected (the same Hfor all the spans, that the moment- carves of successive spans may form a continuous line for the whole girder); since we may easily compute the proper moment ordinate at each support to represent the actual moment, then, for. the H adopted, by (3) 397. The moment-curve of each span, since we know its two extreme points and its pole-distance H, is then constructed by 341. (4.) The shear. Since the last construction involves drawing the special force-diagram for each span, with a ray corresponding to each part of the span between two consecutive loads, the shear at any section of the beam is easily found as being the length of the vertical projection of the " proper ray," interpreted by the force-scale of the force-diagram, as in 389 and 390. With the shears as CONTINUOUS GIRDER BY GRAPHICS. 507 FIG. 449. librium, i.e., ordinates a shear -diagram may now be constructed, if desired, for each span. The directions of the shears should be carefully noted. (5.) Reactions of supports. Let us consider " free " the small por- tion of the girder, at each point of support, included between two sections, one close to the support on each side, Fig. 449. Suppose it is the support (7, and call the reaction, or pressure at that sup- port jR c . Then, for vertical equi- 36), we have (6) and, in general, the reaction at a support equals the (algebraic) sum of the two shears, one close to the support ou the right, the other on the left. The meaning of the subscripts is evident. In applying this rule, however, a free body like that in -Fig. 449 should always be drawn, or conceived ; for the two shears are not always in the same direction ; hence the phrase " algebraic sum." At a terminal support, as A or F, Fig. 446, if the beam is not built in, the reaction is simply equal to the shear (since the beam does not overhang) just as in 241 and 243. Fig. 446 presents the peculiarity that the reaction of the support F is negative, (as compared with R c in Fig. 449); i.e., the support at F must be placed above the beam to prevent its rising (this might also be the case at (7, or D t in Fig. 446, for certain relations between the loads). 403. Numerical Example of Preceding Methods. As illustrat- ing the constructions just given, it is required to investi- gate the case of a rolled wrought-iron " I-beam," [a 15- inch heavy beam of the N. J. Steel and Iron Co.,] extend- ing over four supports at the same level, covering three 508 MECHANICS OF ENGINEERING. spans of 16, 20, and 14 feet respectively, and bearing a single load in each of the extreme spans, but a uniform load over the entire central span. As indicated thus : 20ft. 8ft. 6ft. 30 tons 40 tons 32 tons [This is a practical case where W" is the weight of a brick wall, and P' and P'" are loads transmitted by col- umns from the upper floors of the building ; A and D are simple supports, and the weight of the girder is neglected.] The beam has a moment of inertia / = 707 biquad. inches, and the modulus of elasticity of the iron is E = 25,000,000 Ibs. per sq. in., = 12,500 tons per sq. in. Although with a prismatic continuous girder under given loading, with supports at the same level, it may easily be shown that the moments, shears, and reactions, to be obtained graphically, are the same for all values of I, so long as the elastic limit is not surpassed, still, on account of the necessity of its use in other problems (supports not on a leveD. we shall proceed as if the value of I were essential in this one. Selecting the inch and ton as units for numerical work, we have El= 12500 x 707. = 8,837,500 (sq. in.) (tons) while the respective positive mom.-areas, from eqs. (2) and (3) of 398, are : (M. A.\ f =y 2 x 30 X 84 X 108 = 136,080 (sq. in.) (tons) (M. A.\" = l / a X 40 x 240 2 = 192,000 " " (M. A.\'" = % X 32 x 96 x 72 - 110,592 " " Adopting a scale of 60,000 (sq. in.) (tons) to the linear inch of paper, for mom. -area diagrams, we have for the above mom.-areas 2.27 in., 3.20 in., and 1.84 in., respectively, on the paper, for use in Fig. 448. CONTINUOUS GIRDER BY GRAPHICS. 50!) Having laid off the three spans on a scale of 60 inches to the inch of paper, with A, B, C, and D in the same hori- zontal line, we find by the construction of Fig. 443, that the gravity -vertical of (M. A.\' lies 3.6 in. to the left of the middle in the first span, that of (M. A.\'" 4.8 in. to the right of the middle of the third span ; while that of (M. A.) 2 ", of course, bisects the central span. Hence we draw these verticals ; and also those of the unknown neg- ative mom.-areas through the one-third points ; remember- ing [ 401, (1) and (2)] that (M. A.\' and \M. A.)j" are both zero in this case. Since El = 8,837,500 (sq. in.) (tons), it would require 147.29 in. to represent it, as pole-distance, on a scale of 60,000 (sq. in.) (tons) to the inch ; hence let us take n = 50 for the degree of (vertical) exaggeration of the false poly- FT gons, since the corresponding pole-distance = 2.94 in. of paper is a convenient length for use with the values of (M. A.\', (M. A.\", etc., above given. Following the construction of Fig. 448, except that p' is at A, and pj" is to be joined to D ( 401), (the student will do well to draft the problem for himself, using the pre- scribed scales,) and thus determining the false-polygons, we then construct and compute the neg. mom. -areas according to 402 (1), and the note in 397, obtaining the following results : (M. A.\', 1.43 in. of pap., = 85,800 (sq. in.) (tons) (M. A.\" t 1.77 " " " = 106,200 " (M. A.\" t 1.68 " = 100,800 " " (M. A.\" r ,1.17 " " " = 70,200 " " The remaining results are best indicated by the aid of Fig. 450. Following the items of 402, we find [(3) 397] that the moment at B, using (M. A.\" t is Jf.= gX 106 - 200 = 885 inch-tons. 240 in. 510 MECHANICS OF ENGINEERING. [or, using (M. A.\ f , \M B = 2x85 > 800 =893 in. tons.] X7d Similarly, M = 2x ^ >8 - 840 in. tons, [or, using (M. A.\'", M c = 835.7 in. tons.] Hence, taking means, we have, finally, Jf A =0 ; Jf B =889 in. tons ; Jf c =837.8 ; M D =0. Fig. 450 shows the actual mom.-areas and shear-dia grams, which are now to be constructed. FIG. 450. Selecting a value H 20 tons for the pole-distance of the successive force-diagrams, (the scale of distances being 5 ft. (60 in.) to the inch we have [(3) 397] 20 x ~BG = M B = 889 in. -tons .-. BG = 44.4 in. of actual distance, or 0.74 in. of paper ; also 20 x GK = M G = 837.8 in.-tons .'. GK = 41.89 in., or 0.698 in. of paper. Having thus found G and K, and divided BG into ten equal parts, applying four tons in the middle of each, we construct by 341 an equilibrium polygon which shall pass through and TTand have 20 tons as a pole-distance. (We take a force-scale of 10 tons to the inch.) It will form a (succession of short tangents to a) parabola, and is the moment curve for span BO. Similarly, for the single CONTINUOUS GIRDER BY GRAPHICS. 511 loads P' and P'" in the other two spans, we draw the equilibrium polygons AN'G and KZ"'D, for the same H as before, and passing through A and G, and K and D, re- spectively. Scaling the moment -ordinates NN', QQ", and ZZ"\ reducing to actual distance and multiplying by H, we have for these local moment maxima, M^ = 1008, M^ = 336, and M z = 936, in. tons. Evidently the greatest moment is My and .*. the stress in the outer fibre at ^will be ( 239) = lO.Q tons per sq. inch which is much too large. If we employ a 20-inch heavy beam, with / = 1650 biquad. in., the preceding moments will still be the same (supports att at same level) and we have ^=1008^0= 6.06 tons per sq. in., or nearly 12,000 Ibs. per sq. in., and is therefore safe (183). If three discontinuous beams were to be used, the 20- inch size of beam (heavy) would be much too weak, in each of the three spans, as may be easily shown ; hence the economy of the continuous girder in such a case is readily per- ceived. It will be seen, however, that the cases of conti- nuity and of discontinuity do not differ so much in the shear-diagrams as in the moment curves. By scaling the vertical projection of the proper rays in the special force diagrams (as in 389 and 390) we obtain the shear for any section on AN, as 7 AR (see Fig. 449 for notation) = 12.3 tons ; on NB, J BL = 17.7 tons ; from B to C it varies uniformly from 7 BR = 20.3 tons, through zero at Q, to J CIt 19.7 tons of opposite sign. Also, for CZ, J CR = 18.6 tons ; and for ZD, J" DL = 13.4 tons. Hence, the reactions of the supports are as follows : #A=^AR=12.3 tons ; B E =J^+J KK =38.0 tons. J ff r =7 CR +J r CL =38.3 tons ; jR D =7 DIj =13.4 tons. 512 MECHANICS OF ENGINEERING. [In the shear-diagram, the shear-ordinates are laid off bdow the axis when the shear points down, the "free body " extending to the right of the section considered, (as J" CL in Fig. 449) ; and above, when the shear points upward for the same position of the free body.] If we divide the max. shear, 20.3 tons by the area of the web, 13.75 sq. in., of the 20-inch heavy beam, (256), we obtain 1.5 tons or 3000 Ibs. per sq. in., which is < 4000 ( 183). Notice the points of inflection, i', if', etc., where M is zero. Sufficient bearing surface should be provided at the supports. A swing-bridge offers an interesting case of a continu- ous girder, 404. Continuous Girder of Variable Mom. of Inertia. If / is variable and I denote the mom. of inertia of some con- venient standard section, then we may write 7 = 7 -f- w, when m denotes the number of times I contains /. In a non-prismatic beam, m is different for different sections but is easily found, and will be considered given at each section. In eq. (1) of 391, then, we must put IQ -f- m in place of / and thus write d 2 y_[mMdx] ,-v da?" El, ' ^ ' and (pursuing the same reasoning as there given) may therefore say that in a girder of variable section if each small vertical strip (Mdx) of the moment-area be multiplied by the value of m proper to that section, and these products (or "vir- tual mom.-area strips) considered as loads, the elastic curve is an equilibrium polygon for those loads ivith a pole distance = EI Q . In modifying 400 for a girder of variable section, then, besides taking E!Q -j- n as pole distance, proceed as follows : Construct the positive mom.-area for each span accord- ing to 399 ; for each z of Fig. 447, substitute mz (each z CONTIGUOUS GIRDER BY GRAPHICS. 513 having in general its own m), and thus obtain the " virtual positive mom.-area," and its gravity vertical. Similarly, there will be an unknown " virtual neg. mom.- area" not triangular, replacing each neg. mom.-area of 400. Though it is not triangular, each of its ordi- nates equals the corresponding ordinate of the unknown triangular neg. mom. -area multiplied by the proper m, and its gravity -vertical (ichich is independent of the amount of the unknotvn neg. mom.-area) is found in advance by the process of Fig. 447, using, for z's, a set of ordinates obtained thus : Draw any two straight lines AB and FB, Fig. 445, (for a left-hand trial neg. mom.-area; or FB and GFior a right- hand one) meeting in the end-vertical of the span, divide the span into ten or twenty equal spaces, draw a vertical through the middle of each, noting their intercepts between AB and FB. Add these intercepts and call the sum S. Multiply each intercept by the proper m, and with these new values as z's construct their gravity vertical as in Fig. 447. Add these new intercepts, call the sum S v , and denote the quotient S -s- S v by /?. We substitute the three verticals mentioned, therefore, for the mom.-area verticals of 400, and the " virtual pos. mom.-area " for the pos. mom.-area, in each span ; pro- ceed in other respects to construct the "false polygons " according to 400. Then the result of applying the con- struction in the note 397 will be the " virtual neg. mom.- areas," each of which is to be multiplied by the proper ft to obtain the corresponding triangular neg. mom.-area, with which we then proceed, without further modifica- tions in the process, according to (2), (3), etc. of 402. [The conception of these " virtual mom. -areas " is due to Prof. Eddy ; see p. 36 of his " ^Researches in Graphical Statics," referred to in the preface of this work.] 405. Remarks. It must be remembered that any unequal settling of the supports after the girder has been put in place, may cause considerable changes in the values of the moments, shears, etc., and thus cause the actual stresses to be quite different from those computed without taking 514 MECHANICS OF ENGINEERING. into account a possible change in the heights of the sup- ports. See 271. For example, if some of the supports are of masonry, while others are the upper extremities of high iron or steel columns, the fluctuations of length in the latter due to changes of temperature will produce results of the nature indicated above. If an open-work truss of homogeneous design from end to end (treated as a girder of constant moment of inertia, whose value may be formulated as in 388,) is used as a continuous girder under moving loads, it will be subject to " reversal of stress " in some of its upper and lower hor- izontal members, i.e., the latter must be of a proper de- sign to sustain both tension and compression, (according to the position of the moving loads,) and this may disturb the assumption of homogeneity of design. Still, if / is variable, 404 can be used ; but since the weight of the truss must be considered as part of the loading, several assumptions and approximations may be necessary before 3sta Wishing saiisfactory dimensions. INDEX TC i,lEGH4 NIGS OP Abutment-Line 414 Abutments of Arches 430, 435 Ant i -Re suit ant 402 Anti-Stres-Re*ultant 411 Arches ^inear 386, 396 Arches of Masonry 421, 437 Arch-Ribs 438, 483 /vrch-Ribs, Classification of 458 Arch-Rib of Three Hinges 458, 460 Arch-Rib of Hinged Unas 440, 458, 461 Arch-Rib of Fixed Ends 439, 459, 465 Arch Truss or Braced Arch 478 Autographie Testing Machine 240 Beams , Rectangular, Compar- ative Strength and ntiff- ness 272, 273, 277 Bow's Notation 407 Box-Girder 275, 292 Liraced-Arch 438, 478 Bridge, Arch 430 Buckling of Web-Flat es 383 Built Beams, design ing Sections of 295 built Columns 373 Burr, Prof. . Citations from 224, 229 Butt-Joint 226 Cantilevers 260, 276, 341 Cantilever, Oblique 353, 356 C-aten.'.ry Inverted Catenary, Trans formed 395 Cast Iron 220, 279 Cast Iron, Malleable 224 Centre of Cravity 336 Chrome Ste ^ i224 Jircle as -.stic Curve 262, 343 Circular Arc as Linear Arch 391 Closing line 414 Coblenz, Bridge at 459, 478 Columns, Long 363 Compression of Short Blocks 218 Concurrent forces 397 Continuous Girders, by Analysis 32C 332 Continuous Girders, by Graphics 434 514 Cover-Plates 226 Crane 362 Crank- Shaft, Strength of 314 Crushing, Modulus of 219 424 Dangerous Section 262 332 Deck-Beam 275 )ef lections, (Flexure 342 Derivatives, (Elastic curve) 310 Diagrams, Strain 209 241 Displacement of Joint of /.rch Rib 447 Dove-Tail Joint 269 Eddy, Prof . , Graphic methods , (See Preface) 426 plastic Curve a Circle 262 343 Elastic Curve an Equilibrium Polygon 484 Clastic Curves 245,252, 262 356 Elastic Curves, the Four x-De- rivatives of 316 Elasticity-Line 241 Elasticity, Modulus of 203, 227 Elastic Limit 202 Elliptical Beam 340 Elongation of 7< T rourht-lron Rod 207 Equilibrium Polygon 401 45JD Equilibrium Polygon Through Three Points 418 419 Exaggeration of Vertical Dimensions in Arch- Ribs 470 Examples in Flexure 280 284 Examples in Shearing 231 232 Examples in Tension and Compression 222 223 Examples in Torsion Experiments of an English Llroa; Qommission 314 3 'Experiments of Hodgkinson 207 369 Experiments of Prof. Lanza 280 Experiments on Building Stone 424 Experiments on Columns 378 Ext r ados 421 Euler's Formula for Columns 364 Factor of Safety 223 "False Polygons" 497 501 Fatigue of Metals 224 "Fixed Points" 503 Flexural Stiffness 250 Flexure 244 386 Flexure and Torsion Con- bined 314 Flexure Beams of Uniform Strength 335 ?>. ure, Common Theory 244 Flexure, Eccentric ;>oad 256 301 Flexure, Mastic Curves in 251,252 Flexure, Examples in 280 284 Flexure, Hydrostatic Pressure 308 Flexure, Moving Loads 298 Flexure,Kon-Prismatic Beams 332 335 Flexure of Long Columns 363 Flexure of Prismatic Beams Under Oblique Forces 347 362 Moxura, Safe Loads in 262 284 Flexure, Safe Stress in 279 ^lexure, Shearing Stress in 234 295 Flexure, Special Problems in 295 319 Flexure, Strength in 249 p lexure,the Elastic Forces Flexure, the "Moment" 249 .exure,the "Shear" 243T Flexure, Uniform Load 258,267 302 ',05,307,324,329 340 Flow of Solids 212 Force Diagram 400 "-lygons 'orces, Distributed 197 Friction 422 423 General Properties of Mat -rials 204 Craphic Treatment of Arch 431 .Gravity, cent re of 336 453 4 Gravity, Vertical 453 Graphical Statics, Elements 397 420 Graphical Statics of Vertical Forces 412 420 Kodgkinson's Formulae for Columns 369 -oke's Law 201,203 207 ooks, ^trength of 362 Horizontal Straight Girders by Graphics 479 433 lorse-fower 239 242 I-Beam 275,292 295 337 Inclined Beam 359 Internal Stress, General Problem of 205 Intrados 421 I sot ropes 204 ;iiza, Experiments of Tr^-f. 280 Lateral Contraction 211 229 Lateral Security of Cirders230 298 Linear Arches 506,396,417 425 Live Loads 298 430 Load-Line 413 Locomotive on Ar'-h 430 Locomotice on Girder 298 Malleable Cast Iron 224 Middle Third 423 Modulus of Elasticity 203 227 Modulus of Resilience 213 Modulus of Rupture (Flexure) 278 Modulus of Tenacity 212 Moduli of Compression 219 Mohr's Thoerem 486 Moment- '.re a 435 Moment -diagram 263 Flexure 248,318, : Moment of Inertia by Graphics 454 Moment of Inertia of Kox-Girder 276 :omcnt of Inertia of Built Golumn379 Moment of Inertia of Bui n t Beam 296 Moment of Inert is of Plane Figures 249 274 Moment of Inertia of Truss 478 Moment of Torsion 236 Mortar 422 Moving Loads (Flexure) 298 Ilaperian' kajcariikaix Base 357 337 Navier's Principle 422,436 ireutral nxis 245,247, 347, 355 lion-Concurrent Forces in a Plane 399 Normal Stress 200 Oblique Section of Rod in Te-.sion 200 Parabola as Linear Arch 391 536,344 Phoenix Columns 378 Pier Reactions 404 of Arches 430,435 "Pin- and- Square" Columns 364 Polar-Moment of Inertia 238 Polelin Graphics) 401 Pole-Distance 416,417 Practical Notes 223 Power, Transmission of by Shafts 238 318 Punching Rivit Holes 229 Radius of Curvature 250,353 Radius of Gyration 313,376 Rankine's Formulae for Columns 372,372 Rays of Force Diagram 401 Reaction 4C4 Reduced Load- Contour 429 R&gsdLxfcisra af xxakisisx Resilience 204,213,237,251,3125 Reversal of Stress 514 Riveting fir Built Reams 292 Rivets and Riveted PI 225,292 Robinson, Prof .Integration by 357 Rod in Tension 198,200 Roof Truss 405 Rupture 202 Safe Limit of Stress 202 Safe Loads in Flexure 262^284 St. Louis Bridre 459,467,478 :et,Parmanent 202,209,208,241 Shafts 233 239 Shafts, Non-circular 239 Shear Diagram (Flexure- 265 Shearing 225,232 Shearing Distortion 227 Shear, Distribution of in Flexure 287 Shearing Forces 225 Shearing Stress 200,201,225,234,284 .-hear, the First x-De- rivative of Moment (Flexure) 264 xki&dtxxg Slope (inPlexure) 253 Soffit 421 Spandrel 421 Special Equilibrium Polygon 469,424, 440 Statics, Graphical 397 420 Stiffening of V.'eb-J lutes 383 Stone, Strength of 221 424 "tress Diagram for rch-Ribs 471 Stresses due for Sib Shortening 476 Strain Diagrams 209 241 Strains, t~ o kinds only 196 Stress ' 197 193 Stress and Strain, ''elation Betvcen 201 Stress Couple 253 348 Stress, Nornal and Shearing 200 Strength of Materials 195 Stretching of aTrisra Under its Cwn r ' eight 215 "Sudden" Ap^-lic/ition of a Load 214 255 Summation of Products by Graphics 451 Table for Flexure 279 Table for Shearing 228 Tables for Tension and Com- pression 221 Temperature Stresses 206,217/222,473 Tenacity, Moduli of 212 Testing Machine, autograph! s 240 Theorem of Three Moments ?32 Thrust (in Flexure) 348/350 Torsion 233,243 Tors ion, Angle of Tor s ion, Helix Angle in 233 Tor si on, Moment of 236 Torsional Resilience 237 Torsional stiffness 236 Torsional Strength 235 Transformed Catenary 395 Transmission of Tower by Shafting 238, 318 Trussed Girders 381 Page 'Uniform Strength Beams Uniform Strength Solid of, in Tension 216 Voussoir 386 421 7eb of I-3eam 274 Web of I-Beam,bucklinf of "eb of I -Beam, Shear in 290 W or king- Beam 336 344 Corking Strength 202 This book is DUE on the last date stamped below NOV 6 1946 AUG 1 1 1947 Form L-9-15m-7,'31 TA 350 Church* C47m Mechanics of vv2 engine erirrg. A 001 113692 6 "550 AL1FOBS1*