LIBRARY 
 
 OF THE 
 
 UNIVERSITY OF CALIFORNIA. 
 
 Class 
 
u 
 
MECHANICS 
 
 OF 
 
 INTERNAL WORK 
 
 (OR WORK OF DEFORMATION) 
 
 IN ELASTIC BODIES AND SYSTEMS IN EQUILIBRIUM 
 
 INCLUDING 
 
 THE METHOD OF LEAST WORK 
 
 BY 
 
 TRYING P. CHURCH, C.E. 
 
 Assoc. AM. Soc. C.E. 
 
 Professor of Applied Mechanics and Hydraulics, College of Civil Engineering, 
 Cornell University 
 
 FIRST EDITION 
 
 FIKST THOUSAND 
 
 -- - ' . 
 Of THE 
 
 UNIVERSITY 
 
 OF 
 
 ILJJF 
 
 NEW YORK 
 
 JOHN WILEY & SONS 
 
 LONDON: CHAPMAN & HALL, LIMITED 
 
 1910 
 
Copyight, 1910 
 
 BT 
 IRVING P. CHURCH 
 
 THE SCIENTIFIC PRESS 
 
 IMMONO AND COMPANY 
 BROOKLYN, N. Y. 
 
PREFACE 
 
 AMONG the more modern methods used in dealing with elastic bodies 
 in the Mechanics of Materials, those based on the properties of the 
 derivatives of the work of deformation (or "internal work" as it will be 
 called in the present volume), as first thoroughly established by Cas- 
 tigliano * in 1879, and including the " Method of Least Work," have 
 been but sparingly presented in American text-books; and even in 
 those works which include applications of these methods, such proofs 
 of the methods themselves as may be given are generally so meagre in 
 detail that the student is left much in the dark as to the precise condi- 
 tions which must be observed to render the methods applicable. 
 
 It has therefore seemed desirable to the writer that a text-book be 
 prepared which should deal exclusively with the " Mechanics of Internal 
 Work in Elastic Bodies in Equilibrium" and embody special fullness of 
 explanation and illustration, not only in the demonstrations of the 
 methods to be employed but also in the detail of the applications; and 
 thus indicate clearly the use, scope, and limitations of these valuable 
 methods. 
 
 The following pages are the result of an attempt in this direction; 
 and it is hoped that the book may prove useful not only to students 
 of engineering, but in some degree also to members of the profession. 
 
 In this connection the attention of students who read German is 
 directed to Miiller-Breslau's standard work on "Recent Methods in 
 
 * Castigliano wrote in French, his book being entitled " Theorie de I'Equili- 
 bre des Systemes Elastiques." It was published at Turin in 1879 and has 
 been translated into German. Prof. William Cain's paper on "Determination 
 of the Stresses in Elastic Systems by the Method of Least Work" in the Transac. 
 Am. Soc. Civ. Engs., April, 1891, is a valuable contribution to this subject; 
 as also Prof. George F. Swain's article " On tJie Application of the Principle 
 of Virtual Velocities to the Determination of the Deflection and Stresses of 
 Frames," in the Journal of the Franklin Institute for Feb., March, and 
 April, 1883. 
 
 iii 
 
 211757 
 
iV PREFACE 
 
 Strength of Materials, etc.*" Books in English by Hiroi and Martin 
 are referred to in the present work (see p. 89); and also articles by 
 Tilden, Hudson, and Mensch (see pp. 51, 53, and 93). 
 
 The abbreviation M. of E. will be used in referring to the writer's 
 " Mechanics of Engineering." 
 
 CORNELL UNIVERSITY, ITHACA, N. Y., 
 October 20, 1910. 
 
 * Die neueren Methoden der Festigkeitslehre und der Statik der Baukonstruk- 
 tionen. Leipzig, 1893 (zweite vennehrte und verbesserte Auflage). 
 
CONTENTS 
 
 CHAPTER I. EXTERNAL AND INTERNAL WORK OF ELASTIC 
 
 STRUCTURES 
 
 PAGES 
 
 1-20. Gradual Application of Loads. Structure to Consist of 
 Bars. Successive Introduction of Elastic Properties. 
 
 Limitation to Zero Initial Stresses. Proof that External 
 
 Work Equals Internal Work '. 1-14 
 
 CHAPTER II. THE DERIVATIVES OF INTERNAL WORK 
 
 21-25. Derivative of Internal Work with Respect to an 
 External Force or "Load." Proofs of Castigliano's 
 Theorem, both Analytical and Geometrical. Application 
 to a Frame. FraenkePs Formula. Derivative of Internal 
 Work with Respect to Moment of Couple 15-25 
 
 CHAPTER III. THE "THEOREM OF LEAST WORK"; WITH APPLICA- 
 TIONS TO SYSTEMS OF BARS IN EQUILIBRIUM UNDER LOADS 
 
 26-35. Statically Indeterminate Structures. Example of 
 Frame of Six Bars, One of Them Redundant. Proof of 
 the " Principle of Least Work." Frame with Two Redund- 
 ant Bars; and More than Two. Theorem when a Bar is 
 Originally Too Short, or Too Long, to Fit into Place ; and 
 when there are Several Such Bars. Temperature Stresses. 26-44 
 
 CHAPTER IV. INTERNAL WORK IN BEAIVIS UNDER FLEXURE. 
 APPLICATIONS 
 
 36-56. Internal Work of a Bent Beam; both Straight, and 
 Curved (of Large Radius of Curvature). Examples I to 
 VIII of Straight Prismatic Beams. Straight Horizontal 
 Beams of Variable Moment of Inertia. Continuous Girders; 
 both Prismatic and Tapering. Derivative of Internal 
 Work of Beam with Respect to a Statically Indeter- 
 minate Internal Moment. Proof of the Theorem of 
 Three Moments. Values of the Two Integrals in the 
 Latter Theorem in Particular Cases. Applications of 
 Theorem to Examples 45-77 
 
VI CONTENTS 
 
 CHAPTER V. COMPOSITE SYSTEMS AND CURVED BEAMS 
 
 57-73. Straight Beam Supported by Three Others. Trussed 
 Beam or Brake-beam. Cambered Trussed Beam. Exam- 
 ples. Straight Beam Supported by Two Piers and Parabolic 
 Cable. " Double-knee " Beam. Davit of Uniform Section. 
 Curved Beams of Circular Form of Axis. Semicircular, and 
 Segmental, Hinged at Ends, also Case of No Hinges or 
 Fixed Ends. Simpson's Rule for Approximate Integra- 
 tion. Curved Davit of Variable Section Treated by the 
 Aid of Simpson's Rule 78-122 
 
 APPENDIX 123 
 
 INDEX 125 
 
MECHANICS OF INTERNAL WORK 
 
 CHAPTER I 
 EXTERNAL AND INTERNAL WORK OF ELASTIC STRUCTURES 
 
 1. Gradual Application of Loads. The propositions or theo- 
 rems of this and the two following chapters have to do with 
 elastic bodies, or structures, undergoing slight deformations 
 within the elastic limit. Certain points of the structure are sup- 
 ported on fixed (and inelastic) bodies, while at other points 
 forces or loads are applied, each increasing in value gradually 
 and simultaneously, and proportionately to the others, until 
 their final values are reached at the same instant. During this 
 gradual application of the external or applied forces, gradual, 
 but slight, changes of form occur in the elastic body or struc- 
 ture itself. 
 
 2. Structure to Consist of Bars. For present purposes the 
 body or structure will be conceived to be composed of straight 
 elastic bars, each being pivoted (pin-connected) to the neighbor- 
 ing bars, at its two extremities (and nowhere else); all these 
 bars lying in the same plane (plane of the paper). All external 
 forces or loads are to be applied at joints; that is, at the pins. 
 Each pin is perpendicular to the plane of the bars. 
 
 3. Bars Considered Elastic in Succession. In establishing 
 the theorems, however, it will be convenient to consider that 
 the elastic character of the body is introduced in successive 
 stages; that is, the body will first be considered to be entirely 
 rigid with the exception of one bar, all the remainder of the 
 body being treated as a stiff plate or rigid body. Then, portions 
 of this plate will be considered to be composed of elastic bars; 
 
2 MECHANICS OF INTERNAL WORK 4 
 
 and so on, in successive cases, until a proposition is established 
 for a body composed entirely of straight elastic bars. 
 
 4. Necessary Bars. Redundant Bars. If there are more 
 bars than are necessary for the structure to hold its shape 
 (that is, to hold it, aside from the small elastic deformation 
 which will occur when loads are applied) it is said to contain 
 " redundant bars "; though as to which bars are redundant 
 may be an arbitrary matter, more or less. Thus, if a set of 
 five bars form the four sides and one diagonal of a quadri- 
 lateral, all of these bars are necessary that the framework may 
 hold its shape and be capable of bearing loads applied at certain 
 joints, two other joints being supported; but if a bar be placed 
 
 /7\^ ft in the position of the other diagonal, the frame contains one 
 redundant bar. Any one of the six bars could be treated as 
 the redundant bar and the other five as the " necessary bars "; 
 since, whichever one were omitted, the other five, connecting 
 the original four vertices or joints, would constitute a frame 
 capable of holding its shape. 
 
 5. As a First General Theorem it will be proved that in a 
 structure of elastic bars the " work " of the applied forces or 
 loads is equal to the work expended on stretching or com- 
 pressing the bars, all these loads being gradually applied. Or, 
 briefly, 
 
 External work = Internal work (1) 
 
 It is restricted to cases where all the bars are accurately 
 fitted to each other before loading and are under no initial 
 'stress; that is, under no stress when the application of the 
 applied loads begins; so that all the bars begin to stretch or to 
 shorten at the same instant and the stresses in them increase 
 in direct proportion to their final values. Also, there must be 
 no change of temperature during the application of the applied 
 forces. 
 
 In the case of a bar under compressive stress, though designed 
 by a " long column " formula, its amount of shortening is 
 supposed to be proportional to the stress at any instant. 
 
 Case I. The Structure Contains only One Elastic Bar. 
 Fig. 1 represents a structure in which the plate N is rigid 
 while the bar hb is elastic. M is a fixed rigid support, the 
 
I 5 EXTERNAL AND INTERNAL WORK 3 
 
 straight bar hb being " pin-connected " both to N and M. 
 A load P is applied to the plate at the point e. Originally, at 
 the instant of first application of the load, the point b was in 
 position a, e was at d, and the bar hb was under no stress. As 
 a result of the gradual * application of the load P the bar 
 stretches and finally its extremity b reaches the position of 
 the figure and the load (force) P reaches its final full value P. 
 At this final instant there is a certain tensile stress in the elastic 
 bar which we shall call T. There is also a reaction, R, at the 
 pin m, where the rigid plate is pivoted to the fixed support M. 
 One extremity, h, of the elastic bar is also pivoted to M. We 
 have, therefore, the rigid body N under the action of the three 
 
 bc=Ji 
 
 /k 
 
 FIG. 1. 
 
 forces P, T, and R, and in equilibrium. During the gradual 
 application of the load its point of application has moved 
 from d to its final position e. 
 
 Let a perpendicular now be let fall from d upon the (present) 
 line of action, / . . . k, of the force P. The distance fe is then 
 the projection of de upon the line /. . . k. Similarly, let fall 
 the perpendicular ac from point a upon line bh', be, the projec- 
 tion of ab upon line b . . . h, being thus obtained. 
 
 Since the plate N is a rigid body in equilibrium under the 
 three forces T, P, and R, let us now apply to it the principle 
 of Virtual Velocities (M. of E., p. 68) conceiving it to execute 
 a small displacement of precisely such character and extent as 
 
 * If the force or load P is due to gravity, this gradual application may 
 be supposed to be brought about by the progressive pouring of sand into a 
 pail, for instance. 
 
4 MECHANICS OF INTERNAL WORK 6 
 
 to bring points b and e to their original positions a and d, 
 respectively; that is, a small rotation about point m. It should 
 be remembered that this small displacement is purely imaginary, 
 and that while it is taking place the values of the three forces 
 P, T, and R, remain . constant. For this displacement the 
 " virtual velocity " of T is be and occurs on the positive side 
 of the point b [that is, on the side toward which the force T 
 (arrow) points], while the " virtual velocity " ef of the force P 
 falls on the negative side for that force (on the side opposite 
 from the direction in which P points); the virtual velocity of 
 force R being zero. Giving, therefore, the proper sign to each 
 " virtual moment/' or product, with zero as right-hand member 
 of the equation, we have [from the (Pdw)=0 of p. 68, M. 
 ofE.], 
 
 T.cb-P.ef+R*0 = ...... (2) 
 
 But evidently the distance cb is not only the virtual velocity 
 of the force T, but_also the difference (practically such), X, 
 between its length, lib, under stress and its original length, ha, 
 when under no stress. That is, A is the elongation of the bar. 
 Hence we may write 
 
 Py=TX, ....... (3) 
 
 in which t/, or ef, is the " displacement " of the point d in the 
 direction of the force P (i.e., the projection of de on line of force 
 
 P). 
 
 6. External Work; and Internal Work. While the last 
 result is true as it stands, as a result of the " principle of virtual 
 velocities " applied to a rigid body at rest (the displacement 
 mentioned being only imagined to take place), if we divide both 
 members by 2 we obtain the form 
 
 to be interpreted as follows: P/2 is the average value of the 
 force P during the gradual movement of point d to its final 
 position e (this force having increased from zero to P), while 
 T/2, similarly, is the average stress in the elastic bar during 
 its gradual elongation (the initial stress having been zero). 
 Also, y is the projection upon the line of P of tho " path/' de, 
 
7 EXTERNAL AND INTERNAL WORK 5 
 
 of its point of application; and similarly cb, or the elongation 
 A, is the projection, upon the line of the force T, of the " path/' 
 6, of the point of application of this latter force. Conse- 
 
 p 
 quently the product -^y is called the work done by the force P 
 
 during its gradual application (and this is called " external 
 work "; and if other loads were acting we should add a similar 
 
 T 
 
 term for each one); while the product -~-X is called the " work 
 
 of deformation" or work done in changing the length of the 
 elastic bar from a condition of no stress to its final condition 
 (and the sum of a number of similar terms for a number of 
 bars forming an elastic structure will be called the " internal 
 
 work")- 
 
 7. Case II. The Structure Contains Four Elastic Bars. 
 
 Fig. 2 shows a structure consisting of a rigid plate N, and 
 four elastic bars, namely, 1, 2, 3, and 4, connecting the plate 
 
 FIG. 2. 
 
 to a fixed rigid body M. To the rigid plate are applied two 
 loads PI and P 2 acting (in this final position of the plate) 
 at points I and n. When the loads were first applied these two 
 points were at A; and m, respectively, and the values of the 
 loads were each zero. The pivots b and / where the elastic 
 bars are connected were originally at points a and e, these 
 points forming with points k and m the same rigid configura- 
 tion which they form in their final positions (the plate N being 
 rigid). The four bars comprehend one redundant bar, since 
 
6 MECHANICS OF INTERNAL WORK 7 
 
 three alone would have been sufficient to enable the whole 
 structure to hold its shape and its connection with body M\ 
 but the proposition would be equally true on the supposition 
 that only three bars were present. During the gradual appli- 
 cation of the two loads varying simultaneously in value from 
 zero up (and proportionally) to their final amounts, P\ and P 2 , 
 the point a has moved to some new position b and, likewise, 
 e to/, k to I, and m to n; while the stresses in the four bars, 
 whatever their initial values, attain, respectively, certain final 
 values, T 7 !, T 2 , T 3) and T. The figure, then, shows the rigid 
 body N in this final state of equilibrium and acted on by the 
 six forces, viz., T\, T 2 , TZ, T, P\ and P 2 \ and to this body 
 we shall apply the principle of virtual velocities; regarding, as 
 the small displacement, which is imagined to occur, the very 
 movement which would bring the four points 6, /, /, and n, 
 back to their original positions, a, e, k, and m. If from a we 
 draw the perpendicular ad upon the line of bar 3 (prolonged), 
 thus obtaining the point d, and also a perpendicular from a 
 upon the line of bar 1, obtaining the point c, we note that bd 
 is the virtual velocity of T%, and also the shortening, ^3, of bar 3; 
 while be is the virtual velocity of T\ and also the amount of 
 elongation, X\ y of bar 1. Similarly, drawing at the point e 
 the perpendiculars eh and eg we determine fh, which is at the 
 same time the virtual velocity of T 4 and also the shortening A 4 
 of bar 4 ; and fg as the virtual velocity of T 2 and the elongation 
 X 2 of bar 2. At each of the points k and m, dropping perpen- 
 diculars upon the lines of action of PI and P 2j respectively, 
 we determine the virtual velocity vl( = yi) of PI and also that, 
 on, or 1/2, of P 2 . Forming, then, the algebraic sum of the 
 products of each of the six forces by its virtual velocity, repre- 
 senting the respective elongations or shortenings of the bars 
 by AI, >b, etc., and writing the sum equal to zero, we obtain: 
 
 Pzyz-Q, . . (5) 
 
 or, by transposition, 
 
 4. ... (6) 
 
 At this stage of the present case there is no object in dividing 
 each term of this equation by 2 and calling each term a certain 
 
8 EXTERNAL AND INTERNAL WORK 7 
 
 amount of work done by, or upon, each force during the gradual 
 application of the loads PI and P^ since it may have been 
 necessary, before this application, in putting the bars together, 
 to strain one of the bars (the last one fitted; that is, the 
 " redundant " bar) into the position over the pivots at its 
 extremities; which operation would create in all the four bars 
 certain initial stresses, different from the values which the 
 stresses in these bars will have attained at the end of the 
 gradual application of the loads; therefore, the mean value 
 of the stress in bar 1 (for instance) during the gradual applica- 
 tion would not be one-half of the final value, T\. Hence 
 eq. (6) will be allowed to stand in its present form, as a mere 
 consequence of the principle of virtual velocities, without 
 bringing into play any idea of the " work done " by any force 
 during the gradual application of the loads. 
 
 8. Special Case of the Preceding. No Initial Stress. 
 Let us suppose that although there is one redundant bar, the 
 last bar fitted is of just the proper length to fit over the pivots 
 provided for its extremities without strain; there will then be 
 no initial stress in the bars ; and when the gradual application 
 of the loads begins, all the bars will begin simultaneously either 
 to lengthen or to shorten, and the stress in each will change 
 gradually from its initial value zero to its final value; so that 
 the mean value is one-half the final. In such a case all the 
 terms in the right-hand member of eq. (6) are positive, whether a 
 bar is subjected to a lengthening, or to a shortening. Bar 1, 
 for instance, suffers an elongation, so that the force T\ points 
 to the left and the virtual velocity cb falls on the same side of 
 the point b as that in which the arrow (or force) T\ is directed, 
 so that the product T\\\ is positive in eq. (5). Now con- 
 sidering the case of bar 4, we note from the figure that it is 
 subjected to a shortening, and hence to a compressive force, 
 T. This force, as an action on body N, must therefore point 
 to the right in the figure; where we also note that the virtual 
 velocity for this force, namely, hf, falls on the same side of / 
 as the arrow denoting the force ; so that here again the product 
 (viz., T 4 ^4) will be positive in eq. (5); which is the same result 
 that we reached for the bar in tension. From this it follows,, 
 then, that all the products of the form TX in the right-hand 
 
8 MECHANICS OF INTERNAL WORK 10 
 
 member of eq. (6), in this case of no initial stress in -the bars, 
 are positive; from which fact a very important relation will 
 result, in future propositions. As to the left-hand member of 
 eq. (6), the products of the form Py may be positive, or may 
 bo negative ; according to the positions of the respective applied 
 forces, or loads, PI, P^ etc., since the movement of the rigid 
 body N during the gradual application of these loads may be 
 such, and the forces may be so directed, that the projections, 
 or virtual velocities, 1/1, 2/2, etc., may, or may not, fall on the 
 same side of I (or n, respectively) as the arrow showing the 
 direction of the force. 
 
 9. No Redundant Bars; then All Products TA Positive 
 in eq. (6). If in Case II there has been no redundant bars 
 connecting the rigid bodies M and N (in other words, if no 
 more than the " necessary " bars had been used) there would be 
 no initial stresses in the bars; so that, as the loads are gradually 
 applied, each stress increases gradually from a zero value to its 
 final, and each term on the right in eq. (6) is positive, as already 
 shown for the case of no initial stresses with redundant bars. 
 
 It should therefore be noted, that in both cases, viz., of 
 redundant bars, and of simply necessary bars, the average 
 stress is not equal to one-half the final unless the initial stresses 
 are each zero (which is always so for the case of simply necessary 
 bars, but not necessarily true in a redundant system). 
 
 10. Case III. The Structure Contains Five Elastic Bars, 
 No One of which is Redundant. (N.B. The absence of a 
 redundant bar does not vitiate the truth of the proposition. 
 One or more such bars might be introduced without changing 
 the nature of the demonstration; the object of omitting them 
 is to reduce the amount of detail. See 14.) 
 
 Here, as before, we have (see Fig. 3) a rigid plate TV con- 
 nected to a rigid support M by a number of elastic bars, so 
 arranged as to be five in number and to involve no redundant 
 bars. Of the five joints, or pivots, having to do with these 
 bars, p and q remain stationary during the application of the 
 loads PI and PI, while the other three, now at b, f, and s, were 
 originally at a, e, and r, respectively, when the gradual appli- 
 cation of the loads PI and P% began. At this initial instant 
 each of the five bars was under a zero stress. 
 
12 
 
 EXTERNAL AND INTERNAL WORK 
 
 9 
 
 11. It will be noticed from the figure that the only bars the 
 stresses in which act on points of the rigid body N are bars 3, 
 4, and 5, and that in the special case shown in this figure bar 3 
 has been shortened, so that its action against body N is a force 
 T 3 pointing toward the right. As to whether bars 4 and 5 
 are in tension or in compression does not directly appear in the 
 figure since the movement of joint b would not be thoroughly 
 known except in some particular numerical case. As a supposi- 
 tion to be consistently followed when joint b comes up for 
 consideration, we shall suppose that the action of bar 4 upon 
 .body N is & tension, and hence directed upward at point /; 
 
 FIG. 3. 
 
 .and that the action of bar 5 upon body N is also a tension, 
 .and hence directed to the left at point s. 
 
 12. As before, we shall apply the principle of virtual veloc- 
 ities to the rigid body N in equilibrium under the five forces 
 T 3 , T, T 5 , PI, and P 2 ; imagining by way of a small displace- 
 ment that the points of application (s, /, Z, and n) shift back 
 to their original positions (r, e, k, and m) and letting fall from 
 each of these last points, respectively, the proper perpendiculars 
 upon the lines of action of the respective forces. The five 
 forces concerned remain constant during this shift. In this 
 way we obtain the virtual velocity ts of 7 7 5 ; fh, that of T 3 ', 
 zf, that of T 4 ; vl, that of PI; and wn, that of P 2 . As before, 
 fh, being the elongation of bar 3, will be called ^ 3 ; also vl, y; 
 and wn, y%] whereas, for the present, st and zf, as quantities 
 
10 MECHANICS OF INTERNAL WORK 12 
 
 (lengths), will be written st and zf. Forming the various 
 products, then, or " virtual moments, " we have 
 
 If now we consider the joint or pivot at b as a separate 
 
 " free body," we find the system of concurrent forces, shown 
 
 in Fig. 4, acting on the pivot; 
 and there have been reproduced 
 in this figure, in their proper 
 relative positions, the point o, 
 the four perpendiculars ac, av, 
 ad, and au, as also the prolonga- 
 tions bu and bv of the two bars 
 4 and 2. In other words, we 
 have in this figure the respective 
 virtual velocities be, bd, bu, and 
 
 bv, of the four stresses, or forces, acting on the pin b. 
 
 Now conceiving, as a basis for applying " virtual velocities," 
 
 that the pin b is displaced through the small distance ba, and 
 
 forming the proper products, we derive 
 
 'T4.uo-T 5 . : vb = 0. . . . (8) 
 
 It is now to be noted, that the projection cb is equal to the 
 elongation, ^i, or bar 1, and similarly, that the projection db 
 is the elongation, X 2 , or bar 2, so that eq. (8) may be rewritten 
 
 2 -T 4 .uo-T 5 .vo = ..... (9) 
 We note again, examining both Fig. 3 and Fig. 4 that 
 
 zf-ub= A 4 = the elongation of bar 4; 
 and st vb= >* 5 = the elongation of bar 5. 
 
 These substitutions having been made in the equation resulting 
 from the adding of (7) and (9) we obtain, after transposition, 
 
 Piyi + P 2 y 2 = T^ + T 2 X 2 + 7^3 + Tth + T B l s . . (10) 
 
 13. Attention is now again called to the fact that each 
 term of the form TX in the right-hand member of eq. (10) is 
 positive, and this feature will always occur, in systems having 
 no redundant bars, f<^r reasons previously stated (8); whereas,, 
 
16 EXTERNAL AND INTERNAL WORK 11 
 
 in the left-hand member, where we have to do only with the 
 applied forces and the respective projections y\, etc., some of 
 the terms may be negative; in accordance with the special 
 conditions of any particular case. 
 
 14. In case there had been redundant bars in this case, 
 eq. (10) would have resulted in the general form as already 
 derived; but if the idea of work were to be introduced, we 
 should have to remember that at the beginning of the gradual 
 application of the loads the various elastic bars might be 
 already under stress (that is, there might be " initial stresses ") ; 
 
 rp m 
 
 so that -~-j -~-, etc., would not necessarily be the average values 
 
 of the stresses in bar 1, bar 2, etc., respectively, during the gradual 
 application of the loads. Also the terms of the form TX in the 
 right-hand member of eq. (10) would not all necessarily be 
 positive (though such would be the case if the bars were 
 so accurately fitted originally that there are no initial 
 stresses). 
 
 15. It now becomes evident from the nature of the steps 
 already taken, that eq. (10) is general for any structure of 
 elastic bars which become slightly changed in length by reason 
 of the gradual application of loads PI, etc., and that it is true 
 whether (there being redundant bars) there were already 
 stresses (" initial stresses") existing in some or all of the bars, 
 or not; so long as each symbol T denotes the stress tho ot>oao 
 in a bar at the instant when all the loading is in place and the 
 body at rest, and X denotes the change of length occurring in 
 each bar as due to the gradual loading of the structure (and 
 not the difference between its final length and that which it 
 would have under no stress whatever. 
 
 16. Limitation to Zero Initial Stresses. X in Terms of T. 
 We are now going to limit, very specifically, the conditions of 
 our problems by stipulating that, before the structure is loaded, 
 all redundant bars are of just the proper length, respectively, to 
 fit over their pivots without stress of either kind, tension or 
 compression. Under these circumstances, the stress in each bar 
 is zero before loading takes place, and the quantity X denotes 
 the difference between the length of a bar under its final stress 
 T and its length under no stress. That is, X is the elongation 
 
12 MECHANICS OF INTERNAL WORK 17 
 
 due to T and is proportional to T, bearing to it the relation (see 
 p. 203, M. of E.), 
 
 Tl 
 
 where F is the sectional area (square inches, say) of the bar 
 (which is understood to be a prism in form, and homogeneous in 
 material), I is the length, and E is the modulus of elasticity 
 (pounds per square inch, for instance) of the material. 
 
 Hence, under these special conditions, eq. (10), after division 
 of both members by 2, may be rewritten in the form, 
 
 or [see eq. (11)], 
 
 the form in eq. (13) being of such a nature as to embody certain 
 convenient conceptions, as follows : 
 
 17. " External Work." If we assume that the external 
 forces, or " loads," are applied to the elastic structure simul- 
 taneously, each increasing very gradually but in proportion to 
 the others, from a value of zero to its full (or " final ") value PI 
 (or P 2 , etc.) (this could be done practically by the progressive 
 pouring of sand into pails, for instance) ; their points of appli- 
 cation will be deflected or displaced through the various dis- 
 tances i/j, 2/2, etc., measured in the direction of the corresponding 
 load or "external force"; and the product of JPi (which is 
 the average value of load No. 1 during this motion) by the 
 deflection y\ (in direction of force) is called the "work" done 
 by this force during the gradual loading of the structure. 
 Similarly, \P^^ is the work done by load No. 2, etc. (Some 
 of these products may be negative.) Hence the left-hand 
 member of eq. (13) may be called the " External Work," or 
 work of the applied loads. 
 
 18. Internal Work Equals External Work. During this 
 gradual and progressive application of the external forces, 
 tensile and compressive stresses are created in all the elastic 
 bars, the stress in each bar increasing gradually, and simul- 
 
' 
 
 $19 EXTERNAL AND INTERNAL WORK 13 
 
 taneously with those in all the others, from a value of zero 
 to its full " final value " T (pounds, for instance) which it 
 reaches at the same instant that the stresses in the other bars 
 reach their respective " final " stresses. In this progressive 
 increase of stress, the stress in any bar is always directly pro- 
 portional to the change of length, and hence the average stress 
 in any bar (for purpose of formulating the work done in pro- 
 ducing that change of length) is one-half of its final value T, 
 
 T 
 
 and the product of average stress by elongation, that is, -~-X, 
 
 & 
 
 is called the work done in changing the length of the bar. Now 
 the right-hand member of eq. (12) is seen to consist of the sum 
 of all such products for all the bars of the structure and is called 
 the "Internal Work" of the structure or system of elastic bars, 
 and will be denoted by the symbol [7; that is [see also eq. (13)], 
 U denotes 
 
 . **'"'' (14) 
 
 or, 
 
 and is also called the " work of (elastic) deformation" Eq. (12) 
 or (13) can therefore be read, 
 
 External Work = Internal Work. . . . (16) 
 
 19. Note. It must be repeated that in case there are 
 redundant bars these must be of just such original lengths as 
 to fit into their places without stress before the structure is 
 loaded; or, to express it in another way, the " necessary bars " 
 having been assembled so as to give to the structure its proper 
 form, these necessary bars being without strain since no loads 
 are as yet applied, the other bars, viz., the redundant bars, are 
 supposed to be of just the proper lengths to fit upon their 
 various pivots without subjecting themselves or the necessary 
 bars to any stress. Hence when the loads are now gradually 
 applied to the structure, the initial value of the stress in each 
 bar is zero. It must also be understood that no change of tem- 
 perature occurs in the structure during the application of the 
 loads. 
 
14 MECHANICS OF INTERNAL WORK 20 
 
 20. The arbitrary phrases just adopted (internal and external 
 work, etc.) are convenient, but not really necessary, since 
 eqs. (10) and (12) are mathematically correct for the quantities 
 indicated by the respective symbols, whether any names are 
 given to the products in question, or not; and it is really imma- 
 terial (so far as the final result is concerned), in conceiving 
 the loads to be placed on the structure, whether they are applied 
 in the manner indicated (simultaneously) or whether they are 
 applied in succession, each gradually; the final result is the same. 
 
 For brevity, the fraction ^ , which is dependent onlv on 
 
 r \rj\ 
 
 the dimensions and the material of bar No. 1, may be called 
 Ci (and similarly for the other bars we write C 2 , C 3 , C 4 , etc.; 
 respectively), and hence eq. (14) may be written 
 
 U = internal work, = W l T l 2 + W 2 T 2 2 + etc.', . (17) 
 or, 
 
 U=2(1,CT 2 ); . ..... ... t . i . . (17a) 
 
 while the expression for 
 
 external work is %P\y 1+^22/2 + etc. . . (18) 
 
 Note that each term in (17) is positive; while in (18) some 
 terms may be negative. 
 
21 
 
 DERIVATIVES OF INTERNAL WORK 
 
 15 
 
 CHAPTER II 
 THE DERIVATIVES OF INTERNAL WORK 
 
 21. Castigliano's Theorem. Derivative of Internal Work 
 with Respect to an External Force, or " Load." Statement: 
 In any structure of elastic bars, at rest (under no initial stresses, 
 with or without redundant 6ars;* supported by fixed points, or 
 by fixed and smooth supporting surfaces, or both) the " deriva- 
 tive " (or "first differential coefficient ") of the total internal 
 work with respect to any one of the external forces (Pi, for instance) 
 is equal to the displacement (yi) of the point of application of 
 that force, in the direction of the force (that is, the projection, upon 
 the action line of the force, of the actual displacement of the 
 point of application). 
 
 That is, we are to prove that 
 
 In the next figure, Fig. 5, the network of elastic bars of which 
 the structure is really composed is not shown. The continuous 
 
 FIG. 5. 
 
 straight line OacN represents the structure (for example, a 
 bridge truss) under no load whatever, while the curved con- 
 tinuous line ObeN represents the same structure after the gradual 
 application of the two (vertical) loads PI and P 2 . and N 
 represent two fixed supports. The point of the structure is 
 fixed, but the point (or joint) N bears against a smooth surface 
 * Note the restriction stated in 19. 
 
16 MECHANICS OF INTERNAL WORK 21 
 
 of the fixed support at that end. During the application of 
 the two loads the points of application have moved to final 
 positions b and e respectively, so that their total (vertical) 
 displacements at this stage are ab, or y\, and ce, or y 2) respect- 
 ively. ObeN will be the new position of the truss, and the system 
 of external forces acting upon the truss consists of the two 
 loads PI and P 2 and the two reactions P and P n . Whatever 
 the value of the internal work may be at this stage, we know 
 it to be equal to the external work. 
 
 Note. It is to be noted that the external work done by 
 the reaction PQ is zero, because point does not move; and 
 that the work done by P n is zero, because its point of applica- 
 tion moves at right angles to the action line of P n , and hence 
 the projection of its displacement upon the action line of P n 
 is zero. The reaction of a smooth surface is necessarily in a 
 line normal to that surface. We may therefore write 
 
 P iyi P 2 ?/ 2 PpXO PnXO. 
 
 [/= 7>~+ ; ~ + 2 2 ; ' ' ^ '' 
 that is 
 
 U = lPiyi+lP 2 y2. ....... (20) 
 
 We shall now note the effect, in producing an increment, 
 dU, in the value of U, of giving to one of the external forces 
 (say, PI) an increment, dPi. This increase in PI will cause 
 the truss to descend slightly to a lower position (dotted) 
 Ob'e'N; creating changes dyi and dy 2 in both the displacements 
 2/i and y 2 . (Pz remains of the same value as before.) This 
 will cause a small addition or increment, 
 
 in the external work, and consequently an equal increment 
 (dU) in the internal work; whence, neglecting the term involv- 
 ing the product of two differentials, we have 
 
 . .... (21) 
 
 But we can express dU in another form, as follows : We first 
 conceive the truss to be restored to the unstrained position 
 OacN (by removing all loads), and then consider the pressures 
 (or loads) at a and c to have (Pi + dPi) and P 2 as their respective 
 
21 DERIVATIVES OF INTERNAL WORK 17 
 
 final values, and to be applied gradually and progressively 
 (from zero), while always maintaining to each other the con- 
 stant ratio subsisting between their final values. When these 
 final values are reached, we find the structure in the position 
 (dotted) Ob'e'N', and the whole external work done in this 
 operation of " deformation " (that is, from state OacN to state 
 Ob'SN) is 
 
 ;. .. . (22) 
 
 i.e., neglecting the term involving the product of two differentials, 
 and rearranging terms, 
 
 W = [iPit/i + JP 2 y 2 ] + i(<*Pi)y i + IPidyi + iP*^. (23) 
 
 Now the difference, U' U, must be equal to dU, the incre- 
 ment accruing to the value of U as occasioned by giving to 
 PI an increment dP\\ hence, from eqs. (20) and (23), 
 
 dU = WPi)yi+^Pid yi + ^P2dy 2 . . . . (24) 
 
 But, from eq. (21), \Pifyi>+$Ffy*~fyttT\ and hence 
 eq. (24) becomes 
 
 dU=(dPi)yi; or, ^=7i, .... (25) 
 
 that is, the derivative of U (total internal work) with respect to 
 any one load, .or external force, P, is equal to the displacement 
 (y) in the direction of this P (i.e., the projection of the actual dis- 
 placement upon the line of P) of the point of application of P. 
 
 This is the fundamental theorem or principle in the subject 
 of internal and external work in elastic structures, Castigliano's 
 Theorem. 
 
 If therefore the total internal work, U = ^> or 
 
 (7=(|CT 2 ), of the elastic structure can be expressed as a 
 function of one of the external forces (say, PI), we may find 
 the displacement y\ of the point of application of that load, 
 in the direction of the force, by differentiating the quantity U 
 with respect to PI. 
 
 Although in the foregoing there has been, besides PI, only 
 one other load (P 2 ) [which retains a fixed value during the 
 demonstration] the result would evidently have been the same 
 
18 MECHANICS OF INTERNAL WORK 21 
 
 in case there had been any number of other loads besides PI. 
 (P 2 and other similar loads are independent of the value of 
 PI, so that the derivative of one of them with respect to PI 
 would be zero). The reactions P and P n of Fig. 5 are, of 
 
 /7P 
 
 course, dependent on PI; so that -775-, for instance, would not 
 
 ai i 
 
 be zero. 
 
 It is to be particularly noted in the foregoing demonstration 
 that the mode of support is of such character that the reaction 
 at a supporting point is that of a fixed point or is offered by a 
 smooth and unyielding surface. A rough supporting surface, 
 along which slipping may occur, is precluded. An external force 
 doing zero work may be called a " neutral " force. 
 
 Although for simplicity the loads or external forces PI, P 2 , 
 etc., have been taken in a vertical position any oblique position 
 might have been assumed for them without altering the rigor of 
 the demonstration; nor need the line ON of the structure be 
 taken in a horizontal position necessarily. Also it is understood 
 that the support furnished by the supporting bodies is com- 
 pletely effective whatever change of value may occur in the 
 single load which has been selected as a variable. 
 
 2 la. Theorem for Variable Displacement. Castigliano has 
 proved a second theorem which asserts that if the displacement 
 (y) of the point of application of a load (P) (in its own direction) 
 be made to vary, the displacements of all the other loads remain- 
 ing fixed in value, then, 
 
 d JL- P . 
 2jT 
 
 but this theorem is of little practical utility and has not come 
 into prominence. 
 
 22. Another Proof (Partly Geometrical) of Castigliano's 
 Theorem. In Fig. 6 have been reproduced the various posi- 
 tions a, c, 6, e, etc., of the points of application of the two 
 external forces or loads, PI and P 2 , of Fig. 5, the values of 
 the displacements y\ and 3/2 and their increments dy\ and dy% 
 being indicated. Before PI received its increment dP\, the 
 work done by the two forces, each increasing gradually from 
 zero and in proportion to final values P\ and P 2 , is represented 
 
22 
 
 CASTIGLIANO'S THEOREM 
 
 19 
 
 in Fig. 6 by the two shaded triangles abh and cem, bh being 
 perpendicular to ab and made equal (by scale) to the final 
 value PI. Similarly (on same scale) em is equal to P 2 . The 
 areas of these triangles, therefore, are respectively the work 
 items \P\y\ and \P^y^. If now PI receive its increment, 
 dP\ (a little more sand being poured into that pail), while P 2 
 remains unchanged, the additional work done at b will be 
 represented by the (shaded) trapezoid bhkb' ', and that at e 
 by the (shaded) rectangle emne'; ik being equal to dP\. In 
 this first mode of loading the structure, then (the loads being 
 (P+dPi) and P 2 ), the external work is represented by the 
 shaded areas of Fig. 6. 
 
 Let now the structure be unloaded (pails emptied), and then 
 reloaded ; care being taken, in this reloading, to make the amount 
 of force (amount of sand in pails), at each instant during the 
 whole process, proportional to Pi+dPi, and P 2 , as final values; 
 instead of PI and P 2 . In this second mode of loading, with 
 same final values PI -\-dP\ and P 2 as in the first, the total external 
 work done is represented by the two areas of the two triangles 
 aW and cue'. The former area is seen to be in excess of the 
 shaded parts in the left of the figure, by the area of triangle 
 ahk' } while the latter is less than the shaded part in the right 
 of the figure, by the area of triangle cmn. Now this excess 
 (ahk) on the left must be equal to the deficiency (cnm) on 
 the right; since, whichever of the two modes of loading be 
 conceived to take place, the final state of the component bars, 
 and hence the amount of internal work, C7', are the same in 
 the two cases; and consequently the amounts of the total 
 external work are equal in the two cases. That is, area of 
 
20 
 
 MECHANICS OF INTERNAL WORK 
 
 23 
 
 triangle ahk = area of triangle cnm. Now area ahk= -{-triangle 
 aik triangle aih triangle hik m , i.e., 
 
 +dyi]-Pi . dt/i- 
 
 and 
 
 . (25o) 
 . (256) 
 
 Equating (25a) and (256), we have, neglecting products of 
 two differentials, 
 
 . ... (25c) 
 
 But, by eq. (21), the right-hand member of eq. (25c) is 
 equal to JcH7; and hence, finally, 
 
 dU 
 
 (25) 
 
 which is Castigliano's Theorem. 
 
 23. Example of Application of Castigliano's Theorem. 
 
 The structure in Fig. 7 contains seven bars, three horizontal 
 and four at 45. Pins o and n rest on the smooth horizontal 
 
 FIG. 7. 
 
 surfaces of two fixed piers, so that the reactions P and P n 
 are vertical. The three vertical forces (loads) PI, P 2 , and 
 PS are not equal to each other, necessarily. 
 
 It is required to find the vertical displacement of joint a, 
 where the load PI is applied (i.e., the projection, upon line of 
 PI, of movement of a). (As to whether a moves horizontally, 
 and to what extent, depends on the horizontal sliding of one or 
 both of the pins at o and n. 
 
23 
 
 CASTIGLIANO S THEOREM 
 
 21 
 
 There being no redundant bars, after finding the values of 
 PO and P n by ordinary statics, viz. : 
 
 
 (26) 
 
 we find by ordinary statics the values of the seven stresses in 
 the respective bars (in terms of the given loads and the reac- 
 tions). As to whether any such stress is tensile or compressive, 
 this is immaterial; since the internal work due to each one is of 
 necessity positive (see 9) . The following values are obtained : 
 
 and 
 
 (27) 
 
 Since we now are to consider PI as variable each T must be 
 expressed as a function of the variable P\ and of the con- 
 stants P and P 3 . This is done by the use of eqs. (26); and 
 the derivative of each T with respect to PI is then easily 
 obtained and set down in the proper column. In this way the 
 following values are obtained : 
 
 dPi 4' 
 
 and ^ 4 , 
 dT 3 
 
 4 ' 
 
 V2 
 
 l^L!'x P . 
 
 ~ + ~" + 3 ' 
 
 i 4 
 
 4 
 
 (28) 
 
 OF THE 
 
 UNIVERSITY 
 
 OF 
 
22 
 
 MECHANICS OF INTERNAL WORK 
 
 23 
 
 Before we apply eq. (25) (Castigliano's Theorem) to finding 
 the vertical displacement y\ of joint a we should note that, in 
 general [see eqs. (14), (17), (17a)], since 
 
 the derivative -75- can be written in the form. 
 
 ...,et,. . (29) 
 [Substitution from eq. (28) in this general form gives rise to 
 less complication than in (17).] Hence, since -T- = 2/i, we have 
 
 
 f 3 p+ P * + P 3\ 
 
 ^p,+ y + T ; 
 
 (30) 
 
 for the displacement of the point (or joint) a, where PI is 
 applied, in a vertical direction (Pi being vertical); as due to 
 three loads (i.e., neglecting the weights of the bars) with this 
 mode of support. 
 
 23a. FraenkePs Formula. Since the three loads PI, P 2 , and 
 P 3 may have any values so long as the elastic limit is not passed 
 in any bar, it is evident that by making P\=zero we obtain 
 the deflection of point a as due to the loads P 2 and P 3 only; 
 which leads to the following convenient rule: To find the dis- 
 placement, in a given direction, of a joint (of a loaded truss or 
 elastic structure) where there is no load applied, conceive a load 
 or force, PI, applied at this joint, in the given direction, form 
 an expression for yi, by the same procedure as shown in fore- 
 going example; then make PI = zero (Fraerikd's Formula). 
 N.B. If any of the " loads," or forces, acting on the truss in 
 Fig. 7 were not vertical, either pin o or n would have to be 
 fixed. 
 
24 
 
 CASTIGLIANO S THEOREM 
 
 23 
 
 24. Numerical Case of Foregoing Example (in 23). Let 
 
 the lengths of the bars and the amounts of the loads be as 
 shown in Fig. 7a, viz. : 
 
 Z 1 =Z 4 = / 6 = 20ft. = 240in.; 
 / 2 = Z 3 =Z 5 = Z 7 = 14.14 ft. = 169.7 in.; 
 PI = 8 tons, P% = 2 tons, and P 3 = 4 tons. 
 
 Let the sectional area of each bar be Fi = F 2 = F3, etc., 
 = 2 sq.in., and the value of E the same for all, 30,000,000 Ibs. 
 per sq.in. 
 
 We have, therefore, with the inch and ton as units, 
 
 L 240 
 
 2X15000 
 
 0.008; also 
 
 while 
 
 0.008; 
 
 0.00565. 
 
 Above values in the various terms of eq. (30) give rise to 
 
 L 20- J 2 tons j 
 
 I %- 20- H 
 
 fe 1 * SrZ 6 ? mk: 
 
 1 
 
 FIG. 7a. 
 
 f (0.008) (6 + 1 + 1) = + 0.0480 inches 
 
 |(0.00565)(6 + 1 + 1) =+0.0678 " 
 
 -1(0.00565) (1-2 + 1)= -0.0000 " 
 
 1 (0.008) (4 + 2 + 2) = + 0.0320 ' ' 
 
 1(0.00565) (2 + 1-1)= +0.0056 " 
 
 1(0.008) (2 + 1+3) =+0.0120 " 
 
 i(0.00565)(2 + l +3)= +0.0169 " 
 
 and hence the vertical dis- 
 placement 
 
 ?/i= +0.1823 inches. 
 
24 MECHANICS OF INTERNAL WORK 25 
 
 (N.B. It should be understood that in this case the pins at 
 the two supports are free to slide horizontally on the smooth 
 horizontal surfaces of the fixed piers, where the reactions are 
 therefore vertical in direction; and that the m veinentsof these 
 pins are at right angles to the reactions so that the external 
 work done by the reactions is zero. If the piers could yield 
 or settle during the gradual increase of the reactions (due to 
 gradual application of loads) this would not be true (zero work).) 
 
 25. Derivative of Internal Work with Respect to the 
 Moment of an External Couple. Among the external forces 
 holding an elastic structure in equilibrium let there be a couple 
 consisting of two equal, parallel, but oppositely directed forces, 
 PI and P 2 , applied at the extremities of a rigid cross-bar bd, 
 attached to an elastic structure. Let a denote the perpen- 
 dicular distance between PI and P 2 (" arm " of the couple). 
 
 During the gradual application of the loads or external 
 forces to the structure, the point of application of PI has been 
 displaced from e to b (the projection of this displacement upon 
 the line of P\ may be denoted by y\) ; that of P 2 , from c to d 
 (its projection on the line of P 2 being y 2 ). Then, by eq. (25), 
 we have 
 
 dU ' dU 
 
 _ =2/1 . and _ = 2/2 (31) 
 
 The angle <f>, between the original and the final position of 
 the rigid cross-bar, is so small in any practical case that for 
 tan <f> we may write <j> (i.e., in " radians" or circular measure) 
 and note, after drawing be' parallel to ec, that 
 
 a a '' < 32 > 
 
$ 25 EXTERNAL COUPLE THEOREM 25 
 
 in which for tan </> we now write </>; and, for y\ and 3/2, their 
 values from eq. (31); whence 
 
 dU dU 
 
 
 But a . dPi=d(Pia) =the differential of the moment M, 
 = (Pid) ( = P20>) of the couple; and the left-hand member of 
 >e(|. (33) expresses the complete derivative of U with respect to 
 M: and hence we may write 
 
 Evidently, from the figure, a positive value for 6 refers to 
 a " clockwise " displacement, or turning, of the cross-bar, if 
 the couple itself has a clockwise moment; i.e., in general, a 
 positive </> means an angular displacement of the same nature 
 as that of the moment of the couple; and vice versa. 
 
 For instance, a value of 0.0174 for <j> means an angle of 1. 
 
26 
 
 MECHANICS OF INTERNAL WORK 
 
 26 
 
 CHAPTER III 
 
 THE " THEOREM OF LEAST WORK"; WITH APPLICATIONS TO 
 SYSTEMS OF BARS IN EQUILIBRIUM UNDER LOADS 
 
 26. Statically Indeterminate Structures. A loaded struc- 
 ture containing one or more redundant (or " unnecessary ") 
 bars is called "statically indeterminate"; since the stresses in 
 the bars cannot be determined by ordinary statics. For their 
 determination we must have recourse to the elastic properties 
 of the bars or members. Castigliano's Theorem, bringing into 
 play, as it does, the displacement of any joint and the elastic 
 change of form or " deformation " of each bar, may be availed 
 of to prove a second relation or theorem, the direct outgrowth 
 of the first, called the " Theorem of Least Work," the application 
 of which to practical cases has the great advantage of involving 
 very simple and direct procedures. 
 
 As a preliminary step in the development of this second 
 theorem it will be useful to take up the study and solution of a 
 simple case, from which the general form of the relation is 
 readily inferred ; as follows : 
 
 27. Statically Indeterminate Frame of Six Bars. In 
 
 Fig. 9 is shown a frame of six 
 bars, in a vertical plane, forming 
 the sides of a square and both 
 diagonals, with one (vertical) 
 load, PI, as shown, suspended 
 on the lowest pin at e. Evi- 
 dently there is one redundant 
 bar (bars 1 and 2 are not con- 
 nected in any manner at A). 
 
 The pins at o and n are sup- 
 ported on the horizontal smooth 
 surfaces of the fixed supports 
 
 (at the same level), the reac- 
 
 P 1 
 tions from which are therefore vertical and each equal to -- r 
 
28 STATICALLY INDETERMINATE FRAME 27 
 
 since the only load, PI, is midway between. [Or, pin o might 
 be fixed, and pin n free to slide; the reactions would still be 
 vertical and each would be JPi.] 
 
 It is assumed, as before stipulated ( 19), that all of the 
 six bars are of just the proper lengths to enable them to be 
 fitted together without creating stress in any bar. (For 
 example, if all the bars except bar 1 had been put together 
 and bar 1 were too short or too long for its extremities to fit 
 over the pins at a and e, it could not be placed in position 
 without a forcible stretching or shortening, and when this had 
 been accomplished stresses would be created in all the bars 
 (" initial stresses "). Such a condition is specially precluded 
 here. The load P\ being now gradually applied, it is required 
 to determine the final stress T\ (tension, evidently) in bar 1; 
 and ultimately T 2 , T 3) etc. 
 
 The method of solution to be used depends on the geomet- 
 rical fact that the amount of elongation produced in the 
 (vertical) bar 1 must be equal to the difference between the 
 (downward) vertical displacements of its extremities, i.e., of 
 joints a and e] since if these displacements were equal it would 
 imply that no change had occurred in the length of bar 1. We 
 shall therefore find an expression for the downward vertical dis- 
 placement 2/0 of joint a in terms of the unknown T\, using the 
 method of 21 and 23 (Castigliano's Theorem); and, similarly, 
 an expression for the downward vertical displacement, y ej 
 of joint e in terms of T\] and then subtract the former from 
 
 ] rp 
 
 the latter and place this difference equal to J J . or CiT\. 
 
 iSifi 
 
 which expresses the elongation AI of bar 1 in terms of T\. 
 The resulting equation, 
 
 y.-y <* = &!, (35) 
 
 will contain but one unknown quantity, viz., TI, which is 
 then easily determined. 
 
 28. Detail of Solution. Conceive bar 1 to be removed, 
 but that its influence on the remainder of the loaded structure 
 is fully and completely maintained by the vertical down- 
 ward force TI (unknown) applied at a and an equal vertical 
 upward force T\ applied at e. For clearness, however, in 
 
28 
 
 MECHANICS OF INTERNAL WORK 
 
 28 
 
 subsequent mathematical work, let us denote the T\ at a by T, 
 and that at e by T'. We now have Fig. 10 in place of Fig. 
 
 9 and note that it shows a struc- 
 ture containing no redundant bars ; 
 T and T' playing the part of (un- 
 known) external forces, like PI 
 and the two reactions; each of 
 these reactions being now ex- 
 pressed in the form 
 
 4(P+T-T') 
 
 J-(p+T-T') 
 
 Now the downward vertical dis- 
 placement of a (y a , say) is 
 
 dU 
 y a== ' ) the external force T being vertical and downward, so 
 
 that the desired y a is the projection, on the line of T, of the 
 actual displacement of joint, or pin, a. 
 
 The bars of our present structure being 2 to 6 inclusive, 
 we have [see eq. (29)], 
 
 Similarly, the displacement of pin e in the direction of the 
 external force T f is given by the relation 
 
 (37) 
 
 But since the force T f points vertically upward and y e 
 denotes the downward vertical displacement of pin e, we have 
 
 (38) 
 
 From ordinary statics we obtain the values of T 2 , T%, etc., 
 in terms of T and T' and of the load PI, as shown in the follow- 
 ing table (in which the second and third columns, containing 
 certain derivatives to be used later, have been obtained by 
 obvious means) : 
 
28 
 
 STATICALLY INDETERMINATE FRAME 
 
 29 
 
 T Pt- 
 
 \/2 
 
 dT 4 _ I 
 dT ~ 
 
 dT, 1 
 
 dT V2 
 
 8 = 
 dT 
 
 dT' 
 
 If now we write T' equal to T 7 and each equal to T\, and 
 substitute in eqs. (36) and (37) the values just given in the 
 table, we obtain 
 
 (39) 
 
 and 
 
 i-rj.j. (40) 
 
 Hence the value of yey a will be, after substituting y f for y 
 [see eq. (38)], 
 
 and this should = AI, i.e., 
 is, 
 
 = elongation of bar 1). That 
 
 =0;. . (41) 
 
 and finally, solving, 
 
 2Ci + 2C 2 + C 3 + C 4 + C 5 + C Q ' 
 
 (42) 
 
30 MECHANICS OF INTERNAL WORK 28 
 
 (For instance, if the coefficients Ci, C 2 , C%, etc., were all 
 equal we should obtain 7 7 i = f PI.) 
 
 But we shall now show that eq. (41) is nothing more than 
 
 what may be obtained by writing -77^- = 0: where U is the 
 
 al i 
 
 total " internal work " of the whole framework, that is, of all 
 the six bars, including bar 1 itself. 
 
 Since U = JCi T 1 2 + W 2 T 2 2 + . . .+iC 6 7 7 6 2 , what we under- 
 
 IK du 
 stand by - is 
 
 dT 
 
 (43) 
 
 Hence, utilizing the table following eq. (38), where we find T 2 , 
 T 3 , etc., as functions of T and T ', each of which will now be 
 
 replaced by its real value, TI, so that = "+3^7; etc v 
 we find 
 
 , (44) 
 
 V2 V V2 
 and if this expression be written equal to zero, we obtain 
 
 -C 6 ~^; (45) 
 
 which is identical with eq. (41) and will yield the correct 
 value of TI. 
 
 But the value of TI obtained from eq. (45) is a value which, 
 from the principles of the differential calculus, would cause U, 
 the total internal work of the structure, to be a minimum; 
 that is, of all values that could be proposed for TI, that one 
 is the true and only possible one which makes the expression 
 for U a minimum.* This is called the principle of least work, 
 
 * Sine dU/aT,= 0. See Note C, Appendix. 
 
29 
 
 PRINCIPAL OF LI AST WORK 
 
 31 
 
 \ 
 
 a" 
 
 first established by Castigliano in 1878 ; and it will now be proved 
 in its general form. 
 
 29. Principle of Least Work. General Proof. (For a 
 loaded elastic structure containing one redundant bar; all 
 bars closely fitted originally without initial stress.) Select the 
 bar to be considered as the redundant bar and let T\ denote 
 the final stress in it. In Fig. 11 ae represents this 
 bar before the structure is loaded. As a result 
 of the loading the joint a has moved to a" and 
 the other extremity of the bar, viz., joint e, has 
 moved to e" . Perpendiculars having been let fall 
 from a" and e" upon the original position, ae, of 
 the bar (when unstrained), we note that y a , = aa', 
 is the displacement of joint a in the direction of 
 the bar, and that y ej = ee', is the 'displacement of 
 joint e in the direction of the bar. If we suppose 
 the bar to have stretched as a result of the ap- 
 plication of the loading evidently the difference 
 
 j m 
 
 y e y a is the elongation, = AI, =-= r ^ rj =CiTi, i.e., 
 
 where T\ is the final stress in the bar. 
 
 Now conceive the bar to be removed, the forces which it 
 exerts on the joints a and e at its extremities being applied at 
 those joints, respectively, to preserve the remaining bars in the 
 same position and state of stress as before. These forces are 
 equal, and act in the same line, but point in opposite direc- 
 tions. For this case (bar in tension) they point toward each 
 other (see Fig. 11) and will be called T and T' respectively, 
 although they are actually equal, each being equal to T\. For 
 the remaining bars T and T f play the part of external forces; 
 i.e., either T or T f might be conceived to vary, i.e., change 
 its value, the other remaining constant, without destroying the 
 equilibrium of the structure formed by the remaining bars 
 (note that such a statement as this could not be made if the 
 ibar in question were not a redundant bar). 
 
 The remaining bars constituting a statically determinate 
 
32 MECHANICS OF INTERNAL WORK 29' 
 
 system, then, with T and T' playing the part of independent 
 external forces, and the internal work of this system being 
 
 dU" 
 
 denoted by 17", the result of performing the operation -^ 
 
 would be an expression for the displacement of joint a in the 
 direction of T, i.e., for y a , and no change of sign is necessary, 
 since joint a has been displaced in the direction in which T 
 
 dU" 
 points; but the expression given by the operation -^ would 
 
 have to be taken with a contrary sign to give the displacement 
 y e , since joint e has been displaced in a direction (measured 
 parallel to T') opposite to that in which T' points. That is,, 
 otherwise expressed, we have 
 
 dV" dU" 
 
 ya= ~' ye= ~~'' ' '*' ' 
 
 dT 
 Hence eq. (46) becomes 
 
 dU" dU" dU" dU 
 
 ~~dT ==Cl i; Le '' ~d7 r+ 
 
 dU" dU" 
 Now since T = T' = Ti, the terms "^+^7 are the same 
 
 dU" 
 thing as -r^r, i.e., the complete derivative of V" with respect 
 
 to Ti, U" being the internal work of all the bars except 
 
 d [GiTi 2 ~] 
 bar.l; while C\T\ is nothing more than -JTJT} ^ , i.e, the 
 
 derivative of the internal work of bar 1 itself, with respect to 
 TI. Hence the left-hand member of eq. (48) is simply the 
 derivative, with respect to stress T 7 ! in bar 1, of the internal 
 work of all the bars, including bar 1, i.e., of the total internal 
 work, U, of the original complete system of bars. Eq. (48) 
 then, takes the form 
 
 dU 
 
 dfT' ' ' ..... (49) 
 
 as the condition or equation the solution of which will give the 
 value of the stress T\ in bar 1. 
 
30 TWO REDUNDANT BARS IN FRAME 33- 
 
 In other words, of all values that could be proposed for T t 
 the actual or true value is the one which would make the func- 
 tion U ' , or total internal work, a minimum. Hence the term, 
 Principle, or "Method, of Least Work." (Footnote, p. 30.) 
 
 N.B. If the extremities of bar 1, supposed to be in tension 
 in above proof, were displaced in directions contrary to those 
 assumed; or if the bar were in compression, no matter what 
 the actual movement of the extremities, in space, due to the 
 application of the loading; it is easily shown that the same 
 result would be reached [eq. (49)]. This result is therefore 
 general. 
 
 29a. Theorem of " Least Work " a Particular Case of a 
 More General Relation. The use of eq. (49) has already been 
 illustrated in solving the statically indeterminate frame of 
 28. It should be noted, however, that the idea of " Least 
 Work " which has given rise to the name of the method is 
 purely an incidental matter, of scientific, rather than practical, 
 interest; since the proof of eq. (49) does not depend upon this 
 idea, but is based on the ordinary relations of geometry and 
 elasticity in connection with Castigliano's Theorem for displace- 
 ments of the joints of a statically determinate structure. This 
 same mode of proof will be employed later ( 32) in establish- 
 ing a method of dealing with a structure containing redundant 
 bars which are initially too long, or too short, to fit the joints 
 already determined by the " necessary " bars of the structure; 
 and it will be seen that eq. (49) is merely a particular case of 
 this more general relation. 
 
 30. Statically Indeterminate Structure Containing Two 
 Redundant Bars. The truss in Fig. 12, resting on two smooth 
 horizontal piers at the same level, consists of one square panel 
 and one rectangular panel, involving eleven bars fitted origi- 
 nally (i.e., before the loading) without strain. There are three 
 equal loads, each equal to P, applied at the three upper joints. 
 Evidently there are two redundant or " unnecessary " bars; 
 that is, no more than two bars could be omitted without collapse. 
 For example, bars 5 and 9 could be omitted; or bars 5 and 2; 
 and the remainder of the structure would hold its shape and 
 form a " statically determinate " structure. (In this particular 
 case it also happens that, since bars 6 and 11 are in the same 
 
34 
 
 MECHANICS OF INTERNAL WORK 
 
 30 
 
 line and there is no load at their junction, the three bars 1, 2, 
 and 7 might be omitted without collapse.) 
 
 Let us imagine bars 1 and 2 removed and the (unknown) 
 stresses in them (suppose each to be a tension), TI and T 2 , 
 to act at the proper joints on the remainder of the truss, this 
 
 T 
 
 FIG. 12. 
 
 remainder being now a " statically determinate " structure under 
 the nine external forces (see Fig. 13), TI in two places, T 2 in 
 two places, P in three places (loads), and the two vertical 
 reactions fP and P. 
 
 We can now by simple statics determine the values of the 
 internal stresses T 3 , T 4 , etc., up to T\\ (inclusive) in terms of 
 these " external forces " just mentioned. These values are 
 assembled in the following table and also (for a future purpose) 
 
 FIG. 13. 
 
 values of the derivatives of each of these internal stresses with 
 respect to T\ and to TV We here denote cos a by m, and 
 sin a by n (a itself is 33 40'). 
 
30 
 
 TWO REDUNDANT BARS IN FRAME 
 
 1 
 
 35 
 
 
 T 10 = 
 
 _ -IZp-wfc 
 
 = 
 
 = 
 
 
 dT 2 
 dT 7 
 
 dT^ 
 dT 2 
 
 dTio 
 
 
 n 
 
 dT 
 
 n 
 
 (50) 
 
 Let us now restore bar 2 to the truss in Fig. 11. The result 
 is a new truss with one redundant bar (taken as bar 2) the 
 stress in which (T 2 ) can therefore be found in terms of TI 
 and P by 29; since for this new truss the external forces 
 are the two TI'S, the three P's, the fP and the P; (Ti is yet 
 unknown, of course). 
 
 We now put Vj ' n =0, with the understanding that 
 dJ. 2 
 
 U 2 . . . n, denotes the total internal work of this new truss, 
 consisting of bars 2 to 11 inclusive; that is, 
 
 U 
 
 2 
 
 11 = 
 
 . 
 
 I-...H 
 
 ,, n 
 ., (51) 
 
 whence 
 
 dU 2 ... 
 
 dTo 
 
 +O n Tir%r 9 =0. (52) 
 
 CH 2 
 
36 MECHANICS OF INTERNAL WORK 30 
 
 Filling out details of eq. (52) by the aid of the table, or 
 eqs. (50), there is obtained, after reduction, 
 
 - +[C7 2 +n*C 7 
 
 ] = 0, . . ... . V . (53) 
 
 which contains two unknowns, viz., TI and T 2 . 
 
 Similarly, restore bar 1 to the truss in Fig. 13, and we again 
 have a new truss with only one redundant bar (taken as bar 1). 
 The new truss consists of bar 1 and bars 3 to 11 inclusive, the 
 forces external to it being the two TYs, the three P's and the 
 fP and P. Its total internal work, which may be denoted 
 by Ui, 3 . . . 11, 
 
 2 2 2 
 
 and by putting its first derivative with respect to T\ equal to 
 zero we obtain [by eq. (49)] an expression for T\ in terms of 
 the external forces just mentioned. Now 
 
 dU i>*--"_ CT dT * lC T ^J 
 
 *dT. 4 . , dTi 
 
 ... (55) 
 
 and this is to be placed equal to zero. Therefore, substituting 
 details from eqs. (50), we have, after simplification, 
 
 0. . (56) 
 
 We now have two equations, viz., (53) and (56), from which by 
 ordinary algebra both TI and T 2 can be determined. 
 
 If we take, as a particular instance of this example, the 
 value of 45 for a, we must put \/2 -=-2 both for m and n, while 
 JP replaces the fP and also the fP in eqs. (50), TI will also 
 be equal to T%, from symmetry; and .*, eq. (56) alone will 
 
32 MORE THAN TWO REDUNDANT BARS 37 
 
 serve to determine T\, which in this way is found to be (with 
 all the O's taken equal) 
 
 (57) 
 
 31. Statically Indeterminate Structure with More than 
 Two Redundant Bars. (The usual stipulation is made, that 
 all the bars are closely fitted originally, without strain.) With 
 three redundant bars, say, bars 1, 2, and 3, we have simply 
 to replace them by the pairs of (unknown) forces which they 
 exert on the joints at their extremities, and by applying ordi- 
 nary statics obtain the stresses T, T&, etc., in all the other 
 bars in terms of the unknown T\, T 2 , and T%, and the external 
 forces of the original truss. The derivatives of each of the 
 stresses T, T&, etc., are then obtained with respect to T\, to 
 T 2 , and to !T 3 , and set down in a table. 
 
 Bar 1 is then replaced in the truss, which thus becomes a 
 truss of one redundant member (viz., bar 1) its external forces 
 consisting of the two TVs, the two TS'S, and the external forces 
 of the original truss. By the use of eq. (49) an expression is 
 obtained for T\ in terms of T 2 , T 3 , and the original external 
 forces. 
 
 By replacing bar 2 in the truss instead of bar 1 and using 
 eq. (49), a second equation is obtained giving T^ in terms of 
 Ti and jP 3 , and the original external forces. Similar treatment 
 with bar 3 gives a third equation furnishing T s in terms of 
 T\, T%, and the original external forces. From these three 
 independent equations the values of the three quantities T\, 
 TI, and TS are then easily determined. 
 
 The procedure is now evident for the case of .any number 
 of redundant bars, but manifestly involves a large amount of 
 detail when there are more than two such bars. 
 
 32. Statically Indeterminate Structure with One Redun- 
 dant Bar, this Bar being Originally Too Short, or Too Long, 
 to Fit into its Place. In a structure of one redundant bar 
 let us suppose that the " necessary " bars have first been fitted 
 together and that is it then found that the remaining (redundant) 
 bar is ^ inches longer, or shorter, than the distance between 
 the joints which it is to connect; so that to adjust it to its 
 
38 
 
 MECHANICS OF INTERNAL WORK 
 
 32 
 
 place it is necessary by external means to force the two joints 
 nearer to, or further from, each other, and either to stretch or 
 shorten the bar itself, to cause its extremities to fit over the 
 pins at these joints. This being done, and the external con- 
 straint removed, the structure is provided with some character 
 of support, and is loaded at one or more joints. The small 
 distance ^ (surplus or deficiency, in length; of the bar in 
 question) is supposed to be small; so that when the bar is 
 forced into position and the loads placed, the elastic limit is 
 not passed in any bar. We are now to prove, denoting by U 
 the total internal work of the structure (all the bars) in its 
 final condition under load, and by TI the stress in the redundant 
 bar (bar 1), that 
 
 -*,, ...,.. . (57) 
 
 for the case where bar 1 is assumed to be in tension and is 
 originally too short. 
 
 A specific case will be taken. In Fig. 14 the five " neces- 
 sary " bars (2, 3, 4, 5, and 6) when fitted together form the 
 
 d" 
 
 FIG. 14. 
 
 quadrilateral figure (dotted lines) abed] bar 4 being a diagonal, 
 bd. Bar 1, whose unstrained length is oc', is to form the other 
 diagonal; that is, it is _tp_ connect the joints a and c; but is 
 too short, by an amount c'c, or ^ inches. Joint b is fixed on the 
 support A, while the pin of joint d is free to slide along the 
 
32 REDUNDANT BAR TOO SHORT 39 
 
 smooth hard surface of the fixed support /). Let now the 
 lower extremity of bar 1 (oc') be forcibly fitted over the pin 
 of joint c and the load P 2 applied at the latter joint. This 
 special external constraint having been removed, and the load 
 P2 being in place, we find joint d in a new position, d" ', on 
 the supporting surface of Z); while joints a and c are now found, 
 respectively, at a" and c" '. Bar 1 is in a state of tension, the 
 stress in it being Ti, unknown. Its unstrained length being 
 ac' = li, its present length is a"c" = Z + ^i, where AI is its elonga- 
 tion. 
 
 Let us now consider bar 1 to be removed, its action on the 
 two joints a" and c" being fully represented by the forces T 
 at a and T' at c", these forces being in the same line a"c", 
 each equal to the unknown TI, and pointing toward each other, 
 as shown. They may now be considered as external forces 
 (and independent of each other), acting on the remaining bars, 
 viz., 2, 3, 4, 5, and 6; these bars forming a statically deter- 
 minate structure which has been changed from the form abed 
 (dotted lines) into the form a"bc"d" (full lines) by the action 
 of the external forces T, T', and P 2 ; the supports being of 
 the nature described. 
 
 Hence, if U" denote the internal work for all the bars except 
 bar 1, we have, from Castigliano's Theorem ( 21) the dis- 
 placement of a in the direction of force T 7 , viz., the projection 
 marked y in the figure is 
 
 and similarly, the displacement of c in direction of force T', 
 or the projection y' of the figure, is 
 
 (Note that for this figure both of the above signs are + r 
 since each of the displacements concerned takes place in the 
 direction in which the corresponding force points.) 
 
 But, from the geometry of the figure, oc' + ^ = 2/+a"c" + ?/;. 
 
 Mi+V'. ( ( >) 
 
40 MECHANICS OF INTERNAL WORK 32 
 
 / T 
 
 Introducing into (60) the relation *i gr^K =Ci7V and 
 
 " 1^1 
 
 the values of y and y f from (58) and (59), we have 
 
 7 !^. . .'". .... (61) 
 
 Now T and T' each =T\ and CiTi is nothing more than 
 the derivative of the internal work of bar 1, viz., JCiTi 2 , 
 with respect to T\. Hence the left-hand member of eq. (61) is 
 the complete derivative of the internal work, U, of all the bars 
 (including bar 1) and (61) may therefore be written 
 
 ;#r=4>- .... j- - (62) 
 
 The value of the stress TI may therefore be determined by 
 filling out eq. (62) and solving for T\', i.e., by first expressing 
 the stress in each bar in terms of T\ (use being made of the 
 " free body " (full Hnes) in Fig. 14; the values T and T' being 
 now replaced by jPi); and then writing out the separate deriv- 
 atives needed (besides the stresses thesmelves) for substitution 
 in the six terms in the left-hand member of eq. (62), i.e., in 
 the expression 
 
 =* . . (63) 
 
 While it is evident that in case bar 1 is originally too short 
 to fit into its place it will be found in a state of tension after 
 it has been forcibly fitted into place and the truss left to itself 
 without load; nevertheless, its place in the truss and the 
 position of the load or loads may be such that afterwards, when 
 the truss is loaded, it will be found in a state of compression, 
 instead of tension. This, result would be evidenced by the 
 obtaining of a negative result for the stress TI in the final 
 solution of eq. (63); that is, TI is thus found to be a " negative 
 tension," or compressive stress. 
 
 Bar 1 Originally too Long. If the redundant bar (bar 1) 
 is originally too long, by ^ inches, to connect joints a and c 
 of Fig. 14, point c' will be below c" . We have, therefore, with 
 bar 1 in its final position and the truss loaded, the bar being 
 
33 REDUNDANT BAR TOO SHORT 41 
 
 assumed in a state of compression, and its final shortening being 
 
 ^ClTl, 
 
 ac, =l-*o,=y + (l-h)+y'. . , . (64) 
 
 If we now suppose bar 1 to be removed, the forces T and T f 
 (at a" and c") taking its place for equilibrium, these two forces 
 must be inserted pointing away from, instead of toward, each 
 other; so that we now have 
 
 dV" dU" 
 y = ftf and t/' = ^ (65) 
 
 Hence, by combining, we reach the same result as in the case 
 of the bar being too short, viz., 
 
 dU" dU" 
 
 ~~^~^^~ r ^ 1== ^ 7 
 
 or 
 
 ^r=*>; . (67) 
 
 but it must be noted, that in the use of this relation for the 
 case of bar 1 too long, the stress in the bar must be assumed to be 
 compressive and the determination of stresses in the other 
 bars in terms of T\ must be made on this basis. (If the actual 
 stress in bar 1 is tensile the fact will be brought out by the 
 obtaining of a negative value for T\ in the final numerical 
 result.) 
 
 It will be noted that when ^ = zero for the redundant bar 
 eqs. (62), (63), and (67) reduce to what has been called the 
 Theorem of " Least Work " (see 29a). 
 
 33. Example of Single Redundant Bar, when Originally 
 Too Short. For this we may take the frame of six bars already 
 shown in Fig. 9 (p. 26), 27, and suppose that originally 
 bar 1 was too short, by ^ inches, to connect joints a and e. 
 In the treatment of the problem in 27 the stresses in the other 
 bars were found in terms of T\ (there called T and T', tem- 
 porarily) on the assumption that the final state of bar 1 is 
 tension] which is suitable for present purposes in the use of 
 eq. (57), or (63), according to which we have only to set the 
 
42 MECHANICS OF INTERNAL WORK 34 
 
 expression in eq. (44) equal to fa, instead of zero. This being 
 done, the result is 
 
 ~ 
 
 
 which may now be compared with the value in eq. (42) for the 
 case where bar 1 could be fitted in place originally without 
 strain. 
 
 34. Statically Indeterminate Structure with Two or More 
 Redundant Bars ; Each of which is Too Long, or Too Short, 
 to Fit into Place. A case of two such bars will suffice to 
 indicate the method for any other number. Let the structure 
 contain nine bars, of which only seven are " necessary " bars, 
 the other two being redundant. Let the bars to be taken as the 
 the redundant bars be 4 and 5; bar 4 being originally (^ ) 4 
 inches too short, and bar 5 (fa) 5 inches too long, to fit into 
 place. Assume, therefore, that the final state of 4 is tension; 
 and that of 5, compression. 
 
 Let now the redundant bars 4 and 5 be removed from the 
 truss, and the forces which they exerted on the joints at their 
 extremities put in; T at two joints, and T$ at two joints, 
 pointing in proper directions to correspond with the assumption 
 of tension or compression in the corresponding bars. We now 
 have a statically determinate structure consisting of the seven 
 bars 1, 2, 3, 6, 7, 8, and 9; the stresses in which are now 
 obtained in terms of the loads and of the unknown T and T$ r 
 by ordinary statics. Now conceive bar 4 to be again in place 
 (but not bar 5), and we have a statically indeterminate struc- 
 ture, with only one redundant bar (bar 4), the external forces 
 for which are the two TYs, the loads, and the reactions of 
 supports; in terms of which forces, T may now be obtained by 
 the use of eq. (62), viz. : 
 
 d[Ui, 2, 3, 4; 6, 7, 8, Q] , . v , aQ . 
 
 - 
 
 where the left member consists of terms of the type 
 
 dT 
 
 -m-, one for each of the original bars, except 5. 
 
35 REDUNDANT BARS TOO SHORT OR TOO LONG 43 
 
 Again, let bar 5 be put back in the truss, and bar 4 taken 
 out; and the structure thus obtained contains only one redun- 
 ( lant bar (which now is bar 5) and is acted on by external forces, 
 consisting of the two TYs, the loads, and the reactions of 
 supports. Hence from eq. (67), 
 
 d[Ul, 2, 3; 5, 6, 7, 8, Q] , . . 
 
 ..... (59) 
 
 where the left-hand member consists of terms of the 
 
 dT 
 type CT-T^r, for one each of the bars of the original truss, 
 
 except 4. Eq. (69) will give the stress T 5 in terms of the 
 external forces just mentioned. 
 
 Equations (68) and (69), then, are two simultaneous equa- 
 tions containing the two unknown quantities T ' 4 and T^ ; and 
 the values of the latter are now obtained by ordinary elimina- 
 tion. Should a negative number be obtained as the value of 
 cither T or T$ it indicates that the character of the stress in 
 the bar is the opposite of that originally assumed. 
 
 The procedure to be adopted in case there are more than 
 two redundant bars is now evident. There will be as many 
 independent equations like (68) or (69) as there are redundant 
 bars. As a basis for each one of these equations we have under 
 consideration a truss with only one redundant bar, the forces 
 external to this truss being the original loads, the reactions of 
 supports, and the forces exerted (on the joints at their extrem- 
 ities) by all the other (original) redundant bars. Attention is 
 again called to the necessity of assuming each of the original 
 redundant bars to be in tension, or in compression, according 
 to whether it is too short, or too long (respectively) to fit into 
 place after the " necessary '" bars have been put together. 
 
 35. Temperature Stresses in a System of Elastic Bars. 
 The foregoing treatment finds application in the following 
 circumstances: A change of temperature from that common to 
 all the bars at the time of assembling the truss, if we suppose 
 an original close fit, without strain, at that temperature, may 
 create stresses in the bars without any load being placed on the 
 truss. 
 
44 MECHANICS OF INTERNAL WORK 35 
 
 There may be redundant bars, or the character of the 
 supports may be such, that when the temperature rises (or 
 falls), the lengths natural to the bars at the new temperature 
 (with no load on truss) are not those that they are constrained 
 to have in their connection with each other; so that if the 
 truss were reassembled at the new temperature the bars would 
 not fit into place without special external (temporary) constraint ; 
 that is, after the " necessary " bars had been put together at 
 the new temperature each of the others would be found to be 
 too short, or too long, to fit into place without such external 
 constraint. 
 
 Hence we may determine the stresses in these latter (redun- 
 dant) bars (and ultimately in all the bars), after noting by 
 how much each such bar is too short, or too long, to fit into its 
 place in the assemblage of the " necessary " bars, by applying 
 the methods of 32, 33, and 34. Stresses thus induced are 
 called temperature stresses. These stresses are usually obtained 
 independently of the action of loads, and combined later with 
 those due to the loading. 
 
36 FLEXUKE OF BEAMS. INTERNAL WORK 45 
 
 CHAPTER IV 
 
 INTERNAL WORK IN BEAMS UNDER FLEXURE. APPLICATIONS 
 
 36. Internal Work of a Bent Beam. Straight Beams; 
 and Curved Beams of Large Radius of Curvature. When a 
 beam originally straight, or curved (and in that case the radial 
 thickness must be small compared with the radius of curvature), 
 is slightly deformed by being made the means of holding in 
 equilibrium a system of applied forces and reactions (all acting 
 in the same plane, containing the axis of the beam), each slice 
 or lamina (of the beam), bounded by two imaginary cu'tting 
 planes at right angles to the axis of the beam and a certain 
 small distance apart denoted by ds, measured along that axis, 
 consists of a bundle of fibers (so called) on the extremities of 
 which (in the two cutting planes), by the common theory of 
 flexure, we find elastic forces acting, viz., tensile, or com- 
 pressive, and shearing. In general [see Fig. 15, (A)], we find 
 at each end of the lamina, or block, a total shear represented 
 by J (Ibs.) (or /'), a total thrust, T (Ibs.) (or T'), consisting of 
 a set of uniformly distributed compressive or tensile forces of 
 uniform intensity p\ (Ibs. per sq.in.) ; and a set of tensions and 
 compressions forming a " stress-couple " whose moment (" bend- 
 
 P2^ 
 
 ing moment ") is denoted by M and is equal to (in which p 2 
 
 t/ 
 
 denotes the stress, Ibs. per sq.in., in the outer fiber at a distance 
 e from the center of gravity, or axis, of the cross-section, and / 
 the " moment of inertia " of the plane figure (formed by the 
 cross-section) with reference to the gravity axis G, perpendicular 
 to the paper. dF denotes an elementary area of the cross- 
 section and z is its distance from the gravity axis G. (See 
 p. 354 of the author's " Mechanics.") 
 
 Before stress, the block had the unstrained shape NHH Q No. 
 Under stress, however, its shape is N n H' f H Q N^ which now shape 
 we may consider to have come about thus: The uniformly 
 distributed thrust T causes all the fibers to shorten equally by 
 
46 
 
 MECHANICS OF INTERNAL WORK 
 
 36 
 
 an amount cUi, thus bringing the face NH to a parallel position 
 N'H'; then the couple acts, causing the fibers to shorten or 
 lengthen unequally, but proportionally to the distance z from 
 the gravity axis G of the cross-section (in which operation the 
 outer fiber NN- shortens by an amount dX 2 ) ; then the shearing 
 stress comes into play to distort slightly the corner angles of 
 the small blocks into which each fiber may be conceived to be 
 
 T' 
 
 J \ ,1 ^ 
 
 VKsJ*" 
 
 (B) 
 
 subdivided (no attempt has been made to show this distortion 
 in the figure). 
 
 We have therefore, 
 
 The work of thrust on d, or one fiber, = \p\dF . ab = [%p\dF}dX\ ; 
 
 but, from definition, -~ = Q ; 
 
 hence the work of thrust (one fiber), = ^p\dF- = (? 
 
 
 hence the whole work of thrust for all the fibers composing the 
 lamina is (with F denoting whole area of section N"H"), 
 
 (70) 
 
 As to the work of the stress-couple, each fiber shortens an 
 of d^ due to a force p-2dF', and hence 
 
 fiber =(%.-pd,F\(-dx2\. But ^r|r, 
 
 \ w /> * / \ p I r\ Q H. 
 
 \ O / \t/ / (*o /^ 
 
 amount be, = of dfa, due to a force p-2dF] and hence the 
 
 6 & 
 
 work on one 
 
 and 
 
36 FLEXURE. INTERNAL WORK 47 
 
 therefore the work on one fiber, as due to the stress-couple, is 
 
 p 2 2 ds dFz 2 f 
 
 J . 2 . p ; and therefore the 
 
 Whole work due to stress-couple 1 _ 1 P^ds C N " , 2 _ ds p# 
 for the whole lamina or block J = * ' ~~E#~ ) H ,, d = ~2 ' Ee* ' 
 
 which can be written pr^rf { - ) . But - is M. the moment of 
 ZHjl \ e / e 
 
 the stress-couple (or the " bending moment ") and hence the last 
 expression becomes 
 
 As for the work of the shear, J, if the whole block were 
 gradually distorted from a square-cornered condition to a 
 rhombus [d being the change of angle at the corner. See 
 Fig. 15 (B)] by the shearing couple whose moment is Jds we 
 should have the work of J (all the other forces being neutral) 
 
 = J . Jab=*y(ds)d, where d is the angle of shearing distortion. 
 
 Now (by definition) d = -p~E 8 , where E s is the modulus of 
 elasticity for shearing (p. 227, M. of E.), and therefore the whole 
 
 J2 
 
 work due to shear for this lamina or block would be i . TTTT . ds. 
 
 But since the shear is not uniformly distributed over the cross- 
 section in any case of flexure, this last expression must be 
 modified by the use of a coefficient, or multiplier, A, to suit 
 different forms of cross-section. For example, for a solid 
 rectangular cross-section with two bases parallel to the " force 
 plane " or plane of the paper, it can be proved that A=f ; while 
 for a solid circular section A=^-. For an ordinary I-beam 
 with a thin web, if the web be considered as bearing most of 
 the shear, A becomes practically unity (but then F is the area 
 of the section of the web alone). Hence, the work of the shear 
 for one lamina, of length ds, would be 
 
 - ^ (72) 
 
 2 - ........ (72) 
 
48 MECHANICS OF INTERNAL WORK 37 
 
 The forces shown in Fig. 15 (A), as acting on the slice or 
 lamina situated between two neighboring cross-sections, play 
 the part of external forces for this elastic body. But since this 
 elastic body may be considered to be made up of an immense 
 number of elastic bars pivoted to each other, these bars being 
 gradually stretched or shortened during the gradual application 
 of external forces, the total internal work performed in this 
 operation will be equal in value to the sum of the three terms 
 just found, viz. : 
 
 T 2 ds M 2 ds A J 2 ds 
 
 Now an entire beam, straight or curved, may be considered 
 to be made up of a great number of consecutive slices or 
 laminae like that in Fig. 15 (A), each having a length ds along 
 the axis of the beam and subjected to a thrust T at each end; 
 to two stress-couples, one at each end, of a moment M; and to 
 two shears, one at each end. Consequently, the total internal 
 work of such a beam can be expressed as 
 
 T*ds M 2 ds rAJ 2 ds 
 
 (the integration being extended along the whole length of the 
 beam) and may be placed equal to the sum of the amounts of 
 external work of the external forces applied to the beam, or 
 " applied forces " (including reactions) to which the state of 
 stress of the beam is due. 
 
 37. Displacement of a Point in Axis of a Straight Prismatic 
 Beam in Flexure. The expression for U in eq. (74), having been 
 once found in any given case, may be used inCastigliano's Theorem 
 [21,eq. (19)] for determining deflections or displacements of any 
 point of the axis of a beam where there is a load applied (or 
 where a load may be imagined to be applied and then reduced 
 to zero-, as indicated in 23a). The term due to shear is 
 generally of very small consequence and may therefore be 
 omitted in many cases without sensible error. The term 
 involving the thrust is also sometimes negligible, according to 
 circumstances. The beams concerned being usually of homo- 
 geneous material, the quantity E may then be taken outside 
 
38 FLEXURE. INTERNAL WORK 49 
 
 the integral sign; and, furthermore, if all cross-sections are 
 alike and similarly placed, the " moment of inertia," /, of the 
 cross-section will be the same at all sections and consequently 
 may in such cases be placed outside the sign of integration. 
 Frequently when it is wished to differentiate this expression 
 for U with reference to some quantity which it will eventually 
 contain (say, one of the external forces P), the operation may 
 be shortened by differentiating the general expression under the 
 integral sign; thus (in the term involving M) 
 
 d r i M 2 ds l 2M dM\ ds 
 
 Hence, when (later) M is expressed as a function of P it will 
 simply be necessary to find M, and (-jpO> m terms of P and 
 
 substitute in eq. (75) ; thus dispensing with the squaring of the 
 M as called for literally in eq. (74). A similar remark applies 
 to the other general terms, involving the thrust and the shear, 
 respectively. 
 
 38. Example I. Horizontal, Straight, Prismatic Beam. 
 End Supports. Load in Middle. Central Deflection Re- 
 quired. The (originally) straight, prismatic, homogeneous beam 
 is placed on two supports at the same level and loaded in the 
 middle with a concentrated load P, vertical. Evidently the 
 reaction 'at each pier is V = P + 2 (Fig. 16) and is vertical 
 
 1-f 
 
 1" 
 
 dx i 1 c 
 
 1 
 
 \ ! 1 
 
 M- t a.____ 
 
 ? 
 
 1 
 
 L ^ 
 
 IE 
 
 [ 
 
 i 
 
 | 7 
 
 
 FIG. 16. 
 
 (smooth horizontal surfaces); and the thrust is zero at all sec- 
 tions; while the work of the shear will be neglected, as being 
 very small. 
 
 It is required to find the vertical deflection y\ for the middle 
 point B of the axis of the beam, as due to the gradual applica- 
 tion of the one load P, applied at B. Neglect the weight of 
 
50 MECHANICS OF INTERNAL WORK 38 
 
 the beam itself. Evidently the deflection desired is the dis- 
 placement of the point of application of P in the direction of 
 P itself and therefore can be obtained by an application of 
 eq. (25) (i.e., Castigliano's Theorem). 
 
 In this case we shall measure the abscissa, x, of any dx 
 in the left-hand half of the beam, from the extremity 0; and 
 since the beam is straight, dx will take the place of ds in the 
 general eq. (74). The " free body " of length x is acted on 
 by the reaction JP at one end and by the shear J and the 
 tensions and compressions forming the stress-couple of moment 
 M at the other. From this free body, taking moments about 
 the right-hand extremity, we have M = JPz, and hence the 
 internal work for all the vertical laminae of the left-hand half 
 OB of the beam : 
 
 (76) 
 
 SE1 \ 3 
 
 From symmetry, we may double this for the internal work 
 of the whole beam as due to flexure (that due to shear being 
 neglected and that due to thrust being zero on account of the 
 zero value of the thrust at all sections). That is, the total 
 internal work for the beam is 
 
 E>273 
 
 (77) 
 
 Now the deflection of point B below its original position in the 
 horizontal line drawn through and C, is the displacement of 
 the point B in the direction of the external force P, and hence, 
 by eq. (25), 
 
 dU . d 
 
 P being now regarded as a variable. That is, 
 / Z 3 \ P/ 3 
 
 *"=W) 2P; or *-j&Bi> (78) 
 
 as we know by another method. (See page 254, Mech. of Eng.) 
 
39 FLEXURE. INTERNAL WORK 51 
 
 39. Influence of Shear in Preceding, if the Beam is a 
 Steel I-Beam. If the beam in foregoing problem in a steel 
 I-beam with narrow web the effect of shear in producing deflec- 
 tion may not be negligible. Considering the internal work of 
 shear as well as that of the stress-couples, we have, since the 
 total vertical shear in this case is / = JP at all sections, for 
 the internal work [see eqs. (74) and (77)], 
 
 
 
 A(P\* 
 2V27 ' 
 
 in which F w denotes the sectional area of the web alone while 
 the coefficient A may be taken as unity. Therefore 
 
 dU PI 3 PI 
 
 As a numerical instance, let us take the case of a 10-inch 
 Cambria steel I-beam, weighing 25 Ibs. per foot, and of a length 
 of 10 feet; to be supported as in Fig. 16 and loaded at middle 
 with P = 10,000 Ibs.; (the web placed in the usual vertical 
 position). 
 
 For the cross-section of this beam we find from the Cambria 
 Co.'s hand-book that I (about neutral axis perpendicular to 
 the web) =122.1 in. 4 while ^ = 7.85X0.31 = 2.43 in. 2 Young's 
 modulus may be taken as 7 = 30,000,000 lbs./in. 2 ; and E 8 , the 
 modulus of elasticity for shearing, as E 8 = l 1,500,000 lbs./in. 2 
 With these values eq. (80) gives 
 
 t/i = 0.0982 + 0.0107 = 0. 1089 inch. 
 
 The portion (viz., 0.0107) due to shearing action is seen to 
 be about ten per cent of the whole.* (The weight of the beam 
 itself has been neglected.) The effect of the shear would be 
 much less marked with a beam of rectangular section; and also 
 in the present case of the I-beam if the span were considerably 
 increased. 
 
 * An interesting case of this nature is worked out by Professor C. J. 
 Tilden in the Engineering News of Feb. 24, 1910, p. 228. 
 
52 
 
 MECHANICS OF INTERNAL WORK 
 
 40 
 
 40. Example II. Deflection of Straight Prismatic Beam, 
 Ends Supported, Load Uniform. (Fig. 17.) An (originally) 
 straight homogeneous, prismatic beam rests on two supports 
 
 UlllllllllilJ! 
 
 Si 
 
 FIG. 17. 
 
 at its extremities at the same level, bearing a uniformly dis- 
 tributed load over its whole length of an amount W = wl (w 
 load per linear unit; I being the whole length). It is required 
 to find the vertical deflection of the middle point B. Since 
 there is no finite vertical load at this point let us imagine one 
 (=P, say) to be placed there and finally reduce it to zero. We 
 then have the reaction at each support V = $(W+P). Now 
 for any vertical lamina of thickness dx of the beam the moment 
 of the stress-couple is 
 
 --*-, (81) 
 
 x being measured from the left-hand support 0. There is no 
 thrust. Neglect the work of shear. To secure greater brevity 
 let .us use the form 
 
 fo fdM\ds 
 Jo M (dp)Tl> 
 
 mstead of 
 
 d f C c M*ds 
 
 for the deflection of the point of application of the load P. 
 Since the summation having to do with the dx's in the right- 
 hand half of the beam is, from symmetry, equal to that in the 
 left-hand half, we may double the expression applying to the 
 left-hand half. Hence, noting that in this case ds = dx, and 
 
 that, for any dx on OB, M = 
 have [see eq. (75)]: 
 
 wx 
 
 dM x 
 
 -; and = ; we 
 
41 DEFLECTION OF BEAMS 53 
 
 2 
 
 , 
 
 dp dx 
 
 ---+ (82) 
 
 This gives the deflection of B as due to the distributed load 
 W and the concentrated load P at the middle. We have only 
 to make P equal to zero to obtain the deflection B for the actual 
 
 5 Wl 3 
 case of the distributed load alone, that is, 2/i = ooj ~gT- 
 
 Note. The " dummy " force P having served its purpose 
 
 dM x 
 in enabling us to get -jp-, =^, we might have put P equal to 
 
 zero in the expression for M above, before substituting in the 
 integration, and it is also interesting to note that the value 
 
 of -Tp- is the same as the moment at any section due to a load 
 
 of unity applied vertically at the point B (thus leading to 
 " Fraenkel's Formula." * Of course, also, by putting W= zero 
 in eq. (82) we may obtain the deflection due to P alone and 
 thus check the result already obtained in eq. (78). 
 
 41. Example III. Deflection. Beam with Overhang. A 
 continuous prismatic beam (homogeneous and originally 
 straight) rests on two unyielding supports at the same level 
 as shown in Fig. 18. The vertical force P is applied at D, 
 midway between and B, while the extremity of the over- 
 hanging portion carries a vertical load Q. The length BC 
 equals OB, =2a. Required the vertical deflection of the 
 point D, neglecting the work of shear (that due to thrust being 
 
 zero, as before). By eq. (25) this deflection y\ -jp, where U 
 
 *See Mr. C. W. Hudson's paper on "Deflection of Beams of Variable 
 Moment of Inertia," in Vol. LI (Dec., 1903) of the Transac- Am. Soc. Civ. 
 Engineers, p. 1. 
 
54 
 
 MECHANICS OF INTERNAL WORK 
 
 41 
 
 is the total internal work 
 
 . This summation will be 
 
 f c M*dx 
 'Jo 2EI 
 
 separated into the three integrals (reckoning x for each seg- 
 ment in the way shown in Fig. 18) for OD, DB, and BC y 
 respectively, 
 
 -I 
 
 FIG. 18. 
 
 In this case we note that from ordinary statics the reaction 
 
 p 
 at is V^-jr Q> an d obtain the following expressions for M 
 
 and its various derivatives in the three segments, viz., 
 
 =-e x; and = 
 
 For OD, M 
 for points on DB, 
 
 M=V(a+x)-Px,=(^~Q)(a+x)-Px', and 
 while on CB, M = Qx and ~J' = ^ 
 
 Before substitution, however, we may write (differentiating 
 under the integral sign), 
 
41 DEFLECTION OF BEAMS 55 
 
 in which, wnen values are substituted from above relations for 
 
 M, etc., there results 
 
 (84) 
 
 Performing the integrations and reducing, we obtain 
 
 [P ~ 
 j- 
 
 In this equation, if we now make Q equal to zero, we have 
 the same result as in eq. (78) ; while if P be made equal to zero 
 we have the deflection (upward; note the negative sign) of the 
 point D as due to the single load Q at C (in which case, of 
 course, the extremity must be " latched down," and the 
 reaction V is downward). 
 
 It should be carefully noted that in the above solution P 
 and Q are independent loads; that is, when P is conceived to 
 vary Q remains constant; in other words, Q is not a function 
 of P and hence (as above) d(Qx)/dP = Q. But the reactions, 
 V and V, depend on both P and Q; and neither V nor V can 
 be considered constant if either P or Q is to vary, but must be 
 expressed in terms of the variable (P; above) before dM/dP 
 can be obtained (on OD or DB). We note also that the external 
 work done by these two reactions, when the loads P and Q are 
 gradually applied, is zero; since the supports have smooth 
 horizontal surfaces and the points of the beam in contact with 
 them are thus free to slide (one or both) during the loading; 
 thus justifying us in the application of eq. (25), or (19). [See 
 note following eq. (19)]. 
 
 In case the vertical deflection (y 2 ), of extremity C were 
 desired, there being a vertical load Q at that point, we 
 should have 
 
 -; .... - - ''.' (86) 
 
 the expansion of which would give us eq. (83) with P replaced 
 by Q. In such case, of course, d(Qx)/dQ is not zero, but =x. 
 (Let the student work, out the detail of finding y 2 .) 
 
56 
 
 MECHANICS OF INTERNAL WORK 
 
 42 
 
 42. Example IV. Equation to Curve of Straight Beam. 
 Two End Supports. Single Eccentric Load. It is proposed 
 to use our present methods for determining the equation of 
 the elastic curve OD of the beam OC in Fig. 19 (homogeneous, 
 
 FIG. 19. 
 
 prismatic, and originally straight) resting on two end supports 
 (smooth, horizontal surfaces) and loaded with a concentrated 
 load Q not in the middle but at a distance a from the left-hand 
 support greater than that, 6, from the other support. (Denote 
 a + b by I.) We shall first find the vertical deflection of any 
 point B (between and D) whose distance from is x\, by 
 the process just illustrated, involving the use of a " dummy " 
 vertical force P applied at B. With such a force conceived to 
 
 act, the reactions of the supports are evidently Vo = jQ + 
 
 I 
 
 and V c =jQ+-j-P, respectively. The distance of any point n 
 from will be called x, and the moment of stress-couple at any 
 
 while 
 
 Again, for any section between B and D, we have 
 
 -P(x-x l); and - 
 
 For cross-sections on portion DC it will be more convenient 
 to reckon x (calling it now x') from the support C; whence, 
 for any. section between D and (7, 
 
 A 
 and =y 
 
42 EQUATION TO CURVE OF BENT BEAM 57 
 
 =-^T I M \Jpj 
 
 Now the deflection of the point B is y I=-^T I M \Jpjdx, 
 
 the summation extending over all the dx's (or dx f 's) of the axis 
 of the beam and can be written equal to the sum of three 
 integrals [applying to the segments OB, BD, and DC, respect- 
 ively], viz. : 
 
 or for brevity, 
 
 ..... (87) 
 
 Before substituting details in eq. (87) we may put P = in 
 each expression for M, since in this way the final result is the 
 same as if the P had been retained and had not been placed 
 qual to zero until the final expression had been reached; but 
 
 -each -jp-, of course, is not equal to zero. In detail, therefore, 
 bearing in mind these substitutions, we obtain, successively, 
 
 bQ Xl * 
 
 that is, 
 
 while 
 
 i- x ' dx '=~r) *' w= P 'f- 
 
 Since B is arbitrary in position we may regard it as any 
 point whatever on the elastic curve OD; and hence x\ and yi 
 become abscissa and ordinate of any such point. Therefore, 
 substituting the values of Z\, Z 2 , and Z 3 just found, in eq. (87), 
 
58 MECHANICS OF INTERNAL WORK 43 
 
 and with y for y\ and x for x\ (for simplicity), we have, as the 
 equation to the elastic curve OD, being the origin, 
 
 Q y r "' 3 
 
 Or, after reduction, 
 
 IEI [abl 
 
 ** ' '"' ' ( } 
 
 as the equation to elastic curve OD. 
 
 The abscissa of the lowest point of this curve (note that 
 OD is >DC', i.e., a>b) which will thus locate the point of 
 maximum deflection, may be obtained by writing 
 
 dy Q \abl a?b bx^ 
 
 I- ' i.e,byputting M [ T -^- 2 -J=0; 
 
 which being solved, gives for the required abscissa x, x m , 
 
 which can also be written 
 
 This value for x m being substituted in eq. (88) gives for 
 maximum deflection, 
 
 - (a - a) - (90> 
 
 or, in another form, 
 
 (These results for x m and y m are seen to check with those on 
 pp. 258, 494, and 506 of M. of E.) 
 
 43. Examples for Practice. Straight Prismatic Beams. 
 
 The following involve horizontal straight beams of constant 
 moment of inertia: 
 
 Example V. For the cantilever beam shown at (A) in 
 Fig. 20, find the vertical displacement (i.e., deflection) of the 
 extremity N, the load W being uniformly distributed over the 
 
43 
 
 DEFLECTION OF BEAMS 
 
 59 
 
 whole length; w is the load per running foot and I the length; 
 / constant. 
 
 Ans. 
 
 ' El' 
 
 Example VI. The cantilever beam at (B) in Fig. 20 is 
 built in, horizontally, at the left-hand extremity and carries 
 a variably distributed load, W] the rate of distribution, w, 
 is proportional to the distance x, from the free end, N. Find 
 the vertical deflection of point N 
 
 WP 
 ' El' 
 
 Ans. 
 
 FIG. 20. 
 
 Example VII. Find the vertical .displacement of any point 
 on the elastic curve of the beam at (A) } Fig. 20, in terms of 
 the horizontal distance x\, from extremity N. 
 
 w 
 
 Ans. 
 
 Example VIII. Find the vertical displacement of any 
 point on the elastic curve of the beam at (B), Fig. 20, in terms 
 of the distance x\ from the extremity N. 
 
 w r -~ .1 
 
 Ans. 
 
 44. Horizontal Straight Beam of Variable Moment of 
 Inertia (i.e., a Non-Prismatic Beam). Resting on Two 
 Terminal Supports. As a case presenting this feature let us 
 take the beam shown in Fig. 21. All its sections are rectan- 
 gular and the height is constant throughout, =h. The two 
 extreme quarters, however, are wedge-shaped, the width u 
 
60 
 
 MECHANICS OF INTERNAL WORK 
 
 44 
 
 being proportional to the distance x from the end, while the 
 middle half DA is prismatic, of constant width 6. A single 
 concentrated load, PI, being placed at the middle point B, it 
 is required to find the vertical displacement of that point; or y\. 
 From eq. (25), 
 
 dU c M dM 
 
 that is, E being constant, 
 2 C B M dM 2 
 
 dM 
 
 >~ 8 ~t -. 
 
 I 
 FIG. 21. 
 
 With x measured from support 0, we have, for the dx's 
 along AB, I being constant, =/o, 
 
 _. PI dM x , , bh? . 
 M=-^-x; -7p- = ; and /=TO, = A>; 
 
 Zi dL \ 1- 
 
 while along OA (/ variable), 
 
 ^i am x un* x on* x 
 
 ~5~ x '> ^P" = O an(i ** = T9~ == 7T T9" = ^ ^' 
 air i J LZ Q, LZ & 
 
 Hence eq. (92) becomes 
 2[ T a a Fix x 
 
 L Piz x 
 
 " ~17~ 77 < 
 
 ^ " 
 
 S To * P7 ~ 2^kiT3 * WW 
 
 45. Horizontal Non-Prismatic Beam. I Constant through- 
 out Each of Several Portions of the Length. (Fig. 22.) 
 In the case here shown the cross-section has different moments 
 
46 
 
 STATICALLY INDETERMINATE BEAMS 
 
 61 
 
 of inertia (7) in the different portions OA, AB, etc., of the 
 length (an I-beam with strengthened flanges, for instance), but 
 the value of / is constant through any one such portion. With 
 a symmetrical arrangement as indicated in Fig. 22, and with I\ 
 denoting the moment of inertia along AO and CE, 7 2 that along 
 AB and DE, and 7 3 along BH and HD, we obtain yi of the 
 middle point H by using a relation like that in eq. (92) but 
 with three terms in the bracket instead of two; and in the 
 three integrations concerned a different, but constant, 7 would 
 occur in each. (See foot-note in 41.) 
 
 
 
 ' 
 
 1 
 
 1 
 
 
 l 
 
 / 
 
 t 
 
 B H | 
 
 D 1 
 
 1 , 
 
 1 
 
 
 
 1 
 
 
 Ed 
 
 
 
 
 | 
 
 
 ,-4 
 
 1 
 
 
 
 1 
 
 
 i | 
 
 
 
 
 
 
 
 FIG. 22. 
 
 46. Straight Horizontal Beams on More than Two Sup- 
 ports (Statically Indeterminate). If the supported points 
 of the beam are all free to slip along smooth fixed surfaces 
 (except that one point must be fixed) and there are more than 
 two points supported, the beani is statically indeterminate; i.e. 
 the reactions (and consequently the internal stresses) cannot 
 be determined by simple statics. Our present methods, how- 
 ever (Castigliano's Theorem and the Theorem of Least Work), 
 enable solutions to be obtained by very simple and straight- 
 forward steps. A beam on more than two supports is called 
 a " continuous girder." 
 
 It must be recalled that, in accordance with Castigliano's 
 Theorem ( 21), if a beam is supported at two points (one fixed; 
 the other free to slide on a smooth fixed surface) the displace- 
 ment y\ of the point of application of some one load PI, con- 
 sidered as a variable, in the direction of pointing of this load, 
 
 will be given by 2/i=-Tp-, provided each of the other external 
 forces either is independent of P\, or, on account of being the 
 
62 MECHANICS OF INTERNAL WORK 47 
 
 reaction of a smooth fixed support, does no external work when 
 the beam is loaded. 
 
 If, therefore, a beam has two points supported and is to 
 carry several loads, P, Q, etc., and we desire that the point 
 of application of some one of them, say, Q, shall, during the 
 gradual loading of the beam, execute no movement whatever 
 in the direction of Q; in other words, that its displacement 
 
 shall be zero; we have simply to write rr = zero and solve the 
 
 equation for Q. That is, if the point of application of Q is 
 prevented from moving in the direction of Q, during the loading 
 of the beam, by bearing against a fixed smooth surface at 
 right angles to the direction of Q, the reaction of that surface 
 
 7TT 
 
 will be Q, and must have the value computed from -^ = 0. 
 
 d*t 
 Hence if, before loading, the point of application of Q be 
 
 chosen as a third point of support for the beam (with a fixed 
 bearing surface provided at right angles to Q) the value of the 
 reaction after loading may be found as indicated. 
 
 It is evident that the same reasoning applies to any elastic 
 structure (without initial stresses) as well as to a beam, and 
 also applies when there are more than two points of the beam 
 or structure already provided with supports (of the nature 
 indicated above) ; but, in any case, all the external forces which 
 are dependent on the unknown Q must be expressed as functions 
 
 of Q before the derivative in question is taken; then -377 = 
 
 will give Q in terms of other external forces. 
 
 47. Continuous Girder. Case I. (Fig. 23.) A straight, 
 prismatic, homogeneous beam C'OC is placed on three unyield- 
 ing equidistant supports, at the same level. The two spans 
 C'O and OC, equal in length, each length being Z = a + 6, are 
 loaded symmetrically with two loads, each = P; one in each 
 span, at distance a from the middle support. 
 
 This beam is a " continuous girder " for which at the outset 
 the reactions of the three supports are unknown. The reactions 
 at C and C" are equal, from symmetry, each = F; that at 
 support is V. They cannot be found by ordinary statics 
 and it is now proposed to determine them by the principle 
 announced in the last paragraph. 
 
:47 
 
 CONTINUOUS GIRDERS 
 
 63 
 
 Selecting V as the reaction to be found by the method of 
 46 we are to put 377 = 0- For the present beam there is no 
 
 thrust, and the work of shear will be neglected; so that, by 
 
 . (74) and (75), 38, putting dx for ds, 
 
 dU d 
 
 1 
 
 dM 
 
 extended over the whole beam. The moments (M) of the 
 stress-couples along the beam will aU be obtained in terms of 
 the reaction V, at C (regarded as a variable), and of the known 
 (constant) load P. On account of symmetry it will only be 
 necessary to deal with the internal work, C7, of the right-hand 
 
 half, OC', of the beam; the expression involving which can then 
 be doubled, for the total internal work. 
 
 For any dx on the portion DC (the abscissa x being meas- 
 ured as in the figure) we have 
 
 M = Vx-, and 'jy' ==x > 
 
 while for any dx on OD (and for these dx's we reckon x from 
 point D), 
 
 M = V(b+x)-Px; and jr- 
 Hence, substituting in eq. (94), we obtain 
 
 ' (95) 
 
64 
 
 MECHANICS OF INTERNAL WORK 
 
 48 
 
 The integration being performed and the limits inserted, 
 there results 
 
 2 ' 
 
 
 (When once V is found the other reactions easily follow from 
 ordinary statics) (viz., at C', V; and at 0, 2P-2V). 
 In the particular case where b = a this gives 
 
 (97) 
 
 and hence 
 
 F'-- tf-P. 
 
 48. Continuous Girder. Case II. (Fig. 24.) Straight, 
 prismatic, homogeneous beam on three equidistant and unyield- 
 ing supports, 0, B, and C, at same level. There is but one 
 
 FIG. 24. 
 
 load, viz., P, in middle of left-hand span. Evidently the 
 support at must be placed above the beam and the reaction, 
 Vj at that extremity must point downward, since there is no 
 load between and B. V will be selected as the variable. 
 The beam must be divided into three portions OB, BD, and 
 DC for summation purposes, but first the other two reactions, 
 V and V", since they depend on V, must be expressed as func- 
 tions of V. This is done by ordinary statics, the whole beam 
 being the " free body"; first taking moments about C, and 
 then about B', whence 
 
 F' = 27 + iP and V" = \P -V . . . . (98) 
 
 Reckoning x as shown in figure, for the three respective 
 portions of beam, we have 
 
 onOJ5: 
 
 and 
 
49 
 
 i.e., 
 
 CONTINUOUS GIRDERS 
 
 ,65 
 
 and 
 
 dM 
 dV 
 
 dM 
 
 on DC: M=F"=(iP-F)z; and 
 Hence, substituting in eq. (94), 
 
 f\V(2a-x)-%Px](2a-x)dx 
 
 JQ 
 
 whence, solving, 
 
 ' = and 
 
 49. Continuous Girder. Case III. Non-Prismatic 
 Beam. The straight homogeneous beam of Fig. 25 is sup- 
 ported at three equidistant points at same level. Supports 
 firm. Two equal loads, each = P, one in middle of each span. 
 
 FIG. 25. 
 
 The middle half DA is prismatic, having a rectangular section 
 with constant moment of inertia, /o, =bh 3 -t-12. The extreme 
 quarters, however, CD and AO, have rectangular sections of 
 constant height h and variable width u, u being proportional 
 to distance x from 0, or (7; i.e., u= (x+a)b, so that the moment 
 of inertia along these portions of the beam is variable and is 
 I = uh* + l2 = (x + a)I . 
 
 The unknown reactions being V, V, and V, if V be chosen 
 
 as the variable its value is found by placing -^ I -j-. 
 
66 
 
 MECHANICS OF INTERNAL WORK 
 
 50 
 
 the work of shear being neglected and advantage being taken of 
 symmetry. 
 
 OnOA, M=Vx- r = x; and 7 = / . 
 
 On AB, M=Vx-P(x-a}; j^ = x; and I = I . 
 Hence 
 
 2o 
 
 Va* 
 
 7Pa* 3Pa 3 
 
 . , (98a) 
 
 = AP; and 7' = ffP (986) 
 
 and 
 
 50. Continuous Girder. Case IV. Symmetrically 
 Placed, with Uniform Loading. (Fig. 26.) As in the preced- 
 ing cases (except Case III), the beam is prismatic (I constant 
 along whole length), with two equal spans; supports on a 
 
 Iv |V v 
 
 lUiuiiuliiiimiul 
 
 level; but the load is uniformly distributed over the whole 
 length, at rate of w Ibs. per foot. (Fig. 26.) The load on 
 one span is W = wl, and the reactions are V, V and V. As 
 before, taking F as a variable, and doubling the result obtained 
 from a consideration of the dx's along one span 0(7, or half 
 the length of beam, we are to put 
 
 2 C 
 Eljc 
 
51 CONTINUOUS GIRDERS 67 
 
 The M for any dx is the same function of x for all points of 
 CO; i.e., from the free body of length x, we have along CO, 
 
 x T7 wx 2 dM 
 
 M=Vxwx.-==Vx ~; and ~Jy' ==x 
 
 w 
 
 VI 3 wl* 
 
 -- -- = 0; or V = lwl = lW, and V' = *fW. . (98c) 
 
 51. Continuous Girder. Case V. (Fig. 27.) Beam pris- 
 matic, and loaded uniformly over whole length. Three points 
 of support at same level. Length of right span double that of 
 
 1v v' V"A 
 
 wa 2wa 
 
 .. miimulmuumumiiuuiL 
 
 FIG. 27. 
 
 the left. Rate of loading, w, the same throughout the whole 
 length, which is 3a. E and 7 being constant, we may write, 
 
 (omitting the outside factor I/ El), if V be taken as a variable; 
 the other reactions being V and V" . Measuring x for the 
 dx's on OB, and BC, as shown in the figure, respectively, we 
 have for points on OB: 
 
 ., T7 
 
 M=Vx ^- and -Tf7 = - 
 2 aK 
 
 Taking a free body of length x measured from (7, we obtain, 
 for points on span BC, M=V"xwx 2 ; but since a variation 
 in V would involve a variation of V" also, we must express V" 
 as a function of V before proceeding further. This is done, 
 with the whole beam as a free body, putting by (moms.)==0, 
 about B: whence 
 
 V 3wa 
 ~~ ~~ 
 
68 MECHANICS OF INTERNAL WORK 52 
 
 and 
 
 TUT T*r (V3wa\ wx* dM x 
 
 .-. M on C=^_+_Jz -_; and w =^. 
 
 Hence, substituting above, 
 f/ T , wx 2 \ 
 
 Jo ( F *- ) xdx+ 
 
 i.e., 
 
 Q 
 
 Fa 3 tm 4 27a 3 
 
 383 
 A F = ^; V' = %$wa; and V" = }%wa. . . (99) 
 
 The shear and moment can now be computed for any cross- 
 section of the beam. 
 
 52. Derivative of Internal Work, U, with Respect to MI, 
 the Moment of the Stress-Couple, at Any Section of a Beam 
 in Flexure, when Statically Indeterminate. At any cross- 
 section (1) of a beam, straight or curved, if we imagine the 
 connection of the portions of the beam meeting at this section 
 to consist of a hinge joint at the center of gravity of the cross- 
 section, and two short bars (one on each side of the hinge, 
 and at a proper distance from it) parallel to the axis of the 
 beam at that point, and a distance a apart; then the pressure 
 on the hinge joint will be the resultant of the thrust, T\, and 
 shear, Ji, at that section of the beam, while the stresses PI 
 and PI' in the two bars will be equal to each other, and the 
 product P\a is the moment, MI, of the stress-couple. 
 
 Now if the beam is so supported as to be statically indeter- 
 minate and section 1 is situated between two supports the two 
 short bars above mentioned might be omitted and the beam 
 would still remain in place. In other words, each of the two 
 bars is redundant; and we therefore write 
 
 dU dU 
 
 _ = , and ^-0; 
 
 that is, dividing by a, and adding, 
 
 du "Vo doo) 
 
53 THEOREM OF THREE MOMENTS 69 
 
 But this is simply the complete derivative of U, the internal work 
 of the whole beam with respect to the moment, MI, at cross-section 
 1. (The internal work of the two short bars themselves is 
 considered zero; in other words, they are considered to be 
 infinitesimal in length.) We may therefore write 
 
 ' 
 
 This relation is available for determining the moment of 
 couple MI, at any given section (1) of a statically indeterminate 
 beam, straight or curved (in the latter case the radius of 
 curvature must be relatively large.) 
 
 53. Continuous Girders. Theorem of Three Moments. 
 Although Castigliano's Theorem may still be used for deter- 
 mining the support reactions in case a continuous girder rests 
 on more than three supports, the detail is very burdensome and 
 tedious. The relation just proved in 52, however, may be 
 brought into play with great advantage in such cases, the 
 treatment being very much simplified thereby. The beam 
 being assumed to be homogeneous, straight, and prismatic 
 (constant /), and all the supports to be at the same level and 
 unyielding, a general relation based on eq. (101) will first be 
 proved, known as Clapeyron's Theorem of Three Moments, 
 established as follows : 
 
 In Fig. 28 we have a straight continuous beam with three 
 points of its axis, viz., 0, 1, and 2 (0 and 2 being at the ends) 
 supported on fixed supports at same level (two of these points 
 are free to slide on smooth horizontal surfaces). Various 
 vertical loads, P, P, etc., act on the beam; some in each span; 
 of arbitrary values. Two couples are also applied to the beam, 
 one at each end, of moments MQ and M 2 , respectively; that at 
 the right being clockwise; that at the left, counterclockwise. 
 These couples are virtually arbitrary " loads," like the P's, 
 being independent of them and of each other. It is plain that 
 whatever value might be assigned to any one of these " loads," 
 the beam would still be completely supported. Evidently the 
 reactions V , Vi, and V 2 , at the points of support, are each 
 vertical. 
 
70 
 
 MECHANICS OF INTERNAL WORK 
 
 53 
 
 For any set of given values for the " loads " there will be a 
 corresponding value, MI, of moment of (internal) stress-couple 
 at cross-section 1, over the intermediate support at 1. The 
 beam being statically indeterminate) MI may be found by the use 
 of eq. (101); after rewriting it (neglecting work of shear) thus: 
 
 du d 
 
 2EI 
 
 tfMfSfy* aw 
 
 For sections on span 01 x is reckoned from 0; for those oa 
 span 02, from 2. For section n on 01, taking On as " free body " 
 and n as center of moments, we have, for the moment M, 
 
 (103) 
 
 where (Pa/) denotes the sum of the moments, about n, of all 
 
 ^4 n 
 
 the loads P situated between n and 0. But VQ is a function of 
 MI, obtained by considering 01 as free body and taking moments 
 about point 1, viz.: 
 
 (104) 
 
 Here^jP (Pa;") denotes the sum of the moments, about point 
 
 1, of all the loads P between 1 and 0. 
 
 Now note that if the portion 01 of the beam were cut apart 
 from the remainder and set up horizontally on two supports at 
 its extremities and 1, and loaded with the same loads P as 
 before in same positions (but vnthout the couples acting at 
 
53 THEOREM OF THREE MOMENTS 71 
 
 and 1) the reaction at the support would be R = 
 
 a\ 
 
 and the moment of couple at any section n would be 
 
 ..... (105) 
 
 This ideal moment, M n , which would be occasioned at any sec- 
 tion n if the portion 01 were supported at its extremities as a 
 " simple beam " [bearing the same loads P, but without the 
 end couples] may be called the " normal moment " (as on p. 494, 
 M. ofK). 
 
 As a result of substituting the expression for R in eq. (105) 
 and combination with eqs. (103) and (104) we readily derive 
 for a point n in the left-hand span ... 1, 
 
 M = M -M-(M -Afi)^-, . . . (106) 
 giving M as a function of the one variable MI ; whence 
 
 3IT--. il -. d07) 
 dMi a\ 
 
 Similarly, there are obtained, for any point n in the right span 
 1...2, 
 
 M = M 2 -M n - (M 2 -Mi), (108) 
 
 and -. - > (109) 
 
 a 2 
 
 Substituting in eq. (102), which may be written 
 
 dx = 0,. (110) 
 
 we obtain 
 
 ^-j \M x-M n x-(M -M l )^-\dx 
 
 \ =0. (Ill) 
 
 i C a>i \ x 2 ^ 
 
 + I \M 2 x-M n x-(M 2 -M l )- 
 
 6^2 I ^2 
 
 Of the quantities i, 2 , -M , MI, M 2 , and M n , only M n 
 is a function of x in each span; this ideal moment, or " normal 
 moment," having the meaning just given. Performing the 
 
72 MECHANICS OF INTERNAL WORK 54 
 
 integrations, as far as it can be done in the general case, we 
 have, after reduction, 
 
 6 C ai 6 f a * 
 
 = ) M n xdx + I M n xdx, (112) 
 
 which is Clapeyron's Theorem of Three Moments, giving a 
 relation between the three moments MO, MI, M2, of the couples 
 at the three points 0, 1, and 2; of a homogeneous, prismatic 
 beam supporting a set of vertical loads P, P, etc.; the three 
 points in question resting on three fixed supports at same 
 level. 
 
 Note A. It must be carefully noted that, of the integrals 
 in the right-hand member of (112), the first one deals with the 
 left-hand span and its loads, x being measured from its left end, 
 or point 0; while the second relates to the right-hand span, 
 x being measured from its right end or point 2; and the ideal 
 M n must be a positive number. 
 
 Note B. The three moments, M , M\, and M 2 , have been 
 used in such a way as to imply tension in the upper fibers of 
 the beam, and results must be interpreted on this basis; that 
 is, if a negative number is obtained for one of them it implies 
 that at that section of the beam the upper fibers are in com- 
 pression. 
 
 54. Values of the Two Integrals in the Three-Moment 
 Theorem in Particular Cases. For brevity, the first integral 
 (referring to loads "in the left-hand span) in eq. (112) may be 
 denoted by Zi, and the second (right-hand span) by Z 2 ; and 
 the theorem will then read 
 
 A f\ 
 
 M ai +2M 1(01 + a 2 ) + M 2 a 2 =- . Z l +- . Z 2 . . (113) 
 
 a 0,2 
 
 If the loading on the left-hand span is a uniformly distributed 
 load, Wi covering the whole span (see Fig. 28) a\ at rate w\ 
 Ibs./lin.ft., the value of M n at any point n in portion 01 [con- 
 ceived to be a separate beam resting on two supports at its 
 
 W\ i 
 extremities] would bo x -- ; hence (with W = 
 
54 THEOREM OF THREE MOMENTS 73 
 
 Similarly, if the right-hand span carries a load W 2 , = w 2 a 2 , 
 covering the whole span a 2) we have 
 
 '2=24 
 
 TF 2 a 2 3 . 
 
 With a single concentrated load P in the left-hand span 
 
 Px" 
 
 (Fig. 28) at a distance x" from support 1, we have M n = -- x 
 
 a\ 
 
 for any point n on the left of P, but for any point n between 
 
 Px" 
 
 P and support 1, M n = - x P[x (a\ x"}} ; hence 
 
 tti 
 
 ,-ai-z" /P :r "r\ rai / P'Vr 
 
 Zl= ^LfJ-Hfc+r ( ^-P[; C -( 0l -x 
 
 ~/0 \ a l / */ai-x" \ a l 
 
 i.e., 
 
 s"X'; - '.. - (H5) 
 
 and similarly, for a single concentrated load P in the right- 
 hand span, at distance x" from support 1, 
 
 (If for ai x" we write z" (distance of P from support 0) 
 Zi in (115) becomes 
 
 JP(ai 2 -z" 2 X'; and Z 2 = JP(o 2 2 -2 //2 X / , 
 
 for a load P in right-hand span, with z" = a 2 x".) 
 
 For any number of concentrated loads in the left-hand 
 span, Z\ is the sum of a number of terms of the form given in 
 the right-hand member of eq. (115), one term for each load, 
 and containing the amount, P, of the load and the correspond- 
 ing x" . A similar statement may be made for Z 2 , as regards 
 the right-hand span, if that span carries a number of concen- 
 trated loads; remembering that the x" is measured from support 
 1 (see Fig. 28). 
 
 Of course, if there is no load on the left-hand span, Z\ is 
 zero; and similarly for the right-hand span, with no load on 
 that span, Z 2 = 0. 
 
 For a distributed load over any portion of the left-hand 
 span, the rate of distribution w being constant or variable, we 
 may find the value of Zi by substituting wdx" for P in eq. (115) 
 
74 MECHANICS OF INTERNAL WORK 54 
 
 and summing (i.e., integrating) all such terms between the 
 extremities of the distributed load; that is, by making x" 
 vary between proper limits. If to is variable it must first be 
 expressed in terms of the variable x?'. For example, if a load 
 Wi of constant rate Wi is distributed over the part of the left- 
 hand span which extends c\ ft. toward the left from support 1 
 (so that Wi=wid), we have 
 
 /x"=0 
 
 and similarly if, in the right-hand span, a uniformly distributed 
 load, PP 2 . of rate u> 2 , extends to the right over c 2 ft. from the 
 support 1, we have 
 
 Again, if, in the left-hand span, the uniformly distributed 
 load Wi, =widi, of rate w\, extends d\ ft. toward the right 
 from the support 0, by a similar process we obtain (now making 
 z" vary between the limits GI di and ai), 
 
 while a distributed load Wz, =1020*2, at constant rate u> 2 , cover- 
 ing the portion of right-hand span extending toward the left 
 d 2 ft. from support 2, calls for a value of Z 2 = ^WzdzQa** d#), 
 
 If neither end of a distributed load adjoins a support, results 
 for Zi and Z 2 are obtained by the same methods, by simply 
 using proper limits for the variable x"; (and if to is variable it 
 must first be expressed as a function of x"). 
 
 In conclusion it may be said that, for any land of (vertical) 
 loading whatever, on either span, the value of Z\ (or Z 2 ), is 
 made up of the sum of the separate terms that would apply 
 to each separate part of the loading, concentrated or distributed, 
 carried by the beam in the span hi question. 
 
 55. Continuous Girder: (Prismatic) Resting on an Indefi- 
 nite Number of Fixed Supports at the Same Level. In such a 
 case it is convenient to let the moments hi the sections just over 
 
55 CONTINUOUS GIRDERS BY " THREE MOMENTS 75 
 
 the various supports constitute a set of unknown quantities 
 to be deterimned at the outset, use being made of the Theorem 
 of Three Moments, just proved in 53. 
 
 To apply this method we must note that the beam in 
 Fig. 28 may be considered to be a portion (isolated as a " free 
 body "), comprising any two consecutive spans, of a long con- 
 tinuous girder of several spans. At 0, a section has been made 
 cutting close on the right of a support, MQ being the moment 
 of the couple acting in the section, while VQ may represent the 
 shear in the section; while 2 is a section made close on the 
 left of a support (the second support from the other), M 2 being 
 the moment of couple and Fo representing the shear in the 
 section. We have therefore only to apply this relation to as 
 many lengths, of two consecutive spans each, as circumstances 
 permit, in order to obtain as many equations as there are 
 unknown quantities. 
 
 For example, if there are five spans, i.e., six points of 
 support at same level (the beam simply resting on each sup- 
 port; not clamped or " built-in ") the moments over the six 
 points of supports are the quantities MQ (over the extreme 
 left support; and MQ is either zero if the beam terminates there 
 or there is an unloaded overhang; or is already known if there 
 is a loaded overhang), MI, M 2 , M 9 , M 4 , and finally M 5 
 (which is either zero if there is no overhang, or an unloaded 
 overhang; or is already a known quantity if there is a loaded 
 overhang). Here the theorem may be applied four times: 
 First to the two consecutive spans extending from support 
 to support 2 ; then from 1 to 3 ; then from 2 to 4 ; and finally 
 from 3 to 5. The equations thus obtained are seen to be four 
 in number, containing no unknown quantities except the four 
 moments, MI to M 4 inclusive, which are then determined by 
 ordinary algebra. With these moments known, all reactions, 
 and the moment and shear at any section, are easily found by 
 ordinary statics. 
 
 In case the left-hand end of the beam is clamped, or 
 " built-in/' horizontally, we may consider that the first actual 
 span on the left is the right-hand span of two consecutive 
 spans of which the left-hand span is indefinitely short (ai=0) 
 and carries no load; so that the theorem, being applied to 
 
76 MECHANICS OF INTERNAL WORK 56 
 
 these two spans gives a result obtained by putting a\ and Z\ 
 each equal to zero in eq. (113), 
 
 2Mia 2 + M 2 a2 = (6-ba 2 )Z 2 . . /'--:, (118) 
 
 Similarly, if the right-hand extremity of the girder is fixed, 
 or built-in, horizontally, we consider the last actual span on 
 the right to be the left-hand span of two consecutive spans of 
 which the right-hand one is indefinitely short, and carrying no 
 load; that is, we put a 2 and Z 2 , each equal to zero in eq. (113), 
 and obtain 
 
 M<>ai+2Afiai = (6^ai)Zi. .-.,,. . (119) 
 
 Of course in both of these special cases the built-in section 
 of the beam is represented by the section at point 1 in Fig. 28; 
 with corresponding meaning of the symbols in the above two 
 equations. 
 
 56. Continuous Girders. Case VI. (Fig. 29.) This figure 
 shows a continuous girder extending over three spans and with 
 its extremities resting (not built-in) on the extreme supports, 
 
 f 
 
 wa 
 
 liiUUUU 
 
 II 
 
 wa 
 
 IJIIUi 
 
 ilijiiaiiiimiiiiiiiiii! 
 
 oj 
 
 
 
 i 
 
 
 a 
 
 f 
 
 
 (\}2 "3 1 
 
 j I 
 
 1 u 
 
 
 
 ft 2a - 
 
 i ! 
 
 
 
 and 3. The two spans on the left are each of length a and 
 carry a uniformly distributed load (of the same rate, w, per 
 lin.ft. on both spans) throughout their entire lengths. The 
 other span has double the length of either of the first and 
 carries a uniformly distributed load over its whole length, of 
 rate w'. 
 
 The girder is homogeneous and prismatic and the four 
 supports are at same level, hence the Theorem of Three Moments 
 is available [eqs. (112) and (113)]. Applying it first to the 
 two consecutive spans on the left, viz., ... 1 and 1 ... 2, we 
 find [see eq. (114)], with M = 0, 
 
56 CONTINUOUS GIRDERS BY " THREE MOMENTS " 77 
 
 Next applying it to the (consecutive) spans 1 ... 2 and 
 2 ... 3, we have, with M 3 = Q, 
 
 6 (wa)a* . 6 (2ti/q)(2o) 
 -.-^-+^.- -^- - 
 
 . . .\ . (121) 
 
 By elimination between (120) and (121) we have 
 Mi =Hwa 2 -^w/a 2 ; and M 2 = A-i 
 
 Assuming that w is greater than T 8 rW, we note that both MI 
 and M2 are positive; that is, that the couples acting in the 
 sections at 1 and 2 are of a character producing tension in the 
 upper fibers at those sections (according to Note B of 53). 
 In other words, the curvature of the beam is convex on the 
 upper side; in the way indicated at the points 0, 1, and 2, 
 in Fig. 28. 
 
 The reactions VQ, V\, etc., are now easily found by statics. 
 For example, from the free body ... 1, reaching from sup- 
 port to a cutting plane just short of support 1, we find acting 
 at the right-hand end of the body a shear, J' ', and a couplo 
 (clockwise) whose moment is M\, just obtained. Putting 
 (moms.)=0 about the neutral axis of this section for this 
 body we find 
 
 J' X - V a + wa . - M l = 0, 
 and hence V^=^wa+-^w'a, ..... (122) 
 
 and similarly we may find the other three reactions, taking 
 various "free bodies"; the whole beam being conveniently 
 taken as such, for two of them. 
 
78 MECHANICS OF INTERNAL WORK 57 
 
 CHAPTER V 
 COMPOSITE SYSTEMS; AND CURVED BEAMS 
 
 57. Straight Prismatic Beam Supported by Three Others. 
 Symmetrical Design. In Fig. 30 we have a system of four 
 horizontal straight beams, each prismatic and homogeneous; the 
 upper one, A, resting on the other three, B, B', and B, under- 
 neath; and these latter, on unyielding supports, C" and C". 
 A is at right angles to the others, touching the middle of each. 
 The middle of A touches B'\ and its two extremities touch 
 
 FIG. 30. 
 
 B and B. For A we have the values E and 7; for each of the 
 other beams, E' and /'. There is but one load, viz., P, vertical 
 and applied at the middle of A. Weights of beams neglected. 
 Before the system is loaded the beams are in dose contact with 
 each other and with the supports, without strain; Castigliano's 
 Theorem is therefore applicable; and its corollary (Least 
 Work). 
 
 The nine pressures between the beams and against the 
 supports are all unknown in advance and cannot be deter- 
 mined by statics alone, but they can all be expressed in terms 
 of one of them, say, 7, the pressure under the left-hand 
 extremity of beam A. That at the right extremity of A is 
 also V] and that, V, under the middle of A is readily seen 
 
I 57 COMPOSITE SYSTEMS 79 
 
 to be equal to P 2V, for the equilibrium of /L. Under each 
 end of beams B, B, the pressure is V/2. Under each end of 
 B' it is 772; that is, (P-27)/2. The various half lengths 
 are a and 6; while the x of any point is measured as in Fig. 30. 
 If we consider the left end of beam A to bear on beam B 
 by the interposition of an infinitely short vertical bar the stress 
 in this bar will be the unknown 7, and we may take this bar 
 as the redundant bar of the system; and hence (by 29) write 
 dU/dV = (the internal work of the bar itself being, of course, 
 zero). Here U must denote the total internal work of the six 
 beams, each of which is in a state of flexure, but not under 
 thrust. The work of shear will be neglected. Evidently' the 
 internal work UA, of beam A, is double that of one of its halves, 
 from symmetry (in this case). A similar statement applies to 
 Us', for B'; while for the two beams B and B the internal 
 work UB is four times that of the half length of one of them. 
 It remains to fill in the detail of the relation 
 
 dU d 
 
 . W "37 
 
 1 C^(dM\ , I 
 
 mj M (wH -- 
 
 For beam A, 
 
 2 f*. 
 
 -- \ T/V r/7r 
 
 ~EIJ Q ' ~ 
 
 ' dM dU A 2 *. 27a 3 
 
 r 
 
 dV dV ~EI ~ 3EI' 
 
 For the two beams B, B, 
 
 Vx dM x dU B 4 b Vx x 763 
 
 M 2 5 ^T7 ^O 5 * 
 
 For beam B' ', 
 
 7'x /P-27\ dM 
 2 ~ \~~2~~r 5 d7" = ~ :c; 
 
 and 
 
 dU B ' 2 C 
 ~d^ = ETJ 
 
 b (P-2V)x 2Vb* Pb* 
 
 ~ x > dx ~ 
 
 dV ~E'I'J Q 2 
 
 Substituting these values in eq. (123) we obtain 
 
80 
 
 MECHANICS OF INTERNAL WORK 
 
 58 
 
 If all the beams are of the same material and the same 
 moment of inertia of cross-section, and a = b, this reduces to 
 
 Example I. Find the value of V for the case in Fig. 30 
 in case the weights of the beams are not neglected (weight per 
 running foot being w for beam A, and w f for each of the others). 
 
 Example II. Instead of the concentrated load applied at 
 the middle of beam A, apply a load W distributed uniformly 
 over the whole length of beam A, in Fig. 30; first, neglecting 
 the weights of the beams; then, considering those weights. 
 
 For cases resembling that of Fig. 30, but in which the 
 arrangement is not symmetrical, the mode of treatment will be 
 the same but evidently more detail will be involved. 
 
 58. Trussed-Beam (or Brake-Beam). In Fig. 31 at 
 (A) is shown a straight, prismatic, homogeneous beam with 
 its two extremities resting on two piers at the same level; 
 
 (B) 
 
 provided with a short vertical strut, DK, at its middle point, 
 and also with two oblique tie-rods connecting the lower end 
 of the strut with the two extremities of the beam. It is to be 
 understood that these four members are accurately fitted and 
 secured to each other, before loading, in the position shown, 
 but not so as to create any stress in any member; in other 
 words, the beam is still straight, and stresses will begin to be 
 produced in oil the members as soon as any load is placed 
 upon the beam; that is, there are no " initial stresses." 
 
 In the present case the only load is the concentrated vertical 
 load P, at the point Z); when bearing which, the structure is 
 assumed to be at the same temperature as when the parts were 
 put together. It is required, by the method of " Least Work," 
 to find: The tensile stress, T\, induced in each tie-rod; the 
 
58 TRUSSP:D-BEAMS 81 
 
 compressive force produced in the strut, DK; the thrust, T r 
 produced in the horizontal beam itself; and 4 an expression from 
 which we can compute the " bending moment," M, at any 
 section of the beam. As regards the total internal work, U, 
 of all the members, that of the strut (on account of its short 
 length) will be neglected, that of each tie-rod will be an expres- 
 
 f^ 'T 7 2 
 
 sion for the case of simple tension, viz., ^ , while that for 
 
 CT 2 
 
 the beam will consist of a term ~ for the work of the thrust, 
 
 /M 2 dx 
 9 y , , for the work of the " bending moments "; 
 
 (that due to the shears being neglected). 
 
 Here C\ denotes cri=r (see 20), where l\ is the length and 
 
 FI the sectional area of each tie-rod; E\ being the modulus of 
 elasticity. Also C = l + FE, where I ( = 2a) is the length, and F 
 the sectional area, of the horizontal beam itself; and E the 
 modulus of elasticity of its material. The angle DBK is denoted 
 by a (see Fig. 31). 
 
 At (B) in Fig. "S&, is shown the free body nOnf with all the 
 forces external to it, we, have from simple statics (since P/2 
 is evidently the reaction of each support), 
 
 P 
 - 2 X a y 
 
 dM 
 (so that -Jm-= x sin a); which are to be substituted in the 
 
 expression for internal work of the whole structure, viz. : 
 
 By differentiating this last expression with respect to T^ 
 we have 
 
82 MECHANICS OF INTERNAL WORK 58 
 
 which, after substitution of above relations, is to be put equal 
 to zero, according to eq. (49). We thus progressively derive 
 
 2CiTi+CTi (cos a) cos a 
 
 2_f*-*/Px . \ 
 
 + #/Jo V2 " *)(*:.$ 
 
 or, 
 
 2 sin a C a \P 
 
 Ti[2Ci + C cos 2 a] - ' VJ I -^-x 2 dx - TI (sin 
 ^y J [.2 
 
 that is, 
 
 <^nf _^^ (l 
 
 and finally, solving for 7\, the tension in either tie-rod, noting 
 that a is only the half-length of the beam, we obtain 
 
 T = _ Pa 3 sin , __ 
 +3C cos 2 a) +2o* sin 2 a 
 
 T, and the thrust in DK, are now easily found, and the value 
 of M at any point of beam; and finally the maximum unit 
 stress (which should be below the elastic limit) in the outer 
 fibers of the beam; as due to combined thrust and bending 
 moment. 
 
 If all members of the system are of the same material they 
 expand, or contract, together, in case the temperature changes 
 after the parts are put together; but if the beam were of timber, 
 for example, and the tie-rods iron or steel (as is often the case), 
 the metal would expand so much more than the timber, for a 
 rise of temperature, that the maximum moment, and stresses 
 dependent on it, in the beam, might be much larger than if 
 the temperature had remained at the original value (see next 
 paragraph). Similarly, a fall of temperature might greatly 
 alter the stresses. 
 
 59. Cambered Trussed-Beam ; and Effect of Change in 
 Temperature in an Ordinary Trussed Beam. If it is desired 
 to give a " camber " or slight curvature (convex on upper side) 
 to the beam of the foregoing article, before it is loaded, this 
 may be done by making the vertical strut a little longer than 
 is called for by the exact fitting of all the parts without strain. 
 
59 
 
 TRUSSED BEAMS 
 
 83 
 
 In such a case, to find the stresses in the tie-rods, etc., after 
 loading, use may be made of 32, as follows : 
 
 In Fig. 32 the tie-rods BK and KO are shown in such a 
 position that the distance from B to is horizontal and equal 
 to the length of the beam (between joint centers). The strut 
 KD, of a length a little greater than for no camber, being placed 
 vertically with lower extremity at K, its upper end D projects 
 a small distance D'D, = A', above the line joining B and 0, 
 and hence when the axis of the beam is placed so as to pass 
 through B and K, with one extremity at B, the other extremity 
 will be found at 0'. If now we join KO' and let fall the per- 
 pendicular . . . e upon the line KO', we (practically) deter- 
 mine the distance O'e, = ^, as the amount by which the tie-rod 
 
 FIG. 32. 
 
 on the right is too short to connect the " joints " K and 0'. 
 This is called ^ in 32, the bar in question being a redundant' 
 member. Hence the stress, TI, in the tie-rod, after the parts 
 are forced together and the structure loaded (at the same 
 temperature), may be found from eq. (57) of 32; that is, by 
 putting dU/dTi = ^ , the stress TI being assumed to be tension. 
 From Fig. 32 we have O'e=~00' . sin a = 2Diy . sin a; i.e., 
 ^-S^sincj, 
 
 We may therefore find the stress TI in the tie-rod by putting 
 ^ instead of zero in the right-hand member of eq. (126); thus 
 obtaining 
 
 T 
 
 EI(6d +3C cos 2 a) + 2a 3 sin 2 a 
 
 
 If, with the uncambered trussed-beam, the temperature, t' , 
 after loading, is lower than that, t Q , when the parts were put 
 
84 MECHANICS OF INTERNAL WORK 60 
 
 together, it becomes evident, from a figure drawn to represent 
 the (straight) axes of the beam, the strut, and one of the tie- 
 rods, as assembled (unstrained) at the new temperature t', 
 that the other tie-rod is too short to fit into place by some 
 amount, ^o, T\ may be found from eq. (128) by insertion 
 therein of the value of ^ . For example, if the coefficient of 
 expansion or contraction for the steel tie-rods is >?, while that 
 of the timber beam is negligible in comparison, as also that of 
 short strut, we have, 
 
 If the temperature t f of loading is higher than the original 
 temperature to of " assembling," with steel tie-rods and timber 
 beam, we replace the ^ of eq. (128) by the (negative) quantity : 
 -2y(1f -to)li; for instance, if ' = 35, and fo = 15, C., with 
 Zi = 10 ft. --=120 in. we have 
 
 ^ = -2(0.000011)(35 -15)(120) = -0.053 inch. 
 
 60. Other Cases of Trussed-Beams. Example I. Find 
 the stress in one of the tie-rods of the trussed-beam in Fig. 31 
 in case the loading, instead of being concentrated in the middle 
 D, is a uniformly distributed load, W, covering the whole 
 length (2) of the beam at rate w Ibs. per lin.ft. Disregard 
 changes of temperature. 
 
 5wa 4 sin a 
 l = 24Ci + 12C cos 2 a) +8a 3 sin 2 a 
 
 Example II. In Fig. 31 suppose the single concentrated 
 load P to be placed at a point in the right-hand half of the 
 beam and at a distance = c from support 0. Find T\, the 
 stress in tie-rod OK. 
 
 Example III. Trussed-Beam with Two Struts and Three 
 Tie-rods. Fig. 33 shows this case; the load, W, =3wa, being 
 uniformly distributed over the whole length, 3a, of the beam; 
 which is prismatic and homogeneous and is provided (at points 
 dividing its length into thirds) with two short vertical struts, 
 and three tie-rods 1, 2, and 3, as shown; tie-rod 2 being hori- 
 zontal. The figure shows the notation as to dimensions and 
 angles. The internal work of the struts will be neglected, as 
 also the work of shear in the beam. Rods 1 and 3 are equal in 
 
00 
 
 TRUSSED-BEAMS 
 
 85 
 
 sectional area FI and length /i ( = aseca), their modulus of 
 elasticity being E\ ; corresponding quantities for rod 2 being 
 F 2 , E 2 , and 1 2 ( = a); and for the beam, F, E, and I ( = 3a) 
 (with / as moment of inertia of section) . 
 
 Let 
 
 E 
 r \ 
 
 be denoted by 
 
 
 b Y ^2; and ^^ by C. 
 
 The four members, assumed to be fitted together closely, 
 without strain, before loading, form an elastic system with one 
 redundant bar. Bar (rod) 1 will be taken as the redundant 
 
 w= 3 wa 
 
 ill will-ill 1 1 II 
 
 FIG. 33. 
 
 bar; and the stress in it, T\, is to be determined, for the given 
 loading; change of temperature being disregarded. Hence by 
 29 we write dU/dTi = Q. Let T^the thrust in the beam 
 (evidently the same at all sections), and M the (variable) 
 " moment " in any section; and T 2 , the tensile stress in rod 2. 
 In the present case we have, therefore, 
 
 dU dT * 
 
 dM 
 
 dM 
 
 From the free body situated on the left of a vertical 
 cutting plane, passed anywhere between and H and at a 
 distance x from 0, we find 
 
 and 
 
 wx 2 
 - 
 
 whence 
 
 dT 
 
 dM 
 and -77^"= x sin a. 
 
86 
 
 MECHANICS OF INTERNAL WORK 
 
 61 
 
 Also, from the free body extending from to any vertical 
 cutting plane passed between H and K, x measured from as 
 before, we have 
 
 wx 2 
 T 2 = T, and M = wax- - 
 
 that is, T 2 = 
 whence 
 
 cos a, 
 dT 2 
 
 wx* 
 
 and M = %wax T\a sin a ; 
 
 , dM 
 
 = cos a, and -rsfr= a sin a. 
 al i 
 
 These various functions of TI and corresponding derivatives 
 being substituted in the detailed expression just given for 
 
 -J77T, the various integrations performed, and the result placed 
 al i 
 
 equal to zero; there is obtained, on solving for T\, 
 
 llwa 4 sin a 
 
 6EI[2Ci + (C 2 + C) cos 2 a] + 10a 3 sin 2 a 
 
 (129) 
 
 If the temperature under load is different from the original 
 temperature, the new value of TI may be obtained in the 
 manner already illustrated in 59. 
 
 61. Straight Prismatic Beam Supported by Two Piers and 
 by Parabolic Cable. (Fig. 34.) The extremities of the beam 
 rest on two fixed piers and C, situated at the same level 
 (and capable of preventing upward as well as downward move- 
 
 (B) 
 
 ment of the ends of the beam) and the beam is attached by 
 (very numerous) vertical suspension rods, uniformly spaced 
 along the horizontal, to a cable fastened at its extremities to 
 the two fixed points E and F, which are at a common level 
 
61 BEAM SUSPENDED FROM CABLE 87 
 
 and vertically over and C. It is stipulated that before any 
 load acts on the beam (its own weight and that of rods and 
 cable being neglected) all the suspension rods are barely taut, 
 as also the cable (from one point of attachment of a rod to 
 the next), and that the curve formed by the cable is a common 
 parabola, with its geometrical axis vertical and passing midway 
 between points E and F. R is the vertex of the curve. That 
 is, the beam and the suspension rods and the cable are closely 
 fitted to each other initially, and without strain, so that if 
 a vertical load is gradually applied to the beam the stresses 
 in the beam and rods and the parts of the cable increase pro- 
 gressively, and proportionally, from zero up to their final values. 
 Evidently this presents a statically indeterminate problem; 
 involving but one redundant element, if we consider the bending 
 of the beam under load so slight that the curve of the cable 
 remains practically a parabola. For this character of curve 
 it is known from mechanics that the tensions in the suspension 
 rods are all equal if these rods are very numerous and uniformly 
 spaced along the horizontal, as in the present case. The deforma- 
 tion (stretching) of the cable and suspension rods, when the 
 beam is loaded, will here be neglected, as having but small 
 influence compared with the bending- of the beam. 
 
 Let now a uniformly distributed load, W ', of rate w' per 
 foot run, be placed on the beam, extending from along a 
 portion OD, =a, of the length, Z, of the beam; i.e., W ' = w'a; w' 
 and a being given. This will produce reactions, viz., V at 
 and V at (7, from the piers, and equal stresses in all the 
 vertical rods at same rate w per horizontal foot. The system 
 of forces now acting on the beam, as " free body/' is shown 
 at (E) in Fig. 34. There are three unknowns, viz., V, V ', and 
 w} while statics provides only two independent equations; 
 hence the third must depend on elastic theory. As the single 
 " redundant element " we may take either V, V, or w\ let 
 us choose w; then the others must be expressed as functions 
 of w. From the free body at (B) (whole beam), by moments 
 about C we find 
 
 wl 
 
 < 130 > 
 
88 MECHANICS OF INTERNAL WORK 61 
 
 If the vertical rods are b ft. apart, the stress in some par- 
 ticular rod, which is to be taken as the redundant bar, is ivb] 
 
 and by the principle of least work ( 29) we may put , ^ =0: 
 
 dU 
 
 i.e., 7 = 0. 
 ; dw 
 
 The internal work of shear being neglected, and also that of 
 the cables and vertical rods (as previously implied), we have, 
 therefore (E and 7 constant for beam), 
 
 -- 
 
 dw J D dw 
 
 For the dx at any point n on portion OD of beam, x being 
 measured from 0, taking On as a free body, we have [see also 
 eq. (130)], 
 
 [w'a( a\ wl~\ 
 TV 2/~tj 
 
 wx 2 w'x 2 
 "~2 2~' 
 
 dM I x 2 x 2 -lx 
 and hence ~~[ w ~ = ~2 X+ ~2' = ~~2 ' ' ' * '"- * ' ' 
 
 while for any point n' on portion DB, with x measured from 0, 
 and On' as free body, 
 
 . ... (134) 
 
 [w'a /_ a\ twZ"l wx 2 I a\ 
 
 -rV-2J-yJ x+ -^-2)5 
 
 dM I x 2 x^-lx 
 
 and ' ar'-ff+t- = ^~" ' ' ' I iS ( ^ 
 
 Hence substitution in eq. (131) gives 
 
 w'x 2 ] /x 2 lx 
 
 C a r ^'a 2 .r it?Zx i^a: 2 w'x 
 
 J, r"'ijr -~ + 
 
 w'a?x wlx wx 2 
 -2T -2- + -^ 
 
61 BJ AM SUSPENDED FROM CABLE 89 
 
 . Performing the integrations between respective limits indi- 
 cated, and reducing, we find 
 
 - 2w'a 5 + 5t//a 4 Z - 5w/a 2 Z 3 + 2wl 5 = ; 
 solving which for w, writing m for the ratio a+l, we have 
 
 fy)2 
 
 w = ^-(4m 3 - 5m 2 + 10)w'. (136) 
 
 With w known, V and V, and the moment and shear at any 
 .section of the beam, are easily computed, in a numerical case. 
 
 [Note. Evidently this problem involves the case of one 
 variety of suspension bridge; in a more thorough treatment 
 of which, however, the deformation of the cable should not be 
 neglected; see Professor Hiroi's book on " Statically Indeter- 
 minate Structures " for a fuller investigation. The relation in 
 eq. (136) is confirmed by a statement in Mr. H. M. Martin's 
 small work of same title as Professor Hiroi's; and the results 
 given in the remainder of this paragraph are quoted from Mr. 
 Martin's book.] 
 
 If a single concentrated load P is carried on the beam in 
 Fig. 34, instead of a distributed load, being placed at any 
 distance c from the left support 0, similar analysis (writing n 
 for the ratio c-^-Z) leads to the result 
 
 ; ..... (137) 
 
 while the reaction V at the support C is found to be 
 
 7 / = tn(10n 2 -5n 3 -3)P. .... (138) 
 
 For example, if c = JZ, V = &P (points downward). 
 
 If the beam is built in horizontally both at and 
 (" fixed ends ") instead of merely resting on the piers (" hinged 
 ends "), with original close fit without strain as before, the 
 same treatment may be adopted but the detail is much more 
 cumbersome. We have now three redundant elements; viz., 
 w and the moments M and M c of the two internal couples 
 in the beam at and C. If any one of these, say M , be 
 
90 MECHANICS OF INTERNAL WORK 62 
 
 regarded as a single redundant element, the stresses in the 
 rods and the forces forming the other couple may be treated 
 as independent loads, like P itself (see 53), justifying us in 
 
 placing - =0. Similarly, we have ,,, =0, and -, = 0; and 
 
 ttiVi o CLIVL c (*W 
 
 thus derive three independent equations, aside from the rela- 
 tions of statics. In this way w, M , and M c can be found. 
 
 If in this case ( "fixed ends ") the load is uniformly dis- 
 tributed, viz., W' = w'a, extending a ft. from support 0, we find 
 (with m = a-r-i), 
 
 '\ . : . . (139) 
 
 and the shear in the vertical section of the beam close to 
 support C is 
 
 J c ^m 3 (7m -4 -3m 2 )w'l. . . _. . (140) 
 
 Also the value of M c in section at C is 
 
 . . . . . (141) 
 
 (A positive result indicates tension in upper fibers.) 
 
 Again, if the beam in Fig. 34 (A) has fixed ends and carries 
 a single concentrated load, P, at distance c from left end O 
 by similar process we find (denoting c + l by n), 
 
 P 
 
 , .... (142) 
 
 J c = n 2 (28n-12-15n 2 )P, .... (143) 
 and M c =Jn 2 (3-8tt + 5^ 2 )PZ. .,. . . (144) 
 
 62. Double-Knee Beam ; under Side Pressure. Constant 
 E and I. Fig. 35 shows this " double-knee beam " as it may 
 be called. It is a continuous beam, homogeneous, and of 
 .constant moment of inertia of section (7), consisting of three 
 straight portions, continuous at trie " knees," G and D: two 
 vertical and one horizontal. Its own weight is to be neglected. 
 Its extremities are hinged to two immovable piers, at same 
 level, and B\ and before any load or force is brought upon 
 it, it is under no strain] that is, it does not have to be "sprung" 
 to fit it over the hinge pins. (It must be noted that there 
 are no hinges at the corners G and D.) 
 
62 
 
 ''DOUBLE-KNEE" BEAM 
 
 91 
 
 Let now a horizontal wind pressure, wh, act against the- 
 left-hand side, of intensity of w Ibs. per linear (vertical) foot. 
 It is required to determine the horizontal component H, of the 
 hinge reaction at 0, the vertical component at being V 
 (components at B are H' and V) by the foregoing principles. 
 Notation as in figure. Also let F = common sectional area of 
 
 all parts of beam and let C\ denote -7^77; and C% denote 7 
 
 Denote by T\, Tz, and T 3 , the (unknown) thrusts in por- 
 tions OG, GD, arid DB, respectively; and neglect the internal 
 work due to shears. 
 
 This is a statically indeterminate elastic structure: with 
 only one redundant element, however, and this may be taken 
 as either H, or H f \ say, H. 
 
 We shall therefore put "TJ^O; 
 
 which may be justified in either 
 
 of two ways. Note that of 
 
 the five external forces, wh is 
 
 fixed, i.e., given; and that V 
 
 and V are also fixed and are 
 
 not affected by any imagined 
 
 change in H or H'\ since by 
 
 taking moments about B we 
 
 have V = wh 2 + 2l. If the sup- FIG. 35. 
 
 port at consisted of a fixed 
 
 horizontal smooth surface, a change in H would simply mean 
 
 a shifting of joint of beam along this surface and the beam 
 
 would still be supported; H may therefore be considered as a 
 
 load, independent of P, and must have a value corresponding 
 
 to the fact that in the present case joint of the beam does 
 
 not move horizontally during the gradual application of the 
 
 loads wh and H (H' would simply be a " neutral " reaction 
 
 during this application; see 21); i.e., dU + dH = Q, by 21. 
 
 Again, from another point of view, consider that the beam 
 is connected with the fixed support on the left, not by a hinge 
 joint, but by two inelastic bars, one horizontal, the other ver- 
 tical; then the stress in the former would be H] in the latter, 
 V. These bars must be inelastic, otherwise joint of the beam 
 
92 MECHANICS OF INTERNAL WORK 63 
 
 would move during the gradual application of the load wh\ 
 and are now to be considered as part of the structure. Since 
 the beam is only hinged to the support at B, and not " built-in/' 
 it is evident that the vertical bar is a necessary bar and not 
 redundant, since without it the beam would not be supported. 
 But the horizontal bar is a redundant bar; since, if it were 
 omitted, the beam would still be supported, the joint simply 
 shifting to a position a little to the right of the original. Hence 
 by 29 we may put dU + dH = Q; in which U should include 
 the internal work of the two bars; but these being inelastic, 
 such internal work is zero, so that in this case U simply refers 
 to the internal work of the beam alone. We shall therefore 
 use the relation dU + dH = Q and solve for H; remembering that 
 V does not depend on H and hence need not be replaced by its 
 value in terms of wh, until toward the close of the algebraic 
 work. 
 
 From 37, eq. (74), we have 
 
 and hence 
 
 dU dK dT* dT 1 r" dM 
 
 ^ ClT 
 
 dT 1 r 
 3 dH + EiJ 
 
 For the ds's situated along OG, ds = dy; T\ V\ and 
 
 = Hy-^~. For those on GD, ds = dx-, T 2 =wh-H; and 
 
 wh 2 
 
 = Hh ^ Vx; while along DB, ds = dz; T 3 = V; and 
 
 We have, therefore, 
 
 fl> - ( + y)dy = 
 
 C D * 
 Jo 
 
62 " DOUBLE -KNEE " BEAM 93 
 
 
 Hh* wh* Vlh 2 
 
 12 2 ' 
 
 Noting also that both dTi+dH and dT^^-dH are zero, 
 while dT2+dH= 1, and substituting in the expression 
 dU^dH = 0, we find 
 
 2 * 
 
 A similar problem is treated by Mr. Mensch, on p. 80 of 
 Engineering News of Feb., 1900. Besides the wind pressure 
 on the vertical side Mr. Mensch also considers a uniformly 
 distributed load along the whole length of the horizontal por- 
 tion GD, of intensity of w\ Ibs. per linear foot. He not only 
 neglects the internal work due to shears but also that due to 
 the thrusts. In his problem the moment of inertia of the 
 section of the portion GD is different, viz., /i, from that, /, 
 of the vertical portions of beam. His final result for H is, 
 writing n for the ratio I\ -=- /, 
 
 .. 
 
 The results in eqs. (145) and (146) are seen to agree when in 
 (145) we make C 2 = (implying that the effect of the thrusts 
 is negligible in this connection), and in (146) place w\ equal 
 to zero and I\ equal to /. 
 
 If the horizontal side pressure is P Ibs. concentrated at the 
 corner G, instead of the distributed pressure wh, it may easily 
 be proved that H = H' = $P. 
 
 Example. If, in the case of Fig. 35, the loading consists 
 of one vertical load, P, applied vertically at the middle of GD, 
 find the value of H] and also the vertical displacement of the 
 point of application of P, 
 
MECHANICS OF INTERNAL WORK 
 
 63 
 
 63. Loaded Davit of Uniform Section. Displacement of 
 Extremity. The davit [see (A) in Fig. 36] consists of a straight 
 vertical portion DK, built in vertically in the fixed support at 
 D, and the curved portion KO, which is the quadrant of a circle 
 of radius r. E and 7 are assumed constant throughout the 
 whole length DKO. The tangent at K is vertical (before load- 
 ing). Neglecting the weight of the davit itself, it is required 
 to find the vertical displacement Oa, =?yi, of the extremity 0, 
 as due to the gradual application of the vertical load PI. 
 Before loading, the extremity is at 0; after loading, at 0'. 
 
 ~ 
 
 n /\\ a f T 
 
 r XPS .,1 
 
 d xN ^ PI 
 
 P X \ I / S <1.. 
 
 FIG. 36. 
 
 Since the projection of the displacement 00' upon the line 
 of the external force or load PI is the quantity sought, viz., 
 Oa, =2/1, it may be determined as the derivative of the total 
 internal work, U, of the beam, with respect to PI; i.e., by 
 
 q. (25) of 21, or yi =jp~- Neglecting the work of shear, and 
 
 also in, this case the work of thrust, we have only the internal 
 work due to bending moments, Af; and hence may write 
 [from eqs. (74) and (75), 371, 
 
 dM 
 
 M - 
 
 dM 
 
 (147) 
 
 For the ds's situated along DK we measure z from D; so 
 that ds = dz. For those along the curved part KO, using polar 
 
63 LOADED DAVIT 95 
 
 .coordinates, as shown, with angle measured from radius CK, 
 we have ds r . dd. 
 
 For any point n f on DK we have 
 
 . . . 
 
 M = Pir: -nT- = r: and ds = dz\ 
 
 while for any point n on the curve KO we have, from the free 
 body nO, shown at (B) in Fig. 36, 
 
 M = Pir'cosd: -775- = r cos 6 ; and ds = rd6. 
 
 Substitution in eq. (147) gives (see integral forms in the 
 Appendix), 
 
 _ - (147a) 
 
 To find the horizontal displacement of 0, viz., aO' in (A), 
 Fig. 36, since there is no external force at pointing hori- 
 .zontally we have simply to introduce an imaginary force PI 
 at directed horizontally toward the right, and use the rela- 
 
 tion aO' = 2/2 = -7D~, m which finally P% is to be made equal to 
 
 OIL 2 
 
 :zero; thus following the procedure of 23a and 40. With 
 P2 as well as PI in action we have 
 
 For any point n' on DK : 
 
 For any point n on quadrant KO : 
 
 M = Pir cos d + P 2 r[l -sin 0]; ( r = r(l -sin 
 
 (i/ 2 
 
MECHANICS OF INTERNAL WORK 
 
 64 
 
 Substituting in y 2 = dM/dP2 (making, however, P2 = in 
 each of the expressions for M, before substitution), we derive 
 p r r rh /v/2 n 
 
 y 2 = -L.\ J (h+r-z)<b+i*l (cos -sin cos 
 
 (/. + r-r.V. , . (148). 
 
 64. Semicircular Curved Beam, or Arch Rib ; Hinged at 
 Ends. Single Eccentric Load. The curved beam is of semi- 
 circular form, radius = r, and is assumed to be homogeneous 
 and of constant moment of inertia. That is, E and / are con- 
 
 FIG. 37. 
 
 stant. Its own weight is neglected. It is hinged at its extrem- 
 ities to the fixed supports and K, at the same level, being 
 fitted to the hinges without strain, at temperature to, before load- 
 ing. The temperature remaining constant, a single eccentric 
 vertical load P is then placed at the point m, 60 from tho 
 right-hand horizontal radius, CK. Evidently the line of P is f r 
 from and jr from K. It is required to find the horizontal 
 component, H, of the reaction at each hinge (evidently, in this 
 case, H is the same for both). 
 
 Considering the whole rib as a free body, we easily find by 
 ordinary statics (by moments about and K) that the vertical 
 components of the reactions at and K are }P and fP, 
 respectively (they are independent of H)] that is, H con- 
 stitutes the only redundant element in the problem; and may 
 
 be found by putting -777 = (bringing into play the same method 
 of proof that was used in the fore part of 62). 
 
64 SEMI-CIRCULAR ARCH RIB. END HINGES 97 
 
 Here the work of the thrust, T, will be considered, as well 
 as that of the " bending moment," M, but the work of shear 
 will be neglected. With F as the sectional area of the beam, 
 we have from eqs. (74) and 75), 
 
 j 
 M dff d8 
 
 For any point n between and m, with 6 as variable angle 
 (see Fig. 37), we have, from the free body On (like that at 
 (B) in Fig. 38), 
 
 dM 
 
 M = \Pr(\ -cos 0) -Hr sin 0; --= -r sin 6\ 
 
 dT 
 
 ~ = * [n 6] ds==rd6 '> 
 
 6 
 
 while for any point n' on part Km, reckoning the corresponding 
 variable angle 6' from radius CK, n'K being the free body, 
 
 dT 
 
 = cos 
 
 /=fPsinfl'-#cos 0'. 
 Hence, with ds = rdd, or rdO', we find 
 
 sn + sn cos 
 
 dU r 3 r i20 
 
 dH = in j (~* p si 
 
 r 3 /^60 
 
 + E~1 1 ( ~ 
 
 r rim* 
 
 + I ( \P sin OcosO + H sin 2 
 r /i^o 
 
 sin P + sin ^ cos ^ +r sin2 
 
 sn cos 
 
98 MECHANICS OF INTERNAL WORK 65 
 
 Performing the integrations (see Appendix), collecting terms 
 and reducing, and placing dU/dH = Q, we find 
 
 0. . (149) 
 
 Writing I = Fk 2 , in which k is the radius of gyration of the 
 cross-section of the beam, and solving for H there results 
 
 ' (150) 
 
 H being found, the thrust, shear, and bending moment are 
 easily determined for any section of beam. 
 
 The radial thickness of the beam being small compared 
 with the radius r, the effect of considering the work of thrust 
 in this semicircular form is extremely slight. Disregarding the 
 work of thrust, which simply means dropping the term affected 
 by the factor (r+FE) in eq. (149), or (which is the same thing) 
 neglecting the term k 2 in (150) compared with r 2 , we have 
 
 # = 0.2386P, . . . . . .' (151) 
 
 instead of (150). For example, with A: = 5 in., and r = 20 ft. = 
 240 in., the two results differ by about one-tenth of one per 
 cent, only. Even when the form of rib is a fairly flat circular 
 arc, the effect of the thrust terms is relatively small. 
 
 The effects of a change of temperature will be treated in 
 66. 
 
 65. Semicircular Arch Rib with Tie-Rod. In this case 
 two fixed piers with smooth horizontal surfaces furnish vertical 
 reactions, only, at the ends of the semicircular rib (constant E 
 and I as before) whose extremities are connected by a horizontal 
 bar or tie-rod. Piers at same level. (See (A), Fig. 38.) 
 
 When the load P, to be placed at the " crown/' m, or highest 
 point of the rib, is applied, the ends and K of the rib spread 
 apart slightly, and at the same time the horizontal tie-rod is 
 stretched; the final stress in the latter being 7\, which is to 
 be determined; it being understood that the rib and tie-rod 
 are fitted closely together without strain, at some temperature 
 to, before the load is applied, and that the temperature does 
 
65 
 
 SEMI-CIRCULAR ARCH RIB WITH TIE-ROD 
 
 99 
 
 not change. Let ^i denote the sectional area, li the length, 
 = 2r, and EI the modulus of elasticity, of the tie-rod; C\ repre- 
 senting li~FiEi. 
 
 Evidently, in this elastic system, the tie-rod is a redundant 
 bar; there being but one redundant element in the case; and 
 
 hence we write j^r^O? by 29. The reactions (vertical) of 
 
 the two smooth piers are each P/2, from symmetry, and inde- 
 pendent of TI. Considering the work of thrusts (including TI) 
 
 d f C K M 2 ds C K T^dsl 
 dflljo ~2ET + Jo 2FE\ = 
 2 C m dT 
 
 TTTI I * jrji ^*" '-' 
 
 tijo dl i 
 
 FIG. 38. 
 
 and of moments, but not that of shear, we have from eqs. (74) 
 and (75), etc., 
 
 dU o 
 
 (152) 
 
 (On account of symmetry, we are enabled to put the internal 
 work of the rib equal to twice that of the half between and m.) 
 
 To find moments and thrusts in terms of TI consider the 
 portion of the beam from to any point n in the left-hand half, 
 as a free body; see (B) in Fig. 38. M is found by taking 
 moments about the gravity axis of the section at n\ and T 7 , 
 the thrust, by summing components parallel to the tangent 
 /.-..; thus excluding the shear J from either sum. 
 
 We have, then, for points on the quadrant Om, 
 
 dM 
 M = vVPr(l-cos 0) -TV sin 0; ^r= -rsin 0; 
 
 4^sin0; ds = rdO. 
 
100 MECHANICS OF INTERNAL WORK 66 
 
 Substitution in eq. (152) gives 
 
 9r3 /"/2 
 
 (7 1 y 1 + I ( _ jp s i n e + JP sin cos + TI sin 2 6>)d0 
 ^Vo 
 
 2r A/ 2 
 +-^ J ( JP sin cos + T 7 ! sin 20)d0 = 0. . (153) 
 
 Performing the integrations (see Appendix) and reducing, 
 
 ^ P siT. rP r2V 
 ~/ 2 + 2 EI + 2FE + 2FE u ' 
 
 
 (154) 
 
 66. Temperature Stresses for Semicircular Arch Rib. 
 
 If the rib is hinged to two fixed piers, as in Fig. 37, and the 
 temperature t under load is higher than that fo of " erection " 
 (when the rib was fitted in place without strain), we may con- 
 sider the piers to furnish only the vertical reactions, and that 
 the horizontal reactions are due to an inelastic horizontal rod 
 or bar connecting and K, whose coefficient of expansion is 
 zero, and which was fitted in place without strain at the original 
 lower temperature t Q . Then if the two parts, rib and bar, 
 were to be fitted together at the higher temperature it would 
 be found that the bar is too short by Ao = 2rr)(t to), y being the 
 coefficient of expansion of the material of the rib; since under 
 no constraint the distance between the extremities of the rib 
 has increased by that amount. 
 
 Hence by eq. (57) we may put dU/dH=Ao, and in this, U 
 must include the internal work of the bar; but as the latter 
 is zero U comprises the internal work of the rib alone. There- 
 fore, for the loading of 64 and Fig. 37 we have only to put 
 2rj)(t to) in place of the zero in the right-hand member of 
 eq. (149) and solve for //; obtaining (in case the work of both 
 thrust and shear in the rib is disregarded), 
 
 . . . (155) 
 
66 SEMICIRCULAR ARCH RIB. TEMP. STRESSES 101 
 
 If to is lower than t, the same formula holds but t to is 
 negative. If the rib in Fig. 37 carries no load, the H is entirely 
 due to a change of temperature, if any; having a value 
 
 In Fig. 38, where horizontal constraint is exerted upon the 
 semicircular rib by an actual bar or tie-rod OK, which is elastic 
 and has a coefficient of expansion T?I (different, it may be, from 
 that, )?, of the material of the rib), it would be found that in 
 fitting rib and rod together at a temperature t, higher than 
 that, to, of erection, the rod is too short by an amount 
 io*"2r00 ifc) (t to) to connect the extremities of the rib. Hence 
 we put dU/dTi = fa, U including the internal work of the 
 tie-rod. In other words, we write 2r(r} r n )(t-to} in the place 
 of the zero in the right-hand member of eq. (152) ; and obtain 
 
 (156) 
 
 In case the rib in Fig. 38 is not loaded this reduces to 
 
 (157) 
 
 For a, fall of temperature (t to) would be negative, while if 
 i)i were numerically greater than y, (y yi) would be negative. 
 
 67. Semicircular Curved Beam without Hinges ; E and I 
 Constant. Single Eccentric Load. [See (A) in Fig. 39.] 
 This means that the semicircular arch rib is " built in," or inserted 
 rigidly in the two immovable piers (at same level) at and K 
 without strain before loading, and is continuous between. 
 No change of temperature considered at present. Sectional 
 area of rib is F. A vertical load P is to act on the rib at point m, 
 60 from the horizontal radius CK on the right. Hence, when 
 the load is on, the cross-sectional plane of the rib at (also 
 at K) will be the same as before loading, and the center of gravity 
 of the section at (and at K) will not move as the load is applied. 
 
 At 0, therefore, when the load is on, there will be in action 
 a couple, of moment M ; a horizontal shear, H; a vertical 
 
102 
 
 MECHANICS OF INTERNAL WORK 
 
 67 
 
 thrust, Vj all unknown in advance. Similarly, in the section 
 at K, Mk, V, H'] also unknown. Since, for the whole beam 
 as a free body, statics furnishes only three independent equa- 
 tions (one of which gives immediately H' = H], and there are 
 six unknowns, it is evident that the beam is statically inde- 
 terminate to the extent of three redundant elements; and let 
 V, H, and M be chosen as these three redundant elements. 
 
 If the rib has been duly fitted in place without strain, and 
 then the load P applied, it is evident that, the beam being 
 built in at K, the pier at might be entirely removed and the 
 beam would still be supported, though becoming deformed (it 
 would be a cantilever). The forces of the couple, also V and H, 
 might then be gradually applied as independent loads with the 
 
 f I H 
 
 M t (B) 
 
 r c 
 
 FIG. 39. 
 
 result that the sectional plane at and the center of gravity 
 of that section would finally be brought into the position where 
 the pier constrains them to be under actual conditions under 
 load. In other words, we might remove pier 0, let the rib 
 remain unloaded at first, thus leaving the part of beam in 
 the same position as it will be finally under actual load, and 
 then gradually and simultaneously apply all these " loads," viz., 
 P, V, H, and the forces of the couple, until they reach their 
 respective final values. During this gradual action the 
 points of application of V and H would not be displaced, and the 
 cross-sectional plane of the beam at would not turn, thus justi- 
 fying us in putting, separately [see eqs. (25) and (34)], 
 
 
 - = and 
 
 (158) 
 
67 SEMICIRCULAR ARCH RIB. NO HINGES 103 
 
 It is to be noted that these are three independent equations. 
 For instance, it is not necessary, before performing the differen- 
 tiation implied in dU/dH = Q, to express the other quantities 
 MQ, and V, in terms of H\ since the three quantities, M , V, 
 and H, are independent of each other. For example, we might 
 assume arbitrary values for M Q and V, and then the imposing 
 of the condition dU/dH = would simply give us such a value 
 of H that during the gradual loading just described the point 
 of the beam would not be displaced horizontally; although it 
 would be displaced vertically, and the section would turn 
 through some angle <j> ; as dependent respectively on the arbi- 
 trary choice of V and M . Hence if there is to be neither ver- 
 tical nor horizontal displacement at 0, nor any turning of the 
 plane of the cross-section, all three of the equations (158) must 
 hold, independently. It remains to supply the detail of apply- 
 ing the three equations in (158), in order to determine the 
 three unknown constants, MO, V, and H. 
 
 Neglecting the internal work of shear, but considering that 
 due to thrust T and moment M, we have 
 
 Evidently the rib must be divided into two parts, . . m 
 and m . . . K y for the summations required. For any point n 
 on ... m, taking the free body shown at (B) in Fig. 38, we 
 have 
 
 M = M + Vr(l -cos 6) -Hr sin 6] 
 and hence 
 
 dM dM dM 
 
 T5rr = l; -yf7- = r(l -cos 0); and -jvp= rsm0. 
 
 dM Q dV dH 
 
 Also T = Fc 
 
 dT dT dT 
 
 whence * ~dV = cos ' a ~dff ==sm 
 
104 MECHANICS OF INTERNAL WORK 67 
 
 For any point n' between m and K, with . . . n' as free 
 body, still reckoning angle from radius CO (noting that the 
 distance of P from is fr; from K, %r), we have 
 
 M = M + Vr(l-cos0)-Hrsin + Pr(J+cos 0); 
 hence 
 
 dM dM dM . 
 
 T=V cos + # sin -P cos 
 and therefore 
 
 7 dT 7 dT 
 
 = cos0 and = sm 
 
 First substituting in eq. (159), noting that the second inte- 
 gral disappears since dT/dMo = Q, we have, with ds = rd6, 
 
 r ri2o 
 -^7 (M + Vr - Vr cos -Hr sin 0)dO 
 
 Mi L /n 
 x u 
 
 r r i8o 
 +1T7 I (M + Fr - 7r cos -Ifr sin + P +Pr cos 0)d0 = 0. 
 
 Performing the integrations (see Appendix), noting that 
 sin 120 = 0.8660, cos 120= -0.500, arc 120 = TT, and 
 arc 180 = TT, we obtain after final reduction, 
 
 0; . . (162) 
 
 which is the first of the necessary three simultaneous equations 
 involving the three unknowns M , H, and V. 
 Next we have from eq. (160), 
 
 Substitution in this equation of the values for M, dM/dV, 
 dT/dV already obtained, having regard to the portion of beam 
 within which each applies, with ds = rdO, leads, after integration 
 reduction, to the result, 
 
 =0. (163) 
 
67 SEMICIRCULAR ARCH RIB. NO HINGES 105 
 
 Similarly, the result of substituting in eq. (161), or dU/dH 
 = 0, i.e., in 
 
 .after integration and reduction is found to be 
 
 The three simultaneous eqs. (162), (163), and (164), can now 
 be solved for M , V, and H. 
 
 Work of Thrust Neglected. With such a large angle as 180 
 for a circular arch rib the effect of considering the work of 
 thrust is extremely small; so the terms arising from that 
 source will now be neglected (that is, the terms involving the 
 divisor F). This causes I to disappear from all three equa- 
 tions. By elimination between (162) and (163) we find 
 
 7 = 0.1955P (Ibs.); . , . . . (165) 
 and further combination gives 
 
 ff = 0.312P(lbs.); . , I .. . . (166) 
 ;and 
 
 M = 0.1127Pr (in.4bs.) ..... (167) 
 
 Work of Thrust Retained. If the terms containing the 
 divisor F be retained and eqs. (162), (163), and (164) be com- 
 bined without modification (that is, the work of thrust is now 
 considered; though not that of shear) the result of elimination 
 will be 
 
 0.307, 
 
 (168) 
 
 (0.0466- Y~2 
 
 and H=- ^j ; (169) 
 
 0.1491+?.-^ 
 
106 MECHANICS OF INTERNAL WORK 68 
 
 and, further, regarding both V and H as having been found, 
 
 If in eqs. (168) and (169) we write I = Fk 2 , where k is the 
 " radius of gyration " of the cross-section of rib, the fraction 
 I+Fr 2 reduces to k 2 /r 2 , which ratio is extremely small; and 
 the terms containing it are so small, relatively (in this case 
 of a semicircular rib with radial thickness small compared with 
 the radius r of the semicircle) that the omission of the terms 
 involving this fraction, I + Fr 2 , in the equations mentioned, 
 leads to no practical error and gives rise to the results already 
 obtained in eqs. (165), (166), and (167). For results in case 
 the rib is segments! instead of semicircular, and for tem- 
 perature stresses, see 68 and 69. 
 
 68. Segmental Circular Arch Rib Hinged at Ends. Constant 
 E and I. Fig. 40 shows this case, with a single concentrated 
 load P at any point m, the radius to which, Cm, makes an 
 angle /? with the horizontal radius CO on the left (ft may be 
 < or >90). As before, the rib is homogeneous (constant E) 
 
 Angle 
 
 and its cross-section has a constant 7; the radial thickness 
 being small compared with the span OK. The radius, Cn of 
 any point n of the axis of rib on the left of t m is at some angle 
 with the horizontal measured from CL; while for any point n' 
 on the right the corresponding angle is called 0', measured from 
 CS. The angle SCm will be denoted by /?'; i.e., p^x-p. 
 The axis of the rib is the arc of a circle of radius r, and is 
 symmetrical about the vertical line through center C. With 
 the one vertical load, P, as shown, the reaction of each hinge 
 
68 HINGED SEGMENTAL CIRCULAR ARCH RIB 107 
 
 is oblique (since the rib tends to spread), and will be replaced 
 by its vertical and horizontal components. From S (hor. 
 compon.)=0 the two horizontal components are equal to each 
 other (value H). The vertical reactions PQ and P K are easily 
 found by statics. With the whole rib as free body, moments 
 about hinge K gives, after canceling the r, 
 
 (COS0Q + COS/?)P 
 
 Po= 2 cos 
 
 The piers are immovable and it is assumed that the rib 
 has been fitted in place originally without strain and the load 
 applied later; and that no change in temperature has occurred. 
 The value of H is the statically redundant element (there is 
 but one such element in this problem) and may be found, as 
 already explained in 62, by putting dU/dH = Q. As before, 
 let ds, =rd6 (or rdO', on right of P), bo an elementary portion 
 of the axis of the rib, at any point n (or n r ) ; then, considering 
 the internal work of bending, and of thrust, but not that due 
 to shear, we have [see eq. (75)], 
 
 -* + *-* <172> 
 
 the summation being extended over the whole length of the 
 axis of rib, from to K. Detail as follows: 
 
 For any ds occurring between and m we have, consider- 
 ing portion On as a free body, 
 
 M = P r(cos -cos 0) -Hr(sm -sin ), . (173) 
 
 and -777--= r (sin sin 0o), .... (173a) 
 
 art 
 
 also T = P cos + H sin 0, . . . . (174) 
 
 dT 
 and ' =: +sm ^' ... (174a) 
 
 and the limits for are from = OQ to = /?. 
 
 Similar expressions for M and T will hold for any ds between 
 m and K, if for P , 0, and dd, we write Pk, 0', and dO', respect- 
 ively; and the limits for 0' will be 0o and /?'. 
 
108 MECHANICS OF INTERNAL WORK 68 
 
 Substitution being made in eq. (172) we obtain 
 
 t 
 
 t cos #o sin 6 -sin 6 cos -sin cos +sin cos 6]dO 
 
 -&T I [ sin2 ^ -2 sin sin 6 + sin 2 ]d0 
 ^ 
 
 i sin ^ cos e do + JE ( sin2 ede 
 
 [cos sin 0' - sin 6' cos 0' - sin cos +sm0 cos0']d0' 
 
 rrr I [sin 2 0' -2 sin sin ^ +sin 2 d Q ]dO' 
 
 ^ in ^ cos d ' de ' + 
 
 To effect a solution of eq. (175) for H, we first perform the 
 integrations indicated; which is fairly simple, since each inte- 
 gral contains but one variable, 6 or 6', and the forms may be 
 found in the table of integrals in the Appendix. For P K , 
 PPo is now substituted; and then, for PQ (in each term 
 containing it), the value given in eq. (171). Terms are now 
 collected; and use made, for further reduction, of the following 
 relations: /?' = TT-/?; sin ^ = sin 0; cos /?'= -cos /?; 
 
 sin 2/?'= -sin 2/? and I = Fk?. 
 
 If now we denote by A the expression 
 
 J(sin 2 /9 -3 sin 2 ) +sin [sin 8 - 0i cos +/?i cos /?], 
 
 li 
 result, 
 
 in which Oi=--6 Q and ^=- p, final solution leads to the 
 
 (A; 2 
 A-^.-^[sm 2 t 3-sm 2 
 
 H = - -^- -. . (176) 
 
 0i (1 +2 sin 2 ) -| sin 20 + ^ (0 + sin 20 ) 
 
 With H known, the values of moment, thrust, and shear are 
 easily found for any section of the rib. 
 
69 HINGED SEGMENTAL CIRCULAR ARCH RIB 109 
 
 If the internal work of thrust is disregarded, the result 
 for H is that due to the omission of the terms in (176) contain- 
 ing the factor k 2 /r 2 ; i.e., 
 
 /-t tjn\ 
 
 H== d l (l+2 sin 2 6 ) -I sin 20 ' 
 
 As the arc of the arch rib approaches a semicircle the 
 difference between (176) and (177) becomes very slight. 
 
 69. Temperature Stresses in Circular Segmental Arch Rib of 
 Two Hinges. With load P still in place (in Fig. 40), the length 
 of span being OK = 2r cos OQ, if the temperature rises to a value 
 t higher than that, fo, of erection, the piers being unyielding, 
 we may prove, as in the first part of 66 that the value of H 
 is obtained by putting dU/dH equal to the amount ^ , viz., 
 y(2r cos #o)( fo), that the distance between the hinges and 
 K, of the rib, would have increased, had the rib been free to 
 expand with rising temperature (instead of putting dU/dH = Q). 
 Filling in details in the same manner as in 68 we obtain (for 
 meaning of A, see end of 68), 
 
 -i 4> in V-sin 2 0o])P + -^(2)?cos 6 )(t-to) 
 
 H=- ~ ^ ; (178) 
 
 #i(l+2sin 2 )-f sin 2# -f-^(#i+sin 
 
 j^ 
 
 in which if P be made zero the value of H is that resulting 
 from the change of temperature alone. Of course in that case 
 PO, PK> an d -P ( m Fig. 40) would drop out; and the moment, 
 thrust, and shear at any section, due to the H alone, would be 
 the elements from which the fiber stresses, which now may be 
 called " temperature stresses," would be computed. 
 
 For a fall of temperature below that, to, of erection, the 
 quantity (t to) becomes negative in eq. (178). 
 
 70. Segmental Circular Arch Rib without Hinges. E and I 
 Constant. Ends Fixed in Immovable Piers. Here, as in 67, 
 it is stipulated that the ends of the rib are fixed, or " built in " 
 (or " encastre" ") in the piers or supports without strain before 
 any loading; that is, the rib is put in place originally without 
 initial stresses. Fig. 41 shows this case, a symmetrical arrange- 
 
110 
 
 MECHANICS OF INTERNAL WORK 
 
 70 
 
 ment with extremities and K at the same level; E and I 
 constant. A single vertical load P is placed at any point ra, 
 the radius to which, Cm, makes an angle /? with the left-hand 
 horizontal radius LC. Each of the extreme radii CO and CK 
 makes an angle with the horizontal; and the angle LCK 
 (which = K - OQ) will be called //. 
 
 The variable angle OCn for any point n, or element ds, of 
 the axis of rib on the left of m will be called 0. For any point 
 n', on the right of m, also, the angle will be measured from 
 LC in the present analysis. 
 
 When the load P is in place (weight of rib itself neglected) 
 stresses will be induced in the cross-section at equivalent to 
 
 FIG. 41. 
 
 a couple of moment M , a shear, and a thrust. But the shear 
 and thrust may be replaced by a single resultant passing 
 through the center of gravity of the section at 0, and this 
 resultant may then be decomposed into a horizontal component 
 H and a vertical component T 7 . These are shown in Fig. 41 as 
 acting on that end of the rib. Similarly, in the cross-section 
 at K, the stresses acting are equivalent to a couple of moment 
 MK, a horizontal force HK and a vertical force VK', these latter 
 acting through the center of gravity of the cross-section at K. 
 These two couples, the forces H, HK, V, VK, and the load P, 
 form a system of forces holding the whole rib in equilibrium 
 when considered as a free body. It is seen that (P being 
 given) six unknowns are involved, viz., M , MK, H, HK, V, 
 and VK- Since statics furnishes but three independent equa- 
 tions, the remaining three needed for complete solution must be 
 based on the theory of elasticity. As in 67 we note that there 
 
70 SEGMENTAL CIRCULAR ARCH RIB. NO HINGES 111 
 
 are here three redundant elements involved, which could be 
 any three of the above six quantities. Our present treatment 
 will select MO, H, and F, as the three redundant elements; and 
 the analysis will be so shaped as to exclude the quantities MR, 
 HK, and VK from the equations brought into play for deter- 
 mining M , H, and V. We shall, therefore (see 67), write 
 the three relations 
 
 and from them determine MO, F, and H, in terms of P. 
 
 Details as follows : From the free body . . . n, we have 
 for the moment and thrust at any section n between and m, 
 
 M = Mo + Fr(cos 60 -cos d) -Hr($m 6 -sin ) ; - (180) 
 T=Fcos0+#sin0; . . ........ (181) 
 
 whence 
 
 dM dM dM 
 
 = 1 ; - = r (cos -cos 0) ; -r = -r (sin -sm ) ; 
 
 .and 
 
 dT dT dT 
 
 whereas, for any section n' between m and K we have, from 
 O ... n' as a free body, 
 
 M = MO + Fr(cos cos 0) #r(sin sin ) Pr(cos /? cos 0) ; 
 
 -whence 
 
 dM dM dM 
 
 -jiif = 1; -Tf7 r =r(cos 0o cos 0); ~jrr = r(sm sin 0o); 
 divi o dv tin 
 
 -and 
 
 dT dT dT 
 
112 MECHANICS OF INTERNAL WORK 70' 
 
 First Equation. Expanding the first statement of eq. (179) , 
 neglecting the work of shear, and indicating limits of itegra- 
 tion by limiting values of the variable angle 6 for the two por- 
 tions Om and mK of the rib, we have 
 
 dU I C^-dM 1 r?dT , 
 
 1 C'\,dM i 1 f 
 - I M . d s+ __| 
 tilj* dM FEJ do 
 
 K&MA J_fV^ -n nQtt 
 
 *dMo + Fj, 2 dM dS 
 
 But, from eqs. (182) and (184), dT/dM Q is zero both on 
 . . . m and on m ... K: hence the second and fourth integrals 
 on the right drop out. Substituting, now, the proper values 
 of M and dM/dM , from eqs. (180) to (184), in the first and 
 second integrals, putting ds = r . dd, and denoting the expression 
 forming the right-hand member of eq. (180) by Z, we have, 
 after canceling the r/E7, 
 
 0. (186) 
 
 rz r* C* 
 
 Note. Now I Zdd+ I Zdd is the same thing as I ZdO; 
 
 ^do J ft J o<> 
 
 whence we have 
 
 -cos 6) -Hr(sm d -sin d Q )]dO 
 
 -Pr P(cos.^-cos 0)dO = Q. (187) 
 *$ 
 
 This equation contains but one variable, 0. After inte- 
 gration, with insertion of limits, further reduction is brought 
 about by the use of the relations: /JL = TT #o; sin /* = sin #o; 
 
 cos /== -cos OQ\ we shall also write ^ 00=01, and ^ /?=/?i. 
 
 The final form then becomes 
 
 20i (M + Vr cos ) -2#r[cos -6 l sin 6 ] 
 
 -Pr[(^i+/?i) cos^+sin/?-sin ) = 0; . (188) 
 
 which is the first of the required three equations containing 
 the three unknown constants MO, V, and H (and no others). 
 The given quantities are P, Q , 61, /?, fa, E, 7, F, and r. 
 
70 SEGMENTAL CIRCULAR ARCH RIB. NO HINGES 113 
 
 Second Equation. Expanding dU/dV = [of eqs. (179)] we 
 have (neglecting work of thrust) 
 
 dU__ 
 dV~l 
 
 Values for M, T, dM/dV, and dT/dV, from eqs. (180) to 
 (184), are now inserted in their proper places in (189); and 
 advantage is taken of the relation pointed out in the " Note " 
 following eq. (186). Whence, with ds = rd6, we obtain 
 
 r 
 I 
 
 -ji I [M + Vr (cos -cos 0) -Hr(sm -sin # )](cos 6 -cos d)dO 
 
 r 3 fn 
 
 pl I P ( cos P ~ cos ^)( COS ^o -cos 0)dO 
 
 +jfi\ J (^sin^+Fcos^)cos^^- J Pcos 2 0dO\=0. (190) 
 
 Integration between the limits shown, and the aid of rela- 
 tions already mentioned, reduce this to the form (Fk 2 having 
 replaced /), 
 
 2/^tMo cos #o -2#r(cos 2 -O l sin cos 6 ) 
 
 [( /A T 
 
 20 1 cos 2 #0 + (^1 +j3/(?i -i s in 2 ^o)J 
 
 -Pr ffi +/?i) cos ,5 -sin 6 + sin ^] cos ~ (sin -sin /?) cos 
 
 i+/?i -J-(sin 20o + sin 2^9)] = 0; (191) 
 
 which is the second equation needed; between Af , F, and H. 
 
 Third Equation. Finally, the detail of dU/dH = [see 
 eqs. (179)] is as follows (work of shear omitted) : 
 
 dU 1 r> dM 1 f" dM 
 
 = M - ds + M ' dff ds 
 
114 . MECHANICS OF INTERNAL WORK 70 
 
 Substituting values from eqs. (180) to (184) and utilizing 
 the relation of the " Note " following eq. (186), with ds = r . dO, 
 we have 
 
 r 2 fit 
 
 wj I [M + Fr(cos -cos 6) -Hr(sm d -sin )](sin -sin 0)d0 
 
 &*JOo 
 
 r s r ft 
 
 F7 1 ^(cos p -cos 0)(sin -sin 0)d0 
 tiijB 
 
 +$1 M#sin0 + Fcos0)sin0d0- J Pcos0sin0d0 =0. 
 
 After integration and reduction this becomes * 
 2A/ (0o sin -cos ) +2Fr cos (0i sin -cos ) 
 
 fl +:j)('i +1 sin 20 ) -2 sin 20 -f 20i sin 2 
 
 sin cos/?+sin sin/?-cos/9 (cos + c 
 
 0. ... (194) 
 
 (193) 
 
 This is the third equation needed, of the three between the 
 unknown constants MO, V, and H. 
 
 Elimination. By a fortunate coincidence eqs. (188) and 
 (191) happen to be so constituted that both MO and H may be 
 eliminated by a single operation, viz. : Multiply eq. (188) 
 throughout by cos and add the resulting equation to 
 (191), member to member. The equation thus produced con- 
 tains neither MO nor H\ and solved for'F gives rise to 
 
 cos p (sin p -sin ) + 1 1 + )# \P 
 
 in which the symbol B stands for the expression * 
 
 J(0i +0i -Jam 20 -ism 20). . . . (196) 
 
 Similarly, M and V may be eliminated from the two equa- 
 tions (188) and (194) by one operation, viz. : Solve each of the 
 
 * For meaning of t and ^ see after eq. (187). 
 
70 SEGMENT AL CIRCULAR ARCH RIB. NO HINGES 115 
 
 equations (188) and (194) for M and equate the expressions 
 so derived. The resulting equation contains H, but not MQ 
 nor V. Solving for H, we obtain, using the symbol G to denote 
 the expression 
 
 cos (/?i cos /? + sin /?) -O l cos 2 /?- J sin 20 ), 
 
 (197) 
 
 1 + (0i + 1 sin 20 )0i -2 cos 
 
 Since # and F are now known in terms of P a solution of 
 eq. (188) for MO will give the latter quantity in known terms; 
 thus 
 
 )F 
 + ([0i +/5i] cos /3 + sin /? -sin #o)^-]r. . (198) 
 
 Great complication would result from the insertion in (188) 
 of the algebraic values of H and V from eqs. (195) and (197), 
 and no useful purpose would be served thereby. When H and 
 V have been determined numerically in any given case, placing 
 their values in (198) leads directly to the numerical value of MQ. 
 
 With M , V, and H known, the moment, thrust, and shear 
 (M, T, and /) may be found at any cross-section of the rib 
 (including that at K, where we find MK, VK, and HK) by 
 the conditions of equilibrium of a free body extending from 
 to the section in question (evidently HK = H). Hence all 
 stresses are now determinate, as to a single vertical load P 
 applied at any point m of the rib, Fig. 41. It is understood 
 that the temperature when P is in place is the same as that, 
 fo, of erection. 
 
 If in the foregoing expressions for MO, V , and H, the ratio 
 
 k 2 
 
 -j, wherever occurring, is placed equal to zero, the results are 
 
 those that would be obtained if, besides neglecting the work of 
 shear, we had also neglected the work of thrust. As the form 
 of the rib approaches a semicircle, the work of thrust is of less 
 and less consequence. 
 
116 MECHANICS OF INTERNAL WORK 71 
 
 71. Temperature Stresses in Segmental Circular Arch Rib 
 of No Hinges. Fixed Supports. E and I Constant. (See Fig, 
 41.) With no load whatever on the rib, to being the temperature 
 of placing in position (" erection"), there being then no con- 
 straint, that is, no " initial strains," stresses are induced if the 
 temperature rises to some higher figure ; due to attempted 
 expansion on the part of the rib. At each of the sections O 
 and Kj a couple, a thrust, and a shear, are developed. The 
 thrust and shear being combined into a resultant, and that 
 resultant decomposed along the vertical and horizontal, evi- 
 dently, from the equilibrium of the whole rib (from the con- 
 ditions of statics) each vertical component is zero and the two 
 horizontal components are equal ( = H) while the moments of 
 the two couples are equal ( = M ). There are, therefore, two 
 redundant elements, MQ and //. 
 
 If T) is the coefficient of expansion of the material of the 
 rib, the point of the rib would have moved horizontally 
 toward the left a distance, ^, equal to r)(t to) 2r cos (see 
 Fig. 41) if the rib had been free to change form by the removal 
 of the left-hand pier (the other pier remaining fixed). Also, 
 by the fixity of the left-hand pier, the turning of the plane of 
 the cross-section of rib at 0, which would have occurred if there 
 had been a fixed hinge joint at 0, is prevented. Hence by reason- 
 ing similar to that on p. 102 (see also 66) we must have the con- 
 ditions :* 
 
 =0; and -** '- 
 
 which will serve to determine the two unknowns, MO and H. 
 
 Detail as follows : From the free body . . . n, n being any 
 point of axis of rib between and K, we find 
 
 0-sm# ) - .... (200) 
 and T = #sin0. . . . . . . . . , (201) 
 
 '' ST = 1; ^ = -Ksin0-sin0 ); - - (202) 
 
 dT dT 
 
 -nr = 0; and -r^sin ..... (203) 
 dH 
 
 * See Note D in Appendix. 
 
71 SEGMENTAL CIRCULAR ARCH RIB. NO HINGES 117 
 
 In making each of the summations called for in eqs. (199) 
 advantage can be taken of the symmetry of form of the rib, 
 
 by integrating between = (at 0) and 6 = - (at highest 
 
 point, or " crown," of the rib) and doubling the result; as a 
 simple means of obtaining the summation for all the ds's of 
 the rib axis. 
 
 Expanding dU/dMo = Q, we have (neglecting work of shear), 
 
 * /2 
 
 (204) 
 
 But dT/dMo = and the second integral drops out. Sub- 
 stituting from eqs. (200), etc., we derive 
 
 -sn 
 
 i.e., M (jj -0 ) -ff/|cos - (|-fc) sin J =0, . (205) 
 
 or, putting ^ -0o = 0i, 
 
 0iM -#r(cos 0o-0i sin ) =0. . .. . (206) 
 
 Again, expanding dU/dH=X Q , =rj(t to) 2r cos , we find, 
 putting ds = rdQ, and neglecting work of shear, 
 
 J 1 C* /2 
 
 2 bl 
 
 i.e., 
 
 i r* /2 
 
 ^y I [M -#r(sin -sin )][ -r(sin -sin )]rd0 
 
 i>l , . . (207) 
 
 whence, after integration and reduction, putting -~ 0o = 
 [0i sin -cos ]M + [0i sin 2 -sin 20 ]Fr 
 
 . . (208) 
 
118 MECHANICS OF INTERNAL WORK 
 
 Elimination of M from (206) and (208) gives 
 
 -J3-. y(t-to)eos 
 
 H = 
 
 With H known, M is found from eq. (206), i.e., 
 
 72 
 
 . (209) 
 
 (210) 
 
 If the work of thrust were neglected (besides that of shear) 
 the resulting value of H would be obtained from eq. (209) by 
 writing zero in place of the ratio k 2 /r 2 in the denominator (only) . 
 
 72. Simpson's Rule for Approximate Integration. In 
 applying the methods of internal work it frequently happens 
 that summations must be made which on a strict mathematical 
 basis are either impossible or extremely cumbersome. In such 
 
 cases recourse may be had to Simpson's Rule, or other method 
 of approximate integration, as will be illustrated in the next 
 paragraph ( 73). The present paragraph will consist of a 
 reproduction of the statement and proof of Simpson's Rule 
 taken from the author's Notes and Examples in " Mechanics " 
 (p. 13); as follows: 
 
 If ABCDEFG, Fig. 42, is a smooth curve and ordinates be 
 drawn from its extremities A and G to the axis X, an approxi- 
 mation to the value of the area so inclosed, A...D...G...N 
 . . . ... A, between the curve and the axis X, is obtained by 
 Simpson's Rule, now to be demonstrated. Divide the base ON 
 into an even number, n, of equal parts, each = Jx (so that 
 
72 SIMPSON'S RULE 119 
 
 ON = n . Ax) , and draw an ordinate from each point of division 
 to the curve, the lengths of these ordinates being UQ, u\, u 2 , etc. ; 
 see figure. 
 
 Consider the strips of area so formed in consecutive pairs; 
 for example, CDE&'C" is the second pair in this figure 
 (counting from left to right). Conceive a parabola, with its 
 axis vertical, to be passed through the points C, D, and E. It 
 will coincide with the real curve between C and E much more 
 closely than would the straight chords CD and DE; and the 
 segment CDE, considered as the segment of this parabola, has 
 an area equal to two-thirds of that of the circumscribing paral- 
 lelogram CC'E'E. Hence, since the area of this pair of strips = 
 trapezoid CEE"C" + parabolic segment CDE, we may put 
 
 Area of pair of \ 
 
 rt-nn \ = 2 Ax\$(Uz + ^4) +(^3 i[^2 +^4j)j; 
 
 strips Lilt 
 which reduces to ... JJa:[^2 + 4w 3 +w 4 ]. 
 
 Treating all the ~- pairs of strips in a similar manner, we have 
 finally, after writing Ax= (x n XQ) +n, 
 
 Whole area AG" 
 
 (approx.) [ 3n 
 
 The approximation is closer the more numerous the strips and 
 the more accurate the measurement of the ordinates UQ, Ui, 
 u 2 , etc. 
 
 If the subdivision on the axis X were " infinitely small/' an 
 exact value for the area would be expressed by the calculus form 
 
 f^ 
 udx. Hen.ce for any integral of this form, I udx, if we 
 = XO Jx XQ 
 
 are only able to determine the particular values (UQ, u\, etc.) 
 of the variable u corresponding respectively to the abscissae XQ, 
 XQ + AX, XQ + 2Jx, etc. (where Jx=(x n xo)+n, n being an 
 even number), we can obtain an approximate value of the 
 integral or summation by writing 
 
 x n X n -Xo 
 
 I 
 
 ~ 
 
 Un]. . . , (211) 
 
120 
 
 MECHANICS OF INTERNAL WORK 
 
 73 
 
 As to the meaning of n, note that the first ordinate on the 
 left is not u\, but UQ\ also that while there are n strips, the 
 number of points of division is n + 1, counting the extremities 
 and N. 
 
 73. Davit of Variable Cross-section. Circular Quad- 
 rant. Deflection of Extremity. Use of Simpson's Rule. 
 In Fig. 43 is shown a davit, ... 6, of wrought iron; its axis 
 being a quarter circle. All sections circular, but of varying 
 radii; smaller toward the free extremity (point 6). Radius of 
 circular axis is r, = ... C, = 50 in. 
 
 O....b=l 
 
 75 
 
 At the extremity is " built in," with tangent vertical, in 
 a fixed support. At the other end, 6, a vertical load P, = 150 
 Ibs., is applied; under whose action the extremity 6 (of the 
 axis of davit) is displaced from 6 to 6'. It is required to deter- 
 mine the vertical projection 6 ... a of this displacement, neg- 
 lecting the weight of the davit itself; the value of E being 
 28,000,000 Ibs./inA Let us take seven different points 0, 1, 2, 
 etc., equally spaced along the axis from to 6. The angles (0) 
 of the radii drawn to these points from center C are respect- 
 ively 0, 15, 30, 45, 60, 75, and 90; from the horizontal 
 . . . C. The radii of the cross-sections (circular) at these 
 points are respectively TQ = 1.0 in. ; r\ = 0.98 ; r^ = 0.92 ; r 3 = 0.86 ; 
 r 4 = 0.75; r 5 = 0.62; and r 6 = 0.50 in.; while the corresponding 
 moments of inertia are /o = i^ 4 = 0.7854 in. 4 ; /i= 0.7251 in. 4 ; 
 
73 DEFLECTION OF DAVIT. SIMPSON'S RULE 121 
 
 7 2 = 0.5692; / 3 -0.4298; / 4 = 0.2487; 7 5 = 0.1162: and 7 6 = 
 0.0491 in. 4 . 
 
 As will be seen later, values of cos 2 will also be needed for 
 the seven angles: = 0, 01 = 15, 2 = 30, etc., up to 6 = 90. 
 These values of cos 2 are respectively, 1.0, 0.9330, 0.7500, 
 0.5, 0.25, 0.0670, and 0. 
 
 Since the vertical distance 6 ... a is the projection upon 
 the line of the external force of the displacement 6 ... 6' we 
 may obtain its value by Castigliano's Theorem [ 21] by writing 
 
 jr77 "' = y ' = W' ' ' :> -' (212) 
 
 1 f & M <LM 
 
 i.e. [see 36], y = E) T'dF ds; ' '' ' ' ' (213) 
 
 (neglecting the internal work both of thrust and shear, the radial 
 thickness of the curved beam being small compared with the 
 radius of the circular axis). 
 
 From the free body m . . . 6, a radial section having been 
 made at m, we find by moments about m, 
 
 M = Pr cos ; and -p- = r cos 0. . . (214) 
 These being substituted in eq. (213), with ds = rdd, we obtain 
 
 ..... (215) 
 
 The integral contains two variables, and /. If I were a 
 given algebraic function of 0, this expression could be sub- 
 stituted and the integration performed (provided only known 
 integral forms were encountered). But in the present case, 
 although the curved beam has a smooth taper from to the 
 end 6, and all sections are circles, / is not a known algebraic 
 function of 0. However, the value of / may be computed from 
 the measured radius of section for any assigned value of 0. 
 Simpson's Rule, therefore, offers a means of securing a result 
 accurate enough for practical purposes. 
 
122 MECHANICS OF INTERNAL WORK 73 
 
 Comparing the integral in eq. (215) with that in the left- 
 hand member of (211), we note that dd corresponds to dx', 
 
 n . /cos 2 6\ 
 
 ^ to x n ', to XQ] and I j ) to u. Choosing a value ol o 
 
 for n, we therefore divide the angle (=90) between point 
 and 6 into six equal parts; with corresponding points on the 
 axis of beam (see Fig. 43) and note that 
 
 cos 2 cos 2 15 cos 2 30 cos 2 90 
 
 UQ = j ; ui = j - ; u 2 = j -- ; etc., to UQ = j -- . 
 
 Hence, eq. (215) becomes 
 
 /V 2^p_ 4 cos 2 15 2 cos 2 30 
 E + 
 
 4 cos 2 45 2 cos 2 60 4 cos 2 75 1 
 H T j j -+0 ; . (2K>> 
 
 ^3 *4 ^5 
 
 or, with numerical values, using the inch and pound as units, 
 
 l 4x0933 Q 2X0.7500 
 y ~ 28000000 X 18 L0.7854" f 0.7251 0.5692 
 
 4X0.5000 2x0.2500 4x0.067 
 
 0.06701 
 
 Ti6ir. =L05ln - 
 
 as the value of the vertical deflection ... a. 
 
 Similarly, we may determine the horizontal deflection 
 a ... 6', using the same method as in 64 (involving a " dummy " 
 horizontal force at point 6). 
 
 (It is here supposed that the elastic limit is not passed in 
 the outer fiber of any section of the curved beam. Let the 
 student test this point). 
 
APPENDIX 
 
 Integral Forms. A few integral forms, useful in the mathe- 
 matical work of this book, are here presented. 
 
 /x n+1 
 x n dx= - 
 n+r 
 
 /i'dx 
 J x 
 
 J sin 6 . d6= cos 6; 
 
 fcos 6 . dd= +sin#; 
 
 J sin 6 . cos . dO = J sin 2 6\ 
 
 \0 . cos 0. dd = 6 . sin + cos 0; 
 
 c 
 
 J s 
 
 fcos 2 d . dO = \Q + i sin 26. 
 
 Note C. (See foot of p. 30.) If a curve be plotted from 
 the expression for U, as function of TV in the seventh line 
 from top of p. 30, with U as (vertical) ordinate and T\ as 
 (horizontal) abscissa, this curve is evidently the common parabola; 
 and, moreover, one with its geometrical axis vertical; since it 
 must be a conic section, the highest power of either co-ordinate 
 being the second, and since it is impossible to get more than one 
 value of U for any assigned value of TV These characteristics 
 
 123 
 
124 APPENDIX 
 
 show that the curve cannot be an ellipse, nor an hyperbola, 
 nor a parabola with its axis other than vertical. 
 
 For the angle a, which a tangent line at any point of this 
 
 curve makes with the horizontal, we have tan a =-7-, and 
 
 al i 
 
 this is given in eq. (45); which is an equation of the first 
 degree as respects T^ so that only one value of T\ satisfies 
 
 the condition -7777- = 0. The point of the curve where -77^- = 
 a 1 1 al i 
 
 is, of course, the vertex of the parabola, where the tangent line 
 is horizontal, and here the ordinate U is either a maximum or 
 a minimum. It will be a minimum if the curve on either side 
 of the vertex rises, the final value of U being + infinity; but a 
 maximum, if the curve descends on either side, to a final ordinate 
 of infinity. But the latter value for U is impossible; since, 
 from the nature of the expression for U, each term is positive 
 for all values of T\. Hence the value of TI obtained by putting 
 
 -77-- = corresponds to a minimum value of U. 
 
 Note D. (See p. 116.) To prove eqs. (199) a little more 
 specifically let us suppose that the rib is at first supported 
 (i.e., held) by the right-hand support at K, only, and under 
 no load; but at the higher temperature, t. The point of the 
 axis would then be found a distance A , horizontally to the left 
 from where it is actually constrained to be, and the plane of 
 the cross-section would be parallel to its ultimate position. 
 Let now the system of independent " loads," consisting of H 
 and the forces forming the couple of moment M , be gradually 
 applied- (beginning with zero values) until they reach their 
 respective ultimate and actual values. The horizontal dis- 
 placement of point of the rib will be to, = y(t t )2r cos , 
 
 giving -rjj - h (p. 20) ; while the angle through which the plane 
 of section will have turned is zero, whence "7117" = (p. 25). 
 
 .Q 
 
INDEX 
 
 Angular displacement of arm of 
 couple 24 
 
 Arch rib, or curved beam 96 
 
 Arch rib, segmental, hinged ends 106 
 
 Arch rib, segmental, no hinges, 
 fixed ends. .. 109 
 
 Arch rib, segmental, tempera- 
 ture stresses 109, 116 
 
 Arch rib, semicircular, hinged 
 ends . 96 
 
 Arch rib, semicircular, no hinges, 
 fixed ends.... 101 
 
 Arch rib, semicircular, with tie 
 rod 98 
 
 Arch rib, semicircular, tempera- 
 ture stresses 100 
 
 Bars considered elastic in succes- 
 sion 1 
 
 Bars, necessary 2 
 
 Bars, redundant 2 
 
 Beams, curved 45, 94-122 
 
 Beams in Flexure, Internal work 45 
 Beams, non-prismatic. . . . 60, 61, 65 
 Beam supported by parabolic 
 
 .cable 86 
 
 Beam supported by three others 78 
 
 Beams, straight 49-77 
 
 Beam, trussed 80 
 
 Beam with overhang 53 
 
 Brake-beam 80 
 
 Cable supporting beam 86 
 
 Cambered trussed-beam 82 
 
 Cambria steel I-beam 51 
 
 Cantilever beams 59 
 
 Castigliano's Theorem. 15, 17 
 
 Castigliano's Theorem, geomet- 
 rical proof 18 
 
 Clapeyron's Theorem 72 
 
 Composite systems 78 
 
 Compressive shortening 2 
 
 Continuous girders ... 62, 64, 05-67, 
 
 69-77 
 
 Continuous girder of variable 
 moment of inertia 65 
 
 Couple, Derivative of vork with 
 respect to moment of 24 
 
 Curved beams 45, 94-122 
 
 Davit of uniform section, deflec- 
 tion 94 
 
 Davit of variable cross-section, 
 deflection 120 
 
 Deflection of beams 49, 52, 53, 
 
 59,60 
 
 Deflection of beam with variable 
 moment of inertia 53, 60, 61 
 
 Derivatives of work 17, 20, 24 
 
 Derivative of internal work as to 
 moment of an external couple . 24 
 
 Displacement of joint of frame . 
 
 4, 20 
 
 Displacement of point in axis of 
 beam 48 
 
 Displacement, variable, Theo- 
 rem for 18 
 
 Double-knee beam 90 
 
 Elastic limit not passed 1 
 
 Elongation of bar 12, 13 
 
 Equation to curve found by Cas- 
 tigliano's theorem 56, 59 
 
 Example of deflection of joint of 
 
 frame 20 
 
 External work 4, 11 
 
 Flexure, work of, in beams 45 
 
 Fraenkel's Formula 22, 53 
 
 Frame of one redundant bar . . 5, 26 
 Frame of two redundant bars. . . 33 
 Frame of more than two redun- 
 dant bars 37 
 
 Gradual application of loads ... 1, 3 
 
 Hiroi, Prof., book by 89 
 
 Hudson, C. W., on deflection of 
 
 beams 53 
 
 125 
 
126 
 
 INDKX 
 
 I-beam, work of shear in 47, 51 
 
 Initial stresses 7, 11 
 
 Integral forms (Appendix) 123 
 
 Integrals in the three-moment 
 
 Theorem 72 
 
 Internal work 4, 13 
 
 Internal work equals external 
 
 work. . 13 
 
 Knee-loeam. 
 
 90 
 
 Least work, Principle of 31 
 
 Least work, a particular case of 
 
 more general relation 33 
 
 Limitation to zero initial stresses 
 
 11, 13 
 
 Loads, gradual application of . . . 1 
 
 Martin, H. W. book by 89 
 
 Mensch, problem treated by. ... 93 
 Moment, internal, statically in- 
 determinate, derivative of in- 
 ternal work with respect to. . . 68 
 Moment of inertia, variable, 
 
 beams of 60, 61, 65 
 
 Moment, " normal " 71 
 
 " Necessary " bars 2 
 
 " Neutra) " force 18 
 
 Non-prismatic beams 60,61,65 
 
 " Normal moment " 71 
 
 Overhang, beam with 53 
 
 Parabolic cable sustaining beam 
 
 and load 86 
 
 Projection of displacement 4 
 
 "Redundant bars". . . 2, 5, 8, 11, 13 
 Redundant bar too short or too 
 
 long 37 
 
 Rough supporting surface 18 
 
 Segmental circular arch rib 
 
 hinged at ends 106 
 
 Segmental circular arch rib fixed 
 at ends, no hinges 109 
 
 Shear, work of, in flexure. 47, 48, 51 
 
 Shear, influence on deflection of 
 I-beam 51 
 
 Simpson's Rule for approximate 
 integration 118 
 
 Simpson's Rule applied in prob- 
 lem of davit 120 
 
 Smooth supporting surfaces pos- 
 tulated 18 
 
 Statically indeterminate struc- 
 tures 26 
 
 Surfaces, smooth 18 
 
 Surfaces, rough 18 
 
 Tapering beams 60, 61, 65 
 
 Temperature 2, 13 
 
 Temperature stresses, etc 2, 13, 
 
 43, 84, 100, 109, 116 
 Temperature stresses in elastic 
 
 bars 43 
 
 Temperature stresses in trussed- 
 
 beam 84 
 
 Theorem of three moments 69 
 
 Three moments, Theorem of 69 
 
 Thrust, work of 46 
 
 Tilden, Prof., article by 51 
 
 Trussed-beam 80 
 
 Trussed-beam, cambered 82 
 
 Trussed beam with two struts 
 
 and three tie-rods 84 
 
 Variable displacement theorem . 18 
 Variable moment of inertia . . .60, 61, 
 
 65 
 Virtual velocities 3, 6, 7, 9 
 
 Work of deformation -. . . 5 
 
 Work, external 4,12 
 
 Work, internal 4 
 
 Work of shear considered 51 
 
 Work of shear neglected in 
 
 nearly all cases 48 
 
 Work of thrust neglected. . . 105, 121 
 
Trigonometric Ratios (Natural); including "arc," by which is meant 
 "radians," or "7r-measure, " or "circular measure ;" e.g., arc 100 = 1.7453293, 
 
 arc 
 
 de- 
 gree 
 
 sin cosec 
 
 tan cot an 
 
 sec cos 
 
 
 
 
 
 
 
 I in fin. 
 
 ) infin. 
 
 1.0000 1.0000 
 
 90 
 
 1.5708 
 
 0.0175 
 
 1 
 
 0.0175 57.299 
 
 0.0175 57.290 
 
 L0001 0.9998 
 
 89 
 
 1.5533 
 
 .0349 
 
 2 
 
 .0349 28.654 
 
 .0349 28.636 
 
 1.0006 .9994 
 
 88 
 
 1.5359 
 
 .0524 
 
 3 
 
 .0523 19.107 
 
 .0524 19.081 
 
 1.0014 .9986 
 
 87 
 
 1.5184 
 
 .0698 
 
 4 
 
 .0698 14.336 
 
 .0699 14.301 
 
 1.0024 .9976 
 
 86 
 
 1.5010 
 
 .0873 
 
 5 
 
 .0872 11.474 
 
 .0875 11.430 
 
 1.0038 .9962 
 
 85 
 
 1.4835 
 
 0.1047 
 
 6 
 
 0.1045 9.5668 
 
 0.1051 9.5144 
 
 1.0055 0.9945 
 
 84 
 
 1.4661 
 
 .1222 
 
 7 
 
 .1219 8.2055 
 
 .1228 8.1443 
 
 1.0075 .9925 
 
 83 
 
 1.4486 
 
 .1396 
 
 8 
 
 .1392 7.1853 
 
 .1405 7.1154 
 
 1.0098 .9903 
 
 82 
 
 1.4312 
 
 .1571 
 
 9 
 
 .1564 6.3925 
 
 .1584 6.3138 
 
 1.0125 .9877 
 
 81 
 
 1.4137 
 
 .1745 
 
 10 
 
 .1736 5.7588 
 
 .1763 5.6713 
 
 1.0154 .9848 
 
 80 
 
 1.3963 
 
 0.1920 
 
 11 
 
 0.1908 5.2408 
 
 0.1944 5.1446 
 
 1.0187 0.9816 
 
 79 
 
 1.3788 
 
 .2094 
 
 12 
 
 .2079 4.8097 
 
 .2126 4.7046 
 
 1.0223 .9781 
 
 78 
 
 1.3614 
 
 .2269 
 
 13 
 
 .2250 4.4454 
 
 .2309 4.3315 
 
 1.0263 .9744 
 
 77 
 
 1.3439 
 
 .2443 
 
 14 
 
 .2419 4.1336 
 
 .2493 4.0108 
 
 1.0306 .9703 
 
 76 
 
 1.3264 
 
 .2618 
 
 15 
 
 .2588 3.8637 
 
 .2679 3.7321 
 
 1.0353 .9659 
 
 75 
 
 1.3090 
 
 0.2793 
 
 16 
 
 0.2756 3.6280 
 
 0.2867 3.4874 
 
 1.0403 0.9613 
 
 74 
 
 1.2915 
 
 .2967 
 
 17 
 
 .2924 3.4203 
 
 .3057 3.2709 
 
 1.0457 .9563 
 
 73 
 
 1.2741 
 
 .3142 
 
 18 
 
 .3090 3.2361 
 
 .3249 3.0777 
 
 1.0515 .9511 
 
 72 
 
 1.2566 
 
 .3316 
 
 19 
 
 .3256 3.0716 
 
 .3443 2.9042 
 
 1.0576 .9455 
 
 71 
 
 1.2392 
 
 .3491 
 
 20 
 
 .3420 2.9238 
 
 .3640 2.7475 
 
 1.0642 .9397 
 
 70 
 
 1.2217 
 
 0.3665 
 
 21 
 
 0.3584 2.7904 
 
 0.3839 2.6051 
 
 1.0712 0.9336 
 
 69 
 
 1.2043 
 
 .3840 
 
 22 
 
 .3746 2.6695 
 
 .4040 2.4751 
 
 1.0785 .9272 
 
 68 
 
 1.1868 
 
 .4014 
 
 23 
 
 .3907 2.5593 
 
 .4245 2.3559 
 
 1.0864 .9205 
 
 67 
 
 1.1694 
 
 .4189 
 
 24 
 
 .4067 2.4586 
 
 .4452 2.2460 
 
 1.0946 .9135 
 
 66 
 
 1.1519 
 
 .4363 
 
 25 
 
 .4226 2.3662 
 
 .4663 2.1445 
 
 1.1034 .9063 
 
 65 
 
 1.1345 
 
 0.4538 
 
 26 
 
 0.4384 2.2812 
 
 0.4877 2.0503 
 
 1.1126 0.8988 
 
 64 
 
 1.1170 
 
 .4712 
 
 27 
 
 .4540 2.2027 
 
 .5095 1.9626 
 
 1.1223 .8910 
 
 63 
 
 1.0996 
 
 .4887 
 
 28 
 
 .4695 2.1301 
 
 .5317 1.8807 
 
 1.1326 .8829 
 
 62 
 
 1.0821 
 
 .5061 
 
 29 
 
 .4848 2.0627 
 
 .5543 1.8040 
 
 1.1434 .8746 
 
 61 
 
 1.0646 
 
 .5236 
 
 30 
 
 .5000 2.0000 
 
 .5774 1.7321 
 
 1.1547 .8660 
 
 60 
 
 1.0472 
 
 0.5411 
 
 31 
 
 0.5150 1.9416 
 
 0.6009 1.6643 
 
 1.1666 0.8572 
 
 59 
 
 1.0297 
 
 .5585 
 
 32 
 
 .5299 1.8871 
 
 .6249 1.6003 
 
 1.1792 .8480 
 
 58 
 
 1.0123 
 
 .5760 
 
 33 
 
 .5446 1.8361 
 
 .6494 1.5399 
 
 1.1924 .8387 
 
 57 
 
 0.9948 
 
 .5934 
 
 34 
 
 .5592 1.7883 
 
 .6745 1.4826 
 
 1.2062 .8290 
 
 56 
 
 0.9774 
 
 .6109 
 
 35 
 
 .5736 1.7435 
 
 .7002 1.4281 
 
 1.2208 .8192 
 
 55 
 
 0.9599 
 
 0.6283 
 
 36 
 
 0.5878 1.7013 
 
 0.7265 1.3764 
 
 1.2361 0.8090 
 
 54 
 
 0.9425 
 
 .6458 
 
 37 
 
 .6018 1.6616 
 
 .7536 1.3270 
 
 1.2521 .7986 
 
 53 
 
 0.9250 
 
 .6632 
 
 38 
 
 .6157 1.6243 
 
 .7813 1.2799 
 
 1.2690 .7880 
 
 52 
 
 0.907G 
 
 .6807 
 
 39 
 
 .6293 1.5890 
 
 .8098 1.-2349 
 
 1.2868 .7771 
 
 51 
 
 0.8901 
 
 .6981 
 
 40 
 
 .6428 1.5557 
 
 .8391 1.1918 
 
 1.3054 .7660 
 
 50 
 
 0.8727 
 
 0.7156 
 
 41 
 
 0.6561 1.5243 
 
 0.8693 1.1504 
 
 1.3250 0.7547 
 
 49 
 
 0.8552 
 
 7330 
 
 42 
 
 .6691 1.4945 
 
 .9004 1.1106 
 
 1.3456 .7431 
 
 48 
 
 0.8378 
 
 7505 
 
 43 
 
 .6820 1.4663 
 
 .9325 1.0724 
 
 1.3673 .7314 
 
 47 
 
 0.8203 
 
 .7679 
 
 44 
 
 .6947 1.4396 
 
 .9657 1 0355 
 
 1.3902 .7193 
 
 46 
 
 0.8028 
 
 .7854 
 
 45 
 
 .7071 1.4142 
 
 1.0000 1 0000 
 
 1.4142 .707.1 
 
 45 
 
 0.7854 
 
 
 
 cos sec 
 
 c-otan t a a 
 
 cosec sin 
 
 de- 
 
 arc 
 
 
 
 
 
 
 gree 
 
 
OVERDUE. 
 
 OCT 
 
 DEC V* 
 
 tf* 
 
 MAR 21 ' 946 
 SEP 131946 
 
 f EB ~0 1939 
 
 DEC 6 
 
 D 
 
 oc t 
 
 MAR 2 1942 
 
 1) 
 
 4 1954 LU 
 
 - 
 REC'D LD 
 
 MAR 19 
 
 LD 21A-60m-4,'64 
 (E4555slO)476B 
 
 LD 2l- 
 
 General Library 
 
 University of California 
 
 Berkeley