Cornell University Library TG 260.M38 ENGINEERING LIBRAE. President White Library, CORNELL UNIVERSITY. A U?t9 ///*/96 Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924004022582 STATICALLY INDETERMINATE STRUCTURES PRINCIPLE OF LEAST WORK. HAEOLD MEDWAY MAETIN, Wh. So. Revised and Reprinted from " Engineering.'' London : Offices of " Engineering," 35 and 36, Bedford Street, Strand, W.C. 1895. 5 STATICALLY INDETERMINATE STRUCTURES AND THE PRINCIPLE OF LEAST WORK. Introduction. THE subject of statically indeterminate structures has been much neglected in English engineering text-books. Special cases, such as stiffened suspension bridges and rigid arches, have, it is true, been dealt with by more than one writer, but the general question has only been considered in a few scattered papers by a number of different authors. An excellent treatise on the subject was published in Italy by Castigliano in 1879. In this volume the principle of least work was enunciated and its application to determining the stresses in a structure containing superfluous bars explained. These stresses can, it is true, be determined by other methods which Maxwell seems to have originated, but the method of least work has great practical advantages, and will be adopted as the basis of what follows. Every metallic or wooden structure is elastic, and constitutes a spring. If a spring is loaded by a 4 STATICALLY INDETERMINATE STRUCTURES. weight, it elongates, and a certain amount of work is done in this elongation. This work is stored in the spring in the form of potential energy, and can be reconverted into mechanical work, as is commonly done in clocks and watches. The stiffer the spring the less it is deformed by a given weight, and hence less work is stored in a stiff spring loaded with a 1-lb. weight than in a light one loaded by the same weight. Thus if 1 ton is hung from a steel bar of 2 square inches in section, less work is done in deforming the bar than if it was hung on a steel bar of the same length and of 1 square inch section. If a weight lies on a platform supported by four legs of elastic material, work will be done in deforming the platform and in compressing the legs. If there had been only three legs, the ordinary principles of statics would suffice to determine the weight taken by each leg, which is then quite independent of the comparative stiffness of the legs and the platform. When, however, we have more than three legs, these statical principles no longer suffice, and to determine how much of the weight is carried by each leg it is necessary to introduce other considerations. The one great principle to which such problems can be reduced is known in dynamics as that of least action, and in such problems as we have before us as that of "least work." That is to say that the work stored in an elastic system in stable equilibrium is always the smallest possible. As an illustration, suppose a piece of steel ribbon is bent between two stops A and B, Fig. 1, then there are several possible positions of equilibrium for such INTRODUCTION. 5 a spring. For example, the S-shaped curve A D E B is one possible position of equilibrium, and the three- lobed curve A F G B is another, whilst a third possible position of equilibrium is the single-lobed curve A C B. Of these different possible positions of equilibrium some are more stable than others. Thus a slight shock or displacement will cause either the three- lobed curve or the two-lobed one to pass into a single- lobed one like A B, but this latter cannot by a small shock be made to pass into either of the other forms. Hence it is a case of stable equilibrium whilst the others are unstable. It will be noted that the stable form is the least bent one, and as the work stored in a flat spring increases with the amount of bending, it is obvious that the work stored up as potential energy is less in the stable than in the unstable conditions of equilibrium. Returning now to our four-legged table, there are an infinite number of ways in which the load could be distributed over the four legs, whilst still main- taining equilibrium. But if the equilibrium is to 6 STATICALLY INDETERMINATE STRUCTURES. be stable, the distribution will be such that the work«d stored up as potential energy will be as small as possible. To fix the ideas, let Fig. 2 represent a table 50 in. square, supported by four legs of the same material and of uniform section, and let a weight W be placed as indicated. Numbering the legs 1, 2, 3, and 4, the load taken by each has to be ascertained. If the top of the table is very rigid, compared with the legs, the problem is very easily solved by the method of least work, and we will assume this to be the case. Under these conditions, the work done in deforming the top of the table will be very small compared with that done in compressing the legs, and may, therefore, be neglected without materially affecting the results of the calculation. Now, in aD ordinary spring balance, the extension or INTRODUCTION. 7 compression is twice as great for a load of 2 lb. as it is for one of 1 lb., and for P lb. the extension of the spring is P times as great as it is for 1 lb. Suppose the balance at zero, then, by placing small shot in the scale pan, we can gradually bring up the load in the pan to any desired amount P. Suppose the spring extends £ in. for a weight of 1 lb., then the P . final extension of the spring is P times this, or — in. 8 During the process of adding the shot, the load in the pan has been gradually increased from to P, and P hence the average weight has been -~ . Now the work done by a given force is found by multiplying the average value of the force by the distance through which it has moved. The average force in the present P . P case is — lb-» and the distance it has moved is — j n . 2 8 pa Hence the work stored in the spring is jg inch- pounds. Now, coming back to our table, suppose under a load of 1 lb. each leg is compressed a in. Under a load P lb. it will then compress P a in., and the total work done in the compression will, as shown above, p 2 o be -g- inch-pounds. Let us assume that the weight taken by leg No. 4 is P 4 , and that P 1; P 2 , and P 3 represent the weight taken by each of the other legs. Evidently + P 4 =W (l). 8 STATICALLY INDETERMINATE STRUCTURES. Also, by taking moments, we have 50 (P 4 + P„) = 15. W, or Similarly or 15W-50P 4 _ 3W-10P 4 Fs= 50 ~ 10 • ' ' W 50(P 1 + P 4 ) = 30W, P^ 6 -^^ <*> .,fro m( l)P 2 =^±^. Hence, if P 4 was known, all the others could be determined. To determine P 4 , consider the work done in compressing the legs. Call this U, then U= £(P 1 2 + P 2 i! + P3 !! + P4 !! ), and substituting for P x , P 2 , P s , we find , M j . ( 46 W 2 - 160 W P 4 + 400 P 4 2 ) , and we must choose P 4 so that this is as small as possible. One way of doing this would be to choose a number of values for P 4 , taking it successively equal to £ W, i W, £ W, £ W, £ W, and so on, and to tabulate the values then found for the quantity inside the bracket. We thus get : when P 4 = ^ then 46 W 2 - 160 W P 4 + 400 P 4 2 = 66 W 2 = 37.1 W 2 = 31 W* = 30 W 2 = 30.4W 2 V = , P4 = w 3 p 4 = W 4 p* = W 5 ?4 = W • 6 INTRODUCTION. 9 Now this last value is greater than that imme- diately preceding it, and thus after first steadily diminishing the value begins to increase again. Hence the proper value of P 4 is somewhere W W between -7- , and -^-, and if the numbers were plotted down, and a smooth curve drawn through W them, it would be found that -=- was about the correct value. This method, though quite practic- able, and not difficult of application in the present instance, would in many cases prove exceedingly troublesome, and it is best to make use of the differential calculus, which gives the correct value for P 4 at once, and the differentiations involved are of the very simplest kind. We have u = 26b (46 w " ~ 160 w p 4 + 400 p * s >- Whence f]F 4 = 2 io{- 160W+800I> *} and this is equal to when U has its smallest possible value ; . •. 800P 4 - 160W = 0, or *V _ 5 Having got this, the other reactions are found to be as follows : W + 2W 3 ** - 10 ~ 10 v 6W-2W .4 *i - 10 10 *s - 10 - 10 STATICALLY INDETERMINATE STRUCTURES. This, then, is the distribution of the weight on the four legs when the latter are of uniform section and material. Suppose, however, the legs were of different lengths, and had different sectional areas. In this case the first thing to be done is to deter- mine how much each leg will be compressed by a unit force. If I be the length of leg, D, its area, and P the load on it, the extension will be _ P l A - n • E' The work done in extending it will be, as already- shown, P_\_P 2 _J_ 2 ~ 2 • n . E' Now, returning to our table, suppose legs 1 and 2 are 30 in. long and 2 square inches in area, whilst legs 3 and 4 are 20 in. long and 1 square inch in area. Then U, the work done in compressing the legs, will be : U ~2E - 2 2E • 2 + 2E" 1 2E ' 1 =21: { 15 p i 2 + 15 p s 2 + 20 p 3 2 + 20 P 4 2 } and, as before, we have 6W-10P 4 r >- 10 W + 10P 4 ^~ 10 3W - 10P 4 r3_ 10 LATTICE GIRDERS. 11 Whence tt L /15(6W-]0P 4 ) 2 +15(W + 10P 4 )M < ... 3.33 kd ... 16.05 k I ... 2.56 dm ... 18.02 1 m ... 2.08 mf ... 16.05 m, n ... 1.75 f'o ... 25.50 n o ... 1.75 oh 25.50 op ... 1.75 LATTICE GIRDERS. 13 The bars on the other side of the mid span will have corresponding values of - . In practice, however, the bracing bars only require to be taken into con- sideration as the work done on the top and bottom flanges varies but little with different distributions of load between the two systems of web members, and hence these flanges can and will be entirely neglected in what follows : Let u be the work done in deforming one bar, then the whole work will be equal to the work done in deforming all the bars, or U = s«. But if n = the stress in the supernumerary bar a i, the work done in deforming the whole structure will be a minimum with respect fi. Whence dV „ du but for any bar, as has already been shown above, T 2 L where T is equal to the stress in the bar, and T can always be expressed in the terms of the external d / Ub load on the structure and /i. Whence 2-= — •, can be a n found, and equating this to zero we shall get the value of v.. Suppose, for example, the central apex h only is loaded by a weight W. Let R be the reaction on the left-hand abutment, and fi the stress in the bar a i. Then in this case from symmetry it will be obvious that the stress in the bar corresponding to 14 STATICALLY INDETERMINATE STRUCTURES. aiat the right hand abutment, will also be /*. But on drawing a stress diagram it will be found that for equilibrium the stress in this bar should be — /*. Hence in the case taken fi= o. and the load is carried wholly by one system of bracing although if any bar of this system were removed the structure would still support the load the other system then coming into play. If, however, the load was placed at the apex / instead of at h, then a certain amount of stress will obtain in both systems of bracing. As before let /u, be stress in the bar a i. Then the shear causing stress in the bars i b,b k, &c, will be R — fi where R is the total reaction at the left abutment. The shear causing stress in the bars a j, j c will of course be /j,. Hence the stress in the bars ib,bk, &c, will be 4- (R — /*) sec a up to the apex f, and beyond this point for the remaining bars of this system it will be jf { R — fi — W) sec a. For the other system the stress in the bracing bars will be + /* sec a through- out^ a being the angle between the bars and the vertical. Hence for each of the bracing bars of the first system up to the apex /, L u = 2EQ se ° 2a ' E,- '*) a and beyond this point u = . sec 2 a (R — /*— W) 2 du aeo 2 fl r _, L . . .■.^=-^-. [ ^.R ] _.upto/ J sec 2 a L and= ~^-|> + W-R]-^ LATTICE GIRDERS. 15 for the rest of the bars of the system. Hence taking the value of _ . from the proceeding table we get for the first system of bracing, ^jp 2 | n - R J- 1 13.03 + 13.65 + 16.05 + 18.02 + 16.05 + 1 S6C" Ot f 25 50 + 25.50 j +~jf - W-l 25.50 + 25.50 + 25.50 + 25.50 + 16.05 + 18.02 + 16.05 + 13.65 + 13.03 } or sec 2 a ( i -gr-. 255.6m - 255.6R + 178.8 W. And taking moments, about the right-hand abut- g ment we get, R = — W. Whence on substituting, there results for the sum of-=- , for this first system a it of bracing, sec 2 a f ) -g-{255.6ju- 15.6W . Similarly for the second system of bracing we have throughout for each bar du _ sec 2 a L di, ~ ~W " " " a' and substituting from the table as before and adding we get „.„ „sec 2 o 249.8^g-. M Hence putting U = X u we get . J5 = !^ffj 255.6m + 249.8^ -15.6W}=0, when U is a mininum. . •. 505.4 m =15. 6 W. whence m = -031 W. 16 STATICALLY INDETERMINATE STRUCTURES. The value of /j, for other portions of the load will be as follows : id at Value of n i 011W. a 024W. / 031 W. h 000 W. c 877 W. e 743W. .9 575W. If any three apices such as b, c and d were loaded, each with a load W. the resulting stress in the bar ai. will of course be got by summation, and will in the case given be W ("Oil + -877 + -024) = -912 W. Knowing this the stress in any other bar can be determined by pure statics. In the girder considered the modification in the stresses as obtained on the usual assumption is trivial. The example is of interest, however, as many American Engineers object to the use of double systems of triangulation on the ground that the stresses in them cannot, strictly speaking, be determined by pure statics, the usual assumptions being objected to as being erroneous. From the complete investigation above given, it would appear that the error involved in these assumptions may well be neglected in practice. Trusses with more than one Superfluous Bar. A lattice girder of the type dealt with has one superfluous member only, but the method of least work is also applicable to structures containing more than one such member. As an instance, take TRUSS WITH TWO SUPERFLUOUS BARS. 17 the truss shown in Fig. 4, which, though of a type not to be recommended for adoption in an actual bridge, will, nevertheless, afford a good example of the methods to be employed in calculating such a framework. This truss contains two superfluous members. Thus, if the bars B E and E G were re- moved, the stresses in the various members could be determined by ordinary statics. But these stresses could also be determined if the stresses in these two bars were known. Thus let the stress in B E be P, and in E G be Q. Assume each of these bars is in tension, then their action on the structure will be the same as if the bars Vere Fig. 4-. B D S x X X X ^ Tig. 5. B D C r A/ C 2629. / £ F Ntf x\ taken away altogether and forces applied to the joints B, G, and E, as indicated in Fig. 5. This latter figure represents, it will be seen, a statically determined structure, and the stresses in all the bars can be ex- pressed in terms of the loading and P and Q, whilst these latter can be determined by the method of least work. To apply this principle, however, it is necessary to know the ratio of — for each bar, where L = length 18 STATICALLY INDETERMINATE STRUCTURES. of bar and a its area. These ratios can be determined with sufficient accuracy by the ordinary methods of dealing with such structures, viz., by dividing them up into two separate trusses, each of which is statically determined, and finally combining the results. An approximation to the true stress is thus obtained, permitting the section of the bars to be designed pro- visionally, and the ratio - found. Suppose the ratios determined in this way to be those given in the second column of the annexed Table. Then the third column shows the stresses due to a load of 29 tons at C, the rest of the structure being unloaded. A small slide rule has been used for these calculations, and is quite accurate enough for the purpose. The total stress, obtained by adding up columns 3, 4, and 5, is given in column 6. Then let u be the work done in deform- ing any bar, B D, for example. Then M = afe- {-707P + 21.8} 3 . Hence dn 1 L r _ i ^p=E-n-{- 707P + 21 - 8 } x - 707 > and L -^-forbarBD = 13.5. A v, ■ '■ rfp = 6.75P + 208. In this way columns 7 and 8 are obtained, and finally, if U is the whole work done in deforming the structure, U = 2 u ; and hence ?lF = 2 ^i? and ^? - 3 d u df d? dQ~ dQ' TRUSS WITH TWO SUPERFLUOUS BARS. 19 £ EO OS P £ *H a, O t-5 a S! « S H SI a CD X (NCOtOpH 00 II pt 1 i + l t^ T*Hde* ■« IS CO r-H CO i-t X ira oo aioo -— 3383^ -h|H 1 ++ 1 + . . . . Ph . I> II Cu Ph P-{ Ph £l : : : : jo : s|ph ira »o co^h ira °° i>.t^ w "el's T(i ID O rt ■* C? 0000 IOk^IO SQ J rt rji O •* ™ ™ 8 S2 i +* i + i i + i^d.-. to 3 oo^ooc-i©© £ 1-. r- © l> t>. i>. _j_i>e~ + + | + + + + + a a?© a 1 © ^ f- f- ©"fir Er © 1(5 : : : : : :t-;t^ | i>;t^ ■ • : 1 02 + + + + s PhPh PliPh «p<- . .SSfeSS Ph . •* m m o + + + + 3Q 5 d s Due oad o ns at I~ CO CC 00 CN O U3 CI CO CO CO CN O-*^^©^ "«* ©" 1^ JE> r-i o . . CO nNNMHHOHH rt ■ tres: a L 9 To + 11 + 11 1 ++ 1 1 + 02 CN-HCOCNCOINeOi-KN 53 PQ WOMfiOHaHPHROP=iOHd! i— i «i«!oeHfHOOii ( Q;5& &c. Then the load at a has moved through the distance A a , and has increased uniformly from to W a W a . Hence its average value has been ---. Multi- plying this average value of the force by the distance moved through, the work done by it on the beam W a has been -g-. a Bj and similarly for b, c, d, &c. 22 STATICALLY INDETERMINATE STRUCTURES. Hence the total work, v = W.a, + We a 6 + W c a c +)&o< 2 i 2> Suppose now the force W a is increased by a very- small quantity d W a . Then the deflection at each of the points considered will also be increased by corres- pondingly small quantities. Thus the deflection at S Aft b will be increased by an amount ry^" . d W a , and so on. Hence the total work TJ will be increased by a quantity d U = [Wa + d Wa ] ^~. dVf a + Wft £^-. d W« + W c j-^ . d Wa, + , &c. Then, neglecting the squares of small quantities and dividing by d W a , we get d V 8 Aa „ T S Aft __ S A c rwi = Wa sw a +Wb sw a + Wc rw+' to • w But, as already shown. 2 U = Wa Aa + Wft Aft + &o. „ d U .„ 8 Aa 8 Aft • '• 2 a^ =& + w »nf» + w » s-wT+> &c - < 2 > Then subtracting (1) and (2) we get finally dv and similarly ., _d V . d V . ■ Ab -dW b > A ° = &0 - Hence, finally, the deflection at any point of an elastic structure is equal to the differential coefficient of the work of deformation, with respect to the load applied at that point. Q.E.D. DEFLECTION. 23 The deflection at the point b of an elastic structure caused by load W at the point a is equal to the deflection at a in the direction of the original load, caused by an equal load at the point b applied in the direction of the original deflection. Suppose, as before, a beam loaded at the points a, b, c, &c. Let the deflection at b caused by the load at a be /3± W,, , and similarly for W& , W c , &c, so that the total deflection at b is Aft = 1 Wa + 2 Wo + 3 Wc +, Ac, and similarly deflections at a and c are respectively in = «, Wu + ajWl + a 3 Wc +, Ac, Ac = 7i Wa + 7 2 Wft + 7 3 Wc + , Ac. Then let U be the work done in deforming the structure. Then it has been shown that O = j"[Wa Aa + WS Aj + Wc 4c +,&».] or Then U = ?? [a, W« + o 2 Wo + u 3 Wc +, Ac] + -^-[0! W a + 2 Wb + 3 Wc +, Ac] + Ac, + Ac. d U dWa = i K Wa + a 2 Wo + « 3 Wc +, Ac.] Wa ± Wl. x W '. il, -g" «l + -jj" *i + "J Ti +■. &C - But it has already been shown that 4JE. = Aa = oj Wa + «2 W& + a 3 Wc +, i d Wa and equating these two values of d u d Wa' 24 STATICALLY INDETERMINATE STRUCTURES. it is evident that ft = "*2i 7i = &c. But the deflection caused at the point 6 by a load W at a is by hypothesis fi x W, and similarly the deflection caused at the point a by a load W at b is a 2 W. But yS 1 = a 2 ; which proves the proposi- tion. The Hawkesbury Bridge. As an example of the method, let it be required to find the deflection of the truss shown in Fig. 6, when loaded with 30 tons at each panel point. The truss in question is one of those used for the r h J i i ? \ / / V / V Fuj.6. b ^ i , V \ / A -V- 4 ! '■ "7 '-- -J— \ 7 \ / j — / c ! i e 1 J i 'o I J 3 n > A r \ \ c 3 3 1 j r 7 Si r ' 3 3 'l 1 1 \ 2£?9 £ -- 410' c &' 1 -M Hawkesbury Bridge, N.S.W. The bars shown dotted are counters, stiffening bars, or are other- wise unstrained under a uniform load. To determine the deflection, the length and area of each acting bar must be determined. These are given in the second and third columns of Table II., which extends up to the centre of the bridge only, the truss being symmetrical about mid span. The ratio of these — is given in the fourth column, whilst in the fifth THE HAWKESBURY BRIDGE. 25 column the stresses due to a load of 30 tons at each panel point are written down. As in previous cases, TABLE II. 1. 2. 3. 4. 5. 6. 7. Stress due to (du\ MC'" 1 ° Bar. L n L " n 30 Tons at each Panel Stress due to C. Point. in. sq. in. a i 625.2 175 3.57 + 224.7 + .626C + 502 a e 378.5 60 6 31 - 135.0 _ .376C + 320 eb 504.0 16 3.15 - 30.0 o e 378.5 74 5.11 - 135.0 _ .376C + 259 el 625.2 62 10.11 - 150.3 - .530 C + 805 bd 381.6 138 2.76 + 228.0 + .706 C + 444 d e 552.0 60.7 9.10 + 91.5 + .410C + 341 eg* 378.5 118 3.21 - 226.5 - .706C + 512 dg 558.0 43.7 12.76 - 100.0 _ .460 C + 587 df 381.6 170 2.24 + 294.1 + .962C + 634 gf 600.0 43.5 13.80 + 54.0 + .380 C + 283 9 i 378.5 148 2.56 - 282.0 - .9620 + 694 fi 709.2 28.9 24.50 60.0 — .4180 + 614 fh 381.6 187 2.04 + 317 1 + 1.188C + 762 i h 648.0 28 23.19 + 19.8 + .3500 + 161 i h 378.5 166 2.28 - 315.0 - 1.183C + 843 Tth 750.0 18 41.60 - 20.7 - .3800 + 327 bj 381.6 195 1.95 + 328.5 + 1.380 O + 884 jh 696.0 22 31.60 - 12.0 + .3260 125 k m 378.5 172 2.20 - 325.8 — 1.376 C + 941 j ill 793.2 22.5 35.21 - 32.4 — .5720 + 652 u 378.5 201.4 1.88 + 341.4 + 1.376 C + 884 m I 693.0 22 31.60 + .500 C m n I n lo 378.5 178 2.13 - 341.4 - 1.648 C + 1199 378.5 201.4 1.88 + 341.4 X. 1.975C + 1268 a slide rule has been uged. Now to determine the 26 STATICALLY INDETERMINATE STRUCTURES. deflection. Suppose a force C to be applied at the centre of the bridge as shown. This force will give rise to the stresses given in the sixth column of the Table. But the deflection at the centre of the bridge will, as we have shown, be equal to ay dC' where U is the work done in deforming the struc- ture, and this will be true whatever value has. Hence to find the deflection under the load of 30 tons at each panel point, it is only necessary to determine dC' and then to make C = 0. Let u be the work done in deforming any separate bar of the structure. Then U = 5«, and Now d U _ _ d v, dC c. Coming now to cases of bending, consider a beam E ABD (Fig. 8) supported at A and B, and fitted with scale pans at its extremities in which weights can be placed. Also suppose the weights in the two pans to be always equal. Then the bending moment curve will be as in Fig. 9, being uniform between A and B. Further, let the lengths outside the supports be very rigid, as compared with the remainder of the beam. Then, if equal quantities of shot are gradually poured into the two scale pans, the lower surface of 34 STATICALLY INDETERMINATE STRUCTURES. the beam will take a shape something like that indicated in Fig. 10. The parts A D and B E will each turn through a small angle a, whilst remaining tangents to the under surface of the beam at A and B. Hence perpendiculars to the beam at A and B will make an angle i = 2a with each other. Now, during this bending of the beam, the scale pan at each end has moved through a distance a sin a, and the average value of the load has been W 2' Fig. 8 Hence the total work done by the sinking of the loads will be W . a . sin a. But W . a = the bending moment in the beam between A and B. Therefore the work done will be M sin a = Ma, when a is small, as it always is in practice. But a = — , therefore the work done on a given length of a beam by a uniform bending moment is equal SOLID BEAMS. 35 to that moment multiplied by half the change in the angle between two normal sections of the beam taken at the extremities of the length considered. Now to determine the value of i. The stress on the upper surface of a beam is shown in ordinary works on statics to be : My where I denotes the moment of inertia of the beam, and y the distance of the upper surface from the neutral axis. The upper fibres will be elongated by an amount A _JP*_MyZ Similarly the lower fibres will be compressed by an amount . _ MzZ where z is the distance of the lower fibres from the neutral axis. Then y + z = lb, the depth of the beam, and hence the difference in length of the upper fibres and the lower will after stress be *» + Aa - IT' But if Pi be the radius of the upper surface of the beam and p 2 the radius of the lower, we will have (Pi - Pa)* = *i + *2 = -gj ■ But, in the case considered above, 7 . • MZ ft - fti = * ; ■ •* = gj- 36 STATICALLY INDETERMINATE STRUCTURES. And hence finally the work done in deforming a beam by a uniform moment M will be : 2 2EI' But in most cases which occur in practice the bend- ing moment is not uniform. Nevertheless, we may divide up the whole beam into a number of very small parts, throughout which the bending moment may be considered uniform. Then the total work done in bending the beam may be obtained by cal- culating the work done on each of the parts on the above assumption, and adding the whole together. The greater the number of parts into which the beam is thus considered to be divided, the greater the accuracy of the result. From this it follows that the work done in deform- ing a beam by bending is equal to the length of the M 2 beam multiplied by the average value of 2E1' The Deflection of a Curved Beam. M 2 The average value of — - can, in some few cases, be determined by the integral calculus, but more generally can only be determined by some approxi- mate method, such as Simpson's rule, or those of Cotes or Gauss. A very convenient rule has also been given by Mr. Weddle. We shall make use of this later to determine the deflection of the davit shown in Fig. 11, under a load W applied as indicated. It is assumed that the davit is solidly encastrd at its base, where it CURVED BEAMS. 37 is 4 in. in diameter, which thickness is maintained up to the commencement of the curved portion, after which it tapers down uniformly to a diameter of 2 in. at the extreme end. To determine the deflection of this, we have, as shown on page 21, vertical deflection d a ^ where U is the work done in deforming the beam. But 7 / M 2 \ U = — — x I mean value of — . I 2JS V I / To determine this mean value, divide the centre line 38 STATICALLY INDETERMINATE STRUCTURES. of the beam into six equal parts, as shown, and calcu- late the value of M 2 for each of the points 0, 1, 2, 3, 4, 5, and 6. ' Then if y is the value at 0, y 1 that at 1, and so on, we have, by Mr. Weddle's rule, mean value of Similarly y =jq [y° + y* + »4 + ye + 5 fai + y«) + 6 Vsl dV I ( ., ,dy-\ JW=2E' i mean value of aw J dy = Ay The values of -rrjl for the various sections are given below : M 2 . d y Section. I. M. I dW .7854 1 2.47 27.0 W 295 W 2 590 W 2 6.06 50.4 W 409 W 2 818 W 3 12.57 60.0 VV 286 W 2 572 W 4 12.57 60.0 W 286 W 2 572 W 5 12.57 60.0 W 286 W 2 572 W 6 12.57 60.0 W 286 W 2 572 W Then applying "Weddle's rule, the average value of ^| will be found to be 560 W, and the deflection ^ - ' dW^ l Av = d#=2~E - 56 ° W I = 163". E = 13,000 tons ; V^ . •. A = 3.5 W. Hence if W = 4 ton, A = J". If it were desired to determine the horizontal de- ARCHES. 39 flection of the end of the davit, under the load W, in addition to the vertical one, a force fi may be assumed to act horizontally as indicated in the figure. Then the moment at point 1 is now 27.0 W + .6 /j,. At point 2 it is 50.4 W + 14 /*, and similarly for the other points. Then the -^- can be found for each of these points, and TT l / M *\ U — 2~£; x I average value of ~y I ; and, finally, the horizontal deflection is equal to dU -j — , whatever the value of /a may be. Hence, if, after having obtained -n — , fj, is made 0, the hori- zontal deflection of the davit end under the influence of W acting alone is determined. Using Weddle's rule, the horizontal deflection will be found to be A H = 2.39 W in. where W is the load in tons. Six sections is, perhaps, in this case, rather too small a number of parts in which to divide the beam in question, if any special accuracy were required, but is enough for the purpose of illustration. The device of using an imaginary load to determine deflections, as in the above example, and in that of the Hawkes- bury Bridge, dealt with on page 24, is, we believe, original with Castigliano. Akches. Proceeding now to the discussion of some arch problems, let us suppose A B, Fig. 12, to be the centre line of an arch rib, and let it be loaded by 40 STATICALLY INDETERMINATE STRUCTURES. a weight W as indicated. Assume D E F to represent the line of resistance of this load. Then a certain thrust T, say, acts on the arch along the line D E, o A F B F B and another thrust along the line E F. Let G he the point at which the line D E cuts the vertical through 0. Now the action of the left-hand part of the arch on the right-hand part can be replaced by a force ARCHES. 41 equal to T, acting in the direction indicated in Fig. 14, whilst the action of the right-hand part of the arch on the left-hand part can be replaced by a numeri- cally equal force T acting as indicated in Fig. 13. These forces T may be resolved into their horizontal and vertical components P and S, as shown. Now consider a section H J taken at the right-hand side of the arch. Then the bending moment at this section is evidently P(«+!/) + Sj!-W«i or if Pa= N, the above expression will reduce to N + Py + Si-ffj. Turning now to the left-hand portion of the arch Fig. 12, it will be seen that the bending moment at H'J'is N + Py — S x where P and S now tend to bend the beam in opposite directions. This relation will always hold, and in whatever way the arch is loaded. P and S on one half will tend to bend the arch in the same direction, and on the other half in opposite directions. From the above it will be evident that if N, P, and S can be determined, then the bending stresses at all points of the arch can be found. Thus, if the arch is hinged at A, C, and B, the bending moment at each of these points must vanish, as the arch cannot resist bending there. In that case we should have N = 0, as T must pass through C. Also taking the right-hand half, we have Moment at hinge B = PA + Sa— Wc = 0, 42 STATICALLY INDETERMINATE STRUCTURES. and on the left hand Moment at A = P A — S a = 0. Whence P= 2 h and s = ^f. 2a If, however, there is no hinge at 0, it is no longer justifiable to assume that N is 0; but there being hinges at A and B, we should have and Whence as before, and N + P/t + Sa-Wc = N + P7i-So = 0. N = l 5 -Pl Thus in this case the three unknown quantities reduce to one, and this latter cannot be determined by pure statics. Finally, if there are no hinges at A and B, it is not justifiable to assume that the moments at the abutments are zero ; and hence, in this case, no one of the quantities N, P, and S can be determined by pure statics. We have, however, the general result holding for all elastic bodies, that the work done in deforming them is always the least possible consistent with the maintenance of equilibrium. If, then, in the case of the two- hinged arch we determine P so that the work done TWO-HINGED ARCHES. 43 is a minimum, N can then be found from the equation N = ™-°- P h. 2 In the case of the arch rigid at both abutments and crown, the whole three quantities, N, P, and S, must be chosen so as to make IT a minimum. Two-hinged Aeches. As an example of a two-hinged arch, take the case shown in Fig. 15, which represents diagrammatic- ally the Washington Bridge over the Harlem River, New York. The span in this case is 510 ft. between the centres of hinges, and the rise of the centre line is 90 ft. The dead load on an intermediate rib is said to be 1469.5 tons, and the live load 370 tons. Each rib is 13 ft. deep, and is of I-section, built up of plates and angles. 44 STATICALLY INDETERMINATE STRUCTURES. The work done by the various stresses acting on the rib can be divided under three heads, viz., that due to bending, that due- to direct compression, and finally, that due to the shearing forces. As has been shown, the work done in bending a bar is Ub = ;p= x ( mean value of -^-M, where M is the bending moment at any point of the bar, and I the corresponding moment of inertia of the cross-section ; and similarly d Us i r , c M d M i TV- =E x L meanvalueo£ T-dpJ- To find this mean value, divide up the centre line of the arch as indicated in Fig. 15, and deter- mine the bending moment at each of the points of division. As before, the action of the left-hand half of the arch on the other half may be replaced by a moment N, a horizontal thrust P, and a shear S. Then, similarly to the previous instance, Fig. 14, the moment at any section, say (3), will be M, = K + Pj + Sj;-«1 where w is the load per foot run of the span. Taking lengths in inches, M 3 = N + 277.9 P + 1622.4 S - 395,626 for the right-hand side, = N + 277.9 P - 1622.4 S - 316,049 for the left hand-side. These bending moments are given in Table IV. which has been calculated on the assumption that the live load extends over the right-hand half of the span only. The values of I, as determined from the drawings, will be found in the fourth column. At section 6 there is a hinge, hence here the bend- ing moment and I are both zero. TWO-HINGED ARCHES. 45 In this Table N, P, and S denote, as before, the bending moment, the horizontal thrust, and the shear at the crown, and this notation will be main- tained throughout. Then at section 2, say, on the right hand, we have : M d M _ f N + 124.2 P + 1093.3 S — 179,644 1 I " d P I 1,364,000 I x {Jf + 124.2 } , but ^ = _ h, = — 1080 in. U> ST h being the versed sine of the arch, . M d_M _ / N + 124.2 P + 1093.3 S - 179,644 | ' I' d t " X 1,364,000 i x |— 955.8 j- Similarly at the corresponding section on the left- hand side of the arch, we should have M «[M _ ( N + 124. 2 P— 1093. 3 S — 143,510 \ I " d P I 1,364,000 / x [—955.8 j- From which it will be seen that the terms in S cancel each other when both sides of the arch are considered. Hence, neglecting terms in S, the values of M dM 1' dP' will be found in Table V., the coefficients of S not being calculated as taking both sides into account, they cancel each other, as already mentioned. 46 STATICALLY INDETERMINATE STRUCTURES. M oooooo O O Q © O O oooooo CM WS CO CO CO CO w W !M CO -* "* CO 4H M 13 1 g s a o inch-tons N + 31.1 P — 550.2 S — 36,340 N + 124.2 P — 1093.3 S — 143,510 N + 277.9 P — 1622.4 S — 316,049 N + 490.6 P — 2130.9 S — 545,167 N + 758.9 P — 2612. 1 S — 819,193 T3 ■a (3 o s inch-tons. N N+ 31.1 P+ 550.2 S— 45,490 N + 124.2 P + 1093.3 S — 179,644 N + 277.9 P + 1622.4 S — 395,626 N + 490.6 P + 2130.9 S — 682,434 N + 758.9 P + 2612.1 S — 1,025,454 a o "■£ o o O 11 CJ CO -* "O CO CO *o CO OS IT** OOCOCOCD CO O t-(N OS p-Hi-HOI r-i + + + + + GO GO 02 GG 02 23 i© W p "g II « oa. f*.io w + + + + + PhPwPhPhPh hhimWho uscoosojg w fflOCD'OW id r^ in e! £! 43 ^ V ■ W 00 io o 00 ^h 1 1 II 1 © uscoweoco cn -*r-c»r»^ CKOiaoo MWOIQH-* CO 00 I- »Q ■>* N 1 1 1* l" l.l" > O CO 00 00 00 i_5 ©COWOil> ffl OS 00 CD CN W eS NWOO® CO w |s Mill 1 » PhPjPmPhPh HrlNWJHQ w^b lO CO OS© «D OS © » 55 CS •a • wSi>o --5 c* oi com o go t-H (N >-i a ah 1 1 II 1 fcfcfclzilziiz; O WCQiO CO CO (M t* r- os it* ■— i o^ooooq cowowh^ • oo go r-- w •* cn i I I I I I g o '■§ © H CN CO "* W CO TWO-HINGED ARCHES. 47 Then by Weddle's rule already explained, the mean value of M 9 + 1134 3.018 P — cS + 1557 2.747 P—dS + 1954 2.567 P — eS + 2389 1.877 P — /S + 2219 3,886 P 3.780 P + aS + 500.8 3.367 P + JS + 906.5 3.018 P + cS + 1243 2.747 P + AS + 1562 2.567 P + eS + 1908 1.877 P +/S -t 1772 From Weddle's rule, then, the mean value of T dT q" dP from one end of the arch to the other is mean value of T A T _ 3.086 P + 1319 O dP 1000 TWO-HINGED ARCHES. 49 Hence But axjc _ i d P 1000 K I 3.086 P + 1319). dV _ dV B dUo . dP d P d P .-. 0.5359 N + 110.1 P — 143,091 = 0. Also, since the bending moments at the abutments are zero, we have N + 1080 P — 1,265,689 = by eliminating S from the expression for moments at both abutments. Solving these equations, N = 32,500 inch tons, and P = 1140 tODS. It will be observed that, to calculate the stress arising in the arch, it was necessary to know the dimensions of its various parts, whereas the very object of finding the stresses is to determine these dimensions, and in this way a difficulty arises. Experience shows, however, that for a first ap- proximation the work done in deforming an arch by the normal pressures is small compared with that done by bending. In the case considered it only affects the value of the horizontal thrust by about 1 per cent., hence, in a first approximation, both it, and the variation of I, may be neglected. If this is done it is possible to obtain the provisional value of the horizontal thrust P from suitable tables, such as that given on next page (Table VIII.), which shows tht variation in the thrust as a continuous uniform load moves over a two-hinged arched bridge. The figures give the value of C in the equation P = C w r ; w D 50 STATICALLY INDETERMINATE STRUCTURES. tH © p-H "# "* CO ,_ IO © *0 CO © »o CO IN IO I © CM © © I-H IN IN © OJ GO r~ t- © IO -# t~ 00 IN r~ © © © © © o © ■* >o H* © © CM II © T(l © Tti r~ © © ~H cH © © 00 r~ © © IO T|l a 3 & ra CO IN IO © CO Th © ■* o 00 CM TX r~ IN © (M © d r~ CO 00 H* O !>• 00 1 00 CO J>- c~ © •o ■* CO *43 © to s ■* r-- o © © fn t^» CM i-H 00 IN © IN IN O d to CO 00 IN © © CO > t~ r~ © © IO TP 5 CO o -*» A to r~ CO r* 00 t« © ■* 3 d (M © © © © TK t- •« CO to IO »o r* ■* CO IN *-1 a (— I o £ S !>■ © © r- r~ (N IO CO IO CM p— I CO »■ © f-H CM CM o d 00 © CO © o i-H © -H H «w ■* TH -* CO CO CO IN CM ^ GO W a ^ Eh -*3 IO tK IN © © 00 3 © ■* r- © I-H © © t-» © d CO IN © t>. Tfl i-H CO -# •n CO CO CO IN IN IN l-H p-H <3 o 8 io CO t~ © 1^ CO "O © co IN IN --H t^ © IN IN d o in TO GO © IO CO © © «l-i o 3 00 © CO © CO t~ -* IN (M o r~ © -* © 00 IN co © T« CO i-H IN IN -* d © 00 00 r~ r~ © IO -# o © © © © © © © r~ OJ 00 IO IN IN 00 i-H to 00 r-~ M3 CO IN 00 3 CO i-H © 00 © -* 1— 1 o CM CM -H i-H i-H f-H 1 o o © © © © © © •s^uounnq'E uoau^aq © © o © o © o © pspna^qns tH © 00 © (N Ttl © 00 ojSt 'V RIGID ARCHES. 51 being the weight of the load in tons per foot run, and r the mean radius of the arch in feet. . By taking the value of the horizontal thrust from this table, a provisional design of the rib can be prepared, after which the thrust should be recalculated, taking into account the variation in the moment of inertia of the rib and the work done by the tangential thrust. It should, however, be pointed out that the maximum bending moments will generally occur when the live load is distributed over from -§■ to f of the span, but this of course will not affect the method of calculation which is all we are concerned with here. The value of N will usually be greatest when the live load extends over the central third or fourth of the span only. Eigid Arches. As an example of an arch fixed at the abutments, let us take the recently completed' Pont Morand, Lyons. This bridge is remarkable not only for the low ratio of rise to span, viz., — — , but for the beauty of the completed structure. The span is 67.4 metres, and the rise 4.44 metres. The ribs are of box girder section, .8 metre deep at the crown, and 1 metre deep at the abutments. The dead load on a rib is 3.38 tons per metre run, and the designed live load 1.04 tons per metre run. Dividing as before the medial line, of one-half the rib into six equal parts its moment of inertia^and area 52 STATICALLY INDETERMINATE STRUCTURES. in inch units at each of the points of division are about as follows : Section. I [Inches] 4 . a Sq. In. 31,960 159.5 1 32,960 159.5 2 35,140 144.0 3 37,340 146.0 4 46,220 150.0 5 61,200 170.0 6 84,000 201.0 Let us first find the central bending moment, hori- zontal thrust, and shear for the case in which the arch is loaded from end to end with both its dead and its live load. Then if N, P, and S denote respectively this moment, pressure, and shear, the bending moment M at the other sections will be as follows, S being zero from the symmetry of conditions. Section. 1 2 3 4 5 6 M. In.-Tons. N N + 4.887 P- - 2,806 N + 19.463 P- •11,203 N + 43,756 P- -25,130 N + 77,702 P- -55,995 N+ 121.236 P- ■69,102 N + 174.280 P — 98,816 From this it is easy to get the values of M f dM M dS' I dM 'IP- They are as follows, the values being multiplied by 20,000 simply to avoid long decimals : RIGID ARCHES. 53 Sec- tion. 20,000 M d M 20,000 M d M IdP 1 2 3 4 5 6 .6258 N .6072 N + 2.955 P- .5692N + 11.077 P- .5356 N + 23.437 P- .4327 N + 33.623 P- .3268N + 39.620 P- .2381 N + 41.495 P- - 1,705 - 6,376 -13,460 -24,230 -22,528 -23,528 2.955N+ 14.39 P- 11.077 N + 215.59 P- 23.437 N + 1025.50 P- 33.623 N + 2612.60 P- 39.620 N + 4803.49 P- 41.495 N + 7231.79 P- 8,295 124,104 590,321 1,882,714 2,737,844 •4,100,384 As in previous cases, if TJ B is the work done in bending, dUs If , ,M Ml -r^f = -= . i mean value of — . -^-^ Y dN El I dN) and dV s If , -M dM\ ^P =E-( meanvalueof T-lp) Applying Weddell's formula, we get dVs Ub C I O.OOO^jj = £-. J .4875 N + 21.985P - 12,816 (1) and 20,000-^- = b^- [ 21.985 N + 2015.1 P — 1,168,988 J But account has to be taken of the work U T done in the direct compression of the rib, and a term 20,000 -rw must be added to the last equation before equating it to zero. T is most easily determined as indicated on Fig. 16, where ABC represents the right-hand side of the arch, and O its centre of curvature. On O A take O E = 1, and on this, as 54 STATICALLY INDETERMINATE STRUCTURES. diameter, describe the semicircle E F O. Then the resolved part of P along the arch ring at B is p OF " OE" Hence it is only necessary to scale off F. Simi- larly the resolved part of any central shearing force S acting along the arch ring at B is q EF S -OE' and, finally, the resolved part of the load acting w. 16. along the arch ring at B is equal to the total load E F between A and B multiplied by O E" The total thrust T is the algebraical sum of all these components, it being noted that with unequal loading S acts upwards on the more heavily loaded side of the arch, and downwards on the other. We should thus get for section B : F T = OE P + EF ( OE' 1 ■S for right-hand half, and /r = OF P + EF OE- OV C-^ + B] for left-hand half, where w x = the total load be- RIGID ARCHES. 55 tween the crown and the section considered on the right-hand side, and w 1 x the corresponding quantity on the left-hand side. As before, f T d T ) i mean value of_ . -=— > . I a dPi dT d_U T d P I a T The values of - . ^^ are as below O a r Section. Tangential Thrust. T. T dT 20 ' 000 ^ TP ■ 1 2 3 4 5 6 P .999 P + 1.091 .996 P + 4.364 .991 P + 9.790 .995P + 21.8 .976 P + 26.9 .966 P + 38.5 125.38 P 125.16 P + 137 138.40 P + 607 134.52 P + 1330 129.20 P"+ 2860 112.06 P + 3200 92.88 P + 3702 From this, the value of p T is 20,000 dV j _ i_ r d e E L 123.9 P + 1542 '] But dV d¥ dV v d P + i^. 1 = 0. d P Therefore 21.985 N + 2139.0 P — 1,167,466 = 0. Combining this with (1), and solving, it will be found that N = + 3126 inch-tons, and P = 514 tons. If the live load only extended from one abutment up to the crown of the arch, N and P would be re- duced in the ratio of the new average load on the 56 STATICALLY INDETERMINATE STRUCTURES. arch to the old one. The average load when the live load extends over the whole arch is 4.42 tons per metre, and the average load when the live load extends to the centre of the arch only is 3.38 + 4.42 _ 3 90 tQng metre 2 r Hence the new value of N = ~ (3126^ = 2757 inch-tons. 4.42 \ / P = ?^° ( 5li\ = 454 tons. 4.42 \ ) As, however, the load is no longer symmetrical, there will be a shearing force at the centre, the value of which is unknown, but can be determined by the relation -j-~ = where S denotes this shear at mid span. At any section, say, for example, 3, we should have for the bending moment on one side of the arch, M = N + 43.756 P + 669.1 S — 25,130; and at the corresponding section on the other side, M = N + 43.756 P — 669.1 S — 19,217. This gives, on the one side, M dM 669.1 t 1 I" d~S = ~ T ■ { N + 43,756 P + 669.1 S — 25, 130 1 and, on the other side, M ijM 669.1, „ „ _, T ■ ^S" = T~ i — N — 43 - 756 p + 669 - 1 s + 19 » 217 } " Taking the mean of these, we get M dM Average value of j- - -r-q- at section 3 = 669.1 r , —j- j 669. IS— 2957 \ . Similar results will be obtained for the other RIGID ARCHES. 57 sections, and we thus get the figures in the second column of the annexed Table : Section. M dM I • dS- T dT a'dS- 1 2 3 4 5 6 1.517 S — 1.22 5. 6789 S— 16.76 11.989 S— 52.97 17.144 S— 126.87 20.116 S— 147.38 20.957 S — 183.63 .00001 s — .00005 S — .00012 S — .00020 S — .00027 S — .00033 S — .00007 .00031 .00103 .00296 .00402 .10552 Again, the thrust T at the point 3, say, of the arch will on the one side be .991 P— .1302 S + 9790, and on the other .991 P + .1302 S + 7484. Then the mean value of _T d_T a • d S at the two corresponding points will be .1302 , -, — £- . 1 .1302 S — 1.153}, P cancelling out. The values of this quantity for the different sections are given above. We have then dV d S = = d Ub d Ut d S + d S But and d Ub M > = R2 " " •■■ ••• ■■• ■ ■• = R» And, also, (R + 2 R 2 + 2 R 2 + 2 R s ) = W, W being the load on each rail. Hence the total work done on the troughs will be : 1968 i ^ U t = -^- j R 2 + 2 R x 2 + 2 R 2 2 + 2K,H, or substituting for R , we have 1968 U t = ^-.[(W-2R 1 -2R ! -2 R 3 ) 3 + 2 Ri 2 + 2 R 2 2 + 2 R s 2 ]. Similarly the work done on the rails can be ascertained in terms of R 1; R 2 , and R 3 . Calling this work U r, we have the total elastic work of deformation is Ur + Ut = U, TROUGH FLOORS. 67 say, and U is a minimum, with respect to ~R lt R 3 , and d U dJJ dJJ ■'■ dH 1 = dR 2 = dR s = "■ It is unnecessary to go through the whole work here, but we finally get the three equations : 13,742.8 R 3 + 8000.8 R 2 + 5097.5 Rj = 1968 W. 8,008.8 R 3 + 8226.8 Rj + 4661.8 R 2 = 1968 W. 5,097,5 R 3 + 4661.8 ^ + 6194.3 R x = 1968 W. The solution of these equations, using seven-figure logarithms is Ri = .242.245.2 W. R 2 = .115.406.0 W. R 3 = —.013.838.8 W. Hence R = +.312.375.2 W. From this it appears that in the above case rather less than one-third of the weight directly over a trough is transferred to it, the rest being carried by the adjacent troughs. The results are recorded in Fig. 17. In practice, however, more than one pair of wheels has usually to be reckoned with, and if a second pair of wheels were added at a distance of 7ft. 6in. from the first pair, the resulting reactions would be those given in Fig. 19; and it is the com- bined effect of the two that has to be considered. In this case the maximum reaction is rather greater than one-third the load on one wheel. It will be noted that the negative reaction is less than one- seventieth of the load on one wheel. Another question arises as to the effect of the joint in modifying the above distribution. On this head it may be remarked that with the very stiff 68 STATICALLY INDETERMINATE STRUCTURES. joints now used no diminution in the distributing power of the rail is to be expected, but it may be of interest to ascertain the modification in the dis- tribution when the rail has a perfectly flexible joint Thus suppose the loading to be as in Fig. 20, there being a pin joint at the point shown. On making the calculation it will be found that, as was to be expected, the negative reactions are considerably increased, but the maximum positive reaction is not so much increased as might have been anticipated, the distribution being as shown in Fig. 20. As even the weakest rail joint has some degree of stiffness, it appears, therefore, that the distribution obtained in assuming a continuously stiff rail may be relied upon as that actually existing. If the load rested directly over the joint, the maximum reaction would be .443 W. If the rails are laid directly on the troughs without the intervention of cross sleepers, a much more favourable distribution can be obtained. The distance between the points of support of the rail is now shortened to 20 in., and the maximum reaction for a single load is then .221 of the weight imme- diately above the trough, the distribution being as in Fig. 21. With two sets ,of loads 7 ft. 6 in. apart, the maximum trough reaction would be about .26 the load on one axle. This result shows the advantage of longitudinal sleepers in conjunction with trough flooring. Coming now to the case of a double line of rail, we may take Hobaon's flooring as used in the Liverpool Overhead Railway. This floor consists TROUGH FLOORS. 69 of troughs pitched at 2 ft. 6 in. centres, and 15 in. deep. It is built up of arch plates -f^ in. thick, riveted to j_-i r ons at the bottom. The span between main girders is 25 ft., and the rails are carried by longitudinal sleepers. If the rail was the typical one already used, and fixed direct to the troughing, we find that in the case of the inside pair of rails the distribution would be as in Fig. 22 and for the j_ 8016 rail resting directly \. j on troughs -J !Vj >\\J "W '!V^ M V_y ' \Vj r fV_ + '050 W. +-I3SW. +-I9SW. + '2tlW. +796W. + 135 W + 'OSO W Fig. 21. outside pair as in Fig. 23. Practically the distribu- tion is probably a little better than this, as though the rail actually used is not as stiff, we believe, as the typical one taken, the combined stiffness of the longitudinal sleeper and the rail is probably greater. With two loads 7 ft. 6 in. apart the maximum reactions are .34 W. and .435 W. respectively (Figs. 24 and 25). Just for comparison we have calculated out . liobaon's Flooring ZS 'span. I jw I Distribution of load' on V [ J inside pair of rolls. /|YtY!V!VtVfY|\ -021 W. '07/ W '?S9 w. -36.9 W 'Z6.9 W 071 W - Oil W. Fig. 22. the distribution on this floor, if it had been built of the Lindsay C section, used on the single-line spans referred to above. The results are shown in the diagram (Fig 24). Here it will be seen that the 7-0 STATICALLY INDETERMINATE STRUCTURES. ^ \Hobsons Flooring ZS'&pan f L Distribution of Load on \ J ' outsidt pair of rails. /jYiYiYiYiVfYIV t ! ! ! I ! r -■O+SW -044W. '260 1V