Page:Encyclopædia Britannica, Ninth Edition, v. 4.djvu/344

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300
BRIDGES
[beams.

tinuous girders require less material for the same depth, span, and permanent load than simple girders ; but the difference is hardly worth consideration, except in large spans where the weight of the structure is great as compared with the weight of the passing load. The necessity of al lowing the girder to expand and contract freely with changes of temperature limits the general use of large continuous

girders to two spans.

Fig. 23.

Fig. 24.

Fig. 25.

Consider a continuous girder having an indefinite num ber of supports equally spaced ; let the girder be of constant depth, uniformly loaded, and so proportioned as to be of uniform strength at all sections. The me thod of designing such a girder has not yet been shown, but it may be ad mitted that the design is possible. Then, as we have already seen ( 24), the curvature assumed by the girder will everywhere be constant, i.e., the curves will be circular arcs of constant radius. The points of inversion of flexure A and B (fig. 24) must therefore in this case be at a distance from C and C u , equal to one quarter of the span ; for AA X ? AB ? A x B x &c., and C bisects AA r At B and B! there is no moment of flexure. The girder might here be cut through, and if the ends B and B t were pinned on to the rest of the girder by pins capable of bearing the shearing stress, nothing would be changed in the cur vature of the girder nor in the distri bution of stress. Quite similarly, we may suppose the ends B and B : portion to rest directly on sup ports or piers in troduced for the purpose, the rest of the imaginary girder being wholly removed. We shall then have (fig. 25) the curve assumed by a continuous girder of two spans of constant depth and uniform strength. The points A.A 1 of inversion of flexure will be at a distance CA from the middle pier equal to one- third of the span. The top and bottom members at A and AJ might vanish (if only one distribution of load were to be carried), for at this point there is only a shearing stress and no bending moment. The girder is, as it were, made up of three girders : BA and A 1 B 1 hang on to AAj, which is supported in the middle; half the weight of AB is borne by the pier B ; half the weight of A : B X is borne by the pier B : ; the rest of the weight between B and B x is borne by the middle pier. Thus let L be the length of one span in feet, tv the load per foot run, P the load borne by each of the end piers, and P c the load borne by the centre pier, then we have—

The curve of bending moments can now be calculated for each girder precisely as for a simple girder ; the moment at any section is equal to the sum of the moments of all the external forces on one side of the section ; the beam between A and B will be subject to bending moments equal to those produced by a uniform load on a simple girder if the span BA be similarly loaded. Between A and AJ the moments will be negative, i.e., the left-handed moment produced by the downward action of all the weights to the left will exceed the right-handed moment produced by the upward reaction of the pier at B (or of the two piers B and C if the section considered lies to the right of C). The full black line in fig. 25a shows a curve of bending moments for this case.


Fig. 25a.

The maximum moment is negative arid occurs over the centre pier ; let it be called M c . The maximum positive moment occurs at a distance from B equal to one-third of the span ; let this moment be M d . Then we have—

M e = - ^iv L 2 ,

The shearing stresses F 6 , F d , F a , F c , at the points B D, A, and C, are

5 ...... F = F

Fig. 26 gives a diagram of the shearing stresses for two spans of 60 feet, with a uniform load of 1 ton per foot run.


Fig. 26.
When the beam is of uniform depth and uniform cross

section, the curves in which it deflects are no longer

circular; the defect of strength over the centre pier has