is wl; the reactions at abutments, R1 = R2 = 12wl. The distribution of shear on vertical sections is given by the ordinates of a sloping line. The greatest bending moment is at the centre and = Mc = 18wl2. At any point x from the abutment, the bending moment is M = 12wx(l−x), an equation to a parabola.
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| Fig. 41. |
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| Fig. 42. |
23. Shear due to Travelling Loads.—Let a uniform train weighing w per ft. run advance over a girder of span 2c, from the left abutment. When it covers the girder to a distance x from the centre (fig. 41) the total load is w(c+x); the reaction at B is
R2 = w(c+x) ✕ (c+x)4c = w4c(c+x)2,
which is also the shearing force at C for that position of the load. As the load travels, the shear at the head of the train will be given by the ordinates of a parabola having its vertex at A, and a maximum Fmax. = −12wl at B. If the load travels the reverse way, the shearing force at the head of the train is given by the ordinates of the dotted parabola. The greatest shear at C for any position of the load occurs when the head of the train is at C. For any load p between C and B will increase the reaction at B and therefore the shear at C by part of p, but at the same time will diminish the shear at C by the whole of p. The web of a girder must resist the maximum shear, and, with a travelling load like a railway train, this is greater for partial than for complete loading. Generally a girder supports both a dead and a live load. The distribution of total shear, due to a dead load wl per ft. run and a travelling load wl per ft. run, is shown in fig. 42, arranged so that the dead load shear is added to the maximum travelling load shear of the same sign.
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| Fig. 43. |
24. Counterbracing.—In the case of girders with braced webs, the tension bars of which are not adapted to resist a thrust, another circumstance due to the position of the live load must be considered. For a train advancing from the left, the travelling load shear in the left half of the span is of a different sign from that due to the dead load. Fig. 43 shows the maximum shear at vertical sections due to a dead and travelling load, the latter advancing (fig. 43, a) from the left and (fig. 43, b) from the right abutment. Comparing the figures it will be seen that over a distance x near the middle of the girder the shear changes sign, according as the load advances from the left or the right. The bracing bars, therefore, for this part of the girder must be adapted to resist either tension or thrust. Further, the range of stress to which they are subjected is the sum of the stresses due to the load advancing from the left or the right.
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| Fig. 44. | Fig. 45. |
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| Fig. 46. | |
25. Greatest Shear when concentrated Loads travel over the Bridge.—To find the greatest shear with a set of concentrated loads at fixed distances, let the loads advance from the left abutment, and let C be the section at which the shear is required (fig. 44). The greatest shear at C may occur with W1 at C. If W1 passes beyond C, the shear at C will probably be greatest when W2 is at C. Let R be the resultant of the loads on the bridge when W1 is at C. Then the reaction at B and shear at C is Rn/l. Next let the loads advance a distance a so that W2 comes to C. Then the shear at C is R(n+a)/l−W1, plus any reaction d at B, due to any additional load which has come on the girder during the movement. The shear will therefore be increased by bringing W2 to C, if Ra/l+d > W1 and d is generally small and negligible. This result is modified if the action of the load near the section is distributed to the bracing intersections by rail and cross girders. In fig. 45 the action of W is distributed to A and B by the flooring. Then the loads at A and B are W(p−x)/p and Wx/p. Now let C (fig. 46) be the section at which the greatest shear is required, and let the loads advance from the left till W1 is at C. If R is the resultant of the loads then on the girder, the reaction at B and shear at C is Rn/l. But the shear may be greater when W2 is at C. In that case the shear at C becomes R(n+a)/l+d−W1, if a > p, and R(n+a)/l+d−W1a/p, if a < p. If we neglect d, then the shear increases by moving W2 to C, if Ra/l > W1 in the first case, and if Ra/l > W1a/p in the second case.
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| Fig. 47. |
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| Fig. 48. |
26. Greatest Bending Moment due to travelling concentrated Loads.—For the greatest bending moment due to a travelling live load, let a load of w per ft. run advance from the left abutment (fig. 47), and let its centre be at x from the left abutment. The reaction at B is 2wx2/l and the bending moment at any section C, at m from the left abutment, is 2wx2/(l−m)/l, which increases as x increases till the span is covered. Hence, for uniform travelling loads, the bending moments are greatest when the loading is complete. In that case the loads on either side of C are proportional to m and l-m. In the case of a series of travelling loads at fixed distances apart passing over the girder from the left, let W1, W2 (fig. 48), at distances x and x+a from the left abutment, be their resultants on either side of C. Then the reaction at B is W1x/l+W2(x+a)/l. The bending moment at C is
M = W1x(l−m)/l+W2m{1−(x+a)/l}.
If the loads are moved a distance ∆x to the right, the bending moment becomes
M+∆M = W1(x+∆x)(l−m)/l+W2m{1−(x+∆x+a)/l}
∆m = W1∆x(l−m)/l−W2∆xm/l,
and this is positive or the bending moment increases, if W1(l−m) > W2m, or if W1/m > W2/(l−m). But these are the average loads per ft. run to the left and right of C. Hence, if the average load to the left of a section is greater than that to the right, the bending moment at the section will be increased by moving the loads to the right, and vice versa. Hence the maximum bending moment at C for a series of travelling loads will occur when the average load is the same on either side of C. If one of the loads is at C, spread over a very small distance in the neighbourhood of C, then a very small displacement of the loads will permit the fulfilment of the condition. Hence the criterion for the position of the loads which makes the moment at C greatest is this: one load must be at C, and the other loads must be distributed, so that the average loads per ft. on either side of C (the load at C being neglected) are nearly equal. If the loads are very unequal in magnitude or distance this condition may be satisfied for more than one position of the loads, but it is not difficult to ascertain which position gives the maximum moment. Generally one of the largest of the loads must be at C with as many others to right and left as is consistent with that condition.
This criterion may be stated in another way. The greatest bending moment will occur with one of the greatest loads at the section, and when this further condition is satisfied. Let fig. 49 represent a beam with the series of loads travelling from the right. Let a b be
