119. Beam supported at both Ends and subjected to an Advancing Distributed Load of Uniform Intensity. In fig. 70 suppose an advancing load of w per foot to come on a beam from the right. The maximum positive shearing stress on the section at a distance, x, from the left abutment, occurs when the front of the load is at this section, and the amount of this stress at x, and w (1 − x)2 all points to the left of x, is, F2 = P This shearing 21 stress increases as x diminishes, and becomes a maximum when x = 0, or immediately over the left abutment, where it wl 2' equals = It is easy to see that the maximum shearing stress at the section indicated must occur when the front of the load is just at it, for if the load move to the right, the value of P will diminish, and consequently the shearing stress. If, on the other hand, the load move to the left of the section the value of P is increased by a portion of the load to the left of the section, but the shearing stress is less than the new value of P by the whole load to the left of the section, and consequently it is less than the first value of P. This will hold true whether the moving load is long enough to cover the whole span or only a portion of it. In a similar manner it may be shown that the greatest negative shearing stress occurs at any section when the tail of the load rests over it and that the stress is equal to Q, the supporting force of the right abutment for this position of the load. It will also be apparent from this that the maximum shearing stress at the centre will occur when the beam is loaded over ωι one-half the span, and will be equal to 8 120. Diagram of Shearing Forces when the Length of the Load is equal to, or greater than the Span.-In fig. 70 suppose a load of uniform intensity of w per foot to pass over the beam A B in the direction from B to A. Draw the horizontal at A and is represented by A, A, = (1097) 2 D E Figs. 70 and 70a. 70 B1 70a and occurs when the span is fully loaded. The maximum shearing stress at the centre is C1C2 ωι 8' 3 and occurs when the right half of the beam is loaded. By constructing a similar parabolic curve, A, C, B2, underneath A, B1, we get a diagram for the maximum negative stresses on the beam, these stresses being exactly the same in value as the positive stresses. The shaded portion of the figure is the diagram for the maximum numerical shearing stresses on the beam during the passage of the load. 121. Diagram of Shearing Forces when the Length of the Load is less than the Span. The locus of the maximum shearing stresses in this case is not a parabolic curve, but is made up of a straight line and a parabolic curve. It may be drawn thus Draw the parabolic curve, A, D, B1, fig. 70a, by making A, A, wl = Set off B1 D 2 = a; draw the vertical, D D1, meeting the curve in D1. Through D, draw D, A,, a tangent to the parabola at D1. A practical method of drawing this tangent is to bisect D B1 at E, join E D1, and produce it. The locus of the positive maximum shearing stresses on the beam is a line composed of the curve B1 D1, and the straight line A, D1. Example 10.-A railway train weighing 2 tons per foot passes over a bridge 200 feet span. What are the maximum positive and negative shearing stresses on each of the two main girders at intervals of 20 feet, the train being considered as a uniformly distributed load and longer than the span, and the dead weight of the bridge being neglected? 7 = 200. w = 1 ton for each girder. Dead load on each girder 200 tons. TABLE XVII. = 150 The maximum positive shearing forces may be calculated from the equation, F = the equation, F2 = w (l - x)2 21 wx2 = 21 Q, by giving to x its proper value. It will be seen that the maximum numerical value of the shearing stress occurs at each abutment, and is equal to 100 tons, while the minimum numerical value occurs at the centre of the beam and is equal to 25 tons. Example 11-In the last example, what are the maximum shearing stresses at the different sections of the girder during the passage of the load if the dead weight on the girder, including its own weight, be 150 tons equally distributed? The fourth column in the table gives the shearing stresses arising from the dead weight alone, and the fifth column gives the total maximum shearing arising from both the live and dead loads. These latter stresses are obtained by adding those in column 4 to the corresponding maximum numerical stresses in columns 2 and 3. It will be seen from Table XVII. that whether the live or dead loads, or a combination of them, be taken, the shearing stresses on both halves of the girder are equal, though of opposite sign. Fig. 71 represents graphically the shearing stresses on the girder for the different loads drawn to scale. A B=span=200 feet. Aa=ab=bc, &c. = 20 feet. Draw the ordinates, A A, and B B1, each = 75 tons. Join A, B,; the two triangles, A A, C, B B, C, will consequently represent the diagram of shearing stresses on the girder for the dead weight alone. Next, draw the ordinates, A A2, B B2, each equal to 100 tons. Construct the parabolic curves, A, C, B, B, C, A. The spaces included between these curves and A B will give the diagrams for the maximum positive and maximum negative shearing stresses respectively, arising from the travelling load. Next, by setting off on the ordinates through A, a, b, c, &c., A, A = A A1, ɑ ɑ3 = ɑ ɑ1, b2b2 = b b1, &c., we get the line A, a. . . C„ which is the locus of the maximum positive shearing stresses for both loads taken together. 3 . C2, which is In the same way we can construct B, h3 93 the locus for the maximum negative shearing stresses. CHAPTER IX. CENTRE OF GRAVITY AND MOMENT OF INERTIA OF It will be seen in Chapter X. that, in order to determine the transverse strengths of beams of solid section, it will be necessary in the first instance to know the moments of inertia of the sections of such beams with respect to axes passing through their centres of gravity. It is advisable, therefore, that the student be able to determine the centres of gravity and moments of inertia of the sections of such beams as are usually to be met with. In ordinary language the terms "centre of gravity" and "moment of inertia" have reference to the weight or mass of solid bodies, and belong to the domain of rigid dynamics. As used in this sense they do not concern us here; it is only in their application to plane surfaces that we wish to consider them. 122. Centre of Gravity.-The centre of gravity of a plane simple surface is its geometrical centre. That of a circular surface, for example, is the centre of the circle bounding the surface, while that of a parallelogram is the point where its diagonals intersect each other. |