..-15948072 = (-482335366120+205092 - 15552) + Elc, 15289157 EI I = ΕΙ —15289157 — 482335x — 10170x2 + 158.25x3 − {x}. Greatest deflection occurs in the middle portion, and the point is given by the equation. EXAMPLES OF CONTINUOUS GIRDERS. 1o. Let / uniform moment of inertia of girder. w = load per unit of length uniformly distributed. Find expressions for 1, the bending-moment over each support, 2, the supporting-forces, 3, the greatest bending-moment, 4, the slopes at the supports, 5, the greatest deflection, in each of the following cases :(a) Two equal spans, length (b) Three equal spans, length 7. (c) Four equal spans, length 7. (d) Two spans, lengths /, and /, respectively. (e) Three spans, lengths 7, 4, and /, respectively. 3 (ƒ) Four spans, lengths 1, 2, 3, and 1, respectively. (g) Two equal spans ; loads per unit of length on each span, w, and w, respectively. (h) Three equal spans ; loads per unit of length on each span, w1, w, and w, respectively. 2o. Do the same in the case where each span is loaded with a centre load W, and has no distributed load. 3o. Find greatest bending-moment and greatest deflection for a continuous girder of two spans, uniformly loaded on these two spans with load w per unit of length, and which overhang the outer supports; the overhanging parts having lengths . and respectively, and the same distributed load per unit of length on the overhanging parts. CHAPTER IX. equilibrium curves.—arches and domes. § 248. Loaded Chain or Cord. It has been already shown (§ 126), when the form of a polygonal frame is given, that the loads must be adapted, in direction and magnitude, to that form, or else the frame will not be stable. The same is true of a loaded chain or cord, which would be realized if the frame were inverted. If a set of loads be applied which are not consistent with the equilibrium of the frame under that form, it will change its shape until it assumes a form which is in equilibrium under the applied loads. As to the manner of finding (when a sufficient number of conditions are given) the stresses in the different members, etc., this was sufficiently explained under the head of "Frames," and will not be repeated here, as the figures speak for themselves. In Fig. 257 the polygon fedcbaf (a) F 5 E 3 B с FIG. 257. is the force polygon, while the equilibrium polygon is 123456, an open polygon. A straight line joining e and a would represent the resultant of the loads. (a) CHAIN WITH VERTICAL LOADS. (b) If all the loads are vertical, the broken line edcba becomes a straight line and vertical, as shown in Fig. 2586. Whenever the loads are concentrated at single points, as 2, 3, 4, 5, the form of the chain is polygonal; and when the load is distributed, it becomes a curve, as shown in Fig. 259. F 5 FIG. 258. A CURVED CHAIN WITH A VERTICAL DISTRIBUTED LOAD. Given the form of the chain AOE supported at A and E, is balanced by the two supporting-forces at A and E respectively, as shown in the figure. Hence draw ca parallel to the tangent at E, and ba parallel to that at A, and we have the force polygon abca; the equilibrium curve being the chain AOE itself. Moreover, if the lowest point of the chain be O, then the load must be so distributed that the portion between 0 and A shall be balanced by the tension at O and that at A, and hence that its resultant shall pass through the intersection of the tangents at O and A. Its amount will be found by drawing from a a horizontal line; and then we shall have ao as the tension at o, ab as the tension at A, and bo as the load between A and O. Hence the load between E and O will be oc. |