736
ELASTIC 1
beam, when the bending moment M is given, is identical ■with the catenary or funicular curve passing through the ends of the span under a (fictitious) load per unit length of the span equal to M/EI, the horizontal tension in the funicular being unity. (II.) The directions of the tangents to this funicular curve at the ends of the span are the same for all statically equivalent systems of (fictitious) load. When M is known, the magnitude of the resultant shearing stress at any section is dM/dx, where x is measured along the beam. 11. Let l be the length of a span of a loaded beam (Fig. 9), and M2 the bending moments at the ends, M the bending moment at a section distant x from the end (Md, M' the bending moment at l-cc Vl/t A A Fig. 9. the same section when the same span with the same load is simply supported ; then M is given by the formula l-x . nr , M = M' + M1-y ■s +M 27 and thus a fictitious load statically equivalent to M/EI can be easily found when M' has been found. If we draw a curve (Fig. 10) to pass through the ends of the span, so that its ordinate represents the value of M'/EI, the corresponding fictitious loads are statically equivalent to a single load, of amount represented M 2 by the area of the curve, placed at the point of the span vertically above the centre of gravity of this area. If PN is the ordinate of this curve, and if at the ends of the span we erect ordinates in the proper sense to represent Fig. 10. Mj/EI and Ma/EI, the bending moment at 2 any point is represented by the For a uniformly disr length PQ. tributed load the curve ot M is a parabola — x where w is the load per unit3 EI of length statically laced ;atand the the raiddle omt equivalent tl e fictitious load is TVwZ / P P ,G
span • also the loads statically equivalent to the fictitious loads MfiZ-afi/ZEI and are PM/EI and W/EI placed at the points g, g' of trisection of the span. The funicular polygon for
SYSTEMS 12. When there is more than one span the funiculars in question may be drawn for each of the spans, and, if the bending moments at the ends of the extreme spans are known, the intermediate ones can be determined. This determination depends on two considerations : (1) the fictitious loads corresponding to the bending moment at any support are proportional to the lengths of the spans which abut on that support; (2) the sides of. two funiculars that end at any support coincide in direction. Fig. 11 illustrates the method for the case of a uniform beam on three supports A, B, C, the ends A and C being freely supported. There will be an unknown bending moment M0 at B, and the system3 of fictitious loads is AtwAB^EI at G the middle point of AB, ^wBC3/EI at G' the middle point of BO, - pi0AB/EI at g and -pi0BC/EI at g, where g and g' are the points of trisection nearer to B of the spans AB, BC. The centre of gravity of the two latter is a fixed point independent of M0, and the line VK of the figure is the vertical through this point. We draw AD and CE to represent the loads at G and G' in magnitude ; then D and E are fixed points. We construct any triangle UYW whose sides UV, UW pass through D, B, and whose vertices lie on the verticals g'U, YK, g'W ; the point F where VW meets DB is a fixed point, and the lines EF, DK are the two sides (2, 4) of the required funiculars which do not pass through A, B, or 0. The remaining sides (1, 3, 5) can then Fig. 12. be drawn, and the side 3 necessarily passes through B ; for . n the triangle UYW and the triangle whose sides are 2, 3, 4 are in ^The bending moment M0 is represented in the figure by the vertical line BH where H is on the continuation of the side 4, the scale being given by M(B0 . CE ^BC3 ’ this appears from the diagrams of forces. Fig. 12, m which the oblique lines are marked to correspond to the sides of the funiculars to which they are parallel. , ,, In the application of the method to more complicated cases there are two systems of fixed points corresponding to F, by means of which the sides of the funiculars are drawn. 13. Finite Bending of Thin ifod.—The equation curvature=bending moment4-flexural rigidity may also be applied to the problem of the flexure in a principal plane of a very thin rod or wire, for which the curvature need not be small. When the forces that produce the flexure are applied at the ends only, the curve into which the central line is bent is one of a definite family of curves, to which the name elastica has been given, and there is a division of the family into two species according as the external forces are applied directly to the ends or are applied to rigid arms attached to the ends; the curves of the former species are characterized by the presence of inflexions at all the points at which they cut the line of action of the applied forces. Selecting this case for consideration, the problem of determining the form of the curve (cf. Fig. 13) is mathematically identical with the problem of determining the motion ol a simple circular pendulum oscillating through a finite angle, as is seen by comparing the differential equation of the curve EI
S+Wsin^=0 with the equation of motion of the pendulum
Fig. 11. + F. 13 the fictitious loads can thus be drawn, and the direction of the central line at the supports is determined when the bending The length L of th curve between two moments at the supports are known. inflexions corresponds to the time of oscillation of the pendulum The sign of M is shown by the arrow-heads in Fig. 9, for which, from rest to rest, and we thus have LV(W/EI) = 2K, _ . with y downwards, EI^ + M = 0. where K is the real quarter period of elliptic functions of modulus dx* 3 2 M0 is taken to have, as it obviously has, the opposite sense to that The figure is drawn for a case where the bending moment has the shown in Fig. 9. same sign throughout.
