When M is known, the magnitude of the resultant shearing stress at any section is dM/dx, where x is measured along the beam.

Fig. 20.

58. Let l be the length of a span of a loaded beam (fig. 19), M1 and M2 the bending moments at the ends, M the bending moment at a section distant x from the end (M1), M′ the bending moment at the same section when the same span with the same load is simply supported; then M is given by the formula

M = M′ + M1 l − x+ M2 x,
l l
Fig. 19.

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. 20) 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 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 M1/EI and M2/EI, the bending moment at any point is represented by the length PQ.[2] For a uniformly distributed load the curve of M’ is a parabola M′ = ½wx (l − x), where w is the load per unit of length; and the statically equivalent fictitious load is 1⁄12wl³ / EI placed at the middle point G of the span; also the loads statically equivalent to the fictitious loads M1 (l − x) / lEI and M2x / lEI are ½M1l / EI and ½M2l / EI placed at the points g, g′ of trisection of the span. The funicular polygon for the fictitious loads can thus be drawn, and the direction of the central-line at the supports is determined when the bending moments at the supports are known.

Fig. 21.
Fig. 22.
Fig. 23.

59. 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. 21 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 system[3] of fictitious loads is 1⁄12wAB³/EI at G the middle point of AB, 1⁄12wBC³ / EI at G′ the middle point of BC, −½M0AB / EI at g and −½M0BC / 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 UVW whose sides UV, UW pass through D, B, and whose vertices lie on the verticals gU, VK, 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 C. The remaining sides (1, 3, 5) can then be drawn, and the side 3 necessarily passes through B; for the triangle UVW and the triangle whose sides are 2, 3, 4 are in perspective.