48 Mechanics of Materials §3.5 10 kN/m 20 kN 20kN O kN 40kN C 0 -2m* -3m -2m* 2m 42.5 40 22.5 2.5 S.F diagram (kN) -7.5 -67.5 72.5 87.5 37.565 45 -2.5 B.M.diagram (kN m) Fig.3.9. For complete accuracy one or two intermediate values should be obtained along each u.d.I. portion of the beam, e.g. B.M.midway between A and B=(42.5×1)-(l0×1×) =42.5-5=37.5kNm Similarly,B.M.midway between C and D=45kNm B.M.midway between D and E=-39kNm The B.M.and S.F.diagrams are then as shown in Fig.3.9. 3.5.Points of contraflexure A point of contraflexure is a point where the curvature of the beam changes sign.It is sometimes referred to as a point of infexion and will be shown later to occur at the point,or points,on the beam where the B.M.is zero. For the beam of Fig.3.9,therefore,it is evident from the B.M.diagram that this point lies somewhere between C and D(B.M.at C is positive,B.M.at D is negative).If the required point is a distance x from C then at that point B.M.=(42.55+)-(10×24+x)-203+x)-20x- 20x2 =212.5+42.5x-80-20x-60-20x-20x-10x2 =72.5-17.5x-10x2 Thus the B.M.is zero where 0=72.5-17.5x-10x2 i.e.where x=1.96or-3.748 Mechanics of Materials $3.5 20kN /IOkN 40 kN IO kN/m 2o kN 42 5 S F diagram (kN) .I B M diagram (kN m) Fig. 3.9. For complete accuracy one or two intermediate values should be obtained along each u.d.1. portion of the beam, e.g. B.M. midway between A and B = (42.5 x 1) - (10 x 1 x $) = 42.5 - 5 = 37.5 kNm Similarly, B.M. midway between C and D = 45 kN m B.M. midway between D and E = - 39 kN m The B.M. and S.F. diagrams are then as shown in Fig. 3.9. 3.5. Points of contraflexure A point of contraflexure is a point where the curvature of the beam changes sign. It is sometimes referred to as a point ofinflexion and will be shown later to occur at the point, or points, on the beam where the B.M. is zero. For the beam of Fig. 3.9, therefore, it is evident from the B.M. diagram that this point lies somewhere between C and D (B.M. at C is positive, B.M. at D is negative). If the required point is a distance x from C then at that point 20x2 B.M. = (42.5)(5+~)-(10 x 2)(4+~)-20(3+~)-20~-- 2 = 212.5 + 42.5~ - 80 - 20~ - 60 - 20~ - 20~ - lox2 = 72.5 - 17.5~ - lox2 Thus the B.M. is zero where i.e. where 0 = 72.5 - 17.5~ - lox2 x = 1.96 or -3.7