Question

for the beam shown, draw the shear and moment diagrams. clearly indicate positions where shear us equal to zero and whether each segment is constant, linear, quadratic, or cubic.

85 kN 24 kN/m Hinge A B ob D С 8 m 8 m 3 m

Answer 1

solution ! T 24 kN/m ana 85KN x ता 8m 8м 3m MA=513 8510N Ay 7 64.125 Cy=212.87 64.124 64,125 Đ + 85kn SFD 127.875 s D B. MD KNM 513 1255KNM e parabolic MA -24 kN/m By Dy = 85kw A Bx Ax 8m Bx "By om 3m cy Ay For Right segment: E Fx = 0 -> Bx=0 E fy =0 » By t ly = 24x8 + 85 5277 KN EMB -0 =) 24-48x4 85XL o Cy x 8 cy 1703 8 = 212.875KN - By = 277 - 242-875 = 64.1255 For left Segment! MA By = 641128-KN Axe B 8m Ay EXO - Ax=0 =) Ay = By = 64.125KN E fy zo EMA 20 = MA 8X64.125" so MA = 5 13 KNM.

S.F. at paint A 64.1% KN sofo at point 'B' left- 691129 KN SiF just Right to the point 'B' - decreasing linearly sif at point c (left) (S.Fc = Sif gust Right to a S.F) E 64.125- 24x8 = -127 875 -127.875 +212.875 =85KN BM at A = 513 KNM 1) B 573-64.12548 27 cokNM B.M. at point D=0 KN 8.M at point Gl -(88x3) Right) -255kNm. B.M just left to 'c'- (Bima 85x2 + 212.875(x-3) 24 x(6-3)(2-3) 2 E-85% +212.875 X-638.624 H2(x2+9-6 F-1202? + 199.8 x - 746.621 at point B7 (B.MP4 746.6280 = -12X112 +09+8x11 = 0'825 O at point 'e' (Right) = (Bumib =-108 + 199-883 -786.628 = 255.225 ENM

→ point where shear force zero- 64.125 2 C B 127.875 f 8m 64.125 127.875 8- 2 64.12520 = 127.875 X 8 - 127.875 x = 5.398 CD' hence shear force equal do zero at point which is lies 5.328 m from point'c? In S.FD Constant in the poosion AB and and linearly varying in section Bela - gn B.MiD Linearly varying in the portion AB and CD and parabolic varying in the portion BG