Question

Q.3. [20 pts) The figure below shows the shear force diagram for a loaded beam. a-) Sketch the beam, showing the loading conditions. b-) Draw the bending moment diagram showing the critical values. V (kN) *** 1.6 2.4 36 16 20 -x (m) 1-24 40

Answer 1

Ang shear force from given fig. If we cecwate the area bounded by diagram and compared colth top and bottom Ir. diffrence is zero they no moment (peter acting). Arsea of bottom of shear force diagram: - + x40x2 + x 24 82.4 & 68.8 - noul - zero = 0 Area of top of shear force diagram 36*+4x1641-6 + ļx20x2 =68.8 Diffrence of Top and bottom Hence no external moment act. If we call audte the slope of Indlined SFD then we get the UDL the URL loading intensity. w = 140-0) 20kN/m (Acting downward) Reaction at A - 40+ 36 = 76 KN Point load at B = 36-16 - 20 KN intensity 16 + 24 4 intensity of wz 16kN/m Scanned by TapScanner 2 of w= 40 = lokN/m 20-0 2

Reaction at a - 24+20 44 KN As, loading diagram 20 KN o 20 kN/m W2 = 10 km WO3710 kN/m E 2m th 4m kan RA=76 ai RC=44 loading dregsem? (6) Bending moment at A = w,a? 2 = 20*4 = 40 KNM 2 Bending moment arc = za 2² 2 10x4 – 20 kNm 2 Bending moment et Bending moment at E=0 Bending moment at B = - 20x2x1 +76 xq = ~ 4 KNM Scanned by TapScanner

moment at zero shear= -20x2*3.6 +7682.6 20 X16 - 10X16X16 2 = + 8.8KNM BMD as 88 inim A B C (20vNๆ (4KNM 20 kNm 40kNm zero shear Point BMD Scanned by TapScanner