Question

USE INTEGRATION METHOD

7.81 and 7.82 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment. 250 lb/ft 1250 lb -12 ft - +3f Fig. P7.81

Answer 1

250 lblet 12501b A prima K 12ft À 3fty as pin joints B and c impose two vertical reactions, let them Re and Re To be in static exceilibrium Evertical faces = 0 250 x 12 = 1250 + Potre - PB + R = 1750 i moments about c = 0 1250 X 15 - 250x129(2+3) + Rx3 =0 - f = -2750 [PB = 2750 lb downwards) RE= 1750 + 2750 [Re = 4500 ch

Shear forces at point A SFA = 1250 lb at point B Sff = 1250-250x12 = –1750 lb at point c SFR = -1750-2750 – – 4500 lb 250 WA 1250 1 1250 SED 7-750 max 4500 Кр

let ux take dux at a distance de cross from section left and Mx = 1250 x 250 x DN = 1250 - 250 x 2x - 1250 - 250r q = -250 20 dan (max) He it max where de 20 1250 - 2500 201 2502 = 1250 M = 1250 (5) - 250 (5) 2 x 25 - 3125 max- lb-ft