Question

shesr and moment diagram

SHEAR AND MOMENT DIAGRAMS Determine the shear and moment diagrams for the beam shown below. There is a pin at A and a rocker at B. a. There are 3 sections. Draw the FBD for each section. Also, give the shear and moment equation for each section. у 125 lb/ft 1000 lb A 10 ft 6 ft 10 ft B

Answer 1

JOft 1000 lb 10 Ft * 6 ft B ANO А let Rax & hay be the reaction forces at A acting on the beam 16 Rb be the reaction at B on the beam It's focc body diagram 1000 lb 125 lb) ft Rai oft 1 Ray Rb Foo the beam to be in equilibrium (0) EFFO, (i) 5 =0, x = % 21 = 0 (moment about any point is "o" EF = 0 Rax=0 horizontal force acting on beam) (It is only Raz = 0

Ray+ Rb-1000-125X10 =0 Ray + Ro = 2250 lb ju Taking moment about A to be 2000 >> EMA=0 1- (125x10) (16+ 10 ) + Rp -TODD X ID - + RX 26=0 26 Rb = 10,000+ 1250 x 21 Rb = 1394.230 Lb = 2250 Ray = 2250-RB Ray + Ro Ray = 855.77lb .. Reaction at A, Ra = Ray = 855.77 lb & Rp = 1394,23 lb 125 kb) Ft 1000 lb 6 ft 10 Ft fB A D с soft Ra R = 1394.23 lb 855.771b - 1 In Section A-C 1000 lb A (S.F) = tra 855.77 lb С Ba = 355,11 x (8-11). - Ra* 1 = 855.771 ( lb-ft) 1 In section (-D loft 1000 lb 1 1 (SF), - +Ra-1000 = 855.77-1000 = -144.23 lb D Ra (B,M) = R XI - 1000 (1-10) 855 1-1000(1-10) = -144.23 2 +10,000 (06-H

In section DB 10Ft 1000 lb 6011 ok Ra - 10 0 0 - 125 I-16) (5.7) + + -1000 4125(3-6)) 6-125(1-16) D 1 f Ra Х = -144.23 - 11 (B-M), = +Ra* I-1000(1-10) - 125(2-16) * ( 176 ) • 355.171 -100(1-10) - 125 (1-16) S.F 11 In section A-6, S.F = 855.77 lb CD, S.F = -144 23 lb D-B, SF = -144.23-125[L-16) B.M In section A-C, B.M = 855.771 (B.Me = 855.77 xD = 8557.7 lb ft C-D, B-M = 855.111-1000(2-10) , (B-M), 1.1, 1 = 16 15 3692.32 D-6, B M = 855.111-1000(1-10) - 125 (2-16) (8•M)2 = is IDoolb 125 lb) ft C D > JOEL CE 10 Ft - - 855.77 + S.F in lb 1 -144.23 - 1394.23 Parabola 8557.1 3692.32 + B.M in lb-ft + + 0 -
I hope you understand. Comment if you have any doubt.