Question

3) For the beam shown, find the reactions at the supports and plot the shear-force and bending moment diagrams using singularity functions. Label the diagrams properly and provide values at all key points on the diagram (A, B, C, D, E, and F). Lengths: AB = 3 in, AC = 8 in, AD - 10 in, AE-14 in, and AF-17 in. Use singularity 5lbf 30 lbf in 240 lbf.in functions to find R1 and R2 D

Answer 1

iven 5lbt 240 ebrin wa300belin con ant R2 wasoobelin AB=Bin AC= 8 in AD=Loin AE=14n Af=17in, Consider a cection xx at a distance & from point A & in between EGA eingularity function xta Bending moment at x B.M) x = R1 (2-0) -5(2-3) - 240 (SC-8) + R2 (x-10) - zoko1o) (x-10) + 304-14) ( x4) (B-MAX = R, (x-o) = 5(2-3) - 240 (x-80° +R2 (X-40) - 30 (X4032 +30 (x-14) We know that Bending moment at Alexao i o. at fi.cx=1719-o. B. Maczo to (B. 19) C=1750 (BMD C=1H = R1C17) - 5012-3) - 240 (1) + R2 (12-10) - 30 (17-103² +30 (17-14)²

17 R, -070 - 240 + AR 735 of 135-o 17847R2=9100 be Shear force at & (s.f) ne = R, (x-03 - 5 (x-330 4 R₂(x-103° - 30 (x-los +30 (4-14) ale know that Shear force at salti.e at point &zo (SF) = 17 = R1 (1) - 5(1) + R₂ (1) 30 (17-10) +30 (17-14) so RitR2-5-30 (7) + 30(3) =0 Rit R₂ = 125 lbs / - from equations ① & ② 6- 7x② 17R, +7 R₂ = 910 - (FR1 +7R2 = 875) 10 R, to =35 lbs R = 3.5lbs | R₂ = 121.5 lbs

(s.f) at A (r=0) = 3.5 lbs at B Caight) = 1.5lbs (Sf) at a = -1.5 lbs (Sf) at D caight) = -1.54 121.5=120lbs (Sf) at E = 1200 - (30x40 20 (S.f) f = 0. (OM) = R(x-0)-5(2-3) - 240 (ac-8) + R2(0-10) (B.M) A (x=0) = 0 (B. M) 3 (x-3) = 10.5 lb in (B.Me (x=81 (left = 3 lbin (BM) C (x=8) (right) = -237 lbin B.M) b = - 240 lb in (BM) E (x=14] =0 (BM) f (x = 17) =0

Sher force Draghram 1 2016 3.5ebs - 1.5lbs Bending Moment Diaghram 10.5 lb in 3lbin - 237 lbin -zuolbin AB=3 in A casin AD = loin AE=14in AF=17in.