Question

For the beam (a) (b) using the work-energy method to solve the problem- Compute the slope at A. Compute the slope at B. 3 kips/ft 85 kip. ft B 15 5 15'

Answer 1

z kips/ft A B & 85 kip-ft 151 151 {fy=0; Rat Rc - 3x 15 =0 & Rat Rc = 45 in Eme=0; Rax30 - 3x15% (15 +7.5) +85=0 927.5 30.92 K. Ra= 30 14.08 ki R6 = 45-30.92 Mx = Raxx - wxx² Now; w xx² 30.92x - 3xx2 2 30.92X - 1.5x2 virtual beam: For slope at A & applying unit couple at point A IK ft ERMAPO REX30 - 1-0 Rc = 30 RA = -1 230 for BC Mx IX 30 - 1 x + 1 AB Mx = Raxatl = 30 For slope at B; Ikift 2 Ra= RC = 1280 30 For Bc: I xt1 Mna X + AB mx = 30 30

in Real beam: 24 For portion BC! Mx= - Rca+M doost = -14.08 x + 85 in ytstual eye am: - m Wi = We Now Apply principle energy 1. OA = f Mode 30.92X= 5x2) <tex +1) •dx + f(-14.082 +85) (302).dx Work S 1·DA EI El 30 20 + El 30 30 D 3 24 80 + 30.92 2² _ 1.5x² & OA = EI 1630.92 x² + x3 + 30.922 - 105x²]. du + -14.08x² onal arox.**] → [1159.5 + 632- 81 +3478.5-1687.5 + 637.5-528] 1373. 81 () .30.92 x 90 + DA EI

1.OB em da - EL DB 3 2 1 1 2 (30.92 x - 1.5x²) ( x + 1) dx + 12 (-14-088 +85) (-2+) dz 0 OB= 30 30 30.92 x ² + 1.5 x ² + 30.928 - 1.5x²] dx EI 1715 ЕТ 14.08x² – 85% - 14.08%+85. de. + lo 30 30 + 2 90 .-30.92 x² + 1.5x4 30.92x² 1.5k 30xY 3 + 1 [14.08x² 85x² EI 90 30X2 14.08x² +85x 85x). 2 2- OB=L 30.92(15) 1.5 (15) El 30.92x15 ² + 1.5(15) 90 120 14.08 ( 90 L EI 18 EL -1159.5+ 632.81+ 3478.5-1687.5+ 528-318.75 - 1584 + 1275 1164.56 Aus. EI