HW # 5 solution - HW # 5 SOLUTION = 1- Find the rotation at...

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HW # 5 SOLUTION ============== 1 - Find the rotation at A the vertical deflection at D, and the vertical deflection at B 24 ft-k 8 kips 12' 6' B C D 14 6 8’ V M -24 -48 -48 U S t CA θ A D R D’ D” tan θ A = θ A = t CA /12 tCA = (1/2*8*48/EI)(1/3*8+4)+(24/EI*4)*2+(1/2*4*24/EI)*4/3 = 1536/EI θ A = 1536/EI/12 = 128/EI CCW θ B – θ A = (1/2*8*-48/EI) = -192/EI θ B = -192/EI +128/EI = -64/EI CW θ C – θ B = (-24-48)/2*4/EI) = -144/EI θ C = -144/EI-64/EI = -208/EI CW The vertical displacement at B = RS, RS = RU-SU SU = the tangential deviation at B from a tangent at A. = (1/2*8*48/EI)*8/3 = 512/EI RU = θ A *8 = 1024/EI RS = 1024/EI – 512/EI = 512/EI The vertical deflection at D = DD” = DD’ + D’D” DD’ = θ C*6 = -208/EI*6 = -1248/EI D’D” = the tangential deviation at D from a tangent at C. = (1/2*6*48/EI)(2/3*6) = -576/EI DD” = 1248/EI+576/EI = 1824/EI Now check with the mathematical model. CD: Ym = 8x-48 (origin at C) Ys = 1/EI[4x 2 -48x-208] Yd = 1/EI[4/3x 3 -24x 2 -208x] At x = 6 Yd = 1824/EI
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2 - Find the rotation at B the vertical deflection at D, and the vertical deflection at B 24 ft-k 2k/ft 12' 6' A B C D 8' 17 5 12 V -5 24 7.2 M 16 4.8 E 3.2 36 B’ D θ A B tCA D’ B” D” tc tCA = (1/2*4.8*24/EI)(2/3*4.8+7.2)+(1/2*7.2*-36/EI)*2.4 = 288/EI
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This note was uploaded on 03/29/2011 for the course CIVL 201 taught by Professor Schneider during the Spring '08 term at Manhattan College.

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HW # 5 solution - HW # 5 SOLUTION = 1- Find the rotation at...

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