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predict the maximum stress and deflection in the simply-supported aluminum beam shown below, with h = 1 in, b = 10 in, L = 4 ft, and P = 5 psi.

predict the maximum stress and deflection in the simply-supported aluminum beam shown below, with h = 1 in, b = 10 in, L = 4 ft, and P = 5 psi.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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predict the maximum stress and deflection in the simply-supported aluminum beam shown below, with h = 1 in, b = 10 in, L = 4 ft, and P = 5 psi.
wrong in the process for you to accept them at face value! Using the beam tables at the back of your book, predict the maximum stress and deflection in the simply-supported aluminum beam shown below, with h = 1 in, b = 10 in, L = 4 ft, and P = 5 psi. Record these in your data table. P h b
Transcribed Image Text:wrong in the process for you to accept them at face value! Using the beam tables at the back of your book, predict the maximum stress and deflection in the simply-supported aluminum beam shown below, with h = 1 in, b = 10 in, L = 4 ft, and P = 5 psi. Record these in your data table. P h b
SIMPLY SUPPORTED BEAMS Beam Slope Deflection Elastic Curve 3 Px -(3Ľ² – 4x²) 48EI PL PL V = 6 =-0 : V. max 16EI 48EI for 0<x<½ mas Pb(L – b³) 4 6. Pa’b? Pbx -(Ľ² -b² – x²) 6LEI 6LEI V=- V=- 3LEI Pa(L - a² ) 0, =+ for 0<x<a at x= a 6LEI ML 8 ML 9. 9/3 EI max 3EI Мx V=- 6LEI -(2Ľ² - 3 Lx + x³) ML 0, = + 6EI at x = Lj 1- L. 10 11 12 wĽ 5wL* Wx 0 = -0, -(Ľ’ – 2Lx² +x³) 24EI = - v=- max 24EI 384EI max 13 14 wa? (2L-a) 24LEI -(Lr - 4aLx +2a°xr' + 4a*L 24LEI wa" (4L² -7al +3a²) 24LEI -4a'L+a*) for 0Sxsa V=-. wa? -(2L² -a²) y=-. 24LEI (2x² - 6Lx² +a²x+ 4L²X-a°L) at x =a 0, =+ 24LEI 7- 15 for a<xsL 16 17 18 7w,Ľ w,L W. 360EI V. =-0.00652 Wox %3D -(7L* – 10LX² +3x*) max EI V=- 360LEI 0, =+ 45EI at x = 0.5193L 7-
Transcribed Image Text:SIMPLY SUPPORTED BEAMS Beam Slope Deflection Elastic Curve 3 Px -(3Ľ² – 4x²) 48EI PL PL V = 6 =-0 : V. max 16EI 48EI for 0<x<½ mas Pb(L – b³) 4 6. Pa’b? Pbx -(Ľ² -b² – x²) 6LEI 6LEI V=- V=- 3LEI Pa(L - a² ) 0, =+ for 0<x<a at x= a 6LEI ML 8 ML 9. 9/3 EI max 3EI Мx V=- 6LEI -(2Ľ² - 3 Lx + x³) ML 0, = + 6EI at x = Lj 1- L. 10 11 12 wĽ 5wL* Wx 0 = -0, -(Ľ’ – 2Lx² +x³) 24EI = - v=- max 24EI 384EI max 13 14 wa? (2L-a) 24LEI -(Lr - 4aLx +2a°xr' + 4a*L 24LEI wa" (4L² -7al +3a²) 24LEI -4a'L+a*) for 0Sxsa V=-. wa? -(2L² -a²) y=-. 24LEI (2x² - 6Lx² +a²x+ 4L²X-a°L) at x =a 0, =+ 24LEI 7- 15 for a<xsL 16 17 18 7w,Ľ w,L W. 360EI V. =-0.00652 Wox %3D -(7L* – 10LX² +3x*) max EI V=- 360LEI 0, =+ 45EI at x = 0.5193L 7-
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