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• Derive the following equation of maximum deflection using double integration method. (for type of loading see slide 7-12). Po MI² 6max = (F-') 120EI 2EI ({1* - 48°) at the center, if a>b 48EI PP MI WL3 SEI 48EI 9/3 EI 384EI MI at the center 16EI wL dmax 764EI WL3 6max 30EI at x = 0.475L 192EI Deflection in Simply Supported Beams Px y = -(31 – x) 6EI PI max 3EI Pa 3a-x) for 0

• Derive the following equation of maximum deflection using double integration method. (for type of loading see slide 7-12). Po MI² 6max = (F-') 120EI 2EI ({1* - 48°) at the center, if a>b 48EI PP MI WL3 SEI 48EI 9/3 EI 384EI MI at the center 16EI wL dmax 764EI WL3 6max 30EI at x = 0.475L 192EI Deflection in Simply Supported Beams Px y = -(31 – x) 6EI PI max 3EI Pa 3a-x) for 0

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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• Derive the following equation of maximum deflection using double integration method. (for type of loading see slide 7-12). M 6max at x= 120EI 2EI at x Pb (31* - 46') at the center, if a >b 48EI PP MI? WL3 at Y= SEI 48EI 9/3 EI 384EI at the center 16EI wL Omax 764EI W L3 8max 30EI 192EI at x = 0.475L Deflection in Simply Supported Beams 1. Px (31 – x) 8̟ 6EI max 3EI Pa Pr y=(3a - x) for 0<x<a 6EI Pa y=- (3x-a) for a<x<l 6EI 3 %3D y=- 24EI SEI 120NET 30EI 5. Mx? M² y: 2EI max 2EI 6. Pi 8. 48EI Pr for 0<r<- 12EI
Transcribed Image Text:• Derive the following equation of maximum deflection using double integration method. (for type of loading see slide 7-12). M 6max at x= 120EI 2EI at x Pb (31* - 46') at the center, if a >b 48EI PP MI? WL3 at Y= SEI 48EI 9/3 EI 384EI at the center 16EI wL Omax 764EI W L3 8max 30EI 192EI at x = 0.475L Deflection in Simply Supported Beams 1. Px (31 – x) 8̟ 6EI max 3EI Pa Pr y=(3a - x) for 0<x<a 6EI Pa y=- (3x-a) for a<x<l 6EI 3 %3D y=- 24EI SEI 120NET 30EI 5. Mx? M² y: 2EI max 2EI 6. Pi 8. 48EI Pr for 0<r<- 12EI
7. 0, 10, Pbx (1² -x²-b') for 0<x<a y = Mlx 1 6EI 6IEI Pb [1 y= 6IEI b %3D for a<x<l MI? at x= 9/3 EI at x= N3 IEI MI? Pb -(31 - 46') at the center, if a >b 48EI at the center 16EI 10. OX y = 24EI (P - 2h² +x') }= 360)ET O = 0.00652 at x= 0.5191 EI %3D max 384EI ô=0.00651" at the center EI 11. wL ômax 1 20EI AT at x = wa363 12. 3EI L3 a b=L 13.Total W = wL WL3 384EI 14. w = N/m Smax 764EI at x = 0.475L 15. W W. W 13 &max 192EI R, 16 W L3 &max R1 384EI
Transcribed Image Text:7. 0, 10, Pbx (1² -x²-b') for 0<x<a y = Mlx 1 6EI 6IEI Pb [1 y= 6IEI b %3D for a<x<l MI? at x= 9/3 EI at x= N3 IEI MI? Pb -(31 - 46') at the center, if a >b 48EI at the center 16EI 10. OX y = 24EI (P - 2h² +x') }= 360)ET O = 0.00652 at x= 0.5191 EI %3D max 384EI ô=0.00651" at the center EI 11. wL ômax 1 20EI AT at x = wa363 12. 3EI L3 a b=L 13.Total W = wL WL3 384EI 14. w = N/m Smax 764EI at x = 0.475L 15. W W. W 13 &max 192EI R, 16 W L3 &max R1 384EI
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