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Px (3L – x) PL 3EI PL? 2EI 6EI

Px (3L – x) PL 3EI PL? 2EI 6EI

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 at least 5 cases from the 12 cases of beam loadings that is shown. Derive the deflection at different points of the beam, maximum deflection, and slopes as shown in the table. L -(3L² – 4x²) 0 <x< PL3 Smax = 48EI PL? 04 = Og = 16EI Px Px2 (3L - 6EI PL3 OB = 3EI PL² OB = 2EI A x) 48EI L Pb(L² – b²)³/² 9/3LEI L² – b² Pbx -(L² – x – b²) 6LEI Smax Px2 (За — х) 0<x<a 6 ΕΙ 0<xsa Pab (2L a) 6L EI Ра? (3L – a) 6EI Pa? OB 2EI 8 = Pb (x – a)³ + (L² – b²)x –x at x = Pab (2L – b) 6L EI 3 Pa2 (3х — а) а <х<L 6EI OB = a 6LEI Pb (3L² – 4b²) if a >b a<x<L Scœnter = 48EI Wo Wo wox? (6L² – 4Lx+ x²) 24EI woL3 OB = 6ΕΙ B 5woL4 384EI A Wox dB = 8EI 24ET (L – 2Lx²+x³) Smax O4 = OB = 24EI Wo Wo Wo wox? (6a² – 4ax+x²) 0<xsa 24EI [a²(a – 2L)² +2a(a – 2L)x² + Lx³] dænter = 24L EI (5L* – 12L?b² + 86*) Woa? (a – 2L)² 384 EI A 24L EI Woa3 6ΕΙ 0<xsa A woa3 8 = . if a 2 b 8 = OB = (4L — а)| 0в Woa? (-L+x)(a² – 4Lx+2x²) Woa? -(2L² – a²) OB = 24L EI 24EI woa* (4х — а) 24EI Woa? (3L² – 2a²) if asb 96EI a <x<L 24L EI dcenter =: L a<x<L Wo wo Smax = 0.006 522- EI 7woL3 Wox 360EI Wox? : (10L³ – 10L²X+ 5Lx² – x³) | dB = (7L* – 10L²×² + 3x*) B A A at x = 0.5193L OB = 24 EI 360L EI 120L EI 30EI Sœnter = 0.006 510- EI OB = 45 EI Mo M,L² 8 max 9/3EI M,L Mox? 8 = M,L² SB = 2EI MọL OB at x= V3 %3D A B Mox 6EI A (L² – x²) 6L EI 2EI EI M,L? 16EI MọL OB = 3EI Mo dæenter =
Transcribed Image Text:Derive at least 5 cases from the 12 cases of beam loadings that is shown. Derive the deflection at different points of the beam, maximum deflection, and slopes as shown in the table. L -(3L² – 4x²) 0 <x< PL3 Smax = 48EI PL? 04 = Og = 16EI Px Px2 (3L - 6EI PL3 OB = 3EI PL² OB = 2EI A x) 48EI L Pb(L² – b²)³/² 9/3LEI L² – b² Pbx -(L² – x – b²) 6LEI Smax Px2 (За — х) 0<x<a 6 ΕΙ 0<xsa Pab (2L a) 6L EI Ра? (3L – a) 6EI Pa? OB 2EI 8 = Pb (x – a)³ + (L² – b²)x –x at x = Pab (2L – b) 6L EI 3 Pa2 (3х — а) а <х<L 6EI OB = a 6LEI Pb (3L² – 4b²) if a >b a<x<L Scœnter = 48EI Wo Wo wox? (6L² – 4Lx+ x²) 24EI woL3 OB = 6ΕΙ B 5woL4 384EI A Wox dB = 8EI 24ET (L – 2Lx²+x³) Smax O4 = OB = 24EI Wo Wo Wo wox? (6a² – 4ax+x²) 0<xsa 24EI [a²(a – 2L)² +2a(a – 2L)x² + Lx³] dænter = 24L EI (5L* – 12L?b² + 86*) Woa? (a – 2L)² 384 EI A 24L EI Woa3 6ΕΙ 0<xsa A woa3 8 = . if a 2 b 8 = OB = (4L — а)| 0в Woa? (-L+x)(a² – 4Lx+2x²) Woa? -(2L² – a²) OB = 24L EI 24EI woa* (4х — а) 24EI Woa? (3L² – 2a²) if asb 96EI a <x<L 24L EI dcenter =: L a<x<L Wo wo Smax = 0.006 522- EI 7woL3 Wox 360EI Wox? : (10L³ – 10L²X+ 5Lx² – x³) | dB = (7L* – 10L²×² + 3x*) B A A at x = 0.5193L OB = 24 EI 360L EI 120L EI 30EI Sœnter = 0.006 510- EI OB = 45 EI Mo M,L² 8 max 9/3EI M,L Mox? 8 = M,L² SB = 2EI MọL OB at x= V3 %3D A B Mox 6EI A (L² – x²) 6L EI 2EI EI M,L? 16EI MọL OB = 3EI Mo dæenter =
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