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1. Beam Diagram 26 by CISC Handbook of Steel Construction. (1) using 4th-order differential equation to obtain the deflection equation y (i.e., Ax) for interval 0

1. Beam Diagram 26 by CISC Handbook of Steel Construction. (1) using 4th-order differential equation to obtain the deflection equation y (i.e., Ax) for interval 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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1. Beam Diagram 26 by CISC Handbook of Steel Construction. (1) using 4th-order differential equation to obtain the deflection equation y (i.e., Ax) for interval 0<x<l. (2) Where is the maximum deflection for Ax? And what is the maximum deflection? 26. BEAM OVERHANGING ONE SUPPORT - CONCENTRATED LOAD AT END OF OVERHANG R₁ = V₁ R₂ = V₁ + V₂ V₂. M max. (at R2₂) M, (between supports) M, (for overhang) V R₁ Shear Moment AR, R₂ V₂ M max A max. between supports at x = A max. (for overhang at x₁ = a) A, (between supports) Ax, (for overhang). Pa 1 √3) .... = P (1 + a) Pa Pax 1 = P(a - x₁) = Pal2 9/3 EI Pa² 3 Е! PX <= .06415 (1 + a) Pax 6 Ell (1²-x2) P 6 El Pal2 EI (2al + 3ax, - x₁²)
Transcribed Image Text:1. Beam Diagram 26 by CISC Handbook of Steel Construction. (1) using 4th-order differential equation to obtain the deflection equation y (i.e., Ax) for interval 0<x<l. (2) Where is the maximum deflection for Ax? And what is the maximum deflection? 26. BEAM OVERHANGING ONE SUPPORT - CONCENTRATED LOAD AT END OF OVERHANG R₁ = V₁ R₂ = V₁ + V₂ V₂. M max. (at R2₂) M, (between supports) M, (for overhang) V R₁ Shear Moment AR, R₂ V₂ M max A max. between supports at x = A max. (for overhang at x₁ = a) A, (between supports) Ax, (for overhang). Pa 1 √3) .... = P (1 + a) Pa Pax 1 = P(a - x₁) = Pal2 9/3 EI Pa² 3 Е! PX <= .06415 (1 + a) Pax 6 Ell (1²-x2) P 6 El Pal2 EI (2al + 3ax, - x₁²)
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