Explanation Given information: The point load act at center of the beam ( P ) is 20 kN . The moment at both ends of the beam ( M ) is 10 kN ⋅ m . The modulus of elasticity of the beam ( E ) is 200 GPa The moment of inertia of the beam ( I ) is 42.8 × 10 − 6 m 4 . The length of the beam ( L ) is 6 m . Calculation: Calculate the support reactions A and B ( R A and R B ) : Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 R A + R B − 20 = 0 R A + R B = 20 (1). Calculate the moment about point A by applying the equation of equilibrium: Sum of moments about point A is equal to 0. ∑ M A = 0 ( R B × 6 ) − ( 20 × 3 ) − 10 + 10 = 0 6 R B − 60 = 0 6 R B = 60 R B = 60 6 R B = 10 kN Substitute 10 kN for R B in Equation (1). R A + 10 = 20 R A = 20 − 10 R A = 10 kN Calculate the bending moment about point A : M A = − ( 20 × 3 ) + ( 10 × 6 ) + 10 M A = 10 kN ⋅ m Calculate the bending moment about point B : M B = ( 10 × 0 ) + 10 M B = 10 kN ⋅ m Calculate the bending moment about point C : M C = ( 10 × 3 ) + 10 M C = 40 kN ⋅ m The section is symmetry

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 12.4, Problem 12.11FP

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The maximum deflection of the simply supported beam (υmax).

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Chapter 12.4, Problem 12.10FP

Chapter 12.4, Problem 12.12FP

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The maximum deflection of the simply supported beam ( υ max ) .

Solution Summary: The author calculates the maximum deflection of the simply supported beam (upsilon_mathrmmax).

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The maximum deflection of the simply supported beam (υmax).

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Chapter 12 Solutions