Explanation Given information: The effective length of the column is L e = 5.5 m . The centric dead load is P D = 310 kN . The centric live load is P L = 375 kN . The allowable yield strength of the steel is σ Y = 250 MPa . The modulus of elasticity of the steel is E = 200 GPa . The dead load factor is γ D = 1.2 . The live load factor is γ L = 1.6 . The resistance factor is ϕ = 0.90 . Calculation: Consider W 310 × 60 ; Refer to Appendix C “Properties of Rolled-Steel Shapes” in the textbook. For W 310 × 60 rolled steel shape; The cross sectional area of the section is A = 7 , 550 mm 2 The minimum radius of gyration is r = r y = 49.3 mm . Find the slenderness ratio L r using the equation. L r = 4.71 E σ Y Here, the modulus of elasticity of the material is E and the allowable yield strength is σ Y Substitute 200 GPa for E and 250 MPa for σ Y L r = 4.71 200 GPa × 1,000 MPa 1 GPa 250 = 133.22 Find the ratio of effective length ( L e ) and the minimum radius of gyration ( r ) as follows; L e r = 5.5 m × 1,000 mm 1 m 49.3 = 111.562 < L r = 133.22 Find the effective stress ( σ e ) using the equation. σ e = π 2 E ( L e / r y ) 2 Substitute 200 GPa for E and 111.562 for L e / r y . σ e = π 2 × 200 GPa × 1,000 MPa 1 GPa ( 111.562 ) 2 = 158.598 MPa Find the critical stress ( σ cr ) using the relation. σ cr = [ 0.658 σ Y σ e ] σ Y Substitute 250 MPa for σ Y and 158.598 MPa for σ e . σ cr = [ 0.658 250 158.598 ] 250 = 129.243 MPa Find the ultimate load ( P U ) using the relation. P U = A σ cr Substitute 7 , 550 mm 2 for A and 129.243 MPa for σ cr . P U = 7 , 550 × 129.243 = 975.79 × 10 3 N × 1 kN 1,000 N = 975.79 kN Find the ultimate load using the relation; γ D P D + γ L P L = ϕ P U Here, the dead load factor is γ D , the centric dead load is P D , the live load factor is γ L , the centric live load is P L , and the resistance factor is ϕ . Substitute 1.2 for γ D , 310 kN for P D , 1.6 for γ L , 375 kN for P L , and 0.90 for ϕ

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ISBN: 9780100257061

Author: BEER

Publisher: YUZU

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 10.3, Problem 87P

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Choose the wide-flange shape of W310 mm.

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Chapter 10.3, Problem 86P

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Choose the wide-flange shape of W310 mm.

Solution Summary: The author describes the wide-flange shape of W310 mm in Appendix C "Properties of Rolled-Steel Shapes".

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Chapter 10 Solutions