Explanation Given information: The effective length of the column is L e = 4.5 m . The allowable yield strength of the steel is σ Y = 345 MPa . The modulus of elasticity of the steel is E = 200 GPa . The centric load acting in the column is P = 900 kN . Calculation: Consider the section W 250 × 67 _ . Refer to Appendix C “Properties of Rolled-Steel Shapes” in the textbook. For W 250 × 67 _ rolled-steel shape; The cross sectional area of the section is A = 8 , 580 mm 2 . The minimum moment of inertia of the section is I = I x = 22.2 × 10 6 mm 4 . The minimum radius of gyration is r = 51.1 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 345 MPa for σ Y L r = 4.71 200 GPa × 10 3 MPa 1 GPa 345 = 113.40 Find the ratio of effective length ( L e ) and the minimum radius of gyration ( r ) as follows; L e r = 4.5 m × 1,000 mm 1 m 51.1 = 88.063 < L r = 113.40 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 88.063 for L e / r y . σ e = π 2 × 200 GPa × 1,000 MPa 1 GPa 88

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ISBN: 9780100257061

Author: BEER

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Buckling of columns

The column is described as a vertical member of the structure that carries an axial compressive load, and buckling is a lateral deformation of the column under the application of load.

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Chapter 10, Problem 126RP

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Choose the wide-flange shape of W250.

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Chapter 10, Problem 125RP

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Chapter 10 Solutions

EBK MECHANICS OF MATERIALS

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Choose the wide-flange shape of W250.

Solution Summary: The author calculates the wide-flange shape of W250 using Appendix C "Properties of Rolled-Steel Shapes".

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Chapter 10 Solutions