(a) Explanation Given information: The allowable yield strength in the column is σ Y = 36 ksi . The effective length of the column is L e = 21 ft . The modulus of elasticity of the material is E = 29 × 10 6 psi . The given rolled-steel shape is W 8 × 31 . Calculation: Refer to Appendix C “Properties of Rolled-Steel Shapes” in the textbook. For W 8 × 31 Rolled-steel shape; The cross sectional area is A = 9.12 in . 2 . The minimum radius of gyration in the y -axis is r y = 2.02 in . . 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 stress is σ Y Substitute 29 × 10 6 psi for E and 36 ksi for σ Y . L r = 4.71 29 × 10 6 36 ksi × 1,000 psi 1 ksi = 133.68 Find the ratio of effective length ( L e ) and the minimum radius of gyration ( r y ) as follows; L e r y = 21 ft × 12 in . 1 ft 2.02 = 124.75 < L r = 133.68 Find the effective stress ( σ e ) using the equation. σ e = π 2 E ( L e / r y ) 2 Here, the modulus of elasticity of the material is E . Substitute 29 × 10 6 psi for E and 124 (b) To determine Find the allowable centric load for the column.

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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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 58P

(a)

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Find the allowable centric load for the column.

(b)

To determine

Find the allowable centric load for the column.

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Chapter 10.3, Problem 57P

Chapter 10.3, Problem 59P

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Find the allowable centric load for the column.

Solution Summary: The author calculates the allowable centric load for the column, based on the modulus of elasticity of the material, and the slenderness ratio using the equation.

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Chapter 10 Solutions