Explanation Given information: The effective length of the column is L e = 20 ft . The allowable yield strength of the steel is σ Y = 36 ksi . The modulus of elasticity of the steel is E = 29 × 10 6 psi . The centric load acting in the column is P = 65 kips . Calculation: The outer dimension of the tube is d o = 6 in . . Find the inner dimension of the tube ( d i ) using the relation. d i = d o − 2 t Substitute 6 in. for d o and 1 2 in . for t . d i = 6 − 2 t Find the cross sectional area ( A ) using the equation. A = d o 2 − d i 2 Substitute 6 in. for d o and ( 6 − 2 t ) for d i . A = 6 2 − ( 6 − 2 t ) 2 = 36 − 36 + 24 t − 4 t 2 = 24 t − 4 t 2 Find the moment of inertia ( I ) of the cross section using the equation. I = d o 4 12 − d i 4 12 Substitute 6 in. for d o and ( 6 − 2 t ) for d i . I = 6 4 12 − ( 6 − 2 t ) 4 12 = 1 12 ( 6 4 − ( 6 − 2 t ) 4 ) Find the minimum radius of gyration ( r ) using the relation. r = I A Substitute 1 12 ( 6 4 − ( 6 − 2 t ) 4 ) for I and 24 t − 4 t 2 for A . r = 1 12 ( 6 4 − ( 6 − 2 t ) 4 ) 24 t − 4 t 2 = 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 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 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 ) as follows; L e r = 20 ft × 12 in . 1 ft 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 = 240 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 (1) Consider 240 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 < L r = 133.68 Find the effective stress ( σ e ) using the equation. σ e = π 2 E ( L e / r y ) 2 Substitute 29 × 10 6 psi for E and 240 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 for L e / r y . σ e = π 2 × 29 × 10 6 psi × 1 ksi 10 3 psi ( 240 6 4 − ( 6 − 2 t ) 4 288 t − 48 t 2 ) 2 = 4

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ISBN: 8220100257063

Author: BEER

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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 82P

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Find the thickness of the lightest tube.

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Chapter 10.3, Problem 81P

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Chapter 10 Solutions

EBK MECHANICS OF MATERIALS

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Chapter 10 Solutions