(a) Explanation The length of the steel tube is L = 5 m . The allowable yield stress is σ Y = 250 MPa . The modulus of elasticity of the steel tube is E = 200 GPa . The factor of safety with respect to the permanent deformation is 2.6. The eccentricity of the load is e = 15 mm . The cross sectional area of the cross section is A = 3 , 400 mm 2 . The moment of inertia of the cross section is I = 7.93 × 10 − 6 m 4 . The minimum radius of gyration is r = 48.3 mm . The effective length of the column ( L e ) is equal to the length of the column ( L ). L e = L = 5 m Find the ratio of the effective length to the minimum radius of gyration. L e r = 5 m × 1,000 mm 1 m 48.3 = 103.52 Find the distance between the neutral axis to the extreme fibre ( c ) using the relation. c = a 2 Here, the outer size of the tube is a . Substitute 127 mm for a . c = 127 2 = 63.5 mm Find the value of e c r 2 ; e c r 2 = 15 × 63.5 48.3 2 = 0 (b) To determine Find the allowable load P when the eccentricity is 7.5 mm .

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 10.2, Problem 44P

(a)

To determine

Find the allowable load P when the eccentricity is 15 mm.

(b)

To determine

Find the allowable load P when the eccentricity is 7.5 mm.

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Chapter 10.2, Problem 43P

Chapter 10.2, Problem 45P

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Chapter 10 Solutions

Mechanics of Materials, 7th Edition

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Find the allowable load P when the eccentricity is 15 mm .

Solution Summary: The author calculates the allowable load P when the eccentricity is 15mm and the modulus of elasticity of the steel tube.