________________________________________________________________________ PROBLEM (12.4) An aluminum bar AB is of yield strength 38 ksi and modulus of elasticity E = 10 x 106 psi (see Fig. P12.4). A strain energy of U =150 in.-lb must be stored in the bar when axial load P is applied. Calculate diameter d so that the factor of safety with respect to yielding is ns = 4. SOLUTION U y = 4U = 4(150) = 600 in. ⋅ lb σ y2 (38 × 103 ) 2 = 72.2 in. ⋅ lb in.3 6 2 E 2(10 × 10 ) Uy U y = (U 0 ) y AL or A = (U o ) y L Substituting the given data 600 πd2 A= = , d = 0.383 in. 72.2(6 × 12) 4 (U 0 ) y = = Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.