________________________________________________________________________ PROBLEM (11.41) A structural steel column with pinned ends has a cross-sectional area of width b = 3 in. by depth h = 4 in. Determine the shortest length L min that the column may have using Euler’s formula. Given: σy = 36 ksi, E = 29 x 10 6 psi (from Table B.4) SOLUTION The smallest moment of inertia is 1 1 I min = hb3 = (4)(3)3 = 9 in.4 12 12 and I 9 rmin = min = = 0.866 in. A 3× 4 Euler’s formula (11.9) with σ y = σ cr : Lmin 2 π 2 E π 2 (29 × 103 ) Lmin ) − = = 7,905.519 , = 89.2 36 σ cr r r Hence, Lmin = (89.2)(0.866) = 77.2 in. Comment: If this column is 77.2 or more in length, it will be buckle. ( 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.