________________________________________________________________________ PROBLEM (11.70) A square fixed-pinned end aluminum alloy 6061-T6 column of width a supports a compression load P. Determine the permissible length of the bar. Given: a = 2.2 in., P = 40 kips SOLUTION A = a2 I= a4 12 r= 2.2 I a = = = 0.6351 in. A 2 3 2 3 P 40(103 ) = = 8.26 ksi A (2.2) 2 Assume L r ≥ 66 and use Eq. (11.29) σ all = σ all = 51000(103 ) = 8.26(103 ) 2 (L r) Solving L = 78.58 > 66 r (O.K.) Hence L = 78.57(0.6351) = 49.9 in. ________________________________________________________________________ PROBLEM (11.71) A solid round aluminum strut (alloy 6061-T6) of length L and pinned ends supports an axial load P. Calculate the minimum permissible diameter d of the member. Given: L = 400 mm, P = 60 kN SOLUTION A= πd2 4 I= πd4 64 r= I d = A 4 L ≤ 66 and use Eq.(11.28). r 0.6 60(103 ) σ all = [140 − 0.87( )] = d 4 πd2 4 2 0.0764 , 140d 2 − 1.74d − 0.0764 = 0 140 − (0.87) = 2 d d Assume 9.5 < or 1.74 ± (1.74) 2 + 4(140)(0.0764) d= = 0.0304 m = 30.4 mm 2(140) Since d 30.4 = = 7.6 mm 4 4 Assumption is O.K. r= L 400 = = 52.6 < 66 r 7.6 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.