________________________________________________________________________ PROBLEM (*11.79) An aluminum alloy 6061-T6 compression member AB is fixed at its base and free at its top as shown in Fig. P11.79. Using the interaction formula with (σall ) b = 140 MPa, Calculate the largest permissible eccentric load P. *SOLUTION We have Le = 2 L (from Fig. 11.3a). A = bh = 50(100) = 5000 mm 2 The smallest moment of inertia is 1 1 I = bh3 = (100)(50)3 = 1.042 × 106 mm 4 12 12 1.042 × 103 I r= = = 14.44 5 A The largest slenderness ratio is thus Le 2(1800) = = 249.3 r 14.44 Assume Eq.(11.29) applies (249.3 > 66). 350 ×103 350(109 ) 6 (σ all )c = (10 ) = = 5.63 MPa ( Le r ) 2 (249.3)3 Also P P = = 200 P A 5 × 10−3 Mc Pec P (0.025)(0.05) = 3 = = 300 P I bh 12 (0.05)(0.1)3 12 Equation (11.34) becomes M I P A + c ≤ 1; (σ all )c (σ all )b or P = 26.5 kN 200 P 300 P + ≤1 6 5.63 ×10 140 × 106 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.