________________________________________________________________________ PROBLEM (11.10) Redo Prob. 11.9 for W = 30 kN and α = 40°. SOLUTION FBC ∑F ∑F y 40o FAB x B 30 = 46.67 kN sin 40o = 0 : Pcr = FAB = 46.67 cos 40o = 0 : FBC = = 35.75 kN W=30 kN Pcr = FAB = π 2 EI f s L2 , I= 35.75 ×103 (3)(2.5) 2 = 0.34 ×10−6 m 4 π 2 (200 ×109 ) Therefore, I= π c4 4 = 0.34 × 10−6 , c = 25.6 mm and A = π (25.6)2 = 2066 mm 2 Justification of the formula used: r = I A = 12.8 mm L r = 2500 12.8 = 195 O.K. 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.