________________________________________________________________________ PROBLEM (7.113) Redo Prob. 7.111 for a C4 x 5.4 channel (refer to Table B . 1 0 ) with M = 10 kip ⋅ in. SOLUTION N.A. y Table B.10: C 4 × 5.4 0.184 in. 4 in. z I z = 3.85 in. 4 I y = 0.319 in. 4 Mz C φ z =0.458 in. My A M z = 10 cos 30 o = 8.66 kip ⋅ in. M y = −10 sin 30 o = −5 kip ⋅ in. 1.584 in. 3.85 (a) tan φ = tan(−30o ), 0.319 φ = −81.8o (b) Point A is farthest from N.A. Thus, M z 8.66 × 103 (−2) −5 ×103 (−1.13) M y σ max = σ A = − z A + y A = − + 3.85 0.319 Iz Iy = 4.5 + 17.71 = 22.21 ksi 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.