_______________________________________________________________________ PROBLEM (8.114) Using Mohr's circle, redo Prob. 8.109. SOLUTION ε'= −300 + 420 = 60 μ 2 γ 2 (μ) ε' α o x 20 (-300, -100) x’ α = tan −1 y’ y O C R ε (μ) 10 = 15.5o 36 R = 3602 + 1002 = 373.6 μ γ x ' y ' = 2[373.6 sin( 20o + 15.5o )] = 434 μ ε x ' = −373.6 cos 35.5o + 60 = −304 + 60 = −244 μ ε y ' = 304 + 60 = 364 μ 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.