_______________________________________________________________________ PROBLEM (9.5) Repeat Prob. 9.4 for the case in which the 50-mm-diameter hole in the shaft is eliminated. SOLUTION Critical section is in AB. σx = τ= 4P 2 π d AB σx = 4P 1600 P = 2 π (0.05) π τ =− 16(0.01P) 1280 P =− 3 π (0.05) π 16T 3 π d AB Equation (9.1): 800 P 800 P 2 −1280 P 2 ) σ1 = + ( ) +( π = π π 800 P 1509 P 2309 P + = π π π We have 100 × 106 = 60 × 106 = 2309 P π 1509 P π , , P = 136 kN P = 125 kN Thus, Pall = 125 kN 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.