_______________________________________________________________________ PROBLEM (9.29) A steel shaft. d = 2 in. in diameter and rotating at n = 2400 rpm is acted upon by a bending moment of M = 5 kip-in. Calculate the torque and the horsepower that can also be applied simultaneously to the shaft. Given: τ all = 9 ksi, σ all = 14 ksi SOLUTION Mc 32 M 32(5 × 103 ) σx = = = = 6.37 ksi πd3 π (2)3 I τ (ksi) (637, τ ) R C 6.37 ksi O τ= 3.185 Tc J σ1 σ (ksi) 9 For Mohr's circle: τ = 9 2 − 3.1852 = 8.42 ksi σ 1 = 3.185 + 9 = 12.185 ksi < 14 ksi O.K. Thus, T= τJ = τπ c3 = 8.42 ×103 (π )(1)3 = 13.23 kip ⋅ in. 2 2 c Formula (5.26) TN 13.23 × 103 (2400) = = 504 hp P= 63, 000 63, 000 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.