_______________________________________________________________________ PROBLEM (9.25) A circular bar is fixed at one and loaded as illustrated in Fig. P9.23. Find the required diameter d of the shaft. Given: P = 0, R = 50 π N, T = 120 π N ⋅ m, L = 2 m, σ all = 120 MPa, τ all = 70 MPa SOLUTION D T = 120π N ⋅ m E 2m σx = D 32M πd3 τ = 16T π d 3 50π N 32(2 × 50π ) 3200 = 3 πd3 d −16(120π ) 1920 τt = =− 3 πd3 d σx = Equation (9.1): 120 × 106 = 1600 1600 −1920 + ( 3 )2 + ( 3 )2 3 d d d 120 × 106 = 1 (1600 + 2499.2798), d3 or Equation (9.2): 70 × 10 6 = 2499.2798 d 3 , Similarly, σx = 0 E τ= 1920 +τ d d3 d = 0.0325 m d = 0.0329 m From Example 9.4: VQ 4V τd = =− 3A Ib 4 50π −267 =− = 2 2 3 πd 4 d Equation (9.1): 1 1920 276 2 − 2 ) = 3 (1920 + 267 d ) 3 d d d Solve by trial & error : d = 0.0252 m. Equation (9.2): 1 70 ×106 = 3 (0) 2 + (−1920 − 267d ) 2 d Solve by trial & error : d = 0.0302 m. Thus, d all = 25.2 mm 120 × 106 = 0 + (− 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.