________________________________________________________________________ PROBLEM (11.47) Rework Prob. 11.45 for tube with fixed-pinned ends and a factor of safety of n s = 1.5. SOLUTION Refer to Solution of Prob. 11.45. We now have Le = 0.7 L = 0.7(2.5) = 1.75 m (See Fig. 11.3c) Le 1.75 ×103 = = 75.92 r 23.05 Le = 37.94 2r (a) Equation (1) becomes ⎡ ⎛ σ max = (9.794) ⎢1 + (2.141) sec ⎜ 37.96 ⎜ ⎢⎣ ⎝ 10 × 103 (1021)(210 × 103 ) ⎞⎤ ⎟ ⎥ = 31.5 MPa ⎟⎥ ⎠⎦ (b) Equation (2) Leads to ⎡ ⎛ ⎞⎤ Py ⎢ 1 (2.141) sec 37.96 + ⎜ ⎟⎥ 3 ⎟ ⎜ 1021(10−16 ) ⎢ (1021)(210 10 ) × ⎝ ⎠ ⎥⎦ ⎣ Solving numerically, Py = 68.1 kN . Equation (11.22) results in Py 250(106 ) = Pall = Py ns = 68.1 = 45.4 kN 1.5 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.