________________________________________________________________________ PROBLEM (11.13) A 60-mm outer-diameter and 50-mm inner-diameter steel pipe (E = 20 GPa) 2 m long acts as a spreader in the assembly of Fig. P11.13. Determine for a factor of safety n s = 2, the value of F that will cause buckling of the pipe. SOLUTION Pcr 1 2 2 F I= 4 (c 4 − b 4 ) = Pcr F 1 π 1 2 π 4 F 2 = Pcr 1 (304 − 254 ) = 32.94 × 104 mm 4 Pcr π 2 EI π 2 (200 × 109 )(0.3294 ×10−6 ) = = = 81.28 kN 2(2) 2 ns ns L2 F = 2(81.28) = 162.6 kN Justification of the formula used: A = π (c 2 − b 2 ) = π (302 − 252 ) = 863.94 mm 2 r= I 329, 400 = = 19.5 mm A 863.94 and L 2000 = = 102.6 O.K. r 19.5 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.