________________________________________________________________________ PROBLEMS (*11.45) A structural ASTM-A36 steel tube has length L, outside diameter d o , thickness t, and both ends pinned. A compressive load P acts with an eccentricity e as shown in Fig. P11.43. Calculate: (a) The maximum stress in the tube. (b) The allowable load based on a safety factor n with respect to material yielding. Given: d = 70 mm, t = 5 mm, L = 2.5 m, e = 32.5 mm, P = 10 kN, n s = 3, E = 210 GPa, σy = 250 MPa (from Table B.4) *SOLUTION di = d o − 2t = 70 − 2(5) = 60 mm A= I= π 4 π 64 r= (d o2 − di2 ) = (d o4 − di4 ) = π 4 (702 − 602 ) = 1021 mm 4 π 64 (704 − 604 ) = 542.4 × 103 mm 4 L 2.5 × 103 = = 108.5 28.05 r 542.4(10)3 I = = 23.05 mm 1021 A (a) Use Eq. (11.15b) σ max = ⎛ L P ⎡ ec ⎢1 + 2 sec ⎜⎜ A ⎣⎢ r ⎝ 2r ⎞⎤ ⎟⎟ ⎥ ⎠ ⎦⎥ P AE where 10 ×103 P = = 9.804 MPa , A 1.02 × 10−3 ec (32.5)(35) = = 2.141 r2 (23.05) 2 Thus ⎡ ⎛ ⎜ ⎝ σ max = (9.804) ⎢1 + (2.141) sec ⎜ 54.25 ⎢⎣ = 32.3 MPa ⎞⎤ 10 × 103 ⎟⎥ (1021)(10−6 )(210 ×109 ) ⎟⎠ ⎥ ⎦ (1) (b) Apply Eq.(11.16) σy = ⎛ L Py ⎡ ec ⎢1 + 2 sec ⎜⎜ A ⎢⎣ r ⎝ 2r P AE ⎞⎤ ⎟⎟ ⎥ ⎠ ⎥⎦ ⎡ ⎛ ⎞⎤ Py ⎢ 1 (2.141) sec 54.25 + ⎜ ⎟⎥ 3 ⎟ ⎜ 1021(10−16 ) ⎢ (1021)(210 10 ) × ⎝ ⎠ ⎥⎦ ⎣ Solving numerically: Py = 57.92 kN . Equation (11.22) gives then Py 250(104 ) = Pall = Py ns = (2) 57.92 = 19.31 kN 3 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.