_______________________________________________________________________ PROBLEM (9.44) A simple metallic beam having a box cross section of uniform thickness t = 20 mm supports a concentrated load P as shown in Fig. P9.44. Determine the largest permissible value of P, if the allowable compression stress at point D is not to exceed σ all = 250 MPa. SOLUTION Reaction at B, as found from equilibrium conditions (see Fig. P9.44): RB = 0.375P y At a cross section through the point D: V = 0.375 P M = 0.375 P(0.6) = 0.225 P D Also 100 mm 20 mm 1 1 3 3 N.A. I = (150)(200) − (110)(160) z 12 12 C 100 mm = 62.453(106 ) mm 4 Q = Ay = (150 × 20)(100 − 10) 150 mm = 270(103 ) mm3 So My 0.225 P(0.08) σx = − =− = −288.217 P I 62.453(10−6 ) τ xy = VQ 0.375 P (270 × 10−6 ) = = 40.53P Ib 62.453(10−6 )(2 × 0.02) Hence σ1 = σx 2 + ( σx 2 ) 2 + τ xy2 = σ all 250(10 ) = −144.108 P + (−144.108 P) 2 + (40.53P ) 2 6 Solving Pall = 44.7 kN 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.