_______________________________________________________________________ PROBLEM (9.48) A link with a T cross section is loaded as shown in Fig. P9.48. Determine at section A-B (a) The maximum stress. (b) The location of the neutral axis. SOLUTION y B y 20 B M 47 P yo P=48 70 A 20 mm C z 10 y A 60 70 × 20(45) + 60 ×10(5) = 33 mm 70 × 20 + 60 × 10 y 0 = 20 + y = 53 mm y= A = 20 × 70 + 60 × 10 = 2 × 10 3 mm 3 20(70)3 60(10)3 + 20 × 70(12) 2 + + 60 × 10(28) 2 12 12 6 = 10 (0.572 + 0.202 + 0.005 + 0.47) = 1.249 × 10 6 mm 4 M = 48 × 0.053 = 2.544 kN ⋅ m I= P 48 ×103 (a) σ a = = = 24 MPa A 2 × 10−3 Mc −2544(0.047) (σ b ) B = = = −95.7 MPa I 1.249 ×10−6 33 = 67.2 MPa (σ b ) A = 95.7 47 σ A = 24 + 67.2 MPa σ A = 24 + 67.2 = 91.2 MPa σ B = 24 − 95.7 = −71.7 MPa Continued on next slide 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. (b) Apply Eq.(9.13): 2544 yn P Myn = 24 ×106 − 0= − A I 1.249 × 10−6 y n = 0.0118 m = 11.8 mm σb B N.A x yn A σ 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.