Solution Problem 2-95

MECHANICS OF MATERIALS, 6th Edition RILEY, STURGES AND MORRIS 2-95 The given values for use in drawing Mohr’s circle are 8 ksi x σ = 0 ksi y σ = 5 ksi xy τ =− 3 0 ksi zp σ σ== 80 4.00 ksi 2 a + == 22 4 5 6.403 ksiR =+= 1 1 15 tan 25.670 (CW) 22 4 p φ θ − == = ° (a) 1 4.00 6.40 10.40 ksi (T) p σ =+= ......................................Ans. 25.67° 2 4.00 6.40 2.40 ksi 2.40 ksi (C) p σ =−=− = .................Ans. 64.33° max 6.40 ksi p Rτ τ=== ...........................................................................Ans. (b) 4.00 6.403cos54.920 7.68 ksi (T) ab σ =+ °= ...................................Ans. 6.403sin 54.920 5.24 ksi (CCW) 5.24 ksi ab τ =°==+...................Ans. 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. Leroy D. Sturges 1-1