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01 Solutions 46060 5/6/10 2:43 PM Page 47 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–67. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w 0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for 0 … x 6 a. x a w 0 a Equation of Equilibrium: : + ©F x = 0; -N + 1 2 a w 0 a x + w 0b(a - x) + 1 2 w 0 a = 0 N = w 0 2a A2a2 - x 2 B Average Normal Stress: s = N w 0 A = 2a (2a2 - x 2 ) = w 0 A 2aA A2a2 - x 2 B Ans. *1–68. The prismatic bar has a cross-sectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w 0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for a 6 x … 2a. x a w 0 a Equation of Equilibrium: : + ©F x = 0; -N + 1 2 c w 0 a (2a - x) d(2a - x) = 0 N = w 0 (2a - x)2 2a Average Normal Stress: s = N w 0 A = 2a (2a - x)2 = w 0 (2a - x)2 A 2aA Ans. •1–69. The tapered rod has a radius of r = (2 - x>6) in. and is subjected to the distributed loading of w = (60 + 40x) lb>in. Determine the average normal stress at the center of the rod, B. r w (60 40x) lb/ in. x r = (2 — 6 ) in. A = pa2 - 3 6 b 2 = 7.069 in 2 B x 6 © F x = 0; N - (60 + 40x) dx = 0; N = 720 lb L 3 3 in. 3 in. s = 720 = 102 psi 7.069 Ans. 47

01 Solutions 46060 5/6/10 2:43 PM Page 48 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–70. The pedestal supports a load P at its center. If the material has a mass density r, determine the radial dimension r as a function of z so that the average normal stress in the pedestal remains constant. The cross section is circular. P r 1 z r Require: s = P + W 1 A dW dA = P + W 1 = s A = P + W 1 + dW A + dA P dA + W 1 dA = A dW dA = p(r + dr) 2 - pr 2 = 2p r dr (1) dW = pr 2 (rg) dt From Eq. (1) pr 2 (rg) dz = s 2p r dr r rg dz = s 2 dr z r rg dz = 2s L0 L rg z 2s = ln r r 1 ; r = r 1 e ( p 2a )z However, s = P p r 1 2 r 1 dr r r = r 1 e (p r12 rg ) 2P z Ans. 48

03 Solutions 46060 5/7/10 8:45 AM P

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Magazine: ch01-03 stress & strain & properties