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01 Solutions 46060 5/6/10 2:43 PM Page 21 © 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. 1–31. The column is subjected to an axial force of 8 kN, which is applied through the centroid of the cross-sectional area. Determine the average normal stress acting at section a–a. Show this distribution of stress acting over the area’s cross section. 10 mm 8 kN 75 mm 75 mm 10 mm 10 mm 70 mm 70 mm a a A = (2)(150)(10) + (140)(10) = 4400 mm 2 = 4.4 (10 -3 ) m 2 s = P A = 8 (103 ) 4.4 (10 - 3 = 1.82 MPa ) Ans. *1–32. The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a couple is applied to the lever, determine the average shear stress in the pin between the pin and lever. B A 12 mm 250 mm 250 mm 20 N 20 N a+©M O = 0; -F(12) + 20(500) = 0; F = 833.33 N t avg = V A = 833.33 p 4 ( 6 1000 )2 = 29.5 MPa Ans. 21
01 Solutions 46060 5/6/10 2:43 PM Page 22 © 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. •1–33. The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal and average shear stresses acting over the shaded section, which is oriented at u from the horizontal. Plot the variation of these stresses as a function of u 10 … u … 90°2. P u A P Equations of Equilibrium: R+©F x = 0; V - P cos u = 0 V = P cos u Q+©F y = 0; N - P sin u = 0 N = P sin u Average Normal Stress and Shear Stress: Area at u plane, A¿ = A . sin u s = N A¿ t avg = V A¿ = P sin u A sin u = P cos u A sin u = P A = P A sin2 u P sin u cos u = sin 2u 2A Ans. Ans. 1–34. The built-up shaft consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points. 4 kN A D B 6 kN 6 kN E C 8 kN At D: s D = P A = 4(10 3 ) p 4 (0.0282 - 0.02 2 = 13.3 MPa (C) ) Ans. At E: s E = P A = 8(10 3 ) p = 70.7 MPa (T) 4 (0.0122 ) Ans. 22
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