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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 37 © 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. •1–53. The average shear stress in each of the 6-mm diameter bolts and along each of the four shaded shear planes is not allowed to exceed 80 MPa and 500 kPa, respectively. Determine the maximum axial force P that can be applied to the joint. P Internal Loadings: The shear force developed on each shear plane of the bolt and the member can be determined by writing the force equation of equilibrium along the member’s axis with reference to the free-body diagrams shown in Figs. a. and b, respectively. ©F y = 0; 4V b - P = 0 V b = P>4 ©F y = 0; 4V p - P = 0 V p = P>4 100 mm 100 mm P Average Shear Stress: The areas of each shear plane of the bolts and the members are A and A p = 0.1(0.1) = 0.01m 2 b = p , respectively. 4 (0.0062 ) = 28.274(10 - 6 )m 2 We obtain At allow B b = V b A b ; 80(10 6 ) = P>4 28.274(10 - 6 ) P = 9047 N = 9.05 kN (controls) Ans. At allow B p = V p A p ; 500(10 3 ) = P>4 0.01 P = 20 000 N = 20 kN 37

01 Solutions 46060 5/6/10 2:43 PM Page 38 © 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. 1–54. The shaft is subjected to the axial force of 40 kN. Determine the average bearing stress acting on the collar C and the normal stress in the shaft. 40 kN 30 mm C 40 mm Referring to the FBDs in Fig. a, + c ©F y = 0; N s - 40 = 0 N s = 40 kN + c ©F y = 0; N b - 40 = 0 N b = 40 kN Here, the cross-sectional area of the shaft and the bearing area of the collar are A and A b = p . Thus, 4 (0.042 ) = 0.4(10 - 3 )p m 2 s = p 4 (0.032 ) = 0.225(10 - 3 )p m 2 As avg B s = N s A s = As avg B b = N b A b = 40(10 3 ) 0.225(10 - 3 )p = 56.59(106 ) Pa = 56.6 MPa 40(10 3 ) 0.4(10 - 3 )p = 31.83(106 )Pa = 31.8 MPa Ans. Ans. 38

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