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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 35 © 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. 1–51. During the tension test, the wooden specimen is subjected to an average normal stress of 2 ksi. Determine the axial force P applied to the specimen. Also, find the average shear stress developed along section a–a of the specimen. P a Internal Loading: The normal force developed on the cross section of the middle portion of the specimen can be obtained by considering the free-body diagram shown in Fig. a. 4 in. 2 in. a 1 in. + c ©F y = 0; P 2 + P 2 - N = 0 N = P 4 in. Referring to the free-body diagram shown in fig. b, the shear force developed in the shear plane a–a is P + c ©F y = 0; P 2 - V a - a = 0 V a - a = P 2 Average Normal Stress and Shear Stress: The cross-sectional area of the specimen is A = 1(2) = 2 in 2 . We have s avg = N A ; 2(103 ) = P 2 P = 4(10 3 )lb = 4 kip Using the result of P, V a - a = P 2 = 4(103 ) 2 A a - a = 2(4) = 8 in 2 . We obtain At a - a B avg = V a - a = 2(103 ) = 250 psi A a - a 8 Ans. = 2(10 3 ) lb. The area of the shear plane is Ans. 35

01 Solutions 46060 5/6/10 2:43 PM Page 36 © 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. *1–52. If the joint is subjected to an axial force of P = 9 kN, determine the average shear stress developed in each of the 6-mm diameter bolts between the plates and the members and along each of the four shaded shear planes. 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 - 9 = 0 V b = 2.25 kN ©F y = 0; 4V p - 9 = 0 V p = 2.25 kN 100 mm 100 mm P Average Shear Stress: The areas of each shear plane of the bolt and the member are A and A p = 0.1(0.1) = 0.01 m 2 b = p 4 (0.0062 ) = 28.274(10 - 6 )m 2 , respectively. At avg B b = V b 2.25(10 3 ) = A b 28.274(10 - 6 ) = 79.6 MPa Ans. We obtain At avg B p = V p = 2.25(103 ) = 225 kPa A p 0.01 Ans. 36

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