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# ch01-03 stress &amp; strain &amp; properties

## 01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 43 © 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. •1–61. Determine the maximum magnitude P of the load the beam will support if the average shear stress in each pin is not to allowed to exceed 60 MPa.All pins are subjected to double shear as shown, and each has a diameter of 18 mm. C Referring to the FBD of member AB,Fig.a, a+©M A = 0; F BC sin 30°(6) - P(2) - P(4) = 0 F BC = 2P A 2 m 2 m 30 2 m B : + ©F x = 0; A x - 2P cos 30° = 0 A x = 1.732P + c ©F y = 0; A y - P - P + 2P sin 30° = 0 A y = P Thus, the force acting on pin A is F A = 2 A x 2 + A y 2 = 2 (1.732P) 2 + P 2 = 2P All pins are subjected to same force and double shear. Referring to the FBD of the pin, Fig. b, P P V = F 2 = 2P 2 = P The cross-sectional area of the pin is A = p . Thus, 4 (0.0182 ) = 81.0(10 - 6 )p m 2 t allow = V A ; 60(106 ) = P 81.0(10 - 6 )p P = 15 268 N = 15.3 kN Ans. 43

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