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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 33 © 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. •1–49. The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of the loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins are in double shear as shown, and each has a diameter of 18 mm. C 30 P 4P 4P 2P 0.5m 0.5 m 1 m 1.5 m 1.5 m B A a +©M A = 0; 2P(0.5) + 4P(2) + 4P(3.5) + P(4.5) - (T CB sin 30°)(5) = 0 T CB = 11P : + ©F x = 0; A x - 11P cos 30° = 0 A x = 9.5263P + c ©F y = 0; A y - 11P + 11P sin 30° = 0 A y = 5.5P F A = 2 (9.5263P) 2 + (5.5P) 2 = 11P Require; t = V A ; 80(106 ) = 11P>2 p 4 (0.018)2 P = 3.70 kN Ans. 33

01 Solutions 46060 5/6/10 2:43 PM Page 34 © 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. 1–50. The block is subjected to a compressive force of 2 kN. Determine the average normal and average shear stress developed in the wood fibers that are oriented along section a–a at 30° with the axis of the block. a 50 mm 150 mm 2 kN 2 kN a 30 Force equilibrium equations written perpendicular and parallel to section a–a gives +Q©F x¿ = 0; V a - a - 2 cos 30° = 0 V a - a = 1.732 kN +a©F y¿ = 0; 2 sin 30° - N a - a = 0 N a - a = 1.00 kN The cross sectional area of section a–a is A = a 0.15 b(0.05) = 0.015 m2. Thus sin 30° (s a - a ) avg = N a - a Ans. A = 1.00(103 ) = 66.67(10 3 )Pa = 66.7 kPa 0.015 (t a - a ) avg = V a - a A = 1.732(103 ) 0.015 = 115.47(10 3 )Pa = 115 kPa Ans. 34

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