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

## 01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 45 © 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. *1–64. The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the glue can withstand a maximum average shear stress of 800 kPa, determine the maximum allowable clamping force F. 50 mm 45 F 25 mm glue Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. F : + ©F x = 0; F cos 45° - V = 0 V = 2 2 2 F Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3 )m 2 . We obtain t avg = V A ; 800(103 ) = 2 2 2 F 1.25(10 - 3 ) F = 1414 N = 1.41 kN Ans. •1–65. The triangular blocks are glued along each side of the joint. A C-clamp placed between two of the blocks is used to draw the joint tight. If the clamping force is F = 900 N, determine the average shear stress developed in the glued shear plane. Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the free-body diagram of the triangular block, Fig. a. 50 mm 45 F 25 mm glue F : + ©F x = 0; 900 cos 45° - V = 0 V = 636.40 N Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10 - 3 )m 2 . We obtain t avg = V A = 636.40 1.25(10 - 3 = 509 kPa ) Ans. 45

01 Solutions 46060 5/6/10 2:43 PM Page 46 © 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. 1–66. Determine the largest load P that can be a applied to the frame without causing either the average normal stress or the average shear stress at section a–a to exceed s = 150 MPa and t = 60 MPa, respectively. Member CB has a square cross section of 25 mm on each side. B Analyse the equilibrium of joint C using the FBD Shown in Fig. a, + c ©F y = 0; F BC a 4 5 b - P = 0 F BC = 1.25P Referring to the FBD of the cut segment of member BC Fig. b. : + ©F x = 0; N a - a - 1.25Pa 3 5 b = 0 N a - a = 0.75P 2 m a a + c ©F y = 0; 1.25Pa 4 5 b - V a - a = 0 V a - a = P A C The cross-sectional area of section a–a is A a - a = (0.025)a 0.025 3>5 b = 1.0417(10 - 3 )m 2 . For Normal stress, s allow = N a - a A a - a ; 150(10 6 ) = P = 208.33(10 3 ) N = 208.33 kN For Shear Stress t allow = V a - a A a - a ; 60(10 6 ) = 0.75P 1.0417(10 - 3 ) P 1.0417(10 - 3 ) 1.5 m P P = 62.5(10 3 ) N = 62.5 kN (Controls!) Ans. 46

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