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ch01-03 stress & strain & properties

03

03 Solutions 46060 5/7/10 8:45 AM Page 22 © 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. 3–30. The block is made of titanium Ti-6A1-4V and is subjected to a compression of 0.06 in. along the y axis, and its shape is given a tilt of u = 89.7°. Determine P x , P y , and g xy . y Normal Strain: e y = dL y = -0.06 = -0.0150 in.>in. L y 4 Ans. Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio. e x = -ve y = -0.36(-0.0150) 4 in. u 5 in. x = 0.00540 in. >in. Ans. Shear Strain: b = 180° - 89.7° = 90.3° = 1.576032 rad g xy = p 2 - b = p 2 - 1.576032 = -0.00524 rad Ans. 3–31. The shear stressstrain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 0.75 in. is made of this material and used in the double lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield. Take n = 0.3. The shear force developed on the shear planes of the bolt can be determined by considering the equilibrium of the FBD shown in Fig. a : + ©F x = 0; V + V - P = 0 V = = P 2 P t(ksi) 60 0.00545 P/2 P/2 g(rad) From the shear stressstrain diagram, the yield stress is t y = 60 ksi. Thus, t y = V y A ; 60 = P>2 p 4 A0.752 B P = 53.01 kip = 53.0 kip Ans. From the shear stressstrain diagram, the shear modulus is Thus, the modulus of elasticity is G = G = 60 ksi 0.00545 = 11.01(103 ) ksi E 2(1 + y) ; 11.01(103 ) = E 2(1 + 0.3) E = 28.6(10 3 ) ksi Ans. 22

03 Solutions 46060 5/7/10 8:45 AM Page 23 © 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. *3–32. A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load P is placed on the plug, show that the slope at point y in the rubber is dy>dr =-tan g = -tan1P>12phGr22. For small angles we can write dy>dr = -P>12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = r o . From the result compute the deflection y = d of the plug. r o d P r i r y h y P Shear Stress–Strain Relationship: Applying Hooke’s law with t A = . 2p r h g = t A G = P 2p h G r dy dr P = -tan g = -tan a 2p h G r b (Q.E.D) If g is small, then tan g = g. Therefore, At r = r o , y = 0 Then, y = P 2p h G ln r o r At r = r i , y = d dy dr = - P 2p h G r P dr y = - 2p h G L r y = - 0 = - C = P 2p h G ln r + C P 2p h G ln r o + C P 2p h G ln r o d = P 2p h G ln r o r i Ans. 23

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