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

## 03

03 Solutions 46060 5/7/10 8:45 AM Page 25 © 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. 3–35. The elastic portion of the tension stressstrain s(ksi) diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a 70 diameter of 0.5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Compute the shear modulus G al for the aluminum. 0.00614 From the stressstrain diagram, P (in./in.) E al = s e = 70 = 11400.65 ksi 0.00614 When specimen is loaded with a 9 - kip load, s = P A = 9 p 4 (0.5)2 = 45.84 ksi e long = s E = 45.84 = 0.0040208 in.>in. 11400.65 e lat = d¿-d d = 0.49935 - 0.5 0.5 = - 0.0013 in.>in. e lat -0.0013 V = - = - e long 0.0040208 = 0.32332 G al = E at 2(1 + v) = 11.4(10 3 ) 2(1 + 0.32332) = 4.31(103 ) ksi Ans. *3–36. The elastic portion of the tension stressstrain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is G al = 3.8110 3 2 ksi. s = P A = 10 From the stressstrain diagram E = G = p 4 (0.5)2 = 50.9296 ksi 70 = 11400.65 ksi 0.00614 e long = s E = 50.9296 = 0.0044673 in.>in. 11400.65 E 2(1 + v) ; 3.8(103 ) = 11400.65 2(1 + v) ; v = 0.500 e lat = - ve long = - 0.500(0.0044673) = - 0.002234 in.>in. ¢d = e lat d = - 0.002234(0.5) = - 0.001117 in. s(ksi) 70 0.00614 P (in./in.) d¿ =d +¢d = 0.5 - 0.001117 = 0.4989 in. Ans. 25

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