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

## 03

03 Solutions 46060 5/7/10 8:45 AM Page 8 © 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. 3–11. The stressstrain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 90 ksi, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded. s (ksi) 105 90 75 60 45 30 15 0 P (in./in.) 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 From the stressstrain diagram Fig. a, the modulus of elasticity for the steel alloy is E 1 = 60 ksi - 0 0.002 - 0 ; E = 30.0(103 ) ksi when the specimen is unloaded, its normal strain recovered along line AB, Fig.a, which has a gradient of E. Thus Elastic Recovery = 90 E = Thus, the permanent set is 90 ksi 30.0(10 3 ) ksi = 0.003 in>in Ans. Then, the increase in gauge length is P P = 0.05 - 0.003 = 0.047 in>in ¢L =P P L = 0.047(2) = 0.094 in Ans. 8

03 Solutions 46060 5/7/10 8:45 AM Page 9 © 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. *3–12. The stressstrain diagram for a steel alloy having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material. The Modulus of resilience is equal to the area under the stressstrain diagram up to the proportional limit. Thus, s PL = 60 ksi P PL = 0.002 in>in. (u i ) r = 1 2 s PLP PL = 1 2 C60(103 )D(0.002) = 60.0 in # lb in 3 Ans. The modulus of toughness is equal to the area under the entire stressstrain diagram. This area can be approximated by counting the number of squares. The total number is 38. Thus, C(u i ) t D approx = 38 c15(10 3 ) lb in d a0.05 2 in in b = 28.5(103 ) in # lb in 3 Ans. s (ksi) 105 90 75 60 45 30 15 0 P (in./in.) 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 9

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