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ch01-03 stress & strain & properties

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<strong>03</strong> Solutions 46060 5/7/10 8:45 AM Page 5<br />

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.<br />

3–6. A specimen is originally 1 ft long, has a diameter of<br />

0.5 in., and is subjected to a force of 500 lb. When the force<br />

is increased from 500 lb to 1800 lb, the specimen elongates<br />

0.009 in. Determine the modulus of elasticity for the<br />

material if it remains linear elastic.<br />

Normal Stress and Strain: Applying s = P and e = dL .<br />

A L<br />

s 1 = 0.500<br />

p<br />

4 (0.52 ) = 2.546 ksi<br />

Modulus of Elasticity:<br />

s 2 = 1.80<br />

p<br />

= 9.167 ksi<br />

4 (0.52 )<br />

¢e = 0.009<br />

12<br />

= 0.000750 in.>in.<br />

E = ¢s<br />

¢e<br />

=<br />

9.167 - 2.546<br />

0.000750<br />

= 8.83A10 3 B ksi<br />

Ans.<br />

3–7. A structural member in a nuclear reactor is made of a<br />

zirconium alloy. If an axial load of 4 kip is to be supported<br />

by the member, determine its required cross-sectional area.<br />

Use a factor of safety of 3 relative to yielding. What is the<br />

load on the member if it is 3 ft long and its elongation is<br />

0.02 in.? E zr = 14(10 3 ) ksi, s Y = 57.5 ksi. The material has<br />

elastic behavior.<br />

Allowable Normal Stress:<br />

F.S. =<br />

s y<br />

s allow<br />

3 = 57.5<br />

s allow<br />

s allow = 19.17 ksi<br />

s allow = P A<br />

19.17 = 4 A<br />

A = 0.2087 in 2 = 0.209 in 2<br />

Ans.<br />

Stress–Strain Relationship: Applying Hooke’s law with<br />

e = d L = 0.02 = 0.000555 in.>in.<br />

3 (12)<br />

s = Ee = 14 A10 3 B (0.000555) = 7.778 ksi<br />

Normal Force: Applying equation s = P .<br />

A<br />

P = sA = 7.778 (0.2087) = 1.62 kip<br />

Ans.<br />

5

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