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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 25<br />

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3–35. The elastic portion of the tension <strong>stress</strong>–<strong>strain</strong><br />

s(ksi)<br />

diagram for an aluminum alloy is shown in the figure. The<br />

specimen used for the test has a gauge length of 2 in. and a<br />

70<br />

diameter of 0.5 in. When the applied load is 9 kip, the new<br />

diameter of the specimen is 0.49935 in. Compute the shear<br />

modulus G al for the aluminum.<br />

0.00614<br />

From the <strong>stress</strong>–<strong>strain</strong> diagram,<br />

P (in./in.)<br />

E al = s e = 70<br />

= 11400.65 ksi<br />

0.00614<br />

When specimen is loaded with a 9 - kip load,<br />

s = P A = 9<br />

p<br />

4<br />

(0.5)2<br />

= 45.84 ksi<br />

e long = s E = 45.84<br />

= 0.0040208 in.>in.<br />

11400.65<br />

e lat = d¿-d<br />

d<br />

=<br />

0.49935 - 0.5<br />

0.5<br />

= - 0.0013 in.>in.<br />

e lat<br />

-0.0013<br />

V = - = -<br />

e long 0.0040208 = 0.32332<br />

G al =<br />

E at<br />

2(1 + v) = 11.4(10 3 )<br />

2(1 + 0.32332) = 4.31(1<strong>03</strong> ) ksi<br />

Ans.<br />

*3–36. The elastic portion of the tension <strong>stress</strong>–<strong>strain</strong><br />

diagram for an aluminum alloy is shown in the figure. The<br />

specimen used for the test has a gauge length of 2 in. and a<br />

diameter of 0.5 in. If the applied load is 10 kip, determine<br />

the new diameter of the specimen. The shear modulus is<br />

G al = 3.8110 3 2 ksi.<br />

s = P A = 10<br />

From the <strong>stress</strong>–<strong>strain</strong> diagram<br />

E =<br />

G =<br />

p<br />

4<br />

(0.5)2<br />

= 50.9296 ksi<br />

70<br />

= 11400.65 ksi<br />

0.00614<br />

e long = s E = 50.9296 = 0.0044673 in.>in.<br />

11400.65<br />

E<br />

2(1 + v) ; 3.8(1<strong>03</strong> ) = 11400.65<br />

2(1 + v) ; v = 0.500<br />

e lat = - ve long = - 0.500(0.0044673) = - 0.002234 in.>in.<br />

¢d = e lat d = - 0.002234(0.5) = - 0.001117 in.<br />

s(ksi)<br />

70<br />

0.00614<br />

P (in./in.)<br />

d¿ =d +¢d = 0.5 - 0.001117 = 0.4989 in.<br />

Ans.<br />

25

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