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 />
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