ch01-03 stress & strain & properties
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<strong>03</strong> Solutions 46060 5/7/10 8:45 AM Page 6<br />
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*3–8. The strut is supported by a pin at C and an A-36<br />
steel guy wire AB. If the wire has a diameter of 0.2 in.,<br />
determine how much it stretches when the distributed load<br />
acts on the strut.<br />
A<br />
60<br />
200 lb/ft<br />
Here, we are only interested in determining the force in wire AB.<br />
C<br />
9 ft<br />
B<br />
a+©M C = 0; F AB cos 60°(9) - 1 2 (200)(9)(3) = 0 F AB = 600 lb<br />
The normal <strong>stress</strong> the wire is<br />
s AB = F AB<br />
A AB<br />
=<br />
600<br />
p<br />
4 (0.22 ) = 19.10(1<strong>03</strong> ) psi = 19.10 ksi<br />
Since s AB 6 s y = 36 ksi, Hooke’s Law can be applied to determine the <strong>strain</strong><br />
in wire.<br />
s AB = EP AB ; 19.10 = 29.0(10 3 )P AB<br />
P AB = 0.6586(10 - 3 ) in>in<br />
The unstretched length of the wire is<br />
stretches<br />
L AB = 9(12)<br />
sin 60°<br />
= 124.71 in. Thus, the wire<br />
d AB =P AB L AB = 0.6586(10 - 3 )(124.71)<br />
= 0.0821 in.<br />
Ans.<br />
6
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