ch01-03 stress & strain & properties
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02 Solutions 46060 5/6/10 1:45 PM Page 17<br />
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•2–29. The curved pipe has an original radius of 2 ft. If it is<br />
heated nonuniformly, so that the normal <strong>strain</strong> along its length<br />
is P=0.05 cos u, determine the increase in length of the pipe.<br />
e = 0.05 cos u<br />
¢L =<br />
L<br />
e dL<br />
90°<br />
= (0.05 cos u)(2 du)<br />
L<br />
0<br />
u<br />
2 ft<br />
A<br />
90°<br />
90°<br />
= 0.1 cos u du = [0.1[sin u] 0 ] = 0.100 ft<br />
L<br />
0<br />
Ans.<br />
2–30. Solve Prob. 2–29 if P =0.08 sin u.<br />
dL = 2 due = 0.08 sin u<br />
¢L =<br />
L<br />
e dL<br />
90°<br />
= (0.08 sin u)(2 du)<br />
L<br />
0<br />
= 0.16<br />
L<br />
90°<br />
0<br />
90°<br />
sin u du = 0.16[-cos u] 0<br />
= 0.16 ft<br />
Ans.<br />
u<br />
2 ft<br />
A<br />
2–31. The rubber band AB has an unstretched length of<br />
1 ft. If it is fixed at B and attached to the surface at point A¿,<br />
determine the average normal <strong>strain</strong> in the band.The surface<br />
is defined by the function y = (x 2 ) ft, where x is in feet.<br />
y<br />
y x 2<br />
Geometry:<br />
1 ft<br />
L =<br />
L 0 A 1 + a dy 2<br />
dx b dx<br />
A¿<br />
1 ft<br />
However y = dy<br />
x2 then<br />
dx = 2x<br />
1 ft<br />
L = 21 + 4 x 2 dx<br />
L<br />
0<br />
B<br />
1 ft<br />
A<br />
x<br />
= 1 4 C2x21 + 4 x2 + ln A2x + 21 + 4x 2 1 ft<br />
BD 0<br />
= 1.47894 ft<br />
Average Normal Strain:<br />
e avg = L - L 0<br />
= 1.47894 - 1 = 0.479 ft>ft<br />
L 0 1<br />
Ans.<br />
17
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