ch01-03 stress & strain & properties
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<strong>03</strong> Solutions 46060 5/7/10 8:45 AM Page 11<br />
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3–15. The rigid pipe is supported by a pin at A and an<br />
A-36 guy wire BD. If the wire has a diameter of 0.25 in.,<br />
determine the load P if the end C is displaced 0.075 in.<br />
downward.<br />
B<br />
4 ft<br />
P<br />
A<br />
D<br />
3 ft 3 ft<br />
C<br />
Here, we are only interested in determining the force in wire BD. Referring to the<br />
FBD in Fig. a<br />
a+©M A = 0; F BD A 4 5 B(3) - P(6) = 0 F BD = 2.50 P<br />
The unstretched length for wire BD is L BD = 23 2 + 4 2 = 5 ft = 60 in. From the<br />
geometry shown in Fig. b, the stretched length of wire BD is<br />
L BD¿ = 260 2 + 0.075 2 - 2(60)(0.075) cos 143.13° = 60.060017<br />
Thus, the normal <strong>strain</strong> is<br />
P BD = L BD¿ - L BD 60.060017 - 60<br />
= = 1.00<strong>03</strong>(10 - 3 ) in.>in.<br />
L BD 60<br />
Then, the normal <strong>stress</strong> can be obtain by applying Hooke’s Law.<br />
s BD = EP BD = 29(10 3 )C1.00<strong>03</strong>(10 - 3 )D = 29.01 ksi<br />
Since s BD 6 s y = 36 ksi, the result is valid.<br />
s BD = F BD<br />
A BD<br />
; 29.01(10 3 ) =<br />
2.50 P<br />
p<br />
4 (0.252 )<br />
P = 569.57 lb = 570 lb<br />
Ans.<br />
11