ch01-03 stress & strain & properties
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01 Solutions 46060 5/6/10 2:43 PM Page 28<br />
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•1–41. Solve Prob. 1–40 assuming that pins B and C are<br />
subjected to single shear.<br />
3 ft 3 ft<br />
500 lb<br />
A<br />
3 ft<br />
B<br />
C<br />
Support Reactions: FBD(a)<br />
a +©M g = 0; 500(6) + 300(3) - D y (6) = 0<br />
D y = 650 lb<br />
D<br />
1.5 ft<br />
300 lb<br />
1.5 ft<br />
E<br />
3 ft<br />
; + ©F x = 0; 500 - E x = 0 E x = 500 lb<br />
+ c ©F y = 0; 650 - 300 - E y = 0 E y = 350 lb<br />
From FBD (c),<br />
a+©M B = 0; C y (3) - 300(1.5) = 0 C y = 150 lb<br />
+ c ©F y = 0; B y + 150 - 300 = 0 B y = 150 lb<br />
From FBD (b)<br />
d+©M A = 0; 150(1.5) + B x (3) - 650(3) = 0<br />
B x = 575 lb<br />
From FBD (c),<br />
: + ©F x = 0; C x - 575 = 0 C x = 575 lb<br />
Hence, F B = F C = 2 575 2 + 150 2 = 594.24 lb<br />
Average shear <strong>stress</strong>: Pins B and C are subjected to single shear as shown on FBD (d)<br />
(t B ) avg = (t C ) avg = V A = 594.24<br />
p<br />
4 (0.252 )<br />
= 12106 psi = 12.1 ksi<br />
Ans.<br />
28
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