ch01-03 stress & strain & properties
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01 Solutions 46060 5/6/10 2:43 PM Page 32<br />
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1–47. Determine the average shear <strong>stress</strong> developed in pins<br />
A and B of the smooth two-tine grapple that supports the log<br />
having a mass of 3 Mg. Each pin has a diameter of 25 mm and<br />
is subjected to double shear.<br />
A<br />
20<br />
B<br />
E<br />
C<br />
D<br />
+ c ©F y = 0; 2(F sin 30°) - 29.43 = 0<br />
F = 29.43 kN<br />
0.2 m<br />
a+©M E = 0; P cos 20°(0.2) - (29.43 cos 30°)(1.2) + (29.43 sin 30°)(0.4 cos 30°)<br />
1.2 m<br />
= 0<br />
P = 135.61 kN<br />
t A = t B = V 135.61(10 3 )<br />
A = 2<br />
p<br />
4<br />
(0.025)2<br />
= 138 MPa<br />
Ans.<br />
30<br />
0.4 m<br />
30<br />
*1–48. The beam is supported by a pin at A and a short<br />
link BC. If P = 15 kN, determine the average shear <strong>stress</strong><br />
developed in the pins at A, B, and C. All pins are in double<br />
shear as shown, and each has a diameter of 18 mm.<br />
C<br />
30<br />
P 4P 4P 2P<br />
0.5m<br />
0.5 m<br />
1 m 1.5 m 1.5 m<br />
B<br />
A<br />
For pins B and C:<br />
t B = t C = V A = 82.5 (1<strong>03</strong> )<br />
For pin A:<br />
F A = 2 (82.5) 2 + (142.9) 2 = 165 kN<br />
t A = V A = 82.5 (1<strong>03</strong> )<br />
p<br />
4 ( 18<br />
p<br />
4 ( 18<br />
1000 )2 = 324 MPa<br />
1000 )2 = 324 MPa<br />
Ans.<br />
Ans.<br />
32