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ch01-03 stress & strain & properties

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01 Solutions 46060 5/6/10 2:43 PM Page 32<br />

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently<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

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