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ch01-03 stress & strain & properties

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01 Solutions 46060 5/6/10 2:43 PM Page 42<br />

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

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.<br />

1–59. The open square butt joint is used to transmit a<br />

force of 50 kip from one plate to the other. Determine the<br />

average normal and average shear <strong>stress</strong> components that<br />

this loading creates on the face of the weld, section AB.<br />

50 kip<br />

30<br />

30<br />

Equations of Equilibrium:<br />

a + © F y = 0; N - 50 cos 30° = 0 N = 43.30 kip<br />

+ Q© F x = 0; -V + 50 sin 30° = 0 V = 25.0 kip<br />

A<br />

B<br />

6 in.<br />

2 in.<br />

50 kip<br />

Average Normal and Shear Stress:<br />

2<br />

A¿ =a b(6) = 13.86 in2<br />

sin 60°<br />

s = N = 43.30 = 3.125 ksi<br />

A¿ 13.86<br />

t avg = V = 25.0 = 1.80 ksi<br />

A¿ 13.86<br />

Ans.<br />

Ans.<br />

*1–60. If P = 20 kN, determine the average shear <strong>stress</strong><br />

developed in the pins at A and C. The pins are subjected to<br />

double shear as shown, and each has a diameter of 18 mm.<br />

C<br />

Referring to the FBD of member AB,Fig.a<br />

a+©M A = 0; F BC sin 30° (6) - 20(2) - 20(4) = 0 F BC = 40 kN<br />

: + ©F x = 0; A x - 40 cos 30° = 0 A x = 34.64 kN<br />

A<br />

2 m<br />

2 m<br />

30<br />

2 m<br />

B<br />

+ c ©F y = 0; A y - 20 - 20 + 40 sin 30° A y = 20 kN<br />

Thus, the force acting on pin A is<br />

F A = 2 A x 2 + A y 2 = 2 34.64 2 + 20 2 = 40 kN<br />

Pins A and C are subjected to double shear. Referring to their FBDs in Figs. b and c,<br />

V A = F A<br />

2 = 40<br />

2 = 20 kN V C = F BC<br />

= 40 = 20 kN<br />

2 2<br />

= 81(10 - 6 )p m 2 . Thus<br />

The cross-sectional area of Pins A and C are A A = A C = p 4 (0.0182 )<br />

t A = V A<br />

A A<br />

= 20(1<strong>03</strong> )<br />

81(10 - 6 )p = 78.59(106 ) Pa = 78.6 MPa<br />

t C = V C<br />

A C<br />

= 20(1<strong>03</strong> )<br />

81(10 - 6 )p = 78.59(106 ) Pa = 78.6 MPa<br />

Ans.<br />

Ans.<br />

P<br />

P<br />

42

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