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ch01-03 stress & strain & properties

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01 Solutions 46060 5/6/10 2:43 PM Page 56<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 />

1–83. The three steel wires are used to support the<br />

load. If the wires have an allowable tensile <strong>stress</strong> of<br />

s allow = 165 MPa, and wire AB has a diameter of 6 mm, BC<br />

has a diameter of 5 mm, and BD has a diameter of 7 mm,<br />

determine the greatest force P that can be applied before<br />

one of the wires fails.<br />

A<br />

45<br />

B<br />

30<br />

C<br />

The force in wire BD is equal to the applied load; ie, F BD = P. Analysing the<br />

equilibrium of joint B by referring to its FBD, Fig. a,<br />

: + ©F x = 0; F BC cos 30° - F AB cos 45° = 0<br />

(1)<br />

D<br />

P<br />

+ c ©F y = 0; F BC sin 30° + F AB sin 45° - P = 0<br />

(2)<br />

Solving Eqs. (1) and (2),<br />

F AB = 0.8966 P F BC = 0.7321 P<br />

For wire BD,<br />

s allow = F BD<br />

A BD<br />

; 165(10 6 ) =<br />

For wire AB,<br />

P<br />

p<br />

4 (0.0072 )<br />

P = 6349.94 N = 6.350 kN<br />

s allow = F AB<br />

; 165(10 6 ) = 0.8966 P<br />

p<br />

A AB 4 (0.0062 )<br />

P = 52<strong>03</strong>.42 N = 5.2<strong>03</strong> kN<br />

For wire BC,<br />

s allow = F BC<br />

; 165(10 6 ) = 0.7321 P<br />

p<br />

A BC 4 (0.0052 )<br />

P = 4425.60 N = 4.43 kN (Controls!)<br />

Ans.<br />

56

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