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# ch01-03 stress &amp; strain &amp; properties

## 02 Solutions 46060

02 Solutions 46060 5/6/10 1:45 PM Page 1 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2–1. An air-filled rubber ball has a diameter of 6 in. If the air pressure within it is increased until the ball’s diameter becomes 7 in., determine the average normal strain in the rubber. d 0 = 6 in. d = 7 in. e = pd - pd 0 = 7 - 6 = 0.167 in./in. pd 0 6 Ans. 2–2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip. L 0 = 15 in. L = p(5 in.) e = L - L 0 5p - 15 = = 0.0472 in.>in. L 0 15 Ans. 2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD. D E 4 m P ¢L BD 3 = ¢L CE 7 A B C ¢L BD = 3 (10) 7 = 4.286 mm 3 m 2 m 2 m e CE = ¢L CE L = 10 = 0.00250 mm>mm 4000 Ans. e BD = ¢L BD L = 4.286 4000 = 0.00107 mm>mm Ans. 1

02 Solutions 46060 5/6/10 1:45 PM Page 2 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *2–4. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire. C L œ AC = 2300 2 + 2 2 - 2(300)(2) cos 150° = 301.734 mm 300 mm e AC = e AB = Lœ AC - L AC 301.734 - 300 = = 0.00578 mm>mm L AC 300 Ans. 30 30 A P 300 mm B •2–5. The rigid beam is supported by a pin at A and wires BD and CE. If the distributed load causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD. D 1.5 m E 2 m A 2 m B 3 m C Since the vertical displacement of end C is small compared to the length of member AC, the vertical displacement d B of point B, can be approximated by referring to the similar triangle shown in Fig. a d B 2 = 10 5 ; d B = 4 mm The unstretched lengths of wires BD and CE are L BD = 1500 mm and L CE = 2000 mm. Ae avg B BD = d B L BD = 4 = 0.00267 mm>mm 1500 Ans. w Ae avg B CE = d C = 10 = 0.005 mm>mm L CE 2000 Ans. 2

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