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

## 02 Solutions 46060

02 Solutions 46060 5/6/10 1:45 PM Page 3 © 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. 2–6. Nylon strips are fused to glass plates. When moderately heated the nylon will become soft while the glass stays approximately rigid. Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated. 3 mm y 2 mm P 5 mm 3 mm 5 mm 3 mm x g = tan - 1 a 2 b = 11.31° = 0.197 rad 10 Ans. 2–7. If the unstretched length of the bowstring is 35.5 in., determine the average normal strain in the string when it is stretched to the position shown. 18 in. 6 in. 18 in. Geometry: Referring to Fig. a, the stretched length of the string is L = 2L¿ =2218 2 + 6 2 = 37.947 in. Average Normal Strain: e avg = L - L 0 37.947 - 35.5 = = 0.0689 in.>in. L 0 35.5 Ans. 3

02 Solutions 46060 5/6/10 1:45 PM Page 4 © 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. *2–8. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by u = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched. D u 300 mm P B AB = 2400 2 + 300 2 = 500 mm 300 mm AB¿ =2400 2 + 300 2 - 2(400)(300) cos 90.3° e AB = = 501.255 mm AB¿ -AB AB = = 0.00251 mm>mm 501.255 - 500 500 Ans. A 400 mm C •2–9. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm>mm, determine the displacement of point D. Originally the cable is unstretched. D u 300 mm P AB = 2300 2 + 400 2 = 500 mm B AB¿ =AB + e AB AB 300 mm = 500 + 0.0035(500) = 501.75 mm 501.75 2 = 300 2 + 400 2 - 2(300)(400) cos a A C a = 90.4185° u = 90.4185° - 90° = 0.4185° = p (0.4185) rad 180° ¢ D = 600(u) = 600( p )(0.4185) = 4.38 mm 180° Ans. 400 mm 4

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