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ch01-03 stress & strain & properties

02 Solutions 46060

02 Solutions 46060 5/6/10 1:45 PM Page 5 © 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. 2–10. The corners B and D of the square plate are given the displacements indicated. Determine the shear strains at A and B. A y 16 mm 3 mm D B 3 mm 16 mm x 16 mm C 16 mm Applying trigonometry to Fig. a f = tan - 1 a 13 p rad b = 39.09° a b = 0.6823 rad 16 180° a = tan - 1 a 16 p rad b = 50.91° a b = 0.8885 rad 13 180° By the definition of shear strain, Ag xy B A = p 2 - 2f = p 2 - 2(0.6823) = 0.206 rad Ans. Ag xy B B = p 2 - 2a = p 2 - 2(0.8885) = -0.206 rad Ans. 5

02 Solutions 46060 5/6/10 1:45 PM Page 6 © 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. 2–11. The corners B and D of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonal DB. A y 16 mm 3 mm D B x 3 mm 16 mm 16 mm C 16 mm Referring to Fig. a, L AB = 216 2 + 16 2 = 2512 mm L AB¿ = 216 2 + 13 2 = 2425 mm L BD = 16 + 16 = 32 mm L B¿D¿ = 13 + 13 = 26 mm Thus, Ae avg B AB = L AB¿ - L AB L AB = 2425 - 2512 = -0.0889 mm>mm 2512 Ae avg B BD = L B¿D¿ - L BD 26 - 32 = = -0.1875 mm>mm L BD 32 Ans. Ans. 6

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