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ch01-03 stress & strain & properties

02 Solutions 46060

02 Solutions 46060 5/6/10 1:45 PM Page 11 © 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. 2–18. The piece of plastic is originally rectangular. Determine the shear strain g xy at corners A and B if the plastic distorts as shown by the dashed lines. y 2 mm 2 mm C 5 mm B 4 mm 300 mm Geometry: For small angles, a = c = 2 = 0.00662252 rad 302 D 400 mm A 3 mm 2 mm x b = u = 2 = 0.00496278 rad 403 Shear Strain: (g B ) xy = a + b = 0.0116 rad = 11.6A10 - 3 B rad Ans. (g A ) xy = -(u + c) = -0.0116 rad = -11.6A10 - 3 B rad Ans. 2–19. The piece of plastic is originally rectangular. Determine the shear strain g xy at corners D and C if the plastic distorts as shown by the dashed lines. y 2 mm 2 mm C 5 mm B 4 mm 300 mm D 400 mm A 3 mm 2 mm x Geometry: For small angles, a = c = 2 = 0.00496278 rad 403 b = u = 2 = 0.00662252 rad 302 Shear Strain: (g C ) xy = -(a + b) = -0.0116 rad = -11.6A10 - 3 B rad Ans. (g D ) xy = u + c = 0.0116 rad = 11.6A10 - 3 B rad Ans. 11

02 Solutions 46060 5/6/10 1:45 PM Page 12 © 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. *2–20. The piece of plastic is originally rectangular. Determine the average normal strain that occurs along the diagonals AC and DB. Geometry: y 2 mm 2 mm C 5 mm B 4 mm AC = DB = 2400 2 + 300 2 = 500 mm 300 mm DB¿ =2405 2 + 304 2 = 506.4 mm A¿C¿ =2401 2 + 300 2 = 500.8 mm Average Normal Strain: D 400 mm A 3 mm 2 mm x e AC = e DB = A¿C¿ -AC AC = 0.00160 mm>mm = 1.60A10 - 3 B mm>mm DB¿ -DB DB = = 500.8 - 500 500 506.4 - 500 500 = 0.0128 mm>mm = 12.8A10 - 3 B mm>mm Ans. Ans. •2–21. The force applied to the handle of the rigid lever arm causes the arm to rotate clockwise through an angle of 3° about pin A. Determine the average normal strain developed in the wire. Originally, the wire is unstretched. D 600 mm Geometry: Referring to Fig. a, the stretched length of the consine law, L B¿D can be determined using L B¿D = 2(0.6 cos 45°) 2 + (0.6 sin 45°) 2 - 2(0.6 cos 45°)(0.6 sin 45°) cos 93° A 45 C = 0.6155 m B Average Normal Strain: The unstretched length of wire BD is L BD = 0.6 m. We obtain e avg = L B¿D - L BD 0.6155 - 0.6 = = 0.0258 m>m L BD 0.6 Ans. 12

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