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ch01-03 stress & strain & properties

02 Solutions 46060

02 Solutions 46060 5/6/10 1:45 PM Page 13 © 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. 2–22. A square piece of material is deformed into the dashed position. Determine the shear strain at A. Shear Strain: B y 15.18 mm C (g A ) xy = p 2 - ¢ 89.7° 180° ≤p = 5.24A10 - 3 B rad Ans. A g xy 15 mm D 15 mm 15.24 mm 89.7 15.18 mm x 2–23. A square piece of material is deformed into the dashed parallelogram. Determine the average normal strain that occurs along the diagonals AC and BD. B y 15.18 mm C 15 mm 15.24 mm A 89.7 15 mm D 15.18 mm x Geometry: AC = BD = 215 2 + 15 2 = 21.2132 mm AC¿ =215.18 2 + 15.24 2 - 2(15.18)(15.24) cos 90.3° = 21.5665 mm B¿D¿ =215.18 2 + 15.24 2 - 2(15.18)(15.24) cos 89.7° = 21.4538 mm Average Normal Strain: e AC = e BD = AC¿ -AC AC = 0.01665 mm>mm = 16.7A10 - 3 B mm>mm B¿D¿ -BD BD = = 21.5665 - 21.2132 21.2132 21.4538 - 21.2132 21.2132 = 0.01134 mm>mm = 11.3A10 - 3 B mm>mm Ans. Ans. 13

02 Solutions 46060 5/6/10 1:45 PM Page 14 © 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. *2–24. A square piece of material is deformed into the dashed position. Determine the shear strain at C. g xy B y 15.18 mm C 15 mm 15.24 mm A 89.7 15 mm D 15.18 mm x (g C ) xy = p 2 - ¢ 89.7° 180° ≤p = 5.24A10 - 3 B rad Ans. •2–25. The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt u = 2°. Determine the approximate normal strain in the wire when the frame is in this position. Assume the columns are rigid and rotate about their lower supports. B u 2 u 2 Geometry: The vertical displacement is negligible x A = (1)¢ 2° ≤p = 0.03491 m 180° x B = (4)¢ 2° ≤p = 0.13963 m 180° 3 m 1 m 4 m A x = 4 + x B - x A = 4.10472 m A¿B¿ =23 2 + 4.10472 2 = 5.08416 m AB = 23 2 + 4 2 = 5.00 m Average Normal Strain: e AB = A¿B¿ -AB AB = 5.08416 - 5 5 = 16.8A10 - 3 B m>m Ans. 14

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