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ch01-03 stress & strain & properties

02 Solutions 46060

02 Solutions 46060 5/6/10 1:45 PM Page 15 © 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. 2–26. The material distorts into the dashed position shown. Determine (a) the average normal strains along sides AC and CD and the shear strain g xy at F, and (b) the average normal strain along line BE. 10 mm y 15 mm C B 25 mm D E 90 mm 75 mm A 80 mm F x Referring to Fig. a, L BE = 2(90 - 75) 2 + 80 2 = 26625 mm L AC¿ = 2100 2 + 15 2 = 210225 mm L C¿D¿ = 80 - 15 + 25 = 90 mm f = tan -1 ¢ 25 p rad ≤ = 14.04°¢ ≤ = 0.2450 rad. 100 180° When the plate deforms, the vertical position of point B and E do not change. L BB¿ 90 = 15 100 ; L BB¿ = 13.5 mm L EE¿ 75 = 25 100 ; L EE¿ = 18.75 mm L B¿E¿ = 2(90 - 75) 2 + (80 - 13.5 + 18.75) 2 = 27492.5625 mm Thus, Ae avg B AC = L AC¿ - L AC L AC = 210225 - 100 100 = 0.0112 mm>mm Ans. Ae avg B CD = L C¿D¿ - L CD = 90-80 = 0.125 mm>mm L CD 80 Ans. Ae avg B BE = L B¿E¿ - L BE L BE = 27492.5625 - 26625 26625 = 0.0635 mm>mm Ans. Referring to Fig. a, the angle at corner F becomes larger than 90° after the plate deforms. Thus, the shear strain is negative. 0.245 rad Ans. 15

02 Solutions 46060 5/6/10 1:45 PM Page 16 © 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. 2–27. The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF. The undeformed length of diagonals AD and CF are L AD = L CF = 280 2 + 100 2 = 216400 mm 10 mm y 15 mm C B 25 mm D E The deformed length of diagonals AD and CF are L AD¿ = 2(80 + 25) 2 + 100 2 = 221025 mm 90 mm 75 mm L C¿F = 2(80 - 15) 2 + 100 2 = 214225 mm Thus, Ae avg B AD = L AD¿ - L AD L AD = 221025 - 216400 216400 = 0.132 mm>mm Ans. A 80 mm F x Ae avg B CF = L C¿F - L CF L CF = 214225 - 216400 216400 = -0.0687 mm>mm Ans. *2–28. The wire is subjected to a normal strain that is defined by P=xe - x2 , where x is in millimeters. If the wire has an initial length L, determine the increase in its length. x P xe x2 x L dL = e dx = x e -x2 dx L ¢L = x e -x2 dx L 0 = - c 1 2 e-x2 d L 0 = - c 1 2 e-L2 - 1 2 d = 1 2 [1 - e-L2 ] Ans. 16

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