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

## 02 Solutions 46060

02 Solutions 46060 5/6/10 1:45 PM Page 17 © 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. •2–29. The curved pipe has an original radius of 2 ft. If it is heated nonuniformly, so that the normal strain along its length is P=0.05 cos u, determine the increase in length of the pipe. e = 0.05 cos u ¢L = L e dL 90° = (0.05 cos u)(2 du) L 0 u 2 ft A 90° 90° = 0.1 cos u du = [0.1[sin u] 0 ] = 0.100 ft L 0 Ans. 2–30. Solve Prob. 2–29 if P =0.08 sin u. dL = 2 due = 0.08 sin u ¢L = L e dL 90° = (0.08 sin u)(2 du) L 0 = 0.16 L 90° 0 90° sin u du = 0.16[-cos u] 0 = 0.16 ft Ans. u 2 ft A 2–31. The rubber band AB has an unstretched length of 1 ft. If it is fixed at B and attached to the surface at point A¿, determine the average normal strain in the band.The surface is defined by the function y = (x 2 ) ft, where x is in feet. y y x 2 Geometry: 1 ft L = L 0 A 1 + a dy 2 dx b dx A¿ 1 ft However y = dy x2 then dx = 2x 1 ft L = 21 + 4 x 2 dx L 0 B 1 ft A x = 1 4 C2x21 + 4 x2 + ln A2x + 21 + 4x 2 1 ft BD 0 = 1.47894 ft Average Normal Strain: e avg = L - L 0 = 1.47894 - 1 = 0.479 ft>ft L 0 1 Ans. 17

02 Solutions 46060 5/6/10 1:45 PM Page 18 © 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. *2–32. The bar is originally 300 mm long when it is flat. If it is subjected to a shear strain defined by g xy = 0.02x, where x is in meters, determine the displacement ¢y at the end of its bottom edge. It is distorted into the shape shown, where no elongation of the bar occurs in the x direction. y y 300 mm x Shear Strain: dy dx = tan g xy ; dy dx = tan (0.02 x) L0 ¢y 300 mm dy = tan (0.02 x)dx L 0 300 mm ¢y = -50[ln cos (0.02x)]| 0 = 2.03 mm Ans. •2–33. The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements u A and v B , respectively, determine the normal strain in the fiber when it is in position A¿B¿. y B¿ v B B Geometry: L A¿B¿ = 2(L cos u - u A ) 2 + (L sin u + y B ) 2 A u A A¿ L u x = 2L 3 + u A 2 + y B 2 + 2L(y B sin u - u A cos u) Average Normal Strain: e AB = L A¿B¿ - L L = 1 + u A 2 + y 2 B A L 2 + 2(y B sin u - u A cos u) - 1 L Neglecting higher terms u A 2 and y 2 B e AB = B1 + 2(y B sin u - u A cos u) L Using the binomial theorem: 1 2 R - 1 e AB = 1 + 1 2 ¢ 2y B sin u L - 2u A cos u ≤ + . . . - 1 L = y B sin u L - u A cos u L Ans. 18

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