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ch01-03 stress & strain & properties

03

03 Solutions 46060 5/7/10 8:45 AM Page 20 © 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. 3–26. The short cylindrical block of 2014-T6 aluminum, having an original diameter of 0.5 in. and a length of 1.5 in., is placed in the smooth jaws of a vise and squeezed until the axial load applied is 800 lb. Determine (a) the decrease in its length and (b) its new diameter. 800 lb 800 lb a) s = P A = 800 = 4074.37 psi (0.5)2 p 4 e long = s E = -4074.37 10.6(10 6 ) = -0.0003844 d = e long L = -0.0003844 (1.5) = -0.577 (10 - 3 ) in. Ans. b) V = -e lat e long = 0.35 e lat = -0.35 (-0.0003844) = 0.00013453 ¢d = e lat d = 0.00013453 (0.5) = 0.00006727 d¿ =d +¢d = 0.5000673 in. Ans. 3–27. The elastic portion of the stressstrain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. When the applied load on the specimen is 50 kN, the diameter is 12.99265 mm. Determine Poisson’s ratio for the material. s(MPa) 400 Normal Stress: s = P A = 50(103 ) p = 376.70 Mpa 4 (0.0132 ) 0.002 P(mm/mm) Normal Strain: From the stressstrain diagram, the modulus of elasticity E = 400(106 ) = 200 GPa. Applying Hooke’s law 0.002 e long = s E = 376.70(106 ) 200(10 4 ) = 1.8835A10 - 3 B mm>mm e lat = d - d 0 12.99265 - 13 = = -0.56538A10 - 3 B mm>mm d 0 13 Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio. V = - e lat e long = - -0.56538(10 - 3 ) 1.8835(10 - 3 ) = 0.300 Ans. 20

03 Solutions 46060 5/7/10 8:45 AM Page 21 © 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. *3–28. The elastic portion of the stressstrain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a load of P = 20 kN is applied to the specimen, determine its diameter and gauge length. Take n = 0.4. s(MPa) 400 Normal Stress: s = P A = 20(103 ) p 4 (0.0132 ) = 150.68Mpa 0.002 P(mm/mm) Normal Strain: From the Stress–Strain diagram, the modulus of elasticity E = 400(106 ) = 200 GPa. Applying Hooke’s Law 0.002 e long = s E = 150.68(106 ) 200(10 9 ) = 0.7534A10 - 3 B mm>mm Thus, dL = e long L 0 = 0.7534A10 - 3 B(50) = 0.03767 mm L = L 0 + dL = 50 + 0.03767 = 50.0377 mm Ans. Poisson’s Ratio: The lateral and longitudinal can be related using poisson’s ratio. e lat = -ve long = -0.4(0.7534)A10 - 3 B = -0.3014A10 - 3 B mm>mm dd = e lat d = -0.3014A10 - 3 B(13) = -0.003918 mm d = d 0 + dd = 13 + (-0.003918) = 12.99608 mm Ans. •3–29. The aluminum block has a rectangular cross section and is subjected to an axial compressive force of 8 kip. If the 1.5-in. side changed its length to 1.500132 in., determine Poisson’s ratio and the new length of the 2-in. side. E al 10(10 3 ) ksi. 8 kip 3 in. 1.5 in. 2 in. 8 kip s = P A = 8 = 2.667 ksi (2)(1.5) e long = s E = -2.667 10(10 3 ) = -0.0002667 e lat = 1.500132 - 1.5 1.5 = 0.0000880 v = -0.0000880 -0.0002667 = 0.330 h¿ =2 + 0.0000880(2) = 2.000176 in. Ans. Ans. 21

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