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ch01-03 stress & strain & properties

03

03 Solutions 46060 5/7/10 8:45 AM Page 2 © 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. 3–2. Data taken from a stressstrain test for a ceramic are given in the table.The curve is linear between the origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of resilience. Modulus of Elasticity: From the stressstrain diagram E = 33.2 - 0 0.0006 - 0 = 55.3A103 B ksi Ans. S (ksi) 0 33.2 45.5 49.4 51.5 53.4 P (in./in.) 0 0.0006 0.0010 0.0014 0.0018 0.0022 Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stressstrain diagram (shown shaded). u t = 1 2 (33.2)A103 B ¢ lb in. 2 ≤¢0.0006 in in. ≤ = 9.96 in # lb in 3 Ans. 3–3. Data taken from a stressstrain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine approximately the modulus of toughness.The rupture stress is s r = 53.4 ksi. Modulus of Toughness: The modulus of toughness is equal to the area under the stressstrain diagram (shown shaded). (u t ) approx = 1 2 (33.2)A103 B ¢ lb in. 2 ≤(0.0004 + 0.0010)¢ in in. ≤ + 45.5A10 3 B ¢ lb in. 2 ≤(0.0012)¢ in in. ≤ + 1 2 (7.90)A103 B ¢ lb in. 2 ≤(0.0012)¢ in in. ≤ S (ksi) 0 33.2 45.5 49.4 51.5 53.4 P (in./in.) 0 0.0006 0.0010 0.0014 0.0018 0.0022 + 1 2 (12.3)A103 B ¢ lb in. 2 ≤(0.0004)¢ in in. ≤ = 85.0 in # lb in 3 Ans. 2

03 Solutions 46060 5/7/10 8:45 AM Page 3 © 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. *3–4. A tension test was performed on a specimen having an original diameter of 12.5 mm and a gauge length of 50 mm. The data are listed in the table. Plot the stressstrain diagram, and determine approximately the modulus of elasticity, the ultimate stress, and the fracture stress. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm. Redraw the linear-elastic region, using the same stress scale but a strain scale of 20 mm = 0.001 mm>mm. Stress and Strain: s = P e = dL A (MPa) L (mm/mm) Load (kN) 0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8 Elongation (mm) 0 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380 0 0 90.45 0.00035 259.9 0.00120 308.0 0.00204 333.3 0.00330 355.3 0.00498 435.1 0.02032 507.7 0.06096 525.6 0.12700 507.7 0.17780 479.1 0.23876 Modulus of Elasticity: From the stressstrain diagram (E) approx = 228.75(106 ) - 0 0.001 - 0 = 229 GPa Ans. Ultimate and Fracture Stress: From the stressstrain diagram (s m ) approx = 528 MPa (s f ) approx = 479 MPa Ans. Ans. 3

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