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**03** Solutions 46060 5/7/10 8:45 AM Page 4 © 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. 3–5. A tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. Using the data listed in the table, plot the **stress**–**strain** diagram, and determine approximately the modulus of toughness. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm>mm. Stress and Strain: s = P e = dL A (MPa) L (mm/mm) 0 0 90.45 0.00**03**5 259.9 0.00120 308.0 0.00204 333.3 0.0**03**30 355.3 0.00498 435.1 0.02**03**2 507.7 0.06096 525.6 0.12700 507.7 0.17780 479.1 0.23876 Modulus of Toughness: The modulus of toughness is equal to the total area under the **stress**–**strain** diagram and can be approximated by counting the number of squares. The total number of squares is 187. 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 (u t ) approx = 187(25)A10 6 B ¢ N m ≤ 2 a0.025 m b = 117 MJ>m3 m Ans. 4

**03** Solutions 46060 5/7/10 8:45 AM Page 5 © 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. 3–6. A specimen is originally 1 ft long, has a diameter of 0.5 in., and is subjected to a force of 500 lb. When the force is increased from 500 lb to 1800 lb, the specimen elongates 0.009 in. Determine the modulus of elasticity for the material if it remains linear elastic. Normal Stress and Strain: Applying s = P and e = dL . A L s 1 = 0.500 p 4 (0.52 ) = 2.546 ksi Modulus of Elasticity: s 2 = 1.80 p = 9.167 ksi 4 (0.52 ) ¢e = 0.009 12 = 0.000750 in.>in. E = ¢s ¢e = 9.167 - 2.546 0.000750 = 8.83A10 3 B ksi Ans. 3–7. A structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 4 kip is to be supported by the member, determine its required cross-sectional area. Use a factor of safety of 3 relative to yielding. What is the load on the member if it is 3 ft long and its elongation is 0.02 in.? E zr = 14(10 3 ) ksi, s Y = 57.5 ksi. The material has elastic behavior. Allowable Normal Stress: F.S. = s y s allow 3 = 57.5 s allow s allow = 19.17 ksi s allow = P A 19.17 = 4 A A = 0.2087 in 2 = 0.209 in 2 Ans. Stress–Strain Relationship: Applying Hooke’s law with e = d L = 0.02 = 0.000555 in.>in. 3 (12) s = Ee = 14 A10 3 B (0.000555) = 7.778 ksi Normal Force: Applying equation s = P . A P = sA = 7.778 (0.2087) = 1.62 kip Ans. 5

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