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**Unformatted text preview: **Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.1 At the proportional limit, a 12-inch gage length of a 0.75-in.-diameter alloy bar has elongated 0.0325 in. and the diameter has been reduced 0.0006 in. The total tension force on the bar was 17.5 kips. Determine the following properties of the material: (a) the modulus of elasticity. (b) Poisson’s ratio. (c) the proportional limit. Solution (a) The bar cross-sectional area is 2 2 2 (0.75 in.) 0.441786 in. 4 4 A D ππ = = = and thus, the normal stress corresponding to the 17.5-kip force is 2 17.5 kips 39.611897 ksi 0.441786 in. σ = = The strain in the bar is 0.0325 in. 0.002708 in./in. 12 in. e L ε = = = The modulus of elasticity is therefore 39.611897 ksi 14,626 ksi 14,630 ksi 0.002708 in./in. E = = = = Ans. (b) The longitudinal strain in the bar was calculated previously as long 0.002708 in./in. = The lateral strain can be determined from the reduction of the diameter: lat 0.0006 in. 0.000800 in./in. 0.75 in. D D Δ − = = = − Poisson’s ratio for this specimen is therefore lat long 0.000800 in./in. 0.295421 0.295 0.002708 in./in. ν − = − = − = = Ans. (c) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore, 39.6 ksi PL = Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.2 At the proportional limit, a 20 mm thick × 75 mm wide bar elongates 6.8 mm under an axial load of 480 kN. The bar is 1.6-m long. If Poisson’s ratio is 0.32 for the material, determine: (a) the modulus of elasticity. (b) the proportional limit (c) the change in each lateral dimension. Solution (a) The bar cross-sectional area is 2 (20 mm)(75 mm) 1,500 mm A = = and thus, the normal stress corresponding to the 480-kN axial load is 2 (480 kN)(1,000 N/kN) 320 MPa 1,500 mm σ = = The strain in the bar is 6.8 mm 0.004250 mm/mm (1.6 m)(1,000 mm/m) e L ε = = = The modulus of elasticity is therefore 320 MPa 75,294 MPa 75.3 GPa 0.004250 mm/mm E = = = = Ans. (b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore, 320 MPa PL = Ans. ...

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