01-01ChapGere.0021

01-01ChapGere.0021 - diameter d ? What is the magnitude of...

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Solution 1.5-3 Nylon bar inside steel tube SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 21 C OMPRESSION d 1 5 3.50 in. ± d 1 5 0.01 in. d 2 5 3.51 in. Nylon: E 5 400 ksi n5 0.4 L ATERAL STRAIN e ¿ 5 0.01 in. 3.50 in. 5 0.002857 e ¿ 5 ¢ d 1 d 1 (Increase in diameter) A XIAL STRAIN (Shortening) A XIAL STRESS s 5 E e 5 (400 ksi)( 2 0.007143) 52 2.857 ksi (Compressive stress) Assume that the yield stress is greater than s and Hooke’s law is valid. F ORCE P ( COMPRESSION ) 5 27.5 k P 5 s A 5 (2.857 ksi) ¢ p 4 (3.50 in.) 2 e 52 e ¿ n 52 0.002857 0.4 52 0.007143 d 2 d 1 Problem 1.5-4 A prismatic bar of circular cross section is loaded by tensile forces P (see figure). The bar has length L 5 1.5 m and diameter d 5 30 mm. It is made of aluminum alloy with modulus of elasticity E 5 75 GPa and Poisson’s ratio n 5 1 / 3 . If the bar elongates by 3.6 mm, what is the decrease in
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Unformatted text preview: diameter d ? What is the magnitude of the load P ? d P P L Solution 1.5-4 Aluminum bar in tension L 5 1.5 m d 5 30 mm E 5 75 GPa n 5 1 / 3 d 5 3.6 mm (elongation) A XIAL STRAIN L ATERAL STRAIN 52 0.0008 (Minus means decrease in diameter) e 5 2 ne 5 2 ( 1 3 )(0.0024) e 5 d L 5 3.6 mm 1.5 m 5 0.0024 D ECREASE IN DIAMETER d 5 e 9 d 5 (0.0008)(30 mm) 5 0.024 mm A XIAL STRESS s 5 E e 5 (75 GPa)(0.0024) 5 180 MPa (This stress is less than the yield stress, so Hookes law is valid.) L OAD P ( TENSION ) 5 127 kN P 5 s A 5 (180 MPa) p 4 (30 mm) 2 A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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