01-01ChapGere.0020

01-01ChapGere.0020 - s Y 5 480 MPa L ATERAL STRAIN A XIAL...

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Solution 1.5-1 Steel bar in compression 20 CHAPTER 1 Tension, Compression, and Shear S TEEL BAR d 5 2.00 in. Max. ± d 5 0.001 in. E 5 29 3 10 6 psi n5 0.29 L ATERAL STRAIN A XIAL STRAIN (shortening) e 52 e ¿ n 52 0.0005 0.29 52 0.001724 e ¿ 5 ¢ d d 5 0.001 in. 2.00 in. 5 0.0005 A XIAL STRESS s 5 E e 5 (29 3 10 6 psi)( 2 0.001724) 52 50.00 ksi (compression) Assume that the yield stress for the high-strength steel is greater than 50 ksi. Therefore, Hooke’s law is valid. M AXIMUM COMPRESSIVE LOAD 5 157 k P max 5 s A 5 (50.00 ksi) ¢ p 4 (2.00 in.) 2 Problem 1.5-2 A round bar of 10 mm diameter is made of aluminum alloy 7075-T6 (see figure). When the bar is stretched by axial forces P , its diameter decreases by 0.016 mm. Find the magnitude of the load P . (Obtain the material prop- erties from Appendix H.) d = 10 mm 7075-T6 P P Solution 1.5-2 Aluminum bar in tension d 5 10 mm ± d 5 0.016 mm (Decrease in diameter) 7075-T6 From Table H-2: E 5 72 GPa n5 0.33 From Table H-3: Yield stress
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Unformatted text preview: s Y 5 480 MPa L ATERAL STRAIN A XIAL STRAIN 5 0.004848 (Elongation) e 5 2 2 e ¿ n 5 0.0016 0.33 e ¿ 5 ¢ d d 5 2 0.016 mm 10 mm 5 2 0.0016 A XIAL STRESS s 5 E e 5 (72 GPa)(0.004848) 5 349.1 MPa (Tension) Because s < s Y , Hooke’s law is valid. L OAD P ( TENSILE FORCE ) 5 27.4 kN P 5 s A 5 (349.1 MPa) ¢ p 4 ≤ (10 mm) 2 Problem 1.5-3 A nylon bar having diameter d 1 5 3.50 in. is placed inside a steel tube having inner diameter d 2 5 3.51 in. (see figure). The nylon bar is then compressed by an axial force P . At what value of the force P will the space between the nylon bar and the steel tube be closed? (For nylon, assume E 5 400 ksi and n 5 0.4.) d 2 d 1 Steel tube Nylon bar A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Tech.

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