01-01ChapGere.0023

01-01ChapGere.0023 - Hollow steel cylinder d 1 5 3.9 in. d...

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Problem 1.5-7 A hollow steel cylinder is compressed by a force P (see figure). The cylinder has inner diameter d 1 5 3.9 in., outer diameter d 2 5 4.5 in., and modulus of elasticity E 5 30,000 ksi. When the force P increases from zero to 40 k, the outer diameter of the cylinder increases by 455 3 10 2 6 in. (a) Determine the increase in the inner diameter. (b) Determine the increase in the wall thickness. (c) Determine Poisson’s ratio for the steel. SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 23 Solution 1.5-6 Brass specimen in tension d 5 10 mm Gage length L 5 50 mm P 5 20 kN d 5 0.122 mm ± d 5 0.00830 mm A XIAL STRESS Assume s is below the proportional limit so that Hooke’s law is valid. A XIAL STRAIN e 5 d L 5 0.122 mm 50 mm 5 0.002440 s 5 P A 5 20 kN p 4 (10 mm) 2 5 254.6 MPa (a) M ODULUS OF ELASTICITY (b) P OISSON SRATIO e 95n e ± d 5 e 9 d 5n e d n 5 ¢ d e d 5 0.00830 mm (0.002440)(10 mm) 5 0.34 E 5 s e 5 254.6 MPa 0.002440 5 104 GPa P d 1 d 2 Solution 1.5-7
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Unformatted text preview: Hollow steel cylinder d 1 5 3.9 in. d 2 5 4.5 in. t 5 0.3 in. E 5 30,000 ksi P 5 40 k (compression) d 2 5 455 3 10 2 6 in. (increase) L ATERAL STRAIN (a) I NCREASE IN INNER DIAMETER (b) I NCREASE IN WALL THICKNESS 5 30 3 10 2 6 in. t 5 e t 5 (0.0001011)(0.3 in.) 5 394 3 10 2 6 in. d 1 5 e d 1 5 (0.0001011)(3.9 in.) e 5 d 2 d 2 5 455 3 10 2 6 in. 4.5 in. 5 0.0001011 (c) P OISSON S RATIO Axial stress: (compression) ( s s Y ; Hookes law is valid) Axial strain: 5 0.30 n 5 e e 5 0.0001011 0.000337 5 0.000337 e 5 s E 5 10.105 ksi 30,000 ksi 5 10.105 ksi s 5 P A 5 40 k 3.9584 in. 2 5 3.9584 in. 2 A 5 p 4 [ d 2 2 2 d 1 2 ] 5 p 4 [ (4.5 in.) 2 2 (3.9 in.) 2 ] s 5 P A d 2 d 1 t A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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