01-01ChapGere.0003 - 2 5 4074 psi g L 5 (490 lb / ft 3...

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Solution 1.2-3 Long steel rod in tension Problem 1.2-4 A circular aluminum tube of length L 5 400 mm is loaded in compression by forces P (see figure). The out- side and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction. (a) If the measured strain is e 5 550 3 10 2 6 , what is the shortening d of the bar? (b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P ? Solution 1.2-4 Aluminum tube in compression SECTION 1.2 Normal Stress and Strain 3 5 374.3 psi s max 5 374 psi 1 4074 psi 5 4448 psi Rounding, we get s max 5 4450 psi P A 5 200 lb p 4 (0.25 in.)
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Unformatted text preview: 2 5 4074 psi g L 5 (490 lb / ft 3 )(110 ft) ¢ 1 144 ft 2 in. 2 ≤ s max 5 W 1 P A 5 g L 1 P A d L P = 200 lb P 5 200 lb L 5 110 ft d 5 1 / 4 in. Weight density: g 5 490 lb±ft 3 W 5 Weight of rod 5 g (Volume) 5 g AL Strain gage L = 400 mm P P e 5 550 3 10 2 6 L 5 400 mm d 2 5 60 mm d 1 5 50 mm (a) S HORTENING d OF THE BAR d 5 e L 5 (550 3 10 2 6 )(400 mm) 5 0.220 mm (b) C OMPRESSIVE LOAD P s 5 40 MPa 5 863.9 mm 2 P 5 s A 5 (40 MPa)(863.9 mm 2 ) 5 34.6 kN A 5 p 4 [ d 2 2 2 d 1 2 ] 5 p 4 [ (60 mm) 2 2 (50 mm) 2 ] Strain gage P P 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 Institute of Technology.

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