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 Hooke’s 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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- Spring '08
- Armanios
- 1.5 m, 0.024 mm, 3 3.6 mm
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