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Unformatted text preview: le (txl65) – Elasticity – mohanty – (81440) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Find the minimum diameter of a steel wire 16 . 8 m long that will elongate no more than 9 . 24 mm when a load of 417 kg is hung on the lower end. The acceleration of gravity is 9 . 8 m / s 2 and Young’s modulus for steel is 2 × 10 11 N / m 2 . Correct answer: 6 . 87764 mm. Explanation: Let : L = 16 . 8 m , Δ L = 9 . 24 mm , F = 4086 . 6 N , and Y = 2 × 10 11 N / m 2 . Y = F L A Δ L A = F L Y Δ L = π r 2 r 2 = F L π Y Δ L d 2 = radicalbigg F L π Y Δ L d = 2 radicalbigg F L π Y Δ L = 2 radicalBigg (4086 . 6 N)(16 . 8 m) π (2 × 10 11 N / m 2 )(9 . 24 mm) × 1000 mm 1 m = 6 . 87764 mm . 002 (part 2 of 2) 10.0 points What is the shear stress for such a load? Correct answer: 1 . 1 × 10 8 N / m 2 . Explanation: S = F A = m g π r 2 = 4 m g π d 2 = 4(417 kg)(9 . 8 m / s 2 ) π (6 . 87764 mm) 2 parenleftbigg 1000 mm 1 m parenrightbigg 2 = 1 . 1 × 10 8 N / m 2 . 003 10.0 points Determine the minimum diameter a copper wire can have under a load of 11 . 2 kg if its elastic limit is not to be exceeded. The elas tic limit of copper is 1 × 10 8 N / m 2 and the acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 00118216 m. Explanation: Let : m = 11 . 2 kg , g = 9 . 8 m / s 2 , and S = 1 × 10 8 N / m 2 . Stress = Force Area The maximum stress is S = F A = m g π d 2 4 d 2 = 4 m g π S d = radicalbigg 4 m g π S = radicalBigg 4 (11 . 2 kg) (9 . 8 m / s 2 ) π (1 × 10 8 N / m 2 ) = . 00118216 m . 004 (part 1 of 2) 10.0 points Assume that bone will fracture if a shear stress more than 1 . 23 × 10 8 N / m 2 is exerted....
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This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.
 Spring '11
 sdsd

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