This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Youngs modulus for steel is 2 10 11 N / m 2 . Correct answer: 6 . 87764 mm. Explanation: Let : L = 16 . 8 m , L = 9 . 24 mm , F = 4086 . 6 N , and Y = 2 10 11 N / m 2 . Y = F L A L A = F L Y L = r 2 r 2 = F L Y L d 2 = radicalbigg F L Y L d = 2 radicalbigg F L Y L = 2 radicalBigg (4086 . 6 N)(16 . 8 m) (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 r 2 = 4 m g d 2 = 4(417 kg)(9 . 8 m / s 2 ) (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 d 2 4 d 2 = 4 m g S d = radicalbigg 4 m g S = radicalBigg 4 (11 . 2 kg) (9 . 8 m / s 2 ) (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....
View Full
Document
 Spring '11
 sdsd

Click to edit the document details