1M03 2009 ASSIGNMENT 4-2 ¿¯

1M03 2009 ASSIGNMENT 4-2...

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View Results Name: SOO NAM Attempt: 2 / 3 Out of: 19 Started: March 9, 2009 11:11pm Finished: March 9, 2009 11:27pm Time spent: 15 min. 55 sec. Question 1 (1 point) Consider a brass alloy the stress-strain behavior of which is shown below. A cylindrical specimen of this alloy 20 mm in diameter and 188 mm long is to be pulled in tension. What is the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter? Assume a value of 0.34 for Poisson's ratio. Student response: Percent Value Correct Response Student Response Answer Choices 0.0% a. 120 MPa 100.0% b. 135 MPa 0.0% c. 155 MPa 0.0% d. 160 MPa General feedback: The transverse strain (x or y) for a cylinder of diameter d o is: You can calculate the longitudinal strain by using Poisson's ratio: which is the negative ratio of transverse (x and y) and longitudinal (z) strains. Intercepting the longitudinal strain with the stress-strain curve can show the amount of stress needed to cause the specified reduction in diameter. For more information, see the section on Elastic Properties of Materials in the text. Score: 0 / 1 Question 2 (2 points)
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A specimen of a metal has a rectangular cross section of 10.4 mm x 12.8 mm. It's pulled in tension with a force of 37258 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 72.3 GPa, calculate the resulting strain . Use scientific notation, significant figures in answer: 3 Student response: 3.87e-3 Correct answer: 3.87E-3 (3.87 * 10 -3 ) General feedback: Hooke's law is: The applied stress is equal to: Remember to watch your units. For more information, see the section on Stress-Strain Behaviour in the text. Score:
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This note was uploaded on 04/04/2011 for the course MATLS 1M03 taught by Professor Okon during the Spring '09 term at McMaster University.

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1M03 2009 ASSIGNMENT 4-2...

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