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Unformatted text preview: EMSE 201 – Introduction to Materials Science & Engineering Fall 2010 Case Western Reserve University 1 of 4 Case School of Engineering Solution to Homework #6 Explain how answers were obtained. Give appropriate units for all numerical answers. 1) a) (8 points) A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the [100] direction. If the magnitude of this stress is 4.0 MPa, compute the resolved shear stress in the [1 1 1] direction on each of the (110), (011), and (10 1 ) planes. b) (1 point) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably oriented? (C&R problem 8.8) Solution (modified from the publisher’s solution manual; used with permission) a) For each of these three slip systems, the value of cos λ will be the same—i.e., the cosine of the angle between the direction of the applied stress ([100]) and the slip direction ( [1 11] ). These two diretions can be seen in a (011) plane: From the figure, it can be determined that cos λ = 1/ 3 . Another approach is to use equation 8.6 to determine cos λ and cos φ for the three slip systems. cos λ = u 1 u 2 + v 1 v 2 + w 1 w 2 u 1 2 + v 1 2 + w 1 2 ( ) u 2 2 + v 2 2 + w 2 2 ( ) where (for [100]) u 1 = 1 , v 1 = 0, w 1 = 0, and (for [1 11] ) u 2 = 1, v 2 = –1, w 2 = 1. Therefore, cos λ is determined as cos λ = (1)(1) + (0)( − 1) + (0)(1) (1) 2 + (0) 2 + (0) 2 ⎡ ⎣ ⎤ ⎦ (1) 2 + ( − 1) 2 + (1) 2 ⎡ ⎣ ⎤ ⎦ = 1 3 (Either way of determining cos λ is worth 2 pts .) Now determine cos φ for the angle between the direction of the applied tensile stress — i.e., the [100] direction—and the normal to the (110) slip plane—i.e., the [110] direction. Knowing that these two directions form the edge and diagonal of a (square) face of the unit cell, φ = 45° and cos φ is 1/ 2 . Or, using equation 8.6 where u 1 = 1 , v 1 = 0, w 1 = 0 (for [100]), and u 2 = 1, v 2 = 1, w 2 = 0 (for [110]), cos...
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This document was uploaded on 11/08/2011.
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