HW-04%20Solutions - #1 7.13 We are asked to compute the...

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7.13 We are asked to compute the critical resolved shear stress for Zn. As stipulated in the problem, φ = 65 ° , while possible values for λ are 30 ° , 48 ° , and 78 ° . (a) Slip will occur along that direction for which (cos φ cos λ ) is a maximum, or, in this case, for the largest cos λ . Cosines for the possible λ values are given below. cos(30 ° ) = 0.87 cos(48 ° ) = 0.67 cos(78 ° ) = 0.21 Thus, the slip direction is at an angle of 30 ° with the tensile axis. (b) From Equation 7.4, the critical resolved shear stress is just τ crss = σ y (cos φ cos λ ) max = (2.5 MPa) cos(65 ° ) cos( 30° ) [ ] = 0.90 MPa (130 psi)
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7.24 We are asked to determine the grain diameter for an iron which will give a yield strength of 310 MPa (45,000 psi). The best way to solve this problem is to first establish two simultaneous expressions of Equation 7.7, solve for σ 0 and k y , and finally determine the value of d when σ y = 310 MPa. The data pertaining to this problem may be tabulated as follows: σ y d (mm) d -1/2 (mm) -1/2 230 MPa 1 x 10 -2 10.0 275 MPa 6 x 10 -3 12.91
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This note was uploaded on 09/15/2010 for the course MATSCIE 250 taught by Professor Yalisove during the Fall '08 term at University of Michigan.

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HW-04%20Solutions - #1 7.13 We are asked to compute the...

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