Homework 8 Solutions

# Homework 8 Solutions - ENMA 300 Homework 8 Solutions 1a The...

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ENMA 300 Homework 8 Solutions 1a) The critical resolved shear stress for copper is 0.48 MPa (70 psi). Determine the minimum possible yield strength for a single crystal of Cu pulled in tension. In order to determine the minimum possible yield strength for a single crystal of Cu pulled in tension, we simply employ Equation 8.5 as σ y = 2 τ crss = (2)(0.48 MPa) = 0.96 MPa (140 psi) bi) 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. This part of the problem asks, for a BCC metal, that we compute the resolved shear stress in the [1 1 1] direction on each of the (110), (011), and (10 1 ) planes. In order to solve this problem it is necessary to employ Equation 8.2, which means that we first need to solve for the for angles λ and φ for the three slip systems. For each of these three slip systems, the λ will be the same—i.e., the angle between the direction of the applied stress, [100] and the slip direction, [1 1 1] . This angle λ may be determined using Equation 8.6 λ = cos - 1 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 1 1] ) u 2 = 1, v 2 = –1, w 2 = 1. Therefore, λ is determined as λ = cos - 1 (1)(1) + (0)( - 1) + (0)(1) (1) 2 + (0) 2 + (0) 2 [ ] (1) 2 + ( - 1) 2 + (1) 2 [ ] = cos - 1 1 3 = 54.7 ° Let us now determine 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. Again, 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]), φ is equal to

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φ [100] - [110] = cos - 1 (1)(1) + (0)(1) + (0)(0) (1) 2 + (0) 2 + (0) 2 [ ] (1) 2 + (1) 2 + (0) 2 [ ] = cos - 1 1 2 = 45 ° Now, using Equation 8.2 τ R = σ cos φ cos λ we solve for the resolved shear stress for this slip system as τ R (110) - [1 1 1] = (4.0 MPa) cos (45 ° ) cos(54.7 ° ) [ ] = (4.0 MPa) (0.707)(0.578) = 1.63 MPa Now, we must determine the value of φ for the (011)– [1 1 1] slip system—that is, the angle between the direction of the applied stress, [100], and the normal to the (011) plane—i.e., the [011] direction. Again using Equation 8.6 λ [100] - [ 011] = cos - 1 (1)(0) + (0)(1) + (0)(1) (1) 2 + (0) 2 + (0) 2 [ ] (0) 2 + (1) 2 + (1) 2 [ ] = cos - 1 (0) = 90 ° Thus, the resolved shear stress for this (011)– [1 1 1] slip system is τ R (011) - [1 1 1] = (4.0 MPa) cos (90 ° ) cos (54.7 ° ) [ ] = (4.0 MPa) (0)(0.578) = 0 MPa And, finally, it is necessary to determine the value of φ for the (10 1 ) [1 1 1] slip system —that is, the angle between the direction of the applied stress, [100], and the normal to the (10 1 ) plane—i.e., the [10 1 ] direction. Again using Equation 8.6 λ [100] - [10 1 ] = cos - 1 (1)(1) + (0)(0) + (0)( - 1) (1) 2 + (0) 2 + (0) 2 [ ] (1) 2 + (0) 2 + ( - 1) 2 [ ]
= cos - 1 1 2 = 45 ° Here, as with the (110)– [1 1 1] slip system above, the value of φ is 45 ° , which again leads to τ R (10 1 ) - [1 1 1] = (4.0 MPa) cos (45 ° ) cos (54.7 ° ) [ ] = (4.0 MPa) (0.707)(0.578) = 1.63 MPa

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