A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is \(0.5 \mathrm{MPa}\), calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [101], [10\overline{1} ] \text { , and } [ 0 \overline { 1 1 } ] \text { } directions.

The critical resolved shear stress (CRSS) is the minimum shear stress required to cause slip on a specific crystallographic plane in a specific direction. We need to calculate the applied stress necessary for this situation. The relationship between the applied stress and critical resolved shear stress can be expressed using Schmid's Law, which states: $$ \tau_{R}=\sigma\cos\phi\cos\lambda $$ where \(\tau_{R}\) is the resolved shear stress, \(\sigma\) is the applied stress, \(\phi\) is the angle between the applied stress direction and slip plane normal, and \(\lambda\) is the angle between the applied stress direction and slip direction.

Now, let's calculate the angles \(\phi\) and \(\lambda\) for each slip direction, [101], [10\(\overline{1}\)], and [0\(\overline{1}\)1]. In the FCC crystal structure, the slip plane normal for the (111) plane can be represented as the vector [1, 1, 1]. The tensile stress direction is given as [100]. [101]: - \(\phi\) = angle between [100] and [1, 1, 1]. \(\phi\) = cos\(^{-1}(\frac{\text{dot}(100, 111)}{|\textbf{100}||\textbf{111}|}) = 54.7^{\circ}\) - \(\lambda\) = angle between [100] and [101]. \(\lambda\) = cos\(^{-1}(\frac{\text{dot}(100,101)}{|\textbf{100}||\textbf{101}|}) = 45^{\circ}\) [10\(\overline{1}\)]: - \(\phi\) = angle between [100] and [1, 1, 1]. \(\phi\) = cos\(^{-1}(\frac{\text{dot}(100, 111)}{|\textbf{100}||\textbf{111}|}) = 54.7^{\circ}\) - \(\lambda\) = angle between [100] and [10\(\overline{1}\)]. \(\lambda\) = cos\(^{-1}(\frac{\text{dot}(100,10\overline{1})}{|\textbf{100}||\textbf{10\overline{1}}|}) = 45^{\circ}\) [0\(\overline{1}\)1]: - \(\phi\) = angle between [100] and [1, 1, 1]. \(\phi\) = cos\(^{-1}(\frac{\text{dot}(100, 111)}{|\textbf{100}||\textbf{111}|}) = 54.7^{\circ}\) - \(\lambda\) = angle between [100] and [0\(\overline{1}\)1]. \(\lambda\) = cos\(^{-1}(\frac{\text{dot}(100,0\overline{1}1)}{|\textbf{100}||\textbf{0\overline{1}1}|}) = 90^{\circ}\)

Now we have \(\phi\) and \(\lambda\) for each slip direction. We can plug these values into Schmid's Law and calculate the applied stress for each direction. The critical resolved shear stress is given as 0.5 MPa. [101]: $$ 0.5 \mathrm{MPa}=\sigma\cos(54.7^{\circ})\cos(45^{\circ}) $$ $$ \sigma = \frac{0.5 \mathrm{MPa}}{\cos(54.7^{\circ})\cos(45^{\circ})} = 1.00 \ \mathrm{MPa} $$ [10\(\overline{1}\)]: $$ 0.5 \mathrm{MPa}=\sigma\cos(54.7^{\circ})\cos(45^{\circ}) $$ $$ \sigma = \frac{0.5 \mathrm{MPa}}{\cos(54.7^{\circ})\cos(45^{\circ})} = 1.00 \ \mathrm{MPa} $$ [0\(\overline{1}\)1]: $$ 0.5 \mathrm{MPa}=\sigma\cos(54.7^{\circ})\cos(90^{\circ}) $$ $$ \sigma = \frac{0.5 \mathrm{MPa}}{\cos(54.7^{\circ})\cos(90^{\circ})} \to \text{ Undefined (Slip will not occur)} $$ So, the applied tensile stress necessary to cause slip on the (111) plane are: - 1.00 MPa in the [101] direction - 1.00 MPa in the [10\(\overline{1}\)] direction - Slip will not occur in the [0\(\overline{1}\)1] direction.