Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is \(2.4 \mathrm{MPa}\).

Normalize the direction vector [121] and slip direction vector [111] by dividing each component by the square root of the sum of the squares of the components. $$ \text{Normalized [121]} = \frac{[1,2,1]}{\sqrt{1^2 + 2^2 + 1^2}} = \frac{[1,2,1]}{\sqrt{6}} = [\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}], $$ $$ \text{Normalized [111]} = \frac{[1,1,1]}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{[1,1,1]}{\sqrt{3}} = [\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}]. $$

Calculate the Schmid factor using the normalized direction vector [121] and the normalized slip direction vector [111] by using the formula: $$ \text{Schmid factor} = \text{cos}(\alpha) \cdot \text{cos}(\beta) = \frac{\text{Normalized [121]} \cdot \text{Normalized (101)}}{\|\text{Normalized [121]}\| \cdot \|\text{Normalized (101)}\|} \cdot \frac{\text{Normalized [111]} \cdot \text{Normalized (101)}}{\|\text{Normalized [111]}\| \cdot \|\text{Normalized (101)}\|} $$ where \(\alpha\) is the angle between [121] and (101), and \(\beta\) is the angle between [111] and (101). Since we already have the normalized vectors, we can directly compute the dot products and the Schmid factor. Dot product of Normalized [121] and Normalized (101): \((\frac{1}{\sqrt{6}})(-\frac{1}{\sqrt{2}}) + (\frac{2}{\sqrt{6}})() + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{2}}) = -\frac{1}{\sqrt{12}}\) Dot product of Normalized [111] and Normalized (101): \((\frac{1}{\sqrt{3}})(-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{3}})(0) + (\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{2}}) = 0\) So, the Schmid factor is: $$ \text{Schmid factor} = (-\frac{1}{\sqrt{12}}) \cdot 0 = 0 $$

Use the Schmid factor and the given critical resolved shear stress to calculate the stress at which the crystal yields: $$ \text{Stress at which crystal yields} = \frac{\text{Critical resolved shear stress}}{\text{Schmid factor}} = \frac{2.4 \mathrm{MPa}}{0} $$ However, because the Schmid factor is equal to zero, it means that the applied stress in the [121] direction is not resolved along the (101) slip plane in the [111] direction. As a result, in this case, the stress at which the crystal yields cannot be determined.