(a) 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 \(4.0 \mathrm{MPa}, \mathrm{com}\) pute the resolved shear stress in the \([1 \overline{1} 1]\) di rection on each of the \((110),(011),\) and \((10 \overline{1})\) planes (b) On the basis of these resolved shear stress values, which slip system(s) is (are) most favorably oriented?

First, we need to find the angles between the applied tensile stress in the [100] direction and the slip directions [111]. The dot product of these two directions gives: \(\text{cos}(\alpha) = \frac{[\text{100}] \cdot [\text{1} \overline{\text{1}} \text{1}]}{||\text{[100]}|| ||\text{[1} \overline{1} \text{1]}||} = \frac{0}{\sqrt{1} \cdot \sqrt{3}} = 0\) Thus, \(\alpha = 90^{\circ}\).

Next, we need to find the angles between the slip direction [111] and the normal to the planes (110), (011), and \((10 \overline{1})\). For the (110) plane: \(\text{cos}(\beta_1) = \frac{[\text{1} \text{0} \text{0}] \cdot [\text{1} \text{1} \text{0}]}{||\text{[1} \text{0} \text{0]}|| || [\text{1} \text{1} \text{0}] ||} = \frac{1}{\sqrt{1} \cdot \sqrt{2}} = \frac{1}{\sqrt{2}}\) For the (011) plane: \(\text{cos}(\beta_2) = \frac{[\text{1} \text{0} \text{0}] \cdot [\text{0} \text{1} \text{1}]}{||\text{[1} \text{0} \text{0]}|| || [\text{0} \text{1} \text{1}] ||} = 0\) For the \((10 \overline{1})\) plane: \(\text{cos}(\beta_3) = \frac{[\text{1} \text{0} \text{0}] \cdot [\text{1} \text{0} \overline{\text{1}}]}{||\text{[1} \text{0} \text{0]}|| || [\text{1} \text{0} \overline{\text{1}}] ||} = 1\) Now, let's compute the resolved shear stress using the Schmid's Law.

For the \((110)\) plane: \(\tau_1 = \sigma \text{cos}(\alpha) \text{cos}(\beta_1) = 4\times 10^{6} \text{Pa} \cdot 0 \cdot \frac{1}{\sqrt{2}} = 0\) For the \((011)\) plane: \(\tau_2 = \sigma \text{cos}(\alpha) \text{cos}(\beta_2) = 4\times 10^{6} \text{Pa} \cdot 0 \cdot 0 = 0\) For the \((10 \overline{1})\) plane: \(\tau_3 = \sigma \text{cos}(\alpha) \text{cos}(\beta_3) = 4\times 10^{6} \text{Pa} \cdot 0 \cdot 1 = 0\)

As we can see from our calculations, all three planes have the same resolved shear stress of zero, which means that none of them is prone to slip under the applied tensile stress in the [100] direction.