Let's break down the question: **Question:** > Consider a single crystal loaded with a tensile stress of 10 MPa applied along [010] direction. The slip occurs on (111) plane in [1̅10] direction. Calculate the critical resolved shear stress. --- ### Step 1: Formula for Resolved Shear Stress The resolved shear stress (τ) is given by: \[ \tau = \sigma \cos\phi \cos\lambda \] Where: - \(\sigma\) = applied tensile stress (10 MPa) - \(\phi\) = angle between tensile axis ([010]) and slip plane normal ([111]) - \(\lambda\) = angle between tensile axis ([010]) and slip direction ([1̅10]) --- ### Step 2: Find Cosines #### 1. **Find \(\cos\phi\):** - Tensile axis: [010] - Slip plane normal: [111] \[ \cos\phi = \frac{[010] \cdot [111]}{|[010]||[111]|} \] \[ [010] \cdot [111] = (0*1) + (1*1) + (0*1) = 1 \] \[ |[010]| = \sqrt{0^2 + 1^2 + 0^2} = 1 \] \[ |[111]| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \] \[ \cos\phi = \frac{1}{1 \times \sqrt{3}} = \frac{1}{\sqrt{3}} \] --- #### 2. **Find \(\cos\lambda\):** - Tensile axis: [010] - Slip direction: [1̅10] \[ \cos\lambda = \frac{[010] \cdot [1\bar{1}0]}{|[010]||[1\bar{1}0]|} \] \[ [010] \cdot [1\bar{1}0] = (0*1) + (1*-1) + (0*0) = -1 \] \[ |[1\bar{1}0]| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2} \] \[ \cos\lambda = \frac{-1}{1 \times \sqrt{2}} = -\frac{1}{\sqrt{2}} \] --- ### Step 3: Plug values into the formula \[ \tau = 10 \times \frac{1}{\sqrt{3}} \times \left(-\frac{1}{\sqrt{2}}\right) \] \[ \tau = -10 \times \frac{1}{\sqrt{6}} \] \[ \tau = -10 \times 0.408 = -4.08\, \text{MPa} \] The negative sign just shows direction; the magnitude is what is asked. --- ### **Final Answer:** \[ \boxed{4.083\, \text{MPa}} \] So, the answer is **4.083 MPa** (Option 3).