The question asks about calculating the **critical resolved shear stress** for a single crystal loaded with a tensile stress of 10 MPa applied along the [100] direction, with slip occurring on the (111) plane in the [1̅10] direction. Let's break down the steps: ### Step 1: **Schmid's Law** The resolved shear stress (τ) is given by Schmid’s Law: \[ \tau = \sigma \cdot \cos\phi \cdot \cos\lambda \] Where: - \( \sigma \) = Applied tensile stress (10 MPa) - \( \phi \) = angle between the tensile axis ([100]) and the slip plane normal ([111]) - \( \lambda \) = angle between the tensile axis ([100]) and the slip direction ([1̅10]) --- ### Step 2: **Find the Angles** #### a) **Angle \( \phi \) ([100] and [111])** Cosine of the angle: \[ \cos\phi = \frac{[100] \cdot [111]}{|[100]||[111]|} \] - [100] = (1,0,0) - [111] = (1,1,1) Dot product: \( 1*1 + 0*1 + 0*1 = 1 \) \(|[100]| = 1\) \(|[111]| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\) So, \[ \cos\phi = \frac{1}{\sqrt{3}} \] #### b) **Angle \( \lambda \) ([100] and [1̅10])** - [1̅10] = (1,-1,0) Dot product: \( 1*1 + 0*(-1) + 0*0 = 1 \) \(|[1̅10]| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}\) So, \[ \cos\lambda = \frac{1}{\sqrt{2}} \] --- ### Step 3: **Calculate the Resolved Shear Stress** \[ \tau = \sigma \cdot \cos\phi \cdot \cos\lambda \] \[ \tau = 10 \cdot \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}} \] \[ \tau = 10 \cdot \frac{1}{\sqrt{6}} \] \[ \tau = 10 \cdot 0.4082 = 4.082 \text{ MPa} \] --- ### Step 4: **Select the Closest Answer** The closest answer is: **4.0823 MPa** --- #### **Final Answer:** **4.0823 MPa** is the critical resolved shear stress.