Let's solve the problem in three steps as requested, using Mohr's Circle. --- ### **Step 1: Identify Stresses and Draw Mohr’s Circle** #### **Given:** - Horizontal normal stress, \(\sigma_x = 50\) MPa (tensile, so positive by convention) - Vertical normal stress, \(\sigma_y = 120\) MPa (tensile, so positive) - Shear stress, \(\tau_{xy} = 0\) MPa (since not indicated in the diagram) #### **Mohr’s Circle Construction:** 1. **Center** of Mohr's Circle: \[ C = \frac{\sigma_x + \sigma_y}{2} = \frac{50 + 120}{2} = 85 \text{ MPa} \] 2. **Radius** of Mohr's Circle: \[ R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{\left(\frac{50 - 120}{2}\right)^2 + 0^2} = \sqrt{(-35)^2} = 35 \text{ MPa} \] **Explanation:** This step sets up the normal and shear stresses and computes the center and radius of Mohr’s Circle, which is essential for finding principal and shear stresses. --- ### **Step 2: Find Principal Stresses and Maximum Shear Stresses** #### **a) Principal Stresses (\(\sigma_1\), \(\sigma_2\)):** \[ \sigma_{1,2} = C \pm R = 85 \pm 35 \] - \(\sigma_1 = 120\) MPa - \(\sigma_2 = 50\) MPa #### **b) Maximum Shear Stress (\(\tau_{max}\)):** \[ \tau_{max} = R = 35 \text{ MPa} \] **Planes:** - Principal planes are at angles \(\theta_p\) where shear is zero. - Max shear planes are at \(45^\circ\) to principal planes. **Explanation:** Here, we directly read off the principal stresses and the max shear from the values found in Step 1, using the properties of Mohr’s Circle. --- ### **Step 3: Stresses on a Plane Oriented 45° Clockwise** For an element rotated \(\theta = 45^\circ\) clockwise (which is \(2\theta = 90^\circ\) on Mohr's Circle): \[ \sigma_\theta = C + R\cos(2\theta) \] \[ \tau_\theta = -R\sin(2\theta) \] Plugging in the values: - \(C = 85\) MPa, \(R = 35\) MPa, \(2\theta = 90^\circ\) - \(\cos(90^\circ) = 0\), \(\sin(90^\circ) = 1\) \[ \sigma_\theta = 85 + 35 \times 0 = 85\, \text{MPa} \] \[ \tau_\theta = -35 \times 1 = -35\, \text{MPa} \] **Explanation:** The normal stress on the 45° plane is 85 MPa, and the shear stress is -35 MPa (negative sign indicates direction). --- ## **Final Answers** ### **a) Principal Stresses** - \(\sigma_1 = 120\, \text{MPa}\) - \(\sigma_2 = 50\, \text{MPa}\) ### **b) Maximum Shear Stress** - \(\tau_{max} = 35\, \text{MPa}\) ### **c) Stresses on a Plane Oriented 45° Clockwise** - Normal stress: \(\sigma_\theta = 85\, \text{MPa}\) - Shear stress: \(\tau_\theta = -35\, \text{MPa}\) Let me know if you want to see the Mohr's Circle or a step-by-step sketch!