Certainly! Let's solve **Question 1(a)**, step by step as requested. --- ## **Step 1: Find the principal stresses and directions** Given: - \(\sigma_x = -80 \text{ MPa}\) - \(\sigma_y = 120 \text{ MPa}\) - \(\tau_{xy} = 30 \text{ MPa (cw)}\) The formulas for principal stresses (\(\sigma_1, \sigma_2\)) are: \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] Plug in the values: \[ \frac{\sigma_x + \sigma_y}{2} = \frac{-80 + 120}{2} = 20 \text{ MPa} \] \[ \frac{\sigma_x - \sigma_y}{2} = \frac{-80 - 120}{2} = -100 \text{ MPa} \] Now, \[ \sigma_{1,2} = 20 \pm \sqrt{(-100)^2 + (30)^2} = 20 \pm \sqrt{10000 + 900} = 20 \pm \sqrt{10900} = 20 \pm 104.4 \] So, - \(\sigma_1 = 20 + 104.4 = 124.4 \) MPa - \(\sigma_2 = 20 - 104.4 = -84.4 \) MPa **Explanation:** We used the standard principal stress formula, plugged in the given values, and computed the two principal stresses. --- ## **Step 2: Find the maximum shear stress and the associated normal stress** The maximum shear stress (\(\tau_{max}\)) is: \[ \tau_{max} = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] We already calculated this in the previous step: \[ \tau_{max} = \sqrt{10000 + 900} = \sqrt{10900} = 104.4 \text{ MPa} \] The associated normal stress (\(\sigma_{avg}\)) at this plane is: \[ \sigma_{avg} = \frac{\sigma_x + \sigma_y}{2} = 20 \text{ MPa} \] **Explanation:** The maximum shear stress is the radius of Mohr's circle, and the associated normal stress is the center of Mohr's circle (the average normal stress). --- ## **Final Answers:** ### (i) Principal Stresses: - \(\sigma_1 = 124.4 \) MPa - \(\sigma_2 = -84.4 \) MPa ### (ii) Maximum Shear and Associated Normal Stress: - \(\tau_{max} = 104.4 \) MPa - Associated normal stress = \(20 \) MPa --- If you want to continue with **part (iii) (Mohr's circle)** or part (b), let me know!

Sure! Let's solve all parts of the problem step by step. --- # **Question 1: Stress Analysis** ## **Given:** - \(\sigma_x = -80 \text{ MPa}\) - \(\sigma_y = 120 \text{ MPa}\) - \(\tau_{xy} = 30 \text{ MPa (cw)}\) ## **Part (a): Principal Stresses and Directions** ### **Step 1: Calculate Principal Stresses** The formulas for principal stresses (\(\sigma_1, \sigma_2\)) are: \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] **Calculations:** 1. Calculate the average stress: \[ \frac{\sigma_x + \sigma_y}{2} = \frac{-80 + 120}{2} = 20 \text{ MPa} \] 2. Calculate half the difference in normal stresses: \[ \frac{\sigma_x - \sigma_y}{2} = \frac{-80 - 120}{2} = -100 \text{ MPa} \] 3. Now compute the principal stresses: \[ \sigma_{1,2} = 20 \pm \sqrt{(-100)^2 + (30)^2} = 20 \pm \sqrt{10000 + 900} = 20 \pm \sqrt{10900} = 20 \pm 104.4 \] Thus, we find: - \(\sigma_1 = 124.4 \) MPa - \(\sigma_2 = -84.4 \) MPa **Explanation:** We derived the principal stresses using the formulas for stress transformation, which involved finding the mean and the radius of Mohr's circle. ### **Step 2: Calculate Maximum Shear Stress and Associated Normal Stress** The maximum shear stress (\(\tau_{max}\)) is calculated as: \[ \tau_{max} = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] **Calculations:** 1. Substitute the values: \[ \tau_{max} = \sqrt{(-100)^2 + (30)^2} = \sqrt{10000 + 900} = \sqrt{10900} = 104.4 \text{ MPa} \] 2. The associated normal stress (\(\sigma_{avg}\)): \[ \sigma_{avg} = \frac{\sigma_x + \sigma_y}{2} = 20 \text{ MPa} \] **Explanation:** The maximum shear stress represents the radius of Mohr's circle, while the associated normal stress is the average of the two normal stresses. --- ## **Part (b): Mohr's Circle Construction** ### **Step 1: Plotting Mohr's Circle** 1. **Center of Mohr's Circle**: \[ C = \sigma_{avg} = 20 \text{ MPa} \] 2. **Radius**: \[ R = \tau_{max} = 104.4 \text{ MPa} \] ### **Step 2: Circle Points** 1. **Left Point**: \[ P_1 = \sigma_1 = 124.4 \text{ MPa} \quad \text{(Principal stress)} \] 2. **Right Point**: \[ P_2 = \sigma_2 = -84.4 \text{ MPa} \quad \text{(Principal stress)} \] **Explanation:** Mohr's Circle helps visualize the relationship between normal and shear stresses. The center is at the average normal stress, and the radius corresponds to the maximum shear stress. --- ## **Final Summary:** ### **Principal Stresses:** - \(\sigma_1 = 124.4 \) MPa - \(\sigma_2 = -84.4 \) MPa ### **Maximum Shear Stress:** - \(\tau_{max} = 104.4 \) MPa - Associated normal stress = \(20 \) MPa ### **Mohr's Circle:** - Center: \(20 \text{ MPa}\) - Radius: \(104.4 \text{ MPa}\) If you have any further questions or need additional explanations, feel free to ask!