Let's solve the problem step by step: ### **Given:** - Normal stress on x-face, \(\sigma_x = 90 \) MPa (tension, rightward) - Normal stress on y-face, \(\sigma_y = -60 \) MPa (compression, downward) - Shear stress, \(\tau_{xy} = 30 \) MPa (direction shown) ### **Required:** (a) Principal stresses (\(\sigma_1\) and \(\sigma_2\)) (b) Maximum in-plane shear stress (\(\tau_{max}\)) --- ## **Step 1: Principal Stresses Formula** The principal stresses are calculated as: \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \] --- ## **Step 2: Plug in the Values** - \(\sigma_x = 90\) MPa - \(\sigma_y = -60\) MPa - \(\tau_{xy} = 30\) MPa First, calculate the average normal stress: \[ \frac{\sigma_x + \sigma_y}{2} = \frac{90 + (-60)}{2} = \frac{30}{2} = 15\ \text{MPa} \] Now, calculate the radius (term inside the square root): \[ \frac{\sigma_x - \sigma_y}{2} = \frac{90 - (-60)}{2} = \frac{150}{2} = 75\ \text{MPa} \] \[ \text{So,} \quad \sqrt{(75)^2 + (30)^2} = \sqrt{5625 + 900} = \sqrt{6525} \approx 80.78\ \text{MPa} \] --- ## **Step 3: Calculate Principal Stresses** \[ \sigma_1 = 15 + 80.78 = \boxed{95.78\ \text{MPa}} \] \[ \sigma_2 = 15 - 80.78 = \boxed{-65.78\ \text{MPa}} \] --- ## **Step 4: Maximum In-plane Shear Stress** \[ \tau_{max} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \] \[ \tau_{max} = \sqrt{(75)^2 + (30)^2} = \sqrt{6525} \approx \boxed{80.78\ \text{MPa}} \] --- ## **Final Answers:** ### **Principal Stresses:** - \(\sigma_1 = \boxed{95.78\ \text{MPa}}\) - \(\sigma_2 = \boxed{-65.78\ \text{MPa}}\) ### **Maximum In-Plane Shear Stress:** - \(\tau_{max} = \boxed{80.78\ \text{MPa}}\) --- **Let me know if you need the orientation of the principal planes or a Mohr's circle sketch!**