## Problem 1 ### **Step 1: Identify Stresses and Construct Mohr's Circle** Given: - \(\sigma_x = 15 \) ksi (horizontal) - \(\sigma_y = 6 \) ksi (vertical) - \(\tau_{xy} = \) (no shear shown on the element) We rotate the element **30° clockwise** (\(\theta = -30^\circ\)). #### **Mohr's Circle:** - Center: \(C = \frac{\sigma_x + \sigma_y}{2} = \frac{15 + 6}{2} = 10.5\) ksi - Radius: \(R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{\left(\frac{15-6}{2}\right)^2} = \sqrt{4.5^2} = 4.5\) ksi **Explanation:** We first find the average normal stress (center) and the radius (using the difference in normal stresses and shear). --- ### **Step 2: Find Stresses After Rotation** On Mohr's Circle, a physical rotation of \(\theta\) corresponds to a movement of \(2\theta\) on the circle. For \(\theta = -30^\circ\), move \(-60^\circ\) on the circle. - **Normal stress**: \[ \sigma_\theta = C + R \cos(2\theta) \] \[ \sigma_{-30^\circ} = 10.5 + 4.5 \cos(-60^\circ) \] \[ \cos(-60^\circ) = .5 \] \[ \sigma_{-30^\circ} = 10.5 + 4.5 \times .5 = 10.5 + 2.25 = 12.75\ \text{ksi} \] - **Shear stress**: \[ \tau_\theta = -R \sin(2\theta) \] \[ \tau_{-30^\circ} = -4.5 \sin(-60^\circ) \] \[ \sin(-60^\circ) = -.866 \] \[ \tau_{-30^\circ} = -4.5 \times (-.866) = 3.897\ \text{ksi} \] **Explanation:** We use the Mohr's Circle formulas for normal and shear stress after rotation, plugging in the values. --- ### **Final Answer (Problem 1):** - **Normal stress after 30° clockwise rotation:** \(\boxed{12.75\ \text{ksi}}\) - **Shear stress after 30° clockwise rotation:** \(\boxed{3.90\ \text{ksi}}\) (rounded to two decimals) --- ## Problem 2 ### **Step 1: Identify Stresses and Construct Mohr's Circle** Given: - \(\sigma_x = 60\) MPa - \(\sigma_y = 30\) MPa - \(\tau_{xy} = 45\) MPa Find (a) principal stresses and (b) max in-plane shear stresses. #### **Mohr's Circle:** - Center: \(C = \frac{60 + 30}{2} = 45\) MPa - Radius: \(R = \sqrt{\left(\frac{60-30}{2}\right)^2 + (45)^2} = \sqrt{(15)^2 + (45)^2} = \sqrt{225 + 2025} = \sqrt{225} = 47.43\) MPa **Explanation:** We calculate center and radius for Mohr’s Circle using provided stresses. --- ### **Step 2: Compute Principal and Maximum Shear Stresses** #### **Principal Stresses**: \[ \sigma_{1,2} = C \pm R = 45 \pm 47.43 \] \[ \sigma_1 = 92.43\ \text{MPa} \quad \sigma_2 = -2.43\ \text{MPa} \] #### **Maximum Shear Stress**: \[ \tau_{max} = R = 47.43\ \text{MPa} \] - The corresponding normal stress at maximum shear is \(C = 45\) MPa. **Orientation:** - Principal planes: \(\theta_p = \frac{1}{2} \tan^{-1} \left(\frac{2\tau_{xy}}{\sigma_x - \sigma_y}\right)\) \[ \theta_p = \frac{1}{2} \tan^{-1} \left(\frac{2 \times 45}{60 - 30}\right) = \frac{1}{2} \tan^{-1}(3) = \frac{1}{2} \times 71.56^\circ = 35.78^\circ \] - Maximum shear planes: \(45^\circ\) from principal planes. **Explanation:** Use Mohr's Circle relations to find principal/maximum shear stresses and orientations. --- ### **Final Answer (Problem 2):** #### **(a) Principal Stresses:** - \(\sigma_1 = \boxed{92.43\ \text{MPa}}\) - \(\sigma_2 = \boxed{-2.43\ \text{MPa}}\) - Orientation: \(35.78^\circ\) from original axes #### **(b) Maximum In-Plane Shear Stress:** - \(\tau_{max} = \boxed{47.43\ \text{MPa}}\) - Corresponding normal stress: \(45\) MPa - Orientation: \(35.78^\circ + 45^\circ = 80.78^\circ\) from original axes --- **Draw the rotated elements showing these stresses for each case.** --- *Let me know if you need the drawings or further clarification!*