Let's solve the problem step by step for both (a) and (b): --- ## **General Approach** Given a 2D stress element with normal stresses \(\sigma_x\), \(\sigma_y\) and shear stress \(\tau_{xy}\), the normal (\(\sigma_n\)) and shear (\(\tau_n\)) stresses on an inclined plane at angle \(\theta\) from the x-axis are: \[ \sigma_n = \sigma_x \cos^2\theta + \sigma_y \sin^2\theta + 2\tau_{xy}\sin\theta\cos\theta \] \[ \tau_n = (\sigma_y - \sigma_x)\sin\theta\cos\theta + \tau_{xy}(\cos^2\theta - \sin^2\theta) \] --- ## **(a) Step-by-Step Solution** ### **Given:** - \(\sigma_x = 40\) MPa (right, positive) - \(\sigma_y = 180\) MPa (up, positive) - \(\tau_{xy} = 120\) MPa (CW, so positive in sign convention) - \(\theta = 30^\circ\) ### **1. Calculate \(\sigma_n\):** \[ \sigma_n = \sigma_x \cos^2\theta + \sigma_y \sin^2\theta + 2\tau_{xy}\sin\theta\cos\theta \] Plug in values (\(\cos 30^\circ = \sqrt{3}/2, \sin 30^\circ = 1/2\)): \[ \cos^2 30^\circ = (.866)^2 = .75 \] \[ \sin^2 30^\circ = (.5)^2 = .25 \] \[ 2\sin 30^\circ \cos 30^\circ = 2 \times .5 \times .866 = .866 \] \[ \sigma_n = 40 \times .75 + 180 \times .25 + 120 \times .866 \] \[ = 30 + 45 + 103.92 \] \[ = 178.92 \text{ MPa} \] --- ### **2. Calculate \(\tau_n\):** \[ \tau_n = (\sigma_y - \sigma_x)\sin\theta\cos\theta + \tau_{xy}(\cos^2\theta - \sin^2\theta) \] \[ \sin 30^\circ \cos 30^\circ = .5 \times .866 = .433 \] \[ \cos^2 30^\circ - \sin^2 30^\circ = .75 - .25 = .5 \] \[ \tau_n = (180 - 40) \times .433 + 120 \times .5 \] \[ = 140 \times .433 + 60 \] \[ = 60.62 + 60 \] \[ = 120.62 \text{ MPa} \] --- ### **3. Sketch Explanation** - On the plane at 30°, the normal stress is **178.92 MPa** (pointing away from the plane). - The shear stress is **120.62 MPa** (direction as per positive sign convention, i.e., tangent to the plane). --- ## **(b) Step-by-Step Solution** ### **Given:** - \(\sigma_x = 60\) MPa (right) - \(\sigma_y = 200\) MPa (up) - \(\tau_{xy} = 20\) MPa (CW, so positive) - \(\theta = 45^\circ\) ### **1. Calculate \(\sigma_n\):** \[ \cos^2 45^\circ = (.7071)^2 = .5 \] \[ \sin^2 45^\circ = (.7071)^2 = .5 \] \[ 2\sin 45^\circ \cos 45^\circ = 2 \times .7071 \times .7071 = 1 \] \[ \sigma_n = 60 \times .5 + 200 \times .5 + 20 \times 1 \] \[ = 30 + 100 + 20 \] \[ = 150 \text{ MPa} \] --- ### **2. Calculate \(\tau_n\):** \[ \sin 45^\circ \cos 45^\circ = .7071 \times .7071 = .5 \] \[ \cos^2 45^\circ - \sin^2 45^\circ = .5 - .5 = \] \[ \tau_n = (200 - 60) \times .5 + 20 \times \] \[ = 140 \times .5 + \] \[ = 70 \text{ MPa} \] --- ### **3. Sketch Explanation** - On the plane at 45°, the normal stress is **150 MPa**. - The shear stress is **70 MPa**. --- ## **Final Answers** ### **(a)** - **Normal stress on plane at 30°:** \(\boxed{178.92~\text{MPa}}\) - **Shear stress on plane at 30°:** \(\boxed{120.62~\text{MPa}}\) ### **(b)** - **Normal stress on plane at 45°:** \(\boxed{150~\text{MPa}}\) - **Shear stress on plane at 45°:** \(\boxed{70~\text{MPa}}\) --- ## **Summary Table** | Part | \(\theta\) | \(\sigma_n\) (MPa) | \(\tau_n\) (MPa) | |------|------------|-------------------|-----------------| | (a) | 30° | 178.92 | 120.62 | | (b) | 45° | 150 | 70 | --- If you need sketches of the oriented elements with stress arrows, let me know!