Let's break down **Problem 1** step by step. --- ## **Step 1: Find Stresses on the Face** Given: - Cube dimensions: \( 25.4\, \text{mm} \times 25.4\, \text{mm} \times 25.4\, \text{mm} \) - Forces applied (see diagram). ### **a) Calculate Area** \[ A = 25.4\, \text{mm} \times 25.4\, \text{mm} = 645.16\, \text{mm}^2 = 6.4516 \times 10^{-4} \, \text{m}^2 \] --- ### **b) Find Stresses** Let's assign axes: - \( x \): horizontal right - \( y \): vertical down - \( z \): into the page #### **On the face perpendicular to \( x \) (right face):** - Normal force: \( 3\, \text{kN} \) (compressive, acts into face) - Shear force: \( 2\, \text{kN} \) (downwards) #### **On the face perpendicular to \( y \) (top face):** - Normal force: \( 5\, \text{kN} \) (compressive, acts into face) - Shear force: \( 2\, \text{kN} \) (to the right) #### **Calculate Stresses:** #### **Normal and Shear Stresses** \[ \sigma_x = \frac{3\, \text{kN}}{A} = \frac{300\, \text{N}}{6.4516 \times 10^{-4}} = 4651\, \text{kPa} \] \[ \sigma_y = \frac{5\, \text{kN}}{A} = \frac{500\, \text{N}}{6.4516 \times 10^{-4}} = 7752\, \text{kPa} \] \[ \tau_{xy} = \frac{2\, \text{kN}}{A} = \frac{200\, \text{N}}{6.4516 \times 10^{-4}} = 3101\, \text{kPa} \] --- ## **Step 2: Mohr's Circle Construction** ### **States:** \[ \sigma_x = 4651\, \text{kPa} \quad (\text{compressive, so negative if sign convention is tension +}) \] \[ \sigma_y = 7752\, \text{kPa} \quad (\text{also compressive, negative}) \] \[ \tau_{xy} = 3101\, \text{kPa} \] (If you use compressive as negative, then \(\sigma_x = -4651\) kPa, \(\sigma_y = -7752\) kPa.) --- ## **Step 3: Principal Stresses** \[ \sigma_1, \sigma_2 = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] Plug in negative values (compression negative): \[ \frac{\sigma_x + \sigma_y}{2} = \frac{-4651 + (-7752)}{2} = -6201.5\, \text{kPa} \] \[ \frac{\sigma_x - \sigma_y}{2} = \frac{-4651 - (-7752)}{2} = \frac{3101}{2} = 155.5\, \text{kPa} \] \[ \sqrt{(155.5)^2 + (3101)^2} = \sqrt{ 2,405,051 + 9,620,204 } = \sqrt{12,025,255} \approx 3468.5\, \text{kPa} \] So, \[ \sigma_1 = -6201.5 + 3468.5 = -2733\, \text{kPa} \] \[ \sigma_2 = -6201.5 - 3468.5 = -9669.9\, \text{kPa} \] **Principal stresses** (magnitude): \[ |\sigma_1| = 2733\, \text{kPa},\quad |\sigma_2| = 9669.9\, \text{kPa} \] --- ### **a) Final Answer: Principal Stresses** \[ \boxed{ \sigma_1 = -2733\, \text{kPa},\quad \sigma_2 = -9669.9\, \text{kPa} } \] --- ## **Step 4: Maximum Shear Stress** \[ \tau_{\text{max}} = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] We already calculated this above: \[ \tau_{\text{max}} = 3468.5\, \text{kPa} \] ### **b) Final Answer: Maximum Shear Stress** \[ \boxed{ \tau_{\text{max}} = 3468.5\, \text{kPa} } \] --- ## **Step 5: Normal and Shear Stress on an Inclined Plane** Let plane be inclined at \(\theta = 35^\circ\) clockwise from \(x\). Formulas: \[ \sigma_\theta = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos 2\theta + \tau_{xy} \sin 2\theta \] \[ \tau_\theta = -\frac{\sigma_x - \sigma_y}{2} \sin 2\theta + \tau_{xy} \cos 2\theta \] \[ \cos 2\theta = \cos 70^\circ = .342 \] \[ \sin 2\theta = \sin 70^\circ = .9397 \] \[ \sigma_\theta = -6201.5 + 155.5 \times .342 + 3101 \times .9397 \] \[ = -6201.5 + 530.5 + 2914.5 = -2756.5\, \text{kPa} \] \[ \tau_\theta = -155.5 \times .9397 + 3101 \times .342 \] \[ = -1456.5 + 106.4 = -396.1\, \text{kPa} \] ### **c) Final Answer: Normal and Shear Stress on 35° Plane** \[ \boxed{ \sigma_\theta = -2756.5\, \text{kPa}, \quad \tau_\theta = -396.1\, \text{kPa} } \] --- ## **Summary Table** | Quantity | Value | |------------------------------------|----------------------| | \(\sigma_1\) (Principal) | \(-2733\) kPa | | \(\sigma_2\) (Principal) | \(-9669.9\) kPa | | \(\tau_\text{max}\) (Max Shear) | \(3468.5\) kPa | | \(\sigma_{35^\circ}\) (Normal) | \(-2756.5\) kPa | | \(\tau_{35^\circ}\) (Shear) | \(-396.1\) kPa | --- ## **Mohr Circle (Plotting)** - Center: \(-6201.5\) kPa - Radius: \(3468.5\) kPa --- ### **Pole Method** - On Mohr's circle, find the pole (intersection of lines representing the original stress state). - Draw a line at \(2 \times 35^\circ = 70^\circ\) from the pole; intersection with the circle gives \(\sigma_{35^\circ}\) and \(\tau_{35^\circ}\). --- **Let me know if you want a Python/Matlab script or a hand-drawn Mohr's circle!**

## **Python Code for Mohr's Circle** ```python import numpy as np import matplotlib.pyplot as plt # Given stresses sigma_x = -4651 # kPa sigma_y = -7752 # kPa tau_xy = 3101 # kPa # Calculate center and radius center = (sigma_x + sigma_y) / 2 radius = np.sqrt(((sigma_x - sigma_y) / 2) ** 2 + tau_xy ** 2) # Define angles for the circle theta = np.linspace(0, 2 * np.pi, 100) # Mohr's Circle coordinates circle_x = center + radius * np.cos(theta) circle_y = radius * np.sin(theta) # Plotting Mohr's Circle plt.figure(figsize=(8, 8)) plt.plot(circle_x, circle_y, label='Mohr\'s Circle') plt.plot([sigma_x, sigma_y], [tau_xy, -tau_xy], 'ro', label='Original Stress State') plt.axhline(0, color='black', lw=0.5, ls='--') plt.axvline(0, color='black', lw=0.5, ls='--') plt.title('Mohr\'s Circle') plt.xlabel('Normal Stress (kPa)') plt.ylabel('Shear Stress (kPa)') plt.xlim(center - radius - 1000, center + radius + 1000) plt.ylim(-radius - 1000, radius + 1000) plt.grid() plt.legend() plt.axis('equal') plt.show() ``` ## **MATLAB Code for Mohr's Circle** ```matlab % Given stresses sigma_x = -4651; % kPa sigma_y = -7752; % kPa tau_xy = 3101; % kPa % Calculate center and radius center = (sigma_x + sigma_y) / 2; radius = sqrt(((sigma_x - sigma_y) / 2)^2 + tau_xy^2); % Define angles for the circle theta = linspace(0, 2*pi, 100); % Mohr's Circle coordinates circle_x = center + radius * cos(theta); circle_y = radius * sin(theta); % Plotting Mohr's Circle figure; plot(circle_x, circle_y, 'b', 'LineWidth', 2); hold on; plot([sigma_x sigma_y], [tau_xy -tau_xy], 'ro', 'MarkerSize', 8, 'MarkerFaceColor', 'r'); ax = gca; ax.XAxisLocation = 'origin'; ax.YAxisLocation = 'origin'; grid on; title('Mohr\'s Circle'); xlabel('Normal Stress (kPa)'); ylabel('Shear Stress (kPa)'); xlim([center-radius-1000, center+radius+1000]); ylim([-radius-1000, radius+1000]); axis equal; legend('Mohr\'s Circle', 'Original Stress State'); ``` ### **Instructions for Use** - Copy the code into a Python or MATLAB environment. - Ensure you have `numpy` and `matplotlib` installed for Python. - Running the code will display the Mohr's Circle with the original stress state marked. ### **Note** - Adjust the limits in the plot if necessary to better visualize the circle and stress points.