## Problem Statement **Given:** - **A couple \( M = 1.2 \text{ kN·m} \)** acts in a vertical plane and is applied to a bent beam as shown. - **Cross-sectional properties:** - \( I_y = 1.894 \times 10^6 \ \text{mm}^4 \) - \( I_z = .614 \times 10^6 \ \text{mm}^4 \) - \( I_{yz} = +.800 \times 10^6 \ \text{mm}^4 \) - **Dimensions are given in the diagram.** **Required:** Determine the stress at point **A**. --- ## Solution Steps ### 1. **Convert Units** \[ M = 1.2 \text{ kN·m} = 1.2 \times 10^3 \text{ N} \times 10^3 \text{ mm} = 1.2 \times 10^6 \text{ N·mm} \] --- ### 2. **Locate Point A** From the figure: - Point A is at the top left corner of the section. - Coordinates of A relative to centroid C: \( (y_A, z_A) \) #### Find Centroid location and coordinates of A with respect to C. Suppose \( y_A \) is the vertical distance from C to A, and \( z_A \) is the horizontal distance from C to A. --- ### 3. **Calculate Bending Stress at A** The general formula for bending stress at a point (y, z) due to a moment \( M \) in the y-z plane is: \[ \sigma = -\frac{M_y I_z - M_z I_{yz}}{I_y I_z - I_{yz}^2} y + \frac{M_z I_y - M_y I_{yz}}{I_y I_z - I_{yz}^2} z \] In this case, the moment \( M \) is acting about the **z-axis**, so \( M_z = M \), \( M_y = \): \[ \sigma = \frac{M_z I_y}{I_y I_z - I_{yz}^2} z - \frac{M_z I_{yz}}{I_y I_z - I_{yz}^2} y \] --- ### 4. **Plug in the Values** - \( I_y = 1.894 \times 10^6 \ \text{mm}^4 \) - \( I_z = .614 \times 10^6 \ \text{mm}^4 \) - \( I_{yz} = .800 \times 10^6 \ \text{mm}^4 \) - \( M_z = 1.2 \times 10^6 \ \text{N·mm} \) - \( (y_A, z_A) \): Need to determine from the geometry. #### Denominator: \[ I_y I_z - I_{yz}^2 = (1.894 \times 10^6)(.614 \times 10^6) - (.800 \times 10^6)^2 \] Calculate: \[ I_y I_z = 1.894 \times .614 \times 10^{12} = 1.163 \times 10^{12} \] \[ I_{yz}^2 = .800^2 \times 10^{12} = .640 \times 10^{12} \] \[ I_y I_z - I_{yz}^2 = (1.163 - .640) \times 10^{12} = .523 \times 10^{12} \] --- ### 5. **Coordinates of A (\( y_A, z_A \)):** From the figure, you need to determine the centroid location (C) with respect to A. Let's denote the vertical and horizontal distances from C to A as \( y_A \) and \( z_A \) respectively. Let me know if you want to proceed with the calculation of centroid and coordinates, or if you have them already. --- ## **Summary of Formula** \[ \boxed{ \sigma_A = \frac{M_z I_y}{I_y I_z - I_{yz}^2} z_A - \frac{M_z I_{yz}}{I_y I_z - I_{yz}^2} y_A } \] Plug in the values of \( y_A \) and \( z_A \) depending on your centroid calculations. --- **If you provide the centroid location or coordinates of A with respect to C, I can complete the numeric solution.**