Here's a summarized solution based on the images provided: --- # 1. Vertical Reaction at Support A \[ \sum F_y = \] \[ A_y + B_y = \frac{1}{2} \times 6 \times 4 \] \[ A_y + 6 + 6 = 12\, \text{kN} \] \[ A_y = 6\, \text{kN} \] --- # 2. Horizontal Reaction at Support A \[ \sum F_x = \] \[ A_x - B_x = \] \[ A_x = B_x \] \[ A_x = F_{x} \cos \theta \] Given: \[ A_x = \frac{4}{5} \times 10 = 8\, \text{kN} \] --- # 3. Free Body Diagram - Supports at A and D - Distributed load of \(4\, \text{kN/m}\) on the beam - Point loads at various points - Lengths and distances marked --- # 4. Normal Stress at Point D \[ \sigma_D = \frac{P}{A} \] \[ \sigma_D = \frac{8 \times 10^3}{.1 \times .05} = 160 \times 10^3\, \text{kPa} = 160\, \text{MPa} \] --- # 5. Moment of Inertia \(I\) \[ I = \frac{1}{12} \times .05 \times .13^3 = 4.16 \times 10^{-6}\, \text{m}^4 \] --- # 6. Bending Stress at Point D \[ \sigma_D = \frac{M c}{I} \] \[ \sigma_D = \frac{12 \times 10^3 \times .03}{4.16 \times 10^{-6}} = 86400 \times 10^3\, \text{kPa} = 86400\, \text{kPa} \] --- # 7. Shear Force at Point D \[ V_D = 6 - \left( \frac{1}{2} \times 4 \times 3 \right) = \, \text{kN} \] Shear stress at D: \[ \tau_{z} = \] --- # 8. Normal Stress Due to Bending and Normal Force \[ \sigma_D = \frac{P}{A} + \frac{M c}{I} \] \[ \sigma_D = -160 \times 10^3 - 86400 \times 10^3 = -880 \times 10^3\, \text{kPa} = -88\, \text{MPa} \] --- # 9. Final Result **Stress at point D:** \[ \sigma_D = -88\, \text{MPa} \quad (\text{compression}) \] --- If you'd like any detailed steps or further clarification, please let me know!