Certainly! Here's the solution in a professional, formatted manner with equations rendered using LaTeX: --- # Given Data: - Weight, \( W = 6.7\, \text{kN} \) - Outside diameter, \( D_o = 185\, \text{mm} \) - Wall thickness, \( t = 12.5\, \text{mm} \) - Inner diameter, \( D_i = 13\, \text{mm} \) - Resultant force, \( P = 12\, \text{kN} \) - Support dimensions: \( a=7.6\, \text{m} \), \( b=5.5\, \text{m} \) - \( P_z = 12\, \text{kN} \), \( P_y = 6.7\, \text{kN} \) --- # Step 1: Cross-Sectional Data Inner diameter: \[ D_i = D_o - 2 \times t = 185\, \text{mm} - 2 \times 12.5\, \text{mm} = 160\, \text{mm} \] Cross-sectional area: \[ A = \frac{\pi}{4} (D_o^2 - D_i^2) = \frac{\pi}{4} (185^2 - 160^2) \, \text{mm}^2 \] --- # Step 2: Normal Stress at Point \( H \) ### Normal stress due to axial force \( P \): \[ \sigma_H = \frac{P}{A} \] ### Shear stress due to shear force components: \[ \tau_{xyH} = \frac{P_y}{A} \] --- # Step 3: Bending and Torsional Stresses at Point \( K \): ### Bending moment: \[ M_z = P_z \times b = 12\, \text{kN} \times 5.5\, \text{m} = 66\, \text{kNm} \] ### Shear force: \[ V_y = P_y = 6.7\, \text{kN} \] ### Bending stress at \( K \): \[ \sigma_k = \frac{M_z c}{I} \] where \( c = \frac{D_o}{2} \) and \( I \) is the moment of inertia: \[ I = \frac{\pi}{64} (D_o^4 - D_i^4) \] ### Torsional shear stress: \[ \tau_{yzK} = \frac{T}{J} \] where \( T \) is torsional moment and \( J \) is the polar moment of inertia: \[ T = P_z \times a \] --- # Step 4: Calculations ### Cross-sectional area: \[ A = \frac{\pi}{4} (185^2 - 160^2) \approx 131,946\, \text{mm}^2 \] ### Normal stress at \( H \): \[ \sigma_H = \frac{6.7 \times 10^3\, \text{N}}{131,946\, \text{mm}^2} \approx 50.8\, \text{MPa} \] ### Shear stress: \[ \tau_{xyH} = \frac{6.7 \times 10^3\, \text{N}}{131,946\, \text{mm}^2} \approx 50.8\, \text{MPa} \] --- # Step 5: Bending and Shear Stresses at \( K \) ### Moment of inertia: \[ I = \frac{\pi}{64} (185^4 - 160^4) \approx 2.103 \times 10^{9}\, \text{mm}^4 \] ### Bending stress: \[ \sigma_k = \frac{66 \times 10^3\, \text{N-m} \times c}{I} \] with \( c = 92.5\, \text{mm} \): \[ \sigma_k \approx 332.07\, \text{kPa} \] ### Torsional moment: \[ T = P_z \times a = 12\, \text{kN} \times 7.6\, \text{m} = 91.2\, \text{kNm} \] ### Polar moment of inertia: \[ J = \frac{\pi}{32} (D_o^4 - D_i^4) \approx 4.206 \times 10^{9}\, \text{mm}^4 \] ### Shear stress due to torsion: \[ \tau_{yzK} = \frac{T \times c}{J} \] \[ = \frac{91.2 \times 10^3\, \text{N-m} \times 92.5\, \text{mm}}{4.206 \times 10^{9}\, \text{mm}^4} \approx 2.020\, \text{MPa} \] --- # Final Results: | Quantity | Value | |------------|--------| | \(\sigma_H\) | **50.8 MPa** | | \(\tau_{xyH}\) | **50.8 MPa** | | \(\sigma_K\) | **.332 MPa** | | \(\tau_{yzK}\) | **2.02 MPa** | --- Please replace the approximate values with your detailed calculations for more precision if needed. Let me know if you'd like the exact step-by-step equations in a specific format!