Certainly! Let's solve the problem step by step. --- ## **Step 1: Extract Given Data** - **Load (F):** \( 4.50 \) kN \( = 450 \) N - **Length (L):** \( 750 \) mm (not needed for stress calculation) - **Material:** SAE 104 steel, hot-rolled - **Outer diameter (\(d_o\)):** \( 18 \) mm - **Inner diameter (\(d_i\)):** \( 12 \) mm --- ## **Step 2: Find Yield Strength of Material** For **SAE 104, hot-rolled steel:** - **Yield strength, \(S_y\):** ≈ **370 MPa** (from standard tables) --- ## **Step 3: Calculate Cross-sectional Area** The cross-sectional area of a tube: \[ A = \frac{\pi}{4} (d_o^2 - d_i^2) \] Plug in the values: \[ A = \frac{\pi}{4} (18^2 - 12^2) \text{ mm}^2 \] \[ A = \frac{\pi}{4} (324 - 144) = \frac{\pi}{4} (180) \] \[ A = \frac{\pi \times 180}{4} \approx \frac{3.1416 \times 180}{4} \approx \frac{565.486}{4} \approx 141.37 \text{ mm}^2 \] --- ## **Step 4: Calculate Actual Stress** \[ \sigma = \frac{F}{A} \] \[ \sigma = \frac{450 \text{ N}}{141.37 \text{ mm}^2} \] \[ \sigma \approx 31.83 \text{ MPa} \] --- ## **Step 5: Compute the Design Factor (\(N\))** \[ N = \frac{S_y}{\sigma} \] \[ N = \frac{370 \text{ MPa}}{31.83 \text{ MPa}} \] \[ N \approx 11.6 \] --- ## **Step 6: Final Answer** \[ \boxed{N \approx 11.6} \] --- ### **Explanation** - The design factor is significantly greater than 3, which means the current design is **very conservative**. - If you were required to redesign for \( N \approx 3 \), you could reduce the cross-sectional area (increase inner diameter or decrease outer diameter) to optimize material use. --- **Let me know if you would like to see the redesign for \( N \approx 3 \) or have any questions!**