1. **Problem Statement:** A firm needs a hollow steel tube to carry an axial compressive load of 200000 N (200 kN). Steel yield stress is 250 N/mm², and the design factor of safety (FS) is 2. Available tubes have external diameter $D=101.6$ mm and wall thicknesses $t=3.65$, $4.05$, $4.5$, and $4.85$ mm. We must select the appropriate tube based on strength, manufacturing variability, and safety. 2. **Formulas and Important Rules:** - Allowable stress $\sigma_{allow} = \frac{\sigma_{yield}}{FS}$ - Required cross-sectional area $A_{req} = \frac{P}{\sigma_{allow}}$ - Gross cross-sectional area of hollow tube $A = \pi \left(\frac{D^2}{4} - \frac{(D-2t)^2}{4}\right) = \frac{\pi}{4} (D^2 - (D-2t)^2)$ 3. **Calculate allowable stress:** $$\sigma_{allow} = \frac{250}{2} = 125 \text{ N/mm}^2$$ 4. **Calculate required cross-sectional area:** $$A_{req} = \frac{200000}{125} = 1600 \text{ mm}^2$$ 5. **Calculate gross cross-sectional area for each tube:** For each $t$, compute $$A = \frac{\pi}{4} \left(101.6^2 - (101.6 - 2t)^2\right)$$ - For $t=3.65$ mm: $$A = \frac{\pi}{4} (10323.36 - (101.6 - 7.3)^2) = \frac{\pi}{4} (10323.36 - 8830.09) = \frac{\pi}{4} (1493.27) \approx 1173.5 \text{ mm}^2$$ - For $t=4.05$ mm: $$A = \frac{\pi}{4} (10323.36 - (101.6 - 8.1)^2) = \frac{\pi}{4} (10323.36 - 8760.25) = \frac{\pi}{4} (1563.11) \approx 1226.7 \text{ mm}^2$$ - For $t=4.5$ mm: $$A = \frac{\pi}{4} (10323.36 - (101.6 - 9)^2) = \frac{\pi}{4} (10323.36 - 8640.16) = \frac{\pi}{4} (1683.2) \approx 1321.5 \text{ mm}^2$$ - For $t=4.85$ mm: $$A = \frac{\pi}{4} (10323.36 - (101.6 - 9.7)^2) = \frac{\pi}{4} (10323.36 - 8570.89) = \frac{\pi}{4} (1752.47) \approx 1375.5 \text{ mm}^2$$ 6. **Compare required area and available areas:** - Required area $A_{req} = 1600$ mm² - All available tube areas are less than $1600$ mm², so none meet the required cross-sectional area based on allowable stress. 7. **Recommendation:** Since none of the tubes meet the required area for the given load and safety factor, the design engineer should consider: - Increasing wall thickness beyond 4.85 mm - Using a tube with larger external diameter - Reducing the factor of safety if justified - Using a higher strength steel Among the given options, the tube with $t=4.85$ mm has the largest area and is closest to the requirement, so it is the best choice if no other options are available. 8. **MATLAB formula for area:** ```matlab D = 101.6; % mm t = [3.65, 4.05, 4.5, 4.85]; % mm A = pi/4 * (D^2 - (D - 2*t).^2); ``` 9. **Excel formula for area (assuming D in A1 and t in B1):** `=PI()/4*(A1^2 - (A1 - 2*B1)^2)` 10. **Sensitivity to t and FS:** - Increasing $t$ increases $A$ approximately linearly for small changes. - Increasing FS decreases allowable stress and increases required area. **Final answer:** The tube with $t=4.85$ mm is recommended among the given options but does not fully satisfy the required cross-sectional area for the load and safety factor. Consider design adjustments for safety.