1. **Problem Statement:** Calculate the minimum required shell thickness for a thin-walled cylindrical steel shell under internal gauge pressure $p=6.0$ MPa, with diameter $D=400$ mm, allowable stress $\sigma_{allow}=120$ MPa, and cylinder length $L=1.00$ m. Then find hoop and longitudinal stresses, elastic strains, volume change, factor of safety, and discuss thin-wall assumptions. For the support column of length $L_{col}=3.0$ m and mass $m=850$ kg, select two hollow circular tube sections, calculate Euler buckling load, slenderness ratio, factor of safety, and compare. 2. **Thin-Walled Cylinder Thickness Calculation:** The thin-walled pressure vessel hoop stress formula is: $$\sigma_{hoop} = \frac{pD}{2t}$$ where $t$ is shell thickness. Rearranged for $t$: $$t = \frac{pD}{2\sigma_{allow}}$$ Given: $p=6.0$ MPa = $6.0 \times 10^6$ Pa $D=400$ mm = 0.4 m $\sigma_{allow}=120$ MPa = $120 \times 10^6$ Pa Calculate: $$t = \frac{6.0 \times 10^6 \times 0.4}{2 \times 120 \times 10^6} = \frac{2.4 \times 10^6}{240 \times 10^6} = 0.01 \text{ m} = 10 \text{ mm}$$ 3. **Hoop and Longitudinal Stresses:** Hoop stress: $$\sigma_{hoop} = \frac{pD}{2t} = 120 \text{ MPa}$$ (by design) Longitudinal stress: $$\sigma_{long} = \frac{pD}{4t} = \frac{6.0 \times 10^6 \times 0.4}{4 \times 0.01} = 60 \text{ MPa}$$ 4. **Elastic Strains:** Using Hooke's law for thin-walled pressure vessels: $$\varepsilon_{hoop} = \frac{\sigma_{hoop} - \nu \sigma_{long}}{E}$$ $$\varepsilon_{long} = \frac{\sigma_{long} - \nu \sigma_{hoop}}{E}$$ Given: $E=210$ GPa = $210 \times 10^9$ Pa $\nu=0.30$ Calculate: $$\varepsilon_{hoop} = \frac{120 \times 10^6 - 0.30 \times 60 \times 10^6}{210 \times 10^9} = \frac{120 \times 10^6 - 18 \times 10^6}{210 \times 10^9} = \frac{102 \times 10^6}{210 \times 10^9} = 4.86 \times 10^{-4}$$ $$\varepsilon_{long} = \frac{60 \times 10^6 - 0.30 \times 120 \times 10^6}{210 \times 10^9} = \frac{60 \times 10^6 - 36 \times 10^6}{210 \times 10^9} = \frac{24 \times 10^6}{210 \times 10^9} = 1.14 \times 10^{-4}$$ 5. **Volume Change Estimation:** Initial volume of cylinder: $$V_{cyl} = \pi \left(\frac{D}{2}\right)^2 L = \pi (0.2)^2 \times 1.0 = 0.1257 \text{ m}^3$$ Volume of two hemispherical ends (sphere volume $= \frac{4}{3} \pi r^3$): $$V_{ends} = \frac{4}{3} \pi (0.2)^3 = 0.0335 \text{ m}^3$$ Total initial volume: $$V_0 = 0.1257 + 0.0335 = 0.1592 \text{ m}^3$$ Radial strain $\varepsilon_{hoop}$ causes diameter increase, longitudinal strain $\varepsilon_{long}$ causes length increase. New diameter: $$D_{new} = D (1 + \varepsilon_{hoop}) = 0.4 (1 + 4.86 \times 10^{-4}) = 0.40019 \text{ m}$$ New length: $$L_{new} = L (1 + \varepsilon_{long}) = 1.0 (1 + 1.14 \times 10^{-4}) = 1.00011 \text{ m}$$ New volume: $$V_{new} = \pi \left(\frac{D_{new}}{2}\right)^2 L_{new} + \frac{4}{3} \pi \left(\frac{D_{new}}{2}\right)^3$$ Calculate: $$V_{cyl,new} = \pi (0.200095)^2 \times 1.00011 = 0.1258 \text{ m}^3$$ $$V_{ends,new} = \frac{4}{3} \pi (0.200095)^3 = 0.0335 \text{ m}^3$$ $$V_{new} = 0.1258 + 0.0335 = 0.1593 \text{ m}^3$$ Volume change: $$\Delta V = V_{new} - V_0 = 0.1593 - 0.1592 = 0.0001 \text{ m}^3 = 100 \text{ cm}^3$$ 6. **Factor of Safety (FoS):** Select FoS $= 2$ to account for manufacturing tolerances, material variability, and unexpected loads. 7. **Limitations of Thin-Walled Assumptions:** - Assumes $t < D/10$ (here $t=10$ mm, $D=400$ mm, so $t/D=0.025$ valid). - Neglects stress concentrations at welds or supports. - Assumes uniform stress distribution. - Does not consider local buckling or plastic deformation. 8. **Support Column Selection:** Assume two hollow circular tube sections: - Section 1: Outer diameter $D_o=100$ mm, inner diameter $D_i=80$ mm - Section 2: Outer diameter $D_o=120$ mm, inner diameter $D_i=90$ mm Calculate cross-sectional area $A$ and moment of inertia $I$: $$A = \frac{\pi}{4} (D_o^2 - D_i^2)$$ $$I = \frac{\pi}{64} (D_o^4 - D_i^4)$$ Calculate radius of gyration: $$r_g = \sqrt{\frac{I}{A}}$$ Calculate Euler buckling load: $$P_{cr} = \frac{\pi^2 E I}{(K L)^2}$$ where $K=1$, $L=3.0$ m, $E=210$ GPa. Calculate applied load: $$P = mg = 850 \times 9.81 = 8338.5 \text{ N}$$ **Section 1:** $$A_1 = \frac{\pi}{4} (0.1^2 - 0.08^2) = \frac{\pi}{4} (0.01 - 0.0064) = 2.827 \times 10^{-3} \text{ m}^2$$ $$I_1 = \frac{\pi}{64} (0.1^4 - 0.08^4) = \frac{\pi}{64} (1 \times 10^{-4} - 4.096 \times 10^{-5}) = 2.72 \times 10^{-6} \text{ m}^4$$ $$r_{g1} = \sqrt{\frac{2.72 \times 10^{-6}}{2.827 \times 10^{-3}}} = 0.031 \text{ m}$$ $$P_{cr1} = \frac{\pi^2 \times 210 \times 10^9 \times 2.72 \times 10^{-6}}{(1 \times 3)^2} = 62600 \text{ N}$$ $$\lambda_1 = \frac{3}{0.031} = 96.8$$ $$FoS_1 = \frac{P_{cr1}}{P} = \frac{62600}{8338.5} = 7.51$$ **Section 2:** $$A_2 = \frac{\pi}{4} (0.12^2 - 0.09^2) = \frac{\pi}{4} (0.0144 - 0.0081) = 4.95 \times 10^{-3} \text{ m}^2$$ $$I_2 = \frac{\pi}{64} (0.12^4 - 0.09^4) = \frac{\pi}{64} (2.07 \times 10^{-4} - 6.56 \times 10^{-5}) = 7.04 \times 10^{-6} \text{ m}^4$$ $$r_{g2} = \sqrt{\frac{7.04 \times 10^{-6}}{4.95 \times 10^{-3}}} = 0.038 \text{ m}$$ $$P_{cr2} = \frac{\pi^2 \times 210 \times 10^9 \times 7.04 \times 10^{-6}}{(3)^2} = 162000 \text{ N}$$ $$\lambda_2 = \frac{3}{0.038} = 78.9$$ $$FoS_2 = \frac{162000}{8338.5} = 19.43$$ 9. **Comparison:** Section 2 has higher buckling load and FoS but is heavier and more material costly. Section 1 is lighter, cheaper, but with lower FoS. Choose Section 1 for economy if FoS > 3 is acceptable, else Section 2 for higher safety. 10. **Effects of Imperfections and Eccentric Loading:** - Initial imperfections reduce buckling load significantly. - Eccentric loading causes bending moments, reducing effective load capacity. - Real columns require higher FoS to account for these effects.