Let's solve each part step by step. --- ## (i) Determining the Rankine-Gordon Constant \( a \) ### **Given Data** - **External diameter (\( D \))**: 80 mm - **Wall thickness (\( t \))**: 3 - **Yield load (short length, \( P_{yield} \))**: 392 kN - **Buckling load (2 m length, \( P_{cr} \))**: 230 kN - **Length (\( L \))**: 2 m = 200 mm ### **Step 1: Calculate Internal Diameter** \[ d = D - 2t = 80\,\text{mm} - 2 \times 3\,\text{mm} = 74\,\text{mm} \] ### **Step 2: Calculate Cross-Sectional Area** \[ A = \frac{\pi}{4} (D^2 - d^2) \] \[ A = \frac{\pi}{4} (80^2 - 74^2) = \frac{\pi}{4} (640 - 5476) = \frac{\pi}{4} \times 924 \approx 726.2\,\text{mm}^2 \] ### **Step 3: Yield Stress** \[ \sigma_{yield} = \frac{P_{yield}}{A} \] \[ \sigma_{yield} = \frac{392,000}{726.2} \approx 540\,\text{N/mm}^2 \] ### **Step 4: Moment of Inertia \( I \)** \[ I = \frac{\pi}{64}(D^4 - d^4) \] \[ D^4 = 80^4 = 40,960,000 \] \[ d^4 = 74^4 = 29,986,276 \] \[ I = \frac{\pi}{64}(40,960,000 - 29,986,276) = \frac{\pi}{64}(10,973,724) \approx 538,858\,\text{mm}^4 \] ### **Step 5: Rankine-Gordon Formula** \[ \frac{1}{P_{cr}} = \frac{1}{P_{yield}} + \frac{a}{P_{euler}} \] Where, - \( P_{cr} = \) Buckling load (given, 230 kN) - \( P_{yield} = \) Yield load (given, 392 kN) - \( P_{euler} = \frac{\pi^2 EI}{(L_{eff})^2} \) For pinned ends, \( L_{eff} = L \). ### **Step 6: Calculate \( P_{euler} \)** The modulus of elasticity (\( E \)) is not given, but for steel, typically \( E \approx 210,000\,\text{N/mm}^2 \). \[ P_{euler} = \frac{\pi^2 EI}{L^2} \] \[ P_{euler} = \frac{\pi^2 \times 210,000 \times 538,858}{200^2} \] \[ = \frac{9.870 \times 210,000 \times 538,858}{4,000,000} \] \[ = \frac{3.58 \times 10^{12}}{4,000,000} = 885,510\,\text{N} = 885.5\,\text{kN} \] ### **Step 7: Substitute Values into Rankine-Gordon Formula** \[ \frac{1}{P_{cr}} = \frac{1}{P_{yield}} + \frac{a}{P_{euler}} \] \[ \frac{1}{230,000} = \frac{1}{392,000} + \frac{a}{885,510} \] \[ \frac{1}{230,000} - \frac{1}{392,000} = \frac{a}{885,510} \] Calculate left side: \[ \frac{1}{230,000} = 4.348 \times 10^{-6} \] \[ \frac{1}{392,000} = 2.551 \times 10^{-6} \] \[ \Delta = 1.797 \times 10^{-6} \] So, \[ a = 885,510 \times 1.797 \times 10^{-6} \approx 1.59 \] --- ### **Final Answer (i):** **The value of the constant \( a \) is approximately \( 1.59 \).** --- ## (ii) Euler Buckling Load from Deflection ### **Given Data** - **Span (\( L \))**: 2 m = 200 mm - **Central load (\( P \))**: 900 N - **Deflection (\( \delta \))**: 1.4 mm ### **Step 1: Find \( EI \) from Deflection** For a simply supported beam with a central point load: \[ \delta = \frac{P L^3}{48 EI} \] \[ EI = \frac{P L^3}{48 \delta} \] \[ = \frac{900 \times (200)^3}{48 \times 1.4} \] \[ = \frac{900 \times 8 \times 10^9}{67.2} \] \[ = \frac{7.2 \times 10^{12}}{67.2} \approx 1.07 \times 10^{11}\,\text{N mm}^2 \] ### **Step 2: Euler Buckling Load** \[ P_{euler} = \frac{\pi^2 EI}{L^2} \] \[ = \frac{\pi^2 \times 1.07 \times 10^{11}}{(200)^2} \] \[ = \frac{9.870 \times 1.07 \times 10^{11}}{4 \times 10^6} \] \[ = \frac{1.056 \times 10^{12}}{4 \times 10^6} \] \[ = 264,000\,\text{N} = 264\,\text{kN} \] --- ### **Final Answer (ii):** **The Euler buckling load for the tube is approximately \( 264\,\text{kN} \).** --- ## **Summary Table** | Part | Value/Answer | |------|-------------------------| | (i) | Rankine-Gordon \( a \): \( 1.59 \) | | (ii) | Euler Buckling Load: \( 264 \) kN |