Let's break down the solution step by step for both parts: --- ## **3.1 Calculate the Euler Buckling Load for the Strut** ### **Step 1: Euler Buckling Load Formula** \[ P_{cr} = \frac{\pi^2 E I}{(K L)^2} \] Where: - \( E \) = Young's Modulus for steel \( \approx 200 \text{ GPa} = 200 \times 10^3 \text{ MPa} = 200 \times 10^9 \text{ N/m}^2 \) - \( I \) = Moment of inertia about the relevant axis (use the smaller value for conservative design) - \( L \) = Length of the column \( = 9 \) m \( = 9000 \) mm - \( K \) = Effective length factor (depends on boundary conditions) ### **Step 2: Boundary Conditions** - **About X-X axis:** Fixed at both ends ⇒ \( K = 0.5 \) - **About Y-Y axis:** Fixed at top, pinned at bottom ⇒ \( K = 0.7 \) (approximate, check your course notes or code for precise value) ### **Step 3: Choose the Critical Axis** - \( I_{xx} = 81 \times 10^6 \) mm⁴ (major axis, larger, less likely to buckle) - \( I_{yy} = 5.67 \times 10^6 \) mm⁴ (minor axis, smaller, more likely to buckle) **Use \( I_{yy} \) for buckling.** ### **Step 4: Plug in the Values** Convert units: - \( I_{yy} = 5.67 \times 10^6 \) mm⁴ \( = 5.67 \times 10^{-6} \) m⁴ - \( L = 9 \) m - \( K = 0.7 \) \[ P_{cr} = \frac{\pi^2 \cdot 200 \times 10^9 \cdot 5.67 \times 10^{-6}}{(0.7 \cdot 9)^2} \] Calculate denominator: - \( (0.7 \cdot 9)^2 = (6.3)^2 = 39.69 \) Calculate numerator: - \( \pi^2 \approx 9.87 \) - \( 9.87 \cdot 200 \cdot 5.67 = 9.87 \cdot 1134 = 11192.58 \) - \( 11192.58 \cdot 10^3 = 1.119 \times 10^7 \) So, \[ P_{cr} = \frac{1.119 \times 10^7}{39.69} \times 10^3 \text{ N} = 282,100 \text{ N} = 282.1 \text{ kN} \] **Final Euler Buckling Load:** \[ \boxed{282.1 \text{ kN}} \] --- ## **3.2 Calculate the Failure Load According to Rankine’s Theory** ### **Rankine’s Formula** \[ \frac{1}{P_{Rankine}} = \frac{1}{P_{cr}} + \frac{1}{P_{y}} \] Where: - \( P_{cr} \) = Euler buckling load (from above) - \( P_{y} \) = Yield load = \( A \cdot f_y \) - \( f_y \) = Yield strength of 300W steel = **300 MPa** ### **Plug in the Values** - \( A = 5.76 \times 10^3 \) mm² \( = 5.76 \times 10^{-3} \) m² - \( f_y = 300 \) MPa \( = 300 \times 10^6 \) N/m² \[ P_y = 5.76 \times 10^{-3} \times 300 \times 10^6 = 1.728 \times 10^{6} \text{ N} = 1728 \text{ kN} \] ### **Final Calculation** \[ \frac{1}{P_{Rankine}} = \frac{1}{282.1 \times 10^3} + \frac{1}{1.728 \times 10^6} \] Calculate each term: - \( \frac{1}{282100} = 3.546 \times 10^{-6} \) - \( \frac{1}{1728000} = 5.785 \times 10^{-7} \) Add: - \( 3.546 \times 10^{-6} + 5.785 \times 10^{-7} = 4.125 \times 10^{-6} \) \[ P_{Rankine} = \frac{1}{4.125 \times 10^{-6}} = 242,400 \text{ N} = 242.4 \text{ kN} \] --- ## **Final Answers** ### **3.1 Euler Buckling Load:** \[ \boxed{282.1 \text{ kN}} \] ### **3.2 Rankine Failure Load:** \[ \boxed{242.4 \text{ kN}} \] --- **Let me know if you need step-by-step workings or any clarifications!**