Steel Column Buckling 3D7E81

1. **Problem Statement:** We need to find the minimum diameter $d$ of a round steel column of length $L=8$ ft, fixed at the base and pinned at the top, supporting a load $P=4.5$ kips, such that it will not buckle. 2. **Given Data:** - Length $L = 8$ ft = $8 \times 12 = 96$ in (convert to inches for consistency) - Load $P = 4.5$ kips = $4500$ lbs - Modulus of Elasticity $E = 30 \times 10^6$ psi - Yield stress $\sigma_y = 55$ ksi = $55000$ psi 3. **Formula for Buckling Load (Euler's Buckling Load):** $$ P_{cr} = \frac{\pi^2 E I}{(K L)^2} $$ where: - $P_{cr}$ is the critical buckling load - $I$ is the moment of inertia of the cross-section - $K$ is the effective length factor - $L$ is the length of the column 4. **Effective Length Factor $K$:** For a column fixed at base and pinned at top, $K = \sqrt{2} / 2 \approx 0.7$ (or use $K=0.7$ as standard value) 5. **Moment of Inertia $I$ for a circular cross-section:** $$ I = \frac{\pi d^4}{64} $$ 6. **Maximum allowable stress condition:** The column will not buckle if the critical load $P_{cr}$ is greater than or equal to the applied load $P$. 7. **Calculate $d$ such that $P_{cr} \geq P$:** $$ P \leq \frac{\pi^2 E I}{(K L)^2} = \frac{\pi^2 E \frac{\pi d^4}{64}}{(K L)^2} = \frac{\pi^3 E d^4}{64 (K L)^2} $$ Rearranging for $d$: $$ d^4 \geq \frac{64 P (K L)^2}{\pi^3 E} $$ 8. **Substitute values:** $$ K = 0.7, \quad L = 96 \text{ in}, \quad P = 4500 \text{ lbs}, \quad E = 30 \times 10^6 \text{ psi} $$ Calculate: $$ d^4 \geq \frac{64 \times 4500 \times (0.7 \times 96)^2}{\pi^3 \times 30 \times 10^6} $$ Calculate $(0.7 \times 96)^2$: $$ (0.7 \times 96)^2 = (67.2)^2 = 4515.84 $$ Calculate numerator: $$ 64 \times 4500 \times 4515.84 = 64 \times 4500 \times 4515.84 = 1,301,548,800 $$ Calculate denominator: $$ \pi^3 \times 30 \times 10^6 \approx (31.0063) \times 30 \times 10^6 = 930,189,000 $$ So: $$ d^4 \geq \frac{1,301,548,800}{930,189,000} \approx 1.399 $$ 9. **Calculate $d$:** $$ d \geq \sqrt[4]{1.399} \approx 1.084 \text{ in} $$ 10. **Check yield stress condition:** Calculate cross-sectional area $A$: $$ A = \frac{\pi d^2}{4} = \frac{\pi (1.084)^2}{4} \approx 0.922 \text{ in}^2 $$ Calculate stress due to load: $$ \sigma = \frac{P}{A} = \frac{4500}{0.922} \approx 4882 \text{ psi} < 55000 \text{ psi} $$ Since $\sigma < \sigma_y$, the column will not yield before buckling. **Final answer:** $$ \boxed{d_{min} = 1.08 \text{ in}} $$