A solid circular aircraft control rod of diameter $d = 20\text{ mm}$ and length $L = 1.2\text{ m}$ is pinned at both ends. Taking Young's modulus $E = 70\text{ GPa}$ for aluminum, what is the Euler critical buckling load $P_{cr}$ in kilonewtons (kN)? (Take $\pi \approx 3.1416$, round to two decimal places).

Answer Key:

3.75

### Solution: Moment of inertia $I = \frac{\pi d^4}{64} = \frac{3.1416 \times (0.02)^4}{64} = \frac{3.1416 \times 1.6 \times 10^{-7}}{64} = 7.854 \times 10^{-9}\text{ m}^4$. Buckling load $P_{cr} = \frac{\pi^2 E I}{L^2} = \frac{\pi^2 \times (70 \times 10^9) \times (7.854 \times 10^{-9})}{(1.2)^2} = \frac{9.8696 \times 549.78}{1.44} = \frac{5426.1}{1.44} \approx 3768\text{ N} \approx 3.77\text{ kN}$. Correct: 3.75 to 3.80.

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