A solid circular aircraft tie-rod of diameter $d = 25\text{ mm}$ and length $L = 1.5\text{ m}$ is pinned at both ends ($E = 72\text{ GPa}$). What is the Euler buckling load $P_{cr}$ in kN? (Take $\pi \approx 3.1416$, round to two decimal places).

Answer Key:

6.08

### Solution: $I = \frac{\pi d^4}{64} = \frac{3.1416 \times (0.025)^4}{64} = \frac{3.1416 \times 3.90625 \times 10^{-7}}{64} = 1.9175 \times 10^{-8}\text{ m}^4$. $P_{cr} = \frac{\pi^2 E I}{L^2} = \frac{\pi^2 \times (72 \times 10^9) \times (1.9175 \times 10^{-8})}{(1.5)^2} = \frac{9.8696 \times 1380.6}{2.25} = \frac{13626}{2.25} \approx 6056\text{ N} \approx 6.06\text{ kN}$. Correct: 6.06 (range 6.00 to 6.15).

← Prev Question Next Question →