Tubular column load and eccentric loading

(a) Euler load for the tube

$$I = \frac{\pi}{64}(80^4 - 64^4) = 1.187\times10^6\ \text{mm}^4$$ $$A = \frac{\pi}{4}(80^2 - 64^2) = 1809.6\ \text{mm}^2, \qquad k = \sqrt{\frac IA} = 25.6\ \text{mm}$$

The crippling load of the fixed-free column is one quarter of that of the hinged column. The Euler stresses are $143.9$ MPa and $36.0$ MPa, both below the yield stress and consistent with $\lambda>100$, so Euler's formula is applicable.

(b) Eccentrically loaded short column

Eccentricity is along the 250 mm side ($h = 250$ mm, $b = 150$ mm).

$$A = 150\times250 = 37\,500\ \text{mm}^2, \qquad Z = \frac{bh^2}{6} = \frac{150\times250^2}{6} = 1.5625\times10^6\ \text{mm}^3$$ $$\sigma_0 = \frac PA = \frac{200\times10^3}{37500} = 5.33\ \text{MPa}$$ $$\sigma_b = \frac{Pe}{Z} = \frac{200\times10^3\times40}{1.5625\times10^6} = 5.12\ \text{MPa}$$ $$\sigma_{max} = 5.33 + 5.12 = 10.45\ \text{MPa (compressive)}, \qquad \sigma_{min} = 5.33 - 5.12 = 0.21\ \text{MPa (compressive)}$$

No tension if $\sigma_{min} \ge 0$:

$$\frac{P}{A}\ge\frac{Pe}{Z} \;\Rightarrow\; e\le\frac ZA = \frac h6 = 41.7\ \text{mm}$$

The load must lie within the middle third of the depth (the core, kernel). Here $e = 40 < 41.7$ mm, so there is no tension.

Answer: (a) $P_{cr} = 260.4$ kN ($\lambda=117$) and $65.1$ kN ($\lambda=234$); (b) $\sigma_{max} = 10.45$ MPa, $\sigma_{min} = 0.21$ MPa, limit $e = h/6 = 41.7$ mm.