Columns

Assumptions

1. The column is initially perfectly straight and the load is purely axial.
2. The material is homogeneous, isotropic and obeys Hooke's law (stresses within the proportional limit).
3. The cross-section is uniform and the column bends in the plane of least resistance.
4. The length is large compared with the lateral dimensions; the effect of direct compression is neglected.
5. Self weight is neglected; the end conditions are ideal.

Derivation (both ends hinged)

    P (down)
     |
     v
     o A (hinge)        x measured from A
     |
     |  y (lateral deflection)
     )
     |
     o B (hinge)
     ^
     |
    P (up)

Column of length $L$ with load $P$, buckled into a slightly bent shape. At a section distance $x$ from A the lateral deflection is $y$, so the bending moment is $M = -Py$ (the load tends to increase the deflection):

$$EI\frac{d^2y}{dx^2} = -Py \;\Rightarrow\; \frac{d^2y}{dx^2} + \alpha^2y = 0, \qquad \alpha^2 = \frac{P}{EI}$$

Solution: $y = A\cos\alpha x + B\sin\alpha x$.

• $x = 0,\ y = 0 \Rightarrow A = 0$.
• $x = L,\ y = 0 \Rightarrow B\sin\alpha L = 0$. For a buckled shape $B\neq0$, so $\sin\alpha L = 0$, $\alpha L = n\pi$.

The smallest non-trivial value is $n = 1$: $\alpha^2 = \pi^2/L^2$:

$$\boxed{P_{cr} = \frac{\pi^2EI}{L^2}}$$

$I$ is the least second moment of area, $I_{min} = Ak^2$, $k$ being the least radius of gyration. The buckled shape is a half sine wave.

Other end conditions

$$P_{cr} = \frac{\pi^2EI}{L_e^2}$$

Slenderness ratio and limitation

Slenderness ratio $\lambda = L_e/k$. The Euler stress is

$$\sigma_{cr} = \frac{P_{cr}}{A} = \frac{\pi^2E}{\lambda^2}$$

Euler's formula holds only for long columns, for which $\sigma_{cr}$ is below the proportional limit. For mild steel ($E = 200$ GPa, $\sigma_p \approx 250$ MPa) this needs $\lambda \gtrsim 90$ to $100$. For short and intermediate columns Euler's formula overestimates the strength, and Rankine's or Johnson's formulas are used.

(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.