Straight bar buckling: maximum central deflection

Given: $L = 1.5$ m = 1500 mm, section $12\times5$ mm, hinged ends, $E = 2\times10^5\ \text{N/mm}^2$, yield stress $290\ \text{N/mm}^2$.

The bar buckles about the axis of least resistance (thickness 5 mm).

$$I_{min} = \frac{12\times5^3}{12} = 125\ \text{mm}^4,\qquad A = 60\ \text{mm}^2,\qquad Z = \frac{I}{y} = \frac{125}{2.5} = 50\ \text{mm}^3$$

Euler's buckling load

$$P_{cr} = \frac{\pi^2EI}{L^2} = \frac{\pi^2\times2\times10^5\times125}{1500^2} = 109.66\ \text{N}$$

Direct stress at this load:

$$\sigma_d = \frac{P_{cr}}{A} = \frac{109.66}{60} = 1.83\ \text{N/mm}^2$$

Central deflection at yield

After buckling, the load remains at $P_{cr}$ while the deflection $\delta$ grows. The maximum stress at mid-span is direct stress plus bending stress:

$$\sigma_{max} = \frac{P}{A} + \frac{P\,\delta}{Z}$$ $$290 = 1.83 + \frac{109.66\,\delta}{50}$$ $$\delta = \frac{(290 - 1.83)\times50}{109.66} = 131.4\ \text{mm}$$

Answer: Euler load = 109.7 N; maximum central deflection before yielding $\approx 131$ mm. This is an estimate, because the small-deflection theory is used.