Question

A vertical column fixed at one end is subjected to a compressive axial load at the free end. The column’s section modulus, \(EI = 9.82 \times 10^5 \, \text{Nm}^2\), and the cross-sectional area is \(A = 7.85 \times 10^{-3} \, \text{m}^2\). The length of the column is \(L = 2 \, m\). The yield stress of the material is \(145 \, \text{MPa}\). If the column can fail either in buckling or by Tresca’s criterion, the maximum load that the structure can safely sustain is ............... kN (rounded off to one decimal place).

Hint:

For columns: - Short columns fail by material yielding (\(P_y = \sigma_y A\)). - Long slender columns fail by buckling (\(P_{cr} = \pi^2 EI / (KL)^2\)). The smaller governs safe load.

Answer 1

Correct Answer: 603.5

Step 1: Load by material yielding (Tresca criterion).
Axial stress = \(\sigma = \dfrac{P}{A}\). For yielding: \[ P_y = \sigma_y \cdot A = 145 \times 10^6 \times 7.85 \times 10^{-3} \] \[ P_y = 1.137 \times 10^6 \, N = 1137.0 \, kN \]

Step 2: Load by Euler buckling.
Euler’s critical load: \[ P_{cr} = \frac{\pi^2 EI}{(K L)^2} \] For fixed-free column: \(K = 2\), so effective length = \(L_{eff} = 2L = 4 \, m\). \[ P_{cr} = \frac{\pi^2 (9.82 \times 10^5)}{(4)^2} \] \[ P_{cr} = \frac{9.87 \times 9.82 \times 10^5}{16} \] \[ P_{cr} \approx 6.07 \times 10^5 \, N = 607.0 \, kN \]

Step 3: Governing failure load.
The actual maximum safe load is the lower of the two: \[ P_{safe} = \min(P_y, P_{cr}) = \min(1137, 607) = 607 \, kN \] Final Answer:
\[ \boxed{607.0 \, kN} \]