Question

A force \( F \) is applied at an angle \( \theta = 30^\circ \) on an elastic column as shown in the figure. \(E\) and \(I\) are respectively the Young's modulus and the area moment of inertia. The smallest magnitude of \(F\) needed to cause buckling is:

• A. ( \dfrac{2\pi^2 EI}{\sqrt{3}\,L^2} \)
• B. ( \dfrac{\sqrt{3}\,\pi^2 EI}{2L^2} \)
• C. ( \dfrac{\pi^2 EI}{2L^2} \)
• D. ( \dfrac{2\pi^2 EI}{L^2} \)

Hint:

When a load is inclined, only its axial component contributes to Euler buckling. Always resolve the load before applying the Euler formula.

Answer 1

The Correct Option is A

The axial component of the load causes buckling. The load \(F\) is applied at an angle \(30^\circ\), so its axial component is:
\[ F_{\text{axial}} = F \cos 30^\circ = \frac{\sqrt{3}}{2} F \] For a pinned–pinned column of length \(L\), the Euler buckling load is:
\[ P_{\text{cr}} = \frac{\pi^2 EI}{L^2} \] At the onset of buckling, the axial component equals the critical load:
\[ \frac{\sqrt{3}}{2}F = \frac{\pi^2 EI}{L^2} \] Solving for \(F\):
\[ F = \frac{2}{\sqrt{3}} \cdot \frac{\pi^2 EI}{L^2} \] Thus, the minimum force required is:
\[ \boxed{\frac{2\pi^2 EI}{\sqrt{3}\,L^2}} \]