Question

The Euler buckling load of a column with square cross-section (side length 5 cm) will be _____ times the Euler buckling load of an equivalent column with circular cross-section (diameter 5 cm).

Answer 1

The formula for Euler's buckling load, is:

$P=\frac{\pi^2EI}{L^2}$

Where E is the Young's modulus of the material, I is the cross-sectional area moment of inertia of the column, and L is the length of the column.

If we keep the length and the Young's modulus constant, the ratio between the buckling load of two columns with different cross-sectional area moment of inertia (a square and a circle) will be:

$\frac{P_s}{P_c}=\frac{(\frac{\pi^2EI_s}{L^2})}{(\frac{\pi^2EI_c}{L^2})}=\frac{I_s}{I_c}$

Where the subindex s refers to the square and the subindex c refers to the circle.

The area moment of inertia of a square with side S is given by:

$I_s=\frac{S^4}{12}$

The area moment of inertia of a circle with diameter S is given by:

$I_c=\pi\cdot\frac{S^4}{64}$

Notice that in this case, the side length of the square is the same as the diameter of the circle. Then:

$\frac{I_s}{I_c}=\frac{(\frac{S^4}{12})}{(\frac{\pi S^4}{64})}=\frac{64S^4}{12\pi S^4}=\frac{64}{12\pi}=\frac{16}{3\pi}\approx1.7$

Then:

$\begin{gathered} \frac{P_s}{P_c}\approx1.7 \\ \Rightarrow P_s\approx1.7P_c \end{gathered}$

Therefore, the Euler's buckling load of a column with square cross-section with side length 5cm will be 16/(3π) times the Euler buckling load of an equivalent column with circular cross-section and diameter 5cm.

Approximately, the Euler buckling load of a column with square cross-section will be 1.7 times that of a column with circular cross section.