Understanding Column Buckling Load

The buckling load is a critical parameter in the design of columns, representing the maximum axial compressive load a column can withstand before it becomes unstable and deflects laterally. For long columns, this load is determined by Euler's buckling formula.

Euler's critical buckling load (\(P_e\)) for a column is given by the formula:

\[P_e = \frac{\pi^2 EI_{min}}{(L_e)^2}\]

Where:

• \(P_e\) is the Euler's buckling load.
• \(E\) is the Modulus of Elasticity of the column material. For this problem, it is assumed to be the same for both the original and redesigned columns.
• \(I_{min}\) is the minimum moment of inertia of the column's cross-section. A column tends to buckle about the axis with the minimum moment of inertia.
• \(L_e\) is the effective length of the column. It is given by \(L_e = K L\), where \(L\) is the actual length (or height \(h\) in this question) and \(K\) is the effective length factor. The effective length factor depends on the end conditions of the column and is assumed to be constant for both columns in this problem.

Moment of Inertia Calculation for Rectangular Cross-Sections

For a rectangular cross-section with sides \(b\) and \(d\), the moments of inertia about the centroidal axes are calculated as follows:

• Moment of inertia about the axis parallel to side \(b\): \(I_x = \frac{db^3}{12}\)
• Moment of inertia about the axis parallel to side \(d\): \(I_y = \frac{bd^3}{12}\)

The minimum moment of inertia (\(I_{min}\)) will always be about the axis parallel to the longer side of the rectangle, as this results in the smaller value for the formula \(\frac{base \times height^3}{12}\).

Original Column Parameters and Buckling Load

Let's consider the initial properties of the column:

• Original Height (\(h_1\)): \(h\)
• Original Rectangular Cross-section: \(a \times 2a\)
• Original Buckling Load (\(P_1\)): \(P\)

To find the minimum moment of inertia (\(I_{min1}\)) for the original cross-section \(a \times 2a\):

• If the axis is parallel to the side of length \(2a\), then \(b=2a\) and \(d=a\). The moment of inertia is \(I_x = \frac{a(2a)^3}{12} = \frac{8a^4}{12} = \frac{2a^4}{3}\).
• If the axis is parallel to the side of length \(a\), then \(b=a\) and \(d=2a\). The moment of inertia is \(I_y = \frac{2a(a)^3}{12} = \frac{2a^4}{12} = \frac{a^4}{6}\).

Comparing the two values, the minimum moment of inertia for the original column is \(I_{min1} = \frac{a^4}{6}\).

Applying Euler's formula for the original column:

\[P_1 = P = \frac{\pi^2 E I_{min1}}{(K h_1)^2} = \frac{\pi^2 E \left(\frac{a^4}{6}\right)}{(K h)^2}\]

\[P = \frac{\pi^2 E a^4}{6 K^2 h^2} \quad \ldots(1)\]

Redesigned Column Parameters and Buckling Load

Now, let's look at the properties of the redesigned column:

• Redesigned Height (\(h_2\)): \(1.5h\)
• Redesigned Rectangular Cross-section: \(0.5a \times 3a\)
• Redesigned Buckling Load (\(P_2\)): \(P'\) (this is what we need to calculate)

To find the minimum moment of inertia (\(I_{min2}\)) for the redesigned cross-section \(0.5a \times 3a\):

• If the axis is parallel to the side of length \(3a\), then \(b=3a\) and \(d=0.5a\). The moment of inertia is \(I_x = \frac{0.5a(3a)^3}{12} = \frac{0.5a \times 27a^3}{12} = \frac{13.5a^4}{12} = \frac{9a^4}{8}\).
• If the axis is parallel to the side of length \(0.5a\), then \(b=0.5a\) and \(d=3a\). The moment of inertia is \(I_y = \frac{3a(0.5a)^3}{12} = \frac{3a \times 0.125a^3}{12} = \frac{0.375a^4}{12} = \frac{a^4}{32}\).

Comparing the two values, the minimum moment of inertia for the redesigned column is \(I_{min2} = \frac{a^4}{32}\).

Applying Euler's formula for the redesigned column:

\[P_2 = P' = \frac{\pi^2 E I_{min2}}{(K h_2)^2} = \frac{\pi^2 E \left(\frac{a^4}{32}\right)}{(K \times 1.5h)^2}\]

\[P' = \frac{\pi^2 E a^4}{32 K^2 (1.5)^2 h^2} = \frac{\pi^2 E a^4}{32 K^2 (2.25) h^2}\]

\[P' = \frac{\pi^2 E a^4}{72 K^2 h^2} \quad \ldots(2)\]

Comparing Buckling Loads

To determine the new buckling load \(P'\) in terms of the original load \(P\), we can divide equation (2) by equation (1):

\[\frac{P'}{P} = \frac{\frac{\pi^2 E a^4}{72 K^2 h^2}}{\frac{\pi^2 E a^4}{6 K^2 h^2}}\]

Notice that the terms \(\pi^2 E a^4\), \(K^2\), and \(h^2\) are common in both the numerator and denominator, so they cancel out:

\[\frac{P'}{P} = \frac{1/72}{1/6}\]

\[\frac{P'}{P} = \frac{6}{72}\]

\[\frac{P'}{P} = \frac{1}{12}\]

Therefore, the buckling load of the redesigned column will be:

\[P' = \frac{P}{12}\]

Summary of Buckling Load Factors

The buckling load is directly proportional to the minimum moment of inertia and inversely proportional to the square of the effective length (or height, if K is constant).

The new buckling load \(P'\) can be expressed as:

\[P' = P \times \frac{I_{min2}}{I_{min1}} \times \left(\frac{h_1}{h_2}\right)^2\]

\[P' = P \times \left(\frac{a^4/32}{a^4/6}\right) \times \left(\frac{h}{1.5h}\right)^2\]

\[P' = P \times \left(\frac{6}{32}\right) \times \left(\frac{1}{1.5}\right)^2\]

\[P' = P \times \frac{3}{16} \times \frac{1}{2.25}\]

\[P' = P \times \frac{3}{16 \times 2.25} = P \times \frac{3}{36}\]

\[P' = P \times \frac{1}{12}\]

This calculation confirms that the buckling load of the redesigned column is \(\frac{P}{12}\).