A column with one end fixed and one end free has a critical buckling load of 100 N. For the same column, if the free end is replaced with a pinned end then the critical buckling load will be ___________N (round off to the nearest integer)

The correct answer is 800.

To understand why the buckling load increases to 800 N, we start with Euler's buckling theory for columns. The critical buckling load (P_cr) of a column is given by the formula:
$$P_{cr}=\frac{{\pi }^{2}EI}{L_e^2}$$
where:
E is the Young's modulus of the material,
I is the minimum second moment of area of the column cross-section, and
L_e is the effective length of the column, which depends on its end support conditions.

Since the physical column (including its material properties, cross-section, and actual length L) remains the same, the product π²EI is a constant. Thus, the critical buckling load is inversely proportional to the square of the effective length:
$$P_{cr}\propto \frac{1}{L_e^2}$$

Let us analyze the two support conditions:
1. Case 1: One end fixed and the other end free
For a column of actual length L under these end conditions, the effective length (L₁) is twice the actual length:
L₁ = 2L
The critical buckling load in this condition is given as P₁ = 100 N.

2. Case 2: One end fixed and the other end pinned (hinged)
If the free end is replaced with a pinned end, the boundary conditions change. For a fixed-pinned column, the effective length (L₂) is:
$$L_2=\frac{L}{\sqrt{2}}$$

Now, we can find the new critical buckling load (P₂) by taking the ratio of the two loads:
$$\frac{P_2}{P_1}=\frac{L_1^2}{L_2^2}$$

Substituting the expressions for L₁ and L₂:
$$\frac{P_2}{100}=\frac{{\left(2L\right)}^2}{{\left(\frac{L}{\sqrt{2}}\right)}^2}$$
$$\frac{P_2}{100}=\frac{4L^2}{\left(\frac{L^2}{2}\right)}$$
$$\frac{P_2}{100}=4\times 2=8$$
Solving for P₂:
$$P_2=100\times 8=800\text{ N}$$

Therefore, replacing the free end with a pinned end reduces the effective length of the column, which subsequently increases the critical buckling load to 800 N.