Question

A long steel column of uniform cross section is subjected to a crippling load of $10\text{ kN}$ as per the Euler designed condition. If the length of the column is reduced to half, then what would be the maximum Euler's crippling load?

• A. $50\text{ kN}$
• B. $40\text{ kN}$
• C. $20\text{ kN}$
• D. $10\text{ kN}$

Hint:

Because the column length term is squared in the denominator of Euler's formula, any change in length impacts the buckling load inversely by the square of that change: - Double the length $\rightarrow$ Critical load drops to $\frac{1}{4}$th. - Halve the length $\rightarrow$ Critical load increases by $4$ times.

Answer 1

The Correct Option is B

Concept: According to Euler's buckling theory, the critical crippling or buckling load $P_{\text{cr}}$ of a long, slender column with a uniform cross-section under an axial compressive load is given by the formula: \[ P_{\text{cr}} = \frac{\pi^2 E I}{L_e^2} \] where $E$ is the Young's Modulus of the material, $I$ is the minimum area moment of inertia of the cross-section, and $L_e$ is the effective length of the column, which depends on its end-support conditions. For a column with fixed end conditions, the effective length is directly proportional to its physical length $L$, which means: \[ P_{\text{cr}} \propto \frac{1}{L^2} \]

Step 1: Setting up the initial and final states.
Let the initial state parameters of the steel column be:

• Initial length = $L_1$

• Initial Euler crippling load, $P_1 = 10\text{ kN}$
The problem states that the length of the column is cut in half while keeping the cross-section and material properties exactly the same:

• New length, $L_2 = \frac{L_1}{2}$

• New Euler crippling load = $P_2$

Step 2: Using a ratio equation to solve for the new crippling load $P_2$.
Using the inverse-square relationship between the buckling load and the column length, we can set up the following ratio: \[ \frac{P_2}{P_1} = \left( \frac{L_1}{L_2} \right)^2 \] Substitute the relation $L_2 = \frac{L_1}{2}$ into the ratio: \[ \frac{P_2}{P_1} = \left( \frac{L_1}{\frac{L_1}{2}} \right)^2 = (2)^2 = 4 \]

Step 3: Calculating the numerical value of $P_2$.
Multiply the initial load by the calculated scaling factor: \[ P_2 = 4 \times P_1 = 4 \times 10\text{ kN} = 40\text{ kN} \] Thus, reducing the length of the column by half increases its resistance to buckling by a factor of four, giving a new critical load of $40\text{ kN}$. This matches Option (B).