Simple column buckling problem - pl help

[taureau20]

Knowing that the torsional spring is of constant K and that the rigid bar is of length L, determine the critical load $$P_{cr}$$ beyond which the column would buckle. See pic for the problem.
[As the spring uncurls, it applies moment $$K \theta$$ to the bar.]

The answer is $$P_{cr}=K/L$$

 

[taureau20]

I solved this problem.

 

[Mene]

^2. In order to determine the critical load at which the column would buckle, we need to consider both the torsional spring and the rigid bar. The torsional spring applies a moment of K \theta to the bar as it uncurls, where \theta is the angular displacement of the bar. This moment is directly proportional to the stiffness of the spring, K, and the angular displacement of the bar.

To understand how this moment affects the buckling of the column, we can use Euler's buckling formula, which states that the critical load for a column is given by P_{cr}=\frac{\pi^2EI}{(KL)^2}, where E is the modulus of elasticity and I is the moment of inertia of the column.

In this case, the column is replaced by the rigid bar, and the torsional spring is equivalent to the moment applied at the top of the bar. Therefore, we can substitute the moment K \theta for P_{cr} in the formula, giving us P_{cr}=\frac{\pi^2EI}{(KL)^2}=\frac{K \theta L^2}{(KL)^2}=\frac{K}{L^2}.

This shows that the critical load for buckling is directly proportional to the stiffness of the torsional spring and inversely proportional to the square of the length of the bar. Therefore, for a longer bar or a stiffer spring, the critical load will be higher.

In conclusion, the critical load for buckling in this simple column problem can be determined by using the formula P_{cr}=K/L^2, where K is the stiffness of the torsional spring and L is the length of the rigid bar.