## Column

Consider a 5 m long column with a modulus of elasticity of 10 GPa and a moment of inertia of 400 cm^4 is subjected to buckling. If the column has a fixed-fixed connection, what is the critical axial load in kN?

Hint
Euler’s formula:
$$P_{cr}=\frac{\pi^2EI}{(Kl)^2}$$$where $$P_{cr}$$ is the critical axial load for long columns subjected to buckling, $$E$$ is the elastic modulus, $$I$$ is the moment of inertia, $$K$$ is the effective length factor to account for end supports, and $$l$$ is the unbraced column length. Hint 2 Theoretical effective length factors for columns include: Pinned-pinned - $$K=1.0$$ Fixed-fixed - $$K=0.5$$ Fixed-pinned - $$K=0.7$$ Fixed-free - $$K=2.0$$ Euler’s formula: $$P_{cr}=\frac{\pi^2EI}{(Kl)^2}$$$
where $$P_{cr}$$ is the critical axial load for long columns subjected to buckling, $$E$$ is the elastic modulus, $$I$$ is the moment of inertia, $$K$$ is the effective length factor to account for end supports, and $$l$$ is the unbraced column length. Theoretical effective length factors for columns include:
Pinned-pinned - $$K=1.0$$
Fixed-fixed - $$K=0.5$$
Fixed-pinned - $$K=0.7$$
Fixed-free - $$K=2.0$$
Thus,
$$P_{cr}=\frac{\pi^2(10\cdot 10^9Pa)(4\cdot 10^{-6}m^4)}{(0.5\cdot 5m)^2}=\frac{(\pi^2)40\cdot 10^3N\cdot m^4}{6.25m^2(m^2)}$$$$$=\frac{394,784.2N}{6.25}=63,165.5N=63\:kN$$$
63 kN