## A Column

Consider a 6 m long column with an elastic modulus of 20 GPa and a fixed-free connection is subjected to buckling. If the critical axial load is 50 kN, what is the column’s moment of inertia in cm^4?

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 - $$K=2.0$$
Solving for moment of inertia:
$$I=\frac{P_{cr}(Kl)^2}{\pi^2E}=\frac{50,000N(2\cdot 6m)^2}{\pi^2(20\cdot10^9Pa)}=\frac{50,000N(144m^2)m^2}{\pi^2(20\cdot10^9N)}$$$$$=\frac{50,000(144m^4)}{\pi^2(20\cdot10^9)}=\frac{7.2\cdot 10^6m^4}{1.974\cdot 10^{11}}=3.65\cdot 10^{-5}m^4=3,650\:cm^4$$$
$$3,650\:cm^4$$\$