## 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**

**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**

**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:

$$$I=\frac{P_{cr}(Kl)^2}{\pi^2E}=\frac{50,000N(2\cdot 6m)^2}{\pi^2(20\cdot10^9Pa)}=\frac{50,000N(144m^2)m^2}{\pi^2(20\cdot10^9N)}$$$

$$$=\frac{50,000(144m^4)}{\pi^2(20\cdot10^9)}=\frac{7.2\cdot 10^6m^4}{1.974\cdot 10^{11}}=3.65\cdot 10^{-5}m^4=3,650\:cm^4$$$

$$$3,650\:cm^4$$$