## 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**

**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 -
$$K=2.0$$

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