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?
Expand 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
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