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