Column Modulus
Consider a 6 m long column with a critical axial load of 70 kN and a moment of inertia of 200 cm^4 is subjected to buckling. If the column has a fixed-pinned connection, what is the modulus of elasticity in GPa?
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 modulus of elasticity:
$$$70,000N=\frac{\pi^2E(2\cdot10^{-6}m^4)}{(0.7\cdot 6m)^2}$$$
$$$70,000N=\frac{E(1.97\cdot10^{-5}m^2)}{17.64}$$$
$$$E=\frac{70,000N(17.64)}{1.97\cdot10^{-5}m^2}=6.3\cdot 10^{10}Pa=63\:GPa$$$
63 GPa
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