Column Area
Consider a 10 m long column with an elastic modulus of 17 GPa and a fixed-pinned connection is subjected to buckling. If the critical axial load is 50 kN and the critical buckling stress is 1,000 kPa, what is the column’s cross-sectional area in m^2?
Expand Hint
Critical buckling stress for long columns:
$$$\sigma _{cr}=\frac{P_{cr}}{A}=\frac{\pi ^{2}E}{(KL/r)^{2}}$$$
where
$$P_{cr}$$
is the critical axial load,
$$A$$
is the cross-sectional area,
$$E$$
is the modulus of elasticity,
$$L$$
is the unbraced column length,
$$K$$
is the effective-length factor to account for end supports, and
$$r$$
is the radius of gyration.
Hint 2
Only the critical axial load and critical buckling stress are needed to solve this problem. The other givens simply exist to cause confusion and/or take up time.
Critical buckling stress for long columns:
$$$\sigma _{cr}=\frac{P_{cr}}{A}=\frac{\pi ^{2}E}{(KL/r)^{2}}$$$
where
$$P_{cr}$$
is the critical axial load,
$$A$$
is the cross-sectional area,
$$E$$
is the modulus of elasticity,
$$L$$
is the unbraced column length,
$$K$$
is the effective-length factor to account for end supports, and
$$r$$
is the radius of gyration.
Notice how only the critical axial load and critical buckling stress are needed to solve this problem. The other givens simply exist to cause confusion and/or take up time. Thus,
$$$1,000kPa=\frac{50kN}{A}$$$
$$$A=\frac{50,000N}{1,000,000N/m^2}=0.05\:m^2$$$
$$$0.05\:m^2$$$
Time Analysis
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