Buckling Stress
Consider a 8 m long column with an elastic modulus of 15 GPa and a fixed-pinned connection is subjected to buckling. If the critical axial load is 40 kN and its cross-sectional area is 500 cm^2, what is the column’s critical buckling stress in kPa?
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 area 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 area are needed to solve this problem. The other givens simply exist to cause confusion and/or take up time. Thus,
$$$\sigma _{cr}=\frac{P_{cr}}{A}=\frac{40,000N}{0.05m^2}=800,000N/m^2=800\:kPa$$$
800 kPa
Time Analysis
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