## 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?

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