A square cross-section aluminum column of side b = 20 mm and length L = 1.0 m is pinned at both ends (E = 70 GPa). What is the Euler critical buckling load P_{cr} in kilonewtons (kN)? (Take pi^2 ≈ 9.8696, round to two decimal places).

Answer Key:

9.21

### Solution: Second moment of area I = b^4 / 12 = (0.02)^4 / 12 = 1.6 imes 10^{-7} / 12 = 1.3333 imes 10^{-8} m^4. P_{cr} = \pi^2 EI / L^2 = 9.8696 * (70 imes 10^9) * (1.3333 imes 10^{-8}) / (1.0)^2 = 9.8696 * 933.33 / 1.0 = 9211.6 N ≈ 9.21 kN. Correct: 9.21 (range 9.10 to 9.30).

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