Steel Column Design Compressive Strength
Hardcolumn bucklingcompression
A steel column has no slender elements, KL/r = 70, E = 200,000 MPa, Fy = 250 MPa, Ag = 4,000 mm² and φc = 0.90. Use Fe = π²E/(KL/r)². If Fy/Fe ≤ 2.25, Fcr = 0.658^(Fy/Fe)·Fy; otherwise Fcr = 0.877Fe. Only flexural buckling governs. Find φcPn.
- A694.1 kN
- B1,271.8 kN
- C771.2 kN
- D900.0 kN
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Answer: A. 694.1 kN
Fe = π²(200,000)/70² = 402.84 MPa. Fy/Fe = 0.6206 ≤ 2.25, so the inelastic branch applies: Fcr = 0.6580.6206(250) = 192.81 MPa. Pn = 192.81(4,000) = 771.2 kN; φcPn = 0.90(771.2) = 694.1 kN. Wrong: 1,271.8 kN uses the elastic 0.877Fe branch (353.3 MPa) even though Fy/Fe is below 2.25; 771.2 kN omits φc; 900 kN is the squash load φcFyAg, ignoring buckling.
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