Minimum Steel Area in a Tied Column
Hardcolumnreinforcement-areaaxial-strength
A short tied column has Ag = 160,000 mm², f′c = 30 MPa, fy = 420 MPa, and factored axial load Pu = 3,000 kN. Use Pu ≤ 0.80φ[0.85f′c(Ag − Ast) + fyAst] with φ = 0.65, ignoring slenderness and eccentricity. Find the theoretical minimum total steel area Ast (it should fall within a 1%–8% ratio).
- A4,282 mm²
- B4,022 mm²
- C1,357 mm²
- D2,485 mm²
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Answer: A. 4,282 mm²
Equation: Pu/(0.80φ) = 0.85f′cAg + (fy − 0.85f′c)Ast. Inputs: required P₀ = 3,000,000/0.52 = 5,769,231 N; 0.85f′cAg = 4,080,000 N; fy − 0.85f′c = 394.5 MPa. Substitution: Ast = (5,769,231 − 4,080,000)/394.5 = 4,282 mm² (ratio 2.68%, within 1%–8%). Wrong: 4,022 mm² uses fy alone as the steel increment, ignoring displaced concrete; 1,357 mm² omits the 0.80 cap (Pu/φ only); 2,485 mm² drops the 0.85 factor on f′c.
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