Required Steel Area for a Concrete Beam

Hardflexuresteel-areastrength-design

A singly reinforced rectangular concrete beam has b = 300 mm, effective depth d = 450 mm, f′c = 30 MPa, fy = 400 MPa, and factored moment Mu = 160 kN·m. Use a = Asfy/(0.85f′cb), Mn = Asfy(d − a/2), and φ = 0.90; there is no compression steel, and the steel is assumed to yield. Find the theoretical required tension-steel area As (the small physical root, not a bar layout).

A940 mm²
B16,160 mm²
C1,052 mm²
D988 mm²

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Answer: C. 1,052 mm²

Equation: Mu/φ = Mn = 0.85f′cb·a(d − a/2), with As = 0.85f′cb·a/fy. Inputs: Mn = 160/0.90 = 177.78 kN·m = 177.78×10⁶ N·mm; 0.85f′cb = 0.85(30)(300) = 7,650 N/mm. Substitution: a = d − √[d² − 2Mn/(0.85f′cb)] = 450 − √[202,500 − 46,478] = 450 − 395.00 = 55.00 mm. As = 7,650(55.00)/400 = 1,052 mm². Check: c = a/0.8357 = 65.8 mm, εt = 0.003(450 − 65.8)/65.8 = 0.0175 > 0.005, so yielding and φ = 0.90 are valid, and 1,052 exceeds the minimum 472.5 mm². Wrong: 940 mm² uses Mu as Mn (forgets to divide by φ); 16,160 mm² takes the large quadratic root a ≈ 845 mm, which is not physical; 988 mm² uses the full depth as lever arm, As = Mn/(fyd), ignoring a/2.

Equation:

M_{u} / ϕ

= A_{s} f_{y} (d - \frac{a}{2}), a

= \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b}

\Rightarrow a

= d - d^{2} - \frac{2 M _{n}}{0.85 f _{c}^{'} b}

= 450 - 45 0^{2} - \frac{2 ( 177.78 \times 1 0 ^{6} )}{7650}

= 55.0

; A_{s}

= \frac{7650 ( 55.0 )}{400}

= 1052

^{2}

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Required Steel Area for a Concrete Beam

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