| Symbol | Description |
|---|---|
| \(b\) | breadth of section |
| \(h\) | height of section |
| \(d\) | effective depth of section |
| \(M_{Ed}\) | design bending moment |
| \(f_{cd}\) | design compressive strength of concrete |
| \(f_{yd}\) | design tensile strength of reinforcement |
| \(x\) | neutral axis depth |
| \(A_{s}\) | steel area |
Begin by estimating the neutral axis at the midpoint:
\[ x_{0} = \frac{h}{2} \]
Then use the following recurrence relationship to iterate out:
\[ x_{i+1} = \frac{M_{Ed}}{0.8 f_{cd} b d} + 0.4\frac{x_{i}^{2}}{d} \]
Finally, calculate minimum required area of steel:
\[ A_{s} = \frac{0.8 f_{cd} x b}{f_{yd}} \]
Balance axial forces in Figure 1.
\[\begin{align} A_{s} f_{yd} & = f_{cd} \times 0.8xb \\ A_{s} &= \frac{f_{cd}}{f_{yd}}0.8xb \label{a} \tag{Eq.1} \end{align}\]
Write equation for net moment in Figure 1.
\[\begin{align} M_{Ed} = A_{s} f_{yd} (d - x) + f_{cd} \times 0.8xb \times 0.6 x \end{align}\]
Substitute expression for \(A_{s}\) from \(\ref{a}\)
\[\begin{align} M_{Ed} & = \frac{f_{cd}}{f_{yd}}0.8 x b f_{yd}(d - x) + f_{cd} \times 0.8xb \times 0.6 x \\ \frac{M_{Ed}}{0.8f_{cd} bd} & = (d - x) \frac{x}{d} + 0.6 \frac{x^{2}}{d} \\ \frac{M_{Ed}}{0.8f_{cd} bd} & = x - \frac{x^{2}}{d} + 0.6 \frac{x^{2}}{d} \\ x & = \frac{M_{Ed}}{0.8f_{cd} bd} + 0.4\frac{x^{2}}{d} \end{align}\]
This can be solved for \(x\) using the quadratic equation, but it’s generally fewer calculator button presses if you solve it as a recurrence relationship:
\[ x_{i+1} = \frac{M_{Ed}}{0.8 f_{cd} b d} + 0.4\frac{x_{i}^{2}}{d} \]
Then obtain area of steel from \(\ref{a}\):
\[\begin{align} A_{s} &= \frac{f_{cd}}{f_{yd}}0.8xb \end{align}\]
This is derived using the assumptions of the rectangular stress block method that appears in Figure 3.5 of EN 1992-1-1:2004+A1:2014 where \(f_{ck} \leq 50\textrm{ MPa}\)
Similar rules are present in other design codes: