# Required steel area of a concrete section

## Symbols

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

## Method

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:

$A_{s} = \frac{0.8 f_{cd} x b}{f_{yd}}$

## Derivation

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}

## References

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:

• Figure 3.3 of BS 8110-1:1997