Solution:

Consider a rectangular section of dimension b x h reinforced with Ast amount of steel on tension side with effective cover Ce from tension extreme fiber to C.G of steel.

Then effective depth d=h-ce, measured from extreme compression fiber to C.G of steel strain and stress distribution across the section is shown in.

• The stress distribution is called stress block.

From similar triangle properly applied to strain diagram,

$$ \frac{\varepsilon c u}{x u}=\frac{\varepsilon s}{d-x u} \rightarrow.......(1) \\ $$

$$ \in s=\in c u \times \frac{d-x u}{x u} \rightarrow....(2) \\ $$

For the known value of $x4\ \&\ 6 \mathrm{cu}$ the strain in steel is used to get the value of stress in steel from stress-strain diagram. Equation 4.4-1 can be used to get the value of neutral axis depth as.

$$ x u=\frac{\in c u}{\in s} \times(d-x u)=\frac{\in c u}{\in s} \times d-\frac{\in c u}{\in s} \times x u \\ $$

$$ \begin{aligned} &x u\left(1+\frac{\in c u}{\in s}\right)=\frac{\in c u}{\in s} \times d \\\\ &x u\left(\frac{\in s+\in c u}{\in s}\right)=\frac{\in c u}{\in s} \times d \\\\ &\therefore x u=\frac{\in c u}{\in c u+E s} \times d...(3) \\ \end{aligned} $$

Here, $ \frac{\epsilon c u}{\epsilon c u+\epsilon s} $ is called neutral axis factor,

For equilibrium, $ \mathrm{Cu}=\mathrm{Tu} \\ $.

$ K_{1}, k_{3} f_{c u} b x_{u}=f_{s} A_{s} \\ $

$ \therefore f_{S}=\frac{K 1, k 3 f c u b x u}{A s}=\frac{k 1 k 3 f c u b}{A s} \times \frac{\in c u}{\in c u+\in s} \times d \\ $

$f s=k 1 k 3 f c u \times \frac{E c u}{E c u+E s} \times \frac{b d}{A s}$ Let $p=$ steel raio $=\frac{A s}{b d} \\$ $$ \therefore f s=\frac{k 1 k 3 f c u}{p} \times \frac{E c u}{E c n+E s} \text { or } \frac{E c u}{E c u+E s}=\frac{f s p}{k 1 k 3 f c u}.....(4) \ $$

Value of fs can be graphically computed for a given value of P as shown in Figure.

After getting fs graphically, the ultimate moment or ultimate moment of resistance is calculated as,

$$ \begin{aligned} &M u=T_{u} \times Z=f_{s} A_{s}\left(d-k_{2} \times u\right) \\\\ &M u=C u \times Z=k_{1} k_{2} f_{c u} b x_{u} \times\left(d-k_{2} \times u\right) \\ \end{aligned} $$

Consider,

$$ \begin{aligned} &\mathrm{Mu}=\mathrm{f}_{\mathrm{s}} \mathrm{A}_{s}\left(\mathrm{~d}-\mathrm{k} 2 \times \frac{\epsilon c u}{\epsilon c u+\epsilon s} d\right)=\mathrm{fsAsd}\left(1-\mathrm{kz} \frac{\epsilon c u}{\epsilon c u+\epsilon s}\right) \\\\ &\text { From (4) } \frac{E c u}{E c u+E s}=\frac{\mathrm{f}_{\mathrm{s}} p}{k 1 k 3 f c m} \\\\ &\therefore M u=f s A s d\left(1-\frac{k 2 f s p}{k 1 k 3 f u}\right).....(5) \\ \end{aligned} $$

Here the term, $ 1-\frac{k 2 f s p}{k 1 k 3 f c u} $ is called lever arm factor,

Using As=pbd in (5), the ultimate moment of resistance is computed as,

$$ M u=f s(p b d) d\left(1-\frac{k 2 f s p}{k 1 k 3 f c u}\right) \text { Let } R=\left(1-\frac{f s p}{f c u} \times \frac{k 2}{k 1 k 3}\right) \\ $$

$\frac{M u}{b d^{2}}=p f s x R \quad$ Dividing both sides by fcu we get or,

$$ \frac{M u}{b d^{2}}=p \times \frac{f s}{f c u} \times R.....(6) \\ $$

A graph plotted between $\frac{M u}{f c u b d^{2}}$ as shown in figure and can be used for design.