Solution:

What is stress blocks?

Stress blocks adopted by different codes are based on the stress blocks proposed by different investigators. Among them that proposed by Hog nested and Whitney equivalent rectangular block are used by most of the codes

Case 1: Balanced section:

Balanced section is considered when the ultimate strain in concrete and in steel are reached simultaneously before collapse.

For equilibrium Cu=Tu.

$$ \begin{aligned} &\therefore \quad 0.36 \mathrm{f}_{\mathrm{ck}} \mathrm{x}_{\mathrm{umax}} \mathrm{b}=0.87 \mathrm{f}_{\mathrm{y}} \text { Ast }_{\max } \cdot \\\\ &\mathrm{x}_{\mathrm{umax}}=\frac{0.87 f y}{0.36 f c k} \frac{\text { Ast } \max }{b} \text { dividing both sides by } \\\\ &\frac{x u \max }{d}=\frac{0.87 f y}{0.36 f c k} \frac{\text { Astmax }}{b d} \text { but } \frac{\text { Ast max }}{b d}=\mathrm{pt}_{\operatorname{masc} \cdot} \\\\ &\left.\therefore p t_{\max }=\left(\frac{x u \max }{d}\right) \times \frac{0.36 f c k}{0.87 f y}\right\} \\\\ &\left.p t_{\max }=\frac{x u \max }{d} \times 0.414 \frac{f c k}{f u}-\right\}....(1) \\\\ &\text { From strain diagram } \frac{0.0035}{x u m a c}=\frac{0.002+\frac{0.87 f y}{E s}}{x u m a x-d} \\\\ &\qquad \frac{x u m a x}{d}=\frac{0.0035}{\frac{0.87 f y}{E s}+0.0055}.....(2) \end{aligned} \\ $$

Values of $\frac{x u m a x}{d}$ is obtained from equation $(2)$. This value depends on grade of steel. Based on grade of steel this value is given in note of clause $38.1$ as (pp70).

Fy $\quad$ Xumax $/ d$

$250 \quad 0.53 $

$415 \quad 0.48$

$500 \quad 0.46(0.456)$

$\mathrm{p}_{\text {tmax }}$ given in equation (1) is called limiting percentage steel and denoted as pt lim. To find moment of resistance, the internal moment of $\mathrm{Cu} \& \mathrm{Tu}$ is computed as.

Mulim = Cu, $ \times Z=0.36 f_{c k} x_{\text {ulim }} b\left(d-0.42 x_{\text {ulim }}\right) \\ $

From equation (1), $ \frac{x u m a x}{d}=2.42 \frac{f y}{f c k} \mathrm{p}_{\mathrm{tmax}} \\ $

Mulim = Tu, $x Z=0.87$ fyAst $[d-0.42 x u l i m] $

Mulim, $ =0.87 f y A s t\left[d-0.42 \times 2.42 \frac{f y}{f c k} p t_{\max } d\right] \\ $

$ =0.87$ fyAst [1- $\left.\frac{f y}{f c k} p t m a x\right] \\ $

$$ \frac{\text { Mulim }}{f c k b d^{2}}=0.87 \frac{f y}{f c k} \frac{A s t}{b d}\left(1-\frac{f y}{f c k} p t l i m\right) \\ $$

$$ \frac{\text { Mulim }}{f c k b d^{2}}=0.87 \frac{f y}{f c k} p \text { tlim }\left(1-\frac{f y}{f c k} p t l i m\right)...(3) \\ $$

From equation 4.5-5-2 ptlim can be expressed as

$$ \frac{p t l i m f y}{f c k}=0.414 \frac{x u m a x}{d}-\rightarrow....(4) \\ $$

Values of,

$$ \frac{\text { mulim }}{f c k b d_{2}} \& \frac{p t l i m f y}{f c k} \\ $$

For different grade of Steel is given in Table:

$ \begin{aligned} &\text { Table } 2.1 \text { Limiting Moment resistance \& limiting steel } \\\\ &\begin{array}{llll} \text { Fy } & 250 & 415 & 500 \\\\ \frac{mulim }{f c k b d_{2}} & 0.149 & 0.138 & 0.133 \\\\ \frac{p t l i m f y}{f c k} & 21.97 & 19.82 & 18.87 \\\\ \end{array} \end{aligned} \\ $

Where $p_{\text {tlim }}$ is in\%

Now considering,

$ M_{\text {ulim }}=C_{u} \times Z \\ $.

Mulim,

$ =0.36 f_{c k} x_{\text {ulim }} b \times\left(d-0.42 x_{\text {ulim }}\right) \\ $

$$ \frac{\text { Mulim }}{f c k b d^{2}}=0.36 \times \frac{x \text { ulim }}{d}\left[1-0.42 \frac{\text { xulim }}{d}\right]...(5) \\ $$

Value of $\frac{\text { Mulim }}{f c k b d_{2}}$ is abailalbe in table C of SP16 \& Value of $\frac{\text { Mulim }}{b d_{2}}$ for different grade of concrete and steel is given in Tables.

Value of pt lim for different grade of concrete and steel is given in table E of SP $-{ }^{\prime} 6^{\prime}$.

Term $\frac{M u l i m}{b d_{2}}$ is termed as limiting moment of resistance factor and denoted as Qlim.

$$ \therefore \text { Mulim }=Q_{\text {limbd }} \text {. } \\ $$