A simply supported reinforced concrete beam of rectangular cross-section is 300 $\mathrm{mm}$ wide and $450 \mathrm{~mm}$ deep up to centre of tensile reinforcement which consists of $4-20 \mathrm{~mm}$ dia. bars. Determine depth of neutral axis and maximum tensile stress in steel reinforcement when concrete at extreme compression fibreis stressed to $7 \mathrm{~N} / \mathrm{mm}^{2}$. Take $\mathrm{m}=9$.

So l u t i o n : ~

$$\text { Given: } \mathrm{b}=300 \mathrm{~mm}, \mathrm{~d}=450 \mathrm{~mm}, A_{s t}=4 \times \frac{\pi \times 20^{2}}{4}=1257 \mathrm{~mm}^{2}$$

From the first principle, we have

$$\text { b.n. } \frac{n}{2}=m A_{s t}(d-n)$$

or, $300 \times \frac{n^{2}}{2}=9 \times 1257 \times(450-n)$

or, $\quad n^{2}+75.42 n-33939=0$

Solving the quadratic equation, we get

$$n=150.33 \mathrm{~mm}$$

Using similar triangles of triangular stress distribution diagram (Fig. 2.6), we get

$$\frac{\sigma_{c b c}}{\left(\sigma_{s t} / m\right)}=\frac{n}{d-n} \Rightarrow \sigma_{s t}=\frac{m \sigma_{c b c}(d-n)}{n}=\frac{9 \times 7 \times(450-150.33)}{150.33}=125.6 \mathrm{~N} / \mathrm{mm}^{2}(\mathrm{Ans})$$

So l u t i o n : ~

$$\text { Given: } \mathrm{b}=300 \mathrm{~mm}, \mathrm{~d}=450 \mathrm{~mm}, A_{s t}=4 \times \frac{\pi \times 20^{2}}{4}=1257 \mathrm{~mm}^{2}$$

From the first principle, we have

$$\text { b.n. } \frac{n}{2}=m A_{s t}(d-n)$$

or, $300 \times \frac{n^{2}}{2}=9 \times 1257 \times(450-n)$

or, $\quad n^{2}+75.42 n-33939=0$

Solving the quadratic equation, we get

$$n=150.33 \mathrm{~mm}$$

Using similar triangles of triangular stress distribution diagram (Fig. 2.6), we get

$$\frac{\sigma_{c b c}}{\left(\sigma_{s t} / m\right)}=\frac{n}{d-n} \Rightarrow \sigma_{s t}=\frac{m \sigma_{c b c}(d-n)}{n}=\frac{9 \times 7 \times(450-150.33)}{150.33}=125.6 \mathrm{~N} / \mathrm{mm}^{2}(\mathrm{Ans})$$