A rectangular beam is $250 \mathrm{~mm}$ wide and $500 \mathrm{~mm}$ deep up to centre of reinforcement. Find reinforcement required if it has to resist a moment of $45 \mathrm{KNm}$. Assume that M20 grade concrete mix and HYSD (Grade Fe415) bars have been used

Solution

1. Neutral axis The position of neutral axis shall be found by calculating $\mathrm{k}$ from

$$k=-m p+\sqrt{m^{2} p^{2}+2 m p}$$

Where,

$$\begin{array}{l} m=\frac{280}{3 \sigma_{c b c}}=\frac{280}{3 \times 7}=13.33 \approx 13 \\p=\frac{A_{s t}}{b d}, \quad \text { or } \quad A_{s t}=p b d=p \times 250 \times 500=125000 p \\\therefore k=-13 p+\sqrt{169 p^{2}+26 p} \text { (i) }\end{array}$$

2. Reinforcement required Moment of resistance of beam section with respect to steel

$$=\sigma_{s t} A_{s t} j d=(125000 p) \times 230 \times\left(1-\frac{k}{3}\right) \times 500=45 \times 10^{6}$$

Solving Eqns (i) and (ii) by trial and error, we get $p=0.003424$

$\therefore$ Reinforcement required $=0.003424 \times 250 \times 500=428 \mathrm{~mm}^{2}$

Solution

1. Neutral axis The position of neutral axis shall be found by calculating $\mathrm{k}$ from

$$k=-m p+\sqrt{m^{2} p^{2}+2 m p}$$

Where,

$$\begin{array}{l} m=\frac{280}{3 \sigma_{c b c}}=\frac{280}{3 \times 7}=13.33 \approx 13 \\p=\frac{A_{s t}}{b d}, \quad \text { or } \quad A_{s t}=p b d=p \times 250 \times 500=125000 p \\\therefore k=-13 p+\sqrt{169 p^{2}+26 p} \text { (i) }\end{array}$$

2. Reinforcement required Moment of resistance of beam section with respect to steel

$$=\sigma_{s t} A_{s t} j d=(125000 p) \times 230 \times\left(1-\frac{k}{3}\right) \times 500=45 \times 10^{6}$$

Solving Eqns (i) and (ii) by trial and error, we get $p=0.003424$

$\therefore$ Reinforcement required $=0.003424 \times 250 \times 500=428 \mathrm{~mm}^{2}$