Design simply supported beam, 6 m span

Assumptions: effective span 6 m; $b=300$ mm, $D=600$ mm ($\approx L/10$), effective cover 50 mm, so $d=550$ mm. M20, Fe415. Live load 24 kN/m and dead load 3 kN/m (excluding self weight).

Step 1: Loads

$$\begin{aligned} \text{Self weight}&=0.30\times0.60\times25=4.50\ \text{kN/m}\\ w&=24+3+4.50=31.50\ \text{kN/m},\qquad w_u=1.5w=47.25\ \text{kN/m} \end{aligned}$$

Step 2: Moment

$$M_u=\frac{w_uL^2}{8}=\frac{47.25\times6^2}{8}=212.62\ \text{kN·m}$$

Step 3: Check depth for a singly reinforced section

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times550^2=250.40\ \text{kN·m}>M_u$$

A singly reinforced section is adequate ($d_{req}=\sqrt{M_u/(0.138f_{ck}b)}=507$ mm $<550$ mm).

Step 4: Tension steel

$$M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=1275\ \text{mm}^2$$

Provide 5-20 mm dia bars, $A_{st,prov}=1571$ mm$^2$ ($p_t=0.95\%$). Minimum steel $0.85bd/f_y=338$ mm$^2$ is satisfied; even with the provided steel, $x_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b}=262.6$ mm $<0.48d=264$ mm.

Step 5: Deflection check

$$f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=195.4\ \text{N/mm}^2,\quad k_t\approx1.19\ \text{(Fig. 4)},\quad \left(\frac{L}{d}\right)_{allow}=20k_t=23.8>\frac{6000}{550}=10.91$$

The beam is safe in deflection. Provide 2-12 mm hanger bars and 8 mm stirrups (shear design not required).

Answer: Beam 300 x 600 mm with 5-20 mm bars at the bottom ($A_{st}=1275$ mm$^2$ required).