Beam reinforcement for given loads, 6 m span

Given: $L=6$ m, $b=250$ mm, $D=450$ mm, M20, Fe415. Assume effective cover 40 mm, $d=450-40=410$ mm, $d'=40$ mm.

Step 1: Loads

$$w_u=1.5\times28.812=43.22\ \text{kN/m},\qquad M_u=\frac{w_uL^2}{8}=\frac{43.22\times6^2}{8}=194.48\ \text{kN·m}$$

Step 2: Limiting moment

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times250\times410^2=115.96\ \text{kN·m}<M_u$$

The beam must be doubly reinforced.

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=194.48-115.96=78.53\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.48d=196.8$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{40}{196.8}\right)=0.00279$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=352.1$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{78.53\times10^6}{(352.1-8.92)(410-40)}=619\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=981\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=588\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=1569\ \text{mm}^2$$

Provide: tension 4-25 mm dia ($1963$ mm$^2$, two layers) and compression 4-16 mm dia ($804$ mm$^2$). Use 8 mm stirrups at suitable spacing ($\le16\phi_c$, 300 mm).

Deflection check: $p_t=1.92\%$, $f_s=0.58f_y\dfrac{A_{st,req}}{A_{st,prov}}=192.4$ N/mm$^2$, $k_t\approx0.98$ (Fig. 4); $p_c=0.78\%$, $k_c=1.18$:

$$\left(\frac{L}{d}\right)_{allow}=20\times0.98\times1.18=23.1>\frac{6000}{410}=14.63$$

Safe.

Answer: $A_{st}=1569$ mm$^2$ (4-25 mm) and $A_{sc}=619$ mm$^2$ (4-16 mm).