Design T-beam with 150 mm slab

Assumptions: beam web $b_w=300$ mm, $D=500$ mm (about $L/12$), effective cover 50 mm, $d=450$ mm; floor finish ignored (not given). M20, Fe415.

Step 1: Effective span (IS 456 cl. 22.2): the lesser of $c/c=6+0.25=6.25$ m and clear span $+d=6+0.45=6.45$ m, so $L_{eff}=6.25$ m.

Step 2: Loads per metre length of beam (each beam carries slab load from a 4 m width)

$$\begin{aligned} \text{Slab dead load}&=0.15\times25=3.75\ \text{kN/m}^2;\quad\text{live load}=3\ \text{kN/m}^2\\ \text{Slab load on beam}&=(3.75+3)\times4=27.00\ \text{kN/m}\\ \text{Rib self weight}&=0.30\times(0.50-0.15)\times25=2.625\ \text{kN/m}\\ w&=29.62\ \text{kN/m},\qquad w_u=1.5w=44.44\ \text{kN/m} \end{aligned}$$ $$M_u=\frac{w_uL^2}{8}=\frac{44.44\times6.25^2}{8}=216.98\ \text{kN·m}$$

Step 3: Effective flange width (T-beam)

$$b_f=\frac{l_0}{6}+b_w+6D_f=\frac{6250}{6}+300+6\times150=2242\ \text{mm}\ (<4000\ \text{mm})$$

Step 4: Neutral axis position. Moment capacity when $x_u=D_f$:

$$M_{u,f}=0.36f_{ck}b_fD_f(d-0.42D_f)=936.93\ \text{kN·m}>M_u$$

The neutral axis lies within the flange; design as a rectangular beam of width $b_f$.

Step 5: Tension steel

$$M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{b_fdf_{ck}}\right)\ \Rightarrow\ A_{st}=1374\ \text{mm}^2$$

$x_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b_f}=30.7$ mm $<D_f$ (OK). Minimum steel $0.85b_wd/f_y=277$ mm$^2$ (OK).

Provide 5-20 mm dia bars, $A_{st,prov}=1571$ mm$^2$ ($b_w=300$ mm can hold them in one or two layers).

Step 6: Deflection check. $b_w/b_f=0.13<0.3$, so the flange factor is 0.8. For flanged beams, $p_t$ for Fig. 4 is based on $b_fd$: $p_t=\dfrac{100\times1571}{2242\times450}=0.16\%$, $f_s=0.58f_y\dfrac{A_{st,req}}{A_{st,prov}}=210.5$ N/mm$^2$, so $k_t=2.0$ (maximum).

$$\left(\frac{L}{d}\right)_{allow}=20\times2.0\times0.8=32.0,\qquad\left(\frac{L}{d}\right)_{actual}=\frac{6250}{450}=13.89$$

Safe in deflection.

Answer: T-beam $b_f=2242$ mm, $D_f=150$ mm, $b_w=300$ mm, $D=500$ mm, with 5-20 mm bars ($A_{st,req}=1374$ mm$^2$).