Beam design, 250x475 section

Given: $L=4$ m, $b=250$ mm, $D=475$ mm, effective cover 45 mm, so $d=430$ mm and $d'=45$ mm; superimposed load 45 kN/m and point load 75 kN at mid-span (service), M25. "TMT bars" is taken as Fe500 ($f_y=500$ N/mm$^2$).

Step 1: Loads and moment

$$\begin{aligned} \text{Self weight}&=0.25\times0.475\times25=2.97\ \text{kN/m},\quad w=47.97\ \text{kN/m}\\ w_u&=1.5w=71.95\ \text{kN/m},\quad P_u=1.5\times75=112.5\ \text{kN}\\ M_u&=\frac{w_uL^2}{8}+\frac{P_uL}{4}=143.91+112.50=256.41\ \text{kN·m} \end{aligned}$$

Step 2: Limiting moment (Fe500: $x_{u,max}=0.46d$)

$$M_{u,lim}=0.1336f_{ck}bd^2=0.1336\times25\times250\times430^2=154.40\ \text{kN·m}<M_u$$

The section is too shallow for a singly reinforced design, so it is designed as doubly reinforced.

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=256.41-154.40=102.01\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.46d=197.8$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{45}{197.8}\right)=0.00270$$

From the design stress-strain curve of Fe500 (IS 456 Fig. 23): $f_{sc}=410.2$ N/mm$^2$. Stress in the concrete displaced by the steel: $f_{cc}=0.446f_{ck}=11.15$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{102.01\times10^6}{(410.2-11.15)(430-45)}=664\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1023\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=609\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=1632\ \text{mm}^2$$

Provide: tension 3-25 mm + 1-16 mm dia ($1674$ mm$^2$, one layer in the 250 mm width) and compression 4-16 mm dia ($804$ mm$^2$), with 8 mm stirrups.

Check: $0.87f_yA_{st}=435\times1674=728$ kN, which is less than the compression available at $x_{u,max}$, $0.36f_{ck}bx_{u,max}+A_{sc}(f_{sc}-f_{cc})=445+321=766$ kN, so $x_u<x_{u,max}$, OK. (Providing far more tension steel, such as 6-20 mm = 1885 mm$^2$, would make $x_u>x_{u,max}$.)

Step 3: Deflection check (IS 456 cl. 23.2)

$$p_t=\frac{100A_{st}}{bd}=1.56\%,\quad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=282.7\ \text{N/mm}^2,\quad k_t\approx0.80\ \text{(Fig. 4)}$$

Compression steel $p_c=\dfrac{100A_{sc}}{bd}=0.75\%$, modification factor $k_c=\dfrac{1.6p_c}{p_c+0.275}=1.17$.

$$\left(\frac{L}{d}\right)_{allow}=20\times k_t\times k_c=20\times0.80\times1.17=18.7,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{4000}{430}=9.30$$

Actual $<$ allowable, so the beam is safe in deflection.

Answer: $M_u=256.41$ kN·m; $A_{st}=1632$ mm$^2$ (provide 3-25 + 1-16 mm), $A_{sc}=664$ mm$^2$ (provide 4-16 mm); safe in deflection.