Beam flexure and deflection, 4.5 m span

Given: $L=4.5$ m, superimposed load 45 kN/m, point load 75 kN at mid-span (service), effective cover 45 mm, M25, Fe415.

Step 1: Section (assumed)

Take $b=300$ mm, $D=600$ mm, so $d=600-45=555$ mm.

Step 2: Loads

$$\begin{aligned} \text{Self weight}&=0.30\times0.60\times25=4.50\ \text{kN/m}\\ w&=45+4.50=49.50\ \text{kN/m},\quad w_u=1.5w=74.25\ \text{kN/m}\\ P_u&=1.5\times75=112.5\ \text{kN} \end{aligned}$$

Step 3: Factored moment

$$M_u=\frac{w_uL^2}{8}+\frac{P_uL}{4}=187.95+126.56=314.51\ \text{kN·m}$$

Step 4: Limiting moment (IS 456 cl. 38.1; Fe415: $x_{u,max}=0.48d$)

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times25\times300\times555^2=318.72\ \text{kN·m}\ \ge M_u$$

A singly reinforced section is sufficient.

Step 5: Tension steel (IS 456 Annex G-1.1(b))

$$M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)$$

Solving this quadratic for $M_u=314.51$ kN·m gives $A_{st,req}=1948$ mm$^2$.

Provide 4-25 mm bars, $A_{st,prov}=1963$ mm$^2$ (fits in one layer in the 300 mm width).

• Minimum steel: $0.85bd/f_y=341$ mm$^2$, OK.
• $x_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b}=262.6$ mm $<0.48d=266.4$ mm, OK.

Step 6: Deflection check (IS 456 cl. 23.2, simply supported basic ratio 20)

$$p_t=\frac{100A_{st}}{bd}=1.18\%,\qquad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=238.8\ \text{N/mm}^2$$

Modification factor (IS 456 Fig. 4): $k_t\approx0.96$.

$$\left(\frac{L}{d}\right)_{allow}=20\times0.96=19.2,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{4500}{555}=8.11$$

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

Answer: Beam 300 x 600 mm, $d=555$ mm, $M_u=314.51$ kN·m, $A_{st}=1948$ mm$^2$ (provide 4-25 mm bars); the deflection check is satisfied. Add 2-12 mm hanger bars and 8 mm stirrups (shear design to be done separately).