Beam with torsion, 4.5 m clear span

Assumptions: the imposed load of 35 kN/m and the torsion of 18 kN·m are service values, factored by 1.5. Section $b=300$ mm, $D=600$ mm, effective cover 50 mm, $d=550$ mm. M20, Fe415.

Step 1: Effective span (IS 456 cl. 22.2): the lesser of $c/c=4.5+0.2=4.7$ m and clear span $+d=4.5+0.55=5.05$ m, so $L_{eff}=4.70$ m.

Step 2: Loads

$$\begin{aligned} \text{Self weight}&=0.30\times0.60\times25=4.500\ \text{kN/m}\\ w_u&=1.5(35+4.500)=59.25\ \text{kN/m},\qquad T_u=1.5\times18=27.0\ \text{kN·m}\\ M_u&=\frac{w_uL^2}{8}=\frac{59.25\times4.70^2}{8}=163.60\ \text{kN·m}\\ V_u&=\frac{w_u\times4.5}{2}=133.31\ \text{kN (face)},\quad V_{u,d}=133.31-59.25\times0.55=100.72\ \text{kN} \end{aligned}$$

Step 3: Equivalent forces (cl. 41.3)

$$\begin{aligned} M_t&=T_u\frac{1+D/b}{1.7}=27.0\times\frac{1+600/300}{1.7}=47.65\ \text{kN·m}\\ M_{e1}&=M_u+M_t=163.60+47.65=211.25\ \text{kN·m}\\ V_e&=V_{u,d}+1.6\frac{T_u}{b}=100.72+1.6\times\frac{27.0}{0.3}=244.7\ \text{kN} \end{aligned}$$

$\tau_{ve}=\dfrac{244.7\times10^3}{300\times550}=1.483$ N/mm$^2<2.8$, OK. $M_t<M_u$, so there is no $M_{e2}$.

Step 4: Flexure. $M_{u,lim}=0.138\times20\times300\times550^2=250.40$ kN·m $>M_{e1}$, so the section is singly reinforced.

$$M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=1265\ \text{mm}^2$$

Provide 5-20 mm bars ($1571$ mm$^2$), $p_t=0.95\%$. Minimum steel $0.85bd/f_y=338$ mm$^2$, OK. Provide 2-12 mm bars at the top corners.

Side face bars: $D=600$ mm $>450$ mm, so for torsion (cl. 26.5.1.7(d)) provide $0.1\%\times300\times600=180$ mm$^2$ (90 mm$^2$ per face): 1-12 mm bar on each face at mid-depth ($113$ mm$^2$), giving a spacing of about 250 mm $<300$ mm.

Step 5: Stirrups. $\tau_c=0.61$ N/mm$^2$; $\tau_{ve}>\tau_c$. With $b_1=200$ mm and $d_1=500$ mm:

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_{u}}{2.5d_1(0.87f_y)}=0.748+0.223=0.971\ \text{mm}^2/\text{mm}$$

The code minimum is $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.727$ mm$^2$/mm, so 0.971 governs. For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=104$ mm; limits $x_1=225$ mm, $(x_1+y_1)/4=188$ mm, 300 mm.

Answer: $M_u=163.60$ kN·m; $A_{st}=1265$ mm$^2$ (5-20 mm) for $M_{e1}=211.25$ kN·m; 1-12 mm side face bar on each face; closed 8 mm 2-legged stirrups at 100 mm c/c.