Design 200x400 beam for bending, shear, torsion

Given: $b=200$ mm, $D=400$ mm, $M_u=100$ kN·m, $V_u=80$ kN, $T_u=15$ kN·m, M25, Fe415. Assume effective cover 50 mm: $d=350$ mm. Centre-to-centre distances of the corner bars: $b_1=200-100=100$ mm, $d_1=400-100=300$ mm.

Step 1: Equivalent shear and moment

$$\begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=80+1.6\times\frac{15}{0.2}=200.0\ \text{kN},\qquad \tau_{ve}=\frac{200.0\times10^3}{200\times350}=2.857\ \text{N/mm}^2\\ M_t&=T_u\frac{1+D/b}{1.7}=15\times\frac{1+400/200}{1.7}=26.47\ \text{kN·m}\\ M_{e1}&=M_u+M_t=100+26.47=126.47\ \text{kN·m} \end{aligned}$$

$\tau_{ve}<\tau_{c,max}=3.1$ N/mm$^2$ (M25), so the section is acceptable. $M_t<M_u$, so there is no $M_{e2}$.

Step 2: Longitudinal steel. $M_{u,lim}=0.138f_{ck}bd^2=0.138\times25\times200\times350^2=84.50$ kN·m $<M_{e1}$, so the section is doubly reinforced for $M_{e1}$.

Doubly reinforced design

$$M_{u2}=M_{e1}-M_{u,lim}=126.47-84.50=41.97\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.48d=168.0$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{50}{168.0}\right)=0.00246$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=344.0$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=11.15$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{41.97\times10^6}{(344.0-11.15)(350-50)}=420\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=838\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=387\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=1225\ \text{mm}^2$$

Provide tension 4-20 mm ($1257$ mm$^2$) and compression 3-16 mm ($603$ mm$^2$) bars.

Step 3: Transverse reinforcement. $p_t=\dfrac{100\times1257}{200\times350}=1.80\%$, $\tau_c=0.79$ N/mm$^2$; $\tau_{ve}>\tau_c$.

$$\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=1.385+0.295=1.680\ \text{mm}^2/\text{mm}$$

For 10 mm 2-legged closed stirrups ($A_{sv}=157.1$ mm$^2$): $s_v=93$ mm. Limits: $x_1=125$ mm, $(x_1+y_1)/4=112$ mm, 300 mm.

Answer: Tension steel $A_{st}=1225$ mm$^2$ (4-20 mm), compression steel $A_{sc}=420$ mm$^2$ (3-16 mm), closed 10 mm 2-legged stirrups at 90 mm c/c.