Design beam 300x700 with shear, 7 m span

Given: $L=7$ m (effective), $b=300$ mm, $D=700$ mm, live load 70 kN/m, M20, Fe415. Assume effective cover 50 mm: $d=650$ mm, $d'=50$ mm. Support width assumed 300 mm.

Step 1: Loads and moment

$$\begin{aligned} \text{Self weight}&=0.30\times0.70\times25=5.25\ \text{kN/m},\quad w=70+5.25=75.25\ \text{kN/m}\\ w_u&=1.5w=112.88\ \text{kN/m},\qquad M_u=\frac{w_uL^2}{8}=\frac{112.88\times7^2}{8}=691.36\ \text{kN·m} \end{aligned}$$

Step 2: Limiting moment

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times650^2=349.74\ \text{kN·m}<M_u$$

The beam is doubly reinforced.

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=691.36-349.74=341.62\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.48d=312.0$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{50}{312.0}\right)=0.00294$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=353.4$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{341.62\times10^6}{(353.4-8.92)(650-50)}=1653\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1867\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=1577\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=3444\ \text{mm}^2$$

Provide: tension 8-25 mm dia ($3927$ mm$^2$, in two layers), compression 6-20 mm dia ($1885$ mm$^2$).

Step 3: Shear design

Support shear $V_u=\dfrac{w_uL}{2}=395.1$ kN. Critical section at $d$ from the face of support, i.e. $0.15+0.65=0.80$ m from the support centre:

$$V_{u,d}=w_u\left(\frac{L}{2}-0.80\right)=304.8\ \text{kN}$$ $$\tau_v=\frac{V_{u,d}}{bd}=\frac{304.8\times10^3}{300\times650}=1.56\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2\ \text{(OK)}$$

Taking all tension bars to the support, $p_t=\dfrac{100\times3927}{300\times650}=2.01\%$, so $\tau_c=0.79$ N/mm$^2$ (IS 456 Table 19). As $\tau_v>\tau_c$, shear reinforcement is required:

$$V_{us}=V_{u,d}-\tau_cbd=304.8-154.3=150.5\ \text{kN}$$

Using 10 mm 2-legged stirrups ($A_{sv}=157$ mm$^2$):

$$s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times157\times650}{150.5\times10^3}=245\ \text{mm}$$

Limits: $s_v\le0.75d=487$ mm, 300 mm, and $\dfrac{0.87f_yA_{sv}}{0.4b}=473$ mm. Provide 10 mm 2-legged stirrups at 240 mm c/c near the supports for a length of about 2.13 m (where $V_u>\tau_cbd$). In the remaining central zone provide the minimum: 10 mm at 300 mm c/c. Compression steel is tied by these stirrups (spacing $\le 16\phi$ of compression bar).

Answer: $A_{st}=3444$ mm$^2$ (8-25 mm), $A_{sc}=1653$ mm$^2$ (6-20 mm); stirrups 10 mm 2L at 240 mm c/c.