Design of 250x300 beam on masonry walls

Given: $b=250$ mm, $D=300$ mm, clear span 4 m between walls 300 mm thick, DL = 10 kN/m (including self weight), LL = 8 kN/m, M20, Fe415. Assume an effective cover of 40 mm at top and bottom, so $d=300-40=260$ mm and $d'=40$ mm.

Step 1: Effective span (IS 456 cl. 22.2): the lesser of the centre-to-centre distance of supports ($4+0.3=4.3$ m) and clear span $+\,d$ ($4+0.26=4.26$ m). Take $L_{eff}=4.26$ m.

Step 2: Factored load and moment

$$w_u=1.5(10+8)=27.0\ \text{kN/m},\qquad M_u=\frac{w_uL^2}{8}=\frac{27.0\times4.26^2}{8}=61.25\ \text{kN·m}$$

Step 3: Limiting moment of the section

$$M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times250\times260^2=46.63\ \text{kN·m}$$

Since $M_u>M_{u,lim}$ and the depth is fixed, the beam must be doubly reinforced.

Doubly reinforced design

$$M_{u2}=M_u-M_{u,lim}=61.25-46.63=14.62\ \text{kN·m}$$

Limiting depth $x_{u,max}=0.48d=124.8$ mm. Strain in compression steel (similar triangles):

$$\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{40}{124.8}\right)=0.00238$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=341.6$ N/mm$^2$. Stress in the concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{14.62\times10^6}{(341.6-8.92)(260-40)}=200\ \text{mm}^2$$ $$A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=622\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=184\ \text{mm}^2$$ $$A_{st}=A_{st1}+A_{st2}=806\ \text{mm}^2$$

Provide: tension steel 2-20 mm + 1-16 mm dia ($829$ mm$^2$) and compression steel 2-12 mm dia ($226$ mm$^2$). (Much more tension steel than required, such as 5-16 mm, should not be provided, because it would push $x_u$ beyond $x_{u,max}$ and make the section over-reinforced.)

Check: with the bars provided, $x_u=\dfrac{0.87f_yA_{st}-A_{sc}(f_{sc}-f_{cc})}{0.36f_{ck}b}=\dfrac{299.3-75.2}{1.8}\approx124.5$ mm $\le x_{u,max}=124.8$ mm, OK. $p_t=100\times829/(250\times260)=1.28\%$ $<4\%$, OK. Two-legged 8 mm stirrups at a suitable spacing are provided for shear and to hold the compression bars (spacing $\le 16\phi$ of the compression bar). Shear design to be done separately.

Answer: $M_u=61.25$ kN·m; $A_{st}=806$ mm$^2$ (2-20 + 1-16 mm) and $A_{sc}=200$ mm$^2$ (2-12 mm).