Moment of resistance of doubly reinforced beam

Given: $b=300$ mm, $D=500$ mm, 5-25 mm tension bars ($A_{st}=2454.4$ mm$^2$), 5-12 mm compression bars ($A_{sc}=565.5$ mm$^2$), clear cover 25 mm, M25, Fe415.

Effective depths: $d=500-25-\dfrac{25}{2}=462.5$ mm; $d'=25+\dfrac{12}{2}=31$ mm. (The 4.30 m span is not needed for the moment of resistance.)

Step 1: Neutral axis. Assume the tension steel yields. Equilibrium of forces:

$$0.36f_{ck}b\,x_u+A_{sc}(f_{sc}-f_{cc})=0.87f_yA_{st},\qquad f_{cc}=0.446f_{ck}=11.15\ \text{N/mm}^2$$

Total tension $=0.87\times415\times2454.4=886.2$ kN. Solving by trial (with $f_{sc}$ from the Fe415 design curve at $\varepsilon_{sc}=0.0035(1-d'/x_u)$) gives $x_u=256.28$ mm.

Step 2: Check section type.

$$x_{u,max}=0.48d=0.48\times462.5=222.0\ \text{mm}$$

Since $x_u=256.28$ mm $>x_{u,max}$, the tension steel is more than needed (over-reinforced). The tension steel will not reach $0.87f_y$ and IS 456 limits the neutral axis depth to $x_{u,max}$. So take $x_u=x_{u,max}=222.0$ mm.

Step 3: Compression steel stress

$$\varepsilon_{sc}=0.0035\left(1-\frac{31}{222.0}\right)=0.00301\ \Rightarrow\ f_{sc}=354.0\ \text{N/mm}^2$$

Step 4: Ultimate moment

$$\begin{aligned} M_u&=0.36f_{ck}b\,x_{u,max}(d-0.42x_{u,max})+A_{sc}(f_{sc}-f_{cc})(d-d')\\ &=0.36\times25\times300\times222.0(462.5-0.42\times222.0)+565.5(354.0-11.15)(462.5-31)\\ &=221.33+83.66=304.99\ \text{kN·m} \end{aligned}$$

Answer: $M_u=304.99$ kN·m (limited by concrete, since the section is over-reinforced in tension).