Ultimate moment of L-beam, doubly reinforced

Given: $b_f=925$ mm, $d=450$ mm, $D_f=100$ mm, $b_w=250$ mm; tension steel 4-20 mm, $A_{st}=1256.6$ mm$^2$; compression steel 3-16 mm, $A_{sc}=603.2$ mm$^2$; M20, Fe415. Effective cover to compression steel is not given; assume $d'=40$ mm.

Step 1: Neutral axis. Assume $x_u\le D_f$ so the flange acts as a rectangle of width 925 mm. Force equilibrium $C=T$:

$$0.36f_{ck}b_fx_u+A_{sc}(f_{sc}-f_{cc})=0.87f_yA_{st}$$

where $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$ and $f_{sc}$ depends on the strain $\varepsilon_{sc}=0.0035\left(1-\dfrac{d'}{x_u}\right)$.

Total tension $T=0.87\times415\times1256.6=453.7$ kN.

Trial and error (checking $f_{sc}$ from the Fe415 design curve each time; here the strain is in the elastic range, so $f_{sc}=E_s\varepsilon_{sc}$):

$$x_u=53.20\ \text{mm},\quad \varepsilon_{sc}=0.0035\left(1-\frac{40}{53.20}\right)=0.000868,\quad f_{sc}=2\times10^5\times0.000868=173.7\ \text{N/mm}^2$$

Check: $C=0.36\times20\times925\times53.20+603.2(173.7-8.92)=354.3+99.4=453.7$ kN $=T$.

$x_u=53.20$ mm $<D_f=100$ mm (flange) and $<x_{u,max}=0.48d=216.0$ mm, so the assumption is correct and the section is under-reinforced.

Step 2: Ultimate moment (about the tension steel)

$$\begin{aligned} M_u&=0.36f_{ck}b_fx_u(d-0.42x_u)+A_{sc}(f_{sc}-f_{cc})(d-d')\\ &=0.36\times20\times925\times53.20(450-0.42\times53.20)+603.2(173.7-8.92)(450-40)\\ &=151.52+40.75=192.27\ \text{kN·m} \end{aligned}$$

Answer: $M_u=192.27$ kN·m.