WSM moment of resistance, 250x450 beam

Given: $b=250$ mm, $D=450$ mm, effective cover 40 mm so $d=410$ mm, 4-16 mm dia, $A_{st}=4\times\frac{\pi}{4}(16)^2=804.2$ mm$^2$. M20: $\sigma_{cbc}=7$, $m=13.33$; Fe415: $\sigma_{st}=230$ N/mm$^2$.

Step 1: Constants (M20: $\sigma_{cbc}=7$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 410 = 118.33\ \text{mm} \end{aligned}$$

Step 2: Actual neutral axis (taking moments of areas about the neutral axis, transformed section)

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 804.2\,(410-x)$$ $$x = 149.48\ \text{mm}$$

Since $x = 149.48 > x_c = 118.33$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 149.48\times\left(410-\frac{149.48}{3}\right)\\ &= 47.11\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 162.63$ N/mm$^2$ $<$ 230 N/mm$^2$.

Answer: $M_r=47.11$ kN·m.