WSM moment of resistance, 300x550 beam

Given: $b=300$ mm, $d=550$ mm, 4-12 mm dia, $A_{st}=4\times\frac{\pi}{4}(12)^2=452.4$ mm$^2$. M20: $\sigma_{cbc}=7$ N/mm$^2$, $m=13.33$; Fe415: $\sigma_{st}=230$ N/mm$^2$.

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

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 550 = 158.73\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{300 x^2}{2} = 13.33\times 452.4\,(550-x)$$ $$x = 129.95\ \text{mm}$$

Since $x = 129.95 < x_c = 158.73$ mm, the section is under-reinforced: steel reaches $\sigma_{st}=230$ N/mm$^2$ first, so steel governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \sigma_{st}A_{st}\left(d-\frac{x}{3}\right) = 230\times 452.4\times\left(550-\frac{129.95}{3}\right)\\ &= 52.72\ \text{kN·m} \end{aligned}$$

Stress in concrete at this moment: $\sigma_{c} = \dfrac{\sigma_{st}\,x}{m(d-x)} = 5.34$ N/mm$^2$ $<$ 7 N/mm$^2$ (safe).

Answer: $M_r = 52.72$ kN·m (under-reinforced section).