WSM moment of resistance, 230x350 beam

Given: $b=230$ mm, $D=350$ mm, nominal cover 25 mm, so $d=350-25-\dfrac{16}{2}=317$ mm. $A_{st}=4\times\frac{\pi}{4}(16)^2=804.2$ mm$^2$. $\sigma_{cbc}=7$, $\sigma_{st}=140$ N/mm$^2$, $m=13.33$.

Step 1: Constants ($\sigma_{cbc}=7$, $\sigma_{st}=140$ N/mm$^2$, $m=13.33$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 140} = 0.400\\ x_c &= k d = 0.400\times 317 = 126.78\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{230 x^2}{2} = 13.33\times 804.2\,(317-x)$$ $$x = 131.50\ \text{mm}$$

Since $x = 131.50 > x_c = 126.78$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 230\times 131.50\times\left(317-\frac{131.50}{3}\right)\\ &= 28.92\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 131.62$ N/mm$^2$ $<$ 140 N/mm$^2$.

Answer: $M_r=28.92$ kN·m.