WSM moment of resistance, 250x500 beam

Given: $b=250$ mm, $d=500$ mm, 3-16 mm dia, $A_{st}=3\times\frac{\pi}{4}(16)^2=603.2$ mm$^2$, $\sigma_{cbc}=7$, $\sigma_{st}=230$ N/mm$^2$, $m=13.33$.

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

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 500 = 144.30\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 603.2\,(500-x)$$ $$x = 150.04\ \text{mm}$$

Since $x = 150.04 > x_c = 144.30$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 150.04\times\left(500-\frac{150.04}{3}\right)\\ &= 59.08\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 217.65$ N/mm$^2$ $<$ 230 N/mm$^2$.

Answer: $M_r=59.08$ kN·m.