MoR of beam 250x475 by WSM

Assumption: clear cover 25 mm, so $d=475-25-\dfrac{16}{2}=442$ mm. $b=250$ mm, $A_{st}=3\times\frac{\pi}{4}(16)^2=603.2$ mm$^2$.

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

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 442 = 127.56\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 603.2\,(442-x)$$ $$x = 139.49\ \text{mm}$$

Since $x = 139.49 > x_c = 127.56$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 139.49\times\left(442-\frac{139.49}{3}\right)\\ &= 48.27\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 202.35$ N/mm$^2$ $<$ 230 N/mm$^2$.

Answer: $M_r=48.27$ kN·m (over-reinforced, concrete governs).