WSM moment of resistance, 4.2 m beam

Assumption: clear cover 25 mm, so $d=450-25-\dfrac{20}{2}=415$ mm. $b=230$ mm, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$. Permissible stresses: M20 $\sigma_{cbc}=7$ N/mm$^2$, Fe415 $\sigma_{st}=230$ N/mm$^2$, $m=13.33$. (The span 4.2 m is not needed for $M_r$.)

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

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 415 = 119.77\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{230 x^2}{2} = 13.33\times 1256.6\,(415-x)$$ $$x = 183.59\ \text{mm}$$

Since $x = 183.59 > x_c = 119.77$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 230\times 183.59\times\left(415-\frac{183.59}{3}\right)\\ &= 52.29\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 117.61$ N/mm$^2$ $<$ 230 N/mm$^2$.

Answer: $M_r=52.29$ kN·m.