WSM resistance and stresses, 230x400

Given: $b=230$ mm, $d=380$ mm, $A_{st}=580$ mm$^2$, $\sigma_{cbc}=5$, $\sigma_{st}=140$ N/mm$^2$ (M15 concrete, mild steel), $m=18.67$.

Step 1: Constants ($\sigma_{cbc}=5$, $\sigma_{st}=140$ N/mm$^2$, $m=\dfrac{280}{3\times5}=18.67$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{18.67\times 5}{18.67\times 5 + 140} = 0.400\\ x_c &= k d = 0.400\times 380 = 152.02\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{230 x^2}{2} = 18.67\times 580.0\,(380-x)$$ $$x = 147.85\ \text{mm}$$

Since $x = 147.85 < x_c = 152.02$ mm, the section is under-reinforced: steel reaches $\sigma_{st}=140$ N/mm$^2$ first, so steel governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \sigma_{st}A_{st}\left(d-\frac{x}{3}\right) = 140\times 580.0\times\left(380-\frac{147.85}{3}\right)\\ &= 26.85\ \text{kN·m} \end{aligned}$$

Stress in concrete at this moment: $\sigma_{c} = \dfrac{\sigma_{st}\,x}{m(d-x)} = 4.78$ N/mm$^2$ $<$ 5 N/mm$^2$ (safe).

Actual stresses at $M_r$: concrete $\sigma_c=4.78$ N/mm$^2$, steel $\sigma_s=140.00$ N/mm$^2$.

Answer: $M_r=26.85$ kN·m; actual stresses are $\sigma_c=4.78$ N/mm$^2$ and $\sigma_s=140.00$ N/mm$^2$.