WSM total udl on 300x500 beam

Given: $b=300$ mm, $D=500$ mm, effective cover 50 mm so $d=450$ mm, 4-25 mm dia, $A_{st}=4\times\frac{\pi}{4}(25)^2=1963.5$ mm$^2$, $L=5$ m. M20: $\sigma_{cbc}=7$; Fe415: $\sigma_{st}=230$ N/mm$^2$; $m=13.33$.

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

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 450 = 129.87\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{300 x^2}{2} = 13.33\times 1963.5\,(450-x)$$ $$x = 206.24\ \text{mm}$$

Since $x = 206.24 > x_c = 129.87$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 300\times 206.24\times\left(450-\frac{206.24}{3}\right)\\ &= 82.56\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 110.29$ N/mm$^2$ $<$ 230 N/mm$^2$.

Step 4: Total UDL

$$w=\frac{8M_r}{L^2}=\frac{8\times82.56}{5^2}=26.42\ \text{kN/m}$$

Self weight $=0.30\times0.50\times25=3.75$ kN/m, so the external (superimposed) load is $22.67$ kN/m.

Answer: total UDL including self weight $=26.42$ kN/m.