WSM udl capacity of 25x60 cm beam

Given: $b=250$ mm, $D=600$ mm, $d=600-50=550$ mm, 4-20 mm dia, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$, span $L=6.10$ m, $\sigma_{cbc}=7$, $\sigma_{st}=230$ N/mm$^2$, $m=13.33$ (modular ratio for M20).

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

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\ x_c &= k d = 0.289\times 550 = 158.73\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 1256.6\,(550-x)$$ $$x = 212.63\ \text{mm}$$

Since $x = 212.63 > x_c = 158.73$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 212.63\times\left(550-\frac{212.63}{3}\right)\\ &= 89.14\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 148.05$ N/mm$^2$ $<$ 230 N/mm$^2$.

Step 4: Safe UDL on the simply supported span

$$M_r=\frac{wL^2}{8}\ \Rightarrow\ w=\frac{8M_r}{L^2}=\frac{8\times89.14}{6.10^2}=19.16\ \text{kN/m}$$

Self weight $=0.25\times0.60\times25=3.75$ kN/m, so the superimposed load is $19.16-3.75=15.41$ kN/m.

Answer: total safe UDL $=19.16$ kN/m (including self weight), i.e. superimposed load $=15.41$ kN/m.