WSM beam safety check, 350x650

Given: $b=350$ mm, $d=650$ mm, 5-25 mm dia, $A_{st}=5\times\frac{\pi}{4}(25)^2 = 2454.4$ mm$^2$, $M_{service}=150$ kN·m. M25: $\sigma_{cbc}=8.5$ N/mm$^2$; Fe415: $\sigma_{st}=230$ N/mm$^2$; $m=10.98$.

Step 1: Constants (M25: $\sigma_{cbc}=8.5$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=280/(3\times8.5)=10.98$)

$$\begin{aligned} k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{10.98\times 8.5}{10.98\times 8.5 + 230} = 0.289\\ x_c &= k d = 0.289\times 650 = 187.62\ \text{mm} \end{aligned}$$

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

$$\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{350 x^2}{2} = 10.98\times 2454.4\,(650-x)$$ $$x = 248.62\ \text{mm}$$

Since $x = 248.62 > x_c = 187.62$ mm, the section is over-reinforced: concrete reaches $\sigma_{cbc}=8.5$ N/mm$^2$ first, so concrete governs.

Step 3: Moment of resistance

$$\begin{aligned} M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 8.5\times 350\times 248.62\times\left(650-\frac{248.62}{3}\right)\\ &= 209.73\ \text{kN·m} \end{aligned}$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 150.68$ N/mm$^2$ $<$ 230 N/mm$^2$.

Safety check: $M_r = 209.73$ kN·m $>$ 150 kN·m. The beam is safe.

Stresses at the service moment of 150 kN·m (neutral axis $x=248.62$ mm):

$$\sigma_c=\frac{M}{\frac12 b x(d-x/3)} = 6.08\ \text{N/mm}^2,\qquad \sigma_{s}=\frac{M}{A_{st}(d-x/3)}=107.76\ \text{N/mm}^2$$

Stresses at the limiting (failure) state as per WSM: the section is over-reinforced, so concrete reaches its permissible stress first. At this state concrete stress $= 8.5$ N/mm$^2$ and steel stress $= 150.68$ N/mm$^2$ (less than 230 N/mm$^2$).

Answer: Beam is safe ($M_r=209.73$ kN·m $>150$ kN·m). At failure state: $\sigma_c = 8.5$ N/mm$^2$, $\sigma_s = 150.68$ N/mm$^2$. (At 150 kN·m: $\sigma_c=6.08$, $\sigma_s=107.76$ N/mm$^2$.)