Neutral axis ratio in WSM beam

Given: $M_r=0.85\,b\,d^2$ (N·mm), $\sigma_{cbc}=7$, $\sigma_{st}=140$ N/mm$^2$, $m=18.33$, singly reinforced. Let $n=x/d$.

Step 1: Balanced (critical) neutral axis

$$k=\frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}}=\frac{18.33\times7}{18.33\times7+140}=0.4782$$ $$j=1-\frac{k}{3}=0.8406,\qquad Q=\tfrac12\sigma_{cbc}kj=1.407$$

Balanced moment $=1.407\,bd^2$. The given moment $0.85\,bd^2$ is smaller than this, so the section is under-reinforced and steel stress governs ($\sigma_{st}=140$ N/mm$^2$), with the concrete stress below 7 N/mm$^2$.

Step 2: Equations for an under-reinforced section

Equilibrium of the transformed section: $\dfrac{b x^2}{2}=mA_{st}(d-x)$, so

$$A_{st}=\frac{b\,d\,n^2}{2m(1-n)}$$

Moment about the compression centroid:

$$M_r=\sigma_{st}A_{st}\,d\left(1-\frac{n}{3}\right)=\frac{\sigma_{st}\,b\,d^2\,n^2\left(1-\frac{n}{3}\right)}{2m(1-n)}$$

Step 3: Solve for $n$

Setting $M_r=0.85\,bd^2$:

$$\frac{140\,n^2\left(1-\frac{n}{3}\right)}{2\times18.33\,(1-n)}=0.85$$

Solving numerically (trial and error): $n=0.394$.

Check: concrete stress $\sigma_c=\dfrac{\sigma_{st}\,n}{m(1-n)}=4.97$ N/mm$^2$ $<7$, consistent with an under-reinforced section; $n<k=0.4782$.

Answer: $x/d=0.394$ (neutral axis at about $0.394d$ from the compression fibre).