The shear center doesn't play a role in calculating bending stress, but the location of the neutral axis does. (The neutral axis and the shear center are
two different things).
If you want to calculate the force over a particular portion of the cross section, dF = σ ⋅ dA, and σ = M ⋅ y / I. For a particular distance y from the neutral axis, the bending stress is constant all along that line, just like along the neutral axis, σ = 0. The bending stress σ is proportional to y and the constant of proportionality is the quantity (M / I).
In the section below:
If you wanted to calculate the force acting on the web below the neutral axis, you would proceed thus:
dF = σ ⋅ dA and
σ = M ⋅ y / I
From the previous calculations,
##\bar y = 66.40## mm above the bottom edge of the web
The web has a constant thickness of 8 mm, so dA = 8 ⋅ dy, where dy is the height of a tiny slice of the web.
Then,
dF = (M / I) ⋅ y ⋅ 8 ⋅ dy = 8 ⋅ (M / I) ⋅ y ⋅ dy
To find F, we must integrate both sides of the equation for dF above:
$$\int\, dF = F = \int_0^{66.40} 8 ⋅ (M / I) ⋅ y ⋅ dy$$
$$F = 8 ⋅ (M / I) \int_0^{66.40} y ⋅ dy$$
$$F = 8 ⋅ (M / I) \frac{y^2}{2} \vert_0^{66.40}$$
$$F = 8\,mm ⋅ \frac{150,000\, N⋅mm}{1,202,840\,mm^4} ⋅ \frac{66.40^2\, mm^2}{2}$$
##F = 2,199.30\,N ## (in compression)
You can also compute the average bending stress acting over an area, and the force F = σ(avg) ⋅ Area.