Shear stress distribution in triangular steel profile

[Ieliepielie]

For my job I have to calculate the maximum shear stress in a triangular thin wall steel beam. I know how to calculate this for a retangular shape.
The general formula is:
tau=Vy*Sza/(b*Iz)
with:
tau= shear stress
Vy=force
Sza = statistical moment
b= width
Iz = moment of Inertia

I just don't see how I could calculate the statistical moment for the triangle. It is a closed profile, therefore there is no point where the shear stress is zero and in contrary to the retangular shape the 'legs' don't have equal direction so it can not just be divided by two. I hope somebody can help me.
Thank you!

 

[ank_gl]

What do you know about the neutral axis? its properties?

 

[nvn]

Ieliepielie: My best guess, for maximum shear stress, is currently, tau = 2.0*V/A, where A = s*t, s = triangle perimeter, and t = wall thickness.

 

[curllight]

no idea knowing neutral axis and no properties. Its a random question. Also how to graph the distribution?
Thanks.