If the member is laterally loaded then the flanges will resist the horiziontal shear force not the web (which will resist the vertical shear fotrce from gravity loads).

As dbuzz signals, if we have understood the case well, the flanges will be carrying the shear.

Then the magnitude of the shear stress tau will follow a -if I remember well parabolical- law with a maximum at the center of the flanges -if we dismiss the web presence there, of course- and zero shear at the tips. This tau you evaluate by the equation you refer to, but remember that the static moment corresponds to the area beyond your ordinate in the direction of the load, and all data therein is for such direction of loading.

Bending Stress will follow the usual rule and for this direction will follow

sigma=My/(Iy/c)

By stating these laws as a function of c you may find a particular value of c at which the Von Mises comparison stress

sigmacomparison=SQRT(sigma^2+3·tau^2)

is at maximum.

This is easy to do in Mathcad or akin workspace, or you may chart for the combination.

But if you simply want to design a beam for these circumnstances you may check the on weak axis bending and corresponding shear by LRFD Chapter F, except it falls outside the specifications.

I'm sorry...I indicated that it was a leteral load.  The loading in question is actually a vertical load on the top of the beam flange.

Thanks,

If you have a wide flange beam whereby the upper and lower flanges are in a horizontal position and the beam is loaded horizontally then the average shear stress will be based on  the x-sectional areas of both flanges; however if the load is applied vertically then average shear stress will be based only on the webb x-sectional area as you stated.
Yes use the von misses formula to determine resultant stress from bending and average shear stresses