I understand that for a rectangular c-s the shear stress distribution is parabolic and the max shear stress occurs at the neutral axis and has a value of 1.5V/A. Where V is the 'applied shear force' and A is the cross-sectional area. But this in turn then means that the shear force at this point is equal to 1.5V ( 1.5 times larger than the applied shear force) - which seems a tad strange physically. Is this due to the fact the the average shear force (average shear stress x cs area) is equal to the applied shear force? This is the only way it makes sense to me. Thanks!
You've got your terms confused.
The maximum shear stress at the midpoint is equal to
That is the only viable comparison to be made, stress to stress. And having a maximum stress greater than the average stress is totally reasonable.
Your doubt, however, is that "the shear force at this point is equal to $1.5V$". That is not the case. There is no shear force at any point in the section. There is only a shear stress. The entirety of the shear stress must then be integrated over the area to obtain the shear force.
and prove that the shear force at the midpoint is greater than the applied shear force?" But I already beat you to it. After all, as I mentioned at the start, $\dfrac$ gives you the average stress along the section. So $\dfrac>$ is equivalent to the following stress profile, which clearly isn't the one you're expecting:
