The quick answer is "no". It could change the breakout capacity slightly due to taking the cone from the edge of the plate (but this difference would be minimal at best). This could give a good increase in the pullout capacity as this is directly related to the bearing area of the embedded anchorage.
Assuming the plate is sufficiently thick to fully develop the yield strength of the anchor, there is a non-linear relationship between the radius at the base of the shear cone (at the anchor head) and the surface area of the cone to the free edge. If the geometry is correct, this should provide some assistance.
SA = (pi*r1*s1+pi*r1^2)-(pi*r2*s2+pi*r2^2)
where:
SA = surface area
r1 = radius of shear cone at free edge
r2 = radius of cone at anchor head
s1 = length along side of cone from anchor head to theoretical tip of cone
s2 = length along side of cone from fre edge to theoretical tip of cone
What I typically do is assume my failure cone starts at the centerline of the anchor rod, at the face of the plate/nut. This is conservative, as the cone may start at the edge of the plate/nut.
The next thing I do is size the plate/nut such that I have adequate bearing surface area of the plate or nut to develop the full capacity of the anchor rod, using the ACI equations for bearing on concrete. That way, the rod will yield before the concrete fails in bearing under the head.
For larger rods, a simple nut is not adequate to develop the bearing capacity. For instance, for a 1" rod, the yield strength would be 15 kips, assuming A36 steel. The bearing strength, assuming no edge concerns, would then be 2*0.85*f'c*A. With F'c=3000psi, we would need a bearing area of A=2.94sq.in. So we would need a plate then that was approximately 2"x2". However, a heavy hex nut would only have a bearing area of about 1.5 sq.inch. For 1.5" diameter rod, that plate would need to be 3"x3". This all ignores load factors, which should be used properly, for the sake of simplicity.
I then would check the resulting plate for cantilever bending, assuming 2-way distribution, and shear. Usually I find that using a thickness about equal to the cantilever distance results in a good stiff plate.
Oh ya, and once you have done all this, you don't need to redo it. I simply made a spreadsheet that calculates all this for various rod diameters and strengths. Then I just use whichever table I need on my design drawings.
If the plate is sufficiently large, it can initiate a failure surface. Should be compact like a nut or the head of a bolt. Manitoba Hydro used to use a spiral shaped anchor to reduce the likelihood of a failure caused by a single plate.
