As I said previously, people should use what they want. I have no interest in arguing about it but I think you just made up that percentage. Also, it is important to note that the anvil is most likely going to SHEAR off at the gear. It's not likely to break off 1/3 the way down the anvil. It's not likely to break off 1/2 the way down the anvil. If your model treats the whole anvil equally then it isn't realistic.
I half made up the number. I calculated a while back and assumed the 3/8" anvil was actually a 3/8" circle. I just recalculated the numbers and found it was 4.8% if you assume a 3/8" circle and 3.4% if you assume a 3/8" square.
Note this does look at failure due to torsion which will be a shearing failure. It does not consider stress commentators that may occur at the root of the anvil. The intent isn't to get a perfect answer but to get one close enough to decide that the difference in anvil strength just isn't significant.
Here is where that comes from
https://en.wikipedia.org/wiki/Torsion_constant#Circle
I started with the dimensions of a Husky QR ratchet I have at hand. The anvil is 9.4mm per side with a 4.4mm hole down the center.
If we look at the equations that compare yield strength of a beam under tension we can see that the only factor that is affected by a hole down the center is the torsional constant J. I will say stiffness but J isn't actually stiffness but it's linear with stiffness.
J goes up with the diameter^4. This is why even a small increase in diameter result in big gains in torsional stiffness.
So if we assume a tube then
J=pi*D^4/32 (solid rod) or =pi/32 * (D^4-d^4) where D is the outer diameter and d is the inner diameter. Note the second one is basically the stiffness of a sold rod of diameter D - the stiffness of a sold rod of diameter d.
For a square J=2.25a^4 where a is 1/2 the length of a side of the square.
Here are the numbers:
Solid 3/8 "round" anvil = 766.5 mm^4
With hole = 766.5 - 36.8 = 729.7
4.8% reduction
Solid 3/8 square anvil = 1098 mm^4
With hole = 1098 - 36.8 = 1061.2 mm^4.
3.4% reduction
Since the actual anvils have rounded corners the true value would be somewhere between these two numbers.
The difference is much smaller than people think because the impact of diameter is so big.
puts his weight on a pipe stuck on the end of his ratchet. All of it is moot though since he should have used a breaker bar. Not trying to start a fight. As I said, people should just use whatever they want. Even if it does break, a rebuild kit is often only $10 from the white van if they don't just replace it for free. I don't think it's going to be a big issue for most folks. If something doesn't work for them, they'll switch to something else.