https://www.amazon.com/dp/B01FDAY0DE/?tag=gbody-20
I have some cheapo hubcentric ones off amazon or ebay or whatever on the front of my 2+2. Was to fit c5 17" wheels, 2" spacers. They had to be trimmed to clear the spindle but overall fine with them.
Adding a spacer on a 4.5" BS wheel isn't any difference than just buying a 3.5" wheel (minus unsprung weight). The true wheel centerline is still in the same spot on the wheel. If you slide into a curb having a spacer on isn't going to help or hurt your side load strength.
Also, a lot of 14" wheels aren't hubcentric from the factory anyways. Anything aluminum wheel or post 1990 or truck is sure, but normal pre 90 GM car had a lot of stamped steel lug centric.
Fun fact too, the studs don't actually hold the wheel vertically to the axle. Clamp load of friction between the brake rotor/drum and wheel is what holds the wheel to the car. The hubcentric ring just locates the wheel axially concentric to the axle so there is no runout between the axle and car. It certainly helps, but not the main support function.
Page a-5 on this document goes into clamp load. 7/16-20 bolt has a clamp load of ~10,000 lbs at 80ish ftlbs depending on the friction of the nut and stud. So you have 50,000 lbs of force sandwiching the wheel on.
https://www.fastenal.com/content/documents/FastenalTechnicalReferenceGuide.pdf
And friction between cast iron and steel is .25-.5 (.6 with alum wheels)
https://www.engineeringtoolbox.com/friction-coefficients-d_778.html
So you end up with high of 25,000 lbs to low of 12,500lbs of force holding the wheel to the brake rotor with zero thrust load on the studs itself.
Yay math/engineering!
Think about that, the bolts don't hold the wheel on the car (in an up down force), friction does! If you really want a mind bender, think about a world without friction!
So actual shear stress required to break the bolts off (assuming a non hubcentric, non friction world)
Root diameter of a 7/16-20 screw is ~.375, area of .11 square inches. Shear strength is some % of ultimate tensile, so lets say ~%50 on the conservative side. Grade 8 bolt is 120ksi, so you have 60,000 pounds per square inch (60 ksi), times that .11 square inch. Again, that is about ~6000 lbs of shear stress per bolt. Total of 30k lbs in a frictionless world.
There is fatigue in the picture, but the wheel is going to fail well before.