Here ya go:
Mark’s Standard Handbook for Mechanical Engineers lists Grade 5 fasteners as 120 ksi fasteners. This means the tensile strength is 120,000 lbs per square inch. It also lists Grade 8’s as 150 ksi fasteners meaning the tensile strength is 150,000 lbs per square inch. Also, the ultimate shear strength of a fastener is typically about 60% of its ultimate tension strength. So given a certain diameter (cross-sectional area) and strength rating, someone can figure out how much load that fastener can carry in both tension and shear.
Let’s look at an example of where grade 5 and grade 8 bolts are subjected to single shear loads (winch plate reference).
Using a .250-inch diameter grade 8 fastener gives you the following shear capability:
A = Cross-sectional area of the fastener size (since bolt bodies/shanks have circular cross-sections, use area of a circle) = Pi x r2 where R (radius) = .250/2 = .125, therefore A = Pi x (.125)2 = .0491 square inches (in2)
Capability in shear = 91,000 lbs / in2 x .0491 in2 = 4468 lbs
Using the same .250-inch diameter grade 5 fastener results in the following:
Capability in shear = 75,000 lbs / in2 x .0491 in2 = 3683 lbs
That’s a difference of over 750 lbs or over 1/3 ton. In this example you can clearly see that using a grade 8 fastener has a superior advantage over the grade 5. Therefore the result is if someone is using grade 5 bolts in a shear application like the winch plate example, they will fail almost 800 lbs earlier.