Ok. I respectfully disagree that the flywheel is designed to operate in shear from the dowels.
From Vehicular Engine Design 2015 by Hoag & Dondlinger Chapter 16 (Cranktrain)
"The flywheel connection relies on friction to generate the shear
torque capacity in the joint, but a frictionless condition must also be designed for. "
“If a design is complete and in production, and more torque capacity is required of
the joint, various treatments can be added to increase the capacity of the joint at an
additional cost. The coefficient of friction at the joint surface can be increase by changing
the machined surface finish, by adding abrasive coatings, or by adding additional locating
dowels. Alternately, the fastener grade may be increased to allow greater clampload.”
Torque Capacity of the flange is given by :
Tc = u * Tf/(k*d) * rp * n
Where u = coefficient of friction (Flywheel to crank)
Tf = Fastener Torque
k = nut factor = 0.18 - 0.22 usually assumed 0.2
d = nominal dia of clamp bolt
rp = pitch circle radius of fasteners
n = number of fasteners
In our case, we have 65 ftlb on a 7/16" bolt. 10 bolts. 1 13/16" radius. Assuming 0.15 for the coefficient of friction (which is apparently a safe value in production), then we get 395 ft lb … which is comfortably greater than the 295 ft lb rating of the standard engine. So this results in a factor of safety of 1.3.
This is the primary operating condition of the crank - flywheel flange.
In the case where we have a frictionless condition (a failure has occurred), there are two scenarios recommended :
- Shear is taken on half the bolts. (Min safety factor allowed = 2)
= T/(rp * n * Af)
where T is engine torque, rp is the pitch circle radius of the bolts in shear, n the number of bolts (in this case 5) and Af is the cross sectional area of a single fastener (assumes shear is directed at the unthreaded bolt shank, no on the thread)
Assuming 295 ftlb max engine torque, I make this to result in 2600 psi per fastener.
- Or Shear is taken by one bolt or dowel (Min safety factor allowed = 1)
Results in 13,000 psi on bolt or dowel (using formula above)
Results in a safety factor of 4 (which seems quite high).
(I calc the max allowable shear stress on one 7/16" bolt as being ~52,000 psi.)
I don’t guarantee these numbers. I might have done something dumb. Someone can check my maths - I’m not competent with lbs and ft and things! However it points you in the direction of what is happening and what to look for.
Main issue for me is if you want to tweak an engine and deliver 571 ft lb (which I am) then the 395 ft lb capability of the standard 10 bolt arrangement with a 0.15 coefficient of friction is inadequate. The crank to flywheel interface has to have a coefficient of friction of at least 0.5 (with no factor of safety) or 0.65 to maintain at least a 1.3 factor of safety. Given an ali flywheel has potentially a lower coefficient of friction to the steel crank flange than a steel flywheel, this is another reason why alloy flywheels might not be the best solution when a steel flywheel can be made stronger with the same inertia.
Hope this helps someone.
Cheers
Mark