flywheel interference fit to shaft.... allowance for ID expansion?
flywheel interference fit to shaft.... allowance for ID expansion?
(OP)
Are there any thumbrules for level of interference fit which should be used between flywheel and shaft?
The flywheel of interest to me is keyed to the shaft as well as interference fit.
I have heard that under the influence of centrifugal force the i.d. of the flywheel can expand (if it becomes loose this may allow slight movement resulting in eccentricity or unbalance). Is there any easy way to estimate this expansion?
The flywheel of interest to me is keyed to the shaft as well as interference fit.
I have heard that under the influence of centrifugal force the i.d. of the flywheel can expand (if it becomes loose this may allow slight movement resulting in eccentricity or unbalance). Is there any easy way to estimate this expansion?





RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
I don't have Roark and Young. Is it a textbook?
Any other references?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
I see the formula on page 704.
Thanks.
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
Can anyone check me on the following result.
Steel Flywheel 8" id and 8' od rotating at 1200rpm.
I get 0.020" of expansion which is astonishingly high. Maybe I have made an error? Here is the calculation.
From Roark p704:
deltaR0:=delta*w^2*R0/4/g/E*( (3+nu)*R^2+(1-nu)*R0^2);
where:
> # R0 = inner radius
> # deltaR0 = change in inner radius
> # R = outer radius
> # w = radian speed
> # delta = weight per volum
> # g = acceleration of gravity
> # E = Modulus of elasticity
> # n = poisson's ratio
Substitute values:
R0:=8*inch:R:=96*inch:
> w:=evalf(2*Pi*1200)/minute;
w := 7539.8/minute
> g:=32.2*lbm*ft/sec^2/lbf: #?
> E:=30E6*lbf/inch^2:
> nu:=0.3:
> delta:=7.8*62.4*lbm/ft^3; #SG=7.8
delta := 486.72*lbm/ft^3
Subbing in above gives:
deltaR0 := .17449e7/ft^4/minute^2*inch^5*sec^2
Unit Conversions:
deltaR0:=deltaR0*(ft/12/inch)^4*(minute/60/sec)^2;
This calculation gives:
deltaR0 := .23375e-1*inch
I always get a little confused by the treatment of g in equations like this. Did I mess up that part?
Thx for any assistance.
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
> R0:=4*inch:
> R:=48*inch:
deltaR0 := .29219e-2*inch
0.003" change. More reasonable but still sounds high. Any comments?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
Will you ever want to remove the flywheel in the future?
Manufacturing tolerance - this can be an issue with bore stress levels making selecting ideal interference difficult.
Thermal expansion effects - unlikely to be a problem with a flywheel.
There is an old rule of thumb .001" per inch of shaft diameter. Although I still advocate doing the calculation, my experience is that this guideline usually gives an acceptable result when checked by more sophisticated analysis such as FEA.
Only a small amount of residual interference will make the key redundant. Keep in mind also that with a longish hub and narrow web the stress and deflection in the hub can vary considerably along its axial length.
RE: flywheel interference fit to shaft.... allowance for ID expansion?
Englishmuffin's proposal for Ringfeder is also good. I understand the theory of using the compressive type but I think that the internal ones would work well. I have used both types with success and both types make removal relatively easy.
If there is never going to be reason to remove the flywheel a good heavy shrink fit is least expensive and most reliable. Fit by liquid nitrogen cooling shaft is probably easier than heating flywheel. If you do decide to heat flywheel using gas burners start by heating at rim - don't concentrate heat on hub area.
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
There will be poosible need to remove flywheels in the future.
The flywheel is also keyed to the shaft at 3 locations spaced at 120 degrees apart.
My rough understanding is that the keys play a role in keeping the flywheel centered as follows: Each key allows the flywheel centerline to move only along the line which passes through the center of that key and center of the shaft. There is only one intersection of the three lines. Thus the keys (assuming they have a tight circumferential fit) help to keep the flywheel centered even when the interference disappears.
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
RE: flywheel interference fit to shaft.... allowance for ID expansion?
I have heard of the step increase at a plant in CT. Here is the way I understand the situation: All W machines come with the 3-key design, spaced 120 degrees apart. Originally 2 of those 3 keys are 3-piece key design and the third is a 1-piece key. An upgrade converts the third to 3-piece design. The three-piece key just makes it that much tighter when you assemble the machine initially.
We ourselves have not installed the upgrade and have not experienced any step increases in vib. However during refurbishment it was identified the flywheel was looser than OEM tolerances. That is when I first got dragged into the evaluation. A learning experience for me. I suspect the keys are more critical to preventing vib problems than the flywheel interference, but both play some role.