Keyed Hub Design 1

[BikeDaily]

Greetings,

I am designing a hub that will go over a gearbox output shaft. The key comes on the gearbox output shaft standard, so in this case I don't have to select cross section and length. Although I know the general guidelines for shaft diameter/key size. But what I am having an issue with is how much "meat" needs to be on the hub for sufficient strength.

Any thoughts?

Cheers,

Bike Daily

 

[MikeHalloran]

For a rule of thumb, look at the hub diameter for a mating pulley or sprocket in your Boston or Martin or Browning catalog.





Mike Halloran
Pembroke Pines, FL, USA

 

[BikeDaily]

Mike,

I have used this approach in the past. But often I have large diameter shafts >8" and high torque 500,000 in-lb and greater. And often I need to fit this in a small area. So using higher strength materials is often necessary. Is there a "correction factor" for a keyway in a tube when looking at torsional stress like there is for shafts?

Cheers,
Jeff

 

[MikeHalloran]

You may have to drag out your machine design book for that...


Mike Halloran
Pembroke Pines, FL, USA

 

[Cockroach]

You will need to define what the maximum allowable torque is on the shaft and then use something like 1.50 multiplier on that load. This would give you the extreme peak torque to which the shaft would be designed, thus allowing for uncertainties and alternate variances.

With such a peak torque defined and depending on how it is carried through the shaft, you then look at torque and bending stresses separately, then in a combined state using Mohr's Circle. This will give you the principle stresses in the shaft due to torque and placement. Comparing such a principle stress to the material yield strength will give you a factor of safety, FOS, for your design. It is either acceptable or not.

I would then do a keyway analysis and find the key performance given the geometry of the keyway and configuration of such a key.

These are fairly straight forward mechanics of materials stress computations and can be found in first or second year static textbooks. I would be surprised if you couldn't find an example in such a textbook that would be remarkably similar in nature to your application. So then you would work through that example to get the right answer, apply that example strategy to your situation, an poof....there you have it.

Good luck with it.

Regards,
Cockroach

 

[rb1957]

the key is running along the shaft (parallel with the shaft axis) ?

so the key is removing some of the thickness (so the torsion stresses have to squeeze uner the key, yes ?)

nett area sounds like an approach.

consult Petersen (Stress Concentration Factors) if worried about fatigue.

if you Had to, couldn't you tailor the key section for your new part. I appreciate that it's mating with an existing key-way, though you might assemble it but putting the key in first (if the key in your hub is modified down from the existing key); appreciate this is unusual assembly, but it might take out some meat you don't want or need.

 

[boo1]

A common standard is available from the Mechanical Power Transmission Association, MPTA-B1c-2010.

http://www.mpta.org/

Other useful industry standards are ANSI Standards B17.1 for Keys and Keyseats or 17.2.

Flat keys have two modes of likely failure: shear and crushing.
shear force Fs = T/(d/2)
shear stress key τ = Fs /(L*b)
bearing pressure p = Fs /((h/2)*L)

The required key length can be obtained using either the maximum shear stress theory or by setting the average stress equal to the allowable shear stress. For 8" Shaft Diameter typically the key is 2" W x 1 1/2" D. The key length should be less than about 1.5 times the shaft diameter to ensure a good load distribution over the entire key length when the shaft becomes twisted when loaded in torsion.

REQUIRED KEY LENGTH IN SHEAR
Sys=0.5*Syt
τ_(allow)=0.5*Syt/Nsf
τ_(ave) =2*T/(D*W*L)=0.5*Syt/Nfs

solve for min L
L=4*T*Nfs/(Syt*D*W)

REQUIRED KEY LENGTH IN BEARING
L=4*T*Nfs/(K*Syc*D*H)

triaxial stress
1</=K</=1.5

Note: Keys are generally designed to fail before overloads can cause damage to the shaft or attached component.