# Torque Transmission - Slippage at Press Fit Connection

1. Jul 21, 2015

### TomBanks

Hello everyone. The physics forum is a great online community, and I used to spend a lot of time browsing the philosophy pages. But now I've had to register and post, because I've got a very specific problem with regards mechanical engineering.

My problem is relating to torque transmission through a press fit connection. Basically, I have a design that works and transmits the torque required of it; the design is, however, over-engineered and I wish to optimise it. I wish to be able to specify a torque (say 60Nm) and from that be able to mathematically determine a length of press fit required to transmit that torque.

Step 1.

I believe there are three steps required to reach to get there. The first step requires to determine the contact pressure at the press fit connection. Because there are tolerances on the components, there will be a maximum and minimum pressure. Currently the tolerances on the components are fixed, and I can not change them. The formula I'm using to calculate the pressure is one I've seen referenced in several books and on several websites. The formula I'm using is:

Where:

Vo = Poisson’s ratio of hub = 0.292
Vi = Poisson’s ratio of shaft = 0.292
Dhi = Hub inner diameter = 108.5mm
Dho = Hub outer diameter = 113.0mm
Dsi = Shaft inner diameter = 62.0mm
Dso = Shaft outer diameter = 108.5mm
Eh = Hub elastic modulus = 210000 mpa
Es = Shaft elastic modulus = 210000 mpa
Delta = Size of interferance = 0.119

There is also an excel based online calculator that uses the same equation at:

Using the above formula and values I calculate the pressure (P) to be: 8.475 N/mm^2. Which is also the same answer -- minus rounding errors -- that the excel based calculator gives me.

This leads me to believe that step 1. is the correct approach. Any input would be apprecaited.

Step 2.

Now that I know the pressure at the press fit, I can specify the desired amount of torque I wish to transmit and from that determine the force acting on the press fit.

I wish to transmit = 60 N/m of Torque = 60000 N/mm = T

Shaft Diameter = 108.5mm = D

T = Force x (D/2)

Rearranged to isolate the Force (F):

F = T / (D/2)

F = 60000 / (108.5 / 2)

F = 1106 N

If my thinking is correct, the force exerted on the shaft (and press fit) by transmitting 60Nm of torque is 1106N. From this, and knowing the pressure, I can calculate the length of the press fit required to transmit the torque.

Step 3.

To calculate the length of the press fit required, I'm using the following equation and transposing it to make Length the subject:

F = P x D x pi x L x u

L= F / pi x D x P x u

Where:

u = steel on steel coefficient = 0.4

So:

L = 1106 / pi x 108.5 x 8.475 x 0.4

L = 0,92mm

The required length of press fit necessary to transmit 60Nm of torque I calculate to be 0.92mm, which I feel is well too liberal (I'd expect between 5mm and 10mm). Is there anything that jumps out at anyone as being an obvious error? Perhaps I'm using the wrong equation(s) in step 2 and 3? Perhaps it is a unit error? Or a mistake transposing the equation? Or perhaps it is simply correct. Any input would be appreciated. I've been looking at the problem for a week now and I can't work out where I am going wrong.

Tom

Last edited: Jul 21, 2015
2. Jul 21, 2015

### Nidum

Deleted pro tem

3. Jul 21, 2015

### TomBanks

I think the first equation covers this? It's requires outside and internal diameters for the two components that make up the press fit.

4. Jul 21, 2015

### Nidum

Interface shear stress = radial stress * coefft. friction .

Break away torque = interface shear stress * circumference * shaft radius * length of engagement .

Numbers : Break away torque for a 1 mm length = 8.475 * 0.4 * 108.5 * pi * 108.5/2 = 62695 N-mm .

Your applied torque = 60000 N-mm .

About the same . So about 1 mm engagement gives break point between holding and breaking away .

Now factor that by about 25 for a safe design !

Seriously - an engagement of just 1 mm would be unsafe and unworkable . Applying some sensible factors for safety factor , usage class , shock loading , outer ring creep and practicality of assembly I would not be happy with an engagement of much less than 15 mm . If it doesn't cost too much it is always better anyway to go for good long engagements with press fits because there are always so many uncertainties .

Press fits are rapidly going out of fashion in new designs in favour of positive locking or high strength bonding .

5. Jul 21, 2015

### TomBanks

Nidum, thank you. Not only is this a thorough reply but a speedy one, too.

So I was on the right lines then?

The equation I used in Part 3 is the same as 'Break away torque = interface shear stress * circumference * shaft radius * length of engagement'

I was just shocked that it was giving me such a small value. As well as a press fit, the connection is bonded with locktite. But I just wanted to be able to mathematically determine my limits. The shaft is also subejct to a bending momnet, and I hope to be able to factor that in as a next step. But I first wanted to get to the bottom of the press fit torque transmission. I guess I will just have to accept the results of 1mm?

Last edited: Jul 21, 2015
6. Jul 21, 2015

### Nidum

7. Jul 21, 2015

### Nidum

Case for press fits subject to bending moment proceeds along similar lines to the torque case but you have to sum contributions from many small areas of interface * their individual lever arms to get total holding moment . Can be done by integration or just numerically using say 10 segments /quadrant .

You can't use the interface shear stress twice over - once for torque and once for bending moment . A reasonable apportion has to be given to each applied loading type .

A problem with press fitted joints carrying bending moment is cyclic creep - joints don't nescessarily come apart but they just creep around at very slow speed .

Again best to have generous overdesign.

Last edited: Jul 21, 2015
8. Jul 22, 2015

### TomBanks

I will take the segment aproach to calculating the bending moment's effects upon the press fit.

Is there a way for me, mathmatically, to determine how much of the stress I appropriate to the torque transmission and how much to the bending moment?

9. Jul 22, 2015

### TomBanks

1mm is completely impractical, agreed, but I want to be able to explain why mathmatically. I will try and account for shock loading.

10. Jul 22, 2015

### Nidum

The best way is to estimate the forces acting on each segment due to torque and due to bending and then find the resultant force . This single force can then be checked against maximum available interface shear stress times segment area . That way you don't need to make any guesses .

Situation will of course vary around circumference . If you do checks at worst location then you'll get a very conservative estimate for strength of whole assembly .

11. Jul 22, 2015

### Nidum

Check the hoop stress generated in the outer ring by the press fit .

12. Jul 22, 2015

### TomBanks

The worst case press fit I have is a thin walled cylinder, where the outer ring has an inside diameter of 108.5mm and a wall thickness of 2,25mm (Outisde dimaeter of 113mm).

The pressure I know from step 1 (first post) to be 8.475 N/mm^2.

So the Hoop stress = Pressure x Inside Radius / wall thickness = 8,475 * (108,5/2) / 2,25 = 92.73 N/mm^2.

The material is C45 with a Yield strength of 210000 N/mm^2, so a stress of 92.73 N/mm^2 is neglible.

I have configurated my spreadsheet to account for changes in tolerances, factors of safety, etc, and there is no circumstances under which my press fits are stressed to yield (they don't get close to it). I suspect this will be different once I've calculated the effect of the bending moment. I will try and calculate the twisting moment too, because I believe the effect of torsion will be far higher than the hoop stress dervied from the interferances of press fit.