climbingtech
Mechanical
- Aug 12, 2020
- 7
I've run several calculations for a shrink fit and for likely user error reasons I'm getting very high numbers.
Looking for a 30-40 kn axial force requirement to break a shrink fit on 7075-T6. I ran my calculations with a 0.007" interference not accounting for surface roughness at a frictional coefficient of 1.1 (contact is naked 7075-7075).
The shaft is ID 0.625", OD 0.75"
The collar is ID 0.743", OD 1.125"
Length of engagement is 0.5"
I understand wall thickness and length of engagement are not optimal, just trying to get a feel for where it sits as far as part design is concerned. I just need the press fit to be able to handle >15 kn applied dynamic force to exceed the weakest failure point which is the shaft itself.
My last calculation using Lames equations and conversion from standard to metric yielded 225 kn static frictional force. lol I think I'm making an error somewhere or I personally underestimated the power of the interference.
Any help is appreciated, stay safe!
Looking for a 30-40 kn axial force requirement to break a shrink fit on 7075-T6. I ran my calculations with a 0.007" interference not accounting for surface roughness at a frictional coefficient of 1.1 (contact is naked 7075-7075).
The shaft is ID 0.625", OD 0.75"
The collar is ID 0.743", OD 1.125"
Length of engagement is 0.5"
I understand wall thickness and length of engagement are not optimal, just trying to get a feel for where it sits as far as part design is concerned. I just need the press fit to be able to handle >15 kn applied dynamic force to exceed the weakest failure point which is the shaft itself.
My last calculation using Lames equations and conversion from standard to metric yielded 225 kn static frictional force. lol I think I'm making an error somewhere or I personally underestimated the power of the interference.
Any help is appreciated, stay safe!