Inputs
Joint pressure and relative effective interference at standstill, outside diameter of the outer part, operating speed, plus elastic modulus, Poisson ratio and density; optionally the diameter ratio and yield strengths for the elastic check.
Shaft-hub · DIN 7190-1
Interference fit at speed: centrifugal effect
An interference fit that holds at standstill can let go when it spins: the outer part widens more under centrifugal force than the inner one, the joint pressure falls, and with it everything the fit can transmit. This calculator determines, per DIN 7190-1 section 10.2, the remaining joint pressure, the speed at which the outer part lifts off, and the limit for a pressure drop of at most 10 %.
Valid under the preconditions of section 10.2: a solid inner part, identical elastic constants and densities in both parts, and purely elastic behaviour. The joint pressure and relative effective interference at standstill come from the interference fit design itself; hollow inner parts or differing materials require the extended methods the standard refers to.
pnDIN 7190-1
02
Joint pressure at speed
Set the fit and the speed.
Inputs and method
Interference fit under centrifugal force to DIN 7190-1
Calculate how far centrifugal force reduces the joint pressure of a rotating interference fit and at what speed the outer part lifts off.
Calculation
Per DIN 7190-1 section 10.2, u = π·n·DAa (56) is the circumferential speed of the outer contour and pn = [1 − (u/uab)²]·p (57) the joint pressure at speed, with the lift-off speed uab = 2·√(E·ξw/((3+μ)·ρ)) (58). A pressure drop of at most 10 % requires u ≤ 0.316·uab (59). The purely elastic precondition is checked through the bounds (54) and (55) on ξw.
Example
The standard's own example A.7: ξw = 1.116·10⁻³, DAa = 100 mm, E = 215,000 N/mm², μ = 0.3, ρ = 7.85 kg/dm³ give uab = 192.5 m/s. At 20,000 min⁻¹ the speed is u = 104.7 m/s, and a joint pressure of 90 N/mm² falls to 63.4 N/mm².
Sources and limits: DIN 7190-1:2017-02, section 10.2, formulae (54) to (59); worked example from Annex A.7 of the same standard. Valid only under the preconditions stated there: solid inner part, identical elastic constants and densities in both parts, purely elastic behaviour.
Interference fit under centrifugal force to DIN 7190-1 in detail
Calculate how far centrifugal force reduces the joint pressure of a rotating interference fit and at what speed the outer part lifts off.
Why does an interference fit lose its grip at speed?
An interference fit transmits force purely by friction in the joint, and that friction lives on the joint pressure. When the fit rotates, centrifugal force acts on both parts. Because the outer part moves outwards more than the inner one, part of the interference is consumed and the joint pressure falls – until at a certain speed it disappears entirely and the fit lifts off.
Formula and variables
pn = [ 1 − (u / uab)² ] · p
u = π · n · DAa (n in revolutions per second)uab = 2 · √( E · ξw / ((3 + μ) · ρ) )at most a 10 % drop: u ≤ 0.316 · uabpurely elastic: ξw ≤ (3+μ)/(3+μ+(1−μ)·QA²) · ReLA/E and ξw ≤ ReLI/(QA²·E)
| Symbol / input | Meaning |
|---|---|
| p, pn | Joint pressure at standstill and at the speed considered. |
| ξw | Relative effective interference of the fit, from the standstill design. |
| u, uab | Circumferential speed of the outer contour and the speed at lift-off. |
| E, μ, ρ | Elastic modulus, Poisson ratio and density, identical in both parts. |
Choose the inputs correctly
Joint pressure and relative effective interference at standstill, outside diameter of the outer part, operating speed, plus elastic modulus, Poisson ratio and density; optionally the diameter ratio and yield strengths for the elastic check.
How to use the calculator
Take the joint pressure and the relative effective interference from the interference fit design – the linked calculator reports both. Add the outside diameter of the outer part, not the joint diameter, and the operating speed. Also enter the diameter ratio and the two yield strengths so the calculator can verify the standard's purely elastic precondition; without them that check stays open.
Worked example
The standard's own example A.7: ξw = 1.116·10⁻³, DAa = 100 mm, E = 215,000 N/mm², μ = 0.3, ρ = 7.85 kg/dm³ give uab = 192.5 m/s. At 20,000 min⁻¹ the speed is u = 104.7 m/s, and a joint pressure of 90 N/mm² falls to 63.4 N/mm².
Which value is the informative one?
The remaining joint pressure is the quantity with which the transmission capability has to be recalculated – torque and axial force are directly proportional to it. More informative for judging the situation, though, is the ratio u/uab: because it acts quadratically, it shows immediately how close the fit is to lifting off. A value of 0.5 already means a quarter of the pressure is gone, even though the speed is only half the limit.
Joint pressure and yield strengths in N/mm², diameters in millimetres, speed in min⁻¹, density in kg/dm³. The relative effective interference is dimensionless and typically of the order 10⁻³.
Typical applications
Shrunk-on gear rims, tyres and bandages, rotors of high-speed machines, turbocharger and spindle fits, and any check of whether an interference fit designed at standstill will take its operating speed.
Assumptions, limits and common mistakes
DIN 7190-1:2017-02, section 10.2, formulae (54) to (59); worked example from Annex A.7 of the same standard. Valid only under the preconditions stated there: solid inner part, identical elastic constants and densities in both parts, purely elastic behaviour.
Common mistake: Do not enter the joint diameter instead of the outer part's outside diameter – what governs is the fastest point of the component. Do not confuse ξw with the interference in micrometres: ξw is the effective interference divided by the joint diameter. And do not treat centrifugal force as a small allowance – it enters with the square of the speed, so the joint pressure is already down by a quarter at half the lift-off speed.
Frequently asked questions
Why does the joint pressure fall at all if both parts expand?
Because they expand by different amounts. The outer part has more mass at a larger radius and stretches radially more under centrifugal force than the inner one. The difference between the two expansions eats into exactly the interference that produces the joint pressure – and with the joint pressure, the transmissible torque and axial force fall in the same proportion.
What does the lift-off speed mean in practice?
At it the joint pressure has vanished entirely and the outer part lifts off; the fit transmits nothing. The outer part can additionally move out radially and create large unbalance. DIN 7190-1 therefore explicitly requires an adequate margin against this speed, not only consideration of the optional 10 % criterion.
Why is the limit for a 10 % pressure drop only 0.316 of the lift-off speed?
Because speed enters quadratically: a 10 % drop corresponds to (u/uab)² = 0.1, so u/uab = √0.1 = 0.316. At barely a third of the lift-off speed a tenth of the joint pressure is already gone – the relationship is far less forgiving than a linear intuition suggests.
What if the preconditions are not met?
The formulae of section 10.2 hold for a solid inner part, identical materials in both parts and purely elastic behaviour. If the inner part is hollow or the materials differ, the standard points to extended methods in the literature; elastic-plastic designs have their own reference. The calculator checks the elastic precondition through the bounds (54) and (55) as soon as the values needed for it are supplied.
Does a larger interference help against lift-off?
Yes, and in two ways at once: a larger relative interference raises both the joint pressure at standstill and the lift-off speed, since the latter grows with the square root of ξw. The limit is set by the yield strength: once the interference is large enough to violate condition (54) or (55), the fit leaves the purely elastic range and this calculation no longer applies.