Interference Fit Calculator (Press Fit Pressure and Stress)
Contact pressure, hoop stress and von Mises stress for a press or shrink fit between a shaft and a hub, worked out from the diametral interference and the two materials. The joint is solved as two Lame thick wall cylinders, and the results include the push-out force, the torque the friction ring can carry, and the temperature rise a shrink fit needs.
How to use this calculator
- Enter the joint diameter, the shaft bore diameter and the hub outside diameter. Set the shaft bore to 0 for a solid shaft.
- Enter the diametral interference, which is the shaft diameter minus the hub bore measured before assembly, and the length over which the two parts overlap.
- Pick the shaft and hub materials, or select Custom modulus to type in Young's modulus and Poisson's ratio directly.
- Read the contact pressure, how much the bore opens up, the hub and shaft stresses against yield, the push-out force, the torque capacity and the heating temperature for a shrink fit.
Frequently asked questions
How is press fit pressure calculated?
The interference is shared between the two parts in proportion to their stiffness: delta = p x d x (Ch / Eh + Cs / Es). Ch and Cs are the Lame flexibility coefficients of the hub and the shaft evaluated at the joint diameter, with Ch = (do^2 + d^2) / (do^2 - d^2) + nu and Cs = (d^2 + di^2) / (d^2 - di^2) - nu. For a solid shaft the shaft term collapses to 1 - nu. Rearranged for pressure, p = delta / (d x (Ch / Eh + Cs / Es)).
What interference should I specify for a steel press fit?
Most steel press and shrink fits sit somewhere between 0.04 and 0.15 percent of the joint diameter, which for a 30 mm joint means 12 to 45 microns. Below about 0.02 percent the joint relies almost entirely on friction and can slip under shock loads. Above about 0.25 percent the hub bore risks yielding while it is being pressed on, which destroys the interference you paid for.
Why does the hub bore grow more than the shaft shrinks?
The two parts share the interference in proportion to their stiffness, and a solid shaft is far stiffer than the surrounding hub. In the default case here, a 30 mm joint in a 60 mm hub with 0.03 mm interference, the bore opens up by 22 microns while the shaft closes in by only 8 microns. Those two figures always add back to the original interference.
Why is the torque capacity different from the push-out force?
Both come from friction over the same contact area, but the push-out force is the friction acting on the ring area while the torque is that same friction acting at the radius. The push-out force is mu x p x pi x d x L, and the torque is half of the force times the joint diameter, T = mu x p x pi x d^2 x L / 2.