What is a shrink fit and how is it different from a press fit?

A shrink fit assembles parts by heating the outer sleeve (or hub) so it expands, sliding it over the shaft, and then letting it cool. As it cools the sleeve tries to contract onto the slightly larger shaft, generating a contact pressure p at the interface. A press fit instead pushes the parts together axially with force. Shrink fitting avoids galling of the contact surfaces and allows much larger interferences than press fitting, which is why it is preferred for large-diameter, high-torque joints such as railway wheels on axles or large gears on shafts.

How is the contact pressure p calculated?

From the Lamé thick-cylinder solution. For a solid shaft and an outer sleeve made of the same material, with diametral interference δ, interface radius r_b, sleeve outer radius r_o and Young's modulus E, the contact pressure is p = (E·δ)/(2·r_b) × (r_o²−r_b²)/(2·r_o²). The pressure grows linearly with interference and with E. Making the sleeve thicker raises p, but with diminishing returns: as r_o tends to infinity p approaches a finite limit set by interference alone.

Where is the maximum stress in the sleeve?

The maximum stress in the sleeve is the hoop (circumferential) stress at its inner surface r = r_b, given by σθ_max = p·(r_o²+r_b²)/(r_o²−r_b²). It is tensile and decreases monotonically outward. At the same surface the radial stress equals −p (compressive). In the solid shaft (r_i = 0) the state is uniform σ_r = σ_θ = −p, that is, a pure hydrostatic compression. Designers check that σθ_max stays below the allowable tensile stress of the sleeve material.

How much torque can a shrink fit transmit?

The maximum torque at the slip limit is T = 2π·μ·p·r_b²·L, where μ is the friction coefficient, p is the contact pressure, r_b is the interface radius and L is the engaged length. For dry steel-on-steel μ ≈ 0.15, with oil it drops to about 0.10. In practice a safety factor of 2 is common to account for shock loads and partial slip. To raise the transmissible torque, increase the engaged length first, or increase the interference; the latter trades torque capacity against the risk of bursting the sleeve.

Shrink-Fit Stress Simulator — Free Online Calculator

Parameters

Interface diameter D_b

Sleeve outer diameter D_o

Diametral interference δ

Young's modulus E

GPa

Solid shaft (r_i = 0), engaged length L = 50 mm, friction coefficient μ = 0.15 (dry steel-on-steel) are assumed. Shaft and sleeve share the same material (same E and ν).

Results

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Interface contact pressure p

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Sleeve max σ_θ at bore (tensile)

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Shaft σ_r = σ_θ (compressive)

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Max transmissible torque T (μ=0.15)

Cross-section and radial stress distribution

Left: cross-section (blue = shaft, orange = sleeve) / Right: σ_r(r) (orange) and σ_θ(r) (blue) along the radius; σ_θ jumps at the interface r = r_b

Theory & Key Formulas

The shrink fit between a solid shaft (radius r_i = 0 to r_b) and an outer sleeve (r_b to r_o) of the same material (same E and ν) is described by the Lamé thick-cylinder solution. Here δ is the diametral interference.

Interface contact pressure p (solid shaft):

$$p = \frac{E\,\delta}{2\,r_b}\cdot\frac{r_o^2 - r_b^2}{2\,r_o^2}$$

Radial and hoop stress inside the sleeve (r_b ≤ r ≤ r_o):

$$\sigma_r(r) = \frac{p\,r_b^2}{r_o^2 - r_b^2}\left(1 - \frac{r_o^2}{r^2}\right),\quad \sigma_\theta(r) = \frac{p\,r_b^2}{r_o^2 - r_b^2}\left(1 + \frac{r_o^2}{r^2}\right)$$

Maximum tensile hoop stress at the sleeve bore (r = r_b) and the uniform stress inside the solid shaft:

$$\sigma_{\theta,\max} = p\,\frac{r_o^2 + r_b^2}{r_o^2 - r_b^2},\qquad \sigma_r = \sigma_\theta = -p\ (\text{shaft})$$

Friction-limited transmissible torque (μ is friction coefficient, L is engaged length):

$$T = 2\pi\,\mu\,p\,r_b^2\,L$$

For the defaults (D_b=100, D_o=150, δ=0.1 mm, E=210 GPa, L=50 mm, μ=0.15) we get p ≈ 58.3 MPa, σ_θ,max ≈ 151.7 MPa (tensile), shaft stress −58.3 MPa (compressive) and T ≈ 6.87 kN·m.

What is the Shrink-Fit Stress Simulator

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So a shrink fit is just heating a sleeve and slipping it over a shaft, right? Why does it hold so tightly once it cools down?

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Roughly speaking, the sleeve tries to contract back to its original size but the slightly oversized shaft stops it. That blocked contraction is the interference δ, and it ends up entirely as elastic strain at the interface. The result is a contact pressure p. Slide the "diametral interference δ" from 0.05 mm up to 0.20 mm in the simulator. The p stat card grows almost linearly because in the Lamé solution p is linear in δ.

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So if I want more torque, I just keep cranking up the interference?

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In principle yes — torque is $T = 2\pi\mu p r_b^2 L$, proportional to p. But watch the blue curve on the right, the hoop stress σ_θ at the sleeve bore. Larger interference scales p and σ_θ together. The defaults give σ_θ ≈ 152 MPa, but doubling the interference to 0.20 mm pushes σ_θ above 300 MPa, easily past the allowable tensile stress of plain carbon steel (≈ 200–250 MPa). The sleeve yields, or worse, bursts. It is always a tug-of-war between torque capacity and sleeve failure.

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What if I just make the sleeve thicker? If I slide D_o up, the stress should drop, right?

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Good instinct. Move D_o from 150 to 300 mm and σ_θ drops — but the rate of reduction slows. From $\sigma_{\theta,\max} = p\frac{r_o^2+r_b^2}{r_o^2-r_b^2}$, as r_o → ∞, σ_θ,max approaches p, not zero. No matter how thick the sleeve, the bore hoop stress can never be smaller than the contact pressure. In practice D_o ≈ 1.5 to 2.0 × D_b is the sweet spot between weight and stress. The defaults at D_o/D_b = 1.5 sit right at that economical point.

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And what about Young's modulus E? Say brass (E≈100 GPa) versus steel (E≈210 GPa)?

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p is linear in E. For the same interference, a brass sleeve gives half the contact pressure and half the torque capacity. Interestingly the stress ratio σ_θ/p depends only on geometry (r_o, r_b), so E does not change the "safety margin" relative to the material's own allowable stress. The takeaway is that softer materials need a larger interference to reach the same contact pressure.

Frequently Asked Questions

Both rely on the same Lamé thick-cylinder solution for the final contact pressure, but the assembly methods differ. A press fit pushes parts together with a hydraulic ram, which limits interference and can score the mating surfaces. A shrink fit heats the sleeve (typically 150–300 °C) so it slides on with virtually zero force; on cooling it locks onto the shaft. For large diameters and high transmissible torque, shrink fitting is preferred because the assembly force does not scale with interference.

Required thermal expansion equals "interference δ + an assembly clearance of about 0.05 to 0.10 mm". For steel with α ≈ 12×10⁻⁶ /K, the temperature rise is ΔT = (δ + clearance) / (α·D_b). For example, with D_b = 100 mm, δ = 0.10 mm and 0.05 mm clearance, ΔT ≈ 125 K, so heat the sleeve to about 150 °C. For tempered components, stay well below the tempering temperature — a 200 °C ceiling is a common rule.

This simulator assumes a solid shaft (r_i = 0). For a hollow shaft use the generalized Lamé formula p = (E·δ)/(2·r_b) × (r_o²−r_b²)(r_b²−r_i²) / (2·r_b²·(r_o²−r_i²)). A hollow shaft adds compliance on the inner side, so the same interference produces a lower contact pressure. It also generates a compressive hoop stress at the inner bore (r = r_i), so both the shaft and the sleeve must be checked.

For high-speed components (turbine rotors, flywheels), centrifugal force expands the sleeve and reduces the effective interference. The loss scales like ρω²r_b², so high-speed designs use a larger static interference, or set the "release speed" (the rotational speed at which contact pressure vanishes) well above the design operating speed. This simulator computes static values only; centrifugal correction is required separately for spinning rotors.

Real-World Applications

Railway wheels on axles: The textbook example of shrink fitting. Wheels are shrunk directly onto the axle, transmitting traction and braking torque without keys or welds. A slow "creep" in interference is a known concern, so wheels are inspected periodically with ultrasound and indentation measurements. Modern high-speed rail demands tighter interference and tighter tolerances than freight stock.

Large gears and shaft couplings: Big gears in rolling mills and marine propulsion shafts are mounted keyless to avoid stress concentrations at the keyway. The transmissible torque from the simulator is then divided by a safety factor of 1.5 to 2.0. Hydraulic interference couplings (such as SKF oil-injection couplings) make assembly and disassembly practical for large diameters.

Turbine rotor assembly: Steam and gas turbine discs are shrunk onto a shaft with margin for both operating centrifugal force and thermal expansion. Too much room-temperature interference can crack the disc during startup thermal stress; too little can lift the disc off at high speed, causing vibration. The static stress check in this simulator is the starting point for that design study.

Bearing outer races in housings: Rolling-element bearings press their outer ring into the housing with an interference fit. The resulting hoop stress in the outer race squeezes it inward and reduces internal clearance. International standards (ISO, JIS) tabulate allowable interferences by fit class to prevent this from cutting bearing life prematurely.

Common Misconceptions and Cautions

The most common misconception is treating a shrink fit as if it were a uniform glued joint. The contact pressure is uniform in the idealized analysis, but real fits exhibit "end effects" at the edges of the engagement length L. The hoop stress at the sleeve edge can be 1.3 to 2.0 times the central value, and fatigue cracks nearly always originate there. Practical designs chamfer the sleeve ends or taper the engagement to spread the stress. This simulator gives the stress at the central cross-section only.

The next most common error is mixing up diametral and radial interference. This simulator and JIS/ISO tolerance tables use "diametral δ = D_shaft − D_hole". Some textbooks use "radial δ_r = δ/2"; substituting one for the other gives a factor-of-two error in contact pressure. Always confirm whether your source is diametral or radial; everything here is diametral.

Finally, checking only the contact pressure against yield is dangerously optimistic. The hoop stress σ_θ is tensile and several times the contact pressure: at the default geometry (D_o/D_b = 1.5) σ_θ,max ≈ 2.6 × p. So p = 100 MPa already produces σ_θ,max = 260 MPa. The number that matters is σ_θ,max at the sleeve bore, compared with the allowable tensile stress of the material. For brittle materials like cast iron, also use the maximum principal stress (not von Mises) because they are weak in tension.

Shrink-Fit Stress Simulator — Lamé Thick-Cylinder Solution

What is the Shrink-Fit Stress Simulator

Frequently Asked Questions

Real-World Applications

Common Misconceptions and Cautions

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