Press, Shrink and Expansion Fits

An interference fit holds two parts together by friction alone. The shaft is slightly larger than the hole, so when they are assembled the hub stretches and the shaft squeezes, and the contact pressure that results carries torque and axial load with no key, pin or fastener. This page explains how to get an interference fit together (pressing it, heating the hub, or cooling the shaft), and how to estimate the pressure, stress and holding force it creates. The fit classes themselves are on Fits – Metric (ISO System) (H7/p6, s6, u6) and Fits – Inch (ANSI B4.1) (LN and FN). It is part of the Fits & Tolerances guide.

How an interference fit holds

The difference between the shaft diameter and the hole diameter is the interference, δ. When the parts are assembled the elastic deformation of both parts takes up the interference, and the pressure p at the interface is what holds them together. Three things follow:

• The pressure sets the capacity. Torque capacity and axial holding force are both proportional to the pressure and to the contact area, times the friction coefficient.
• The pressure also stresses the hub. The hub is pulled outward like a pressure vessel; the hoop stress at its bore is the largest stress in the joint, and it must stay below the material’s limit (the Handbook calls this “the most important point to consider”).
• The fit range matters. The parts are made to a tolerance, so the interference is a range. At the loose end the joint must still carry its load; at the tight end the hub must not yield. That is why FN classes keep the spread small.

Forced (press) fits

The Handbook’s traditional guidance for forced fits (assembled with a press or by driving):

• Allowance per inch of diameter usually runs from 0.001 to 0.0025 in, with 0.0015 a fair average. The allowance per inch decreases as the diameter increases: a 2-inch diameter might have a total allowance of 0.004 in, and 8 inches no more than 0.009 or 0.010 in.
• Cylindrical or tapered. The parts are usually cylindrical, though sometimes slightly tapered. A taper needs less pressure, wears the surfaces less and eases removal, but a tapered fit is less reliable: if it loosens, the whole fit is free with only a little axial movement.
• Lubricate the pin and bore before pressing (the Handbook suggests white lead and lard oil mixed to the consistency of paint) to reduce the tendency to abrade and gall.

Estimating the pressing force (Handbook rule)

The Handbook gives a formula for the approximate ultimate pressing force in tons, for a steel shaft in a cast-iron hub whose outside diameter is twice the bore:

P = A × a × F ÷ 2
A = area of the surface in contact (in²), a = total allowance (in), F = pressure factor from the table

Diameter	1 in	2 in	3 in	4 in	5 in	6 in	8 in	10 in	12 in	14 in	16 in	18 in
Pressure factor F	500	240	156	115	91	75	55	43.5	35.9	30.5	26.5	23.4

The Handbook adds that the average ultimate pressure in tons commonly used runs from 7 to 10 times the diameter in inches. Transposing the formula gives the allowance needed for a given tonnage: a = 2P ÷ (A F). This is an old, empirical method: use it for a quick check of press size, not for strength design.

Shrinkage and expansion fits

Instead of forcing the parts together, you can change the size of one of them with temperature. Heat the hub until it has expanded enough to slip over the shaft, or cool the shaft until it has shrunk enough to slip into the hub; when the temperatures equalize, the parts grip. There is no scoring of the surfaces and no press is needed.

Method	How	Notes
Shrinkage fit (heat the outer part)	Heat the hub (oven, oil bath, induction) until the bore has grown by the interference plus some assembly clearance	Do not heat so far that the material is damaged. Steel is often kept to a few hundred degrees C; check the material and any hardening or surface treatment
Expansion fit (cool the inner part)	Cool the shaft: dry ice (about −109 °F, −78 °C) or liquid nitrogen (about −320 °F, −196 °C)	Dry ice will not contract some parts enough; liquid nitrogen shrinks them further. The Handbook notes the cooling equipment ranges from an insulated bucket to an automatic unit
Both together	Heat the hub and cool the shaft	For large interferences, where one alone is not enough, or when heating alone would harm the material

How much a part shrinks on cooling (Handbook)

For a temperature drop from 75 °F to −321 °F (a 220 °C difference), the shrinkage per inch of diameter is about:

Material	Shrinkage per inch of diameter (in)
Steel	0.002 to 0.003
Aluminum alloys	0.0042
Magnesium alloys	0.0046
Copper alloys	0.0033
Monel metal	0.0023
Cast iron (not alloyed)	0.0017

Shrinkage allowances and holding strength

• Shrinkage allowances are generally set a little smaller than forced-fit allowances, although in many shops they are about the same, and for some work they exceed them. They depend a great deal on the form of the part: the thickness and distribution of the metal around the hole is the most important factor.
• Locomotive tires: allowances adopted by the American Railway Master Mechanics Association, for center diameters of 38, 44, 50, 56, 62 and 66 inches, are 0.040, 0.047, 0.053, 0.060, 0.066 and 0.070 in.
• Shrink beats force. Handbook tests comparing shrinkage and forced fits with the same dimensions and allowances found the resistance to slippage for an axial pull was 3.66 times greater for the shrink fit, and in rotation or torsion 3.2 times greater. A forced fit scrapes off some of its interference as the shaft is driven in; a shrink fit does not.
• Choose by situation. Pressing a tire over a wheel center without heating is awkward; pins and similar parts are easily and quickly forced in with a hydraulic press, which also tells you the exact pressure used. There is more uncertainty with a shrink fit unless the stresses are calculated.

Pressure, stress and holding force

The physics of an interference fit is the thick-walled cylinder (Lamé) problem. The equations below are the textbook elastic solution for a solid shaft in a hub, with k = hub outside diameter ÷ bore. They give the contact pressure, the stress, and the holding capacity, and they are what a calculation should check, whatever fit class was chosen.

Interference per unit diameter: δ/D = p × [ ( (k² + 1) / (k² − 1) + ν_hub ) / E_hub + ( 1 − ν_shaft ) / E_shaft ]
Same material, solid shaft: p = (δ/D) × E × (k² − 1) / (2 k²)
Hoop stress at the hub bore: σ_t = p × (k² + 1) / (k² − 1)
Axial holding force: F = π D L p μ
Torque capacity: T = π D² L p μ / 2

• δ is the effective diametral interference: the measured interference minus what the surface roughness flattens off when the parts are assembled. Roughness can use up a meaningful share of a small interference; measure the surfaces, or specify a finish.
• μ is the friction coefficient. Typical values for clean steel on steel run from about 0.1 to 0.2, depending on finish, lubrication and assembly method; treat it as an assumption to test.
• Use the loose end of the fit (the smallest interference) for holding force, and the tight end (the largest) for stress.
• Limits of the model: it is elastic, assumes a uniform fit, and ignores end effects, plastic yielding and temperature. Hubs shorter than the shaft diameter, keyways, thin hubs and mixed materials need more care (or finite-element analysis and a test).

The Handbook’s shrinkage allowance method

The Handbook gives a way of finding the allowance per unit of diameter from the stress you are willing to put in the hub. Take T as the true tangential tensile stress at the inner surface of the outer member, C as a factor from the table, and E as the modulus of elasticity (30,000,000 psi for steel, 15,000,000 for cast iron):

Cast-iron hub on a steel shaft: A = T (2 + C) / 30,000,000
Steel hub on a solid steel shaft: A = T (1 + C) / 30,000,000
A = allowance per inch of diameter, T in psi

The factor C depends on the ratio of the hub’s outside diameter to the shaft diameter, D₂/D₁ (Handbook Table 1, solid steel shaft):

D2 / D1	C, steel hub	C, cast-iron hub
1.5	0.227	0.234
1.6	0.255	0.263
1.8	0.299	0.311
2.0	0.333	0.348
2.2	0.359	0.377
2.4	0.380	0.399
2.6	0.397	0.417
2.8	0.410	0.432
3.0	0.421	0.444
3.2	0.430	0.455
3.4	0.438	0.463
3.6	0.444	0.471
3.8	0.450	0.477
4.0	0.455	0.482

Separate tables (Tables 2 and 3) give C for hollow steel shafts, as a function of the ratios D₂/D₁ and D₁/D₀, for steel and cast-iron hubs. The Handbook’s procedure is: choose the stress T you will allow (it must not exceed the elastic limit, or permanent set results, and in the extreme the hub bursts); find C; then compute the allowance. The thicker the hub, the stronger the grip and the greater the resistance to slippage.

Assembly temperatures

The temperature to which the outer member of a shrinkage fit must be heated depends on the total expansion needed and on the coefficient of linear expansion α of the metal. The total expansion in diameter is the shrinkage allowance plus an added amount for assembly clearance. The Handbook gives α for nickel steel as 0.000007 per °F, for steel in general 0.0000065, and for cast iron 0.0000062.

ΔT = (allowance + assembly clearance) ÷ (α × D)

Worked examples

Example 1: the Handbook’s shrinkage temperature

An outer steel member is to be expanded by 0.005 in per inch of internal diameter (0.001 is the shrinkage allowance; the remainder is clearance). Steel α = 0.0000065 per °F.

• Temperature rise = 0.005 ÷ 0.0000065 = 769 °F above room temperature (the Handbook’s “769 degrees F”).
• In SI that is a rise of about 427 °C. A hub heated this far must be of a material that tolerates it; in practice a smaller rise is used with a smaller interference, or the shaft is cooled as well.

Example 2: pressing force by the Handbook rule

A 2-inch steel shaft is pressed into a cast-iron hub with a contact length of 3 inches. Take the total allowance as 0.004 in (the Handbook’s figure for a 2 inch diameter).

• Contact area A = π × 2 × 3 = 18.85 in².
• Pressure factor for 2 inches: F = 240.
• P = A × a × F ÷ 2 = 18.85 × 0.004 × 240 ÷ 2 = 9.0 tons.
• The Handbook’s 7 to 10 times the diameter in inches gives 14 to 20 tons: the two rules differ, so size the press with generous margin.

Example 3: pressure, stress and holding capacity (SI)

A 40 mm steel shaft in a steel hub of 80 mm outside diameter (k = 2), contact length 50 mm, fit 40H7/s6. E = 207 GPa, friction μ = 0.15 (assumed).

• The ISO limits (see the metric fits page): hole 40.000 to 40.025, shaft 40.043 to 40.059 mm. Interference: 18 to 59 µm.
• Pressure at the loose end: p = (0.018 ÷ 40) × 207000 × 3 ÷ 8 = 34.9 MPa. At the tight end: 114.5 MPa.
• Hoop stress at the bore at the tight end: 114.5 × 5 ÷ 3 = 191 MPa. This must stay safely below the hub material’s yield strength, with margin for stress concentrations.
• Holding capacity at the loose end: axial force = π × 40 × 50 × 34.9 × 0.15 = 32.9 kN; torque = π × 40² × 50 × 34.9 × 0.15 ÷ 2 = 658 N·m.
• The tight end carries about 3.3 times the pressure of the loose end: that is the cost of a wide fit range. Roughness smoothing reduces the effective interference at both ends, so measure the actual finish.

Example 4: how hot, or how cold?

To assemble the 40H7/s6 fit of Example 3, the hub must grow by the largest interference plus an assembly clearance (assume 0.025 mm).

• Growth needed = 0.059 + 0.025 = 0.084 mm.
• Heating the hub: ΔT = 0.084 ÷ (11.7 × 10⁻⁶ × 40) = 179 K, so heat the hub to about 199 °C (assuming steel α = 11.7 × 10⁻⁶ per K, a typical value).
• Cooling the shaft in liquid nitrogen (−196 °C, from 20 °C): contraction = 11.7 × 10⁻⁶ × 40 × 216 = 0.101 mm, compared with the 0.084 mm needed. That is enough, with little margin, so combine a modest heating of the hub with cooling the shaft.

Troubleshooting

Symptom	Likely cause	Try
The joint slips under torque	Interference at the loose end is too small after roughness smoothing, or μ is lower than assumed (oil, coating)	Increase the interference, clean and degrease, or add a key or pin; measure surface finish
The hub cracks or yields	Interference at the tight end too large for the hub thickness or material; thin or cast-iron hubs have low strength	Check hoop stress at the tight end; use a lower FN class or a thicker hub
The shaft scores while pressing	No lubricant, burrs, or misalignment; an interference fit scrapes off some of its allowance	Lubricate, chamfer both parts, align, and consider a shrink fit
A shrink fit will not go together	The hub cooled during assembly, or the temperature rise was too small	Heat higher, assemble faster, or add shaft cooling; check α and the clearance allowed
The fit loosens over time or with temperature	Dissimilar materials: the hub expands more than the shaft when hot	See Temperature, Coatings and Fit Problems; choose materials with similar expansion

Related

• Fits & Tolerances: the overview and quick reference.
• How Fits Work: the three fit types.
• Fits – Metric (ISO System) and Fits – Inch (ANSI B4.1): the fit classes.
• Fits in Practice: bearings, keys and other common fits.
• Temperature, Coatings and Fit Problems: thermal expansion and coatings.

FAQ

Is a shrink fit stronger than a press fit?

For the same dimensions, yes. The Handbook reports tests where a shrink fit resisted axial slip 3.66 times and torsion 3.2 times better than a forced fit, because forcing the shaft in scrapes off some of the interference.

How much interference should I use?

Enough that the loose end of the fit still carries the load, and no more than the hub can take at the tight end. The Handbook’s forced-fit rule of thumb is 0.001 to 0.0025 in per inch of diameter; the fit classes (FN, or H7/s6 and u6) give a controlled range.

Why is the stress in the hub bigger than the contact pressure?

The hub behaves like a pressure vessel. The contact pressure acts on the bore and sets up a hoop stress around the wall that is larger than p, highest at the bore (about 1.67 p for a hub twice the bore diameter).

Does the Handbook’s allowance formula agree with the Lamé equations?

They are the same kind of elastic calculation but use different simplifications, so the numbers differ by around ten percent for a steel hub. Both are estimates; for critical joints use finite-element analysis and test a prototype.

Can I use a keyed fit as well?

Yes, but a key weakens the hub and changes the stress, and it is better to design the interference fit to carry the torque, using the key as a back-up.

Why does the Handbook’s metric formula show a factor of 25.4?

The metric versions of its allowance formulas include a factor of 25.4 (mm per inch). Evaluated literally with stress in MPa and E = 206,843 MPa, T(1 + C)/E is a dimensionless allowance per unit of diameter; check the units before using the metric form, and verify the result with a dimensionless calculation.

References

• Machinery’s Handbook, 31st edition, Industrial Press: “Limits and Fits”: forced fits (allowance, pressure, pressure factors), expansion fits, shrinkage fits, allowances for shrinkage fits (Tables 1 to 3) and temperatures for shrinkage fits.
• Timoshenko, S. P. and Goodier, J. N., Theory of Elasticity; Roark’s Formulas for Stress and Strain (thick-walled cylinders, shrink fits).
• ANSI/ASME B4.1 and B4.2 for the interference fit classes.

Last reviewed: October 2026. The equations are elastic estimates; verify critical joints by analysis and test.