Interference Fit and Shrink Fit Calculator

Interference Fit and Shrink Fit Calculator

A hollow hub on a solid or hollow shaft in two different materials, by the Lamé thick-wall equations: contact pressure, torque capacity, push-out force, hub hoop stress and the assembly temperature — computed at BOTH ends of the ISO 286 interference range, with the roughness the surfaces lose on assembly subtracted the way DIN 7190 does it.

Interference and shrink fits

Fit class and materials → pressure, torque, temperature
The nominal size of the fit. The ISO 286 bands this page carries run from 1 mm to 180 mm and were each verified against four identities from the standard’s own rules.
A thicker hub is stiffer, so the same interference gives more pressure — and the hoop stress in the hub falls. Both effects favour a thick hub.
A hollow shaft collapses more easily, so the same interference gives LESS pressure, and the compressive hoop stress at its bore is much higher than the contact pressure.
The interference is a RANGE, not a number. Every fit here has a minimum the design must work at and a maximum it must survive.
Used only with the last fit option. Diametral, not radial.
Torque and push-out force are both proportional to this. Contact pressure is not.
DIN 7190’s allowance is in Rz, the peak-to-valley height, NOT in Ra. Rz runs roughly four to seven times Ra for a turned or ground surface — using an Ra figure here throws most of the correction away.
Both moduli and both Poisson ratios enter the Lamé solution, and both coefficients of thermal expansion enter the assembly temperature.
DIN 7190’s published slip coefficients: about 0.08 to 0.10 for steel on steel pressed on, 0.12 to 0.14 shrunk on dry. Pressing on destroys asperities; shrinking on does not, which is why a shrunk joint grips better at the same interference.
Not a circuit: a number line in units of the MAXIMUM interference, so the right-hand end is always full scale and the figure is about proportions. The top band is what the two tolerance zones give — its left end is the minimum interference and its right end is the maximum, and the gap between them is the whole difficulty with an interference fit. The band below it is the same interference after DIN 7190's roughness allowance has been taken off both ends, which shifts it bodily to the left. The narrow strip under that is the bite itself: 0.8 times the sum of the two peak-to-valley heights, drawn on the same scale so you can compare it against the minimum interference directly — on a small fit with an ordinary finish it is most of it. The two pressure bars are drawn to one scale with the maximum at full length, so their ratio is visible; the bar at the bottom is the hub's equivalent stress at the maximum interference as a percentage of its yield strength, with the line at 100. Everything is quantised to fiftieths.
11.5N/mm²Example

A steel hub, 90 mm outside diameter and 60 mm long, on a solid 50 mm alloy steel shaft, H7/s6, both surfaces at Rz 6.3

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Lamé for two materials, and a range rather than a number

δ/d_F = p·[K_o/E_o + K_i/E_i]  ·  K_o = (1+Q_o²)/(1−Q_o²) + ν_o, K_i = (1+Q_i²)/(1−Q_i²) − ν_i  ·  T = p·π·d_F²·l_F·μ / 2  ·  F = p·π·d_F·l_F·μ  ·  δ_eff = δ − 0.8(R_z1 + R_z2)  ·  ΔT = (δ_max + IT7) / (α·d_F)
δ
DIAMETRAL interference. Half of it is the radial interference, and mixing the two up is the commonest arithmetic error in press-fit calculations — it gives you twice the pressure you expected
Q_o, Q_i
d_F/d_H for the hub and d_is/d_F for the shaft. A solid shaft has Q_i = 0 and K_i reduces to 1 − ν
E, ν
modulus and Poisson ratio of each part separately. With the same material both sides the two ν terms cancel and this collapses to the single-material form the dowel page uses
R_z
peak-to-valley roughness height, NOT R_a. DIN 7190 subtracts 0.8 of the sum of the two, because the asperities flatten on assembly and that interference is simply gone
μ
slip coefficient. DIN 7190 gives a different one for the axial and the circumferential direction, and the circumferential one is lower
α
coefficient of thermal expansion. The hub’s for a shrink fit, the shaft’s for a cold fit, and the DIFFERENCE between them for what happens in service

Worked example

A steel hub, 90 mm outside diameter and 60 mm long, on a solid 50 mm alloy steel shaft, H7/s6, both surfaces at Rz 6.3
The interference is a RANGE and it comes from the two tolerance bands. At 50 mm, H7 is 0 to +25 µm and s6 is +43 to +59 µm, so the interference runs from 43 − 25 = 18 µm to 59 − 0 = 59 µm. More than three to one
Now take the roughness off. DIN 7190 subtracts 0.8(R_z1 + R_z2) = 0.8 × 12.6 = 10.1 µm — which is 56% of the MINIMUM interference. Effective range: 7.9 to 48.9 µm. Note carefully that the allowance is in R_z and not R_a; if you put an R_a figure of 1.6 in there instead you would subtract 2.6 µm and keep interference you do not have
Lamé, at the minimum. Q_o = 50/90 = 0.5556, so K_o = (1+Q_o²)/(1−Q_o²) + ν = 2.1929 and K_i = 1 − ν = 0.7 for a solid shaft. p = δ/d_F ÷ (K_o/E_o + K_i/E_i) = 11.5 N/mm²
That pressure is what the design has to work with: torque capacity = p·π·d_F²·l_F·μ/2 = 379 N·m, and push-out force p·π·d_F·l_F·μ = 15.17 kN, both at μ = 0.14, DIN 7190's figure for a shrunk steel joint. A pressed-on joint at the same interference gets about 57% of that, because pressing shears the asperities off
At the MAXIMUM interference the pressure is 71.0 N/mm², which is what the hub has to survive. Hoop stress at the hub bore is p(1+Q_o²)/(1−Q_o²) = 134 N/mm² in tension, and the equivalent stress that decides yielding is 2p/(1−Q_o²) = 205 N/mm² — 59% of an ordinary steel hub's yield. It survives, but a thinner hub would not: at a 70 mm outside diameter the same fit would be past yield
To get it on: the hub bore has to open by the maximum interference PLUS enough clearance to pass the shaft, 73.9 µm, so ΔT = δ/(α·d_F) = 126 K above ambient — about 146 °C, which is comfortable for a plain steel hub and too hot for a hardened one. Cooling the shaft needs the same temperature CHANGE the other way: from 20 °C, dry ice gives you about 98 K and liquid nitrogen about 216, so for this fit cooling alone would just do it
Two lessons. The design works at the minimum and survives the maximum, and those are different calculations on the same joint. And a good surface finish is worth real interference here: going from Rz 6.3 to Rz 1.6 on both surfaces recovers 7.5 µm, which is 95% more effective minimum interference than this joint has — it nearly doubles the torque the fit is guaranteed to carry, for the price of a finer finishing pass
Finally, change the hub to aluminium and watch two things happen at once. The pressure at the minimum interference falls to 4.46 N/mm², 39% of the steel hub's, because the softer hub takes up most of the interference itself. And the joint now LOSES 60 µm of interference for every 100 K it warms up, because aluminium expands twice as fast as steel — which is more than the whole effective interference range of an H7/s6. An aluminium hub shrunk onto a steel shaft and then run hot comes off

The interference RANGE for each fit class, in micrometres, computed from ISO 286 rather than copied

Nominal size (mm)IT7IT6H7/p6H7/r6H7/s6H7/t6H7/u6
over 1 to 3106-4 to 120 to 164 to 20—8 to 24
over 3 to 61280 to 203 to 237 to 27—11 to 31
over 6 to 101590 to 244 to 288 to 32—13 to 37
over 10 to 1418110 to 295 to 3410 to 39—15 to 44
over 14 to 1818110 to 295 to 3410 to 39—15 to 44
over 18 to 2421131 to 357 to 4114 to 48—20 to 54
over 24 to 3021131 to 357 to 4114 to 4820 to 5427 to 61
over 30 to 4025161 to 429 to 5018 to 5923 to 6435 to 76
over 40 to 5025161 to 429 to 5018 to 5929 to 7045 to 86
over 50 to 6530192 to 5111 to 6023 to 7236 to 8557 to 106
over 65 to 8030192 to 5113 to 6229 to 7845 to 9472 to 121
over 80 to 10035222 to 5916 to 7336 to 9356 to 11389 to 146
over 100 to 12035222 to 5919 to 7644 to 10169 to 126109 to 166
over 120 to 14040253 to 6823 to 8852 to 11782 to 147130 to 195
over 140 to 16040253 to 6825 to 9060 to 12594 to 159150 to 215
over 160 to 18040253 to 6828 to 9368 to 133106 to 171170 to 235
Every number here is computed from the ISO 286 grades and the standard’s own fundamental-deviation rules, and then checked four ways: p is IT7 plus nought to five micrometres at every band; s is IT8 plus one to four up to 50 mm and IT7 + 0.4·D above it, which reproduces all seven bands above 50 mm to a quarter of a micrometre; r is the geometric mean of p and s at all fifteen bands above 3 mm; and u is IT7 + D at all eleven bands from 18 mm up. Look at the H7/p6 column and notice that its minimum is essentially zero everywhere — exactly zero at 12 mm and one micrometre at 50. That is not an accident: p’s fundamental deviation IS IT7 to within five micrometres, and H7’s upper deviation is IT7, so they cancel by construction. H7/p6 is a locating fit, not a press fit, and a design that needs torque from it has no margin at all. Then look at the ratio of maximum to minimum in the s6 column: at 50 mm it is 59 over 18, more than three to one. Everything else on this page follows from that ratio. These dimensions come from a published standard’s table, not from a formula. The standard itself is cited below and the printed values are attributed to the catalogue they were taken from; a different publisher may round differently in the last digit.

The same 30 µm interference on a 50 mm steel shaft, with the hub in each material

Hub materialE (GPa)να (10⁻⁶/K)Yield (N/mm²)Contact pressure (N/mm²)Hub hoop stress (N/mm²)Equivalent stress as % of yieldTemperature rise to open 30 µm (K)
Steel, general (E 210 GPa)2100.3011.735043.6823651
Alloy steel 42CrMo4 / 4140 (E 210 GPa)2100.3011.765043.6821951
Grey cast iron EN-GJL-250 (E 110 GPa)1100.2610.525026.2503057
Aluminium 6061-T6 (E 69 GPa)690.3323.627616.9321825
Brass CuZn39Pb3 (E 97 GPa)970.3420.025022.8432630
Aluminium bronze CuAl10Ni5Fe4 (E 120 GPa)1200.3216.030027.6522738
Stainless 304 / X5CrNi18-10 (E 193 GPa)1930.2917.321541.0785535
This is the table the dowel-pin page on this site deliberately does not have, because a dowel pin and its hole are the same material and the two Poisson terms cancel. Here they do not, and the consequences are larger than people expect. An aluminium hub on a steel shaft takes the same interference at about two fifths of the contact pressure, because the hub does most of the yielding of the interference — so the joint is much weaker in torque. It also needs less than half the temperature rise to assemble, because aluminium expands twice as fast. And it is the combination that fails in service: an aluminium hub on a steel shaft LOSES interference as the assembly warms up, at about 60 µm per 100 K on a 50 mm fit, which is MORE than the whole interference range of an H7/s6. Every modulus, Poisson ratio and expansion coefficient here is a published room-temperature figure for a representative alloy; use your own material’s data sheet for anything that matters. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.

Friction at the interface, and why it is two numbers

ConditionLongitudinal slip coefficientCircumferential slip coefficientPush-out force at 30 µm on a 50 × 60 mm joint (kN)Torque capacity (N·m)
Steel on steel, dry, pressed on longitudinally0.100.0841.1821
Steel on steel, lightly oiled, pressed on0.080.0732.8718
Steel on steel, shrunk on (heated hub), dry0.140.1257.51,232
Steel on cast iron, dry0.120.1049.31,026
Steel on aluminium alloy, dry0.100.0741.1718
DIN 7190 gives two coefficients for the same interface because sliding along the axis and slipping around it do not happen on the same asperities, and the circumferential figure is the lower one. Two things follow. First, a joint that will not push out may still slip in torsion, so the torque capacity on this page uses the circumferential figure where you enter one and you should too. Second, SHRINKING a joint on grips better than PRESSING it on at the same interference — roughly 0.14 against 0.10 — and the reason is mechanical rather than chemical: pressing on shears the asperities off and they do not grow back, while a shrunk joint’s surfaces come together without relative sliding and keep them. That is the same physics as the roughness allowance, seen from the other side. The same designation can mean different dimensions in different standards families — ANSI against ISO, inch against metric, one national standard against another. The family used here is named beside every figure; check which one your part was made to.

What this page is, and what it is not

QuestionWhere it belongs
A solid pin in a hole, same material both sidesThe dowel pin and hole calculator on this site. It derives the single-material Lamé pressure, and this page’s two-material form reduces to exactly its answer — to twelve significant figures — when both moduli and both Poisson ratios are set equal and the shaft is solid. That agreement is the check that the extra algebra here did not break the simple case
Converting Ra to Rz, or an N grade to eitherThe converters plugin owns surface roughness, and it owns it for a good reason: Ra and Rz are different measurements of different things and the ratio between them depends on the process. Four to seven times is the usual band for turning and grinding, but it is not a conversion factor and this page will not pretend it is. Enter Rz, because that is what DIN 7190’s allowance is in
A keyed hubThe parallel key and keyway dimensions page here, and the key capacity page already on this site. Note that a keyway and an interference fit on the same hub interact: the keyway is a slot in the pressurised bore and the hoop stress runs round it
A shrink disc, a keyless bush, or a tapered adapter sleeveNot here. Those are proprietary geometries with published rating tables, and the maker’s numbers include effects — the taper’s own mechanics, the clamping screws’ preload scatter — that a plain cylindrical Lamé calculation does not contain
Fatigue of the shaft at the end of the fitNot here, and it matters: the step in pressure at the end of an interference fit is a stress raiser, and a rotating bending load cracks shafts exactly there. The stress-concentration page here covers grooves and fillets but not the end of a press fit
Plastic (elasto-plastic) interference fitsNot here. DIN 7190 covers them and they are used deliberately, but once the hub yields the pressure stops following the interference and this page’s elastic solution is simply wrong. The page tells you when the hub yields and then stops
The first row is the important one for anyone who has already used the dowel page: that page does a solid pin in a hole in one material and stops there, on purpose. This one does a hollow hub on a solid or hollow shaft in two different materials, which is the case that arises whenever a gear, sheave, coupling hub, bearing inner ring or impeller goes onto a shaft. They are not the same calculation and neither duplicates the other.

A range not a number, roughness in Rz not Ra, and two materials rather than one

The interference is a range, and the design has to work at one end and survive the other. An H7/s6 fit on a 50 mm shaft is not “38 µm of interference”. It is anything from 18 to 59 µm, because the hole can be anywhere in its 25 µm band and the shaft anywhere in its 16 µm band. The torque the joint carries is set by the MINIMUM, and the stress the hub sees is set by the MAXIMUM, and those two numbers differ by more than three to one. A calculation done at the mean interference is wrong twice over: it promises torque that a third of your production will not deliver and it fails to notice a hub that yields on assembly. This page computes the ISO 286 bands rather than copying them — from the standard’s own defining rules, verified against four internal identities — and reports both ends throughout.

Then take the roughness off, and take it off in R_z. The asperities on the two surfaces flatten as the parts go together, and that interference is simply gone. DIN 7190 subtracts 0.8(R_z1 + R_z2), and the figure it subtracts is the PEAK-TO-VALLEY height, not the arithmetic average. That distinction is not pedantry: R_z runs roughly four to seven times R_a for a turned or ground surface, so putting an R_a figure into that formula throws away most of the correction. On the worked example here the allowance is 10.1 µm against a minimum interference of 18 — more than half of it — and improving both surfaces from Rz 6.3 to Rz 1.6 recovers 7.5 µm, nearly doubling the effective minimum. A better finish is often the cheapest interference you can buy. Converting between R_a, R_z and N grades belongs to the converters plugin on this site and is deliberately not attempted here, because the ratio depends on the process and is not a conversion.

Two materials, which is the case the dowel-pin page does not do. Lamé’s thick-wall solution gives the radial displacement of each cylinder separately, and the interference closes when the hub’s outward movement plus the shaft’s inward movement add up to half the diametral interference. With the same material both sides the two Poisson terms cancel and the algebra collapses to the single-material form the dowel pin and hole calculator already derives — and this page’s two-material form reproduces that answer to twelve significant figures, which is how we know the extra algebra did not break anything. With different materials the results move a long way: an aluminium hub on a steel shaft develops about two fifths of the contact pressure at the same interference, needs less than half the temperature rise to assemble, and loses more interference than an H7/s6 has when the assembly warms 100 K. The whole solution here is asserted against a finite-difference solve of the axisymmetric elasticity equations at five geometries, which knows none of the algebra.

The assembly temperature, and the reason cooling the shaft is sometimes the answer. To drop a hub on, the bore has to open by the maximum interference plus enough clearance to pass the shaft at its largest — so the temperature rise is set by the maximum and not by the nominal. That is often 80 to 150 K for a steel hub, which is comfortable; above about 250 °C it stops being a heating problem and becomes a metallurgical one, because a quenched and tempered steel loses hardness, a hardened bearing ring is damaged (which is why bearing makers cap induction heating at 120 °C) and any seal, cage or coating on the part has its own limit. Cooling the shaft needs the same temperature CHANGE in the other direction, and dry ice reaches −78 °C where liquid nitrogen reaches −196, so for a modest fit it is a real alternative and for a large one it is not enough on its own.

What the fit delivers, and what it must survive. Torque capacity is p·π·d²·l·μ/2 and push-out force is p·π·d·l·μ, both linear in the pressure and therefore both computed at the minimum. DIN 7190 gives two friction coefficients for the same interface, a longitudinal one and a lower circumferential one, so a joint that will not push out can still slip in torsion — and the same source puts a shrunk joint’s coefficient above a pressed one’s, for the mechanical reason that pressing on shears the asperities off. At the other end of the band, the hoop stress at the hub bore is p(1+Q²)/(1−Q²) in tension while the equivalent stress that actually yields is 2p/(1−Q²), because the bore is in hoop tension and radial compression at once. The page reports both and says when the hub yields. The alternative connections are on the spline torque capacity calculator and the key and keyway dimensions page.

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Frequently asked questions

How much interference does an H7/s6 fit give?

A range, not a number. On a 50 mm joint, 18 to 59 µm of diametral interference — because the H7 hole can be anywhere in 25 µm and the s6 shaft anywhere in 16 µm. Subtract DIN 7190’s roughness allowance of 0.8(Rz₁ + Rz₂) and at Rz 6.3 on both surfaces the effective range becomes 7.9 to 48.9 µm. The torque the joint delivers comes from the 8 and the stress the hub survives comes from the 49.

Why subtract surface roughness from the interference?

Because the asperities flatten as the parts go together and that interference no longer exists. DIN 7190 subtracts 0.8 times the sum of the two surfaces’ peak-to-valley heights. It matters more than people expect on small joints: on the 50 mm example here it is more than half the minimum interference. And it is in Rz, not Ra — Rz is roughly four to seven times Ra for turned or ground surfaces, so using an Ra figure understates the loss badly.

What temperature do I heat a hub to for a shrink fit?

Enough to open the bore by the MAXIMUM interference plus a working clearance: ΔT = δ/(α·d). For a 50 mm steel hub at the top of an H7/s6 band that is about 84 K above ambient, so a little over 100 °C. Use the maximum and not the mean, or a proportion of your assemblies will seize halfway on. And check the material limit before the heating method: above about 250 °C a quenched and tempered steel starts losing hardness, and bearing makers cap induction heating of their rings at 120 °C.

Can I cool the shaft instead of heating the hub?

Yes, and it is the right answer whenever the hub carries something that cannot be heated — a seal, a plastic cage, a coating, a hardened surface. The temperature CHANGE needed is the same magnitude, but it is limited by how cold you can get: dry ice reaches about −78 °C and liquid nitrogen about −196, so from a 20 °C ambient you have about 100 K or about 215 K to work with. For a large interference on a large diameter that is not enough on its own and the practical answer is to heat the hub and cool the shaft together, or to press it in.

Does a thicker hub grip better?

Yes, in two ways at once, and it is the cheapest improvement available. A thicker hub is stiffer, so the same interference develops more contact pressure and therefore more torque capacity; and the hoop stress at its bore falls, so it is further from yielding. Both effects saturate: past a hub outside diameter of about twice the shaft diameter, extra wall buys very little. The series on this page plots exactly that curve so you can see where it flattens.

Why is my aluminium hub loose when the machine warms up?

Because aluminium expands about twice as fast as steel, so an aluminium hub on a steel shaft loses interference as the assembly heats. On a 50 mm fit the loss is around 12 µm per 100 K, which is most of the minimum interference of an H7/s6. The calculator reports it explicitly. The fixes are a tighter fit class chosen at the service temperature, a hub material closer to the shaft’s expansion, or a positive drive — a key, a pin, or a spline — rather than relying on friction.

How is this different from the dowel pin calculator on this site?

That page does a solid pin in a hole in one material, which is the case where both Poisson terms cancel and the algebra is short. This one does a hollow hub, on a solid or hollow shaft, in two different materials, which is what a gear, sheave, coupling hub or impeller on a shaft actually is. Setting both materials the same and the shaft solid here reproduces the dowel page’s answer to twelve significant figures — we check that — but the two-material case is what the dowel page deliberately does not attempt.

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References

  1. DIN 7190-1:2017, Interference fits — Part 1: Calculation and design rules for cylindrical self-locking pressfits. Cited by clause; not fetched. The two things this page takes from it, via the eAssistant handbook below, are the roughness allowance and the assembly-temperature relation. The elastic theory itself is Lamé’s and is derived here.
  2. GWJ Technology. eAssistant Handbook, chapter “Interference Fit According to DIN 7190”. The source for the roughness allowance — the effective interference is the nominal interference less s = 0.8(R_zA + R_zI) — and for the assembly-temperature and press-force relations. Note carefully that the standard’s allowance is in R_z, the peak-to-valley height, and NOT in R_a: R_z runs about four to seven times R_a for a turned or ground surface, so using R_a in that formula understates the loss by most of it. It also carries the friction coefficients and the external-load relations p_r = F_r/(d_F·l_F) and p_b = (9/2)·M_b/((2 − Q_W)·D_F·l_F²).
  3. ISO 286-1:2010, Geometrical product specifications (GPS) — ISO code system for tolerances on linear sizes — Part 1: Basis of tolerances, deviations and fits. Cited by clause; not fetched. Its defining relations are what this page computes from: the standard tolerance factor i = 0.45 D^(1/3) + 0.001 D at the geometric mean of each size band, the grade multipliers IT6…IT11 = 10i, 16i, 25i, 40i, 64i, 100i, and the fundamental-deviation rules for the interference letters. The table values were checked against those relations rather than trusted, and the two relations disagree below 6 mm because the standard rounds small sizes to convenient numbers.
  4. Machining Doctor. Engineering Fits & Tolerances — Calculator & Charts, fundamental-deviation-of-shafts chart. The source for the p, r, s, t and u lower deviations over the sixteen bands this batch uses, with the split sub-bands above 10 mm printed separately. Four independent identities from ISO 286-1 check it: p = IT7 + (0 to 5) at every band above 3 mm; s = IT8 + (1 to 4) up to 50 mm and s = IT7 + 0.4 D above it, which reproduces all seven bands above 50 mm to a quarter of a micrometre; r = √(p·s) at all fifteen bands above 3 mm; and u = IT7 + D at all eleven bands from 18 mm up.
  5. Dalloway Precision. ISO 286 Tolerance Grade Chart — IT5 to IT11 in Microns. The source for the IT table. Its column HEADINGS come through in the order IT5 IT6 IT8 IT7 while its VALUES are in the order IT5 IT6 IT7 IT8; the R5 ratio settled which reading is right, because each grade step must be about 10^(1/5) = 1.585 and 10 → 1.67 satisfies that where 14 → 2.33 does not. Every one of the 78 values was then checked against the tolerance-factor formula.
  6. ISO 286-2:2010, Part 2: Tables of standard tolerance classes and limit deviations for holes and shafts. Its freely published sample pages carry Tables 1 to 3 in full, which is where the D-hole fundamental deviation on this page comes from: EI = 20, 30, 40, 50, 65 and 80 µm for the first six bands, reproduced to better than 1.5 µm by 16 D^0.44. The fetch returned the D11 upper-deviation column beside the correct EI column; ES − EI = IT caught it at every row, and only EI, which is grade-independent, was taken.
  7. Schaeffler (INA/FAG). Technical tables: dimension and tolerance symbols, shaft and housing fits. Fetched as the intended cross-check on the p6, r6 and s6 deviations and DISCARDED: the extract came back with three different row offsets in three columns, so its “30–50 mm r6” row carried the 18–30 mm values while its p6 column was offset the other way. ES − EI = IT6 caught it. Nothing on this page is attributed to it.
  8. This plugin’s own dowel-pin-and-hole calculator. The single-material Lamé pressure it derives is the degenerate case of the two-material form used here, and the two agree to twelve significant figures when both moduli and both Poisson ratios are set equal and the shaft is solid. That is the check that this page’s extra algebra did not break the case the site already had right.