Rotating Disk with a Central Bore

stressenergy-storagemass-cost

Verified build 9 relations · 4 identities proven · 2 modeling steps · 6 parity samples

The solid flywheel ended on a fine print: no central hole. But nearly every real rotor needs one — a shaft has to get through. This page pays that bill, and the price is startling. Drill the smallest bore you can machine through the disk’s centre and the peak stress does not creep up by a few percent. It doubles.

$\sigma_{\theta,max} = \frac{\rho\,\omega^2}{4}\Big[(3+\nu)\,R^2 + (1-\nu)\,a^2\Big] \qquad f_{bore} = 2 + \frac{2\,(1-\nu)\,a^2}{(3+\nu)\,R^2}$

The widget’s bore penalty readout $f_{bore}$ is that stress divided by the solid disk’s centre value at identical geometry, speed, and material. Drag the bore radius $a$ down toward nothing and watch it settle on 2.0 — not 1.0. The limit is a genuine discontinuity: a disk with an infinitesimal hole carries twice the stress of a disk with no hole. The resolution is a classic of elasticity: the solid disk’s centre sits in equal biaxial tension ( $\sigma_r = \sigma_\theta$ there), and punching a circular hole into an equibiaxial field multiplies the stress at its edge by exactly the hole’s concentration factor, $K = 2$ . The “paradox” is just stress concentration wearing rotational clothes — and it is why the field in the derivation below has a $1/r^2$ term the solid disk was forced to discard: with material removed at the centre, nothing forbids it anymore, and it bites hardest exactly at the bore edge.

Three consequences worth turning the knobs for:

The speed limit drops by $1/\sqrt{2}$ . Stress doubles at fixed speed, and stress goes as $\omega^2$ — so the yield-onset speed $\omega_y$ falls to about 70.7% of the solid disk’s. The flywheel page’s specific-energy ceiling was a solid-disk privilege; the bored rotor loses roughly half its energy capacity at the limit. That is the real cost of a shaft hole, and why high-end flywheels use integral shafts or profiled hubs instead.
At a fixed speed, the bore is almost free — even helpful. $e = E_k/m$ rises when you bore the disk (526 vs 506 J/kg at this page’s defaults), because the centre you removed was the part storing almost nothing ( $r^2$ weighting). Energy lives at the rim; stress lives at the bore. The knobs let you feel both truths at once.
The radial stress never competes. It must vanish at both free surfaces, so it bulges to a modest peak at the geometric-mean radius $r = \sqrt{aR}$ and stays below the hoop stress everywhere — same verdict as the thick-walled cylinder, and for the same reason: hoop directions carry the burden in every radial-field THING.

One boundary this page deliberately leaves quiet: the bore here is free — nothing presses on it. Put a shaft into that bore with an interference fit and the contact pressure adds a Lamé field on top of this one. The fit mechanics — microns of interference becoming a contact pressure, residual stresses superposing on a working field — now live in the compound cylinder, and they are secretly a chaining story.

Try it

Configuration

Material

T3, bare flat sheet 0.010-0.128 in. thick, AMS 4037 / AMS-QQ-A-250/4 (MIL-HDBK-5J Table 3.2.3.0(b1), p. 3-71)

Bound properties of 2024-T3 aluminum sheet (bare)
nu	0.33 1	typical	mil-hdbk-5j
rho	0.1 lb/inch**3	typical	mil-hdbk-5j
sigma_y	47 ksi	design min.	mil-hdbk-5j

3 materials in the database are not listed here: no published value in our cited sources for every property this THING needs.

Materials modeled here: 2024-T3 aluminum sheet (bare) 304 stainless steel 6061-T6 aluminum 7075-T6 aluminum AISI 1045 medium-carbon steel AISI 4340 low-alloy steel (Ni-Cr-Mo) ASTM A36 structural steel (hot-rolled) C26000 Cartridge Brass (70/30) Nylon 6/6 (PA66), unfilled Ti-6Al-4V

Governing relations

m = \rho\,\pi\,(R^2 - a^2)\,t

Assumes: uniform annular disk — constant thickness, free bore surface, free rim · Valid while: The bore has met (or passed) the rim — no disk remains. Every readout on this page is refused until a < R again.

Source: Hibbeler, R. C., Engineering Mechanics: Dynamics, 14th ed., Pearson, 2016 — §17.1 (mass moments of inertia; the annulus I = m(R² + a²)/2) and ch. 18 (planar kinetics, work and energy: E = ½Iω² for rotation about a fixed axis).

I_z = \tfrac{1}{2} m\,(R^2 + a^2)

Assumes: rigid uniform annulus spinning about its axis of symmetry

E_k = \tfrac{1}{2} I_z \omega^2

Assumes: rigid body in pure spin about a fixed axis — no wobble, no translation

e = \frac{E_k}{m}

Assumes: energy stored per kilogram carried — note the bore RAISES this at a given speed (the removed centre mass was barely storing anything) while lowering the speed limit

Source: Ashby, M. F., Materials Selection in Mechanical Design, 4th ed., Butterworth-Heinemann, 2011 — the "materials for flywheels" case study: kinetic energy per unit mass governed by the material index σ_y/ρ; the bored rotor keeps the index but loses a factor on the coefficient.

\sigma_{\theta,max} = \frac{\rho\,\omega^2}{4}\left[(3+\nu)\,R^2 + (1-\nu)\,a^2\right]

Assumes: linear elastic, isotropic, uniform-thickness annulus in plane stress (thin next to its radius); both surfaces free — no shaft pressing in the bore, no shrink-fitted hub (that story is the compound cylinder's); the hoop stress peaks at the bore edge, where the radial stress is zero · Valid while: The bore edge has reached yield — beyond this speed the elastic field (and every number built on it) stops being the truth. First yield is NOT burst: a ductile disk redistributes stress and survives well past this line, but this model ends here. Thickness is no longer small next to the radius — the plane-stress thin-disk solution is fading; a disk this chunky wants the axisymmetric thick-disk treatment.

Source: Timoshenko, S. P., & Goodier, J. N., Theory of Elasticity, 3rd ed., McGraw-Hill, 1970 — ch. 4 (two-dimensional problems in polar coordinates), §32 "Rotating Disks": the plane-stress annulus with free bore and free rim, peak hoop stress at the bore edge, and the small-hole limit of twice the solid disk's central stress.

\sigma_{r,max} = \frac{3+\nu}{8}\,\rho\,\omega^2\,(R-a)^2 \quad \text{at } r=\sqrt{aR}

Assumes: the radial stress vanishes at BOTH free surfaces and peaks between them, at the geometric mean radius r = √(aR); always smaller than the bore hoop stress — the hoop direction governs everywhere

f_{bore} = \frac{\sigma_{\theta,max}}{\sigma_{solid}} = 2 + \frac{2\,(1-\nu)\,a^2}{(3+\nu)\,R^2}

Assumes: ratio to the SOLID disk's centre stress (3+ν)ρω²R²/8 at the same geometry, speed, material; the vanishing-bore limit is exactly 2 — the equibiaxial hole concentration factor, not a gradual transition

\mathrm{SF} = \frac{\sigma_y}{\sigma_{\theta,max}}

Assumes: margin against first yield at the bore edge, not against burst — burst arrives later, governed by ultimate strength and crack behaviour

Source: Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — ch. 16 (flywheels: energy exchange and sizing) and ch. 1 (design factor against first yield).

\sigma_y = \frac{\rho\,\omega_y^2}{4}\left[(3+\nu)\,R^2 + (1-\nu)\,a^2\right]

Assumes: the spin speed at which the bore edge first reaches yield — 1/√2 of the solid disk's yield-onset speed in the small-bore limit

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

\sigma_{r} = \frac{\omega^{2} \rho \left(\nu + 3\right) \left(- \frac{R^{2} a^{2}}{r^{2}} + R^{2} + a^{2} - r^{2}\right)}{8}

1. The modeling step: same equilibrium-plus-compatibility ODE as the solid flywheel, but the bore changes which solutions are admissible. The solid disk had to DISCARD the 1/r² term to stay finite at the centre; with material removed there, that term returns, and the two integration constants are now fixed by TWO free surfaces — σ_r(a) = 0 and σ_r(R) = 0. This is the classical rotating-annulus field. (The test suite re-solves the boundary-value problem independently.) — elasticity: radial equilibrium + compatibility, free bore and free rim modeling step

\sigma_{t} = \frac{\omega^{2} \rho \left(- r^{2} \left(3 \nu + 1\right) + \left(\nu + 3\right) \left(\frac{R^{2} a^{2}}{r^{2}} + R^{2} + a^{2}\right)\right)}{8}

2. The same solution's hoop component. The 1/r² term that the solid disk forbade is exactly the term that bites: it ADDS to the hoop stress as r shrinks, so the maximum now lives at the bore edge, not the centre — there is no centre anymore. — elasticity: hoop component of the same field modeling step

\frac{\omega^{2} \rho \left(- a^{2} \left(3 \nu + 1\right) + \left(2 R^{2} + a^{2}\right) \left(\nu + 3\right)\right)}{8} = \sigma_{t max}

3. Evaluate the hoop field at the bore edge, r = a (the a²R²/r² term becomes R², so the bracket reads (3+ν)(2R² + a²) − (1+3ν)a², which tidies to 2(3+ν)R² + 2(1−ν)a²): the page's governing stress, σ_θ(a) = ρω²[(3+ν)R² + (1−ν)a²]/4. The radial stress is zero there (free surface), so the bore edge sits in pure uniaxial hoop tension — the simplest possible yield check, at the worst possible place. — evaluate the hoop field at r = a

\frac{\omega^{2} \rho \left(\nu + 3\right) \left(R^{2} - 2 R a + a^{2}\right)}{8} = \sigma_{r max}

4. The radial stress must vanish at both free surfaces and bulge between them; setting its derivative to zero lands the peak exactly at the geometric mean radius r = √(aR), where the a²R²/r² and r² terms both equal aR and the bracket collapses to (R−a)². It never catches the bore hoop stress — the hoop direction governs this THING everywhere, just as it governed the thick-walled cylinder. (The calculus locating √(aR) is re-run in the test suite.) — maximize the radial field — peak at the geometric mean radius

\frac{R^{2} f_{bore} \omega^{2} \rho \left(\nu + 3\right)}{8} = \sigma_{t max}

5. Divide the bore-edge stress by the SOLID disk's centre value (3+ν)ρω²R²/8. The ratio is f = 2 + 2(1−ν)a²/((3+ν)R²) — and as the bore shrinks to nothing, f goes to exactly 2, not to 1. An arbitrarily small hole doubles the stress: this is the equibiaxial stress-concentration factor of a circular hole, K = 2, hiding inside the rotating disk (the solid disk's centre sits in equal biaxial tension, and the vanishing bore is a hole punched in exactly that field). The discontinuity is real and the test suite pins it. — the vanishing-hole limit: f → 2 as a → 0

R^{2} \omega_{y}^{2} \left(\nu + 3\right) + a^{2} \omega_{y}^{2} \left(1 - \nu\right) = \frac{4 \sigma_{y}}{\rho}

6. Run the bore stress up to yield and read it as a speed: in the small-bore limit ω_y ≈ √(4σ_y/((3+ν)ρR²)) — exactly 1/√2 ≈ 70.7% of the solid disk's yield-onset speed. Halving the limit speed quarters... no: the energy at the limit falls by half. The flywheel page's specific-energy ceiling e = 2σ_y/((3+ν)ρ) was a SOLID-disk privilege; the bored rotor pays for its shaft hole in stored energy. — yield-onset speed — the 1/√2 penalty

How it fails

The widget’s margin is first yield at the bore edge of a free-free annulus under steady spin. Real bored rotors inherit every failure mode of the solid flywheel — burst sectors leaving at rim speed, brittle materials skipping the warning yield entirely, overspeed scaling as $\omega^2$ — plus the ones the bore brings itself:

The bore edge is where cracks are born. It carries the peak stress, it is machined (tool marks, keyways, snap-ring grooves all concentrate further), and in a charge–discharge flywheel or a start–stop turbine it is stress-cycled at full amplitude every cycle. Bore-edge fatigue initiation is the canonical life-limiter of bored rotors — which is why critical turbine and compressor disks get their bores honed, cold-worked, and inspected with particular paranoia.
A keyway in the bore stacks concentrations. The factor-of-2 here assumes a smooth circular hole. Cut a square keyway into it and the corner adds its own concentration on top of the already-doubled field — geometric factors multiply. Real high-speed designs avoid keyed bores entirely (friction drives, curvic couplings, integral shafts) for exactly this reason.
First yield at the bore is even less like burst than it was for the solid disk. The overstressed region is a thin annulus at the bore; yielding there redistributes load outward into a large mass of cooler material, and ductile rotors run far past first bore yield before average-stress burst criteria matter. The margin here is honest about what it measures: the end of the elastic model, not the end of the disk.
A pressed-in shaft changes the sign of the problem. This page’s bore is free. An interference-fitted shaft or hub presses outward on the bore — a Lamé pressure field superposes on the rotation field, raising bore hoop stress at standstill and, paradoxically, relieving contact as speed grows (the disk grows away from the shaft). The fit that is tight enough at speed and not yielded at standstill is a genuine two-sided design problem — the compound cylinder carries the fit mechanics (and its failure note carries the release speed), where this page and the thick-walled cylinder meet.
The thin-disk assumption fails for chunky hubs. Plane stress wants $t \ll R$ . Thick bored disks (gear blanks, turbine wheels with deep hubs) develop axial stress variation and need the axisymmetric treatment; the widget’s thickness warning marks where this model starts fading, not where the part fails.

Compound Cylinder (Shrink Fit)
Where the monobloc wall gave up: shrink a jacket over a liner and the interference squeezes the bore into hoop compression before the pressure ever arrives. Service tension must spend that compression first — and at the balanced fit with the interface at √(r_i·r_o), the elastic pressure ceiling approaches DOUBLE the one no solid wall could pass.
- stress
- mass-cost
Thick-Walled Cylinder (Lamé)
Where the thin-wall pressure vessel hands off: the exact elastic field for any wall thickness. The stress piles up at the bore and decays as 1/r² — and the bore shear always exceeds the pressure, so past p = σ_y/2 no amount of thickness can contain it elastically.
- stress
- mass-cost
Thin-Walled Pressure Vessel (Cylinder)
A pressurized tube with closed ends — scuba tank, boiler, rocket stage. The hoop stress is exactly twice the longitudinal stress, which is why sausages split lengthwise; and because the relations are undirected, the same widget runs backwards as a design tool: pick a safety factor and the wall thickness falls out.
- stress
- mass-cost
Cantilever Beam (End Load)
A beam fixed at one end, loaded at the other — the fruit-fly of structures. One widget shows why stiffness (E) and strength (σ_y) are independent axes: swap steel for titanium and deflection goes UP while the safety factor also goes up.
- stress
- mass-cost
Composite Bar (Core + Sleeve)
A solid core inside a concentric sleeve, bonded between rigid end plates and pushed by a centric axial load. The two materials must stretch together, so the load splits in proportion to each member's axial stiffness A·E — and the build solves that coupled 2×2 share exactly. Swap the sleeve's metal and watch the load migrate to the stiffer member.
- stress
- mass-cost
Eccentric Column (Secant Formula)
Load a column even slightly off-axis and the clean buckling story dissolves: it bows from the first newton, stress grows faster than load, and the Euler limit survives only as the asymptote the deflection chases. Because nothing here is linear, the safety factor must be taken on the LOAD — the page solves that transcendental equation live, by bracketed root-finding.
- stability
- stress
- mass-cost

Chains with

Outputs whose SI dimension and quantity kind match another THING's input — the only wires the planner's connectionLegal accepts (invariant 2, computed at build time, not hand-listed). Wire these on the chaining demo.

Impact Loading (Falling Mass, Energy Method)
- m_disk m
Circular Plate under Uniform Pressure (Clamped vs Simply Supported)
- sigma_r_max q
- sigma_r_max sigma_allow
- sigma_t_max q
- sigma_t_max sigma_allow
Simply Supported Beam (Center Load + UDL)
- SF SF
Shaft in Torsion (Solid, Circular)
- omega omega
- omega_y omega
Shaft under Combined Bending + Torsion
- SF SF_t
Thin-Walled Tube in Torsion (Bredt)
- SF SF
Euler Column (Buckling)
- f_bore K
Compound Cylinder (Shrink Fit)
- sigma_r_max p
- sigma_t_max p

+ 12 more THINGs its outputs can legally feed (showing the first 8 in course order).

Sources

Timoshenko, S. P., & Goodier, J. N., Theory of Elasticity, 3rd ed., McGraw-Hill, 1970 — ch. 4 (two-dimensional problems in polar coordinates), §32 "Rotating Disks": the plane-stress annulus with free bore and free rim, peak hoop stress at the bore edge, and the small-hole limit of twice the solid disk's central stress.
Hibbeler, R. C., Engineering Mechanics: Dynamics, 14th ed., Pearson, 2016 — §17.1 (mass moments of inertia; the annulus I = m(R² + a²)/2) and ch. 18 (planar kinetics, work and energy: E = ½Iω² for rotation about a fixed axis).
Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — ch. 16 (flywheels: energy exchange and sizing) and ch. 1 (design factor against first yield).
Ashby, M. F., Materials Selection in Mechanical Design, 4th ed., Butterworth-Heinemann, 2011 — the "materials for flywheels" case study: kinetic energy per unit mass governed by the material index σ_y/ρ; the bored rotor keeps the index but loses a factor on the coefficient.