Rotating Discs and Cylinders – Radial and Hoop Stresses

Rotating discs, rings and cylinders develop internal stresses because each element of material requires centripetal acceleration. The resulting radial and circumferential (hoop) stresses depend on rotational speed, density, Poisson's ratio and component geometry.

This page develops the classical elastic solutions for thin rotating discs and for long rotating cylinders. The equations are applicable to idealised axisymmetric components such as flywheels, rotors and rotating rings where the assumptions stated below are appropriate.

Contents

• Symbols and Units
• Assumptions
• Thin Rotating Disc – General Equations
• Solid Rotating Disc
• Rotating Disc with a Central Hole
• Long Rotating Cylinder – General Equations
• Solid Rotating Cylinder
• Hollow Rotating Cylinder
• Engineering Design Considerations

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Symbols and Units

Tensile stresses are taken as positive.

Symbol	Description	Units
σr	Radial stress	Pa (N/m2)
σθ	Circumferential or hoop stress	Pa (N/m2)
σz	Axial stress	Pa (N/m2)
εr	Radial strain	–
εθ	Circumferential strain	–
εz	Axial strain	–
E	Young's modulus	Pa (N/m2)
ν	Poisson's ratio	–
ρ	Material density	kg/m3
ω	Angular velocity	rad/s
r	Radius at point under consideration	m
R1	Inner radius	m
R2	Outer radius	m
u	Radial displacement	m

If rotational speed is specified as N revolutions per minute:

$$\omega =\frac{2\pi N}{60}$$

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Assumptions

The classical solutions on this page assume:

• axisymmetric geometry and loading;
• constant angular velocity;
• homogeneous, isotropic, linearly elastic material;
• constant density;
• small strains and displacements;
• constant disc or cylinder thickness within the region being analysed;
• no significant thermal stress;
• no internal or external surface pressure unless separately included;
• no local stress concentrations from holes, keyways, blades, spokes or other discontinuities.

The thin-disc and long-cylinder solutions use different axial constraints and must therefore not be interchanged without checking the physical assumptions.

For an isotropic linearly elastic material:

$${\epsilon }_r=\frac{{\sigma }_r-\nu {\sigma }_{\theta }-\nu {\sigma }_z}{E}$$ $${\epsilon }_{\theta }=\frac{{\sigma }_{\theta }-\nu {\sigma }_r-\nu {\sigma }_z}{E}$$ $${\epsilon }_z=\frac{{\sigma }_z-\nu {\sigma }_r-\nu {\sigma }_{\theta }}{E}$$

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Thin Rotating Disc – General Equations

Consider a thin disc of uniform thickness rotating at angular velocity ω. Because the disc is thin and its faces are free, the classical solution treats it as a plane-stress problem:

$${\sigma }_z=0$$

If the radial displacement at radius r is u:

$${\epsilon }_{\theta }=\frac{u}{r}$$

and:

$${\epsilon }_r=\frac{du}{dr}$$

For plane stress:

$$E\frac{u}{r}={\sigma }_{\theta }-\nu {\sigma }_r$$ $$E\frac{du}{dr}={\sigma }_r-\nu {\sigma }_{\theta }$$

Radial Equilibrium

Consider a small rotating annular element.

Radial equilibrium, including the centrifugal body force, gives:

$$\frac{d{\sigma }_r}{dr}+\frac{{\sigma }_r-{\sigma }_{\theta }}{r}+\rho {\omega }^{2}r=0$$

or:

$${\sigma }_{\theta }-{\sigma }_r-r\frac{d{\sigma }_r}{dr}=\rho {\omega }^2{r}^2$$

Combining equilibrium with the plane-stress elastic relations gives:

$${\sigma }_r=A+\frac{B}{r^2}-\frac{\left(3+\nu \right)\rho {\omega }^2{r}^2}{8}$$ $${\sigma }_{\theta }=A-\frac{B}{r^2}-\frac{\left(1+3\nu \right)\rho {\omega }^2{r}^2}{8}$$

The constants A and B are determined from the radial boundary conditions.

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Solid Rotating Disc

For a solid disc, the stresses must remain finite at r = 0. Therefore:

$$B=0$$

At the free outside surface r = R₂:

$${\sigma }_r\left(R_2\right)=0$$

Hence:

$$A=\frac{\left(3+\nu \right)\rho {\omega }^2{R}_2^2}{8}$$

Radial Stress

$${\sigma }_r=\frac{\left(3+\nu \right)\rho {\omega }^2}{8}\left(R_2^2-r^2\right)$$

Circumferential Stress

$${\sigma }_{\theta }=\frac{\rho {\omega }^2}{8}\left(\left(3+\nu \right)R_2^2-\left(1+3\nu \right)r^2\right)$$

Maximum Stress

At the centre of the disc:

$${\sigma }_{r,\text{max}}={\sigma }_{\theta ,\text{max}}=\frac{\left(3+\nu \right)\rho {\omega }^2{R}_2^2}{8}$$

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Rotating Disc with a Central Hole

For an annular disc having free inner and outer cylindrical surfaces:

$${\sigma }_r\left(R_1\right)=0$$ $${\sigma }_r\left(R_2\right)=0$$

Solving for the integration constants:

$$A=\frac{\left(3+\nu \right)\rho {\omega }^2}{8}\left(R_1^2+R_2^2\right)$$ $$B=-\frac{\left(3+\nu \right)\rho {\omega }^2{R}_1^2{R}_2^2}{8}$$

Radial Stress

$${\sigma }_r=\frac{\left(3+\nu \right)\rho {\omega }^2}{8}\left(R_1^2+R_2^2-\frac{R_1^2{R}_2^2}{r^2}-r^2\right)$$

Circumferential Stress

$${\sigma }_{\theta }=\frac{\rho {\omega }^2}{8}\left(\left(3+\nu \right)\left(R_1^2+R_2^2+\frac{R_1^2{R}_2^2}{r^2}\right)-\left(1+3\nu \right)r^2\right)$$

Maximum Circumferential Stress

The maximum circumferential stress occurs at the bore, r = R₁:

$${\sigma }_{\theta ,\text{max}}=\frac{\rho {\omega }^2}{4}\left(\left(1-\nu \right)R_1^2+\left(3+\nu \right)R_2^2\right)$$

Maximum Radial Stress

Differentiating the radial-stress equation shows that the maximum radial stress occurs at:

$$r=\sqrt{R_1{R}_2}$$

and:

$${\sigma }_{r,\text{max}}=\frac{\left(3+\nu \right)\rho {\omega }^2}{8}{\left(R_2-R_1\right)}^2$$

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Long Rotating Cylinder – General Equations

A long rotating cylinder requires a different axial constraint from the thin-disc plane-stress solution. The classical solution used here assumes that the axial strain is uniform through the radius:

$$\frac{d{\epsilon }_z}{dr}=0$$

This is a generalised plane-strain condition. From:

$$E{\epsilon }_z={\sigma }_z-\nu {\sigma }_r-\nu {\sigma }_{\theta }$$

constant axial strain requires:

$$\frac{d{\sigma }_z}{dr}=\nu \left(\frac{d{\sigma }_r}{dr}+\frac{d{\sigma }_{\theta }}{dr}\right)$$

Combining this relation with radial equilibrium gives:

$${\sigma }_r=A-\frac{B}{r^2}-\frac{\left(3-2\nu \right)\rho {\omega }^2{r}^2}{8\left(1-\nu \right)}$$ $${\sigma }_{\theta }=A+\frac{B}{r^2}-\frac{\left(1+2\nu \right)\rho {\omega }^2{r}^2}{8\left(1-\nu \right)}$$

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Solid Rotating Cylinder

At r = 0 the stresses must remain finite, therefore:

$$B=0$$

For a free outer cylindrical surface:

$${\sigma }_r\left(R_2\right)=0$$

Radial Stress

$${\sigma }_r=\frac{\left(3-2\nu \right)\rho {\omega }^2}{8\left(1-\nu \right)}\left(R_2^2-r^2\right)$$

Circumferential Stress

$${\sigma }_{\theta }=\frac{\rho {\omega }^2}{8\left(1-\nu \right)}\left(\left(3-2\nu \right)R_2^2-\left(1+2\nu \right)r^2\right)$$

Maximum Stress

At the centre:

$${\sigma }_{r,\text{max}}={\sigma }_{\theta ,\text{max}}=\frac{\left(3-2\nu \right)\rho {\omega }^2{R}_2^2}{8\left(1-\nu \right)}$$

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Hollow Rotating Cylinder

For a hollow cylinder with free inner and outer surfaces:

$${\sigma }_r\left(R_1\right)={\sigma }_r\left(R_2\right)=0$$

Define:

$$C=\frac{\left(3-2\nu \right)\rho {\omega }^2}{8\left(1-\nu \right)}$$

Then:

$$A=C\left(R_1^2+R_2^2\right)$$ $$B=C{R}_1^2{R}_2^2$$

Radial Stress

$${\sigma }_r=C\left(R_1^2+R_2^2-\frac{R_1^2{R}_2^2}{r^2}-r^2\right)$$

Circumferential Stress

$${\sigma }_{\theta }=\frac{\rho {\omega }^2}{8\left(1-\nu \right)}\left(\left(3-2\nu \right)\left(R_1^2+R_2^2+\frac{R_1^2{R}_2^2}{r^2}\right)-\left(1+2\nu \right)r^2\right)$$

Maximum Radial Stress

The maximum radial stress occurs at:

$$r=\sqrt{R_1{R}_2}$$

and is:

$${\sigma }_{r,\text{max}}=C{\left(R_2-R_1\right)}^2$$

Maximum Circumferential Stress

The maximum circumferential stress occurs at the inner surface r = R₁:

$${\sigma }_{\theta ,\text{max}}=\frac{\rho {\omega }^2}{4\left(1-\nu \right)}\left(\left(1-2\nu \right)R_1^2+\left(3-2\nu \right)R_2^2\right)$$

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Engineering Design Considerations

The equations above describe ideal elastic stress distributions. Real rotating components normally require additional checks.

• Equivalent stress: radial, circumferential and, where applicable, axial stresses should be combined using an appropriate failure criterion such as von Mises or Tresca.
• Stress concentrations: bores, keyways, grooves, bolt holes, changes in section, blade attachments and other discontinuities can produce stresses substantially greater than the axisymmetric solution.
• Material behaviour: yielding, fatigue, creep and fracture behaviour may govern before the ideal elastic stress reaches a simple limiting value.
• Residual stress: manufacturing processes, heat treatment, interference fits and overspeeding can alter the final stress field.
• Temperature: temperature gradients and differential expansion can introduce substantial thermal stresses in high-speed rotors.
• Dynamic behaviour: critical speeds, vibration, imbalance and rotor dynamics are separate from the static rotational-stress calculation.
• Geometry: the equations on this page assume constant thickness. Variable-thickness rotating discs require a different solution.

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Related RoyMech References

• Flywheel Design – energy storage, inertia and rotating stresses.
• Thick-Walled Cylinders – radial and hoop stresses caused by internal and external pressure.
• Stress and Strain – elastic stress-strain relationships and Poisson's ratio.
• Mohr's Circle – principal stresses and stress transformation.
• Failure Modes – yielding and failure criteria.
• Torsion – torsional stresses in shafts.
• Stress-Strain Curve – tensile behaviour and material yielding.

Other Mechanics References

• Vectors
• Statics
• Kinematics
• Kinetics and Dynamics
• Mechanical Linkages
• Simple Machines
• Component Loading
• Buckling
• Loaded Flat Plates

External References

1. CodeCogs – Rotating Discs and Cylinders – additional reference material on rotating-disc and cylinder stresses.
2. DANotes – Thick-Walled Cylinders – detailed notes on related thick-cylinder stress analysis.
3. ExcelCalcs – Stresses in Rotating Discs and Rings – spreadsheet calculation associated with the original RoyMech notes.

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