Introduction..... Symbols..... Assumptions..... Basic Equations for a Rotating Disk..... Basic Equations for a Rotating Cylinder.....

Introduction These notes relate to the stresses and strains existing in thick walled cylinders when rotated at speed they are generally applicable to design of flywheels. The primary assumption is that the cylinders are not subject to internal or external prssure. A basic review of solid disks, rings and cylinders is carried out. Members of excelcalcs.com can upload a spreadsheet including all of the derived equations below at ExcelCalcs.com calculation Stresses in Rotating Disks & Rings ... Symbols / Units
Tensile stresses are considered positive and compressive stresses are negative.
Initial Assumptions
The following relationships are assumed for
the strains ε
ε Thick Disk basics
Consider a "disk"/ "thin ring" subject internal stresses resulting from the inertial forces as a result of its rotational speed.
Under the action of the inertial forces only, the three principal
stress will be σ Consider a microscopically small area under stress as shown. u is the radial displacement at radius r . The circumferential (Hoop) strain due to the internal pressure is At the outer radius of the small section area (r + δr ) the radius will increase to (u + δ ). The resulting radial strain as δr -> 0 is Referring to the stress/strain relationships as stated above. The following equations are derived.
Basis of equations...
Eq. 1)......E ε _{a} =0 - υσ _{t} - υσ _{r}Eq. 2)......E.ε _{t} = E.u/r = σ _{t} - υσ _{r}Eq. 3)......E.ε _{r} = E.du/dr = σ _{r} - υσ _{t} - Multiplying 2) x r Eu = r ( σ Differentiating Edu/dr = σ Eq. 4)........(σ Now considering the radial equilibrium of the element of the section. Forces based on unit length of cylinder 2.σ In the limit this reduces to
Substitute for σ ( r dσ Therefore dσ Integrating
Subtract equation 5... 2.σ This is the same as (1/r).d (σ Integrating σ
σ
σ Solid Disk At the centre the B/r σ σ σ The maximum stress is at the centre σ Disk with a hole At the outside edge r = R Therefore
Solving
2) Cylinders The primary difference between a long rotating cylinder and a thin one is that it the axial stress σ
Basis of equations... Refer to introductory notes at the top of webpage Multiplying 2) x r Eu = r ( σ differentiating Edu/dr = σ
substituting for dσ (σ (σ Now considering the radial equilibrium of the element of the section as shown in the notes above Eq. 5 results
Substitute for (σ
r.(1-υ)(dσ (dσ Integrating This is similar to the equation 6 for the Rotating Disk analysis completed above. In fact the rotating disks equation can apply for the long cylinder if (1 + υ ) in the disk equations are replace by 1 / (1 - υ ) . Or if υ is replaced by (υ / (1- υ).. Now eliminating σ Integrating Now substituting for σ Solving for A and B for a solid cylinder
At the centre of the cylinder R B = 0 At the outside diameter of the cylinder r = R σ The maximum radial stress and tangential stress are equal at r = 0 = Solving for A and B for a hollow cylinder
At r = R Solving for A 0 = σ Resulting in.. σ σ σ σ |

- Rotating Disks and Cylinders Similar Notes to mine using different (american?) terminology.
- Dan Notes Thick Walled Cylinders Excellent source of information on this topic