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.
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ExcelCalcs.com calculation Stresses in Rotating Disks & Rings
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Symbols / Units
Tensile stresses are considered positive and compressive stresses are negative.
p _{1} = Internal pressure (N/m^{2})
p _{2} = External pressure (N/m^{2})
σ _{r} = radial (N/m^{2})
σ _{t} = tangential stress (N/m^{2})
σ _{a} = axial/longitudonal stress (N/m^{2})
E = Young's moudulus of elasticity (N/m^{2})
ρ = density. ( kg/m^{2})
υ = Poisson's ratio
r = radius at point or analysis (m).
R _{1}= inside radius of cylinder (m).
R _{2}= outside radius of cylinder (m).
ε_{r} = radial strain
ε_{t} = tangential strain
ε_{a} = axial /longitudinal strain
u = radial deflection (m)

Initial Assumptions
The following relationships are assumed for
the strains ε_{1},ε_{2},
ε_{3} associated with the stress σ _{1},
σ _{2} and σ _{3}. υ = Poisson's ratio.
Reference Notes on stress and strain
ε_{1} = σ _{1} /E  υσ _{2} /E  υσ _{3} /E
ε_{2} = σ _{2} /E  υσ _{1} /E  υσ _{3} /E
ε_{3} = σ _{3} /E  υσ _{1} /E  υσ _{2} /E
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 _{r} tensile radial stress, σ _{t} tensile
tangential stress and σ _{a} and axial stress
which is generally also tensile. The stress conditions occur throughout the section
and vary primarily relative to the radius r.
It is assumed that the axial stress σ _{a} is
constant along the length of the section and because the disk is thin compared to its diameter the axial stress throughout the
section is assumed to be zero.It is also assumed that the internal pressure (P_{1} ) and the external pressure ( P_{2} ) = 0.
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...
σ _{r} is equivalent to σ _{1}...
σ _{t} is equivalent to σ _{2}...
σ _{a} is equivalent to σ _{3}...
derived equations
Strictly the following equations apply
Eq. 1)......E ε _{a} = σ _{a}  υσ _{t}  υσ _{r}
Eq. 2)......E.ε _{t} = E.u/r = σ _{t}  υσ _{a}  υσ _{r}
Eq. 3)......E.ε_{r} = E.du/dr = σ _{r}  υσ _{t}  υσ _{a}
However because of the assumption that σ _{a} = 0 the equations are modified as follow.

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 _{t}  υσ _{r} )
Differentiating
Edu/dr = σ _{t}  υσ _{r} +
r. [ dσ _{t} /dr  υ.( dσ _{r} / dr ) ]
= σ _{r}  υσ _{t} .. from 3 above )
Simplifying by collecting terms.
Eq. 4)........(σ _{t}  σ _{r} ). ( 1 + υ )
+ r.(dσ _{t}/ dr ) )  υ.r.(dσ _{r} / dr) = 0
Now considering the radial equilibrium of the element of the section. Forces based on unit length of cylinder
2.σ _{t}. δ _{r}.sin(1/2.δθ) +
σ_{r}δθ  (σ _{r} +
δσ _{r}) (r + σ r )θδθ =
ρr^{2} ω^{2}δr δθ
In the limit this reduces to
Eq 5 ).....
σ _{t}  σ _{r}  r dσ _{r} / d r
= ρr^{2} ω^{2}
Substitute for σ _{t}  σ _{r} in Equation 4 results in
( r dσ _{r} / d r + ρr^{2} ω^{2} ). ( 1 + υ )
+ r.(dσ _{t}/ dr ) )  υ.r.(dσ _{r} / dr) = 0
Therefore
dσ _{t}/dr + dσ _{r}/dr
=  ρrω ^{2}(1+υ)
Integrating
Eq 6 )..... σ _{t} + σ _{r}
=  ρ r^{2} ω ^{2}(1+υ)/2 + 2A
Subtract equation 5...
2.σ _{r} + r.dσ _{r}/dr
=  ρr^{2}ω ^{2}(3+υ)/2 + 2A
This is the same as
(1/r).d (σ _{r}.r^{2})dr
=  ρ r^{2} ω ^{2}(3+υ)/2 + 2A
Integrating
σ _{r}.r^{2}
=  (ρ r^{4} ω ^{2}(3+υ)/8 + Ar^{2} + B
Eq 7 ...
σ _{r} = A + B/r^{2}  (3 + υ)ρ r^{2} ω ^{2}/8
Eq 8 ......combining Eqn. 7 with Eqn. 6
σ _{t} = A  B/r^{2}  ( 1+ 3υ)ρ r^{2} ω ^{2}/8
1) Solid Disk
At the centre the B/r^{2} term implies infinite stresses which are clearly not credible and therefore B must equal 0.
At r = R_{2} on the outside edge of the disk. the radial stress is equal to the surface stress which is equal to 0.
Therefore at R_{2}
σ _{r}= 0 = A  (3 + υ)ρ R_{2} ^{2} ω ^{2}/8
Therefore A = ρ R_{2} ^{2} ω ^{2}/8
B = 0
σ _{t} = (ρ ω ^{2}/8)[ (3 + υ)R_{2} ^{2}
 (1 + 3υ) r^{2}]
σ _{r} = (ρ ω ^{2}/8)[ (3 + υ)( R_{2} ^{2}
 r^{2})]
The maximum stress is at the centre
σ _{t_max} = σ _{r_max}
= (ρ ω ^{2}/8).(3 + υ)R_{2} ^{2}
2) Disk with a hole
At the outside edge r = R_{2} and at the hole radius r = R_{1} the radial surface stress is asssumed to be 0
Therefore
σ _{r}= 0 = A + B/R_{2}^{2}  (3 + υ)ρ R_{2}^{2} ω ^{2}/8
σ _{r}= 0 = A + B/R_{1}^{2}  (3 + υ)ρ R_{1}^{2} ω ^{2}/8
Solving
B = (3 + υ)ρ ω ^{2}/8.(R_{1}^{2} . R_{2}^{2})
A = (3 + υ)ρ ω ^{2}/8.(R_{1}^{2} + R_{2}^{2})
σ _{r} = (3 + υ)ρ ω ^{2}/8)
(R_{1}^{2} + R_{2}^{2}  R_{1}^{2}.R_{2}^{2}/ r^{2}  r^{2} )
σ _{t} = ρ ω ^{2}/8)
[(3 + υ)(R_{1}^{2} + R_{2}^{2}
+ R_{1}^{2}.R_{2}^{2}/ r^{2})
 (1+3υ) r^{2} )]
The maximum tangential stress σ _{t} is at the inside hole surface and equals
σ _{t_max} = ρ ω ^{2}/4) [ (1 υ)R_{1}^{2} + (3 + υ)R_{2}^{2})]
The maximum radial stress σ _{r} is at r = sqrt (R _{1}. R_{2} ) and equals .
σ _{r_max} = (3+υ) (ρ ω ^{2}/8)(R_{2}  R_{1})^{2}

2) Cylinders
The primary difference between a long rotating cylinder and a thin one is that it the axial stress σ _{a} is
not equal to 0. The assumption in this case is that the longitudonal strain ε _{a} is constant: cross sections remain plane.
Basis of equations... Refer to introductory notes at the top of webpage
σ _{r} is equivalent to σ _{1}...
σ _{t} is equivalent to σ _{2}...
σ _{a} is equivalent to σ _{3}...
derived equations
Eq. 1)......E ε _{a} = σ _{a}  υσ _{t}  υσ _{r}
Eq. 2)......E.ε _{t} = E.u/r = σ _{t}  υσ _{a}  υσ _{r}
Eq. 3)......E.ε_{r} = E.du/dr = σ _{r}  υσ _{t}  υσ _{a}
Multiplying 2) x r
Eu = r ( σ _{t}  υσ _{a}  υσ _{r} )
differentiating
Edu/dr = σ _{t}  υσ _{a}  υσ _{r} +
r. [ dσ _{t} /dr  υ.( dσ _{a} / dr )  υ.( dσ _{r} / dr ) ]
= σ _{r}  υσ _{t}  υσ _{a} ..( from 3 above )
Simplifying by collecting terms.
Eq. 4)........(σ _{t}  σ _{r} ). ( 1 + υ ) + r.(dσ _{t}/ dr )  υ.r.(dσ _{a} / dr )  υ.r.(dσ _{r} / dr) = 0
Now from 1) above since ε _{a} is constant then dσ _{a} /dr = υ (dσ _{t}/dr + dσ _{r} /dr )
substituting for dσ _{a}/dr in Eq 4)
(σ _{t}  σ _{r} ). ( 1 + υ ) + r.(dσ _{t}/ dr )  υ.r.(υ (dσ _{t}/dr + dσ _{r} /dr )  υ.r.(dσ _{r} / dr) = 0
(σ _{t}  σ _{r} ). ( 1 + υ )
+ r.(1υ^{2})(dσ _{t}/ dr )
 υ.r.(1 + υ ) (dσ _{r}/dr)
= 0
(σ _{t}  σ _{r} )
+ r.(1υ)(dσ _{t}/ dr )
 υ.r (dσ _{r}/dr)
= 0
Now considering the radial equilibrium of the element of the section as shown in the notes above Eq. 5 results
Eq 5........
σ _{t}  σ _{r}  r dσ _{r} / d r
= ρr^{2} ω^{2}
Substitute for (σ _{t}  σ _{r} )
r.(1υ)(dσ _{t}/ dr )
+ (1  υ).r (dσ _{r}/dr)
=  ρr^{2} ω^{2}
Therefore
(dσ _{t}/ dr )
+ (dσ _{r}/dr)
=  ρ r ω^{2} /(1υ)
Integrating
Eq.6)... σ_{t} + σ_{r}
= ρr^{2}ω^{2} /2(1υ) + 2A
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 _{t} by substituting into Eq 5)
2σ _{r} + r dσ _{r} / d r =
2A  ρr^{2}ω^{2}  ρr^{2}ω^{2} /2(1υ)
Therefore
(1/r).d(σ_{r}.r^{2})/dr = 2A  ρr^{2}ω^{2}(3 2υ ) /2(1υ)
Therefore
d(σ_{r}.r^{2})/dr = 2Ar  ρr^{3}ω^{2}(3 2υ ) /2(1υ)
Integrating
Eq.7).... σ_{r} =  ρr^{2}ω^{2}(3 2υ ) /8(1υ)
+ A  B / r^{2}
Now substituting for σ_{r} in Equation 7 to obtain σ_{t}
Eq.8).... σ_{t} =  ρr^{2}ω^{2}(1 + 2υ ) /8(1υ)
+ A + B / r^{2}
Solving for A and B for a solid cylinder
At the centre of the cylinder R_{1} = 0 the stresses cannot be infinite so B is cleraly equal to 0.
B = 0
At the outside diameter of the cylinder r = R_{2} .
σ _{r} = 0 = σ_{r} =  ρr^{2}ω^{2}(3 2υ ) /8(1υ)
+ A
Therefore
A = ρω^{2}R_{2}^{2}(3 2υ ) /8(1υ)
Eq.9).... .σ_{r} = ρω^{2}(3 2υ ) /8(1υ ) (R_{2} ^{2}  r ^{2})
Eq.10).... σ_{t} = ρω^{2} /8(1υ) [(3 2υ )R_{2}^{2} 
(1 + 2υ ) r^{2} ]
The maximum radial stress and tangential stress are equal at r = 0 =
σ_{r_max} = σ_{t_max} =
ρω^{2}(3 2υ ) /8(1υ ))R_{2}^{2}
Solving for A and B for a hollow cylinder
At r = R_{1} and at r = R_{2} the radial stress σ_{r} = 0
Therefore 0 = σ_{r} =  ρ R_{1}^{2}ω^{2}(3 2υ ) /8(1υ)
+ A  B / R_{1}^{2}
Therefore 0 = σ_{r} =  ρ R_{2}^{2}ω^{2}(3 2υ ) /8(1υ)
+ A  B / R_{2}^{2}
Therefore ...
ρ R_{1}^{2}ω^{2}(3 2υ ) /8(1υ)
+ B / R_{1}^{2} =
ρ R_{2}^{2}ω^{2}(3 2υ ) /8(1υ)
+ B / R_{2}^{2}
ρ (R_{1}^{2}  R_{2}^{2})ω^{2}(3 2υ ) /8(1υ)
= ρ B / R_{2}^{2}  B / R_{1}^{2}
Therefore B = R_{1}^{2}.R_{2}^{2}.ρ ω^{2}(3 2υ ) /8(1υ)
Solving for A
0 = σ_{r}
=  ρR_{1}^{2}ω^{2}(3 2υ ) /8(1υ)
+ A 
R_{2}^{2}.ρ ω^{2}(3 2υ ) /8(1υ)
Therefore
A = (R_{2}^{2} + R_{1}^{2}).ρ ω^{2}(3 2υ ) /8(1υ)
Resulting in..
σ_{r} = ρω^{2}(3 2υ ) /8(1υ )
[  r^{2}
+ (R_{2}^{2} + R_{1}^{2})
 (R_{1}^{2}.R_{2}^{2})/r^{2} ]
σ_{r_max} = ρω^{2}(3 2υ ) /8(1υ )
((R_{2}^{2}  R_{1}^{2})... Is located at at r = Sqrt (R_{1}.R_{2} )
σ_{t} = ρω^{2} /8(1υ )
[  r^{2} (1 + 2υ ) + (3 2υ ) {( R_{2}^{2} + R_{1}^{2}) +
(R_{1}^{2}.R_{2}^{2})/r^{2}} ]
σ_{t_max} = ρω^{2} /4(1υ )
[ (1  2υ )R_{1}^{2} + (3 2υ )R_{2}^{2}]
... Is located at r = R_{1}.
