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Thick Walled Cylinders

Introduction..... Symbols..... Assumptions..... Basic derivations..... General Equations..... Interference Fit.....

Introduction

These notes relate to the stresses and strains existing in thick walled cylinders when subject to internal and external pressures.   The notes are directed towards the analysis two or more cylindrical parts assembled by press fitting or shrinking to resulting in an interference fit between the parts.  The equations resulting enable estimates of the forces need to assemble and separate the parts and the maximum torque which can be transmitted by the assembly



Symbols / Units
p = radial stress (compressive) (N/m2)
p 1 = Internal pressure (N/m2)
p 2 = External pressure (N/m2)
σ r = radial (normally compressive)(N/m2)
σ t = tangential stress (normally tensile)(N/m2)
σ a = axial/longitudonal stress (normally tensile)(N/m2)
E = Young's moudulus of elasticity (N/m2)
υ = Poissan's ratio
υ h = Poissan's ratio hub
υ s = Poissan's ratio shaft
d= diameter at point or analysis (m).
d 1= inside diameter of cylinder (m).
d 2= outside diameter of cylinder (m).
r = radius at point or analysis (m).
r 1= inside radius of cylinder (m).
r 2= outside radius of cylinder (m).
εr = radial strain
εt = tangential strain
εa = axial /longitudinal strain
u = radial deflection (m)
us = radial deflection of shaft (m)
uh = radial deflection of hub(m)
ut = radial deflection of hub and shaft(m)
δr 1h = Radial increase in hole (m)
δr 1s = Radial decrease in hole (m)



Initial Assumptions

The following relationships are assumed for the strains ε1,ε2, ε3 associated with the stress σ 1, σ 2 and σ 3. υ = Poisson's ratio.

Reference
Notes on stress and strain

ε1 = σ 1 /E   -   υσ 2 /E    -    υσ 3 /E
ε2 = σ 2 /E   -   υσ 1 /E    -    υσ 3 /E
ε3 = σ 3 /E   -   υσ 1 /E    -    υσ 2 /E



Thick Cylinder basics

Consider a thick cylinder subject to internal pressure p 1 and an external pressure p2.    Under the action of radial pressures on the surfaces the three principal stress will be σ r compressive radial stress, σ t tensile tangential stress and σ a 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...This condition generally applies away from the ends of the cylinder and away for discontinuities.

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 ( compressive) is equivalent to - σ 1...
σ t ( tensile ) is equivalent to σ 2...
σ a ( tensile ) 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.εr = 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 r + σ t ). ( 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

dσ a / dr = υ.((dσ t / dr - σ r / dr )
Substituting this into equation 4)
(σ r + σ t )(1 + υ ) + r( 1 - υa ).(dσ t / dr )+ υ.r.(1 + υ )(dσ r / dr ) =0
This reduces to..

Eq. 5).....σ r + σ t + r( 1 - υ ).(dσ t / dr ) + υ.r.(dσ r / dr ) = 0

Now considering the radial equilibrium of the element of the section

2σ t δ r.sin [(1/2) δ θ ] + (σ r + δσ r)(r + span>δ r ) - σ r δ θ = 0
In the limit .sin [(1/2) δ θ ] - > [(1/2) δ θ ] and neglecting small products the equation reduces to..

Eq.6).....σ t + σ r + r.dσ r / dr = 0

Now subtracting eq 6 from eq. 5

r(1 - υ)(dσ t / dr )+ υr (dσ r / dr) - r(dσ r / dr) = 0
Which results in
dσ t/dr - dσ r/dr = 0...
Which on integrating gives

Eq 7)....σ t - σ r = 2.a (2.a = assumed constant of integration).

Subtracting Eq.7) from Eq.6)

Integrating this equation

σ r.r 2 = - a.r 2 + B
σ r = - a +B / r 2

Eq. 8)   σ r = - a + b / d 2
Eq.9)   σ t = a + b / d 2

d = 2r , b = 4B, a and b are constants which depend on the dimensions and loading



General Equation for Thick Walled Cylinder

The general equation for a thick walled cylinder subject to internal and external pressure can be easily obtained from eq)8 and eq) 9 as follows.

Consider a cylinder with and internal diameter d 1, subject to an internal pressure p 1.  The external diameter is d 2 which is subject to an external pressure p 2.    The radial pressures at the surfaces are the same as the applied pressures therefore

p1 = -a + b / d 12
p2 = -a + b / d 22

The resulting general equations are therefore as follows

If the external pressure is zero this reduces to

If the internal pressure is zero this reduces to



Interference Fit

Consider a press fit of a shaft inside a hole.  The compression of the shaft and the expansion of the hub result in a compressive pressure at the interface.   The conditions are shown in the figures below

The radial interference δr 1 = the sum of the shaft deflection δr 1s and the hole deflection δr 1h
The longitundonal pressure and hence σ 2 are assumed to be zero and the internal pressure in the shaft hole and the external pressure outside the hub are also assumed to be zero. (ref. to equation 2)

Radial Increase in Hole diameter = u. 1h = ( r f / E h ) (σ t + υ h.σ r )
Radial decrease in shaft diameter = u 1s = - ( r f / E s ) (σ t - υ s.σ r )
Total interference u t = u 1h + u 1s

The displacement of the hole u 1h and the shaft u 1s are as follows.
( σ r = p f , d1 = d f )

The total interference is therefor equal to

If the hub and the shaft are the same material with the same E and σ the equation simplifies to

The normal engineering application is when the shaft is solid i.e. r1 = zero therefore the equation further simplifies to

It is often required to determine the interface pressure when the radial interference u t is known (This is half the shaft interference) i.e to determine the torque which can be transmitted or the force require to make or separate the interference joint.



Example calculation of torque transmitted by and interference fit

Consider a steel shaft 100mm dia. pressed into a hole. The length of the hole is 50mm .  The interference = 0,1mm  The assumed coefficient of friction μ = 0,15.  The hub and shaft are both steel with E = 210.109 N/m2.  Poissens ratio υ = 0,3.


Notes to be added



Relevant Links
  1. Dan Notes Thick Walled Cylinders Excellent source of information on this topic
  2. Lecture 21 Thick Walled Cyliders Very informative pdf download
  3. Mecheng Calculations..Thick Walled Cylinder..Calculator + Associated formulea
  4. EngineersEdge -Torque transmitted by a press fit.. Useful calculator


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Last Updated 03/02/2008