These notes relate to the stresses and strains existing in thick walled
cylinders when subject to internal and external pressures.
The notes include the analysis of two or more cylindrical parts,
assembled by press fitting or shrinking, 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

Tensile stresses are considered positive and compressive stresses are negative.

The pressures p, p _{1}, p _{2} & p _{f} are negative

**Thick Cylinder basics**
Consider a thick cylinder subject to internal pressure p _{1} and an external pressure p_{2}.
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 from 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} 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

dσ _{a} / dr = υ.(dσ _{t} / dr + dσ _{r} / dr )

Substituting this into equation 4)

( σ _{t} - σ _{r} )( 1 + υ ) + r ( 1 - υ ^{2} ).( dσ _{t}/dr ) - υ.r.( 1 + υ )( dσ _{r} / dr ) = 0

This reduces to..

Eq. 5).....σ _{t} - σ _{r} + r( 1 - υ ).(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

(σ _{r} + δ σ _{r} )(r + δ r)δθ
- σ _{r }r δ θ - 2σ _{t} δ r.sin [(1/2) δ θ ] = 0

In the limit .sin [(1/2) δ θ ] - > [(1/2) δ θ ]
and neglecting small products the equation reduces to..

σ_{r} δ r + δσ_{r}.r
-σ _{t} δ r = 0... and as δ r --> 0 then σ_{r}. d r + dσ_{r}.r
-σ _{t}d r = 0...as δ and this results in..

Eq.6)....σ_{r} + r. (d σ_{r}/dr ) = σ _{t}

Now eliminating σ _{t} by substituting eq 6 into eq 5)

σ_{r} + r. (d σ_{r}/dr ) - σ _{r}
+ r( 1 - υ ).(dσ _{t} / dr ) - υ.r.(dσ _{r} / dr ) = 0

r( 1 - υ )(d σ_{r}/dr )
+ r( 1 - υ ).(dσ _{t} / dr ) = 0

simpifying to

(d σ_{r}/dr )
+ (dσ _{t} / dr ) = 0

Integration of this equation results in

Eq.7)....
σ_{r} + σ _{t} = 2.A (constant of integration.)

substituting this into equation 6 to eliminateσ_{t}

σ_{r} + r. (d σ_{r}/dr ) = 2.A - σ_{r}

Therefore

2.A = 2. σ_{r} + r. (d σ_{r}/dr )
which is equivalent to 2.A = (1 /r ). d ([ r^{2}. σ _{r} ]/dr )

therefore 2.A.r = (d[r^{2}.σ_{r}]/dr )

Which on integrating gives σ_{r}.r^{2} = A..r^{2} + B

with resulting

Eq 8) **σ**_{r} = A + B / r ^{2} and substituting this into Eq. 7)

Eq 9) **σ**_{t} = A - B / r ^{2}

**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

σ _{r} = A + B / r ^{2}

σ _{t} = A - B / r ^{2}

The radial pressures at the surfaces are the same as the applied pressures therefore

- p_{1} = A + B / r _{1}^{2}

-p_{2} = A + B / r _{2}^{2}

__ The resulting general equations are known as __** Lame's Equations** and are shown as follows

**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 _{s}
and the hole deflection δr _{h}

The longitundonal pressure and hence σ _{a} 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)

** Eq. 2)......E.ε **_{t} = E.u/r =
σ _{t} - υσ _{a} - υσ _{r}
= σ _{t} - υσ _{r}

Radial Increase in Hole diameter = u. _{h} = ( r _{f} / E _{h} ) (σ _{t}
- υ _{h}.σ _{r} )
... The condition is equivalent to a thick cylinder with zero external pressure

Radial decrease in shaft diameter = u _{s} = - ( r _{f} / E _{s} ) (σ _{t}
- υ _{s}.σ _{r} ) ... The condition is equivalent to a thick cylinder with zero internal pressure

Total interference u _{t} = u _{h} + u _{s}

The displacement of the hole u _{h} and the shaft u _{s} are as follows.

( (Hole r_{1} = r _{f} p _{f} = p _{1} ) ( Shaft r_{2} = r _{f} p _{f} = p _{2} )

The total interference is therefore 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. r_{1} = 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 an interference fit**

Consider a steel shaft 100mm dia. pressed into a ring which has and OD of 200mm. 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 = 210.10^{9} N/m^{2}. Poissens ratio υ = 0,3.

## Notes to be added