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
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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
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.
