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
ε1 = σ 1 /E - υσ 2 /E - υσ 3 /E 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... 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 )
= -σ r - υσ t - υσ a ... ( from 3 above )
Eq. 4)........(σ r + σ t ). ( 1 + υ ) + r.(dσ t/ dr ) - υ.r.(dσ a / dr ) + υ.r.(dσ r / dr) = 0 dσ a / dr = υ.((dσ t / dr - σ r / dr ) Substituting this into equation 4) ( σ r + σ t )( 1 + υ ) + r ( 1 - υ 2 ).( 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. Forces based on unit length of cylinder 2σ t δ r.sin [(1/2) δ θ ] + (σ r + δ σ r )(r + δ r)δθ - σ r r δ θ = 0 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).
Integrating this equation σ r.r 2 = - a.r 2 + B Eq. 8) σ r = - a + b / d 2 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. p1 = -a + b / d 12 If the external pressure is zero this reduces to If the internal pressure is zero this reduces to 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 Radial Increase in Hole diameter = u. 1h = ( r f / E h ) (σ t + υ h.σ r )
The displacement of the hole u 1h and the shaft u 1s are as follows. 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 an interference fit
ε2 = σ 2 /E - υσ 1 /E - υσ 3 /E
ε3 = σ 3 /E - υσ 1 /E - υσ 2 /E
Thick Cylinder basics
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
Simplifying by collecting terms.
Now from 1) above since ε a is constant
In the limit .sin [(1/2) δ θ ] - > [(1/2) δ θ ]
and neglecting small products the equation reduces to..
Subtracting Eq.7) from Eq.6)
σ r = - a +B / r 2
Eq.9) σ t = a + b / d 2...Using equation 7
d = 2r , b = 4B, a and b are constants which depend on the dimensions and loading
General Equation for Thick Walled Cylinder
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
p2 = -a + b / d 22
The resulting general equations are therefore as follows
Interference Fit
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 decrease in shaft diameter = u 1s = - ( r f / E s ) (σ t - υ s.σ r )
Total interference u t = u 1h + u 1s
( σ r = p f , d1 = d f )
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