Thick Walled Cylinders
Consider a thick walled cylinder having an inner radius = a; outer radius = b. Let the cylinder is
subjected to internal pressure pi and outer pressure po. For the purpose of analysis, thick walled
cylinder can be considered to consist of a series of thin rings (Figure 1a.).
Consider a typical ring located at a radius r having a thickness dr
*
As the result of internal and external pressure loading, a radial stress σr would develop at the
interface between rings located at a radial distance r.
*
A slightly different radial stress (σr + dσr) would develop at a radial position (r + dr)
*
These stresses would be uniformly distributed over the inner and outer surface of the ring.
*
Shear stress would not develop on the surfaces, since the pressure loading do not tend to force
the rings to rotate with respect to one another.
*
A tangential or hoop stress develops when a pressure difference exists between the inner and
outer surface.
*
The planes on which these tangential stresses act can be exposed by considering only a small
part of a ring
*
The tangential stress is assumed to be uniformly distributed through the thickness of the ring,
since the thickness of the ring is very very small.
*
A relationship between σr and σt can be derived by considering the equilibrium of a small
element (Fig. 1b)
*
Axial stress σa, that may be present is omitted, since it does not contribute to equilibrium of the
element in radial and tangential directions.
Fig 1b
Fig 1a
1
Considering the equilibrium in radial direction i.e.. the sum of all the force in radial direction is zero
∑F
r
=0
0
(σ r + dσ r )( r + dr ) dθ dl − σ r rdθ dl − 2σ t drdl sin ( dθ / 2 ) =
(1)
By neglecting higher order terms and noting sin ( dθ / 2 ) = dθ / 2
The equation (a) can be reduced to
r
dσ r
+σr −σt =
0
dr
(2)
The equation (2) can be integrated since σr and σt are functions of radial position r.
For the case of thick cylinders, the axial strain εa at any point in the wall can be expressed as
εa =
σ a − ν (σ r + σ t )
(3)
E
This means plane transverse section before remain plane after loading.
So far as the axial stress σa is concerned, two cases are of interest in a wide variety of design
applications.
i)
Axial load induced by pressure not carried by the walls of the cylinders (σa = 0)
eg. Gun barrels hydraulic cylinders.
ii)
Walls of the cylinder carry the loads for example, pressure vessels with closed ends.
Regions of the cylinders away from the ends, axial stress are uniformly distributed. Hence εa, σa,
E ν are constant, therefore
σt )
(σ r +=
σ a − Eε a
= 2C1
ν
(4)
Where C1 is a constant
Substituting σt into differential equation, we obtain
r
dσ r
+ 2σ r =
2C1
dr
(5)
2
d 2
( r σ r ) = 2C1r
dr
(6)
Integrating this equation further yields
2
r=
σ r C1r 2 + C2
(7)
Where C2 is a constant of integration.
Thus
σ=
C1 +
r
C2
r2
(8)
C
σ=t C1 − 22 r 2
r
Values of the constants C1 and C2 can be obtained from the known values of internal and external
pressures.
σ r = − pi
σ r = − po
at r = a
at r = b
(- sign indicates that the pressure produces compressive radial stresses).
Using these boundary conditions, we can solve for C1 and C2 as
C1 =
a 2 pi − b 2 po
b2 − a 2
a b ( pi − po )
b2 − a 2
Then the radial and tangential stresses can be obtained as
(9)
2 2
C2 = −
 a 2 pi − b 2 po   a 2b 2 ( pi − po )  1
=
σr 
 2
−
2
2
b2 − a 2
 b −a
 
r
 a 2 pi − b 2 po   a 2b 2 ( pi − po )  1
=
σt 
 2
+
2
2
b2 − a 2
 b −a
 
r
3
(10)
(11)
Radial and circumferential deformations play important roles in press-fit and shrink-fit problems.
Change in circumference δc of the thin ring when internal and external pressures are applied can be
expressed in terms of radial displacement of a point in the ring as
δ=
c
{2π ( r + dr ) − 2π r=}
2πδ r ;
where δr is change in radius of the ring
Circumferential deformation can also be expressed in terms of circumferential strain
δ c = ε t 2π r
∴
δr = εt r
(12)
For most applications σa = 0. Tangential strain εt can be expressed in terms of radial and tangential
stresses by Hooke’s law as:
εt =
(σ t −νσ r )
(13)
E
Thus the radial displacement (or change in radius) at any radius r is given by
δr =
(σ t −νσ r ) r
(14)
E
Where the radial and tangential stresses are calculated at radius r.
This change in radius can be expressed in terms of internal and external pressures as
 1 −ν
=
δr 
 E
2
2
  a pi − b po   1 +ν
r +

2
2
 b − a
  E
2 2
  a b ( pi − po )  1


b2 − a 2

r
(15)
The maximum numerical value of σr is found at r = a to be pi, provided that pi exceeds po.
If po > pi, the maximum σr occurs at r = b and equals po.
On the other hand, the maximum σt occurs at either inner or outer radius according to the pressure
ratio.
The maximum shearing stress at any point equals one-half the algebraic difference between the
maximum and minimum principal stresses. At any point in the cylinder, we may therefore state that
4
=
τ max
σt −σr )
(=
2
 a 2b 2 ( pi − po )  1

 2
b2 − a 2

r
(16)
The largest value of τmax is found at r = a, the inner surface. The effect of reducing po is clearly to
increase τmax. Consequently, the greatest τmax corresponds to r= a and po =0 is given by
τ max =
pi b 2
( b2 − a 2 )
(17)
Since σr and σt are principal stresses, τmax occurs on planes making an angle of 45o with the plane on
which σr and σt act.
Special cases
Internal Pressure only: If only internal pressure acts equations for stresses and change in radius
reduce to
a2 pi  b2
σr = b2 - a2 1 - r2 


(18)
a2 pi 
b2
σt = b2 - a2 1 + r2 


(19)
a2 pi r 
b2
δr = E (b2 - a2) (1 - ν) + (1 + ν) r2 


Fig. 2
b2
Since r2 ≥ 1, σr is negative (compressive) for all r except r = b,
in which case σr = 0. The maximum stress occurs at r = a.
σt, it is positive (tensile) for all radii, and also has maximum at r
= a. This is illustrated in the figure 3.
5
(20)
External Pressure only: In this case, pi = 0, and the equations are given as
b2 po  a2
σr = - b2 - a2 1 - r2 


(21)
b2 po 
a2
σt = - b2 - a2 1 + r2 


b2 po r 
a2
δr = E (b2 - a2) (1 - ν) + (1 + ν) r2 


(22)
(23)
Fig. 3
Figure 4
The maximum radial stress occurs at r = b and is compressive for all r.
The maximum σt is found at r = a, and is likewise compressive. And this is illustrated in the figure 4.
Closed End cylinder: In the case of closed-end cylinder subjected to internal and external pressures,
longitudinal stress exists in addition to the radial and tangential stresses. For a transverse section some
distance from the ends, this stress may be assumed uniformly distributed over the wall thickness. The
magnitude of σl is then determined by equating the net force acting on one end attributable to pressure
loading, to the internal directed force in the cylinder wall:
piπ a 2 − poπ b 2 =π ( b 2 − a 2 ) σ l
Fig. 4
6
The resulting expression for longitudinal stress, applicable only away from the ends, is
σl =
pi a2 - po b2
b2 - a2
(23)
Now consider a thick cylinder subjected to inner pressure only. Then radial and tangential stresses are
given by
a2 pi  b2
σr = b2 - a2 1 - r2 


a2 pi 
b2
σt = b2 - a2 1 + 2 
r 

Both tangential and radial stresses are maximum, at inner surface
σt
max
 a 2 + b2 
= pi  2
2 
b −a 
Let K be the ratio of outer radius to inner radius, K = b/a.
Then
σt
max
σt
min
 K 2 +1
= pi  2

 k −1 
 2 
= pi  2  or σ tmin = σ tmax − pi
 k −1 
The ratio of σ tmax to average tangential stress or tangential stress obtained by membrane equation is
shown in the table below
K = b/a = 1 + h/a
σt /σt
max
min
1.1
1.2
1.4
1.6
1.8
2.0
1.05
1.1
1.23
1.37
1.51
1.67
7
We can observe that for small ratios of thickness to inner radius, there is a little difference in tangential
stress. For instance wall thickness 20% of inner radius, maximum stress is only 10% larger.
From membrane equation for cylindrical shell σt is given by
a
pi a
pi
σt = pi h = b - a = K - 1
A modification for this equation for cylindrical shell appears in Sections I and VIII of the ASME
Code, for thickness range, h ≤ 0.5 Ri (inner radius).
8
Compound Cylinders (Laminated Cylinders)
From the sketch of the stress distribution (Figure 3) it is clear that there is a large variation in
tangential (hoop) stress across the wall of a cylinder subjected to internal pressure. The material of the
cylinder is not therefore used to its best advantage.
 To obtain a more uniform tangential stress distribution, cylinders are often built up by
shrinking one tube on to the outside of another.
 When the outer tube contracts on cooling, the inner tube is brought into state of compression
and the outer tube will conversely be brought into a state of tension.
 If this compound cylinder is now subjected to internal pressure the resultant tangential stress
will be algebraic sum of those resulting from shrinkage and those resulting from internal
pressure.
 The compound cylinder increases the load carrying capacity. The outer cylinder is usually
called hoop, jacket or shell. The inner cylinder is called cylinder or tube.
The compound cylinders are designed such that
•
•
•
The jacket has inside diameter slightly smaller than the outside diameter of the tube.
The difference in diameter at the “common” surface is normally termed the shrinkage or
interference allowance or simply interference.
Normally, the outer cylinder or the jacket is heated until it will freely slide over the tube thus
exerting a shrinkage pressure (ps ) on cooling
Fig. 5
9
Compound cylinder
The method of solution for compound cylinders constructed from similar materials is to break the
problem down into three separate effects:
a) Shrinkage pressure only on the inside cylinder;
b) Shrinkage pressure only on the outer cylinder;
c) Internal pressure only on the compound cylinder
For each of the resulting load conditions there are two known values of radial stress from which the
stresses can be determined in each case.
Condition (a): Consider inner cylinder subjected to shrinkage pressure only
σr = 0
σr = - ps
at r= a
at r = c
(Note: the internal cylinder is subjected to an external
pressure of ps)
The tangential and radial stresses for this condition can be obtained from the equations 21 and 22
σria
c2 ps  a2
= - c 2 - a 2  1 - r2 


(24)
σtia
c2 p s 
a2
= - c2 - a2 1 + r2 


(25)
Note: These are the stresses in the inside cylinder due to shrinkage pressure only
10
Condition (b): Consider outer cylinder (jacket) due to shrinkage pressure only
σr = 0
σr = - ps
at r= b
at r = c
(Note: the external cylinder is subjected to an
internal pressure of ps)
The tangential and radial stresses for this condition can be obtained from the equations 18 and 19
c2 ps  b2
= b2 - c2 1 - r2 


(26)
c2 ps 
b2
σtob = b2 - c2 1 + r2 


(27)
σrob
Note: These are the stresses in the outer cylinder due to shrinkage pressure only.
Condition (c): internal pressure acting on compound cylinder
σr = 0 at r= b
σr = - pi at r = a
the tangential and radial stresses for this condition can be obtained from the equations 18 and 19
a2 pi  b2
= b2 - a2 1 - r2 


(28)
a2 pi 
b2
σtc = b2 - a2 1 + r2 


(29)
σrc
11
Combining all these three cases the stresses in inner or outer cylinder can be obtained.
Stresses in inner cylinder (Total) due to internal pressure
a2 pi  b2
c2 ps  a2
σri = - c2 - a2 1 - r2  + b2 - a2 1 - r2 




c2 ps 
a2
a2 pi 
b2
σti = - c2 - a2 1 + r2  + b2 - a2 1 + r2 




(30)
(31)
Stresses in outer cylinder (total) due to internal pressure
σro
c2 ps  b2
a2 pi  b2
= b2 - c2 1 - r2  + b2 - a2 1 - r2 




(32)
σto
c2 ps 
b2
a2 pi 
b2
= b2 - c2 1 + r2  + b2 - a2 1 + r2 




(33)
Shrinkage or interference allowance
Consider a compound cylinder made up of two different materials.
Let the pressure set up at the junction of the cylinder owing to the shrink fit be pressure ps.
Let the tangential stresses set up on the inner and outer tubes resulting from the pressure ps be σti
(compressive) and σto (tensile) respectively at the common radius of these tubes ‘c’.
Let δo = radial shift of outer cylinder and δi = radial shift of inner cylinder at radius ‘c’
Since, circumferential strain = diametrical strain
Circumferential strain at radius ‘c’ on outer cylinder = δo/c = εto
Circumferential strain at radius ‘c’ on inner cylinder = δi/c = - εti
(negative since it is a decrease in diameter)
Total interference or shrinkage (I) = δo + δi = c(εto - εti)
(34)
Now assuming open ends, i.e. σa = 0,
σto ν1
εto = E - E (-ps)
1
1
since σro = - ps
12
(35)
σti ν2
εti = E - E (-ps) since σri = - ps
2
2
(36)
where E1, and ν1, E2 and ν2 are the elastic modulus and poisson’s ratio of the outer and inner
cylinders respectively.
Therefore total interference or shrinkage allowance (based on radius)
1

1
I=
r
 (σ to +ν 1 ps ) − (σ ti +ν 2 ps )  c
E2
 E1

(37)
Where ‘c’ is the initial nominal radius of the mating surfaces.
Note: σto and σti are evaluated at radius c of outer and inner cylinders respectively. σti being
compressive will change the negative sign to a positive one when its value is substituted.
Shrinkage allowance based on diameter will be twice this value.
Generally, however, if the tubes are of the same material
∴
E1 = E2 = E and ν1 = ν2 = ν
The values of σto and σti may be determined from the equations in terms of shrinkage or
interference allowance.
13
14
15
Press Fits
In a press fit, the shaft is compressed and the hub is expanded. Press fits, or interference fits, are
similar to pressurized cylinders in that the placement of an oversized shaft in an undersized hub
results in a radial pressure at the interface.
Characteristics of Press Fits
1)
The shaft is compressed and the hub is expanded.
2)
There are equal and opposite pressures at the mating surfaces.
3)
The relative amount of compression and expansion depends on the stiffness (elasticity
and geometry) of the two pieces.
4)
The sum of the compression and the expansion equals the interference introduced.
5)
The critical stress location is usually the inner diameter of the hub, where maximum
tensile hoop stress occurs.
2
2
2
2


−
−
b
c
c
a
)(
)
Eδ r (


ps =
2
2
2
c  2c ( b − a ) 


Where δr is the radial interference for hub
and shaft of the same material, with modulus of elasticity, E.
Where δr is the radial interference for hub and shaft of the same material, with modulus of
elasticity, E.
If the shaft is solid, a = 0 and ps is given by
16
Eδ r
ps
=
2c
 c2 
1 − b 2 


If the shaft and hub are of different materials
ps =
δr
 c  a2 + c2

c  b2 + c2
+
+
−
ν
ν
o
i
 c2 − a2
Eo  b 2 − c 2
E


i 
Tangential stress at the internal diameter of the hub is given by
b2 + c2
σ to = ps 2 2
b −c
Tangential stress at the outer diameter of the shaft is given by
a2 + c2
σ io = − ps 2 2
c −a
σ io = − ps if shaft is soilid
17