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1
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 r would develop at
the
interface between rings located at a radial distance r. * A
slightly different radial stress (r + dr) 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 r and t can be derived by considering the
equilibrium of a small
element (Fig. 1b) * Axial stress a, that may be present is
omitted, since it does not contribute to equilibrium of the
element in radial and tangential directions.
Fig 1a
Fig 1b
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2
Considering the equilibrium in radial direction i.e.. the sum of
all the force in radial direction is zero 0rF =
( )( ) ( )2 sin / 2 0r r r td r dr d dl rd dl drdl d + + = (1)
By neglecting higher order terms and noting ( )sin / 2 / 2d d = The
equation (a) can be reduced to
0r r tdrdr
+ = (2)
The equation (2) can be integrated since r and t are functions
of radial position r. For the case of thick cylinders, the axial
strain a at any point in the wall can be expressed as
( )r tE
aa
+= (3)
This means plane transverse section before remain plane after
loading. So far as the axial stress a 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 (a =
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 a, a,
E are constant, therefore
( )r t 12a aE C
+ = = (4)
Where C1 is a constant Substituting t into differential
equation, we obtain
12 2r rdr Cdr
+ = (5)
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3
( )2 12rd r C rdr = (6) Integrating this equation further
yields
2 21 2rr C r C = + (7) Where C2 is a constant of integration.
Thus
21 2rCCr
= +
(8)
221 2tCC rr
=
Values of the constants C1 and C2 can be obtained from the known
values of internal and external pressures.
at r = a at r = b
r i
r o
pp
=
=
(- sign indicates that the pressure produces compressive radial
stresses). Using these boundary conditions, we can solve for C1 and
C2 as
2 2
1 2 2i oa p b pC
b a
=
(9)
( )2 2
2 2 2i oa b p pC
b a
=
Then the radial and tangential stresses can be obtained as
( )2 22 22 2 2 2 2
1i oi or
a b p pa p b pb a b a r
= (10)
( )2 22 2
2 2 2 2 2
1i oi ot
a b p pa p b pb a b a r
= + (11)
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4
Radial and circumferential deformations play important roles in
press-fit and shrink-fit problems. Change in circumference c 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
( ){ }2 2 2c r dr r r = + = ; where r is change in radius of the
ring Circumferential deformation can also be expressed in terms of
circumferential strain 2c t r =
r tr = (12) For most applications a = 0. Tangential strain t can
be expressed in terms of radial and tangential stresses by Hookes
law as:
( )t rt E
= (13)
Thus the radial displacement (or change in radius) at any radius
r is given by
( )t r rrE
= (14)
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
( )2 22 22 2 2 2
1 1 1i oi o a b p pa p b pr rE b a E b a r
+ = + (15)
The maximum numerical value of r is found at r = a to be pi,
provided that pi exceeds po. If po > pi, the maximum r occurs at
r = b and equals po. On the other hand, the maximum t 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
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5
( ) ( )2 2
max 2 2 2
12
t r i oa b p pb a r
= =
(16)
The largest value of max is found at r = a, the inner surface.
The effect of reducing po is clearly to increase max. Consequently,
the greatest max corresponds to r= a and po =0 is given by
( )
2
max 2 2ip b
b a =
(17)
Since r and t are principal stresses, max occurs on planes
making an angle of 45o with the plane on which r and t act. Special
cases Internal Pressure only: If only internal pressure acts
equations for stresses and change in radius reduce to
r = a2 pi
b2 - a2
1 -
b2
r2 (18)
t = a2 pi
b2 - a2
1 +
b2
r2 (19)
r = a2 pi r
E (b2 - a2)
(1 - ) + (1 + )
b2
r2 (20)
Since b2
r2 1, r is negative (compressive) for all r except r = b,
in which case r = 0. The maximum stress occurs at r = a. t, it
is positive (tensile) for all radii, and also has maximum at r = a.
This is illustrated in the figure 3.
Fig. 2
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6
External Pressure only: In this case, pi = 0, and the equations
are given as
r = - b2 po
b2 - a2
1 -
a2
r2 (21)
t = - b2 po
b2 - a2
1 +
a2
r2 (22)
r = b2 po r
E (b2 - a2)
(1 - ) + (1 + )
a2
r2 (23)
Figure 4
The maximum radial stress occurs at r = b and is compressive for
all r. The maximum t 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 l 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:
( )2 2 2 2i o lp a p b b a =
Fig. 4
Fig. 3
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The resulting expression for longitudinal stress, applicable
only away from the ends, is
l = 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
r = a2 pi
b2 - a2
1 -
b2
r2
t = a2 pi
b2 - a2
1 +
b2
r2
Both tangential and radial stresses are maximum, at inner
surface
max
2 2
2 2t ia bpb a
+
=
Let K be the ratio of outer radius to inner radius, K = b/a.
Then max
2
2
11t i
Kpk
+
=
min 2
21t i
pk
= or
mint =
maxt ip
The ratio of
maxt to average tangential stress or tangential stress obtained
by membrane equation is
shown in the table below
K = b/a = 1 + h/a
1.1 1.2 1.4 1.6 1.8 2.0
max min/t t 1.05 1.1 1.23 1.37 1.51 1.67
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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 t is given
by
t = pi ah =
pi ab - a =
piK - 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).
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9
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
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10
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
r = 0 at r= a r = - ps 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
ria = - c2 ps
c2 - a2
1 - a2r2 (24)
tia = - c2 ps
c2 - a2
1 + a2r2 (25)
Note: These are the stresses in the inside cylinder due to
shrinkage pressure only
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11
Condition (b): Consider outer cylinder (jacket) due to shrinkage
pressure only
r = 0 at r= b r = - ps 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
rob = c2 ps
b2 - c2
1 -
b2
r2 (26)
tob = c2 ps
b2 - c2
1 +
b2
r2 (27)
Note: These are the stresses in the outer cylinder due to
shrinkage pressure only. Condition (c): internal pressure acting on
compound cylinder
r = 0 at r= b r = - pi at r = a
the tangential and radial stresses for this condition can be
obtained from the equations 18 and 19
rc = a2 pi
b2 - a2
1 -
b2
r2 (28)
tc = a2 pi
b2 - a2
1 +
b2
r2 (29)
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Combining all these three cases the stresses in inner or outer
cylinder can be obtained. Stresses in inner cylinder (Total) due to
internal pressure
ri = - c2 ps
c2 - a2
1 -
a2
r2 + a2 pi
b2 - a2
1 -
b2
r2 (30)
ti = - c2 ps
c2 - a2
1 +
a2
r2 + a2 pi
b2 - a2
1 +
b2
r2 (31)
Stresses in outer cylinder (total) due to internal pressure
ro = c2 ps
b2 - c2
1 -
b2
r2 + a2 pi
b2 - a2
1 -
b2
r2 (32)
to = c2 ps
b2 - c2
1 +
b2
r2 + a2 pi
b2 - a2
1 +
b2
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 ti (compressive) and to (tensile)
respectively at the common radius of these tubes c. Let o = radial
shift of outer cylinder and i = radial shift of inner cylinder at
radius c Since, circumferential strain = diametrical strain
Circumferential strain at radius c on outer cylinder = o/c = to
Circumferential strain at radius c on inner cylinder = i/c = - ti
(negative since it is a decrease in diameter) Total interference or
shrinkage (I) = o + i = c(to - ti) (34) Now assuming open ends,
i.e. a = 0,
to = toE1 -
1E1 (-ps) since ro = - ps (35)
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ti = tiE2 -
2E2 (-ps) since ri = - ps (36)
where E1, and 1, E2 and 2 are the elastic modulus and poissons
ratio of the outer and inner cylinders respectively.
Therefore total interference or shrinkage allowance (based on
radius)
( ) ( )1 21 2
1 1r to s ti sI p p cE E
= + +
(37)
Where c is the initial nominal radius of the mating
surfaces.
Note: to and ti are evaluated at radius c of outer and inner
cylinders respectively. ti 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 1 = 2 =
The values of to and ti may be determined from the equations in
terms of shrinkage or interference allowance.
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15
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16
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
2 2 22r
s
b c c aEpc c b a
=
Where r is the radial interference for hub and shaft of the same
material, with modulus of elasticity, E. Where r 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
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17
2
212r
sE cp
c b
=
If the shaft and hub are of different materials
2 2 2 2
2 2 2 2
rs
o io i
pc b c c a cE b c E c a
=
+ ++ +
Tangential stress at the internal diameter of the hub is given
by
2 2
2 2to sb cpb c
+=
Tangential stress at the outer diameter of the shaft is given
by
2 2
2 2io sa cpc a
+=
if shaft is soilidio sp =
Fig 1bSpecial casesFig. 3Fig. 5