THICK CYLINDERS
Intoduction
Cylinders are also call tanks or pressure vessels and they
are used to store fluids under pressure. The fluid being
stored may undergo a change of state inside the cylinder, as
in the case of steam boiler or combine with other reagents,
as in chemical plant.
The cylinders are designed with great care because rapture
of pressure means an explosion which may cause loss of life
and property. The material of the cylinder may be brittle
such as cast iron or ductile such as steel.
A cylinder may be considered as thin cylinder or thick
cylinder. Some of the differences between thin and thick
cylinders are given below.
Thin cylinder Thick cylinder
Wall thickness is smaller than Wall thickness is larger than
1/20 of tube or cylinder 1 / 20 of internal diameter
diameter.
Hoop and longitudinal stresses The hoop and longitudinal
are uniformly distributed all the stresses are not uniformly
way through the wall. distributed throughout the wall.
Thin cylinder is only subjected Thick cylinder is subjected
to the internal pressure to internal as well as external
pressure.
Stress is distributed uniform Stress is not distributed
throughout thickness of the uniformly throughout
cylinder thickness of the cylinder
Stress is constant throughout Maximum stress in inner side
cylinder and minimum stress in outer
side
Low stress consuming capacity More stress consuming capacity
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Difference in treatment between thin and thick
cylinders
The theoretical treatment of thin cylinders assumes the
following:
Stresses are uniformly distributed across the
thickness of the cylinder wall
In calculating the volumetric strain on the contents of a
thin cylinder under pressure, its radial stress is
assumed to be negligible in relation to the
circumferential and longitudinal stresses
There is no pressure gradient across the wall
Neither of these assumptions is justified for thick cylinders
for which hoop and radial stresses vary across the wall of
the cylinder and dependent on the radius of the section of
element under consideration.
If a thick cylinder subjected to internal pressure 𝑝, the hoop
stress (or circumferential stress), 𝜎𝐻 and radial stress, 𝜎𝑟 at
any point in the wall cross-section at radius 𝑟 are given by
expressions called Lame’ equations as follows:
Here, 𝐴 and 𝐵 are constants whose values depend on the
dimensions of the loading conditions.
The variations of these stresses across the thickness of the
cylinder wall is given as under.
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The hoop stress or circumferential stress acts in the
direction tangential to the circumference. In other words, it
is a tensile stress on longitudinal section (or on the cylinder
walls).
We see that the hoop stress is maximum at the inner surface
and minimum at the outer surface of the cylinder.
The radial stress is maximum at the inner surface and zero
at the outer surface of the cylinder.
Development of Lame’ equations
In developing Lame’ equations, we use the following
assumptions.
The material of the cylinder is homogeneous and
isotropic
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Plane cross-section perpendicular to the longitudinal
axis of the cylinder remain plane even after the
application of the internal pressure. This implies that:
- longitudinal strain, 𝜀𝐿 is constant across the wall
of the cylinder
- longitudinal stress, 𝜎𝐿 set up in the cylinder wall is
uniform across a cross-section of the cylinder
remote from the ends.
All longitudinal fibres of the material equally strained
and are free to expand or contract independently
without being confined by the adjacent fibers
Consider the thick cylinder shown in figure (a). The stresses
acting on an element of unit length at radius 𝑟 are as shown
in figure (b), the radial stress increasing from 𝜎𝑟 to 𝜎𝑟 + 𝑑𝜎𝑟
over the element thickness 𝑑𝑟 (all stresses are assumed
tensile).
Resolving forces on the element radially over a unit length
of cylinder, we have:
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𝑑𝜃 𝑑𝜃
For small angles (in radians), then sin ≈ . Therefore,
2 2
neglecting second-order small quantities,
Assuming now that plane sections remain plane, i.e. the
longitudinal strain 𝜀𝐿 is constant across the wall of the
cylinder, then we write:
Also, basing on the assumption that the longitudinal stress
𝜎𝐿 is constant across the cylinder walls at points remote
from the ends, we have:
From (i) and (ii), substitution yields
Multiplying all through by 𝑟 and rearranging,
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Therefore, integrating gives
And from (ii), we get
Equations (A) and (B) are known as Lame' Equations.
For any pressure condition there will always be two known
conditions of stress (usually radial stress) sufficient to solve
for the constants 𝐴 and 𝐵 and radial and circumferential
stresses or hoop stress at any radius 𝑟 can then be
evaluated.
Thick cylinder subjected to internal pressure only
Consider now the thick cylinder shown in the figure below
subjected to an internal pressure 𝑃, the external pressure
being zero.
It is important to note that the internal pressure is considered
as a negative radial stress since it will produce a radial
compression (i.e. thinning) of the cylinder walls and the
normal stress convention takes compression as negative.
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Thus, in the common case of the cylinder with internal
pressure only, the two known conditions of stress which
enable the Lame’ constants 𝐴 and 𝐵 to be determined are:
Substituting these in the Lame’ equation gives
Solving these simultaneously gives the constants 𝐴 and 𝐵
as
Accordingly,
Similarly,
The equations (3) and (4) yield the stress distributions given
earlier, with maximum values of both 𝜎𝑟 , and 𝜎𝐻 at the inside
radius.
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Longitudinal stress
Consider now the cross-section of a thick cylinder with
closed ends subjected to an internal pressure 𝑃1 and an
external pressure 𝑃2 , as shown under.
It is assumed in Lame’ Theory that the longitudinal stress,
𝜎𝐿 set up in the cylinder walls is uniform across a cross-
section and its value is obtained by considering the
equilibrium of the forces exerted on the end of the cylinder.
For horizontal equilibrium,
This equation reveals the following:
The longitudinal stress is a constant or uniform, as
expected.
It can be shown that the constant has the same value as
the constant 𝐴 of the Lame’ equations. This can be
verified for the internal pressure only, by
substituting 𝑃2 = 0.
Hence, for combined internal and external pressures, the
relationship 𝜎𝐿 = 𝐴 also applies.
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Maximum shear stress
In the absence of any external shearing force or torque,
𝜎𝑟 , 𝜎𝐻 and 𝜎𝐿 at any point in the cylinder wall are principal
stresses and hence the maximum shear stress at any point
is half the difference of the maximum and minimum principal
stresses.
Remembering that 𝜎𝑟 is compressive, while 𝜎𝐻 and 𝜎𝐻 are
usually tensile, the maximum shear stress at any point is
given by
By substitution,
From the equation, the greatest value of 𝜏𝑚𝑎𝑥 thus normally
occurs at the inside radius where 𝑟 = 𝑅1
Question
The internal and external diameters of a thick cylinder are
80 𝑚𝑚 and 120 𝑚𝑚 respectively. It is subjected to an external
pressure of 40 𝑁/𝑚2 and internal pressure of 120 𝑁/𝑚2 .
Calculate the;
(a) circumferential stress at the external and internal
surfaces
(b) radial and circumferential stresses at the mean radius
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Solution
(a) Information given:
𝑟 = 40 𝑚, 𝜎𝑟 = −120 𝑁/𝑚2
𝑟 = 60 𝑚, 𝜎𝑟 = −40 𝑁/𝑚2
𝐵
We now use Lame’ equation 𝜎𝑟 = 𝐴 − .
𝑟2
𝐵
−120 = 𝐴 − (1)
402
𝐵
−40 = 𝐴 − (2)
602
1 1
80 = 𝐵 ( 2 − ) ∴ 𝐵 = 230400
40 602
From (2),
230400
−40 = 𝐴 − ∴ 𝐴 = 24
602
We now find circumferential stress (hoop stress) from
𝐵
𝜎𝐻 = 𝐴 + at the two radii.
𝑟2
230400
𝑟 = 40 𝑚𝑚, 𝜎𝐻 = 24 + 2
= 168 𝑁/𝑚2
40
230400
𝑟 = 60 𝑚𝑚, 𝜎𝐻 = 24 + 2
= 88 𝑁/𝑚2
60
(b) The mean radias is (40 + 60)/2 = 50 𝑚𝑚. Hence we have:
230400
𝑟 = 50 𝑚𝑚, 𝜎𝑟 = 24 − 2
= −68.16 𝑁/𝑚2
50
230400
𝑟 = 50 𝑚𝑚, 𝜎𝐻 = 24 + 2
= −116.15 𝑁/𝑚2
50
Question
A thick cylinder of 100 𝑚𝑚 internal radius and 150 𝑚𝑚
external radius is subjected to an internal pressure of
60 𝑀𝑁/𝑚2 and an external pressure of 30 𝑀𝑁/𝑚2 . Determine
the hoop and radial stresses at the inside and outside of the
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cylinder together with the longitudinal stress if the cylinder
is assumed to have closed ends.
Solution
The internal and external pressures both have the effect of
decreasing the thickness of the cylinder; the radial stresses
at both the inside and outside radii are thus compressive,
i.e. negative, as shown above.
Now,
Substitution these in Lame’ equations gives
Solving the simultaneous equations gives the values of
Lame’ constants 𝐴 and 𝐵.
Accordingly,
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The longitudinal stress is calculated as under.
Change of cylinder dimensions
(a) Change of diameter
From thin cylinder, we know that:
change of diameter
diametral strain =
original diameter
change of diameter = diametral strain × original diameter
The diametric strain on a cylinder equals the hoop or
circumferential strain. Hence,
change of diameter = hoop strain × original diameter
With the principal stress system of hoop, radial and
longitudinal stresses, all assumed tensile, the
circumferential strain is given by
Thus, the change of diameter at any radius 𝑟 of the
cylinder is given by
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(b) Change of length
Similarly, the change of length of the cylinder is given by
Question
An external pressure of 10 𝑀𝑁/𝑚2 is applied to a thick
cylinder of internal diameter 160 𝑚𝑚 and external
diameter 320 𝑚𝑚.
(a) If the maximum stress permitted on the inside wall of
the cylinder is limited to 30 𝑀𝑁/𝑚2 , what maximum
internal pressure can be applied assuming the cylinder
has closed ends?
(b) What will be the change in outside diameter when this
pressure is applied?
𝐸 = 207 𝐺𝑁/𝑚2 , 𝑣 = 0.29.
Solution
(a) From the question, corresponding values are:
We now substitute the corresponding known values in
Lame’ equations and solve the resulting equations
simultaneously to determine Lame’ constants as under.
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At 𝑟 = 0.08 𝑚, we find allowable maximum internal pressure
𝑃 accordingly.
(b) We know that change of diameter is given by
We now first find the values of hoop stress and
longitudinal stress as follows:
Hence, by substitution, we finally have
Compound cylinders
From the variation of stresses across the thickness of the
thick cylinder wall, it is evident that there is a large variation
in hoop stress across the wall of a cylinder subjected to
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internal pressure. The material of the cylinder is not
therefore used to its best advantage.
To obtain a more uniform hoop stress distribution, the outer
cylinder is shrunk fit over the inner cylinder by heating and
cooling. The combination of the two cylinders becomes
concentric cylinders, are is called a compound cylinder.
Both longitudinal-section and cross-section of a compound
cylinder are shown below.
The following should be noted about the compound
cylinder.
(a) On cooling, the contact pressure is developed at the
junction of the two cylinders, which induces compressive
tangential stress in the material of the inner cylinder and
tensile tangential stress in the material of the outer
cylinder.
(b) When the cylinder is loaded, the compressive stresses are
first relieved and then tensile stresses are induced. Thus,
a compound cylinder is effective in resisting higher
internal pressure than a single cylinder with the same
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overall dimensions. The principle of compound cylinder is
used in the design of gun tubes.
(c)If the compound cylinder is subjected to internal pressure,
the stress distribution in the inner and outer cylinders are
as shown below.
The resultant hoop stresses will be the algebraic sum of
those resulting from internal pressure and those resulting
from shrinkage as drawn in the figure below.
Method of solution for compound cylinder
A compound cylinder can be constructed from same
materials (homogenous) or different materials. In either of
the cases, the method of solution will be different. We
consider only same materials here.
The method of solution for compound cylinders constructed
from similar materials is to break the problem down into
three separate effects as shown in the figure under.
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For each of the resulting load or boundary conditions, there
are two known values of radial stress which enable the Lame'
constants to be determined in each case. These conditions
are given under, in each case.
Shrinkage pressure only on the inside cylinder (internal
cylinder);
Shrinkage pressure only on the outside cylinder;
Internal pressure only on the complete cylinder;
Question
A compound cylinder is formed by shrinking a tube of
250 mm internal diameter and 25 mm wall thickness onto
another tube of 250 mm external diameter and 25 mm wall
thickness, both tubes being made of the same material. The
stress set up at the junction owing to shrinkage is 10 MN/m2 .
The compound tube is subjected to an internal pressure of
80 MN/m2 . If the compound cylinder acts as a single cylinder
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of 300 mm external diameter and 50 mm thickness subjected
to the same pressure, determine the:
(i) Hoop stresses in the outer and inner tubes due to
shrinkage only
(ii) Hoop stresses in the complete cylinder due to internal
pressure only
(iii) resultant Hoop stresses due to combined shrinkage and
internal pressure.
Solution (i)
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Solution (ii)
Solution (iii)
Consider Hoop stresses due to combined shrinkage and
internal pressure.
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