Thick cylinders- Stresses due to internal

and external pressures.

Instructional Objectives:
At the end of this lesson, the students should have the knowledge of:

• Stresses in thick cylinders.

• Lame’s equation for radial and circumferential stresses.
• Distribution of radial and circumferential stresses for different boundary conditions.
• Methods of increasing elastic strength of thick cylinders by prestressing.

9.2.1 Stresses in thick cylinders

For thick cylinders such as guns, pipes to hydraulic presses, high pressure
hydraulic pipes the wall thickness is relatively large and the stress variation across the
thickness is also significant. In this situation the approach made in the previous section is
not suitable. The problem may be solved by considering an axisymmetry about z-axis and
solving the differential equations of stress equilibrium in polar co-ordinates. In general
the stress equations of equilibrium without body forces can be given as
∂σr 1 ∂τrθ ∂τrz σr − σθ
+ + + =0
∂r r ∂θ ∂z r
∂τθr 1 ∂τθ ∂τθz τ
+ + + 2 θr = 0 (1)
∂r r ∂θ ∂z r
∂τzr 1 ∂τzθ ∂σz τzr
+ + + =0
∂r r ∂θ ∂z r
∂
For axisymmetry about z-axis = 0 and this gives
∂θ
∂σr ∂τrz σr − σθ
+ + =0
∂r ∂z r
∂τθr ∂τθz τ
+ + 2 θr = 0 (2)
∂r ∂z r
∂τzr ∂σz τzr
+ + =0
∂r ∂z r
In a plane stress situation if the cylinder ends are free to expand σz = 0 and due to
uniform radial deformation and symmetry τrz = τθz = τrθ = 0. The equation of equilibrium
reduces to

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 ∂σr σr − σθ
+ =0
∂r r
This can be written in the following form:
∂σr
r + σ r = σθ (3)
∂r

If we consider a general case with body forces such as centrifugal forces in the case of a
rotating cylinder or disc then the equations reduce to
∂σr σr − σθ
+ + ρω2 r = 0 which may be written as
∂r r
∂σr
r − σθ + σ r + ρω2 r 2 = 0 (4)
∂r
It is convenient to solve the general equation so that a variety of problems may be solved.
Now as shown in figure- 9.2.1.1, the strains εr and εθ may be given by
∂u r 1
εr = = [ σr − νσθ ] since σ z = 0 (5)
∂r E
( r + u r ) Δθ − rΔθ = u r 1
εθ = = [σθ − νσr ] (6)
rΔθ r E
∂u r
ur + δr
ur ∂r

' '
A A B B
'
A
A

Δθ
r

B
'
B
9.2.1.1F- Representation of radial and circumferential strain.

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Combining equation (5) and (6) we have
∂σθ ∂σ
r − νr r + (1 + ν ) ( σθ − σ r ) = 0 (7)
∂r ∂r
Now from equation (4) we may write
∂σθ ∂ 2σ ∂σ
= r 2 r + 2 r + 2ρω2 r and combining this with equation (7) we
∂r ∂r ∂r
may arrive at
∂ 2σr ∂σ
r + 3 r + ( 3 + ν ) ρω2 r = 0 (8)
∂r 2
∂r
For a non-rotating thick cylinder with internal and external pressures pi and po we
substitute ω = 0 in equation (8) and this gives

∂ 2σr ∂σ
r 2 +3 r = 0 (9)
∂r ∂r
A typical case is shown in figure- 9.2.1.2. A standard solution for equation (9) is
σr = c rn where c and n are constants. Substituting this in equation (9) and also
combining with equation (3) we have
c2
σr = c1 +
r2 (10)
c
σθ = c1 − 22
r
where c1 and c2 are constants.

ro
ri

pi po

9.2.1.2F- A thick cylinder with both external and internal pressure.

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Boundary conditions for a thick cylinder with internal and external pressures pi and po
respectively are:
at r = ri σr = -pi
and at r = ro σr = -po
The negative signs appear due to the compressive nature of the pressures. This gives
2 2 2 2
pi ri − po ro ri ro ( po − pi )
c1 = 2 2
c2 = 2 2
ro − ri ro − ri

The radial stress σr and circumferential stress σθ are now given by

2 2 2 2
pi ri − p o ro ri ro ( po − pi ) 1
σr = 2 2
+ 2 2 2
ro − ri ro − ri r
(11)
2 2 2 2
pi ri − po ro ri ro ( po − pi ) 1
σθ = 2 2
− 2 2 2
ro − ri ro − ri r

It is important to remember that if σθ works out to be positive, it is tensile and if it is

negative, it is compressive whereas σr is always compressive irrespective of its sign.
Stress distributions for different conditions may be obtained by simply substituting the
relevant values in equation (11). For example, if po = 0 i.e. there is no external pressure
the radial and circumferential stress reduce to

pi ri⎛ r22 ⎞
σ r = 2 2 − 2 + 1⎟
⎜ o

ro − ri ⎜⎝ r ⎟
⎠
(12)
2 ⎛ 2 ⎞
pr r
σθ = 2 i i 2 ⎜ o2 + 1⎟
ro − ri ⎜⎝ r ⎟
⎠
The stress distribution within the cylinder wall is shown in figure- 9.2.1.3.

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 ro
ri σθ

σr pi

9.2.1.3F- Radial and circumferential stress distribution within the cylinder

wall when only internal pressure acts.

It may be noted that σr + σθ = constant and hence the deformation in z-direction is

uniform. This means that the cross-section perpendicular to the cylinder axis remains
plane. Hence the deformation in an element cut out by two adjacent cross-sections does
not interfere with the adjacent element. Therefore it is justified to assume a condition of
plane stress for an element in section 9.2.1.
If pi = 0 i.e. there is no internal pressure the stresses σr and σθ reduce to

po ro ⎛ ri ⎞
2 2

σ r = 2 2 2 − 1⎟
⎜
ro − ri ⎜⎝ r ⎟
⎠
(13)
po ro ⎛ ri ⎞
2 2

σ θ = − 2 2 ⎜ 2 + 1⎟
ro − ri ⎜⎝ r ⎟
⎠
The stress distributions are shown in figure-9.2.1.4.

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 po

ro
ri

σr
(negative)
σθ (negative)

9.2.1.4F- Distribution of radial and circumferential stresses within the

cylinder wall when only external pressure acts.

9.2.2 Methods of increasing the elastic strength of a thick cylinder

by pre-stressing

In thick walled cylinders subjected to internal pressure only it can be seen from equation
(12) that the maximum stresses occur at the inside radius and this can be given by

2 2
σ r ( max ) = − pi ro + ri
r = ri
σθ(max) = pi 2 2
r = ri
ro − ri

This means that as pi increases σθ may exceed yield stress even when pi < σyield.
Furthermore, it can be shown that for large internal pressures in thick walled cylinders
the wall thickness is required to be very large. This is shown schematically in figure-
9.2.2.1. This means that the material near the outer edge is not effectively used since the
stresses near the outer edge gradually reduce (Refer to figure- 9.2.1.3).

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