CET 1:

Stress Analysis &
Pressure Vessels

Lent Term 2005

Dr. Clemens Kaminski
Telephone: +44 1223 763135
E-mail:
URL:

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Synopsis
1

Introduction to Pressure Vessels and Failure Modes

2

3-D stress and strain

3

Thermal Effects

4

Torsion.

1.1
1.2
1.3
2.1
2.2
2.3
2.4

3.1
3.2
3.3
4.1
4.2
4.3

Stresses in Cylinders and Spheres
Compressive failure. Euler buckling. Vacuum vessels
Tensile failure. Stress Stress Concentration & Cracking

Elasticity and Strains-Young's Modulus and Poisson's Ratio
Bulk and Shear Moduli
Hoop, Longitudinal and Volumetric Strains
Strain Energy. Overfilling of Pressure Vessels
Coefficient of Thermal Expansion
Thermal Effect in cylindrical Pressure Vessels
Two-Material Structures

Shear Stresses in Shafts - τ/r = T/J = Gθ/L
Thin Walled Shafts
Thin Walled Pressure Vessel subject to Torque


5

Two Dimensional Stress Analysis

6

Bulk Failure Criteria

7

Two Dimensional Strain Analysis

5.1
5.2
5.3
5.4
6.1
6.2

7.1
7.2
7.3
7.4

Nomenclature and Sign Convention for Stresses
Mohr's Circle for Stresses
Worked Examples
Application of Mohr's Circle to Three Dimensional Systems
Tresca's Criterion. The Stress Hexagon

Von Mises' Failure Criterion. The Stress Ellipse

Direct and Shear Strains
Mohr's Circle for Strains
Measurement of Strain - Strain Gauges
Hooke’s Law for Shear Stresses


Supporting Materials
There is one Examples paper supporting these lectures.
Two good textbooks for further explanation, worked examples and exercises are
Mechanics of Materials (1997) Gere & Timoshenko, publ. ITP [ISBN 0-534-93429-3]
Mechanics of Solids (1989) Fenner, publ. Blackwell [ISBN 0-632-02018-0]
This material was taught in the CET I (Old Regulations) Structures lecture unit and was examined
in CET I (OR) Paper IV Section 1. There are consequently a large number of old Tripos questions
in existence, which are of the appropriate standard. From 1999 onwards the course was taught in
CET1, paper 5. Chapters 7 and 8 in Gere and Timoshenko contain a large number of example
problems and questions.

Nomenclature
The following symbols will be used as consistently as possible in the lectures.
E
G
I
J
R
t
T

Young’s modulus

Shear modulus
second moment of area
polar moment of area
radius
thickness

α
ε
γ
η
ν
σ
τ

thermal expansivity
linear strain
shear strain
angle
Poisson’s ratio
Normal stress
Shear stress

τορθυε


A pressure vessel near you!


Ongoing Example
We shall refer back to this example of a typical pressure vessel on several

occasions.
Distillation column

2m

P = 7 bara
carbon steel
t = 5 mm
18 m


1. Introduction to Pressure Vessels and Failure Modes
Pressure vessels are very often
• spherical (e.g. LPG storage tanks)
• cylindrical (e.g. liquid storage tanks)
• cylindrical shells with hemispherical ends (e.g. distillation columns)
Such vessels fail when the stress state somewhere in the wall material exceeds some failure criterion. It is thus important to be able to be able to understand and quantify
(resolve) stresses in solids. This unit will concentrate on the application of stress analysis to bulk failure in thin walled
vessels only, where (i) the vessel self weight can be neglected and (ii) the
thickness of the material is much smaller than the dimensions of the vessel
(D » t).

1.1. Stresses in Cylinders and Spheres
Consider a cylindrical pressure vessel
L

External diameter
D

internal gauge pressure P


r
L
h

wall thickness, t

The hydrostatic pressure causes stresses in three dimensions.
1.
Longitudinal stress (axial) σL
2.
Radial stress
σr
3.
Hoop stress
σh
all are normal stresses.
SAPV LT 2005
CFK, MRM

1


σ

r

σ

r

L

L
h

σ

h

a, The longitudinal stress σL
P

σ

L

Force equilibrium

π D2
P = π D t σL
4
if P > 0, then σ L is tensile
σL =

PD
4t

b, The hoop stress σh
P


σ

h

σ
P

h

P

Force balance, D L P = 2 σ h L t
σh =
SAPV LT 2005
CFK, MRM

2

PD
2t


c, Radial stress

σ
σ

r

varies from P on inner surface to 0 on the

outer face

r

σr ≈ o ( P )
D

σh , σL ≈ P (
).
2t
thin walled, so D >> t
so σ h , σ L >> σ r
so neglect σ r
Compare terms

d, The spherical pressure vessel

P

σ

h

SAPV LT 2005
CFK, MRM

π D2
P
= σh π D t
4

PD
σh =
4t

P

3


1.2. Compressive Failure: – Bulk Yielding & Buckling
– Vacuum Vessels
Consider an unpressurised cylindrical column subjected to a single load W.
Bulk failure will occur when the normal compressive stress exceeds a yield
criterion, e.g.

W

σ bulk =

W
= σY
πDt

Compressive stresses can cause failure due to buckling (bending instability).
The critical load for the onset of buckling is given by Euler's analysis. A full
explanation is given in the texts, and the basic results are summarised in the
Structures Tables. A column or strut of length L supported at one end will
buckle if

π 2 EI

W= 2
L
Consider a cylindrical column. I = πR3t so the compressive stress required
to cause buckling is

σ buckle

W
π 2 EπD3t 1
π 2 ED 2
=
=
⋅
=
2
πDt
8L
πDt
8L2

σ buckle

π2 E
=
2
8( L D)

or

SAPV LT 2005

CFK, MRM

4


where L/D is a slenderness ratio. The mode of failure thus depends on the
geometry:
σ
stress

Euler buckling locus
σ

y

Bulk yield

Short

Long
L /D ratio

SAPV LT 2005
CFK, MRM

5


Vacuum vessels.
Cylindrical pressure vessels subject to external pressure are subject to compressive hoop stresses


∆PD
2t
Consider a length L of vessel , the compressive hoop force is given by,
σh =

∆P D L
2
If this force is large enough it will cause buckling.
σh L t =

length

Treat the vessel as an encastered beam of length πD and breadth L

SAPV LT 2005
CFK, MRM

6


Buckling occurs when Force W given by.
4π 2 EI
W=
(π D) 2

I=

∆P D L 4π 2 EI
=

2
(π D )2

b t3 L t3
=
12
12

SAPV LT 2005
CFK, MRM

∆p buckle

7

=

2E ⎛ t ⎞
⎜ ⎟
3 ⎝D⎠

3


1.3. Tensile Failure: Stress Concentration & Cracking
Consider the rod in the Figure below subject to a tensile load. The stress distribution across the rod a long distance away from the change in cross section (XX) will be uniform, but near XX the stress distribution is complex.
D

X


X
d
W

There is a concentration of stress at the rod surface below XX and this value
should thus be considered when we consider failure mechanisms.
The ratio of the maximum local stress to the mean (or apparent) stress is described by a stress concentration factor K

K=

σ max
σ mean

The values of K for many geometries are available in the literature, including
that of cracks. The mechanism of fast fracture involves the concentration of
tensile stresses at a crack root, and gives the failure criterion for a crack of
length a

σ πa = Kc
where Kc is the material fracture toughness. Tensile
stresses can thus cause failure due to bulk yielding or due
to cracking.
SAPV LT 2005
CFK, MRM

8


σ crack =


Kc 1
⋅
√ π √a

stress

failure locus

length of crack. a

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CFK, MRM

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2.

3-D stress and strain

2.1. Elasticity and Yield
Many materials obey Hooke's law

σ = Eε

σ
E

ε


applied stress (Pa)
Young's modulus (Pa)
strain (-)

σ
failure
Yield
Stress

ε
Elastic
Limit

up to a limit, known as the yield stress (stress axis) or the elastic limit (strain
axis). Below these limits, deformation is reversible and the material eventually
returns to its original shape. Above these limits, the material behaviour depends
on its nature.
Consider a sample of material subjected to a tensile force F.
2
F

F

1
3

An increase in length (axis 1) will be accompanied by a decrease in dimensions
2 and 3.
Hooke's Law


ε1 = (σ1 ≡ F / A ) / E
10


The strain in the perpendicular directions 2,3 are given by

ε2 = −ν

σ1
E

;ε3 = − ν

σ1
E

where ν is the Poisson ratio for that material. These effects are additive, so for
three mutually perpendicular stresses σ1, σ2, σ3;
σ2
σ1
σ3

Giving

ε1 =

σ1
E

ε2 = −

ε3 = −

νσ 2

−

νσ1
E

νσ1
E

E

−

νσ 3
E

+

σ2

−

νσ 2

E

E


−

νσ 3
E

+

σ3
E

Values of the material constants in the Data Book give orders of magnitudes of
these parameters for different materials;

Material
Steel
Aluminum alloy
Brass

E
(x109 N/m2)
210
70
105

11

ν
0.30
0.33

0.35


2.2 Bulk and Shear Moduli
These material properties describe how a material responds to an applied stress
(bulk modulus, K) or shear (shear modulus, G).
The bulk modulus is defined as

Puniform = − Kε v

i.e. the volumetric strain resulting from the application of a uniform pressure. In
the case of a pressure causing expansion

so

σ1 = σ 2 = σ 3 = −P
−P
1
σ1 − νσ 2 − νσ 3 ] =
(1 − 2 ν )
[
E
E
−3P
ε v = ε1 + ε 2 + ε 3 =
(1− 2 ν )
E
E
K=
3(1 − 2 ν )


ε1 = ε 2 = ε 3 =

For steel, E = 210 kN/mm2, ν = 0.3, giving K = 175 kN/mm2
For water, K = 2.2 kN/mm2
For a perfect gas, K = P (1 bara, 10-4 kN/mm2)
Shear Modulus definition

τ = Gγ

γ - shear strain

12


2.3. Hoop, longitudinal and volumetric strains
(micro or millistrain)
Fractional increase in dimension:
ε L – length
ε h – circumference
ε rr – wall thickness
(a)

Cylindrical vessel:

Longitudinal strain

εL =

σL


-

υσ h

-

υσ L

E

E

υσ r

-

E

=

PD
δL
(1 - 2υ ) =
4tE
L

Hoop strain:

εh =


σn
E

E

=

δR
δD
PD
=
(2 - υ ) =
R
D
4tE

radial strain

εr =

δt
3PDυ
1
=
σ r - υσ h - υσ L ] = [
t
4ET
E


[fractional increase in wall thickness is negative!]

13


[ONGOING EXAMPLE]:

εL =

1
(σ L - υσ n )
E

[

=

1
60 x 10 6 - (0.3)120 x 10 6
9
210 x 10

=

1.14 x 10-4 -≡ 0.114 millistrain

ε h = 0.486 millistrain
ε r = -0.257 millistrain
Thus: pressurise the vessel to 6 bar: L and D increase: t decreases
Volume expansion


Cylindrical volume:

⎛ πD 2 ⎞
Vo = ⎜ o ⎟ Lo
⎝ 4 ⎠

New volume

V =

π

(Do
4

(original)
2

+ δD) (Lo + δL)

2

L π Do
2
1 + ε h ] [1 + ε L ]
= o
[
4
δV

Define volumetric strain ε v =
V
V - Vo
2
∴ εv =
= (1 + ε h ) (1 + ε L ) - 1
Vo

(

)

= 1 + 2ε h + ε h2 (1 + ε L ) - 1

ε v = 2 ε h + ε L + ε h2 + 2ε h ε L + ε L ε n2
Magnitude inspection:
14

]


ε max (steel) =

6

σy

190 x 10
−3
=

∴ small
9 = 0.905 x 10
210 x 10

E

Ignoring second order terms,
εv = 2ε h + ε L
(b)

Spherical volume:

εh =
so

1
[σ h - υσ L - υσ r ] = PD (1 - υ )
4Et
E

{

π (Do + δD )3 - Do3
εv =
6
πD 3o 6

}

= (1 + εh)3 – 1 ≈ 3εh + 0(ε2)

(c)

General result

εv = ε1 + ε2 + ε3
εii are the strains in any three mutually perpendicular directions.
εL = 0.114 mstrain

{Continued example} – cylinder

εn = 0.486 mstrain
εrr = -0.257 mstrain
εv = 2εn + ε L
∴ new volume = Vo (1 + εv)
Increase in volume =

π D2 L
4

ε v = 56.55 x 1.086 x 10-3
= 61 Litres

Volume of steelo = πDLt = 0.377 m3
εv for steel = εL + εh + εrr = 0.343 mstrains
increase in volume of steel

= 0.129 L

Strain energy – measure of work done


Consider an elastic material for which F = k x
15


Work done in expanding δx
dW = Fδx

F

A=area

work done

L0
x

Work done in extending to x1

2
kx1
1
x1
x1
w = ∫o Fdx = ∫o k x dx =
=
Fx
2
2 1 1

Sample subject to stress σ increased from 0 to σ1:

Extension Force:

x1 = Lo ε1 ⎫
AL oε 1σ1
⎬ W =
F1 = Aσ 1 ⎭
2

(no direction here)

Strain energy, U = work done per unit volume of material, U =
⇒ U =

Al o ⎛ 1 ⎞
⎟ ε σ
⎜
2 ⎝ Al o ⎠ 1 1

U =

ε1σ 1
2

ALo ε1σ 1
2(ALo )

σ 12

=


2E

1
[ε1σ1 + ε2σ2 + ε3σ3]
2
υσ 3
υσ 2
σ
etc
Now ε1 = 1 E
E
E

In a 3-D system, U =

So U =

[

]

1
σ 12 + σ 2 2 + σ 32 - 2ν (σ 1σ 2 + σ 2σ 3 + σ 3σ 1)
2E

Consider a uniform pressure applied: σ1 = σ2 = σ3 = P
2

2


3P
P
∴U =
(1 - 2 υ) =
2E
2K

energy stored in system (per unit vol.

16


For a given P, U stored is proportional to 1/K → so pressure test using liquids
rather than gases.
{Ongoing Example}

P – 6 barg

.

δV = 61 x 10-3 m3

increase in volume of pressure vessel

Increasing the pressure compresses the contents – normally test with water.
5

6 x 10
∆P
= - 0.273 mstrains

= −
∆V water? ε v (water ) = −
K
2.2 x 109
∴ decrease in volume of water = -Vo (0.273) = -15.4 x 10 –3 m3
Thus we can add more water:
Extra space = 61 + 15.4 (L)
= 76.4 L water
extra space

p=0

p=6

17


3. Thermal Effects
3.1. Coefficient of Thermal Expansion
ε = αL∆T Linear

Definition: coefficient of thermal expansion

Coefficient of thermal volume expansion εv = αT
Steel: αL = 11 x 10-6 K-1
reactor

Volume

∆T = 10oC


∴ εL = 11 10-5

∆T = 500oC

εL = 5.5 millistrains (!)

Consider a steel bar mounted between rigid supports which exert stress σ

Heat
σ

σ

ε = α∆T -

σ
E

If rigid: ε = 0 ⇒

so

σ = Eα∆T
(i.e., non buckling)

steel:

σ = 210 x 109 x 11 x 10-6 ∆T = 2.3 x 106 ∆T
σy = 190 MPa:


failure if ∆T > 82.6 K

18


{Example}: steam main, installed at 10oC, to contain 6 bar steam (140oC)
if ends are rigid, σ = 300 MPa→ failure.
∴ must install expansion bends.

3.2. Temperature effects in cylindrical pressure vessels

.
steel construction
L = 3 m . full of water t = 3 mm
D=1m
Initially un pressurised – full of water: increase temp. by ∆T: pressure rises
to Vessel P.

The Vessel

Wall stresses (tensile)

σL =

PD
= 83.3 P
4t

σn = 2σL = 166.7 P

19


Strain (volume)

εL =

υ

=

E
α

= 210 x 10 ⎬
= 11 x 10 −6 ⎪⎭

0.3

σ L νσ h
+ α l ∆T
E
E

⎫

9⎪

εL = 1.585 x 10-10 P + 11 x 10-6 ∆ T


Similarly
→

εh = 6.75 x 10-10 P + 11 10 –6 ∆T
εv = εL + 2εn = 15.08 x 10-10 P + 33 x 10-6 ∆T = vessel vol. Strain

vessel expands due to temp and pressure change.
The Contents, (water)

Expands

due to T

increase

Contracts

due to P

increase:

εv, H2O = αv∆T – P/K
∴

H2O = αv = 60 x 10-6 K-1

εv = 60 x 10-6 ∆T – 4.55 x 10-10 P

Since vessel remains full on increasing ∆T:
εv (H20) = εv (vessel)

Equating

→

P = 13750 ∆T

pressure, rise of 1.37 bar
Now

per 10°C increase in temp.

σn = 166.7 P = 2.29 x 106 ∆T
∆σn = 22.9 Mpa per 10°C rise in Temperature
20