CET 1:
Stress Analysis &
Pressure Vessels
Lent Term 2005
Dr. Clemens Kaminski
Telephone: +44 1223 763135
E-mail: clemens_kaminski@cheng.cam.ac.uk
URL:
http://www.cheng.cam.ac.uk/research/groups/laser/
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
SAPV LT 2005
CFK, MRM
9
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
∴
Failure does not need a large temperature increase.
Very large stress changes due to temperature fluctuations.
MORAL: Always leave a space in a liquid vessel.
(εv, gas = αv∆T – P/K)
21
3.3. Two material structures
Beware, different materials with different thermal expansivities
can cause difficulties.
{Example} Where there is benefit. The Bimetallic strip -
temperature
controllers
a=
4 mm (2 + 2 mm)
b=
10 mm
a
d
Cu
Fe
L = 100mm
Heat by ∆T:
b
Cu expands more than Fe so the strip will bend: it will
bend in an arc as all sections are identical.
22
Cu
Fe
The different thermal expansions, set up shearing forces
in the strip, which create a bending moment. If we apply a sagging bending
moment of equal [: opposite] magnitude, we will obtain a straight beam and
F
Cu
Fe
F
can then calculate the shearing forces [and hence the BM].
Shearing force F
compressive in Cu
Tensile
in Fe
F
F
= α Fe ∆T +
bdE cu
bdE Fe
Equating strains:
α cu ∆T -
So
1 ⎫
F ⎧ 1
+
⎨
⎬ = (α cu - α Fe ) ∆T
bd ⎩ E cu E Fe ⎭
23
bd = 2 x 10-5 m2
Ecu = 109 GPa
αcu 17 x 10-6 k-1
∆T = 30°C
EFe = 210 GPa
αFe = 11 x 10-6 K-1
F = 387 N
(significant force)
F acts through the centroid of each section so BM = F./(d/2) = 0.774 Nm
Use data book to work out deflection.
ML2
δ=
2 EI
This is the principle of the bimetallic strip.
24
Consider a steel rod mounted in a upper tube – spacer
Analysis – relevant to Heat Exchangers
Cu
Fe
assembled at room temperature
.
increase ∆T
Data: αcu > αFe
:
copper expands more than steel, so will generate
a TENSILE stress in the steel and a compressive
stress in the copper.
Balance forces:
Tensile force in steel
|FFe| = |Fcu| = F
Stress in steel
= F/AFe = σFe
“
= F/Acu = σcu
“ copper
Steel strain:
εFE = αFe ∆T + σFe/EFe
= αFe∆T + F/EFeAFe
copper strain
εcu = αcu∆T – F/EcuAcu
25
(no transvere forces)
Strains EQUAL:
⎡ 1
1 ⎤
⇒ F⎢
+
=
Acu Ecu ⎥⎦
Fe E Fe
⎣1A4
442444
3
(α cu - α Fe ) ∆T
∆d
sum of strengths
So you can work out stresses and strains in a system.
26
4. Torsion – Twisting – Shear stresses
4.1. Shear stresses in shafts –τ/r = T/J = G θ/L
Consider a rod subject to twisting:
Definition : shear strain γ ≡ change in angle that was originally Π/2
Consider three points that define a right angle and more then:
Shear strain
γ = γ1 + γ2
[RADIANS]
γ1
A
B
Hooke’s Law
C
τ=Gγ
B
A
γ
2 C
G – shear modulus =
27
E
2(1 + υ)
Now consider a rod subject to an applied torque, T.
T
2r
Hold one end and rotate other by angle θ
.
B
B
θ
γ
B
B
L
Plane ABO was originally to the X-X axis
Plane ABO is now inclined at angle γ to the axis: tan γ ≈ γ =
Shear stress involved τ = Gγ =
Grθ
L
28
rθ
L
Torque required to cause twisting:
τ
τ
r
dr
.δT = τ 2Π r.δr r
T = ∫ τ 2Π r 2 dr
or
A
T
cf
∫ τ r.dA
A
=
Gθ
L
=
Gθ
{J}
L
∫r
2
dA
so
T Gθ τ
=
=
J
L
r
M E σ
= =
I R y
DEFN: J ≡ polar second moment of area
29
Grθ ⎫
⎧
⎨τ =
⎬
⎩
L ⎭
Consider a rod of circular section:
R
π
o
2
J = ∫ 2π .r r 2 dr =
R4
y
r
x
J=
πD 4
32
Now
r2 = x2 + y2
It can be shown that J = Ixx + Iyy
[perpendicular axis than]→ see Fenner
this gives an easy way to evaluate Ixx or Iyy in symmetrical geometrics:
Ixx = Iyy = πD4/64 (rod)
30
Rectangular rod:
d
b
I xx
I yy
bd 3 ⎫
=
⎪
12 ⎪
⎬
3⎪
db
=
⎪
12 ⎭
J=
[
bd 2
b +d2
12
31
]
Example: steel rod as a drive shaft
D = 25 mm
L = 1.5 m
Failure when τ = τy = 95 MPa
G = 81 GPa
τ max
rmax
Now
J=
πD 4
32
T 95 x 10 6 ⎫
= =
⎪
J
0.0125 ⎪
⎪
⎬ so T = 291 Nm
= 383 x 10 −8 m 4 ⎪
⎪
⎪⎭
(
)
Gθ T
81 x 109 θ
9
= ⇒ 7.6 x 10 =
⇒
From
L
J
1.5
Say shaft rotates at 1450 rpm: power
32
θ = 0.141 rad = 8.1°
=
Tω
=
291 x
=
45 kW
2π
x 1450
60
4.2. Thin walled shafts
(same Eqns apply)
Consider a bracket joining two Ex. Shafts:
T = 291
Nm
D
min
0.025m
What is the minimum value of D for connector?
rmax = D/2
J = (π/32){D4 – 0.0254}
τy
rmax
6
32
⎛ T ⎞ 95 x 10 x 2 291
=⎜ ⎟⇒
=
4
D
π D - 0.0254
⎝J⎠
D4 – 0.0254 = 6.24 x 10-5 D
(
)
D ≥ 4.15 cm
33
4.3. Thin walled pressure vessel subject to torque
τ T
=
r J
now cylinder
J=
=
≈
so
τ2
D
τ=
=
[(D + 2t )
32
π
4
- D4
[8D t + 24 D t
32
π
π
4
3
D 3t
4T
πD 3 t
2T
πD 2 t
34
]
2 2
]
+ ...
CET 1, SAPV
5. Components of Stress/ Mohr’s Circle
5.1 Definitions
Scalars
tensor of rank 0
Vectors
tensors of rank 1
r
r
F = ma
hence :
F1 = ma1
F2 = ma2
F3 = ma3
or :
Fi = mai
Tensors of rank 2
3
pi = ∑ Tij q j
i, j = 1,2,3
j =1
or :
p1 = T11q1 + T12 q2 + T13 q3
p2 = T21q1 + T22 q2 + T23 q3
p3 = T31q1 + T32 q2 + T33 q3
Axis transformations
The choice of axes in the description of an engineering problem is
arbitrary (as long as you choose orthogonal sets of axes!). Obviously the
physics of the problem must not depend on the choice of axis. For
example, whether a pressure vessel will explode can not depend on how
we set up our co-ordinate axes to describe the stresses acting on the
34
CET 1, SAPV
vessel. However it is clear that the components of the stress tensor will
be different going from one set of coordinates xi to another xi’.
How do we transform one set of co-ordinate axes onto another, keeping
the same origin?
x1
x2
x3
a11
a12
a13
x2 ' a21
x3 ' a31
a22
a32
a23
a33
x1 '
... where aij are the direction cosines
Forward transformation:
3
xi ' = ∑ aij x j
“New in terms of Old”
j =1
Reverse transformation:
3
xi = ∑ a ji x j
“Old in terms of New”
j =1
We always have to do summations in co-ordinate transformation and it is
conventional to drop the summation signs and therefore these equations
are simply written as:
xi ' = aij x j
xi = a ji x j
35
CET 1, SAPV
Tensor transformation
How will the components of a tensor change when we go from one coordinate system to another? I.e. if we have a situation where
pi = ∑ Tij q j = Tij q j (in short form)
j
where Tij is the tensor in the old co-ordinate frame xi, how do we find the
corresponding tensor Tij’ in the new co-ordinate frame xi’, such that:
pi ' = ∑ Tij ' q j ' = Tij ' q j ' (in short form)
j
We can find this from a series of sequential co-ordinate transformations:
p' ← p ← q ← q'
Hence:
pi ' = aik pk
pk = Tkl ql
ql = a jl q j '
Thus we have:
36
CET 1, SAPV
pi ' = aik Tkl a jl q j '
= aik a jl Tkl q j '
= Tij ' q j '
For example:
Tij ' = ai1 a jl T1l
+ ai 2 a jl T2l
+ ai 3 a jl T3l
= ai1 a j1 T11 + ai1 a j 2 T12 + ai1 a j 3 T13
+ ai 2 a j1 T21 + ai 2 a j 2 T22 + ai 2 a j 3 T23
+ ai 3 a j1 T31 + ai 3 a j 2 T32 + ai 3 a j 3 T33
Note that there is a difference between a transformation matrix and a 2nd
rank tensor: They are both matrices containing 9 elements (constants)
but:
Symmetrical Tensors:
Tij=Tji
37
CET 1, SAPV
38
CET 1, SAPV
We can always transform a second rank tensor which is symmetrical:
Tij
→
Tij '
such that :
⎡T1 0
Tij ' = ⎢ 0 T2
⎢
⎢⎣ 0 0
0⎤
0⎥
⎥
T3 ⎥⎦
Consequence? Consider:
pi = Tij q j
then
p1 = T1 q1 ,
p2 = T2 q2 , p3 = T3 q3
The diagonal T1, T2, T3 is called the PRINCIPAL AXIS.
If T1, T2, T3 are stresses, then these are called PRINCIPAL STRESSES.
39
CET 1, SAPV
Mohr’s circle
Consider an elementary cuboid with edges parallel to the coordinate
directions x,y,z.
y
Fxy
y face
Fx
x
z face
Fxz
Fxx
z
x face
The faces on this cuboid are named according to the directions of their
normals.
There are thus two x-faces, one facing greater values of x, as shown in
Figure 1 and one facing lesser values of x (not shown in the Figure).
On the x-face there will be some force Fx. Since the cuboid is of
infinitesimal size, the force on the opposite side will not differ
significantly.
The force Fx can be divided into its components parallel to the coordinate
directions, Fxx, Fxy, Fxz. Dividing by the area of the x-face gives the
stresses on the x-plane:
σ xx
τ xy
τ xz
It is traditional to write normal stresses as σ and shear stresses as τ.
Similarly, on the y-face:
τ yx , σ yy , τ yz
40
CET 1, SAPV
and on the z-face we have:
τ zx , τ zy , σ zz
There are therefore 9 components of stress;
⎡σ xx τ xy
⎢
σ ij = ⎢τ yx σ yy
⎢ τ zx τ zy
⎣
τ xz ⎤
⎥
τ yz ⎥
σ zz ⎥⎦
Note that the first subscript refers to the face on which the stress acts and
the second subscript refers to the direction in which the associated force
acts.
σ yy
y
x
τyx
τ xy
σ xx
σ xx
τ xy
τ yx
σ yy
But for non accelerating bodies (or infinitesimally small cuboids):
and therefore:
⎡σ xx τ xy τ xz ⎤ ⎡σ xx τ xy τ xz ⎤
⎢
⎥ ⎢
⎥
σ ij = ⎢τ yx σ yy τ yz ⎥ = ⎢τ xy σ yy τ yz ⎥
⎢ τ zx τ zy σ zz ⎥ ⎢ τ xz τ yz σ zz ⎥
⎣
⎦ ⎣
⎦
Hence σij is symmetric!
41
CET 1, SAPV
This means that there must be some magic co-ordinate frame in which all
the stresses are normal stresses (principal stresses) and in which the off
diagonal stresses (=shear stresses) are 0. So if, in a given situation we
find this frame we can apply all our stress strain relations that we have set
up in the previous lectures (which assumed there were only normal
stresses acting).
Consider a cylindrical vessel subject to shear, and normal stresses (σh, σl,
σr). We are usually interested in shears and stresses which lie in the
plane defined by the vessel walls.
Is there a transformation about zz which will result in a shear
Would really like to transform into a co-ordinate frame such that all
components in the xi’ :
σ ij
→
σ ij '
So stress tensor is symmetric 2nd rank tensor. Imagine we are in the coordinate frame xi where we only have principal stresses:
0⎤
⎡σ 1 0
σ ij = ⎢⎢ 0 σ 2 0 ⎥⎥
⎢⎣ 0 0 σ 3 ⎥⎦
Transform to a new co-ordinate frame xi’ by rotatoin about the x3 axis in
the original co-ordinate frame (this would be, in our example, z-axis)
42
CET 1, SAPV
The transformation matrix is then:
⎛ a11
⎜
aij = ⎜ a21
⎜a
⎝ 31
a13 ⎞ ⎛ cos θ
⎟ ⎜
a23 ⎟ = ⎜ − sin θ
a33 ⎟⎠ ⎜⎝ 0
a12
a22
a32
sin θ
cosθ
0⎞
⎟
0⎟
1 ⎟⎠
0
Then:
σ ij ' = aik a jl σ kl
⎛ cos θ
⎜
= ⎜ − sin θ
⎜ 0
⎝
sin θ
cosθ
0
0 ⎞⎛ cos θ
⎟⎜
0 ⎟⎜ sin θ
1 ⎟⎠⎜⎝ 0
− sin θ
cosθ
0
⎡
σ 1 cos 2 θ + σ 2 sin 2 θ
⎢
= ⎢− σ 1 cosθ sin θ + σ 2 cosθ sin θ
⎢
0
⎣
43
0 ⎞ ⎡σ 1 0
0⎤
⎟⎢
0⎟ 0 σ 2 0 ⎥
⎢
⎥
⎟
1 ⎠ ⎢⎣ 0
0 σ 3 ⎥⎦
σ 1 cosθ sin θ + σ 2 cosθ sin θ
σ 1 cos 2 θ + σ 2 sin 2 θ
0
0⎤
⎥
0⎥
σ 3 ⎥⎦
CET 1, SAPV
Hence:
σ 11 ' = σ 1 cos 2 θ + σ 2 sin 2 θ
1
1
= (σ 1 + σ 1 ) − (σ 2 − σ 1 ) cos 2θ
2
2
σ 22 ' = σ 1 sin 2 θ + σ 2 cos 2 θ
=
1
1
(σ 1 + σ 1 ) + (σ 2 − σ 1 ) cos 2θ
2
2
σ 12 ' = −σ 1 cosθ sin θ + σ 2 cosθ sin θ
=
1
(σ 2 − σ 1 ) sin 2θ
2
44
Yield conditions. Tresca and Von Mises
Mohrs circle in three dimensions.
y
z
x
Shear stresses τ
y,z plane
x, y plane
normal
stresses
σ
x,z plane
1
We can draw Mohrs circles for each principal plane.
6. BULK FAILURE CRITERIA
Materials fail when the largest stress exceeds a critical value. Normally we
test a material in simple tension:
P
P
σy =
Pyield
A
τ
σ
σ = σ
σ=σ
This material fails under the stress combination (σy, 0, 0)
τmax = 0.5 σ y = 95 Mpa
for steel
We wish to establish if a material will fail if it is subject to a stress
combination (σ1, σ2, σ3) or (σn, σL, τ)
2
Failure depends on the nature of the material:
Two important criteria
(i)
Tresca’s failure criterion: brittle materials
Cast iron: concrete: ceramics
(ii)
Von Mises’ criterion: ductile materials
Mild steel + copper
6.1. Tresca’s Failure Criterion; The Stress Hexagon (Brittle)
A material fails when the largest shear stress reaches a critical value, the
yield shear stress τy.
Case (i)
Material subject to simple compression:
σ1
σ1
Principal stresses (-σ1, 0, 0)
τ
-σ
3
M.C: mc passes through (σ1,0), (σ1,0) , ( 0,0)
σ1
τ max
σ1
τmax = σ1/2
occurs along plane at 45° to σ1
Similarly for tensile test.
Case (ii)
σ2 < 0 < σ1
τ
-σ
σ
σ2
σ
1
σ1 - σ2
M.C.
2
Fails when
= τ max
= τy =
σy
2
σ1 - σ2 = σy
i.e., when
material will not fail.
4
Lets do an easy example.
5
6.2 Von Mises’ Failure Criterion; The stress ellipse
(ductile materials)
Tresca’s criterion does not work well for ductile materials. Early hypothesis
– material fails when its strain energy exceeds a critical value (can’t be true
as no failure occurs under uniform compression).
Von Mises’: failure when strain energy due to distortion, UD, exceeds a
critical value.
UD = difference in strain energy (U) due of a compressive stress C equal to
the mean of the principal stresses.
C =
1
[σ + σ2 + σ 3 ]
3 1
UD =
1 ⎧ 2
⎨σ +
2E ⎩ 1
1
=
(σ1
12G
{
[
]
1
⎫
σ 22 + σ 32 + 2υ (σ1σ 2 + σ 2 σ 3 + σ 3σ 1 ) 3C 2 + 6υC 2 ⎬
⎭
2E
2
2
2
- σ 2 ) + (σ 2 - σ 3 ) + (σ 3 - σ1 )
}
M.C.
τ
σ3
σ
σ
σ
2
1
6
Tresca → failure when max (τI) ≥ τy)
Von Mises → failure when root mean
square of {τa, τb, τc} ≥ critical value
τ
σ
σ
y
Compare with (σy, σ, σ) . simple tensile test – failure
Failure if
{
}
{
1
(σ1 - σ2 )2 + (σ2 + σ3 )2 + (σ3 - σ1 )2 > 1 σ2y + 02 + σ2y
12G
12G
{(σ - σ )
2
1
2
}
+ (σ 2 + σ3 ) + (σ3 - σ1 ) > 2σ 2y
2
2
7
}
8
Lets do a simple example.
9
Example Tresca's Failure Criterion
The same pipe as in the first example (D = 0.2 m, t' = 0.005 m) is subject to
an internal pressure of 50 barg. What torque can it support?
PD
= 50 N / mm 2
4t'
PD
σh =
= 100 N / mm 2
2t'
σL =
Calculate stresses
and σ3 = σr ≈0
Mohr's Circle
(100,τ)
τ
50
100
σ
s
(50,−τ)
Circle construction
s = 75 N/mm2
t = √(252+τ2)
The principal stresses
σ1,2 = s ± t
Thus σ2 may be positive (case A) or negative (case B). Case A occurs if τ is
small.
10
Case A
(100,τ)
τ
100 σ1
σ2 50
σ3
s
(50,−τ)
Case B
τ
(100,τ)
2β
σ2
σ3
100
50
σ1
σ
s
(50,−τ)
We do not know whether the Mohr's circle for this case follows Case A or
B; determine which case applies by trial and error.
Case A; 'minor' principal stress is positive (σ2 > 0)
Thus failure when
τ max = 12 σ y = 105N / mm 2
11
For Case A;
τ max =
⇒
σ1
2
=
1
2
[75 +
(252 + τ 2 )]
τ 2 = 135 − 252
τ = 132.7N / mm 2
σ1 = 210 N / mm2 ; σ 2 = −60 N / mm2
⇒
Giving
Case B;
We now have τmax as the radius of the original Mohr's circle linking our
stress data.
Thus
τ max = 252 + τ 2 = 105 ⇒
τ = 101.98N / mm2
Principal stresses
σ1,2 = 75 ± 105
⇒ σ 1 = 180N / mm2 ; σ 2 = −30N / mm 2
Thus Case B applies and the yield stress is 101.98 N/mm2. The torque
required to cause failure is
T = πD2 t' τ / 2 = 32kNm
Failure will occur along a plane at angle β anticlockwise from the y (hoop)
direction;
102
⇒ 2 λ = 76.23° ;
75
2β = 90 - 2λ ⇒ β = 6.9°
tan(2 λ ) =
12
Example
More of Von Mises Failure Criterion
From our second Tresca Example
σ h = 100N / mm 2
σ L = 50 N / mm 2
σ r ≈ 0 N / mm2
What torque will cause failure if the yield stress for steel is 210 N/mm2?
Mohr's Circle
(100,τ)
τ
50
100
σ
s
(50,−τ)
Giving
σ1 = s + t = 75 + 252 + τ 2
At failure
UD =
σ 2 = s − t = 75 − 252 + τ 2
σ3 = 0
1
(σ 1 − σ 2 ) 2 + ( σ 2 − σ 3 ) 2 + ( σ 3 − σ 1 ) 2 }
{
12G
13
Or
(σ1 − σ 2 ) 2 + ( σ 2 − σ 3 ) 2 + ( σ 3 − σ1 ) 2 = ( σ y )2 + (0)2 + (0 − σ y ) 2
4t 2 + (s − t)2 + (s + t) 2 = 2σ 2y
2s2 + 6t 2 = 2 σ y2
s 2 + 3t 2 = σ 2y
⇒ 752 + 3(252 + τ 2 ) = 210 2
τ = 110 N / mm 2
The tube can thus support a torque of
πD2 t' τ
T=
= 35kNm
2
which is larger than the value of 32 kNm given by Tresca's criterion - in this
case, Tresca is more conservative.
14
7. Strains
7.1. Direct and Shear Strains
Consider a vector of length lx lying along the x-axis as shown in Figure
1. Let it be subjected to a small strain, so that, if the left hand end is fixed
the right hand end will undergo a small displacement δx. This need not be
in the x-direction and so will have components δxx in the x-direction and
δxy in the y-direction.
δx
γ1
δxy
δxx
lx
Figure 1
We can define strains εxx and εxy by,
ε xx =
δ xx
;
lx
ε xy =
δ xy
lx
εxx is the direct strain, i.e. the fractional increase in length in the direction
of the original vector. εxy represents rotation of the vector through the
small angle γ1 where,
γ1 ≅ tan γ1 =
δ xy
l x + δ xx
≅
δ xy
lx
= ε xy
Thus in the limit as δx→ 0, γ1 → εxy.
Similarly we can define strains εyy and εyx = γ2 by,
ε yy =
δ yy
ly
;
ε yx =
δ yx
ly
1
δ yx
Figure 2
δ
yy
δy
ly
γ
2
γ1
as in Figure 2.
lx
Or, in general terms:
ε ij =
δ ij
li
where i, j = 1,2,3
;
The ENGINEERING SHEAR STRAIN is defined as the change in an
angle relative to a set of axes originally at 90°. In particular γxy is the
change in the angle between lines which were originally in the x- and ydirections. Thus, in our example (Figure 2 above):
γ xy = ( γ 1 + γ 2 ) = ε xy + ε yx or
γ xy = −(γ 1 + γ 2 )
depending on sign convention.
τ yx
A'
A
τ
B
xy
C
B'
C'
Figure 3b
Figure 3a
Positive values of the shear stresses τxy and τyx act on an element as
shown in Figure 3a and these cause distortion as in Figure 3b. Thus it is
sensible to take γxy as +ve when the angle ABC decreases. Thus
2
γ xy = +( γ 1 + γ 2 )
Or, in general terms:
γ ij = (ε ij + ε ji )
and since
τij = τji,
we have
γij = γji.
Note that the TENSOR SHEAR STRAINS are given by the averaged
sum of shear strains:
1
1
1
1
γ ij = (ε ij + ε ji ) = (γ 1 + γ 2 ) = γ ji
2
2
2
2
7.2 Mohr’s Circle for Strains
The strain tensor can now be written as:
⎡
⎢ ε 11
⎢1
ε ij = ⎢ y 21
⎢2
⎢1 y
⎢⎣ 2 31
1
y12
2
ε 22
1
y32
2
1 ⎤ ⎡
y13
ε 11
2 ⎥ ⎢
⎥ ⎢1
1
y 23 ⎥ = ⎢ y12
2
⎥ ⎢2
1
ε 33 ⎥ ⎢ y13
⎥⎦ ⎣⎢ 2
1
y12
2
ε 22
1
y 23
2
1 ⎤
y13
2 ⎥
⎥
1
y 23 ⎥
2
⎥
ε 33 ⎥
⎥⎦
where the diagonal elements are the stretches or tensile strains and the
off diagonal elements are the tensor shear strains.
Thus our strain tensor is symmetrical, and:
3
ε ij = ε ji
This means there must be a co-ordinate transformation, such that:
ε ij '
→
ε ij
such that :
⎡ε 1 0
ε ij = ⎢⎢ 0 ε 2
⎢⎣ 0 0
0⎤
0⎥
⎥
ε 3 ⎥⎦
we only have principal (=longitudinal) strains!
Exactly analogous to our discussion for the transformation of the stress
tensor we find this from:
ε ij ' = aik a jl ε kl
⎛ cos θ
⎜
= ⎜ − sin θ
⎜ 0
⎝
sin θ
cosθ
0
0 ⎞⎛ cos θ
⎟⎜
0 ⎟⎜ sin θ
1 ⎟⎠⎜⎝ 0
− sin θ
cosθ
0
⎡
ε 1 cos 2 θ + ε 2 sin 2 θ
⎢
= ⎢− ε 1 cosθ sin θ + ε 2 cosθ sin θ
⎢
0
⎣
And hence:
4
0 ⎞ ⎡ε1 0
⎟
0 ⎟⎢ 0 ε 2
⎢
1 ⎟⎠ ⎢⎣ 0 0
0⎤
0⎥
⎥
ε 3 ⎥⎦
ε 1 cosθ sin θ + ε 2 cosθ sin θ
ε 1 cos 2 θ + ε 2 sin 2 θ
0
0⎤
⎥
0⎥
ε 3 ⎥⎦
ε11 ' = ε1 cos 2 θ + ε 2 sin 2 θ
1
1
= (ε1 + ε1 ) − (ε 2 − ε1 ) cos 2θ
2
2
ε 22 ' = ε1 sin 2 θ + ε 2 cos 2 θ
=
1
1
(ε1 + ε1 ) + (ε 2 − ε1 ) cos 2θ
2
2
1
2
ε12 ' = γ 12 ' = −ε1 cos θ sin θ + ε 2 cos θ sin θ
=
1
(ε 2 − ε1 ) sin 2θ
2
For which we can draw a Mohr’s circle in the usual manner:
5
Note, however, that on this occasion we plot half the shear strain against
the direct strain. This stems from the fact that the engineering shear
strains differs from the tensorial shear strains by a factor of 2 as as
discussed.
7.3 Measurement of Stress and Strain - Strain Gauges
It is difficult to measure internal stresses. Strains, at least those on a
surface, are easy to measure.
•
Glue a piece of wire on to a surface
•
Strain in wire = strain in material
•
As the length of the wire increases, its radius decreases so its
electrical resistance
increases and can be readily measured.
In practice, multiple wire assemblies are used in strain gauges, to
measure direct strains.
strain gauge
_
_
_
45°
Strain rosettes are employed to obtain three measurements:
7.3.1 45° Strain Rosette
Three direct strains are measured
ε
C
εB
θ
Mohr’s circle for strains gives
6
εA
ε1
principal strain
120°
A
γ/2
B
ε
2θ
εB
εA
εC
C
radius t
so we can write
circle, centre s,
ε A = s + t cos(2θ )
ε B = s + t cos(2θ + 90) = s − t sin( 2θ )
ε C = s − t cos(2θ )
3 equations in 3 unknowns
Using strain gauges we can find the directions of Principal strains
γ/2
ε
7
7.4 Hooke’s Law for Shear Stresses
St. Venant’s Principle states that the principal axes of stress and strain are
co-incident. Consider a 2-D element subject to pure shear (τxy = τyx =
τo).
τo
y
Y
The Mohr’s circle for stresses is
τo
Q
τo
P
τo
X
x
where P and Q are principal stress axes and
σ pp = σ1 = τ o
σ qq = σ 2 = − τ o
τ pq = τ qp = 0
Since the principal stress and strain axes are coincident,
ε pp = ε1 =
=
τo
E
σ1
E
−
νσ 2
E
(1+ ν )
ε qq = ε 2 =
σ2
E
−
νσ1
E
=
−τ o
(1+ ν )
E
and the Mohr’s circle for strain is thus
γ/2
the Mohr’s circle shows that
Y
Q
P
εqq
εpp
ε
ε xx = 0
− γ xy
2
X (0,−γ )
xy
8
=−
τo
E
(1 + ν )
So pure shear causes the shear strain γ
τo
τo
γ/2
γ=
γ/2
and
2τ o
(1+ ν )
E
But by definition τo = Gγ
so
G=
τo
E
=
γ 2(1+ ν )
Use St Venants principal to work out principal stress values from a
knowledge of principal strains.
Two Mohrs circles, strain and stress.
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
9
So using strain gauges you can work out magnitudes of principal strains.
You can then work out magnitudes of principal stresses.
Using Tresca or Von Mises you can then work out whether your
vessel is safe to operate. ie below the yield criteria
10