MATERIALS SCIENCEMATERIALS SCIENCE C4:...
92
Natural Sciences Tripos Part II MATERIALS SCIENCE MATERIALS SCIENCE C4: Tensors Dr E. G. Bithell Michaelmas Term 2013 14 Name............................. College.......................... Michaelmas Term 2013-14 II
Transcript of MATERIALS SCIENCEMATERIALS SCIENCE C4:...
Natural Sciences Tripos Part II
MATERIALS SCIENCEMATERIALS SCIENCE
C4: Tensors
Dr E. G. Bithell
Michaelmas Term 2013 14
Name............................. College..........................
Michaelmas Term 2013-14
II
PART II MATERIALS SCIENCE C4: Tensors
EGB/MT13
Contents
Reference list ....................................................................................................................... 3
Formulae .............................................................................................................................. 4
1 Introduction .................................................................................................................. 5
1.1 Reasons for using tensors ....................................................................................... 5
1.2 Material properties represented by tensors ............................................................. 7
1.3 Anisotropy and symmetry ........................................................................................ 8
2 Stress and Strain .......................................................................................................... 9
2.1 Failure modes of materials and structures ............................................................... 9
2.2 Choice of volume element and coordinate system .................................................10
2.3 Alternative notations ...............................................................................................12
2.4 Examples of stress tensors ....................................................................................13
2.5 Einstein summation convention ..............................................................................14
2.6 Transformation of axes ...........................................................................................17
2.7 Transformation of components ...............................................................................18
2.8 Transformation laws for tensors .............................................................................19
2.9 Resolution of stresses ............................................................................................20
2.10 Principal axes .........................................................................................................25
2.11 Definition of strain at a point ...................................................................................31
2.12 Physical meaning of eij ...........................................................................................32
2.13 Separating shape changes from volume changes ..................................................35
3 Experimental Study of Stresses and Strains .............................................................37
3.1 Strain gauges .........................................................................................................37
3.2 Photoelasticity ........................................................................................................41
3.3 Brittle coating methods ...........................................................................................44
3.4 X-ray diffraction ......................................................................................................46
3.5 Ultrasonics .............................................................................................................46
3.6 Optical interferometry .............................................................................................47
3.7 Residual Stresses ..................................................................................................47
3.8 Example of stress and strain analysis: elasticity in thin films ..................................48
C4/Tensors EGB/MT13
- 2 -
4 Other tensor properties ..............................................................................................51
4.1 The representation surface for second rank tensors ...............................................51
4.2 Optical properties ...................................................................................................52
4.3 Piezoelectricity .......................................................................................................53
4.4 Effects of crystal symmetry .....................................................................................54
4.5 Ferroelectricity ........................................................................................................54
4.6 Applications of piezoelectricity ................................................................................56
4.7 Pyroelectricity .........................................................................................................57
4.8 Electrostriction ........................................................................................................57
5 Elasticity ......................................................................................................................58
5.1 Elasticity in isotropic media ....................................................................................58
5.2 Elasticity in general anisotropic media ....................................................................62
5.3 Crystal Symmetry ...................................................................................................66
5.4 Elastic stress distributions ......................................................................................70
5.5 Elastic waves .........................................................................................................72
5.6 Saint-Venant’s principle ..........................................................................................75
5.7 Examples of stress distributions .............................................................................76
5.8 Dislocations ............................................................................................................80
Lecture Examples ...............................................................................................................84
Preparatory Questions .......................................................................................................87
Question Sheet 1 ................................................................................................................88
Question Sheet 2 ................................................................................................................89
Question Sheet 3 ................................................................................................................90
Past Tripos Questions .......................................................................................................91
EGB/MT13 C4/Tensors
- 3 -
Reference list
Principal textbooks
Nye: Physical Properties of Crystals, chapters 1, 2, 5, 6 & 8 (Lj13b)
We use Nye's notation throughout.
Lovett: Tensor Properties of Crystals, all chapters (NbA 73b)
Less comprehensive than Nye, but the content corresponds closely to that of the course.
Kelly and Knowles: Crystallography and Crystal Defects, chapters 5, 6 & A6
(NbA99,a,b)
Tensors, stress and strain, piezoelectricity, elasticity
Newnham, Properties of Materials, chapters 1, 5, 8 – 13, 18, 25, 27 (AB200)
Tensors, stress and strain, physical properties
Supplementary textbooks
Cottrell: Mechanical Properties of Matter, chapters 4, 5 & 6 (Lj18a)
Elasticity, stress distributions, elastic waves
Dieter: Mechanical Metallurgy, chapters 1 & 2 (Kz1a,c,e)
Analysis of mechanical failures, stress and strain, elasticity
Gere & Goodno: Mechanics of Materials, chapters 1, 7, 8 & 9 (Ka110a,b)
Stress and strain distributions, engineering design
Holister: Experimental Stress Analysis, chapters 1, 2 & 4 (Ky2)
Strain gauges, photoelasticity
Knott: Fundamentals of Fracture Mechanics, chapters 1 & 2 (Ke45a)
Modes of failure, stress concentrations
Le May: Principles of Physical Metallurgy, chapters 1, 2 & 4 (Kz31b)
Stress and strain, criteria for failure, dislocations
Lovell, Avery and Vernon: Physical Properties of Materials, ch. 8 & 10 (AB59)
Pyroelectricity, piezoelectricity, ferroelectricity, electro-optics, non-linear optics
Wilson and Hawkes: Optoelectronics: An Introduction, chapter 3 (LcD7)
Electro-optic effect, non-linear optics
Wyatt and Dew-Hughes: Metals, Ceramics and Polymers, ch. 5 & 6 (Ab14b)
Tensile testing, yield point phenomena, elasticity
Web resources
DoITPoMS TLPs: Introduction to Anisotropy; Tensors; Dislocations
C4/Tensors EGB/MT13
- 4 -
Formulae
This is a list of formulae and equations appearing in the Tensors course which are sufficiently
fundamental to the subject area that you may be expected to derive them, or to reproduce
them from memory in order to carry out further derivations or calculations.
Einstein summation convention (example using two indices):
( )
Total, normal and shear stresses on a plane:
| | (
)
⁄
( )
⁄
Hydrostatic stress:
Dilatation:
Elasticity relationships (tensor notation):
Superposition of normal stresses in an isotropic medium:
( )
Important note
This course goes beyond the mathematics which can reasonably be done as a ‘pencil and
paper’ exercise, and you will also learn (in outline) the background to computational
approaches to tensor calculations. You should therefore be prepared to answer essay-type
questions on this course, as well as calculations, and be able to illustrate your answers with
appropriate mathematics in the same way as you would with diagrams. In these cases, key
equations will not normally be provided.
EGB/MT13 C4/Tensors
- 5 -
1 Introduction
1.1 Reasons for using tensors
We are concerned with the quantitative description and analysis of the properties of
materials.
There are two reasons why we need tensors:
1.1.1 Anisotropy (matter tensors)
A materials property such as density is just a number – direction has no meaning. But
properties such as electrical conductivity:
can depend upon direction
are anisotropic
need tensors for a proper description
These are “matter tensors”:
the properties have a particular orientation with respect to the material
they must conform to the material's symmetry (Neumann’s Principle)
Anisotropy is most readily associated with single crystals but this is not necessarily the only
possibility.
1.1.2 External influences and stimuli (field tensors)
Even if a material is isotropic, its properties relate specific quantities i.e. cause → effect, for
example:
electric field → current (via conductivity)
stress → strain (via stiffness or compliance)
Field tensors:
can have any orientation with respect to the material
can exist in isotropic materials
need not conform to the symmetry of the material
C4/Tensors EGB/MT13
- 6 -
We have two particular areas of interest:
1.1.3 Crystal physics
This is concerned with the behaviour of solid state devices, for example:
optical properties (birefringence is a strong indicator for anisotropy)
optoelectronic devices
magneto-optic devices
elasto-optic devices
1.1.4 Elasticity
This is concerned with the analysis of stress and strain:
at the atomic level – relevant to dislocations and cracks
at the microscopic level – e.g. thin films and micro devices
at the macroscopic level – components and structures
EGB/MT13 C4/Tensors
- 7 -
1.2 Material properties represented by tensors
Many materials properties can be described by matter tensors – properties which relate two
other quantities (see also the Tensors TLP).
Property Relating the quantities
Density Mass Volume
Heat capacity Temperature Energy
Pyroelectricty Polarisation Temperature change
Electrocaloric effect Entropy change Electric field
Electrical conductivity Current density Electric field
Thermal conductivity Heat flow Temperature gradient
Permittivity Dielectric displacement Electric field
Permeability Magnetic induction Magnetic field
Thermal expansion Strain Temperature change
Direct piezoelectric effect Polarisation Stress
Electro-optic effect Change in dielectric
permeability Electric field
Elastic compliance Strain Stress
Piezo-optical effect Change in dielectric
impermeability Stress
Electrostriction Strain 2 electric field components
C4/Tensors EGB/MT13
- 8 -
1.3 Anisotropy and symmetry
These material properties are matter tensors and must obey Neumann’s Principle:
The symmetry elements of a material property must include
the symmetry elements of the point group of the material
Our treatment will rely upon Neumann’s Principle. First we need to understand how to
classify the symmetry found in different types of macroscopic materials:
An isotropic material
o Has infinite symmetry (rotation axes/mirrors in all directions)
o Is found not only in a macroscopic continuum but in a real amorphous solid
composed of atoms
Lower symmetry
o Is clearly seen in single crystals
o Is described by the 32 point groups, none of which have infinite symmetry
Polycrystals
o Are macroscopically isotropic if crystal shape and orientation are random
o Otherwise have preferred orientation or texture → anisotropy
Amorphous solids
o These may be isotropic but we can induce anisotropy
e.g. by annealing under a tensile stress
The property may have greater symmetry than the material (depending upon the type of
property), so for example second rank tensors (optical properties, electrical conductivity) are
isotropic for cubic crystals but cubic crystals are not isotropic in general and this is apparent
in higher order tensor properties such as elasticity.
EGB/MT13 C4/Tensors
- 9 -
2 Stress and Strain
2.1 Failure modes of materials and structures
There are three modes of mechanical failure:
elastic
plastic
brittle
Analysis of each of these requires a treatment of stress and strain.
2.1.1 Elastic deformation
This is recoverable - remove the load and the body returns to its original state. In our course
we will deal only with linear time-independent elasticity.
In fact the stress-strain relationship need not be linear (anelasticity), and the strain response
need not be immediate (viscoelasticity).
The ability to analyse elastic deformations is important in many cases, such as …
2.1.2 Plastic deformation
Permanent and irrecoverable
Strains may be large
Often the desired failure mode e.g. energy absorption on car impact, because it is
slower and more predictable than ...
2.1.3 Brittle failure
Catastrophic propagation of a crack, with very little or no macroscopic plastic
deformation
2.1.4 Complex stress states
In a general, complex stress state, we need to consider also:
shear stress
that the stress in a real object can be very complicated
C4/Tensors EGB/MT13
- 10 -
So we need to:
describe the stress at a point
be able to deal with a general (complicated) stress
This will require several force/area terms. To define stress at a point we need to:
take a volume element surrounding the point
measure the forces transmitted across the faces
resolve forces along the coordinate axes
shrink the volume to a point
force / unit area = stress
2.2 Choice of volume element and coordinate system
We can choose the one which is most convenient for the given problem. Usually this means
that the symmetry of the coordinate system will reflect the symmetry of the body and/or the
stress state, and will normally be either Cartesian or cylindrical polar coordinates.
2.2.1 Cartesian coordinates
use a right-handed coordinate system
define axes 1, 2, 3
the volume element is a cube
We visualise the components of the stress tensor acting on a cubic volume element:
EGB/MT13 C4/Tensors
- 11 -
i corresponds to the direction in which the stress (or force) acts
j is used for the face (or its normal) on which the stress (or force) acts
e.g. 13 represents a force acting in the x1 direction on a face normal to the x3 direction. This
notation is a convention (and others exist).
This means that we can identify two kinds of stress:
normal: ij with i = j
shear ij with i ≠ j
There must be no net force on the cube, otherwise it would accelerate (linearly), so:
a) Stresses on equal and opposite faces of the cube are equal and opposite, and we only
need to consider three faces. This means that 9 quantities are sufficient to describe the
stress at a point, and we can write these as a 3x3 array:
[
]
b) We can simplify this further by considering rotational as well as linear acceleration. For
example, considering moments about the x axis, and only showing the relevant stresses, the
total moment must be zero, so
In general, σij = σji for all i,j so our 3x3 array is symmetrical and there are only 6 independent
quantities. This array of 9 numbers is a second rank tensor, so we have also demonstrated
that stress is a symmetrical second rank tensor.
[
]
C4/Tensors EGB/MT13
- 12 -
2.2.2 Cylindrical coordinates
These are useful in describing cylindrically symmetrical geometries (tubes, disclocations ...)
σzr = force in z direction on r face, and as before σzr = σrz etc.
2.3 Alternative notations
The notation used is a matter of convention and convenience: in particular the use of τ for
shear stresses is common, which gives equivalent quantities as:
Equivalent notations
Normal stresses
(e.g. stress parallel to x1 on face 1) σ11 σx
Shear stresses
(e.g. stress parallel to x2 on face 1) σ21 τxy
We will use the numerical suffix notation in this course because it allows us to keep to the
same conventions when considering tensors other those for elastic properties, and because
it allows us to use the simplifications offered by the Einstein summation convention. Other
courses may use other conventions and it is important that you become accustomed to these
differences, which are widespread in both textbooks and the research literature.
EGB/MT13 C4/Tensors
- 13 -
2.4 Examples of stress tensors
2.4.1 Uniaxial tension
Choosing sensible axes, with one (x1) along the axis of the rod, σ11 = σ, and all other σij = 0:
[
]
(σ > 0 for tension)
2.4.2 Uniaxial compession
[
]
(σ < 0 for compression)
2.4.3 Hydrostatic compression
(e.g. under water)
This time there are no shear stresses:
[
] ( )
C4/Tensors EGB/MT13
- 14 -
2.4.4 Pure shear
σ12 = σ21 = τ and all other σij = 0, so we have:
[
]
2.5 Einstein summation convention
We will use electric field, current and conductivity as an example to illustrate a more
generally applicable notation convention.
If we apply an electric field, a current will flow. Both of these quantities are vectors, and they
are related by electrical conductivity:
electric field E = (E1, E2, E3) [units V m–1] potential gradient
current density J = (J1, J2, J3) [units A m–2] current density
For a general anisotropic material the relationship between J and E can be expressed as a
series of equations relating their components:
This describes the conductivity as a tensor of the second rank, writing the above equations
as:
[
] [
] [
]
EGB/MT13 C4/Tensors
- 15 -
The components of the tensor are 11
, 12
, etc, and the number of suffices indicates the rank
of the tensor.
Like conductivity, many other materials properties can be described by matter tensors, and
these properties then relate two other quantities:
Rank Number of
components Property Relating the quantities (with rank)
0 1 Density Mass (0) Volume (0)
Heat capacity Temperature (0) Energy (0)
1 3 Pyroelectricty Polarisation (1)
Temperature change (0)
Electrocaloric effect Entropy change (0) Electric field (1)
2 9
Electrical conductivity Current density (1) Electric field (1)
Thermal conductivity Heat flow (1) Temperature gradient (1)
Permittivity Dielectric displacement (1)
Electric field (1)
Permeability Magnetic induction (1)
Magnetic field (1)
Thermal expansion Strain (2) Temperature change (0)
3 27
Direct piezoelectric effect
Polarisation (1) Stress (2)
Electro-optic effect Change in dielectric permeability (2)
Field (1)
4 81
Elastic compliance Strain (2) Stress (2)
Piezo-optical effect Change in dielectric impermeability (2)
Stress (2)
Electrostriction Strain (2) 2 electric field components (2 × 1)
C4/Tensors EGB/MT13
- 16 -
We can rewrite the previous equations for the tensor components as summations, so these
equations:
Become:
∑
∑
∑
or:
∑
( )
This last equation can be written in contracted form using the Einstein Summation
Convention:
( )
A suffix occurring twice in a product (j in this case) implies summation with respect to that
suffix and is called the dummy suffix.
EGB/MT13 C4/Tensors
- 17 -
2.6 Transformation of axes
We need to go from ‘old’ axes (x1, x
2 and x
3) to ‘new’ axes (x
1', x
2' and x
3').
Axis x2', for example, makes angles
12,
22 and
23 with the x
1, x
2 and x
3 axes. We define
direction cosines, a21
= cos 21
, a22
= cos 22
, a23
= cos 23
etc, completely specifying the
angular relationships between the axes. The nine direction cosines form a transformation
matrix:
Old axes
x1 x2 x3
New
axe
s
x'1 a11 a12 a13
x'2 a21 a22 a23
x'3 a31 a32 a33
So we have:
x'2 = x1 cos θ21 + x2 cos θ22 + x3 cos θ23
and so on ...
If one of the axes is unchanged this reduces to a 2x2 matrix which should look familiar!
C4/Tensors EGB/MT13
- 18 -
2.7 Transformation of components
When different reference axes are chosen, the tensor components change but the material
property must be the same. We need to know how the components change.
Using our conductivity example again:
Old axes:
New axes:
Transforming J → J' one component at a time:
or:
which is the 'old' to 'new' transformation.
Similarly for the 'new' → 'old' transformation
and so on, i.e.
So we have:
New from old
Old from new
But we want to relate J' to E':
EGB/MT13 C4/Tensors
- 19 -
and
and
Therefore:
But also
So:
This gives us the ‘new from old’ transformation, and similarly for the ‘old from new’
transformation:
2.8 Transformation laws for tensors
These laws are the basis for defining tensors of a given rank.
Transformation law
Rank of tensor New from old Old from new
0 (scalar) ' = = '
1 (vector) pi' = aijpj pi = ajipj'
2 Tij' = aikajlTkl Tij = akialjTkl'
3 Tijk' = ailajmaknTlmn Tijk = aliamjankTlmn'
4 Tijkl' = aimajnakoalpTmnop Tijkl = amianjaokaplTmnop'
C4/Tensors EGB/MT13
- 20 -
2.9 Resolution of stresses
Force is a vector and is easily resolved into its components on Cartesian axes:
Resolution of stresses is not as simple as resolution of forces because we have to take into
account the plane on which the forces act (i.e. whether it is a normal or a shear stress).
Consider a body in which the stress is homogeneous. Suppose we are interested in the
stress acting within that body on planes in a particular orientation.
We consider the action of a vector force P per unit area (or "stress") on a plane ABC whose
orientation is specified by its unit normal vector n. The vector n makes angles 1, 2,3 with
the axes x1, x2, x3, and the direction of n is defined by its direction cosines.
EGB/MT13 C4/Tensors
- 21 -
λ1, λ2 and λ3 are also the components of the unit vector n.
In order to work out the stress on the plane ABC due to P, we can start by taking any
element which has ABC as one of its faces, and it is simplest to take the tetrahedron OABC.
In equilibrium the net force on this element must be zero: the force per unit area P acting on
face ABC can be resolved into its components P1, P2, P3 and each of these should be
balanced by the stress components acting on the other faces of the volume element OABC.
We write the area of triangle ABC as ΔABC (and other triangles similarly).
P is a force per unit area, so the force on ABC is:
( )
We can resolve P. ΔABC because it is a vector:
But P. ΔABC is balanced by the other forces acting on the other faces, and we can take
these one direction at a time. In the x1 direction we have:
This can be simplified by noting that the volume of the tetrahedron OABC is:
Volume
And:
C4/Tensors EGB/MT13
- 22 -
Therefore:
Likewise:
Now dividing through by ΔABC:
And similarly for P2 and P3:
This set of 9 numbers relates a vector P (with components P1, P2, P3) to a vector n (with
components λ1, λ2, λ3), and defines the σij as a second rank tensor relating force and plane
orientation:
Remembering that:
∑
EGB/MT13 C4/Tensors
- 23 -
2.9.1 Total, normal and shear stresses
The total stress acting on a plane ABC is:
| | (
)
⁄
We want to calculate the normal and shear stresses separately.
2.9.2 Normal stress
This is important in (for example) tensile fracture
We need Pn, the component of P normal to the plane
We resolve each Pi in the direction of n:
But:
C4/Tensors EGB/MT13
- 24 -
2.9.3 Shear stress
This matters for slip
We want Ps, the component of P lying in the plane
We know that:
And we have expressions for P and Pn, so we can find Ps from:
( )
⁄
EGB/MT13 C4/Tensors
- 25 -
2.10 Principal axes
All symmetrical second rank tensors possess three orthogonal principal axes. “Diagonalising”
a tensor to write its components referred to these axes leads to a simplified form of the
tensor:
[
] [
]
2.10.1 Principal axes of the stress tensor
A principal axis of the tensor is one along which the resultant stress is purely a normal stress
There are no shear stresses in the plane perpendicular to a principal axis
So whatever the state of stress, however complicated, there is always a set of axes such that
the stresses on the cubic volume element are:
And the stress tensor is:
[
]
So:
This is a diagonal tensor
There are no shear stresses
By convention, a single suffix denotes a principal stress
C4/Tensors EGB/MT13
- 26 -
2.10.2 Diagonalisation of a tensor
We need to find the orientation of a plane normal n (specified by direction cosines λ1, λ2, λ3)
for which the total force per unit area P is parallel to the normal.
Let the magnitude of P be ξ, so:
But:
So:
( )
( )
( ) }
These three simultaneous equations have non-trivial solutions only if their determinant is
equal to zero:
|
|
This gives a single cubic equation in ξ – the “secular equation” – with three roots (values of ξ)
corresponding to the three principal stresses.
2.10.3 Orientation of the principal axes
We need to find λ1, λ2, λ3 for each principal axis.
Take each value of ξ and substitute it back into the 3 simultaneous equations
Check:
EGB/MT13 C4/Tensors
- 27 -
Then we have:
[
]
where σ1, σ2, σ3 are the three values of ξ
By convention the principal axes are (nearly always!) labelled such that:
2.10.4 2D diagonalisation
In many cases diagonalisation is much easier because one of the principal axes is already
known. We then simply have to rotate about that axis – effectively diagonalising in 2D only.
If one principal axis as already known, the tensor to be diagonalised is:
[
]
Remembering that σ12 = σ21
The diagonalisation required is in two dimensions, not three, and is comparatively
straightforward. The determinant becomes:
|
|
Multiplying out to get the secular equation:
( ){( )( ) }
One solution is ξ = σ3 for the principal axis already known. The other principal stresses are
obtained by solving the remaining quadratic:
( )
( ) √( )
( )
( )
√(
)
C4/Tensors EGB/MT13
- 28 -
2.10.5 The Mohr circle construction
Tha above solutions can also be represented geometrically. We start with the same stress
tensor, and take σ11 > σ22:
[
]
We construct the circle using:
From which:
Points to note:
The x-axis represents normal stresses.
The y-axis represents shear stresses.
The circle is centred on the x-axis and intercepts it at the principal stresses (which are
on the x-axis because there is no shear).
The circle represents the stress state of the volume we are interested in. The
coordinates of the two ends of any diameter give the normal and shear components
of the stress tensor for one particular orientation of the axes.
EGB/MT13 C4/Tensors
- 29 -
The angle of rotation of the diameter (= 2θ) is twice the angle of rotation of the axes in
real space (= θ). The sense of rotation is the same.
The same principal stresses could arise from different combinations of σ11, σ22 and θ,
and these would all correspond to different diameters DC of the same circle.
The procedure works for any second rank tensor (and thus also for strain).
2.10.6 Uses of the Mohr circle
To find principal axes and strains.
To find the maximum shear stress.
Knowing the stress tensor for one orientation, we can find the tensor for any other
orientation.
2.10.7 Representation of a 3D stress state
This can be done by plotting 3 Mohr circles on one diagram, with one circle for plane
perpendicular to a principal axis.
For example, consider again the pure shear case:
[
]
x1x2 plane
[
]
x1x3 plane
[
]
x2x3 plane
[
]
C4/Tensors EGB/MT13
- 30 -
Note that the maximum shear stress still occurs in the x1x2 plane at 45° to the principal axes
– also that this is one case where we commonly do not observe the σ1 > σ2 > σ3 convention!
2.10.8 Triaxial tension
This time we will specify σ1 > σ2 > σ3, and the stress tensor is then:
[
]
Note that although we can represent the stress state of a 3D system using Mohr circles when
we already know the principal axes, we cannot diagonalise a general 3D tensor using Mohr
circles.
EGB/MT13 C4/Tensors
- 31 -
2.11 Definition of strain at a point
Starting in one dimension, imagine an extensible string:
And now extend it:
The strain in OP is:
And the strain at the origin is defined as:
(
)
Now we do the same in 3 dimensions. After deformation of a body, suppose that point
P (x1, x2, x3) has moved to P' (x'1, x'2, x'3) with displacements u1, u2, u3.
So:
If the ui are the same for all points, then we have a rigid body displacement i.e. zero strain.
For there to be a strain, the ui must vary through the body.
( )
In one dimension:
C4/Tensors EGB/MT13
- 32 -
So:
And in 3 dimensions:
Which can be written as:
The form of the summation convention defines eij as a second rank tensor.
2.12 Physical meaning of eij
𝜕
𝜕
𝜕
𝜕
𝜕
𝜕
These are simply the tensile strains along the axes x1, x2, x3. Note that the signs are the
same as for stress: positive in tension and negative in compression.
Now consider what would happen for deformation in the x1x2 plane of a rectangular block,
when:
( )
For small angles, very much exaggerated, this corresponds to:
EGB/MT13 C4/Tensors
- 33 -
Given that:
𝜕
𝜕 ( )
𝜕
𝜕 ( )
Consider possible combinations of e12 and e21:
1. e12 = e21 is “pure shear”
2. e12 = – e21 is “pure rotation” (with no deformation)
3. e12 ≠ 0 and e12 = 0 is “simple shear”
Note the distinction between pure shear and simple shear.
In general e12 and e21 describe both shear and rotation together. Note that displacements for
which (e21 + e12) is constant produce the same shape change but different rotations.
We can separate the effects of shear and rotation by writing:
Or in general:
C4/Tensors EGB/MT13
- 34 -
Where:
( )
And:
( )
Note that εij is symmetric with respect to i and j, and is a measure of shape change, whereas
ωij is antisymmetric with respect to i and j and is a measure of rotation:
And:
( )
So ωij has only 3 independent tensor elements, giving 3 independent rotations – one about
each coordinate axis.
Note the form of the strain tensor written in terms of the shear strain angle in simple shear:
[
]
[
⁄
⁄
] [
⁄
⁄
]
From now on, “the strain tensor” will mean εij
In this way, both stress and strain can be written as symmetrical second rank tensors
EGB/MT13 C4/Tensors
- 35 -
2.13 Separating shape changes from volume changes
2.13.1 Hydrostatic and deviatoric components of stress
We can separate out two components of stress:
1. Hydrostatic volume changes
2. Deviatoric shape changes
For example under compressive pressure the stress tensor is:
[
] where
And plotting this on a Mohr circle:
The Mohr circle is simply a point
The stress tensor is the same for any orientation of the tensor axes i.e. it is an
isotropic tensor
There is never any shear stress
So in general we can write:
[
] [
] [
( )
( )
( )]
i.e. the full stress tensor is the sum of the hydrostatic component (volume change) and the
deviatoric component (shape change).
The normal components of the stress exert a pressure on the volume element, and the
hydrostatic component of the stress is their average:
C4/Tensors EGB/MT13
- 36 -
The hydrostatic stress is also known as the mean stress.
Plastic flow of metals depends only on the deviatoric component of the stress tensor and not
on the hydrostatic component, because it occurs without volume change.
2.13.2 Dilatational and deviatoric components of strain
This is exactly analogous to the separation of the stress components.
In this case, for small strains we can also write the volume of the unit cube:
( )( )( )
Where Δ is known as the “dilatation”.
So:
[
]
[
]
[ (
)
(
)
(
)]
i.e. the full strain tensor is the sum of the dilatational component (volume change) and the
deviatoric component (shape change).
EGB/MT13 C4/Tensors
- 37 -
3 Experimental Study of Stresses and Strains
3.1 Strain gauges
These are devices which exploit the change of electrical resistance resulting from strain.
Start with a long straight wire of length L, and stretch it to L +dL
The resistance R is given by:
[ ρ =resistivity, A = area of cross-section; r = radius of wire]
Take logarithms:
Differentiate:
If the elongation is elastic, then:
Experimentally we find:
[ C ≈ 1 for metals; V = volume = πLr2]
So we have:
[
]
And therefore:
( ) ( )
The exact proportionality depends on the material (via the constants C and ν), so we
introduce a “gauge factor”:
To measure the strain, we:
Fix the wire onto the object
Allow it to deform
C4/Tensors EGB/MT13
- 38 -
Measure the change in R
Calculate the corresponding strain
However the strains are usually small ( 1%) so dR/R is difficult to measure. But we want to
map out the stress distribution across an object, so we cannot just use a very long wire to
make the absolute value of ΔR bigger – we need to be able to measure the strain over a
small area.
We deal with this by folding the wire, which gives us a small actual gauge length (typically 1
– 10mm) but a large effective length, large R and a large, easily measured R.
Gauges are made from wire or – more commonly – from foil etched to give an
appropriate pattern
A common material is “constantan” (55Cu + 45Ni), which gives a good linear
response
Semiconductor strain gauges are also available with C 1 and gauge factors ≈ 100
In the simplest case, the strain gauge can be connected to a Wheatstone Bridge
circuit to measure R
3.1.1 Thermal effects on strain gauges
The resistivity ρ and the gauge factor k are both temperature-dependent, which can lead to
false strain readings
It is better to use thermally compensated gauges, in which the thermal expansion is matched
to common substrate materials
3.1.2 Analysis of strain gauge results
A strain gauge is sensitive only to strains parallel to its length i.e. to normal strains, and not
to shear strains
If we know the directions of the principal stresses, we can measure the strains directly
EGB/MT13 C4/Tensors
- 39 -
For example, the principal axes may be fixed by the symmetry of the body and the loading
geometry. In this case they are known at A but not at B:
But if the principal axes are unknown, the standard method is to use a rosette – an array of
gauges – from which the strain in any direction can be calculated.
The direction of the principle strains can then be deduced from the three readings, for
example using a Mohr circle.
3.1.3 Example of a strain gauge calculation: a 60° rosette
C4/Tensors EGB/MT13
- 40 -
Suppose we measure normal strains (the shear strains are unknown from the gauges):
As a vector diagram this is:
But on the Mohr circle the angles are doubled (and note that this is an arbitrary orientation
with respect to the principal axes):
The centre of the circle is at a distance:
And we need the principal strains εx and εy:
EGB/MT13 C4/Tensors
- 41 -
( )
( )
So:
3.2 Photoelasticity
This technique allows the strain field in a transparent model to be visualised using stress-
induced birefringence: when we take a suitable transparent, isotropic material and apply a
stress, it can become optically anisotropic.
In an optically anisotropic material, light propagates with two different velocities and two
perpendicular planes of polarisation, such that:
( )
Where n1 and n2 are the refractive indices and c is the “stress-optical coefficient”.
C4/Tensors EGB/MT13
- 42 -
3.2.1 Analysis of plane stress distributions using a plane polariscope
We imagine looking towards the polariser, through a sample which is a plate cut in the x1x2
plane of the stress tensor.
At a particular point, the incident wave will be:
This wave splits into two components parallel to σ1 and σ2:
( )
( )
The wave travels through a plate of thickness d, which shifts the phase of each wave by:
So on the exit side of the plate our two waves become:
( ) (
⁄ )
( ) (
⁄ )
There is now a phase difference between the two waves. The wave passing through the
analyser is the sum of the resolved components:
( ) (
⁄ ) (
⁄ )
(
) (
)
This is a travelling wave, with amplitude:
| | (
)
EGB/MT13 C4/Tensors
- 43 -
We get extinction (zero amplitude) when:
σ1 or σ2 is parallel to either the polariser or the analyser, which gives:
This gives black fringes joining the places where the principal stresses are parallel to
the polariser and the analyser. These are isoclinic fringes and are independent of
wavelength.
The two waves are in phase with each other:
(
) ( )
Given:
We get:
( )
( )
And:
( )
C is the stress-optical coefficient; m comesfrom counting the fringes. We get
extinction when the optical path difference is a whole number of wavelengths. These
are isochromatic fringes and:
o Depend on wavelength
o Join points with the same value of (σ1 – σ2)
o Are black in monochromatic light
o Are coloured in white light
Isoclinics orientation of principal stresses
Isochromatics value of (σ1 – σ2)
3.2.2 3D stress analysis
For 3-dimensional stress analysis we can:
Apply a stress to a hot, 3D transparent model
Cool to ‘freeze in’ the birefringence
Remove the external stresses
Section and view slices in a plane polariscope
C4/Tensors EGB/MT13
- 44 -
3.2.3 Birefringent coatings
These can be applied to the surface of an object under stress
A reflection polariscope (still using cross polars) is needed
Possible coatings need to have high values of the stress-optical coefficient C, for
example:
o Glass
o Epoxy resin
o Polycarbonate
o Perspex (PMMA)
3.3 Brittle coating methods
These can be used to determine the stresses and strains in the surface of a body
We apply a thin coating of a brittle material with a well-defined fracture strain
If it is thin enough and well-bonded, we can assume that the strain in the coating is
equal to the strain in the surface to which it has been applied
The coating will crack under tensile strains
To a first approximation, cracks appear when:
σ1(specimen) = E(specimen) εcrit(coating)
A pattern of cracks develops, oriented perpendicular to the maximum tensile stress in
the surface
3.3.1 Tension in one dimension
In uniaxial tension we will see:
The orientation of the cracks gives the direction of σ1
The load at which the cracks appear gives the magnitude of σ1 at that load
EGB/MT13 C4/Tensors
- 45 -
3.3.2 Tension in two dimensions
If σ1 > σ2 > 0 and both are increasing, then the first set of cracks will appear perpendicular to
σ1 and a second set will appear perpendicular to σ2.
If σ1 = σ2 >0 then the direction of the cracks is indeterminate, and a crazed surface results.
This is commonly seen in:
Dried mud
Old china
Surfaces of varnished oil paintings
In these cases the crazing is the result of differential expansion or contraction.
C4/Tensors EGB/MT13
- 46 -
Brittle coatings can also be applied deliberately as an analytical technique:
Coatings are 50 - 200μm thick
Applied as a spray of resin
The solvent evaporates and the resin cures on heating leaving a brittle coating
Ceramic coatings can also be used (these need firing to melt, but can then be used at
high temperatures)
3.3.3 Advantages
Simple to use on a real structure
Useful to determine the directions of the principal stresses
Can then apply strain gauges for accurate measurements
3.3.4 Problems
Ceramic coatings are needed for high temperature use
Failure strains need calibrating
3.4 X-ray diffraction
This technique allows measurements of changes in lattice parameter (and is
especially good for thin films)
X-rays are truly non-destructive and non-contact
Useful for residual surface strains
A back reflection method is commonly used
But there are drawbacks:
Inconvenient to set up
Small penetration depth (tens of microns)
Only measures surface strains
3.5 Ultrasonics
Wave velocity depends on modulus and density:
√
EGB/MT13 C4/Tensors
- 47 -
The modulus and density both depend on stress and strain (given that Poisson’s ratio
≠ ½ for most materials)
The appropriate modulus depends on what type of wave is being used
But again there are drawbacks:
Real materials are non-linearly elastic
Interpretation is not straightforward
3.6 Optical interferometry
This includes both holography and laser speckle interferometry. In general optical methods
are attractive because they are intrinsically non-contact and very sensitive.
3.7 Residual Stresses
These are internal stresses which are not a result of external forces, and which must be
retained by interlocking (for example, a residual compressive stress cannot exist by itself and
must be balanced by a tensile stress somewhere).
There are three main origins of residual stresses in macroscopic bodies:
3.7.1 Thermal stresses due to differential expansion
Consider a plate which has been roll-bonded at high temperature and is then cooled to room
temperature. αth = 1210-6 K-1 for Fe and αth = 1.710-6 K-1 for Cu so they apportion the
thermal strain between the layers differently:
C4/Tensors EGB/MT13
- 48 -
If the sheet is not symmetric, the residual stresses will cause bending: when this can be
calibrated (e.g. a bimetallic strip) it can be used for temperature measurement.
3.7.2 Plastic flow
Consider bending a bar so that the surfaces (but only the surfaces) are stressed beyond their
elastic limit. The surface regions flow, so when the stress is released the bar will remain
slightly bent in order to accommodate the residual stresses:
This means that residual stresses can be found as a result of forming processes such as
shot peening (surface hardening) and machining.
3.7.3 Changes in volume
Residual stresses can also result from non-uniform or localised changes in volume, such as
occur in:
Phase transformation
Carburising and nitriding (in which interstitials cause a change in lattice parameter)
3.8 Example of stress and strain analysis: elasticity in thin films
This is important for:
Pottery glazes
Optical and other coatings
Electronic devices
etc …
In addition, devices are often multilayered e.g. GaAs/AlAs solid state lasers, W/C x-ray
mirrors.
EGB/MT13 C4/Tensors
- 49 -
3.8.1 Stresses in thin films
There cannot be any forces on a free surface, so σ33 = 0 and σi3 = 0 and this is plane stress:
[
]
In almost all cases σ11 = σ22 ( = σ ) so we have isotropic biaxial tension (or compression).
These are now principal stresses, and we should expect to find both biaxial in-plane strains,
and a normal strain.
[
] [
]
The biaxial stress σ and the biaxial strain ε in the (x1,x2) plane of the film are related by:
( )
( )
The normal strain in the film can be determined (for example) by x-ray diffractometry, and
is given by:
( )
C4/Tensors EGB/MT13
- 50 -
Residual stresses of this type can arise from:
thermal stresses due to differential contraction between film and substrate
deposition stresses – a film which is bombarded with gas atoms or ions during
deposition will normally develop a compressive stress
epitaxial stresses – a single crystal film in a fixed orientation with respect to its
substrate can develop a strain as a result of the difference in lattice parameters
3.8.2 Epitaxy in multilayers
Consider a stack of alternate layers of (100) Cu and (100) Pd on a (100) NaCl substrate.
Would the interfaces remain coherent?
1. If the multilayer stack is coherent (i.e. there are no dislocations) there is a 7.6%
difference in lattice parameter but the lattices are forced to match.
For Cu: a = 3.6148Å
For Pd: a = 3.8908Å
So the Cu must be in tension and the Pd in compression, with equal and opposite
stresses for equal layer thicknesses.
2. If the stack is not coherent, there is no strain energy due to matching but a 2D
network of dislocations is needed at each interface.
So we have two scenarios:
For thin layers (d < dcrit) the interfaces will be fully coherent with no dislocations
For thick layers (d > dcrit) there will be few interfaces but these will contain dislocations
EGB/MT13 C4/Tensors
- 51 -
4 Other tensor properties
4.1 The representation surface for second rank tensors
For second rank tensors there exists a simple geometrical representation which is useful, for
example, in deriving the magnitude of a material property in a given direction. Taking
electrical conductivity as an example, the property referred to its principal axes is:
[
]
The applied field E has components (λ1E, λ2E, λ3E).
The resulting current density J therefore has components (σ1λ1E, σ2λ2E, σ3λ3E).
Thus the component of J resolved parallel to E is:
( )
So the conductivity σ parallel to E is:
This conductivity can be represented by the surface:
Which will be an ellipsoid if all the coefficients are positive. A radius in a given direction
(λ1, λ2, λ3) will have length r and intersect the surface at a point (x1, x2, x3) given by:
With these coordinates:
In general for a second rank tensor property, the radius length in any direction is the
(property)–1/2 in that direction. For an applied vector quantity parallel to the radius (e.g. field),
the normal to the representation surface at the point where the radius intersects the surface
C4/Tensors EGB/MT13
- 52 -
is parallel to the resultant vector (e.g. current density). [For discussion of this radius-normal
property, see Nye p28]
The representation surface (or quadric) for a second rank tensor property,
shown in this case for conductivity
4.2 Optical properties
For an isotropic material the electric displacement D is related to the electric field strength E
by:
where κo is the permittivity of vacuum and K is the dielectric constant. The real refractive
index is (K)1/2.
For an anisotropic material, the electric displacement and field strength are not necessarily
parallel. Their vector components are related by:
Where Kij is the tensorial dielectric constant, and Bij is the relative dielectric impermeability.
On principal axes therefore, the real refractive index is n = (K)1/2 = (B)–1/2. The representation
surface for Bij is therefore a plot of refractive index – this is the optical indicatrix.
EGB/MT13 C4/Tensors
- 53 -
4.3 Piezoelectricity
4.3.1 Direct piezoelectric effect
This is the name given to the situtation where we apply a stress to a crystal, which develops
(or changes) its electric moment as a result.
The effect is linear, i.e. moment ∝ stress and when the stress is reversed, so is the moment.
So we need to relate the polarisation vector Pi (moment per unit volume, a first rank tensor)
and the stress σij.
This requires a relationship of the form:
and similarly for P2 and P3.
In general we have:
where dijk is a third rank tensor composed of the 27 piezoelectric moduli.
dijk = dikj always, so we have only 18 independent coefficients.
4.3.2 Converse piezoelectric effect
In this case the crystal develops a strain in response to an applied electric field.
So:
dijk must have the same components as before, otherwise we can make a perpetual motion
machine!
C4/Tensors EGB/MT13
- 54 -
4.4 Effects of crystal symmetry
Consider a crystal with a centre of symmetry.
Apply a stress e.g. σ11 (and this is inherently centrosymmetric)
The result must also be centrosymmetric
So any resultant dipole will be cancelled by its opposite.
Thus centrosymmetric crystals (belonging to the 11 centrosymmetric point groups, out of 32)
cannot be piezoelectric.
But we can expect to make piezoelectric devices from non-centrosymmetric single crystals
e.g. quartz, point group 32.
We can also have piezoelectric polycrystals provided that:
The individual grains are piezoelectric
The polycrystal has texture (preferred orientation)
4.5 Ferroelectricity
Texturing of piezoelectric polycrystalline ceramics makes use of ferroelectric properties, as in
BaTiO3.
Below 120°C the Ti atom moves towards one of the oxygens in the coordinating octahedron.
This causes:
A cubic → tetragonal transformation on cooling
EGB/MT13 C4/Tensors
- 55 -
Spontaneous polarisation
The direction of polarisationcan be influenced by cooling in an electric field (“poling”).
High T cubic BaTiO3 polycrystal
(no texture)
↓
Cool in electric field E
↓
Low T tetragonal BaTiO3 polycrystal
(with texture and a net dipole parallel to E)
Ferroelectricity is only possible in the 10 polar classes:
1, 2, m, mm2, 3, 3m, 4, 4mm, 6, 6mm
These are a subset of the piezoelectric classes, and the final direction of polarisation in a
single crystal will be a unique direction.
C4/Tensors EGB/MT13
- 56 -
4.6 Applications of piezoelectricity
These are generally as transducers, which convert a mechanical signal into an electrical
signal, e.g.:
Sonar (e.g. ultrasonic testing)
Microphones
Earphones
Resonators
Gas lighters
4.6.1 Scanning tunnelling microscope
[CC-SA from http://commons.wikimedia.org/wiki/File:ScanningTunnelingMicroscope_schematic.png]
The tip is scanned laterally and the height adjusted to maintain a constant tunnelling current,
with control to better than 1Å, and thus atomic resolution. The tip has to be moved laterally
with extreme precision, and this is achieved with piezoelectric transducers.
EGB/MT13 C4/Tensors
- 57 -
4.7 Pyroelectricity
Ferroelectricity is the existence of a spontaneous polarisation. If this polarisation varies with
temperature then we have pyroelectricity, according to:
ΔT also causes a change in volume ΔV, which also contributes to ΔPi.
If V is held constant (i.e. ΔV=0) then ΔT causes primary pyroelectricity; if it varies, we also
measure secondary pyroelectricity. An experiment will normally measure the sum of both
primary and secondary effects.
Pyroelectric materials come from the same 10 polar classes as the ferroelectric materials.
They are primarily used for temperature measurement and the detection of infra-red
radiation. Examples include BaTiO3 (doped), and triglycerine sulphate (TGS).
4.8 Electrostriction
Magnetostriction is a more important and useful effect, but electrostriction is an example of a
second order effect which illustrates the treatment of non-linearity. Whereas centrosymmetric
crystals cannot show piezoelectricity, they can show electrostriction.
We start from converse piezoelectricity:
If the effect is non-linear we require an additional term:
( )
The first term ( ) is linear in applied field (if the field is reversed, the strain is reversed),
and there can be no effect if the crystal is centrosymmetric.
It can also be useful to write this expression as:
( )
Where the “correction term” γiljk El is a third rank tensor which represents electrostriction.
The effect is quadratic in applied field (if the field is reversed, the strain is the same) and it
can appear for centrosymmetric crystals.
C4/Tensors EGB/MT13
- 58 -
5 Elasticity
We need to relate two second rank tensors: stress σij and strain εij.
Each component of σ can in general depend upon all nine components of ε, so we need a
fourth rank tensor:
5.1 Elasticity in isotropic media
In the perfectly general case, Cijkl could have 81 coefficients, so first we will consider the
simpler case of isotropic materials include:
Polycrystalline materials and ceramics (if texture-free) – these are isotropic on a
macroscopic scale although individual crystallites or grains will not be
Amorphous materials (glasses and glassy polymers)
Anisotropy arises from:
Texture (e.g. rolled metals)
Structure (e.g. composite materials, single crystals, wood …)
But we can often assume isotropy as an approximation.
5.1.1 Isotropy
Isotropy has two important, simplifying consequences:
1. The strain depends only on the stress state in the material and not on the orientation
2. The principal axes of stress and strain coincide
EGB/MT13 C4/Tensors
- 59 -
5.1.2 Linear elasticity theory
This makes the dual assumptions of linearity and superposition.
1. Linearity – Hooke’s Law
This assumes a linear relationship (proportionality) between stress and strain e.g.
E is a scalar: the “Modulus of Elasticity” or “Young’s Modulus” and is the same
in both tension and compression.
2. Superposition
Suppose a stress σA produces a strain εA and σB produces εB. Then:
( ) ( )
This is a consequence of linearity and holds only for small strains.
5.1.3 Poisson’s ratio
Consider again the rod in uniaxial tension:
But there is a lateral contraction in the x2x3 plane:
Experimentally we find:
Where ν is Poisson’s ratio, and is constant for any one material (note the sign convention).
Typically it takes values of 0.25 – 0.33 for most materials in the elastic range.
If there were no volume change, then:
This requires ν = 0.5, and is true for the elastic deformation of rubber (caused by coils
unwinding rather than bonds stretching).
C4/Tensors EGB/MT13
- 60 -
Note that ν does not have to be positive, and for example cork has a negative Poisson’s ratio
because of its cell structure, as does α-cristobalite. These are known as auxetic materials.
5.1.4 General expression for strain
Using superposition, we can see that each principal strain is the sum of a tensile extension
due to the parallel stress, and two lateral contractions due to the perpendicular stresses, for
example:
( )
And similarly:
( )
( )
There are no shear stresses in these expressions because the tensors are referred to their
principal axes.
Even for arbitrary cases we still have equivalent expressions:
( )
For simple shear, shear stresses are related to shear strains by τ = μγ:
Or equivalently (given that γ = 2εij):
( )
EGB/MT13 C4/Tensors
- 61 -
μ is the “shear modulus” or “modulus of rigidity” (alternatively sometimes written as G).
5.1.5 Other elastic constants
In isotropic media there are just two independent elastic constants (we can specify any two),
and so there are algebraic relationships between them. For example, to describe volume
changes due to hydrostatic pressure:
Dilatation Δ is given by:
And hydrostatic stress σH by:
( )
K is the bulk modulus and β is the compressibility.
We can find relationships between E, ν, μ, K and β by algebraic manipulation. For example
we can add the three equations relating principal stresses and strains:
( )
( )
Which gives:
And so:
( )
C4/Tensors EGB/MT13
- 62 -
5.1.6 Lamé’s constants
We have expressions for strains given known stresses, but we also need expressions for
stresses given known strains. Manipulating the above equations gives:
( )
( )( )( )
If we define:
( )( )
Then we have:
We call μ and λ Lamé’s constants.
For shear stresses σ12 = 2με12 etc as before. μ and λ are often quoted as the two elastic
constants but the choice is arbitrary.
5.2 Elasticity in general anisotropic media
We want to relate the two second rank tensors stress (σij) and strain (εij). So we write:
∑ ∑
The Cijkl are the 81 stiffness coefficients.
We can also write the reverse:
∑ ∑
Where the Sijkl are the compliances.
Again there are 81 of them, forming a 4th rank tensor.
⁄
EGB/MT13 C4/Tensors
- 63 -
C → stiffness constants (elastic constants, Young’s modulus etc)
S → compliances (elastic moduli)
Both σij and εij are symmetrical and only have 6 independent components., so for stiffness
and compliance the number of independent coefficients is reduced to 66 = 36
i.e. Cijkl = Cijlk = Cjikl = Cjilk
5.2.1 Matrix notation
We have:
independent stress components σ1 → σ6
independent strain components ε1 → ε6
We would like to be able to write expressions like:
( ∑
)
( ∑
)
However going from tensor to matrix notation is not straightforward and must be done
carefully. First we replace the stress notation like this:
[
] [
]
Note that the σ1 → σ6 are not principal stresses.
In order for our equation σi = Cijεj to work, we have to define ε6 = 2ε12 etc, so the 6 strain
terms become:
[
]
[
]
Note that this makes ε4 the simple shear angle γ23 (and similarly for ε5 and ε6).
C4/Tensors EGB/MT13
- 64 -
So we can now write:
( ∑
)
Where Cijkl → Cmn for all m,n (appropriately ordered) i.e. the stiffness constants are
unchanged.
The compliances are more complicated. To write:
We require:
[
] [
] [
]
So that:
Sijkl = Smn when m and n are 1,2,3
Sijkl = ½ Smn when either m or n are 4,5,6
Sijkl = ¼ Smn when both m and n are 4,5,6
Defiend in this way, we can now write:
( ∑
)
Note that the two arrays of numbers Cij and Sij are matrices but not tensors.
5.2.2 Elastic strain energy
Considering the elastic energy stored in a body allows us to simplify the equations further.
Consider a bar in uniaxial tension, and remember that the elastic strain energy stored in a
body must be equal to the work done by the external forces in deforming it:
EGB/MT13 C4/Tensors
- 65 -
At one instant the bar is under a load F with strain ε. Suppose it elongates by dL, then the
work done is:
Where A is the cross-sectional area, and:
So the work done per unit volume is:
[Compare this with σ = Eε for uniaxial stress]
In the general case, and using matrix notation, the work done per unit volume will be:
If the deformation is isothermal and reversible, the work done per unit volume is also equal to
the increase in free energy dψ, and so:
Now we can differentiate twice, giving:
𝜕
𝜕 (𝜕
𝜕 )
Similarly, we can derive: 𝜕
𝜕 (𝜕
𝜕 )
But ψ is a function of the state of the body and defined by the strains, so the order of
differentiation must be immaterial, so:
𝜕
𝜕 (𝜕
𝜕 )
𝜕
𝜕 (𝜕
𝜕 )
And therefore:
C4/Tensors EGB/MT13
- 66 -
Similarly
In the same way as the 1D case, this gives us the strain energy per unit volume = ½Cijεiεj.
Since Cij = Cji, the 36 coefficients reduce to 21 independent coefficients for a perfectly
general anisotropic solid.
5.3 Crystal Symmetry
The 21 independent coefficients are all that we need to describe the properties of a crystal of
minimum symmetry (i.e. triclinic), but for higher symmetry the number of independent
coefficients is lower still.
5.3.1 The cubic system
We should choose axes which reflect the symmetry of the lattice, so in the cubic
system we choose the cube axes (i.e. x1 is parallel to [100], x2 is parallel to [010] and
x3 is parallel to [001]). For the most general result, we choose the least symmetric
cubic point group – 23. This point group has:
Diads parallel to <100>
Triads parallel to <111>
No mirror planes
No 4-fold axes
EGB/MT13 C4/Tensors
- 67 -
The four triads require the three cube axes to be equivalent to each other, so:
(and similarly for the compliances)
The diads require all the remaining constants to be zero (see Nye p137/8 for a proof).
So for all cubic crystals there are only 3 independent elastic constants:
C11
C12
C44
(or S11, S12 and S44)
The stiffness matrix is therefore:
[ ]
So, for example:
And similarly for the other axes.
5.3.2 Deformation modes of a cubic crystal
Each independent elastic constant is associated with a fundamental mode of deformation.
For an isotropic material there are two modes: dilatation and shear.
The three independent constants in a cubic crystal require there to be three independent
modes of deformation
C4/Tensors EGB/MT13
- 68 -
a) Dilatation by hydrostatic stress
Under a hydrostatic pressure p:
Then there are two shear modes:
b) Shear on a cube face parallel to a cube axis
e.g. on the (010) plane in the [001] direction:
C44 is a shear modulus, which by convention we name μo.
c) Shear at 45° to a cube axis
e.g. on the (110) plane in the ̅ direction. In this case the shear modulus is:
( )
EGB/MT13 C4/Tensors
- 69 -
So we can define an anisotropy factor:
( )
For an isotropic material:
And:
This explains why cubic crystals are not in general isotropic.
W Al Fe Cu Na
1.0 1.2 2.4 3.2 7.5
Structure bcc fcc bcc fcc bcc
5.3.3 The Cauchy Relation
Where atoms lie at centres of symmetry and where interatomic forces act entirely
along the line joining atom centres, then:
This is roughly true for ionic crystals (e.g. KCl, NaBr) but does not work for metals,
where bonding is not localised.
C4/Tensors EGB/MT13
- 70 -
5.4 Elastic stress distributions
Additional equations are needed to describe stresses or strains which vary through a body.
We need to take account of:
Stress equilibrium
Strain compatibility
The stress-strain relationship
5.4.1 Stress equilibrium
The net forces on an element must balance, but the stress may change across a volume
element.
For example in direction x1:
𝜕
𝜕 𝜕
𝜕 𝜕
𝜕
may not necessarily equal zero. However it can be shown that:
𝜕
𝜕
𝜕
𝜕
𝜕
𝜕
So we have three equations which (using the summation convention) can be written as:
𝜕
𝜕 [∑
]
EGB/MT13 C4/Tensors
- 71 -
5.4.2 Strain compatibility
The displacements ui must vary smoothly through the body with no gaps and no
discontinuities, which means that the ui components must be single-valued, continuous
functions of xi.
This leads to six “compatibility relations”:
𝜕
𝜕 𝜕 (
𝜕
𝜕
𝜕
𝜕
𝜕
𝜕 ) ( )
And:
𝜕
𝜕
𝜕
𝜕
𝜕
𝜕 𝜕 ( )
5.4.3 Stress-strain relationship
For linear elasticity we have:
( )
which gives another six equations.
So in general we have 3 + 6 + 6 = 15 equations with 15 unknowns: six stress components,
six strain components and three displacements.
5.4.4 The Airy Stress Function
The mathematical treatment of the general condition is formidable unless it is simplified by
symmetry. For example, plane stress gives σ1 and σ2 ≠ 0 and σ3 = 0. In this case we can use
the Airy Stress Function χ. This is defined such that:
𝜕
𝜕
𝜕
𝜕
𝜕
𝜕 𝜕
C4/Tensors EGB/MT13
- 72 -
The Airy stress function allows all 15 equations to be combined into the single biharmonic
equation:
( )
5.4.5 Computational procedure
Solve the differential equation for appropriate boundary conditions to obtain χ. This
can be done analytically in some cases (such as holes)
Differentiate χ to obtain σij
Obtain the strains from the stresses using Sijkl
5.5 Elastic waves
These are important in analysing the dynamic response of materials and structures (for
example impact/explosive loading can give a very different fracture pattern from static
loading). They are also important in the measurement of mechanical properties at high strain
rates.
Ultrasonic waves in particular are used for:
Flaw detection (non-destructive testing)
Medical applications
Microstructure refinement during casting and welding
Piezoelectric transducers in scanning probe microscopes
We will consider only isotropic materials (because anisotropic behaviour is much more
complex).
5.5.1 Longitudinal waves in a rod
There is a uniform stress across the planar cross-section.
EGB/MT13 C4/Tensors
- 73 -
The net force on an element is:
𝜕
𝜕
If the displacement of an element from its rest position is u, we can use:
( )
And:
𝜕
𝜕
To give:
𝜕
𝜕
𝜕
𝜕 ( )
This is a 1-dimensional wave equation for a longitudinal wave propagating with velocity vo,
where:
√
These are rod waves. Similarly, for pure shear we get torsion waves according to:
√
5.5.2 Waves in infinite media
In this case the relevant strain is ε1, and the relevant modulus is C11, so we get compression
waves with a velocity given by:
√
√
Similarly we get transverse waves (shear distortion waves) according to:
C4/Tensors EGB/MT13
- 74 -
√
In general for a bulk, isotropic material there will be two waves, corresponding to the two
independent deformation modes.
For a fluid, μ = 0 (there is no shear), so only longitudinal waves are possible.
5.5.3 Surface waves
Displacements perpendicular to a surface are Rayleigh waves. On the surface of the Earth,
these are responsible for causing earthquake damage.
( )√
( )
( )
So in general, vs < v2, so the Rayleigh (surface) waves arrive after the bulk waves, and also
lose least energy because they are propagating in only two dimensions instead of three.
Elastic wave velocities (ms-1)
Steel Aluminium Rubber
Rod waves vo 5190 5090 46
Bulk, longitudinal v1 5940 6320 1040
Bulk, shear v2 3220 3100 27
Surface vs 2980 2920 26
EGB/MT13 C4/Tensors
- 75 -
5.6 Saint-Venant’s principle
Any geometrical discontinuity (holes, cracks, notches …) will create a non-uniform stress
distribution, but far away from the discontinuity the stresses will remain uniform. This is an
example of Saint-Venant’s principle in operation, which can be stated as:
The stress distribution in a body depends only locally
on discontinuities or on the method of loading
Or more formally:
If the forces acting on a small part of the surface of a body are replaced
by statically equivalent forces (i.e. same resultant and couple),
the stress state is negligibly changed at large distance.
For example, the stress in the middle of a plate does not depend upon the method of
gripping:
This is an important principle which is implicit in all mechanical testing.
C4/Tensors EGB/MT13
- 76 -
5.7 Examples of stress distributions
5.7.1 Holes in plates
In general, the solutions for the stresses are:
{
(
) }
(
)
{
(
) }
Examining the limiting cases illustrates how the stresses vary:
At r = ro, we have σrr = 0 and σrθ = 0 because there is no stress perpendicular to a free
surface. But σθθ = σ ( 1 – 2 cos 2θ), so
θ = 0 σθθ = – σ
θ = ±π/2 σθθ = 3σ
EGB/MT13 C4/Tensors
- 77 -
So we have stress magnification (3σ) beside the hole, which can lead to crack propagation
and catastrophic failure and stress reversal (–σ) above and below the hole. In brittle
materials (e.g. masonry) this can lead to failure above and below holes where a compressive
stress is reversed to become tensile.
At θ = π/2 ( cos 2θ = – 1 ) and large r, the stress falls off as 1/r2. As r → ∞
which is as expected from Saint-Venant’s principle.
C4/Tensors EGB/MT13
- 78 -
5.7.2 Stresses at notches and cracks
[See Cottrell p145 for an approximate treatment of notches and Knott p29 for a more
sophisticated treatment]
We can use the results for a hole to get a rough idea of the effect of a notch.
If there are n holes in total, after the nth and last hole the stress is:
Suppose that the radii decrease by a factor of 10 from one hole to the next, with the largest
radius being a and the smallest ρ. Then:
(
)
Taking logs of the first equation we have:
(
)
(
)
Taking exponentials of both sides gives:
√
EGB/MT13 C4/Tensors
- 79 -
This is a good order magnitude estimate – for comparison, the exact solution for the stress at
the edge of an elliptical hole is:
( √
)
Note also (and don’t confuse with) the result for a sharp crack:
√
5.7.3 General features of stress distributions around holes and notches
C4/Tensors EGB/MT13
- 80 -
The general features of the stress distribution ahead of a notch are similar to those ahead of
a hole. The tensile stress in the radial direction is important in understanding how cracks are
stopped in composite materials:
If the fibre/matrix interfaces are weak in tension, they can open up ahead of a crack resulting
in crack blunting and stopping.
5.8 Dislocations
5.8.1 Screw dislocations
For the volume element on the surface of the cylinder surrounding the core of the dislocation:
In a finite medium this causes a problem at the ends of the cylinder because the forces do
not balance. This causes an Eshelby twist along the length.
Consider an annular element of a rod of total length L:
EGB/MT13 C4/Tensors
- 81 -
Shear stress
Area
Tangential force
Moment
Total torque ∫
Suppose that the torque T causes a twist φ, which is the same everywhere. For the annular
element:
Shear strain
Shear stress (
)
Tangential force (
)
Moment (
)
Total torque ∫ (
)
(
)
So for this dislocation:
(
)
which gives the Eshelby twist as :
C4/Tensors EGB/MT13
- 82 -
Note that this depends on b and R, but is independent of μ.
5.8.2 Edge dislocations
In this case the expressions for the stresses are:
( )
( )
This is plane strain, so:
( )
( )
Exploring the special cases:
⁄ ( )
( )
⁄ ( )
( )
EGB/MT13 C4/Tensors
- 83 -
The region below the dislocation half plane is therefore a desirable site for solute atoms.
PART II MATERIALS SCIENCE C4: Tensors
EGB/MT13
Lecture Examples
Lectures 4 and 8 will be in Examples Class format, during which you will have the
opportunity to work through the following questions. These are closely based upon
the content of the immediately preceding lectures.
1. A single crystalline orthorhombic material has an electrical conductivity tensor ij:
[
]
when referred to its principal axes x1, x2, x3 which are parallel to the crystallographic axes x, y
and z. A thin plate is cut from the crystal according to ‘new’ axes x1', x2', x3', with x1' being
the normal to the plate and x2', x3' lying in the plate. The axes are such that x2' is parallel to
x2, and x1' lies between x1 and x3 at 30˚ to x1. Calculate 11'.
Electrodes are plated on to the faces of the thin plate and a potential gradient E (V m–1) is
applied to it. The electric field vector will be perpendicular to the plate faces. What are the
components of E, referred to the 'old' (principal) axes? Using the diagonalised tensor ij
calculate the current density vector (referred to the old axes).
2. The orientation of a vector (x1, x2, x3) can be described by nine direction cosines (aij)
relative to the co-ordinate axes. These direction cosines are not independent, and are related
as follows:
( )
( )
Why is this the case?
3. A body is subjected to principal stresses of magnitudes 5, –3 and 0 respectively along
the x1, x2, and x3 axes. Calculate the normal and shear stresses acting on a plane which is
perpendicular to the x1x2-plane and inclined at 55˚ to the x1x3-plane.
4. Almost all second rank tensors of interest, whether of the matter or field type, are
symmetrical, i.e. Tij = Tji. Tensors with Tij = –Tji are called antisymmetrical. Separate the
following tensors into symmetrical and antisymmetrical components.
[
] [
]
EGB/MT13 C4/Tensors
- 85 -
If these tensors were describing general strains, what would be the significance of the
symmetrical and antisymmetrical components?
5. Separate the following stress tensors into hydrostatic and deviatoric components:
[
] [
] [
]
6. The strains measured by three strain gauges arranged in a 60˚ rosette on the surface
of a steel plate are 1.2 10–3, 2.0 10–3 and –1.0 10–3. Determine the orientation and
magnitude of the principal strains in the surface. What is the maximum shear strain?
7. A 45° strain gauge rosette attached to the surface of an aluminium alloy panel
measures the following strains:
εa = –0.0025; εb = 0.0020; εc = –0.0040
The orientation of the gauges is as shown:
a) What are the normal stresses in the a, b and c directions?
b) What are the principal stresses in the plane of the surface?
The Young’s modulus of the aluminium alloy is 70GPa and its Poisson’s ratio is 0.33
8. Derive the following relationships:
( )
( )
Using these relationships, show that the bulk modulus K is given by:
( )
C4/Tensors EGB/MT13
- 86 -
9. A body of an isotropic elastic material is subjected to principal stresses along the
axes x1, x2 and x3 of such magnitudes that the elongations along x2 and x3 are zero. In a
second experiment it is stressed along these same axes in such a way that the elongations
along x2 and x3 are equal and that along x1 is zero. Show that the effective modulus (stress-
strain ratio) along x1 in the first case is (1 – ) times that along x2 or x3 in the second case.
PART II MATERIALS SCIENCE C4: Tensors
EGB/MT13
Preparatory Questions
You should be able to complete these questions from your knowledge and understanding of
the IB Mechanical Properties course. You may need to review your notes from that course,
and the recap at the start of the Part II Tensors course will also be helpful. Start (and finish)
this sheet as soon as you can – you should not need help from your supervisor with these
questions!
1. Construct a Mohr circle diagram for each of the following stress states and deduce
the values of the principal stresses. Describe each stress state in words. Then diagonalise
each tensor using the secular equation and check that you obtain the same values of
principal stress.
[
] [
] [
]
2. Construct Mohr circles for the following stress tensors. Find the principal stresses
and the angle between the tensor axes and the principal axes. Then calculate the same
quantities from the secular equation.
[
] [
]
3. A cylindrical tube of diameter 80 mm and wall-thickness 2 mm is pressurised to
10 MPa above atmospheric pressure. It bears a tensile load of 12.5 kN and is subjected to a
torque of 1 kN m at its ends. What are the principal stresses in this system, and at what
angles to the tube axis do they occur?
Answers
1. (a) σ1 = 2; σ2 = 0; σ3 = 0
(b) σ1 = +1; σ2 = -1; σ3 = 0
(c) σ1 = 0; σ2 = 0; σ3 = -4
2. (a) √ √
(b) √ √
3. (a) MPa MPa
PART II MATERIALS SCIENCE C4: Tensors
EGB/MT13
Question Sheet 1
1. A three-dimensional state of stress is given by
11 = + 80 MPa 12 = + 20 MPa
22 = – 40 MPa 23 = + 30 MPa
33 = + 60 MPa 31 = – 50 MPa
Show that the total stress on a plane with direction cosines:
√
is 106.6 MN m–2 and that the normal and shear stresses on this plane are 79.5 MPa and
71.0 MPa respectively.
2. A three-dimensional state of strain is given by:
11 = 4 10–4 12 = 3 10–4
22 = –6 10–4 23 = –3 10–4
33 = 2 10–4 31 = –5 10–4
Show that the greatest axial strain is 9.27 10–4 and that there is no volume expansion.
3. Tests are made in combined tension and shear on molybdenum single crystals to
decide whether cleavage fracture is controlled by a critical value of the normal stress across
the most highly stressed {100} cleavage plane. One such experiment is made on a crystal
whose tensile axis lies along [123]. If the crystal fractures under a combination of a tensile
stress of 400 MPa and a shear stress, acting on the (123) plane in the [2–1 0] direction, of 300
MPa, which is the {100} plane with the highest normal stress, and what is the critical value of
this stress?
4. A transformation of axes should not, of course, affect the property of a material or a
stress state, but the description in terms of tensor components will be changed. Show,
however, that a symmetrical tensor (of the second rank) will always remain symmetrical on
transformation of the reference axes.
PART II MATERIALS SCIENCE C4: Tensors
EGB/MT13
Question Sheet 2
1. (From Tripos 1996, Qu. 11) A cylindrical strut 20 mm in diameter is made from a
titanium alloy. A strain gauge is fixed to the surface of the strut at an angle of 30˚ to the axis
of the strut. What strain is registered by the gauge when the strut is subjected to an axial
tensile load of 5 kN? An axial torque is then applied to the strut, producing an additional
surface shear strain of 5 10–5. The sense of the torque is such as to decrease the strain
registered by the gauge. What value of strain is recorded by the gauge under these
conditions?
[For the titanium alloy, Young’s modulus E = 120 GPa, Poisson’s ratio = 0.30]
2. Two strain gauges are arranged to measure the principal strains on the surface of a
thin-walled tube which is subject to a tensile load P and a torque T. When the tube is on the
point of yielding the tensile strain has just twice the magnitude of the compressive strain.
Show that at yield T = 0.56 Pr, where r is the radius of the tube. [Poisson's ratio = 1/3]
3. In the manufacture of a test circuit, a polycrystalline thin film of aluminium is
deposited on to a thick silicon substrate at 120˚C. Subsequently, the silicon wafer with
aluminium layer uppermost is mounted in an X-ray diffractometer (in the standard setting with
the normal to the wafer bisecting the angle between the incident and diffracted beams). At
room temperature (20˚C) the diffractometer trace has aluminium peaks showing a lattice
parameter decrease of 0.167%, compared to the unstressed state at 20˚C. Assuming that
the aluminium was stress-free during deposition, what is the thermal expansion coefficient of
aluminium?
[The thermal expansion coefficient for silicon is 7.6 10–6 K–1. Poisson’s ratio for aluminium
is 0.345.]
4. A thin multilayer of many alternating (100) GaAs layers 10 unit cells thick and (100)
AlAs layers 4 unit cells thick is deposited on to a thick (100) GaAs substrate.
(a) If all the interfaces are fully coherent, what are the diagonalised strain tensors
representing the states in the GaAs layers and in the AlAs layers?
(b) If the multilayer were to lose coherency with the substrate but to retain it between the
GaAs and AlAs layers, the net force in the multilayer would be zero, i.e. the tensile and
compressive forces in alternate layers would balance. In that case, what would be the strain
tensors?
(c) If all coherency were to be lost, what would be the tensors?
[The lattice parameter, a, for GaAs is 0.5563 nm and for AlAs is 0.5571 nm. Poisson's ratio
is 0.429 for both GaAs and AlAs. Assume that GaAs and AlAs have equal elastic moduli.]
PART II MATERIALS SCIENCE C4: Tensors
EGB/MT13
Question Sheet 3
1. Poisson’s ratio for rubber is 0.5. Show that a bicycle inner tube can be inflated with
zero change in circumferential length. (A small length of tube can be treated as a thin-walled
cylinder.)
2. The stress distribution in a rotating circular disc is of practical importance in the
design of gas turbine engines. For a disc containing a central hole of radius a, the principal
stresses in the plane of the disc depend on its density , its angular velocity and its
maximum radius b as follows:
{
}
{
[
] }
For a disc containing a central hole of radius a = b/10, plot the variation of r and with r.
Assume that Poisson's ratio is 1/3. Where are the maximum radial and tangential stresses?
Compare the maximum value of tangential stress with the maximum value for a disc with no
hole (a = 0) and note the stress-concentrating effect of the hole.
If a rotating transparent disc of suitable material is viewed in a polariscope, isochromatic
fringes will be seen joining all points with the same value of (r – ). Sketch the
appearance of these fringes.
3. An infinitely long edge dislocation of Burgers vector b in an
elastically isotropic medium, with shear modulus μ and Poisson’s ratio
ν, gives rise to plane strain and, via the Airy stress function, the
following stress tensor components are found:
where r and θ are defined with respect to the dislocation in the figure, and:
( )
(continued overleaf)
EGB/MT13 C4/Tensors
- 91 -
a) Determine
b) Determine the strain tensor
c) Derive an expression for the dilatation and determine at what angle this is a
maximum for r > 0
d) Using the Mohr circle construction or otherwise, determine expressions for the
principal strains
[from Tripos 2010 q2(b)]
Past Tripos Questions
This list covers those Part II Tripos questions set during the last five years which were
directly based upon the Tensors course. It does not include questions drawn from other
courses which nevertheless require a sound knowledge of tensors.
Whilst the content of the Tensors course itself has remained relatively constant over recent
years, there have been changes in the structure, content and examination of the Part II
course as a whole, and also in the coverage of tensor-related material in Parts IA and IB. All
of the questions on this list could be attempted by a student from the current Part II Class,
but slight changes in coverage mean that some of the problems may appear more difficult
(or, conversely, easier!) than when they were first set. This is particularly likely to be true for
questions older than these which you may find in bound collections of Tripos papers in
libraries.
2013 Paper II questions 4; 5
2012 Paper I questions 3; 4
2011 Paper I questions 3; 4
2010 Paper I questions 2; 3; 4
2009 Paper I questions 2; 3; 4
