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EGB / MT 2012 C4 Tensors / C4-H1

Part II Materials Science

TENSORS

Course C4 (9 + 2 lectures) Dr E G Bithell

SYNOPSIS

This course will be given in 11 one-hour blocks. It does not have a separate examples class, but the allocated time will be divided between lecture-style delivery and short blocks of time for you to spend on worked examples.

Introduction What is a tensor? Tensor quantities and properties (field and matter tensors).

Stress and strain Key principles (brief recall of IB Mechanical Properties course). Definition of stress at a point. Notation and coordinate systems. Einstein summation convention; resolution of stresses and transformation of axes; principal axes; diagonalisation of a general tensor; the Mohr circle construction. Varying stress and strain fields. Definitions of strain at a point. Distinction between rigid body displacement and rotation, and shear.

Methods of experimental stress analysis Strain gauges: fundamentals underlying gauge factor. Practical details. Analysis of strain gauge results. Photoelasticity: birefringence, stress-optical co-efficients, isochromatic and isoclinic fringes, applications. Other methods of stress measurement: brittle coatings, X-ray diffraction, ultrasonics. Residual stresses: origin and measurement. Stresses and strains in thin films: origin, measurement, epitaxy.

Tensor properties other than elasticity The representation surface for second rank tensors. Optical properties and the indicatrix. Piezoelectricity: tensor notation. Applications of the piezoelectric effect, electrical transducers in a variety of devices. Effects of symmetry. Ferroelectricity, pyroelectricy, polycrystalline piezoelectric transducers. Introduction to higher order properties.

Elasticity Isotropic medium: linear elasticity theory, principle of superposition, interrelationships of elastic constants. General anisotropic medium: stiffness and compliance tensors. Simplification by symmetry: matrix notation. Strain energy density: symmetry of general stiffness and compliance


matrices. Effects of crystal symmetry. Physical interpretation of three deformation modes in a cubic crystal. Anisotropy factor for several materials.

Elastic stress distributions General elasticity theory: stress equilibrium, strain compatibility, stress-strain relationship. St. Venant's principle. Examples of elastic stress distributions: circular hole in plate, notch in plate, dislocations.

Elastic waves Dynamic fracture behaviour, ultrasonics. Wave equation for longitudinal wave in rod. Other waves: torsion in rod, dilatation and distortion in an infinite medium, Rayleigh waves. Comparison of wave velocities for steel, aluminium and rubber.


REFERENCE LIST

Principal textbooks Nye: Physical Properties of Crystals, chapters 1, 2, 5, 6 & 8 (library code: Lj13b, ref)

We use Nye's notation throughout. Fundamental basis for the description of material properties using tensors.

Lovett: Tensor Properties of Crystals, all chapters (NbA 73b) Less comprehensive than Nye, but the content of this book corresponds very closely to that of the course. Relatively cheap in paperback.

Other textbooks Cottrell: Mechanical Properties of Matter, chapters 4, 5 & 6 (Lj18a, ref)

Elasticity, stress distributions, elastic waves Dieter: Mechanical Metallurgy, chapters 1 & 2 (Kz1a,c,e, ref)

Analysis of mechanical failures, stress and strain, elasticity Holister: Experimental Stress Analysis, chapters 1, 2 & 4 (Ky2)

Strain gauges, photoelasticity Kelly and Groves: Crystallography and Crystal Defects, chapters 4 & 5 (Ll30)

Tensors, stress and strain, elasticity Knott: Fundamentals of Fracture Mechanics, chapters 1 & 2 (Ke45a, ref)

Modes of failure, stress concentrations Le May: Principles of Physical Metallurgy, chapters 1, 2 & 4 (Kz31b, ref)

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, ref)

Tensile testing, yield point phenomena, elasticity

Web resources DoITPoMS TLPs: Introduction to Anisotropy; Tensors; Dislocations


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 reproduce them from memory in order to carry out derivations or calculations.

Einstein summation convention (example):

= , = 1,2,3 Total, normal and shear stresses on a plane:

|| = + + =

= Hydrostatic stress:

= 3 = + + 3 Dilatation:

= = + + Elasticity relationships (tensor notation):

= !"!" [stiffnesstensor] = -!"!"[compliancetensor] Superposition of normal stresses in an isotropic medium:

= 3 + etc

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.


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 (Neumanns 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


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

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

Example 1: annotate this table with M for matter tensors and F for field tensors


1.3 Anisotropy and symmetry

These material properties are matter tensors and must obey Neumanns 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 Neumanns Principle. First we need to understand how to classify the symmetry found in different types of macroscopic materials:

An isotropic material

Has infinite symmetry (rotation axes/mirrors in all directions) Is found not only in a macroscopic continuum but in a real amorphous solid composed of atoms

Lower symmetry

Is clearly seen in single crystals

Is described by the 32 point groups, none of which have infinite symmetry

Polycrystals

Are macroscopically isotropic if crystal shape and orientation are random Otherwise have preferred orientation or texture anisotropy

Amorphous solids These may be anisotropic but we can induce isotropy 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.


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).

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

But we need to consider also: shear stress

that the stress in a real object can be very complicated

So we need to describe the stress at a point

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be able to deal with a general (complicated) stress

This will require several force/area terms.

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 cood

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:

i corresponds to the direction

be able to deal with a general (complicated) stress

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

Choice of volume element and coordinate system

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 cood

handed coordinate system

the volume element is a cube


direction in which the stress (or force) acts

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e need to:

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 coodinates.


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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 x

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 opposto 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 rotatioexample, considering moments about the x axis, and only showing the relevant stresses, the total moment must be zero

So:

In general, ij = ji for all i,j so our 3x3 array is symmetrical and there are onlquantities. This array of 9 numbers is a second rank tensor, so we have also demonstrated that stress is a symmetrical second rank tensor.

(or its normal) on which the stress (or force) acts

represents a force acting in the x1 direction on a face normal to the x

(and others exist).

This means that we can identify two kinds of stress:


a) Stresses on equal and opposite faces of the cube are equal and opposite, and we only need . This means that 9 quantities are sufficient to describe the stress at a

point, and we can write these as a 3x3 array:

4 5 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

= for all i,j so our 3x3 array is symmetrical and there are onl

quantities. This array of 9 numbers is a second rank tensor, so we have also demonstrated that second rank tensor.

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(or its normal) on which the stress (or force) acts

direction on a face normal to the x3 direction. This


ite, and we only need . This means that 9 quantities are sufficient to describe the stress at a

nal as well as linear acceleration. For

example, considering moments about the x axis, and only showing the relevant stresses, the

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

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2.2.2 Cylindrical coordinates

These are useful in describing

zr = force in z direction on r face, and as before

2.3 Alternative notations

The notation used is a matter of conventionshear stresses is common, which gives equivalent quantities as:

Normal stresses

(e.g. stress parallel to x

Shear stresses

(e.g. stress parallel to x

We will use the numerical suffix

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.

4 5

Cylindrical coordinates

These are useful in describing cylindrically symmetrical geometries (tubes, disclocations ...)

= force in z direction on r face, and as before zr = rz etc.

is a matter of convention and convenience: in particular the use of , which gives equivalent quantities as:

Equivalent notations

Normal stresses

(e.g. stress parallel to x1 on face 1) 11

Shear stresses

(e.g. stress parallel to x2 on face 1) 21

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

se other conventions and it is important that you become accustomed to these differences, which are widespread in both textbooks and the research literature.

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cylindrically symmetrical geometries (tubes, disclocations ...)

: in particular the use of for

Equivalent notations

x

xy

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

se other conventions and it is important that you become accustomed to these differences, which are widespread in both textbooks and the research literature.

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2.4 Examples of stress tensors

2.4.1 Uniaxial tension

Choosing sensible axes, with one (x

( > 0 for tension)

2.4.2 Uniaxial compession

( < 0 for compression)

Examples of stress tensors

Choosing sensible axes, with one (x1) along the axis of the rod, 11 = , and all other

4 0 00 0 00 0 05

4 0 00 0 00 0 05

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, and all other ij = 0:

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2.4.3 Hydrostatic compression

This time there are no shear stresses:

2.4.4 Pure shear

12 = 21 = and all other ij = 0, so we have:

Hydrostatic compression (e.g. under water)

are no shear stresses:

4 0 00 00 0 5 7 0

= 0, so we have:

40 8 08 0 00 0 05

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2.3 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 m1] potential gradient

current density J = (J1, J2, J3) [units A m2] current density

For a general anisotropic material the relationship between J and E can be expressed as a series of equations relating their components:

J1=11E1+12E2+13E3 J2=21E1+22E2+23E3 J3=31E1+32E2+33E3

This describes the conductivity as a tensor of the second rank, written as:

4 5The 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)


We can rewrite the previous equations for the tensor components as summations, so these

equations:

J1=11E1+12E2+13E3 J2=21E1+22E2+23E3 J3=31E1+32E2+33E3

Become:

=

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2.4 Transformation of axes

We need to go from old axes (x

Axis x2', for example, makes angles

direction cosines, a21 = cos 21angular relationships between the axes. The nine direction cosines form a transformation matrix:

So we have:

x'2

and so on ...

If one of the axes is unchanged this reduces to a 2x2 matrix which should look famil

axes (x1, x2 and x3) to new axes (x1', x2' and x3').

', for example, makes angles 21, 22 and 23 with the x1, x2 and x3 axes. We define

21, a22 = cos 22, a23 = cos 23 etc, completely specifying the

angular relationships between the axes. The nine direction cosines form a transformation

Old axes

x1 x2 x3

New

ax

es x'1 a11 a12 a13

x'2 a21 a22 a23

x'3 a31 a32 a33

2 = x1 cos 21 + x2 cos 22 + x3 cos 23

If one of the axes is unchanged this reduces to a 2x2 matrix which should look famil

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axes. We define

specifying the

angular relationships between the axes. The nine direction cosines form a transformation

If one of the axes is unchanged this reduces to a 2x2 matrix which should look familiar!


Example 2

The nine direction cosines are not independent, because:

>!>! = 1 = >!>! = 0

Why is this the case?

2.5 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:

A = > + > + > A = > + > + > A = > + > + >

or:

A = > which is the 'old' to 'new' transformation.

Similarly for the 'new' 'old' transformation

= > + > + > and so on, i.e. = >


So we have:

New from old A = > Old from new = >

But we want to relate J' to E':

=>!! and ! =!"" and " =>" Therefore: =>!!">" But also =

So:

A = >!>"!" This gives us the new from old transformation, and similarly for the old from new transformation: = >!>"!"


2.6 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'

Example 3(a) A single crystalline orthorhombic material has an electrical conductivity tensor ij:

422 0 00 10 00 0 75 10DFmF

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'.

Example 3(b) Electrodes are plated on to the faces of the thin plate and a potential gradient E (V m1) 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).

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2.7 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

account the plane on which the forces act (i.e

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.


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.

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because we have to take into . 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

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We consider the action of a vector force

orientation is specified by its unit normal vector

the axes x1, x2, x3, and the direction of

1, 2 and 3 are also the components of the unit vector

In order to work out the stress on the plane ABC due to 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 face ABC can be resolved into its components P

balanced by the stress components acting on the other faces of the volume element OABC.

We write the area of triangle ABC as

P is a force per unit area, so the force on ABC is:

We can resolve P. ABC because it is a vector:

. HI

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

, and the direction of n is defined by its direction cosines.

= JKL = JKL = JKL

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 which has ABC as one of its faces, and it is simplest to take the tetrahedron OABC. In

he net force on this element must be zero : the force per unit area face ABC can be resolved into its components P1, P2, P3 and each of these should be


We write the area of triangle ABC as ABC (and other triangles similarly).

is a force per unit area, so the force on ABC is:

>NO>KPHI = . HI ABC because it is a vector:

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ea (or "stress") on a plane ABC whose makes angles 1, 2, 3 with

is defined by its direction cosines.

, 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

he net force on this element must be zero : the force per unit area P acting on and each of these should be


triangles similarly).


But P. ABC 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:

. HI = TI + TH + THI This can be simplified by noting that the volume of the tetrahdron OABC is:

KUVWO = 13TX. HI = 13TH. TI = 13TI. TH = 13T . THI

And:

TX = T . cos Y = T . = TI. = TH.

Therefore:

THI = TXT . HI = . HI

Likewise:

TI = . HI TH = . HI


Now dividing through by ABC:

= + + 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 1, 2, 3), and defines the ij as a second rank tensor relating force and plane orientation:

= Remembering that:

=

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2.8.1 Normal stress

This is important in (for example) tensile fracture

We need Pn, the component of

We resolve each Pi in the direction of

But:

This is important in (for example) tensile fracture

, the component of P normal to the plane

in the direction of n:

= + + =

= = KN =

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2.8.2 Shear stress

This matters for slip

We want Ps, the component of

We know that:

And we have expressions for P and P

Example 4

A body is subjected to principal stresses of magnitudes 5, 0 and x2, and x3 axes. For a plane which is perpendicular to the xthe x1x2-plane, show that the 3.46.

, the component of P lying in the plane

= + And we have expressions for P and Pn, so we can find Ps from:

=

A body is subjected to principal stresses of magnitudes 5, 0 and 3 respectively along the xaxes. For a plane which is perpendicular to the x1x3-plane and inclined at 30

plane, show that the normal and shear stresses acting on it are respectively

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3 respectively along the x1, plane and inclined at 30 to

normal and shear stresses acting on it are respectively 1 and

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2.9 Principal axes

All symmetrical second rank tensors possess three orthogonal principal axes. a tensor to write its components referred to these axes leads to a simpl

4\\\

2.9.1 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

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

All symmetrical second rank tensors possess three orthogonal principal axes. a tensor to write its components referred to these axes leads to a simplified form of the tensor:

\ \ \ \ \ \5 ^\ 0 00 \ 00 0 \_

.1 Principal axes of the stress tensor


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:

4 0 00 00 0 5

; ; This is a diagonal tensor

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All symmetrical second rank tensors possess three orthogonal principal axes. Diagonalising ified form of the tensor:


the plane perpendicular to a principal axis

So whatever the state of stress, however complicated, there is always a set of axes such that


There are no shear stresses

By convention, a single suffix denotes a principal stress

2.9.2 Diagonalisation of a tensor

This is an important procedure ...

We need to find the orientation of a plane normal n (specified by direction cosines 1, 2, 3) for which the total force per unit area P is parallel to the normal.

Let the magnitude of P be , so:

= b; = b; = b

But:

= = + + = b = + + = b = + + = b

So:

c b + + = 0 + b + = 0 + + b = 0d These three simultaneous equations have non-trivial solutions only if their determinant is equal to zero:

e b b be = 0 This gives a single cubic equation in the secular equation with three roots (values of ) corresponding to the three principal stresses.


2.9.3 Orientation of the principal axes

We need to find 1, 2, 3 for each principal axis.

Take each value of and substitute it back into the 3 simultaneous equations

Check: + + = 1 Then we have:

4 0 00 00 0 5 where 1, 2, 3 are the three values of

By convention the principal axes are (nearly always!) labelled such that: > > 2.9.4 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:

4 0 00 0 5 Remembering that 12 = 21 The diagonalisation required is in two dimensions, not three, and is comparatively straightforward. The determinant becomes:

e b 0 b 00 0 be = 0 Multiplying out to get the secular equation:

bg b b h = 0

One solution is = 3 for the principal axis already known. The other principal stresses are obtained by solving the remaining quadratic:

b + b + = 0

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b = += +2

2.9.5 The Mohr circle construction

Tha above solutions can also be represented geometrically. tensor, and take 11 > 22:

We construct the circle using:

From which:

+ i j + 4 2 + 2 i lm 2 n +

ircle construction

Tha above solutions can also be represented geometrically. We start with the same stress

4 0 00 0 5

I = H = 2 H = =

tan 2Y = 2

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We start with the same stress


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.

The angle of rotation of the diameter (= 2) is twice the angle of rotation of the axes in real space (= ). The sense of rotation is the same.

The same principal stresses could arise from different combinations of 11, 22 and , 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).

2.9.6 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.

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2.9.7 Representation of a 3D stress state

This can be done by plotting 3perpendicular to a principal axis.

For example, consider again the pure shear case:

x1x2 plane

o+ 00 p

Note that the maximum shear stress still

also that this is one case where we commonly do not observe the

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:

4+ 0 00 00 0 05 x1x3 plane

o+ 00 0p x

o

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 1 > 2 >

C4 Tensors / C4-H32

Mohr circles on one diagram, with one circle for plane

x2x3 plane

o 00 0p

to the principal axes

> 3 convention!

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2.9.8 Triaxial tension

This time we will specify 1 >

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 circles.

> 2 > 3, and the stress tensor is then:

4 0 00 00 0 5

8qrs = 2 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

C4 Tensors / C4-H33

Note that although we can represent the stress state of a 3D system using Mohr circles when

3D tensor using Mohr

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2.10 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 x2, x3) has moved to P' (x'1, x'2

So:

If the ui are the same for all points, then we have a

For there to be a strain, the ui must vary through the body.

In one dimension:

So:

Definition of strain at a point

Starting in one dimension, imagine an extensible string:

TT = Vt And the strain at the origin is defined as:

O = limsu mVtn Now we do the same in 3 dimensions. After deformation of a body, suppose that point P (x

2, x'3) with displacements u1, u2, u3.

t = t + V are the same for all points, then we have a rigid body displacement

must vary through the body.

V = Pt

O = vVvt

V = Ot C4 Tensors / C4-H34

se that point P (x1,

displacement i.e. zero strain.

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And in 3 dimensions:

Which can be written as:

The form of the summation convention defines e

2.9 Physical meaning of eij

O = wVwt These are simply the tensile strains along the axes x

as for stress: positive in tension

Now consider what would happen for deformation in the xwhen:

For small angles, very much exaggerated, this corresponds to:

Given that:

O x

V = Ot + Ot + Ot V = Ot + Ot + Ot V = Ot + Ot + Ot

V = Ot onvention defines eij as a second rank tensor

O = wVwt OThese 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,

O 0;O 0 For small angles, very much exaggerated, this corresponds to:

wVwt x >yzUO;O x wVwt x >yzUO

C4 Tensors / C4-H35

second rank tensor.

O = wVwt . Note that the signs are the same

plane of a rectangular block,

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Consider possible combinations of e

1. e12 = e21 is pure shear

2. e12 = e21 is pure rotation

3. e12 0 and e12 = 0 is simple shear

Note the distinction between pure

In general e12 and e21 describe both shear which (e21 + e12) is constant produce the same shape change but different rotations.

We can separate the effects of shear and

Or in general:

Where:

Consider possible combinations of e12 and e21:

pure shear

pure rotation (with no deformation)

simple shear

pure shear and simple shear.

describe both shear and rotation together. Note that displacements for ) is constant produce the same shape change but different rotations.

We can separate the effects of shear and rotation by writing:

O = + {

O = + {

= 12 |O + O} C4 Tensors / C4-H36

rotation together. Note that displacements for ) is constant produce the same shape change but different rotations.

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And:

Note that ij is symmetric with respect to i and j, and is a measure of shape change, whereas ij is antisymmetric with respec

And:

So ij has only 3 independent tensor elements, giving 3 independent rotations each coordinate axis.

Note the form of the strain tensor written in terms

From now on,

In this way, both stress and strain

{ = 12 |O O} is symmetric with respect to i and j, and is a measure of shape change, whereas

is antisymmetric with respect to i and j and is a measure of rotation:

= ; { = {

{ = 0 = has only 3 independent tensor elements, giving 3 independent rotations

Note the form of the strain tensor written in terms of the shear strain angle in simple shear:

O = 40 ~ 00 0 00 0 05 = +{ From now on, the strain tensor will mean ij

stress and strain can be written as symmetrical second rank tensors

C4 Tensors / C4-H37

is symmetric with respect to i and j, and is a measure of shape change, whereas

has only 3 independent tensor elements, giving 3 independent rotations one about

of the shear strain angle in simple shear:

can be written as symmetrical second rank tensors

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Example 5 Almost all second rank tensors of interest, whether of the symmetrical, i.e. Tij = Tji. Tensors with Tfollowing tensors into symmetrical and antisymmetrical components.

If these tensors were describing general strains, what would be the significance of the symmetrical and antisymmetrical components?

2.10 Separating shape changes from volume changes

2.10.1 Hydrostatic and deviatoric components of

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:

400And 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

Almost all second rank tensors of interest, whether of the matter . Tensors with Tij = Tji are called antisymmetrical. Separate the

following tensors into symmetrical and antisymmetrical components.

42 3 75 2 53 11 2545 1 145 4 60 4 2 5

If these tensors were describing general strains, what would be the significance of the symmetrical and antisymmetrical components?

Separating shape changes from volume changes

Hydrostatic and deviatoric components of stress

We can separate out two components of stress:

volume changes

shape changes

under compressive pressure the stress tensor is:

0 0 00 5where = = = And plotting this on a Mohr circle:

Mohr circle is simply a point


There is never any shear stress

C4 Tensors / C4-H38

or field type, are are called antisymmetrical. Separate the

If these tensors were describing general strains, what would be the significance of the



So in general we can write:

4 5 = 4 0 00 00 0 5 + ^

_ 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:

= 3 = + + 3 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.

2.10.2 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:

= 1 + 1 + 1 + x 1 + + + = + + =

Where is known as the dilatation.

So:

4 5 =

13 0 00 13 00 0 13

+

13 13 13

i.e. the full strain tensor is the sum of the dilatational component (volume change) and the deviatoric component (shape change).


Example 6

The stress tensor is symmetrical. Separate the following stress tensors into hydrostatic and deviatoric components:

41 0 00 2 00 0 3540 1 11 3 11 1 054

72 17 3917 5 2439 24 405


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:

= H = N[ =resistivity, A = area of cross-section; r = radius of wire]

Take logarithms: log = log + log 2 log N + constant Differentiate: v = v + v 2vNN If the elongation is elastic, then: vNN = 3 v Experimentally we find: v = v

[ C 1 for metals; V = volume = piLr2]

So we have:

v = v + 2vNN = v . [1 23]

And therefore:

v = v [ 1 23 + 1 + 23] x 2formostmetals3~0.5

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The exact proportionality depends on the material (viaa gauge factor:

vTo measure the strain, we:

Fix the wire onto the object Allow it to deform

Measure the change in R

Calculate the corresponding strain

However the strains are usually small (map out the stress distribution across an object, so we cannot just use a very long wire to make the absolute value of R bigger 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

Gauges are made from wire or appropriate pattern

A common material is constantan (55Cu + 45Ni), which gives a good linear response

Semiconductor strain gauges are also available with C

In the simplest case, the strain gauge can be connected to a Wheatston

to measure R

The exact proportionality depends on the material (via the constants C and

v = v = x 2 normalstrain

Fix the wire onto the object

Measure the change in R

Calculate the corresponding strain

ually 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

R bigger we need to be able to measure the strain over a small

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

Gauges are made from wire or more commonly from foil etched to give an

A common material is constantan (55Cu + 45Ni), which gives a good linear

Semiconductor strain gauges are also available with C 1 and gauge factors In the simplest case, the strain gauge can be connected to a Wheatston

C4 Tensors / C4-H42

the constants C and ), so we introduce

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

we need to be able to measure the strain over a small

We deal with this by folding the wire, which gives us a small actual gauge length (typically 1 R.

from foil etched to give an

A common material is constantan (55Cu + 45Ni), which gives a good linear

1 and gauge factors 100

In the simplest case, the strain gauge can be connected to a Wheatstone Bridge circuit

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3.1.1 Thermal effects on strain gauges

The resistivity 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 thermalmatched 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

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 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

.1.1 Thermal effects on strain gauges

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 thermalmatched to common substrate materials

.1.2 Analysis of strain gauge results


to shear strains

If we know the directions of the principal stresses, we can measure the strains directly

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 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

C4 Tensors / C4-H43

and the gauge factor k are both temperature dependent, which can

It is better to use thermally compensated gauges, in which the thermal expansion is


stresses, we can measure the strains directly

For example, the principal axes may be fixed by the symmetry of the body and the

to use a rosette an

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

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3.1.3 Example of a strain gauge calculati

Suppose we measure normal strains

As a vector diagram this is:

.1.3 Example of a strain gauge calculation: a 60 rosette

strains (the shear strains are unknown from the gauges)

A = 2.12 10F B = 3.53 10F C = 3.17 10F

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(the shear strains are unknown from the gauges):

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But on the Mohr circle the angles are doubled (and note that this is an with respect to the principal axes):

The centre of the circle is at a distance:

And we need the principal strains

N coscos

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:

3.53 + 2.12 + 3.173 = 2.94 And we need the principal strains x and y:

N cos 2Y = 3.53 2.94 = 0.59 cos60 2Y = 2.94 2.12 = 0.82

cos 2Ycos60 2Y = 0.590.82 2Y = 45.7

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arbitrary orientation


So:

Y = 22.8; N = 0.845 s = 2.94 + 0.845 = 3.785 10F = 2.94 0.845 = 2.095 10F

Example 7 The strains measured by three strain gauges arranged in a 120 rosette on the surface of a steel plate are 1.2 103, 2.0 103 and 1.0 103. Determine the orientation and magnitude of the principal strains in the surface. What is the maximum shear strain?

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3.2 Photoelasticity

This technique allows the strain field in a transparent model to be visualised using induced birefringence: when we take

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

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 xplane of the stress tensor.

At a particular point, the incident wave will be:

This wave splits into two components parallel to

The wave travels through a plate of thickness d, which shifts

So on the exit side of the plate our two waves become:

This technique allows the strain field in a transparent model to be visualised using : 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:

y y = are the refractive indices and c is the stress-optical coefficient.

.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 x

At a particular point, the incident wave will be:

H = > cos{ This wave splits into two components parallel to 1 and 2:

H = > cos cos{ H = > sin cos{

The wave travels through a plate of thickness d, which shifts the phase of each wave by:

2v = 2vP = {v So on the exit side of the plate our two waves become:

C4 Tensors / C4-H47

This technique allows the strain field in a transparent model to be visualised using stress-a suitable transparent, isotropic material and apply a

In an optically anisotropic material, light propagates with two different velocities and two

optical coefficient.

We imagine looking towards the polariser, through a sample which is a plate cut in the x1x2

the phase of each wave by:


H = > cos cos{m v n H = > sin cos{m v n

There is now a phase difference between the two waves. The wave passing through the analyser is the sum of the resolved components:

HAanalyser = > sin cos cos{m v n + > cos sin cos{m v n = > sin 2 sin{ v2 v2 sin{ v2 v2

This is a travelling wave, with amplitude:

H = > sin 2 sin{ v2 v2 We get extinction (zero amplitude) when:

1 or 2 is parallel to either the polariser or the analyser, which gives: sin 2 = 0 = 0or90 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:

{ v2 v2 = WwhereWisaninteger Given: {J = 2 and = Jy We get: v y y = W vy y = W And:

= Wv 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 (1 2) o Are black in monochromatic light o Are coloured in white light

Isoclinics orientation of principal stresses

Isochromatics value of (1 2)

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

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 thing 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 strin in the surface to which it has been applied

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The coating will crack under tensile strains

To a first approximation, cracks appear when:

1(specimen) = E(specimen)

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

The load at which the cracks appear gives the magnitude of

3.3.2 Tension in two dimensions

If 1 > 2 > 0 and both are increasing, then the first set of cracks will appear perpendicular to 1 and a second set will appear perpendicular to

If 1 = 2 >0 then the direction of the cracks is indeterminate, and a crazed surface results.

The coating will crack under tensile strains

To a first approximation, cracks appear when:

(specimen) = E(specimen) crit(coating)

A pattern of cracks develops, oriented perpendicular to the maximum tensile stress in

in one dimension

In uniaxial tension we will see:

The orientation of the cracks gives the direction of 1

The load at which the cracks appear gives the magnitude of 1 at that load

Tension in two dimensions

0 and both are increasing, then the first set of cracks will appear perpendicular to and a second set will appear perpendicular to 2.

>0 then the direction of the cracks is indeterminate, and a crazed surface results.

C4 Tensors / C4-H50

A pattern of cracks develops, oriented perpendicular to the maximum tensile stress in

at that load

0 and both are increasing, then the first set of cracks will appear perpendicular to

>0 then the direction of the cracks is indeterminate, and a crazed surface results.

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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.

Brittle coatings can also be applied deliberately as an analytical technique:

Coatings are 50 - 200m 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

Useful to determine the directions of the principal stresses

Can then apply strain gauges for accurate measurements

Surfaces of varnished oil paintings


Brittle coatings can also be applied deliberately as an analytical technique:

200m thick

Applied as a spray of resin


Ceramic coatings can also be used (these need firing to melt, but can then be used at

Simple to use on a real structure

Useful to determine the directions of the principal stresses

Can then apply strain gauges for accurate measurements

C4 Tensors / C4-H51



Ceramic coatings can also be used (these need firing to melt, but can then be used at


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:

= lmodulus The modulus and density both depend on stress and strain (given that Poissons ratio

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

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3.6 Optical interferometry

This includes both holography and laser speckle interferometry. In general optical methods are attractive because they are intrinsically non

3.7 Residual Stresses

These are internal stresses which are not a result of external forces, and which muretained 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 Consider a plate which has been roll

temperature. th = 1210-6 K-1

thermal strain between the layers differently:

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) 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:

holography and laser speckle interferometry. In general optical methods are attractive because they are intrinsically non-contact and very sensitive.

These are internal stresses which are not a result of external forces, and which muretained 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:

Thermal stresses due to differential expansion Consider a plate which has been roll-bonded at high temperature and is then cooled to room

for Fe and th = 1.710-6 K-1 for Cu so they apportion the

thermal strain between the layers differently:

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.

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:

C4 Tensors / C4-H53

holography and laser speckle interferometry. In general optical methods

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

then cooled to room

so they apportion the

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.

are stressed beyond their elastic limit. The surface regions flow, so when the stress is released the bar will remain

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This means that residual stresses can be found as a result of forming procepeening (surface hardening) and machining.

3.7.3 Changes in volume

Residual stresses can also result from nonoccur in:

Phase transformation

Carburising and nitriding (in which

This means that residual stresses can be found as a result of forming processes such as shot peening (surface hardening) and machining.

Residual stresses can also result from non-uniform or localised changes in volume, such as

Carburising and nitriding (in which interstitials cause a change in lattice parameter)

C4 Tensors / C4-H54

sses such as shot

uniform or localised changes in volume, such as

interstitials cause a change in lattice parameter)

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4. EXAMPLES OF STRESS A

4.1 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 xmirrors.

4.1.1 Stresses in thin films

There cannot be any forces on a free surface, so

In almost all cases 11 = 22 ( =

These are now principal stresses, and we should expect to find both biaxial inand a normal strain.

EXAMPLES OF STRESS AND STRAIN ANALYSIS

Optical and other coatings

multilayered e.g. GaAs/AlAs solid state lasers, W/C x

There cannot be any forces on a free surface, so 33 = 0 and i3 = 0 and this is

4 0 00 0 05 ( = ) so we have isotropic biaxial tension (or compression).

These are now principal stresses, and we should expect to find both biaxial in

4 0 00 00 0 05 4 0 00 00 0 5

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ND STRAIN ANALYSIS

multilayered e.g. GaAs/AlAs solid state lasers, W/C x-ray

= 0 and this is plane stress:

(or compression).

These are now principal stresses, and we should expect to find both biaxial in-plane strains,


The biaxial stress and the biaxial strain in the (x1,x2) plane of the film are related by:

= 1 3 = 1 3

The normal strain in the film can be determined (for example) by x-ray diffractometry, and is given by:

= 231 3 Example 8 Derive the above relationships for this plane stress, thin film condition, i.e.

= 1 3 and = 231 3

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

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4.1.2 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.

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,

of dislocations is needed at each interface.

So we have two scenarios:

For thin layers (d < dcrit For thick layers (d > dcrit

dislocations

Consider a stack of alternate layers of (100) Cu and (100) Pd on a (100) NaCl substrate. Would the interfaces remain coherent?

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. ForCu: > = 3.6148 So the Cu must be in tension and the Pd in compression, with equal and opposite stresses for equal layer thicknesses.

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.

crit) the interfaces will be fully coherent with no dislocationscrit) there will be few interfaces but these will contain

C4 Tensors / C4-H57

Consider a stack of alternate layers of (100) Cu and (100) Pd on a (100) NaCl substrate.

If the multilayer stack is coherent (i.e. there are no dislocations) there is a 7.6%

So the Cu must be in tension and the Pd in compression, with equal and opposite

there is no strain energy due to matching but a 2D network

) the interfaces will be fully coherent with no dislocations be few interfaces but these will contain


5. OTHER TENSOR PROPERTIES

5.1 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:

4 0 00 00 0 5 The applied field E has components (1E, 2E, 3E).

The resulting current density J therefore has components (11E, 22E, 33E).

Thus the component of J resolved parallel to E is:

parallel = + + So the conductivity parallel to E is:

= + + = This conductivity can be represented by the surface:

= 1 Which will be an ellipsoid if all the coefficients are positive. A radius in a given direction (1, 2, 3) will have length r and intersect the surface at a point (x1, x2, x3) given by:

t = N; t = N; t = N With these coordinates:

N = 1 N = 1 N = F/

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 is

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parallel to the resultant vector (e.g. current density).property, see Nye p28]

The representation surface (or quadric) for a second rank tensor property,shown in this case for conductivity

5.2 Optical properties

For an isotropic material the electric displacement by:

where o 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)surface for Bij is therefore a plot of refractive index

parallel to the resultant vector (e.g. current density). [For discussion of this radius

ntation surface (or quadric) for a second rank tensor property,shown in this case for conductivity

For an isotropic material the electric displacement D is related to the electric field strength

= u is the permittivity of vacuum and K is the dielectric constant. The real refractive

For an anisotropic material, the electric displacement and field strength are not necessarily parallel. Their vector components are related by:

X = u or = IXu 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

is therefore a plot of refractive index this is the optical indicatrix

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[For discussion of this radius-normal

ntation surface (or quadric) for a second rank tensor property,

is related to the electric field strength E

is the permittivity of vacuum and K is the dielectric constant. The real refractive

For an anisotropic material, the electric displacement and field strength are not necessarily

Where Kij is the tensorial dielectric constant, and Bij is the relative dielectric impermeability. . The representation

optical indicatrix.


5.3 Piezoelectricity

5.3.1 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 ij.

This requires a relationship of the form:

= v + v + v + v + v + v+ v + v + v and similarly for P2 and P3.

In general we have:

= v!! where dijk is a third rank tensor composed of the 27 piezoelectric moduli. dijk = dikj always, so we have only 18 independent coefficients.

5.3.2 Converse piezoelectric effect

In this case the crystal develops a strain in response to an applied electric field.

So:

! = v! dijk must have the same components as before, otherwise we can make a perpetual motion machine!

5.4 Effects of crystal symmetry

Consider a crystal with a centre of symmetry.

Apply a stress e.g. 11 (and this is inherently centrosymmetric) The result must also be centrosymmetric

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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 none.g. quartz, point group 32.

We can also have piezoelectric polycrystals provided that:

The individual grains are piezoelectric

The polycrystal has texture (preferred orientation)

5.5 Ferroelectricity

Texturing of piezoelectric polycrystalline ceramics makes use of ferroelectric properties, as in

BaTiO3.

Above 120C the Ti atom moves towards one of the oxygens in the coordinating octahedron. This causes:

A cubic tetragonal transformation on cooling

Spontaneous polarisation

The direction of polarisationcan be influenced by cooling in an electric field (poling).

So any resultant dipole will be cancelled by its opposite.

Thus centrosymmetric crystals (belonging to the 11 centrosymmetric point groups, out of 32)

But we can expect to make piezoelectric devices from non-centrosymmetric single crystals

We can also have piezoelectric polycrystals provided that:

The individual grains are piezoelectric

The polycrystal has texture (preferred orientation)

polycrystalline ceramics makes use of ferroelectric properties, as in

C the Ti atom moves towards one of the oxygens in the coordinating octahedron.

tetragonal transformation on cooling

Spontaneous polarisation

e direction of polarisationcan be influenced by cooling in an electric field (poling).

High T cubic BaTiO3 polycrystal (no texture)

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Thus centrosymmetric crystals (belonging to the 11 centrosymmetric point groups, out of 32)

rosymmetric single crystals

polycrystalline ceramics makes use of ferroelectric properties, as in

C the Ti atom moves towards one of the oxygens in the coordinating octahedron.

e direction of polarisationcan be influenced by cooling in an electric field (poling).


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.

5.6 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

5.6.1 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 1, and thus atomic resolution. The tip has to be moved laterally with extreme precision, and this is achieved with piezoelectric transducers.

5.7 Pyroelectricity

Ferroelectricity is the existence of a spontaneous polarisation. If this polarisation varies with

temperature then we have pyroelectricity, according to:

= \ T also causes a change in volume V, which also contributes to Pi.

If V is held constant then T 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).

5.8 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:

! = v! If the effect is non-linear we require an additional term:

! = v! + ~"!" |v! = v!at = 0}

The first term (d E) 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:

! = |v! + ~"!"} Where the correction term iljk 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.


6. ELASTICITY

We need to relate two second rank tensors: stress ij and strain ij.

Each component of can in general depend upon all nine components of , so we need a fourth rank tensor:

= !"!"

6.1 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.

6.1.1 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

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6.1.2 Linear elasticity theory

This makes the dual assumptions of linearity and

1. Linearity Hookes LawThis assumes a linear relationship (proportionality) between stress and strain e.g.

E is a scalar: the Modulus of Elasticityboth tension and compression.

2. Superposition Suppose a stress A produces a strain

This is a consequence of linearity and holds only for small strains.

6.1.3 Poissons ratio

Consider again the rod in uniaxial tension:

But there is a lateral contraction in the x

Experimentally we find:

Where is Poissons ratio, and is constant for any one material (note the sign convention). Typically it takes values of 0.25

If there were no volume change, then:

.1.2 Linear elasticity theory

dual assumptions of linearity and superposition.

s Law

This assumes a linear relationship (proportionality) between stress and strain e.g. = Modulus of Elasticity or Youngs Modulus and is the same in

both tension and compression.

produces a strain A and B produces B. Then: + + This is a consequence of linearity and holds only for small strains.

Consider again the rod in uniaxial tension:

= = 0 lateral contraction in the x2x3 plane:

= ? 0

= = 3 = 3 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:

+ + = 0

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This assumes a linear relationship (proportionality) between stress and strain e.g.

and is the same in

s ratio, and is constant for any one material (note the sign convention). he elastic range.


This requires = 0.5, and is true for the elastic deformation of rubber (caused by coils unwinding rather than bonds stretching).

Note that does not have to be positive, and for example cork has a negative Poissons ratio because of its cell structure, as does -cristobalite. These are known as auxetic materials.

6.1.4 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:

= 3 + And similarly:

= 3 + = 3 +

There are no shear stresses in these expressions because the tensors are referred to their principal axes.

Example 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.

Even for arbitrary cases we still have equivalent expressions:

= 3 + etc[seeNyeforaproof,

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For simple shear, shear stresses are related to shear strains by

Or equivalently (given that = 2

is the shear modulus or modulus of rigidity

6.1.5 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 is given by:

And hydrostatic stress H by:

For simple shear, shear stresses are related to shear strains by = :

= 2ij): = 2 ?

modulus of rigidity (alternatively sometimes written as G).

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:

= V = =

= + +

= 13 + +

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(alternatively sometimes written as G).

are just two independent elastic constants (we can specify any two), and so there are algebraic relationships between them. For example, to describe volume


K is the bulk modulus and is the compressibility.

We can find relationships between E, , , K and by algebraic manipulation. For example we can add the three equations relating principal stresses and strains:

+ + = 1 23 + + Which gives:

= 1 23 . 3 And so:

= 31 23 Example 9 Derive the following relationships:

= 21 + 3 = 9 + 3 3 = 3 26 + 2

6.1.6 Lams constants

We have expressions for strains given known stresses, but we also need expressions for stresses given known strains. Manipulating the above equations gives:

= 1 + 3 + 31 + 31 23 + + If we define:

= 31 + 31 23 Then we have:

= 2 + We call and Lams constants.

For shear stresses 12 = 212 etc as before. and are often quoted as the two elastic constants but the choice is arbitrary.


6.2 Elasticity in general anisotropic media

We want to relate the two second rank tensors stress (ij) and strain (ij). So we write:

= !"!" =


We would like to be able to write expressions like:

= = < =

= - =

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So that:

Sijkl = Smn when m and n are 1,2,3

Sijkl = Smn when either

Sijkl = Smn when both

Defiend in this way, we can now write:

Note that the two arrays of numbers C

6.2.2 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

At one instant the bar is under a load F with strain work done is:

Where A is the cross-sectional area, and:

So the work done per unit volume is:

[Compare this with = E for uniaxial stress]

when m and n are 1,2,3

either m or n are 4,5,6

both m and n are 4,5,6

Defiend in this way, we can now write:

= - =


In the general case, and using matrix notation, the work done per unit volume will be:

v = v If the deformation is isothermal and reversible, the work done per unit volume is also equal to the increase in free energy d, and so:

v = v Now we can differentiate twice, giving:

ww ww = Similarly, we can derive: ww ww = But is a function of the state of the body and defined by the strains, so the order of differentiation must be immaterial, so:

ww ww = ww ww And therefore:

= Similarly:

- = - In the same way as the 1D case, this gives us the strain energy per unit volume = Cijij.

Since Cij = Cji, the 36 coefficients reduce to 21 independent coefficients for a perfectly general anisotropic solid.

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6.2.3 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 fcoefficients is lower still.

6.2.4 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], x[001]). For the most general result, we choose the least symmetric cubic point group This point group has:

Diads parallel to

Triads parallel to

No mirror planes

No 4-fold axes

The four triads require the three cube axes to be equivalent to each other, so:

(and sim

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