ucsd nano106 - 08 - principal directions and representation quadrics

Principal Directions and Representation Quadrics

Shyue Ping OngDepartment of NanoEngineeringUniversity of California, San Diego

Summary of rank 2 tensors for all crystal systems

¡ Note that rank 2 tensors take these forms only when the measurement axes are based on the crystal axes as per the analysis. In a general direction, a rank 2 tensor has 9 non-zero components (though considerably fewer independent components, as we shall see later)

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 8

2

σ ij =

σ11 σ12 σ13σ 21 σ 22 σ 23

σ 31 σ 32 σ 33

!

"

####

$

%

&&&&

Triclinic

σ ij =

σ11 0 σ130 σ 22 0σ 31 0 σ 33

!

"

####

$

%

&&&&

Monoclinic

σ ij =

σ11 0 00 σ 22 00 0 σ 33

!

"

####

$

%

&&&&

Orthorhombic

σ ij =

σ11 0 00 σ11 00 0 σ 33

!

"

####

$

%

&&&&

Tetragonal, Trigonal, Hexagonal Cubic

σ ij =

σ11 0 00 σ11 00 0 σ11

!

"

####

$

%

&&&&

Principal Directions¡ A general rank 2 tensor is simply a matrix with 9 non-zero

elements.

¡ However, in a certain basis, a rank 2 tensor will take the form

¡ These basis vectors are known as the principal axes. Note that these axes are not always orthogonal, unlike our assumption thus far.

¡ When a “force” is applied in the principal directions, the response is always parallel to the applied “force”.


3

σ ij =

σ11 0 00 σ 22 00 0 σ 33

!

"

####

$

%

&&&&

Finding the Principal Axes¡ Finding the principal axes is essentially an eigenvalue

problem. You are trying to find vectors where:

¡ Where we have used T to denote a general tensor and λis the eigenvalue. Rewriting the above equation, we have

¡ To obtain useful solutions, we look for values satisfying the characteristic equation:


4

Tv = λv

(T −λI )v = 0

det(T −λI ) = 0

Example¡ Consider the following tensor with the form:


5

σ ij =5 1 11 3 11 1 3

!

"

###

$

%

&&&

Blackboard

Magnitude of Tensor in a Given Direction¡ Consider the electrical conductivity:

¡ A typical experimental measurement may involve applying the electric field (voltage / unit length) in a particular direction (e.g., [111]) and then measuring the current density (charge / time / area) in the same direction. Note that j is in general not parallel to E, and the value of the tensor is not given by

¡ Instead, we define the value of the tensor along that direction as the component of j in the direction of E, divided by the magnitude of E, i.e.


6

ji =σ ikEk

jE

j111E

Magnitude of Tensor in a Given Direction


7

Consider j=σEThe experiment measures the component of j in the direction of E. This is given by the dot product of j with the unit vector in the direction of E. Hence, the tensorσ in the direction of E is given by

σ [E ] =j⋅EE

⋅1E=σE ⋅EE 2 =

σ ikEkEi

E 2

Now, if E = E(l1, l2, l3), where li are the direction cosinesfor the vector E. Therefore,σ [E ] =σ iklilk The magnitude of a tensor in a given direction

is specified by the direction cosines of that direction.

Example¡ Let’s say we are interested in the value of a tensor along

the [100] direction. We have

¡ Hence, the diagonal elements of the tensor is always the value of the tensor along the three orthogonal axis directions.


8

Blackboard

The representation quadric¡ The representation quadric is a useful geometric

representation of a rank 2 tensor property.

¡ Consider the following equation:

¡ This equation can be plotted as a 3-dimensional surface that completely describes the tensor T.

¡ To make this easier to handle, let us consider the form of the tensor in the principal axes, in which case the off-diagonal elements of T are zero.


9

Tij xix j =1

T11x12 +T22x2

2 +T33x32 + (T12 +T21) ⋅ x1x2 + (T13 +T31) ⋅ x1x3 + (T23 +T32 ) ⋅ x2x3 =1

The representation quadric¡ In the principal directions, the representation

quadric equations becomes:

¡Which can be rewritten as:


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T11x12 +T22x2

2 +T33x32 =1

x12

1 /T11+

x22

1 /T22+

x32

1 /T33=1

Forms of the representation quadric


11

Ellipsoid, T11,T22,T33 > 0



12

Hyperboloid of 1 sheet, T11 < 0, T22,T33 > 0



13

Hyperboloid of 2 sheets, T11,T22 < 0, T33 > 0

Interpreting the representation surface¡ Recall that the representation surface is given by:

¡ Let us now write , where li is the ith direction cosine for xand r is the radius. We get

¡ Recall that we earlier showed that the value of a tensor in a given direction is given by ! Hence, using the representation surface, we can geometrically obtain the value of a tensor in a direction simply by measuring the radius!


14

Tij xix j =1

xi = rli

Tijrlirl j =1

Tijlil j =1r2

Tijlil j

The radius normal property¡ Besides just the magnitude, we can also geometrically determine the

direction of the response. We will do this analysis again in the principal axes to simplify the math. Consider the following:


15

p = Tq or pi = TijqjLet qj = qljConsider a point on the representation surface such that OQ is parallel to q.This point is given by r(l1, l2, l3).The tangent plane to this point is given by (substitute r(l1, l2, l3) into the quadric eqn).:T11x1rl1 +T22x2rl2 +T33x3rl3 =1The normal to this plane is therefore r(T11l1,T22l2,T33l3), which is parallelto p=q(T11l1,T22l2,T33l3)!

Example¡ We will use an ellipsoid quadric surface (all principal values >0, which

is the most common form for property tensors). The analysis will be done in 2D (assume qz component is 0) to facilitate visualization.


16

R

Let's say the generalized force q is in the direction indicated by the bluearrow, with magnitude q.Using the representation quadric, we can determine that the magnitudeof the tensor in the direction of q is given by

T//q =1R2

The magnitude of the response p in the direction of q is therefore

p//q =qR2

The direction of the actual response is given by the purple arrow.

Example¡Consider the following tensor:

¡Draw the projection of the representation quadric in the X1-X2 plane.

¡Determine from the representation quadric the value of the tensor in the [110] direction and the direction of the response.


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σ ij =5 0 10 3 11 1 3

!

"

###

$

%

&&&

Stress & strain¡ Before embarking on the journey beyond rank 2 tensors,

we will now go in-depth into stress (force / unit area) and strain (change in length / unit length). Both of these are rank-2 tensors, but they are not property tensors. Typically, they are known as field tensors to distinguish them from property tensors.

¡ Unlike property tensors, crystal symmetry restrictions do not apply to stress and strain.

¡ We will subsequently discuss higher order tensors that relate stress and strain to other measurements.


18

Stress¡ The stress tensor describes the force acting on a

specimen.

¡ First subscript refers to the direction of the force, the second to the normal to the face on which the force acts. To prevent translational motion, each force is balanced by an equal and opposite force on the reverse side of the specimen.

¡ Xii are tensile stresses in which both the force and the normal are along Zi, and Xij are shear stresses in which a force along Zi acts on a face normal to Zj.

¡ Static equilibrium requirement (balanced torques) => Stress tensor must be symmetric with Xij = Xji.

¡ Stress state is specified by six independent components: three tensile stresses X11, X22, and X33, and three shear components X12, X13, and X23.


19

σ ij =

σ11 σ12 σ13σ12 σ 22 σ 23

σ13 σ 23 σ 33

!

"

####

$

%

&&&&

Stress Quadric¡ As stress is a rank-2 tensor, it can be represented as a

quadric.

¡ Everything we have derived about the representation quadric for rank-2 tensors applies to the stress quadric as well:¡ The square of the radius to a point is the reciprocal of the value of

stress in that direction, given by . Note that for the stress tensor, the direction cosines is based on the normal to the surface of the generalized “displacement”.

¡ The radius normal property applies as well.


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σ ij xix j =1

σ ijlil jli

Forms of Stress Tensor


21

σ ij =

σ11 0 00 0 00 0 0

!

"

####

$

%

&&&&

Uniaxial tension/compression

σ ij =

σ11 0 00 σ 22 00 0 0

!

"

####

$

%

&&&&

Biaxial stress

σ ij =

σ11 0 00 σ 22 00 0 σ 33

!

"

####

$

%

&&&&

Triaxial stress

σ ij =

−p 0 00 −p 00 0 −p

"

#

$$$$

%

&

''''

Hydrostatic pressure

σ ij =σ 0 00 −σ 00 0 0

"

#

$$$

%

&

'''

Pure shear

σ ij =0 !σ 0!σ 0 00 0 0

"

#

$$$

%

&

'''

Strain¡ Strain is the fractional change in shape of a specimen.

¡ Like stress, strain is represented by a symmetric rank 2 tensor.

¡ The symmetry of the strain tensor arises from the elimination of pure rotations that do not involve any deformation of the object.


22

εij =δuiδx j

=

ε11 ε12 ε13ε12 ε22 ε23ε13 ε23 ε33

!

"

####

$

%

&&&&

Displacement tensor¡ To illustrate the symmetry of the strain tensor, let us

consider a generalized fractional displacement tensor denoted by e (note that this is different from the symbol used for strain!). e is in general not symmetric.

¡ For any non-symmetric tensor, we can always write it in terms of the sum of a symmetric tensor ε and an anti-symmetric tensor w as follows:


23

eij = εij +wij

where εij =12

(eij + eji ) and wij =12

(eij − eji )

Rigid body rotations¡ For a rigid body rotation, we have a displacement vector u

that is always perpendicular to the vector r everywhere.


24

u

r Blackboard

Definition of strain¡Rigid body rotations correspond to anti-symmetric

displacement tensors.


25

eij = εij +wij

rotationActual strain

Graphical example using pure shear strain

Dilation¡ What is the fractional volume change of a strained

specimen?

¡ For simplicity, we will again consider a diagonalized strain tensor and work in the principal axes.


26

Blackboard

Forms of the strain tensor


27

εij =

ε11 0 00 0 00 0 0

!

"

####

$

%

&&&&

Tensile or compressive strain

εij =

ε11 0 00 ε22 00 0 0

!

"

####

$

%

&&&&

Plane strain

εij =ε 0 00 −ε 00 0 0

"

#

$$$

%

&

'''

Shear strain (special case of plane strain!)Note that pure shear has no dilation, in line with our intuition

εij =0 !ε 0!ε 0 00 0 0

"

#

$$$

%

&

'''

Voigt notation¡ Method of representing symmetric tensors by reducing its

order.

¡ Instead of rank 2 tensors (matrices), stress and strain can be represented as a 6-element vectors.


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σ ij =

σ11 σ12 σ13

σ12 σ 22 σ 23

σ13 σ 23 σ 33

!

"

####

$

%

&&&&→

σ1

σ 2

σ 3

σ 4

σ 5

σ 6

!

"

########

$

%

&&&&&&&&

=

σ11

σ 22

σ 33

σ 23

σ13

σ12

!

"

########

$

%

&&&&&&&&

, εij =ε11 ε12 ε13

ε12 ε22 ε23

ε13 ε23 ε33

!

"

####

$

%

&&&&→

ε1

ε2

ε3

γ4

γ5

γ6

!

"

########

$

%

&&&&&&&&

=

ε11

ε22

ε33

2ε23

2ε13

2ε12

!

"

########

$

%

&&&&&&&&

Voigt notation, contd.¡ Note that the off-diagonal strain elements has a factor of 2

in the vector notation. γij = 2εij are known as the engineering shear strains.

¡ The benefit of this representation is that the following scalar invariance is preserved:


29

E =σ ij ⋅εij =

σ1σ 2

σ 3

σ 4

σ 5

σ 6

"

#

$$$$$$$$

%

&

''''''''

⋅

ε1ε2ε3γ4γ5γ6

"

#

$$$$$$$$

%

&

''''''''

Voigt notation for higher order tensors

¡The same principle can be (and in fact is frequently) applied to higher order tensors.

¡For example, the rank 3 piezoelectric tensor is represented as a 6x3 matrix (or 3x6 in the case of the converse piezoelectric effect), while the elastic and compliance moduli tensor are represented as 6x6 matrices.


30

Electric polarization¡ The electric polarization is the vector field that expresses

the density of permanent or induced electric dipole moments in a dielectric material. When a dielectric is placed in an external electric field, its molecules gain electric dipole moment and the dielectric is said to be polarized.


31

+-

Electric polarization


32

+-d

Electric dipole moment p = qdwhere d is the displacement vector

Polarization P =pV

(average dipole moment per unit volume)

¡ Electric polarization is a rank 1 field tensor.

ucsd nano106 - 08 - principal directions and representation quadrics

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