Elasticity of Anisotropic Bodies at Structural Phase Transitions

FOREWORD

If you are interested in scientific research and discoveries, in the popular media

these days, the majority of what you hear is about particle accelerators / particle

physics, high energy physics, cosmic particles, space travel, but mostly something

about cosmological topics like the Big Bang, the expansion of the universe, black

holes, wormholes etc., and of course climate change. All this is currently quite 'in

vogue'.

Nevertheless, there are numerous 'classical' physics topics that - although

seemingly relegated to the background - are being intensively researched and not

infrequently produce important, unexpected and impressive results.

In my opinion, this includes the field of solid state physics, in particular crystal

physics. Powerful mathematical tools, such as group theory, as applied in the

theory of the phase transitions (Landau’s Theory), not seldom lead to new

theoretical insights. Since crystals are touchable, material substances, one can

concretely measure their properties and compare these with the predictions. In this

way, for example, materials with specific properties can be developed.

The background of this paper lies mainly in the description and determination of

the elasticity of crystals that have undergone a phase transition. Certain crystals

exhibit so-called ferroelastic properties (by analogy with ferromagnetic and

ferroelectric crystals). This group often shows different and also extreme elastic

constants and deformations, which have to be measured and described. At this

point I also point to a collection of articles I wrote in 2017. There it is about Landau’s

Theory in general and applied to a number of concrete phase transitions.

In the following article, in addition to the basics of anisotropic elasticity, various

investigation possibilities were discussed in detail in connection with selected

crystals.

Hempel

December 2021

PS: Interested parties shall feel free to make use of the attached considerations,

but referencing to this source is highly appreciated.

Contact: [email protected]

Elasticity of Anisotropic Bodies

Hempel 2020/21 Page 1 of 104

Table of Contents

1. Displacements in Continua………………

2. Stress and Elasticity……………………...

3. Hooke’s Law / Elastostatics / Energy….

4. Applications (selection)…………………..

a: Uniform Compression…………...

b: Uniaxial Extension…………….….

c: Shear………………………………..

d: Bending of Rods………………….

e: Torsion of Rods…………………...

5. Ferroelastics…………………………….…..

a: Hysteresis Loop…………………..

b: Domain Switching (and more).…

c: Torsion Pendulum (I. & II.)………

d: Practical Consideratins………….

6. Literature…………………………………….

2

8

10

14

14

16

18

20

28

54

54

56

76

99

103



1. Displacements in Continua

• Every point of a continuum shall have moved from a position, described

by to it new position, described by

• The displacement of that point can be described by

(1)

• Generally, displacements can be decomposed in a pure translation, rotation

and distortion

• Further it is assumed that all displacements are small, and located closely

to the origin of the System of Coordinates (SoC)*)

• The shift of the origin of SoC ( ) is described by

(2)

Note: The contents of chapters 1.-3. base mainly on dedicated textbooks (e.g. /1/,/2/)

*) Note: SoC means the (Cartesian) crystal-physical system of coordinates (x, y, z) – particularly defined for each crystal classes

r

r

rrrs

−=)(

0)0()0( ssr

==

0=r



• Taylor-Series’ development of at yields

(3)

with: = 2nd rank tensor of displacement

= anti-symmetric part

= symmetric part

shift of the origin of SoC

Antisymmetric Tensor – Representation of the Rotation

• Having done the (‘standard’) decomposition of tensor , for the anti-

symmetric part follows explicitly:

(4)

*) Note: Expressions of the kind are called dyad ( ͦ )and represent a tensor, not a scalar product!

s

sa

r

UrUrsUrssr

rsrs

++=+=

+=

=

00

0

0)(

=0s

0=r

0,,

,02

1

2

1

−

==

−

=

k

i

i

kk

kii

akik

kii

a

x

s

x

seeUee

rss

rU

sUaU

U

U

s

r



• Looking closer at the tensor’s components we find:

• is indeed anti-symmetric: ,

• has 3 independent components, which means that it can be

equivalently represented by an (axial) vector

• The rotational displacement can be written as (using (3), (4)):

(5)

whereby denotes the infinitesimal small (axial) rotation vector.

• Comparison yields:

, and in components: (6)

• In particular we obtain, invoking eqs. (6) and (4):

(7a)

Note: is the so called Levi-Civita symbol

aa UU

~−=

aki

aik UU −=

rrsrr

ssr

rUrrs arot

=

−

==

002

1

2

1)(

almklmk U

2

1=

0

0

)(2

1

2

1srots

r

=

=

klm

m

nimni

x

s

= 2

1

azy

ayz

yzx UU

z

s

y

s−==

−

==

0

12

1

aU

aU



(7b)

(7c)

• The resulting (infinitesimal) overall rotation calculates as the sum acc. to:

with being the unit vectors along the x-,y-,z-axes (8)

of the SoC

Symmetric Tensor – Representation of the Distortion (Strain)

• The symmetric part of tensor is named deformation-, respectively strain

tensor and is often denoted as (capital S - for strain)

(9)

axz

azx

zxy UU

x

s

z

s−==

−

==

0

22

1

ayx

axy

xyz UU

y

s

x

s−==

−

==

0

32

1

=

33

22

11

e

e

e

321 ,, eee

0,,02

1

2

1

+

==

+

==

k

i

i

kk

kiiikk

kii

S

x

s

x

seeSee

rss

rSU

U

sU

S



• Tensor has 6 independent components

• The 3 diagonal components stand for the relative change of length along

x-, y- and z-directions:

(10)

• The relative change of volume due to a strain calculates as the sum of the

diagonal components:

(11)

• The Trace is independent of the SoC.

• Taking eq. (9) for i=k (trace components) we get:

(12)

whereby stands for the displacement vector (ref. to eq. (1))

332211 Sz

zS

y

yS

x

x=

=

=

==++=

iii STrSSSS

V

V

332211

sdivr

s

x

s

x

s

x

seeSTr

i i

i

i

i

i

ii

ii

=

=

=

+

=

0002

1

s

STr

S



• The 3 non-diagonal components characterise the position of the tensor

ellipsoid in the SoC

• For subscripts I≠ k we have*): (13)

• After some calculation it follows (as approximation):

(14)

The non-diagonal strain

components represent one half

of the angles between the ‘old

and new’ directions (i.e. before

and after distortion) – see left

*) Note: vectores are not unit vectors anymore!

kikikiik eeeeeSeS

===

ik ee ,

+

=

+

=

+

=

=========

z

s

y

sS

z

s

x

sS

y

s

x

sS

SSSSSS

yzxzxy

2

1,

2

1,

2

1

22,

22,

22

231312

3232

2323

3131

1313

2121

1212

1e

1e

2e 2e

221

212



• Usually the strain tensor components are denoted with one subscript

(Voigt’s Notation /2/) instead of two – to shorten the expressions.

(15)

2. Stress and Elasticity

• The stress of a body is defined via the substitute forces that

maintain exactly the present deformation state (at the ‘cutting’ surfaces

at point )

(16)

(17)

with being the stress tensor.

222456

321

322331132112332211

SSSSSS

SSSSSSSSS

===

)(),( rTAdAdrFdFd

==

Fd

r

ikkki

i TeeT

=,T



• The stress tensor is a symmetric tensor of 2nd rank (see also later)

(18)

and its components have the following meaning:

Tik (i=k) are denoted as the

normal stresses

Tik (i≠k) are denoted as the

tangential stresses

TT

~= kiik TT =



• Likewise as the stains, the stress tensor’s components are denoted with one

subscript (Voigt’s Notation /2/) instead of two – to shorten the expressions.

(19)

3. Hook’s Law / Elastostatics / Energy

• Hooke’s Law describes the linear and unique relation between Stress and

Strain:

(in Voigt’s Notation (20)

• Shorter *):

Note 1: are the so called elastic stiffness coefficients

Note 2: Hooke’s Law is typically named to be the linear relation between and , whereby non-linear terms (of )

can be added. This is not in the focus here.

*) Note: It has to be summed up if subscripts appear several times in one expression

456321

322331132112332211

TTTTTT

TTTTTTTTT

===

==

ScTScTml

lmiklmik,

cc iklm,

ScTScT lmiklmik ==

T

S

S

)6...1:,



• Since and are symmetric tensors, the tensor of the elastic stiffness

has at maximum 21 independent components, when taking into

account the tensor’s symmetry (21)

• Crystals with lowest symmetry (triclinic) do have those 21 coefficients.

• Elastic coefficients for all crystallographic Point Groups (PGs) are listed

in respective textbooks (e.g. /2/).

• Furthermore the invers of eq. (20) is also valid, namely

(22)

• It has to be noticed that means the inverse stiffness tensor

component, denominated as the elastic compliance*) tensor component:

(23)

Note: The conversion from stiffness to compliance requires the calculation of the complete tensor inversion.

*) Note: The elastic compliance- and the elastic stiffness tensors exhibit the same symmetry.

c

cc =

( ) lmiklmik TcS 1−= ( ) TcS 1−

=

( ) 1−= cs

( ) 1−c

S

T



Static Equilibrium Conditions

1. Forces:

The sum of all acting forces has to be zero.

(24)

is the density of the external volume forces, and are the

area-proportional forces.

This can be reformed to be:

(25)

2. Torques:

The sum of all acting torques has to be zero.

(26)

0=+ AdTdVf

00 , =+=

+ klkl TfTr

f

f

AdT

0=− rTAdfrdV



Reforming of eq. (26) leads to *):

↓ ↓

(27)

Finally the equilibrium of the torques results in the symmetry of the

stress tensor (see e.g. /1/). It arises:

(28)

Energy of Elastic Continua

• As shown is respective textbooks (e.g. /1/) the elastic energy density

can be easily written in Voigt’s Notation**) (making use of eqs. (20) & (23)):

(29)

*) Note: Arrows indicate the targets of the action of the differential operators

**) Note: It has to be summed up if subscripts appear several times in one expression

00 =

=

milklmk

i

xTex

rTr

TT

~= kiik TT =

TsTTSScS2

1

2

1

2

1===



• The elastic coefficients are now calculated as the 2nd derivatives:

and (30)

• Comment: for the elastic stiffness components applies:

(30a)

and for the elastic compliance components applies:

(30b)

4. Applications (selection)

a: Uniform Compression

• If a pure hydrostatic pressure p acts on a surface of a body then these

compression forces are directed anti-parallel*) with respect to the normal

vectors , characterising the surface

*) Note: In case of a uniform expansion the directions will be parallel (minus sign turns into plus sign)

SSc

SSc

lmik

iklm

=

=

22

TTs

TTs

lmik

iklm

=

=

22

n

)6...1;6...1( === klijcc ijkl

)6,5,4;6,5,4(4

)3,2,1;6,5,4(2

)6,5,4;3,2,1(2

)3,2,1;3,2,1(

===

===

===

===

klijss

klijss

klijss

klijss

ijkl

ijkl

ijkl

ijkl



• The related boundary condition can be formulated like:

(31)

• This can be satisfied by:

(32)

Comment: is the unit tensor (symbolically: ) and is the Kronecker Symbol,

providing the values: ‘1’ if i=j and ‘0’ if i≠k.

• Turning now to the strains:

(33)

• From eq. (33) follows: with k=1,2,3 (34)

• The elastic compliances (i.e. the Compressibility Coefficients) for all the

crystallographic classes can be easily calculated or simply looked up in /2/.

ijij pnnTnpnT −=−=

ijij pTEpT −=−=

E

ij

100

010

001

klijklklijklij psTsSEpsTsS −==−==

psS ijkkij −=

ijkks



b: Uniaxial Extension

• A rod with an unchanging (but arbitrary) cross-section (of area A)

along the length shall experience uniformly distributed tensile forces

at the ends.

• Having a unit vector along the rod axis, the stress tensors

reads:

(35)

• For the resulting strain parallel to direction of follows:

(36)

• The elongation of the rod along can be obtained:

(37)

• The expression:

(38)

is called the reciprocal Young’s Modulus for direction

jiij qTqTqqTT ==

)1( =qq

lkijklij qqTsSqqsTS ==

lkjiijkljiij qqqqTsqqSl

l==

lkjiijkl qqqqsqE =− )(1

q

q

q

AT



Comment: Vector can be depicted as follows:

(39)

• To ease laborious calculations of eq. (38) Voigt’s Notation will be useful:

(40)

with acc. to : 11→1; 22→2; 33→3; 12→6; 13→5; 23→4

• Rewriting stress and strain (ref. to eqs. (35), (36)) this yields*):

and (41)

*) Note: The expression is called dyad and represents a tensor, not a scalar product!

)1( =qq

)cos(),cos(

)cos(),cos(

)cos(),cos(

3

2

1

===

===

===

zzz

yyy

xxx

eqeqqq

eqeqqq

eqeqqq

)( qqqq ji

=

→ji ,

z

x

y

xq

q

zq

yq

)()( qqTsSqqTT

==

)( qq



• Finally the reciprocal Young’s Modulus can be expressed. It is listed

for the crystallographic classes e.g. in /2/.

(42)

c: Shear

• Let’s have a rectangular rod with uniformly distributed tangential forces

to the side faces. The rod shall be in equilibrium of forces.

• shall be a unit vector normal to the glide-planes and a unit vector

toward the glide-direction (e.g. ref. to eqs. (18), (19), and the pic there).

Both vectors shall be mutually perpendicular.

(43)

• Now the strain can be expressed*):

(44)

*) Note: Factor 2 in eq. (44) originates from the fact that it holds: (because of the symmetry of

regarding subscripts k and l)

)()()(1 qqqqsqE

=−

q

p

)()( jijiij pqqpTTpqqpTT +=+=

lkijklij qpTsSqpsTS 2)(2 ==

qppqqp

2=+ijkls



• Similar to eq. (37) the following shear component of is of interest:

(45)

with being the shear-angle, and use has been made of eqs. (13),(14)

• From eq. (45) the Shear Modulus is given:

(46)

which reads in Voigt’s Notation:

(47)

where (48)

Comment: It is worth to mention (/2/) that:

- The Shear Modulus is symmetric:

- Anisotropic bodies experience volume changes under shear stresses.

( ) ( )2

22

=== lkjiijkl qpqpTsqpsqpTqSp

S

( ) ( )T

qpqpsqpsqpqpG lkjiijkl

==− 44),(1

( ) ( )T

pqqppqqpsqpG

=++=−

),(1

),(),( 11 pqGqpG

−− =

( ) ( ) 6,...,1=+=+ ijpqqppqqp jiji



d: Bending of Rods

• Let’s assume a rectangular rod (length: c, width: 2a, thickness: 2b,

with c>>a,b),

fixed on its

left side.

• The vectors are unit vectors and belong to the special

(non-crystal-physical), but orthonormal SoC

• The bending moment (torque) is created by an external force

applied at the end : (49)

• The equilibrium condition for the torques demands the that sum of

the ‘replacement torque’ (which is uniform along axis ) and

the externally applied (also not depending on ) must be zero)

qnm,,

( )cx ='3

( )'3'2

'1 XXX

0F

0M

SM

'3X

0M

ncFFqcM

000 −==

b2

c

m

n

q

F0

→'3x

MS

→neutral plane

M0

→

'3X



(50)

• Assuming forces (per unit area) across the rod’s cross

section S like: (51)

we get:

and on the other hand:

• With eq. (50): we obtain (52)

• Eq. (52) expresses the boundary conditions regarding the

torques at the rod’s ends

03

4 30

=

+− nkabcF

30

4

3

ab

cFk =

00 =+ SMM

qkxTFS'1

' −==

nkabdxdxqkxmxdSFmxMa

a

b

b

S

S

S

3'

2'1

'1

'1

'1

3

4=−==

+

− −

ncFFqcM

000 −==

cxx == '3

'3 ,0

qxab

cFqT

'13

0'

4

3)( =



• Yielding a stress tensor like:

(53)

• In the special SoC only one component is non-zero:

(54)

• With Hooke’s Law follows:

(55)

• That strain results in a gradual bending of the rod along axis which can

be easily described by a rotation angle around whereby

(accord. to eq. (6)):

and in components (56)

)(4

3 '13

0' qqxab

cFT

=

'3

'13

0'33

4

3Tx

ab

cFT ==

'13

0'3

'3

'3

''33

'33

'

4

3x

ab

cFsTsSTsS ikik ==→=

'2X)( '

3'2 x

0

0

)(2

1

2

1srots

r

=

=

m

nimni

x

s

= 2

1



• To calculate we use the formulae:

(57)

Proof:

With eq. (56) we start with

whereby the last term, which has been arbitrarily added, is identical to zero and therefore

no mistake. Summarizing the expression and using definition of eq. (9) leads to:

Finally we verify eq. (57):

Comment:

It has to be carefully distinguished between (displacement vector) and (strain tensor)

The apostrophes always indicate that the given quantities are referred to the special SoC which

does not necessarily coincide with the crystal-physical SoC.

Elastic coefficients as listed in the textbooks are referred to the respective crystal-physical SoCs

and may need to be transformed according to current application.

'3

'2

x

''

'2

''

'2

''

'2

'

'

'

'

' 2

1

2

1

2

1

2

1

mn

jimn

mj

nimn

mj

nimn

m

nimn

jj xx

s

xx

s

xx

s

x

s

xx i

+

=

=

=

'

''

'

'

'

'''

'2

''

'2'

' 2

1

2

1

2

1nj

m

imn

n

j

j

n

m

imn

mn

jimn

mj

nimn

j

Sxx

s

x

s

xxx

s

xx

s

x i

=

+

=

+

=

ijnj

m

imni

j

SrotSxx

)( ''

'

'

'

=

=

ijnj

m

imni

j

SrotSxx

)( ''

'

'

'

=

=

'S

's



• Execution of eq. (57) results in:

(58)

• with eq. (55) follows:

(59)

and finally we get:

(60)

• Next step is the calculation of the dependence

(61)

• Constant C equals to zero, because must be zero

'1

'3

'3

'5

'1

'33

'3

'13

'1

'33

213'3

'13

231'3

'2

2

1

x

S

x

S

x

S

x

S

x

S

x

S

x

−

=

−

=

+

=

30'

33'1

'3

'3

'5

4

30

2

1

ab

cFs

x

S

x

S−=

−=

30'

33'3

'2

4

3

ab

cFs

x−=

−== '33

0'33

'2

'3

'2

4

3)( dx

ab

cFsdx

)( '3

'2 x

Cab

cxFsx +−=

3

'30'

33'3

'2

4

3)(

)0( '3

'2 =x



• Having now the bending angle of the rod we can calculate the

maximum deviation ( ) of the rod from axis at position

• Angle is defined like:

• Therefore we finally have:

(62)

with (reciprocal Young’s Modulus) along the rod axis (see

also eqs. (38, 42)) - within the special SoC.

Note 1: The arrangement of the special SoC (meaning: “how the crystal rod is cut”) in relation to the crystal-physical SoC might

require certain transformations of the elastic coefficients, which have been described in previous sections (e.g. eqs. (38-42)).

Note 2: Above made considerations also apply to other rod cross sections – just eqs. (51,52) need to be adjusted.

Note 3: The body’s material Is extended above the neutral plane (see pic.) and compressed underneath.

Note 4: The body’s volume doesn’t change during bending.

'3

'1

'3

'1'

2dx

dx

x

x=

Eab

cF

ab

csFdxxcxxl

c

X 3

30

3

3'330

0

'3

'3

'2

'3

'1

8

3

8

3)()('

1

−=−====

'2

cx ='3'3X

'2

'3x

'1x

'2

1'33

−= Es

'1X

l



Question: Are there other deformations when an anisotropic rod is bent?

• Similar like in the preceding considerations of this chapter we have

to check

(63)

and (64)

• With eq. (55) we get:

(65)

(66)

'2

'5

'1

'4

'2

'13

321'1

'23

312'3

'3

2

1

2

1

x

S

x

S

x

S

x

S

x

−

=

+

=

'3

'4

'2

'3

'3

'23

132'2

'33

123'3

'1

2

1

x

S

x

S

x

S

x

S

x

−

=

+

=

'33

0'34

'3

'3

8

3)( x

ab

cFsx =

0)( '3

'1 =x

2

2'340'

2'3

'3

'3

'1

4

3)()('

1 ab

csFdxcxcxxl

b

bX

===== −

00)(0

'3

'3

'2'

2

==== c

Xdxcxxl



Answer: Yes, there are more deformations – in particular twisting around

the rod’s axis! Not happens to isotropic bodies!

• The responsible elastic coefficient within the special SoC can be

expressed by the tabulated elastic constants (which refer to the

crystal-physical SoC):

(67)

• And for eq. (65) follows:

(68)

Comment:

The denote the projections (cosines) of vectors and onto the directions of

the respective crystal-physical SoC, as shown in eq. (39).

• Even more compact

eq. (68) reads: (69)

with

)()(22 '3332

'34 nqsqqss

==

lkjiijklXnqqqs

ab

cFl

2

20

4

3'2

=

lkji nqqq ,,, q

n

)()(4

32

20

'2

qnnqqqsab

cFlX

+=

)6,...,1()(

)6,...,1()(

=+=+

==

klqnnqqnnq

ijqqqq

lklk

ji



e: Torsion of Rods

• Let’s assume a round rod (length: l, radius: R) with l>R, fixed on its lower side.

(70)

• The torsion moment (torque) is created by an external force per unit area

applied at the end of the rod

• Beside the special (non-crystal-

physical) SoC we intro-

duce a local SoC at

each point of the rod

• The relationship between both

systems writes:

( )qnm,,R

0

'3X

M

ekrF

=

m

n

q

lx ='3

M→

'1X

'2X

l

re

q

e

( )qee r,,

'3

'3

'2

'1

sin

cos

xx

rx

rx

=

=

=

area S



• Calculating the torque we get:

with: (71)

• The boundary conditions can be expressed as follows:

→ no forces on the lateral surface of the rod

and (72)

→ only forces on the end-surface of the road

and are satisfied by the stress tensor:

(73)

• shall be now expressed in terms of the special (Cartesian) SoC

4

2

R

Mk

=

qkRddrrekrerdSFrMR

r

S

4

2

0 02

1

===

erR

MqT

)

2(

4

' =

( )

eqqerR

MT

+=

4

' 2

0' = reT

er



• Acc. to the pic above we can write:

(74)

• stands perpendicular to and →

(75)


(76)

• Introducing the following indexation for the special SoC:

leads to

2'2

2'1

'2

'1 xxrwithnxmxerr r +=+==

nxmx

xx

qnm

reqer r

'1

'2

'2

'1 0

100 +−===

er

req

( ) ( ) mxnxqqmxnxR

MT

'2

'1

'2

'14

' 2−+−=

( ) ( ) mqqmxnqqnxR

MT

+−+= '

2'14

' 2

3,2,1 qnm



(77)

and finally to:

(78)

• Now we can write down all the strains within the special SoC

(79)

+

−

+

=

001

000

000

000

000

100

010

000

000

000

100

0002 '

2'14

' xxR

MT

−

−

=

0

00

002

'1

'2

'1

'2

4

'

xx

x

x

R

MT

( )

( )

( )

( )

( )

( )'1'64

'2

'654

'6

'4

'64

'5

'65

'6

'1

'54

'2

'554

'5

'4

'54

'5

'55

'5

'1

'44

'2

'454

'4

'4

'44

'5

'45

'4

'1

'34

'2

'354

'3

'4

'34

'5

'35

'3

'1

'24

'2

'254

'2

'4

'24

'5

'25

'2

'1

'14

'2

'154

'1

'4

'14

'5

'15

'

2

2

2

2

2

21

xsxsR

MSTsTsS

xsxsR

MSTsTsS

xsxsR

MSTsTsS

xsxsR

MSTsTsS

xsxsR

MSTsTsS

xsxsR

MSTsTsS

+−=→+=

+−=→+=

+−=→+=

+−=→+=

+−=→+=

+−=→+=



• The axial torsion (per unit length) is acc. to eq. (57)

defined as:

(80)

and results together wit eq. (79) in: (81)

Comment: Other contributions from eq. (79) are not valid because of eq. (80)

• Next step is the calculation of the dependence

(82)

• Constant C equals to zero, because must be zero

−

=

−

=

= '

5'2

'4'

1

'13'

2

'23'

1

'3'

3

'

2

13

Sx

Sx

Sx

Sxx

( )'55

'444

'

3ss

R

M+=

'3

)( '3

'3 x

( ) +== '34

'55

'44

'3

'3

'2 )( dx

R

Mssdx

CxssR

Mx ++= '

3'55

'444

'3

'3 )()(

)0( '3

'3 =x



• Having now the torsion angle we have the maximum torsion

angle of the rod around axis

(83)1)

• Eq. (83) is valid within the special SoC.

• Rewriting the elastic compliances yield:

• This expression still refers to the special SoC. Using the direction cosines

of the rod’s vector it can be referred to the crystal-physical SoC

(see eq.(39)).

(84)

Note 1: Quantity C is called the torsion stiffness of the rod. This is the formula to be used to calculate in experiments.

Note 2: The arrangement of the special SoC (meaning: “how the crystal rod is cut”) in relation to the crystal-physical SoC might

require certain transformations of the elastic coefficients, which have been described in previous sections .

Note 3: Above made considerations do not apply to other rod cross sections.

)()()(

)()( '3

'3

'3

'3'

55'44

4'55

'444

'3

'3 lx

l

Clx

ssl

RMss

R

lMlx ===

+=+

==

)( '3

'3 x

'3X)( '

3'3 lx =

( ) ( ) ( )'3333

'33

'3333

'3333

'1313

'2323

'1313

'2323

'55

'44 444)( ssssssssss kk −=−++=+=+

( ) ( )( )ljkiikijkllkjiijkllikjkl qqqqsqqqqsqqs −=− 44

q

'55

'44 ss +



• With eq. (84) C can be written now:

• Further we obtain: (85)

and

This is called the reciprocal of the shear modulus and can be found

tabulated for the crystallographic classes e.g. in /2/

Question: Are there other deformations when an anisotropic rod is twisted?

•

Similar like in the preceding considerations (eqs. (57,80)) we have to check

whether it bends around axis respectively . With eq. (79) we get:

(86)

( )( )ljkiikijkl qqqqsR

qCqG

−=

=

2

1)(2)(

4

'354'

3

'4

'2

'3

'3

'23

132'2

'33

123'3

'1' 2

2

11

sR

M

x

S

x

S

x

S

x

S

x

−=

−

=

+

=

=

'344'

1

'3

'3

'5

'1

'33

213'3

'13

231'3

'2' 2

2

12

sR

M

x

S

x

S

x

S

x

S

x

=

−

=

+

=

=

'2X

'1X

( )( )ljkiikijkl qqqqs

RqC

−=

4)(

4

−= −− )(2)( 11 qEqZqqG



• There is an axis of bend parallel to with a magnitude of

(87)

• The radikand can be rewritten to be:

followed by the transformation to the crystal physical SoC which yields:

• This expression can be shortened, thus finally we have:

Answer: Yes, there are more deformations – in particular bending around the

rod’s axes perpendicular to torsion axis! Not happens to isotropic bodies!

nsms '

34'35 +

( ) ( )2'34

2'354

' 2ss

R

M+=

( ) ( )'3333

'3333

'333

'333

'3333

'3333

'3333

'3333

'3323

'3323

'3313

'3313

2'34

2'35 44)( ssssssssssssss kk −=−++=+

( ) ( )lkjiijkllkjiijklljiijklljiijklkk qqqqsqqqqsqqqsqqqsssss −=− 44 '3333

'3333

'333

'333

qqqqssqqqqqqqssqqqR

M

−=4

'

qqqsqqIsqqqR

M

−= )(4

'



• Let’s now focus on rods with an arbitrary cross-section, fixed on its

lower side. Details can be found in the literature (e.g. /8/,/9/)

• For sake of better illustration we use (in the beginning) the same pic as

done for rods with circular cross-section.

• We have the special (non-crystal-

physical) SoC

• The torsion angle per length is

constant and calculates:

(88)

• The displacements vector varies

along the rod’s axis:

• The radius vector is represented by:

( )qnm,,

0

'3X

m

n

q

M→

'1X

'2X

l

3

r

'3

area S

'3

''3

'3

'3'

33)( xx

dl

d==

rqxs

= )( '3

'3

'

2'2

2'1

'2

'1 xxrwithnxmxr +=+=

s→



• Calculation of the displacement yields:

(89)

• It has to be considered that - in general - every cross-section plane

along the rod’s axis can become, as a result of the torsion, curved, i.e.

is not planar anymore.

• Therefore the so called ‘torsion function’ has to be added to eq. (89)

and we get

(90)

• Next step is the calculation of the strain tensor components. According to

eqs. (9,14) we obtain:

and (91)

• All other strains are identical to zero.

nxxmxx

xx

x

qnm

s

+−== )()(

0

)(00 '3

'3

'1

'3

'3

'2

'2

'1

'3

'3

'

qxxnxxmxxs

++−= ),()()( '2

'13

'3

'3

'1

'3

'3

'2

'

)(22

1 '2'

1

'3'

13 xx

Sx

s

x

sS

k

i

i

kik −

=

+

=

)(2

'1'

2

'3'

23 xx

S +

=



• Next step is the calculation of the resulting stress components,

according to Hooke’s Law (see eq. (20)):

• In Voigt’s Notation this reads:

(92)

• Within the current formalism we have to stipulate that only certain

stresses and strains are present (here exclusively T’4, T’5, S’4, S’5 ).

• Therefore we have to demand that in the special SoC following elastic

coefficients are identical to zero:

+

+−

=+= )()(22 '

1'2

'1323

'2'

1

'1313

'3

'23

'1323

'13

'1313

'13 x

xcx

xcScScT

+

+−

=+= )()(22 '

1'2

'2323

'2'

1

'2313

'3

'23

'2323

'13

'2313

'23 x

xcx

xcScScT

+

+−

=+= )()( '

1'2

'54

'2'

1

'55

'3

'4

'54

'5

'55

'5 x

xcx

xcScScT

+

+−

=+= )()( '

1'2

'44

'2'

1

'45

'3

'4

'44

'5

'45

'4 x

xcx

xcScScT

'65

'35

'25

'15

'64

'54

'34

'24

'14 ,,,,,,,, ccccccccc

3,2,1,)()(22 '1'

2

'23

'2'

1

'13

'3

'23

'23

'13

'13

' =

+

+−

=+= kiwithx

xcx

xcScScT ikikikikik

'''lmiklmik ScT =

'''mimi ScT =

6,3,2,1)()( '1'

2

'4

'2'

1

'5

'3

'4

'4

'5

'5

' =

+

+−

=+= mwithx

xcx

xcScScT mmmmm



• If the special SoC and the crystal-physical SoC coincide, this is NOT(!) the

case for the Point Groups (PGs): triclinic, monoclinic, and trigonal /2/.

• In other cases – when the SoCs don’t coincide - the elastic coefficients of

the crystal-physical SoC have to be transferred into the special SoC –

reflecting the orientation of the rod as cut. If then, in the special SoC,

the above mentioned coefficients are not equal to zero (or at least very

small), the forthcoming formalism doesn’t apply precisely.

• Now we introduce the auxiliary function

and (93)

• Together with eqs. (91,92) we have:

and (94)

• Conversion of the equations lead to:

( )'2'1, xx

'2

'3

'5

xT

'1

'3

'4

xT

−

)()( '1'

2

'44'

1

'2'

1

'55'

2

xx

cx

xx

cx

+

−=

−

=

'1'

1'44

'2

'2'

2'55

'1

11x

xcxx

xcx−

−=

+

=



• Through formation of the difference we get the differential equation:

(95)

• The boundary conditions require special attention. We don’t want to

have forces on the lateral surface of the rod (also in accordance with the

experiment):

with


(96)

• Let’s now focus on the vector which stands perpendicular

on the surface, whereby we have

211

2'1

2

'44

2'2

2

'55

−=

+

xcxc

,0,0,0 '2

''1

''3

''''2

'1

'3

===== vvvvvvdSvTdFXXXkiki

11

11

2'1

2

'44

'2

'1

2'2

2

'55

'2

'1

−

−=

+

=

xcxxxcxx

00 '2

'4

'1

'5

'2

'32

'1

'31 =+=+ vTvTvTvT

0'2'

1

'3

'1'

2

'3 =

−

v

xv

x

0,0,0 '2

'1

'3 = vvv

nvmvv '

2'1

' +=



• We start with a vector along the circumference of the cross-section:

• This vector stands always perpendicular to which yields

• Normalisation means and leads together with

to and (97)

• Inserting of eq. (97) into eq. (96) we get:

• Multiplication by (-dl) leads finally to :

(98)

which equals the total differential (d) of the function

ndxmdxld

'2

'1

' +=

nvmvv '

2'1

' +=

'1

'2

'2'

1'2

'2

'1

'1

'' 0dx

dxvvdxvdxvldv −==+=

( ) ( )2'2

2'1

' 1 vvv +==

( ) ( ) ( )2'2

2'1

2' dxdxdl +=dl

dxv

dl

dxv

'1'

2

'2'

1 =−=

0'1

'1

'2

'2

=

−

−

dl

dx

xdl

dx

x

0'1'

1

'2'

2

=

+

dx

xdx

x



• Since d=0 on the circumference, itself must be constant there.

• In the differential equation (see eq. (95)) only derivatives of appear, thus

we can put 0 on the circumference.

• Now we must select a specific cross-section of the rod – and we are going

to choose a rectangular cross-section (view along the rod axis).

(99)

• Using the Fourier ‘Ansatz’: the (100)

boundary conditions in direction of X’1 (i.e. 1. and 2.) are already fulfilled.

'1X

'2X

'3X

a

b

• We do have 4 boundary conditions:

0),(.20),0(.1 '2

'1

'2

'1 ==== xaxxx

0)2

,(.40)2

,(.3 '2

'1

'2

'1 ===−=

bxx

bxx

( )

=

=0

'1

'2

'2

'1 sin),(

kk x

a

kxWxx



• We want now to set up a differential equation for the Fourier coefficients

• The scalar product looks like:

• Since we get:

(101)

• Finally, applying the definition of the scalar product, we obtain:

(102)

Note: To ease the scripting, until further notice the apostrophes are omitted.

)( '2xWk

( )

=

=0

'1

'2

'1

'2

'1

'1 sinsin),(sin

kk x

a

kxWx

a

mxxx

a

m

( ) '1

'1

0

'2

'2

'1

'1 sinsin),(sin x

a

kx

a

mxWxxx

a

m

kk

=

=

=

==mkif

amkif

ax

a

kx

a

mmk

2

0

2sinsin '

1'1

kma

xWxxxa

mm =

2)(),(sin '

2'2

'1

'1

'1

'2

'1

0

'1

'2 ),(sin

2)( dxxxx

a

k

axW

a

k

=



• The first integration by parts leads to:

• Due to the boundary conditions (see eq. (99)) the 1st term vanishes

• The second integration by parts leads to:

• Again the 1st term vanishes and we have:

(103)

• The differential equation (see eq. (95)) can be reformed to be:

and is inserted into eq. (103) →

+

=

=−=

a

k dxx

xxx

a

k

k

a

x

axxxx

a

k

k

a

axW

01

1

211

1

1

2112

),(cos

0),(cos

2)(

−

=

=

=

a

k dxx

xa

k

k

a

x

ax

xx

a

k

k

a

axW

012

1

2

122

2

1

1

1

122

2

2 sin0

sin2

)(

−=

a

k dxx

xa

k

k

axW

012

1

2

1222 sin2

)(

442

2

2

552

1

2 12 c

xcx

−−=

−−−=

a a

k dxc

cx

a

k

dx

ddxcx

a

k

k

axW

0 01

55

4412

2

2

1441222 sin2sin2

)(



• Acc. to eq. (101) the value of the last integral is and yields:

• Further reformation finally leads to the differential equation for

(104)

• First we go now for the homogeneous solution of eq. (104) and use the

‘Ansatz’ and get:

• The solution can be written down as:

)(2

2xWa

k

+=aa

k dxxa

k

dx

d

kc

acdxx

a

k

k

acxW

0112

2

2

2255

44

01122

442 sin

2sin

4)(

)(sin4

)( 22255

442

01122

442 xW

kc

cadxx

a

k

k

acxW k

a

k

+=

)( 2xWk

−=−a

kk dxxa

k

a

cxW

ac

kcxW

011

5522

44

2255

2 sin4

)()(

2)( 2x

k exW −=44

255

22

ca

ck =

244

255

22

244

255

22

212 )(x

ca

ckx

ca

ck

k ececxW

−+

+=



• Now we go for the particular solution of eq. (104) and use the ‘Ansatz’

• Insertion into eq. (104) yields

• The integral amounts to:

and is unequal to zero only if k=1,3,5,… and amounts then

• Consequently we get for A:

• The general solution of the differential equation is the sum of the

homogeneous and particular solutions and can be written down as:

AxWk =)( 2

−=−a

dxxa

k

a

cA

ca

kc0

1155

442

2255 sin

4

3344

28

k

caA =

( )1cos0

cossin1

1

10

11 −−==

=−=

k

k

a

x

axx

a

k

k

adxx

a

ka

k

a2

3344

2

212

8)(

244

255

22

244

255

22

k

caececxW

xca

ckx

ca

ck

k ++=−+



• Adjustment of the solution to the boundary conditions 3. and 4. results

for the constants c1 and c2 in:

• The general solution now reads:

bca

ckb

ca

ck

b

ca

ckb

ca

ck

ee

eeAccc

442

5522

442

5522

442

5522

442

5522

22

21

−+

−+

−

−−==

AececxWx

ca

ckx

ca

ck

k +

+=−+ 2

442

5522

244

255

22

22 )(

3344

22

2

2

1

3344

22

2

2

1

80

80

442

5522

442

5522

442

5522

442

5522

k

caecec

k

caecec

b

ca

ckb

ca

ck

b

ca

ckb

ca

ck

++=

++=

+−

−+



+

−= 1cosh2

sinh

2sinh

)( 2

442

5522

442

5522

442

5522

2 xca

ck

bca

ck

b

ca

ck

AxWk

• Making use of the identity

• Finally we get with eq. (100) the auxiliary stress function to be:

(105)

• Next task is the calculation of the torsion stiffness (see also eq. (83))

xxx coshsinh22sinh =

−=

2cosh

cosh

18

)(

442

5522

2

442

5522

3344

2

2b

ca

ck

xca

ck

k

caxWk

1

44

55

2

44

55

,...5,3,133

442

22 sin

2cosh

cosh

118

),( xa

k

bc

c

a

k

xc

c

a

k

k

caxx

k

−=

=



• Starting with the energy of the twisted rod we detail eq. (29)

• Inserting the strains from eq. (92) and the auxiliary stress function

(see eq. (93)) we get:

(106)

• Torsion energy per length requires the integration over the rod’s

cross-section S

( )

( ) ( )445544445555

3232232331311313

2

1

22

12

1

2

1

TSTSTSTSTSTS

TSTSTSTSTS ikik

+=+++

=

+++==

2

1

23

44

2

2

23

55

4

44

45

55

5

2

1

2

1

2

1

+

=

+=

xcxcT

c

TT

c

T

21

2

144

2

255

23

11

2

1dxdx

xcxcS

+

=

−

=

=

+

−

−=

=

=

−−

2

2

20

121

2

441

1

144

1

0

2

2

222

2

552

2

255

23

1

0

11

2

21

2

1dxdx

xcx

ax

xcdxdx

xcbx

bx

xc

b

b

aab

b



• The function is identical to zero on the circumference, and

consequently the terms are also zero.

• Further we get:

• The expression in brackets represents the differential equation (eq.(95))

and amounts exactly to -2 (depends only on the cross-section).

(107)

• To calculate the total energy of the entire rod, integration along the

rod’s axis is required. Use was also made of eq. (107).

(108)

Note: Now we are going to use the apostrophes again, in order to express that we are still moving

within the special SoC

2121

2

44

23122

2

2

55

23

1

2

11

2

1dxdx

xcdxdx

xc SS

−

−=

21

,xx

2121

2

4422

2

55

23

11

2

1dxdx

xcxcS

+

−=

212321

23 2

2

1dxdxCCdxdx

SS

===

l

Cl

l

Cl

Cdx

Cdx

ll

rod

2'3

'

2

2'3

'2'3

''3

0

2'3

'

0

'3

2222

=====



• The potential energy of a road twisted by an external torque M (M=D) (109)

can be generally expressed as follows:

(110)

• Comparing eqs. (108) & (110) we find: with l being the (111)

length of the rod. C is called torsion stiffness (see also eq. (83)).

• Now we need to calculate composition of C’, which is acc. to eq. (107)

road axis (112)

parallel to X3’

2'3

'

0

'3

'3

'

0

'3

2

1''

=== DdDdMpot

'

''

M

l

CD ==

−

=

−==2

20

'2

'1

'1

'44

'55

'

'44

'55

,...5,3,133

'44

2'2

'1

' sin

2cosh

cosh

118

222

b

b

a

kS

dxdxxa

k

bc

c

a

k

xc

c

a

k

k

cadxdxC

−==

=

bc

c

a

k

c

c

k

ab

k

caCC

kx '

44

'55

'55

'44

,...5,3,134

'44

3'

2tanh

2132'3



• Likewise we find:

road axis (113)

parallel to X1’

road axis (114)

parallel to X2’

• It needs to be mentioned that the case of a square cross-section is covered,

if we put simply: a=b

• Putting everything together (eqs. (108)-(114)) we can imagine to have the

following experimental approach / setup:

• The rod (dimensions: a, b, l) with defined orientation is excited

by a certain external torque M

• The respective twisting angle is measured

−==

=

bc

c

a

k

c

c

k

ab

k

caCC

kx '

66

'44

'44

'66

,...5,3,134

'66

3'

2tanh

2132'2

−==

=

bc

c

a

k

c

c

k

ab

k

caCC

kx '

55

'66

'66

'55

,...5,3,134

'55

3'

2tanh

2132'1

'i



• The value of is calculated according to

• Since the - depending on the PG – frequently consist of

two different elastic coefficients, the procedure for the

calculation of the single coefficients might be:

• to use a numerical iteration procedure, or

• to cut / do an approximation of the series expression

for the such way, that the elastic coefficients can be

approximatively calculated

• Special attention has to be paid to cases where the special SoC

is different to the crystal-physical SoC (i.e. in cases of special

cuts of the rod). Here the elastic coefficients have to be

transformed acc. to the recipe (see. eq. (122)):

and the applicability of the above outlined formalism has to be

checked – i.e. the non-existence of :

''

ix

MlC

i

=

'ix

C

'ix

C

'ix

C

nopqmqlpkoiniklm cDDDDc ='

'65

'35

'25

'15

'64

'54

'34

'24

'14 ,,,,,,,, ccccccccc



5. Ferroelastics

a: Hysteresis Loop

• A ferroelastic material is characterized by a non-zero spontaneous

strain (SS) – analogue to the spontaneous polarization (in ferroelectrics)

or to the spontaneous magnetization (in ferromagnetics) – see e.g. the

comprehensive overview papers /3/, /4/, /5/.

• A full ferroelastic crystal has at least two Orientation States (OS, domains)

which differ in their SS-tensors (/6/).

• Ferroelasticity is the result of a (real or fictive) Phase Transition (PT) from

a Paraelastic- (PP) to a Ferroelastic Phase (FP).

• Main (and obvious) differentiator between elastic and ferroelastic behaviour

is the dependence of the strain S on stress T.

• Ferroelastics (in contrast to classical elastic materials that follow Hooke’s

Law) are characterized by a non-linear S=S(T) dependence which is called

Ferroelastic Hysteresis Loop.



• Aizu (/6/) derived the specific forms of the SS-Tensors for all possible ferro-

elastic PTs within the framework of Point Groups.

• As outlined above, in the FP the crystal usually exists in a multi-domain

state, whereby the different OSs are separated by domain walls.

• Sapriel (/7/) provided tables with the permissible domain walls for all

ferroelastic PTs.

• Ferroelastic domains can be normally observed with the aid of polarizing

microscopy because through the ferroelastic PT the crystal system changes

(if the hexagonal and trigonal systems are treated as one system)

S

T

S

T

SS

Tc

Linear Elasticity Ferroelasticity

TC: Coercive

stress

SS: Spontaneous

strain



b: Domain Switching (and more)

• Under application of specific external stresses (Text) the OSs behave

differently – they can grow, disappear or even remain as they have been.

• The criterion for a potential change is the value of the interaction energy

Eint (see eq. (29) and /9/):

(115)

• Under the action of the stress the crystal tends to achieve the state with

the lowest interaction energy (the minimum).

• Wadhawan (/3/) proposed a simple scheme how to calculate the directions

of (compressive) stress to facilitate/generate a certain OS.

• In many cases the stress directions are obvious (see examples below), but

sometimes it requires indeed detailed calculations.

• Let’s now consider to use the Bending of Rods (see above ch. 4.d:) to record

hysteresis loops, and possibly to measure simultaneously elastic coefficients

iexti SSTE −=int



• With external bending forces F0 acting up and down, tensile and

compressive stresses interchange in the upper, respectively in the lower

part of the rod.

• That means we have got a ‘vehicle’ that can produce alternating pure

tensile and compressive stresses.

• The stresses are not homogeneously distributed across the rod’s cross-

section, but vary linearly along axis

m

n

q

F0

→

neutral plane

(zero stress)

Tensile stress

in upper part

Compressive

stress in lower part

( )mX'

1



(116)

• The coercive stress TC is the applied stress, where an OS switch to another

(ref. to the pic above in this chapter, showing the hysteresis loop)

• TC cannot be realized on the neutral plane, but it should be located in the

intervals 0< and 0< , respectively 0> and 0>

• This means, that within a stress-range of , which corresponds

acc. to eq. (116) to a certain band in the rod with regards to , no switching

of OSs can be initiated. This situation is depicted in the following drawing:

represents the neutral

plane of the rod (stress-free)

b ( )mX'

1

'13

0'1

'

4

3)( x

ab

cFxT =

'T

'

0

3'''

1'

3

4)(' CCT

TcF

abTTxx

C===

-b

0'1 =x

bx '1''bC TT bx −'1

''bC TT −

'''CC TTT −

Tension

Compression

'1X

'CT

'CT−

'

0

3'''

1'

3

4)(' CCT

TcF

abTTxx

C−=−==

−

''CT

x

''CT

x−

'bT

'bT −−



• Remember: In an elastically deformed rod (see chapter: 4.d:, eq. (62)) the

deviation l of the bar at its end calculates as:

(117)

• But, it must be noted that in the FP we have several OSs, which differ in

their orientation with regard to the crystal-physical SoC as set up in the PP.

• Expecting that trough the tensile stress in the upper and through the

compressive stress in the lower parts of the rod the domain situation is

unambiguously defined, but in the grey marked band (see pic above) not!

• In that area a randomized mixture of all possible OSs must be expected

m

n q

0

''CT

x

''CT

x−

No OS switching in

the grey marked area

cx ='3

Eab

cF

ab

csFle 3

30

3

3'330

8

3

8

3−=−=



• This means that for Ferroelastics in the FP the Young Module E, as

used in eq. (117), must be treated as an effective , stemming from:

1. The E-Module of the upper part with defined OS-structure

2. The E-Module of the lower part with defined OS-structure,

but (generally) different from 1.

3. The E-Module of the grey band with unknown OS-structure

• To this end we need to adjust eq. (117) using :

(118)

• In Ferroelastics in addition we have to consider the spontaneous strains

(SSs), that might occur along the rod’s axis, and consequently will contribute

to the overall deviation/bending of the rod.

• Those contributions will (generally) differ in the upper and lower parts, and

of course depend on the strength and on the direction of the externally

applied force.

E

Eab

cF

ab

csFle 3

30

3

3'330

8

3

8

3−=−=

E



• It is required to calculate those contributions to l of eqs. (117,118).

• lu and ll are the lengths of the rod’s upper and lower parts, and

varies between 0 and c

• Using the same procedure as previously we get the rod deviation

caused by strains (subscript S):

(119)

Note 1: stands for the strain in the upper part and in the lower part, respectively

( )nSSb

x

l

lb

r

nbr

ln

r

l

lu

l

u

ul

−−=→

−

=→

+−=−=

2

1

22

'3'

'

'3x

'3

'1

'3

'1

2dx

dx

x

x=

2

'3x

'1x

( ) ( )luc

luc

S SSb

cdxSS

b

xdxxcxxl −−=−−==== 42)()(

2

0

'3

'3

0

'3

'3

'2

'3

'1

uS lS



• As already mentioned above, in the FP the particular arrangement of the

OSs results in different Young’s Moduli in the different parts of the rod.

(This creates a significant difference to the considerations made in chapter

4.d:, where no spatial distribution of the E-Module is present)

• From eq. (55):

• Denoting the elastic coefficients acc. to the rod’s parts, we get

upper part

lower part

• The strains on the edges of the parts are written:

upper part

lower part

• Making use of eq. (119) we have:

(120)( )28

3 '33

'33

3

30

lu

e

ss

ab

cFl

+−=

'13

0'33

'3

'13

0'3

'

4

3

4

3x

ab

cFsSx

ab

cFsS =→=

l

u

s

s'33

'33

→

→

bab

cFsbxS

bab

cFsbxS

ll

uu

30'

33'1

'3

30'

33'1

'3

4

3)(

4

3)(

−=−=→

==→



• Finally the total deviation from the rod’s axis is obtained when the

contribution from the SSs are added ( in eq. (119) just need to

be replaced by the SSs: ):

(121)

Example I: ferroelastic species 4mmFmm2

• For the lattice parameters in the FP (PG mm2) it is assumed a || x is greater

than a of the tetragonal PP, and b || y is smaller than a of the PP. c || z is not

relevant here.

• Therefore schematically the 2 OSs are arranged as follows regarding the

crystal-physical SoC. They differ by a rotation of 90° around the z-axis.

( ) ( )lulu

SSetot SSSSb

css

ab

cFlll −−

+−=+=

428

3 2'33

'33

3

30

ye

xe

ze

OS 1 OS

2

tetragonal PP

lu SS ,lu SSSS ,



• The transformation of the tensors of the spontaneous strains was done

with the aid of the rotation matrix (angle ) around the z-axis:

acc. to the recipe: (122)

• Since in this particular case, tension favours OS 1 and

compression OS 2 (because the interaction energy is minimal, respectively)

)(2

1

000

00

00

000

00

00

2121 SSww

w

SSw

w

SS OSOS −=

−

=

−= rotation by

90° around z

−=

100

0cossin

0sincos

klD ( ) ( )imlmkikl SSDDSS =

F0

→

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1

OS

2

OS

2

OS

2

OS

2

OS

2 F0

→

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1OS

2

OS

2

OS

2

OS

2

OS

2

ye

xe

ze

0)(2

121 −= SSw

Case A Case B



• According to eq. (29,115) the interaction energy is defined as:

(123)

• Goal is to find the minimum, whereby tension is counted positive,

compression negative and shear positive, if the shear angle decreases

• First we have to achieve a defined (preferable mono-domain) state

in the upper and lower parts of the rod.

• This can be materialized via an applied force F0 of suitable strength that

initiates the switching of domains into the desired state(s).

• Afterwards the force shall be removed and the domains are expected to

remain in the defined status.

• According to eq. (124)

the bar shall be afterwards excited by F0 and the rod’s deviation le as

linear function of the force allows to calculate the elastic coefficient

along the rod axis.

• Switching of OSs must not happen when F0 is applied, i.e. the force to

measure the elastic coefficients must be small.

( ) ( )lulu

SSetot SSSSb

css

ab

cFlll −−

+−=+=

428

3 2'33

'33

3

30

( )lu ss '33

'33 +

iexti SSTE −=int



• In relation to the crystal-physical SoC we have:

and for Case A (125)

whereby for Case B just the superscripts u and l need to be interchanged.

• Finally we have: for Case A

and (126)

for Case B

• But again with the remark, that not the whole rod unambiguously

contributes to the elastic coefficients (because the grey band in the

pics above isn’t in a well defined state with regard to the OSs).

• Next undertaking is the consideration of the SSs.

• Assuming – as already mentioned above - that an applied force F0

which e.g. increases gradually and as a consequence, if the stress exceeds

the coercive stress in the upper und lower parts of the rod, the OSs should

switch and both parts shall enter (ideally) into a mono-domain (or at least

well defined) state.

( )28

3 22113

30 ss

ab

cFle

+−=

( )28

3 11223

30 ss

ab

cFle

+=

22222222'3311111111

'33 ssssssss lluu ======



• If then the switching force F0 is removed the rod are expected to remain

bent by the SSs exclusively.

• Using eq. (121) we calculate for Case A and

for Case B

• Thus the deviation lSS of the rod at zero forces (F0=0) is a direct measure

for the spontaneous strain w

• Moreover, this setup should allow to record Hysteresis Loops (also like the

forthcoming 2 examples)

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1

OS

2

OS

2

OS

2

OS

2

OS

2

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1OS

2

OS

2

OS

2

OS

2

OS

2ye

xe

ze

Case A Case BSSu=w

SSl=-w SSu=w

SSl=-w

( )wb

clSS 2

4

2

−= ( )wb

clSS 2

4

2

=



0)(2

1

000

022

3

02

3

2

000

022

3

02

3

2

000

00

00

21

321

−=

−

=

−

−−

=

−=

SSw

ww

ww

SSw

w

ww

SSw

w

SS OSOSOS

Example II: (NH4)2SO4 - (assumed) ferroelastic Species 6/mmmFmmm (/10/,/11/)

• Lattice parameters at room temperature (in FP with Point Group (PG) mmm)

• c = 7,728 Å

a = 10.636 Å (|| x)

b = 5,993 Å (||y) which is larger than the hexagonal

• Therefore schematically the 3 OSs are arranged as follows regarding the

SoC. They differ by a rotation of 120° around the y-axis.

7320,137746,1 ==b

a

xe

ze

ye

OS 1

rotation by

120°

around z

again

rotation by

120°

around z

hexagonal PP



• Since in this particular case, tension favours OS 1 and

compression OS 2 and OS 3 (because the interaction energy is minimal,

respectively) – see also eq. (123) above.

• Again looking at: (127)

we first focus on , using the transformation of 120° around z →

• OS 1: OS 2, OS 3:

• These are the results for Case A, whereby for Case B the superscipts l

and u need to be interchanged.

F0

→

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1

F0

→

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1

0)(2

121 −= SSw

Case A Case B

xe

ze

ye

( ) ( )lulu

SSetot SSSSb

css

ab

cFlll −−

+−=+=

428

3 2'33

'33

3

30

( )lu ss '33

'33 +

11111111'33 ssss uu === ( )6612221111

'33 369

16

1ssssss uu +++==



• Finally we get:

for Case A

and

for Case B

• Now step is to investigate the spontaneous strains:

• Using eq. (127) we calculate for Case A and

for Case B

( )32

36917

8

3 661222113

30 ssss

ab

cFle

+++−=

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1

OS 1 OS 1 OS 1

Case A Case B

xe

ze

ye

SSu=w

SSu=w

SSl=-w/2

SSl=-w/2

( )32

36917

8

3 661222113

30 ssss

ab

cFle

+++=

−= wb

clSS

2

3

4

2

= w

b

clSS

2

3

4

2



Example III: Rochelle Salt - ferroelastic Species 222F2 (/12/)

• Lattice parameters at 323 K (in PP with Point Group (PG) 222)

b =11,924 Å(||y); a=14,298 Å (||x); c=6,230 Å (||z)

• Lattice parameters at 274 K (in FP with Point Group (PG) 2)

b = 11,869 Å; a = 14,316 Å; c = 6,223; (a,c)= 89,26°

(the 2-fold b-axis being also a ferroelectric axis)

• In the FP we shall have 2 OSs as follows regarding the SoC. They

differ by a 2-fold rotation around the x- or z-axis

*)

5212

1

00

000

00

00

000

00

Sw

w

w

SS

w

w

SS OSOS =

−

−

=

= 2-fold rotation

around x or z

rhombic PP

ye

xe

ze

OS 1 OS 2

*) Note: Shear components of the strain tensor are counted positive if the shear angle decreases under

action of a shear stress.



• Since the SS-Tensors of the two OSs do not posses components in

x-direction, no interaction with the tensile nor compressive stress shall

be possible in the bar, and therefore also no domain switching.

• The way forward would be to try to rotate the crystal around y, thus the

desired SS-strain components would appear.

• To this end let’s rotate the crystal with regard to the SoC by 45° counter

clock-wise around y

• Applying the related rotation matrix and the recipe (see eq. (122)):

we get the rotated strain tensors, which now posses components in

x-direction (i.e. along the rod axis)

5212

1

00

000

00

00

000

00

Sw

w

w

SS

w

w

SS OSOS =

−

=

−

=

−

=

−

=

2

20

2

20102

20

2

2

cos0sin

010

sin0cos

klD( ) ( )imlmkikl SSDDSS =



• It is obvious that under the assumption , tension (Case A:

upper part) favours OS 1 and compression (Case A: lower part) OS 2

(because the interaction energy has to be at minimum – see eq. (115).

• Again looking at: (128)

we first focus on , using the transformation of 45° around y

and for OS 2 additionally of 180° around z we get:

for OS 2:

for OS 1:

( ) ( )lulu

SSetot SSSSb

css

ab

cFlll −−

+−=+=

428

3 2'33

'33

3

30

( )lu ss '33

'33 +

02

1513 == SSw

( )55351513331111'33 222

4

1ssssssss uu +++++==

F0

→

F0

→Case A Case B

( )55351513331111'33 222

4

1ssssssss ll +−−++==

ye

xe

ze



• Finally we have:

for Case A

and (129)

for Case B

• But again with the remark that not the whole rod unambiguously

contributes to the elastic coefficients (because the grey band in the

pics above isn’t in a defined state with regard to the OSs).

• Next undertaking is the consideration of the SSs.

• Assuming – as already mentioned above - that an applied force F0

which e.g. increases gradually and as a consequence, if the stress

exceeds the coercive stress in the upper und lower parts of the rod,

the OSs should switch and both parts shall enter (ideally) into a mono-

domain (or at least well defined) state.

( )4

2

8

3 551333113

30 ssss

ab

cFle

+++−=

( )4

2

8

3 551333113

30 ssss

ab

cFle

+++=



• If then the switching force F0 is removed the rod is expected to remain

bent by the SSs exclusively.

• Taking into account that and using eq. (128) we calculate

for Case A:

and

for Case B:

• Thus the deviation lSS of the rod at zero forces (F0=0) is a direct

measure for the spontaneous strain w=S13

Case

A

Case BSSu=w

SSu=w

SSl=-w

SSl=-w

( ) wb

cw

b

clSS

22

4

22

−=−=

02

1513 == SSw

( ) wb

cw

b

clSS

22

4

22

==

ye

xe

ze



c: Torsion Pendulum I.

• A pendulum shall be a device where an oscillating torque excites a (round)

rod to get it (alternating) twisted by a certain angle.

• Based on the considerations made in chapter 4. (e: Torsion of Rods)

an elastic pendulum (e.g. /13/) can be the means to measure shear moduli

as well as spontaneous strains SSs (latter of ferroelastics).

• At a first glance it becomes obvious that crystals are promising candidates

that exhibit ferroelastic behaviour accompanied by SSs that are pure shear

strains.

• According to /6/,/7/ following ferroelastic species are candidates (selection

only): 422F222, 4mmFmm2, 4/mmmFmmm, 422F2, 222F2, …

• Let’s look at Rochelle Salt - see chapter 5. (b: Domain Switching (and more),

example III).

• Having the torsion angle (see eq. (83)) at the end of the round rod of

length l in dependence on the exciting torque and of the elastic compliance

moduli:

(130))()( '

55'444

'3

'3 ss

R

lMlx +

==



• The orientation of the crystal regarding the rod’s special SoC

shall be as follows:

• We have for OS 2: and

and for OS 1: and

2323'2323

'44 44 sss ==

0

'3X

'1X

'2X

m

n

q

xe

ye

ze

1313'1313

'55 44 sss ==

2323'2323

'44 44 sss == 1313

'1313

'55 44 sss ==

crystal-

physical SoC

qnm,,

special SoC



• Since the crystal behaves uniformly regarding the torsion (i.e. OS 1 and

OS 2 do not differ in their the elastic coefficients), the sum can

be precisely measured (see eq. (83)), even in a not-mono-domain state

(but in latter case addt’l. contributions to the elastic coefficients from the

domain walls should be expected).

• If the position of the crystal in the rod is another one than depicted, a

corresponding transformation of the elastic coefficients (from one to the

other SoC) has to be performed

• The situation is getting more complicated if we now look at the spontaneous

strains (SSs)

• When the ferroelastic PT in Rochelle Salt occurs (species 222F2) and we

have the crystal’s orientation as shown in the pic above, then the crystal

exists typically in a mixture of the two OSs (with unknown spatial distribution).

• Reminder: the stress tensor in cylindrical coordinates (ref. to eq. (73))

reads:

(131)

Note: The stress tensor doesn’t depend on the rod’s axis coordinate

5544 ss +

( )

eqqerR

MT

+=

4

' 2

'3x



• This yields that 1) its value depends only on the radius coordinate r and

2) any interaction with the (fixed) crystal depends on the angle coordinate

➢ 1) means that, if we have a coercive stress TC (as discussed in chapter 5.

a: Hysteresis Loop, eq. (116)), with eq. (132) we can calculate a ‘coercive

radius’, which is independent from crystal’s orientation.

(132)

In the region the persisting stress (by value) is smaller than TC

which results in the fact that no switching of OSs shall be possible there.

BTW: The coercive radius will decrease if the exciting torque (M) increases.

➢ 2) means that we have to look at the mutual orientation of the stress tensor

ellipsoid (see eq. (78)) and the crystal (mono-domain state with OS 2 shown):

CC TM

Rr

2

4=

Crr

1

2

5

4

3OS 1 OS 2

rC )( '2Xn

)( '1Xm

)( '3Xq



• Exploration in which orientation of the crystal (1…6 in the pic above) the

interaction energy between the respective SS-Tensor and the stress tensor

becomes minimal, results in the OS that will be preferred and switched into.

• Next step is the calculation of the interaction energy for OS 1 and OS 2

(see eq. (115)).

We use the stress tensor (from eq. (78))

and the SS-Tensors for OS 1 and OS 2 (see previously) are:

• The interaction energies amount:

for OS 1 (133)

for OS 2 (134)

4'1

'2

'1

'2

4

' 2

0

00

002

R

MCwith

xx

x

x

R

MT

−

−

=

02

1

00

000

00

00

000

00

521 =

−

−

=

= Swwith

w

w

SS

w

w

SS OSOS

( )wxCSSTE OSi

extiOS

'2

1int1 2−−=−=

( )wxCSSTE OSi

extiOS

'2

2int2 2−=−=



• Now we’re going to investigate the individual position 1…6 for OS 1 and OS 2

• The positions differ in their coordinates which enter the stress tensor

• We do have for 1: for 2: for 3:

for 4: for 5: a.s.o. (135)

• It can be easily shown the same way that in the other (left) side of the rod

(i.e. the region where ) OS 1 is energetically preferred.

'2

'1, xx

0, '2

'1 =−= xxx

xxxx2

2,

2

2 '2

'1 == xxx == '

2'1 ,0

xxxx2

2,

2

2 '2

'1 =−=

0, '2

'1 == xxx

No. / angle Eint (OS 1) Eint (OS 2)Comment

(mind that w>0)

1 / 0° 0 0 OS 1 / OS 2 - no preference

2 / 45° OS 2 preferred

3 / 90° OS 2 preferred (with deeper energy

than in pos. 2 and 4)

4 / 135° OS 2 preferred

5 / 180° 0 0 OS 1 / OS 2 - no preference

xwC 2 xwC 2−

xwC 2 xwC 2−

xwC 2 xwC 2−

0'2 x



• All above made consideration base on the fact the exciting torque shall

be directed parallel to of the special SoC.

• Should the torque be directed anti-parallel to then OS 1 and OS 2

swap their roles (“switching”). Those circumstances are depicted on the

following picture.

• Looking at the pics it can be summarised:

• If the torque is positive (counter-clockwise twist as shown

on the right hand pic) than the rod is expected to exist in

two mono-domain regions (left, right acc. to the special SoC)

q

q

( )M

+

OS 1 OS 2

M

+M

−

( )M

−



• If the torque is negative (clockwise twist as shown on the

left hand pic) than the rod is expected also to exist in

two mono-domain regions – but swapped

• As discussed earlier, in the rod there is a region where

no OS-switching can be possible, because the value of

the stress doesn’t reach TC.

• Regarding the angle-dependence of the stress tensor around

the rod’s axis it has to be accepted that in pos. 1 & 5 no

relevant stress prevails (means stress smaller than TC).

• The stress increases from pos. 1 over pos. 2 to 3 where it

becomes maximal, followed by a decrease towards pos. 5

• To identify the regions of the rod where switching can be possible at all,

we need to check where the effective stress locally exceeds TC (taking into

consideration not only the absolute value but the value caused by the

mutual orientation (angle) between the stress and the crystal’s OSs).

• Having the ‘stress vector’ eqs. (72,75):

Crr

( )nxmxR

Mer

R

MT

'1

'244

' 22+−==



• The relevant component lies parallel to :

(136)

and depends only on .

• From that we have: (137)

• Combination of eqs. (136) and (137) yields:

(138)

'24

' 21

xR

MT X

−=

)( '2Xn

)( '1Xm

RR

MTRxTr

R

MTrxT RXCCCX 4

''2

'

4

''2

' 2)(,

2)(

11 ======

'2x

)( '1Xm

'24

'2

'' 21

xR

Mx

r

TT

C

CX

==

)( '2Xn

)( '1Xm

R

Cr Cr R

'CT

'RT

'

1XT



• Within the special SoC the stress

component equals, respectively

exceeds, the coercive stress in the

blue bordered regions.

• Therefore switching among the two

OSs (incl. monodomainisation) shall

be possible exactly only there.

• Remaining parts are idle.

)( '2Xn

)( '1Xm

'

1XT

'CT

• Now we want to investigate by which angle the rod will remain in a twisted

state - caused by the acting/prevailing spontaneous strains (SSs) – that

means without externally applied torque. Precondition would be of course

that the blue regiones above have been transferred into a monodomain state

beforehand (by a suitable torque, which has to be removed afterwards).

• In pic above we see 4 prominent points, characterised by the coordinates:

(139)

1

2

4

6

CCCC

CCCC

rxrRxRxxrxrRx

rxrRxRxxrxrRx

=−−====−=

−=−=−==−=−−=

'2

22'1

'2

'1

'2

22'1

'2

22'1

'2

'1

'2

22'1

,:6,0:5,:4

,:3,0:2,:1

'3

3

5



• The displacement vector of the unit cell can be simply drawn for OS 2.

• With lattice parameters a, b, c, and we have:

• (140)

• At the end of the rod (length l) we calculate the total displacement to be:

(141)

Remark

- this displacement value is only

locally correct at positions &

- In other positions projections are

to be considered and yield for the

component of along the circum-

ference:

OS 2

ye

xe

ze

mc

ec

s x

tantan' −=−=

OS 2

OS 2

OS 2

ml

s tot

tan' −=

l

c

'tots

's

2 5

'2

''

tanx

R

less totcircum

−==

'tots



• Finally the maximum displacements, with +M that twists the rod, are:

for the right hemisphere:

and likewise (142)

for the left hemisphere:

• The existing situation is depicted here:

)( '2Xn

)( '1Xm

'3

• The torsion angle of the rod can

be now expressed:

=

R

s tot'

'3 arctan

ml

s tot

tan' −=

ml

s tot

tan' =

tan' l

s tot =



• Finally we get with and with

(143)

and

• Thus through the measurement of the twisting angle after respective

switching of the OSs the SS (S13) can be determined and Hysteresis

Loops recorded – e.g. for Rochelle Salt.

• Analogue to eq. (121), it should be recognised that the torsion angle

consists of an elastic part (see eq. (83)) and of a portion stemming

from the spontaneous strain (eq. (143):

(144)

=

R

Sl

SR

l 13

13

'3

2arctan

2cotarctan

1329090 SSS −=−=

)2cot()290tan( 1313 SS =−

'313

2l

RS

'3

++

=+=

13

55444

'3

'3

'3

2cotarctan)(

SR

lss

R

lMSSetot



Torsion Pendulum II.

• In this chapter we want to investigate the behaviour of a rod with a

rectangular cross-section, which shall be excited by an oscillating torque.

• Because of easier preparation of specimens with quadratic/rectangular-

compared to circular-cross-sections, respective shapes and experiments

are clearly preferred.

• Based on the considerations made in chapter 4. (e: Torsion of Rods)

we want to concentrate on the elastic/ferroelastic properties of

(NH4)2SO4 (Species 6/mmmFmmm). This crystal was discussed already

in chapter 5. Ferroelastics, b: Domain Switching (and more), Example II.

• The orientation of the crystal (meaning OS 1…3) from page 68 have

been (via transformation of the coordinate system) referred to the rod’s

special SoC (see overleaf)

• This also concerns the SS-tensors, where we have now:

(145)

qnm,,

0)(2

1

20

2

30002

30

2

20

2

30002

30

2

00

000

00

21321 −=

−−

−

=

−

=

−

= SSww

w

ww

SSw

w

ww

SS

w

w

SS OSOSOS



'3X

'1X

'2X

m

n

q

xe

ye

ze

crystal-

physical SoC

special SoC

a

b

lOS 1

OS 3

OS 2

• The same transformation has to be performed for the elastic stiffnesses of

the orthorhombic OSs 1…3 to express (see eq. (112) through the

relevant coefficients referred to the crystal-physical SoC

• According to the ‘recipe’ (eq. (122)) we get after some cumbersome calculations:

'55

'44, cc



(146)

• Since the actual composition of the rod isn’t known regarding the OSs, we

would – if measuring the torsion stiffness – obtain a mixture of the coefficients

as shown in the table above.

• Therefore we should strive to bring the crystal (via suitable external stresses)

into a well defined domain status.

• Investigation of the interaction energy (see eq. (115)) together with definition

of the stress components (see eq. (92)) and the SS-Tensors (see eq. (145))

(147)

yields:

OS 1

OS 2

OS 3

='44c ='

55c

4

3 4455 cc +

66c55c

4

3 4455 cc +

16

4633 66122211 cccc +−+

16

4633 66122211 cccc +−+

'25

'2'

1

'55

'3

'5 )( xCx

xcT −−

=

'

14'1'

2

'44

'3

'4 )( xCx

xcT +

=



(148)

• The essence is that with stresses T’5 a monodomainization towards OS 2,

respectively OS 3 should be feasible in the two hemispheres!

(149)

Sign of x‘2 Eint (OS 1) Eint (OS 2) Eint (OS 3)Comment

(mind that w>0)

>0 0 >0 <0 OS 3 is preferred

=0 0 0 0 no OS preferred

<0 0 <0 >0 OS 2 is preferred

( ) 00'25

1int1 =−−=−= xCSSTE OS

iextiOS

( )

−−=−= wxCSSTE OS

iextiOS

2

3'25

2int2

( )

−−−=−= wxCSSTE OS

iextiOS

2

3'25

3int3

OS 3 OS 2

M

+M

−

Note:

T‘4 doesn‘t interact with the SS.



• As also in the previous chapter discussed the stress depends on the

coordinates (see eq. (132)) and in a certain region its value is

smaller than the coercive stress TC

Getting the stress vector (comp. eqs.

(72,75) and eq. (92)):

Assuming that C5 C4=C we have:

(150)

• This means that in the grey circle area no switching among the OSs shall

be possible. Consequently the domain situation there is undefined!

• Question: Taking into account the specific orientation of the (NH4)2SO4

crystal, what are the regions where switching can be possible at all?

• Answer: Only stress initiates switching.

Crr

)( '2Xn

Cr

)( '1Xm

nxCmxCT

+− '14

'25

'

CCC rCrrTT

rCxxCxCxCT

==

=++

)('

2'1

2'2

2'1

2

4

2'2

2

5'

'25

'5 xCT −



• Looking now at a potential measuring strategy:

• If the torque is positive (counter-clockwise twist as shown

on the right hand pic above, see pic (149)) than the rod is

expected to exist in two mono-domain regions (left, right

acc. to the special SoC)

• If the torque is negative (clockwise twist as shown on the

left hand pic) than the rod is expected also to exist in two

mono-domain regions – but swapped

)( '2Xn

Cr

)( '1Xm

• Within the special SoC the stress

component T’5 equals, respectively

exceeds, the coercive stress TC in

the blue bordered regions.

• Therefore switching among the two

OSs (incl. monodomainisation) shall

be possible exactly only there.

• Remaining parts are idle, as well

as OS 1.



• But, in the rod’s middle there is a region (surrounded in blue)

where no OS-switching can be possible, because the

value of the stress doesn’t reach TC. The domain situation

there is undefined.

• Initially the rod shall be increasingly twisted by a torque +M

(or –M) with the aim to achieve mono-domain states left and

right.

• Subsequently taking away the torque, it is assumed that the

domain states regarding OS 2 and OS 3 are conserved.

• If now the rod is subjected to twisting oscillations (small torque,

in order not to amend the domain structures), the elastic

coefficients can be measured.

• As outlined in table (146) we have the coefficients expressed

within the crystal-physical SoC

• This way we can measure the coefficients (which are

identical for OS 2 and OS 3), but have to accept, that

the contribution of the grey marked area (eq. (150)) is unknown.



• In particular the amount of OS 1 in the grey area cannot be

quantified, and leads to a certain/small measuring error.

• Ultimately with eqs. (111,112) and eq.(146) we get:

• Next we shall check the spontaneous strain (SS). We focus on OS 2 and

OS 3 since OS 1 cannot contribute. According to eqs. (14,145) we have the

relevant SS-Tensor components:

(151)

with

• The displacement vector of the unit cell can be schematically drawn

for OS 2, OS 3 as follows.

+

+−+

+−+

+−

+

==

=

bcc

cccc

a

k

cccc

cc

k

ab

k

ccal

MC

k

x

4455

66122211

66122211

4455

,...5,3,13

44455

3

'3

3

4633

4tanh

4633

341

)3(8'3

0)(2

12

3

22

3

2

21

'313'

3133

'213'

2132

−=

−====== −−

−−

SSw

wSSSwSSS OSOSOS

OSOSOS



• At the end of the rod (length l) we calculate the total displacement for

OS 2 and for OS 3 to be:

OS 2:

OS 3: (152)

• Finally the maximum displacements,

with applied +M that twists the rod, are:

for the right hemisphere:

and likewise

for the left hemisphere:

ye

xe

ze

mlmls tot

= '13

'13

' tan

l

'tots

's's

'13

mls tot

− '13

'

mls tot

'13

'

'13

'13

mlmls tot

−−= '13

'13

' tan



• The existing situation is depicted here:

• The torsion angle of the rod

can be now expressed:

(153)

• With eq. (151) we can write and with eq. (153) we

get the dependence of the torsion angle on the SS:

(154)

• Thus through the measurement of the twisting angle - after respective

switching of the OSs - the SS be determined

and Hysteresis Loops can be recorded – e.g. for (NH4)2SO4.

)( '2Xn

)( '1Xm

'3

=

b

s tot'

'3

2arctan

'13

' ls tot

'13

'13

' 2Slls =

'3

'13

'3

'13

'13'

34

tan4

4arctan

l

bS

l

bS

b

Sl

'3

)(4

3

2

321

'13 SSwS −==



d: Further Considerations

• The considerations in the previous chapters follow the ‘thread’:

Force → Stress → Strain >>>> Elastic Coefficients (compliances sijkl)

• Only for the torsion of non-square rods the derivation scheme differs:

• Strain → Stress >>>> Elastic Coefficients (stiffnesses cijkl), but certain

limitations apply.

• In the theoretical modelling of ferroelastic PTs (see e.g. /14/) the spon-

taneous strains play a central role (partly even as order parameter), as

well as the elastic stiffnesses (being the inverse tensor components to

the elastic compliances)

• At ferroleastic PTs specific (combinations of) elastic stiffnesses show typical

dependences on the temperature (and sometimes also on pressure).

• Moreover, to describe a ferroelastic PT with a suitable free energy function,

the state parameters are strains (Sij) and polarizations (Pk). Therefore the

elastic stiffnesses of interest are those at constant polarization (see e.g. /14/)



• Also, if – experimentally - elastic compliances are measured, the need

arises to convert them into elastic stiffnesses, with the aim to compare with

the theoretical models

• This isn’t a simple undertaking at all. Exemplarily, here the conversion is

provided for cubic and hexagonal crystals (/2/,/15/)

Cubic:

(3 components)

Hexagonal:

(5 components)

( )( ) ( )( ) 44

44

12111211

1213

12111211

121111

1

sc

ssss

sc

ssss

ssc =

+−

−=

+−

+=

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )121166

44

44

2

13121133

13132

13121133

121133

12112

13121133

3312

12112

13121133

3311

2

11

22

1

22

11

22

1

cccs

c

ssss

sc

ssss

ssc

ssssss

sc

ssssss

sc

−==

−+−=

−+

+=

−−

−+=

−+

−+=



• Also it must be taken into account that elastic coefficients:

• depend on the (measuring) frequency, and there are

numerous measuring techniques available

• are complex numbers, which consists of real and imaginary

parts (focus was here entirely on the real part)

• are related to specific boundary/measuring conditions:

▪ isothermal measurement (in thermal equilibrium)

▪ adiabatic measurement (fast measurement)

▪ at constant electric polarization (superscript P) –

measured at an open circuit. (In the high-frequency

limit the elastic coefficients at constant polarization

are investigated (see e.g. /16/))

▪ at constant electric field (superscript E) – measured

at a short circuit (e.g. via an electric generator)

Comment: in non-piezoelectric crystals applies: and Eij

Pij

Eij

Pij ccss ==



• At the end a simple experimental setup is suggested to investigate the

findings/predictions as discussed beforehand.

• Taking into account all pros & cons of the methods, bending of bars seems

to be the most uncomplicated approach (incl. easy sample preparation)

• During the quasi-static bending, simultaneously the crystal could be observed

with the aid of a polarizing microscope, thus potentially watching the movement

of the domain walls within the process of domain switching, a.s.o.

• A respective device could look like this:

mirror

magnet

(neodymium)

bar

polarizer analyzer

magnet coil



6. Literature

/1/ Macke, W., “Mechanik der Teilchen, Systeme und Kontinua”, Akademische

Verlagsgesellschaft Geest & Portig K.-G., 1967

/2/ Sirotin, Yu. I., Shaskolskaya, M. P., “Fundamentals of Crystal Physics”,

Mir Publishers, Moscow, 1982

/3/ Wadhawan, V. K., Phase Transitions, 3, 3 (1982)

/4/ Wadhawan, V. K., Bull. Mater. Sci., 6, 733 (1984)

/5/ Wadhawan, V. K., Phase Transitions, 34, 3 (1991)

/6/ Aizu, K., J. of the Phys. Soc. of Japan, 28, 706 (1970)

/7/ Sapriel, J., Phys. Rev., B12, 5128 (1975)

/8/ Landau, L. D., Lifshitz, E. M., “Theory of Elasticity”, Pergamon Press Ltd., 1970

/9/ Pakulski, G., et al., Ferroelectrics, 48, 259 (1983)

/10/ Makita, Y., Sawada, A., Takagi, Y., J. of the Phys. Soc. of Japan, 41, 167 (1976)

/11/ Sawada, A., Makita, Y., Takagi, Y., J. of the Phys. Soc. of Japan, 41, 174 (1976)

/12/ Solans, X., et al., J. of Solid State Chemistry, 131, 350 (1997)

/13/ Gridnev, S. A., et al., J. Phys. D: Appl. Phys., 40, 856 (2007)



/14/ Hempel, H., “Structural Phase Transitions and Landau Theory”, 2018

/15/ Rabiei, M., et al., Materials 2020, 13, 4380

/16/ Burfoot, J. C., Taylor, G. W., “Polar Dielectrics and Their Applications”, The

Macmillan Press LTD, 1979

Elasticity of Anisotropic Bodies at Structural Phase Transitions

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