Defining sign of stress tensorKittel’s Fig. 15 may be confusing about sign of Txx (which he calls Xx )

Stress tensor component Txx is defined as the x-component of force, transmitted in –x direction. The compressive force shown (labeled Xx) is in the –x direction (note that the x axis points left!) and is transmitted in the –x direction, so the sign of Txx is negative. Tension would be positive.

Elasticity (Chapter 3)

– only 6 of 9 elements are independent.Replace 3x3 matrix by a 6-component column vector:

Lecture 21Oct. 8, 2010PH 481/581

zzzyzx

yzyyyx

xzxyxx

TTT

TTT

TTT

Recall Young’s modulus Y:

L

LY

A

F

Stress = elasticity constant x strain

In general, stress and strain are 3x3 matrices, so the most general relation would be

,

MT

Where M has 4 indices (3x3x3x3 =81 elements)

Fortunately, not all elements of T and are independent: these matrices are symmetric:

←L+L →

← L →

FF

Elasticity (continued)

T is symmetric matrix – only 6 of 9 elements are independent.Replace 3x3 matrix By 6-element

column vector:

zzzyzx

yzyyyx

xzxyxx

TTT

TTT

TTT

6

5

4

3

2

1

T

T

T

T

T

T

T

T

T

T

T

T

xy

zx

yz

zz

yy

xx

z

y

x

z

y

x

no

no

no

has

has

has

How about strain tensor? Again replace the 3x3 symmetric matrix

By a 6-element column vector:

zzzyzx

yzyyyx

xzxyxx

sss

sss

sss1

2

3

4

5

6

2

2

2

xx xx

yy yy

zz zz

yz yz zy

zx zx xz

xy xy yx

se

se

se

se

se

se

Note factor of 2, only in off-diagonal elements. (Later, makes energy formula simpler.)

We will use and to represent the composite indices 1..6, and and to represent x,y,z.

Elasticity (continued)

6

5

4

3

2

1

66

24232221

161514131211

6

5

4

3

2

1

...

...

T

T

T

T

T

T

S

SSSS

SSSSSS

e

e

e

e

e

e

Now we can write the most general linear relationship between stress and strain as a matrix equation e = S T:

xy

zx

yz

zz

yy

xx

xyxyxxxy

zzzzyyzzxxzz

zzyyyyyyxxyy

xyxxzzxxyyxxxxxx

xy

zx

yz

zzx

yy

xx

T

T

T

T

T

T

SS

SSS

SSS

SSSS

e

e

e

e

e

e

))(())((

))(())(())((

))(())(())((

))(())(())(())((

)(

)(

)(

)(

)(

)(

...

....

...

Recalling that the 1..6 indices are short for (xx),...(xy), this is

Elasticity (continued)

11 12 12

12 11 12

12 12 11

44

44

44

0 0 0

0 0 0

0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

S S S

S S S

S S SS

S

S

S

In the case of cubic symmetry, the axes are equivalent, so that S(xx)

(xx) = S(yy)(yy) , i.e. S11 = S22. Similarly, S(xy)(xy) = S(yz)(yz) , so S66 = S44 . Also, S13 = S12 -- all the off-diagonal elements in the upper left 3x3 matrix are equal. Also, anything like S(xx)(xy) that has a single y index must vanish due to the y↔ -y symmetry. This leads to

We call S (in e = S T) the compliance matrix. We denote the inverse matrix (“stiffness matrix”) by C = S-1, so that T = C e.In a system with no symmetry, all 36 components of C and S are independent.

Cubic symmetry only

The stiffness matrix C has the same form -- a cubic material has 3 independent stiffness coefficients C11, C12, and C44.

Elasticity (continued)

Important special case: uniaxial stress, Txx >0, other T = 0. Then

This experiment defines Young’s modulus Y = (F/A)/(L/L) = Txx / xx = 1/S11, as well as Poisson’s ratio P = -yy /xx = -S12 /S11. Note that yy < 0 when stress is applied along x only, so S12 < 0.

1 11 12 13 14 15 16 11 11

2 21 22 23 24 21 12

3 31 12

4 41

5 51

6 66 61

... 0

... 0

0 0

0 0

0 0

xxxx xx

xx xx

xx xx

xx

xx

xx

e S S S S S S S TT S T

e S S S S S T S T

e S T S T

e S T

e S T

e S S T

0

0

0

xx

yy

zz

if cubic symm

etry

←L+L →

← L →

FF

Elastic Energy: energy/volume U is

1 12 2

,

1 11 1 6 62 2

a a xx xx xy xy yx yxU T T T T

T e T e T e

1 2(using e , 2 )xx xy yx xye s

12

,

C e e

(C can be chosen to be symmetric)

using T C e

Elastic Waves (Chapter 3)Lecture 23Oct. 13, 2010PH 481/581 Write equation of motion for a

volume element from position r, now at displaced position r + R(t):

r

RrR

that so or εrR

becomes

F

F

But displacement R is related to the strain tensor, which is related to the stress tensor T. Expressing everything in terms of R,

aF m 2

2

t

RT

r

Origin

r

R

r

RCCT

)()(

2

2

)( t

R

r

RC

r

So F=ma becomes

Like a wave equation.

Elastic Waves (continued)Derived a wave equation for displacement R:

rr

RC

t

R2

)(2

2

Solutions: try a plane wave )(0),( tiet rkRrR

2

2

66

2

12

2

11

2

2

)(

2

)(2

2

)(2

2

y

x

yx

y

xx

x

y

xxyxy

yx

yyyxx

x

xxxxx

x

r

RC

rr

RC

rr

RC

r

RC

rr

RC

r

RC

t

R

Writing part of the =x component explicitly,

This is Eq. (57a) in Kittel.

xyyyxxxx RkCRkkCRkCR 02

6601202

1102

Try k = (K, 0, 0) & R along x: xx RKCR 02

1102

So and the wave speed is 2

112 KC 2/1

11 )/(/ CK

so

Elastic Waves (Chapter 3)Lecture 24Oct. 13, 2010PH 481/581

Review: becomes

in terms of displacement R – wave equation.

aF m

rr

RC

t

R2

)(2

2

Insert a plane wave )(

0),( tiet rkRrRTry k along x, i.e. (100) & R along 100:v2 = C11

& k along (100) & R along 010: v2 = C44

& k along (110) & R along 110: v2 = ½(C11 + C12 + 2C44) & k along (110) & R along 001: v2 = C44

& k along (110) & R along (1-10):v2 = ½(C11 - C12)Condition for isotropy: C11 - C12 = 2C44

Kittel gives table – W is close to isotropicLamé constants: T = 2 + tr 1; derive C11=2C12= C44=