Phys463.nb
11
4
Elasticity and Stiffness
Ref. Theory of elasticity by Landau and Lifshitz.
Q1: Soft or hard, how should we describe it?
Q2: Why do we use cork oak to make bottle stoppers?
Q3: Why the sound velocity of the LA mode is always larger than the TA mode?
4.1. Hooke’s law
For a spring
F = C DL
(4.1)
The elastic energy:
1
E=
C DL2
(4.2)
2
Q: How to generalize this Hooke’s law to a 3D solid?
4.2. Strain tensor
4.2.1. l.h.s. of the Hooke’s law
Let’s take a closer look at the r.h.s. of Eq. (4.1). DL is the change of length of the spring. In other words, it is the change distance between point A
and B, where A and B are the two ending points of the spring.
Q: For a 3D solid, how should we describe the change of distance between to points A and B?
A: the strain tensor
For a solid, we label every points in a crystal using its coordinates r = Hx, y, zL. If we deform the solid, point r will be moved to a new position
®
®
®
and we call the new position r '. The deformation of this point is
®
®
®
(4.3)
u = r' - r
Notice that for each point r , we will have a deformation vector u . So u is a function of r , u Jr N.
®
® `
® `
® `
u Jr N = ux Jr N x + u y Jr N y + uz Jr N z
® ®
The strain tensor is defined as:
®
®
® ® ®
(4.4)
Phys463.nb
12
exx exy exz
eyx eyy eyz
(4.5)
exz eyz ezz
where
exx =
¶ uy
¶ ux
eyy =
¶x
exy = eyx =
1 ¶ ux
2
ezz =
¶y
¶ uz
(4.6)
¶z
¶ uy
eyz = ezy =
+
¶y
¶x
This tensor is a symmetric tensor Heab = eba L.
1 ¶ uz
2
¶ uy
ezx = exz =
+
¶y
¶z
1 ¶ uz
2
+
¶x
¶ ux
(4.7)
¶z
There are six independent component for a 3×3 symmetric tensor.
In the text book, the off-diagonal terms are defined with out the fact 1/2. (In that case, one can NOT write these es as a tensor).
Remark: a more precise definition of the strain tensor is
exx =
¶ ux
+
¶x
1 ¶ ul ¶ ul
2 ¶x ¶x
¶ uy
eyy =
+
¶y
1 ¶ ul ¶ ul
ezz =
2 ¶y ¶y
¶ uz
¶z
+
1 ¶ ul ¶ ul
(4.8)
2 ¶z ¶z
exy =
eyx =
1 ¶ ux
2
¶ uy
+
¶y
+
¶x
¶ ul ¶ ul
¶x ¶ y
eyz = ezy =
1 ¶ uz
2
¶y
¶ uy
+
+
¶z
¶ ul ¶ ul
¶ y ¶z
ezx = exz =
1 ¶ uz
2
¶x
+
¶ ux
¶z
+
¶ ul ¶ ul
(4.9)
¶x ¶z
We typically assumes that the second order derivatives are much smaller than the first order derivatives, so that we can ignore them. As a result, it
recovers the definition used above.
4.2.2. Review on vectors and tensors
Q: What is a vector?
A: two criteria.
Ÿ Write three components in a row
Ÿ Rotational same as the coordinates. R v
Q: What is a tensor?
A: two criteria.
Ÿ Write nine components as a three by three matrix
Ÿ Rotation: R M RT
Q: What are symmetric and anti-symmetric tensors? Why should we care about them?
A: symmetric tensor: Mij = M ji. Anti-symmetric tensor: Mij = -M ji. For any tensor Mij, we can write it as the sum of a symmetric tensor and an
any symmetric tenor.
M = MS + M A
(4.10)
Here M is an arbitrary tensor, M S is a symmetric tensor and M A is an anti-symmetric tensor. Here,
MS =
M + MT
(4.11)
2
MA =
M - MT
(4.12)
2
Where M T is the transpose matrix of the matrix M.
Q: If I have a vector field u Jr N, can I get a tensor from it?
® ®
Phys463.nb
13
A: Yes. The first order derivatives ¶i u j gives you a tensor.
¶x ux ¶x u y ¶x uz
U = ¶ y ux ¶ y u y ¶ y uz
¶z ux ¶z u y ¶z uz
(4.13)
This tensor can be written as the sum of a symmetric tensor and an anti-symmetric tensor
¶x ux
¶ x u y +¶ y u x
¶ x u z +¶ z u x
2
2
¶ x u y +¶ y u x
U=
¶ y u z +¶ z u y
¶y uy
2
0
+ -
2
¶ x u z +¶ z u x
¶ y u z +¶ z u y
2
2
¶z uz
-
¶ x u y -¶ y u x
2
¶ x u y -¶ y u x
0
2
¶ x u z -¶ z u x
2
-
¶ y u z -¶ z u y
2
¶ x u z -¶ z u x
2
¶ y u z -¶ z u y
2
(4.14)
0
The symmetric part is the strain tensor.
4.2.3. Physical meaning of the strain tensor
The strain tensor measures the change of distance between any two points A and B in a solid.
®
`
`
`
Let’s consider two nearby points A and B. Before we deform the solid, the distance between A and B is â r = â x x + â y y + â z z.
®
After deformation, the distance between A and B changes into â r ' . One can show that
â r '2 = â r2 + 2 eij â ri â r j + OIâ r3 M
(4.15)
4.2.4. Elastic energy
For a spring, the elastic energy only depends on the change of length (distance)
1
U=
C DL2
(4.16)
2
For a solid, the elastic energy depends on the change of length between any two points too, which is described by the strain tensor
U =à âr â
ijlk
®
1
2
Cij;kl eijJr N elk Jr N
®
®
(4.17)
Cij;kl is a rank-4 tensor, which is known as the elastic modulus tensor. Each of the component is known as an elastic stiffness Constant (or simply
an elastic Constant). They measures how “hard”this solid is. A large elastic stiffness constant means that it cost more energy to deform this solid.
In other words, the solid is “hard”. Smaller elastic Constant means that the solid is “soft”.
For a rank 4 tensor, there are four indices, i, j, k and l, each of which takes three possible values x, y and z. So we have There are
3 ´ 3 ´ 3 ´ 3 = 81 different Cij;lk (81 elastic constants). But in reality, we just need to know 21 of them, and the value of the other 60 can be
determined from these 21 elastic constants.
This is because eij = e ji, we can prove that Cij;kl = C ji;kl = Cij;lk . In addition one can prove that Cij;kl = Ckl;ij, so we have
Cij;kl = C ji;kl = Cij;lk = Ckl;ij
(4.18)
For a spring, there is only 1 elastic constant (the spring constant). For a solid, there are 21 elastic constants.
Remark: if we choose proper axes (choosing the direction of the x, y and z axes), we can make three of the 21 elastic constants 0. So, in reality,
we have only 18 independent elastic constants.
4.2.5. Another way to write down the elastic energy
Define eΛ with Λ=1,2,… 6. Here e1 = exx , e2 = eyy , e3 = ezz , e4 = exy = eyx , e5 = eyz = ezy , e6 = exz = ezx
U =à âr â
âΛ=1
Λ=1
®
6
6
1
2
CΛΜ eΛJr N e Μ Jr N
®
®
Here, one can prove that CΛΜ = C ΜΛ . In other words, CΛΜ is a 6×6 symmetric matrix.
For a 6×6 symmetric matrix, there are 21 components. So we have 21 elastic constants (same number as using Cij;kl ).
(4.19)
14
Phys463.nb
4.2.6. Elastic constants in crystals
We know crystals have their rotational symmetry (point group). These rotational symmetry enforce extra constrain on the elastic constants, which
will further reduce the number of independent elastic constants.
Triclinic: 18
Monoclinic: 12
Rhomic: 9
Tetragonal: 6
Hexagonal: 5
Cubic: 3
Isotropic: 2
Example: Cubic system
® 1
U = à â r B Cxxxx Iexx 2 + eyy 2 + ezz 2 M + Cxxyy Iexx eyy + exx ezz + eyy ezz M + 2 Cxyxy Iexy 2 + exz 2 + eyz 2 MF
2
(4.20)
4.2.7. Elastic constants for an isotropic solid
For an isotropic system, Cxxxx = Cxxyy + 2 Cxyxy
U = à â rB
®
= à â rB
®
= à â rB
®
= à â rB
®
Cxxyy + 2 Cxyxy
2
Cxxyy
2
Iexx 2 + eyy 2 + ezz 2 M + Cxxyy Iexx eyy + exx ezz + eyy ezz M + 2 Cxyxy Iexy 2 + exz 2 + eyz 2 MF
Iexx 2 + eyy 2 + ezz 2 + 2 exx eyy + 2 exx ezz + 2 eyy ezz M + Cxyxy Iexx 2 + eyy 2 + ezz 2 + 2 exy 2 + 2 exz 2 + 2 eyz 2 MF
Iexx + eyy + ezz + 2 exx eyy + 2 exx ezz + 2 eyy ezz M + GIexx + eyy + ezz + 2 exy + 2 exz + 2 eyz MF
Λ
2
2
Λ
2
2
2
I exx + eyy + ezz M + GIeij eijMF = à â r B
®
2
2
Here we use the Einstein notation eij eij = Úi Új eij eij
Λ
2
2
2
2
2
(4.21)
2
H eiiL2 + GIeij eijMF
We can rewrite this elastic energy as
U = à â rB
®
where B = Λ +
1
2
2
3
B eii2 + G eij -
1
3
∆ij el l F
2
G and B is called the bulk modulus and G is called the shear modulus. Here we only have 2 elastic constants B and G.
Q: Why we use this complicated form, instead of just using Λ and G?
A: B and G have important physical meanings.
(4.22)
Phys463.nb
15
The first term measures the energy cost for changing the volume of a solid. The second term measure the energy cost for share deformation
(without changing the volume).
Fig. 1. Bulk modulus B = -V
âP
(if we want to change the volume by âV, how strong the pressure one need to apply). Figure from wikipedia.
âV
Fig. 2. Shear modulus G = -
Fl
A Dx
. Figure from wikipedia.
For typical solids, B~1-400 GPa. For typical liquids, B~1 GPa. For gases, B~0.1MPa
For many solids, B ~ G. For rubbers, G<<B.
Rubber is “soft”. The B of rubber is comparable with other solids (~1GPa), but the G of rubber is much smaller H ~ 10-3 GPaL. For rubber, it is
hard to change the volume, but if we fix its volume, it is very easy to deform without changing its volume.
4.2.8. Poisson’s ratio
Poisson’s ratio (Ν) measures: when we squeeze (stretch) an object along one direction, how much it expands (shrink) in the perpendicular
directions.
Ν » DL '  DL
(4.23)
The Poisson’s ratio can be calculated using B and G as:
3B-2G
Ν=
2 H3 B + GL
(4.24)
16
Phys463.nb
Typically solids have B ~ G or B >> G, so that Ν>0 (positive Poisson’s ratio). But there are some materials has B << G, so that Ν<0 (negative
Poisson’s ratio).
Materials with negative Poisson’s ratio are known as “auxetic materials”or “auxetics”. They are very useful.
Ÿ Wine cork.
Ÿ Bullet proof materials.
4.2.9. Equations of motion and sound waves
Three types of sound waves: 1 longitudinal mode and 2 transverse modes. Dispersion relation:
Ω = v L k Hlongitudinal modesL
(4.25)
Ω = vT k Htransverse modesL
(4.26)
The velocity of the longitudinal sound
B+
vL =
4
3
G
(4.27)
Ρ
The velocity of the transverse sound
G
vT =
(4.28)
Ρ
Notice that v L > vT . Longitudinal sound always have a larger velocity.
(The rest part of this section is note required)
U = à â rB
+ G Iexx 2 + eyy 2 + ezz 2 M + ΛIexx eyy + exx ezz + eyy ezz M + 2 GIexy 2 + exz 2 + eyz 2 MF
Λ
®
2
Remember that
exx =
¶ uy
¶ ux
eyy =
¶x
1 ¶ ux
exy = eyx =
2
ezz =
¶y
¶ uz
(4.30)
¶z
¶ uy
¶y
¶x
+G B
¶ ux
1 ¶ uz
eyz = ezy =
+
(4.29)
2
¶ uy
ezx = exz =
+
¶y
¶z
1 ¶ uz
2
+
¶x
¶ ux
(4.31)
¶z
So
U = à â rB
®
Λ
Λ
2
¶ ux ¶ u y
= à â rB
B
+
2
G
2
Kinetic energy
K=à âr
2G
®
1
2
3
B
¶ ux
¶ uy
+
+
¶y
¶ ux ¶ uz
+
¶x ¶z
B
¶ ux
2
2
¶ uy
¶ uz
¶ uy
¶z
¶ u y ¶ uz
2
+
+
2
+
¶ uz
2
2
2
¶ uy
+
Ρ AH¶t ux L2 + I¶t u y M + H¶t uz L2 E
where Ρ is the mass density.
G
¶z
¶y
2
F+
+
¶ uz
¶y
¶x
2
¶ y ¶z
¶x
+
¶y
2
+
¶x
¶x ¶ y
®
2
B
+
¶z
+
¶y
F+ B2
¶ uy
¶ ux
¶ uz
+
¶x
2
G
3
¶ uz
¶x
2
¶y
¶ ux ¶ u y
+
¶ ux
¶z
2
+2
2
+
¶ uz
¶z
+
¶x ¶ y
2
¶ uy
+
¶ ux ¶ uz
¶x
+
¶x ¶z
¶ ux ¶ u y
¶ y ¶x
+
¶ ux
2
¶z
¶ u y ¶ uz
F
(4.32)
+
¶ y ¶z
+2
¶ uz ¶ u y
¶ y ¶z
+
¶ uz ¶ ux
¶x ¶z
F
(4.33)
Phys463.nb
17
Lagrangian is
L=K-U
(4.34)
L is depends on ux , u y , uz , ¶t ux , ¶t u y and ¶t uz .
We can find the equation of motion for ux ,u y and uz as
∆L
¶t
∆L
=
∆ ¶t ux
∆L
¶t
∆L
=
∆ ¶t u y
(4.36)
∆ uy
∆L
¶t
(4.35)
∆ ux
∆L
=
∆ ¶t uz
(4.37)
∆ uz
2 Ρ ¶t 2 u x = -
∆U
4G
∆ ux
2
¶x 2 ux + B -
= B+
3
3
G I¶x ¶ y u y + ¶x ¶z uz M +
GA¶ y ux + ¶z ux + ¶x ¶ y u y + ¶x ¶z uz E = B B +
2
2
4G
3
Similarly, we have
2 Ρ ¶t 2 u y = -
2 Ρ ¶t 2 u z = -
∆U
¶ y ¶x ux + B B +
G
= B+
∆ uy
3
∆U
G
∆ uz
¶z ¶x ux + B +
3
3
3
Assuming that
2
2
¶ y 2 u y + GI¶x 2 u y + ¶z 2 u y MF + B +
4G
G
= B+
¶x ux + GI¶ y ux + ¶z ux MF + B +
2
¶z ¶ y u y + B B +
4G
3
G
3
(4.38)
G
¶x ¶ y u y + B +
3
¶x ¶z uz
G
¶ y ¶z uz
3
(4.39)
¶z 2 uz + GI¶x 2 uz + ¶ y 2 uz MF
(4.40)
ux = Ax expIä kx x + ä k y y + ä kz z - ä Ω tM
(4.41)
u y = A y expIä kx x + ä k y y + ä kz z - ä Ω tM
(4.42)
uz = Az expIä kx x + ä k y y + ä kz z - ä Ω tM
(4.43)
(plane wave solutions)
The three equations above can be written in terms of a matrix form
Ρ Ω2
Ax
1
Ay =
2
Az
IB +
4G
3
M kx 2 + GIk y 2 + kz 2 M
IB +
G
IB +
3
G
3
M kx k y
IB +
M kx kz
IB +
4G
3
G
3
M kx k y
M k y 2 + GHkx 2 + kz 2 L
IB +
G
3
M k y kz
IB +
IB +
IB +
4G
3
G
3
G
3
M kx kz
M k y kz
Ax
Ay
M kz + GIkx + k y M
2
2
2
This is an eigen value problem for the 3×3 matrix in the equation above. Eigenvalues of the matrix is G k 2 , G k 2 and IB +
So, we have
G
Ω1 =
B+
G
k
Ρ
and Ω2 =
k
Ρ
and Ω3 =
4
3
(4.44)
Az
4G
3
M k2
G
(4.45)
k
Ρ
The amplitude of these three modes are
A1 = H-kz , 0, kx L ,
®
and A2 = I-k y , kx , 0M
®
®
A3 = Ikx , k y , kz M ,
®
and
® ® ®
(4.46)
®
So the first two modes are transverse modes with A perpendicular to k (A × k = 0), while the third mode is a longitudinal mode with A parallel to
®
®
®
k (A ´ k = 0).
18
Phys463.nb
FullSimplifyBEigensystemB
IB +
4G
3
M kx2 + G Iky2 + kz2 M
IB +
IB +
G
3
G
3
M kx ky
M kx kz
IB +
4G
3
IB +
kz
kx
, 0, 1>, :-
ky
kx
, 1, 0>, :
kx
IB +
ky
,
kz
3
M kx ky
M ky + G Ikx + kz M
::G Ikx2 + ky2 + kz2 M, G Ikx2 + ky2 + kz2 M,
::-
G
2
G
3
1
3
2
M ky kz
2
IB +
4G
3
IB +
G
3
G
3
M kx kz
M ky kz
M kz2 + G Ikx2 + ky2 M
H3 B + 4 GL Ikx2 + ky2 + kz2 M>,
, 1>>>
kz
IB +
FF