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Defining sign of stress tensor. Kittel’s Fig. 15 may be confusing about sign of T xx (which he calls X x ). - PowerPoint PPT Presentation
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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=