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```Stress, Strain, and Elasticity, 2006
Rick Aster
February 1, 2006
In seismology, a fundamental goal is to understand the vector equation of
motion
F = ma
(1)
for force, F, mass, m, and acceleration, a, and its many consequences in continuous elastic and nearly elastic media.
In a continuous medium of density ρ, we can write an equation of force
balance at point x and time, t, as
f (x, t) = ρ
∂ 2 u(x, t)
∂t2
(2)
or, in index form as
∂ 2 ui (x, t)
(3)
∂t2
where f is a force density (a force per unit volume) and the displacement of the
media from its equilibrium state in space and time is u(x, t).
Stress is the continuum analog of pressure. It has units of force per area, and
the total force on a real or imagined surface within a volume can be calculated
by integrating the stress over that surface. Stress can thus be thought of as a
surface force of contact force, in contrast with a volume force, such as gravity,
where a total force calculation would involve integrating over a volume.
The traction at some point x with a medium is the resolved force per unit
area on a surface defined by the normal n̂
fi (x, t) = ρ
δf
(4)
δS
where S is a small surface element defined by n̂. Note that τ and n̂ will not,
in general, be parallel; the traction will typically have a nonzero normal and a
nonzero tangential component.
A complete equation of motion with both surface and body force densities,
fb , is thus
Z Z
Z Z Z
∂2u
ftotal =
τ (n̂) dS +
fb (n̂) dV = m 2 .
(5)
∂t
S
V
τ (n̂) =
lim
δS→0
1
Figure 1: Traction Continuity
Now, consider a thin wafer of material. As its thickness decreases to zero,
the volume, and the volume integrals above, go to zero, but the surface area and
the surface integral remain finite. It is thus clear that the integrated surface
force must go to zero for a thin waver of any orientation (otherwise we would
have a finite force applied to a zero mass!). Thus, the traction on each side must
be equal and opposite, i.e.
τ (n̂) = −τ (−n̂)
(6)
everywhere within a connected medium. Traction is thus a continuous quantity
in such a medium; it cannot have discontinuities.
The stress tensor is simply a matrix of traction components


  (1)
(2)
(3)
τ1
τ1
τ1
σ11 σ12 σ13


σij =  σ21 σ22 σ23  =  τ2(1) τ2(2) τ2(3) 
(7)
(1)
(2)
(3)
σ31 σ32 σ33
τ3
τ3
τ3
(j)
where τi is the ith component of traction on the surface of an infinitesimal
cube defined by the outward normal in the j th direction.
2
Figure 2: Traction Component Decomposition
To see how the stress tensor can be easily used to find the traction on an
arbitrary surface, consider a right angle tetrahedron acted upon only by surface
forces, with three of its faces, dSi , lying in the three coordinate planes.
As the volume of the tetrahedron in Figure 2 goes to zero, so must the net
force (otherwise we would have a finite force operating on a zero mass and a
resulting infinite acceleration). Thus
lim τ (n̂)dS + τ (−n̂1 )dS1 + τ (−n̂2 )dS2 + τ (−n̂3 )dS3 = 0
or
∆V →0
(8)
−τ (−n̂1 )dS1 − τ (−n̂2 )dS2 − τ (−n̂3 )dS3
.
(9)
dS
The components of n̂ are just the projections onto the coordinate axes, or
equivalantly the correspondin area ratios of the sizes of the right angle tetrahedron
τ (n̂) =
n̂ = (n̂ · n̂1 , n̂ · n̂2 , n̂ · n̂3 ) = (dS1 /dS, dS2 /dS, dS3 /dS)
so that the traction on dS is
τ (n̂) = −τ (−n̂1 )n̂ · n̂1 − τ (−n̂2 )n̂ · n̂2 − τ (−n̂3 )n̂ · n̂3
= τ (n̂1 )n̂ · n̂1 + τ (n̂2 )n̂ · n̂2 + τ (n̂3 )n̂ · n̂3
= τ (n̂i )(n̂ · n̂i ) = σij n̂j .
3
(10)
Figure 3: Traction Torques
We thus see that traction for an arbitrary surface defined by the normal
n̂ is given by the matrix multiplication of σ and n̂; the stress tensor is the
mathematical machine that transforms one vector (the normal) into another
(the traction).
The diagonal terms of the stress tensor define the normal stresses in the coordinate system in which σ is expressed. The off-diagonal terms define the shear
stresses. Normal stresses act parallel to n̂ while shear stresses act perpendicular
to n̂. In a viscosity-free liquid, where the flow of the material acts to eliminate
shear tractions, the stress tensor is diagonal, and defines the pressure. Indeed,
the stress tensor in such a medium has the very simple form:
σij = −P δij
(11)
where P is the pressure and δ is the Kronecker delta. The minus sign arises
because of our outward normal convention; tractions that push inward are negative (be aware that this convention is sometimes reversed in engineering applications). Our convention is convenient, as we shall see, because positive stresses
produce positive strains.
Next, consider a small stressed rectangular prism (dx × dy in length/width)
of in the (x, y) plane.
The stresses on the faces in the (x, z) and (y, z) planes are approximately
given by the values of the stress tensor at O, σij , Because the cube is small, we
can approximate the stresses on the other faces at x = dx and y = dy using the
4
first-order approximation
σxi + δσxi = σxi +
∂σxi
· dx
∂x
(12)
σyi + δσyi = σyi +
∂σyi
· dy
∂y
(13)
and
The net torque, Tz (counterclockwise) on the prism in Figure 3 about ẑ at O
for a prism of thickness dz is a sum of terms consisting of contact forces (stress
times area), times their corresponding perpendicular moment arms (there are
only six terms instead of eight because σxy acting on the −x̂ face and σyx
operating on the −ŷ face both have associated lever arms of zero length and
hence contribute zero torque)
Tz = (σxy + δσxy ) · dy dz dx − (σyx + δσyx ) · dx dz dy − 1/2(σxx + δσxx ) · dy dz dy
(14)
+1/2(σxx ) · dy dz dy + 1/2(σyy + δσyy ) · dx dz dx − 1/2(σyy ) · dx dz dx
where the 1/2 factors arise because the normal forces operate on lever arms of
1/2 dy or 1/2 dx.
To keep the prism in equilibrium (and from undergoing an infinite angular
acceleration as its dimensions and volume, ∆V = dx · dy · dz, go to zero), the
net torque must go to zero faster than the volume and moment of inertia, so
that
0 = lim
∆V →0
Tz
= (σxy + δσxy ) − (σyx + δσyx ) − 1/2(σxx + δσxx ) · dy/dx (15)
∆V
+1/2(σxx ) · dy/dx + 1/2(σyy + δσyy ) · dx/dy − 1/2(σyy ) · dx/dy
As dx, dy, and dz all go to 0, each of the δσij terms go to zero due to (13). The
remaining σii terms cancel, giving
0 = lim
∆V →0
τ
= σxy − σyx .
∆V
(16)
We could have just as easily have performed this derivaton in the (x, z) or
(y, z) planes, which would have given the results
0 = σxz − σzx
(17)
0 = σyz − σzy .
(18)
and
Because of these symmetries, the stress tensor is a symmetric tensor, and
thus has only 6 independent elements.
As stress is the first tensor object that we will encounter in this course, this
is a good time to review tensor properties. A tensor is an object that specifies
some physical or abstract condition (e.g., stress, strain, conductivity), and can
5
be expressed in an arbitrary coordinate system. Tensors are just linear operators
converting one object (such as a vector) to another. Recall that a vector is an
object specifying a magnitude and direction. To represent this quantity in a new
coordinate system, we simply multiply by the appropriate matrix of direction
cosines
A = ê0i · êj
(19)
where the êj are the original coordinate axes and the ê0i are the new coordinate
axes. The coordinate transformation for an arbitrary vector is thus
v0 = Av
(20)
To see how to express a tensor such as stress in a new coordinate system, consider
traction and unit normal vectors in two coordinate systems
τ 0 = Aτ
(21)
n̂0 = An̂
(22)
As A is an orthogonal matrix, the inverse transformation is
τ = AT τ 0
(23)
n̂ = AT n̂0
(24)
In the primed coordinate system, we thus have
τ 0 = σ 0 n̂0 = Aτ = Aσn̂ = AσAT n̂0
(25)
Thus, the stress tensor in the new coordinate system, which is the tensor relating
n̂0 and τ 0 , is given by the transformation
σ 0 = AσAT
(26)
Because it is symmetric, the stress tensor is diagonalizable, meaning there is
always a coordinate system where the off-diagonal elements are all zero. In this
coordinate system the stress tensor can be written as
σ 0 = Λ = δij λi = VT σV
(27)
where the eigenvalues λi are normal principal stresses, and the transforming
matrix V consists of columns of eigenvectors corresponding to each eigenvalue
in the eigenvalue-eigenvector decomposition
σ = VΛVT .
(28)
Thus, the coordinate system where σ is diagonalized is defined by the columns
of V, and the tractions in each of these directions (called the principal stress
directions) are λi v̂i .
6
To examine how shear and normal tractions vary with coordinate system,
consider a biaxial stress


σxx
0 0
σyy 0  .
(29)
σ= 0
0
0 0
A clockwise rotation of the coordinate system about the ẑ axis by an angle θ
is characterized by the rotation matrix of direction cosines between coordinate
axes, ê0i and êj


cos θ − sin θ 0
Aij = ê0i · êj =  sin θ
cos θ 0  .
(30)
0
0
1
so that the stress tensor in the new coordinate system is

 
cos θ − sin θ 0
σxx
0
cos θ 0  ·  0
σ 0 = AσAT =  sin θ
σyy
0
0
1
0
0


cos θ
 − sin θ
0
sin θ
cos θ
0

0
0 =
1
σxx cos2 θ + σyy sin2 θ (σxx − σyy ) sin θ cos θ
 (σxx − σyy ) sin θ cos θ σyy cos2 θ + σxx sin2 θ
0
0
 0

0
0
σxx σxy σxz
0
0
0
 .
σyy
σyz
=  σyx
0
0
0
σzx
σzy
σzz

0
0 ·
0
(31)

0
0 
0
0
If σxx and σyy are of opposite sign, there will be two angles where σxx
=0
0
(no normal traction on the surface defined by the x̂ direction), when
σxx cos2 θ + σyy sin2 θ = 0
(32)
tan2 θ = −σxx /σyy ,
(33)
or
which has a real solution when σxx σyy ≤ 0 of
q
θ = ± arctan −σxx /σyy .
(34)
0
Similarly if σxx σyy ≤ 0, then there are two angles for which σyy
= 0 (no normal
0
traction on the surface defined by the x̂ direction), when σxx σyy ≤ 0
q
θ = ± arctan −σyy /σxx .
(35)
7
Figure 4: A Simple Stress Coordinate System Stress Rotation
We can thus rotate the stress tensor so that two of the four faces have only
shear tractions. For example, if σxx = −σyy , we can get all diagonal terms
of the stress tensor to be zero through a ±45◦ coordinate system rotation, as
illustrated below. In this case, the pre- and post-rotation stress tensors are just


−P 0 0
P 0 
(36)
σ= 0
0
0 0
and

0
σ 0 =  −P
0
−P
0
0

0
0  .
0
(37)
It is very important to realize that the physical state of stress has not changed,
only the numbers and coordinate system that we chose to express it in!
There is, as you would expect, an important link between stresses and faulting, as tractions provide the force that makes faults slip. In general, fault
movement will be more likely to occur near planes that resolve maximum shear
stress, although other factors such as fluid pressure or other processes acting to
8
reduce fault stress, as well as the long-term structural evolution and rotation
of fault zones can produce active fault systems that are oriented at appreciable
angles from these maximum shear-traction planes.
The mean stress is the average of the stress diagonal
M = (σxx + σyy + σzz )/3 = σii /3 .
(38)
M is invariant with respect to the coordinate system, as we can see by noting
that
0
= Aik σkl ATlj = Aik σkl Ajl = Aik Vkl Λlm Vjm Ajm
σij
(39)
where V is the matrix of eigenvectors that rotates σ into the diagonal matrix of
eigenvalues, λn , Λ. The double rotation effected by A and V can be expressed
as a single rotation by the orthogonal matrix
so that
Bil = Aik Vkl
(40)
0
σij
= Bil Λlm Bjm = Bil δln λn δmn Bjm .
(41)
The trace of the stress tensor in any coordinate system thus the sum of the
eigenvalues
X
0
σii
= λn Bin Bin =
λn .
(42)
n
The deviatoric stress tensor is
Dij = σij − M δij
(43)
Stresses are usually measured in the MKS units of Newtons per square meter,
or Pascals (Pa), but you may frequently see it expressed in bars, were 1 bar = 0.1
MPa. Within the Earth, where the mean stress at shallow depths is essentially
Z z
P = −g
ρ(z) dz ,
(44)
0
P is typically much greater than the off-diagonal deviatoric stress tensor elements. Nevertheless, the deviatoric stresses are frequently of the most interest
in seismology, as they drive faulting and deformation that is not purely volumetric. Deviatoric stresses can range up to 100’s of MPa (kbars) in the crust,
but appear to be much lower on most active faults. Lithostatic stresses, on the
other hand, reach up approximately 20 GPa in the outermost approximately
600 km of the Earth where earthquakes occur.
Note that any matrix, K, can be written as the sum of a symmetric part, S
and an antisymmetric part, A
Kij = 1/2(Kij + Kji ) + 1/2(Kij − Kji ) ≡ Sij + Aij
9
(45)
It is instructive to examine the geometric significance of antisymmetric and
symmetric matrix operators. An antisymmetric tensor can be closely related to
a vector. Consider an arbitrary antisymmetric tensor


0
a12 a13
Aij =  −a12
0
a23 
(46)
−a13 −a23 0
Left multiplication of an arbitrary vector x by A gives

 

a12 x2 + a13 x3
w2 x3 − w3 x2
vi = Aij xj =  −a12 x1 + a23 x3  =  w3 x1 − w1 x3  ≡ w × x
−a13 x1 + a32 x2
w1 x2 − w2 x1
where

−a23
w =  a13 
−a12
(47)

(48)
Thus, multiplication by an antisymmetric tensor is equivalent to a cross-product
multiplication by w, which introduces a 90◦ rotation and a scaling by |w| sin θ
(where θ is the angle between w and x).
A symmetric tensor operator, S, applied to an arbitrary vector, x has another
geometric interpretation which can be seen by looking at the dot product of the
input and output vectors of the operation, which is
C(x) = xi (Sij xj ) = s11 x21 +s22 x22 +s33 x23 +2(s12 x1 x2 +s13 x1 x3 +s23 x2 x3 ) . (49)
The derivatives of the scalar function C with respect to xi are
∂C
= 2sij xj
∂xi
(50)
Thus, the gradient of C, which is everywhere normal to the surface where C is
some constant (say c0 ), is proportional to the result of the tensor operation on
x. the C = c0 surface is thus described by the equation
c0 = xi Sij xj = s11 x21 + s22 x22 + s33 x23 + 2(s12 x1 x2 + s13 x1 x3 + s23 x2 x3 ) . (51)
If we select the principal coordinate system, where S is diagonalized, then
c0 = xi Sij xj = s11 x21 + s22 x22 + s33 x23
(52)
which is the equation for a quadric surface. If all of the eigenvalues of S are
positive or negative, and c0 has the same sign as the
p eigenvalues, then this is the
equation of an ellipsoid with semiaxes of length c1 /λi . A symmetric tensor
operation thus scales and rotates the set of all unit vectors to trace out a quadric
surface in 3-space.
10
Figure 5: Strain
We are now prepared to introduce the deformational counterpart of stress,
strain. Simply put, elastic strain (which we are most interested in seismology)
is a normalized measure the elastic (reversible) deformation of a medium from
equilibrium. If we have a deformation field, ui (x), the first order linearization
for this deformation near a point x is
ui (x + δx) = ui (x) +
∂ui(x)
δxj = ui (x) + δui
∂xj
(53)
The quantities of interest, the products of the δxj and the spatial derivatives of
the displacement field
∂ui (x)
δui =
δxj ,
(54)
∂xj
can be written as a sum of symmetric and an antisymmetric matrices
∂ui (x) ∂uj (x)
δui = 1/2
+
δxj
∂xj
∂xi
∂ui (x) ∂uj (x)
+1/2
−
δxj ≡ (ij + Wij )δxj .
∂xj
∂xi
(55)
First, let’s consider the antisymmetric component; from our previous discus-
11
Figure 6: Rotating a Vector
sion, it is clear that left multiplication of δx by W is equivalent to
 
∂u3 (x)
∂u2 (x)
−
∂x3 
 ∂x2
 ∂u1 (x) ∂u3 (x) 
1/2 
− ∂x1
 × δx ≡ w × δx
 ∂x3

∂u2 (x)
∂u1 (x)
∂x1 − ∂x2
(56)
For the first-order strain expansion (53) to be valid, kwk and kδxk must be
sufficiently small. To see what the cross product corresponds to in this case,
consider the arbitrary clockwise rotation of a vector r about some axis n̂ (Figure
6).
We have
r0 = ON + N V + V Q = n̂(n̂ · r) + (r − n̂(n̂ · r)) cos θ + (r × n̂) sin θ
(57)
= r cos θ + n̂(n̂ · r)(1 − cos θ) + (r × n̂) sin θ
If the rotation is small, then cos θ ≈ 1 and sin θ ≈ θ, so that
r0 ≈ (r × n̂) · θ + r
(58)
The change in the coordinate vector to the particle originally at r is thus
δr = r0 − r = (r × n̂) · θ = −(n̂ × r) · θ
(59)
We thus see that, if the terms in w are small, then left multiplication by w
corresponds to a counterclockwise rotation about the ŵ axis by an angle |w|.
As pure rotations of materials do not release or store elastic energy, we do not
12
Figure 7: Basic strains and rotations: a)
∂ux
∂y
=
∂uy
∂x
=> 0; e)
∂ux
∂y
=
∂u
− ∂xy
∂ux
∂x
> 0; b)
∂ux
∂x
< 0; c)
∂ux
∂y
> 0; d)
=> 0
usually worry about them in seismology (just as we do not generally concern
ourselves with rigid translations). We are principally interested in the symmetric
component, the strain tensor
∂ui (x) ∂uj (x)
ij = 1/2
+
.
(60)
∂xj
∂xi
First, note that the elements of are dimensionless. The diagonal elements
specify changes in lengths (contractions or extensions or longitudinal strains),
while the off-diagonal elements specify displacements perpendicular to the directions of interest, and are thus shear strains.
Figure 7 shows the basic types of strains. In the longitudinal strains, the
medium is either extended or compressed along one axis. It thus undergoes a
change in both volume and shape. Under simple shear, the shape changes (but
not, to first order, the volume), but this deformation requires an antisymmet13
ric (rotational) component of the strain tensor. If we remove the rotational
component , we are left with pure shear.
It is important and useful to differentiate between changes in volume and
changes in shape. The trace of the strain tensor is the dilatation
Θ = ii =
∂uz
∂ux ∂uy
+
+
=5·u
∂x
∂y
∂z
(61)
which is just the divergence of the displacement field. To see the physical
significance of Θ, consider the first-order change in volume that occurs when a
material of volume V = dx · dy · dz is strained to a volume V + ∆V
∂uy
∂uz
∂ux
V + ∆V = 1 +
1+
1+
dx dy dz
(62)
∂x
∂y
∂z
∂uz
∂ux ∂uy
≈ 1+
+
+
dx dy dz = V + ΘV .
∂x
∂y
∂z
We thus see that Θ ≈ ∆V /V is the volumetric strain.
The relationship between stress and strain, called the constitutive equation
is fundamental to seismology. In one dimension, this relationship for an elastic
system is given by Hooke’s law
fx = k · δx
(63)
where fx is the force applied to the system, k is the spring constant, and δx is
the resulting displacement from equilibrium. Hooke’s law is linear ; a doubling of
force produces a doubling of displacement, and so forth. If we look at a cartoonish relationship between displacement and force in the simple spring system, in
Figure 8 we see that there is a range of forces and corresponding strains for
which we expect linear behavior, but more complicated behavior outside of this
range. If the force gets too large, the metal in the spring will exceed its elastic
limit, and become permanently deformed. On the other hand, if the magnitude
of a negative force is large enough to push coils together, it suddenly becomes
much harder to deform it, as the spring constant of a hollow metal cylinder,
rather than a coil spring, will become appropriate. The slope of the curve is a
measure of the stiffness of the system, so there is a large slope increase when
this occurs. In between these extremes, we see that there is a linear relationship
between force and displacement. To understand the relationship between stress
and strain in elastic media, we must generalize Hooke’s law for continuous elastic media. In seismology, strains are commonly small (displacements of microns
and wavelengths of km), which makes much of seismology a linear science (this
is fortunate, especially considering the many other complications that we must
deal with!). The most general linear relationship between stress and strain is
σij = cijkl kl
(64)
The elements of c which relate stress and strain are called elastic moduli. As
strain is dimensionless, the moduli have units of stress (Pa). Although c has 34 =
14
Figure 8: Linear and Nonlinear Elasticity in the Spring System
15
81 elements, it is easy to see that the actual number must be considerably less,
as both the stress and strain tensors are symmetric. This, and thermodynamic
considerations that we won’t elaborate on here, reduce the maximum physically
possible number of independent elements in c to 21 (still a daunting number).
Fortunately, much of seismology can be done using the simplest possible c for
solid media, where there are only two independent moduli
cijkl = λδij δkl + µ(δik δjl + δil δjk )
(65)
where the moduli, λ and µ are called the Lamé parameters. Thus, for this
simplest case of an isotropic medium, where elastic properties vary in no way
with coordinate system, the constitutive equation is
σij = (λδij δkl + µ(δik δjl + δil δjk ))kl

(66)
= λkk δij + 2µij = λΘδij + 2µij
λΘ + 2µxx
2µyx
=
2µzx
2µxy
λΘ + 2µyy
2µzy

2µxz

2µyz
λΘ + 2µzz
λ has no common name (we’ll see its physical interpretation later). µ is an
important quantity in seismology called the rigidity, it is the ratio of pure shear
strain due to an applied shear stress component in an isotropic medium
µ=
σij
2ij
(67)
where i 6= j.
The bulk modulus or incompressibility describes the “volume stiffness” of
an isotropic medium. For a medium under isotropic stress characterized by a
pressure, P , we have
σij = −P δij = cijkl kl
(68)
= λΘδij + 2µij .
Evaluating the trace of the stress tensor for isotropic media by summing the
three equations for the diagonal stress tensor terms gives
−3P = 3λΘ + 2µΘ = (3λ + 2µ)Θ .
(69)
The bulk modulus is defined as the ratio of pressure to volume change
κ=−
P
2
=λ+ µ .
Θ
3
(70)
In an ideal (zero viscosity) fluid, mu = 0, thus, the Lamé parameter λ is the
incompressibility of an ideal fluid.
Consider a stretching experiment where tension is applied to an isotropic
medium along a principal axis (say x̂).
16
Figure 9: A uniaxial stress test
The stress state is

σxx
σ= 0
0
0
0
0

0
0  .
0
(71)
Using the isotropic cijkl (66) gives the stress-strain relationships
σxx = λΘ + 2µxx
(72)
σyy = σzz = 0 = λΘ + 2µyy = λΘ + 2µzz
(73)
σxy = σyx = σxz = σzx = σyz = σzy = 0
(74)
= 2µxy = 2µyx = 2µxz = 2µzx = 2µyz = 2µzy .
These expressions imply
yy = zz
(75)
(as would be expected from symmetry), so that
λ(xx + 2yy ) + 2µyy = 0
(76)
or
yy = −
λ
xx ≡ −νxx
2(λ + µ)
(77)
where ν, the negative of the dimensionless ratio of radial to longitudinal strain
is Poisson’s Ratio.
The ratio of longitudinal stress to strain in the x̂ direction in the stretching
experiment is found by substituting −νxx for yy and zz in (72)
σxx = λ(xx − 2νxx ) + 2µxx
17
(78)
to obtain the expression for Young’s Modulus
µ(3λ + 2µ)
σxx
=
≡E .
xx
λ+µ
(79)
Note that all of these moduli (µ, λ, ν, κ, E) are positive and have units of
stress (except for the dimensionless ν). For many Earth materials λ ≈ µ. If we
take λ = µ, we have what is called a Poisson solid, where ν = 0.25, E = 5µ/2,
and κ = 5µ/3.
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