# 4.StressStrain

```Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Defining Stress in a Solid
Imagine applying
external forces to a solid.
Make an imaginary cut inside,
and apply a set of forces
carefully chosen so that the
rest of the solid does not feel
the cut.
These forces will depend on
the position and orientation of
the surface.
These forces that we must apply to the cuts are
exactly the forces that act within the original
solid.
In general these forces depend on:
• the position within the solid
• the orientation of the surface
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Stress and Strain Analysis
Why are surfaces important?
• When we push on something, we apply forces
to a surface.
• When we break something, we create surfaces.
Things break along surfaces.
• When we repair something, we reattach or glue
surfaces.
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Stress and Strain Analysis
Look more closely at the forces acting on a surface
tangential
component
normal
component
ν
Force F
unit normal
area dA
F
ν
The orientation of the surface is
specified by the outward normal
vector ν
The normalized force per unit area is called the
Traction Vector or Stress Vector.
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Stress and Strain Analysis
Tractions and Stresses
We define the traction vector as the force per unit
area acting on a surface.
We define the stress as the set of components of
tractions acting on the various surfaces. In cartesian
coordinates:
σ zz
σ zy
σ zx
σ yz
σ xz
σ xx
z
σ yy
y
σ yx
σ xy
x
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Units of Stress
1 bar = 106 dyne/cm2 = 14.50 psi
10 bar = 1 MPa = 106 N/m2
1 psi/ft = 144 lb/ft3
= 19.24 lb/gal
= 22.5 kPa/m
1 lb/gal = 0.052 psi/ft
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Traction on a Surface of
Arbitrary Orientation
z
T
ν = (βx, βy, βz)
y
x
Cauchy’s Formula:
Tx
σ xx σ xy σ xz
Ty = σ yx σ yy σ yz
σ zx σ zy σ zz
Tz
≈
T = σν
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νx
νy
νz
Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
The Stress Tensor is Symmetric
We must also balance the torques applied to
the cube and we find that:
σxy = σyx
σxz = σzx
σyz = σzy
σ xx σ xy σ xz
σ yx σ yy σ yz
σzx σzy σzz
Stress tensor is symmetric.
There are only 6 independent terms.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Einstein Summation Convention
It is customary in stress and strain analysis to use
a shorthand notation, sometimes called the Einstein
summation convention. Any repeated index within
a term implies summation over that index with a
range of 1 to 3.
For example:
c ijkl σ kl
3 3
c ijkl σkl
Σ
Σ
k=1l=1
and
are equivalent. Here, k is repeated so it implies a
summation; l is also repeated, so it also implies a
summation. i and j are not repeated so there are
no summations.
There is no summation in this expression:
σ ij + σ jk
because there is no repeated summation in any of
the terms. The two occurrances of the index j are
in two different terms.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Transformation of Coordinates
We found it convenient to define the elements of the stress
tensor in a given coordinate system. We did so by labeling
the various components of forces that act on the faces of
orthogonal planes in that coordinate system.
We now want to be able to write the stress tensor in terms of
another coordinate system. That is, for the same physical
configuration of forces, how do we simply label or describe
these in terms of another set of axes? Or, how do we find the
components of force acting on the orthogonal planes in the
new coordinate system?
We do it in two steps:
1. First find the coordinate transformation that relates the
two sets of axes.
2. Then use it to re-express the stress tensor in the new
coordinate system.
z
r
z'
x'
y
x
y'
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Coordinate Transformations
A real physical vector can be expressed in any two
coordinate systems:
r = x 1e 1 + x 2e 2 + x 3e 3
r = x ′1e ′1 + x ′2e ′2 + x ′3e ′3
It is important to remember that we are simply giving
the same physical vector different names in different
coordinate systems.
x ′1
β 11 β 12 β 13
x ′2 = β 21 β 22 β 23
β 31 β 32 β 33
x ′3
Or in index notation:
3
x ′i =
j=1
Σ
x1
x2
x3
β ijx j = β ijx j
where β ij is the direction cosine of the new ith axis
relative to the old jth axis.
So for any vector we can write:
′
≈
r = βr
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
So now go back to Cauchy’s formula
Told = σoldν old
Now transform T and ν to the new coordinates
β ij TTnew = σold β ij Tν new
Rewrite as:
Tnew = β ij σ old β ij T ν new
Rename term in brackets:
Tnew = σ new ν new
β ij σold β ij T = σnew
Formula for coordinate transformation
σij = β ikβ jlσ kl
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
The hydrostatic stress or mean stress is the
average of the three normal stresses.
σxx + σ yy + σ zz
σ αα
=
σ0 =
3
3
This is an invariant. The mean stress is the
same in any coordinate system.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Principal Stresses
Stress is a real physical state or quantity: we have
forces, surfaces, and stress whether or not we choose a
coordinate system. Now, if we do define a coordinate
system, then we assign numerical values to the
components σ ij in the matrix:
σ ij
In general, different choices of coordinate systems
cause the numerical values in the matrix (stress tensor)
to vary.
Now recall Cauchy’s formula:
T = σ ij ν
Can we find directions such that the traction vector is
normal to the surface?
T = σ ij ν = σν
where T is the magnitude of the traction vector, ν is
the direction, and σ is a scalar.
This is a standard problem of finding the eigenvalues
and eigenvectors of a matrix. We call the three
eigenvalues the principal stresses, and the three eigen
vectors the principal directions.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Principal Stresses
Let’s emphasize the physical interpretation:
The traction T is parallel to ν , which is always
perpendicular to the surface. If T is normal to the
surface then there is no component of T parallel to
the surface. Therefore, there is no shear stress.
The simplest and most important thing to remember
about the principal axes is that there are no shear
stresses on surfaces perpendicular to the three
principal axes.
If we find these directions and make them axes then
we know how to transform the stress tensor to these
axes and we will find:
σ1 0 0
0 σ2 0
0 0 σ3
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Infinitesimal strain
P'
P
Q'
Q
We can describe the deformation of a continuum with
a vector displacement field U . This displacement
gives the vector motion of each point in the body.
It is important to note that the displacement can vary
with position.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Rigid Body Translation
The vector displacement field is a constant.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Rigid Body Rotation
P'
P
Q'
Q
u=ω&times;r
ω = rotation vector
• direction is along axis of rotation
• ω is the angle of rotation
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Strain
If a body undergoes a displacement that cannot be
simulated point-by-point as a rigid body translation or
rotation, then we say it is strained.
With strain we have deformation or distortion. The
size or shape is changed.
U(x,y,z)
U(x + dx, y + dy, z + dz)
(x,y,z)
(x + dx, y + dy, z + dz)
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Ux
Uy
Uz
∂U x
∂x
∂U y
∂x
∂Uz
∂x
Ux
= Uy
Uz
r + dr
∂Ux
∂y
∂Uy
∂y
∂Uz
∂y
∂U x
∂z
∂U y
∂z
∂U z
∂z
+
r
∂U x
∂x
∂Uy
∂x
∂Uz
∂x
∂U x
∂y
∂U y
∂y
∂U z
∂y
∂Ux
∂z
∂Uy
∂z
∂Uz
∂z
dx
dy
dz
0 –ω 3 ω2
ε xx ε xy ε xz
= ε yx ε yy ε yz + ω3 0 –ω 1
ε zx ε zy ε zz
–ω 2 ω1 0
strain=
symmetric part
rotation=
antisymmetric
part
We usually work in the range of infinitesimal strain
• mathematically
• geophysically
∂Ux
&lt;&lt; 1
∂x
∂Ux
&lt;&lt; 10 –4
∂x
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Exploring the Physical Interpretation of Strain
1. Extensional Strain
U(x)
U(x+dx)
L
Lo
The relative change in length can be described as
∆L = L – L 0 = U x + dx – U x
≈ ∂U L 0
∂x
∆L ≈ ∂U ≈ ε
xx
L 0 ∂x
The strain terms along the diagonal describe
relative changes in length.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Exploring the Physical Interpretation of Strain
2. Volumetric Strain
The original volume is
V0 = L x0L y 0L z0
The volume after straining is
V0 + ∆V = L x 0 + ∆L x L y 0 + ∆L y L z0 + ∆L z
V0 + ∆V ≈ L x 0L y0L z0
∆L x ∆L y ∆L z
1+
+
+
L x0
L y0
L z0
∆V ≈ ε + ε + ε
xx
yy
zz
V0
The volumetric strain is the trace of the strain tensor
or is the sum of the three extensional strains. This is
sometimes called the dilatation.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Exploring the Physical Interpretation of Strain
3. Pure Shear Strain
x
α
α
y
When we deform the square like this, notice that the
corner changes from a 90˚ angle to a lesser angle
90˚ - 2α. Note that
∂Ux ∂U y
tan α =
=
∂y
∂x
For small angles:
∂Ux ∂Uy
1
tan α ≈ α ≈
+
∂x
2 ∂y
α = ε xy
The shear strain describes a change in angle.
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Stanford Rock Physics Laboratory - Gary Mavko
Stress and Strain Analysis
Exploring the Physical Interpretation of Strain
4. Simple Shear Strain
x
2α
y
In this case the angle change is again exactly 2α.
But now:
∂U
∂U
x
∂y
Therefore:
y
=0 ,
∂x
= 2α
∂Ux ∂U y
1
ε xy =
+
≈α
∂x
2 ∂y
∂Uy ∂Ux
1
ωxy =
–
≈α
∂y
2 ∂x
Hence the strain in this example is identical to the
previous case (pure shear). However, the rotation α is
superimposed.
α
2α
=
+
α
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α
```