defining stress in a solid - stanford earth stress in a solid these forces that we must apply to the...

Stanford Rock Physics Laboratory - Gary Mavko

Stress and Strain Analysis

30

Defining Stress in a Solid

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

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.



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• 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.

Why are surfaces important?



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Look more closely at the forces acting on a surface

The normalized force per unit area is called the

Traction Vector or Stress Vector.

The orientation of the surface is

specified by the outward normal

vector ν

F

Force F

tangentialcomponent

normalcomponent

area dA

ν

unit normal

ν



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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:

x

z

y

σzz

σzx

σzy σyz

σyy

σyx

σxx

σxy

σxz



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Units of Stress

1 bar = 106 dyne/cm2 = 14.50 psi10 bar = 1 MPa = 106 N/m2

Mudweight to Pressure Gradient1 psi/ft = 144 lb/ft3

= 19.24 lb/gal = 22.5 kPa/m1 lb/gal = 0.052 psi/ft



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Traction on a Surface ofArbitrary Orientation

Cauchy’s Formula: TxTyTz

=σxx σxy σxzσyx σyy σyzσzx σzy σzz

νxνyνz

≈ T = σν

T

x

y

z

ν = (βx, βy, βz)



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σxx σxy σxzσyx σyy σyzσzx σzy σzz

σxy = σyxσxz = σzxσyz = σzy

The Stress Tensor is Symmetric

We must also balance the torques applied tothe cube and we find that:

Stress tensor is symmetric.

There are only 6 independent terms.



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Einstein Summation Convention

It is customary in stress and strain analysis to usea shorthand notation, sometimes called the Einsteinsummation convention. Any repeated index withina term implies summation over that index with arange of 1 to 3.

For example:

cijkl σkl Σ

k=1

3cijkl σklΣ

l=1

3and

are equivalent. Here, k is repeated so it implies asummation; l is also repeated, so it also implies asummation. i and j are not repeated so there areno summations.

σ ij + σ jk

There is no summation in this expression:

because there is no repeated summation in any ofthe terms. The two occurrances of the index j arein two different terms.



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Transformation of CoordinatesWe 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.

zz'

y

y'

x'

x

r



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r = x1e1 + x2e2 + x3e3

r = x1′ e1

′ + x2′ e2

′ + x3′ e3

′

Coordinate Transformations

A real physical vector can be expressed in any two

coordinate systems:

It is important to remember that we are simply giving

the same physical vector different names in different

coordinate systems.

Or in index notation:

x1′

x2′

x3′

=β 11 β 12 β 13β 21 β 22 β 23β 31 β 32 β 33

x1x2x3

xi

′ = β ijxj = β ijxjΣj = 1

3

where is the direction cosine of the new ith axis

relative to the old jth axis.

So for any vector we can write:

β ij

r ′ = βr ≈



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So now go back to Cauchy’s formula

Now transform T and ν to the new coordinates

Rewrite as:

Rename term in brackets:

Formula for coordinate transformation

σij = β ikβ jlσkl

Told = σoldνold

β ij

TTnew = σold β ijTνnew

Tnew = β ij σold β ijT νnew

Tnew = σnew νnew β ij σold β ij

T = σnew



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The hydrostatic stress or mean stress is theaverage of the three normal stresses.

This is an invariant. The mean stress is thesame in any coordinate system.

σ0 =σxx + σyy + σzz

3 = σαα

3



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where is the magnitude of the traction vector, isthe 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.

T

σ ij σ ij

Principal StressesStress 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 in the matrix:

In general, different choices of coordinate systems

cause the numerical values in the matrix (stress tensor)

to vary.

Now recall Cauchy’s formula:

Can we find directions such that the traction vector is

normal to the surface?

ν

T = σ ij ν

T = σ ij ν = σν



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Principal Stresses

Let’s emphasize the physical interpretation:

σ1 0 00 σ2 00 0 σ3

The traction is parallel to , which is always

perpendicular to the surface. If is normal to the

surface then there is no component of 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:

ν T T

T



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P

P'

Q

Q'

Infinitesimal strain

We can describe the deformation of a continuum with

a vector displacement field . This displacement

gives the vector motion of each point in the body.

It is important to note that the displacement can vary

with position.

U



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Rigid Body Translation

The vector displacement field is a constant.



46

P Q

P' Q'

Rigid Body Rotation

ω

u = ω × r

= rotation vector • direction is along axis of rotation • is the angle of rotation

ω



47

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)

(x,y,z)

U(x + dx, y + dy, z + dz)

(x + dx, y + dy, z + dz)



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strain=symmetric part

rotation=antisymmetricpart

We usually work in the range of infinitesimal strain

• mathematically

• geophysically

∂Ux∂x << 1 ∂Ux∂x << 10–4

UxUyUz r + dr

=UxUyUz r

+

∂Ux∂x

∂Ux∂y

∂Ux∂z

∂Uy∂x

∂Uy∂y

∂Uy∂z

∂Uz∂x

∂Uz∂y

∂Uz∂z

dxdydz

∂Ux∂x

∂Ux∂y

∂Ux∂z

∂Uy∂x

∂Uy∂y

∂Uy∂z

∂Uz∂x

∂Uz∂y

∂Uz∂z

=εxx εxy εxzεyx εyy εyzεzx εzy εzz

+0 –ω3 ω2

ω3 0 –ω1–ω2 ω1 0



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Exploring the Physical Interpretation of Strain

1. Extensional Strain

The relative change in length can be described as

∆L = L – L 0 = U x + dx – U x≈ ∂U∂x L0

∆LL0

≈ ∂U∂x ≈ εxx

The strain terms along the diagonal describe

relative changes in length.

U(x) U(x+dx)

Lo L



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2. Volumetric Strain

The original volume is

The volume after straining is

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.

V0 = Lx0Ly0

Lz0

V0 + ∆V = Lx0+ ∆Lx Ly0

+ ∆Ly Lz0+ ∆Lz

V0 + ∆V ≈ Lx0

Ly0Lz0

1 + ∆LxLx0

+∆LyLy0

+ ∆LzLz0

∆VV0

≈ εxx + εyy + εzz



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3. Pure Shear Strain

α = εxy

When we deform the square like this, notice that the

corner changes from a 90˚ angle to a lesser angle90˚ - 2α. Note that

For small angles:

The shear strain describes a change in angle.

x

y

α

α

tan α = ∂Ux

∂y =∂Uy∂x

tan α ≈ α ≈ 1

2∂Ux∂y +

∂Uy∂x



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4. Simple Shear Strain

In this case the angle change is again exactly 2α.

But now:

Therefore:

Hence the strain in this example is identical to theprevious case (pure shear). However, the rotation α is

superimposed.

∂Ux∂y = 0 ,

∂Uy∂x = 2α

εxy = 1

2∂Ux∂y +

∂Uy∂x ≈ α

ωxy = 1

2∂Uy∂x – ∂Ux

∂y ≈ α

x

y

2α

α

= + αα

2α

defining stress in a solid - stanford earth stress in a solid these forces that we must apply to the...

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