basic concepts of strain and tilt - unavco

1

Basic Concepts of Strain and Tilt

Evelyn Roeloffs, USGS

June 2008

Basics of Strain, Tilt, and Poroelasticity June 2008

of 24

CoordinatesRight-handed coordinate system, with positions along the three axes specified by

€

x,y,z .

€

x,y will usually be horizontal, and

€

z will usually be up.

We will often use the coordinate system

€

x,y,z=East, (geographic) North, Up.Sometimes we will use 1,2,3 instead of x,y,z to allow equations to be written more compactly.

The distinctions among the reference frames used for GPS are not important for the interpretation of strainmeter ortiltmeter data.


of 24

Displacements

Displacement of a point is described by the three displacements in these respective directions denoted by

€

uE ,

€

uN ,and

€

uUP .


of 24

Stress and Strain in One Dimension

If

€

L is the length of the rod and the force stretches it by

€

dL , then strain,

€

ε, is the dimensionless quantity

€

ε = dL /LThe stress,

€

σ , is the force per unit area, in this case

€

σ = F /πr2

Since this is a “linearly elastic” rod, the stress is proportional to the strain:

€

F /πr2 = E(dL /L) or

€

σ = Eε

a) we consider

€

ε positive when the rod gets longer; we consider

€

σ to be positive when the rod is in tension.

b)

€

E is a material property called “Young’s modulus” with dimensions of force per unit area, units hPa, MPa, GPa

c) The stress

€

σ is present throughout the length of the rod, not just at the end where the force is applied.

d) Strain is uniform throughout the length of the rod; displacement increases linearly from the fixed to the free end.


of 24

Typical Values of Young’s Modulus

Elastic strains are small for many realistic situations.

Material Young's modulus (GPa) Strain caused by a 0.1MPa (apx 14.5 psi) loadparallel to axis of a rod

Rubber 0.007 14E-3

Spider thread 3 .03E-3

Hair 10 .01E-3

Brick 14 7E-6=7 microstrain

Oak 14 7E-6=7 microstrain

Concrete 27 4E-6=4 microstrain

Marble 50 2E-6=2 microstrain

Aluminum 65 1.5E-6=1.5 microstrain

Granite 70 1.4E-6 = 1.4 microstrain

Iron 190 0.5E-6=0.5 microstrain

Table. Typical values of Young’s modulus (from http://www.physics.usyd.edu.au/teach_res/db/d0004c.htm)


of 24

Strain in Three DimensionsThe strains are the spatial gradients of displacement.If displacement is uniform over a body (ie., every part of the body moves same distance in the same direction),then there is no strain.The body can be picked up, moved, or rotated and there will not necessarily be any strain.

The equation for obtaining strain from displacements, in three dimensions, is

€

εij = 12 ∂ui /∂x j + ∂u j /∂xi[ ] or

€

εij = 12 Δui /Δx j + Δu j /Δxi[ ] for

€

i, j = 1,2,3 (or

€

x,y,z , or East-North-Up). (1)

Writing out a few of them:

€

εzz = 12 ∂uz /∂z + ∂uz /∂z[ ] = ∂uz /∂z ≈ Δuz /Δz

€

εxy = 12 ∂ux /∂y + ∂uy /∂x[ ]

€

εNN = 12 ∂uN /∂xN + ∂uN /∂xN[ ] = ∂uN /∂xN ≈ ΔuN /ΔxN

€

εNE = 12 ∂uN /∂xE + ∂uE /∂xN[ ] = 1

2 ∂uE /∂xN + ∂uN /∂xE[ ] = εEN (a shear strain)

Engineering shear strain:

€

γ xy = 2εxy = ∂ux /∂y + ∂uy /∂x[ ]


of 24

3D Strain, continued

The 3 displacements can each vary in any of 3 directions, so there are 9

€

i, j combinations.

However, there are only 6 independent strains, because if

€

i and

€

j are different, then

€

εij = ε ji

Example: for the rod, strain in the direction parallel to the rod is the slope of the plot of displacement vs. distancealong the rod:

€

εxx = ∂ux /∂x = dL /L


of 24

Some basic assumptions:

1) “Small” region:

ie., region in which it is OK to use the value of displacement at a point and its spatial gradients to estimate thedisplacementSize of region depends on the spatial distribution of displacement.For example, slip on the San Andreas fault produces a displacement field that has a jump across the fault.Can’t use displacement and strain at a point SW of the fault to estimate displacement NE, even if points are only100 m apart.For two points on the same side of the fault, the region might be about 100 m or more.For two sites several km from the fault, the displacement field is much smoother and observations one or more kmapart are sufficient.

2) “Small” deformation:

Generally we will be speaking of strains in range 10-10 (0.1 nanostrain) to 10-4 (100 microstrain).This ranges goes from resolution of GTSM21 borehole strainmeter to approximate strain within 10’s of meters of aM7 fault rupture.

3) Only changes matter:

For example, we will consider vertical stress changes caused by atmospheric pressure fluctuations, but we will notexplicitly worry about the more or less constant overburden pressure.


of 24

“Units” of Strain

Strain is dimensionless but is often referred to as if it had units

A 1% change in volume is a volumetric strain of 0.01 = 10,000 microstrain = 10,000 parts-per-million (ppm)

A 1-mm change in a 1-km baseline is a linear strain of 10-6=1 microstrain1 microstrain is sometimes written 1

€

µε

1 nanostrain=10-9 =0.001 microstrain, also called 1 part-per-billion (ppb)

1 nanostrain is sometimes written 1

€

nε

We will mostly be considering strains that range from about 0.1 nanostrain (an approximate lower bound onresolution for borehole strainmeters) to 1000 microstrain. 10 microstrain is a ballpark value for the coseismicstrain within 5 km of a M7 earthquake.


of 24

Sign Conventions for Strain

In mathematical descriptions, increases of length or volume are always considered to be positive strains.

There is also an unambiguous sign convention for shear strain based on its defining equation.

€

εxy = 12 ∂ux /∂y + ∂uy /∂x[ ]


of 24

Example: Strain Near Transition from Creeping to Locked on a Strike-Slip Fault

The strain-displacement equations unambiguously define a sign convention for shear strain. However, an arbitrarysign convention is sometimes used in reporting data. It’s always a good idea to check.

If fault creep decreases from 30 mm/year to 0 over a 30 km reach of the fault, then

€

εyy increases by 1microstrain/year.If the creeping zone is 100 m wide, then at the creeping end

€

εxychanges by -300 microstrain/year.


of 24

Matrix NotationIt’s convenient to represent strain using a matrix:

€

ε3×3 =

ε11 ε12 ε13ε12 ε22 ε23ε13 ε23 ε33

or

€

ε3×3 =

εxx εxy εxzεxy εyy εyzεxz εyz εzz

or

€

ε3×3 =

εEE εEN εEZεEN εNN εNZεEZ εNZ εZZ

The term “strain tensor” is often used. This is because strain can be expressed in different coordinate systems(“transformed”) according to certain rules.

We will often work in 2 horizontal dimensions, where

€

ε2×2 =εxx εxyεxy εyy

The three independent strain components in 2D are usually written in vector form:

€

εxxεyyεxy

3×1

or

€

εxx + εyyεxx −εyy2εxy

3×1


of 24

Form of the Horizontal Strain Tensor that We Will Use in this Workshop

We will mostly use the form on the right and refer to its entries as follows:

€

εxx + εyy “Areal Strain”

€

εa

€

εxx −εyy “Differential Extension”

€

γ1

€

2εxy “Engineering Shear Strain”

€

γ 2

Dashed line = initial shape

Solid line = deformed shape

All of the diagrams show positive strains.


of 24

StressStress has dimensions of force per unit area.

Inside a deforming body, forces act on every small planar region. Force has magnitude and direction, and 3components: one perpendicular to the surface, and two parallel to the surface, at right angles to each other.

There can be 3 directions of force on each plane, and 3 orthogonal planes, each perpendicular to one of thecoordinate axes, giving rise to 9 components of stress. We do not distinguish between

€

σ ij and

€

σ ji , because stresses arise from the forces that do not move (accelerate) thebody. Mathematically, the requirement that

€

σ ij =σ ji results from the equilibrium equations (see any elasticity text).Intuitively, the picture can be used to visualize what would happen if these shear stresses were not equal.

Because

€

σ ij =σ ji , there are only 6 independent stress components in 3 dimensions, similar to the situation for strain.

€

σ 3×3 =

σ xx σ xy σ xz

σ xy σ yy σ yz

σ xz σ yz σ zz


of 24

Types of Stress and Stress States

The three stress components with two equal subscripts are referred to as “normal stresses”. They apply tension orcompression along a specified coordinate axis. They act parallel to the normal to the face of the cube.

Stress components with unequal subscripts are “shear stresses”. They apply equal and opposite forces on opposingfaces of a cube of material, acting parallel to those faces.

The average of the three normal stresses

€

13 σ xx +σ yy +σ zz( ) is called the “mean stress”, or sometimes the “average

stress”. It is frequently written

€

σ kk /3, using the repeated index to imply the sum.

A simple and important stress state is that in which the three normal stresses are equal. This is referred to ashydrostatic or isotropic stress. Pressure in a liquid is equivalent to a hydrostatic stress state.

See Engelder (1993) for discussions of stress states relevant to the lithosphere.


of 24

Stress-strain Relations

The way in which stress and strain are coupled for a particular material is described by “constitutive equations”.For a linearly elastic medium, the constitutive relations basically say that strain is proportional to stress, in 3dimensions. An isotropic material is one which has the same mechanical properties in all directions.

The constitutive equations for normal stresses and strains, in an isotropic linearly elastic material, are:

€

2Gεxx =σ xx −ν1+ ν

σ xx +σ yy +σ zz( ) ,

€

2Gεyy =σ yy −ν1+ ν

σ xx +σ yy +σ zz( ) ,

€

2Gεzz =σ zz −ν1+ ν

σ xx +σ yy +σ zz( ) (2a,b,c)

and the constitutive equations for shear stresses and strains are:

€

2Gεxy =σ xy ,

€

2Gεxz =σ xz ,

€

2Gεyz =σ yz (2d,e,f)

All 6 constitutive equations can be written compactly as:

€

2Gεij =σ ij −ν1+ ν

σ kkδij (another way to write 2a-f)

using the summation convention and

€

δij =1 for

€

i = j , and 0 for

€

i ≠ j .

Two material properties appear:

€

G, the shear modulus (with units of force per unit area); and

€

ν , the Poisson ratio(dimensionless).More elastic moduli are needed if the material is not isotropic, but that situation is beyond the scope of this shortcourse.


of 24

Relationships Among Elastic Moduli

Although only two material properties are needed to relate stress and strain in an isotropic elastic material, thereare many ways to represent these properties. The inter-relationships useful in this short course are:

Young’s modulus,

€

EBulk modulus,

€

KShear modulus,

€

G

€

E

€

3(1− 2ν )K

€

2G(1+ ν )

€

K

€

E3(1− 2ν )

€

2(1+ ν )G3(1− 2ν )

€

G

€

E2(1+ ν )

€

3(1− 2ν )K2(1+ ν )

Table. Relationships between elastic moduli

We will use

€

K and

€

G frequently. They are examples of “moduli”, with dimensions of force/unit area.

The Poisson ratio couples extension in one direction to contraction in the perpendicular directions. It is always >0and <0.5, taking on the upper limit of 0.5 for liquids. The Poisson ratio is dimensionless and is not a modulus.

To see the role of the bulk modulus, add up the equations relating the three normal stresses and strains:

€

ε11 + ε22 + ε33( ) =3(1− 2ν )2G(1+ ν )

σ11 +σ 22 +σ 33

3

=

1K

σ11 +σ 22 +σ 33

3

The bulk modulus is the coefficient of proportionality between volumetric strain (ie., expansion or contraction),and mean stress change (for example, pressure).


of 24

Typical Values of Elastic Moduli and Poisson ratio

Material BulkModulus(GPa)

Shearmodulus(GPa)

Poissonratio

Air (0°C, 0.1MPa)

0.142x10-3 0 0.5

Water (25°C) 2.23 0 0.5PolycrystallineIce

4.41 0.16 0.2

Plagioclase 63.1 7.47 0.29Quartz Crystal 37.9 6.34 0.08Calcite 74.8 8.45 0.32Olivine 131.3 31.9 0.24Berea Sandstone 8 6 0.2WesterlyGranite

25 15 0.25

TennesseeMarble

40 24 0.25

Table. Typical values of bulk modulus, shear modulus, and Poisson ratio


of 24

Sizes of relevant and/or familiar stresses and strains:Material immediately adjacent to a fault that experiences a seismic stress drop of 10 MPa undergoes strains orderof 1000 microstrain.

Barometric pressure drop during a typical storm is 20 millibars = 0.02 bars = 0.002 MPa = 2 kPa =20 hPa

Pressure underwater increases at rate of 2.3 psi/foot = 1 bar/10 meters. = 1MPa/100 meters; pressure differencefrom bottom to top of 10-foot-deep pool is 23 psi = apx 0.3 bar = 0.03 MPa.


of 24

Rotating CoordinatesWe’ll need to switch back and forth between coordinate systems that have different horizontal orientations (thevertical direction remains the same in these notes; See any text on elasticity for the 3D formulas).

€

εx 'x 'εy 'y 'εx 'y '

3×1

=12

1+ cos2θ 1− cos2θ 2sin2θ1− cos2θ 1+ cos2θ −2sin2θ−2sin2θ 2sin2θ 2cos2θ

3×3

εxxεyyεxy

3×1

€

εx 'x ' + εy 'y'εx 'x ' −εy 'y'2εx 'y'

3×1

=

1 0 00 cos2φ sin2φ0 −sin2φ cos2φ

3×3

εxx + εyy 'εxx −εyy2εxy

3×1

(3a,b)

The method for converting stresses to a rotated coordinate system is exactly the same as that for converting strains.


of 24

Invariants of Strain and Stress

€

εx 'x ' + εy 'y'εx 'x ' −εy 'y'2εx 'y'

3×1

=

1 0 00 cos2φ sin2φ0 −sin2φ cos2φ

3×3

εxx + εyy 'εxx −εyy2εxy

3×1

(3b again)

Inspecting equation (3b) shows that :

€

εx'x ' + εy 'y ' = εxx + εyy for any value of

€

φ (4)

This is also true of the volumetric strain in 3D, and of mean stresses. In other words, the areal and volumetricstrain, and the mean stress, are the same (invariant) no matter what Cartesian coordinates they are expressed in.

In 2D there is one other invariant of strain, which is

€

εxy

2− εxxεyy

Principal Strains and Principal Stresses

By setting

€

εx'y '=0 in equation (3b),

€

φ can be found such that the shear stresses (and strains) vanish. In thiscoordinate system, the e stress and strain tensors have only diagonal elements, called principle stresses and strains.,related by:

€

εI 0 00 εII 00 0 εIII

=

1/E −ν /E −ν /E−ν /E 1/E −ν /E−ν /E −ν /E 1/E

σ I 0 00 σ II 00 0 σ III

(5)

Near the Earth’s surface, the vertical stress is usually a principle stress, so the other two principle stresses are in thehorizontal plane.


of 24

Reducing to Two Dimensions using the Plane Stress AssumptionPlane stress: one of the principle stresses is zero, so that only the stress components in one plane are nonzero.

This is the situation when the vertical stress is unchanging (for example, it equals the overburden) and cantherefore be assumed to be zero. Shear strains in vertical planes also vanish, so the only nonzero strains are

€

εxx ,

€

εyy ,

€

εzz , and

€

εxy .

However, we can express

€

εzz in terms of

€

εxx and

€

εyy . To see this, write out the constitutive equations for the normalstrains:

€

2Gεzz = −ν1+ ν

(σ xx +σ yy )

€

2G(εxx + εyy ) =1−ν1+ ν

(σ xx +σ yy )

Comparing these two equations shows that

€

εzz =−ν1−ν

(εxx + εyy ) (6)

€

εzz is of opposite sign from

€

εxx + εyy ; when there is contraction in the horizontal plane, the vertical strain isextensional. Consequently, for plane stress the volumetric strain is smaller than the areal strain:

€

εxx + εyy + εzz =1− 2ν1−ν

(εxx + εyy ) for plane stress (7)


of 24

TiltTilt is defined simply as the change in inclination with respect to either the horizontal or the vertical.

Tilt is conceptually different from strain - it does not necessarily entail deformation. That is, displacements can bedistributed spatially such that there is no strain, but there are nonzero tilts.

Tilt can be measured using vertical sensors in boreholes or horizontal sensors on the earth’s surface. As forborehole strainmeters, tiltmeters in boreholes are usually considered to be at the surface for mathematical purposes.

Two tiltmeter components are needed to characterize the change in inclination of a plane. Tilt is a vector.

Figure. (A) Tilt without strain. (B) Tilt with strain.If one component is oriented East-West and the other North-South, tilts are related to displacements by:

€

tiltEW = ∂ux /∂z = −∂uz /∂x and

€

tiltNS = ∂uy /∂z = −∂uz /∂y


of 24

Borehole Strainmeters as Elastic Inclusions

basic concepts of strain and tilt - unavco

Documents