evelyn roelo s u. s. geological survey earthquake science ... · pbo 4-component gtsm: gauge con...

Introduction to Strain and Borehole Strainmeter Data

Evelyn RoeloffsU. S. Geological Survey

Earthquake Science Center

March 28, 2016

Evelyn Roeloffs, USGS ESC Strainmeters: Introduction March 28, 2016 1 / 28

Plate Boundary Observatory (PBO) BoreholeStrainmeter Network

Funded by NSF as partof the Earthscopeiniative

78 Gladwin TensorStrainmeters

Installed 2004-2013

Depths 500-800 feet(150-250 m)


UNAVCO Engineers Installing B201 (Mount St. Helens)


A typical PBO borehole strainmeter installation

Strainmeter is grouted into anuncased section of borehole

Every PBO borehole alsocontains a 3-componentseismometer

23 of the boreholes alsocontain pore pressure sensors

PBO boreholes at Mount St.Helens and Yellowstone alsocontain tiltmeters

Circular diagram shows BSMgauge orientations


Gladwin Tensor Strain Meter

Developed in Australia byMichael Gladwin

Four ”gauges” measure innerdiameter of steel housing

Three gauges (CH0, CH1, andCH2) are 120◦ apart aroundthe borehole axis

The fourth gauge (CH3) isperpendicular to CH1


Gladwin Tensor Strain Meter - Capacitive SensingElement

Gauge changes length inresponse to strain along axis

Gap d1 is fixed

Strain changes capacitanceacross gap d2

Capacitance changes aremeasured using a bridgecircuit whose other arms areat the surface


Strainmeters fill ”gap” between Seismology and GPS

Strain = spatial derivative of displacement

Seismometer: measures time derivative of displacement; needarray to measure strain

GPS: measures displacement; need array to measure strain


Strainmeter ”niche” is hours to days

7.5 yearsdaily averages

996 1000 1004 1008

07/18/09 07/25/09 08/01/09 08/08/09 08/15/09 08/22/09 08/29/09

B201 barometer (hPa)

0 4 8

12

mm B201 rainfall

-40-20 0 20 40

nano

stra

in

B201 CH3 N 48.0E cc.tb

-40-20 0 20 40 B201 CH2 N 78.0E c

c.tb

-40-20 0 20 40 B201 CH1 N138.0E c

c.tb

-40

0

40

80B201 coldwt201bwa2007 Coldwater Visitor Center Mount St Helens


45 days30-minute samples

detrended

4 minutes20 samples per second


Basic GTSM data processing steps

7.5 yearsdaily averages 996

1000 1004 1008

07/18/09 07/25/09 08/01/09 08/08/09 08/15/09 08/22/09 08/29/09

B201 barometer (hPa)

0 4 8

12

mm B201 rainfall

-40-20 0 20 40

nano

stra

in


-40-20 0 20 40 B201 CH2 N 78.0E c

c.tb

-40-20 0 20 40 B201 CH1 N138.0E c

c.tb

-40

0

40

80B201 coldwt201bwa2007 Coldwater Visitor Center Mount St Helens


45 days30-minute samples

detrended

Applied to figures shown here:

”Clean” the dataRemove long-term trendsCorrect for atmosphericpressureRemove tidal variations

To be discussed:

Optionally, develop yourown calibrationsObtain strain components aslinear combinations of gaugeelongations


What is ”strain”?

”Strain” is a change in one or more dimensions of a solid body,relative to a reference state

Size may changeShape may change

We assume here that strains are small, so ”infinitesimal straintheory” applies


Coordinate Systems

Right-handed Cartesian coordinate system

Various sets of names for coordinate axes (examples above)

Strainmeters do not care about:

Curvature of the earthGeodetic reference frames


Displacements

Material at a point can move in three directions, e.g. (ux, uy, uz)

Various sets of names for components of displacement

e.g., 1,2,3 or x, y, zHorizontal axes will not always be East and North

Strain is a result of spatially varying displacement


Spatial derivatives of displacement

Displacements can vary in three coordinate directions

9 partial derivatives ∂ui∂xj

where i, j = 1, 2, 3

Strain components: εij = 12 [ ∂ui

∂xj+

∂uj

∂xi]

”Normal” strains have i = j: εii = ∂ui∂xi

(no summation implied)

”Shear” strains have i 6= j, note that εij = εji


Horizontal strain components

unstrainedmaterial

+x

+y

εxxε εyy εxy

contraction in the x-direction

(a negative strain)extension

in the y-direction(a positive strain)

xy shear(a positive strain becausey-displacement increases

with increasing x)

εxx = ∂ux∂x εyy =

∂uy

∂y εxy = 12 [∂ux

∂y +∂uy

∂x ]


”Units” of strain

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

εxx = 0.01 might be called ”1% strain” or ”10,000 microstrain” or”10,000 ppm strain”1 mm shortening of a 1-km baseline is a strain of −10−6 = -1microstrain = -1 ppm strain1 mm lengthening of a 1000-km baseline is a strain of 10−9 = 1nanostrain = 1 ppb strain


Sign conventions

The following sign conventions are used here to minimizemathematical confusion:

Increasing length (”extension”) is positive strain; decreasing length(”contraction”) is negative strainIncreasing area or volume (”expansion”) is positive strainStresses that produce positive strains are positive (ie., tension ispositive)Stresses that produce negative strains are negative (ie., compressionis negative)

...but note that some publications do not use these signconventions:

In geotechnical literature, contraction and compressional stress arereferred to as positivePublished work on volumetric strainmeter data describescontraction as positive


Example: Locked vs. creeping strike-slip fault

Map view


GTSM gauge output is proportional to inner diameterchange of housing

Empty Borehole

Goal: Measure the strain of

the formation

Deforms much more than formation

Borehole filledwith rigid material

Does not deform at all

Strainmeter Groutedinto Borehole

Deforms more than formation but less

than empty boreholeDeformation depends on relative moduli of

strainmeter, grout, and formationEvelyn Roeloffs, USGS ESC Strainmeters: Introduction March 28, 2016 18 / 28

Elongation of a single ideal gauge in response to strain

Gauge elongation, ei, is a linear combination of strainparallel and perpendicular to the gauge

ex = Aεxx −BεyyA and B are positive scalars with A > B


Areal strain, differential extension, engineering shear

unstrainedmaterial

+x

+y

xxε yy+ε2εxy

areal contraction (a negative strain)

differential extension(a positive strain) engineering shear

(a positive strain becausey-displacement increases

with increasing x)

xxε yy−ε

Areal strain εxx + εyy does not change if axes are rotatedDifferential extension (εxx − εyy) and engineering shear 2εxy areshear strain components

Neither shear strain component changes areaShear strains depend on coordinate system


Elongation of ideal gauge: Areal strain and differentialextension

ex = Aεxx −Bεyy can be re-written as linear combination of arealstrain and differential extension

Rearrange: ex = 0.5(A−B)(εxx + εyy) + 0.5(A+B)(εxx − εyy)

Define C = 0.5(A−B) and D = 0.5(A+B)

ex = C(εxx + εyy) +D(εxx − εyy)

NOTE: εxy does not change length for ideal gauge parallel to x or y


PBO 4-component GTSM: Gauge configuration

CH0

CH1

CH2

CH3

North

φ0

60o

60o

30o

This is the azimuthon the PBO web pagegiven CW from North

=e1

=e3

x

y

CH0,CH1, and CH2 are equally spacedCH3 is perpendicular to CH1

Blue dots: end of gauge whose azimuth is givenRed circles: ends of CH2 and CH0 that are -120o and +120o from CH1

1

1

CH2 =e0

CH0 =e2

East

θ

x1 and y1 are Cartesiancoordinates parallel andperpendicular to CH1

Azimuths are clockwise fromNorth; Polar coordinate anglesare counterclockwise from East

Polar coordinates (r, θ) areused for math

The polar angle of y1 is +90from the polar angle of x1


2 strain components from 2 perpendicular ideal gauges

x, y = directions ofperpendicular gauges CH1and CH3e1 = C(εxx+εyy)+D(εxx−εyy)e3 = C(εxx+εyy)−D(εxx−εyy)

Solve for areal strain and differential extension:(εxx + εyy) = (e1 + e3)/2C(εxx − εyy) = (e1 − e3)/2DAreal strain is proportional to average of gauge elongations

Differential extension is proportional to difference between gaugeelongations


Transforming horizontal strains to rotated coordinates

Horizontal strain tensor canbe expressed in a coordinatesystem rotated about thevertical axis

εx′x′ + εy′y′

εx′x′ − εy′y′2εx′y′

=

1 0 00 cos 2θ sin 2θ0 − sin 2θ cos 2θ

εxx + εyyεxx − εyy

2εxy

Areal strain is invariant under rotation:

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


Expressing gauge elongations in rotated coordinates

Each gauge has its owngauge-parallel coordinates, e.g.:e1 = C(εxx + εyy) +D(εxx − εyy)e0 = C(εx′x′ +εy′y′)+D(εx′x′−εy′y′)

Express CH2 elongation in CH1-parallel coordinates:

εx′x′ + εy′y′ = εxx + εyyεx′x′ − εy′y′ = cos 2θ(εxx − εyy) + sin 2θ(2εxy)

e0 = C(εxx + εyy) +D cos 2θ(εxx − εyy) +D sin 2θ(2εxy)e0 = C(εxx + εyy)− 0.5D(εxx − εyy) + 0.866D(2εxy) for θ = 60◦

NOTE: e0 does not respond to 2εx′y′ , but does respond to 2εxy


3 gauge elongations to 3 strain components

x, y are parallel and perpendicular to CH1 = e1

3 identical gauges 120◦ apart (CH2, CH1, CH0) = (e0, e1, e2)

Express elongations in CH1-parallel coordinates:e0 = C(εxx + εyy) +Dcos(−240◦)(εxx − εyy) +Dsin(−240◦)(2εxy)e1 = C(εxx + εyy) +D(εxx − εyy)e2 = C(εxx + εyy) +Dcos(240◦)(εxx − εyy) +Dsin(240◦)(2εxy)

Solve for strain components:(εxx + εyy) = (e0 + e1 + e2)/3C(εxx − εyy) = [(e1 − e0) + (e1 − e2)]/3D2εxy = (e0 − e2)/[2(0.866D)]

Areal strain is proportional to average of outputs from equallyspaced gauges

Shear strains are proportional to differences among gauge outputs


Gauge elongations to strain components: Example


Any questions?


evelyn roelo s u. s. geological survey earthquake science ... · pbo 4-component gtsm: gauge con...

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