Copyrighted Material

1 Deformation Stress and Conservation Laws

In this chapter we will develop a mathematical description of deformation Our focus is onrelating deformation to quantities that can be measured in the field such as the change indistance between two points the change in orientation of a line or the change in volume ofa borehole strain sensor We will also review the Cauchy stress tensor and the conservationlaws that generalize conservation of mass and momentum to continuous media Last wewill consider constitutive equations that relate the stresses acting on a material element tothe resultant strains andor rates of strain This necessarily abbreviated review of continuummechanics borrows from a number of excellent textbooks on the subject to which the readeris referred for more detail (references are given at the end of the chapter)

If all the points that make up a body (say a tectonic plate) move together without a changein the shape of the body we will refer to this as rigid body motion On the other hand ifthe shape of the body changes we will refer to this as deformation To differentiate betweendeformation and rigid body motion we will consider the relative motion of neighboringpoints Tomake this mathematically precise we will consider the displacement u of a point atx relative to an arbitrary origin x0 (figure 11) Note that vector u appears in boldface whereasthe components of the vector ui do not

Taking the first two terms in a Taylorrsquos series expansion about x0 dxj + middot middot middot i = 1 2 3 (11)partuiui (x) = ui (x0) +partxj x0

in equation (11) and in what follows summation on repeated indices is implied The firstterm ui (x0) represents a rigid body translation since all points in the neighborhood of x0

share the same displacement The second term gives the relative displacement in terms ofthe gradient of the displacements partuipartxj or nablau in vector notation The partial derivativespartuipartxj make up the displacement gradient tensor a second rank tensor with nine independentcomponents

partu1partx1 partu1partx2 partu1partx3 partu3partx1 partu3partx2 partu3partx3

The displacement gradient tensor can be separated into symmetric and antisymmetric partsas follows

(12)partu2partx1 partu2partx2 partu2partx3

1 1partui partuj partui partujui (x) = ui (x0) + dxj + dxj + middot middot middot (13)minus+2 2partxj partxi partxj partxi

If the magnitudes of the displacement gradients are small |partuipartxj | laquo 1 the symmetric andantisymmetric parts of the displacement gradient tensor can be associated with the smallstrain and rotation tensors as in equations (14) and (15) Fortunately the assumptionof smallstrain and rotation is nearly always satisfied in the study of active crustal deformationwith theimportant exceptions of deformation within the cores of active fault zones where finite strainmeasures are required

2

Copyrighted Material

Chapter 1

x0+dx u(x0+dx)

dx

x0u(x0)

Figure 11 Generalized displacement of a line segment dx Point x0 displaces by amount u(x0)whereas the other end point x0 + dx displaces by u(x0 + dx)

Define the symmetric part of the displacement gradient tensor with six independentcomponents to be the infinitesimal strain tensor Ei j

1 partui partuj (14)Ei j equiv + 2 partxj partxi

The antisymmetric part of the displacement gradient tensor is defined to be the rotation ωi j which includes the remaining three independent components

1 partui partuj (15)ωi j equiv minus 2 partxj partxi

By substituting equations (14) and (15) into (13) and neglecting terms of order (dx)2 we seethat the displacement u(x) is composed of three componentsmdashrigid body translation strainand rigid body rotation

u(x) = u(x0) + E middot dx + ω middot dx ui (x) = ui (x0) + Ei jdxj + ωi jdxj (16)

We must still show that Ei j is sensibly identified with strain and ωi j with rigid body rotationThis will be the subject of the next two sections

11 StrainTo see why Ei j is associated with strain consider a line segment with end points x and x + dx(figure 12) Following deformation the coordinates of the end points are ξ and ξ + dξ Thelength of the line segment changes from dl0 equiv |dx| to dl equiv |dξ |

The strain can be defined in terms of the change in the squared length of the segmentmdashthatis dl2 minus dl0

2 The squared length of the segment prior to deformation is

dl02 = dxidxi = δi jdxidxj (17)

where the Kronecker delta δi j = 1 if i = j and δi j = 0 otherwise Similarly the squared lengthof the segment following deformation is

dl2 = dξmdξm = δmndξmdξn (18)

3

Copyrighted Material

Deformation Stress and Conservation Laws

x+dx ξ+dξ

|dξ|=dl|dx|=dl0 Deformation

ξx

Figure 12 Strained line segment Line segment with endpoints x and x + dx in the undeformed statetransforms to a line with endpoints ξ and ξ + dξ in the deformed state

We can consider the final coordinates to be a function of the initial coordinates ξi = ξi (x) orconversely the initial coordinates to be a function of the final coordinates xi = xi (ξ ) Adoptingthe former approach for the moment and writing the total differential dξm as

dξm = partξmdxj (19)partxj

we find

partξm partξndl2 = δmn dxidxj (110)partxi partxj

Thus the change in the squared length of the line segment is

partξm partξndl2 minus dl02 = δmn minus δi j dxidxj equiv 2Ei jdxidxj (111)

partxi partxj

Ei j as defined here is known as theGreenrsquos strain tensor It is a soshycalled Lagrangian formulationwith derivatives of final position with respect to initial position

Employing the inverse relationship xi = xi (ξ ) which as described here is an Eulerian formushylation yields the soshycalled Almansi strain tensor ei j

partxm partxndl2 minus dl02 = δi j minus δmn dξidξ j equiv 2ei jdξidξ j (112)

partξi partξ j

Note that both equations (111) and (112) are valid whether or not |partuipartxj | laquo 1 We do notyet have the strains described in terms of displacements To do so note that the displacementis the difference between the initial position and the final position ui = ξi minus xi Differentiating

partξi partui= + δi j (113)partxj partxj

Substituting this into equation (111) gives

partum partun2Ei j = δmn + δmi + δnj minus δi j (114)partxi partxj

partun partum partum partun= δmn δmi + δnj + + δmiδnj minus δi j (115)partxj partxi partxi partxj

Copyrighted Material

4 Chapter 1

Thus the Green strain is given by

Ei j = 12

partui

partxj+ partuj

partxi+ partun

partxi

partun

partxj (116)

The corresponding Almansi (Eulerian) strain is given by

1 partui partuj partun partunei j = + minus (117)2 partξ j partξi partξi partξ j

Notice that the third terms in both equations (116) and (117) are products of gradients indisplacement and are thus second order These terms are safely neglected as long as thedisplacement gradients are small compared to unity Furthermore in this limit we do notneed to distinguish between derivatives taken with respect to initial or final coordinates Theinfinitesimal strain is thus given by equation (14)

More generally the Lagrangian formulation tracks a material element or a collectionof particles It is an appropriate choice for elasticity problems where the reference state isnaturally defined as the unstrained configuration The Eulerian formulation refers to theproperties at a given point in space and is naturally used in fluid mechanics where one couldimagine monitoring the velocity density temperature and so on at fixed points in spaceas the flowing fluid moves by In contrast the Lagrangian formulation would track theseproperties as one moved with the flow For example fixed weather stations that measurewind velocity barometric pressure and temperature yield an Eulerian description of theatmosphere Neutrally buoyantweather balloonsmovingwith thewind yield a correspondingLagrangian description

Geodetic observing stations are fixed to the solid earth suggesting that a Lagrangian formushylation is sensible Furthermore constitutive equations which refer to a parcel of material areLagrangian However the familiar Cauchy stress to be defined shortly is inherently Eulerian(Dahlen and Tromp 1998 give a more complete discussion of these issues) These distinctionsturn out not to be important as long as the initial stresses are of the same order as or smallerthan the incremental stresses (and any rotations are not much larger than the strains) Forthe most part in this book we will restrict our attention to infinitesimal strains and will notfind it necessary to distinguish between Lagrangian and Eulerian formulations In the earthhowever the initial stresses at least their isotropic components are likely to be large and amore thorough accounting of the effects of this prestress is required This arises in chapter 9where we will consider the role of gravitation in the equilibrium equations

An important measure of strain is the change in the length of an arbitrarily oriented linesegment In practice linear extension is measured by a variety of extensometers rangingfrom strain gauges in laboratory rock mechanics experiments to field extensometers and laserstrainmeters To the degree that the strain is spatially uniform geodimeters or ElectronicDistanceMeters (EDM)measure linear extension Consider the change in length of a referenceline dx with initial length l0 and final length l In the small strain limit we have

l2 minus l02 = (l + l0)(l minus l0) = 2Ei jdxidxj

2l0(l minus l0) 2Ei jdxidxj

l d d Ei jdxidx j (118)l0

where ddxi are the components of a unit vector in the direction of the baseline dx The changein length per unit length of the baseline dx is given by the scalar product of the strain tensor

5

Copyrighted Material

Deformation Stress and Conservation Laws

x0+dxdx

du

x0

Figure 13 Extension and rotation of a line segment The extension is the dot product of the relativedisplacement duwith the unit vector in the direction of the line segment dx = dx|dx| The rotation ofthe line segment is the cross product of the relative displacement with the same unit vector

dand the unit vectors dxd This can also be written as dxd middot E middot dx This result has a simplegeometric interpretation The change in length of the baseline is the projection of the relativedisplacement vector du onto the baseline directionl = duiddxi (figure 13) In an irrotationalfield the relative displacement between nearby points is given by dui = Ei jdxj (equation [16])Thusl = Ei jddxidxj Dividing both sides by l0 yields the last equation in equation (118)

Geodimeters which measure distance with great accuracy can be used to determine thechange in distance l between two benchmarks separated by up to several tens of kilometersIf the baseline lies in a horizontal plane dxT = (dx1 dx2 0) and the strain is spatially uniformbetween the two end points the extensional strain is given by

l 2= E11cos α + 2E12cos α sin α + E22sin2 α (119)l0

where α is the angle between the x1 axis and the baseline Equation (119) shows how theelongation varies with orientation In the small strain limit we can relate the rate of linelength change to the strain rate by differentiating both sides of equation (119) with respectto time to firstshyorder changes in orientation partαpartt can be neglected The exact relationshipcorresponding to equation (118) valid for finite strain is (1 l)dldt = ddξ middot D middot ddξ (eg Malvern1969) where the rate of deformation tensor is defined as

1 partvi partv jDi j equiv + (120)2 partξ j partξi

where v is the instantaneous particle velocity with components vi The rate of deformationtensor inwhich derivatives are takenwith respect to the current configuration is not generallyequivalent to the time derivative of a strain tensor in which derivatives are taken with respectto the initial configuration The distinction is negligible however if one takes the initialconfiguration to be that at the time of the first geodetic measurement not an unrealizableundeformed state so that generally |partuipartxj | laquo 1

If the rate of elongation is measured on a number of baselines with different orientationsαi it is possible to determine the components of the average strainshyrate tensor by least squaressolution of the following equations

(1li )dlidt cos2 αi 2 cos αi sin αi sin2 αiE11 = E12 (121) E22

where the superimposed dot indicates differentiation with respect to time

6

Copyrighted Material

Chapter 1

200

150

100

50

0

ndash50

ndash100

ndash150

ndash2000

Azimuth(degrees)

Figure 14 Linear extension rate as a function of azimuth for precise distance measurements from theEastern California Shear Zone Data from two networks are shown The Garlock network is located atthe eastern end of the Garlock fault The estimated strain-rate fields for the two networks (shown bythe continuous curves) are very similar From Savage et al (2001)

In figure 14 linear extension rates vary as a function of azimuth in a manner consistentwith a uniform strainshyrate field within rather large errors given the magnitude of the signalThese data were collected in the Southern California Shear Zone using precise distancemeasuring devices Data from the US Geological Survey (USGS) Geodolite program whichmeasured distance repeatedly using an Electronic Distance Meter (EDM) were analyzed byJohnson et al (1994) to determine the average strain rate in southern California Figure 15shows the principal strainshyrate directions and magnitudes for different polygons assumingthat the strain rate is spatially uniformwithin each polygon

Borehole dilatometers measure volume change or dilatational strain in the earthrsquos crust(figure 16) The volume change of an element is related to the extension in the threecoordinate directions Let V0 be the initial volume and V be the final volume and define thedilatation as the change in volume over the initial volume equiv (V minus V0)V0 If xi representthe sides of a small cubical element in the undeformed state then V0 = x1x2x3 Similarly if ξirepresent the sides in the deformed state then V = ξ1ξ2ξ3 The final dimensions are relatedto the undeformed dimensions by ξ1 = x1(1 + E11) ξ2 = x2(1 + E22) and ξ3 = x3(1 + E33)so that

x1x2x3(1 + E11)(1 + E22)(1 + E33) minus x1x2x3 = (122)

x1x2x3

To first order in strain this yields = E11 + E22 + E33 = Eii = trace(E) which can be shownto be independent of the coordinate system In other words the dilatation is an invariantof the strain tensor Note further that Eii = partuipartxi so volumetric strain is equivalent to thedivergence of the displacement field = nabla middot u

Exte

nsio

nra

te(n

anos

trai

nyr

)

BarstowGarlockBarstowfitGarlockfit

30 60 90 120 150 180

2

Copyrighted Material

Chapter 1

x0+dx u(x0+dx)

dx

x0u(x0)

Figure 11 Generalized displacement of a line segment dx Point x0 displaces by amount u(x0)whereas the other end point x0 + dx displaces by u(x0 + dx)

Define the symmetric part of the displacement gradient tensor with six independentcomponents to be the infinitesimal strain tensor Ei j

1 partui partuj (14)Ei j equiv + 2 partxj partxi

The antisymmetric part of the displacement gradient tensor is defined to be the rotation ωi j which includes the remaining three independent components

1 partui partuj (15)ωi j equiv minus 2 partxj partxi

By substituting equations (14) and (15) into (13) and neglecting terms of order (dx)2 we seethat the displacement u(x) is composed of three componentsmdashrigid body translation strainand rigid body rotation

u(x) = u(x0) + E middot dx + ω middot dx ui (x) = ui (x0) + Ei jdxj + ωi jdxj (16)

We must still show that Ei j is sensibly identified with strain and ωi j with rigid body rotationThis will be the subject of the next two sections

11 StrainTo see why Ei j is associated with strain consider a line segment with end points x and x + dx(figure 12) Following deformation the coordinates of the end points are ξ and ξ + dξ Thelength of the line segment changes from dl0 equiv |dx| to dl equiv |dξ |

The strain can be defined in terms of the change in the squared length of the segmentmdashthatis dl2 minus dl0

2 The squared length of the segment prior to deformation is

dl02 = dxidxi = δi jdxidxj (17)

where the Kronecker delta δi j = 1 if i = j and δi j = 0 otherwise Similarly the squared lengthof the segment following deformation is

dl2 = dξmdξm = δmndξmdξn (18)

3

Copyrighted Material

Deformation Stress and Conservation Laws

x+dx ξ+dξ

|dξ|=dl|dx|=dl0 Deformation

ξx

Figure 12 Strained line segment Line segment with endpoints x and x + dx in the undeformed statetransforms to a line with endpoints ξ and ξ + dξ in the deformed state

We can consider the final coordinates to be a function of the initial coordinates ξi = ξi (x) orconversely the initial coordinates to be a function of the final coordinates xi = xi (ξ ) Adoptingthe former approach for the moment and writing the total differential dξm as

dξm = partξmdxj (19)partxj

we find

partξm partξndl2 = δmn dxidxj (110)partxi partxj

Thus the change in the squared length of the line segment is

partξm partξndl2 minus dl02 = δmn minus δi j dxidxj equiv 2Ei jdxidxj (111)

partxi partxj

Ei j as defined here is known as theGreenrsquos strain tensor It is a soshycalled Lagrangian formulationwith derivatives of final position with respect to initial position

Employing the inverse relationship xi = xi (ξ ) which as described here is an Eulerian formushylation yields the soshycalled Almansi strain tensor ei j

partxm partxndl2 minus dl02 = δi j minus δmn dξidξ j equiv 2ei jdξidξ j (112)

partξi partξ j

Note that both equations (111) and (112) are valid whether or not |partuipartxj | laquo 1 We do notyet have the strains described in terms of displacements To do so note that the displacementis the difference between the initial position and the final position ui = ξi minus xi Differentiating

partξi partui= + δi j (113)partxj partxj

Substituting this into equation (111) gives

partum partun2Ei j = δmn + δmi + δnj minus δi j (114)partxi partxj

partun partum partum partun= δmn δmi + δnj + + δmiδnj minus δi j (115)partxj partxi partxi partxj

Copyrighted Material

4 Chapter 1

Thus the Green strain is given by

Ei j = 12

partui

partxj+ partuj

partxi+ partun

partxi

partun

partxj (116)

The corresponding Almansi (Eulerian) strain is given by

1 partui partuj partun partunei j = + minus (117)2 partξ j partξi partξi partξ j

Notice that the third terms in both equations (116) and (117) are products of gradients indisplacement and are thus second order These terms are safely neglected as long as thedisplacement gradients are small compared to unity Furthermore in this limit we do notneed to distinguish between derivatives taken with respect to initial or final coordinates Theinfinitesimal strain is thus given by equation (14)

More generally the Lagrangian formulation tracks a material element or a collectionof particles It is an appropriate choice for elasticity problems where the reference state isnaturally defined as the unstrained configuration The Eulerian formulation refers to theproperties at a given point in space and is naturally used in fluid mechanics where one couldimagine monitoring the velocity density temperature and so on at fixed points in spaceas the flowing fluid moves by In contrast the Lagrangian formulation would track theseproperties as one moved with the flow For example fixed weather stations that measurewind velocity barometric pressure and temperature yield an Eulerian description of theatmosphere Neutrally buoyantweather balloonsmovingwith thewind yield a correspondingLagrangian description

Geodetic observing stations are fixed to the solid earth suggesting that a Lagrangian formushylation is sensible Furthermore constitutive equations which refer to a parcel of material areLagrangian However the familiar Cauchy stress to be defined shortly is inherently Eulerian(Dahlen and Tromp 1998 give a more complete discussion of these issues) These distinctionsturn out not to be important as long as the initial stresses are of the same order as or smallerthan the incremental stresses (and any rotations are not much larger than the strains) Forthe most part in this book we will restrict our attention to infinitesimal strains and will notfind it necessary to distinguish between Lagrangian and Eulerian formulations In the earthhowever the initial stresses at least their isotropic components are likely to be large and amore thorough accounting of the effects of this prestress is required This arises in chapter 9where we will consider the role of gravitation in the equilibrium equations

An important measure of strain is the change in the length of an arbitrarily oriented linesegment In practice linear extension is measured by a variety of extensometers rangingfrom strain gauges in laboratory rock mechanics experiments to field extensometers and laserstrainmeters To the degree that the strain is spatially uniform geodimeters or ElectronicDistanceMeters (EDM)measure linear extension Consider the change in length of a referenceline dx with initial length l0 and final length l In the small strain limit we have

l2 minus l02 = (l + l0)(l minus l0) = 2Ei jdxidxj

2l0(l minus l0) 2Ei jdxidxj

l d d Ei jdxidx j (118)l0

where ddxi are the components of a unit vector in the direction of the baseline dx The changein length per unit length of the baseline dx is given by the scalar product of the strain tensor

5

Copyrighted Material

Deformation Stress and Conservation Laws

x0+dxdx

du

x0

Figure 13 Extension and rotation of a line segment The extension is the dot product of the relativedisplacement duwith the unit vector in the direction of the line segment dx = dx|dx| The rotation ofthe line segment is the cross product of the relative displacement with the same unit vector

dand the unit vectors dxd This can also be written as dxd middot E middot dx This result has a simplegeometric interpretation The change in length of the baseline is the projection of the relativedisplacement vector du onto the baseline directionl = duiddxi (figure 13) In an irrotationalfield the relative displacement between nearby points is given by dui = Ei jdxj (equation [16])Thusl = Ei jddxidxj Dividing both sides by l0 yields the last equation in equation (118)

Geodimeters which measure distance with great accuracy can be used to determine thechange in distance l between two benchmarks separated by up to several tens of kilometersIf the baseline lies in a horizontal plane dxT = (dx1 dx2 0) and the strain is spatially uniformbetween the two end points the extensional strain is given by

l 2= E11cos α + 2E12cos α sin α + E22sin2 α (119)l0

where α is the angle between the x1 axis and the baseline Equation (119) shows how theelongation varies with orientation In the small strain limit we can relate the rate of linelength change to the strain rate by differentiating both sides of equation (119) with respectto time to firstshyorder changes in orientation partαpartt can be neglected The exact relationshipcorresponding to equation (118) valid for finite strain is (1 l)dldt = ddξ middot D middot ddξ (eg Malvern1969) where the rate of deformation tensor is defined as

1 partvi partv jDi j equiv + (120)2 partξ j partξi

where v is the instantaneous particle velocity with components vi The rate of deformationtensor inwhich derivatives are takenwith respect to the current configuration is not generallyequivalent to the time derivative of a strain tensor in which derivatives are taken with respectto the initial configuration The distinction is negligible however if one takes the initialconfiguration to be that at the time of the first geodetic measurement not an unrealizableundeformed state so that generally |partuipartxj | laquo 1

If the rate of elongation is measured on a number of baselines with different orientationsαi it is possible to determine the components of the average strainshyrate tensor by least squaressolution of the following equations

(1li )dlidt cos2 αi 2 cos αi sin αi sin2 αiE11 = E12 (121) E22

where the superimposed dot indicates differentiation with respect to time

6

Copyrighted Material

Chapter 1

200

150

100

50

0

ndash50

ndash100

ndash150

ndash2000

Azimuth(degrees)

Figure 14 Linear extension rate as a function of azimuth for precise distance measurements from theEastern California Shear Zone Data from two networks are shown The Garlock network is located atthe eastern end of the Garlock fault The estimated strain-rate fields for the two networks (shown bythe continuous curves) are very similar From Savage et al (2001)

In figure 14 linear extension rates vary as a function of azimuth in a manner consistentwith a uniform strainshyrate field within rather large errors given the magnitude of the signalThese data were collected in the Southern California Shear Zone using precise distancemeasuring devices Data from the US Geological Survey (USGS) Geodolite program whichmeasured distance repeatedly using an Electronic Distance Meter (EDM) were analyzed byJohnson et al (1994) to determine the average strain rate in southern California Figure 15shows the principal strainshyrate directions and magnitudes for different polygons assumingthat the strain rate is spatially uniformwithin each polygon

Borehole dilatometers measure volume change or dilatational strain in the earthrsquos crust(figure 16) The volume change of an element is related to the extension in the threecoordinate directions Let V0 be the initial volume and V be the final volume and define thedilatation as the change in volume over the initial volume equiv (V minus V0)V0 If xi representthe sides of a small cubical element in the undeformed state then V0 = x1x2x3 Similarly if ξirepresent the sides in the deformed state then V = ξ1ξ2ξ3 The final dimensions are relatedto the undeformed dimensions by ξ1 = x1(1 + E11) ξ2 = x2(1 + E22) and ξ3 = x3(1 + E33)so that

x1x2x3(1 + E11)(1 + E22)(1 + E33) minus x1x2x3 = (122)

x1x2x3

To first order in strain this yields = E11 + E22 + E33 = Eii = trace(E) which can be shownto be independent of the coordinate system In other words the dilatation is an invariantof the strain tensor Note further that Eii = partuipartxi so volumetric strain is equivalent to thedivergence of the displacement field = nabla middot u

Exte

nsio

nra

te(n

anos

trai

nyr

)

BarstowGarlockBarstowfitGarlockfit

30 60 90 120 150 180

3

Copyrighted Material

Deformation Stress and Conservation Laws

x+dx ξ+dξ

|dξ|=dl|dx|=dl0 Deformation

ξx

Figure 12 Strained line segment Line segment with endpoints x and x + dx in the undeformed statetransforms to a line with endpoints ξ and ξ + dξ in the deformed state

We can consider the final coordinates to be a function of the initial coordinates ξi = ξi (x) orconversely the initial coordinates to be a function of the final coordinates xi = xi (ξ ) Adoptingthe former approach for the moment and writing the total differential dξm as

dξm = partξmdxj (19)partxj

we find

partξm partξndl2 = δmn dxidxj (110)partxi partxj

Thus the change in the squared length of the line segment is

partξm partξndl2 minus dl02 = δmn minus δi j dxidxj equiv 2Ei jdxidxj (111)

partxi partxj

Ei j as defined here is known as theGreenrsquos strain tensor It is a soshycalled Lagrangian formulationwith derivatives of final position with respect to initial position

Employing the inverse relationship xi = xi (ξ ) which as described here is an Eulerian formushylation yields the soshycalled Almansi strain tensor ei j

partxm partxndl2 minus dl02 = δi j minus δmn dξidξ j equiv 2ei jdξidξ j (112)

partξi partξ j

Note that both equations (111) and (112) are valid whether or not |partuipartxj | laquo 1 We do notyet have the strains described in terms of displacements To do so note that the displacementis the difference between the initial position and the final position ui = ξi minus xi Differentiating

partξi partui= + δi j (113)partxj partxj

Substituting this into equation (111) gives

partum partun2Ei j = δmn + δmi + δnj minus δi j (114)partxi partxj

partun partum partum partun= δmn δmi + δnj + + δmiδnj minus δi j (115)partxj partxi partxi partxj

Copyrighted Material

4 Chapter 1

Thus the Green strain is given by

Ei j = 12

partui

partxj+ partuj

partxi+ partun

partxi

partun

partxj (116)

The corresponding Almansi (Eulerian) strain is given by

1 partui partuj partun partunei j = + minus (117)2 partξ j partξi partξi partξ j

Notice that the third terms in both equations (116) and (117) are products of gradients indisplacement and are thus second order These terms are safely neglected as long as thedisplacement gradients are small compared to unity Furthermore in this limit we do notneed to distinguish between derivatives taken with respect to initial or final coordinates Theinfinitesimal strain is thus given by equation (14)

More generally the Lagrangian formulation tracks a material element or a collectionof particles It is an appropriate choice for elasticity problems where the reference state isnaturally defined as the unstrained configuration The Eulerian formulation refers to theproperties at a given point in space and is naturally used in fluid mechanics where one couldimagine monitoring the velocity density temperature and so on at fixed points in spaceas the flowing fluid moves by In contrast the Lagrangian formulation would track theseproperties as one moved with the flow For example fixed weather stations that measurewind velocity barometric pressure and temperature yield an Eulerian description of theatmosphere Neutrally buoyantweather balloonsmovingwith thewind yield a correspondingLagrangian description

Geodetic observing stations are fixed to the solid earth suggesting that a Lagrangian formushylation is sensible Furthermore constitutive equations which refer to a parcel of material areLagrangian However the familiar Cauchy stress to be defined shortly is inherently Eulerian(Dahlen and Tromp 1998 give a more complete discussion of these issues) These distinctionsturn out not to be important as long as the initial stresses are of the same order as or smallerthan the incremental stresses (and any rotations are not much larger than the strains) Forthe most part in this book we will restrict our attention to infinitesimal strains and will notfind it necessary to distinguish between Lagrangian and Eulerian formulations In the earthhowever the initial stresses at least their isotropic components are likely to be large and amore thorough accounting of the effects of this prestress is required This arises in chapter 9where we will consider the role of gravitation in the equilibrium equations

An important measure of strain is the change in the length of an arbitrarily oriented linesegment In practice linear extension is measured by a variety of extensometers rangingfrom strain gauges in laboratory rock mechanics experiments to field extensometers and laserstrainmeters To the degree that the strain is spatially uniform geodimeters or ElectronicDistanceMeters (EDM)measure linear extension Consider the change in length of a referenceline dx with initial length l0 and final length l In the small strain limit we have

l2 minus l02 = (l + l0)(l minus l0) = 2Ei jdxidxj

2l0(l minus l0) 2Ei jdxidxj

l d d Ei jdxidx j (118)l0

where ddxi are the components of a unit vector in the direction of the baseline dx The changein length per unit length of the baseline dx is given by the scalar product of the strain tensor

5

Copyrighted Material

Deformation Stress and Conservation Laws

x0+dxdx

du

x0

Figure 13 Extension and rotation of a line segment The extension is the dot product of the relativedisplacement duwith the unit vector in the direction of the line segment dx = dx|dx| The rotation ofthe line segment is the cross product of the relative displacement with the same unit vector

dand the unit vectors dxd This can also be written as dxd middot E middot dx This result has a simplegeometric interpretation The change in length of the baseline is the projection of the relativedisplacement vector du onto the baseline directionl = duiddxi (figure 13) In an irrotationalfield the relative displacement between nearby points is given by dui = Ei jdxj (equation [16])Thusl = Ei jddxidxj Dividing both sides by l0 yields the last equation in equation (118)

Geodimeters which measure distance with great accuracy can be used to determine thechange in distance l between two benchmarks separated by up to several tens of kilometersIf the baseline lies in a horizontal plane dxT = (dx1 dx2 0) and the strain is spatially uniformbetween the two end points the extensional strain is given by

l 2= E11cos α + 2E12cos α sin α + E22sin2 α (119)l0

where α is the angle between the x1 axis and the baseline Equation (119) shows how theelongation varies with orientation In the small strain limit we can relate the rate of linelength change to the strain rate by differentiating both sides of equation (119) with respectto time to firstshyorder changes in orientation partαpartt can be neglected The exact relationshipcorresponding to equation (118) valid for finite strain is (1 l)dldt = ddξ middot D middot ddξ (eg Malvern1969) where the rate of deformation tensor is defined as

1 partvi partv jDi j equiv + (120)2 partξ j partξi

where v is the instantaneous particle velocity with components vi The rate of deformationtensor inwhich derivatives are takenwith respect to the current configuration is not generallyequivalent to the time derivative of a strain tensor in which derivatives are taken with respectto the initial configuration The distinction is negligible however if one takes the initialconfiguration to be that at the time of the first geodetic measurement not an unrealizableundeformed state so that generally |partuipartxj | laquo 1

If the rate of elongation is measured on a number of baselines with different orientationsαi it is possible to determine the components of the average strainshyrate tensor by least squaressolution of the following equations

(1li )dlidt cos2 αi 2 cos αi sin αi sin2 αiE11 = E12 (121) E22

where the superimposed dot indicates differentiation with respect to time

6

Copyrighted Material

Chapter 1

200

150

100

50

0

ndash50

ndash100

ndash150

ndash2000

Azimuth(degrees)

Figure 14 Linear extension rate as a function of azimuth for precise distance measurements from theEastern California Shear Zone Data from two networks are shown The Garlock network is located atthe eastern end of the Garlock fault The estimated strain-rate fields for the two networks (shown bythe continuous curves) are very similar From Savage et al (2001)

In figure 14 linear extension rates vary as a function of azimuth in a manner consistentwith a uniform strainshyrate field within rather large errors given the magnitude of the signalThese data were collected in the Southern California Shear Zone using precise distancemeasuring devices Data from the US Geological Survey (USGS) Geodolite program whichmeasured distance repeatedly using an Electronic Distance Meter (EDM) were analyzed byJohnson et al (1994) to determine the average strain rate in southern California Figure 15shows the principal strainshyrate directions and magnitudes for different polygons assumingthat the strain rate is spatially uniformwithin each polygon

Borehole dilatometers measure volume change or dilatational strain in the earthrsquos crust(figure 16) The volume change of an element is related to the extension in the threecoordinate directions Let V0 be the initial volume and V be the final volume and define thedilatation as the change in volume over the initial volume equiv (V minus V0)V0 If xi representthe sides of a small cubical element in the undeformed state then V0 = x1x2x3 Similarly if ξirepresent the sides in the deformed state then V = ξ1ξ2ξ3 The final dimensions are relatedto the undeformed dimensions by ξ1 = x1(1 + E11) ξ2 = x2(1 + E22) and ξ3 = x3(1 + E33)so that

x1x2x3(1 + E11)(1 + E22)(1 + E33) minus x1x2x3 = (122)

x1x2x3

To first order in strain this yields = E11 + E22 + E33 = Eii = trace(E) which can be shownto be independent of the coordinate system In other words the dilatation is an invariantof the strain tensor Note further that Eii = partuipartxi so volumetric strain is equivalent to thedivergence of the displacement field = nabla middot u

Exte

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4 Chapter 1

Thus the Green strain is given by

Ei j = 12

partui

partxj+ partuj

partxi+ partun

partxi

partun

partxj (116)

The corresponding Almansi (Eulerian) strain is given by

1 partui partuj partun partunei j = + minus (117)2 partξ j partξi partξi partξ j

Notice that the third terms in both equations (116) and (117) are products of gradients indisplacement and are thus second order These terms are safely neglected as long as thedisplacement gradients are small compared to unity Furthermore in this limit we do notneed to distinguish between derivatives taken with respect to initial or final coordinates Theinfinitesimal strain is thus given by equation (14)

More generally the Lagrangian formulation tracks a material element or a collectionof particles It is an appropriate choice for elasticity problems where the reference state isnaturally defined as the unstrained configuration The Eulerian formulation refers to theproperties at a given point in space and is naturally used in fluid mechanics where one couldimagine monitoring the velocity density temperature and so on at fixed points in spaceas the flowing fluid moves by In contrast the Lagrangian formulation would track theseproperties as one moved with the flow For example fixed weather stations that measurewind velocity barometric pressure and temperature yield an Eulerian description of theatmosphere Neutrally buoyantweather balloonsmovingwith thewind yield a correspondingLagrangian description

Geodetic observing stations are fixed to the solid earth suggesting that a Lagrangian formushylation is sensible Furthermore constitutive equations which refer to a parcel of material areLagrangian However the familiar Cauchy stress to be defined shortly is inherently Eulerian(Dahlen and Tromp 1998 give a more complete discussion of these issues) These distinctionsturn out not to be important as long as the initial stresses are of the same order as or smallerthan the incremental stresses (and any rotations are not much larger than the strains) Forthe most part in this book we will restrict our attention to infinitesimal strains and will notfind it necessary to distinguish between Lagrangian and Eulerian formulations In the earthhowever the initial stresses at least their isotropic components are likely to be large and amore thorough accounting of the effects of this prestress is required This arises in chapter 9where we will consider the role of gravitation in the equilibrium equations

An important measure of strain is the change in the length of an arbitrarily oriented linesegment In practice linear extension is measured by a variety of extensometers rangingfrom strain gauges in laboratory rock mechanics experiments to field extensometers and laserstrainmeters To the degree that the strain is spatially uniform geodimeters or ElectronicDistanceMeters (EDM)measure linear extension Consider the change in length of a referenceline dx with initial length l0 and final length l In the small strain limit we have

l2 minus l02 = (l + l0)(l minus l0) = 2Ei jdxidxj

2l0(l minus l0) 2Ei jdxidxj

l d d Ei jdxidx j (118)l0

where ddxi are the components of a unit vector in the direction of the baseline dx The changein length per unit length of the baseline dx is given by the scalar product of the strain tensor

5

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Deformation Stress and Conservation Laws

x0+dxdx

du

x0

Figure 13 Extension and rotation of a line segment The extension is the dot product of the relativedisplacement duwith the unit vector in the direction of the line segment dx = dx|dx| The rotation ofthe line segment is the cross product of the relative displacement with the same unit vector

dand the unit vectors dxd This can also be written as dxd middot E middot dx This result has a simplegeometric interpretation The change in length of the baseline is the projection of the relativedisplacement vector du onto the baseline directionl = duiddxi (figure 13) In an irrotationalfield the relative displacement between nearby points is given by dui = Ei jdxj (equation [16])Thusl = Ei jddxidxj Dividing both sides by l0 yields the last equation in equation (118)

Geodimeters which measure distance with great accuracy can be used to determine thechange in distance l between two benchmarks separated by up to several tens of kilometersIf the baseline lies in a horizontal plane dxT = (dx1 dx2 0) and the strain is spatially uniformbetween the two end points the extensional strain is given by

l 2= E11cos α + 2E12cos α sin α + E22sin2 α (119)l0

where α is the angle between the x1 axis and the baseline Equation (119) shows how theelongation varies with orientation In the small strain limit we can relate the rate of linelength change to the strain rate by differentiating both sides of equation (119) with respectto time to firstshyorder changes in orientation partαpartt can be neglected The exact relationshipcorresponding to equation (118) valid for finite strain is (1 l)dldt = ddξ middot D middot ddξ (eg Malvern1969) where the rate of deformation tensor is defined as

1 partvi partv jDi j equiv + (120)2 partξ j partξi

where v is the instantaneous particle velocity with components vi The rate of deformationtensor inwhich derivatives are takenwith respect to the current configuration is not generallyequivalent to the time derivative of a strain tensor in which derivatives are taken with respectto the initial configuration The distinction is negligible however if one takes the initialconfiguration to be that at the time of the first geodetic measurement not an unrealizableundeformed state so that generally |partuipartxj | laquo 1

If the rate of elongation is measured on a number of baselines with different orientationsαi it is possible to determine the components of the average strainshyrate tensor by least squaressolution of the following equations

(1li )dlidt cos2 αi 2 cos αi sin αi sin2 αiE11 = E12 (121) E22

where the superimposed dot indicates differentiation with respect to time

6

Copyrighted Material

Chapter 1

200

150

100

50

0

ndash50

ndash100

ndash150

ndash2000

Azimuth(degrees)

Figure 14 Linear extension rate as a function of azimuth for precise distance measurements from theEastern California Shear Zone Data from two networks are shown The Garlock network is located atthe eastern end of the Garlock fault The estimated strain-rate fields for the two networks (shown bythe continuous curves) are very similar From Savage et al (2001)

In figure 14 linear extension rates vary as a function of azimuth in a manner consistentwith a uniform strainshyrate field within rather large errors given the magnitude of the signalThese data were collected in the Southern California Shear Zone using precise distancemeasuring devices Data from the US Geological Survey (USGS) Geodolite program whichmeasured distance repeatedly using an Electronic Distance Meter (EDM) were analyzed byJohnson et al (1994) to determine the average strain rate in southern California Figure 15shows the principal strainshyrate directions and magnitudes for different polygons assumingthat the strain rate is spatially uniformwithin each polygon

Borehole dilatometers measure volume change or dilatational strain in the earthrsquos crust(figure 16) The volume change of an element is related to the extension in the threecoordinate directions Let V0 be the initial volume and V be the final volume and define thedilatation as the change in volume over the initial volume equiv (V minus V0)V0 If xi representthe sides of a small cubical element in the undeformed state then V0 = x1x2x3 Similarly if ξirepresent the sides in the deformed state then V = ξ1ξ2ξ3 The final dimensions are relatedto the undeformed dimensions by ξ1 = x1(1 + E11) ξ2 = x2(1 + E22) and ξ3 = x3(1 + E33)so that

x1x2x3(1 + E11)(1 + E22)(1 + E33) minus x1x2x3 = (122)

x1x2x3

To first order in strain this yields = E11 + E22 + E33 = Eii = trace(E) which can be shownto be independent of the coordinate system In other words the dilatation is an invariantof the strain tensor Note further that Eii = partuipartxi so volumetric strain is equivalent to thedivergence of the displacement field = nabla middot u

Exte

nsio

nra

te(n

anos

trai

nyr

)

BarstowGarlockBarstowfitGarlockfit

30 60 90 120 150 180

5

Copyrighted Material

Deformation Stress and Conservation Laws

x0+dxdx

du

x0

Figure 13 Extension and rotation of a line segment The extension is the dot product of the relativedisplacement duwith the unit vector in the direction of the line segment dx = dx|dx| The rotation ofthe line segment is the cross product of the relative displacement with the same unit vector

dand the unit vectors dxd This can also be written as dxd middot E middot dx This result has a simplegeometric interpretation The change in length of the baseline is the projection of the relativedisplacement vector du onto the baseline directionl = duiddxi (figure 13) In an irrotationalfield the relative displacement between nearby points is given by dui = Ei jdxj (equation [16])Thusl = Ei jddxidxj Dividing both sides by l0 yields the last equation in equation (118)

Geodimeters which measure distance with great accuracy can be used to determine thechange in distance l between two benchmarks separated by up to several tens of kilometersIf the baseline lies in a horizontal plane dxT = (dx1 dx2 0) and the strain is spatially uniformbetween the two end points the extensional strain is given by

l 2= E11cos α + 2E12cos α sin α + E22sin2 α (119)l0

where α is the angle between the x1 axis and the baseline Equation (119) shows how theelongation varies with orientation In the small strain limit we can relate the rate of linelength change to the strain rate by differentiating both sides of equation (119) with respectto time to firstshyorder changes in orientation partαpartt can be neglected The exact relationshipcorresponding to equation (118) valid for finite strain is (1 l)dldt = ddξ middot D middot ddξ (eg Malvern1969) where the rate of deformation tensor is defined as

1 partvi partv jDi j equiv + (120)2 partξ j partξi

where v is the instantaneous particle velocity with components vi The rate of deformationtensor inwhich derivatives are takenwith respect to the current configuration is not generallyequivalent to the time derivative of a strain tensor in which derivatives are taken with respectto the initial configuration The distinction is negligible however if one takes the initialconfiguration to be that at the time of the first geodetic measurement not an unrealizableundeformed state so that generally |partuipartxj | laquo 1

If the rate of elongation is measured on a number of baselines with different orientationsαi it is possible to determine the components of the average strainshyrate tensor by least squaressolution of the following equations

(1li )dlidt cos2 αi 2 cos αi sin αi sin2 αiE11 = E12 (121) E22

where the superimposed dot indicates differentiation with respect to time

6

Copyrighted Material

Chapter 1

200

150

100

50

0

ndash50

ndash100

ndash150

ndash2000

Azimuth(degrees)

Figure 14 Linear extension rate as a function of azimuth for precise distance measurements from theEastern California Shear Zone Data from two networks are shown The Garlock network is located atthe eastern end of the Garlock fault The estimated strain-rate fields for the two networks (shown bythe continuous curves) are very similar From Savage et al (2001)

In figure 14 linear extension rates vary as a function of azimuth in a manner consistentwith a uniform strainshyrate field within rather large errors given the magnitude of the signalThese data were collected in the Southern California Shear Zone using precise distancemeasuring devices Data from the US Geological Survey (USGS) Geodolite program whichmeasured distance repeatedly using an Electronic Distance Meter (EDM) were analyzed byJohnson et al (1994) to determine the average strain rate in southern California Figure 15shows the principal strainshyrate directions and magnitudes for different polygons assumingthat the strain rate is spatially uniformwithin each polygon

Borehole dilatometers measure volume change or dilatational strain in the earthrsquos crust(figure 16) The volume change of an element is related to the extension in the threecoordinate directions Let V0 be the initial volume and V be the final volume and define thedilatation as the change in volume over the initial volume equiv (V minus V0)V0 If xi representthe sides of a small cubical element in the undeformed state then V0 = x1x2x3 Similarly if ξirepresent the sides in the deformed state then V = ξ1ξ2ξ3 The final dimensions are relatedto the undeformed dimensions by ξ1 = x1(1 + E11) ξ2 = x2(1 + E22) and ξ3 = x3(1 + E33)so that

x1x2x3(1 + E11)(1 + E22)(1 + E33) minus x1x2x3 = (122)

x1x2x3

To first order in strain this yields = E11 + E22 + E33 = Eii = trace(E) which can be shownto be independent of the coordinate system In other words the dilatation is an invariantof the strain tensor Note further that Eii = partuipartxi so volumetric strain is equivalent to thedivergence of the displacement field = nabla middot u

Exte

nsio

nra

te(n

anos

trai

nyr

)

BarstowGarlockBarstowfitGarlockfit

30 60 90 120 150 180

6

Copyrighted Material

Chapter 1

200

150

100

50

0

ndash50

ndash100

ndash150

ndash2000

Azimuth(degrees)

Figure 14 Linear extension rate as a function of azimuth for precise distance measurements from theEastern California Shear Zone Data from two networks are shown The Garlock network is located atthe eastern end of the Garlock fault The estimated strain-rate fields for the two networks (shown bythe continuous curves) are very similar From Savage et al (2001)

In figure 14 linear extension rates vary as a function of azimuth in a manner consistentwith a uniform strainshyrate field within rather large errors given the magnitude of the signalThese data were collected in the Southern California Shear Zone using precise distancemeasuring devices Data from the US Geological Survey (USGS) Geodolite program whichmeasured distance repeatedly using an Electronic Distance Meter (EDM) were analyzed byJohnson et al (1994) to determine the average strain rate in southern California Figure 15shows the principal strainshyrate directions and magnitudes for different polygons assumingthat the strain rate is spatially uniformwithin each polygon

Borehole dilatometers measure volume change or dilatational strain in the earthrsquos crust(figure 16) The volume change of an element is related to the extension in the threecoordinate directions Let V0 be the initial volume and V be the final volume and define thedilatation as the change in volume over the initial volume equiv (V minus V0)V0 If xi representthe sides of a small cubical element in the undeformed state then V0 = x1x2x3 Similarly if ξirepresent the sides in the deformed state then V = ξ1ξ2ξ3 The final dimensions are relatedto the undeformed dimensions by ξ1 = x1(1 + E11) ξ2 = x2(1 + E22) and ξ3 = x3(1 + E33)so that

x1x2x3(1 + E11)(1 + E22)(1 + E33) minus x1x2x3 = (122)

x1x2x3

To first order in strain this yields = E11 + E22 + E33 = Eii = trace(E) which can be shownto be independent of the coordinate system In other words the dilatation is an invariantof the strain tensor Note further that Eii = partuipartxi so volumetric strain is equivalent to thedivergence of the displacement field = nabla middot u

Exte

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nra

te(n

anos

trai

nyr

)

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7

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Deformation Stress and Conservation Laws

344

342

340

338

336

334

332

330

328

326

Nor

thla

titud

e

SanDiego

02micro yrndash1

contraction

Pureshear

Puredilation

BajaCalifornia

B

B

A

A ε

ndash1180 ndash1175 ndash1170 ndash1165 ndash1160 ndash1155 ndash1150Eastlongitude

Figure 15 Strain-rate distribution in southern California determined from the rate of linear extensionmeasured by the US Geological Survey Geodolite project Maximum and minimum extension ratesare average for the different quadrilaterals Scale bar is located in the upper-right corner From Johnsonet al (1994)

We have seen that the three normal strains Ei j i = j represent stretching in the threecoordinate directions The offshydiagonal components Ei j i j or shear strains represent a=change in shape It can be shown that the shear strain is equal to the change in an initiallyright angle In section 12 you will see how horizontal angle measurements determined bygeodetic triangulation can be used tomeasure crustal shear strain

111 Strains in Curvilinear CoordinatesDue to the symmetry of a particular problem it is often convenient to express the strains incylindrical polar or spherical coordinates We present the results without derivation detailsare given in for example Malvern (1969) In cylindrical polar coordinates the strains are

parturEr r =

partr

Eθθ = 1rpartuθpartθ

+ urr

Ezz = partuz

partz

Er θ = 12

1rparturpartθ

+ partuθpartr

minus uθr

Er z = 12

partuz

partr+ partur

partz

Ezθ = 12

partuθpartz

+ 1rpartuz

partθ (123)

8

Copyrighted Material

Chapter 1

Figure 16 Cross section of a Sacks-Evertson borehole dilatometer The lower sensing volume is filledwith a relatively incompressible fluid The upper reservoir is partially filled with a highly compressibleinert gas and is connected to the sensing volume by a thin tube and bellows The bellows is attachedto a displacement transducer that records the flow of fluid into or out of the sensing volume As theinstrument is compressed fluid flows out of the sensing volume The output of the displacementtransducer is calibrated to strain by comparing observed solid earth tides to theoretically predictedtides After Agnew (1985)

In spherical coordinates the strains are given by

parturEr r =

partr1 partuθ ur

Eθθ = + r partθ r

1 partuφ ur cot θEφφ = + + uθ r sin θ partφ r r

9

Copyrighted Material

Deformation Stress and Conservation Laws

Er θ = 12

1rparturpartθ

+ partuθpartr

minus uθr

Erφ = 12

1r sin θ

parturpartφ

+ partuφpartr

minus uφr

Eθφ = 12

1r sin θ

partuθpartφ

+ 1rpartuφpartθ

minus cot φr

uφ (124)

12 RotationThere are only three independent components in the rotation tensor as defined by equashytion (15) so the infinitesimal rotationmay be represented as a vector Q

1Qk equiv minus eki jωi j (125)

2

where ei jk is the permutation tensor ei jk vanishes if any index is repeated is equal to +1for e123 or for any cyclic permutation of the indices and is equal to minus1 for e321 or for anycyclic permutation of the indices Because it is a rankshythree tensor the permutation tensoris easily distinguished from the Almansi strain From the definition of ωi j (15) we can writethe rotation Q as

1 1 partui 1 partujQk = minus eki j + eki j (126)

2 2 partxj 2 partxi

Now minuseki j = ekji so (noting that i and j are dummy indices)

1 1 partui 1 partujQk = ekji + eki j (127)

2 2 partxj 2 partxi

1 partujQk = eki j (128)

2 partxi

1Q = (nabla times u) (129)

2

That is the rotation vector is half the curl of the displacement fieldEquation (125) can be inverted to write ωi j in terms of Qk To do so we make use of the

soshycalled eshyδ identity

ekmneki j = δmiδnj minus δmjδni (130)

Multiplying both sides of equation (125) by emnk

1emnkQk = minus emnkeki jωi j 2

1 = minus δmiδnjωi j minus δmjδniωi j 21= minus (ωmn minus ωnm) = minusωmn (131)2

Copyrighted Material

10 Chapter 1

x3 Ω

dxu

Figure 17 Rotational deformation

We can now see clearly themeaning of the termωi j dxj in the Taylorrsquos series expansion (16) forthe displacements

dui = ωi j dxj = minusei jkQkdxj = eikjQkdxj (132)

which in vector notation is du = Q times dx This is illustrated in figure 17Consider now how an arbitrarily oriented baseline rotates during deformation This could

represent the change in orientation of a tiltmeter or the change in orientation of a geodeticbaseline determined by repeated triangulation surveys We will denote the rotation of abaseline as e and recognize that the rotation of a line segment is different from the rotationalcomponent of the deformation Q Define the rotation of a line segment e as

dx times dξe equiv (133)

dlo2

(Agnew 1985) Note that this definition is reasonable since in the limit that the deformationgradients are small the magnitude of e is

|dl||dlo||e| = sin e sim e (134)dlo2

wheree is the angle between dx and dξ In indicial notation equation (133) is written as

ei jkdxjdξkei = (135)

dxndxn

As before we write the final position as the initial position plus the displacement ξi = xi + ui so

partξk partukdξk = dxm = + δkm dxm (136)partxm partxm

11

Copyrighted Material

Deformation Stress and Conservation Laws

Substituting equation (136) into (135) yields

dxndxnei = ei jkdxjpartuk + δkm dxmpartxm

= ei jkpartuk dxjdxm + ei jkdxjdxkpartxm

= ei jk(Ekm + ωkm)dxjdxm (137)

Note that the term ei jkdxjdxk is the cross product of dx with itself and is therefore zero Wehave also expanded the displacement gradients as the sum of strain and rotation so finally

dei = ei jk(Ekm + ωkm)ddx jdxm (138)

where as before the ddx j are unit vectorsConsider the rotational term in equation (138)

ei jkωkmddx jddxm = minusei jkekmnQnddx jddxm= minus(δimδ jn minus δinδ jm)Qnddx jddxm= Qi minusQ jddx jddxi (139)

where in the first equation we have made use of equation (131) and in the last thatddxmddxm = 1

Thus the general expression for the rotation of a line segment (138) can be written as

ei = ei jkEkmddx jddxm +Qi minusQ jddx jddxi (140)

or in vector form as

d d d (141)e = dx times (E middot dx)ddx) + Q minus (Q middot dx

The strain component of the rotation has a simple interpretation Recall from figure 13that the elongation is equal to the dot product of the relative displacement and a unit vectorparallel to the baseline The rotation is the cross product of the unit vector and the relativedisplacement dui = Ei jddx j

To understand the dependence of e on Q geometrically consider two endshymember caseswhen the strain vanishes In the first case the rotation vector is normal to the baselinedIn this case Q middot dx = 0 so e = Q the rotation of the line segment is simply Q (figure 18)(Notice from the rightshyhand rule for cross products that the sign of the rotation due to strain isconsistent with the sign of Q) If on the other hand the rotation vector is parallel to the linesegment then (Q middot dx)d dx = Q so that ed dx = |Q|d = 0 When the rotation vector is parallel tothe baseline the baseline does not rotate at all (figure 18) For intermediate cases 0 lt e lt Q

Consider the verticalshyaxis rotation of a horizontal line segment dxT = (dx1 dx2 0) In thiscase equation (140) reduces to

e3 = E12(cos2 α minus sin2 α) + (E22 minus E11)cos α sin α +Q3 (142)

Copyrighted Material

12 Chapter 1

Ω

Ω

dx

dx

Θ

Figure 18 Rotation of a line segment In the left figure the rotation vector is perpendicular to thebaseline and e = Q In the right figure the rotation vector is parallel to the baseline and e = 0

x2

x1

α1

α21

α2

Figure 19 Angle α21 between two direction measurements α1 and α2

where α is the angle between the x1 axis and the baseline Note that the rotation e3 isequivalent to the change in orientation dα With geodetic triangulation surveyors were ableto determine the angle between three observation points with a theodolite Triangulationmeasurements were widely made during the late nineteenth and early twentieth centuriesbefore the advent of laser distance measuring devices Surveying at different times it waspossible to determine the change in the angle Denoting the angle formed by two baselinesoriented at α1 and α2 as α21 = α2 minus α1 (figure 19) then

E22 minus E11δα21 = δα2 minus δα1 = E12 (cos 2α2 minus cos 2α1) + (sin 2α2 minus sin 2α1) (143)

2

Note that the rotationQ3 is common to both δα1 and δα2 and thus drops out of the differenceThe change in the angle thus depends only on the strain field Defining engineering shearstrains in terms of the tensor strains as γ1 = (E11 minus E22) and γ2 = 2E12

1 1δα21 = γ2 (cos 2α2 minus cos 2α1) minus γ1 (sin 2α2 minus sin 2α1) (144)

2 2

Notice from figure 110 that γ1 is a northndashsouth compression and an eastndashwest extensionequivalent to rightshylateral shear across planes trending 45 degrees west of north γ2 is rightshylateral shear across eastndashwest planes

Determinationof shear strain using the change in anglesmeasured during a geodetic surveywas first proposed by Frank (1966) Assuming that the strain is spatially uniform over some

13

Copyrighted Material

Deformation Stress and Conservation Laws

γ2

γ1 γ1

γ2

Figure 110 Definition of engineering shear strains

area it is possible to collect a series of angle changes in a vector of observations related to thetwo shear strains

minus 1 1δαi j 2 sin 2αi minus sin 2α j 2 cos 2αi minus cos 2α j γ1 = (145) γ2

Given a system of equations such as this the two shear strains are easily estimated by leastsquares As expected the angle changes are insensitive to areal strain A significant advantageof trilaterationusing geodimeters whichmeasure distance over triangulationwhichmeasuresonly angles is that with trilateration it became possible to determine areal strain as well asshear strain

Savage and Burford (1970) used Frankrsquos method to determine the strain in small quadrilatshyerals that cross the San Andreas fault near the town of Cholame California The triangulationnetwork is shown in figure 111 The quads are designated by a letter (A to X ) from southwest tonortheast across the fault The shear strain estimates with error bars are shown in figure 112The data show a positive γ1 centered over the fault as expected for a locked rightshylateral faultThe γ2 distribution shows negative values on the southwest side of the fault and positive valueson the northeast side of the fault Savage and Burford (1970) showed that these strains werecaused by the 1934 Parkfield earthquake which ruptured the San Andreas just northwest ofthe triangulation network The data are discussed inmore detail by Segall and Du (1993)

13 StressWhile this text focuses on deformation parameters that can be measured by modern geoshyphysical methods (Global Positioning System [GPS] strainmeters Interferometric SyntheticAperture Radar [InSAR] and so on) it is impossible to develop physically consistent modelsof earthquake and volcano processes that do not involve the forces and stresses acting withinthe earth It will be assumed that the reader has been introduced to the concept of stress in

Copyrighted Material

14 Chapter 1

33deg45

0 5 10

X

W

VUT

SR

Q

PO

15km SanAndreas

NL

K M

D

A

B

H

FG J

I

CE

120deg15

Figure 111 Cholame triangulation network

the mechanics of continuous deformation We will distinguish between body forces that actat all points in the earth such as gravity from surface forces Surface forces are those that acteither on actual surfaces within the earth such as a fault or an igneous dike or forces thatone part of the earth exerts on an adjoining part The traction vector T is defined as the limit ofthe surface force per unit area dF acting on a surface element dA with unit normal n as thesize of the area element tends to zero (figure 113) The traction thus depends not only on theforces actingwithin the body but also on the orientation of the surface elements The tractioncomponents acting on three mutually orthogonal surfaces (nine components in all) populatea secondshyrank stress tensor σi j

σ11 σ12 σ13

σ = σ21 σ22 σ23

σ31 σ32 σ33

We will adopt here the convention that the first subscript refers to the direction of the forcewhereas the second subscript refers to the direction of the surface normal (figure 114) Alsowe will adopt the sign convention that a positive stress is one in which the traction vector actsin a positive direction when the outwardshypointing normal to the surface points in a positivedirection and the traction vector points in a negative direction when the unit normal pointsin a negative direction Thus unless otherwise specified tensile stresses are positive

Note that this definition of stressmakes nomention of the undeformed coordinate systemso this stress tensor known as the Cauchy stress is a Eulerian description Lagrangian stress

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

19

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

8

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Chapter 1

Figure 16 Cross section of a Sacks-Evertson borehole dilatometer The lower sensing volume is filledwith a relatively incompressible fluid The upper reservoir is partially filled with a highly compressibleinert gas and is connected to the sensing volume by a thin tube and bellows The bellows is attachedto a displacement transducer that records the flow of fluid into or out of the sensing volume As theinstrument is compressed fluid flows out of the sensing volume The output of the displacementtransducer is calibrated to strain by comparing observed solid earth tides to theoretically predictedtides After Agnew (1985)

In spherical coordinates the strains are given by

parturEr r =

partr1 partuθ ur

Eθθ = + r partθ r

1 partuφ ur cot θEφφ = + + uθ r sin θ partφ r r

9

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Deformation Stress and Conservation Laws

Er θ = 12

1rparturpartθ

+ partuθpartr

minus uθr

Erφ = 12

1r sin θ

parturpartφ

+ partuφpartr

minus uφr

Eθφ = 12

1r sin θ

partuθpartφ

+ 1rpartuφpartθ

minus cot φr

uφ (124)

12 RotationThere are only three independent components in the rotation tensor as defined by equashytion (15) so the infinitesimal rotationmay be represented as a vector Q

1Qk equiv minus eki jωi j (125)

2

where ei jk is the permutation tensor ei jk vanishes if any index is repeated is equal to +1for e123 or for any cyclic permutation of the indices and is equal to minus1 for e321 or for anycyclic permutation of the indices Because it is a rankshythree tensor the permutation tensoris easily distinguished from the Almansi strain From the definition of ωi j (15) we can writethe rotation Q as

1 1 partui 1 partujQk = minus eki j + eki j (126)

2 2 partxj 2 partxi

Now minuseki j = ekji so (noting that i and j are dummy indices)

1 1 partui 1 partujQk = ekji + eki j (127)

2 2 partxj 2 partxi

1 partujQk = eki j (128)

2 partxi

1Q = (nabla times u) (129)

2

That is the rotation vector is half the curl of the displacement fieldEquation (125) can be inverted to write ωi j in terms of Qk To do so we make use of the

soshycalled eshyδ identity

ekmneki j = δmiδnj minus δmjδni (130)

Multiplying both sides of equation (125) by emnk

1emnkQk = minus emnkeki jωi j 2

1 = minus δmiδnjωi j minus δmjδniωi j 21= minus (ωmn minus ωnm) = minusωmn (131)2

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10 Chapter 1

x3 Ω

dxu

Figure 17 Rotational deformation

We can now see clearly themeaning of the termωi j dxj in the Taylorrsquos series expansion (16) forthe displacements

dui = ωi j dxj = minusei jkQkdxj = eikjQkdxj (132)

which in vector notation is du = Q times dx This is illustrated in figure 17Consider now how an arbitrarily oriented baseline rotates during deformation This could

represent the change in orientation of a tiltmeter or the change in orientation of a geodeticbaseline determined by repeated triangulation surveys We will denote the rotation of abaseline as e and recognize that the rotation of a line segment is different from the rotationalcomponent of the deformation Q Define the rotation of a line segment e as

dx times dξe equiv (133)

dlo2

(Agnew 1985) Note that this definition is reasonable since in the limit that the deformationgradients are small the magnitude of e is

|dl||dlo||e| = sin e sim e (134)dlo2

wheree is the angle between dx and dξ In indicial notation equation (133) is written as

ei jkdxjdξkei = (135)

dxndxn

As before we write the final position as the initial position plus the displacement ξi = xi + ui so

partξk partukdξk = dxm = + δkm dxm (136)partxm partxm

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Deformation Stress and Conservation Laws

Substituting equation (136) into (135) yields

dxndxnei = ei jkdxjpartuk + δkm dxmpartxm

= ei jkpartuk dxjdxm + ei jkdxjdxkpartxm

= ei jk(Ekm + ωkm)dxjdxm (137)

Note that the term ei jkdxjdxk is the cross product of dx with itself and is therefore zero Wehave also expanded the displacement gradients as the sum of strain and rotation so finally

dei = ei jk(Ekm + ωkm)ddx jdxm (138)

where as before the ddx j are unit vectorsConsider the rotational term in equation (138)

ei jkωkmddx jddxm = minusei jkekmnQnddx jddxm= minus(δimδ jn minus δinδ jm)Qnddx jddxm= Qi minusQ jddx jddxi (139)

where in the first equation we have made use of equation (131) and in the last thatddxmddxm = 1

Thus the general expression for the rotation of a line segment (138) can be written as

ei = ei jkEkmddx jddxm +Qi minusQ jddx jddxi (140)

or in vector form as

d d d (141)e = dx times (E middot dx)ddx) + Q minus (Q middot dx

The strain component of the rotation has a simple interpretation Recall from figure 13that the elongation is equal to the dot product of the relative displacement and a unit vectorparallel to the baseline The rotation is the cross product of the unit vector and the relativedisplacement dui = Ei jddx j

To understand the dependence of e on Q geometrically consider two endshymember caseswhen the strain vanishes In the first case the rotation vector is normal to the baselinedIn this case Q middot dx = 0 so e = Q the rotation of the line segment is simply Q (figure 18)(Notice from the rightshyhand rule for cross products that the sign of the rotation due to strain isconsistent with the sign of Q) If on the other hand the rotation vector is parallel to the linesegment then (Q middot dx)d dx = Q so that ed dx = |Q|d = 0 When the rotation vector is parallel tothe baseline the baseline does not rotate at all (figure 18) For intermediate cases 0 lt e lt Q

Consider the verticalshyaxis rotation of a horizontal line segment dxT = (dx1 dx2 0) In thiscase equation (140) reduces to

e3 = E12(cos2 α minus sin2 α) + (E22 minus E11)cos α sin α +Q3 (142)

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12 Chapter 1

Ω

Ω

dx

dx

Θ

Figure 18 Rotation of a line segment In the left figure the rotation vector is perpendicular to thebaseline and e = Q In the right figure the rotation vector is parallel to the baseline and e = 0

x2

x1

α1

α21

α2

Figure 19 Angle α21 between two direction measurements α1 and α2

where α is the angle between the x1 axis and the baseline Note that the rotation e3 isequivalent to the change in orientation dα With geodetic triangulation surveyors were ableto determine the angle between three observation points with a theodolite Triangulationmeasurements were widely made during the late nineteenth and early twentieth centuriesbefore the advent of laser distance measuring devices Surveying at different times it waspossible to determine the change in the angle Denoting the angle formed by two baselinesoriented at α1 and α2 as α21 = α2 minus α1 (figure 19) then

E22 minus E11δα21 = δα2 minus δα1 = E12 (cos 2α2 minus cos 2α1) + (sin 2α2 minus sin 2α1) (143)

2

Note that the rotationQ3 is common to both δα1 and δα2 and thus drops out of the differenceThe change in the angle thus depends only on the strain field Defining engineering shearstrains in terms of the tensor strains as γ1 = (E11 minus E22) and γ2 = 2E12

1 1δα21 = γ2 (cos 2α2 minus cos 2α1) minus γ1 (sin 2α2 minus sin 2α1) (144)

2 2

Notice from figure 110 that γ1 is a northndashsouth compression and an eastndashwest extensionequivalent to rightshylateral shear across planes trending 45 degrees west of north γ2 is rightshylateral shear across eastndashwest planes

Determinationof shear strain using the change in anglesmeasured during a geodetic surveywas first proposed by Frank (1966) Assuming that the strain is spatially uniform over some

13

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Deformation Stress and Conservation Laws

γ2

γ1 γ1

γ2

Figure 110 Definition of engineering shear strains

area it is possible to collect a series of angle changes in a vector of observations related to thetwo shear strains

minus 1 1δαi j 2 sin 2αi minus sin 2α j 2 cos 2αi minus cos 2α j γ1 = (145) γ2

Given a system of equations such as this the two shear strains are easily estimated by leastsquares As expected the angle changes are insensitive to areal strain A significant advantageof trilaterationusing geodimeters whichmeasure distance over triangulationwhichmeasuresonly angles is that with trilateration it became possible to determine areal strain as well asshear strain

Savage and Burford (1970) used Frankrsquos method to determine the strain in small quadrilatshyerals that cross the San Andreas fault near the town of Cholame California The triangulationnetwork is shown in figure 111 The quads are designated by a letter (A to X ) from southwest tonortheast across the fault The shear strain estimates with error bars are shown in figure 112The data show a positive γ1 centered over the fault as expected for a locked rightshylateral faultThe γ2 distribution shows negative values on the southwest side of the fault and positive valueson the northeast side of the fault Savage and Burford (1970) showed that these strains werecaused by the 1934 Parkfield earthquake which ruptured the San Andreas just northwest ofthe triangulation network The data are discussed inmore detail by Segall and Du (1993)

13 StressWhile this text focuses on deformation parameters that can be measured by modern geoshyphysical methods (Global Positioning System [GPS] strainmeters Interferometric SyntheticAperture Radar [InSAR] and so on) it is impossible to develop physically consistent modelsof earthquake and volcano processes that do not involve the forces and stresses acting withinthe earth It will be assumed that the reader has been introduced to the concept of stress in

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14 Chapter 1

33deg45

0 5 10

X

W

VUT

SR

Q

PO

15km SanAndreas

NL

K M

D

A

B

H

FG J

I

CE

120deg15

Figure 111 Cholame triangulation network

the mechanics of continuous deformation We will distinguish between body forces that actat all points in the earth such as gravity from surface forces Surface forces are those that acteither on actual surfaces within the earth such as a fault or an igneous dike or forces thatone part of the earth exerts on an adjoining part The traction vector T is defined as the limit ofthe surface force per unit area dF acting on a surface element dA with unit normal n as thesize of the area element tends to zero (figure 113) The traction thus depends not only on theforces actingwithin the body but also on the orientation of the surface elements The tractioncomponents acting on three mutually orthogonal surfaces (nine components in all) populatea secondshyrank stress tensor σi j

σ11 σ12 σ13

σ = σ21 σ22 σ23

σ31 σ32 σ33

We will adopt here the convention that the first subscript refers to the direction of the forcewhereas the second subscript refers to the direction of the surface normal (figure 114) Alsowe will adopt the sign convention that a positive stress is one in which the traction vector actsin a positive direction when the outwardshypointing normal to the surface points in a positivedirection and the traction vector points in a negative direction when the unit normal pointsin a negative direction Thus unless otherwise specified tensile stresses are positive

Note that this definition of stressmakes nomention of the undeformed coordinate systemso this stress tensor known as the Cauchy stress is a Eulerian description Lagrangian stress

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

Copyrighted Material

24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

9

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Deformation Stress and Conservation Laws

Er θ = 12

1rparturpartθ

+ partuθpartr

minus uθr

Erφ = 12

1r sin θ

parturpartφ

+ partuφpartr

minus uφr

Eθφ = 12

1r sin θ

partuθpartφ

+ 1rpartuφpartθ

minus cot φr

uφ (124)

12 RotationThere are only three independent components in the rotation tensor as defined by equashytion (15) so the infinitesimal rotationmay be represented as a vector Q

1Qk equiv minus eki jωi j (125)

2

where ei jk is the permutation tensor ei jk vanishes if any index is repeated is equal to +1for e123 or for any cyclic permutation of the indices and is equal to minus1 for e321 or for anycyclic permutation of the indices Because it is a rankshythree tensor the permutation tensoris easily distinguished from the Almansi strain From the definition of ωi j (15) we can writethe rotation Q as

1 1 partui 1 partujQk = minus eki j + eki j (126)

2 2 partxj 2 partxi

Now minuseki j = ekji so (noting that i and j are dummy indices)

1 1 partui 1 partujQk = ekji + eki j (127)

2 2 partxj 2 partxi

1 partujQk = eki j (128)

2 partxi

1Q = (nabla times u) (129)

2

That is the rotation vector is half the curl of the displacement fieldEquation (125) can be inverted to write ωi j in terms of Qk To do so we make use of the

soshycalled eshyδ identity

ekmneki j = δmiδnj minus δmjδni (130)

Multiplying both sides of equation (125) by emnk

1emnkQk = minus emnkeki jωi j 2

1 = minus δmiδnjωi j minus δmjδniωi j 21= minus (ωmn minus ωnm) = minusωmn (131)2

Copyrighted Material

10 Chapter 1

x3 Ω

dxu

Figure 17 Rotational deformation

We can now see clearly themeaning of the termωi j dxj in the Taylorrsquos series expansion (16) forthe displacements

dui = ωi j dxj = minusei jkQkdxj = eikjQkdxj (132)

which in vector notation is du = Q times dx This is illustrated in figure 17Consider now how an arbitrarily oriented baseline rotates during deformation This could

represent the change in orientation of a tiltmeter or the change in orientation of a geodeticbaseline determined by repeated triangulation surveys We will denote the rotation of abaseline as e and recognize that the rotation of a line segment is different from the rotationalcomponent of the deformation Q Define the rotation of a line segment e as

dx times dξe equiv (133)

dlo2

(Agnew 1985) Note that this definition is reasonable since in the limit that the deformationgradients are small the magnitude of e is

|dl||dlo||e| = sin e sim e (134)dlo2

wheree is the angle between dx and dξ In indicial notation equation (133) is written as

ei jkdxjdξkei = (135)

dxndxn

As before we write the final position as the initial position plus the displacement ξi = xi + ui so

partξk partukdξk = dxm = + δkm dxm (136)partxm partxm

11

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Deformation Stress and Conservation Laws

Substituting equation (136) into (135) yields

dxndxnei = ei jkdxjpartuk + δkm dxmpartxm

= ei jkpartuk dxjdxm + ei jkdxjdxkpartxm

= ei jk(Ekm + ωkm)dxjdxm (137)

Note that the term ei jkdxjdxk is the cross product of dx with itself and is therefore zero Wehave also expanded the displacement gradients as the sum of strain and rotation so finally

dei = ei jk(Ekm + ωkm)ddx jdxm (138)

where as before the ddx j are unit vectorsConsider the rotational term in equation (138)

ei jkωkmddx jddxm = minusei jkekmnQnddx jddxm= minus(δimδ jn minus δinδ jm)Qnddx jddxm= Qi minusQ jddx jddxi (139)

where in the first equation we have made use of equation (131) and in the last thatddxmddxm = 1

Thus the general expression for the rotation of a line segment (138) can be written as

ei = ei jkEkmddx jddxm +Qi minusQ jddx jddxi (140)

or in vector form as

d d d (141)e = dx times (E middot dx)ddx) + Q minus (Q middot dx

The strain component of the rotation has a simple interpretation Recall from figure 13that the elongation is equal to the dot product of the relative displacement and a unit vectorparallel to the baseline The rotation is the cross product of the unit vector and the relativedisplacement dui = Ei jddx j

To understand the dependence of e on Q geometrically consider two endshymember caseswhen the strain vanishes In the first case the rotation vector is normal to the baselinedIn this case Q middot dx = 0 so e = Q the rotation of the line segment is simply Q (figure 18)(Notice from the rightshyhand rule for cross products that the sign of the rotation due to strain isconsistent with the sign of Q) If on the other hand the rotation vector is parallel to the linesegment then (Q middot dx)d dx = Q so that ed dx = |Q|d = 0 When the rotation vector is parallel tothe baseline the baseline does not rotate at all (figure 18) For intermediate cases 0 lt e lt Q

Consider the verticalshyaxis rotation of a horizontal line segment dxT = (dx1 dx2 0) In thiscase equation (140) reduces to

e3 = E12(cos2 α minus sin2 α) + (E22 minus E11)cos α sin α +Q3 (142)

Copyrighted Material

12 Chapter 1

Ω

Ω

dx

dx

Θ

Figure 18 Rotation of a line segment In the left figure the rotation vector is perpendicular to thebaseline and e = Q In the right figure the rotation vector is parallel to the baseline and e = 0

x2

x1

α1

α21

α2

Figure 19 Angle α21 between two direction measurements α1 and α2

where α is the angle between the x1 axis and the baseline Note that the rotation e3 isequivalent to the change in orientation dα With geodetic triangulation surveyors were ableto determine the angle between three observation points with a theodolite Triangulationmeasurements were widely made during the late nineteenth and early twentieth centuriesbefore the advent of laser distance measuring devices Surveying at different times it waspossible to determine the change in the angle Denoting the angle formed by two baselinesoriented at α1 and α2 as α21 = α2 minus α1 (figure 19) then

E22 minus E11δα21 = δα2 minus δα1 = E12 (cos 2α2 minus cos 2α1) + (sin 2α2 minus sin 2α1) (143)

2

Note that the rotationQ3 is common to both δα1 and δα2 and thus drops out of the differenceThe change in the angle thus depends only on the strain field Defining engineering shearstrains in terms of the tensor strains as γ1 = (E11 minus E22) and γ2 = 2E12

1 1δα21 = γ2 (cos 2α2 minus cos 2α1) minus γ1 (sin 2α2 minus sin 2α1) (144)

2 2

Notice from figure 110 that γ1 is a northndashsouth compression and an eastndashwest extensionequivalent to rightshylateral shear across planes trending 45 degrees west of north γ2 is rightshylateral shear across eastndashwest planes

Determinationof shear strain using the change in anglesmeasured during a geodetic surveywas first proposed by Frank (1966) Assuming that the strain is spatially uniform over some

13

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Deformation Stress and Conservation Laws

γ2

γ1 γ1

γ2

Figure 110 Definition of engineering shear strains

area it is possible to collect a series of angle changes in a vector of observations related to thetwo shear strains

minus 1 1δαi j 2 sin 2αi minus sin 2α j 2 cos 2αi minus cos 2α j γ1 = (145) γ2

Given a system of equations such as this the two shear strains are easily estimated by leastsquares As expected the angle changes are insensitive to areal strain A significant advantageof trilaterationusing geodimeters whichmeasure distance over triangulationwhichmeasuresonly angles is that with trilateration it became possible to determine areal strain as well asshear strain

Savage and Burford (1970) used Frankrsquos method to determine the strain in small quadrilatshyerals that cross the San Andreas fault near the town of Cholame California The triangulationnetwork is shown in figure 111 The quads are designated by a letter (A to X ) from southwest tonortheast across the fault The shear strain estimates with error bars are shown in figure 112The data show a positive γ1 centered over the fault as expected for a locked rightshylateral faultThe γ2 distribution shows negative values on the southwest side of the fault and positive valueson the northeast side of the fault Savage and Burford (1970) showed that these strains werecaused by the 1934 Parkfield earthquake which ruptured the San Andreas just northwest ofthe triangulation network The data are discussed inmore detail by Segall and Du (1993)

13 StressWhile this text focuses on deformation parameters that can be measured by modern geoshyphysical methods (Global Positioning System [GPS] strainmeters Interferometric SyntheticAperture Radar [InSAR] and so on) it is impossible to develop physically consistent modelsof earthquake and volcano processes that do not involve the forces and stresses acting withinthe earth It will be assumed that the reader has been introduced to the concept of stress in

Copyrighted Material

14 Chapter 1

33deg45

0 5 10

X

W

VUT

SR

Q

PO

15km SanAndreas

NL

K M

D

A

B

H

FG J

I

CE

120deg15

Figure 111 Cholame triangulation network

the mechanics of continuous deformation We will distinguish between body forces that actat all points in the earth such as gravity from surface forces Surface forces are those that acteither on actual surfaces within the earth such as a fault or an igneous dike or forces thatone part of the earth exerts on an adjoining part The traction vector T is defined as the limit ofthe surface force per unit area dF acting on a surface element dA with unit normal n as thesize of the area element tends to zero (figure 113) The traction thus depends not only on theforces actingwithin the body but also on the orientation of the surface elements The tractioncomponents acting on three mutually orthogonal surfaces (nine components in all) populatea secondshyrank stress tensor σi j

σ11 σ12 σ13

σ = σ21 σ22 σ23

σ31 σ32 σ33

We will adopt here the convention that the first subscript refers to the direction of the forcewhereas the second subscript refers to the direction of the surface normal (figure 114) Alsowe will adopt the sign convention that a positive stress is one in which the traction vector actsin a positive direction when the outwardshypointing normal to the surface points in a positivedirection and the traction vector points in a negative direction when the unit normal pointsin a negative direction Thus unless otherwise specified tensile stresses are positive

Note that this definition of stressmakes nomention of the undeformed coordinate systemso this stress tensor known as the Cauchy stress is a Eulerian description Lagrangian stress

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

Copyrighted Material

16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

10 Chapter 1

x3 Ω

dxu

Figure 17 Rotational deformation

We can now see clearly themeaning of the termωi j dxj in the Taylorrsquos series expansion (16) forthe displacements

dui = ωi j dxj = minusei jkQkdxj = eikjQkdxj (132)

which in vector notation is du = Q times dx This is illustrated in figure 17Consider now how an arbitrarily oriented baseline rotates during deformation This could

represent the change in orientation of a tiltmeter or the change in orientation of a geodeticbaseline determined by repeated triangulation surveys We will denote the rotation of abaseline as e and recognize that the rotation of a line segment is different from the rotationalcomponent of the deformation Q Define the rotation of a line segment e as

dx times dξe equiv (133)

dlo2

(Agnew 1985) Note that this definition is reasonable since in the limit that the deformationgradients are small the magnitude of e is

|dl||dlo||e| = sin e sim e (134)dlo2

wheree is the angle between dx and dξ In indicial notation equation (133) is written as

ei jkdxjdξkei = (135)

dxndxn

As before we write the final position as the initial position plus the displacement ξi = xi + ui so

partξk partukdξk = dxm = + δkm dxm (136)partxm partxm

11

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Deformation Stress and Conservation Laws

Substituting equation (136) into (135) yields

dxndxnei = ei jkdxjpartuk + δkm dxmpartxm

= ei jkpartuk dxjdxm + ei jkdxjdxkpartxm

= ei jk(Ekm + ωkm)dxjdxm (137)

Note that the term ei jkdxjdxk is the cross product of dx with itself and is therefore zero Wehave also expanded the displacement gradients as the sum of strain and rotation so finally

dei = ei jk(Ekm + ωkm)ddx jdxm (138)

where as before the ddx j are unit vectorsConsider the rotational term in equation (138)

ei jkωkmddx jddxm = minusei jkekmnQnddx jddxm= minus(δimδ jn minus δinδ jm)Qnddx jddxm= Qi minusQ jddx jddxi (139)

where in the first equation we have made use of equation (131) and in the last thatddxmddxm = 1

Thus the general expression for the rotation of a line segment (138) can be written as

ei = ei jkEkmddx jddxm +Qi minusQ jddx jddxi (140)

or in vector form as

d d d (141)e = dx times (E middot dx)ddx) + Q minus (Q middot dx

The strain component of the rotation has a simple interpretation Recall from figure 13that the elongation is equal to the dot product of the relative displacement and a unit vectorparallel to the baseline The rotation is the cross product of the unit vector and the relativedisplacement dui = Ei jddx j

To understand the dependence of e on Q geometrically consider two endshymember caseswhen the strain vanishes In the first case the rotation vector is normal to the baselinedIn this case Q middot dx = 0 so e = Q the rotation of the line segment is simply Q (figure 18)(Notice from the rightshyhand rule for cross products that the sign of the rotation due to strain isconsistent with the sign of Q) If on the other hand the rotation vector is parallel to the linesegment then (Q middot dx)d dx = Q so that ed dx = |Q|d = 0 When the rotation vector is parallel tothe baseline the baseline does not rotate at all (figure 18) For intermediate cases 0 lt e lt Q

Consider the verticalshyaxis rotation of a horizontal line segment dxT = (dx1 dx2 0) In thiscase equation (140) reduces to

e3 = E12(cos2 α minus sin2 α) + (E22 minus E11)cos α sin α +Q3 (142)

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12 Chapter 1

Ω

Ω

dx

dx

Θ

Figure 18 Rotation of a line segment In the left figure the rotation vector is perpendicular to thebaseline and e = Q In the right figure the rotation vector is parallel to the baseline and e = 0

x2

x1

α1

α21

α2

Figure 19 Angle α21 between two direction measurements α1 and α2

where α is the angle between the x1 axis and the baseline Note that the rotation e3 isequivalent to the change in orientation dα With geodetic triangulation surveyors were ableto determine the angle between three observation points with a theodolite Triangulationmeasurements were widely made during the late nineteenth and early twentieth centuriesbefore the advent of laser distance measuring devices Surveying at different times it waspossible to determine the change in the angle Denoting the angle formed by two baselinesoriented at α1 and α2 as α21 = α2 minus α1 (figure 19) then

E22 minus E11δα21 = δα2 minus δα1 = E12 (cos 2α2 minus cos 2α1) + (sin 2α2 minus sin 2α1) (143)

2

Note that the rotationQ3 is common to both δα1 and δα2 and thus drops out of the differenceThe change in the angle thus depends only on the strain field Defining engineering shearstrains in terms of the tensor strains as γ1 = (E11 minus E22) and γ2 = 2E12

1 1δα21 = γ2 (cos 2α2 minus cos 2α1) minus γ1 (sin 2α2 minus sin 2α1) (144)

2 2

Notice from figure 110 that γ1 is a northndashsouth compression and an eastndashwest extensionequivalent to rightshylateral shear across planes trending 45 degrees west of north γ2 is rightshylateral shear across eastndashwest planes

Determinationof shear strain using the change in anglesmeasured during a geodetic surveywas first proposed by Frank (1966) Assuming that the strain is spatially uniform over some

13

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Deformation Stress and Conservation Laws

γ2

γ1 γ1

γ2

Figure 110 Definition of engineering shear strains

area it is possible to collect a series of angle changes in a vector of observations related to thetwo shear strains

minus 1 1δαi j 2 sin 2αi minus sin 2α j 2 cos 2αi minus cos 2α j γ1 = (145) γ2

Given a system of equations such as this the two shear strains are easily estimated by leastsquares As expected the angle changes are insensitive to areal strain A significant advantageof trilaterationusing geodimeters whichmeasure distance over triangulationwhichmeasuresonly angles is that with trilateration it became possible to determine areal strain as well asshear strain

Savage and Burford (1970) used Frankrsquos method to determine the strain in small quadrilatshyerals that cross the San Andreas fault near the town of Cholame California The triangulationnetwork is shown in figure 111 The quads are designated by a letter (A to X ) from southwest tonortheast across the fault The shear strain estimates with error bars are shown in figure 112The data show a positive γ1 centered over the fault as expected for a locked rightshylateral faultThe γ2 distribution shows negative values on the southwest side of the fault and positive valueson the northeast side of the fault Savage and Burford (1970) showed that these strains werecaused by the 1934 Parkfield earthquake which ruptured the San Andreas just northwest ofthe triangulation network The data are discussed inmore detail by Segall and Du (1993)

13 StressWhile this text focuses on deformation parameters that can be measured by modern geoshyphysical methods (Global Positioning System [GPS] strainmeters Interferometric SyntheticAperture Radar [InSAR] and so on) it is impossible to develop physically consistent modelsof earthquake and volcano processes that do not involve the forces and stresses acting withinthe earth It will be assumed that the reader has been introduced to the concept of stress in

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14 Chapter 1

33deg45

0 5 10

X

W

VUT

SR

Q

PO

15km SanAndreas

NL

K M

D

A

B

H

FG J

I

CE

120deg15

Figure 111 Cholame triangulation network

the mechanics of continuous deformation We will distinguish between body forces that actat all points in the earth such as gravity from surface forces Surface forces are those that acteither on actual surfaces within the earth such as a fault or an igneous dike or forces thatone part of the earth exerts on an adjoining part The traction vector T is defined as the limit ofthe surface force per unit area dF acting on a surface element dA with unit normal n as thesize of the area element tends to zero (figure 113) The traction thus depends not only on theforces actingwithin the body but also on the orientation of the surface elements The tractioncomponents acting on three mutually orthogonal surfaces (nine components in all) populatea secondshyrank stress tensor σi j

σ11 σ12 σ13

σ = σ21 σ22 σ23

σ31 σ32 σ33

We will adopt here the convention that the first subscript refers to the direction of the forcewhereas the second subscript refers to the direction of the surface normal (figure 114) Alsowe will adopt the sign convention that a positive stress is one in which the traction vector actsin a positive direction when the outwardshypointing normal to the surface points in a positivedirection and the traction vector points in a negative direction when the unit normal pointsin a negative direction Thus unless otherwise specified tensile stresses are positive

Note that this definition of stressmakes nomention of the undeformed coordinate systemso this stress tensor known as the Cauchy stress is a Eulerian description Lagrangian stress

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

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Deformation Stress and Conservation Laws

Substituting equation (136) into (135) yields

dxndxnei = ei jkdxjpartuk + δkm dxmpartxm

= ei jkpartuk dxjdxm + ei jkdxjdxkpartxm

= ei jk(Ekm + ωkm)dxjdxm (137)

Note that the term ei jkdxjdxk is the cross product of dx with itself and is therefore zero Wehave also expanded the displacement gradients as the sum of strain and rotation so finally

dei = ei jk(Ekm + ωkm)ddx jdxm (138)

where as before the ddx j are unit vectorsConsider the rotational term in equation (138)

ei jkωkmddx jddxm = minusei jkekmnQnddx jddxm= minus(δimδ jn minus δinδ jm)Qnddx jddxm= Qi minusQ jddx jddxi (139)

where in the first equation we have made use of equation (131) and in the last thatddxmddxm = 1

Thus the general expression for the rotation of a line segment (138) can be written as

ei = ei jkEkmddx jddxm +Qi minusQ jddx jddxi (140)

or in vector form as

d d d (141)e = dx times (E middot dx)ddx) + Q minus (Q middot dx

The strain component of the rotation has a simple interpretation Recall from figure 13that the elongation is equal to the dot product of the relative displacement and a unit vectorparallel to the baseline The rotation is the cross product of the unit vector and the relativedisplacement dui = Ei jddx j

To understand the dependence of e on Q geometrically consider two endshymember caseswhen the strain vanishes In the first case the rotation vector is normal to the baselinedIn this case Q middot dx = 0 so e = Q the rotation of the line segment is simply Q (figure 18)(Notice from the rightshyhand rule for cross products that the sign of the rotation due to strain isconsistent with the sign of Q) If on the other hand the rotation vector is parallel to the linesegment then (Q middot dx)d dx = Q so that ed dx = |Q|d = 0 When the rotation vector is parallel tothe baseline the baseline does not rotate at all (figure 18) For intermediate cases 0 lt e lt Q

Consider the verticalshyaxis rotation of a horizontal line segment dxT = (dx1 dx2 0) In thiscase equation (140) reduces to

e3 = E12(cos2 α minus sin2 α) + (E22 minus E11)cos α sin α +Q3 (142)

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12 Chapter 1

Ω

Ω

dx

dx

Θ

Figure 18 Rotation of a line segment In the left figure the rotation vector is perpendicular to thebaseline and e = Q In the right figure the rotation vector is parallel to the baseline and e = 0

x2

x1

α1

α21

α2

Figure 19 Angle α21 between two direction measurements α1 and α2

where α is the angle between the x1 axis and the baseline Note that the rotation e3 isequivalent to the change in orientation dα With geodetic triangulation surveyors were ableto determine the angle between three observation points with a theodolite Triangulationmeasurements were widely made during the late nineteenth and early twentieth centuriesbefore the advent of laser distance measuring devices Surveying at different times it waspossible to determine the change in the angle Denoting the angle formed by two baselinesoriented at α1 and α2 as α21 = α2 minus α1 (figure 19) then

E22 minus E11δα21 = δα2 minus δα1 = E12 (cos 2α2 minus cos 2α1) + (sin 2α2 minus sin 2α1) (143)

2

Note that the rotationQ3 is common to both δα1 and δα2 and thus drops out of the differenceThe change in the angle thus depends only on the strain field Defining engineering shearstrains in terms of the tensor strains as γ1 = (E11 minus E22) and γ2 = 2E12

1 1δα21 = γ2 (cos 2α2 minus cos 2α1) minus γ1 (sin 2α2 minus sin 2α1) (144)

2 2

Notice from figure 110 that γ1 is a northndashsouth compression and an eastndashwest extensionequivalent to rightshylateral shear across planes trending 45 degrees west of north γ2 is rightshylateral shear across eastndashwest planes

Determinationof shear strain using the change in anglesmeasured during a geodetic surveywas first proposed by Frank (1966) Assuming that the strain is spatially uniform over some

13

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Deformation Stress and Conservation Laws

γ2

γ1 γ1

γ2

Figure 110 Definition of engineering shear strains

area it is possible to collect a series of angle changes in a vector of observations related to thetwo shear strains

minus 1 1δαi j 2 sin 2αi minus sin 2α j 2 cos 2αi minus cos 2α j γ1 = (145) γ2

Given a system of equations such as this the two shear strains are easily estimated by leastsquares As expected the angle changes are insensitive to areal strain A significant advantageof trilaterationusing geodimeters whichmeasure distance over triangulationwhichmeasuresonly angles is that with trilateration it became possible to determine areal strain as well asshear strain

Savage and Burford (1970) used Frankrsquos method to determine the strain in small quadrilatshyerals that cross the San Andreas fault near the town of Cholame California The triangulationnetwork is shown in figure 111 The quads are designated by a letter (A to X ) from southwest tonortheast across the fault The shear strain estimates with error bars are shown in figure 112The data show a positive γ1 centered over the fault as expected for a locked rightshylateral faultThe γ2 distribution shows negative values on the southwest side of the fault and positive valueson the northeast side of the fault Savage and Burford (1970) showed that these strains werecaused by the 1934 Parkfield earthquake which ruptured the San Andreas just northwest ofthe triangulation network The data are discussed inmore detail by Segall and Du (1993)

13 StressWhile this text focuses on deformation parameters that can be measured by modern geoshyphysical methods (Global Positioning System [GPS] strainmeters Interferometric SyntheticAperture Radar [InSAR] and so on) it is impossible to develop physically consistent modelsof earthquake and volcano processes that do not involve the forces and stresses acting withinthe earth It will be assumed that the reader has been introduced to the concept of stress in

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14 Chapter 1

33deg45

0 5 10

X

W

VUT

SR

Q

PO

15km SanAndreas

NL

K M

D

A

B

H

FG J

I

CE

120deg15

Figure 111 Cholame triangulation network

the mechanics of continuous deformation We will distinguish between body forces that actat all points in the earth such as gravity from surface forces Surface forces are those that acteither on actual surfaces within the earth such as a fault or an igneous dike or forces thatone part of the earth exerts on an adjoining part The traction vector T is defined as the limit ofthe surface force per unit area dF acting on a surface element dA with unit normal n as thesize of the area element tends to zero (figure 113) The traction thus depends not only on theforces actingwithin the body but also on the orientation of the surface elements The tractioncomponents acting on three mutually orthogonal surfaces (nine components in all) populatea secondshyrank stress tensor σi j

σ11 σ12 σ13

σ = σ21 σ22 σ23

σ31 σ32 σ33

We will adopt here the convention that the first subscript refers to the direction of the forcewhereas the second subscript refers to the direction of the surface normal (figure 114) Alsowe will adopt the sign convention that a positive stress is one in which the traction vector actsin a positive direction when the outwardshypointing normal to the surface points in a positivedirection and the traction vector points in a negative direction when the unit normal pointsin a negative direction Thus unless otherwise specified tensile stresses are positive

Note that this definition of stressmakes nomention of the undeformed coordinate systemso this stress tensor known as the Cauchy stress is a Eulerian description Lagrangian stress

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

12 Chapter 1

Ω

Ω

dx

dx

Θ

Figure 18 Rotation of a line segment In the left figure the rotation vector is perpendicular to thebaseline and e = Q In the right figure the rotation vector is parallel to the baseline and e = 0

x2

x1

α1

α21

α2

Figure 19 Angle α21 between two direction measurements α1 and α2

where α is the angle between the x1 axis and the baseline Note that the rotation e3 isequivalent to the change in orientation dα With geodetic triangulation surveyors were ableto determine the angle between three observation points with a theodolite Triangulationmeasurements were widely made during the late nineteenth and early twentieth centuriesbefore the advent of laser distance measuring devices Surveying at different times it waspossible to determine the change in the angle Denoting the angle formed by two baselinesoriented at α1 and α2 as α21 = α2 minus α1 (figure 19) then

E22 minus E11δα21 = δα2 minus δα1 = E12 (cos 2α2 minus cos 2α1) + (sin 2α2 minus sin 2α1) (143)

2

Note that the rotationQ3 is common to both δα1 and δα2 and thus drops out of the differenceThe change in the angle thus depends only on the strain field Defining engineering shearstrains in terms of the tensor strains as γ1 = (E11 minus E22) and γ2 = 2E12

1 1δα21 = γ2 (cos 2α2 minus cos 2α1) minus γ1 (sin 2α2 minus sin 2α1) (144)

2 2

Notice from figure 110 that γ1 is a northndashsouth compression and an eastndashwest extensionequivalent to rightshylateral shear across planes trending 45 degrees west of north γ2 is rightshylateral shear across eastndashwest planes

Determinationof shear strain using the change in anglesmeasured during a geodetic surveywas first proposed by Frank (1966) Assuming that the strain is spatially uniform over some

13

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Deformation Stress and Conservation Laws

γ2

γ1 γ1

γ2

Figure 110 Definition of engineering shear strains

area it is possible to collect a series of angle changes in a vector of observations related to thetwo shear strains

minus 1 1δαi j 2 sin 2αi minus sin 2α j 2 cos 2αi minus cos 2α j γ1 = (145) γ2

Given a system of equations such as this the two shear strains are easily estimated by leastsquares As expected the angle changes are insensitive to areal strain A significant advantageof trilaterationusing geodimeters whichmeasure distance over triangulationwhichmeasuresonly angles is that with trilateration it became possible to determine areal strain as well asshear strain

Savage and Burford (1970) used Frankrsquos method to determine the strain in small quadrilatshyerals that cross the San Andreas fault near the town of Cholame California The triangulationnetwork is shown in figure 111 The quads are designated by a letter (A to X ) from southwest tonortheast across the fault The shear strain estimates with error bars are shown in figure 112The data show a positive γ1 centered over the fault as expected for a locked rightshylateral faultThe γ2 distribution shows negative values on the southwest side of the fault and positive valueson the northeast side of the fault Savage and Burford (1970) showed that these strains werecaused by the 1934 Parkfield earthquake which ruptured the San Andreas just northwest ofthe triangulation network The data are discussed inmore detail by Segall and Du (1993)

13 StressWhile this text focuses on deformation parameters that can be measured by modern geoshyphysical methods (Global Positioning System [GPS] strainmeters Interferometric SyntheticAperture Radar [InSAR] and so on) it is impossible to develop physically consistent modelsof earthquake and volcano processes that do not involve the forces and stresses acting withinthe earth It will be assumed that the reader has been introduced to the concept of stress in

Copyrighted Material

14 Chapter 1

33deg45

0 5 10

X

W

VUT

SR

Q

PO

15km SanAndreas

NL

K M

D

A

B

H

FG J

I

CE

120deg15

Figure 111 Cholame triangulation network

the mechanics of continuous deformation We will distinguish between body forces that actat all points in the earth such as gravity from surface forces Surface forces are those that acteither on actual surfaces within the earth such as a fault or an igneous dike or forces thatone part of the earth exerts on an adjoining part The traction vector T is defined as the limit ofthe surface force per unit area dF acting on a surface element dA with unit normal n as thesize of the area element tends to zero (figure 113) The traction thus depends not only on theforces actingwithin the body but also on the orientation of the surface elements The tractioncomponents acting on three mutually orthogonal surfaces (nine components in all) populatea secondshyrank stress tensor σi j

σ11 σ12 σ13

σ = σ21 σ22 σ23

σ31 σ32 σ33

We will adopt here the convention that the first subscript refers to the direction of the forcewhereas the second subscript refers to the direction of the surface normal (figure 114) Alsowe will adopt the sign convention that a positive stress is one in which the traction vector actsin a positive direction when the outwardshypointing normal to the surface points in a positivedirection and the traction vector points in a negative direction when the unit normal pointsin a negative direction Thus unless otherwise specified tensile stresses are positive

Note that this definition of stressmakes nomention of the undeformed coordinate systemso this stress tensor known as the Cauchy stress is a Eulerian description Lagrangian stress

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

Copyrighted Material

18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

19

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

13

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Deformation Stress and Conservation Laws

γ2

γ1 γ1

γ2

Figure 110 Definition of engineering shear strains

area it is possible to collect a series of angle changes in a vector of observations related to thetwo shear strains

minus 1 1δαi j 2 sin 2αi minus sin 2α j 2 cos 2αi minus cos 2α j γ1 = (145) γ2

Given a system of equations such as this the two shear strains are easily estimated by leastsquares As expected the angle changes are insensitive to areal strain A significant advantageof trilaterationusing geodimeters whichmeasure distance over triangulationwhichmeasuresonly angles is that with trilateration it became possible to determine areal strain as well asshear strain

Savage and Burford (1970) used Frankrsquos method to determine the strain in small quadrilatshyerals that cross the San Andreas fault near the town of Cholame California The triangulationnetwork is shown in figure 111 The quads are designated by a letter (A to X ) from southwest tonortheast across the fault The shear strain estimates with error bars are shown in figure 112The data show a positive γ1 centered over the fault as expected for a locked rightshylateral faultThe γ2 distribution shows negative values on the southwest side of the fault and positive valueson the northeast side of the fault Savage and Burford (1970) showed that these strains werecaused by the 1934 Parkfield earthquake which ruptured the San Andreas just northwest ofthe triangulation network The data are discussed inmore detail by Segall and Du (1993)

13 StressWhile this text focuses on deformation parameters that can be measured by modern geoshyphysical methods (Global Positioning System [GPS] strainmeters Interferometric SyntheticAperture Radar [InSAR] and so on) it is impossible to develop physically consistent modelsof earthquake and volcano processes that do not involve the forces and stresses acting withinthe earth It will be assumed that the reader has been introduced to the concept of stress in

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14 Chapter 1

33deg45

0 5 10

X

W

VUT

SR

Q

PO

15km SanAndreas

NL

K M

D

A

B

H

FG J

I

CE

120deg15

Figure 111 Cholame triangulation network

the mechanics of continuous deformation We will distinguish between body forces that actat all points in the earth such as gravity from surface forces Surface forces are those that acteither on actual surfaces within the earth such as a fault or an igneous dike or forces thatone part of the earth exerts on an adjoining part The traction vector T is defined as the limit ofthe surface force per unit area dF acting on a surface element dA with unit normal n as thesize of the area element tends to zero (figure 113) The traction thus depends not only on theforces actingwithin the body but also on the orientation of the surface elements The tractioncomponents acting on three mutually orthogonal surfaces (nine components in all) populatea secondshyrank stress tensor σi j

σ11 σ12 σ13

σ = σ21 σ22 σ23

σ31 σ32 σ33

We will adopt here the convention that the first subscript refers to the direction of the forcewhereas the second subscript refers to the direction of the surface normal (figure 114) Alsowe will adopt the sign convention that a positive stress is one in which the traction vector actsin a positive direction when the outwardshypointing normal to the surface points in a positivedirection and the traction vector points in a negative direction when the unit normal pointsin a negative direction Thus unless otherwise specified tensile stresses are positive

Note that this definition of stressmakes nomention of the undeformed coordinate systemso this stress tensor known as the Cauchy stress is a Eulerian description Lagrangian stress

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

14 Chapter 1

33deg45

0 5 10

X

W

VUT

SR

Q

PO

15km SanAndreas

NL

K M

D

A

B

H

FG J

I

CE

120deg15

Figure 111 Cholame triangulation network

the mechanics of continuous deformation We will distinguish between body forces that actat all points in the earth such as gravity from surface forces Surface forces are those that acteither on actual surfaces within the earth such as a fault or an igneous dike or forces thatone part of the earth exerts on an adjoining part The traction vector T is defined as the limit ofthe surface force per unit area dF acting on a surface element dA with unit normal n as thesize of the area element tends to zero (figure 113) The traction thus depends not only on theforces actingwithin the body but also on the orientation of the surface elements The tractioncomponents acting on three mutually orthogonal surfaces (nine components in all) populatea secondshyrank stress tensor σi j

σ11 σ12 σ13

σ = σ21 σ22 σ23

σ31 σ32 σ33

We will adopt here the convention that the first subscript refers to the direction of the forcewhereas the second subscript refers to the direction of the surface normal (figure 114) Alsowe will adopt the sign convention that a positive stress is one in which the traction vector actsin a positive direction when the outwardshypointing normal to the surface points in a positivedirection and the traction vector points in a negative direction when the unit normal pointsin a negative direction Thus unless otherwise specified tensile stresses are positive

Note that this definition of stressmakes nomention of the undeformed coordinate systemso this stress tensor known as the Cauchy stress is a Eulerian description Lagrangian stress

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

19

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

Copyrighted Material

22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

Copyrighted Material

26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

15

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Deformation Stress and Conservation Laws

40

30

20

10

0

ndash10

ndash2020

10

0

ndash10

ndash20

ndash30

ndash40

DistancefromSanAndreas(km)

Figure 112 Shear strain changes in the Cholame network between 1932 and 1962 After Savage andBurford (1970) The positive values of γ1 centered on the fault at quadrilateral S result from the 1934Parkfield earthquake northwest of the triangulation network Slip in this earthquake also caused fault-parallel compression on the east side of the network (positive γ2) and fault-parallel extension on thewest side (negative γ2)

γ 2(micros

trai

n)

γ 1(micros

trai

n)

1932ndash1962

A B CDE F GHI J K LMN O P Q R S T U V W X

ndash60 ndash40 ndash20 0 20

dA

n

dF

Figure 113 Definition of the traction vector Force dF acts on surface dA with unit normal n Thetraction vector is defined as the limit as dA tends to zero of the ratio dFdA

tensors known as Piola-Kirchoff stresses can be defined where the surface element and unitnormal are measured relative to the undeformed state (eg Malvern 1969) It is shown inall continuum mechanics texts and thus will not be repeated here that in the absenceof distributed couples acting on surface elements or within the body of the material thebalance of moments requires that the Cauchy stress tensor be symmetric σi j = σ j i This isnot necessarily true for other stress tensorsmdashin particular the first PiolashyKirchoff stress is notsymmetric

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

19

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

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16 Chapter 1

x2

x1

σ11

σ21

σ31

σ22

σ33

σ23

σ13

x3

Figure 114 Stresses acting on the faces of a cubical element All of the stress components shownare positivemdashthat is the traction vector and outward-pointing surface normal both point in positivecoordinate directions

It is also shown in standard texts by considering a tetrahedron with three surfaces normalto the coordinate axes and one normal to the unit vector n that equilibrium requires that

(146)Ti = σi j nj T = σ middot n

This is known as Cauchyrsquos formula it states that the traction vector T acting across a surface isfound by the dot product of the stress and the unit normal

The mean normal stress is equal to minus the pressure σkk3 = minus p the negative signarising because of our sign convention The mean stress is a stress invariantmdashthat is σkk3 isindependent of the coordinate system It is often useful to subtract the mean stress to isolatethe soshycalled deviatoric stress

τi j = σi j minus σkkδi j (147)

3

There are in fact three stress invariants The second invariant is given by II = 12 (σi jσi j minusσiiσ j j )

The second invariant of the deviatoric stress often written as J2 = 21 (τi jτi j ) measures the

squaredmaximum shear stress It can be shown that

2 2 2J2 = 1 (σ11 minus σ22)2 + (σ22 minus σ33)2 + (σ11 minus σ33)2

+ σ12 + σ23 + σ13 (148)6

The third invariant is given by the determinant of the stress tensor although it is not widelyused

14 Coordinate TransformationsIt is often necessary to transform stress or strain tensors from one coordinate system toanothermdashfor example between a local North East Up system and a coordinate system thatis oriented parallel and perpendicular to a major fault To begin we will consider the familiarrotation of a vector to a new coordinate system For example a rotation about the x3 axis from

jj

j j

j j

17

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

jj

j j

j j

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Deformation Stress and Conservation Laws

x2 θ

θ

x2

x1

x1

Figure 115 Coordinate rotation about the x3 axis

the original (x1 x2) system to the new (x1j x2

j ) system (figure 115) can be written in matrixform as

vj cos θ sin θ 0 v11 (149)minussin θ cos θ 0=v v22

0 0 1v v33

or compactly as [v]j = A[v] where [v]j and [v] represent the vector components with respect tothe primed and unprimed coordinates respectively In indicial notation

vij = ai jv j (150)

The elements ai j are referred to as direction cosines It is easy to verify that the inverse transforshymation is given by [v] = AT[v]j

vi = ajivjj (151)

The matrix rotating the components of the vector from the primed to the unprimedcoordinates is the inverse of the matrix in equation (149) Clearly ai j = aT

ji but also comparshyminus1 minus1ing equations (150) and (151) ai j = aji Therefore aT

ji = aji so that the transformation isorthogonal

To determine the ai j let ei represent basis vectors in the original coordinate system and eji

represent basis vectors in the new coordinates One can express a vector in either system

x = xje j = xjje

jj (152)

Taking the dot product of both sides with respect to ei we have

xj (e j middot ei ) = xj (e j middot ei )xjδi j = xj (e j middot ei )

xi = xjj (e

jj middot ei ) (153)

so that aji = (ejj middot ei ) A similar derivation gives ai j = (e j middot eij ) For example for a rotation about

the x3 axis we have

a11 = e1 middot e1j = cos(θ)

a12 = e2 middot ej1 = sin(θ)

a21 = e1 middot e2j = minus sin(θ)

a22 = e2 middot e2j = cos(θ) (154)

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

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18 Chapter 1

We can now show that secondshyrank tensors transform by

Ti jj = aikajlTkl (155)

Ti j = akial j T j (156)kl

To do so consider the secondshyrank tensor Tkl that relates two vectors u and v

vk = Tklul (157)

For example u could be pressure gradient v could be fluid flux and T would be the pershymeability tensor The components of the two vectors transform according to

vji = aikvk (158)

and

ul = ajlujj (159)

Substituting equations (158) and (159) into (157) we have

jvi = aikvk = aikTklulj jvi = aikTklajlu j

j jv = aikajlTklu j (160)i

j = T jThis can be written as vi i j ujj if we define T in the rotated coordinate system as

T j (161)i j equiv aikajlTkl

or inmatrix notation as

[T]j = A[T]AT (162)

where [T j] refers to the matrix of the tensor with components taken relative to the primedcoordinates whereas [T] are the components with respect to the unprimed coordinates Notethat all secondshyrank tensors including stress transform according to the same rules Thederivation can be extended to show that higher rank tensors transform according to

j = aiαajβakγ αβγ (163)i jkl

15 Principal Strains and StressesThe strain tensor can be put into diagonal form by rotating into proper coordinates in whichthe shear strains vanish

E1 0 0 0 E2 0 (164)

0 0 E3

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

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22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

19

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Deformation Stress and Conservation Laws

The strain tensor itself has six independent components In order to determine the strain wemust specify not only the three principal values E(k) k = 1 2 3 but also the three principalstrain directions so that it still takes six parameters to describe the state of strain The principaldirections are mutually orthogonal (see the following) Thus while it requires two angles tospecify the first axis the second axis must lie in the perpendicular plane requiring only oneadditional angle The third axis is uniquely determined by orthogonality and the rightshyhandsign convention

The principal strains are also the extrema in the linear extension l l0 For simplicity weshow this for the case of irrotational deformation ω = 0 however see problem 5 In the smallstrain limit the linear extension is proportional to the dot product of the relative displacementvector and a unit vector parallel to the baseline

l duidxi du middot dx= = (165)l0 dxkdxk dx middot dx

(equation [118]) The maximum and minimum extensions occur when du and dx areparallelmdashthat is du = Edx where E is a constant or

dui = Edxi = Eδi jdx j (166)

From equation (16) the relative displacement of adjacent points in the absence of rotation isdui = Ei jdx j Equating with (166) yields

Ei jdxj = Eδi jdx j (167)

or rearranging

Ei j minus Eδi j dx j = 0 (168)

Thus the principal strains E in equation (168) are the eigenvalues of the matrix Ei j and thedxj are the eigenvectors or principal strain directions Similarly the principal stresses and theprincipal stress directions are the eigenvalues and eigenvectors of thematrix σi j

To determine the principal strains and strain directions we must find the eigenvalues andthe associated eigenvectors Note that the equations (168) have nontrivial solutions if andonly if

det Ei j minus Eδi j = 0 (169)

In general this leads to a cubic equation in E The strain tensor is symmetric whichmeans thatall three eigenvalues are real (eg Strang 1976) They may however be either positive negashytive or zero some eigenvaluesmaybe repeated For thedistinct (nonrepeated) eigenvalues thecorresponding eigenvectors are mutually orthogonal With repeated eigenvalues there existssufficient arbitrariness in choosing corresponding eigenvectors that they can be chosen to beorthogonal

For simplicity in presentation wewill examine the twoshydimensional case which reduces to

E11 minus E E12det = 0 (170)E21 E22 minus E

which has the following two roots given by solution of the preceding quadratic

E11 + E22 E11 minus E222

+ E2E = plusmn 12 (171)2 2

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20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

Copyrighted Material

22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

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24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

20 Chapter 1

Figure 116 Left Two-dimensional strain state given by equation (172) where the arrows denoterelative displacement of the two sides of the reference square Right Same strain state in the principalcoordinate system

The principal strain directions are found by substituting the principal values back intoequation (168) and solving for the dx(k) As an example consider the simple twoshydimensionalstrain tensor illustrated in figure 116

1 1Ei j = (172)

1 1

From equation (171) we find that E(1) = 2 and E(2) = 0 Now to find the principal straindirection corresponding to themaximumprincipal strain we substitute E(1) = 2 into equation(168) yielding

1 minus 2 1 dx1 = 0 (173)1 1 minus 2 dx2

These are two redundant equations that relate dx1 and dx2

minusdx1 + dx2 = 0 (174)

dx1 minus dx2 = 0 (175)

The eigenvectors are usually normalized so that radic1 minusradic1

dx(1) = 2 or 2 (176)

radic1 minusradic12 2

For this particular example the principal direction corresponding to themaximum extensionbisects the x1 and x2 axes in the first and third quadrants (figure 116)We know that the secondprincipal direction has to be orthogonal to the first but we could find it also by the sameprocedure Substituting E(2) = 0 into equation (168) yields

1 minus 0 1 dx1 = 0 (177)1 1 minus 0 dx2

or dx1 = minusdx2 Again we normalize the eigenvector so that 1 minus 12 2dx(2) =

radicor

radic (178)minusradic1 radic1

2 2

showing that the second eigenvector corresponding to the minimum extension is also45 degrees from the coordinate axes but in the second and fourth quadrants

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

Copyrighted Material

22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

Copyrighted Material

24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

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Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

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28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

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Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

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30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

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Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

21

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Deformation Stress and Conservation Laws

16 Compatibility EquationsRecall the kinematic equations relating strain to displacement (14) Given the strains theserepresent six equations in only three unknowns the three displacement components Thisimplies that there must be relationships between the strains These relations known ascompatibility equations guarantee that the displacement field be single valued To be moreconcrete this means that the displacement field does not admit gaps opening or materialinterpenetrating

In two dimensions we have

Exx = partux

partx (179)

Eyy = partuy

party (180)

1 partux partuyExy = + (181)

2 party partx

Notice that

part2Exy part3ux part3uy part2Exx part2Eyy2 = + = + (182)partxparty party2partx partx2party party2 partx2

An example of a strain field that is not compatible is as follows Assume that the onlynonzero strain is Exx = Ay2 where A is a constant Note that such a strain could not occurwithout a shear strain nor in fact without gradients in shear strain This follows fromequation (182)

In three dimensions the compatibility equations can be written compactly as

(183)Ei jkl + Ekli j minus Eik jl minus E jlik = 0

where we have used the notation i equiv partpartxi The symmetry of the strain tensor can be usedto reduce the 81 equations (183) to 6 independent equations Furthermore there are threerelations among the incompatibility elements so there are only three truly independent comshypatibility relations The interested reader is referred toMalvern (1969)

When do you have to worry about compatibility If you solve the governing equationsin terms of stress then you must guarantee that the implied strain fields are compatibleIf on the other hand the governing equations are written in terms of displacementsthen compatibility is automatically satisfiedmdashassuming that the displacements are contishynuous (true except perhaps along a fault or crack) one can always derive the strains bydifferentiation

17 Conservation LawsThe behavior of a deforming body is governed by conservation laws that generalize theconcepts of conservation of mass and momentum Before detailing these we will examinesome important relations that will be used in subsequent derivations

The divergence theorem relates the integral of the outward component of a vector f over aclosed surface S to the volume integral of the divergence of the vector over the volume V

Copyrighted Material

22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

Copyrighted Material

24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

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26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

Copyrighted Material

Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

Copyrighted Material

28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

22 Chapter 1

enclosed by S

part fifini dS = dV (184)

S V partxi f middot n dS = nabla middot f dV (185)

S V

The time rate of change of a quantity f is different in a fixed Eulerian frame from a movingLagrangian frame Any physical quantity must be the same regardless of the system Thisequivalence can be expressed as

f L(x t) = f E [ξ (x t) t] (186)

where the superscripts L and E indicate Lagrangian and Eulerian descriptions respectivelyHere f can be a scalar vector or tensor Recall also that x refers to initial coordinates whereasξ refers to current coordinates Differentiating with respect to time and using the chain ruleyields

d f L part f E part f E partξi= + (187)dt partt partξi partt

The first term on the rightshyhand side refers to the time rate of change of f at a fixed point inspace whereas the second term refers to the convective rate of change as particles move to aplace having different properties The Lagrangian derivative is often written where using anEulerian description as DfDt and is also known as thematerial time derivative

Df part f E= + v middot nabla f E (188)Dt partt

A linearized timeshyintegrated form of equation (188) valid for small displacements

δ f L = δ f E + u middot nabla f E (189)

relates the Lagrangian description of a perturbation in f to the Eulerian description (Dahlenand Tromp 1998 equation 316) The spatial derivative in equation (189) is with respect to thecurrent coordinates although this distinction can be dropped in a linearized analysis

We will also state here without proof the Reynolds transport theorem which givesthe material time derivative of a volume integralmdashthat is following a set of points involume Vmdashas the region moves and distorts The theorem states (eg Malvern 1969) that fora propertyA

D DAρAdV = ρ dV (190)

Dt V V Dt

where ρ is the Eulerianmass densityWe are now ready to state the important conservation laws of continuummechanics Mass

must be conserved during deformation The time rate of change of mass in a current element with volume V V(partρpartt)dV must be balanced by the flow of material out of the element

23

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Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

Copyrighted Material

24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

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Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

Copyrighted Material

26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

Copyrighted Material

Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

Copyrighted Material

28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

23

Copyrighted Material

Deformation Stress and Conservation Laws

S ρv middot ndS where S is the surface bounding V Making use of the divergence theorem (185)

partρdV + ρv middot ndS = 0

parttV S partρ + nabla middot ρv dV = 0 (191)parttV

This must hold for every choice of volume element which yields the continuity equation

partρ partρvi (192)+ = 0partt partxi

Note that for an incompressible material with uniform density the continuity equationreduces to nabla middot v = 0

Conservation of linear momentum for the current volume element V surrounded bysurface S requires that the rate of change of momentum be balanced by the sum of surfaceforces and body forces

DρvdV = TdS + ρfdV (193)

Dt V S V

HereT is the traction acting on the surface S and f are body forces per unitmass such as gravshyity or electromagnetic forces The traction is related to the stress viaT = σ middot n (equation [146])Application of the divergence theorem (185) yields

Dρv dV = [nabla middot σ + ρf] dV (194)

Dt V V

Application of the Reynolds transport theorem (190) yields Dvnabla middot σ + ρf minus ρ dV = 0 (195)DtV

Again this must hold for all volume elements so that

Dv (196)nabla middot σ + ρf = ρ Dt

For small deformations DvDt = uuml Furthermore if accelerations can be neglected the equashytions of motion (196) reduce to the quasi-static equilibrium equations

partσi j + ρ fi = 0partxj (197)nabla middot σ + ρ f = 0

Copyrighted Material

24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

Copyrighted Material

Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

Copyrighted Material

26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

Copyrighted Material

Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

Copyrighted Material

28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

24 Chapter 1

Note that this is an Eulerian description In chapter 9 we will write Lagrangian equations ofmotion for perturbations in stress and displacement relative to a gravitationally equilibratedreference state In this formulation an advective stress term appears which from equation(189) is of the form u middot nablaσ 0 where σ 0 is the stress in the reference state

171 Equilibrium Equations in Curvilinear CoordinatesIn cylindrical polar coordinates

1 part 1 partσr θ partσr z σθθ(rσr r ) + + minus = 0r partr r partθ partz r

1 part 2 1 partσθθ partσθz(r σr θ ) + + = 0r 2 partr r partθ partz

1 part 1 partσθz partσzz(rσzr ) + + = 0 (198)r partr r partθ partz

In spherical coordinates

partσr r 1 partσr θ 1 partσrφ 1+ + + (2σr r minus σθθ minus σφφ + σr θ cot φ) = 0partr r partθ r sin θ partφ r

partσr θ 1 partσθθ 1 partσθφ 1+ + + [3σr θ + (σθθ minus σφφ) cot θ ] = 0partr r partθ r sin θ partφ r

partσrφ 1 partσθφ 1 partσφφ 1+ + + (3σrφ + 2σθφ cot θ) = 0 (199)partr r partθ r sin θ partφ r

All equations are shown without body forces (Malvern 1969)

18 Constitutive LawsIt is worth summarizing the variables and governing equations discussed to this pointThere are three displacement components ui six strains Ei j six stresses σi j and densityρ for a total of 16 variables (all of which are generally functions of position and time) Interms of the governing equations we have the six kinematic equations that relate strain anddisplacement (14) the three equilibrium equations (197) and the single continuity equation(192) for a total of 10 At this point we have six more unknowns than equations whichmakes sense because we have not yet defined the material behavior The soshycalled constitutiveequations provide the remaining six equations relating stress and strain (possibly includingtime derivatives) so that we have a closed system of equations It should be noted that thisdiscussion is limited to a purely mechanical theory A more complete thermomechanicaldescription including temperature entropy and conservation of energy is discussed brieflyin chapter 10

For a linear elastic material the strain is proportional to stress

(1100)σi j = Ci jklEkl

where Ci jkl is the symmetric fourthshyrank stiffness tensor Equation (1100) is referred to asgeneralized Hookersquos law The symmetry of the strain tensor makes it irrelevant to considercases other than for which the Ci jkl are symmetric in k and l The stress is also symmetric so

25

Copyrighted Material

Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

Copyrighted Material

26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

Copyrighted Material

Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

Copyrighted Material

28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

25

Copyrighted Material

Deformation Stress and Conservation Laws

the Ci jkl must be symmetric in i and j Last the existence of a strain energy function requiresCi jkl = Ckli j as follows The strain energy per unit volume in the undeformed state is defined as

dW= σi jdEi j (1101)

so that σi j = partWpartEi j From themixed partial derivatives part2WpartEi jpartEkl it follows that partσi jpartEkl =partσklpartEi j Making use of equation (1100) leads directly to Ci jkl = Ckli j

It can be shown that for an isotropic mediummdashthat is one in which the mechanicalproperties do not vary with directionmdashthe most general fourthshyrank tensor exhibiting thepreceding symmetries is

Ci jkl = λδi jδkl + micro(δikδ jl + δilδ jk) (1102)

(eg Malvern 1969) Substituting equation (1102) into (1100) yields the isotropic form ofHookersquos law

(1103)σi j = 2microEi j + λEkkδi j

where micro and λ are the Lame constants micro being the shear modulus that relates shear stress toshear strain

Other elastic constants which are convenient for particular applications can be definedFor example the bulk modulus K relates mean normal stress σkk3 to volumetric strain Ekk

σkk = 3K Ekk (1104)

Alternatively equation (1103) can be written in terms of Youngrsquos modulus E and Poissonrsquosratio ν as

EEi j = (1 + ν)σi j minus νσkkδi j (1105)

Youngrsquos modulus relates stress and strain for the special case of uniaxial stress Applying stressin one direction generally leads to strain in an orthogonal direction Poissonrsquos ratio measuresthe ratio of strain in the orthogonal direction to that in the direction the stress is appliedNotice that while there are five widely used elastic constants micro λ K ν and E only two areindependent Sometimes the symbol G is used for shear modulus but we reserve this forthe universal gravitational constant There are a host of relationships between the five elasticconstantsmdashsome prove particularly useful

2micro(1 + ν) 2K = = λ+ micro

3(1 minus 2ν) 3

2microνλ =

(1 minus 2ν)

E = 2micro(1 + ν)micro = 1 minus 2ν

λ+ micro

λ = ν (1106)

λ+ 2micro 1 minus ν

Copyrighted Material

26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

Copyrighted Material

Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

Copyrighted Material

28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

26 Chapter 1

The existence of a positive definite strain energy requires that both micro and K are nonnegativeThis implies from the first of equations (1106) that ν is bounded byminus1 le ν le 05 For ν = 051K = 0 and thematerial is incompressible

It is possible to write the compatibility equations in terms of stresses using Hookersquos lawInverting equations (1103) to write strain in terms of stress and substituting into equation(183) yields

σi jkl + σkli j minus σik jl minus σ jlik = νσnnklδi j + σnni jδkl minus σnn jlδik minus σnnikδ jl (1107)

1 + ν

As before only six of these equations are independent Setting k = l and summing

σi jkk + σkki j minus σik jk minus σ jkik = νσnnkkδi j + σnni j (1108)

1 + ν

This result can be further simplified using the equilibrium equations (197) yielding theBeltrami-Michell compatibility equations

1 νσi jkk + σkki j minus σkkmmδi j = minus(ρ fi ) j minus (ρ f j )i (1109)

1 + ν 1 + ν

Auseful relationship can be obtained by further contracting equation (1109) setting i = j andsumming which gives

1 + νnabla2σkk = minus nabla middot (ρf) (1110)1 minus ν

There will be occasion to discuss fluidlike behavior of rocks at high temperature within theearth The constitutive equations for a linear Newtonian fluid are

σi j = 2ηDijj minus pδi j (1111)

where η is the viscosity p is the fluid pressure and Dijj are the components of the deviatoric

rate of deformation defined by equation (120) and

DkkDi jj = Dij minus δi j (1112)

3

Equation (1111) is appropriate for fluids that may undergo arbitrarily large strains Inchapter 6 we will consider viscoelastic materials and implicitly make the approximation thatdisplacement gradients remain small so that the rate of deformation tensor can be approxshy

imated by the strain rate Dij asymp Ei j Constitutive equations for viscoelastic and poroelasticmaterials will be introduced in chapters 6 and 10

We will now return to the question of the number of equations and the number ofunknowns for an elasticmedium If it is desirable to explicitly consider the displacementsmdashforexample if the boundary conditions are given in terms of displacementmdashthen the kinematicequations (14) can be used to eliminate the strains from Hookersquos law leaving 10 unknowns(three displacement six stresses and density) and 10 equations (three equilibrium equationssix constitutive equations and continuity) If on the other hand the boundary conditions aregiven in terms of tractions it may not be necessary to consider the displacements explicitly Inthis case there are 13 variables (six stresses six strains and density) The governing equationsare equilibrium (three) the constitutive equations (six) continuity (one) and compatibility(three) leaving 13 equations and 13 unknowns

27

Copyrighted Material

Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

Copyrighted Material

28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

27

Copyrighted Material

Deformation Stress and Conservation Laws

fi f i

TiTi

uiui

Figure 117 Two sets of body forces and surface tractions used to illustrate the reciprocal theorem LeftUnprimed system Right Primed system

In general solving 13 coupled partial differential equations can be a formidable taskFortunately we can beginmdashin the next chaptermdashwith a far simpler twoshydimensional problemthat does a reasonably good job of describing deformation adjacent to long strikeshyslip faults

19 Reciprocal TheoremThe Betti-Rayleigh reciprocal theorem is an important tool fromwhichmany useful results can bederived As one example in chapter 3 the reciprocal theorem is employed to derive Volterrarsquosformula which gives the displacements due to slip on a general planar discontinuity in anelastic medium We will present a derivation of the reciprocal theorem here following forexample Sokolnikoff (1983) Imagine two sets of selfshyequilibrating body forces and surfacetractions fi Ti and fi

j Tij acting on the same body (figure 117) The unprimed forces and

tractions generate displacements ui whereas the primed generate uji both unique to within

rigid body motions We will consider the work done by forces and tractions in the unprimedstate acting through the displacements in the primed state

j j j jTiuidA+ fiuidV = TiuidAminus σi j j uidV (1113)A V A V

where we have rewritten the volume integral making use of the quasishystatic equilibriumequations (197) Use of the divergence theorem on the first integral on the rightshyhand sideleads to

j j j jTiuidA+ fiuidV = (σi j ui ) jdV minus σi j j uidVA V V V

= σi j uji jdV

V

j= Ci jkluklui jdV (1114)V

the last step resulting from application of Hookersquos law (1100) You have seen that theCi jkl havethe symmetry Ci jkl = Ckli j implying that the leftshyhand side of equation (1114) is symmetricwith respect to primed and unprimed coordinates In other words we can interchange theprimed and unprimed states

j j T j f jTiuidA+ fiuidV = i uidA+ i uidVA V A V

(1115)

Copyrighted Material

28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

28 Chapter 1

Equation (1115) is one form of the reciprocal theorem It states that the work done bythe primed body forces and tractions acting through the unprimed displacements is equalto the work done by the unprimed body forces and tractions acting through the primeddisplacements

110 Problems1 (a) Derive expression (112) for the Almansi strain

(b) Show that in terms of the displacement ui the Almansi strain tensor is given byequation (117)

2 Consider a line segment ddx that lies in the (x1 x2) plane Show that the elongation isgiven by

l 2= E11cos θ + 2E12cos θ sin θ + E22sin2 θl

where θ is the orientation of the baseline with respect to the x1 axis Given the strainE11 E12 01 01 = E21 E22 01 01

graphl l between 0 and π What is the meaning of themaximum andminimum

3 Prove that a pure rotation involves no extension in any directionHint Replace ωi j for Ei j in equation (118) and note the antisymmetry of ωi j

4 Consider the rotation of a line segment due to strain Derive an expression fore3 in termsof the strain components For simplicity take dx to lie in the (x1 x2) plane

dxT = (dx1 dx2 0)

Draw diagrams showing that the sign of the rotation you compute for each strain composhynent makes sense physically

5 Show formally that the principal strains are the extrema in the linear extension Recallfrom equation (118) that for small deformations

l Ei j xi xj= l0 xkxk

To find the extrema set the gradient ofl l0 with respect to xn to zero First show that thegradient is given by

part (l l0) 2Einxi Ei j xi xj 2xn= minus (1116)partxn xkxk xkxk xmxm

Notice that the quantity in parentheses Ei j xi xjxkxk is a scalar It is the value ofl l0 at theextremum which we denote E Show that the condition that the gradient (1116) vanishleads to

(Ein minus Eδin)xi = 0

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

29

Copyrighted Material

Deformation Stress and Conservation Laws

Mapview

σ1

uh

uv

θδ

σ3

LPF SAF LPFSAF

Figure 118 Loma Prieta fault geometry SAF = San Andreas fault LPF = Loma Prieta fault

thus demonstrating that the maximum andminimum elongations are the eigenvalues ofthe strain

6 In two dimensions show that the principal strain directions θ are given by

2E12tan 2θ = E11 minus E22

where θ is measured counterclockwise from the x1 axis Hint if you use an eigenvector approach the following identity may be useful

2 tan θtan 2θ =

1 minus tan2 θ

However other methods also lead to the same result

7 This problem explores Hookersquos law for twoshydimensional deformation For plane straindeformation all strains in the 3shydirection are assumed to be zero E13 = E23 = E33 = 0

(a) Show that for this case σ33 = ν(σ11 + σ22)

(b) Show that for plane strain Hookersquos law (1103) for an isotropic medium reduces to

2microEαβ = σαβ minus ν(σ11 + σ22)δαβ α β = 1 2 (1117)

(c) Show that for plane stress conditions σ13 = σ23 = σ33 = 0 Hookersquos law reduces to thesame form as in (b) but with ν replaced by ν(1 + ν)

8 The M 69 1989 Loma Prieta California earthquake occurred with oblique slip on a faultthat strikes parallel to the SanAndreas fault (SAF) and dips to the southwest approximately70 degrees (figure 118) We believe that the SAF last slipped in a purely rightshylateral eventin 1906 on a vertical fault plane The problem is to find whether a uniform stress statecould drive pure rightshylateral strike slip on a vertical fault and the observed oblique slipon the parallel Loma Prieta fault (LPF)The important points to note are that the SAF is vertical the LPF dips to the southwest

(δ asymp 70 degrees) slip in 1906 was pure rightshylateral strike slip and slip in 1989 wasuh = βuv where uh is horizontal displacement uv is vertical displacement and β asymp 14

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

Copyrighted Material

30 Chapter 1

Assume that the fault slips in the direction of the shear stress acting on the fault plane sothat τh = βτv where τh and τv are respectively the horizontal and vertical components ofthe resolved shear stress on the LPF You can assume that the intermediate principal stressσ2 is verticalHint Solve for τh and τv in terms of the principal stresses The constraint that τh = βτv

will allow you to find a family of solutions (stress states) parameterized by

σ2 minus σ3φ equiv

σ1 minus σ3Note that

φ rarr 1 as σ2 rarr σ1

φ rarr 0 as σ2 rarr σ3

and thus 0 lt φ lt 1 Express the family of admissible stress states in terms of φ as a functionof θ

9 Use the strainshydisplacement equations (14) and the isotropic form of Hookersquos law (1103)to show that the quasishystatic equilibrium equations (197) can be written as

partukkmicronabla2ui + (λ+ micro) + ρ fi = 0 (1118)

partxi

These equations are known as the Navier form of the equilibrium equations in elasticitytheory

10 Consider a solid elastic bodywith volumeV and surface S with unit normal ni Tractions Tiare applied to the S in general they are nonuniform and may be zero in places We wantto find the volume change

V = uinidsS

where ui is the displacement Use the reciprocal theorem to show that the volume changecan be computed by

1V = Tixids3K S

where K is the bulk modulus

111 ReferencesAgnew D C 1985 Strainmeters and tiltmeters Reviews of Geophysics 24(3) 579ndash624Dahlen F A and J Tromp 1998 Theoretical global seismology Princeton NJ Princeton University

PressFrank F C 1966 Deduction of earth strains from survey data Bulletin of the Seismological Society of

America 56 35ndash42Johnson H O D C Agnew and F K Wyatt 1994 Presentshyday crustal deformation in Southern

California Journal of Geophysical Research B 99(12) 23951ndash23974Malvern L E 1969 Introduction to the mechanics of continuous media Englewood Cliffs NJ Prentice

Hall

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220

31

Copyrighted Material

Deformation Stress and Conservation Laws

Savage J C and R O Burford 1970 Accumulation of tectonic strain in California Bulletin of theSeismological Society of America 60 1877ndash1896

Savage J C W Gan and J L Svarc 2001 Strain accumulation and rotation in the EasternCalifornia Shear Zone Journal of Geophysical Research 106 21995ndash22007

Segall P and Y Du 1993 How similar were the 1934 and 1966 Parkfield earthquakes Journal ofGeophysical Research 98 4527ndash4538

Sokolnikoff I S 1983 Mathematical theory of elasticity New York McGrawshyHill reprinted MalabarFL Robert E Krieger Publishing pp 390ndash391

Strang G 1976 Linear algebra and its applications New York Academic Press pp 206ndash220