elasticity - terrapub

Chapter 7

Elasticity

Elastic strain of rocks is very small, so that it is usually difficult to detect the strainfrom geologic structures. However, it is possible in some cases, allowing us to infermechanical properties of rocks at depths. A simple model, linear elasticity, is introducedin this chapter to discuss the states of stress. It should be noted that models in thischapter are ideal ones, because elastic strain energy tends to be released, over millionsof years, in shallow levels in the Earth by weathering, diagenesis, fracturing, and plasticflow.

7.1 Linear elasticity

7.1.1 Constitutive Equation

Let us derive the constitutive equation for isotropic elastic materials based on the representationtheorem (Eq. (5.5)). The equation is used in Chapters 7 and 8 to construct models for tectonics.

Elastic strains are very small so that the tensor is linearized (Eq. (2.6)) as V = I+E. Substitutingthis into Eq. (5.5), we obtain the theorem for an infinitesimal strain (E ≈ O). Namely,

S = φ0 I + φ1(I + E) + φ2(I + E)2

≈ (φ0 + φ1 + φ2)I + (φ1 + 2φ2)E, (7.1)

where the approximation (I + E)2 ≈ I + 2E is used, and the coefficients φ0, φ1 and φ2 are thefunctions of basic invariants of E. Note that EI = trace E is similar to E in order of magnitude. BothEII ∼ O(E2) and EIII ∼ O(E3) are negligible compared to EI. Therefore, the geometrical linearityallows us to assume that the functions depend only on EI = trace E.

Experiments show that elastic stress and strain have a linear relationship for rocks. If we definethe unloaded shape of a rock for a reference configuration, Eq. (7.1) should be homogeneous—zerostress must correspond to zero strain. Consequently, we have S = a(trace E)I + bE, where a and bare scalar constants. Therefore, a linear elastic body should have a constitutive equation of the form

S = λ (trace E) I + 2GE, (7.2)

153

154 CHAPTER 7. ELASTICITY

where λ and G are material constants known as Lame’s constants. As EI represents volume change,λ indicates the resistance to the changes. Deformations without volume change are associated withshear strains that induce stresses due to the second term of Eq. (7.2). G indicates the resistance toshear strains, and is called the shear modulus. Equation (7.2) is rewritten with r and e as1

r = λ (trace e) I + 2G e. (7.3)

The tensorial equations of the linear elasticity (Eqs. (7.2) and (7.3)) relate not only to the mag-nitude of stress and strain but also to their principal axes. We have assumed isotropy to use therepresentation theorem. The assumption results in the correspondence—principal stress axes areparallel to principal strain axes. Therefore, the relation of the principal values is often enough fortectonics, provided that the axial directions are known a priori. If we take the axes of Cartesian coor-dinates as parallel to the principal axes, the diagonal components of S and E are identical with theirprincipal values. Using the values, the constitutive equations (Eqs. (7.2) and (7.3)) are rewritten as⎧⎨⎩S1

S2

S3

⎫⎬⎭ =

⎧⎨⎩λ + 2G λ λλ λ + 2G λλ λ λ + 2G

⎫⎬⎭⎧⎨⎩E1

E2

E3

⎫⎬⎭⎧⎨⎩σ1

σ2

σ3

⎫⎬⎭ =

⎧⎨⎩λ + 2G λ λλ λ + 2G λλ λ λ + 2G

⎫⎬⎭⎧⎨⎩ε1

ε2

ε3

⎫⎬⎭ . (7.4)

The column vectors and coefficient matrices are put in braces here instead of being parenthesized,because they are not first- or second-order tensors that are transformed as Eq. (C.34), but the prin-cipal values are invariant to coordinate rotations and the components of the matrices are materialconstants that do not depend on the rotation, either.

Sometimes we need to determine strain from stress. They are linearly related, so that the linearequations

E =1 + νY

S − ν

YSII, e =

1 + νY

r − ν

YσII (7.5)

meet the necessity, where Y and ν are material constants called Young’s modulus2 and Poisson’sratio. Table 7.1 shows those constants of typical rocks. Equation (7.5) is rewritten by the relationbetween the principal values as⎧⎨⎩E1

E2

E3

⎫⎬⎭ =1Y

⎧⎨⎩ 1 −ν −ν−ν 1 −ν−ν −ν 1

⎫⎬⎭⎧⎨⎩S1

S2

S4

⎫⎬⎭ ,

⎧⎨⎩ε1

ε2

ε3

⎫⎬⎭ =1Y

⎧⎨⎩ 1 −ν −ν−ν 1 −ν−ν −ν 1

⎫⎬⎭⎧⎨⎩σ1

σ2

σ3

⎫⎬⎭ . (7.6)

1The signs of stresses S and r and of strains E and e are defined not to affect the form of constitutive equations as thecase of Eqs. (7.2) and (7.3).

2Most literature assign the symbol E to Young’s modulus. However, we have appropriated the symbol to infinitesimalstrain, so that Y is used instead.

7.1. LINEAR ELASTICITY 155

Table 7.1: Elastic property of typical rocks [245].

ρ Y G ν×103 kg m−3 GPa GPa

Sedimentary Rockshale 2.1–2.7 10–30 14sandstone 2.2–2.7 10–60 4–30 0.2–0.3limestone (non-crystalline) 2.2–2.8 60–80 20–30 0.25–0.3limestone (crystalline) 2.2–2.8 30–90 20–35 0.1–0.4

metamorphic rockgneiss 2.7 4–70 10–35 0.4–0.15amphibolite 3.0 50–100 0.4

igneous rockbasalt 2.95 60–80 30 0.25granite 2.65 40–70 20–30 0.1–0.25gabbro 2.95 60–100 20–35 0.15–0.2diorite 2.80 60–80 30–35mantle material* 3.359 160 60 0.3

*Inferred values at a depth of 100 km by seismological observations.

It is obvious that Eqs. (7.4) and (7.5) are the inverse of each other. Inverse of the coefficient matrixof Eq. (7.6) is

Y

(ν + 1)(2ν − 1)

⎧⎨⎩ν − 1 −ν −ν−ν ν − 1 −ν−ν −ν ν − 1

⎫⎬⎭ .

This should be equal to the coefficient matrix of Eq. (7.4), so that we have the following relationsamong material constants:

λ =Y ν

(1 − 2ν)(1 + ν)(7.7) ν =

λ

2(λ + G)(7.8)

Y =G(3λ + 2G)λ + G

(7.9) G =Y

2(1 + ν). (7.10)

7.1.2 Physical Interpretation of Y, ν and G

Suppose a uniaxial compression where σ1 �= 0, σ2 = σ3 = 0. Combining these conditions and Eq.(7.6), we obtain

σ1 = Y ε1. (7.11)

The subscripts in this equation indicate that stress is proportional to strain in the extending directionand Y is the constant of proportionality. This is compared to the behavior of springs that obeys theequation F = −kx where F and x are force and length change of the spring and k is the spring


constant. Linear elasticity is the three-dimensional extension of Hooke’s law3, and Y is the stiffnesscomparable to the spring constant.

Let us estimate elastic strains with using Eq. (7.11). Tectonic stresses are in the orders of 10–100 MPa in tectonically active regions, and Young’s modulus of rocks is about 60 GPa (Table 7.1).Therefore, elastic strain is σ/Y ≈ 10−4–10−3. Strains may be less than this in inactive regions.The lithosphere cannot support stresses larger than the magnitude, but brittle or plastic deforma-tions occur to create permanent deformations that may be observable as geologic structures. Thedeformation relieves tectonic stresses, giving rise to the upper limit of the stresses.

Suppose a cylindrical sample is subject to the uniaxial compression σ1 > 0 and σ2 = σ3 = 0. Ifthe sample is made of a linear elastic material, combining Eqs. (7.6) we have ε1 = σ1/Y and

ε2 = ε3 = −νσ1/Y = −νε1. (7.12)

Although O-2 and -3 directions are stress free, the cylindrical sample becomes thicker in thosedirections. This is called the Poisson effect. It should be pointed out that plain stress does not resultin plain strain due to the effect. Plane strain does not cause plain stress.

Poisson’s ration is constrained in the range from 0 to 1/2. Equation (7.12) shows that ν indicateshow much Poisson effect appears. If ν = 0, no such effect occurs. Common materials including rockshave the positive Poisson effect that a sample thickens in directions perpendicular to the direction ofuniaxial compression. This is not satisfied if ν < 0. Therefore, it is permitted to assume that ν ≥ 0. Inaddition, elastic strains are infinitesimal, so that volume change is δV/V = −εI = −(ε1+ε2+ε3) (Eq.(2.11)). Therefore, combining Eq. (7.12) we obtain δV/V = −(1 − 2ν) ε1. We have assumed thatthe sample is compressed, therefore the ratio should be negative in sign. In this case (ε1 > 0), so thatPoisson’s ratio must satisfy the inequality ν ≤ 1/2. If ν = 1/2, the elastic strain is incompressible—no volume change occurs. Consequently, the ratio has a value of between 0 and 1/2.

Poisson’s ratio is often assumed at 0.25 in tectonic models because most rock types have similarratios (Table 7.1). Seismology shows that the representative Poisson ratio for the crust is 0.25 ±0.04 on average, and at about 0.27 and 0.29 under Cenozoic orogenic belts and shields, respectively[274]. In addition, the ratio at 0.25 is convenient for modeling. The reason is that although theconstitutive equation of linear elasticity (Eq. (7.2)) has two material parameters, λ and G, they havea common value if ν = 1/4. In this case, Eq. (7.8) becomes 1/2 = λ/(λ+G), indicating that λ = G.In the following sections, we shall often use the value ν = 0.25 as the representative Poisson ratiofor rocks.

The shear modulus G indicates the resistance of a material to simple shear (Exercise 7.1).

3In general, linear elasticity is expressed by the equation S = C : E, where C = (Cijkl) is a fourth-order tensor called thetensor of elastic constants and represents the anisotropy of a material. This equation indicates proportionality between stressand strain, called the generalized Hooke’s law. For isotropic materials, C is the fourth-order isotropic tensor, and the aboveequation reduces to Eq. (7.2). See [61] for detail.

7.2. EARTH PRESSURE AT REST 157

7.1.3 Two-dimensional elasticity

Orogenic belts and rift zones have linear belts in which vigorous deformations take place. Accord-ingly, it is sometimes convenient to model the deformations on the vertical section perpendicular tothe belts by means of two-dimensional elasticity, where the plane strain or plane stress condition iscombined with the linear elasticity.

First, let us derive the constitutive equations for plane-strain problems. Consider plane strain onthe O-13 plane, where the O-1 and -3 axes are taken to be parallel to the principal strain axes. In thiscase we have ε2 = 0, so that the constitutive equation of linear elasticity (Eq. (7.4)) becomes⎧⎨⎩σ1

σ2

σ3

⎫⎬⎭ =

⎧⎨⎩(λ + 2G)ε1 + λε3

λ(ε1 + ε3)λε1 + (λ + 2G)ε3

⎫⎬⎭ . (7.13)

Eliminating ε1 and ε3 from Eq. (7.13) and exchanging Lame’s constants by Poisson’s ratio usingEq. (7.8), we obtain the constitutive equation for the plane strain problem,

σ2 = ν (σ1 + σ3) . (7.14)

The deformation is incompressible if ν = 1/2. In this case, we have σ2 = (σ1 + σ3)/2, i.e., theincompressible plane strain of a linear elastic body results in the stress ratio, Φ = 1/2.

The state of plane stress parallel to the O-12 plane is sometimes assumed to model the elasticbehavior of the lithosphere, later. The state is represented by the principal stresses, S1 �= 0, S2 �= 0and S3 = 0. Substituting these into Eq. (7.6), we obtain the constitutive equation for this states⎧⎨⎩E1

E2

E3

⎫⎬⎭ =1Y

⎧⎨⎩ S1 − νS2

−νS1 + S2

−(S1 + S2)

⎫⎬⎭ . (7.15)

It should be noted that plane stress does not necessarily result in plane strain. This discordance isdue to the Poisson effect which is explained in the following section.

7.2 Earth Pressure at Rest

We have introduced the lithostatic state of stress in Chapter 3 as a reference state, where rocks behaveas fluid at depths. Now we study the state of stress with the assumption that rocks behave as linearelastic materials.

For simplicity, we derive the state of stress for a linearly elastic rock mass on five assumptions:(1) we assume isotropy for rocks at depth: (2) we assume a layered structure: density below thehorizontal surface, ρ, is horizontally homogeneous but has vertical variations. We take the origin ofthe Cartesian coordinates O-xyz at the surface and the z-axis downward. (3) We assume that gravityis the only origin of stress. If so, one of the principal stress axes must be vertical and the rest liehorizontal. Since the system has no horizontal variation at all, we expect axial symmetry around the


vertical for the state of stress. Therefore, σxx, σyy and σzz equals principal stresses with σxx = σyy. Ifrocks are isotropic, the strain field must have the same symmetry: εxx, εyy and εzz are the principalstrains and εxx = εyy. (4) Rocks are confined at depth in the crust, horizontal expansion is restrictedby adjacent rock. Therefore, let us assume that the layers are horizontally constrained so as not toallow horizontal displacement: ux = uy = 0. Therefore, we have

εxx =∂ux∂x

= 0, εyy =∂uy

∂y= 0. (7.16)

(5) Rocks behave as linear elastic materials. This allows us to use Eq. (7.4), in which Eq. (7.16) issubstituted. The result is the equation⎧⎨⎩σxxσyy

σzz

⎫⎬⎭ =

⎧⎨⎩λ + 2G λ λλ λ + 2G λλ λ λ + 2G

⎫⎬⎭⎧⎨⎩ 0

0εzz

⎫⎬⎭ =

⎧⎨⎩ λλ

λ + 2G

⎫⎬⎭ εzz.

It is found from this equation that

σxx = σyy =

(λ

λ + 2G

)σzz. (7.17)

Using Eqs. (7.7) and (7.10), we have

λ

λ + 2G=

Y ν(1−2ν)(1+ν)

Y ν(1−2ν)(1+ν) +

Y(1+ν)

=ν

1−2νν

1−2ν + 1=

ν

ν + (1 − 2ν)=

ν

1 − ν .

Using Eq. (7.17), we obtain

σxx = σyy =( ν

1 − ν)σzz =

( ν

1 − ν)pL, (7.18)

where pL stands for overburden (Eq. (3.26)). This is called earth pressure at rest which is the stateof stress that is induced by gravity in a horizontally layered rock mass. Hence, let us write the stateas

rgravity = pL

⎛⎝ν/(1 − ν) 0 00 ν/(1 − ν) 00 0 1

⎞⎠ . (7.19)

We are able to use earth pressure at rest as the reference for the state of stress at depths instead ofthe lithostatic state of stress. It is necessary to specify the reference state to discuss crustal stresses.

Since Poisson’s ratio is bounded as 0 < ν ≤ 1/2, the coefficient ν/(1 − ν) in Eqs. (7.18) and(7.19) ranges between 0 and 1. Therefore, the inequality 0 < (σxx, σyy) ≤ σzz holds, indicating thathorizontal stress is less than the overburden. Rocks are vertically constricted by overburden but arehorizontally extended by the Poisson effect. In the extreme case of ν = 1/2, Eq. (7.18) becomesσxx = σyy = σzz. This is equivalent to the lithostatic state of stress, though the media is not assumedto be fluid but elastic.

7.3. STABILITY OF ELASTIC ROCK MASSES 159

Since Poisson’s ratio of representative rocks can be approximated as ν ≈ 1/4 (p. 156), thecoefficient K = σHmean/σv is approximately equal to 1/3. Horizontal stress is about one-third ofoverburden. In-situ stress measurements conducted in various regions of the world show that a fewregions show K ≈ 1/3 [24]. The ratio is considerably scattered in the shallow levels of the crust,but seems to converge with increasing depth to ∼ 1, the lithostatic state of stress. The above modelpredicts the differential stress Δσ = (1 − K)pL, which piles up unlimitedly with increasing depth.However, real rocks have their own yield stress. The yielding may relief the differential stress tocause convergence in the ratio. This is perhaps at least one of the reasons for the convergence. Thestress state discussed above is an idealized state, convenient for theoretical considerations, but weshould be careful to apply the model to real tectonics. Rocks behave as elastic materials in a humantimescale. However, elastic stress may be released in a geological timescale.

7.3 Stability of elastic rock masses

Is a rock stable if its behavior is linear elastic? Let us examine whether the Earth pressure at rest isunder the frictional strength of rocks. If it is not so, rocks may collapse through normal faulting oninclined pre-existing fractures without tectonic stress. The Earth pressure at rest has the principalstresses

σv = pL > σH = σh =( ν

1 − ν)pL. (7.20)

Figure 7.1a shows this stress state.In one extreme case, ν = 0, this is a uniaxial stress σH = σh = 0. The Mohr circles corresponding

to various depths pass the origin of the Mohr diagram so that the line of frictional strength alwayscuts the circles, i.e., the elastic rock body with ν = 0 is unstable. In another extreme case, ν = 1/2,we have lithostatic states σv = σH = σh = pL. Then no deformation occurs at all and the elastic rockbody is always stable.

For the general case where 0 < ν < 1/2, the Mohr circles has a linear envelope passing theorigin of the Mohr diagram for the proportionality of all the principal stresses with pL (Fig. 7.1(a)).The slope of the line is given by means of the Pythagorean theorem about the triangle OCD. Namely,using Eq. (7.20), we have

OC =12

(pL +

ν

1 − ν pL

)=

12(1 − ν)

pL, CD =12

(pL − ν

1 − ν pL

)=

2ν − 12(1 − ν)

pL,

∴ OD =√c2 − r2 =

( ν

1 − ν)1/2

pL.

Consequently, we obtain the slopeCD

OD=

2ν − 1

2√ν(1 − ν)

.

Figure 7.1(b) shows the variation of this slope. If we use the value ν = 0.25 for the representativePoisson ratio for rocks, we have a slope of 0.58. This is slightly smaller than the coefficient of


Figure 7.1: (a) Mohr diagram showing the Earth pressure at rest. (b) Relationship between ν andthe slope in (a).

friction. Consequently, the rocks are stable under the Earth pressure at rest. Rocks with smallPoisson ratios may be unstable.

We have neglected the effect of pore pressure in the above argument. However, pore pressureplays a critical role in the stability because increased pore pressure shifts the Mohr circle to the leftand destabilizes rocks with ordinary Poisson ratios. Therefore, earthquakes can occur on pre-existingfracture surfaces only through a building up of pore pressure.

7.4 Thermal stress

Changes of temperature leads to thermal stress which is as large as a stress increment resulting fromother factors. Since temperature increases generally with depth, depth changes accompanied byexhumation or basin subsidence affect the state of stress.

Basic equation

Rocks expand when heated. Consider a strain E of a rod accompanied by the temperature changeof ΔT . Experiments show that the proportionality E = α�ΔT holds for a rod of rock if the lengthof the rod is unconstrained. The constant of proportionality α� is known as the coefficient of linearexpansion. Rocks have a coefficient of about 10−5 K−1. The temperature change results in anincreased volume of a cube given by

ΔVV

=(L + ΔL)3 − L3

L3=

(L + ΔLL

)3

− 1 = 3ΔLL

+ 3(ΔLL

)2

+

(ΔLL

)3

≈ 3ΔLL,

where L is the initial length of the sides of the cube and higher-order terms are neglected. UsingΔL = LE, we have

ΔVV

≈ 3E = 3α�ΔT.

7.4. THERMAL STRESS 161

Accordingly, we define the coefficient of volume expansion or simply thermal expansion coefficientα so as to satisfy the relationship ΔV /V = αΔT . Obviously, we have

3αl ≈ α. (7.21)

Rocks have the thermal expansion coefficients at around 3 × 10−5 K−1.

The coefficient of linear expansion of a single crystal has anisotropy resulting from its crystal-lographic structure. However, a rock mass consists of a polycrystalline material made of mineralsusually with random crystallographic orientations. Therefore, we assume that α� is isotropic whenwe consider the stress or strain of rocks. This is formulated as

ε1 = ε2 = ε3 = −αlΔT. (7.22)

Incremental deformation against viscous forces dissipates kinetic energy (Eq. (3.41) to affectthe temperature field, i.e., mechanical and thermal processes are generally coupled. However, inmany situations their interaction can be neglected. The resulting analysis is known as the uncoupledthermoelastic theory of continua. In this case, strain is considered to be the sum of the strains causedby the incremantal stress and temperature. The constitutive equation of the former is given by Eq.(7.5) or (7.3) and the latter is given by Eq. (7.22), so that we have

e =1 + νY

r −(νσI

Y+ αlΔT

)I, (7.23)

r = 2Ge +

(λεI +

Y αlΔT1 − 2ν

)I. (7.24)

Using principal stresses and strains, Eq. (7.23) is rewritten as

ε1 =1Yσ1 −

ν

Yσ2 −

ν

Yσ3 − αlΔT,

ε2 = − νYσ1 +

1Yσ2 −

ν

Yσ3 − αlΔT , (7.25)

ε3 = − νYσ1 −

ν

Yσ2 +

1Yσ3 − αlΔT .

Prior to using Eq. (7.25), the principal orientations should be specified.

When magma cools down to be solidified, a temperature change of several hundreds degreesleads to a stress change of Y α1ΔT ∼ 102 MPa, as the representative Young’s modulus of rocksis about Y = 60 GPa. Since the tensile strength of rocks is in the order of 101 MPa, this thermalstress easily forms fractures in the solidified igneous body, called cooling joints. Cooling of a tabularintrusive body splits the body into columns to form columnar joints perpendicular to the intrudingplane (Fig. 7.2).


Figure 7.2: Doleritic dike penetrating purple granite and overlying volcanic rocks. The dike hascolumnar joints. Oga peninsula, Northeast Japan.

Comparison with overburden stress

The state of stress in a linear elastic material at depth is described by the Earth pressure at rest,which indicates a normal fault regime of stress. If a lava flow that was consolidated at the surfaceis gradually buried in a sedimentary basin, the layer experiences horizontal extension. On the otherhand, the temperature of the layer increases with depth, leading to horizontal compression if thelayer is horizontally constrained as the condition from which the earth pressure at rest was derived.Let us consider these competing effects on the stress state.

For this purpose, we assume a level surface and no horizontal density or tempeature variation.Under these conditions, stress and strain have a vertical principal orientation. In addition, we as-sume uncouple mechanical and thermal processes so that incremental stresses accompanied by anincreases of temperature and overburden are considered separately.

The layer is horizontally constrained but can freely displace the overlying strata by its thermalexpansion, i.e., the layer is not vertically constrained. Therefore, the thermal expansion does notresult in an increase of the vertical stress component, i.e., σzz = 0 for a contribution from the

7.4. THERMAL STRESS 163

temperature increase. Substituting this vertical stress into Eq. (7.25), we have

εxx =1Y

(σxx − νσyy) − αlΔT

εyy =1Y

(−νσxx + σyy) − αlΔT (7.26)

εzz =ν

Y(−σxx − σyy) − αlΔT

Combining the condition of horizontal constraints εxx = εyy = 0, we obtain

σxx = σyy =Y αlΔT1 − ν (7.27)

for the incremental stress components due to the temperature increase. Let Γ be a geothermal gradi-ent, then the temperature at depth z is given by T = T0+Γz, where T0 is the surface temperature. Thelayer had no strain at all at the surface, therefore ΔT = Γz. Substituting this temperature differenceinto Eq. (7.27), we obtain

σxx = σyy =Y αlΓz1 − ν . (7.28)

The differential stress due to the temperature difference equals

Δσ =

(Y αlΓ1 − ν

)z. (7.29)

Geothermal gradients are usually a few degrees per 100 m except for geothermal areas. Using thevalues Γ = 0.01 K m−1, Y = 60 GPa, ν = 0.25, and α� = 10−5 K−1, we obtain Δσ/z = 8 kPa m−1,and the gradient of overburden stress ρg = 2.4 kPa m−1. Consequently, the thermal stress is as greatas othe verburden stress for a layer that was consolidated at the surface4.

Polygonal fracture pattern on Venus

Venusian plain regions have polygonal terrains that are characterized polygonal fracture network byrelatively uniform spacing (Fig. 7.3). The random orientations of the fractures suggest isotropicsmall-strain deformations, probably reflecting the thermal stress included by decreased surface tem-perature over a billion years [93].

Venus was resurfaced by global massive volcanism a few billion years ago. Greenhouse effectgases such as H2O and SO2 emitted by the volcanism raised the surface temperature by more than100 K. Later atmospheric cooling contracted rocks in the crust. Just like the half-space coolingmodel of the oceanic lithosphere (§3.12), rocks at shallow levels in the crust are cooled swifter thanthose at deep levels. The time lag induces a variation in the magnitude of horizontal contraction to

4This model assumes that a horizontally extensive elastic layer subsided neither with jointing nor faulting in a geologicaltime scale. This is not probable in reality. Fractures between blocks may allow horizontal displacements, violating thecondition of horizontal constraints.


Figure 7.3: Polygonal terrain on Venus [5].

eventually form the polygonal fractures eventually [5]. Let us estimate the contraction and associatedthermal stress by the climate change.

Consider the temperature changes with time as T = A sinωt at the surface z = 0. The tempera-ture in the half-space z ≥ 0 is calculated by the heat conduction equation ∂T/∂t = κ∂2T/∂z2. Here,T represents the temperature anomaly. Substituting the solution of the form T = u(z)eiωt, we haved2u/dz2 = (iω/κ)u. Using the condition that the solution does not diverge to infinity at z = ∞, weobtain

T = Ae−Bz cos(ωt − Bz),

where B =√ω/2κ [36]. Therefore, the temperature change at z is delayed by Bz. The vertical

temperature variation induces thermal stresses. We have the variation

∂T

∂z= ABe−Bz

[sin(ωt − Bz) − cos(ωt − Bz)

].

Hence, the near-surface geothermal gradient is represented by Γ ≈ AB. Using the values A = 100K, ω = 1 rad/Ga ≈ 3 × 10−16 s−1, and κ = 1 × 10−5 K−1, we have Γ ≈ AB ≈ 10 K/km. Assumingconstrained horizontal displacement, the thermal stress between different depths Δz is evaluated asEq. (7.29), i.e., we have

σ ≈(Y α�Γ1 − ν

)Δz.

Using Y = 60 GPa and ν = 0.25, we have Y α�Γ/(1 − ν) ≈ 1000 Pa m−1. On the other hand, thegravitational acceleration at on the Venusian surface is g ≈ 9 ms−1 so that the gradient of overburdenstress is ρg ≈ 3000 Pa m−1. Therefore, the thermal stress is comparable and if the surface tempera-ture drop was a few hundred degrees, the thermal stress becomes large enough to produce map-scalefractures5.

5See [5] for detail.

7.5. GLOBAL THERMAL CHANGES AND SURFACE STRESS FIELD 165

7.5 Global thermal changes and surface stress field

Terrestrial planets and satellites experienced drastic global cooling in their early history. Except forthe Earth and Venus which have active geological processes including tectonism and erosion, tec-tonic features are clues to their early thermal history. For example, it is suggested that the cessationof extensional tectonics some 3.6 billion years ago was the surface manifestation of global cooling[127]. Thrust faults on Mercury are thought to be the result of global cooling, also. In this section,we investigate the linkage between the cooling and tectonics.

Consider a self-gravitating solid spherical body with a radius of R, in which all physical prop-erties have spherical symmetry. The symmetry leads to a vertically axisymmetric stress field. LetSV and SH be the vertical and horizontal stresses. Here, we assume a solid planetary body, so thatthe vertical and horizontal stresses can be different. The unit volume in the body is subject to thegravitational force −ρg, where the density ρ and gravitational acceleration g are functions of theradial coordinate r. The body force and the stress components shown in Fig. 7.4 are found to leadto an equillibrium equation in the form6

dSV

dr+

2r

(SV − SH) − ρg = 0. (7.30)

Thermal expansion of the body is very small compared to the radius of the planetary body R, sothat the former does not depend on temperature. Therefore, let us utilize the uncoupled thermoelastictheory to investigate thermal stresses in the body. Stresses due to gravity and to temperature changecan be treated separately. Hence, the gravity term −ρg is deleted from Eq. (7.30) to estimate thestress field due to temperature changes. Consequently, the equation of thermal stress (Eq. (7.25)) isrewritten as

EV − αΔT =1Y

(SV − 2νSH) , (7.31)

EH − αΔT =1Y

[SH − ν(SV + SH)

], (7.32)

where SV and SH are vertical and horizontal infinitesimal strain components due to temperaturechange ΔT . These strain components are related to the vertical displacement u by the equations

EV =dudr, EH =

u

r. (7.33)

Rearranging Eqs. (7.31) and (7.32), we have

SV =Y[

(1 − ν)EV + 2νEH − (1 + ν)αΔT]

(1 + ν)(1 − 2ν), SH =

Y[EH + νEV − (1 + ν)αΔT

](1 + ν)(1 − 2ν)

. (7.34)

6Equation (3.46) that describes the acceleration due to gravity as a function of r lacks the second term of Eq. (7.30).When we derived the former equation, we implicitly assumed that the planetary body is composed of fluid at rest. Therefore,the differential stress SV − SH vanished.


Figure 7.4: Force balance of a small circular truncated cone with a thickness of dr and an apicalangle of dφ.

Subsitituting these equations into Eq. (7.30), we have the differential equation of u,

d2u

dr2+

2r

dudr

− 2ur

= α(1 + ν

1 − ν)dΔT

dr.

Rearranging this equation, we obtain

ddr

[1r

ddr

(r2u)]

= α(1 + ν

1 − ν)dΔT

dr,

which is integrated to give

u =α

r2

(1 + ν1 − ν

) ∫ rr0

ΔTr2dr + C1r +C2

r, (7.35)

where C1 and C2 are constants of integration and r0 designates the base of a spherical shell. Com-bining this differential equation and Eqs. (7.33) and (7.34), we obtain the equations [246],

SV = − 2αY(1 − ν)r3

∫ rr0

ΔTr2dr +Y C1

1 − 2ν− 2Y C2

(1 + ν)r3, (7.36)

SH =αY

(1 − ν)r3

∫ rr0

ΔTr2dr +Y C1

1 − 2ν+

Y C2

(1 + ν)r3− αYΔT

1 − ν . (7.37)

7.5. GLOBAL THERMAL CHANGES AND SURFACE STRESS FIELD 167

The constant C2 is determined through Eq. (7.35) by the vanishing displacement at the center r = 0as C2 = 0 because of

limr→0

1r2

∫ r0ΔTr2dr = 0.

Let ΔT0 be the temperature change at the center, then we have

limr→0

1r3

∫ r0ΔTr2dr =

ΔT0

3.

Therefore, the other constant is constrained by assuming a free surface at r = R as

Y C1

1 − 2ν=

2αY1 − ν

1R3

∫R0ΔTr2dr.

Consequently, Eqs. (7.36) and (7.37) become

SV =2αY1 − ν

(1R3

∫R0ΔTr2dr − 1

r3

∫ r0ΔTr2dr

),

SH =αY

1 − ν

(2R3

∫R0ΔTr2dr +

1r3

∫ r0ΔTr2dr − ΔT

).

Now consider a planetary body being composed of n spherical shells, where the top is the 0thlayer. The shells have different material constants but the constants are uniform within each layer.Let Yi and νi be Young’s modulus and Poisson’s ratio of the ith layer. Then, we have

u(r) = rI (r) + Air + BiR3

r2, (7.38)

SV(r) = −2YiI (r)1 + νi

+YiAi

1 − 2νi− 2YiBiR3

(1 + νi)r3,

SH(r) = −YiI (r)1 + νi

+YiAi

1 − 2νi+

YiBiR3

(1 + νi)r3− Yiα(r)ΔT (r)

3(1 − νi), (7.39)

where

I (r) =1 + ν

3(1 − ν)r3

∫ r0α(r)ΔT (r)r2dr (7.40)

and Ai and Bi are dimensionless numbers corresponding to the following three conditions on thestress field [218]. Namely, (1) there is no singularity at the center, (2) no discontinuity betweenlayers, and (3) the surface of the planetary body is free, SV = 0. The integral in Eq. (7.40) equalsthe thermal expansion within the radius of r.

The goal of this argument is to compare the surface horizontal stress SH∣∣r=R

and observedtectonic features. For brevity, we assume that the surface temperature is constant through time,ΔT (R) = 0. Combining the free surface condition, we obtain the equation for the surface layer,

−2Y0I (R)1 + ν0

+Y0A0

1 − 2ν0− 2Y0B0

1 − ν0= 0.


Rearranging this equation, we get

B0 =12

1 + ν0

1 − 2ν0A0 − I (R). (7.41)

Combining Eqs. (7.38), (7.39) and (7.41), we obtain the displacement and horizontal stress of thesurface layer (r = R),

u(R) =32R(1 − ν0)A0

1 − 2ν0, (7.42)

SH(R) =32Y0A0

1 − 2ν0. (7.43)

The increase of the radius ΔR equals the upward displacement of the top layer u(R). Therefore,combining Eqs. (7.42) and (7.43), we obtain

SH(R) =Y0

1 − ν0

ΔRR. (7.44)

Consequently, the surface horizontal stress determined only by the stretch of the radius ΔR/R andthe elastic constants of the surface layer, independent from the physical properties of the inside. Theratio ΔR/R reflects the internal changes.

The activity of thrust faults (Fig. 2.14) contracted Mercury 3–4 billion years ago. The orientationand distribution of these thrust faults appear to be rondom, the contraction in the radius ΔR wasroughly estimated via Eq. (2.56) at minus 1–2 km, and is interpreted as the surface manifestationof global cooling [232]7. |ΔR| is much smaller than the present radius of Mercury at 2439 km.Therefore, the initial radius can be replaced by the present one, and we have the ratio −ΔR/R =(4–9)×10−4. Substituting this ratio with Y = 60 GPa and ν = 0.25 as the representative values ofrocks into Eq. (7.44), we obtain the surface horizontal stress at −SH = 30–60 MPa.

According to the theory of thermal evolution, that amount of contraction is explained well bythe heat loss just as the amount of heat production by core formation alone, and the total freezing ofthe core can lead to an 8 km contraction of the radius [217]. The contraction estimated from thrustfaults may be the lower bound, so that Mercury probably has a large inner core [208].

The Moon has tectonic features indicating both horizontal contraction and expansion, which areevidenced by wrinkle ridges and linear rilles, respectively. In addition, those features appear toreflect regional tectonics characteristic of mare basins (§8.8). Global cooling left a less conspicuouspattern of tectonic features on the Moon than on Mercury. The lack of a pattern suggests the globalexpansion as small as |ΔR| < 1 km for the Moon [129]. This corresponds to |SH| < 50 MPa.However, the wrinkle ridges still suggest global cooling, as extensional tectonics terminated at ∼3.6Ga on the Moon, and horizontal compression prevailed thereafter. Too hot initial conditions leadto a large contraction, which is inconsistent with the tectonic features on the Moon. Using theseconstraints, Solomon et al. [219] estimated the depth of the magma ocean at about 200 km at thebeginning of the history of the Moon.

7Several researchers raise objections to the randomness, as the surface of the Mercury has not been globally photographedwith enough resolution [147, 241].

7.6. EXERCISES 169

7.6 Exercises

7.1 Demonstrate that the shear modulus G indicates the resistance of a linearly elastic material tosimple shear.

7.2 The Venusian surface has polygonal fractures, each of which has a diameter of several kilo-meters (Fig. 7.3). The spacing between the fractures is thought to reflect the depth extent of crackpropagation. Derive the equation

ΔSH =ΔΓαY1 − ν

( te2− z),

describing the change in the surface horizontal stress due to the change in the geothermal gradientΔΓ, where te is the thickness of the surface elastic layer. Assuming the Coulomb-Navier criterion,estimate the maximum depth for fracture propagation due to thermal stress [93].

elasticity - terrapub

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