Elasticity. Chapter Linear elasticity Constitutive Equation

Transcription

1 Chapter 7 Elasticity Elastic strain of rocks is very small, so that it is usually difficult to detect the strain from geologic structures. However, it is possible in some cases, allowing us to infer mechanical properties of rocks at depths. A simple model, linear elasticity, is introduced in this chapter to discuss the states of stress. It should be noted that models in this chapter are ideal ones, because elastic strain energy tends to be released, over millions of years, in shallow levels in the Earth by weathering, diagenesis, fracturing, and plastic flow. 7.1 Linear elasticity Constitutive Equation Let us derive the constitutive equation for isotropic elastic materials based on the representation theorem (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. Substituting this into Eq. (5.5), we obtain the theorem for an infinitesimal strain (E O). Namely, S = φ 0 I + φ 1 (I + E) + φ 2 (I + E) 2 (φ 0 + φ 1 + φ 2 )I + (φ 1 + 2φ 2 )E, (7.1) where the approximation (I + E) 2 I + 2E is used, and the coefficients φ 0, φ 1 and φ 2 are the functions of basic invariants of E. Note that E I = trace E is similar to E in order of magnitude. Both E II O(E 2 ) and E III O(E 3 ) are negligible compared to E I. Therefore, the geometrical linearity allows us to assume that the functions depend only on E I = trace E. Experiments show that elastic stress and strain have a linear relationship for rocks. If we define the unloaded shape of a rock for a reference configuration, Eq. (7.1) should be homogeneous zero stress must correspond to zero strain. Consequently, we have S = a(trace E)I + be, where a and b are scalar constants. Therefore, a linear elastic body should have a constitutive equation of the form S = λ (trace E) I + 2GE, (7.2) 153

2 154 CHAPTER 7. ELASTICITY where λ and G are material constants known as Lamé s constants. AsE I represents volume change, λ indicates the resistance to the changes. Deformations without volume change are associated with shear strains that induce stresses due to the second term of Eq. (7.2). G indicates the resistance to shear strains, and is called the shear modulus. Equation (7.2) is rewritten with r and e as 1 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 magnitude of stress and strain but also to their principal axes. We have assumed isotropy to use the representation theorem. The assumption results in the correspondence principal stress axes are parallel to principal strain axes. Therefore, the relation of the principal values is often enough for tectonics, provided that the axial directions are known a priori. If we take the axes of Cartesian coordinates as parallel to the principal axes, the diagonal components of S and E are identical with their principal values. Using the values, the constitutive equations (Eqs. (7.2) and (7.3)) are rewritten as S 1 λ + 2G λ λ E 1 S 2 = λ λ+ 2G λ E 2 S 3 λ λ λ+ 2G E 3 σ 1 λ + 2G λ λ ε 1 σ 2 = λ λ+ 2G λ ε 2. (7.4) λ λ λ+ 2G σ 3 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 principal values are invariant to coordinate rotations and the components of the matrices are material constants that do not depend on the rotation, either. Sometimes we need to determine strain from stress. They are linearly related, so that the linear equations E = 1 + ν Y S ν Y S II, e = 1 + ν Y r ν Y σ II (7.5) meet the necessity, where Y and ν are material constants called Young s modulus 2 and Poisson s ratio. Table 7.1 shows those constants of typical rocks. Equation (7.5) is rewritten by the relation between the principal values as E 1 E 2 = 1 1 ν ν S 1 ε 1 ν 1 ν S 2 Y, ε 2 = 1 1 ν ν σ 1 ν 1 ν σ 2 Y. (7.6) ν ν 1 ν ν 1 E 3 S 4 1 The signs of stresses S and r and of strains E and e are defined not to affect the form of constitutive equations as the case of Eqs. (7.2) and (7.3). 2 Most literature assign the symbol E to Young s modulus. However, we have appropriated the symbol to infinitesimal strain, so that Y is used instead. ε 3 ε 3 σ 3

3 7.1. LINEAR ELASTICITY 155 Table 7.1: Elastic property of typical rocks [245]. ρ Y G ν 10 3 kg m 3 GPa GPa Sedimentary Rock shale sandstone limestone (non-crystalline) limestone (crystalline) metamorphic rock gneiss amphibolite igneous rock basalt granite gabbro diorite mantle material* *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 matrix of Eq. (7.6) is Y ν 1 ν ν ν ν 1 ν (ν + 1)(2ν 1). ν ν ν 1 This should be equal to the coefficient matrix of Eq. (7.4), so that we have the following relations among material constants: Yν λ λ = (7.7) ν = (7.8) (1 2ν)(1 + ν) 2(λ + G) G(3λ + 2G) Y Y = (7.9) G = λ + G 2(1 + ν). (7.10) Physical Interpretation of Y, ν and G Suppose a uniaxial compression where σ 1 0, σ 2 = σ 3 = 0. Combining these conditions and Eq. (7.6), we obtain σ 1 = Yε 1. (7.11) The subscripts in this equation indicate that stress is proportional to strain in the extending direction and Y is the constant of proportionality. This is compared to the behavior of springs that obeys the equation F = kx where F and x are force and length change of the spring and k is the spring

4 156 CHAPTER 7. ELASTICITY constant. Linear elasticity is the three-dimensional extension of Hooke s law 3, and Y is the stiffness comparable to the spring constant. Let us estimate elastic strains with using Eq. (7.11). Tectonic stresses are in the orders of MPa in tectonically active regions, and Young s modulus of rocks is about 60 GPa (Table 7.1). Therefore, elastic strain is σ/y Strains may be less than this in inactive regions. The lithosphere cannot support stresses larger than the magnitude, but brittle or plastic deformations occur to create permanent deformations that may be observable as geologic structures. The deformation relieves tectonic stresses, giving rise to the upper limit of the stresses. Suppose a cylindrical sample is subject to the uniaxial compression σ 1 > 0 and σ 2 = σ 3 = 0. If the sample is made of a linear elastic material, combining Eqs. (7.6) we have ε 1 = σ 1 /Y and ε 2 = ε 3 = νσ 1 /Y = νε 1. (7.12) Although O-2 and -3 directions are stress free, the cylindrical sample becomes thicker in those directions. This is called the Poisson effect. It should be pointed out that plain stress does not result in 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 indicates how much Poisson effect appears. If ν = 0, no such effect occurs. Common materials including rocks have the positive Poisson effect that a sample thickens in directions perpendicular to the direction of uniaxial compression. This is not satisfied if ν<0. Therefore, it is permitted to assume that ν 0. In addition, elastic strains are infinitesimal, so that volume change is δv/v = ε I = (ε 1 +ε 2 +ε 3 ) (Eq. (2.11)). Therefore, combining Eq. (7.12) we obtain δv/v = (1 2ν) ε 1. We have assumed that the sample is compressed, therefore the ratio should be negative in sign. In this case (ε 1 > 0), so that Poisson 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 similar ratios (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 the constitutive equation of linear elasticity (Eq. (7.2)) has two material parameters, λ and G, they have a 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 ratio for rocks. The shear modulus G indicates the resistance of a material to simple shear (Exercise 7.1). 3 In general, linear elasticity is expressed by the equation S = C : E, where C = (C ijkl ) is a fourth-order tensor called the tensor of elastic constants and represents the anisotropy of a material. This equation indicates proportionality between stress and strain, called the generalized Hooke s law. For isotropic materials, C is the fourth-order isotropic tensor, and the above equation reduces to Eq. (7.2). See [61] for detail.

5 7.2. EARTH PRESSURE AT REST Two-dimensional elasticity Orogenic belts and rift zones have linear belts in which vigorous deformations take place. Accordingly, it is sometimes convenient to model the deformations on the vertical section perpendicular to the belts by means of two-dimensional elasticity, where the plane strain or plane stress condition is combined with the linear elasticity. First, let us derive the constitutive equations for plane-strain problems. Consider plane strain on the O-13 plane, where the O-1 and -3 axes are taken to be parallel to the principal strain axes. In this case we have ε 2 = 0, so that the constitutive equation of linear elasticity (Eq. (7.4)) becomes σ 1 (λ + 2G)ε 1 + λε 3 σ 2 = λ(ε 1 + ε 3 ). (7.13) σ 3 λε 1 + (λ + 2G)ε 3 Eliminating ε 1 and ε 3 from Eq. (7.13) and exchanging Lamé s constants by Poisson s ratio using Eq. (7.8), we obtain the constitutive equation for the plane strain problem, σ 2 = ν (σ 1 + σ 3 ). (7.14) The deformation is incompressible if ν = 1/2. In this case, we have σ 2 = (σ 1 + σ 3 )/2, i.e., the incompressible 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 elastic behavior of the lithosphere, later. The state is represented by the principal stresses, S 1 0, S 2 0 and S 3 = 0. Substituting these into Eq. (7.6), we obtain the constitutive equation for this states E 1 E 2 = 1 S 1 νs 2 νs 1 + S 2 Y. (7.15) (S 1 + S 2 ) E 3 It should be noted that plane stress does not necessarily result in plane strain. This discordance is due 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 behave as fluid at depths. Now we study the state of stress with the assumption that rocks behave as linear elastic 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 the horizontal surface, ρ, is horizontally homogeneous but has vertical variations. We take the origin of the Cartesian coordinates O-xyz at the surface and the z-axis downward. (3) We assume that gravity is the only origin of stress. If so, one of the principal stress axes must be vertical and the rest lie horizontal. Since the system has no horizontal variation at all, we expect axial symmetry around the

6 158 CHAPTER 7. ELASTICITY vertical for the state of stress. Therefore, σ xx, σ yy and σ zz equals principal stresses with σ xx = σ yy.if rocks are isotropic, the strain field must have the same symmetry: ε xx, ε yy and ε zz are the principal strains and ε xx = ε yy. (4) Rocks are confined at depth in the crust, horizontal expansion is restricted by adjacent rock. Therefore, let us assume that the layers are horizontally constrained so as not to allow horizontal displacement: u x = u y = 0. Therefore, we have ε xx = u x x = 0, ε yy = u y y = 0. (7.16) (5) Rocks behave as linear elastic materials. This allows us to use Eq. (7.4), in which Eq. (7.16) is substituted. The result is the equation σ xx λ + 2G λ λ 0 λ σ yy = λ λ+ 2G λ 0 = λ ε zz. λ λ λ+ 2G λ + 2G σ zz ε zz It is found from this equation that σ xx = σ yy = ( ) λ σ zz. (7.17) λ + 2G Using Eqs. (7.7) and (7.10), we have λ λ + 2G = Yν (1 2ν)(1+ν) Yν (1 2ν)(1+ν) + Y (1+ν) = ν 1 2ν ν 1 2ν + 1 = ν ν + (1 2ν) = ν 1 ν. Using Eq. (7.17), we obtain ( ν ) ( ν ) σ xx = σ yy = σ zz = p L, 1 ν 1 ν (7.18) where p L stands for overburden (Eq. (3.26)). This is called earth pressure at rest which is the state of stress that is induced by gravity in a horizontally layered rock mass. Hence, let us write the state as ν/(1 ν) 0 0 r gravity = p L 0 ν/(1 ν) 0. (7.19) We are able to use earth pressure at rest as the reference for the state of stress at depths instead of the 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 < (σ xx,σ yy ) σ zz holds, indicating that horizontal stress is less than the overburden. Rocks are vertically constricted by overburden but are horizontally extended by the Poisson effect. In the extreme case of ν = 1/2, Eq. (7.18) becomes σ xx = σ yy = σ zz. This is equivalent to the lithostatic state of stress, though the media is not assumed to be fluid but elastic.

7 7.3. STABILITY OF ELASTIC ROCK MASSES 159 Since Poisson s ratio of representative rocks can be approximated as ν 1/4 (p. 156), the coefficient K = σ Hmean /σ v is approximately equal to 1/3. Horizontal stress is about one-third of overburden. In-situ stress measurements conducted in various regions of the world show that a few regions 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 model predicts the differential stress Δσ = (1 K)p L, which piles up unlimitedly with increasing depth. However, real rocks have their own yield stress. The yielding may relief the differential stress to cause convergence in the ratio. This is perhaps at least one of the reasons for the convergence. The stress state discussed above is an idealized state, convenient for theoretical considerations, but we should be careful to apply the model to real tectonics. Rocks behave as elastic materials in a human timescale. 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 is under the frictional strength of rocks. If it is not so, rocks may collapse through normal faulting on inclined pre-existing fractures without tectonic stress. The Earth pressure at rest has the principal stresses ( ν ) σ v = p L >σ H = σ h = p L. (7.20) 1 ν Figure 7.1a shows this stress state. In one extreme case, ν = 0, this is a uniaxial stress σ H = σ h = 0. The Mohr circles corresponding to various depths pass the origin of the Mohr diagram so that the line of frictional strength always cuts the circles, i.e., the elastic rock body with ν = 0 is unstable. In another extreme case, ν = 1/2, we have lithostatic states σ v = σ H = σ h = p L. Then no deformation occurs at all and the elastic rock body is always stable. For the general case where 0 <ν<1/2, the Mohr circles has a linear envelope passing the origin of the Mohr diagram for the proportionality of all the principal stresses with p L (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 = 1 2 ( p L + ν ) 1 ν p L = 1 2(1 ν) p L, CD = 1 ( p L 2 ν ) 1 ν p L = 2ν 1 2(1 ν) p L, ( ν ) 1/2 OD = c 2 r 2 = pl. 1 ν Consequently, we obtain the slope CD 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 representative Poisson ratio for rocks, we have a slope of This is slightly smaller than the coefficient of

8 160 CHAPTER 7. ELASTICITY Figure 7.1: (a) Mohr diagram showing the Earth pressure at rest. (b) Relationship between ν and the slope in (a). friction. Consequently, the rocks are stable under the Earth pressure at rest. Rocks with small Poisson ratios may be unstable. We have neglected the effect of pore pressure in the above argument. However, pore pressure plays a critical role in the stability because increased pore pressure shifts the Mohr circle to the left and destabilizes rocks with ordinary Poisson ratios. Therefore, earthquakes can occur on pre-existing fracture 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 from other factors. Since temperature increases generally with depth, depth changes accompanied by exhumation 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 change of ΔT. Experiments show that the proportionality E = α l ΔT holds for a rod of rock if the length of the rod is unconstrained. The constant of proportionality α l is known as the coefficient of linear expansion. Rocks have a coefficient of about 10 5 K 1. The temperature change results in an increased volume of a cube given by ΔV V = (L +ΔL)3 L 3 L 3 = ( L +ΔL L ) 3 1 = 3 ΔL L + 3 ( ) 2 ΔL + L ( ) 3 ΔL 3 ΔL L L, where L is the initial length of the sides of the cube and higher-order terms are neglected. Using ΔL = LE,wehave ΔV V 3E = 3α lδt.

9 7.4. THERMAL STRESS 161 Accordingly, we define the coefficient of volume expansion or simply thermal expansion coefficient α so as to satisfy the relationship ΔV/V = αδt. Obviously, we have 3α l α. (7.21) Rocks have the thermal expansion coefficients at around K 1. The coefficient of linear expansion of a single crystal has anisotropy resulting from its crystallographic structure. However, a rock mass consists of a polycrystalline material made of minerals usually with random crystallographic orientations. Therefore, we assume that α l is isotropic when we consider the stress or strain of rocks. This is formulated as ε 1 = ε 2 = ε 3 = α l ΔT. (7.22) Incremental deformation against viscous forces dissipates kinetic energy (Eq. (3.41) to affect the temperature field, i.e., mechanical and thermal processes are generally coupled. However, in many situations their interaction can be neglected. The resulting analysis is known as the uncoupled thermoelastic theory of continua. In this case, strain is considered to be the sum of the strains caused by 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 + ν Y ( r νσi Y r = 2Ge + ) + α lδt ( λε I + Yα lδt 1 2ν Using principal stresses and strains, Eq. (7.23) is rewritten as I, (7.23) ) I. (7.24) ε 1 = 1 Y σ 1 ν Y σ 2 ν Y σ 3 α l ΔT, ε 2 = ν Y σ Y σ 2 ν Y σ 3 α l ΔT, (7.25) ε 3 = ν Y σ 1 ν Y σ Y σ 3 α l ΔT. 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 degrees leads to a stress change of Yα 1 ΔT 10 2 MPa, as the representative Young s modulus of rocks is about Y = 60 GPa. Since the tensile strength of rocks is in the order of 10 1 MPa, this thermal stress easily forms fractures in the solidified igneous body, called cooling joints. Cooling of a tabular intrusive body splits the body into columns to form columnar joints perpendicular to the intruding plane (Fig. 7.2).

10 162 CHAPTER 7. ELASTICITY Figure 7.2: Doleritic dike penetrating purple granite and overlying volcanic rocks. The dike has columnar 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 surface is gradually buried in a sedimentary basin, the layer experiences horizontal extension. On the other hand, the temperature of the layer increases with depth, leading to horizontal compression if the layer 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 assume uncouple mechanical and thermal processes so that incremental stresses accompanied by an increases of temperature and overburden are considered separately. The layer is horizontally constrained but can freely displace the overlying strata by its thermal expansion, i.e., the layer is not vertically constrained. Therefore, the thermal expansion does not result in an increase of the vertical stress component, i.e., σzz = 0 for a contribution from the

11 7.4. THERMAL STRESS 163 temperature increase. Substituting this vertical stress into Eq. (7.25), we have ε xx = 1 Y (σ xx νσ yy ) α l ΔT ε yy = 1 Y ( νσ xx + σ yy ) α l ΔT (7.26) ε zz = ν Y ( σ xx σ yy ) α l ΔT Combining the condition of horizontal constraints ε xx = ε yy = 0, we obtain σ xx = σ yy = Yα lδt 1 ν (7.27) for the incremental stress components due to the temperature increase. Let Γ be a geothermal gradient, then the temperature at depth z is given by T = T 0 +Γz, where T 0 is the surface temperature. The layer had no strain at all at the surface, therefore ΔT =Γz. Substituting this temperature difference into Eq. (7.27), we obtain σ xx = σ yy = Yα lγz 1 ν. (7.28) The differential stress due to the temperature difference equals Δσ = ( ) Yαl Γ z. (7.29) 1 ν Geothermal gradients are usually a few degrees per 100 m except for geothermal areas. Using the values Γ=0.01 K m 1, Y = 60 GPa, ν = 0.25, and α l = 10 5 K 1, we obtain Δσ/z = 8kPam 1, and the gradient of overburden stress ρg = 2.4kPam 1. Consequently, the thermal stress is as great as othe verburden stress for a layer that was consolidated at the surface 4. Polygonal fracture pattern on Venus Venusian plain regions have polygonal terrains that are characterized polygonal fracture network by relatively uniform spacing (Fig. 7.3). The random orientations of the fractures suggest isotropic small-strain deformations, probably reflecting the thermal stress included by decreased surface temperature over a billion years [93]. Venus was resurfaced by global massive volcanism a few billion years ago. Greenhouse effect gases such as H 2 O and SO 2 emitted by the volcanism raised the surface temperature by more than 100 K. Later atmospheric cooling contracted rocks in the crust. Just like the half-space cooling model of the oceanic lithosphere ( 3.12), rocks at shallow levels in the crust are cooled swifter than those at deep levels. The time lag induces a variation in the magnitude of horizontal contraction to 4 This model assumes that a horizontally extensive elastic layer subsided neither with jointing nor faulting in a geological time scale. This is not probable in reality. Fractures between blocks may allow horizontal displacements, violating the condition of horizontal constraints.

12 164 CHAPTER 7. ELASTICITY Figure 7.3: Polygonal terrain on Venus [5]. eventually form the polygonal fractures eventually [5]. Let us estimate the contraction and associated thermal stress by the climate change. Consider the temperature changes with time as T = A sin ωt at the surface z = 0. The temperature in the half-space z 0 is calculated by the heat conduction equation T/ t = κ 2 T/ z 2. Here, T represents the temperature anomaly. Substituting the solution of the form T = u(z)e iωt,wehave d 2 u/dz 2 = (iω/κ)u. Using the condition that the solution does not diverge to infinity at z =, we obtain 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 = 100 K, ω = 1 rad/ga s 1, and κ = K 1,wehaveΓ AB 10 K/km. Assuming constrained horizontal displacement, the thermal stress between different depths Δz is evaluated as Eq. (7.29), i.e., we have ( ) Yαl Γ σ Δz. 1 ν Using Y = 60 GPa and ν = 0.25, we have Yα l Γ/(1 ν) 1000 Pa m 1. On the other hand, the gravitational acceleration at on the Venusian surface is g 9ms 1 so that the gradient of overburden stress is ρg 3000 Pa m 1. Therefore, the thermal stress is comparable and if the surface temperature drop was a few hundred degrees, the thermal stress becomes large enough to produce map-scale fractures 5. 5 See [5] for detail.

13 7.5. GLOBAL THERMAL CHANGES AND SURFACE STRESS FIELD Global thermal changes and surface stress field Terrestrial planets and satellites experienced drastic global cooling in their early history. Except for the Earth and Venus which have active geological processes including tectonism and erosion, tectonic features are clues to their early thermal history. For example, it is suggested that the cessation of 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 properties have spherical symmetry. The symmetry leads to a vertically axisymmetric stress field. Let S V and S H be the vertical and horizontal stresses. Here, we assume a solid planetary body, so that the vertical and horizontal stresses can be different. The unit volume in the body is subject to the gravitational force ρg, where the density ρ and gravitational acceleration g are functions of the radial coordinate r. The body force and the stress components shown in Fig. 7.4 are found to lead to an equillibrium equation in the form 6 ds V dr + 2 r (S V S H ) ρg = 0. (7.30) Thermal expansion of the body is very small compared to the radius of the planetary body R,so that the former does not depend on temperature. Therefore, let us utilize the uncoupled thermoelastic theory to investigate thermal stresses in the body. Stresses due to gravity and to temperature change can be treated separately. Hence, the gravity term ρg is deleted from Eq. (7.30) to estimate the stress field due to temperature changes. Consequently, the equation of thermal stress (Eq. (7.25)) is rewritten as E V αδt = 1 Y (S V 2νS H ), (7.31) E H αδt = 1 [ ] S H ν(s V + S H ), (7.32) Y where S V and S H are vertical and horizontal infinitesimal strain components due to temperature change ΔT. These strain components are related to the vertical displacement u by the equations E V = du dr, E H = u r. (7.33) Rearranging Eqs. (7.31) and (7.32), we have S V = [ ] Y (1 ν)e V + 2νE H (1 + ν)αδt, S H = (1 + ν)(1 2ν) [ ] Y E H + νe V (1 + ν)αδt (1 + ν)(1 2ν). (7.34) 6 Equation (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 S V S H vanished.

14 166 CHAPTER 7. ELASTICITY Figure 7.4: Force balance of a small circular truncated cone with a thickness of dr and an apical angle of dφ. Subsitituting these equations into Eq. (7.30), we have the differential equation of u, d 2 u dr + 2 du 2 r dr 2u ( 1 + ν ) dδt r = α 1 ν dr. Rearranging this equation, we obtain [ d 1 d ( r 2 u )] 1 + ν ) dδt = α( dr r dr 1 ν dr, which is integrated to give u = α ( 1 + ν ) r ΔTr 2 dr + C r 2 1 r + C 2 1 ν r 0 r, (7.35) where C 1 and C 2 are constants of integration and r 0 designates the base of a spherical shell. Combining this differential equation and Eqs. (7.33) and (7.34), we obtain the equations [246], S V = 2αY (1 ν)r 3 S H = αy (1 ν)r 3 r r r 0 ΔTr 2 dr + YC 1 1 2ν 2YC 2 (1 + ν)r 3, (7.36) ΔTr 2 dr + YC 1 r 0 1 2ν + YC 2 αy ΔT (1 + ν)r3 1 ν. (7.37)

15 7.5. GLOBAL THERMAL CHANGES AND SURFACE STRESS FIELD 167 The constant C 2 is determined through Eq. (7.35) by the vanishing displacement at the center r = 0 as C 2 = 0 because of 1 r lim ΔTr 2 dr = 0. r 0 r 2 0 Let ΔT 0 be the temperature change at the center, then we have 1 r lim ΔTr 2 dr = ΔT 0 r 0 r Therefore, the other constant is constrained by assuming a free surface at r = R as YC 1 1 2ν = 2αY 1 1 ν R 3 Consequently, Eqs. (7.36) and (7.37) become S V = 2αY 1 ν S H = αy 1 ν ( 1 R R 3 0 R ( 2 R 3 0 R 0 ΔTr 2 dr. ΔTr 2 dr 1 r ) ΔTr 2 dr, r 3 0 ΔTr 2 dr + 1 r ) ΔTr 2 dr ΔT. r 3 0 Now consider a planetary body being composed of n spherical shells, where the top is the 0th layer. The shells have different material constants but the constants are uniform within each layer. Let Y i and ν i be Young s modulus and Poisson s ratio of the ith layer. Then, we have R 3 u(r) = ri(r) + A i r + B i r, (7.38) 2 S V (r) = 2Y ii(r) + Y ia i 2Y ib i R ν i 1 2ν i (1 + ν i )r, 3 S H (r) = Y ii(r) + Y ia i + Y ib i R ν i 1 2ν i (1 + ν i )r Y iα(r)δt (r), (7.39) 3 3(1 ν i ) where 1 + ν r I(r) = α(r)δt (r)r 2 dr (7.40) 3(1 ν)r 3 0 and A i and B i are dimensionless numbers corresponding to the following three conditions on the stress field [218]. Namely, (1) there is no singularity at the center, (2) no discontinuity between layers, and (3) the surface of the planetary body is free, S V = 0. The integral in Eq. (7.40) equals the thermal expansion within the radius of r. The goal of this argument is to compare the surface horizontal stress S r=r H and observed tectonic 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, 2Y 0I(R) 1 + ν 0 + Y 0A 0 1 2ν 0 2Y 0B 0 1 ν 0 = 0.

16 168 CHAPTER 7. ELASTICITY Rearranging this equation, we get B 0 = ν 0 A 0 I(R). (7.41) 2 1 2ν 0 Combining Eqs. (7.38), (7.39) and (7.41), we obtain the displacement and horizontal stress of the surface layer (r = R), u(r) = 3 R(1 ν 0 )A 0, (7.42) 2 1 2ν 0 S H (R) = 3 Y 0 A 0. (7.43) 2 1 2ν 0 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 S H (R) = Y 0 ΔR 1 ν 0 R. (7.44) Consequently, the surface horizontal stress determined only by the stretch of the radius ΔR/R and the elastic constants of the surface layer, independent from the physical properties of the inside. The ratio ΔR/R reflects the internal changes. The activity of thrust faults (Fig. 2.14) contracted Mercury 3 4 billion years ago. The orientation and distribution of these thrust faults appear to be rondom, the contraction in the radius ΔR was roughly estimated via Eq. (2.56) at minus 1 2 km, and is interpreted as the surface manifestation of 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) Substituting this ratio with Y = 60 GPa and ν = 0.25 as the representative values of rocks into Eq. (7.44), we obtain the surface horizontal stress at S H = MPa. According to the theory of thermal evolution, that amount of contraction is explained well by the heat loss just as the amount of heat production by core formation alone, and the total freezing of the core can lead to an 8 km contraction of the radius [217]. The contraction estimated from thrust faults 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 are evidenced by wrinkle ridges and linear rilles, respectively. In addition, those features appear to reflect regional tectonics characteristic of mare basins ( 8.8). Global cooling left a less conspicuous pattern of tectonic features on the Moon than on Mercury. The lack of a pattern suggests the global expansion as small as ΔR < 1 km for the Moon [129]. This corresponds to S H < 50 MPa. However, the wrinkle ridges still suggest global cooling, as extensional tectonics terminated at 3.6 Ga on the Moon, and horizontal compression prevailed thereafter. Too hot initial conditions lead to a large contraction, which is inconsistent with the tectonic features on the Moon. Using these constraints, Solomon et al. [219] estimated the depth of the magma ocean at about 200 km at the beginning of the history of the Moon. 7 Several researchers raise objections to the randomness, as the surface of the Mercury has not been globally photographed with enough resolution [147, 241].

17 7.6. EXERCISES Exercises 7.1 Demonstrate that the shear modulus G indicates the resistance of a linearly elastic material to simple shear. 7.2 The Venusian surface has polygonal fractures, each of which has a diameter of several kilometers (Fig. 7.3). The spacing between the fractures is thought to reflect the depth extent of crack propagation. Derive the equation ΔS H = ΔΓαY 1 ν ( te 2 z ), describing the change in the surface horizontal stress due to the change in the geothermal gradient ΔΓ, where t e 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].