Plasticity - TERRAPUB · They include plastic body, Bingham, and power-law ﬂuids. The equations...

Chapter 10

Plasticity

Fluids with a non-linear constitutive equation are introduced in this chapter.They include plastic body, Bingham, and power-law fluids. The equations ofplasticity are applied to the deformation of impact craters. Rocks behave aspower-law fluids whose constitutive equation is used in Chapter 12 to investi-gate the mechanics of the lithosphere.

10.1 Quasilinear Fluids

The constitutive equation of Newtonian fluid (Eq. (9.3)) has a viscosity η that does not depend onfluid motions. Let us broaden this limitation, i.e., we assume isotropy, incompressibility (DI = 0),and the dependency of viscosity on D (Fig. 10.1). This fluid is used to model tectonic deformationsin Chapter 10. The function η(D) must satisfy the representation theorem so that the viscosity is afunction of basic invariants DII and DIII. The second law of thermodynamics applies to this fluidflow. It is known that the law demands that the function does not include DIII [276]. Consequently,the variable viscosity fluids have a constitutive equation of the form

T = 2η(DII)D (10.1)

and is spoken of as quasilinear fluids. In this case, η is called shear-rate dependent viscosity. Quasi-linear fluids include power-law fluids, Bingham fluids, and plastic bodies which are collectivelycalled viscoplastic bodies.

Attention must be paid to the sign of DII. We have assumed incompressibility (DI = trace D =0), so that we can presume D as a deviatoric tensor (see Appendix C.6). The same is true forthe strain rate tensor Ė, because D is identical with Ė (Eq. (2.27)). Therefore, their second basic

237

238 CHAPTER 10. PLASTICITY

Figure 10.1: Deviatoric stress T versus velocity gradient D for quasilinear fluids. The inclinationof the segment OP equals twice the effective viscosity, which decreases with increasing velocitygradient for pseudoplastic fluids. That of dilatant fluids increases. SY is the yield stress.

invariants satisfy the inequalities (Eq. (C.50)):

DII =12

(D : D) ≥ 0, ĖII =12

(Ė : Ė

) ≥ 0. (10.2)Consider a simple shearing motion on the O-12 coordinate plane to see the basic behavior of

quasilinear fluids. The stretching tensor for this flow is

D =

⎛⎝ 0 D 0D 0 00 0 0

⎞⎠ . (10.3)Substituting Eq. (10.3) into (10.1), we find that the two components T12 = T21 = T are the onlynon-zero components of T. Accordingly, the equation

T = 2aD +2kπ

tan−1D

b(10.4)

is a constitutive equation for this flow, and the graph for the equation is given by the convex upwardcurve in Fig. 10.1. Note that Eq. (10.4) includes the constitutive equation of Newtonian fluid asthe case where k = 0. In the case of a = 0 and b → 0, Eq. (10.4) is a constitutive equation of arigid-perfectly-plastic body.

The shear-rate-dependent viscosity is half of the slope of the line segment OP in this figure, sothat we have

2η(D) = 2a +2kπD

tan−1D

b. (10.5)

This viscosity increases with D. This is spoken of as shear-thinning viscosity. Fluids with this typeof viscosity are called pseudoplastic fluids. By contrast, shear-thickening viscosity increases withD, and fluids with this type of viscosity are called dilatant fluids.

10.2. PRINCIPAL STRESS SPACE 239

Taking the limit b → ∞ and replacing k by SY, Eq. (10.4) becomes

T = 2aD + SY, (10.6)

the graph of which is shown in the two line segments that meet on the vertical axis at S = SY. Thisis called Bingham fluid, and SY is of that fluid. The fluid has a viscosity

2η(D) = 2a + SY/D (10.7)

that goes to infinity withD → 0. Therefore, the decrease ofD increases the viscosity. The fluid flowis consequently rapidly decelerated. Bingham fluids start to flow when a driving force overcomesthe yield stress. Under the critical stress, the fluid behaves as a rigid or elastic body.

For the simple shear flow, DII = 12 D : D = 2D2. Accordingly, the expression for the viscosity of

Bingham fluids (Eq. (10.7)) may be broadened for general flow cases as

2η(DII) = 2a +SY√DII/2

. (10.8)

The constitutive equation of Bingham fluids is therefore

T = 2

(a +

SY√2DII

)D

(D �= O) . (10.9)

Let us consider the simple shear flow (Eq. (10.3)), again, to derive the relationship between SYand T. Substituting DII = 2D2 into Eq. (10.9), we have

T =(

2a +SYD

)D =

⎛⎝ 0 2aD + SY 02aD + SY 0 00 0 0

⎞⎠ , T2 =⎛⎝T 212 0 00 T 212 0

0 0 0

⎞⎠ .The second basic invariant of the deviatoric stress is TII = − 12

[(trace T)2 − trace T2], so that

limD→0

TII = S2Y.

This indicates the fluid yields when TII reaches a critical level1.

10.2 Principal stress space

If a rock is isotropic, its mechanical properties are the same in any direction, only the magnitudeof the principal stresses plays a role in describing the yield or failure behavior. Hence, we use athree-dimensional stress space where three principal stresses are the coordinate axes O-σ1σ2σ3 orO-S1S2S3 (Fig. 10.2(a)). The states of hydrostatic and lithostatic stresses S1 = S2 = S3 are

1This is the yield criterion of von Mises, which is introduced in Section 10.3.


Figure 10.2: (a) Principal stress space. A point on the line S1 = S2 = S3 represents a hydrostaticstate of stress with the pressure S0. The distance from the line indicates deviatoric stress . Perpen-dicular to the line is called a deviatoric plane. The equation S1 + S2 + S3 = const. represents theplane. If S0 = 0, the plane is called a π-plane. (b) Θ is an angle on the π-plane and represents thesymmetry of Lamé’s stress ellipsoid.

represented by the points that make up a line parallel to the unit vector e⊥ = (1, 1, 1)/√

3. Planesperpendicular to the line are called deviatoric planes. The point P in Fig. 10.2 is an example. Thosepoints designate isotropic stresses, whereas points out of the line designate anisotropic stresses. Thepoint Q represents those points, and has the position vector

−−→OQ = (S1, S2, S3)T. If the line PQ is

perpendicular to the line OP, we have∣∣−→OP∣∣ = −−→OQ · e⊥ = SI√3=√

3S0. (10.10)

Hence, we have−→OP =

∣∣−→OP∣∣ e⊥ = √3S0 · (1, 1, 1)T/√3 = (S0, S0, S0)T, and further−→PQ =

−−→OQ − −→OP =

⎛⎝S1 − S0S2 − S0S3 − S0

⎞⎠ =⎛⎝T1T2T3

⎞⎠ . (10.11)Therefore, the length

∣∣−→PQ∣∣ is related to the second basic invariant of T and the octahedral shearstress by the equation ∣∣−→PQ∣∣ =√T 21 + T 22 + T 23 =√2TII =√3 SoctS , (10.12)where Eqs. (4.14) and (4.17) are used. Namely, TII and SoctS indicate the magnitude of anisotropy.

The plane passing through the origin O, and perpendicular to the line OP, is known as the π-plane in the theory of plasticity. Let O-S ′i be the orthogonal projection of the coordinate axes O-Si

10.2. PRINCIPAL STRESS SPACE 241

onto the plane (Fig. 10.2(b)). Obviously, the S ′1-, S′2- and S

′3-axes meet at 120

◦. Let the point Q′

be the foot of the base of the perpendicular dropped from Q (S1, S2, S3) onto the π-plane, thenQ′ has the coordinates (T1, T2, T3) (Eq. (10.11)). Let Θ be the angle between OQ′ and the S ′1-axis. This is known as Lode angle. The orthogonal projection has the projector (I − e⊥e⊥). TheS1-axis has the base vector e = (1, 0, 0)T, so that the unit vector e′ parallel to the S ′1-axis satisfiese′ ∝ (I − e⊥e⊥) · e = e − (e⊥ · e) e⊥ = (2, −1, −1)T/3. Therefore, we have

e′ =1√6

(2, −1, −1)T, (10.13)

and−−→OQ′ · e′ =

∣∣∣−−→OQ′∣∣∣ cosΘ = ∣∣∣−→PQ∣∣∣ cosΘ =√2TII cosΘ. Substituting Eq. (10.11), we have−−→OQ′ · e⊥ =

−→PQ · (2, −1, −1)√

6=

(T1, T2, T3) · (2, −1, −1)√6

=2T1 − T2 − T3√

6=

√3√2T1,

∴ cosΘ =√

32

S1TII. (10.14)

Using the formula cos 3Θ = 4 cos3 Θ − 3 cosΘ, this is rearranged as

cos 3Θ =3√

32

(TII)−3/2(T 31 + T1TII

). (10.15)

Eliminating T1 from Eq. (10.15) with the three equations,

T1 + T2 + T3 = 0, TII = −(T1T2 + T2T3 + T3T1), TIII = T1T2T3,

we obtain

cos 3Θ =3√

32TIII (TII)−3/2 (0 ≤ Θ ≤ 60◦). (10.16)

Rearranging Eq. (10.14), we obtain

T1 =2√TII√3

cosΘ, T2 =2√TII√3

cos(120◦ − Θ), T3 =2√TII√3

cos(120◦ + Θ) (10.17)

and

S1 =SI3

+2√TII√3

cosΘ, S2 =SI3

+2√TII√3

cos(120◦ − Θ), S3 =SI3

+2√TII√3

cos(120◦ + Θ),

(10.18)where 0 ≤ Θ ≤ 60◦. Principal axes exchange their orientations when Θ value exceeds the multiplesof 60◦. Stresses with these Θ values are axial stresses (Fig. 10.3(a)). Φ and Θ have a one-to-onecorrespondence (Fig. 10.3(b)),

Φ =2 sinΘ√

3 cosΘ + sinΘ, (10.19)

which is derived from Eq. (10.17), provided that Θ is in the range [0, 60◦].


Figure 10.3: (a) Diagram showing the relationship between Mohr circles and Lode angle. (b) Graphof Eq. (10.19).

10.3 Yield conditions

At the limits of a → 0 and b → 0 in Eq. (10.4), the pseudoplastic fluid becomes a perfectly plasticfluid that yields when TII = 2k2 is satisfied. In this section, a phenomenological theory of yieldingis introduced.

For simplicity, we assume an isotropic and incompressible plastic material. The material yieldswhen the state of stress satisfies a yield condition, which is represented by the function of stresstensor F (S) = 0, called the yield function. Since an isotropic material is assumed, this function isdepends only on the principal stresses. That is, the function is rearranged as F (S1, S2, S3) = 0,which is a function of the points in the stress space and defines a surface in the space, known as theyield surface. Using Eqs. (10.16) and (10.18), the principal stresses can be replaced by TII and TIIIso that

F (S0, TII, TIII) = 0 (10.20)

also represents the yield condition.Laboratory experiments show that the yield conditions of metals do not depend on mean stress.

However, mean stress is not negligible for porous materials including rocks, indicating that overbur-den stress is significant for yielding. S0 in Eq. (10.20) indicates this dependency. For simplicity,

10.3. YIELD CONDITIONS 243

Figure 10.4: Yield surfaces for yield criteria in three-dimensional stress space. Dot-bar lines desig-nate the line σ1 = σ2 = σ3.

we first introduce the yield conditions that have no such dependence. In that case, the yield functionbecomes F (TII, TIII) = 0.

Tresca yield criterion

The Tresca yield criterion assumes that plastic yielding occurs when the maximum shear stressreaches the critical value cY. Since the maximum shear stress equals half the differential stress (Eq.(4.18)), we have the yield condition

12ΔS =

12

(S1 − S3) = cY. (10.21)

When ΔS/2 < cY, deformation of the material is accommodated by elasticity. The Tresca crite-rion does not include the intermediate principal stress in contrast to the following von Mises yieldcriterion. Neglecting the inequality S3 ≤ S2 ≤ S1, we obtain the yield condition

max(∣∣S1 − S2∣∣, ∣∣S2 − S3∣∣, ∣∣S3 − S1∣∣) − 2cY = 0.

This is rearranged as∣∣S1 − S2∣∣ = 2cY or ∣∣S2 − S3∣∣ = 2cY or ∣∣S3 − S1∣∣ = 2cY. (10.22)Each of these equations designates a plane in the stress space. The plane represented by

∣∣S1−S2∣∣ =2cY is perpendicular to the unit vectors ±(1, −1, 0)T/

√2, which with the vector e′ in Eq. (10.13)

make an angle of

cos−1[± (1, −1, 0)

T

√2

· (2, −1, −1)T

√6

]= 30◦ or 150◦.


Figure 10.5: Yield locus on the π-plane of three yield criteria.

The vector e′ is parallel to the S ′1-axis. The same angles are obtained from the equations∣∣S2 −

S3∣∣ = 2cY and ∣∣S3 − S1∣∣ = 2cY in Eq. (10.22), so that the solid figure indicated by this equation

is a hexagonal prism with the axis perpendicular to the π-plane (Fig. 10.4(a)). If a stress stateis designated by a point within the prism, the deformation of the material is accommodated byelasticity. The figure is the yield surface of the Tresca criterion. The regular pentagon in Fig. 10.5shows the yield locus of the criterion on the π-plane.

Von Mises yield criterion

The condition known as von Mises yield criterion assumes that plastic yielding occurs when elasticstrain energy reaches a critical value. Consider a linear spring under a tensile force F , which isrelated to the displacement u and the constant k as F = ku. The energy stored in the spring is

U =∫ u

0Fdu =

∫ u0ku du =

12ku2 =

12Fu

which is generalized for a material with linear elasticity as

U =12

S : E. (10.23)

This is divided into two parts corresponding to the changes of volume and shape. Substituting

E =1 + νY

S − νY

(trace S) I

into Eq. (10.23), we obtain

U =12

S :[

1 + νY

S − νY

(trace S) I]=

1 + ν2Y

S : S − ν2Y

(trace S)2.

10.3. YIELD CONDITIONS 245

Combining S = T + S0I = T + (trace S) I/3 we have

U =1 + ν2Y

(T + S0I) : (T + S0I) −ν

2Y

[trace (T + S0I)

]2.

Since TI = 0 (Eq. (4.12)), we obtain

U =

US︷︸︸︷1 + ν2Y

T : T +

UV︷︸︸︷3(1 − 2ν)

2YS0, (10.24)

where US and UV are strain anergy corresponding to changing the shape and volume, respectively.Using the second basic invariant TII = 12 T : T, we have

US =1 + νY

TII. (10.25)

Using Eq. (4.17), this is rewritten as

US =3(1 + ν)

2Y

(SoctS

) 2. (10.26)The von Mises criterion assumes that plastic yielding occurs when this energy reaches a critical

value. Namely, the yield condition for this criterion is

TII − c2Y or32

(SoctS

) 2 = c2Y. (10.27)Combining Eqs. (10.12) and (10.27), we obtain |−−→PQ| = √2cY, which does not depend on Θ, i.e.,the yield surface is a cylinder with a radius of

√2cY (Fig. 10.4).

Unlike the Tresca criterion, the yield condition of the von Mises criterion includes the inter-mediate principal stress via TII (Fig. 4.3). If the Tresca and von Mises criteria are assumed for amaterial, the critical differential stresses predicted by the criteria should be equal to each other foraxial stresses. Figure 10.5 shows the yield loci for the criteria on the π-plane. The critical differ-ential stress predicted from the Tresca criterion is smaller than that from the von Mises criterion.The loci of the former is represented by a regular hexagon inscribing the circle that represents thelatter. The difference reaches the maximum at Θ = 30◦ or, equivalently, Φ = 1/2. At that time,∣∣−→PQ∣∣ for the criteria is different by 1 − cos 30◦ ≈ 15%. Since ∣∣−→PQ∣∣ is proportional to SoctS (Eqs.(10.12)), which is further proportional to ΔS, the difference in ΔS is about 15 %, also. Laboratoryexperiments demonstrate that the von Mises criterion is more appropriate for rocks than the Trescacriterion. However, the maximum difference is small and, therefore, is sometimes neglected.

Coulomb-Navier yield condition

The Coulomb-Navier criterion is designated by an inclined line in the Mohr diagram (Fig. 10.6),and its yield condition is represented by

σ1 − σ3 = 2τ0 cosφ + (σ1 + σ3) sinφ. (10.28)


Figure 10.6: Schematic picture showing the Coulomb-Navier criterion.

Using Θ and basic invariants, this is rewritten as

−σI sinφ +12√τII

[√3(3 − sinφ) cosΘ + 3(1 + sinφ) sinΘ

]− 3τ0 cosφ = 0, (10.29)

where 0 ≤ Θ ≤ π/3. The yield loci and yield surface corresponding to this criterion are representedby a hexagon (Fig. 10.5) and a hexagonal pyramid (Fig. 10.4(c)), respectively.

Extended Tresca and extended von Mises criteria

Unlike metals, porous materials including rocks have yield a stress depending on the confining pres-sure. An extended Tresca yield criterion assumes that the critical values is a function of the meanstress as

ΔS = f (S0). (10.30)

Likewise, the yield condition for the extended von Mises yield criterion is represented by√TII = g(S0). (10.31)

If the function g(S0) is linear, it is called a Drucker-Prager yield criterion. Laboratory experimentsdemonstrate the S2 dependence of the yield conditions, and the extended von Mises criterion is betterthan the extended Tresca criterion [154]. It is shown that the function g in Eq. (10.31) is not linear[212].

10.4 Plastic yielding by folding

Hinge zone of a fold is the most highly curved zone near the hinge line of the fold. Under Kirchhoff’shypothesis, layer-parallel normal stress is proportional to curvature (Eqs. (8.9) and (8.12)) (Fig.10.7(a)). Accordingly, plastic yielding can occur in the hinge zone when a thin elastic-plastic plateis folded. In order to consider the yielding, we firstly calculate stress in the zone. Let us use thesame coordinates with those in Section 8.2. Namely, O-xy plane is defined by the middle surface

10.4. PLASTIC YIELDING BY FOLDING 247

of the plate when it was flat. Plane strains on O-xz plane is assumed. The displacement of themiddle surface is assumed to be small (w � 1), so that strain and stress have the principal axesapproximately parallel to the coordinate axes. Consider the correspondence, ε1 = εxx, ε2 = εyy andε3 = εzz. Neglecting the effect of overlying and underlying layers, we have σzz = 0. The curvatureis approximated as K = w′′, so that we have

εxx = −w′′z, (10.32)

where z is the distance from the middle surface and w′′ = d2w/dx2. In this case, Eq. (8.9) isrewritten as σxx = [Y/(1 − ν2)]εxx. Substituting Eq. (10.32) into this equation, we have

σxx = −Y z

1 − ν2w′′. (10.33)

Equation (7.13) is expressed in the present coordinates as

σxx = (λ + 2G)εxx + λεzz, σyy = λ(εxx + εzz), σzz = λεxx + (λ + 2G)εzz.

Substituting σzz = 0 and Eq. (10.33) into the above three equations, and combining Eqs. (7.7) and(7.10) to eliminate λ and G, we obtain the intermediate principal stress

σyy =Y νz

1 − ν2w′′ = −νσxx.

Now we know all the principal stresses and have the octahedral shear stress(σoctS

)2 = 19

[(σx − σy)2 + (σy − σz)2 + (σz − σx)2

]=

29

(1 + ν + ν2

)σ2x.

The yield condition for the von Mises criterion (Eq. (10.27)) is indicated by 2S2Y = (σxx − σyy)2 +(σyy − σzz)2 + (σzz − σxx)2, where SY is the critical stress. Then, yielding occurs when σxx satisfies

σxx =SY√

1 + ν + ν2.

Let H be the thickness of the plate, then the normal stress at its base (z = H/2) is given by Eq.(10.33), i.e.,

σxx

∣∣∣z=H/2

= − YHw′′

2(1 − ν2) . (10.34)

The zones of plastic deformation appear at the top and base of the plate along the hinge line, andgradually expand toward the middle surface (Fig. 10.7(b)). The elastic core is the central part wheredeformation is accommodated by elasticity. The elastic core is eroded at both sides and eventuallydisappears. Let w′′0 be the curvature when the yielding begin, and cY be σxx

∣∣z=H/2. Then, Eq. (10.34)

is rearranged as

w′′0 = −2(1 − ν2)cY

YH, (10.35)


Figure 10.7: (a) Stress states in the hinge zone of a fold. (b) Plastic yielding of a thin elastic-plasticplate.

indicating the critical curvature for the yielding to begin.The elastic core is attenuated with increasing curvature w′′. If a vertical load exerted on the

lithosphere increases or abruptly appears, the lithosphere is firstly bent as an elastic plate. However,the elastic core is gradually thinned. Whole lithosphere failure is said to occur if the elastic core inthe lithosphere eventually disappears. However, it takes a time to thin the elastic core so that yieldingmay be negligible for outer trench swells that are accompanied by rapid subduction.

10.5 Slip line theory

If a rock mass has numberless fracture surfaces with various orientations, faulting with very smalldisplacements on the surfaces gives rise to a deformation of the mass that can be regarded as a plasticdeformation. In this case, which surfaces are activated as faults? Slip line theory gives a clue to thisproblem.

Strain increments

Unlike elastic strain, plastic strain can be very large. It is not easy to relate stress and finite strain.To this end, we have to know deformation history quantitatively. Instead, strain incremental theory

10.5. SLIP LINE THEORY 249

assumes that an incremental plastic strain dEp is proportional to deviatoric stress

dEp = Tdλ, (10.36)

where dλ is the constant of proportionality. It is assumed that the total strain is the sum of the plasticand elastic strains

dE = dEp + dEe, (10.37)

where the superscripts “p” and “e” stand for the plastic and elastic parts, respectively.When mechanical properties are measured, stresses are often applied to a test piece. If the mate-

rial has isotropic mechanical properties, the resultant strain has the same axial symmetry. Therefore,it is convenient to make some formulas to convert triaxial stress and strain to axial stress and strain.For this purpose, equivalent stress is defined based on the von Mises yield criterion:

Te =√

3TII =

√32

T : T =3√2SoctS . (10.38)

For a uniaxial stress (S1 �= S2 = S3 = 0), we have the mean and principal deviatoric stressesS0 = S1/3, T1 = 2S1/3 and T2 = T3 = −S1/3. In this case, therefore, the equivalent stress equalsthe non-zero principal stress (S1 = Te).

In relation to the energy dissipation by plastic strain dW p, the equivalent plastic strain incrementdEe is defined as dW e = TedEe, i.e., equivalent plastic strain increment is given by

dEe =

√32

(dE) : (dE). (10.39)

An equivalent strain rate is defined as

Ėpe =

√32

Ė : Ė. (10.40)

Equivalent strain and stress are sometimes defined as

TE =

√12

T : T =√TII, ĖE =

√12

Ė : Ė (10.41)

instead of as in Eqs. (10.38) and (10.40). These quantities are convenient for comparing the resultsof a simple shear test that measures mechanical properties under simple shear. Consider a simpleshear

F =

⎛⎝1 2q 00 1 00 0 1

⎞⎠ .Then, the stress tensor has only non-zero components S12 = S21 if the test piece is isotropic, andhas the principal stresses −S12, 0 and S12. Hence, we have ΔS = 2|S12|. In this case, we have thestress ratio Φ = 0, so that Eq. (4.15) reduces to TII = |S12|. In addition, from Eq. (10.41) we obtainTE = |S12|. Therefore, TE equals the shear stress for simple shear tests.


Figure 10.8: Slip lines and Mohr circles.

Slip line field

�extitslip line field is introduced by assuming incompressible plane strain on theO-13 plane. Becauseof the plane strain, the intermediate principal strain vanishes. Therefore, we have

dEp1 = −dEp3 , dE

p2 = 0. (10.42)

Substituting Eq. (10.42) into (10.36), we have

dEp2 = T2dλ =[S2 −

13

(S1 + S2 + S3)]

dλ =13

(−S1 + 2S2 − S3) dλ = 0.

∴ S0 = S2 =S1 + S3

2. (10.43)

In this case, we have Φ = 1/2 so that the Tresca and von Mises criteria predict the same yieldcondition (Fig. 10.4), i.e., yielding occurs on the surfaces with the maximum shear stress. Theymake angles with the principal axes at 45◦ (Fig. 10.8(a)). The intersection between the surfaces andthe O-12 plane is known as a slip line. The slip line that makes a clockwise angle of 45◦ with theS1-axis is called an α-line. A slip line perpendicular to this line is called a β-line (Fig. 10.8(a)). Thetrajectories of the α- and β-lines are called slip line field.

In this case, we have S12 = S23 = 0. Therefore, the yield condition for the von Mises criterionis given by

(S11 − S33

)2+ 4S213 = 4k

2 or(S11 −

S11 + S332

)2+ S213 = k

2.

The latter designates the Mohr circle for the slip lines. Let φ be the angle shown in Fig. 10.8(b),then we have

S1 = S0 + k, S2 = S0, S3 = S0 − k, S11 = S0 − k sin 2φ, S33 = S0 + k sin 2φ

10.6. COLLAPSE OF IMPACT CRATERS 251

Figure 10.9: (a) Mosaic photograph around the south pole of the Moon (Clementine BJ90A000).The upper half of this picture shows the near side. The Schödinger basin is about 320 km in diameter.(b) The multi-ringed basin Orientale on the Moon (Lunar Orbiter IV194M). The Cordillera ring isabout 900 km in diameter.

and

tan 2φ =S22 − S11

2S13. (10.44)

The trends of the slip lines are obtained through this equation.The slip-line theory can be applied not only to plane strains but also to axisymmetric strain fields.

The theory is applied in the next section to the stability of impact craters. It should be noted that thetheory is not valid for triaxial strains. Therefore, the theory is not useful in regions where crustalthickening or thinning is significant.

10.6 Collapse of impact craters

Impact craters have various shapes. However, the shapes can be classified by diameter [145]. Smallcraters are bowl-shaped with raised rims, whereas larger craters have complex shapes. Large cratersare classified as central peaks (Fig. 1.8), peak-ring and multi-ring craters. The Schöredinger basinis a peak-ring crater, and the Orientale basin has multiple rings (Fig. 10.9).

The depthH versus diameterD for lunar craters is shown in Fig. 10.10. Obviously, lunar cratershave two trends that intersect at about D = 15 km. Craters smaller than this are bowl-shaped andcalled simple craters. Larger ones are called complex craters. On the Moon, craters greater thanabout 15 km in diameter usually have landslides on their rims to reduce the H/D ratio. Craters with


Figure 10.10: Depth H versus diameter D of craters in highland (dots) and maria (open circles) onthe Moon [182].

the sizes of D ∼ 101 km initially had bowl-shapes, which are called transient craters, but they wereimmediately deformed by landslides.

The intersection at D = 15 km marks the critical diameter for a bowl-shaped basin to keepits topography against gravity. A large crater has a deep basin that generates a large difference inoverburden stresses around the crater. The imbalance causes gravitational collapse of the basin.Therefore, the critical diameter enables an order-of-magnitude estimation of the strength of the ma-terials under craters [145]. To this end, the topography of a transient crater with the depth H anddiameter D is approximated by a paraboloid of revolution

z = H

[1 −

(r

D/2

)2], (10.45)

where r is the horizontal distance from the crater center and z is depth. The mass defect of the basinis the source of force for deformation. Using Eq. (10.45), the force is roughly estimated to be of theorder of

F = ρg∫H

0πr2 dz =

18πρgHD2.

The surface of the hemisphere with a radius of D/2 is S = πD2/2. Therefore,

σ =F

S=

14ρgH

is the representative stress under the crater. The transient crater has a D/H ratio of around 2.7, sothat a critical diameter of about 15 km corresponds to a critical depth of H ≈ 5.6 km. Using theparameters ρ ≈ 3 × 103 kg m−3, g ≈ 1.5 ms−1, we obtain the critical stress σY ≈ 6 MPa. This is oneorder of magnitude smaller than the tensile strength of rocks, and two orders of magnitudes smaller

10.6. COLLAPSE OF IMPACT CRATERS 253

Figure 10.11: Central peak transition diamter D and surface gravity g on icy satellites and terrestrialplanets [205].

than their compressive strength. The small strength suggests that the crashed rocks were fluidized byimpact cratering so that they behave as a plastic fluid. If it was the case, the critical diameter shouldhave inverse correlation with surface gravity g. Impact craters on icy satellites and terrestrial planetsdemonstrate the inverse correlation. Figure 10.11 suggests that the fluid made from the icy crust wasweaker than that from rocks. The peak-rings and multi-rings are considered to be the waves on theplastic fluid that were quenched when the fluidization came to an end (§10.7).

Width of landslide blocks

The strength can be estimated from the width of landslide blocks at the rim of small complex cratersusing the slip line theory [143]. Among complex craters, smaller ones were merely subject to land-slides (Fig. 1.8). The slopes of those craters are stepped by the blocks that slid down along thearcuate fractures surfaces. In order for the estimation, we assume that the sliding was very slowrelative to freefall, so that inertia forces are negligible. Secondly, the materials under the crater ofa are rigid-perfectly-plastic body following the Tresca yield criterion (σ1 − σ3)/2 = cY. Thirdly,Assuming axial symmetry about the vertical for the transient crater whose shape is approximated bythe paraboloid of revolution (Eq. (10.45)), we define the cylindrical coordinates O-rφz. Fourthly,


Figure 10.12: Slip line fields under a transient crater whose shape is approximated by a paraboloid ofrevolution (Eq. (10.45)) with aD/H ratio of 0.2 [143]. Dot-bar lines indicate the axis of revolution.Note the convergence of β-lines to a thick line near the crater rim. Stress state is under the yieldcondition in the gray region in (a). Slip lines are not drawn under the thick lines in (b) and (c)because this model fails there.

the Haar-von Kármàn hypothesis σφφ = σ1 = σ2 is employed to constrain the hoop stress σφφ [45].Figure 10.12 shows the slip line fields obtained through numerical calculations under these con-

ditions2, where ρgH/cY is the dimensionless depth of the transient crater. It was found that the stressstate near the crater rim reaches the yield condition for a dimensionless depth of greater than 5. Inaddition, if the dimensionless depth is greater than 7, β-lines show convergence under the rim to thethick lines in Fig. 10.12. Those lines may be utilized for landslides. If ρgH/cY is ∼10, the areawhere the maximum shear stress exceeds cY becomes extensive under the rim, suggesting an entirecollapse of the slope and resultant upheaval of the fractured crater center.

Landslides may occur along the surface that is designated by thick lines in Fig. 10.12. Accord-ingly, the stability of the parallelogram in Fig. 10.13 constrains the strength of the materials underthe rim. The inclination of the slope at its top φ0 is obtained by differentiating Eq. (10.45). Theresult is tanφ0 = |dz/dr| = 4H/D. Let h be the height of the block that is shown in Fig. 10.13. Theshear stress at the base of the gray parallelogram is σS = ρgh cosφ0 sinφ0. When this shear stressreaches the critical value cY, the block slides down. Namely, the failure condition is

ρgh cosφ0 sinφ0 = cY.

The height is replaced by the width w = h/ tanφ0, so that we obtain the critical width of the landslideblock

w =cYρg

[1 +

116(H/D)2

]. (10.46)

Equation (10.46) allows us to estimate the critical value cY. Lunar craters have those blocks withwidths of several kilometers [177]. Since simple craters have an H/D ratio of 0.2, we assume theorder of magnitude to be H/D ∼ 10−1 for the ratio of complex craters. Then we have cY ∼ 100MPa. This is consistent with the critical stress estimated on p. 252.

2See [143] for details.

10.7. BINGHAM MODEL FOR COMPLEX CRATERS 255

Figure 10.13: Parallelogram approximating the cross-sectional shape of the landslide block under acrater rim. The thick line approximates the line to which the β-lines converge in Fig. 10.12.

Stability of caldera walls

The above model can be applied to calderas. Assuming a cylindrical depression for the caldera,the slip line field is shown in Fig. 10.14. The vertical wall is stable for shallow depressions withρgH/cY < 5. When this dimensionless depth reaches 6.8, the failed region reaches the center of thedepression, indicating an uplift of the center to accommodate the subsidence of the rim.

Figure 10.14(c) shows the large caldera called Creidne on Io. There are few landslides in thecaldera, although its wall is as high as 1 km. The surface of Io is covered by various sulphur-rich compounds. However, this cover is probably thin because the compounds do not have enoughstrength to support the wall [40].

10.7 Bingham model for complex craters

The shape of a complex crater depends on the diameter of the crater D. Lunar craters with D greaterthan 20 km have central peaks and those with D � 140 km have peak rings. Central peaks, peakrings, and multirings are thought to be formed by the quench of waves of fluidized materials excitedby impacts. A moving Bingham fluid can exhibit rapid deceleration (§10.1) so that the final stateof the fluid is quenched. Assuming the damped oscillation of Bingham fluids, let us estimate themechanical properties of the fluids from those geological structures [144]. Quenching at half theoscillation cycle leads to a central peak. If the oscillation was quenched at the end of the first cycle,a peak ring is the result. A multiring structure is the result of cessation after a few cycles. If thishypothesis is valid, the crater diameters for the onset of central peaks and that of peak rings shoulddepend on surface gravity g, because gravity is the driving force of the waves. If gravitational forcedoes not overcome the critical stress of the Bingham fluid, oscillation does not occur. Figure 10.15demonstrates the existence of this correlation.

We are going to determine the effective values of the Bingham parameters. Instead of using theparameters in considering the quenching, we utilize the critical depth of the transient crater to beunstable. The critical dimensionless depth was presented as ρgH/cY ≈ 5 on p. 254. Accordingly,


Figure 10.14: (a) Slip line field for a cylindrical depression with H/D = 0.05 [143]. (b) Modesof gravitational collapse of the depression for three-dimensionless depth ρgH/cY. (c) Surface ofJovian satellite Io (Voyager image 0145J1+000). The caldera called Creidne has a depth of � 1 kmrelative to the surrounding plain and a major axis of 100 km. Montes Haemus is about 10 km high.

let us use the critical amplitude

A∗ =5cYρg

. (10.47)

The oscillation is frozen when the amplitude of oscillation falls to A∗.

Consider a hemisphere with a radius of L in which fluidization and oscillation occur. The hemi-sphere includes a transient crater. Let A, ω and t be the representative amplitude, angular frequency,and the time since the beginning of the oscillation. Then the representative vertical displacement isexpressed as z = A cosωt, and P = 2π/ω is the period of oscillation. Let h be the root-mean-square

10.7. BINGHAM MODEL FOR COMPLEX CRATERS 257

Figure 10.15: The minimum diameter of central peak craters [205]. That of Venus is after [77].Gray lines have an inclination of −1. Open circles: rocky planets and satellites; closed circles: icysatellites.

of z over a cycle:

h2 =ω

2π

∫ 2π/ω0

(A cosωt)2 dt = A21

2π

∫ 2π0

12

(cos 2τ + 1) dτ =A2

2π

∫ 2π0

dτ2

=A2

2.

Taking damping into account, the root-mean-square is somewhat smaller than above. However,h ≈ A/√2 can be used as a rough estimate. This mean displacement occurs over the surface areaπL2, so that the potential energy is represented by

V ≈ − (ρgh2) × (πL2) = −π2ρgA2L2. (10.48)

The minus sign is attached here for the convenience of later arrangement. On the other hand, ifȦ sinωt is the vertical velocity, we similarly obtain the representative velocity v ≈ Ȧ/√2. The massof the hemisphere is M = (4π/6)L3ρ, where ρ is the density of the Bingham fluid. Therefore, thekinetic is roughly estimated as

U ≈ 12Mv2 =

π

6ρL3Ȧ2. (10.49)

In order to deal with the damped oscillation, we utilize the Laglangian [117]

L = U − V = π6ρL3Ȧ2 +

π

2ρgA2L2, (10.50)

where Eqs. (10.48) and (10.49) are used. Let η be the viscosity of the fluid. The velocity of thecenter is represented by Ȧ, but the velocity vanishes at a distance L from the center. Hence, we have


the representative velocity gradient Ȧ/L and viscous resistance 2ηȦ/L. The energy dissipationwithin a unit volume per unit time is 2ηȦ2/L2. Therefore, we have the energy dissipation within thehemisphere

D =2πL3

3× 2ηȦ

2

L2=

4π3ηȦ2L. (10.51)

Since Laglange’s equation of motion with dissipation is [117]

ddt

(∂L∂Ȧ

)− ∂L∂A

+12∂D

∂Ȧ= 0

so that substituting Eqs. (10.50) and (10.51), we obtain the equation of damped oscillation

mÄ + bȦ + kA = 0, (10.52)

wherem =

π

3ρL3, b =

4π3ηL, k = πρgL2. (10.53)

Note that these parameters do not include the assumed angular frequency ω. We define the angularfrequency for the free oscillation (b = 0),

ω0 =

√k

m=

√3gL

(10.54)

and the magnitude of damping

γ =b

2m=

2ηρL2

. (10.55)

Using parameters in Eqs. (10.54) and (10.55), we have the solution of Eq. (10.52) in the form of

A(t) = A0e−γt cosΩt, (10.56)

where A0 is the initial amplitude and Ω =√ω02 − γ2.

The initial phase angle is assumed to be zero. The number of cycles until freezing depends onthe initial potential energy which determines A0. The initial potential energy is obtained from theshape of the transient crater that is approximated by a paraboloid of revolution. From Eq. (10.45)we have the shape r = a

√(H − z)/H , where a is the radius of the transient crater. Then, we have

the initial potential energy3

V0 = −∫H

0ρgz

(πr2

)dz = −

∫H0πρga2

z(H − z)H

dz = −16πρga2H2. (10.57)

Combining Eqs. (10.48) and (10.57), we obtain the initial amplitude

A0 =aH√3L

. (10.58)

3In reality, gravitational collapse begins while the transient crater is excavated [146]. However, the initial kinetic energyis neglected here for brevity.

10.8. POWER-LAW FLUID 259

If this is lower than A∗ in Eq. (10.47), the oscillation is not excited.The number of cycles while the amplitude of the damped oscillation A(t) is greater than A∗

determines the number of rings of the complex crater. Let t∗ be the duration of oscillation. This isdetermined by solving the equation A(t∗) = A∗ or

A∗ = A0e−γt∗. (10.59)

Substituting Eqs. (10.47), (10.54) and (10.57) into Eq. (10.59), we obtain the duration

t∗ =ρL2

2ηlog

(ρgaH

5√

3cYL

).

The number of zero-crossing for A(t) is N = Ωt∗/π, therefore we obtain4.

N =1π

(3ρg2L3

4η2− 1

)1/2log

(1

5√

3

ρgH

cY

a

L

). (10.60)

Among complex craters, smaller ones have only landslides at their rims that characterize the com-plexity. In those cases, it is possible to estimate the radius of transient crater a and the extent Lto which landslides occur from present crater morphology. Detailed analysis indicates L/a ≈ 1.5[144]. On the other hand, H ≈ 0.4a is the depth/diameter ratio of the transient cavity of a crater.Unknowns are the Bingham parameters cY and η in Eq. (10.60). These values can be estimated fromthe crater morphology (Table 10.1).

Values in the two columns † and ‡ in Table 10.1 indicate observed smallest radii for central peakand peak-ring craters on Earth, Moon, and other terrestrial planets and icy satellites. The densityvalues in the table are assigned for rock 3.0 × 103 kg m−3 and ice 0.9 × 103 kg m−3. Assuming thatthe radii correspond to a in Eq. (10.60) for N = 1 and 2, the values cY and η are evaluated andshown in the table. Values in the column T designate the period of oscillation when the minimumcentral peak craters were formed.

The estimated Bingham parameters in Table 10.1 show that the critical values cY of rocky plan-etary bodies are 1–3 MPa. This is consistent with those estimated on p. 252 and p. 254, supportingthe Bingham model for the formation of complex craters. Large impacts gave rise to the fluids byacoustic fluidization [144]. Interestingly, the critical values for icy satellites are one or two orders ofmagnitude smaller than those of the rocky planets.

10.8 Power-law fluid

10.8.1 Definition

Rocks get fluidity at temperatures higher than about half their melting temperatures on a geologi-cal timescale. Laboratory experiments have determined the rheological properties of rocks at high

4The numeral “5” of the denominator in the parentheses of Eq. (10.60) comes from the numeral “5” in Eq. (10.47), whichdepends on the distribution of the topographic load.


Table 10.1: Bingham properties of planets and satellites [205]. T is the period of oscillation for theformation of a minimum central-peak crater. †, the diameter at the onset of central peaks (N = 1);‡, diameter onset of central rings or pits (N = 2).

g ρ a (km) cY η T

(ms−2) (kg m−3) † ‡ (MPa) (Pa s) (minute)Earth

Crystalline rocks 9.8 3.0 × 103 1.9 4.8 2.23 1.6 × 107 1.6Sedimentary rocks 9.8 3.0 × 103 1.9? 2.3 ca. 2.23 ca. 1.4 × 108 1.1

MoonHighlands 1.62 3.0 × 103 10.9 27 2.12 1.5 × 108 9.6Maria 1.62 3.0 × 103 8.6 19 1.68 2.9 × 108 8.0

Mercury 3.78 3.0 × 103 4.7 13 2.13 1.0 × 108 4.3Mars 3.72 3.0 × 103 3.1 6 1.38 1.4 × 108 3.0Icy satellites

Ganymede 1.43 0.9 × 103 < 10 ca. 4 < 0.33 < 7.8 × 107 < 4.0Callisto 1.25 0.9 × 103 < 10 < 10 < 0.29 −Titania 0.372 0.9 × 103 > 1 < 20 > 0.01 −Rhea 0.285 0.9 × 103 8.6 15 0.06 6.3 × 107 17Ariel 0.251 0.9 × 103 17.0 15 0.10 1.5 × 108 18Dione 0.224 0.9 × 103 17.0 20 0.09 1.6 × 108 22Tethys 0.185 0.9 × 103 26.0 27.5 0.11 2.7 × 108 28Miranda 0.083 0.9 × 103 > 25 > 25 > 0.05 −Mimas 0.079 0.9 × 103 14.1 27.5 0.03 5.9 × 108 46

temperatures and very low strain rates. It was found that rocks with these conditions behave aspower-law fluids, equivalent strain rate and equivalent stress of which obey the equation

ĖE = ATn

E, (10.61)

where A > 0 and n > 1. Namely, the equivalent strain rate is proportional to some power of equiv-alent stress. The parameter n is known as the power-law exponent. The power-law fluid includesNewtonian fluids when n = 1. Differentiating both sides of Eq. (10.61) by TE, we obtain

dĖEdTE

= AnT n−1E =n

TE

(ATnE

)=nĖETE

. (10.62)

Since the viscosity of Newtonian fluids was defined as half the constant of proportionality betweenthe strain rate and stress (Eq. (9.3)),

η =TE

2ĖE(10.63)

gives the effective viscosity. Combining Eqs. (10.62) and (10.63), the power-law exponent is relatedto the effective viscosity as

n = 2ηdĖEdTE

. (10.64)


Figure 10.16: Equivalent stress versus equivalent strain rate for a power-law fluid with n > 1.Effective viscosity η decreases with an increasing strain rate. The slope μ is smaller than 2η for thisfluid.

Pseudoplastic fluids includes power-law fluids with n > 1, and the effective viscosity decreases withincreasing ĖE, i.e., the fluids show shear-thinning behavior. Let μ = dTE/dĖE be the slope of thegraph in Fig. 10.16 and we have

n = 2η/μ (10.65)

from Eq. (10.64).In order to apply the power-law rheology to tectonics, the constitutive equation (Eq. (10.61))

should be rewritten with tensors to account for various stress states. For this purpose and for simplic-ity, we assume anisotropy in the rheological properties of rocks. Then, Eq. (10.36) is transformedto proportionality Ė = λT. The equivalent strain rate and equivalent stress should satisfy this pro-portionality so that we have ĖE = λTE. The constant is given by comparing this and Eq. (10.61):λ = ATn−1E . A depends on lithology and temperature. Consequently, power-law fluids have thetensorial constitutive equation

Ė = ATn−1E T. (10.66)

According to the strain incremental theory, strain rate and deviatoric stress have proportionalityTij/Ėij = TE/ĖE (Eq. (10.36)). Combining this and Eq. (10.61), we obtain

T = T EĖ−1E Ė = A

1n Ė

1n−1E Ė. (10.67)

Non-linear fluids have the effective viscosity that is defined through the equation T = 2η(D)D =2η(Ė)Ė (Eq. (9.3)). Comparing this with Eq. (10.67), we obtain the effective viscosity of power-lawfluid

η =12A

1n Ė

1n−1E . (10.68)

Definition of equivalent stress is based on the von Mises yield criterion (§10.5). When rocksbehave as power-law fluids, they obey the criterion. This is evidenced by the observation that ex-perimental deformations by triaxial shear tests and shear tests show consistent relationships betweenequivalent stresses and equivalent strain rates.


Figure 10.17: Velocity profiles for pressure-driven flow of incompressible power-law fluids betweenstationary parallel planes. The horizontal axis is the ratio vx/vx, where the denominator is the meanvelocity.

10.8.2 Velocity profiles for power-law fluids

Consider the velocity profile for the pressure-driven flow of an incompressible power-law fluid withn > 1 between stationary parallel plates (Fig. 10.17). We take the x-axis in the flow direction, so thatthe velocity vector has the velocity components vx > 0 and vy = vz = 0. The flow has the boundarycondition v = 0 at the walls z = ±h. The velocity profile for the fluid with n ≈ 1 may resemblethat of Newtonian fluids (Fig. 9.2), i.e., vx is the maximum and the velocity gradient dvx/dz is theminimum at z = 0. Hence, the effective viscosity of the power-law fluid decreases towards the wallsfrom the flow center z = 0. Therefore, the velocity gradient near the walls may be greater than thatof Newtonian fluids, but that near the center may be smaller.

For the verification of the above argument, we firstly consider the force balance equation (Eq.(3.23))

∂Sxx∂x

+∂Sxz∂z

= 0. (10.69)

The first term in this equation equals the pressure gradient Δp/L, where L is the length along theflow and Δp is the pressure difference between the ends of the length. The second term in Eq.(10.69) equals Szx, which is the shear stress SS working at planes parallel to the flow. Therefore,Eq. (10.69) is rewritten as

dSSdz

= −ΔpL. (10.70)

According to Eq. (10.61), the power-law fluid has v′x = ASnS. Hence, we have

SS = A1n

(v′x) 1n , (10.71)

where v′x = dvx/dz. Substituting Eq. (10.71) into (10.70), we obtain

A− 1n (v′x) 1n = −ΔpL .


Combining the boundary conditions vx∣∣z=±h = 0, we have

vx =A

n + 1

(ΔpL

)n (hn+1 − |z|n+1) .

Figure 10.17 shows the graphs of this equation for various n. Velocity gradient increases toward thewalls. There is a zone at the flow center in which the velocity gradient is small. In the case of n� 1,the zone more or less behaves like a rigid body. This is called a plug, and the velocity gradient isconcentrated in the vicinity of the walls. Consequently, faults are sometimes simulated by thosenarrow zones of high velocity gradients.

10.8.3 Temperature and other dependence

Laboratory experiments have demonstrated that when rocks behave as power-law fluids, their rheo-logical parameters change significantly with temperature, obeying Dorn’s equation

ε̇ = AD(Δσ)n exp(− QRT

), (10.72)

Δσ = BDε̇1n exp

(Q

nRT

), (10.73)

where R is the gas constant, T the absolute temperature, and Q the activation energy. Q depends onlithology. AD and BD are constants and are related to each other through the equation

BnD = 1/AD. (10.74)

AD and BD indicate the predisposition and difficulty for flow, respectively.Power-law parameters have some pressure p-dependence. If that is significant, the scalar consti-

tutive equation is expressed as

ε̇ = A′(Δσ)n exp(−H

∗ + pV ∗

RT

), (10.75)

where H∗ and V ∗ are known as activation enthalpy and activation volume, respectively.Unlike dislocation creep by the gliding motion along crystalline dislocations, diffusion creep by

the diffusion of individual atoms across individual crystals is easy for fine grained rocks, i.e., flowby the diffusion has grain-size d-dependence of the form

ε̇ = A′′(Δσ)ndm exp(−H

∗ + pV ∗

RT

), (10.76)

where m is a material constant slightly larger than unity5.

5See [97] for further reading on the rheology of rocks.


Table 10.2: Rheological parameter values of selected rock types and minerals. †, abbreviations inFig. 10.18. ‡, MPa−ns−1.

† AD n Q or H∗ V ∗ Reference(GPa−ns−1) (kJ mol−1) (cm3mol−1)

Granite

dry Gr 5.0 3.2 123 [99, 192]wet Gr(w) 100 1.9 137 [99, 192]

Quartzitedry Qz 100 2.4 156 [99, 192]wet Qz(w) 2.0 × 103 2.3 154 [99, 192]

Albite rock Ab 1.3 × 106 3.9 234 [99, 192]Anorthosite An 1.3 × 106 3.2 238 [99, 192]Granodiolite Qd 2 × 104 2.4 219 [99, 192]Diabase Db ‡190, ‡8 4.7 487 [130]Olivine

dry Ol ‡6.3 × 104 3.5 533 [99]wet Ol(w) ‡1.9 × 103 3.0 420 [98]

Halite (rock salt) RS 5.0 × 1016 5.3 102 [99, 192]Ice

≤ 195 K I 1.6 × 10−7 6.0 39 –13 [53]195–240 K I 126 4.0 61 –13 [53]240–258 K I 6.3 × 108 4.0 91 –13 [53]

Power-law parameters of several rock types are listed in Table 10.2. The values of AD are differ-ent in order of magnitude, indicating that creep strengths are different for rock types by several ordersof magnitude. This further suggests that prediction of tectonics requires detailed knowledge aboutthe subsurface distribution of lithology. Continental crust has an intricate structure shaped throughits long history so that continental tectonics is more difficult to understand than oceans where thecrust has a simpler constitution.

Ductile strength is obtained by eliminating BD from Eqs. (10.74) and (10.73). Consequently,we have

Δσ = A− 1nD ε̇

1n exp

(Q

nRT

), (10.77)

designating the viscous resistance to exerted strain rate ε̇. The rate is more or less in the order of10−15 s−1 for the tectonics of the present Earth. Substituting ε̇ = 10−15 s−1 as a representative strainrate into Eq. (10.77), Fig. 10.18 shows the strength of the rocks listed in Table 10.2.

Olivine is the most abundant mineral in the upper mantle. The figure designates that the mantleis much stronger than crustal rocks if they are at the same temperature. However, Olivine has largertemperature dependence than the crustal rocks, so that the mantle lithosphere is more weakened byheating than the crust. This leads to the mechanical instability of the lithosphere (§12). Recent


Figure 10.18: Ductile strength and effective viscosity of rocks and minerals when they behave aspower-law fluids at the strain rate of 10−15 s−1 under axial stresses. The strengths (Δσ) are calculatedwith Eq. (10.77). Abbreviations designating rock types and minerals are listed in Table 10.2.

experiments showed that diabase, a constituent of the crust, has a similar rheology to olivine [130].

On the other hand, rock salt is commonly deposited in thick layers of evaporite. It is not a rareconstituent of the crust, but has peculiar properties. Namely, it is much weaker than rocks by about10 orders of magnitude, and has a smaller density (2.2 × 103 kg m−3). Hence, rock salt often makediapirs and domes. Due to the sharp difference in the mechanical properties of salt and silicate rocks,a salt dome can disturb the stress field around it [22].

Ice is also a very weak material compared to rocks. However, ice is as strong as some kind rocksincluding granite at the surface temperatures of icy satellites, Outer Solar System6. The temperatureis slightly over 100 K on Jovian moons, and some 70 K on Saturnian moons [119]. At those verylow temperatures, ice scarcely flows thus maintaining geological structures over billions of years.

6The ice listed in Table 10.2 had grain sizes of ∼100 μm. Ice made of grains smaller than several microns has a smallerstrength than that illustrated in Fig. 10.18.


10.8.4 Stability

Let us derive equations for describing the growth of perturbations in incompressible power-lawfluids. The fluids can be regarded as Newtonian fluids with the effective viscosity η(DII) and withthe constitutive equation S = −pI + 2η(DII)D. Dividing variables in this equation into means andperturbations, we have

S + S̃ = −(p + p̃)I + 2η(DII + D̃II) (D + D̃) . (10.78)Substituting Taylor expansion η

(DII)= η

(DII)+ η′

(DII)D̃II + · · · into Eq. (10.78), we have

S + S̃ ≈ −(p + p̃)I + 2 [η(DII) + η′(DII)D̃II] (D + D̃)≈ −(p + p̃)I + 2η(DII)D + 2η(DII)D̃ + 2η′(DII)D̃IID, (10.79)

where higher-order terms are neglected and η′ = dη/dDII. Combining Eq. (10.79) and the constitu-tive equation satisfied by the mean flow S = −pI + 2η(DII)D, we obtain

S̃ = −p̃I + 2η(DII)D̃ + 2η′(DII)D̃IID. (10.80)Now we assume incompressible plane strains on the O-12 plane, i.e., the stretching tensor has

the form

D =

⎛⎝D11 D12 0D21 D22 00 0 0

⎞⎠ , D12 = D21.The flow is incompressible so that D is a deviatoric tensor. Hence, the second basic invariant of Dsatisfies the equation with the same form as Eq. (4.13). Accordingly, we have

DII =12

(D211 +D

222

)+D212. (10.81)

Incompressibility is designated in this case by the equation, trace D = D11 + D22 = 0, so that theabove equation is transformed to

DII = D211 +D

212. (10.82)

From (10.81), we have

2(DII + D̃II

)=(D11 + D̃11

)2+ 2

(D12 + D̃12

)2+(D22 + D̃22

)2.

Rearranging this equation,

DII + D̃II =12

(D

211 + 2D

212 +D

222

)+(D11D̃11 + 2D12D̃12 +D22D̃22

)+

12

(D̃211 + 2D̃

212 + D̃

222

). (10.83)


The content of the last pair of parentheses is composed of second-order terms of perturbations, sothat they are negligible. On the other hand, from Eq. (10.82) we have

2DII = D211 +D

212 +D

222. (10.84)

Comparing Eqs. (10.83) and (10.84), we have

D̃II = D11D̃11 + 2D12D̃12 +D22D̃22. (10.85)

If pure-shear is the mean flow, we take the coordinate axes parallel to the principal axes of thepure shear. Then, we have D12 = 0. Therefore, Eq. (10.85) is simplified to

D̃II = D11D̃11 +D22D̃22. (10.86)

Substituting Eq. (10.86) into (10.80), we have

S̃ = −p̃I + 2η(DII)D̃ + 2 [η′(DII) · (D11D̃11 +D22D̃22)]D, (10.87)or, equivalently, T̃11 = 2μD̃11, T̃22 = 2μD̃22 and S̃12 = 2η

(DII)D̃12, where

μ = η(DII)+ 2η′

(DII)D

211. (10.88)

Using these equations, the biharmonic equation in Eq. (9.11) is rewritten as

∂4ψ̂

∂x41+ 2

(2μη

− 1)

∂4ψ̂

∂x21x22

+∂4ψ̂

∂x42= 0 (10.89)

for power-law fluid7 [216].Using these equations, the solution of single layer folding in Section 9.4 was extended to power-

law fluids [57, 216]. Consequently, the dominant dimensionless wavelength λ/H was found to havestrong dependence on the ratio of power-law exponents of the layer and its matrix, and that shortwavelength folds can be formed (Fig. 10.19). The flow stability of power-law fluids was employedto explain an imbricate thrust system [68].

Exercises

10.1 Locate a point for the octahedral plane on the Mohr diagram for a triaxial stress (0 < σ3 <σ2 < σ3).

10.2 To what extent does a dense rigid body sink within a low density matrix for each of the caseswhere the matrix behaves as an elastic, Newtonian and Bingham fluids?

10.3 Estimate the effective elastic thickness of the folded thin elastic-plastic plate shown in Fig.10.7.

7The parameters η and μ are oppositely defined in [216].


Figure 10.19: Dominant dimensionless wavelength λ/H of single-layer folding for power-law fluids[216]. The horizontal axis is the ratio of the effective viscosity of the layers (1) and (2), which areillustrated in Fig. 9.10. H is the thickness of the layer (2).

Plasticity - TERRAPUB · They include plastic body, Bingham, and power-law ﬂuids. The equations...

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Transcript of Plasticity - TERRAPUB · They include plastic body, Bingham, and power-law ﬂuids. The equations...