Rheology III. Ideal materials Laboratory tests Power-law creep The strength of the lithosphere The role of micromechanical defects in power-law creep
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1 Rheology III Ideal materials Laboratory tests Power-law creep The strength of the lithosphere The role of micromechanical defects in power-law creep
2 Ideal materials fall into one of the following categories: ü Elastic Viscous Plastic
3 Viscous materials Δσ n = 2η e n where: σ s = 2η e s η is the coefficient of viscosity The strain is non-recoverable The larger the applied stress the faster the fluid flow The amount of strain is not related to the magnitude of the applied stress Given enough time, any (!) stress can produce any arbitrarily large strain
4 Example: Marble bench Deformed marble bench that sagged and locally fractured under the influence of gravity (and occasionally also users).
5 Plastic materials σ s K where: K is the yield stress The strain is non-recoverable Undergo no strain if the applied stress is smaller than the yield stress, but flow readily if it is equal to it
6 The behavior of real materials is better described by combining simple models in series or parallel Fig. from ic.ucsc.edu/~casey/eart150/lectures/rheology/13rheology.htm
7 Firmo-viscous (Kelvin or Voight) materials It is elastic to the extent that the equilibrium strain is a linear function of the applied stress, but the strain rate is governed by the viscous response The strain is recoverable An everyday application of the mechanical analogue is an automobile suspension system, which damps out the elastic oscillations of the spring
8 Example: Post-glacial rebound
9 Example: Post-glacial rebound Fennoscandian British Isle
10 Exhibit a combination of recoverable elastic strain and permanent plastic deformation The plastic does not begin until the yield stress is reached The elastic-plastic system is believed to be a suitable analogue for the loading of crustal faults by tectonic plate motion Elastic-plastic (Prandtl) materials
11 The spring-slider and the seismic cycle preseismic Stress postseismic V plate represents the constant plate velocity k represents the elastic stiffness of the tectonic plate Time
12 Displays linear viscous behavior only above the the plastic yield stress Non-recoverable Visco-plastic (Bingham) materials
13 It makes a big difference whether the viscous and the elastic components are connected in series or parallel Visco-elastic (or Maxwell) materials Firmo-viscous (Kelvin or Voight) materials
14 The extrapolation from lab scale to geological scales is non-trivial Rock Mechanics Lab at BGU Usually, rock samples are subject to either constant stress or constant strain rate Fig. from
15 Experimental strain rates are 10-7 S -1 or higher how are these rates compared with natural geological rates? Given that the: Half-spreading of oceanic lithosphere is 5 cm/year Assume that this displacement is accommodated by shear distributed linearly with depth down to the base of the asthenosphere that is 200 km thick We get: 5 Cm/year 200 km = s 1 We expect the geologic strain rates to proceed at strain rates between s -1 and s -1, that is 5 to 10 orders of magnitude lower than the rates of laboratory tests. 200 km 5 cm/year
16 Increasing the temperature reduces rock s strength Fig. From Fowler Note the resemblance to the elastoplastic model
17 Rocks are weaker under lower strain rates Note that: Strength is strain rate dependent Stress-strain curves are similar to that of the elasto-plastic model Fig. From Fowler
18 Rocks are weaker under lower strain rates Slow deformation allows diffusional crystal-plastic processes to more closely keep up with applied stresses
19 The constitutive equation that accounts for most moderate-stress, steady-state deformation is the power-law equation. e = C 0 ( σ 1 σ 3 ) n exp( Q ) RT, where: e C 0 is the strain rate is a constant σ 1 σ 3 is the differential stress n Q R is a constant is the activation energy is the universal gas constant T is the temperature (bold text indicate experimentally determined parameters)
20 Steady-state strain rate is plotted against the differential stress for n=1, 3, 5 and 7. As the stress exponent increases, the power-law behavior approaches the idealized plasticity When the stress exponent, n, is close to 1, the material exhibits viscous behavior Fig. From Fowler
21 e = C 0 ( σ 1 σ 3 ) n exp( Q ) RT Question 1: Given the above constitutive law and the differential stress vs. strain rate curve, compute the differential stress exponent, n.
22 e = C 0 ( σ 1 σ 3 ) n exp( Q ) RT Question 2: Given the above constitutive law and the differential stress vs. strain rate curve, compute the differential stress exponent, n, the coefficient C 0 and the activation energy, Q.
23 Based on the above equation, it is possible to construct differential stress versus temperature for a variety of rock types.
24 Given a geothermal gradient, temperature may be translated into depth. Based on the above equation, it is possible to construct differential stress versus temperature for a variety of rock types.
25 The geothermal gradient can vary significantly from one tectonic setting to another. hot continental cold continental An important conclusion is that only olivine would have significant strength at the base of continental crust!
26 Here is how we think the differential stress (needed for maintaining a strain rate of s -1 ) varies with depth mostly Qz moho mostly Ol
27 Here is how we think the differential stress (needed for maintaining a strain rate of s -1 ) varies with depth mostly Qz What about the top 15 km? moho mostly Ol
28 We assume x and y are principal directions of the stress tensor, and s yy is equal to the lithostatic pressure. Then: and: for a thrust: for a normal fault: σ yy = ρgh σ xx = ρgh + Δσ Δσ > 0 Δσ < 0
29 the effective normal stress is: 1. σ n = ρgh P + Δσ 2 ( 1+ cos2θ ) and the shear stress is: 2. τ = ± Δσ 2 sin2θ plugging 1 and 2 into the Coulomb criterion we get: 2µ ( ρgh P) 3. Δσ = ±sin2θ µ 1+ cos2θ ( ). At the top 15 (or so) kilometers, deviatoric stresses increase with depth.
30 Here is how we think the differential stress (needed for maintaining a strain rate of s -1 ) varies with depth mostly Qz moho mostly Ol
31
32 A fault near the surface becomes mylonite at great depth:
33 The strength of rocks is strongly dependent on the temperature and the strain rate. Curve 1 is appropriate to low strain rates or high geotherms. Curve 2 is appropriate to lower geotherms or higher strain rates. The brittle-ductile transition can be viewed as the depth at which failure mode switches, and coincides approximately with the base of the seismogenic zone.
34 What are the implications of this model on earthquake depth-distribution?
35 Maggi et al., Geology, 2000.
36
37 In CA, the deformation throughout the upper km is accommodated by brittle deformation Earthquake depth versus heat flow for over 140,000 earthquakes in California between 1985 and The maximum depth of earthquakes generally decreases with increasing heat flow. The 400 C isotherm is shown (dashed line) calculated using average values of heat generation and thermal conductivity with the shown surface heat flow. Fig. from Bonner et al., BSSA, 2003
38 Landers post seismic relaxation Figs. from Savage and Svarc, JGR, 1997.
39 A common view is that the brittle-ductile transition marks the boundary between seismic and aseismic deformation. Fig. from Rolando et al., GRL, 2004
40 Micromechanics defects play a central role in crystal plasticity. Vacancies are missing atoms. Point (0-D) defects Interstitials are atoms that are sitting not in their regular place.
41 Point (0-D) defects Point defects generally are mobile - at least at high temperatures. The migration of point defects in crystalline solids are referred to as creep deformation. Interstitial diffusion Vacancy diffusion
42 Line (1-D) defects Dislocations are linear defects in lattice structure. These are the most important defects for the understanding of deformation under crustal conditions - including the rupture of earthquakes. STM (Scanning Tunneling Microscope) image of a dislocation.
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