Geodynamics Lecture 5 Basics of elasticity

Transcription

1 Geodynamics Lecture 5 Basics of elasticity Lecturer: David Whipp david.whipp@helsinki.fi! Geodynamics 1

2 Goals of this lecture Introduce linear elasticity! Look at the applications of elasticity to several simple stress/ strain scenarios! Provide the background needed to consider flexure of the lithosphere 2

3 Toward rheology At this point, we ve talked about how to describe stress and strain and a few examples of measurements that can be made in rocks For the rest of this lecture and the following two, our focus will turn to the connections between stress and strain The relationship between stress and strain (or strain rate) are given by constitutive equations that describe how rock deforms when a given force is applied (also known as rheological laws) Today, we start with elasticity 3

4 What parts of the Earth are elastic? Essentially all rocks are elastic under the right conditions Elastic: Low temperature, low deviatoric stress Brittle (fracture): Low temperature, high deviatoric stress Lecture 11 (7.10) Viscous (flow): High temperature Lecture 12 (9.10) 4

5 Flexure of the lithosphere in foreland basins BEARTOOTH MOUNTAINS 3- S N 2- I- M//\/ V ~* rtiary / (Fort Union FmJ 0- KM -I- Mesozoic -2- i l l Paleozoic -3- Ä -4- /$rèçàmbriqny,v;v/ ; V/^ A. NORTHERN BIGHORN BASIN 10 km Base of Upper Cretaceous Mesaverde Formation KM -10- w I OVERTHRUST Absaroka Darby BELT Prospect Thrust Wind River Thrust Tertiary S Mesozoic Mesozoic.Pgle.Qzoic; -20- Precambrian E I V WIND RIVER MOUNTAINS B. NORTHERN GREEN RIVER BASIN Modified from Royse, Warner, and Reese, km Top of Upper Cretaceous Mesaverde Formation Hagen et al.,

6 Flexure of the lithosphere Hawaii Watts et al.,

7 Inter-seismic strain accumulation ru ptu 38 re Plate tectonic motions lead to the gradual build up of elastic strain near fault zones g pin n ee tio cr sec ! rup tur e 34 Rather than being localized to the fault, this deformation is distributed over large LETTERS areas (>10 km from the fault zones)! 57 Pacific Ocean 33 Fialko, NATURE 245 Figure S1. Shaded relief map of California. Pink lines denote sections of the S fault that ruptured in great earthquakes in 1857 and A red line denotes t part of SAF that did not produce a major earthquake in historic times. Black show other geologically mapped faults. A white box outlines the study area shown in the main text. During an earthquake, slip on the faults leads to unrecoverable deformation and releases (some of) the stored elastic stress 4 7 Figure 2 Average LOS velocities (grey dots) and GPS/EDM data (coloured 0 accumulation due to a deep slip below the mapped tra

8 Elasticity Normal stress Shear stress n or s n or s Twiss and Moores, 2007! / " Stress is proportional to strain For 1-D normal stress xx = E" xx E : Young s modulus (1D) μ : Shear modulus (1D) If stress 0, strain 0 (recoverable) xx = E" xx E or 2µ " n or " s " n or " s 8

9 Elasticity Normal stress Shear stress n or s n or s Twiss and Moores, 2007! / " Stress is proportional to strain For 1-D normal stress xx = E" xx E : Young s modulus (1D) G : Shear modulus (1D) If stress 0, strain 0 (recoverable) xx = E" xx E or 2G " n or " s " n or " s 9

10 Elasticity Normal stress Shear stress n or s n or s Twiss and Moores, 2007! / " Stress is proportional to strain For 1-D normal stress xx = E" xx E : Young s modulus (1D) G : Shear modulus (1D) If stress 0, strain 0 (recoverable) xx = E" xx E or 2G " n or " s " n or " s 10

11 Elasticity Normal stress Shear stress n or s n or s Twiss and Moores, 2007! / " Stress is proportional to strain For 1-D normal stress xx = E" xx E : Young s modulus (1D) G : Shear modulus (1D) If stress 0, strain 0 (recoverable) xx = E" xx E or 2G " n or " s " n or " s 11

12 Linear elasticity in terms of stress For linear elasticity, stress is linearly proportional to strain 1 =( +2G)" 1 + " 2 + " 3 2 = " 1 +( +2G)" 2 + " 3 3 = " 1 + " 2 +( +2G)" 3 where λ and G are known as the Lamé parameters G is also known as the modulus of rigidity or shear modulus Note, we assume the material properties are isotropic 12

13 Linear elasticity in terms of stress For linear elasticity, stress is linearly proportional to strain 1 =( +2G)" 1 + " 2 + " 3 2 = " 1 +( +2G)" 2 + " 3 3 = " 1 + " 2 +( +2G)" 3 where λ and G are known as the Lamé parameters Principal strain ε produces a stress (λ + 2G)ε in the same direction and stresses λε in the perpendicular directions 13

14 Linear elasticity in terms of strain We can also formulate the equations of linear elasticity in terms of principal strains " 1 = 1 E 1 E 2 E 3 " 2 = E E 2 E 3 " 3 = E 1 E E 3 where E is known as Young s modulus and ν is Poisson s ratio In this case, principal stress σ produces a strain of σ/e in the direction it acts and strains of -(νσ/e) in the perpendicular directions 14

15 Material properties of common rock types Density E G k α kg m Pa Pa ν Wm --1 K K --1 Sedimentary Shale Sandstone Limestone Dolomite Metamorphic Gneiss Amphibole Marble Igneous Basalt Granite Diabase Gabbro Diorite Pyroxenite Anorthosite Granodiorite Mantle Peridotite Dunite Miscellaneous Ice E : Pa (typical values: GPa) G : Pa ν : (typical values ) 15

16 Uniaxial stress Uniaxial stress occurs when only one of the principal stresses is nonzero (σ1 for this example) If σ2 = σ3 = 0, the equations for linear elasticity in terms of strain reduce to " 2 = " 3 = E 1 = " 1 In this case, the equation above can be simplified further to yield Hooke s law 1 = E" 1 16

17 Uniaxial stress We can also find the change in volume of the rock parcel or dilatation Δ from the principal strains =" 1 + " 2 + " 3 = " 1 (1 2 ) 17

18 Uniaxial stress We can also find the change in volume of the rock parcel or dilatation Δ from the principal strains =" 1 + " 2 + " 3 = " 1 (1 2 ) What is Poisson s ratio for an incompressible material? 18

19 Quartzite under uniaxial compression 19

20 Quartzite under uniaxial compression What is happening here? 20

21 Stresses as a result of burial How does elastic stress change in sedimentary rocks as a result of burial?! What stress/strain conditions are appropriate for this scenario? 21

22 Uniaxial strain Uniaxial strain occurs when only one component of the principal strains is nonzero (ε1 in this example) In this case, if we consider ε2 = ε3 = 0, the equations for linear elasticity reduce to 2 = 3 = (1 ) 1 1 = (1 )E" 1 (1 + )(1 2 ) 22

23 Uniaxial strain Let s consider a parcel of rock initially at the surface that now has been buried by sediments of density ρ to a depth h In this case, we can assume σ1 is vertical and equal to the weight of the overburden, σ1 = ρgh From the equations on the previous slide, we find 2 = 3 = (1 ) gh 23

24 Uniaxial strain Let s now consider the effects on the deviatoric stress, the principal stresses minus pressure p! p = 1 3 ( (1 ) )= 3(1 ) gh Which results in deviatoric stresses 0 2(1 2 ) 1 = 1 p = 3(1 ) gh 0 2 = 2 p = 0 3 = 3 p = (1 2 ) 3(1 ) gh 24

25 Uniaxial strain Under tension! Let s now consider the effects on the deviatoric stress, the principal stresses minus pressure p! p = 1 3 ( (1 ) )= 3(1 ) gh Which results in deviatoric stresses 0 2(1 2 ) 1 = 1 p = 3(1 ) gh 0 2 = 2 p = 0 3 = 3 p = (1 2 ) 3(1 ) gh 25

26 Pure shear and simple shear Plane stress occurs when only one principal stress is zero (σ3 = 0) Pure shear is a special case of plane stress Consider the case with σ3 = 0 and σ1 = -σ2 In the case on the left with an angle θ = -45 between σ1 and the x axis, we find In this case, the equations for plane stress conditions yield " 1 = xx = yy = 0 and xy = 1 (1 + ) E 1 = (1 + ) E xy = " 2 26

27 Pure shear and simple shear Plane stress occurs when only one principal stress is zero (σ3 = 0) Pure shear is a special case of plane stress Consider the case with σ3 = 0 and σ1 = -σ2 In the case on the left with an angle θ = -45 between σ1 and the x axis, we find In this case, the equations for plane stress conditions yield " 1 = xx = yy = 0 and xy = 1 (1 + ) E 1 = (1 + ) E xy = " 2 27

28 Pure shear and simple shear Similar to the stresses, with an angle θ = -45 between σ1 and the x axis, we find " xx = " yy = 0 and " xy = " 1 From the equation on the previous slide for we see xy = E! 1+ " xy If we recognise that the modulus of rigidity G can be found from E G = 2(1 + ) then we finally see the simple relationship xy =2G" xy 28

29 Pure shear and simple shear Similar to the stresses, with an angle θ = -45 between σ1 and the x axis, we find " xx = " yy = 0 and " xy = " 1 From the equation on the previous slide for we see xy = E! 1+ " xy If we recognise that the modulus of rigidity G can be found from E G = 2(1 + ) then we finally see the simple relationship xy =2G" xy 29

30 Pure shear and simple shear Similar to the stresses, with an angle θ = -45 between σ1 and the x axis, we find " xx = " yy = 0 and " xy = " 1 From the equation on the previous slide for we see xy = E! 1+ " xy If we recognise that the modulus of rigidity G can be found from E G = 2(1 + ) then we finally see the simple relationship xy =2G" xy Note this works for both pure and simple shear 30

31 Isotropic stress If all three principal stresses are equal, σ1 = σ2 = σ3 p and the state of stress is isotropic In this case, the principal strains are also equal, ε1 = ε2 = ε3 = 1/3Δ From the equations of elasticity in terms of stress we find 3 +2G p = K 1 3 where K is the bulk modulus and its reciprocal β is the compressibility 31

32 Isotropic stress If all three principal stresses are equal, σ1 = σ2 = σ3 p and the state of stress is isotropic In this case, the principal strains are also equal, ε1 = ε2 = ε3 = 1/3Δ From the equations of elasticity in terms of stress we find 3 +2G p = K 1 3 where K is the bulk modulus and its reciprocal β is the compressibility 32

33 Isotropic stress Any change in volume of rock must conserve mass For a parcel of rock with volume V a change in volume δv will result in a change in density δρ, or ( V )=0 In terms of dilatation of the rock parcel Δ we can say V = = V assuming Δ is small Thus, the density change as a function of pressure, compressibility and initial density is simply = p 33

34 Isotropic stress Any change in volume of rock must conserve mass For a parcel of rock with volume V a change in volume δv will result in a change in density δρ, or ( V )=0 In terms of dilatation of the rock parcel Δ we can say V = = V assuming Δ is small Thus, the density change as a function of pressure, compressibility and initial density is simply = p 34

35 Isotropic stress Finally, both the bulk modulus and compressibility can be found from the elastic properties of rock K = 1 = E 3(1 2 ) 35

36 Isotropic stress Finally, both the bulk modulus and compressibility can be found from the elastic properties of rock! K = 1 E = 3(1 2 ) What does this suggest about rock (in)compressibility as a function of Poisson s ratio ν? 36

37 Recap Linear elasticity describes the linear relationship between stress and strain in rocks under low deviatoric stress and at low temperatures! The elastic properties of rock, combined with assumptions about the stress acting on the rock parcel can be used to predict elastic deformation! The elastic properties of rock suggest rocks are actually reasonably compressible (ν = ) 37

38 References Fialko, Y. (2006). Interseismic strain accumulation and the earthquake potential on the southern San Andreas fault system. Nature, 441(7096), doi: /nature04797! Hagen, E. S., Shuster, M. W., & Furlong, K. P. (1985). Tectonic loading and subsidence of intermontane basins: Wyoming foreland province. Geology, 13(8), 585. doi: / (1985)13<585:tlasoi>2.0.co;2! Watts, AB, Brink, Ten, U. S., Buhl, P., & Brocher, T. M. (1985). A multichannel seismic study of lithospheric flexure across the Hawaiian Emperor seamount chain. Nature, 315(6015),