Geodynamics Lecture 3 Forces and stresses

Transcription

1 Lecture 3 Forces and stresses Lecturer: David Whipp david.whipp@helsinki.fi

2 Goals of this lecture Define forces and stresses Discuss how stresses can be represented graphically and mathematically Look at various useful stress relationships and their geological value 2

3 Rocks record strain, so who cares about stress? Folded micaceous marble in the Sawtooth Mountains, Idaho, USA These folds are beautiful examples of rock deformation (strain), but can they tell us anything about how they formed? 3

4 Rocks record strain, so who cares about stress? Forces applied to geologic materials produce deformation (strain) Plate tectonics is driven by gravitational body forces (which can be affected by temperature) So, if we want to understand what drives geodynamic behavior, we need to understand stress 4

5 Forces Force: A push or pull applied to a body. Force = mass x acceleration (Newton s second law) Units: Newtons [N]; 1 N = 1 kg m s - 2 Representation: Vector Example: Gravity (directed toward center of Earth) 5

6 Forces Force: A push or pull applied to a body. Force = mass x acceleration (Newton s second law) Units: Newtons [N]; 1 N = 1 kg m s - 2 Representation: Vector Example: Gravity (directed toward center of Earth) Force x F y after Twiss and Moores, 26 6

7 Forces Force: A push or pull applied to a body. Force = mass x acceleration (Newton s second law) Units: Newtons [N]; 1 N = 1 kg m s - 2 Representation: Vector Example: Gravity (directed toward center of Earth) Force Component forces x x Fx F Fy F y y after Twiss and Moores, 26 7

8 Body forces versus surface forces Body force: Forces that act throughout the volume of a solid. Proportional to its volume or mass. Example: Slab pull (gravity) Surface force: Forces that act on the surface area bounding an element or volume. Proportional to the area upon which the force acts. Example: Friction along a fault plane 8

9 Body and surface forces Consider a column of rock in the crust and the force acting on its basal surface (σyy A) Fig. 2.1, Turcotte and Schubert, 214 It is subject to gravity, so the base must support the weight of the overlying rock column 9

10 Body and surface forces Consider a column of rock in the crust and the force acting on its basal surface (σyy A) The rock column has a thickness, y, and cross-sectional area, A Thus, the weight is simply ρgy A, the density times gravitational acceleration time the volume of the column Since the force at on the basal surface must equal the column weight, σyy A = ρgy A, or σyy = ρgy Fig. 2.1, Turcotte and Schubert, 214 Note, σyy is acting vertically upward, balancing the downward weight of the column 1

11 Body and surface forces Consider a column of rock in the crust and the force acting on its basal surface (σyy A) The rock column has a thickness, y, and cross-sectional area, A Thus, the weight is simply ρgy A, the density times gravitational acceleration time the volume of the column Since the force at on the basal surface must equal the column weight, σyy A = ρgy A, or σyy = ρgy Fig. 2.1, Turcotte and Schubert, 214 Note, σyy is acting vertically upward, balancing the downward weight of the column 11

12 Body and surface forces Why is this true? Fig. 2.1, Turcotte and Schubert, 214 Consider a column of rock in the crust and the force acting on its basal surface (σyy A) The rock column has a thickness, y, and cross-sectional area, A Thus, the weight is simply ρgy A, the density times gravitational acceleration time the volume of the column Since the force at on the basal surface must equal the column weight, σyy A = ρgy A, or σyy = ρgy Note, σyy is acting vertically upward, balancing the downward weight of the column 12

13 Surface stresses Stress: A force per unit area transmitted through a material by interatomic force fields Surface stress: A pair of equal and opposite forces acting on the area of a surface in a specific orientation Units: Pascals [Pa]; 1 Pa = 1 N m - 2 Representation: Pair of vectors with a specified surface area/orientation Example: Hand pushing on table, table pushing back 13

14 Surface stresses Stress: A force per unit area transmitted through a material by interatomic force fields Surface stress: A pair of equal and opposite forces acting on the area of a surface in a specific orientation Units: Pascals [Pa]; 1 Pa = 1 N m - 2 Representation: Pair of vectors with a specified surface area/orientation Example: Hand pushing on table, table pushing back Surface stress σ area A - σ Component surface stresses after Twiss and Moores, 26 14

15 Surface stresses Stress: A force per unit area transmitted through a material by interatomic force fields Surface stress: A pair of equal and opposite forces acting on the area of a surface in a specific orientation Units: Pascals [Pa]; 1 Pa = 1 N m - 2 Representation: Pair of vectors with a specified surface area/orientation Example: Hand pushing on table, table pushing back Surface stress σn σ area σs A - σ Component surface stresses Normal stress - σs - σn Shear stress after Twiss and Moores, 26 15

16 Surface stress at the Moho Assuming that the continental crust is 35 km thick and has an average density of 275 kg m -3, how large is the vertical surface stress at its base? You may want to report your value in megapascals (MPa, 1 million Pa). σyy =??? Fig. 2.1, Turcotte and Schubert,

17 Crustal materials at mantle conditions typical 3 µm thin sections, microdiamonds may appear to be yellow due to small amounts of nitrogen (FIG. 3B), and their occurrence at different focal depths within the host mineral (FIG. 3C) dispels the possibility of their introduction during sample preparation. Petrographic identification of microdiamond requires reflected-light microscopy to observe its very high reflectivity and internal reflection. Graphite replacing diamond characteristically appears optically in dull (crystallographically disordered) and brighter (crystallographically ordered) varieties. Graphite is often formed when fluid infiltrates along fractures and reacts with the apparently protected diamond inclusions. Graphite inclusions alone, however, are not a reliable indicator of the former presence of diamond. However, radiating polishing scratch patterns (FIG. 3D) around raised diamonds or, more commonly, around graphite-lined holes remaining after the diamonds were plucked during polishing may indicate the presence of microdiamond. Microdiamond has generally been interpreted to form from the reduction of dense, supercritical COH + silicate fluids trapped during UHP metamorphism, and thus tiny multiphase, graphite-bearing inclusions are a good pointer for more detailed investigation. Taken to high pressures and temperatures, crustal materials undergo transitions from one crystal form to another What happens to these minerals when they undergo phase transitions? Schertl diagram and O Brien, Pressure temperature showing 213 the reaction boundaries for andalusite sillimanite kyanite (Holdaway 1971), calcite aragonite (Hacker et al. 25), quartz coesite (Bose and Ganguly 1995), coesite stishovite (Yong et al. 26), graphite diamond (Kennedy and Kennedy 1976), rutile-type FIGURE 1 With this in mind, do you see any problems with the plot? Experimental studies on the HP behavior of TiO2 suggest that the natural occurrence of -PbO2 -type TiO2 (e.g. Withers et al. 23) requires formation pressures that considerably exceed the graphite diamond phase boundary (FIG. 1). Thus -PbO2 -type TiO2 can be an important indicator for formation pressures between the graphite diamond boundary and the stishovite field. If an -PbO2 type TiO2 phase is found only as nanometer-size, epitaxial slabs between twinned rutile crystals and not as true relics, caution is required before using these nanocrystals for routine geobarometry and as an unequivocal proof of 17

18 Continental buoyancy Fig. 2.2, Turcotte and Schubert, 214 Consider a block of continental crust floating on the mantle The crust floats because a typical crustal density ρc is ~275 kg m -3 whereas the mantle density ρm is ~33 km m -3 (i.e., its buoyant) In hydrostatic equilibrium the weight of a crustal column of thickness h is equal to that of the mantle beside it at the depth of its base b, or c h = m b 18

19 Continental buoyancy Fig. 2.2, Turcotte and Schubert, 214 Consider a block of continental crust floating on the mantle The crust floats because a typical crustal density ρc is ~275 kg m -3 whereas the mantle density ρm is ~33 km m -3 (i.e., its buoyant) In hydrostatic equilibrium the weight of a crustal column of thickness h is equal to that of the mantle beside it at the depth of its base b, or c h = m b How high does a 35-km-thick continent stick out above the mantle? 19

20 Continental buoyancy Continents are mostly near sea level Oceans are mostly 4 km below 2

21 Horizontal surface stresses The horizontal surface stresses σxx and σzz act normal (or perpendicular) to vertical planes in the Earth Combined σxx, σyy and σzz are the three normal stresses In hot/weak rocks all three may be equal to the weight of the overlying rock or lithostatic pressure pl p L xx = yy = zz = gy Fig. 2.7, Turcotte and Schubert, 214 Note: We will use the orientation of the coordinate axes show on the left in subsequent lectures 21

22 Tangential surface stresses Forces and stresses can act both normal and parallel (or tangent) to surfaces, such as in strike-slip faults These stresses, such as σxz are called shear stresses Fig. 2.1, Turcotte and Schubert, 214 In the convention for labeling these stresses, the first letter is the direction normal to the surface element and the second letter is the direction of the shear force 22

23 Tangential surface stresses Forces and stresses can act both normal and parallel (or tangent) to surfaces, such as in strike-slip faults These stresses, such as σxz are called shear stresses Fig. 2.1, Turcotte and Schubert, 214 In the convention for labeling these stresses, the first letter is the direction normal to the surface element and the second letter is the direction of the shear force What is the orientation of σyx? 23

24 Stress in two dimensions Surface forces in 2D In two dimensions, we consider forces acting on four faces of an infinitesimal cube of dimension δx δy δz Here we assume no forces act or vary in the z direction Normal stresses: σxx, σyy Shear stresses: σxy, σyx At equilibrium we can state σxy = σyx Fig. 2.13, Turcotte and Schubert, 214 Why? 24

25 Stress in two dimensions Surface forces in 2D In two dimensions, we consider forces acting on four faces of an infinitesimal cube of dimension δx δy δz Here we assume no forces act or vary in the z direction Normal stresses: σxx, σyy Shear stresses: σxy, σyx At equilibrium we can state σxy = σyx Fig. 2.13, Turcotte and Schubert, 214 Why? 25

26 Stress in two dimensions Surface forces in 2D In two dimensions, we consider forces acting on four faces of an infinitesimal cube of dimension δx δy δz Here we assume no forces act or vary in the z direction Normal stresses: σxx, σyy Shear stresses: σxy, σyx At equilibrium we can state σxy = σyx Fig. 2.13, Turcotte and Schubert, 214 Why? 26

27 Changing coordinate reference frames The normal and shear stresses in the coordinate system x, y can also be represented in another coordinate system xʹ, yʹ that has been rotated by angle θ Knowing σxx, σyy and σxy, their equivalent values in the rotated coordinate system xʹ, yʹ are x x = xx cos 2 + yy sin 2 + xy sin 2 y y = xx sin 2 + yy cos 2 xy sin 2 x y = 1 2 ( yy xx)sin2 + xy cos 2 27

28 Changing coordinate reference frames Why would we want to do this? As it turns out, for any state of stress σxx, σyy and σxy it is possible to find a stress state with no shear stresses This can be found by setting σxʹyʹ equal to zero and solving for θ in the equation below x y = 1 2 ( yy xx)sin2 + xy cos 2 which yields tan 2 = xx 2 xy The orientation of θ is the principal axis of stress, as is θ + π/2 yy 28

29 Changing coordinate reference frames Why would we want to do this? As it turns out, for any state of stress σxx, σyy and σxy it is possible to find a stress state with no shear stresses This can be found by setting σxʹyʹ equal to zero and solving for θ in the equation below x y = 1 2 ( yy xx)sin2 + xy cos 2 which yields tan 2 = xx 2 xy The orientation of θ is the principal axis of stress, as is θ + π/2 yy 29

30 Changing coordinate reference frames Why would we want to do this? As it turns out, for any state of stress σxx, σyy and σxy it is possible to find a stress state with no shear stresses This can be found by setting σxʹyʹ equal to zero and solving for θ in the equation below x y = 1 2 ( yy xx)sin2 + xy cos 2 which yields tan 2 = xx 2 xy The orientation of θ is the principal axis of stress, as is θ + π/2 yy 3

31 Principal stresses The normal stresses in the principal axis coordinate system are know as the principal stresses σ1 and σ2 The principal stresses can be found from stresses in coordinate system x, y using apple 1,2 = xx + yy ( xx yy ) 2 1/2 ± + xy Conversely, we can get the stresses in coordinate system x, y from the principal stresses using xx = ( 1 2) 2 xy = 1 2 ( 1 2)sin2 yy = ( 1 2 ) 2 cos 2 cos 2 31

32 Stress in three dimensions In three dimensions, we consider forces acting on all six faces of an infinitesimal cube of dimension δx δy δz Normal stresses: σxx, σyy, σzz Shear stresses: σxy, σyx, σxz, σzx, σyz, σzy At equilibrium we can state σxy = σyx, σxz = σzx, σyz = σzy 32

33 Stress in three dimensions As before, we can also determine the principal stresses in three dimensions The convention is that σ1 σ2 σ3 Isotropic stress is when σ1 = σ2 = σ3 = p The hydrostatic state of stress is when the normal stresses equal p and there are no shear stresses The lithostatic stress is the hydrostatic state of stress where stress increases in proportion to the density of the overlying rock 33

34 Stress in three dimensions As before, we can also determine the principal stresses in three dimensions The convention is that σ1 σ2 σ3 Isotropic stress is when σ1 = σ2 = σ3 = p The hydrostatic state of stress is when the normal stresses equal p and there are no shear stresses The lithostatic stress is the hydrostatic state of stress where stress increases in proportion to the density of the overlying rock 34

35 Stress in three dimensions As before, we can also determine the principal stresses in three dimensions The convention is that σ1 σ2 σ3 Isotropic stress is when σ1 = σ2 = σ3 = p The hydrostatic state of stress is when the normal stresses equal p and there are no shear stresses The lithostatic stress is the hydrostatic state of stress where stress increases in proportion to the density of the overlying rock 35

36 Stress in three dimensions As before, we can also determine the principal stresses in three dimensions The convention is that σ1 σ2 σ3 Isotropic stress is when σ1 = σ2 = σ3 = p The hydrostatic state of stress is when the normal stresses equal p and there are no shear stresses The lithostatic stress is the hydrostatic state of stress where stress increases in proportion to the density of the overlying rock 36

37 Stress in three dimensions When the principal stresses are not equal p = 1 3 ( ) = 1 3 ( xx + yy + zz ) 37

38 Deviatoric stresses We often subtract pressure from normal stresses to determine the deviatoric stresses (indicated by primes) xx = xx p xy = xy yy = yy p xz = xz zz = zz p yz = yz The same can be done for the principal stresses 1 = 1 p 2 = 2 p 3 = 3 p What is the sum of the deviatoric principal stresses? For rock deforming by viscous flow, all deformation is the result of a nonzero deviatoric stress 38

39 Deviatoric stresses We often subtract pressure from normal stresses to determine the deviatoric stresses (indicated by primes) xx = xx p xy = xy yy = yy p xz = xz zz = zz p yz = yz The same can be done for the principal stresses 1 = 1 p 2 = 2 p 3 = 3 p What is the sum of the deviatoric principal stresses? For rock deforming by viscous flow, all deformation is the result of a nonzero deviatoric stress 39

40 Deviatoric stresses We often subtract pressure from normal stresses to determine the deviatoric stresses (indicated by primes) xx = xx p xy = xy yy = yy p xz = xz zz = zz p yz = yz The same can be done for the principal stresses 1 = 1 p 2 = 2 p 3 = 3 p What is the sum of the deviatoric principal stresses? For rock deforming by viscous flow, all deformation is the result of a nonzero deviatoric stress 4

41 Deviatoric stresses We often subtract pressure from normal stresses to determine the deviatoric stresses (indicated by primes) xx = xx p xy = xy yy = yy p xz = xz zz = zz p yz = yz The same can be done for the principal stresses 1 = 1 p 2 = 2 p 3 = 3 p What is the sum of the deviatoric principal stresses? For rock deforming by viscous flow, all deformation is the result of a nonzero deviatoric stress 41

42 Recap A surface stress is a pair of equal and opposite forces applied to an area Stating the stress in two-dimensions requires knowing three stress values (2 normal, 1 shear); in 3D, we must know six stress values (3 normal, 3 shear) Deformation of the lithosphere occurs as the result of applied stresses and understanding these stresses is integral to understanding geodynamic processes in the Earth 42

43 References Schertl, H. P., & O'Brien, P. J. (213). Continental Crust at Mantle Depths: Key Minerals and Microstructures. Elements, 9(4), doi:1.2113/gselements Twiss, R. J., & Moores, E. M. (26). Structural Geology, 2nd Edition. W.H. Freeman Co. 43