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# force Stress strain deformation

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### force Stress strain deformation

1. 1. Subjects Force Stress Strain Deformation Rheology
2. 2.  Mass: Dimension: [M] Unit: g or kg  Length: Dimension: [L] Unit: cm or m  Time: Dimension: [T] Unit: s Velocity, v = distance/time = dx/dt  Change in distance per time) v =[L/T] or [LT-1] units: m/s or cm/s Acceleration (due to gravity): g = velocity/time  Acceleration is change in velocity per time (dv/dt). g = [LT-1 ]/[T] = LT-2, units: m s -2 Force: F = mass . acceleration  F = mg F = [M][LT-2]  units: newton: N = kg m s-2
3. 3.  A property or action that changes or tends to change the state of rest or velocity or direction of an object in a straight line  In the absence of force, a body moves at constant velocity, or it stays at rest  Force is a vector quantity; i.e., has magnitude, direction
4. 4.  Gravitational force  Acts over large distances and is always attractive  Ocean tides are due to attraction between Moon & Earth  Thermally-induced forces  e.g., due to convection cells in the mantle.  Produce horizontal forces (move the plates)  The other three forces act only over short ranges (atomic scales). May be attractive or repulsive  Electromagnetic force  Interaction between charged particles (electrons)  Nuclear or strong force  Holds the nucleus of an atom together.  Weak force  Is responsible for radioactivity
5. 5.  Any part of material experiences two types of forces:  surface & body  Body Force: Results from action of a field at every point within the body  Is always present  Could be due to gravity or inertia  e.g., gravity, magnetic, centrifugal  Its magnitude is proportional to the mass of the body
6. 6.  Act on a specific surface area in a body  Are proportional to the magnitude of the area  Reflect pull or push of the atoms on one side of a surface against the atoms on the other side  e.g., force of a cue stick that hits a pool ball
7. 7.  Forces applied on a body do either or both of the following:  Change the velocity of the body  Result in a shape change of the body  A given force applied by a sharp object (e.g., needle) has a different effect than a similar force applied by a dull object (e.g., peg). Why?  We need another measure called stress which reflect these effects
8. 8.  A force acting on a small area such as the tip of a sharp nail, has a greater intensity than a flat-headed nail! s = [MLT-2] / [L2]=[ML -1T-2] s = kg m-1 s-2 pascal (Pa) = newton/m2  1 bar (non-SI) = 105 Pa ~ 1 atmosphere  1 kb = 1000 bar = 108 Pa = 100 Mpa  1Gpa = 109 Pa = 1000 Mpa = 10 kb  P at core-mantle boundary is ~ 136 Gpa (at 2900 km)  P at the center of Earth (6371 km) is 364 Gpa
9. 9.  Tension: Stress acts _|_ to and away from a plane  pulls the rock apart  forms special fractures called joint  may lead to increase in volume  Compression: stress acts _|_ to and toward a plane  squeezes rocks  may decrease volume  Shear: acts parallel to a surface  leads to change in shape
10. 10.  Force (F) across any of these planes can be resolved into two components: Shear stress: Fs , & normal stress: Fn, where: Fs = F sin θ Fn = F cos θ tan θ = Fs/Fn  Smaller θ means smaller Fs
11. 11.  Stress on an arbitrarily-oriented plane through a point, is not necessarily perpendicular to the that plane  The stress (s) acting on a plane can be resolved into two components:  Normal stress (sn)  Component of stress perpendicular to the plane, i.e., parallel to the normal to the plane  Shear stress (ss) or t  Components of stress parallel to the plane
12. 12.  The average overburden pressure (i.e., lithostatic P) at the base of a 1 km thick rock column (i.e., z = 1 km), with density (r) of 2.5 gr/cm3 is 25 to 30 MPa P = rgz [ML -1T-2] P = (2670 kg m-3)(9.81 m s-2)(103 m) = 26192700 kg m-1s-2 (pascal) = 26 MPa   The geopressure gradient: dP/dz  30 MPa/km  0.3 kb/km (kb = 100 MPa)  i.e. P is  3 kb at a depth of 10 km
13. 13.  Physical quantities, such as the density or temperature of a body, which in no way depend on direction  are expressed as a single number  e.g., temperature, density, mass  only have a magnitude (i.e., are a number)  are tensors of zero-order
14. 14.  Some physical quantities are fully specified by a magnitude and a direction, e.g.:  Force, velocity, acceleration, and displacement  Vectors:  relate one scalar to another scalar  have magnitude and direction  are tensors of the first-order  have 1 subscript (e.g., vi) and 21 and 31 components in 2D and 3D, respectively
15. 15.  Some physical quantities require nine numbers for their full specification (in 3D)  Stress, strain, and conductivity are examples of tensor  Tensors:  relate two vectors  are tensors of second-order  have 2 subscripts (e.g., sij); and 22 and 32 components in 2D and 3D, respectively
16. 16.  The stress tensor matrix: | s11 s12 s13 | sij = | s21 s22 s23 | | s31 s32 s33 |  Can be simplified by choosing the coordinates so that they are parallel to the principal axes of stress: | s1 0 0 | sij = | 0 s2 0 | |0 0 s3 |  In this case, the coordinate planes only carry normal stress; i.e., the shear stresses are zero  The s 1 , s2 , and s 3 are the major, intermediate, and minor principal stress, respectively  s1>s3 ; principal stresses may be tensile or compressive
17. 17.  A component of deformation dealing with shape and volume change  Distance between some particles changes  Angle between particle lines may change  The quantity or magnitude of the strain is given by several measure based on change in:  Length (longitudinal strain) - e  Angle (angular or shear strain) -   Volume (volumetric strain) - ev
18. 18.  Extension or Elongation, e: change in length per length e = (l´-lo) / lo = Dl/ lo [dimensionless]  Where l´ and lo are the final and original lengths of a linear object  Note: Shortening is negative extension (i.e., e < 0)  e.g., e = - 0.2 represents a shortening of 20% Example:  If a belemnite of an original length (lo) of 10 cm is now 12 cm (i.e., l´=12 cm), the longitudinal strain is positive, and e = (12-10)/10 * 100% which gives an extension, e = 20%
19. 19.  Stretch: s = l´/lo = 1+e = l [no dimension] X =  l1 = s1 Y =  l2 = s2 Z =  l3 = s3  These principal stretches represent the semi-length of the principal axes of the strain ellipsoid. For Example: Given lo = 100 and l´ = 200 Extension: e = (l´-lo)/ lo = (200-100)/100 = 1 or 100% Stretch: s = 1+e = l´/lo = 200/100 = 2 i.e., The line is stretched twice its original length!
20. 20.  Gives the change of volume compared with its original volume  Given the original volume is vo, and the final volume is v´, then the volumetric stain, ev is: ev =(v´-vo)/vo = dv/vo [no dimension]
21. 21.  Any deformed rock has passed through a whole series of deformed states before it finally reached its final state of strain  We only see the final product of this progressive deformation (finite state of strain)  Progressive strain is the summation of small incremental distortion or infinitesimal strains
22. 22.  Incremental strains are the increments of distortion that affect a body during deformation  Finite strain represents the total strain experienced by a rock body  If the increments of strain are a constant volume process, the overall mechanism of distortion is termed plane strain (i.e., one of the principal strains is zero; hence plane, which means 2D)  Pure shear and simple shear are two end members of plane strain
23. 23.  Distortion during a homogeneous strain leads to changes in the relative configuration of particles  Material lines move to new positions  In this case, circles (spheres, in 3D) become ellipses (ellipsoids), and in general, ellipses (ellipsoids) become ellipses (ellipsoids).  Strain ellipsoid  Represents the finite strain at a point (i.e., strain tensor)  Is a concept applicable to any deformation, no matter how large in magnitude, in any class of material
24. 24.  Series of strain increments, from the original state, that result in final, finite state of strain  A final state of quot;finitequot; strain may be reached by a variety of strain paths  Finite strain is the final state; incremental strains represent steps along the path
25. 25.  We can think of the strain ellipse as the product of strain acting on a unit circle  A convenient representation of the shape of the strain ellipse is the strain ratio Rs = (1+e1)/(1+e3) = S1/S3 = X/Z  It is equal to the length of the long axis over the length of the short axis
26. 26.  If a line parallel to the radius of a unit circle, makes a pre-deformation angle of  with respect to the long axis of the strain ellipse (X), it rotates to a new angle of ´ after strain  The coordinates of the end point of the line on the strain ellipse (x´, z´) are the coordinates before deformation (x, z) times the principal stretches (S1, S 3)
27. 27. .
28. 28.  3D equivalent - the ellipsoid produced by deformation of a unit sphere  The strain ellipsoids vary from axially symmetric elongated shapes – cigars and footballs - to axially shortened pancakes and cushions
29. 29.  If the strain axes have the same orientation in the deformed as in undeformed state we describe the strain as a non-rotational (or irrotational) strain  If the strain axes end up in a rotated position, then the strain is rotational
30. 30.  An example of a non-rotational strain is pure shear - it's a pure strain with no dilation of the area of the plane  An example of a rotational strain is a simple shear
31. 31. 1. Axially symmetric extension  Extension in one principal direction (l1) and equal shortening in all directions at right angles (l2 and l3) l1 > l2 = l3 < 1  The strain ellipsoid is prolate spheroid or cigar shaped 2. Axially symmetric shortening  This involves shortening in one principal direction (l3) and equal extension in all directions at right angles (l1 and l2 ). l1 = l2 > 1 > l3  Strain ellipsoid is oblate spheroid or pancake-shaped
32. 32. The sides of the parallelogram will progressively lengthen as deformation proceeds but the top and bottom surfaces neither stretch nor shorten. Instead they maintain their original length, which is the length of the edge of the original cube
33. 33.  In contrast to simple shear, pure shear is a three- dimensional constant-volume, irrotational, homogeneous flattening, which involves either plane strain or general strain.  Lines of particles that are parallel to the principal axes of the strain ellipsoid have the same orientation before and after deformation  It does not mean that the principal axes coincided in all increments!  During homogeneous flattening a sphere is transformed into a pancake-like shape and a box is changed into a tablet or book-like form.
34. 34.  During pure shear the sides of the cube that are parallel to the z-axis are shortened, while the lengths of the sides that are parallel to the x-axis increase. In contrast, the lengths of the sides of the cube that are parallel to the y-axis remain unchanged.  When such geometrical changes occur during the transformation of a rock body to a distorted state then the mechanism of distortion is termed plane strain.
35. 35.  Collective displacements of points in a body relative to an external reference frame  Deformation describes the transformations from some initial to some final geometry  Deformation of a rock body occurs in response to a force
36. 36.  Deformation involves any one or a combination of the following four components:  Ways that rocks respond to stress: 1. Rigid Body Translation 2. Rigid Body Rotation 3. Distortion or Strain 4. Dilation
37. 37. .
38. 38.  Distortion is a non-rigid body operation that involves the change in the spacing of points within a body of rock in such a way that the overall shape of the body is altered with or without a change in volume  Changes of points in body relative to each other  Particle lines may rotate relative to an external coordinate system  Translation and spin are both zero  Example: squeezing a paste  In rocks we deal with processes that lead to both movement and distortion
39. 39.  Dilation is a non-rigid body operation involving a change in volume  Pure dilation:  The overall shape remains the same  Internal points of reference spread apart (+ev) or pack closer (-ev) together  Line lengths between points become uniformly longer or shorter
40. 40.  Though commonly confused with each other, strain is only synonymous with deformation if there has been distortion without any volume change, translation, or rotation  Strain represents only one of four possible components involved in the overall deformation of a rock body where it has been transformed from its original position, size, and shape to some new location and configuration  Strain describes the changes of points in a body relative to each other, or, in other words, the distortions a body undergoes  The reference frame for strain is thus internal
41. 41.  Originally straight lines remain straight  Originally parallel lines remain parallel  Circles (spheres) become ellipses (ellipsoids)
42. 42. .
43. 43. Heterogeneous strain affects non-rigid bodies in an irregular, non-uniform manner and is sometimes referred to as non-homogeneous or inhomogeneous strain Leads to distorted complex forms
44. 44.  Viscous deformation is a function of time  This means that strain accumulates over time  Hence deformation is irreversible, i.e. strain is  Non-recoverable  Permanent • Flow of water is an example of viscous behavior. • For a constant stress, strain will increase linearly with time (with slope: s/h) • Thus, stress is a function of strain and time! s = he/t
45. 45.  The terms elastic and plastic describe the nature of the material  Brittle and ductile describe how rocks behave.  Rocks are both elastic and plastic materials, depending on the rate of strain and the environmental conditions (stress, pressure, temperature), and we say that rocks are viscoelastic materials.
46. 46.  Plasticity theory deals with the behavior of a solid.  Plastic strain is continuous - the material does not rupture, and the strain is irreversible (permanent).  Occurs above a certain critical stress (yield stress = elastic limit)  where strain is no longer linear with stress  Plastic strain is shear strain at constant volume, and can only be caused by shear stress
47. 47.  Brittle rocks fail by fracture at less than 3-5% strain  Ductile rocks are able to sustain, under a given set of conditions, 5-10% strain before deformation by fracturing
48. 48.  Confining pressure, Pc  Effective confining pressure, Pe  Pore pressure, Pf is taken into account  Temperature, T .  Strain rate, e
49. 49. – Increasing T increases ductility by activating crystal-plastic processes – Increasing T lowers the yield stress (maximum stress before plastic flow), reducing the elastic range – Increasing T lowers the ultimate rock strength •Ductility: The % of strain that a rock can take without fracturing in a macroscopic scale
50. 50. The time interval it takes to accumulate a certain amount of strain Change of strain with time (change in length per length per time). Slow strain rate means that strain changes slowly with time – How fast change in length occurs per unit time
51. 51.  Shear strain rate: . .  =2e [T-1]  Typical geological strain rates are on the order of 10-12 s-1 to 10-15 s-1  Strain rate of meteorite impact is on the order of 102 s-1 to 10-4 s-1
52. 52.  Decreasing strain rate:  decreases rock strength  increases ductility .  Effect of slow e is analogous to increasing T  Think about pressing vs. hammering a silly putty  Rocks are weaker at lower strain rates  Slow deformation allows diffusional crystal-plastic processes to more closely keep up with applied stress
53. 53. • Increasing confining pressure: • Greater amount of strain accumulates before failure • i.e., increases ductility –increases the viscous component and enhances flow –resists opening of fractures •i.e., decreases elastic strain
54. 54. • Increasing pore fluid pressure – reduces rock strength – reduces ductility • The combined reduced ductility and strength promotes flow under high pore fluid pressure • Under ‘wet’ conditions, rocks deform more readily by flow – Increasing pore fluid pressure is analogous to decreasing confining pressure
55. 55.  Rupture Strength (breaking strength)  Stress necessary to cause rupture at room temperature and pressure in short time experiments  Fundamental Strength  Stress at which a material is able to withstand, regardless of time, under given conditions of T, P and presence of fluids without fracturing or deforming continuously