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Rheology I. Rheology Part of mechanics that deals with the flow of rocks, or matter in general Deals with the relationship of the following: (in terms.

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Presentation on theme: "Rheology I. Rheology Part of mechanics that deals with the flow of rocks, or matter in general Deals with the relationship of the following: (in terms."— Presentation transcript:

1 Rheology I

2 Rheology Part of mechanics that deals with the flow of rocks, or matter in general Deals with the relationship of the following: (in terms of constitutive equations): stress,  strain, e strain rate e. (hence time, t) material properties other external conditions Rocks flow given time and other conditions!

3 Linear Rheologies The ratios of stress over strain or stress over strain rate is constant, e.g.: Elastic behavior:  = Ee Viscous behavior:  = ηe.

4 Rheology Explains Behavior Drop onto a concrete floor four objects: a gum eraser a cube of halite a ball of soft clay one cm 3 of honey When they fall, they behave the same by following the Newton’s Second Law (F = mg) Their difference is when they reach the ground: The eraser rebounds and bounces (elastic) The clay flattens and sticks to the floor (ductile) The halite fractures and fragments scatter (brittle) The honey slowly spreads on the floor (viscous)

5 Material Parameters Rheology depends on: Extrinsic (external) conditions such as: P, T, t, chemistry of the environment Intrinsic (internal) material properties such as: rock composition, mass, density

6 Material Parameters Are actually not purely “material constants” Are related to the rheological properties of a body, e.g.: rigidity compressibility viscosity, fluidity elasticity These depend on external parameters Are scalars in isotropic material and tensors of higher order in anisotropic material

7 Constitutive Equations Mechanical state of a body is specified by: Kinematic quantities such as: strain, e displacement, d velocity, v acceleration, a Dynamic quantities such as: force, F stress, σ

8 Constitutive Equations, Example F = ma  = E e The constitutive equations involve both mechanical and material parameters: f (e, e., ., ……, M ) = 0 M is material property depending on P, T, etc.

9 Law of Elasticity - Hooke’s Law A linear equation, with no intercept, relating stress (  to strain (e) For longitudinal strain:  = E e(  e/  t = 0) The proportionality constant ‘E’ between stress and longitudinal strain is the Young’s modulus Typical values of E for crustal rocks are on the order of 10 -11 Pa Elasticity is typical of rocks at room T and pressures observed below a threshold stress (yield stress)

10 Characteristics of Elasticity Instantaneous deformation upon application of a load Instantaneous and total recovery upon removal of load (rubber band, spring) It is the only thermodynamically reversible rheological behavior Stress and strains involved are small Energy introduced remains available for returning the system to its original state (internal strain energy) It does not dissipate into heat; i.e., strain is recoverable Typically, elastic strains are less than a few percents of the total strain

11 Law of Elasticity.

12 Shear Modulus For shear stress and strains  s = G  The proportionality constant G between stress and shear strain is the shear modulus (rigidity)

13 Bulk Modulus For volume change under pressure: P = Ke v K = P/e v is the bulk modulus; e v is dilation K is the proportionality constant between pressure and volumetric strain The inverse of the bulk modulus is the compressibility: k = 1/K

14 Units of the proportionality constants The proportionality constants ‘E’, ‘G’, and ‘K’ are the slope of the line in the  -e diagram (slope =  /e) Since ‘E’,,’G’, and K’ are the ratio of stress over strain (  /e), their units are stress (e.g., Pa, Mpa, bar) because ‘e’ is dimensionless

15 Poisson Ratio,  nu  Under uniaxial load, an elastic rock will shorten under compression while expanding in orthogonal direction Poisson ratio: The ratio of the elongation perpendicular to the compressive stress (called: transverse, e t, or lateral strain, e lat ) and the elongation parallel to the compressive stress (longitudinal strain, e l ) = e lat /e long = e t /e l [no dimension] It shows how much a core of rock bulges as it is shortened

16 http://silver.neep.wisc.edu/~lakes/PoissonIntro.html

17 http://en.wikipedia.org/wiki/Poisson's_ratio | ½ e l _ ½ e t = e t /e l

18 Poisson Ratio … Because rocks expand laterally in response to an axially applied stress, they exert lateral stress (Poisson effect) on the adjacent material If no lateral expansion is allowed, such as in a confined sedimentary basin or behind a retaining wall, the tendency to expand laterally produces lateral stress

19 Poisson Ratio  = e t /e l By setting the lateral (i.e., transverse, e t ) strains to zero, and loading a column of earth, describing its tendency to expand by Poisson's ratio and translating these lateral strains into stresses by Young's modulus we can show that (assume  1 is vertical):  2 =  3 =  lateral =  vertical /(1- ) or  h =  v /(1- ) (h =horizontal, v =vertical) For a material that expands as much as it is compressed (fully incompressible), for example a fluid ( = 0.5), this leads to:  h =  v (hydrostatic response) The second equation is used by engineers in calculating stresses behind retaining walls to estimate lateral stresses in mine shafts or in sedimentary basins. This is an elastic model, other options can be used to estimate stress at plastic failure

20 http://silver.neep.wisc.edu/~lakes/PoissonIntro.html Material become narrower when they are stretched!

21 Poisson ratio, =e t /e l ranges between 0.0 and 0.5 = 0.0 for fully compressible material, i.e., those that change volume under stress without extending laterally (i.e., e t =0 ): if e t = 0.0  =e t /e l = 0.0 Note: Sponge has a low  = 0.5 for fully incompressible material (e.g., fluid) which maintain constant volume irrespective of stress (material extends laterally): i.e., =e t /e l = 0.5  e t = 0.5 e l Note: lead cylinder a high 

22 Values of the Poisson ratio in natural rocks range between 0.25 and 0.35 (  0.25 for most rocks) For a  0.25, the magnitude of lateral stress (  h =  2 =  3 ) for most rocks (i.e., the Poisson effect) is  1/3 of the greatest principal stress (  l is vertical), i.e.,  3 = 1/3  l because:  h =  v /(1- )or  3 = /(1- )  l  3 = 0.25/(1-0.25)   3 = 1/3  l


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