Download presentation

Presentation is loading. Please wait.

Published byFrida Leeper Modified over 5 years ago

1
Distribution of Microcracks in Rocks Uniform As in igneous rocks where microcrack density is not related to local structures but rather to a pervasive process such as cooling. Variable as a function of distance with respect to local structure such as joints, shear fractures, and larger faults.

2
Mechanical Processes and Microcracking The tensile stress necessary for propagation of microcracks arise either from thermal or mechanical processes. Mechanical processes include: stress concentrations at grain boundaries and around cavities. elastic mismatch-induced cracking kink band and deformation lamellae-associated microcracking. twin-induced microcracking. Mechanically-induced microcracks tend to have a non- uniform distribution in association with local structures or discontinuities.

3
Thermal Stresses and Microcracking Arise from differential and incompatible expansion or contraction between: Grains of different thermoelastic properties Similar but mis-aligned anisotropic grains. Experimental work shows that a: T < 50 o C can generate thermally-induced microcracks. Such changes can occur during burial and uplift; so they should be common within uplifted rocks. Thermally-induced microcracks are more likely to have a uniform distribution in large rock volume.

4
Loading History Loading refers to history of applied forces (over time), displacement, and temperature, which control the net deformation of a rock. There are three types of loading: Gravitational Thermal Tectonic

5
Gravitational Loading Is the most important, and varies with time by erosion. Is due to the weight of the overlying rocks: z = g z It affects horizontal stress because rocks are elastic and tend to expand transverse to compressive vertical stress Poisson ratio: = e t /e l where = 0.25 for most rocks when material is free to expand laterally. If rock is not free to expand (e t = 0), a transverse stress t is created which is given by: t = /(1- z which for a = 0.25 becomes: t = z /3 = gz/3 This means the horizontal stress, induced by gravitational load, is 1/3 the vertical load.

6
Uplift and Denudation Many joints form by the release of tensile stresses generated during uplift and denudation. These stresses change the pre-existing state of stress ( ) in three ways: 1.Horizontal stretching through the geometry of uplift, s 2.Expansion through the release of the gravitational load, g 3.Contraction through cooling, t

7
Uplift and Denudation, cont’d If a body of rock with mean density, , is lithified at some depth (h) in the earth, it is subjected to an isotropic stress, P: P = z = x = y = gh The state of stress after a given amount of erosion (negative z ) is: z = g(h+ z ) The new horizontal stress is determined by the sum of the effects due to : horizontal stretching expansion by the release of the gravitational load. contraction through cooling. x = y = gh + s + g + t

8
Horizontal Stretching When a segment of the crust is uplifted; it may stretch horizontally. The amount of stretching is inversely proportional to the radius of curvature of the flexure, i.e.: The strongly curved parts of folds formed under brittle upper crustal conditions commonly display more concentrations of joints and small faults.

9
Release of the Gravitational Load Release of gravitational load by erosion causes expansion of compressed rock. Since rocks are not free to expand horizontally, much of the horizontal compressive stress remains after erosion.

10
Thermal Contraction A change in horizontal stress is produced during erosion as a result of cooling. Strain and temperature are related by: e = T where is the coefficient of thermal expansion. If, however, the rock is not free to expand or contract (i.e., e t = 0), a stress, called thermal loading, is set up, given by: t = - E /1- T

11
Example The surface of a lava flow is in tension because it cools faster than inside it; so it fractures. If an uplift leads to a T=100 o C in a confined rock, we get: t = - E /1- T t = (10 -6 / o C) (10 5 Mpa) / (1-0.25) 100 o C Which gives a t = -13 Mpa; equal to the tensile strength of most rocks, so joints will form.

12
Thermal Contraction … For slow uplift, the change in temperature is determined by the equilibrium thermal gradient. t = T/ z z For very rapid uplift the temperature change may be substantially higher In this case, given a T= T/ z z, the maximum thermal estimate for stress is: t = E /1- T/ z z

13
The Net Effect of Uplift The stress generated during uplift and erosion is the net effect of the horizontal stretching, expansion by the release of the gravitational load, and contraction through cooling. x = gh + s + g + t The horizontal stress, x, at a depth z =(h+ z ) after some erosion (negative z ) is given by: x = g[ z -(1-2 )/(1- ) z]+ E/1- dT/dz z

14
The Net Effect of Uplift Notice that the resulting stress is a function of several material properties ( ,, , E ) and the thermal gradient (dT/dz). Thus, under constant thermal stress, some rocks with different properties may form joints. Joints form by the release of tensile stresses generated by uplift and denudation.

15
Residual Stress A stress system that acts in a body of rock at equilibrium when external forces and temperature gradients do not exist. Builds up in rocks normal to compressive stresses when rocks cannot accommodate strain in transverse direction Note: e t /e l, e t = 0

16
Residual Stress Caused by change in T and P in homogeneous rocks with different thermoelastic properties or their spatial relationship (e.g., granite surrounded by schist). Residual stresses generated by pressure release, and cooling during uplift and erosion, play an important role in the formation of some joints.

17
Differential Stress for Jointing The shape and size of the Mohr circle places an upper limit for the deviatoric stress and hence depth of joint formation. The maximum deviatoric stress ( 1 - 3 ) for a tensile effective normal stress ( n < 0) is about 6 times the tensile strength (T o ). ( 1 - 3 ) < -6 T o This is even smaller (4 times the tensile strength ) for true joints: ( 1 - 3 ) < |4T o |

18
Depth of Jointing Joints require tensile effective normal stress ( n < 0 ) But the vertical stress due to gravitational load is compressive ( z = gz) Therefore, formation of joints at deep levels require both small deviatoric stress and high fluid pressure ( )

19
For transitional joint, the estimate for the maximum depth, z, is given: z = | (2+2 o 2)T o / g(1- ) | For true tensile joints, the maximum effective stress ( 1 ) is limited to 3 times the tensile strength ( 1 < |3T o |). The maximum depth for true joints is given by (is 60% of z for transitional joints): z < | 3T o / g(1- ) |

20
Orientation of Joints Fluid-solid interfaces such as earth surface and open fluid-filled joints are principal planes of stress (since fluid cannot support shear stress). Thus, the principal stress must be perpendicular to these interfaces.

21
Orientation of Joints Since joints form perpendicular to the least principal stress ( = 0 o ; = 90 o ), we expect joints to be either: parallel ( 3 vertical) or perpendicular ( 1 or 2 vertical) to the earth surface (dikes, veins, sills). Tensile joints propagating near a preexisting open joint will bend into perpendicular or parallel orientation with respect to the preexisting joint.

22
Joints in Bedded Sedimentary Rocks Assume a multi-layer of different rock type (1, 2, …, n) with their own individual Young’s modulus (E), tensile strength (T), and thickness (d): F and e are the horizontal stress & strain. _____________________________ E 1, T 1, d 1, F 1 = E 1 e 1 layer 1 _____________________________ E 2, T 2, d 2, F 2 = E 2 e 2 layer 2 _____________________________ E n, T n, d n, F n = E n e n layer n ____________________________

23
Joints vs. Rock type in Bedded Rocks The new horizontal stress in each of the layers after a uniform horizontal stretching (e x ; e.g., due to epeirogenic warping) is: F 1 = E 1 (e 1 - e x ) F 2 = E 2 (e 2 - e x ) F 3 = E 3 (e 3 - e x ) F n = E n (e n - e x ) Notice that even though the e x is uniform, the horizontal stresses in each layer will be different. Thus some layers will fracture first if their tensile strength is reached. Therefore, fracture spacing is a function of rocks type (T, E).

24
Joint Spacing and Bed Thickness Cracks release significant stress only within a radius of about one crack length. Thus, a fracture has little effect on the state of stress in its own bed more than one bed thickness away. The rest of the bed is very close to the stress for tensile failure. With the small increase in strain, other joints will form, each only relaxing the stress within one bed thickness spacing. Thus, fracture spacing is related to bed thickness. With continued strain, the entire bed will be jointed with spacing equal to bed thickness.

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google