Structural geology Flashcards

Question

Harmony

Answer 1

The extend to which folding of multilayer sequence is consistent through the sequence. - disharmonic fold - dies out within a couple of half wavelengths - harmonic fold - continuous along its axial trace for many multiples of its half wavelength (each bed doing the same thing) - polyharmonic fold - harmony on various multiples of half wavelengths within the multilayer sequence.

Answer 2

Polyharmonic folding generates parasitic folds. Occurs on many scales. Can describe the collective geometry of parasitic folds with respect to their enveloping surface. Parasitic fold geometries are predictable in relation to associated main fold (map vergence)

Answer 3

Characteristic of parasitic folds Indicates the direction of movement and rotation during deformation. Can identify the location of major folds using minor folds. Z (LHS of fold) M (top of fold) S (RHS of fold) Powerful field mapping tool, especially in areas that are relatively poorly exposed.

Answer 4

Axial planar cleavage - cleavage that is parallel to the axial plane Planes defined by fractures or mineral alignment parallel to a folds axial plane -bedding steeper than cleavage = overturned limb -cleavage steeper than bedding = upright limb Assume: -bedding folded and right way up -cleavage parallel to axial plane -axial plane bisects interlimb angle

Answer 5

Alignment of minerals, forming cleavage or schistosity. Usually the plane of flattening xy plane of strain ellipsoid.

Answer 6

Brittle deformation - buckling, bending, flexural slip and kinking Ductile deformation - passive folding, flexural shear/flow, oblique shear

Answer 7

Horizontal forces Sinusoidal (smooth wave) folding of a single strong layer within a weaker matrix by lateral compression (layer maintains thickness). Tends to form parallel folds with uniform orthogonal layer thickness. Shallow crust. Strain accommodated parallel to layering (tangential longitudinal strain). Most strain accommodated in hinge (nothing happening in limbs). Outer arc extended; normal faults, extension veins, boudinage, horizontal cleavage Inner arc compressed and shortened; thrust faults, stylolites, vertical cleavage. Between them is a neutral surface where no deformation is taking place.

Answer 8

Often hard to separate from buckling Vertical forces Achieved when F act across the layer (at 90) and may involve more than one mechanism. Look at context - diapir (salt or intrusion rises into layers and pushes them up), faults (beds respond to vertical motion - reverse and normal), boudinage *layers drop into folds)

Answer 9

Where many thick, strong and still layer in between which is some material with low cohesion (weak boundary). Length of beds does not change, slip between (eg: bending a book). All action in limbs and nothing in hinge. Sense of shear changes across hinge zones - consistent between anticlinal and synclinal limbs. Produces parallel folds - constant bed thickness. Slickensides and striations are lineations and oriented at right angles to fold axis. Slip at layer boundaries form bedding-plane faults (thrust) characterised by grooving and mineral growth parallel to movement direction (slickensides or striations). Within the competent layer, veins often open at 45 angle to the shear directions of individual beds. Sigmoidal shape of cracks due to progressive lengthening of veins with rotation.

Answer 10

Produces folds with straight limbs and very sharp hinges - chevron folds or kink bands. Occurs in strongly layered or laminated sequences where have strong mechanical anisotropy between very competent layers and thin incompetent layers (associated with thin bedded sandstones interbedded with shales; slates and schists). Resultant folds show flexural slip and are parallel folds. MIX of buckle and flexural. Straight limbs, angular hinge Usually parallel folds - shallow crust.

Answer 11

Kinking See very intense layer-parallel slip on incompetent layers (discontinuities between layers) gaps in hinge zones = saddle reefs (can fill with mineralisation) lock up when inter-limb angle ~60

Answer 12

``` Kinking Smaller scale (mm) than chevron Two parallel sided, very well defined axial surface. Layering deflects suddenly across the 1st kink plane then returns to its original orientation across the 2nd kink plane Ideally, each kink has an inter-limb angle of 120 and is symmetric around axial plane ```

Answer 13

Produces harmonic folds where the layering plays no mechanical role and has no influence on the fold shape. Can form in response to any kind of ductile strain (simple shear, subsimple shear, transpression, pure shear).

Answer 14

``` Distortion of layering by shear on an infinite number of closely spaced parallel planes that are inclined perpendicular/at a high oblique angle to layering. Displacement on shear planes varies along the layer so it folds. Deformation = heterogeneous simple shear -> class 2 (similar) folds. Rocks must be very ductile and have low viscosity contrast - deep in crust or in particularly weaker rocks (salt). ```

Answer 15

Modification of folding layer by simple shear that occurs parallel to the layer and is evenly distributed within the layer. Occurs in layers with an initial uniform fabric (shale) and is geometrically similar to flexural slip folding thousands of infinitesimally thin layers. Distortion in limbs, no strain in hinge. Strain becomes more evenly distributed in limbs, shear increases down limbs (change in orientation), pure flexural folds have no neutral surface and strain increases away from hinge. Do not see individual slip surface.

Answer 16

Two main controls - viscosity and thickness (both proportional to wavelength but t more significant due to 2t vs cube rooted n) Viscosity = resistance to flow, high viscosity (most competent rocks). Increased viscosity ratio = longer wavelength (proportional) -high viscosity ratio = high viscosity layer encased in low viscosity rock -low viscosity ratio - folds are poorly developed (short wavelength), layer thickens by pure shear Thicker layers = longer wavelength

Answer 17

Interplay between layer thickness and viscosity contrast on intensity of folding (wavelength, A, number of folds) explains how parasitic folds might form. Thinner beds with lower viscosity contrasts -> short wavelength; incorporated within larger wavelength folds defined by thicker layers

Answer 18

To form parasitic folds, polyharmonic folding must occur. Controls on fold harmony relate to layer competency and thickness, but also to the relative abundance of and distance between competent layers. -closely spaced with same properties -> harmonic folds -competent layers are at intermediate distances get polyharmonic folds -few similarary competent layers which are far apart -> disharmonic folding

Answer 19

``` Multiple deformations can produce their own fold sets (F1, F2...). Superimposition can produce complex geometries (complex to distinguish in 2D). Ramsay and Huber - orientation of fold axis and axial surface of first generation folds Type 0 (redundant superimposition) - exact same orientation, add together to get greater amplitude Type 1 (dome-basin or egg-carton) - folds perpendicular to each other, forming dome-basin interference pattern. -dome = anticline + anticline -basin = syncline + syncline Type 2 (dome-crescent or mushroom) - axial surface has low dip fold add to upright fold Type 3 (convergent-divergent or rippled fold) - isoclinal recumbent fold + upright fold (produces synformal anticline etc) ```

Answer 20

Gravity and lithostatic stress | -unloading - removal of overburden and corresponding lateral stresses, allowing expansion of rock by fracturing.

Answer 21

Deformation in which one of the principal strain axes does not change length.

Answer 22

By meteorites, catastrophic event responsible for large local stresses (craters).

Answer 23

Changes in T can change stress conditions by 1) increasing P of fluids in rock 2) causing expansion (heating) or contraction (cooling) 3) causing mineralogical phase changes

Answer 24

The most important forces arise from plate motions - drives geological change. Structures usually at plate boundaries (high stress).

Answer 25

Related to P. Force per unit area (planar surface within rock or a surface partly containing volume of rock). Stress is a pair of equal and opposite pressures acting on a surface = a pair of equal and opposite forces acting on a unit area of a body. Stress is not seen 1Pa=1N/m^2 1bar = 0.1 MPa Stress gradient with depth is 250bar/km or 25MPa/km

Answer 26

Measure of the response of a rock to stress. Relates to change of shape and/or volume.

Answer 27

The removal of the deforming stress causes an immediate return of the body to its original shape. Strain is temporary and recoverable. Straight line Hooke's law: σ=Ee -linear relationship between stress (σ) and strain (e). Not seen in rocks as rocks only record the permanent strain history since they formed. Eventually get an earthquake. Shallow crust Temporary distortions of the lattice of individual crystals. Atomic spacing changed proportional to stress (relation to inter-actomic bonding force which is characteristic of each mineral -> different E).

Answer 28

The rock does not fracture and there is no return to original body shape after removal of deforming stress Strain is permanent and unrecoverable. Curved line Viscosity = material's resistance to deform Viscous strain is seen in rocks as it is permanent. Viscous deformation = material flow without failure Ductile deformation = permanent strain without visible fracturing Occurs at depths (high PT) or in weak rocks with lots of clay close to the surface Accommodated by 3 principal deformation mechanisms (microscopic scale) 1) cataclasis - fracture and sliding of solid particles, individual crystals fractured 2) Intracrystalline Plasticity - dislocation creep and dislocation glide, crystals change shape 3) Diffusive mass transfer - molecular scale diffusion of material in solid state or in solution.

Answer 29

The rock fractures and there is no return to original body shape after removal of deforming stress. Strain permanent and unrecoverable. Associated with joints, veins and faults.

Answer 30

Beyond this point, the sample will change shape permanently.

Answer 31

Elasticity of the material E=σ/e Different rocks have different elasticities which can change at different conditions.

Answer 32

Ideal viscous fluid. σ=ne. n=viscosity (steeper line = more viscous) e.=strain rate Relationship between stress and strain rate: with constant stress, strain increases linearly with time: e=σt/n -higher the applied stress, the faster the material deforms Total strain depends on the magnitude of the stress and the time over which stress acts. A shearing stress must be constantly applied to maintain movement More strain = lower viscosity

Answer 33

Perfectly elastic materials show linear stress-strain curves until they reach their failure strength and fracture - brittle failure Perfectly viscous materials never fails, continue to flow as long as a constant stress is applied; flat-line curve - ductile failure

Answer 34

A material that behaves elastically overall, but takes time to reach limiting value. Material instantly behaves elastically with applied stress, but the straining continues after this, decreasing exponentially with time until reaching elastic limit. Stress below failure strength -> instantaneous elastic response and remaining strain decays to 0.

Answer 35

Elastic + visco-elastic + viscous. Elastic immediately recovered, viscoelastic recovers over time. Viscous is permanent.

Answer 36

``` P T Rock properties Strain rate (e.) Fluid pressure ```

Answer 37

Effects of depth and T on deformation give rise to the brittle-ductile transition (BDT), below which all rocks deform in a ductile fashion. BDT moves up for higher T, faster rate of strain and more fluids Typical depth ~10-12km Brittle upper zone - faulting and earthquakes BDT Ductile lower zone - shear zone and fault creep

Answer 38

Due to the density of overlying water at rest

Answer 39

Due to the density of overlying rock

Answer 40

With increasing confining P a rock gets stronger (yield stress/strength increases) Behaviour has viscoelastic properties and becomes more ductile (inhibiting failure)

Answer 41

Increase in T decreases yield strength (stress) and Young's modulus. Increased T introduces viscoelastic behaviour and becomes more viscous. BDT ~ 300 degrees

Answer 42

Such as strength of rock (cohesion) - related to mineralogy, texture, degree of fracturing... Mudstones/salts weak and more prone to ductile flow Granites strong and more prone to fracturing

Answer 43

Lower strain rates, decreasing yield strength and increasing ductility. Long-term strain behaviour is creep. On long time scales, viscous strain can arise under low stresses that would produce only elastic effects on short time scales.

Answer 44

Pore fluids reduce the resistance to slip along internal planes/grains. Usually water-based, the elastic properties of rocks are reduced (water is incompressible). Fluids weaken a rock and decrease Young's modulus, causing ductile deformation.

Answer 45

Joints - fractures in rocks with no shear displacement Veins - fracture which is filled with precipitated material Faults - rock deformed brittlely with shear displacement

Answer 46

1) deformation of objects (3D). Sphere -> ellipsoid 2) Deformation of layers -> folding 3) Foliation: alignment of minerals by rotation, recrystallization of both.

Answer 47

Rigid body translation - object shape constant, movement in 1D Rigid body rotation - object shape constant, body rotated about an axis Distortion - shape change by squeezing or squashing Volume change - deforming object to change V

Answer 48

Strain usually heterogeneous, but on some scales (

Answer 49

Measures the amount of ductile deformation in 3D. Matches the changed shape. X-axis - long axis; Z-axis - short axis. -flattening strain - y-axis extended -constrictional strain - y-axis shortened -plane strain - Y-axis unchanged, all deformation in x-z plane.

Answer 50

Irrotational strain. Major axes of strain ellipse unchanged through deformation (ie. axes remain where they are and get longer/shorter) - coaxial strain. 2D coaxial strain measured by aspect ratio of strain ellipse (X/Z). All deformation in this dimension No change in volume

Answer 51

Rotational strain. The major axes of strain ellipse rotate through deformation. 2D non-coaxial strain measured by aspect ratio of strain ellipse (X vs Z) and by angular shear and shear strain (Y=tan(a)) No change in volume

Answer 52

1) measure the length changes of lines 2) measure changes of the angle between lines Shear strain (Y) = tan (angle) = change X/I

Answer 53

e=I-I0/I0 -shortening with be negative strain Strain (%) = e x 100 Stretch = I/I0

Answer 54

``` Quantitative classification of strain ellipsoid shapes. y-axis - ratio of x/y axes x-axis - ratio of y/z axes k=(a01)/(b-1) = tan (q) 3 end members types of 3D homogeneous strain. -uniaxial prolate, k=infinity -plane strain, k=1 -uniaxial oblate, k=0 ```

Answer 55

Using change in angles and length requires you know the original shapes and/or dimensions of the deformed objects = strain markers i) originally circular - oolite, reduction spots, lapilli tuffs ii) originally elliptical - conglomerate pebbles iii) original known size - belemnites, acicular minerals, boudinage. e=I-I0/I iv) objects originally possessed bilateral symmetry - fossils (tribolites)

Answer 56

strain ratio: x/y=tan(a) | Measure long and short axes dimension and plot. Angle (a) is angle from line of best fit to short axis (x-axis)

Answer 57

Centre-to-centre method Higher strain occurs in matrix and strain reflected by changes in distance between randomly arranged particles (original shape required). Measure distance between 2 centres and the angle between that line and a reference line. Undeformed rock - horizontal line More deformation - more extreme sine curve

Answer 58

If the principal direction unknown, use Wellman graphical method. i) transfer hinges and symmetry lines ii) draw arbitrary line A-B iii) draw lines parallel to hinge and symmetry line from points A and B iv) draw best fit strain ellipse through all corners

Answer 59

Stress ellipsoid: σ1>σ2>σ3, all perpendicular to one another. Perpendicular to strain ellipse. Origins of stress - crustal loading, crustal unloading, thermal effect and plate tectonics Geostatic stresses - stresses caused by water or rock pgz=σv=σ1=σ2=σ3

Answer 60

σ1=σ2=σ3 - spherical stress ellipse | Newly solidified magma chambers; young, undeformed sediments

Answer 61

2 axes = 0. σ1>0, σ2=σ3=0. Condition of mine pillars.

Answer 62

One axis = 0 σ1>0 and σ3>0, σ2=0. Implicit in many 2D cross-sections 2D stress ellipse

Answer 63

Most common All stresses >0. Represented by strain ellipsoid

Answer 64

Usually results from extension or compression and if have moderate confining pressure (σ3 +ve (compressional)) Conjugate faults, where sigma 1 bisects the acute angle

Answer 65

Normal faults - σ1 vertical and σ3 horizontal Reverse faults - σ1 horizontal and σ3 vertical -folds Strike-slip faults - σ1 horizontal and σ3 horizontal

Answer 66

Usually result from extension and if have low confining pressure (σ3 low or negative (tensile)). σ1 parallel to joints

Answer 67

Represents the stress state at the point of failure (relationship between shear stress, normal stress, principal stresses and angle of failure). Made from experimental data. Plot s1 and 3 Midpoint = mean stress (s1+s3/2) Radius = Deviatoric stress (s1-s3/2) Differential stress = s1-s3 (drives failures - larger circle = more likely failure will occur) P - point of failure B is the angle of internal friction = 90-2a C is the cohesion/strength = y-int (minimum shear stress value required to cause failure when normal stress is 0) Failure envelope - by subjecting rock to different s1 and s3 values and noting the shear and normal stresses at which failure occurred. Divided fields of stable stress from unstable stress. Mohr circles toughing failure envelope = critical states of stress Curved line Increase confining P, increase shear stress needed for failure (bigger mohr circle)

Answer 68

weaken rocks envelope moves - less stress required to fail. Veins may strengthen a rock

Answer 69

T=c+uσ(n) Positive linear relationship - predicts linear Mohr failure -c=cohesion u=coefficent of internal frication

Answer 70

Rocks in experiments do not fail according to Coulomb criterion when σ3 is low/negative (tensile) Griffiths failure criterion - idea that failure occurs from the opening and linkage of micro cracks. Rounded end to failure envelope, failure closer to origin than predicted. Rocks are stronger in compression than tension.

Answer 71

Low confining pressure - Griffiths criterion Higher confining pressure - Coulomb criterion -ve numbers - tension failure (steep dip) - surface +ve numbers - shear failure very +ve numbers - ductile (30-40) - depth Faults are steep at the surface and shallow out with depth - Mohr circle

Answer 72

Pore fluid P's decrease values of σ1/3. Thus, elevated pore fluid pressures make it possible to have movement on very low angle faults. Shifts Mohr circle to less differential stress needed to overcome shear resistance on plane. More fluid = more chance of failure (closer to vertical axis)

Answer 73

Use pore fluids in oil/gas industry to weaken rocks are force Mohr circle to move.

Answer 74

Joints often systematic - parallel, similar spacings; discrete sets Can be 100s m in length Extensional and have no measurable shear displacement Common Tensile failure (steeply dipping), low stress and strain (low confining P, -ve) Joints are planes. Shape controlled by bed thickness (thicker units more circular, thinner units start and stop at edge of bed) Joints often occur in association with folds. Timing of joints can be synchronous with folding or can be uplift and unloading related structures which form much later than faults. Transverse joints Diagonal joints Longitudinal joints

Answer 75

Master/systematic joints - long, pronounced, older than others, regular spacing Cross-joints - shorter and occur between systematic joints, less regular Orthogonal join formation - strain during development of set 1 joints causes σ1 and σ3 to flip. Only possible if differential stress is low (shallow) Joints can be dissolved out by meteoric water to provide fissures (limestone). They can have major impacts on flow within flows.

Answer 76

Decreases with layer thickness and strain More joints in stiffer layers Joints often confined to beds Each joint relieves part of total stress acting on rock. Each joint acts within its own bed. Joint in thicker bed has longer length, relieves more stress Join in thinner bed has thinner length, so relieves less stress Thus, to relieve total stress evenly, more joints with closer spacing required in thin beds relative to thick beds.

Answer 77

Joints start at origin and propagate away. | Ribbing if front stops, before more stress applied and propagates out again

Answer 78

Like unpeeling onion Pluton forms, lifted to surface, stresses relieved and form parallel to pluton Removal of overburden allows expansion of rock by fracturing

Answer 79

Hot magma hits air and cools slowly from outside in. Become stressed and fracture.