Solutions to Fossen Structural Geology

Structural Geology: Solutions to problems On these pages you will find solutions to the extra problem set prepared for several of the book chapters.

Version date: 01 Feb. 2011

Structural Geology/Fossen

Deformation (Chapter 2) Solution to problem 2-1

a) To calculate the extension or elongation, simply measure the present length (l) from one tip to the other, using a ruler, and then estimate the original length (l0) of the belemnite. This can be done by restoration (Figure SP2.1) or by adding the lengths of individual belemnite segments. Elongation = (l-l0)/ l0 = (11.56 – 6.22)/6.22 ≈ 0.86 = 86%. 86% extension means that the belemnite has been stretched by a factor (stretch s) of 1.46. Note that some of the boudins are barrel shaped, which introduces an uncertainty. If we assume that this shape is an expression of ductile deformation along the margins or corners of the boudins, then restoring a line through the center of the boudinaged belemnite would be the way to go. In the restoration shown here (Figure SP2.1a) I have chosen to balance the gaps and overlaps.

6.22 cm

2 cm

11.56 cm

Figure SP2.1a Restored and stretched belemnite of Problem 2-1.

b) This one is similar to what we did above: Measure the current length (l) and then the restored length (l0Tr and l0B) and calculate the extension (=elongation) as above. Top Triassic extension: (l-l0Tr)/ l0Tr = (236.6 – 210)/210 ≈ 0.127 = 12.7%. Top Basement extension: (l-l0B)/ l0B = (236.6 – 178.5)/178.5 ≈ 0.326 = 32.6% Is the extension evenly distributed in the two cases? For the Top Triassic marker, there is clearly more extension (gaps) between the Gullfaks Field and the Viking Graben then elsewhere. For the Top Basement marker the extension is more evenly distributed in the central – eastern part of the section, and lower in the western part (Shetland Platform). For the North Sea section, how do the two extension estimates compare? Basement is stretched more than the Top Triassic marker. The difference (~20% or 31.5 km extension) is a strain estimate for the late Permian–mid Triassic phase of rifting in this part of the North Sea rift. It tells us that this first phase of rifting involved considerably more extension than the second (Jurassic) and last phase. How much extension is taken up by the largest 4-5 faults? To answer this question, restore only the largest 4-5 faults and do the same calculation over again. This shows that the 5 largest faults take up 11/26.6 = 41% of the post-Triassic stretching, and 37.6/58.1 = 65% of the total basement extension estimated above (Figure SP2.1c). This tells us that small faults play a role in extension estimates. Since there must be small faults that are not shown in the interpretation (because of limited seismic resolution), there is a component of fault extension that is missing. Hence all of our results underestimate the real extension.

2

Problem Set: Solutions A

20 km

Top Triassic

Top Basement

Horda Platform

A’

l = 236.6 km

(a)

l0Tr = 210 km (26.6 km extension)

(b)

(c)

Viking Graben

Gullfaks Field

Shetland Platform

l0B =178.5 km (58.1 km extension)

Figure P2.1b (a) Cross-section through the northern North Sea, where post-Triassic strata have been removed. Based on deep seismic line NSDP84-1. (b) Top Triassic restored (fault offsets removed without any rotation of the layering). (c) Top Basement restored.

1

2

3

4

(a)

5

l = 236.6 km 20 km

(b)

l0Tr = 225.6 km (11 km extension)

(c)

l0B =199 km (37.6 km extension)

Figure P2.1c Restoration of 5 of the largest fault displacements only.

Is there any other way that we could estimate the extension along the North Sea section? If we knew top MOHO, we could do a constant area restoration, assuming that the prerift crustal thickness was constant and equal to the present thickness at the rift margins. A comment on uncertainties: The restored line should ideally be more or less planar and horizontal, since this is how sedimentary layers are deposited. We will see from Chapter 20 that this can be fixed, but it requires a choice of deformation such as vertical shear (which does not change our estimate of l0) or rigid rotation of dipping line segments or fault blocks (which increases l0).

3


Problem 2-2.

Bedding is horizontal, and we trace the orientation of the two fossils and measure their angle to bedding (bedding is taken to represent the shear plane). It appears that the two trace fossils have somewhat different orientations, so we get two estimates for the angular shear (38.5 and 53). The shear strains (g) are tan 38.5° ≈ 0.8 and tan 53°≈1.3. The difference in strain estimates may be due to different original orientations. However, the skolitos tend to have a fairly consistent bedding-perpendicular initial orientation. Hence it is likely that the shear strain is heterogeneous at the scale of observation. The fact that the lower fossil seems to be tighter may support this interpretation. However, we need to analyze more strain markers in this rock to see if there is a systematic variation in the orientations and thus in strain.

38.5°

Angular shear

53°

Figure SP2.2 Drawing of the essential features of the pictire and angular observations.

Problem 2-3.

To find the new positions of the four points we have to use the deformation matrix. This easy for simple shear and pure shear (use Equations 2.15 and 2.16 in the textbook), but more difficult for subsimple shear. We then have to calculate G from Equation 2.17 (note that the general formula for G has a printing error, so use the one for no area change), which in this case (g=2 and kx=2) becomes: G = g(kx-1/kx)/2(lnkx) = 2(2-0.5)/2ln(0.5) = 3/(-1.386)=-2.164. The deformation matrices and calculations of new points are shown in Figure SP2.3.

4

Problem Set: Solutions

y

y

Simple shear, γ=2 1 0

3

x x+2y = y y

2 1

Subsimple shear, kx=2, γ=2 2 2.164 0 0.5

3

for x=1, y=2:

for x=1, y=2: 1 0

1

2

3

4

5

6

2 2.164 0 0.5

1

i.e., x’=5, y’=2 1

y

1 5 = 2 2

2 1

1

2

2 0 x 2x = 0 0.5 y 0.5y

2 2 0 1 = 1 0 0.5 2

2

3

4

5

6

4

5

6

x

for x=2, y=1: x’=(4+2.164)=6.164 y’=0.5

Figure SP2.3 Graphical illustration of the three deformations. An ellipse has been added for clarity.

i.e., x’=2, y’=1. 1

3

for x=0, y=1: x’=2.164 y’=0.5

for x=1, y=2:

1

1 6.328 = 2 1

i.e., x’=6.328, y’=1.

x

Pure shear, kx=2

3

x 2x+2.164y = 0.5y y

x

Problem 2-4. The matrix

 3 0 0.25  0  0 0.5 0.5  0 0

   

describes a three-dimensional deformation where the off-diagonal term 0.25 reoates to a simple shear strain and the diagonal terms (3, 0.5, 0.5) describe a three-dimensional coaxial strain. The pure shear involves extension along the x-axis of the coordinate system, and shortening in the plane orthogonal to x (i.e. along the y- and z-axes of the coordinate system). The simple shear affects the z-value The fact that the simple shear and coaxial strain are contained in a single matrix tells us that, mathematically, they are applied simultaneously. The determinant of this matrix equals the product of the diagonal elements, which represent the pure shear components (geologically this is consistent with the fact that simple shear preserves volume or area): Det D=3x0.5x0.5= 0.75. The 3x extension along the x-axis is less than the combined shortening along y and z, which causes a reduction in volume by 25%, all caused by the coaxial component.

z

z

x

x y

Simple shear

y

Coaxial strain

Figure SP2.4 The two components of deformation represented by the deformation matrix in Problem 2-4 in a specified coordinate system.

5


Problem 2-5.

Since the deformation is taking place in the x-y plane, this is the section we want to study. To write the deformation matrices, we need to remember the premultiplication thing about matrices, which means that the last deformation is represented by the first matrix:

k kγ (1 + ∆ )  0  1 γ  1 0   k 1+ ∆  0 1 / k  0 1  0 1 + ∆  =   0       k Problem 2-6

All cases except case (e) are homogeneous, because we see from Figure P2.6 that straight lines remain straight and parallel lines remain parallel. Case (b) does not involve strain, because all the displacement are of equal length and parallel (translation). Neither does (a) involve strain, although this is a little harder to see. It becomes obvious once we realize that the displacement is that of rigid rotation. Remember: Deformation = Rigid rotation + Translation + Strain. The deformation matrix for each of the deformations are shown in Figure SP2-6. y

y Rotation ω=-22.5°

6

Translation

6

cosω -sinω sinω cosω

4

2.25 0

0 2.25

4

(a)

2

6

4

y

1.5 0

Pure shear 6 kx=1.5, ky=0.667

x

(b)

2

y

0 0.67

Simple shear Angular shear=22.5

6

4

6

4

1 0

x

0.41 1

4

(c)

2

4

6

x

(d)

2

y

y

Heterogeneous strain: 6 nonlinear transformation

6

4

4

4

6

x

Simple shear (tan22.5), then rotation (11.25°). Equal to rotating grid 11.25, then performing ss along the line dipping 11.25°

2

(e)

2

4

6

x

(f)

2

Figure SP2.6 Displacement fields and information about the deformation for Problem 2-6.

6

4

6

x


Problem 2-7

a) A rock with vertical foliation (strike/dip= 000/90) and vertical lineation (000/90):

z

z

l0=1

l1 x

x

45° y

Undeformed

y

Simple shear (γ=1)

Figure SP2.7 The initial situation and the effect of a shear strain of 1.

b) Because of the special orientation of this lineation (parallel to the z-axis) its angle to z will always be the angular shear strain, which for g=1 is 45°, and for g=10 becomes tan-110= 84.3°, which means that it makes an acute angle of 5.7° to the horizontal shear plane. To use the deformation matrix D to find the line rotation, we first need to find D. It is a matrix with no coaxial strain or volume change, which means that the diagonal elements are unity. Then there is one simple shear element, which we denote g. This element is off-diagonal, and we have to place it where it affects the x-value of any vector that D is applied to. Trying and failing shows that this is the upper right-hand corner. We then need to multiply a unit vector l representing the undeformed lineation with the deformation matrix D (see Appendix A.4). l=(001) and the calculation is simple:

 1 0 γ   0  γ  0 1 0  0  =  0        0 0 1   1   1  With g=1 and 10 we get the new vectors (1,0,1) and (10,0,1). We find the angles that these vectors make with the x-axis by using the tangent relationship. For g=1 we get tan-1(1/1)=45°, and for g=10 we get tan-1(10/1)=5.7°. c) To find the elongation of a line of unit length parallel to the lineation, we use what we found in b) above, which was that the unit vector (0,0,1) changes to (1,0,1) and (10,0,1) for the two deformations. Since we started out with a vector of unit length the new lengths give us the elongation. For g=1, the length becomes:

12 + 0 2 + 12 = 2 and for g=10, we get a new length that is close to 10 times the original one:

10 2 + 0 2 + 12 = 101 ≈ 10

7


d) Nothing happens to the orientation of the foliation during these deformations: It remains vertical and parallel to the x-z plane. You can use Equation A.16 to check this. The pole to the plane is represented by the normal vector p=(0,1,0), and we get the new orientation by multiplying this vector with the inverse of the matrix D: −1

1 0 γ   1 0 −γ   0    [ 0 1 0 ]  0 1 0  = [ 0 1 0 ]  0 1 0  =  1   0 0 1   0 0 1   0  The vector p does not change, hence there is no change in the orientation of the plane.

Problem 2-8

a) For subsimple shear with Wk=0.5 (with g being 1 and 10) we are looking at more complicated calculations. The general matrix for this type of deformation is:

k 0 Γ  D =  0 1 0   0 0 1 / k  But how do we find k and G? We have to pull k out of the expression for Wk given by Equation 2.29 or A.22. This is a twodimensional formula, which is fine since we can exclude the y-direction (we have a case of plane strain in the x-z plane). For g=1 and Wk=0.5 we get : 0.5 = cos[tan-1(2lnk)] cos-10.5 = tan-1(2lnk) 60 = tan-1(2lnk) tan60 = 2lnk 0.866 = lnk k=e0.866 k≈2.377 Similarly, for g=10 we get k ≈ 5769. We can get these k-values from the downloadable Excel-file as well, in which case we would use the sheet called “Wk-based 2D-Strain” and set Wk=0.5 and check the k-values calculated for g=1 and 10. Now we can calculate G like we did for Problem 2.3: For g =1: G = g(k-1/k)/[2(lnk)] = 1(2.377-1/2.377)/[2(ln2.377)] = 1.956/1.7317 = 1.13 For g =10: G = g(k-1/k)/2(lnk) = 10(5769-1/5769)/2[8.66] = 57690/17.32= 3330.8 The new line orientation and length for the two cases can now be calculated by multiplying l and D. For g =1:

z

1.13   0   1.13   2.377 0  Dl =  0 1 0   0  =  0   0 0 1 / 2.377   1   0.421

tanφ = 0.421/1.13 φ= 20.43° l1 φ 1.13

8

length of l1= 0.4212+1.132 = 1.206 0.421

x


For g =10:

z

3331   0   3331   5769 0   Dl =  0 1 0   0  =  0   0 0 1 / 5769   1   0.0001733

tanφ = 0.00017/3331 φ= 0.000003° l1 3331

length of l1= 33312+0.000172 = 3331

0.00017

x

Why does the last case give such a long low-angle lineation (almost parallel with the coordinate axis x)? Because for g =10 requires a huge coaxial component to fulfill Wk=0.5. In this sense, the coaxial component accumulates faster than the simple shear component for a constant-Wk deformation. As for plane rotations, multiplying the deformation matrices with the pole p to the foliation plane has no effect, so the foliation remains vertical:

 k 0 Γ  0  0  pD = [ 0 1 0 ]  0 1 0   0  =  1   0 0 1 / k   1   0  . b) The angle a between the flow apophyses in this subsimple shear is governed by Wk and therefore the same for any strain value (it is a flow parameter, not a strain parameter). We can use Equation 2.25: a = cos-1 Wk = cos-1 0.5 = 60° or we can use Figure 2.24 for Wk = 0.5 and read off a = 60°. The angle q‘ between the long axis (X) of the strain ellipsoid and the x-axis (and shear direction) can be found from Figure 5.12b for g=1 and k=2.377, which gives us an angle close to 5. g =10 is outside the range of this graph. We then have to use the Excel spread sheet called “Wk-based 2D-Strain” (which also gives us more precise values for g=1) and read off q‘, which for g=1 and 10 becomes 4.0° and 7.4x10-7.

Oblique flow apophysis

X γ=0°, Wk=0.5

z

X γ=1°, Wk=0.5 X γ=10° x

α= 60°

Figure SP2-8 Illustration of flow apophyses (blue lines) and the orientation of the finite strain ellipse (X, red lines) for Wk=0.5 and g=0, 1, and 10.

9


Problem 2-9

a) The amount of compaction is 20% (porosity is reduced by 50%, while the total volume is reduced by 20%, from 100% to 80%). The vertical elongation D therefore becomes -0.2, and the deformation matrix is:

0  1 0 0  1 0  D = 0 1 0  =  0 1 0   0 0 1 + ∆   0 0 0.8  b) The strain ellipsoid has two horizontal axes of equal length (X=Y) and a vertical axis (Z) that is 0.8 times the length of the other two. R is the ratio between two principal axes, and there is one R-value for each section containing these axes: Rxy=X/Y=1/1=1, Ryz=X/Y=1/0.8=1.25, Rxy=X/Z=1/0.8=1.25.

c) The strain ellipsoid is oblate (“hamburger-shaped”) and plot in the lower part of the Flinn diagram (Figure SP2.9)

1

X/Y

0

Y/Z

0

1

1.25

Figure SP2.9 Plot of the compactional strain in the Flinn diagram (linear).

10


Stress (Chapter 4-5) Problem 4-1

a) Planes 1 and 2 in the Mohr diagram both have q = 60°, q being the angle between s1 and the poles to the two planes. We don’t know if the two planes cross. We don’t even know their actual dips, only their angular relation to the principal stress axes. This is really a two-dimensional problem, but a three-dimensional interpretation is given in Figure SP4.1a, showing the orientations of the planes for a vertical s1 (extensional tectonic regime). The values of sn and ss are found directly from the Mohr diagram: sN ≈ 2.17 MPa, sS ≈ ±2.05 MPa. b) A force of 100 N (Newtons) acts normal to a 0.1 m2 plane. sN = 100N/0.1m2 =1000N/m2 = 1 kPa. sS = 0 on the plane (since the force is acting normal to the plane). c) A plane that makes 45 degrees to sN is represented by the star at the top of the Mohr circle in Figure SP4.1b, and the normal stress and shear stress across the plane becomes 0.5 kPa. Using Equation 4.2 gives: sN = 1 kPa cos245 = 0.5 kPa sS = (1 kPa sin90)/2=0.5 kPa

θ=60°

σs

σ1

kPa

1

0.5

σ3 σ2

60°

σ3 θ=60°

2θ=90°

1

Figure SP4.1b Mohr circle for problem 4-1c.

Figure SP4.1a Geometric interpretation of the two planes and their relation to the principal stress axes for Problem 4-1.

11

σ1 kPa


Problem 4-2

a) The states of stress are presented in Figure SP4.2a. b) We know from Box 7.2 that even though a plane oriented at 45° to s1 has the largest resolved shear stress of any plane orientation, a somewhat more steeply dipping plane (i.e. one that makes a smaller angle to s1, meaning that q>45) is more easily reactivated. This is so because even if the shear stress is a bit smaller, the normal stress across that plane is considerably smaller. We can calculate this by means of Equation 4.2 if we wish. These relationships depend on the frictional properties of the material (rock or soil). Consider the Mohr diagram. The critical angle of failure is given by the tangent point between the Mohr circle and the line representing the Mohr-Coulomb criterion. This line has the slope m = tanf, so that f is the angle of that line with the horizontal axis (see Figure 7.10). Further, the orientation of the critical shear fracture is given by 2q. Remember that the radius to the tangent point has to be perpendicular to the line representing the Mohr-Coulomb criterion. This tells us that: 2q = ±(90+f), q = ±(45+f/2) Since for most rocks, f ≈30° (or m ≈ tan30 ≈ 0.6), we can expect q ≈ ±(45° + 15°)= ±60°. Hence, it is likely that the weak planes dipping 60° will more easily activated in all three cases. The sense of slip will be normal since the largest principal stress is vertical.

100

100

σs

75

MPa

50

75

Mean stress: 25/2 MPa =12.5 MPa Deviatoric stress: 25 MPa

25

i)

100

σs

25

θ θ

50

60°

50

Mean stress: 100/2 MPa =50 MPa Deviatoric stress: 100 MPa

75

MPa

25

2θ=90° 2θ=120°

σ3

MPa

σs

75

MPa

σn

100

ii)

50

Mean stress: (100+50)/2 MPa =75 MPa Deviatoric stress: 50 MPa

25

σ3 25

50

45°

75

MPa

60°

σn

100

iii)

σ3 25

45°

50

75

MPa

σn

100

60°

θ=45°

Figure SP4.2a Mohr circle illustrations of the three stress states, mean and deviatoric stress, and illustration of the two weak plane orientations (bottom) for Problem 4-2.

12


Problem 4-3

a) The stress ellipse contains information about the three-dimensional state of stress (magnitudes of s1, s2 and s3) and the orientations of the principal stress axes (s1, s2 and s3). That means that we can find the normal and shear stresses across/ along any plane orientation that goes through the point or volume to which the ellipse applies. We can also find the maximum normal and shear stress that can occur and the planes for which these stresses occur (which we already know are the vertical and 45° inclined planes). All principal stress axes must be either positive (compressive) or negative (tensile). The Mohr circle tells us the magnitudes (not the orientations) of the principal stresses, whether they are compressive or tensile. From that information we can find the mean stress (center of the Mohr circle) and the differential stress (the radius). Any point on the circle represents a plane orientation, and using the Mohr diagram we can find the shear and normal stress across that plane. b) See Figure SP4.3(a). This is a triaxial state of stress. c) See Figure SP4.3(b). This is a case of plane stress since one of the principal stresses is zero.

σs

150

σ1

100

50 σ3

σs

50

σ2 100

MPa

25 σ2

MPa

50

(a)

50 σ1

25

50

σ3

100

σ2

150

σ1

σn

200

(b)

MPa

σ3

25

σ2

50

σ1

75

σn

MPa

Figure SP4.3 Mohr circle illustrations of the two stress states, and drawing of the stress ellipsoids. Because the second case is one of plane stress, the ellipsoid degenerates into an ellipse.

13


Problem 4-4

a) For 1 km depth: sv = rgh=2.7g/cm3 x 9.8m/s2 x 1000m = 26.46 MPa. If we have a lithostatic state of stress, then by definition s1 = s2 = s3 = 26.46 MPa, and the Mohr diagram presentation becomes a single point on the horizontal axis at this value (leftmost red filled circle in Figure SP4.4). For 5 km depth: sv = 26.46 x 5 = 132.3 MPa For 10 km depth: sv = 26.46 x 10 = 264.6 MPa b) The uniaxial-strain model gives the same value for sv as the lithostatic model, but sH becomes less (Figure SP4.4(b)). For a Poisson’s ration of 0.3 we get: sH =[ 0.3/ (1–0.3)] x 26.46 MPa = 0.43 x 26.46 MPa = 11.4 MPa sH =[ 0.3/ (1–0.3)] x 132.3 MPa = 0.43 x 132.3 MPa = 56.8 MPa sH =[ 0.3/ (1–0.3)] x 264.6MPa = 0.43 x 264.6 MPa = 113.8 MPa The mean stress becomes considerably less than the corresponding lithostatic stress, and we have plane stress.

σs

100

MPa 50

σV

(a)

10

30

σV 50

100

σV 150

200

MPa

250

σV 150

200

MPa

250

σn

σs

100

MPa 50

σH

(b)

10

σ

30 V

σH 50

σH

100

σV

σn

Figure SP4.4 Mohr circle illustrations of the lithostatic cases (a) and uniaxial-strain stress models (Mohr circles) (b) for Problem 4.4.

14


Fracture (Chapter 7)

Problem 7-1

The stress data are plotted in the Mohr diagram by plotting corresponding values for sS and sN, shown as points in Figure SP7.1. Since we know the confining pressure in each case we can also construct a Mohr circle for each of the 8 experiments, where the lower intercept with the sN-axis represents the confining pressure. The data points can be fitted to a straight line with slope m around 0.67, making an angle f with the horizontal axis of 33.7°. The intersection C is close to 10.7 MPa, and the Coulomb fracture criterion becomes sS = 10.7 + 0.67sN. For a confining pressure of 250 MPa (which is very high) we utilize the graphical representation of the Coulomb fracture criterion that we have drawn in Figure SP7.1 and construct the Mohr circle for the given confining pressure. That is a circle whose left side intersects the sN-axis at 250 MPa. The top of the circle should touch the failure criterion, because this gives us the stress conditions at which a plane inclined at 45° would fail. From this point we read off sN ≈ 730 MPa and sS ≈ 495 MPa.

σs

500

φ 450

φ=33.7° µ=tan 33.7°=0.667

400

350

300

250

200

MPa

150

100

50

C≈10.7 MPa

50

100

150

200

250

300

350

400

450

MPa

500

550

600

650

700

750

σn

Figure SP7.1 Experimental results from Problem 7.1 plotted in Mohr space. A straight line defines a fracture criterion that seems to fit the data very well.

15


Faults (Chapter 8) Problem 8-1

The data are plotted as poles to planes in Figure SP8.1a. They plot along a great circle whose pole is 080/30. This means that the fault measurements fit with a cylindrical model. Think of the fault surface as being N curved, just like a fold or a corrugated metal roof plate: the pole represents the “fold axis” or the direction of the corrugations (Figure SP8.1b). This direction is the direction that the hanging wall can move without geometrical complications and is therefore the most likely slip direction, which gives us the extension direction when projected onto the horizontal plane, since they are normal faults from the North Sea Rift system.

080/30

Figure SP8.1a Measurements of fault orientations plotted in an equal-area stereonet. The best fit great circle is shown.

080/30

Figure SP8.1b Geometric interpretation of the fault orientation data: Cylindrical fault surfaces.

16


Folds and folding (Chapter 11)

Problem 11-1

a) The clue here is that the non-folded vein is cross-cut by the folded one. This means that only the folded vein was in the field of finite shortening. Since the two veins are more or less perpendicular, and since the folds are quite symmetric, we can assume that the folded vein is oriented along the principal shortening direction (the Z-axis of the strain ellipse). The amount of shortening can be estimated by assuming constant length (our second assumption), which gives about 45% shortening (see Figure SP11.1). To construct the strain ellipse, deform a circle so that it its horizontal axis is 0.55 times its original length, and the vertical axis is 1/0.55=1.82 times its original length.

e=(l-l0)/l0=-0.45=-45%, which means 45% shortening L0=79.29 cm A

A’

1 cm

X

?

L=43.4 cm

20 cm

Z

1 cm

Figure SP11.1a Solution to Problem 11-1a).

b) There are two things we can do that give us a quick idea about the class of fold. One is to measure the hinge thickness and move it around the hinge. If it is a class 2 fold, this thickness measured parallel to the axial trace should be constant. It is not for our fold. Then let’s check if it is a Class 1B fold, in which case the orthogonal thickness should be constant. It is not quite constant, and for the most part it seems that the limbs are thinner than the hinges. Hence the folds are Class 1C. It is likely that this is a Class 1B fold that has been modified, which happens when shortening exceeds something like 36%.

1 cm Figure SP11.1b Checking the layer thickness around the folds. In general, the layer seems to have thinner limbs.

c) To find the dominant wavelength Ld we measure a number of wavelengths. The average is around 6.24 cm (Figure SP11.1c), but the distribution is somewhat bimodal with peaks at 2.5-5 and 4-4.5. Ideally we would like to see one peak: in this case some parts of the layer has been more tightly folded than other parts.

17


Interval 0-1 1-2 2-3 3-4 4-5 5-6 6-7 7-8 9-10 10-11 11-12 12-13 13-14 14-15

cm 7.75 4.59 6.1 3.53 7.1 6,37 7.43 7,81 6.43 5.18 2.55 8.73 8.1 5.72

Average: 6.24 cm

0

6

5

2

1

3

4

l=43.4 cm

8

7

?

9 10

1 cm

6 5

11

12

13 14

4

15

3 2 1 8.5-9

8-8.5

7.5-8

7-7.5

6.5-7

6-6.5

5.5-6

5-5.5

4.5-5

4-4.5

3.5-4

3-3.5

2.5-3

0

Figure SP11.1c Measuring wavelength between red points (numbered). Table and histogram shown. Note that you have to convert to cm.

The dominant wavelength is calculated in Figure SP11.1c. What can we say about viscosity at the time of deformation, using Equation 11.2 in the textbook? We know the thickness (1 cm) and the wavelength (about 6.24 cm), and we solve for mL/mM to get mL/mM = 6(6.24/2p)3 ≈ 6. Hence the viscosity contrast is around 6:1, i.e. the vein being six times as viscous as the matrix. This is not a very high contrast, but obviously enough to cause buckling of the layer.

18


Problem 11-2

a) The dominant wavelength (Ld) is found as for Problem 11-1, and plotted against the estimated layer thickness (h). The results, shown in Figure SP11-2, nicely illustrates how thicker layers show longer dominant wavelengths. In this case we can define a linear relationship (straight line in the graph) where the wavelength is approximately 1.5 times the layer thickness. b) I estimated the amount of shortening by measuring the upper boundary of each folded layer, where l is the distance from each end point (straight line) and l0 is the distance measured along the folded boundary (which I did on a computer, but it can also be done by means of a thread of cotton. Using e=(l-l0)/ l0 I got the values shown in Figure SP11-2. The amount of shortening is very similar for layer A, C and D, but is somewhat higher for layer E (layer B is too short and to close to C to be considered). We usually expect to find this kind of shortening estimates to be lower for thicker layers because they tend to start to fold at a later stage than thin layers. In this case we must look for another explanation, probably that strain is heterogeneous and higher in the lowermost part of the sample.

shortening: 44.5% Ld=1.71 cm, h=0.24

.19 .3

A

.28 .21

.27

.22

.28

shortening: 38.9%, Ld=0.46 , h=0.09 B

.09

.14

C

.21

.12 .28 .24

.19 .2

.24

.13 .14

.22

.28

.19

.24

shortening: 47.7% Ld=1.50, h=0.19 cm

.25

D

.17

.13

.16

.11

.24

.11

.17

.17

shortening: 44.6% Ld=0.86, h=0.14 cm

.41

E

.22

.22

.27

.44

.29 .31

.22

1 cm

h (cm)

0.3

Ld=1.5h

0.2

0.1 0

B 0

D

1

Ld (cm)

C

shortening: 51.6% Ld=1.70, h=0.29

E A

2

Figure SP11.2 Measuring wavelength between red points (numbered). Table and histogram shown. Note that you have to convert to cm.

19


Problem 11-3

Analyze folds that have fairly regular limb shapes. When plotted in the diagram (Figure SP11.3) it appears that the folds are close to Class 2 (similar) folds. This is typical for strongly sheared quartzites. I chose layers 1, 2 and 3, and I rotated the image to make the axial traces vertical. I then drew lines dipping 20, 40, 60 and 80° in pairs that touched the inner and outer arcs of the fold, respectively. The perpendicular distance ta between these pairs of parallel lines were measured and normalized. The results are shown in Figure SP11.3. These layers seem to have had uneven thickness before folding, which accounts for at least some of the deviation from the Class 2 trend in the t'a- a plot. Do the results give us information about the mechanical properties of the layers during folding? Yes, if the viscosity contrast had been great we would expect some degree of buckling, which would have produced alternating Class 1B and Class 3 folds. Class 2 folds indicate that viscosity contrast is low or zero.

Class 1A Layer 2

1B

1.0

Layer 1 Layer 2 Layer 2

Class 1C s2 as Cl

t'α 0.5

Class 3

α

tα t0

1 yer La

0.0 Layer 3

0°

45°

Local dip of layer (α)

Figure SP11.3 Answer to Problem 11.3.

20

90°

t'α=tα/t0


Problem 11-4

The black layers, particularly layer B, show collapsed hinge zones, which gives Class 3 folds when plotted in a t’a- a plot (Figure SP11.4). The white layers are Class 1C folds, i.e. they show some hinge thickening, but less so than the black layers. The white layers are more competent, and developed local reverse faults in a couple of the hinge zones during the shortening.

F H

B

C

E D

I G

A

Class 1A

1.0

Competent Layer C Layer A Layer A

1B Class 1C

Incompetent Layer B Layer B Layer B

t'α s2 as Cl

0.5

Class 3

0.0 0°

45°

Local dip of layer (α)

Figure SP11.4 Answer to Problem 11.4. Symbols on drawing indicate where the measurements were taken.

21

90°


Shear zones and mylonites (Chapter 15)

Problem 15-1

a) Shear strain profiles are shown in Figure SP15.1a, assuming simple shear (g). g is found by rearranging Equation 15.4: g=2/(tan2q’).

34.6° 35.4°

A

B 45°

25 °

31.5°

27.2° 19.2° 5.4° 20° 21.8° 33.7°

31.2° 10° 4° 9.4° 25.7° 37°

34.6°

43°

1 cm

4.0 3.5 3.0

cm

2.5 2.0 1.5 1.0 0.5

A, B

0

2

4

6

8

10

12

14

16

γ Figure SP15.1a Measurements of the orientations of the foliation (q’) at various locations across the zone. Bottom: shear strain graphs found by calculating g from q.

22


b) Offset across the zone can be found by calculating the area under the shear strain graph, as shown in Figure SP15.1b (graphs). They show fairly consistent results. One of the curves (A) has a lower peak, which is compensated for by a somewhat wider zone. An alternative way of finding the offset is to calculate y from g using the relationship g =tany. This gives angles that tell us the local displacement, and if we add these displacements we can construct how a line perpendicular to the zone would change orientation, as shown in Figure SP15-1b (bottom). c) The maximum shear strain value is difficult to constrain, because it occurs in the central part of the shear zone where the foliation is difficult to estimate. If we use the graphs in Figure SP15.1a, we get gmax of 10-14, which from Figure 15.11 implies a R around 200-275, i.e. very high ellipticities.

4.0 3.5

4.0 1/2

3.5

3/4

3.0

Sum: 16 units=8 cm

cm

2.0

1/2

1/2

1/2

1/2

1/2

1/2

1/4 1/4

1/4 1/4

1/4 1/4

1/4 1/4

Sum: 16.75 units=8.4 cm

1/2

2.5

cm

1/2

2.5

1/4

3.0

2.0

1/4

1.5

3/4

3/4

1/2

1/2

1/2

1/2

1/2

1/2

3/4

3/4

1/2

1/2

1/2

1/2

1/2

1/2

1/4 1/4

1/4 1/4

1/4 1/4

1.5 1/2

1.0

1.0

1/2

0.5

A

1/2

0.5 0

2

4

6

γ

8

10

B

12

0

2

4

6

γ

8

10

12

14

B ψ ψ

8 cm

Figure SP15.1b Finding offset from the area under the shear strain graph (profile B), and by using offset across discrete element based on y, which is found from the measured orientations of the foliation (q’) at various locations across the zone.

23


Problem 15-2

a) A shear strain profile from A to B is shown in Figure SP15.2. b) The area covered by the graph gives a displacement of ca. 60 cm, which is close (but not identical) to the 55 cm offset of the marker. c) This may indicate a deviation from simple shear, as can the variations in shear zone orientation and thickness.

10 cm

60° B 55°

6.2° 10.7° 13.2° 12.8° 16 15° 14 15.5° 12 26.3°

51° 49.2° 32 45.3° 30 39.9° 28 41.1° 26 15.5° 20

42° 53.8° 6 53.8° 4 β’=55° β=60°A 2 0 cm

22

Outer shear zone

Inner shear zone

β=60°

40 cm

36

D ≈ 55 cm

Outer shear zone

'

40

B

35

1.25

cm

30 25

2.5 3.75

20 5

5

5

5

5

5

5

3.75

2.5

5

2.5

1.25

15 10

2.5

5 0

A

0

Sum: 60 cm

2

4

γ

6

=cot '-cot 60 55 53.8 53.8 42 26.3 15.5 15 12.8 13.2 10.7 6.2 15.5 41.1 39.9 45.3 49.2 51 55 60

8

Figure SP12.1 Analysis of the shear zone (Problem 15-2).

24

0 0.122857269 0.154539092 0.154539092 0.533262246 1.445995976 3.02853324 3.154700538 3.82416617 3.686171487 4.715000187 8.627806164 3.02853324 0.568971253 0.618636312 0.412232206 0.285826576 0.232433764 0.122857269 0

cm 1 2 4 6 8 9.7 9.8 10 12 14 16 18 20 22 24 26 28 31 35 37


Strike-slip, transpression and transtension (Chapter 18) N

Problem 18-1

a) The stereoplot is shown in Figure SP18.1a.

Direction of rotation

ISA3

b) The orientation of the various structures are consistent with sinistral sense of shear, based on the information that the zone is N-S striking and the relative orientations of contractional and extensional structures.

ISA1


The type of deformation: A simple shear deformation would have ISA at 45° to the zone. When we look at the orientations of the axial planes, we see that they have a lower angle to the shear zone, more like 35° for axial planes of open folds and 25° for tight folds. This could indicate a component of shortening (pure shear) across the shear zone. However, the reason why open and tight folds have different orientations can be explained by the longer history of rotation for tight folds: they formed earlier and have experienced more rotation. Any fold in a simple shear-dominated system rotates from the very moment it forms, so we don’t know the initial orientation of the axial planes, only that they made a higher angle to the shear zone. A similar argument could be made for the faults (small faults having experienced less rotation). This all fits with a deformation type that is close to simple shear. Whether there was a small component of shortening across the zone we don’t know.


ISA1 ISA3 Direction of rotation

Contraction/extension direction, from small structures/open folds Contraction/extension direction, from large structures/tight folds

Axial planes, open folds Axial planes, tight folds Normal faults, small Normal faults, large Reverse faults, small Reverse faults, large Strike-slip surfaces

c) The ISA are oriented at 45° to the shear zone for simple shear. Figure SP18.1a Plot of the structures in Problem 18-1. Very approximate d) Figure SP18.1b shows small-scale structures that could be found in the shear zone.

R’

Veins R

en

Stylolite

P

ula

tio

ns

N

Stylolite

Cr

Figure SP18.1b Structures in a fictive strike-slip shear zone (map scale).

25

Vein systems

Shear zone

contraction and extension directions are added, based on the orientations of the structures. ISA for simple shear are shown as blue arrows.


Balancing and restoration (Chapter 20) Problem 20-1

To construct the hanging-wall roll-over we must figure out how the lower part of the pulled-out hanging wall will move. For vertical shear it will move straight down as far as it can, that is to the detachment fault. The material above will follow the same path (vertical paths for vertical shear, 45° paths antithetic to the fault in the other case). Think of the hanging wall as consisting of piles of sand grains, vertical or inclined, depending on the angle of shear. The resulting hanging-wall geometry is different for the two cases. The rollover is wider and dips are shallower for antithetic shear. Note that the heave (horizontal component of fault displacement) is smaller for the case of antithetic shear, although the total extension applied is the same. This makes extension estimates more difficult or uncertain than if everything was vertical shear. Similarly the throw is different for the two cases.

extension (e)

Vertical shear

e

heave = e throw

e

Antithetic shear (45°)

e

heave < e throw

Figure SP20.1 Solution to Problem 20-1. Vertical shear at the top, antithetic shear below.

26


Problem 20-2

Rigid block restoration means that blocks can rotate and translate. It is not possible to do a perfect restoration of this section unless ductile deformation is accounted for, but we can get a rough result that can be useful for some purposes. a) Extension is 17 and 13% for top Statfjord and top Teist, respectively, according to Figure SP20.2. b) When restoring to top Statfjord Fm. time, there seems to be a residual fault displacement on the leftmost major fault. This may be evidence for pre-Statfjord activity on this fault. True, fault displacement changes along faults, and many faults in rift basins have propagated upward (causing upward-decreasing fault displacement), but the displacement gradient would be unusually steep to explain the difference in this way. There is a clear stratigraphic westward thickening of the Lunde Fm. that bevomes obvious after restoration. This may indicate that a major fault is located to the west of our cross-section that was active during deposition of the Lunde Fm. (late Triassic). c) The restored versions have some problems. There are some bumps in the restored top Statfjord Formation that should not be there, and the graben area where E and W-dipping faults interact is problematic. In that area we have reverse offsets at Teist level when the Statfjord Fm. is restored. This results in a lower extension estimate for the Teist Fm., and it is possible that the interpretation should be revised in this structurally challenging area. d) The restorations indicate that the initial dip of the western domino-style faults was around 43°. This is a very low initial dip, which is commonly around 60° or a bit more for normal faults. In contrast, the eastern horst-bounding faults have restored dips around 60°. This may suggest that soft deformation (such as vertical shear) has taken place in the western domino area. e) An indirect indication of ductile or “soft” deformation was mentioned in d). In addition, the fact that it is impossible to restore the layering to a perfectly horizontal layer indicates the presence of such deformation.

27

Top Statfjord

1 km

Fm

Figure SP20.2 Gullfaks cross-section, showing present and restored lengths of the top Statfjord and Upper Teist formations, and the extensions involved. .

28

Upper Teist Fm

Top Teist

l0, Statfjord Fm.

l0, Upper Teist Fm.

m

ist F

e er T Upp

e Fm Lund

j. Statf

r.

Brent G

l, Statfjord Fm.

?

1040

e = (l-l0)/l0 =0.127=12.7%

e = (l-l0)/l0 =0.172=17.2%

960

l, Upper Teist Fm. 1120

1200

1280

Line 736 1360

5000 mbsl

4000

3000

2000

1000



Problem 20-3

The answer to this exercise is given in Figure 20.13 in the textbook. The displacement field (Figure 20.13c), which is found by connecting points that can be identified in the present and restored maps, is close to plane strain in the northern part of the area and shows the extension direction. The extension varies somewhat locally, and it can be found from the map as shown in b) in Figure SP20.3. Any balanced section must be parallel to the displacement vectors. In general the restoration looks reasonable, but there are some areas of overlap/gaps that need local attention.

l

l0

e=0.36

N

a)

Gullfaks Field

1km

b)

Restored 42% area difference

Orientation of lines for restoration

c)

Displacement field

Figure SP20.3 Original (a) and restored (b) map. The displacement field is shown in c).

29


Pc

O C

Ramp

S

S

C D

b) A balancd version is shown in Figure SP20.4b. The “loose ends” on the left-hand side line up reasonably well. The restoration exercise made me realize that there was a horizontal fault at the base of the Cambrian that was later affected by a deeper flat fault (lower left-hand part of the section).

Hanging-wall anticline (faultbend fold)

a) An interpretation is shown in Figure SP20.4a. The first thing to note is irregularities in the stratigraphy encountered in two of the wells. In the middle well, the Cambro-Ordovician section is repeated, and we can put in a thrust fault at the base of the upper Cambrian unit. In the leftmost well the Silurian is repeated, and we have another instance of a thrust fault. Then we use the geometric relations between the kinkfolds and ramps, as shown in Figure 16.17 etc. to construct the section.

Ramp

Problem 20-4.

Ramp

Pc

Flat

O C O C

Forelimb

S

30

Figure SP20.4a Interpretation of the cross-section in Problem 20-4.

Ramp

Pc

O C

S

C D

2 km

Fault-bend fold

Hanging-wall anticline

Flat

Backlimb

C D

Hanging-wall anticline (faultbend fold)

c) From the balanced section we calculate shortening to be around 11% .

Figure SP20.4b Restored section.

e = (l-l0)/l0 =-0.11=11% shortening

l l0

Eroded


31

Solutions to Fossen Structural Geology

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