1

Accepted in In: Fagereng A, Billi A. (Eds) Problems and Solutions in Structural 1 Geology and Tectonics. Developments in Structural Geology and Tectonics Book Series. 2

Series Editor: Mukherjee S. Elsevier ISSN: 2542-9000. 3

4

Kinematics of pure shear ductile deformation within rigid walls: new 5

analyses 6

7

Soumyajit Mukherjee 8

Department of Earth Sciences, Indian Institute of Technology Bombay, Powai, Mumbai 400 9

076, Maharashtra, INDIA 10

e-mail: soumyajitm@gmail.com 11

12

Abstract: Pure shear deformation/flow of rocks in ductile regime is quite important in 13

structural geology and tectonics. This work deduces time dependent spatial flow profiles of 14

2D pure shear of an incompressible Newtonian viscous fluid with non-deformable walls. 15

During the deformation history, the line parallel to the compression direction that bisects one 16

of the sides of the shear zone with finite length remains straight, and parabolic profiles away 17

from this line towards the free end of the shear zone taper more and more. At every time 18

instants, shear strain is ideally zero at the vertices of the parabolic profiles with maximum 19

curvature. Towards the shear zone boundaries, shear strain increases and curvature of the 20

flow profile reduces. Thus, shear strain maximizes at the shear zone margins. The locus of the 21

vertices of the flow profiles is the line that bisects and is perpendicular to the width of the 22

shear zone at every instant. Shear sense flips across this line. (In case the abstract will be 23

published as online only version, then I would request to also present the abstract as the 24

first para of the Introduction, under the new title/subtitle as “Summary of this 25

chapter”) 26

27

2

Keywords: Ductile shear, structural geology, tectonics, Newtonian viscous fluid 28

29

1. Introduction: 30

Pure shear deformation is one of the end members of ductile shearing phenomenon in 31

structural geology and tectonics that work at widely different tectonic domains and from a 32

depth exceeding 8 or 15 km in the ductile regime (Passchier and Trouw, 2005). In such a 33

deformation, the two long rigid boundaries/margins of a ductile shear zone, considered 34

parallel in most of the cases, move towards (or away from) each other perpendicular to their 35

lengths. Secondary synthetic and antithetic shears in ductile shear zones in meso- and micro-36

scales probably indicate a pure shear component (Mukherjee, 2013). Natural ductile shear 37

zones have been reported to usually have one more deformation component, viz., simple 38

shear/Couette flow (= pure shear of Gere and Goodno, 2012; see Xypolias 2010 and Bose et 39

al. 2018 as examples). Besides, pure shear kinematics is one of the ways to explain rifting 40

(Fig. 1a) and nappe movement/gravitational spreading (Fig. 1b) (e.g. Fossen, 2016). 41

42

{Insert Fig. 1 about here} 43

44

Analyzing ductile shear in terms of pure- and simple shear components have been in practice 45

for more than 100 years in geoscience (Howarth, 2003). There has also been significant 46

discussion regarding pure shear (and simple shear) in material science (Segal 2002). In 47

particular, velocity at any point in a shear zone, pure-, simple- or general- has been presented 48

in Jaeger (1969). Simple- and pure shear, if operate together, is called sub-simple or general 49

shear. Understanding kinematics of shear zones is of great importance in deciphering plate 50

3

tectonics, rheology and seismicity (e.g., Regenauer-Lieb and Yuen, 2003; Hobbs and Ord, 51

2014; Mulchrone and Mukherjee, 2016; Fossen and Cavalcante, 2017). 52

53

Pure shear mechanism can work on some flowing lavas as well (Manga, 2005). Such a shear 54

kinematics has been worked out from different perspectives in mechanics/rheology works 55

(e.g., Segal, 2002; “Stephan’s squeezing flow” in Papanstasiou et al. 2000; Malkin and 56

Isayev, 2006). The deformation involves parabolic flow path of fluid particles (Schlichting, 57

1960). However, how it would deform straight marker layers was not analysed properly, even 58

though pure shear is referred frequently in structural geological texts (e.g., Dennis, 1987; 59

Reitan, 1987; Twiss and Moores, 2007; Fossen, 2016). The other detail of pure shear of two 60

media has been available. For example, Treagus and Lan (2000) studied the mode of 61

deformation of square objects within another medium under pure shear. 62

63

This work elaborates kinematics of pure shear presuming ductile rocks to be a single 64

incompressible Newtonian viscous fluid. Such a consideration has yielded other first order 65

understanding of ductile shear zones (e.g., Ramsay and Lisle, 2000; Mukherjee, 2012). Pure 66

shear dominated ductile shear zones do exist in nature (e.g., Bailey et al. 2007) where this 67

analysis could approximately apply. However, a natural shear zone with only pure shear and 68

zero simple shear has not yet been reported. 69

70

{Insert Fig. 2 about here} 71

72

4

A. Instruction to the Instructor: The prerequisite knowledge for students here would be 73

elementary calculus. We are going to do an area balancing exercise typically what is 74

done for cross-section balancing (e.g., Lopez-Mir, submitted). Our aim is to find out 75

equation of the velocity profile for pure shear. Step-wise present materials below to 76

students. Give them sufficient time to solve equations independently. 77

78

2. Pure shear kinematics: 79

2.a. One boundary stationary and the other moves: 80

Consider a rectangle ABCD with a rigid base BC. Side AD is compressed with a constant 81

velocity V. After instant ‘t’, AD comes to A/D/. The incompressible Newtonian viscous fluid 82

expels at the two free sides as parabolic profiles P1 and P2. With the chosen coordinates as in 83

Fig. 2a, AA/=V.t, therefore area AA/DD/= L.V.t. P1 is represented by: 84

x=Ay2+By+C (1) 85

As this parabola passes through the origin (0,0), therefore C=0 (2) 86

Or, x=Ay2+By (3) 87

88

B. Instruction to the Instructor: Ask students what will be the problem if the origin is 89

chosen differently. Also, one needs to inform the students that the profile need not be 90

parabolic for other rheology of the fluid. 91

92

5

The origin is chosen in this way in order to simplify eqn 1. This does not modify the physical 93

process of deformation, rather the representation is simplified. And as it passes through point 94

A/[0, (y0-Vt)], putting x=0 and y=(y0-Vt) in eqn (3) and simplifying: 95

B=-A(y0-Vt) (4) 96

Now, the area bound by the line AM (or the Y-axis) and the parabola P1: 97

(y0-Vt) 98

A1= ∫x dx (5) 99

0 100

Putting the expression for ‘x’ from eqn (3) into eqn (5) and integrating: 101

A1=(y0-Vt)2[0.33*A(y0-Vt)+0.5*B] (6) 102

Considering that the area lost by compression (i.e. area AA/DD/= L.V.t) would be equally 103

distributed at the two parabolic sides, i.e., A1= 0.5LVt (7) 104

Equating A1 from eqns (6) and (7), and using eqn (4): 105

A= -3VtL (y0-Vt)-3 (8) 106

B=3*VtL(y0-Vt)-2 (9) 107

Using eqns (3), (8) and (9): 108

x=3*VtLy(y0-Vt)-2[1-y(y0-Vt)-1] (10) 109

Therefore velocity-wise, 110

Vx= x t-1=3*VLy(y0-Vt)-2[1-y(y0-Vt)-1] (11) 111

Writing Vx henceforth as x, eqn (3) can be written as, by writing (y0-Vt)=ht, 112

(y-0.5ht)2=-0.33*L-1V-1ht(x-0.75*LVht

-1) (12) 113

6

Coordinate of its vertex is deduced (also recall ht=y0-Vt): 114

{0.75LV(y0-Vt)-1, 0.5(y0-Vt)} (expression 1) 115

Note (i) the x-ordinate of the vertex of the parabola as in eqn (11) represents highest speed. 116

(ii) The vertex is equidistant from the two walls at every moment. 117

The parabolic profile that would be produced by the line MN passing through any point M (-118

m,0) lying on the X-axis (Fig. 2a) will be deduced now. If only the rectangle A/BMN were 119

there (Fig. 2a), the distance ‘M1’ would have been, by replacing ‘m’ in place of ‘L’ in the x-120

ordinate of expression (1), 121

M1=0.75*mV(y0-Vt)-1 (13) 122

If only the rectangle MNDC were there (Fig. 2a), the distance ‘M2’ would have been, by 123

replacing in place of ‘L’ in the x-ordinate of expression (1), 124

M2=0.75*(L-m)V(y0-Vt)-1 (14) 125

The resultant of these two parabolic profiles, in case both the rectangles MNDC and A/BMN 126

exist side by side and in contact with each other along the line MN: 127

ΔM=M1-M2=0.75*(2m-L)V(y0-Vt)-1, if M1>M2 (15) 128

ΔM=M2-M1= 0.75*(L-2m)V(y0-Vt)-1, if M1<M2 (16) 129

And, in specific, ΔM=0, if m=0.5*L (17) 130

131

Eqn (17) indicates that at the middle of the rectangle ABCD, the linear inactive marker simply 132

shrinks yet continues as a straight line. Eqns (15) and (16) indicate that as one goes at either 133

margins, AB and CD, one gets parabolic markers with increasing tapering, or in other words, 134

7

with increasing curvature at their vertices. Secondly choosing different values of m (≤ L), one 135

can constrain the exact parabolic profile passing through the corresponding point M. 136

C. Instruction to the Instructor: Instruct the student at this point what are the meanings 137

of curvature, strain, shear strain and tangential shear strain. 138

139

Curvature of P1 is deduced using the formula: 140

ρ={1+(dx/dy)2}3/2(d2x/dy2)-1 (18) 141

In the present case, dx/dy = 3*LV(1-2*yht-1)ht

-2 (19) 142

And d2x/dy2= -6*LVht-3 (20) 143

Using eqns (19) and (20) in eqn (18), curvature at the vertex (ρv), for which y=0.5ht is given 144

by: 145

ρv= -0.17ht3L-1V-1 (21) 146

This is the highest curvature on the parabola P1. 147

Shear strain at any point on the P1 parabola is studied as follows. As per Fig. 2a, Φ is the 148

angular shear strain at point T. Line TZ is a tangent on the parabola P1 drawn at the point T. 149

Note tanΦ=Cotθ=dx/dy (22) 150

From eqn (11), dx/dy=3*LVht-2(1-2*yht

-1) (23) 151

Therefore tanΦ=3*LVht-2 (1-2*yht

-1) (24) 152

Thus, Φ and tanΦ vary with time inside a pure shear zone. The only exception would be the 153

points on the line that bisects the length of the rectangle (line GH in Fig. 2a), where in fact 154

Φ=tanΦ=0 remains time independent. Note, at vertex, i.e., for y=0.5ht in eqn (24): 155

8

Φvertex=0 (25) 156

This matches with the intuition: at vertex the tangent drawn on the parabola parallels the Y-157

axis, therefore their angle becomes zero. 158

At origin (0,0), putting y = 0 in eqn (24), 159

tanΦat origin=3*LVht-2 (26) 160

And at point A/, putting y = ht in eqn (24), 161

tanΦatA/ =- 3*LVht-2 (27) 162

The difference in sign in eqns (26) and (27) indicate opposite ductile shear sense. While at 163

origin it is top-to-right, at A/ it is top-to-left. The shear sense flips across the vertex of the 164

parabola. 165

166

Strain at any point inside the pure shear shear zone can also be deduced. With respect to the 167

new co-ordinate system (Fig. 2b) where the X-axis passes through the vertex of the parabola, 168

the equation of the (resultant) parabolic profile can be written as, since it passes through 169

points N: [0, 0.5*(y0-V.t)] and M[0, - 0.5*(y0-V.t)]: 170

x=H[y2-0.25*(y0-V.t)2] (28) 171

At y=0, 172

x = -0.25*H(y0-V.t)2 (29) 173

This has to equal ΔM of either eqn (15) or eqn (16), therefore, 174

H= -4ΔM(y0-V.t)-2 (30) 175

9

Putting back H in eqn (28), the parabola is now given by: 176

x=ΔM[1-4y2(y0-vt)-2] (31) 177

Therefore, in the new coordinate system: 178

tanΦ1=dx/dy= - 8ΔM.y(y0-vt)-2 (32) 179

Putting back the expression of ΔM from eqn (15) into eqn (32): 180

tanΦ1= - 6*(2m-L)V(y0-Vt)-3y (33) 181

Choose different magnitudes of ‘m’, ‘y’ and ‘t’ to get tangential shear strain tanΦ1(and 182

therefore shear strain Φ1) at any point inside the pure shear zone. For example, for y = 0, i.e., 183

for the new X-axis bisecting the pure shear zone at any instant (i.e. for any ‘t’), tanΦ1 = 0. 184

This is as expected since at the vertex of the parabolic profiles, shear strain is zero. Secondly, 185

for y= 0.5*(y0-vt), and m = 0, i.e., at point A/: 186

tanΦ1 = 3*L.V.(y0-Vt)-2 (34) 187

This exactly is the same as eqn (26) deduced in a simplified way. 188

189

{Insert Fig. 3 about here} 190

191

2.b. Both the boundaries move (Fig. 3a): 192

Consider the boundaries of shear zones move towards each other at velocities V1 and V2(V1 > 193

V2). Let AD and AB are respectively L and 2y0 units long. Points A (0,y0) and B (0,-y0) on the 194

two walls, after time ‘t’, comes at A/ (0, y0-V1t) and B/ (0, V2t-y0), respectively. The vertex of 195

10

the parabola will be equidistant from these two points. Therefore, its y-ordinate would be -196

0.5*t(V1-V2). Let the x-ordinate of the vertex be p. Thus the complete coordinate is {p, -197

0.5*t(V1-V2)}. Let the equation of this parabola is: 198

Y2+Dx+Ey+F=0 (35) 199

200

D. Instruction to the Instructor: At this point onward, ask the students to proceed 201

themselves. It will be very interesting to see whether they can deduce the velocity 202

profiles! 203

Putting A/and B/coordinates in eqn (35), E and F are solved: 204

E=(V1-V2)t (36) 205

F=(y0-V1t) (V2t-y0) (37) 206

Putting them back in eqn (35): 207

y2+D.x+(V1-V2)ty+(y0-V1t)(V2t-y0)=0 (38) 208

Area bound by this parabola with the Y-axis between the coordinates {0,(y0-V1t)} and {0,(V2t-209

y0)}: 210

(y0-V1t) 211

A1=∫x dx (39) 212

(V2t-y0) 213

214

Putting x in eqn (39) from eqn (38), and integrating: 215

A1= D-1{2*y0-t(V1+V2)}[0.33*{(V2-V1)2t2+2(y0-V1t)(V2t-y0)}-0.5*t(V1+V2)] (40) 216

217

11

Area lost due to compression equals area gained at the two sides, being an incompressible 218

fluid under consideration inside the rectangle ABCD. Therefore, 219

A1=0.5*(V1+V2)tL (41) 220

Eliminating A1 from eqns (40) and (41): 221

D= 2*t-1 L-1(V1+V2)-1{2*y0-t(V1+V2)}[0.33*{(V2-V1)

2t2+2(y0-V1t)(V2t-y0)}-0.5*t(V1+V2)]222

(42) 223

Eliminating D from eqns (40) and (42), one gets the resultant parabolic flow profile P1. In 224

this case also, the vertex will mark the point of minimum shear strain and maximum 225

curvature. 226

E. Instruction to the Instructor: One can easily locate in the classroom by now students 227

who solved correctly and enjoyed taking the challenge. Refer to that small group of 228

students Mukherjee (2012, 2014), and ask them to find out velocity profile for general 229

shear deformation. This would be a step forward from the title of this article. 230

231

3. Discussions 232

Wittenberg (2003) presented briefly shear strain for a sphere pure sheared to an ellipsoid. But 233

unlike the present work, point to point variation of shear strain for a pure sheared object, here 234

a rectangle, was not presented by any. Allmendinger et al. (2012)and Zeb et al.(2013) 235

presented kinematic analyses of pure shear zones that especially track fluid particles over 236

time. They did not however provide equations of parabolic velocity profiles produced by such 237

an ideal deformation. Eqn (12) of flow profile for one of the boundaries movement (Case 2.a) 238

matches with that presented straightway by Spurk (1999). However, Spurk (1999) did not 239

12

give the deduction, and to author’s knowledge it does not exist in any publications. The 240

second deduction of profile for both the boundaries move can be obtained by eliminating ‘D’ 241

from eqns (40) and (42). This does not exist any previous publications. Note parabolic flow 242

profile also develops for Poiseuille flow of Newtonian viscous fluid through parallel-sided 243

boundaries. However, in that case, the profile becomes time-independent (Mukherjee, 2012). 244

This is unlike pure shear where the curvature at every moment at the vertex is a (cubic) 245

function of time (such as eqn 21). Changing shear strain with time inside shear zones is also 246

known in real cases (e.g., Grasemann et al.1999). Note that, by putting V2 = 0 in eqns (40) 247

and (42) and then eliminating ‘D’, one would not simply go back to eqn (12): the case of 248

movement of a single boundary. This is because in the two cases the origins (0,0) were 249

selected differently. 250

251

In case one of the sides of the rectangular pure sheared rectangle is completely rigid and 252

precludes flow materials outward due to the presence of some massive geological body, eqns 253

(7) and (41) alter respectively to: 254

A1= LVt (43) 255

A1= (V1+V2)tL (44) 256

Accordingly the flow profiles (eqns 12, 40 and 42 after elimination of ‘D’), shear strain 257

pattern (eqn 24) inside the pure shear zone, and the curvature of the profile (eqn 21) would 258

change. 259

260

In case both the boundaries move with equal velocity U (=V1=V2), eqns (40) and (42) 261

simplify, respectively to: 262

13

y2+D.x- (y0-Ut)2=0 (45) 263

D= - 4L-1(y0-tU)3 (46) 264

Eliminating ‘D’ from eqns (45) and (46), the parabolic profile becomes: 265

y2– 4L-1(y0-tU)3.x- (y0-tU)2= 0 (47) 266

The Y-ordinate of its vertex is ‘y0’. This means that the vertex is not only equidistant 267

temporally from both the moving walls, but maintains its initial position. This is unlike the 268

situations when (i) one of the walls is static (case 2.a), and (ii) both the walls move with 269

unequal velocities (V1 ≠ V2situation in case 2.b). Note eqn (47) cannot be compared 270

straightway with eqn (12) by merely considering 2U = V, since the coordinate systems of 271

cases 2.a and 2.b differ. In case the boundaries move away from each other in a pure shear 272

manner, the co-ordinates of the points A/, B/ etc. should be obtained first (Fig. 3b) and then 273

proceed in the same way to deduce flow profiles. In this case the convex sides of the two 274

parabolic margins are towards the center of the rectangle. 275

276

The deductions presented here work obviously under several presumptions. Since rocks have 277

quite low compressibility (2.55*10-11 Pa-1 for sediments: Turgut, 1997), considering it to be 278

incompressible looks justified. However, in a bit different context, layer perpendicular 279

compression of sediments in basins due to their own weights (“gravitational compaction”: 280

Wangen, 2010) can reduce volume (Mukherjee and Kumar, submitted). Shear zone 281

boundaries would be rigid if the ratio of viscosity between the shear zone and its 282

surroundings is ≤ 10−7. Melting and other fluid activities during ductile shearing can 283

significantly reduce volume of the shear zones. Ultra-high-pressure rocks under dissolution– 284

precipitation creep and granular flow at low stress act as Newtonian fluids (review in 285

14

Mukherjee, 2012; Mukherjee and Mulchrone, 2012). Likewise, natural granitic melt can be a 286

Newtonian fluid, but this is not necessarily true for all kinds of crustal melts. Natural shear 287

zones can obviously contain more than one lithology. Viscous dissipation might elevate 288

temperature that might modify the parabolic profile. Unlike constant V considered in case 289

2.aand constant V1 and V2 in case 2.b, the movement rate of boundaries of shear zones can 290

vary in nature (e.g., Janecke et al. 2010). 291

292

This work deduces kinematics of a horizontal pure shear zone (Figs. 2a, 3a). Horizontal 293

(Mies, 1991) and sub-horizontal ductile shear zones (Lister and Davis, 1989) have been 294

referred by many from nature. In reality, shear zones can dip (Jegouzo, 1980), and in that 295

case the effect of gravity needs to be considered in the kinematic analysis. 296

297

Elliptical symmetric quartz blebs, symmetric pressure shadows and fringes indicate pure 298

shear, although asymmetric fabrics can also be produced by simple shear (Mukherjee, 2017 299

and references therein). Myrmekitization induced augen in ductile shear zones indicate 300

volume loss (e.g., Menegon et al. 2008), therefore the presented model for constant area (and 301

volume) deformation would not work there. Also, augens can indicate heterogeneous 302

deformation (Dell’Angelloand Tullis, 1989) that would restrict the use of the presented 303

model. 304

305

Acknowledgements: A research sabbatical for the year 2017 and a CPDA grant provided by 306

IIT Bombay are thanked. This work bases preliminary analysis of pure shear in Mukherjee 307

(2007). Thanks to Amy Shapiro, Tasha Frank and the proofreading team (Elsevier), and Ake 308

15

Fagereng (University of Cardiff) and Andrea Billi (Sapienza Universita di Roma) for 309

reviewing and handling chapters. I dedicate this article to Late Profs. Gaurishankar Ghatak 310

and Malay Bhushan Chakrabarti for their exhaustive teaching of fundamentals of geology 311

during SM's B.Sc. programme in the then Presidency College, Kolkata during 1996-1999. 312

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Figure captions 441

Fig. 1a. Pure shear induced rifting. The uprising magma body exerts pure shear on the 442

overlying rock body. b. gravitational spreading by its own exerts pure shear on the spreading 443

mass. 444

445

Fig. 2a. Pure shear on rectangle ABCD on side AD with a velocity ‘V’. Side BC places rigidly 446

on the X-axis. Line AD moves to A/D/ after time ‘t’. Had A/BMN be the only rectangle 447

present, line NG would have been deformed as the green parabolic profile. Had ND/CM be 448

the only rectangle present, line NG would have been deformed as the orange parabola. When 449

both the rectangles are present, i.e., when we consider the rectangle A/BCD/, Line NG would 450

deform to the red parabola. Note (QS-QW)=QR. b. Shear strain for the red parabolic profile 451

on point T1 is Φ1. 452

453

Fig. 3. Pure shear on sides AD with a velocity V3 and BC with V4, of the rectangle ABCD. 454

After time ‘t’, A reaches A/ and B to B/. a. Compression. b. Extension. 455

456