kinematics of pure shear ductile deformation within rigid walls: … · 32 structural geology and...
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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
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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: [email protected] 11
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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
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Keywords: Ductile shear, structural geology, tectonics, Newtonian viscous fluid 28
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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