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Prediction of Fracturing and Dynamic Roof Failure inPlatinum Mines
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, TankiMotsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe
Ramanna, Charlene Chipoyera
Moderators: Prof. Mason, Prof. Fowkes, Ashleigh Hutchinson
January 20, 2014
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 1 / 31
Introduction
Roof collapses occur occasionally in platinum mines and aredevastating events
When roof collapse occurs, large slabs of the rock fracture
These roof collapses compromise miner safety and are very costly tothe mine
By understanding how a roof collapse occurs, we can predict itsoccurrence as well as attempt to mitigate the risks, thus avoidingunnecessary loss
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 2 / 31
Introduction
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 3 / 31
Introduction
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 4 / 31
Crack Formation
Given slabbing, what factors contribute to/determine the uniformthickness of the slabs in non-sedimentary rock?
Definitions
Slabbing: A phenomenon whereby a rock mass peels off in uniformlayers
Thickness: Distance from free surface to the fracture
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 5 / 31
Crack Formation
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 6 / 31
Crack Formation
Considerations
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 7 / 31
Crack Formation
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 8 / 31
Crack Formation
Possible Slabbing Explanations
P ∝ H, so the pressure closer to the lower free surface is the greatest.This is why we have roof collapse.
The geometric distribution of pre-existing cracks may influence themanner in which the rock fractures.
Seismicity induced by drilling, explosions and natural effects causes are-distribution of stresses which contributes to crack extension.
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 9 / 31
Exfoliation Problem
Exfoliation of surface rocks is a complementary phenomenon that has beenobserved. We believe it to be a buckling beam problem and that it alsocorresponds to the eigenvalue problem, as the effect of gravity was seen tobe negligible.
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 10 / 31
Exfoliation Problem
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 11 / 31
Exfoliation Problem
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 12 / 31
Exfoliation Problem
The beam equation is given by
EId4w
dx4+ P
d2w
dx2= q,
where q is the sum of the body force and the surface traction per unitlength.
For the exfoliation problem, q=0 thus resulting in
d4w
dx4+ B2 d
2w
dx2= 0.
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 13 / 31
Exfoliation Problem
Finding the general solution to our equation yields
w(x) = A cos(Bx) + C sin(Bx) +D
B2x +
F
B2,
subject to the boundary conditions for a beam clamped at both ends:
w(0) = 0,
w(1) = 0,
w ′(0) = 0,
w ′(1) = 0.
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 14 / 31
Exfoliation Problem
In order to produce non-trivial solutions, we impose our boundaryconditions to obtain a homogeneous system of equations in the formHx = 0:
1 0 0 B−2
cos(B) sin(B) B−2 B−2
0 B B−2 0−B sin(B) B cos(B) B−2 0
ACDF
=
0000
We want the determinant of the matrix to be equal to 0.
det(H) = 4B5 sin
(B
2
)(sin
(B
2
)− B
2cos
(B
2
))= 0
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 15 / 31
Exfoliation Problem
Case 1: sin
(B
2
)= 0
Solving for B gives B = 2nπ n ∈ Z.
We substitute this into the matrix H:
1 0 01
(2nπ)2
1 01
(2nπ)21
(2nπ)2
0 2nπ1
(2nπ)20
0 2nπ1
(2nπ)20
.
Solving the resulting matrix system with B = 2nπ gives
w(x) = A(cos(2nπx)− 1),
where A is an arbitrary constant.
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 16 / 31
Exfoliation Problem
Plot of the beam deflection
0.2 0.4 0.6 0.8 1.0x
-0.5
-1
-1.5
-2
wHxL
n=3
n=2
n=1
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 17 / 31
Exfoliation Problem
Plot of the beam curvature
0.2 0.4 0.6 0.8 1.0x
50
100
150
200
250
300
350
Curvature
B=6Π, n=3
B=4Π, n=2
B=2Π, n=1
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 18 / 31
Exfoliation Problem
Case 2:
(sin
(B
2
)− B
2cos
(B
2
))= 0⇒ tan
(B
2
)− B
2= 0
Π 2 Π 3 Π 4 Π 5 Π 6 Π 7 Π
B
-20
-10
10
20
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 19 / 31
Exfoliation Problem
From tan
(B
2
)=
B
2, we can derive
sinB =B
B2
4+ 1
cosB =
B2
4− 1
B2
4+ 1
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 20 / 31
Exfoliation Problem
We substitute these expressions into the matrix H, and apply rowoperations, we obtain
1 0 0 1
1− B2
4B 1 +
B2
41 +
B2
40 B 1 0
−B2 B(1− B2
4) 1 +
B2
40
.
Solving the resulting matrix system gives
w(x) = D(−B
2cos(Bx) + sin(Bx)− Bx +
B
2),
where D is an arbitrary constant.
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 21 / 31
Exfoliation Problem
Plot of the beam deflection, for varying B values
0.2 0.4 0.6 0.8 1.0x
-20
-10
10
20
wHxL
B=28.1
B=21.8
B=15.45
B=8.9
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 22 / 31
Exfoliation Problem
Plot of the beam curvature, for varying B values
0.2 0.4 0.6 0.8 1.0x
2000
4000
6000
8000
10 000
curvature
B=28.1
B=21.8
B=15.45
B=8.9
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 23 / 31
Slanting Beam Problem
Blasted tunnels have roofs that seem to behave as beams; however, theyare at an angle, say θ. We thus felt that examining how the roof in aplatinum mine fractured would be equivalent to examining the slantedbeam equation.
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 24 / 31
Slanting Beam Problem
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 25 / 31
Slanting Beam Problem
Once resolving the components of force, we deal with the followingequation
EId4w
dx4+ P cos θ
d2w
dx2= q cos θ.
Upon non-dimensionalising this, we obtain
d4w
dx4+ B2 cos θ
d2w
dx2= cos θ.
This equation governs the slanting roof beams of interest.
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 26 / 31
Slanting Beam Problem
The graphical representation of our deflection, for different Beam numbervalues with constant θ, is given by the following plot:
0.2 0.4 0.6 0.8 1.0x
0.0005
0.001
0.0015
0.002
Deflection
B=Π
B=Π�2
B=Π�3
B=Π�4
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 27 / 31
Slanting Beam Problem
The graphical representation of our deflection, for different θ values withconstant B, is given by the following plot:
0.2 0.4 0.6 0.8 1.0x
0.0005
0.001
0.0015
0.002
0.0025
0.003
Deflection
Θ=Π�3
Θ=Π�6
Θ=Π�12
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 28 / 31
Slanting Beam Problem
The graphical representation of the curvature, for different Beam numbervalues with constant θ, is given by the following plot:
0.2 0.4 0.6 0.8 1.0x
0.01
0.02
0.03
0.04
0.05
0.06
Curvature
B=Π
B=Π�2
B=Π�3
B=Π�4
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 29 / 31
Slanting Beam Problem
The graphical representation of the curvature, for different θ values withconstant B, is given by the following plot:
0.2 0.4 0.6 0.8 1.0x
0.02
0.04
0.06
0.08
Curvature
Θ=Π�3
Θ=Π�6
Θ=Π�12
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 30 / 31
Conclusions and Further Research
We need to find a mathematical model to explain the formation andpropagation of fractures
For the exfoliation problem, we need to improve our model in order tofind the arbitrary constant
Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera[0.2cm]Moderators: Prof. Mason, Prof. Fowkes, Ashleigh HutchinsonMISG 2014 January 20, 2014 31 / 31