prediction of fracturing and dynamic roof failure in ... · platinum mine fractured would be...

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