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Prediction of Fracturing and Dynamic Roof Failure in Platinum Mines Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Thebe Ramanna, Charlene Chipoyera Moderators: Prof.

SLIDE 1

Prediction of Fracturing and Dynamic Roof Failure in Platinum Mines

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, 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, Theb MISG 2014 January 20, 2014 1 / 31

SLIDE 2

Introduction

Roof collapses occur occasionally in platinum mines and are devastating events When roof collapse occurs, large slabs of the rock fracture These roof collapses compromise miner safety and are very costly to the mine By understanding how a roof collapse occurs, we can predict its

ccurrence as well as attempt to mitigate the risks, thus avoiding

unnecessary loss

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 2 / 31

SLIDE 3

Introduction

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 3 / 31

SLIDE 4

Introduction

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 4 / 31

SLIDE 5

Crack Formation

Given slabbing, what factors contribute to/determine the uniform thickness of the slabs in non-sedimentary rock? Definitions Slabbing: A phenomenon whereby a rock mass peels off in uniform layers Thickness: Distance from free surface to the fracture

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 5 / 31

SLIDE 6

Crack Formation

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 6 / 31

SLIDE 7

Crack Formation

Considerations

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 7 / 31

SLIDE 8

Crack Formation

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 8 / 31

SLIDE 9

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 the manner in which the rock fractures. Seismicity induced by drilling, explosions and natural effects causes a re-distribution of stresses which contributes to crack extension.

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 9 / 31

SLIDE 10

Exfoliation Problem

Exfoliation of surface rocks is a complementary phenomenon that has been

bserved. We believe it to be a buckling beam problem and that it also

corresponds to the eigenvalue problem, as the effect of gravity was seen to be negligible.

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 10 / 31

SLIDE 11

Exfoliation Problem

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 11 / 31

SLIDE 12

Exfoliation Problem

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 12 / 31

SLIDE 13

Exfoliation Problem

The beam equation is given by EI d4w dx4 + P d2w dx2 = q, where q is the sum of the body force and the surface traction per unit length. For the exfoliation problem, q=0 thus resulting in d4w dx4 + B2 d2w dx2 = 0.

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 13 / 31

SLIDE 14

Exfoliation Problem

Finding the general solution to our equation yields w(x) = A cos(Bx) + C sin(Bx) + D B2 x + 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, Theb MISG 2014 January 20, 2014 14 / 31

SLIDE 15

Exfoliation Problem

In order to produce non-trivial solutions, we impose our boundary conditions to obtain a homogeneous system of equations in the form Hx = 0:     1 B−2 cos(B) sin(B) B−2 B−2 B B−2 −B sin(B) B cos(B) B−2         A C D F     =         We want the determinant of the matrix to be equal to 0. det(H) = 4B5 sin B 2 sin B 2

− B

2 cos B 2

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 15 / 31

SLIDE 16

Exfoliation Problem

Case 1: sin B 2

Solving for B gives B = 2nπ n ∈ Z. We substitute this into the matrix H:            1 1 (2nπ)2 1 1 (2nπ)2 1 (2nπ)2 2nπ 1 (2nπ)2 2nπ 1 (2nπ)2            . 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, Theb MISG 2014 January 20, 2014 16 / 31

SLIDE 17

Exfoliation Problem

Plot of the beam deflection

0.2 0.4 0.6 0.8 1.0 x 0.5 1 1.5 2 wx

n3 n2 n1 Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 17 / 31

SLIDE 18

Exfoliation Problem

Plot of the beam curvature

0.2 0.4 0.6 0.8 1.0 x 50 100 150 200 250 300 350 Curvature

B6Π, n3 B4Π, n2 B2Π, n1 Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 18 / 31

SLIDE 19

Exfoliation Problem

Case 2:

B 2

− B

2 cos 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, Theb MISG 2014 January 20, 2014 19 / 31

SLIDE 20

Exfoliation Problem

From tan B 2

2 , we can derive sin B = B B2 4 + 1 cos B = B2 4 − 1 B2 4 + 1

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 20 / 31

SLIDE 21

Exfoliation Problem

We substitute these expressions into the matrix H, and apply row

perations, we obtain

       1 1 1 − B2 4 B 1 + B2 4 1 + B2 4 B 1 −B2 B(1 − B2 4 ) 1 + B2 4        . Solving the resulting matrix system gives w(x) = D(−B 2 cos(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, Theb MISG 2014 January 20, 2014 21 / 31

SLIDE 22

Exfoliation Problem

Plot of the beam deflection, for varying B values

0.2 0.4 0.6 0.8 1.0 x 20 10 10 20 wx

B28.1 B21.8 B15.45 B8.9

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 22 / 31

SLIDE 23

Exfoliation Problem

Plot of the beam curvature, for varying B values

0.2 0.4 0.6 0.8 1.0 x 2000 4000 6000 8000 10 000 curvature

B28.1 B21.8 B15.45 B8.9 Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 23 / 31

SLIDE 24

Slanting Beam Problem

Blasted tunnels have roofs that seem to behave as beams; however, they are at an angle, say θ. We thus felt that examining how the roof in a platinum mine fractured would be equivalent to examining the slanted beam equation.

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 24 / 31

SLIDE 25

Slanting Beam Problem

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 25 / 31

SLIDE 26

Slanting Beam Problem

Once resolving the components of force, we deal with the following equation EI d4w dx4 + P cos θd2w dx2 = q cos θ. Upon non-dimensionalising this, we obtain d4w dx4 + B2 cos θd2w 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, Theb MISG 2014 January 20, 2014 26 / 31

SLIDE 27

Slanting Beam Problem

The graphical representation of our deflection, for different Beam number values with constant θ, is given by the following plot:

0.2 0.4 0.6 0.8 1.0 x 0.0005 0.001 0.0015 0.002 Deflection

BΠ BΠ2 BΠ3 BΠ4

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 27 / 31

SLIDE 28

Slanting Beam Problem

The graphical representation of our deflection, for different θ values with constant B, is given by the following plot:

0.2 0.4 0.6 0.8 1.0 x 0.0005 0.001 0.0015 0.002 0.0025 0.003 Deflection

ΘΠ3 ΘΠ6 ΘΠ12

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 28 / 31

SLIDE 29

Slanting Beam Problem

The graphical representation of the curvature, for different Beam number values with constant θ, is given by the following plot:

0.2 0.4 0.6 0.8 1.0 x 0.01 0.02 0.03 0.04 0.05 0.06 Curvature

BΠ BΠ2 BΠ3 BΠ4 Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 29 / 31

SLIDE 30

Slanting Beam Problem

The graphical representation of the curvature, for different θ values with constant B, is given by the following plot:

0.2 0.4 0.6 0.8 1.0 x 0.02 0.04 0.06 0.08 Curvature

ΘΠ3 ΘΠ6 ΘΠ12 Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 30 / 31

SLIDE 31

Conclusions and Further Research

We need to find a mathematical model to explain the formation and propagation of fractures For the exfoliation problem, we need to improve our model in order to find the arbitrary constant

Kedy Mazibuko, Kirsten Louw, Despina Zoras, Emile Meoto, Tanki Motsepa, Yachna Bharath, Sanelisiwe Bingo, Vuyelwa Makibelo, Theb MISG 2014 January 20, 2014 31 / 31