MISG 2011, Problem 1:
Coal Mine pillar extraction
Group 1 and 2
January 14, 2011
Group 1 () Coal Mine pillar extraction January 14, 2011 1 / 30
Group members
• C. Please, D.P. Mason, M. Khalique, J. Medard.
• A. Hutchison, R. Kgatle , B. Matebesse, K.R. Adem, E. Yohana, A.R. Adem, B. Bekezulu, M. Hamese, B. Khumalo.
• Industry representative: Halil Yilmaz
Group 1 () Coal Mine pillar extraction January 14, 2011 2 / 30
1 Introduction
2 Model for pillar stress calculation
3 Buckling of a Strut
4 Euler-Bernoulli Beam
5 Combined Beam-Strut
6 Conclusion
Group 1 () Coal Mine pillar extraction January 14, 2011 3 / 30
Introduction
A small coal mining developpement
Group 1 () Coal Mine pillar extraction January 14, 2011 4 / 30
Introduction
Introduction
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• Cutting pillars to make snooks
• QUESTION: How do snooks collapse in order to have a safe andcontrolled collapse of the rock?
Group 1 () Coal Mine pillar extraction January 14, 2011 5 / 30
Introduction
Two approaches
• How do snooks collapse? “ The pillar problem”
• How stable is the roof when snooks collapse ’strut-beam analysis’
Group 1 () Coal Mine pillar extraction January 14, 2011 6 / 30
Model for pillar stress calculation
The set up
0
z = −H
z = H
τ(x)τ(x)
τ(x)τ(x)
τzx (x, H)
ROOF
FLOOR
τxx (x,−H)
x
z
τzz (x,−H)
τzz (x, H)
1 2
τxx
Figure: The stresses in a pillar
Group 1 () Coal Mine pillar extraction January 14, 2011 7 / 30
Model for pillar stress calculation
Governing equations
The stresses are considered in the xz-plane in the form
τxx = τxx(x , z), τxz = τxz(x , z), τzz = τzz(x , z)
and the average stress is defined as
τ̄ik(x) =1
2H
∫ H
−H
τik(x , z)dz .
The equations of static equilibrium in 2D reads
∂
∂xτxx +
∂
∂zτzx = 0 (1)
∂
∂xτxz +
∂
∂zτzz = 0 (2)
Taking the averages in the previous equations, we arrive at
d
dxτ̄xx +
1
2H[τzx(x ,H) − τzx(x ,−H)] = 0 (3)
d
dxτ̄xz +
1
2H[τzz(x ,H) − τzz(x ,−H)] = 0 (4)
Group 1 () Coal Mine pillar extraction January 14, 2011 8 / 30
Model for pillar stress calculation
Governing equations and BCs for Region 1
• For −1 ≤ x̄ ≤ 0,
d2
dx̄2τ̄
(1)xx +
3
µ
L
H
d
dx̄τ̄
(1)xx + 3m
(
L
H
)2
τ̄(1)xx = −3
(
L
H
)2
(5)
with
m =[
(1 + µ2)12 + µ
]2> 1 (6)
and µ the coefficient of internal friction, x̄ = xL.
τ̄zz = c0 + µτ̄xx .
and the boundary condition
τ̄(1)xx (−1) = 0
d τ̄(1)xx
dx̄(0) = 0. (7)
Group 1 () Coal Mine pillar extraction January 14, 2011 9 / 30
Model for pillar stress calculation
Governing equations and BCs for Region 2
• For 0 ≤ x̄ ≤ 1,
d2
dx̄2τ̄
(2)xx − 3
µ
L
H
d
dx̄τ̄
(2)xx + 3m
(
L
H
)2
τ̄(2)xx = −3
(
L
H
)2
(8)
with the boundary condition
τ̄(2)xx (1) = 0. (9)
Moreover, the following compatibility conditions hold
τ̄(1)xx (0) = τ̄
(2)xx (0) (10)
d τ̄(2)xx
dx̄(0) = 0. (11)
Group 1 () Coal Mine pillar extraction January 14, 2011 10 / 30
Model for pillar stress calculation
Solution of the Mohr-Coulomb constitutive equations
From the characteristic equations of the ODEs, we get the critical value ofµ,
µcrit
.=
√
3
4(1 +√
3)≈ 0.52394565828751.
• µ < µcrit,
τ̄(1)xx =
1
m
λ2exp(−λ1x̄) − λ1exp(−λ2x̄)
λ2exp(λ1) − λ1exp(λ2)− 1
m, (12)
and
τ̄(2)xx =
1
m
λ1exp(λ2x̄) − λ2exp(λ1x̄)
λ1exp(λ2) − λ2exp(λ1)− 1
m, (13)
Group 1 () Coal Mine pillar extraction January 14, 2011 11 / 30
Model for pillar stress calculation
Solution of the Mohr-Coulomb constitutive equations
• If µ = µcrit,
τ̄(1)xx =
1 + αx̄
m(1 − α)e−α(1+x̄) − 1
m, (14)
and
τ̄(2)xx =
1 − αx̄
m(1 − α)e−α(1−x̄) − 1
m, (15)
• If µ > µcrit,
τ̄(1)xx =
β̃ cos β̃x̄ + α sin β̃x̄
m(β̃ cos β̃ − α sin β̃)e−α(1+x̄) − 1
m, (16)
and
τ̄(2)xx =
β̃ cos β̃x̄ − α sin β̃x̄
m(β̃ cos β̃ − α sin β̃)e−α(1−x̄) − 1
m. (17)
Group 1 () Coal Mine pillar extraction January 14, 2011 12 / 30
Model for pillar stress calculation
Analytical solutionsCase 1
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1
−0.5
0
0.5
1
1.5
x
stre
ss
Group 1 () Coal Mine pillar extraction January 14, 2011 13 / 30
Model for pillar stress calculation
Analytical solutionsCase 2
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1.2
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
x
stre
ss
Group 1 () Coal Mine pillar extraction January 14, 2011 14 / 30
Model for pillar stress calculation
Analytical solutionsCase 3
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
x
stre
ss
Group 1 () Coal Mine pillar extraction January 14, 2011 15 / 30
Buckling of a Strut
Buckling of a Strut
v
p p
x
y
• Two equations for the bending moment at point x along the strut
EId2v
dx2= M(x), M(x) = −pv(x). (18)
• These equations gives
d2v
dx2+
p
EIv(x) = 0. (19)
• Solving (19) with the corresponding boundary conditions
v(0) = 0, v(L) = 0,
gives
v(x) = A sin(nπ
Lx), n = 1, 2, 3, · · · (20)
Group 1 () Coal Mine pillar extraction January 14, 2011 16 / 30
Buckling of a Strut
Buckling of a StrutNumerical results
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
defle
ctio
n
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
x
defle
ctio
n
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
x
defle
ctio
n
Group 1 () Coal Mine pillar extraction January 14, 2011 17 / 30
Euler-Bernoulli Beam
Euler-Bernoulli Beam
w
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At x = 0 At x = L
x = 0 x = L
w = 0 w = 0
wdx
= 0 wdx
= 0q
d2
dx2(EI
d2w
dx2) = q (21)
Assumptions
• Mass is uniformly distributed.
• The beam is composed of an isotropic material.
− EId2w
dx2= M Bending moment (22)
− d
dx(EI
d2w
dx2) = Q Shear force on the beam (23)
Group 1 () Coal Mine pillar extraction January 14, 2011 18 / 30
Euler-Bernoulli Beam
Euler-Bernoulli Beam
• Now the deflection of the beam satisfies the equation
d4w
dx4=
q
EI(24)
• subject to the boundary conditions
w(0) = w(L) = 0dw
dx(0) =
dw
dx(L) = 0 (25)
• The solution is found as
w(x) =q
24EIx2(x − L)2. (26)
• The curvature is given by
C = | − q L2 − 6 q x L + 6 q x2
12|
• A plot of C reveals that the beam always breaks at the end points.
Group 1 () Coal Mine pillar extraction January 14, 2011 19 / 30
Combined Beam-Strut
A combined beam and strutMotivation
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pp
q
x = 0 x = L
Figure: A combined beam and strut
Group 1 () Coal Mine pillar extraction January 14, 2011 20 / 30
Combined Beam-Strut
A combined beam and strutGoverning equations
• Use the Euler-Bernoulli equation and the theory of Euler strut.
• For a unified model, we use the previous two models to have theequation
d4w
dx4+
p
EI
d2w
dx2= − q
EI. (27)
• with the boundary conditions
w(0) = 0, dwdx
(0) = 0 at x = 0;
w(L) = 0, dwdx
(L) = 0 at x = L.(28)
• The nondimensional form of the model is
d4w̄
d x̄4+
pL2
EI
d2w̄
d x̄2= −qL4
EIs. (29)
Group 1 () Coal Mine pillar extraction January 14, 2011 21 / 30
Combined Beam-Strut
A combined beam and strutGoverning equations
• Define the so called Mason number M and s such that
M2 =pL2
EI, s =
qL4
EI(30)
• The problem reduces to
d4w̄
d x̄4+ M2 d2w̄
d x̄2= −1. (31)
• with the boundary conditions
w̄(0) = 0, dw̄dx̄
(0) = 0 at x = 0;
w̄(1) = 0, dw̄dx̄
(1) = 0 at x = 1.(32)
Group 1 () Coal Mine pillar extraction January 14, 2011 22 / 30
Combined Beam-Strut
A combined beam and strutSolution to the Governing equations
• The solution of the previous equation is found as
w̄(x) =1
2M2
(
1
4− (x − 1
2)2
)
+1
2M3 sin M2
(
cosM
2− cos(M(x − 1
2))
)
.
(33)
• The curvature is mesured with the help of
|w̄ ′′(x)| =
∣
∣
∣
∣
∣
− 1
M2+
cos(M(x − 12))
2M sin M2
∣
∣
∣
∣
∣
.
Group 1 () Coal Mine pillar extraction January 14, 2011 23 / 30
Combined Beam-Strut
A combined beam and strutNumerical Solution to the Governing equations
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−3.5
−3
−2.5
−2
−1.5
−1
−0.5
0x 10
−3
x
defle
ctio
n
Figure: A plot of the deflection for different values of M , M = 0.3 − 3.
Group 1 () Coal Mine pillar extraction January 14, 2011 24 / 30
Combined Beam-Strut
A combined beam and strutNumerical Results
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
x
curv
atur
e
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
xcu
rvat
ure
Figure: A plot of the curvature |w ′′(x)| for different values of M .
Group 1 () Coal Mine pillar extraction January 14, 2011 25 / 30
Combined Beam-Strut
A combined beam and strutNumerical Results
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
x
curv
atur
e
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
xcu
rvat
ure
Figure: A plot of the curvature |w ′′(x)| for different values of M .
Group 1 () Coal Mine pillar extraction January 14, 2011 26 / 30
Combined Beam-Strut
A combined beam and strutNumerical Results
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
x
curv
atur
e
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−3
−2.5
−2
−1.5
−1
−0.5
0
0.5x 10
−3
x
defle
ctio
n
Figure: A plot of the deflection and the curvature M = 0.8 and M = 8.
Group 1 () Coal Mine pillar extraction January 14, 2011 27 / 30
Combined Beam-Strut
A combined beam and strutImportant results
• It appear that for small (M < 2π) values of M, the beam break at theend points.
• We never recover the first mode of the solution of the Euler strutequation.
• This is due to the fact that for the beam we have 4 boundaryconditions and only 2 for the Euler strut. The equivalent fourth orderODE associated with the Euler strut has different boundaryconditions.
Group 1 () Coal Mine pillar extraction January 14, 2011 28 / 30
Conclusion
Conclusions
• Snook collapse• Stress is dependent of the quantity L
H. This ration of height to width of
the pillar is crucial. In general mining practices, LH≥ 5; but for
secondary mining operations, we can have LH
< 5.• In this later case, the stress is confined to boundary layers and the
problem correspond to singularly perturbed problem.
• Roof stability• We found a critical ratio, M , for the roof to buckle or to break.• When M < 2π, what happen in most of the cases, the roof always
break next to the snook.
Group 1 () Coal Mine pillar extraction January 14, 2011 29 / 30