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 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 Cutting pillars to make snooks QUESTION: How do snooks collapse in order to have a safe and controlled collapse of the rock? Group 1 () Coal Mine pillar extraction January 14, 2011 5 / 30
Two approaches Introduction 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
The set up Model for pillar stress calculation τ zx(x, H) z τ zz(x, H) τ(x) τ(x) ROOF z = H 1 2 0 x τ xx FLOOR z = H τ(x) τ(x) τ xx(x, H) τ zz(x, H) 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 τ xx = τ xx (x,z), τ xz = τ xz (x,z), τ zz = τ zz (x,z) and the average stress is defined as τ ik (x) = 1 2H H H τ ik (x,z)dz. The equations of static equilibrium in 2D reads x τ xx + z τ zx = 0 (1) x τ xz + z τ zz = 0 (2) Taking the averages in the previous equations, we arrive at d dx τ xx + 1 2H [τ zx(x,h) τ zx (x, H)] = 0 (3) d dx τ xz + 1 2H [τ zz(x,h) τ zz (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 x 0, d 2 d x 2 τ(1) xx + 3 µ L H ( ) d L 2 ( ) d x τ(1) xx + 3m τ xx (1) L 2 = 3 (5) H H with m = [ (1 + µ 2 ) 1 2 + µ ] 2 > 1 (6) and µ the coefficient of internal friction, x = x L. and the boundary condition τ zz = c 0 + µ τ xx. xx ( 1) = 0 d τ(1) xx (0) = 0. (7) d x τ (1) 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 x 1, d 2 d x 2 τ(2) xx 3 µ L H with the boundary condition ( ) d L 2 ( ) d x τ(2) xx + 3m τ xx (2) L 2 = 3 (8) H H Moreover, the following compatibility conditions hold τ (2) xx (1) = 0. (9) τ (1) xx (0) = τ (2) xx (0) (10) d τ xx (2) (0) = 0. (11) d x 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 of µ,. 3 µ crit = 4(1 + 3) 0.52394565828751. µ < µ crit, τ xx (1) = 1 λ 2 exp( λ 1 x) λ 1 exp( λ 2 x) 1 m λ 2 exp(λ 1 ) λ 1 exp(λ 2 ) m, (12) and τ xx (2) = 1 λ 1 exp(λ 2 x) λ 2 exp(λ 1 x) 1 m λ 1 exp(λ 2 ) λ 2 exp(λ 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 = µ crit, and If µ > µ crit, τ (1) xx = 1 + α x m(1 α) e α(1+ x) 1 m, (14) τ (2) xx = 1 α x m(1 α) e α(1 x) 1 m, (15) τ (1) xx = β cos β x + αsin β x m( β cos β αsin β) e α(1+ x) 1 m, (16) and τ (2) xx = β cos β x αsin β x m( β cos β αsin β) e α(1 x) 1 m. (17) Group 1 () Coal Mine pillar extraction January 14, 2011 12 / 30
Model for pillar stress calculation Analytical solutions Case 1 1.5 1 0.5 stress 0 0.5 1 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 x Group 1 () Coal Mine pillar extraction January 14, 2011 13 / 30
Model for pillar stress calculation Analytical solutions Case 2 0.8 0.6 0.4 0.2 0 stress 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 1 x Group 1 () Coal Mine pillar extraction January 14, 2011 14 / 30
Model for pillar stress calculation Analytical solutions Case 3 1 0.8 0.6 0.4 stress 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 Group 1 () Coal Mine pillar extraction January 14, 2011 15 / 30
Buckling of a Strut Buckling of a Strut p p v x y Two equations for the bending moment at point x along the strut EI d2 v = M(x), dx2 M(x) = pv(x). (18) These equations gives d 2 v dx 2 + p v(x) = 0. EI (19) Solving (19) with the corresponding boundary conditions v(0) = 0, v(l) = 0, gives v(x) = A sin( nπ x), L n = 1,2,3, (20) Group 1 () Coal Mine pillar extraction January 14, 2011 16 / 30
Buckling of a Strut Buckling of a Strut Numerical results 1 1 1 0.9 0.8 0.8 0.8 0.6 0.6 0.7 0.4 0.4 0.6 0.2 0.2 deflection 0.5 deflection 0 deflection 0 0.4 0.2 0.2 0.3 0.4 0.4 0.2 0.6 0.6 0.1 0.8 0.8 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x Group 1 () Coal Mine pillar extraction January 14, 2011 17 / 30
Euler-Bernoulli Beam Euler-Bernoulli Beam Assumptions w At x = 0 At x = L w = 0 w = 0 w dx = 0 q dx w = 0 x = 0 x = L d 2 dx 2(EI d2 w dx 2 ) = q (21) Mass is uniformly distributed. The beam is composed of an isotropic material. EI d2 w = M dx2 Bending moment (22) d dx (EI d2 w ) = Q dx2 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 subject to the boundary conditions w(0) = w(l) = 0 The solution is found as The curvature is given by w(x) = d 4 w dx 4 = q EI dw dx (24) dw (0) = (L) = 0 (25) dx q 24EI x2 (x L) 2. (26) C = q L2 6q x L + 6q x 2 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 strut Motivation p x = 0 q x = L p Figure: A combined beam and strut Group 1 () Coal Mine pillar extraction January 14, 2011 20 / 30
Combined Beam-Strut A combined beam and strut Governing equations Use the Euler-Bernoulli equation and the theory of Euler strut. For a unified model, we use the previous two models to have the equation d 4 w dx 4 + p d 2 w EI dx 2 = q EI. (27) with the boundary conditions w(0) = 0, w(l) = 0, dw dx dw The nondimensional form of the model is (0) = 0 at x = 0; dx (L) = 0 at x = L. (28) d 4 w d x 4 + pl2 d 2 w EI d x 2 = ql4 EIs. (29) Group 1 () Coal Mine pillar extraction January 14, 2011 21 / 30
Combined Beam-Strut A combined beam and strut Governing equations Define the so called Mason number M and s such that The problem reduces to M 2 = pl2 EI, s = ql4 EI (30) d 4 w d x 4 + w M2d2 = 1. (31) d x 2 with the boundary conditions w(0) = 0, w(1) = 0, d w d x d w (0) = 0 at x = 0; d x (1) = 0 at x = 1. (32) Group 1 () Coal Mine pillar extraction January 14, 2011 22 / 30
Combined Beam-Strut A combined beam and strut Solution to the Governing equations The solution of the previous equation is found as w(x) = 1 ( 1 2M 2 4 (x 1 ) 1 2 )2 + (cos M2 2M 3 sin M cos(m(x 12 ) )). 2 (33) The curvature is mesured with the help of w (x) = 1 M 2 + cos(m(x 1 2 )) 2M sin M. 2 Group 1 () Coal Mine pillar extraction January 14, 2011 23 / 30
Combined Beam-Strut A combined beam and strut Numerical Solution to the Governing equations 0 x 10 3 0.5 1 deflection 1.5 2 2.5 3 3.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 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 strut Numerical Results 0.18 0.9 0.16 0.8 0.14 0.7 0.12 0.6 curvature 0.1 0.08 curvature 0.5 0.4 0.06 0.3 0.04 0.2 0.02 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 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 strut Numerical Results 0.09 0.09 0.08 0.08 0.07 0.07 0.06 0.06 curvature 0.05 0.04 curvature 0.05 0.04 0.03 0.03 0.02 0.02 0.01 0.01 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 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 strut Numerical Results 0.09 0.5 x 10 3 0.08 0 0.07 0.06 0.5 curvature 0.05 0.04 deflection 1 1.5 0.03 2 0.02 0.01 2.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 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 strut Important results It appear that for small (M < 2π) values of M, the beam break at the end points. We never recover the first mode of the solution of the Euler strut equation. This is due to the fact that for the beam we have 4 boundary conditions and only 2 for the Euler strut. The equivalent fourth order ODE associated with the Euler strut has different boundary conditions. Group 1 () Coal Mine pillar extraction January 14, 2011 28 / 30
Conclusions Conclusion 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, L H 5; but for secondary mining operations, we can have L H < 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
Thank you Conclusion Thank you THANK YOU Group 1 () Coal Mine pillar extraction January 14, 2011 30 / 30