Fold catastrophe model of dynamic pillar failure in asymmetric mining

Received 20 May 2008; accepted 11 July 2008 Projects Y2005-A03 supported by the Natural Science Foundation of Shandong Province of China and G04D15 by the Educational Committee of Shandong Province of China Corresponding author. Tel: +86-532-86879624; E-mail address: [email protected]

Fold catastrophe model of dynamic pillar failure in asymmetric mining

PAN Yue, LI Ai-wu, QI Yun-song College of Civil Engineering, Qingdao Technological University, Qingdao, Shandong 266520, China

Abstract: A rock burst disaster not only destroys the pit facilities and results in economic loss but it also threatens the life of the miners. Pillar rock burst has a higher frequency of occurrence in the pit compared to other kinds of rock burst. Understanding the cause, magnitude and prevention of pillar rock burst is a significant undertaking. Equations describing the bending moment and displacement of the rock beam in asymmetric mining have been deduced for simplified asymmetric beam-pillar systems. Using the symbolic operation software MAPLE 9.5 a catastrophe model of the dynamic failure of an asymmetric rock-beam pillar system has been established. The differential form of the total potential function deduced from the law of conservation of energy was used for this deduction. The critical conditions and the initial and final positions of the pillar during failure have been given in analytical form. The amount of elastic energy released by the rock beam at the instant of failure is determined as well as. A diagrammatic form showing the pillar failure was plotted using MATLAB software. This plot contains a wealth of information and is important for understanding the behavior during each deformation phase of the rock-beam pillar system. The graphic also aids in distinguishing the equivalent stiffness of the rock beam in different directions. Keywords: pillar; fold catastrophe; asymmetric mining; energy importing rate; energy releasing amount.

1 Introduction

Sometimes the rock body bursts by dynamic failure during deep mining work. The rock burst disaster destroys the pit facilities and results in economic loss but, worse, also threatens the life of the miners. Because pillar rock burst occurs more frequently in the pit[1–3], research into its prevention, generation and magnitude is of great significance.

Catastrophe theory is an effective way to study problems of catastrophic dynamic failure. A catastrophe model for narrow pillar failure in symmetric mining was established for rock beams fixed at the ends. The behavioral characteristics of prototype pillar failures, including system stability, were also described[4]. Many factors limit excavation in a way that they are usually executed under conditions of asymmetric mining. These factors include the geological environment.

The stability of the mined area has a direct relation to both safe production in the mine and to the excavation plans for the rock body. Li et al., have examined the law predicting pillar rock burst under a rigid roof using computer models based on symmetric

mining and the idea that the un-excavated coal bed is an elastic foundation[5–6]. In this article the stability of the pillar during asymmetric mining is analyzed by using catastrophe theory.

2 An analytic model of asymmetric mining

The stope layout is shown in Fig. 1. There is a solid rock stratum above the coal layer. When rock burst occurs in the pillar the roof of the rock stratum participates only in energy release instead of being destroyed[1]. If the base plate does not deform and the working surface is very long then the unit length at section I-I can be taken as the subject under investigation. The roof is then regarded as a rock beam and both the self-weight of the rock beam and the action of the rock stratum above it are simplified to a uniformly distributed load, q. The action of a unit length on the pillar can also be regarded as a concentrated force, F, acting at x=a. For small beam the turning angle from the rock beam to the deviations of the pillar from the midpoint of the rock pillar can be assumed to be zero. Because the pillar is relatively narrow and its compression is far greater

Mining Science and Technology 19 (2009) 0049–0057

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Mining Science and Technology Vol.19 No.150

than that of the un-excavated coal layer. To simplify the analysis it is supposed that the un-excavated coal layer is incompressible and that the rock beam is located within it as shown in Fig. 2. The rock beam deforms elastically until fracture of the pillar; its bending stiffness is EI.

Fig. 1 Layout of the stope

1. Working surface; 2. Unexcavated mine bed; 3. Narrow pillar; 4. Tunnel; 5. Rock beam

Fig. 2 A simplified model of a rock beam and pillar

The stress-strain relationship of the pillar is: [7]

oo

expm

E εσ εε

= − (1)

where oE is the initial elastic modulus, m is the homogeneous exponent of the curve and the relation between oε and the strain cε corresponding to

peak stress cσ is: 1/

o c1 m

mε ε=

where o cε ε= , while 1=m . The action of the pillar on point C of the rock beam, or the load-displacement relationship of the pillar (Fig. 3) is then:

Fig. 3 Load-displacement curves for a narrow pillar

o

( ) expm

uF u uu

λ= − (2)

where o /E B Hλ = is the initial stiffness of the pillar and B and H are the width and original height of the pillar, respectively. Also, u is the deformation of the pillar, or displacement of its top end, i.e.; the deflection, at ax = , of the rock beam (Fig. 2) for

o ou Hε= .

3 Equation for the curvature of the rock beam in asymmetric mining

If the support pillar shown in Fig. 2 is removed, the rock beam will sink to a static equilibrium position, shown in Fig. 4. This results from a balance between the weight of the rock and the elastic properties of the rock. The deflection at the middle of the rock beam is

)24/(~~ 4max EIlquc = . The supporting force of the

pillar, ( )F u in Fig. 2, restricts the displacement of the rock beam at each point. But under the action of gravity each point along the curve has a tendency toward the static equilibrium position shown in Fig. 4. This is the tendency that results in the failure rupture of the pillar.

Fig. 4 Deflection curve of a rock beam without pillar support

Assuming the angle at ax = on the beam is zero, a uniform load, q, is opposed by a counterforce

)(uF from the pillar, which is treated as a beam fixed at both ends (Fig. 2). When the displacement of the pillar is u the bending moment and shear forces at the beam ends can be divided into two parts as shown in Figs. 5a and b.

Fig. 5 Counter force at the pillar end of the rock-beam

From structural mechanics[8] we know that if a beam fixed at both ends undergoes uniform load, q , there will be a clockwise lateral motion, u , at its

PAN Yue et al Fold catastrophe model of dynamic pillar failure in asymmetric mining 51

right end (Fig. 5a). The bending moment and shear force at the ends A and C are:

3

6 6, 12 12

12 12,2 2

2 2

AC CA2 2

AC CA3

qa EI qa EIM u M ua a

qa EI qa EIQ u Q ua a

= − − = − +

= + = − + (3)

where the convention is that the bending moment at the underside of the rock beam is drawn as positive.

In Fig. 5b the uniform load, q, causes an anticlockwise lateral motion, u, at the left end. The bending moment and shear force at the ends C and Bare:

2 2

2 2

3 3

6 6, 12 12

12 12,2 2

CB BC

CB BC

qb EI qb EIM u M ub b

qb EI qb EIQ u Q ub b

= − + = − −

= − = − − (4)

Taking point A as the origin, from Eq.(3) the equation for the bending moment in section AC is:

2

1

2 2

3 2

( )2

12 62 2 12

AC ACqxM x M Q x

qx qa EI qa EIu x ua a

= + −

= − + + − + (5)

Taking B as the origin, Eq.(4) gives the bending moment in section BC as:

2

2

2 2

3 2

( )2

12 62 2 12

BC BCqxM x M Q x

qx qb EI qb EIu x ub b

= − −

= − + + − + (6)

From Eq.(5) and the differential equation for displacement, )(xMyEI −=′′ , and taking A as the origin, the equations for the turning angle and the displacement in section AC are:

3 2

1

2

126 2 2

6 12

3

2

qx qa EI xEI y ua

qa EI u xa

′ = − +

+ +

(7)

4 3

1 3

2 2

2

1224 2 6

6 12 2

qx qa EI xEI y ua

qa EI xua

= − +

+ +

(8)

From Eq.(6), in a similar way taking B as the origin, the turning angle and displacement in section BC are:

3 2

2 3

2

2

126 2 2

6 12

qx qb EI xEI y ub

qb EI u xb

′ = − +

+ +

(9)

4 3

2 3

2 2

2

1224 2 6

6 12 2

qx qb EI xEI y ub

qb EI xub

= − +

+ +

(10)

From Eqs.(7)–(10) it is easy to verify that the displacement satisfies these boundary conditions: for points A and B 0)0( =y , 0)2( =ly , 0)0( =′y and

0)2( =′ ly ; For point C 0)('1 =ay , 0)('

2 =by and ubyay == )()( 21 .

From Eqs.(3) and (4), it is also seen that for a deformation in the pillar of u the rock beam undergoes a counterforce from the pillar. The expression relating F, q and u at point C is:

33 3

( )1 112

CA CB CA CBF u Y Y Q Q

ql l k ua b

= + = − +

= − + (11)

where

3lEIk = (12)

4 A catastrophe theory analysis of pillar failure

4.1 A fold catastrophe model of pillar failure in asymmetric mining

In the following passage the sign d expresses a differential with respect to position and the sign δexpresses a differential with respect to the variables.

The dynamic failure of a pillar occurs on the softened region of the load-displacement curve, ( )F u , after the peak value has occurred. On this region of the ( )F u curve when the equilibrium position of the beam-pillar system has quasi-static displacement uδ , i.e.; the deformation of the pillar caused by compression of its top end is uδ , the rock beam will release the elastic deformation energy e ( 0)Uδ < as each point on its deformation curve approaches the equilibrium position shown in Fig. 4. The pillar dissipates energy p ( 0)Uδ > because internal minute

cracks extend and connect. If e pU Uδ δ− < , the external force q replenishes the energy through work, )0(>Wδ , during displacement, which makes the pillar deform. Being in a quasi-stationary state and without effect from kinetic energy the law of conservation of energy requires that:

e p 0U U Wδ δ δ+ − = (13) This is the equilibrium relationship between the

work and energy increments of the system for a small


displacement of the pillar )0(>uδ . The system is in a quasi-stationary state, i.e.; the differential form of the total potential function, established by taking the condition that the upper end of the pillar has been displaced to u, is now the reference state.

When the upper end of the pillar has the displacement increment uδ the dissipated energy is:

pU F uδ δ= (14) Accordingly, the energy released by the rock beam

is: 2

e 0

0

2

1 10

2 20

( ) ( )d

1 ( ) ( )d

1 ( ) ( )d

1 ( ) ( )d

1 ( ) ( )d

l

a

l

a

a

b

U M x x x

M x M x xEI

M x M x xEI

M x M x xEI

M x M x xEI

δ δκ

δ

δ

δ

δ

=

=

+

=

′ ′ ′+

(15)

where x′ is the position coordinate (taking B as the origin on the left). Then EIxMx /)()( =κ is the curvature at point x on the displacement curve of the rock beam and the deflection at point C is u; this is the displacement of the pillar. And )(xκδ is the differential in the curvature at point x on the displacement curve so the deflection increment is

uδ at point C. The work increment of the external force is given

by:

xxyqxxyqxxyqWl

a

ald)(d)(d)(

2

0

2

0 +== δδδδ

xxyqxxyqba

d)(d)(0 20 1 += δδ (16)

where )(xyδ is the deflection increment at any point x on the displacement curve for a deflection increment at point C of uδ . Substituting Eqs.(5) and (6) into Eq.(15) gives:

22

e 30

22

3

22

30

22

3

1 6 (2 )2 6

6 (2 ) d2 6

1 6 (2 )2 6

6 (2 ) d2 6

a

b

q a EIU x a x u x aEI a

q a EIx a x u x a xa

q b EIx b x u x bEI b

q b EIx b x u x b xb

δ

δ δ

δ δ

= − − + + −

× − − + + ⋅ −

+ − − + + ⋅ −

× − − + + ⋅ −

where q is a function of F and u, and qδ is a function of Fδ and uδ , see Eq.(11). Integrating

and simplifying this equation with respect to x by using the symbolic-operation software MAPLE 9.5, it can be found that:

e1

360

( 24 ) 24 ( 384 )

UB

FF Aku Ak F Cku uu

δ

δ δδ

= ×

+ + + (17)

where 3 3

33 3

5

5 5

4 2 27

3 3 5 5

,22

( ) 42 ( )

a bA la b

lBa ba b a bC l

a b a b

+=

=+− +=

+

(18)

Substituting (17) into the differential form of the total potential function (13) and then dividing both sides of the equality by uδ gives:

+++ )384(24)24(360

1 kuCFkAuFkuAF

B δδ

0=−+ JF (19) It should be pointed out that Eq.(19) is the

equilibrium equation when the upper end of the pillar has a positive incremental displacement, ( 0)uδ > , and the system is in a quasi-stationary state. This state is established by taking the state where the upper end of the pillar has displaced to u as the reference state where:

uWJ

δδ

= (20)

where J is the required energy change of the system for a unit displacement of the pillar. Because this energy comes from the surroundings, this is termed the energy importing rate[1]. From the derivation and physical meaning of Eq.(13) we know that in Eq.(19) the pillar has a quasi-static displacement if 0>J . 0=J , the displacement of the pillar, can increase automatically only by virtue of energy transfer within the system without applying work from the external force, q, i.e.; the elastic strain energy accumulating in the apical plate or in the rock beam transfers into the pillar. This shows that the system is in critical state, so

0=J (21) can be regarded as the critical condition for the system leaving the quasi-stationary state: the onset of failure rupture of the pillar.

In Fig. 3 the displacement, tu , at the inflexion point, t, on the softened region of the )(uF curve, which is given by Eq.(2), satisfies 0)( =′′ tuF . The


relation between tu and ou is: 1/

o

1 mtu m

u m+= (22)

This shows that the bigger the curve exponent, m, is the smaller o/tu u is. That is to say, the steeper the softened region of the )(uF curve the stronger the impact from movement in the pillar media. Rock burst occurs in the softened region of the )(uF curve after the peak value. The burst starts at a certain point before the inflexion point, t, and ends at another point after t. Using Eq.(22), the Taylor expansion of Eq.(2), the displacement, tu , at the inflexion point can be calculated as:

2 32

2 2 43

( ) ( ) (1 ) ( )6

(3 11 14)(1 ) ( )24

t t tt

tt

F u u m u u m m u uu

m m m u uu

ββ β

β

= − − + + −

+ − + + −

+ (23)

where exp[ (1 ) / ]m mβ λ= − + . Substituting Eq.(23) into the equilibrium equation, Eq.(19) gives:

22

3

(1 ) ( 24 ) ( )2

(1 ) ( 24 )( ) (1 )( 24 )

360 ( ) 0

tt

t

t

t

m m Ak u uu

m K m Ak u um K Ak u

Bk J O u u

β β

β ββ β

+ + −

+ − × − −− − +

− + − =

(24)where

)(384384

tuFCk

mCkK

′−==

β (25)

In the following passage, the value of K is assumed to be around 1. Because the quadratic term of

)( tuu − is the lowest power term whose coefficient is not zero in Eq.(24) the principle of determinacy states that Eq.(24) corresponds to the equilibrium equation of a fold catastrophe model[9–10]. The stability of the rock-beam pillar system will be discussed with terms higher than the cubic term being ignored. Simplifying Eq.(24) to a dimensionless form and collating gives:

2

2

(1 )( 24 )(1 ) ( 24 )

t

t

u u K m Aku m Ak

ββ

− − −++ +

2

2 2

2

(1 )( 24 ) 2(1 )(1 ) ( 24 ) (1 )

15 08 (1 ) (1 24 ) ( )t

K m Ak Km Ak m

BKJC m Ak F u

ββ

β

− − −− −+ + +

− =+ +

(26)

This is deduced from ( )t tF u uβ= and Eq.(25). Substitution of variables gives:

2

2

2 2

21

(1 )( 24 )(1 ) ( 24 )

(1 )( 24 ) 2(1 )(1 ) ( 24 ) (1 )

158 (1 ) (1 24 ) ( )

t

t

u u K m Akwu m Ak

K m Ak Km Ak m

BK JC m k F u

ββ

ββη

β

− − −= ++ +

− − −++ + += −

⋅++ + ⋅

(27)

and allows simplification of Eq. (26) to the canonical form[7] of the equilibrium equation of a fold catastrophe model,

02 =+ηw (28) where w is a state variable and η is a control

variable. If 0≤η , a plot of Eq.(28) gives a parabola. The straight line 0η = (or the axis 1K − ) divides the parabola into upper and lower branches as shown in Fig. 6.

Fig. 6 Equilibrium surface of the fold catastrophe model

Remembering that:

tuAkmAkmKu

)24()1()24)(1(1 2

*

++−−−=

ββ

(29)

The first equality in Eq.(27) maps the point *u of the ( )F u curve to the axis 1K − with 0w = , in

Fig. 6. Apparently *u changes according to K:

tuu =* while 1=K . Substituting Eqs.(8) and (10) into Eq.(16), and

using Eq.(11), qδ can be expressed in terms of Fδand uδ . The work increment of the external force is:

[ ]

4 3 2 23 2

30

4 3 2 23 2

30

(2 3 ) d24 12 24

(2 3 ) d24 12 24

384 (30)360

a

b

q x ax a x EI uW q x ax xEI a

q x bx b x EI uq x bx xEI b

ql F Ck uBk

δδ δ

δδ

δ δ

= − + − −

+ − + − −

= +

where B and C are given by Eq.(18). Substituting Eq.(30) into Eq.(20) gives the energy importing rate:


384360

W ql FJ Cku Bk u

δ δδ δ

= = + (31)

0>J is true in some parts of the softened region

of the ( )F u curve. In these regions Eq.(26) can be written as the following two equations:

2

12

(1 )( 24 ) 1 (1 )( 24 ) 152(1 )(1 ) ( 24 ) 1 (1 )( 24 ) 248 1 ( )

t

tt

u u K m Ak K m Ak BKJw Ku m Ak m m Ak AkC F u

β ββ β

β

− − − − − −+ = = + − ++ + + + +

+ (32)

2

22

(1 )( 24 ) 1 (1 )( 24 ) 152(1 )(1 ) ( 24 ) 1 (1 )( 24 ) 248 1 ( )

t

tt

u u K m Ak K m Ak BKJw Ku m Ak m m Ak AkC F u

β ββ β

β

− − − − −+ = = + − ++ + + + +

+ (33)

Eqs.(32) and (33) are the upper and lower branches of the quasi-static equilibrium path of the rock-beam pillar system. These correspond to the sections above and below the inflexion point, t, of the softened region of the ( )F u curve. The equilibrium position of the rock-beam pillar system will finally reach branch 2 in Fig. 6. For a η that is not zero the system has two equilibrium positions corresponding to branch 2. For a certain K the equilibrium position can move to branch 2 from branch 1 via the original point and/or the axis K 1. In-other-words, the two quasi-static equilibrium paths can be extended to the axis K 1. If this happens the pillar will fracture in the form of a quasi-stationary state. Otherwise, the equilibrium position will reach branch 2 by way of jumping and the pillar will break up in the form of a dynamic failure.

4.2 Characteristics of pillar failure in asymmetric mining

For certain values of K Eqs.(27), (31) and (32) show that, when *uu < , as u increases due to pillar compression ( )F u′ and J decrease. Thus ηand 1w change from negative values to zero and the equilibrium position, 1( , )wη , of the system moves to the right along branch 1 in Fig. 6. From Eqs.(27), (31) and (33) we know that as u increases when *uu >then ( )F u′ and J also increase and η becomes more negative while 2w becomes more positive: The equilibrium position, ),( 2wη , moves left along branch 2 in Fig. 6.

In the following passage, several situations will be discussed. When K has different values uapproaches and deviates from the point *u as in Eq.(29). The equilibrium position ( , )wη approaches and deviates from the axis K 1.

1) 384C ( )tk F u′< − , i.e.; K is smaller than 1. When the equilibrium position ),( 1wη moves

right along branch 1 it can be proved that there exists a certain point on branch 1 where 0jw < . There is a

certain point, ju , given the condition that *uu < , that satisfies:

( ) 384 0jF u Ck′ + = (34)

At the point ju , in Eq.(31):

0Wu

δδ

= or 0J = (35)

and Eq.(21) is satisfied. This shows that the Cooks rigidity criterion is met[11]. Thus the system is in a critical state. From Eq.(16) we see:

2

0 ( )d 2

l

mq y x x ql yW

u u u

δ δδδ δ δ

= = (36)

where 2

0 ( )d

2

l

m

q y x xy

l

δδ = (37)

is the mean value of the deflection increment of each point on the displacement curve, for deflection increment uδ at point ax = . Between points Aand B there is a ( ) 0y xδ > at each point on the displacement curve of the rock beam when the deformation of the pillar 0uδ > . Hence the mean deflection increment is always greater than zero;

0myδ > . When the equilibrium position of the system moves along branch 1 of the quasi-static equilibrium path myδ and uδ are same order of magnitude. From Eq.(36) and the critical conditions

0J W uδ δ= = or / 0my uδ δ =

∞→myu

δδ

(38)

showing that the displacement u has an abrupt and

limited variation at ju . Except for the point on the axis at 1K − , for the same η w has two corresponding states. Hence the equilibrium position jumps from point jw on branch 1 to point sw on branch 2, in Fig. 6. At the point sw the following equation, which has the same form as Eq.(23), holds:


( ) 384 0sF u Ck′ + = (39) This is shown in Fig. 3. Because both points j and shave J=0, from Eqs.(32)–(33), the failure amplitude of the pillar, in terms of the dynamic force, is:

2 (1 )( 24 ) 2(1 )1 (1 )( 24 )

s j

t

u u u

K m Ak K um m Ak

ββ

Δ = − =

− − + − ⋅+ + +

(40)

Note that ( ) ( ) [ ( ) ]F u u F u u F u uδ δ δ+ = ⋅ in Eq.(17). Now integrating the sum of Eqs.(17) and (14) from ju to su (in this phase J=0) we obtain the elastic strain energy released by the rock beam during failure. This exceeds the energy dissipated by the pillar by the extension of its internal cracks. The energy is given by:

2 21 ( ) 48 ( )720 s j s s j jE F F Ak F u F u

BkΔ = − + −

s2 2 224 384 ( ) ( )j

u

s j uA Ck u u F u uδ× × − + (41)

whose geometric form is the shadowed area enclosed by the F(u) curve and the oblique straight line ojtranslated to point j on the right in Fig. 3. Since this is an energy release 0<ΔE . The excess energy turns into kinetic energy, T, of the rock-beam pillar system:

ET Δ−=Δ (42) A small part of this kinetic energy is dissipated by dynamic expansion of the cracks in the pillar. Most of it transforms into kinetic energy of the rock beam itself and thus causes an elastic wave.

2) 384 ( )tC k F u′= − i.e.; 1K = . When 1K = ,

tuu =* in Eq.(29) corresponds to the points on the axis 1−K in Fig. 6. Eqs.(32) and (33) become:

11 15

1 248 1 ( )

t

tt

u u BJwu m AkC F u

β

− −= =+

+ (43)

21 15

1 248 1 ( )

t

tt

u u BJwu m AkC F u

β

−= =

++

(44)

The equilibrium position ),( 1wη moves to the right along branch 1 and J decreases from a positive value. From t( ) 384 F u C k′− = and Eq.(31) we know that the critical condition 0J = is satisfied when *uuu t == because then 021 == ww in Eqs.(43) and (44). Here branch 1 and branch 2 of the quasi-static equilibrium path of the rock-beam pillar system in Fig. 6 join at the axis k–1. Thus the equilibrium position transitions smoothly (non- jumping) from branch 1 to branch 2 and then moves

left along branch 2. This analysis indicates that the pillar will fracture continuously in a progressive form as the working face moves forward and the distribution of force on the pillar changes. These changes weaken the supporting effect of the rock beam.

3) t384 ( )C k F u′> − i.e.; K is larger than 1. The equilibrium position transitions smoothly from

branch 1 to branch 2. The pillar will fracture in a progressive manner. The details will not be given here because of length limits.

5 Equivalent stiffness and energy gradient curve of the rock beam

Eq.(34) shows that Cook’s stiffness criterion comes into existence at the starting point of pillar failure. The equivalent stiffness of the rock beam in asymmetric mining is:

4 2 27

e 3 3 5 5

( ) 4384 3842 ( )a b a bK Ck l k

a b a b− += =

+ (45)

From Eq.(45) it is seen that as the pillar deviates minutely from the center of the rock beam the relationship between the equivalent stiffness of the rock beam, e 384K Ck= , and the span length, lq / , can be plotted as shown in Fig. 7. Here, we notice that when lba == the equivalent stiffness of the rock beam reaches the maximum emax 384K k= . The value of eK decreases while ba ≠ . When 1.2a l=and 0.8b l= we have from Eq.(18):

24 30.38, 360 255.68384 30.3 255.68 286.06

A BC= =

= + = (46)

Fig. 7 Relationship between equivalent stiffness of the rock beam and asymmetry

The asymmetric position of the pillar not only reduces the equivalent stiffness of the rock beam,

eK , but also reduces the stiffness of the pillar. The combined effect of these two factors is that the dynamic failure strength of the rock-beam pillar system is less than if the pillar were in a symmetrical position.

When lba == the mine layout in Figs. 1 and 2


is symmetrical. It is obvious that when lba == : 1=A , 1=B , 1=C

in Eq.(18). Now all the Eqs. mentioned above can describe pillar failure in asymmetric mining.

The expression for the elastic potential energy gradient of the rock beam with respect to displacement, u, at ax = , the system state variable, is:

e 1360

( 24 ) 24 ( 384 )

Uu Bk

FF Aku Ak F Ckuu

δδ

δδ

= ×

+ + +

(47)

where 0F = when 0u = and so e / 0U uδ δ = . Substituting the initial and end points at j and s

on the )(uF curve in Fig. 3, the system failure condition CkuF 384)( =′ from Eqs.(34) and (39) when put into Eq.(47) give:

e ( ) 0jU u F uδ δ = − < , e ( ) 0jU u F uδ δ = − < (48)

Substituting Eq.(2) into Eq.(47) and dividing by e o o384 K u C ku= gives the gradient expression for

the dimensionless elastic potential energy of the rock beam:

e e

o

2

o o o

o o

o o

1384 360 384

1 exp 2

24 2

exp 24 384

m m

m

m

U u UC ku u B C

u u umk u u u

u uA mk u u

u uA Cu u

δ δ δδ

λ

λ

= =×

− −

+ −

× − + ×

(49)

where ou u u= . When 1.2a l= and 0.8b l=Eq.(46) is / 11 384k Cλ = × for 1=m . Eq.(49) can be plotted using the software MATLAB. This function, e /U uδ δ ~ u , is shown in the plot in Fig. 8. When the deformation of the pillar d1 c( )u u> then

e / 0U uδ δ = in Eq.(47), which corresponds to the zero point in Fig. 8. On the left side of d1u

e / 0U uδ δ > and the curve is above the axis u . This shows that the rock beam is absorbing energy. On the right side of d1u e / 0U uδ δ < and the curve is below the axis u . This shows that the rock beam is releasing energy.

From Eq.(14) we know that for the pillar at the points j and s it is true that:

/ ( ) 0P jU u F uδ δ = > , / ( ) 0P sU u F uδ δ = > (50)

Eq.(2) divided by o384Cku is:

o o

1 exp384

m

PU u uFu C k u u

δ λδ

= = − (51)

Fig. 8 Dimensionless rate of change in the elastic potential energy of the rock beam and in the

energy dissipated by the pillar

The dimensionless expression in Eq.(51) is plotted in Fig. 8 from MATLAB. The )(uF curve is always above the axis u because the pillar absorbs and dissipates energy, deforming until the point of fracture. After turning the )(uF curve around axis u by an angle of 180 it can be easily seen from Eqs. (48) and (50) that the two curves intersect at the points ))(,( jj uFu − and ))(,( ss uFu − . On the

left side of jue /U uδ δ− < uU P δδ so the pillar

deforms to balance the work applied by the external force, i.e.; 0>>>>J ). At the point ju where

e PU u U uδ δ δ δ− = the system is in a critical state.

On the right side of jue PU u U uδ δ δ δ− > , i.e., the

strain energy released by the rock beam is greater than the energy dissipated by the pillar. Therefore, the deformation of the pillar increases significantly without additional applied work from the external force. Thus the initial position of pillar dynamic failure is at the point j in Fig. 3. ( ) 0J u = along the

path continuing until su . The area of the shaded region EΔ in Fig. 8 multiplied by 2

e oK u is EΔ . This is the energy released by the rock beam that exceeds the energy dissipated by the pillar and is given by:

2 2e o o384E K u E Cku EΔ = Δ = Δ (52)

Since 0EΔ < the shadowed area is located under the axis u in Fig. 8. EΔ is transformed into


kinetic energy TΔ , which leads to dynamic failure of the pillar—rock burst. The end point of the pillar dynamic failure is the point s in Fig. 3. There

e /U uδ δ− < ( )F u after point su . So the condition ( ) 0J u > is required for pillar deformation 0uδ > .

This depends on the work applied by the external force. That is to say, the point in Fig. 3 is the stopping point of dynamic failure of the pillar. If the pillar fractures completely the rock beam will lose the support of the pillar and the displacement curve of the rock beam moves to the lower side of that shown in Fig. 4. This is caused by the action of the gravity load and the elastic potential energy of the rock beam increases as it deforms. This is expressed in Fig. 8 where e 0U uδ δ > while d2u u> i.e.; the gradient of the elastic potential energy of the rock beam extends above the axis u where d2 ou u u= .

Fig. 8 illustrates all the mathematical descriptions of pillar failure discussed in Section 4 in a geometric form. Notice that the angles of inclination of ( )F uat the points j and s are equal to each other, i.e.,

j sα α= . Compare this to Fig. 3 where the angles of inclination of ( )F u at j and s are also equal to each other: j sα α= . This intuitively affirms that the mathematical operations from Eqs.(3–52) are accurate and that the reasoning of the article is coherent and consistent. In addition, the area of the shaded region EΔ , expressed by Eq.(41), in Fig. 3 is equal to the energy, EΔ , released by the system upon dynamic failure of the pillar. The latter EΔ is expressed by Eq.(52), which is deduced from the area of the shaded region, EΔ , in Fig. 8. The former represents deformations of the rock beam and the pillar, while the later is represents the deformation, u, of the pillar.

6 Conclusions

1) The equations for the bending moment and the deflection curve of the rock beam in asymmetric mining have been accurately deduced using catastrophe theory. This allows the amplitude and positions of the starting and ending points of pillar failure to be ascertained in mathematical form. The elastic energy released by the rock beam at failure can also be computed.

2) The rate of elastic potential change of the rock beam and the load-displacement relationship of the

pillar can be plotted on a dimensionless scale, as shown in Fig. 8. This plot contains a wealth of information and provides insight into the behavior rules in every phase of deformation in the rock-beam pillar system. The equivalent stiffness of the rock beam in a certain direction can also be determined from the plot.

Acknowledgements

Financial support for this work provided the Natural Science Foundation of Shandong Province of China (Project Y2005-A03) and Educational Committee of Shandong Province of China (Project G04D15), are gratefully acknowledged. Here we express our sincere appreciation to Mr Li Chengjun of Journal of CUMT for his help of revision this article.

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Fold catastrophe model of dynamic pillar failure in asymmetric mining