research article methods to analyze flexural...

Research ArticleMethods to Analyze Flexural Buckling of the ConsequentSlabbed Rock Slope under Top Loading

Hongyan Liu Guihe Wang and Feng Huang

College of Engineering amp Technology China University of Geosciences Beijing Beijing 100083 China

Correspondence should be addressed to Hongyan Liu lhyan1204126com

Received 6 May 2016 Accepted 26 June 2016

Academic Editor Paolo Maria Mariano

Copyright copy 2016 Hongyan Liu et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The consequent slabbed rock slope is prone to flexural buckling failure under its self-weight and top loading However nearly noneof the existing studies consider the effect of the top loading on the slope flexural critical buckling height (CBH) Therefore on thebasis of Eulerrsquos Method and the flexural buckling failure mode of the consequent slabbed rock slope the calculation method of theCBH of the vertical slabbed rock slope under the self-weight is firstly proposed and then it is extended to that of the consequentslabbed rock slope The effect of slope dip angle friction angle and cohesion between the neighboring rock slabs and rock elasticmodulus on the slope CBH is discussed Secondly the calculation method of the CBH of the consequent slabbed rock slope underits self-weight and top loading is proposed according to the superposition principle Finally on the basis of the hypothesis that therock mechanical behavior obeys the statistical damage model the effect of the rock mechanical parameters 119899 and 120576

0on the slope

CBH is studied The results show that the rock strength has much effect on the slope CBH If the rock is supposed to be a linearelastic body without failure in Eulerrsquos Method the result from it is the maximum of the slope CBH

1 Introduction

Rock masses are often intersected by a single set of steeplydipping discontinuities such as regular bedding planes orjoints forming a slab or slabbed structure Meanwhile theloads on the top of the slope are often encountered in thepractical engineering such as the vehicle load in the trans-portation engineering and the building load in the civil engi-neering When the slope surface is parallel to the discontinu-ities under the action of driving force due to the self-weightof rock slabs and top loading failure by buckling may be ini-tiated namely the slabs near the toe of slope buckle graduallyand correspondingly the slabs above the buckle slide along aweak interlayer (Figure 1) In the condition of the length andwidth of the rock slab being far larger than its thickness thebuckling of rock slabs may be simplified as a beam stabilityproblem

Because buckling of rock slabs is a common phenomenonof slopemovements in sedimentary rocks lots of studies havebeen conducted to understand the slip-buckling slope failuremechanism Harrison and Falcon [1] called some naturally

formed buckles roof and wall structures and cascades John-son [2] discussed buckling phenomena related to formationof sheet structures in New England quarry floors Nemcoket al [3] and Radbruch-Hall [4] attributed some buckles toldquocreepsrdquo Kutter [5] briefly discussed buckling in British openpit coal mines Cavers [6] analyzed mechanisms of single-slab buckling for Euler buckles and three hinged buckles forplanar or curved slabs Also on the basis of Eulerrsquos theory Sun[7] obtained the buckling critical height of the rock slab withan energy equilibriumprinciple Hu andCruden [8] regardedthe notion that the modes of buckling were determinedby bedding thickness and accordingly they proposed threebuckling modes Euler buckles block-flexure buckles andblock buckles Pant and Adhikary [9] and Adhikary et al [10]conducted an explicit and implicit finite element numericalstudy on the mechanism of flexural buckling failure of foli-ated rock slopes Qin et al [11] proposed a catastrophe cuspmodel to study the failure mechanisms of the slip-bucklingslope Pereira and Lana [12] analyzed a buckling failure occur-ring in an open pit mine with the numerical method Suppos-ing that the rock element strength obeysWeibull distribution

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 3402547 8 pageshttpdxdoiorg10115520163402547

2 Mathematical Problems in Engineering

model Zhang et al [13] introduced the rock strength intothe calculation of the critical buckling height (CBH) of theconsequent slabbed rock slope Qi et al [14] presented an ana-lytical solution on slip buckling slope failure which fully con-siders both the effect of earthquake and pore water pressurebased on energy equilibrium theory

However it can be found from the existing research thatalthough many researchers have conducted rather profoundwork on the buckling failure of the consequent slabbed rockslope the following aspects are still to be studied further Firstof all the load acting on the slope is not perfectly consideredin the existing studies Although the loads such as self-weightof rock slabs groundwater pressure and the seismic force areconsidered in the models proposed by many researchers [1114] the slope is often subjected to the top loading such as thetransportation and building loads which are not consideredin the calculation of the CBH of the consequent slabbed rockslope Moreover the rock is assumed to be elastic in theexisting studies and cannot fail in strengthwhich is not alwaystrue in the practical engineering Although Zhang et al [13]introduced the rock strength into the calculation of the CBHof the consequent slabbed rock slope they did not take intoaccount the top loading of the slope and the cohesion betweenthe neighboring rock slabs and the buckling deformationdeflection curve equation of the rock slab they adopted is alsounreasonable

Therefore the present paper is aimed at presenting a com-prehensive theoretical model for the flexural buckling of theconsequent slabbed rock slope under top loading Firstlysuppose the rock to be a perfect elastic material the calcula-tion method of the CBH of the vertical slabbed rock slope isproposed and then it is generalized to that of the consequentslabbed rock slope Secondly assume the rock strength obeysWeibull distribution and then the effect of its strengthproperty on the CBH of the vertical slabbed rock slope isdiscussed In all the proposedmethod provides a way to con-sider the effect of rock strength and top loading on theCBHofthe consequent slabbed rock slope

2 Flexural Buckling Model forthe Consequent Slabbed Rock Slope underIts Self-Weight Assuming the Rock toBe a Perfect Elastic Material

21 Flexural BucklingModel for the Vertical Slabbed Rock Slopeunder Its Self-Weight Because the vertical one is a specialcase of the consequent slabbed rock slope the vertical slabbedrock slope is firstly studied and then its result can be gen-eralized to the consequent slabbed rock slope Zhang and Li[15] summarized the failure mode of the vertical slabbed rockslope into the following two namely buckling failure and top-pling failure shown in Figures 2(a) and 2(b) and the formeris only discussed here The classical buckling theory (EulerrsquosMethod) makes the following assumptions in the derivationof the formulae [16] (1) The column is elastic and obeysHookersquos Law (2) The slope of the deflection curve can beapproximated by a linear function (3)The column is weight-less (4) The column is perfectly straight

y

t

px

120572

l

cl

G

f

Figure 1 Buckling failure model for the consequent slabbed rockslope

According to the assumptions above the critical bucklingload is

119875cr =1205872119864119868

(120583119897)2 (1)

where119875cr is the critical buckling loadN119864 is Youngrsquosmodulusof the material Pa 119868 is the moment of inertia m4 119897 is thelength of the column m and 120583 is a constant describing endconditions for pinned ends 120583 = 1

The buckling instability of the vertical slabbed rock slopeis triggered by the gravity of the rock slope and the cohesionand frictional force between the neighboring rock slabs andthe loads above are linear ones along the height of the rockslab Therefore it cannot be solved with the classic bucklingtheory (Eulerrsquos Method) which will otherwise lead to theengineering accident or unnecessary engineering cost

The instability model for the vertical slabbed rock slopeunder its self-weight should satisfy the following hypothesisA only the self-weight of the rock slab and the cohesionbetween the neighboring rock slabs are included here and thetop loading of the rock slab is not consideredB the bottomof the rock slope is embedded into the ground so it can beregarded as the fixed constraintThe buckling failure mode ofthe vertical slabbed rock slope is shown as Figure 2(a) It canbe seen that the upper rock slab of the slope slips down alongthe bedding plane at the initial stage of the slope instabilityand its normal deformation on the bedding plane is restrictedbecause of the neighboring rock slabTherefore the bucklinginstability failure mode of the vertical slabbed rock slope canbe simplified as the mechanical model with the top slidingand bottom fixation constraints shown in Figure 2(c) withthe corresponding coordinate system Assume the depth ofthe rock slab perpendicular to the plane is the unit the criticalbuckling load of the slope can be solved with the energymethod It can be seen from Figure 2(c) that the axial part ofthe rock slab will deviate from the original position namely

Mathematical Problems in Engineering 3

t

l

(a) (b)

y

x

x

dx

(c)

Figure 2 Failure modes of the vertical slabbed rock slope and the pressure bar model (a) Buckling failure (b) toppling failure (c) thecalculation model of the pressure bar

119909 direction Assume V is the deflection of the rock slab itsgeometrical and mechanical boundary conditions are

V|119909=0 = 0

V|119909=119897 = 0

V101584010038161003816100381610038161003816119909=0

= 0

(2)

According to the experiments of rock slab under uniaxialcompression by Sun [7] it is reasonable to assume thedeflection curve is

V = 120575 (1 minus cos 2120587119909119897

) (3)

where 120575 is the constant to determine It is easily verified that(3) satisfies the boundary condition

According to elastic theory [17] the strain energy 119880stored in the rock slab is

119880 = int

119897

0

119864119868

2

(V10158401015840)2

119889119909 =

411986411986812058741205752

1198973

(4)

As shown in Figure 2(c) assume the differential lengthalong the deflection curve is 119889119904 when 119909 increases from 119909 to119909+119889119909 the distance119889120582(119909) that the load above119909-sectionmovesis

119889120582 (119909) = 119889119904 minus 119889119909 = (radic1 + (V1015840)2 minus 1) 119889119909 (5)

Making Taylor extension for (5) and ignoring the high-order items (5) is abbreviated

119889120582 (119909) =

1

2

(V1015840)2

119889119909 (6)

Therefore the displacement 120582(119909) that the rock slabmovesalong 119909-axis is as follows

120582 (119909) = int

119909

0

1

2

(V1015840)2

119889119909 =

12058721205752

1198972(119909 minus

119897

4120587

sin 4120587119909119897

) (7)

The corresponding potential energy119882 of the rock slab is

119882 = int

119897

0

(119902 minus 119888) 120582 (119909) 119889119909 =

12058721205752

2

(119902 minus 119888) (8)

where 119902 = 120574119905 is the gravitational load intensity Nm2 120574 is theunit weight of the rock slab Nm3 119905 is the slab thickness (jointspacing) m and 119888 is the cohesion between the neighboringrock slabs Pa

The overall potential energy of the bending rock slab isΠ = 119880minus119882 and then according to the principle of minimumpotential energy 120597Π120597120575 = 0 we obtain

119875cr = (119902 minus 119888) 119897cr0 =81205872119864119868

1198972

cr0 (9)

where 119875cr is the critical force Nm and 119897cr0 is the CBH of thevertical slabbed rock slope under its self-weight 119868 = (112)1199053is the moment of inertia m4

In order to compare with the result obtained with EulerrsquosMethod (9) can be changed into

(119902 minus 119888) 119897cr0 =1205872119864119868

(035119897cr0)2 (10)

Therefore the CBH 119897cr0 of the vertical slabbed rock slopeunder its self-weight is

119897cr0 =3radic

1205872119864119868

012 (119902 minus 119888)

=3radic

12058721198641199053

144 (120574119905 minus 119888)

(11)

where the meaning of all the parameters is stated as above

22 Flexural Buckling Model for the Consequent Slabbed RockSlope under Its Self-Weight However the completely verticalslabbed rock slope is rare in practical engineering almost allof the slabbed rock slops are consequent or anticonsequent


Therefore in order to make the study result more applicablethe study on the CBH of the consequent slabbed rock slopeshown in Figure 1 is done next Because the self-weight cohe-sion and friction force will all exist in the consequent slabbedrock slope the residual driving force 119865

119909of the rock slab along

the interlayer is

119865119909= 120574119897 sin120572 minus 120574119897 cos120572 sdot tan120601 minus 119888119897 (12)

where 120572 is the slope dip angle (ie the slab dip angle) 120601 isthe friction angle between the neighboring rock slabs and theother parameters are as stated above

Therefore according to the calculation result of the verti-cal slabbed rock slope namely substituting 120574119905 into (11) with120574119905(sin120572minuscos120572sdottan120601) the CBH 119897cr1 of the consequent slabbedrock slope under its self-weight is

119897cr1 =3radic

12058721198641199053

144 [120574119905 (sin120572 minus cos120572 sdot tan120601) minus 119888] (13)

It can be seen that the vertical slabbed rock slope is a specialcase of the consequent one

23 Calculation Examples The calculation example by Xiaoand Yang [18] is discussed here The rock slope is divided bya set of consequent joints and the thickness of the rock slabis 063m The rock slope will be prone to flexural bucklingfailure under its self-weightTheCBH 1198971015840cr of the slope obtainedby Xiao and Yang [18] with Eulerrsquos Method is (ignoring thecohesion between the neighboring rock slabs)

1198971015840

cr =3radic

12058721198641199053

6120574 (sin120572 minus cos120572119905119892120601) (14)

It can be seen that the result obtained from (13) is 161times that obtained from (14) when the cohesion betweenthe neighboring rock slabs is not considered So it can beregarded that the result obtained from (14) is too conser-vative which will lead to the unnecessary engineering costThe reason leading to this result is that Xiao and Yang [18]assumed the gravity of the rock slab to be a concentrated loadacting on its top which will reduce the CBH of the slopeMeanwhile the cohesion between the neighboring rock slabsis also neglected in their study

The effect of the parameters such as the slope dip angle onthe CBH is discussed here Assume 119864 120574 119905 120572 120601 119888 are 10GPa25300Nm3 063m 70∘ 15∘ and 5 kPa respectively and theother parameters remain unchanged when one of them isstudied The variation of the slope CBH with 120572 119888 120601 and 119864obtained from (3) is shown as in Figures 3ndash6 It can be foundthat the slope CBH gradually decreases with increasing theslope dip angle but with the increase in the friction angle andcohesion between the neighboring rock slabs and the rockelastic modulus the slope CBH increases

Criti

cal b

uckl

ing

50 60 70 80 9040Slope dip angle (degree)

110130150170190210

heig

ht (m

)

Figure 3 Variation of slope CBH with the slope dip angle

Criti

cal b

uckl

ing

5 10 15 20 250Friction angle between the neighboring rock slabs (degree)

115

120

125

130

135

heig

ht (m

)

Figure 4Variation of the slopeCBHwith the friction angle betweenthe neighboring rock slabs

3 Flexural Buckling Model forthe Consequent Slabbed Rock Slope underSelf-Weight and Top Loading Assumingthe Rock to Be a Perfect Elastic Material

Although the flexural buckling model for the consequentslabbed rock slope under its self-weight is only discussed inSection 2 there are many other loads such as the transporta-tion and building loads on the slope top in many conditionsTherefore the instability of the consequent slabbed rock slopeunder the slope top loading and its self-weight should bestudied

When only the top loading shown in Figure 1 is consid-ered the CBH of the slope can be solved with EulerrsquosMethodnamely (1) According to boundary condition of the slope 120583should be equal to 07 Therefore the CBH 119897cr2 of the slabbedrock slope is

119897cr2 = radic12058721198641199052

588119901 sin120572 (15)

where 119901 is the top loading intensity of the slope Nm2 andthe other parameters are as stated above

Assume the critical buckling load of the slabbed rockslope under the following three load conditions such as itsself-weight and top loading its self-weight and top loading is119875cr 119875cr1 and 119875cr2 Then we can obtain

119875cr = 119875cr1 + 119875cr2 (16)

If it is assumed that the critical height of the slabbedrock slope under its self-weight and top loading is 119897cr andaccording to Eulerrsquos Method we can obtain

1205872119864119868

(120583119897cr)2=

1205872119864119868

(120583119897cr1)2+

1205872119864119868

(120583119897cr2)2 (17)

Mathematical Problems in Engineering 5Cr

itica

l buc

klin

g

3 6 9 120Cohesion between the neigboring rock slabs (kPa)

100130160190220250

heig

ht (m

)

Figure 5 Variation of the slope CBHwith the cohesion between theneighboring rock slabs

Criti

cal b

uckl

ing

3 6 90Rock elastic modulus (GPa)

55

69

83

97

111

125

heig

ht (m

)

Figure 6 Variation of the slope CBHwith the rock elastic modulus

Then we can obtain

1

1198972

cr=

1

1198972

cr1+

1

1198972

cr2 (18)

Also take the calculation example in Section 23 as anexample the variation of the slope CBH with the top loadingis studiedThe parameters in Section 23 namely119864 = 10GPa120574 = 25300Nm3 119905 = 063m 120572 = 70∘ 120601 = 15∘ and 119888 = 5 kPaare also adopted we can obtain

1

1198972

cr=

119901

711198909

+

1

121892 (19)

It can be seen from Figure 7 that the slope CBH almostlinearly decreases with increasing top loading

4 Flexural Buckling Model for the ConsequentSlabbed Rock Slope regardingthe Rock to Be the Damage Material

41 The Statistical Damage Constitutive Model for a RockWith Eulerrsquos Method the rock is assumed to be a perfect lin-ear elastic body and it does not fail So only the deformationalparameter such as the rock elastic modulus is consideredwhile its strength one is not However the rock is a kind ofnatural damage geological body and it contains many ran-domly distributed microcracks which will have effect on therock strengthThe statistical damage mechanics is a powerfultool to study the occurrence propagation and coalescenceprocesses of these microcracks and their effect on rockmechanical behaviors By means of it the distribution law ofthese microcracks in rock such as normal or Weibull distri-bution is assumed so as to build up the mesoscopic elementswith strength in a rock and to determine its damage state

80

90

100

110

120

130

Criti

cal b

uckl

ing

heig

ht (m

)

02 04 060Pressure acting on the slope top (MPa)

Figure 7 Variation of the slope CBHwith the load on the top of theslope

Thus a damage statistical constitutivemodel for a rock can beset up Till nowmuch progress in the study of damage statisti-cal constitutivemodels has beenmade [19ndash22]The establish-ment of a rock damage statistical constitutivemodel is mainlybased on the following two aspects (1) choose the strengthcriteria for the rock mesoscopic element for instance themaximum principle strain criterion Mohr-Coulomb crite-rion and Drucker-Prager criterion and (2) determine thedistribution law of the rock mesoscopic element strength forexample power function distribution and Weibull distribu-tion The studies show that the damage constitutive modelbased on Weibull distribution is better than that based onpower function distribution and its calculation process iseasier Therefore the damage constitutive model based onWeibull distribution and the maximum principle straincriterion is adopted here

The strength of mesoscopic elements obeys the followingWeibull distribution function [19]

119875 (120576) =

119899

1205760

(

120576

1205760

)

119899minus1

exp [minus( 1205761205760

)

119899

] 120576 gt 0

0 120576 le 0

(20)

where 120576 is an elemental strength parameter or stress level andbecause the strain strength theory is adopted here it denotesstrain 119899 and 120576

0are the distribution parameters which can

be obtained by fitting with the test stress-strain curve of therock and 119875(120576) is the percentage of damaged ones out of thetotal number of the mesoscopic elements in the rock

Therefore the damage119863 of the rock can be defined as

119863 = 1 minus

119864

1198640

= 1 minus exp [minus( 1205761205760

)

119899

] (21)

where 119863 takes a value between 0 and 1 corresponding todamage states of the rock from undamaged to fully damaged1198640and 119864 are the elastic moduli of the rock without any

damage and with some damage respectivelyAssume the mechanical behavior of the rock mesoscopic

elements obeys Hooke law its constitutive law is

120590 = 119864120576 (1 minus 119863) (22)

where 120590 and 120576 are the stress and strain for the rock respec-tively


Criti

cal b

uckl

ing

70

80

90

100

110

heig

ht (m

)

05 1 15 20n

Figure 8 Variation of the slope CBH with 119899 (1205760= 002)

Criti

cal b

uckl

ing

70

80

90

100

110he

ight

(m)

00005 000101205760

Figure 9 Variation of the slope CBH with 1205760(119899 = 2)

42 Flexural Buckling Model for the Consequent Slabbed RockSlope under Its Self-Weight and Top Loading Based on the RockStatistical Damage Constitutive Model As stated above therock is a kind of natural damage geological body in whichthe damage will increase even under little load Therefore itis unsuitable to assume the rock to be the linear elastic bodywithout any damage

So in order to discuss the effect of the rock damage onthe CBH of the slope the flexural buckling model for theconsequent slabbed rock slope based on Weibull statisticaldamage constitutive model for the rock is set up From (21)it can be known that the rock damage and its evolution aremainly reflected by the variation of the rock elastic modulusTherefore substituting (21) into (18) leads to

1

1198972

cr=

1

3radic(120587211990531198640exp [minus (120576120576

0)119899] 144 [120574119905 (sin120572 minus cos120572 sdot tan120601) minus 119888])

2

+

1

12058721198640exp [minus (120576120576

0)119899] 119905248119901 sin120572

(23)

where

120576 =

119901 sin120572 + 120574119905 (sin120572 minus cos120572 sdot tan120601) minus 119888119864

(24)

Because when the buckling failure of the slope occurs therock damage is rather little so 119864 can be replaced with 119864

0

namely

120576 asymp


(25)

The effect of 119899 and 1205760on the CBHof the slope is discussed

The parameters in Section 23 namely 119864 = 10GPa 120574 =

25300Nm3 119905 = 063m 120572 = 70∘ 120601 = 15∘ and 119888 = 5 kPaare also adopted at the same time the parameters of 119899 and 120576

0

are assumed to be 20 and 002 respectively by experience Ifwe assume the pressure on the top of the slope is 02MPa theCBH of the slope is 10503m with (23) By comparison thecorresponding result is 10234m when the rock is assumed tobe a perfect elastic body If ignoring the calculation error they

are assumed to be equalThe effects of 119899 and 1205760on the CBH of

the slope are discussed It can be seen from Figure 8 that theCBH of the slope increases from 7101m to 10503m when 119899increases from 0 to 2 which indicates that 119899 has much effecton the calculation result However from the increase extentthe CBH of the slope increases much more when 119899 increasesfrom0 to 05 and then it increases rather less It shows that theeffect of 119899 on the calculation result is little when it increasesto a certain value and the similar result can also be found inFigure 9

In order to illustrate the effect of 119899 and 1205760on the rock

stress-strain curve the results are given in Figures 10 and 11 Itcan be seen that the rock elastic modulus and climax strengthwill both vary with them and in some cases the variationof the rock strength is more obvious Therefore it can beregarded that the effect of the rock strength on the CBHof theslope ismuchWhen the rock is assumed to be a perfect elasticbody namely its climax strength is infinite the CBH of theslopewill reach itsmaximumwhich can be seen fromFigures8 and 9 So the rock strength and deformational behavior


0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

n = 1

n = 05

n = 2

n = 15

Figure 10 Variation of the rock stress-strain with 119899 at 1205760= 002

0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

1205760 = 001

1205760 = 0005

1205760 = 002

1205760 = 0015

Figure 11 Variation of the rock stress-strain with 1205760at 119899 = 2

should be considered at the same time in order to accuratelyobtain the CBH of the slope

5 Conclusions

Assuming the rock to be a perfect elastic material we firstlyinvestigate the calculation method of the CBH of the verticalslabbed rock slope under its self-weight based on the energyprinciple and then extend it to that of the consequent slabbedrock slope The effect of slope dip angle friction angle andcohesion between the neighboring rock slabs and rock elasticmodulus on the slope CBH is discussed It can be seen thatthe slope CBH gradually decreases with increasing the slopedip angle but it will increase with the increase in the frictionangle and cohesion between the neighboring rock slabs andthe rock elastic modulus

According to the superposition principle of EulerrsquosMethod we deduce the calculation method of the CBH ofthe consequent slabbed rock slope under its self-weight andtop loading Meanwhile the effect of top loading on the CBHof the consequent slabbed rock slope is also discussed It canbe seen that the slope CBH almost linearly decreases withincreasing top loading

Thirdly assuming the rockmechanical behavior obeys thestatistical damage model we establish the corresponding cal-culation method of the CBH of the consequent slabbed rockslope under its self-weight and top loading We also discussthe effect of the rock strength characteristic parameters 119899 and

1205760on the slope CBH and the results show that their effect on

the slope CBH is large when 119899 and 1205760are little and then when

they increase to a certain value their effect will become verylittle Overall the proposed method provides a way to calcu-late the CBH of the consequent slabbed rock slope under itsself-weight and top loading

Finally it is noted that the proposed method is moresuitable to the case that the top loading cannot be ignoredcomparingwith the self-weight of the rock slope for examplethere are heavy building and transportation load on the top ofthe rock slope

Competing Interests

The authors declare that they have no competing interests

Acknowledgments

This study is supported by ldquothe Fundamental ResearchFunds for the Central Universities (26520140192652015263)of Chinardquo and ldquoChina Scholarship Funds (2016)rdquo

References

[1] J V Harrison and N L Falcon ldquoGravity collapse structures andmountain ranges as exemplified in south-western Iranrdquo Quar-terly Journal of the Geological Society of London vol 92 no 1ndash4pp 91ndash102 1936

[2] A M Johnson Physical Processes in Geology Freeman Cooperamp Co San Francisco Calif USA 1970

[3] A Nemcok J Pasek and J Rybar ldquoClassification of landslidesand other mass movementsrdquo Rock mechanics vol 4 no 2 pp71ndash78 1972

[4] D H Radbruch-Hall ldquoGravitational creep of rock masses onslopesrdquo in Rockslides and Avalanches 1 Natural Phenomena BVoight Ed pp 607ndash657 Elsevier New York NY USA 1978

[5] H K Kutter ldquoMechanisms of slope failure other than pureslidingrdquo in Rock Mechanics International Center for MechanicalSciences Courses and Lectures L Muller Ed vol 165 SpringerNew York NY USA 1974

[6] D S Cavers ldquoSimple methods to analyze buckling of rockslopesrdquo Rock Mechanics and Rock Engineering vol 14 no 2 pp87ndash104 1981

[7] G Z Sun Rock Mass Structure Mechanics Science PressBeijing China 1988 1988 (Chinese)

[8] X-Q Hu and D M Cruden ldquoBuckling deformation in theHighwood Pass Alberta Canadardquo Canadian Geotechnical Jour-nal vol 30 no 2 pp 276ndash286 1993

[9] S R Pant and D P Adhikary ldquoTechnical note implicit andexplicit modelling of flexural buckling of foliated rock slopesrdquoRockMechanics and Rock Engineering vol 32 no 2 pp 157ndash1641999

[10] D P Adhikary H-BMuhlhaus andA V Dyskin ldquoA numericalstudy of flexural buckling of foliated rock slopesrdquo InternationalJournal for Numerical and Analytical Methods in Geomechanicsvol 25 no 9 pp 871ndash884 2001

[11] S Q Qin J J Jiao and S J Wang ldquoA cusp catastrophe modelof instability of slip-buckling sloperdquo Rock Mechanics and RockEngineering vol 34 no 2 pp 119ndash134 2001


[12] L C Pereira and M S Lana ldquoStress-strain analysis of bucklingfailure in phyllite slopesrdquoGeotechnical and Geological Engineer-ing vol 31 no 1 pp 297ndash314 2013

[13] L M Zhang S R Lv J H Zhang and H Y Liu ldquoInstabilityanalysis of bedding rock slope based on the statistical constitu-tive damage modelrdquo Geotechnical Investigation amp Survey no 9pp 7ndash29 2014 (Chinese)

[14] S W Qi H X Lan and J Y Dong ldquoAn analytical solution toslip buckling slope failure triggered by earthquakerdquo EngineeringGeology vol 194 pp 4ndash11 2015

[15] T J Zhang and Y P Li ldquoLinear viscoelasticity stability analysisof bluff rock sloperdquoMechanics in Engineering vol 25 no 6 pp51ndash54 2003 (Chinese)

[16] A Chajes Principles of Structural Stability Theory Civil Engi-neering and Engineering Mechanics Series Prentice HallEnglewood Cliffs NJ USA 1974

[17] D O Brush and B P Almroth Buckling of Bars Plates andShells McGraw-Hill New York NY USA 1975

[18] S F Xiao and S B YangRockMassMechanics Geological PressBeijing China 1987 (Chinese)

[19] W Weibull ldquoA statistical distribution function of wide applica-bilityrdquo Journal of Applied Mechanics vol 18 pp 293ndash297 1951

[20] D Krajcinovic and M A G Silva ldquoStatistical aspects of thecontinuous damage theoryrdquo International Journal of Solids andStructures vol 18 no 7 pp 551ndash562 1982

[21] Z-L Wang Y-C Li and J G Wang ldquoA damage-softening sta-tistical constitutive model considering rock residual strengthrdquoComputers amp Geosciences vol 33 no 1 pp 1ndash9 2007

[22] H Y Liu and X P Yuan ldquoA damage constitutive model for rockmass with persistent joints considering joint shear strengthrdquoCanadian Geotechnical Journal vol 52 no 8 pp 1136ndash11432015

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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Applied MathematicsJournal of


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Journal of


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OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


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Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


model Zhang et al [13] introduced the rock strength intothe calculation of the critical buckling height (CBH) of theconsequent slabbed rock slope Qi et al [14] presented an ana-lytical solution on slip buckling slope failure which fully con-siders both the effect of earthquake and pore water pressurebased on energy equilibrium theory

However it can be found from the existing research thatalthough many researchers have conducted rather profoundwork on the buckling failure of the consequent slabbed rockslope the following aspects are still to be studied further Firstof all the load acting on the slope is not perfectly consideredin the existing studies Although the loads such as self-weightof rock slabs groundwater pressure and the seismic force areconsidered in the models proposed by many researchers [1114] the slope is often subjected to the top loading such as thetransportation and building loads which are not consideredin the calculation of the CBH of the consequent slabbed rockslope Moreover the rock is assumed to be elastic in theexisting studies and cannot fail in strengthwhich is not alwaystrue in the practical engineering Although Zhang et al [13]introduced the rock strength into the calculation of the CBHof the consequent slabbed rock slope they did not take intoaccount the top loading of the slope and the cohesion betweenthe neighboring rock slabs and the buckling deformationdeflection curve equation of the rock slab they adopted is alsounreasonable

Therefore the present paper is aimed at presenting a com-prehensive theoretical model for the flexural buckling of theconsequent slabbed rock slope under top loading Firstlysuppose the rock to be a perfect elastic material the calcula-tion method of the CBH of the vertical slabbed rock slope isproposed and then it is generalized to that of the consequentslabbed rock slope Secondly assume the rock strength obeysWeibull distribution and then the effect of its strengthproperty on the CBH of the vertical slabbed rock slope isdiscussed In all the proposedmethod provides a way to con-sider the effect of rock strength and top loading on theCBHofthe consequent slabbed rock slope

2 Flexural Buckling Model forthe Consequent Slabbed Rock Slope underIts Self-Weight Assuming the Rock toBe a Perfect Elastic Material

21 Flexural BucklingModel for the Vertical Slabbed Rock Slopeunder Its Self-Weight Because the vertical one is a specialcase of the consequent slabbed rock slope the vertical slabbedrock slope is firstly studied and then its result can be gen-eralized to the consequent slabbed rock slope Zhang and Li[15] summarized the failure mode of the vertical slabbed rockslope into the following two namely buckling failure and top-pling failure shown in Figures 2(a) and 2(b) and the formeris only discussed here The classical buckling theory (EulerrsquosMethod) makes the following assumptions in the derivationof the formulae [16] (1) The column is elastic and obeysHookersquos Law (2) The slope of the deflection curve can beapproximated by a linear function (3)The column is weight-less (4) The column is perfectly straight

y

t

px

120572

l

cl

G

f

Figure 1 Buckling failure model for the consequent slabbed rockslope

According to the assumptions above the critical bucklingload is

119875cr =1205872119864119868

(120583119897)2 (1)

where119875cr is the critical buckling loadN119864 is Youngrsquosmodulusof the material Pa 119868 is the moment of inertia m4 119897 is thelength of the column m and 120583 is a constant describing endconditions for pinned ends 120583 = 1

The buckling instability of the vertical slabbed rock slopeis triggered by the gravity of the rock slope and the cohesionand frictional force between the neighboring rock slabs andthe loads above are linear ones along the height of the rockslab Therefore it cannot be solved with the classic bucklingtheory (Eulerrsquos Method) which will otherwise lead to theengineering accident or unnecessary engineering cost

The instability model for the vertical slabbed rock slopeunder its self-weight should satisfy the following hypothesisA only the self-weight of the rock slab and the cohesionbetween the neighboring rock slabs are included here and thetop loading of the rock slab is not consideredB the bottomof the rock slope is embedded into the ground so it can beregarded as the fixed constraintThe buckling failure mode ofthe vertical slabbed rock slope is shown as Figure 2(a) It canbe seen that the upper rock slab of the slope slips down alongthe bedding plane at the initial stage of the slope instabilityand its normal deformation on the bedding plane is restrictedbecause of the neighboring rock slabTherefore the bucklinginstability failure mode of the vertical slabbed rock slope canbe simplified as the mechanical model with the top slidingand bottom fixation constraints shown in Figure 2(c) withthe corresponding coordinate system Assume the depth ofthe rock slab perpendicular to the plane is the unit the criticalbuckling load of the slope can be solved with the energymethod It can be seen from Figure 2(c) that the axial part ofthe rock slab will deviate from the original position namely


t

l

(a) (b)

y

x

x

dx

(c)



V|119909=0 = 0

V|119909=119897 = 0

V101584010038161003816100381610038161003816119909=0

= 0

(2)


V = 120575 (1 minus cos 2120587119909119897

) (3)



119880 = int

119897

0

119864119868

2

(V10158401015840)2

119889119909 =

411986411986812058741205752

1198973

(4)




119889120582 (119909) =

1

2

(V1015840)2

119889119909 (6)


120582 (119909) = int

119909

0

1

2

(V1015840)2

119889119909 =

12058721205752

1198972(119909 minus

119897

4120587

sin 4120587119909119897

) (7)


119882 = int

119897

0

(119902 minus 119888) 120582 (119909) 119889119909 =

12058721205752

2

(119902 minus 119888) (8)



119875cr = (119902 minus 119888) 119897cr0 =81205872119864119868

1198972

cr0 (9)



(119902 minus 119888) 119897cr0 =1205872119864119868

(035119897cr0)2 (10)


119897cr0 =3radic

1205872119864119868

012 (119902 minus 119888)

=3radic

12058721198641199053

144 (120574119905 minus 119888)

(11)






the interlayer is




119897cr1 =3radic

12058721198641199053




1198971015840

cr =3radic

12058721198641199053

6120574 (sin120572 minus cos120572119905119892120601) (14)



Criti

cal b

uckl

ing


110130150170190210

heig

ht (m

)


Criti

cal b

uckl

ing


115

120

125

130

135

heig

ht (m

)





119897cr2 = radic12058721198641199052

588119901 sin120572 (15)



119875cr = 119875cr1 + 119875cr2 (16)


1205872119864119868

(120583119897cr)2=

1205872119864119868

(120583119897cr1)2+

1205872119864119868

(120583119897cr2)2 (17)


itica

l buc

klin

g


100130160190220250

heig

ht (m

)


Criti

cal b

uckl

ing


55

69

83

97

111

125

heig

ht (m

)


Then we can obtain

1

1198972

cr=

1

1198972

cr1+

1

1198972

cr2 (18)


1

1198972

cr=

119901

711198909

+

1

121892 (19)




80

90

100

110

120

130

Criti

cal b

uckl

ing

heig

ht (m

)





119875 (120576) =

119899

1205760

(

120576

1205760

)

119899minus1

exp [minus( 1205761205760

)

119899

] 120576 gt 0

0 120576 le 0

(20)





119863 = 1 minus

119864

1198640

= 1 minus exp [minus( 1205761205760

)

119899

] (21)




120590 = 119864120576 (1 minus 119863) (22)



Criti

cal b

uckl

ing

70

80

90

100

110

heig

ht (m

)

05 1 15 20n


Criti

cal b

uckl

ing

70

80

90

100

110he

ight

(m)

00005 000101205760




1

1198972

cr=

1

3radic(120587211990531198640exp [minus (120576120576


2

+

1

12058721198640exp [minus (120576120576

0)119899] 119905248119901 sin120572

(23)

where

120576 =


(24)


0

namely

120576 asymp


(25)




0







0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

n = 1

n = 05

n = 2

n = 15


0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

1205760 = 001

1205760 = 0005

1205760 = 002

1205760 = 0015



5 Conclusions








Competing Interests


Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










t

l

(a) (b)

y

x

x

dx

(c)



V|119909=0 = 0

V|119909=119897 = 0

V101584010038161003816100381610038161003816119909=0

= 0

(2)


V = 120575 (1 minus cos 2120587119909119897

) (3)



119880 = int

119897

0

119864119868

2

(V10158401015840)2

119889119909 =

411986411986812058741205752

1198973

(4)




119889120582 (119909) =

1

2

(V1015840)2

119889119909 (6)


120582 (119909) = int

119909

0

1

2

(V1015840)2

119889119909 =

12058721205752

1198972(119909 minus

119897

4120587

sin 4120587119909119897

) (7)


119882 = int

119897

0

(119902 minus 119888) 120582 (119909) 119889119909 =

12058721205752

2

(119902 minus 119888) (8)



119875cr = (119902 minus 119888) 119897cr0 =81205872119864119868

1198972

cr0 (9)



(119902 minus 119888) 119897cr0 =1205872119864119868

(035119897cr0)2 (10)


119897cr0 =3radic

1205872119864119868

012 (119902 minus 119888)

=3radic

12058721198641199053

144 (120574119905 minus 119888)

(11)






the interlayer is




119897cr1 =3radic

12058721198641199053




1198971015840

cr =3radic

12058721198641199053

6120574 (sin120572 minus cos120572119905119892120601) (14)



Criti

cal b

uckl

ing


110130150170190210

heig

ht (m

)


Criti

cal b

uckl

ing


115

120

125

130

135

heig

ht (m

)





119897cr2 = radic12058721198641199052

588119901 sin120572 (15)



119875cr = 119875cr1 + 119875cr2 (16)


1205872119864119868

(120583119897cr)2=

1205872119864119868

(120583119897cr1)2+

1205872119864119868

(120583119897cr2)2 (17)


itica

l buc

klin

g


100130160190220250

heig

ht (m

)


Criti

cal b

uckl

ing


55

69

83

97

111

125

heig

ht (m

)


Then we can obtain

1

1198972

cr=

1

1198972

cr1+

1

1198972

cr2 (18)


1

1198972

cr=

119901

711198909

+

1

121892 (19)




80

90

100

110

120

130

Criti

cal b

uckl

ing

heig

ht (m

)





119875 (120576) =

119899

1205760

(

120576

1205760

)

119899minus1

exp [minus( 1205761205760

)

119899

] 120576 gt 0

0 120576 le 0

(20)





119863 = 1 minus

119864

1198640

= 1 minus exp [minus( 1205761205760

)

119899

] (21)




120590 = 119864120576 (1 minus 119863) (22)



Criti

cal b

uckl

ing

70

80

90

100

110

heig

ht (m

)

05 1 15 20n


Criti

cal b

uckl

ing

70

80

90

100

110he

ight

(m)

00005 000101205760




1

1198972

cr=

1

3radic(120587211990531198640exp [minus (120576120576


2

+

1

12058721198640exp [minus (120576120576

0)119899] 119905248119901 sin120572

(23)

where

120576 =


(24)


0

namely

120576 asymp


(25)




0







0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

n = 1

n = 05

n = 2

n = 15


0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

1205760 = 001

1205760 = 0005

1205760 = 002

1205760 = 0015



5 Conclusions








Competing Interests


Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












the interlayer is




119897cr1 =3radic

12058721198641199053




1198971015840

cr =3radic

12058721198641199053

6120574 (sin120572 minus cos120572119905119892120601) (14)



Criti

cal b

uckl

ing


110130150170190210

heig

ht (m

)


Criti

cal b

uckl

ing


115

120

125

130

135

heig

ht (m

)





119897cr2 = radic12058721198641199052

588119901 sin120572 (15)



119875cr = 119875cr1 + 119875cr2 (16)


1205872119864119868

(120583119897cr)2=

1205872119864119868

(120583119897cr1)2+

1205872119864119868

(120583119897cr2)2 (17)


itica

l buc

klin

g


100130160190220250

heig

ht (m

)


Criti

cal b

uckl

ing


55

69

83

97

111

125

heig

ht (m

)


Then we can obtain

1

1198972

cr=

1

1198972

cr1+

1

1198972

cr2 (18)


1

1198972

cr=

119901

711198909

+

1

121892 (19)




80

90

100

110

120

130

Criti

cal b

uckl

ing

heig

ht (m

)





119875 (120576) =

119899

1205760

(

120576

1205760

)

119899minus1

exp [minus( 1205761205760

)

119899

] 120576 gt 0

0 120576 le 0

(20)





119863 = 1 minus

119864

1198640

= 1 minus exp [minus( 1205761205760

)

119899

] (21)




120590 = 119864120576 (1 minus 119863) (22)



Criti

cal b

uckl

ing

70

80

90

100

110

heig

ht (m

)

05 1 15 20n


Criti

cal b

uckl

ing

70

80

90

100

110he

ight

(m)

00005 000101205760




1

1198972

cr=

1

3radic(120587211990531198640exp [minus (120576120576


2

+

1

12058721198640exp [minus (120576120576

0)119899] 119905248119901 sin120572

(23)

where

120576 =


(24)


0

namely

120576 asymp


(25)




0







0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

n = 1

n = 05

n = 2

n = 15


0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

1205760 = 001

1205760 = 0005

1205760 = 002

1205760 = 0015



5 Conclusions








Competing Interests


Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










itica

l buc

klin

g


100130160190220250

heig

ht (m

)


Criti

cal b

uckl

ing


55

69

83

97

111

125

heig

ht (m

)


Then we can obtain

1

1198972

cr=

1

1198972

cr1+

1

1198972

cr2 (18)


1

1198972

cr=

119901

711198909

+

1

121892 (19)




80

90

100

110

120

130

Criti

cal b

uckl

ing

heig

ht (m

)





119875 (120576) =

119899

1205760

(

120576

1205760

)

119899minus1

exp [minus( 1205761205760

)

119899

] 120576 gt 0

0 120576 le 0

(20)





119863 = 1 minus

119864

1198640

= 1 minus exp [minus( 1205761205760

)

119899

] (21)




120590 = 119864120576 (1 minus 119863) (22)



Criti

cal b

uckl

ing

70

80

90

100

110

heig

ht (m

)

05 1 15 20n


Criti

cal b

uckl

ing

70

80

90

100

110he

ight

(m)

00005 000101205760




1

1198972

cr=

1

3radic(120587211990531198640exp [minus (120576120576


2

+

1

12058721198640exp [minus (120576120576

0)119899] 119905248119901 sin120572

(23)

where

120576 =


(24)


0

namely

120576 asymp


(25)




0







0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

n = 1

n = 05

n = 2

n = 15


0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

1205760 = 001

1205760 = 0005

1205760 = 002

1205760 = 0015



5 Conclusions








Competing Interests


Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Criti

cal b

uckl

ing

70

80

90

100

110

heig

ht (m

)

05 1 15 20n


Criti

cal b

uckl

ing

70

80

90

100

110he

ight

(m)

00005 000101205760




1

1198972

cr=

1

3radic(120587211990531198640exp [minus (120576120576


2

+

1

12058721198640exp [minus (120576120576

0)119899] 119905248119901 sin120572

(23)

where

120576 =


(24)


0

namely

120576 asymp


(25)




0







0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

n = 1

n = 05

n = 2

n = 15


0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

1205760 = 001

1205760 = 0005

1205760 = 002

1205760 = 0015



5 Conclusions








Competing Interests


Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

n = 1

n = 05

n = 2

n = 15


0

20

40

60

80

100

Stre

ss (M

Pa)

001 002 003 004000Strain

1205760 = 001

1205760 = 0005

1205760 = 002

1205760 = 0015



5 Conclusions








Competing Interests


Acknowledgments


References































Volume 2014




Journal of











Journal of


Function Spaces






Algebra




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article methods to analyze flexural...

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