CN109885980A - Determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference - Google Patents
Determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference Download PDFInfo
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Abstract
Determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference the invention discloses a kind of, S1, setting joint thin layer microscopic element body load-bearing enter the linear elasticity stage, joint thin layer microscopic element body is isotropic continuous media, microscopic element body is converted into the state of damaging by nondestructive state and instantaneously completes, and process is irreversible;S2, it is based on Weibull distribution function, defines outer lotus threshold values F*, obtain the failure probability density function of microscopic element body;S3, the Statistical Damage Constitutive Model at joint under shear action is obtained using the combine analog microscopic element body load-bearing situation of spring and friction plate based on rheological model mechanical elements;S4, become damage after damaging according to rock-soil material load-bearing and do not damage two parts, determine outer lotus threshold values F*, obtain the shear-deformable Damage Constitutive Model of damage evolution model and joint at shear-deformable process joint;S5, shear-deformable Damage Constitutive Model parameter m, u in the joint is determined0、us;S6, the correctness for verifying constitutive model.
Description
Technical field
The invention belongs to the technical fields of rock shearing face characteristic, and in particular to a kind of to determine yield point based on stress difference
Shear Complete Damage Process constitutive model in joint.
Background technique
Natural rock mass experienced complicated geologic process in forming process, inside there are distinct all kinds of sections
Reason, the failure mechanism of engineering rock mass is largely controlled by joint, and the mechanical property at joint is mainly by its shearing strength control
System, the shear property research to joint are key scientific problems and hot issue in ROCK MECHANICS RESEARCH.All the time, joint
The emphasis of shearing principle research is all to establish one to can accurately reflect joint mechanical characteristic and shear the constitutive model of overall process.State
Inside and outside scholar has carried out a large number of experiments analysis and theoretical research for joint shear property, proposes some classical constitutive models and retouches
State the shearing mechanics deformation characteristic at joint.For example, Bandis[1]According to direct shear test as a result, proposing to be fitted using hyperbolic function
Shear stress-the displacement curve in peak value proparea.S.Saeb and B.Amadei[2]Shear stress-displacement the overall process at joint is analyzed,
It proposes purely linear shearing constitutive model, force-displacement relationship is answered using piecewise linear function description.G.Grasselli[3,4]To 37
Group joint direct shear test result is analyzed, it is believed that is partially used linear function fit before peak, is partially used hyperbola letter behind peak
Number fitting.In fact, a large amount of Keeping Shear Experiment Data[4]Shear stress-the displacement relation for showing pre-and post-peaking is not simple line
Sexual intercourse.Since the 1970s, the rise of damage mechanics provides new approaches for research joint shear property.Damage force
It learns and regards existing natural microfissures a large amount of in rock as a kind of damage[5,6], numerous scholars[7-13]Rock constitutive model is carried out
Research.C.S.Desai[14]From the random distribution of defect, this structure mould is sheared based on the DSC that defect theory establishes joint
Type.Wang Z L[15]It was found that microfissure is a continuous process from fracture is damaged to, in relatively Drucker-Prage criterion
After Mohr Coulomb's criteria, propose it is a kind of based on unit intensity follow Weibull distribution damage of rock softening count this structure
Model.Simon[16]It was found that joint it is shear-deformable reach will appear nonlinear softening phenomenon after peak value, and it is bent using exponential function
Line is described.
Above for the research in terms of the shear property model constitutive model of joint, defect theory is effectively promoted in rock mechanics
In development.However, existing constitutive model mainly for the purpose of with trial curve adhered shape, imports parameter excessively and object
It is true to manage interrogatory, it is relatively deficient for the research of Joint Element mechanics response pattern, joint is sheared before peak and post-peak deformation
The research carried out as a whole is even more to be rarely reported.It is as follows above with reference to document:
[1]Bandis SC,Lumsden AC,Barton NR.Fundamentals of rock joint
deformation[J].International Journal of Rock Mechanics&Mining Sciences&
Geomechanics Abstracts,1983,20(6):249-268.
[2]Saeb S.Modelling of rock joints under shear and normal loading:
Saeb,S;Amadei,B Int J Rock Mech Min SciV29,N3,May 1992,P267–278[J]
.International Journal of Rock Mechanics&Mining Sciences&Geomechanics
Abstracts,1992,29(6):344.
[3]Grasselli G,Egger P.Constitutive law for the shear strength of
rock joints based on three-dimensional surface parameters[J].International
Journal of Rock Mechanics&Mining Sciences,2003,40(1):25-40.
[4]GRASSELLI.Shear strength of rock joints based on quantified
surface description[J].Rock Mechanics&Rock Engineering,2006,39(4):295.
[5]Lemaitre J.How to use damage mechanics☆[J].Nuclear Engineering&
Design,1984,80(2):233-245.
[6] thank to peaceful rock-concrete damage mechanics [M]: publishing house, China Mining University;1990.
[7]Kachanov ML.A microcrack model of rock inelasticity part I:
Frictional sliding on microcracks[J].Mechanics of Materials,1982,1(1):19-27.
[8]Kachanov ML,Kachanov ML.A microcrack model of rock inelasticity
part II:Propagation of microcracks[J].Mechanics of Materials,1982,1(1):29-41.
[9] Xu Jingnan, Zhu Weishen compression-shear stress act on mechanical characteristic-fracture damage EVOLUTION EQUATION of lower Jointed rock masses
And verification experimental verification [J] rock-soil mechanics, 1994, (2): 1-12.
[10] the CT dynamic test of three axis mesoscopic numerical test rule of Ge Xiurun, Ren Jianxi, Pu Yibin, et al. coal petrography
[J] Chinese Journal of Rock Mechanics and Engineering, 1999, (05): 497-502.
[11] Ren Jianxi, Ge Xiurun uniaxial compression damage of rock, which develop, carefully sees mechanism and its Constitutive Models Study [J] rock
Stone mechanics and engineering journal, 2001,20 (4): 425-425.
[12] Du Xiuli, Huang Jingqi, Jin Liu, et al. rock the research of D elastic-plastic Damage Constitutive Model [J] ground work
Journey journal, 2017,39 (6): 978-985.
[13] Ling Jianming, damage Mechanical Analysis [J] the rock mechanics of Jiang's rank of nobility light intermittent fracture rock mass mechanics characteristic with
Engineering journal, 1992,11 (4): 373-373.
[14]Desai CS,Ma Y.Modeling of joints and interfaces using the
disturbed state concept[J].International Journal for Numerical&Analytical
Methods in Geomechanics,2010,16(9):623-653.
[15]Wang ZL,Li YC,Wang JG.A damage-softening statistical constitutive
model considering rock residual strength[J].Computers&Geosciences,2007,33(1):
1-9.
[16]Simon R,Aubertin M,Mitri HS.A non-linear constitutive model for
rock joints to evaluate unstable slip[J].1999.
Summary of the invention
It is an object of the invention to be directed to above-mentioned deficiency in the prior art, one kind is provided based on stress difference and determines yield point
Joint shear Complete Damage Process constitutive model, it is above-mentioned to solve the problems, such as or improve.
In order to achieve the above objectives, the technical solution adopted by the present invention is that:
It is a kind of to determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference comprising:
S1, setting joint thin layer microscopic element body load-bearing enter the linear elasticity stage, and joint thin layer microscopic element body is each
To the continuous media of the same sex, microscopic element body is converted into the state of damaging by nondestructive state and instantaneously completes, and process is irreversible;
S2, it is based on Weibull distribution function, defines outer lotus threshold values F*, obtain the failure probability density letter of microscopic element body
Number;
S3, based on the mechanical elements in rheological model, list is carefully seen using the combine analog joint thin layer of spring and friction plate
First body load-bearing situation, obtains the Statistical Damage Constitutive Model at joint under shear action;
S4, become damage after damaging according to rock-soil material load-bearing and do not damage two parts, determine outer lotus threshold values F*, obtain
The shear-deformable Damage Constitutive Model of damage evolution model and joint at shear-deformable process joint;
S5, the shear-deformable Damage Constitutive Model form parameter m in the joint, scale parameter u are determined0It is displaced and joins with yield point
Number us;
S6, several joint shear test datas are substituted into the shear-deformable Damage Constitutive Model in joint, the section verified
Managing shear-deformable Damage Constitutive Model accurately can shear full deformation process in simulation joint, and reflect the stage of shear history
Feature.
Preferably, the method for probability density function is destroyed in step S2 are as follows:
It is greater than threshold values F when a certain stage micro unit institute is loaded*When, rock material damages, and microscopic element body is broken
Bad probability density function P (F) are as follows:
Wherein, m, F0Respectively Weibull profile shape parameter and scale parameter.
Preferably, in step S3 joint Statistical Damage Constitutive Model are as follows:
S3.1, using the combine analog microscopic element body load-bearing situation of spring and friction plate, joint thin layer microscopic element body
Mechanical response in shear history regards that by stiffness coefficient be k as1iSpring and stiffness coefficient be k2iSpring friction plate
Two parts element in parallel composition;Since damaged portion and non-damaged portion are mixed in together, i.e., the displacement coordination in deformation process
Unanimously and it is equal to macroscopical shear displacemant, it is assumed that P1si、P2siIt is the suffered shearing of two parts composition respectively, their summation is equal to thin
See the suffered shearing P of cell cubesi, phase shift etc. because two parts element in parallel is ascended the throne is known by stress relationship, when microscopic element position moves
U is less than critical value usWhen, Psi=P1si+P2si=k1iu+k2iu;Displacement components u is greater than critical value usWhen, microscopic element body enters damage shape
State, Psi=P2si=PniTan φ, numerically PniTan φ=k2ius;It is mixed in together with non-damaged portion due to damaging, by becoming
Shape coordination principle knows that two parts displacement is equal and is equal to macroscopical shear displacemant, so the displacement critical value u of cell cubesAs save
The yield point of reason is displaced;When joint is by shearing PsWhen effect, if Joint Element total number is N, the unit number of destroyed at this time
Mesh is Nf, the damage of material is exactly as caused by the continuous destruction of these microscopic element bodies, then shearing may be expressed as:
Wherein, u is shear displacemant, usFor yield point displacement, N indicates the thin sum for seeing micro unit, NfTo indicate a certain stage
The thin sight micro unit number of destroyed;
S3.2, according to stiffness coefficient of two springs after in parallel it is Psi-u picture lines elastic stage slope, obtains shear surface
Upper shearing PsAre as follows:
Ps=(N-Nf)ksu+Nfk2ius
Wherein, ks=k1i+k2i, i.e. shearing rigidity;
S3.3, according to shear surface area ANBy damaged area ANfNot yet damaged portion A composition, obtains damaging parameter D
Are as follows:
S3.4, the shear surface in step S3.1 is sheared into expression formula both sides simultaneously divided by AN, and substitute into step S3.2 and obtain
To apparent shear stress τ:
τ=ksu(1-D)+k2iusD
Wherein, apparent shear stress τ includes the τ that non-damaged portion undertakes*=ksThe τ that u and damaged portion undertaker=k2iusGroup
At then the expression formula in step S3.4 becomes:
τ=τ*(1-D)+τrD
S3.5, Weibull are distributed lower damaging parameter DIn conjunction with apparently cutting
Stress τ obtains the Statistical Damage Constitutive Model at joint:
Preferably, the shear-deformable damage of damage evolution model and joint at shear-deformable process joint is obtained in step S4 originally
The method of structure model are as follows:
S4.1, using a certain stress state when effective stress shear stress τ corresponding with yield pointsDifference τ*-τsMeasurement
It is thin to see whether micro unit intensity reaches faulted condition:
F-F*=τ*-τs=ksu-τs
S4.2, work as F-F*When >=0,Have at this point for yield point:
τs=ksus
S4.3、usCorresponding shear displacemant when having just enter into the strain hardening stage for sample, i.e. yield point are displaced, similarly,
Weibull parameter F0Expression are as follows:
F0=ksu0
S4.4, the expression formula in step S4.1, S4.2 and S4.3 is substituted into equationIn, obtain the damage evolution model at shear-deformable process joint:
S4.5, the failure probability density that step S4.1, S4.2, S4.3 and S4.4 are substituted into the microscopic element body in step S2
In function, the shear-deformable Damage Constitutive Model in joint is obtained:
Preferably, shear-deformable Damage Constitutive Model parameter m, u in joint is determined0Method are as follows:
In peak strength point function derivative it is zero according to joint shear deformation breaks down curve, can obtains:
At peak point, by u=uf, τ=τfIt substitutes into the shear-deformable Damage Constitutive Model in joint still to set up, then by step
S5.1 substitutes into the shear-deformable Damage Constitutive Model in joint, obtains determining Parameters of constitutive model m, u0Calculation formula:
Preferably, the shear-deformable Damage Constitutive Model parameter u in the joint is determinedsMethod are as follows:
The data in the linear elasticity stage that test data is located at peak value proparea are chosen, fitting obtains the oblique of linear elasticity stage
Rate, i.e. shearing rigidity ks;
Reference line τ=k is drawn in a coordinate systemsU calculates the shear stress values τ of measured data valueActual measurementWith relevant shear position
Move τ on reference line corresponding to uWith reference toDifference, difference is denoted as τ*;
τ is drawn in a coordinate system*- u image and reference line τ*=0, it is required bend that inflection point is intuitively read from image
It takes a little, the corresponding u of yield point coordinate is model parameter us。
It is provided by the invention to determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference, have following
The utility model has the advantages that
1, model of the present invention describes all deformed characteristic and its damage evolution law behind before the peak of joint and peak, model form
Simply, parameter explicit physical meaning, model prediction result and test result have the quite high goodness of fit, it was demonstrated that the mould established
Type is reasonable;The problem of shear-deformable Damage Constitutive Model in existing joint can be efficiently solved with it is insufficient.
2, the shear stress-displacement curve at joint usually by before peak hardening phase and the peak after-tack stage constitute, it is existing
Model generally using segmentation empirical function description and mostly only consider peak before Temperature measurement, the present invention proposition can reflect peak after
Constitutive model is sheared in shear stress-displacement of softening feature entirely, has Keeping Shear Experiment Data using the model analysis, higher quasi-
Close the reasonability that precision shows new model.
3, model of the present invention is using shear stress and shear displacemant as basic parameter, only comprising conventional rock power in model expression
Learn parameter, explicit physical meaning, it is thus necessary to determine that fitting parameter it is few, precision more wins previous model.
4, the present invention is based on infinitesimal hypothesis, and the damage evolution model of proposition being capable of the shear-deformable full mistake in Unify legislation joint
The injuring rule of journey, and damage equation form is simple, parameter explicit physical meaning.
5, it on the basis of the existing yield strength of analysis determines the advantage and disadvantage of method, proposes based on this paper damage evolution model
Yield strength establish new method, this method had not only theoretically illustrated the present position of yield point, but also was avoided as much as people
For the interference of factor, also having the characteristics that ease for operation, the yield point that context of methods is established substitutes into established constitutive model,
Demonstrate the reasonability and feasibility of method.
Detailed description of the invention
Fig. 1 is to determine that Complete Damage Process constitutive model joint shear stress-shearing is sheared at the joint of yield point based on stress difference
Relative displacement relationship.
Fig. 2 is to determine that the damage of Complete Damage Process constitutive model ideal joint thin layer is sheared at the joint of yield point based on stress difference
Microscopic element body.
Fig. 3 is to determine that the thin sight body infinitesimal element of Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference
Model.
Fig. 4 is to determine that the element simulation model of Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference.
Fig. 5 is to determine that Complete Damage Process constitutive model difference u is sheared at the joint of yield point based on stress difference0To shear-deformable
Curve influences.
Fig. 6 is to determine that Complete Damage Process constitutive model difference m is sheared to shear-deformable in the joint of yield point based on stress difference
Curve influences.
Fig. 7 is to determine that the joint of yield point shears Complete Damage Process constitutive model and determines that us method is illustrated based on stress difference
Figure.
Fig. 8 is to determine that the joint of yield point shears Complete Damage Process constitutive model and tests measured value and model based on stress difference
Calculated curve compares.
Fig. 9 is to determine that the joint of yield point shears Complete Damage Process constitutive model and tests measured value and model based on stress difference
Calculated curve compares.
Figure 10 is to determine that Complete Damage Process constitutive model measured value and model meter are sheared in the joint of yield point based on stress difference
Curve is calculated to compare.
Figure 11 is to determine that the shearing examination of Complete Damage Process constitutive model JM1 type joint is sheared at the joint of yield point based on stress difference
Measured value is tested compared with model calculated curve.
Figure 12 is to determine that the shearing examination of Complete Damage Process constitutive model JM2 type joint is sheared at the joint of yield point based on stress difference
Measured value is tested compared with model calculated curve.
Figure 13 is to determine that the shearing examination of Complete Damage Process constitutive model JM3 type joint is sheared at the joint of yield point based on stress difference
Measured value is tested compared with model calculated curve.
Figure 14 is to determine that the shearing examination of Complete Damage Process constitutive model JM4 type joint is sheared at the joint of yield point based on stress difference
Measured value is tested compared with model calculated curve.
Figure 15 is to determine that Complete Damage Process constitutive model direct stress σ is sheared at the joint of yield point based on stress differencenWith parameter μ0
Relationship.
Figure 16 is to determine that Complete Damage Process constitutive model direct stress σ is sheared at the joint of yield point based on stress differencenWith parameter m
Relationship.
Figure 17 be based on stress difference determine yield point joint shear Complete Damage Process constitutive model JM1 type test damage because
Sub- trend chart.
Figure 18 be based on stress difference determine yield point joint shear Complete Damage Process constitutive model JM2 type test damage because
Sub- trend chart.
Figure 19 be based on stress difference determine yield point joint shear Complete Damage Process constitutive model JM3 type test damage because
Sub- trend chart.
Figure 20 be based on stress difference determine yield point joint shear Complete Damage Process constitutive model JM4 type test damage because
Sub- trend chart.
Figure 21 is to determine that Complete Damage Process constitutive model typical case joint staight scissors damage is sheared at the joint of yield point based on stress difference
Hurt evolutionary process.
Specific embodiment
A specific embodiment of the invention is described below, in order to facilitate understanding by those skilled in the art this hair
It is bright, it should be apparent that the present invention is not limited to the ranges of specific embodiment, for those skilled in the art,
As long as various change is in the spirit and scope of the present invention that the attached claims limit and determine, these variations are aobvious and easy
See, all are using the innovation and creation of present inventive concept in the column of protection.
According to one embodiment of the application, determine that the joint of yield point is sheared based on stress difference with reference to Fig. 1 this programme
Complete Damage Process constitutive model, comprising:
S1, ideal joint thin layer element body is established as shown in Fig. 2, setting joint thin layer element body load-bearing enters linear elasticity
Stage, joint thin layer microscopic element body are isotropic continuous medias, and joint thin layer microscopic element body is converted by nondestructive state
The state of damaging is instantaneously to complete, and process is irreversible;
S2, it is based on Weibull distribution function, defines outer lotus threshold values F*, obtain the failure probability density letter of microscopic element body
Number;
S3, rheological model mechanical elements are based on, using the combine analog joint thin layer microscopic element body of spring and friction plate
Load-bearing situation obtains the Statistical Damage Constitutive Model at joint under shear action;
S4, become damage after damaging according to rock-soil material load-bearing and do not damage two parts, determine outer lotus threshold values F*, obtain
The shear-deformable Damage Constitutive Model of damage evolution model and joint at shear-deformable process joint;
S5, the shear-deformable Damage Constitutive Model form parameter m in the joint, scale parameter u are determined0It is displaced and joins with yield point
Number us;
S6, several joint shear test datas are substituted into the shear-deformable Damage Constitutive Model in joint, the section verified
Managing shear-deformable Damage Constitutive Model accurately can shear full deformation process in simulation joint, and reflect the stage of shear history
Feature.
Above-mentioned steps are described in detail below
S1, ideal joint thin layer element body is established as shown in Fig. 2, setting joint thin layer element body load-bearing enters linear elasticity
Stage, joint thin layer microscopic element body are isotropic continuous medias, and joint thin layer microscopic element body is converted by nondestructive state
The state of damaging is instantaneously to complete, and process is irreversible;
With reference to Fig. 1, typical joint shear history can be divided into 4 stages, be divided into linear elasticity stage (OA), strain hardening rank
Section (AB), strain softening stage (BC), residual strength stage (CD), wherein us、τsIt is corresponding when respectively entering yielding stage to cut
Cut displacement, shear stress, uf、τfThe respectively corresponding shear displacemant of peak point, shear stress, τrFor residual strength.
In order to establish reflection jointed rock mass shear deformation characteristic damage model, make following setting:
1, material load-bearing enters the linear elasticity stage;
2, joint thin layer microscopic element body is isotropic continuous media, from macroscopically saying that its size is sufficiently small, can be seen
Work is a material particle, and from thin see, size is sufficiently large, contains the essential information of material damage, countless microscopic elements
Body constitutes the whole of damage body, embodies the statistical average property of material;
3, microscopic element body is converted into the state of damaging by nondestructive state and instantaneously completes, and process is irreversible.
S2, it is based on Weibull distribution function, defines outer lotus threshold values F*, obtain the failure probability density letter of microscopic element body
Number;
The defects of there is the microcracks of a large amount of random distributions, gap inside rock material, the presence of these defects makes
Numerous infinitesimal shape and intensity are different, and the destruction of micro unit is random under external load function, based on Weibull points
Cloth[19]The infinitesimal failure probability density function of function are as follows:
Wherein, F is material micro-unit strength metric form: as micro-unit strength F < 0, not being damaged at this time, infinitesimal damage
Variable perseverance is zero, and infinitesimal is in linear elasticity state;
When micro-unit strength is greater than 0, micro unit is in different degrees of faulted condition, damaging parameter D value be located at section [0,
1], damage variable is directly related with stress-strain state locating for micro unit at this time, and therefore, F=0 can be understood as micro unit
Failure criteria, the logical starting point of infinitesimal bulk damage.
At this point, there are an outer lotus threshold values F*, F*Institute is loaded when being numerically equal to micro unit destruction, when a certain stage
Micro unit institute is loaded to be less than threshold values F*When, material does not damage, and original state can be restored to by shedding outer lotus material at this time;When
A certain stage micro unit institute is loaded to be greater than threshold values F*When, material damages, then the failure probability density letter of microscopic element body
Number are as follows:
Wherein, m, F0Respectively Weibull form parameter and scale parameter, wherein that m reflection is form parameter, F0It represents
Macroscopical mean intensity of material.
S3, rheological model mechanical elements are based on, using the combine analog microscopic element body load-bearing situation of spring and friction plate,
Obtain the Statistical Damage Constitutive Model at joint under shear action;
The damage of rock material is exactly to be indicated carefully to see the total of micro unit with N as caused by the continuous destruction for carefully seeing micro unit
Number, NfIt indicates the thin sight micro unit number of a certain stage destroyed, defines the micro unit that damaging parameter D is a certain moment destroyed
The ratio between number and micro unit sum, i.e. D are as follows:
External load is by F*When being loaded into F, the thin sight micro unit sum N that destroys in the processfAre as follows:
Equation (4) substitution equation (3) is had:
Bullet is used based on the method for commonly using primary element and combinations thereof reflection constitutive model in rheological model with reference to Fig. 3
Joint thin layer microscopic element body load-bearing situation is simulated in the combination of spring and friction plate:
Mechanical response of the microscopic element body in shear history regards that by stiffness coefficient be k as1iSpring and rigidity system
Number is k2iSpring friction plate two parts element in parallel composition;Since damaged portion and non-damaged portion are mixed in together, that is, exist
Displacement coordination is consistent in deformation process and is equal to macroscopical shear displacemant, it is assumed that P1si、P2siIt is to be cut suffered by two parts composition respectively
Power, their summation are equal to the suffered shearing P of microscopic element bodysi, phase shift etc. because two parts element in parallel is ascended the throne, by stress relationship
Know, when microscopic element body displacement components u is less than critical value usWhen, Psi=P1si+P2si=k1iu+k2iu;Displacement components u is greater than critical value usWhen,
Cell cube enters faulted condition Psi=P2si=PniTan φ, numerically PniTan φ=k2ius;Due to damage and non-damaged portion
It is mixed in together, know that two parts displacement is equal and is equal to macroscopical shear displacemant by principle of deformation consistency, so the position of cell cube
Move critical value usThe as yield point displacement at joint.When joint is by shearing PsWhen effect, if Joint Element total number is N, at this time
The number of unit of destroyed is Nf(Fig. 2 dash area is destroyed unit), the damage of material is exactly by these microscopic elements
Caused by the continuous destruction of body, then shearing may be expressed as:
Wherein, u is shear displacemant, usFor yield point displacement, N indicates the thin sum for seeing micro unit, NfTo indicate a certain stage
The thin sight micro unit number of destroyed.
With reference to Fig. 4, the stiffness coefficient after two springs are in parallel is Psi-u picture lines elastic stage slope, i.e. ks=k1i+
k2i, therefore formula (7) becomes:
Ps=(N-Nf)ksu+Nfk2ius (8)
According to shear surface area ANBy having damaged (destruction) area ANfNot yet damaged portion A composition, damaging parameter D are as follows:
To (7) formula both sides simultaneously divided by AN, substitute into (8) Shi Ke get:
τ=ksu(1-D)+k2iusD (10)
The τ that apparent shear stress τ is undertaken by non-damaged portion*=ksThe τ that u and damaged portion undertaker=k2iusComposition, then
(10) formula becomes:
τ=τ*(1-D)+τrD (11)
Simultaneous (6), (11) can obtain the Statistical Damage Constitutive Model at joint under shear action:
However want to reflect joint failure by shear overall process, it must be determined that in Statistical Damage Constitutive Model formula
F*, determination will be analyzed it below.
S4, become damage after damaging according to rock-soil material load-bearing and do not damage two parts, determine outer lotus threshold values F*, obtain
The shear-deformable Damage Constitutive Model of damage evolution model and joint at shear-deformable process joint;
The prior art becomes damage after thinking the damage of rock-soil material load-bearing and does not damage two parts, from damage factor D=0
Consider with 1 two extreme cases, it is believed that when infinitesimal enters the initial phase that yielding stage is damage, as D=0, line bullet at this time
Sexual stage effective stress is less than τs, degree of impairment does not occur, when stress reaches threshold values τsWhen think damage have occurred, i.e. D=1, because
This, use is a certain
Corresponding shear stress τ when effective stress when stress state is with strain hardeningsDifference τ*-τsMeasurement is thin to see infinitesimal
Whether body intensity reaches faulted condition, it may be assumed that
F-F*=τ*-τs=ksu-τs (13)
Work as F-F*When >=0, thenFor separation, that is, yield point of linear elasticity and hardening phase, have at this point:
τs=ksus (14)
usIt is that sample has just enter into corresponding shear displacemant when the strain hardening stage, for Weibull parameter F0, equally can be with
It is expressed as:
F0=ksu0 (15)
Formula (13), (14), (15) formula substitution (6) formula can be obtained to the damage evolution model at shear-deformable process joint:
Formula (13), (14), (15) (16) formula are substituted into (10) formula, it is as follows to obtain the shear-deformable damage Constitutive Equation in joint:
S5, the shear-deformable Damage Constitutive Model form parameter m in the joint, scale parameter u are determined0It is displaced and joins with yield point
Number us;
Weibull random distribution parameter u0The geometric scale and form that the shear-deformable curve in joint will be will affect with m, for
Formula (17), other conditions are constant, change u respectively0, m, with reference to Fig. 5 and Fig. 6, Weibull distribution parameter m, u0With shear-deformable song
There are following rules for the variation relation of line:
1, in shear stress-displacement curve peak front elastic stage, curve shape is not influenced substantially by parameter;Joint reaches
After yield strength, with u0Increase with the value of m, curve slightly moves up;In the strain softening stage, parameter u0Curve is influenced with m
It is larger.
2, peak after-tack stage, parameter u0Bigger, the peak strength and residual strength at joint are bigger, reach peak strength
When shear displacemant it is also bigger, curve integrally moves right upwards, u0When increasing to 3.02 from 1.02, peak strength increases
56.21%, corresponding shear displacemant value increases by 31.86%, different parameters u0In the case of, curve in post-peak area group global approximation parallel
Column.Wherein, u0Mainly reflect the size of rock macroscopic statistics mean intensity.
3, with the increase of parameter m, peak strength is in increase tendency, the integral inclined increase of curve in post-peak area.Different parameters m feelings
Under condition, there are a common intersection between peak strength and residual strength, the variation of intersection point anterioposterior curve becomes curve in post-peak area group
Gesture is on the contrary.Parameter m from 2.53 increase to 4.53 when, peak strength increase by 10.42%, m mainly reflect rock brittleness spy
The distribution intensity of sign and material internal micro-unit strength.
Determine shear-deformable Damage Constitutive Model parameter m, u in joint0Comprising:
It in peak strength point function derivative is zero according to joint plane shear deformation breaks down curve, it may be assumed that
And at peak point, by u=uf, τ=τf(17) formula of substitution is still set up, and (18) are substituted into (17) formula at this time, are obtained
Determine Parameters of constitutive model m, u0Calculation formula:
By formula (19) it is found that determining m, u0Or even the key of entire shearing constitutive model is to determine linear elasticity and hardens rank
The separation u of sections。
Determine the shear-deformable Damage Constitutive Model parameter u in joints:
In the shear history of joint, there are an important stress thresholds for online elastic stage and nonlinear phase (peak value proparea)
Value, i.e. yield strength.Yield strength point is joint macroscopic deformation by linearly switching to nonlinear separation, yield limit
Determination is to study the important link of joint shear history.
There are two types of methods for yield strength point in existing determining shear history:
(1) by directly establish shear stress-displacement curve on by linear transition be nonlinear turning point.
(2) it directly determines yield stress and carries out linear fit with Mohr Coulomb's criteria again and obtain yield strength index (cs、
ψs)。
Above two method has its defect: the determination of turning point is strongly depend on the subjective judgement of experimenter, by artificial
Factor is affected, and the determination of yield stress also depends on artificial judgement, how objectively to determine that shear stress-displacement is bent
Line is by nonlinear conversion being linearly the accurate key point for determining yield strength.
It is poor that the present invention is based on shear stresses, and surrender is determined by trial curve and with reference to the shear stress difference between straight line
Point specific value, principle is:
Positioned at the data point in linear elasticity stage, corresponding practical shear stress and the aforementioned difference for calculating theoretical shear stress are in base
It being fluctuated about directrix τ=0, this difference is the function of u, it is denoted as τ (u), and it is located at the data point after yield point, with u's
Increase, difference τ increases, because yield point there are following characteristics:
(1) it is located at reference line to fluctuate up and down, intersects with reference line or almost intersect;
(2) after the point, all data points are more and more remoter apart from reference line, so we are finding out each measured data
Point with reference to the stress difference between straight line after, with τ=0 for benchmark line, the yield point almost τ (u) on reference line, after yield point
Reduce (negative, bigger negative of absolute value itself are smaller) with the increase of u, corresponding bend can be intuitively found from coordinate system
It takes a little.
Specific steps are as follows:
With reference to Fig. 7, the data in the linear elasticity stage that experimental data is located at peak value proparea are chosen, fitting obtains linear elasticity
The slope in stage, i.e. shearing rigidity ks。
Reference line τ=k is drawn in a coordinate systemsU calculates the shear stress values τ of measured data valueActual measurementWith relevant shear position
Move τ on reference line corresponding to uWith reference toDifference, difference is denoted as τ*。
τ is drawn in a coordinate system*- u image and reference line τ*=0, determine inflection point, which is
Required yield displacement us。
So far, m, u0、usIt determines completely, and model parameter explicit physical meaning, is suitable for different normal stress shapes
Situation under state, however its feasibility must also be verified by test examples.
S6, several joint shear test datas are substituted into the shear-deformable Damage Constitutive Model in joint, the section verified
Managing shear-deformable Damage Constitutive Model accurately can shear full deformation process in simulation joint, and reflect the stage of shear history
Feature, specific steps include:
Keeping Shear Experiment Data analysis based on T.Papaliangas
According to Keeping Shear Experiment Data (examination of the T.Papaliangas jointed rock mass under the normal stress of 0.10 and 0.05MPa
Test that the results are shown in Table 1), it is fitted using constitutive relation model of the invention,
1 T.Papaliangas direct shear test result of table
It can be obtained by test data, work as σnWhen=0.05MPa, ks=0.024, τr=0.047, it is asked by method in step S5
Obtain uS=1.637;Work as σnWhen=0.1MPa, ks=0.026, τr=0.077, it can be in the hope of u3=3.045.
Two groups of parameters are substituted into respectively in (17) formula, matched curve effect is as shown in figure 8, as can be seen from Figure 8, this hair
Bright model can accurately simulate joint and shear full deformation process, to preferably reflect the phasic characteristics of shear history.
Based on S.Bandis[26]Keeping Shear Experiment Data analysis
In well-established law under stress condition, S.Bandis etc. has done a series of direct shear test, joint size to natural joint
For 9cm × 5cm, the shear displacemant of application is 6~7mm, and the uniaxial compressive strength that joint is faced the wall and meditated is 2MPa, and direct shear test result is such as
Shown in table 2.It is analyzed for the Keeping Shear Experiment Data that normal stress is 0.09MPa, fitting result is as shown in Figure 9:
2 S.Bandis direct shear test result of table
It can be obtained by test data, work as σnWhen=0.09MPa, ks=0.131, τr=0.077, acquire us=0.547, it will join
Number substitutes into (17) formula, and matched curve is as shown in figure 9, goodness of fit R2Numerical value be in close proximity to 1, show the sheet established herein
Structure model can not only characterize shear-deformable overall process well, can also protrude the deformation characteristics of joint shear stage.
Based on Liu Yuanming, Xia Caichu[27]Keeping Shear Experiment Data analysis
Liu Yuanming, Xia Caichu are to the non across jointed rock mass at 3 kinds of joints of simulation material containing tooth form in different normal stress items
Direct shear test is carried out under part, is analyzed for the experimental data that direct stress is 0.5MPa, test result is shown in Table 3:
3 Liu Yuanming, Xia Caichu direct shear test result of table
It can be obtained by test data, work as σnWhen=0.5MPa, ks=1.512, τr=1.278, acquire us=1.04, by parameter
In (17) formula of substitution, matched curve effect is as shown in Figure 10, the model established as can be seen from Figure 10 and the test result goodness of fit
It is high.
The present invention is for parameter m and u0Discussion
Parameter m, u0It is related with external load with joint itself, to study influence form and degree of each factor to parameter,
Present invention combination document[28]Middle similar materials analysis of experiments, the physical and mechanical parameter of test specimen are as shown in table 4:
4 test specimen physical parameter table of table
The direct shear test result at 4 kinds of non-through class joints is analyzed, using 4 kinds of different Undulating angle simulation sections in test
Reason, joint type are as shown in table 5:
5 joints characteristics of table
Formula (17) are substituted into the shearing test at four seed type joints of verifying as a result, being fitted image as shown in figures 11-14:
Figure 11-14 reflects constitutive model proposed by the present invention and the test goodness of fit is good, and it is same to investigate different methods respectively
Stress lower die shape parameter and direct stress σnRelationship, data are as shown in table 6:
6 Weibull distribution parameter of table and direct stress relationship
The above results are fitted, parameter u0, m and σnFitting image it is as shown in Figure 15,16:
It can be seen that two parameter u0, m and direct stress σnThere is preferable linear relationship, and summarize following empirical equation:
u0=a σn+b (17)
M=c σn+d (18)
Wherein a, b, c, d are trial curve fitting coefficient, and formula (17), (18) are substituted into formula (14) available difference and just answered
The parameter determination method of joint shearing Complete Damage Process constitutive model under power.
The damage factor evolution curve that four class joint test specimens are drawn according to formula (16), as shown in Figure 17,18,19,20:
Figure 17-20 illustrates the damage variable evolution curve that joint is sheared under different operating conditions, obtains according to this paper model
Damage evolution model can be corresponded with joint failure by shear overall process.It is limited to length, is only in direct stress with JM1 type joint
Staight scissors data instance under 0.5MPa analyzes joint damage evolution law during direct shear test.Damage process can be divided into
3 stages as shown in figure 21:
(1) it damages the holding stage, which generates almost without damage, and the value of damage variable is held essentially constant non-
Very close in 0 level.Joint is in the initial stage cut, and shear deformation is smaller, shear stress increases comparatively fast, and shearing is answered
Power and displacement growth in a linear relationship;
(2) it damages the incipient stage, this stage is that joint fissure generates, extends and penetrate through the stage, with the continuous of microfissure
Extension extends, and damage variable is approached from 0 to 1;
(3) damage and failure stage, shearing process enter the remaining stage, and damage variable tends to be flat in this stage with change in displacement
It is slow, stablize about 1.
The present invention is based on stress differences to determine that Complete Damage Process constitutive model is sheared compared with the existing technology in the joint of yield point,
All deformed characteristic and its damage evolution law behind before the peak of joint and peak are described, model form is simple, parameter physical significance is bright
Really, model prediction result and test result have the quite high goodness of fit, it was demonstrated that the model established is reasonable;It can effectively solve
The problem of shear-deformable Damage Constitutive Model in certainly existing joint with it is insufficient.
Although being described in detail in conjunction with specific embodiment of the attached drawing to invention, should not be construed as to this patent
Protection scope restriction.In range described by claims, those skilled in the art are without creative work
The various modifications and deformation made still belong to the protection scope of this patent.
Claims (6)
1. a kind of determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference characterized by comprising
S1, setting joint thin layer microscopic element body load-bearing enter the linear elasticity stage, and joint thin layer microscopic element body is each to same
The continuous media of property, microscopic element body is converted into the state of damaging by nondestructive state and instantaneously completes, and process is irreversible;
S2, it is based on Weibull distribution function, defines outer lotus threshold values F*, obtain the failure probability density function of microscopic element body;
S3, based on the mechanical elements in rheological model, using the combine analog microscopic element body load-bearing situation of spring and friction plate,
Obtain the Statistical Damage Constitutive Model at joint under shear action;
S4, become damage after damaging according to rock-soil material load-bearing and do not damage two parts, determine outer lotus threshold values F*, sheared
The shear-deformable Damage Constitutive Model of the damage evolution model at deformation process joint and joint;
S5, the shear-deformable Damage Constitutive Model form parameter m in the joint, scale parameter u are determined0With yield point displacement parameter us;
S6, several joint shear test datas are substituted into the shear-deformable Damage Constitutive Model in joint, the joint verified is cut
Shear deformation Damage Constitutive Model accurately can shear full deformation process in simulation joint, and reflect that the stage of shear history is special
Point.
2. according to claim 1 determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference,
It is characterized in that, the method for probability density function is destroyed in the step S2 are as follows:
It is greater than threshold values F when a certain stage micro unit institute is loaded*When, rock material damages, the failure probability of microscopic element body
Density function P (F) are as follows:
Wherein, m, F0Respectively Weibull profile shape parameter and scale parameter.
3. according to claim 1 determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference,
It is characterized in that, the Statistical Damage Constitutive Model at joint in the step S3 are as follows:
S3.1, using the combine analog microscopic element body load-bearing situation of spring and friction plate, joint thin layer microscopic element body is being cut
Mechanical response during cutting regards that by stiffness coefficient be k as1iSpring and stiffness coefficient be k2iSpring friction plate two
Subelement composes in parallel;Since damaged portion and non-damaged portion are mixed in together, i.e., displacement coordination is consistent in deformation process
And it is equal to macroscopical shear displacemant, it is assumed that P1si、P2siIt is the suffered shearing of two parts composition respectively, their summation is single equal to carefully seeing
First body is suffered to shear Psi, phase shift etc. because two parts element in parallel is ascended the throne is known by stress relationship, when microscopic element body displacement components u is small
In critical value usWhen, Psi=P1si+P2si=k1iu+k2iu;Displacement components u is greater than critical value usWhen, microscopic element body enters faulted condition,
Psi=P2si=PniTan φ, numerically PniTan φ=k2ius;It is mixed in together with non-damaged portion due to damaging, by deforming
Coordination principle knows that two parts displacement is equal and is equal to macroscopical shear displacemant, so the displacement critical value u of cell cubesAs joint
Yield point displacement;When joint is by shearing PsWhen effect, if Joint Element total number is N, the number of unit of destroyed at this time
For Nf, the damage of material is exactly as caused by the continuous destruction of these microscopic element bodies, then shearing may be expressed as:
Wherein, u is shear displacemant, usFor yield point displacement, N indicates the thin sum for seeing micro unit, NfTo indicate that a certain stage has broken
Bad thin sight micro unit number;
S3.2, according to stiffness coefficient of two springs after in parallel it is Psi-u picture lines elastic stage slope, obtains cutting on shear surface
Power PsAre as follows:
Ps=(N-Nf)ksu+Nfk2ius
Wherein, ks=k1i+k2i, i.e. shearing rigidity;
S3.3, according to shear surface area ANBy damaged area ANfNot yet damaged portion A composition, obtains damaging parameter D are as follows:
S3.4, the shear surface in step S3.1 is sheared into expression formula both sides simultaneously divided by AN, and substitute into step S3.2 and obtain apparently
Shear stress τ:
τ=ksu(1-D)+k2iusD
Wherein, apparent shear stress τ includes the τ that non-damaged portion undertakes*=ksThe τ that u and damaged portion undertaker=k2iusComposition, then
Expression formula in step S3.4 becomes:
τ=τ*(1-D)+τrD
S3.5, Weibull are distributed lower damaging parameter DIn conjunction with apparent shear stress τ
Obtain the Statistical Damage Constitutive Model at joint:
4. according to claim 1 determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference,
It is characterized in that, the shear-deformable Damage Constitutive of damage evolution model and joint at shear-deformable process joint is obtained in the step S4
The method of model are as follows:
S4.1, using a certain stress state when effective stress shear stress τ corresponding with yield pointsDifference τ*-τsMeasurement is thin to be seen
Whether micro unit intensity reaches faulted condition:
F-F*=τ*-τs=ksu-τs
S4.2, work as F-F*When >=0,Have at this point for yield point:
τs=ksus
S4.3、usCorresponding shear displacemant when having just enter into the strain hardening stage for sample, i.e. yield point are displaced, similarly, Weibull
Parameter F0Expression are as follows:
F0=ksu0
S4.4, the expression formula in step S4.1, S4.2 and S4.3 is substituted into equation
In, obtain the damage evolution model at shear-deformable process joint:
S4.5, the failure probability density function that step S4.1, S4.2, S4.3 and S4.4 are substituted into the microscopic element body in step S2
In, obtain the shear-deformable Damage Constitutive Model in joint:
5. according to claim 1 determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference,
It is characterized in that, determines shear-deformable Damage Constitutive Model parameter m, u in the joint0Method are as follows:
In peak strength point function derivative it is zero according to joint plane shear deformation breaks down curve, can obtains:
At peak point, by u=uf, τ=τfIt substitutes into the shear-deformable Damage Constitutive Model in joint still to set up, then by step S5.1
The shear-deformable Damage Constitutive Model in joint is substituted into, obtains determining Parameters of constitutive model m, u0Calculation formula:
6. according to claim 1 determine that Complete Damage Process constitutive model is sheared at the joint of yield point based on stress difference,
It is characterized in that, determines the shear-deformable Damage Constitutive Model parameter u in the jointsMethod are as follows:
The data in the linear elasticity stage that test data is located at peak value proparea are chosen, fitting obtains the slope in linear elasticity stage,
That is shearing rigidity ks;
Reference line τ=k is drawn in a coordinate systemsU calculates the shear stress values τ of measured data valueActual measurementWith relevant shear displacement components u institute
τ on corresponding reference lineWith reference toDifference, difference is denoted as τ*;
τ is drawn in a coordinate system*- u image and reference line τ*=0, it is required surrender that inflection point is intuitively read from image
Point, the corresponding u of yield point coordinate are model parameter us。
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