CN105426660B - A kind of Forecasting Methodology of machine components table planar residual stress field - Google Patents
A kind of Forecasting Methodology of machine components table planar residual stress field Download PDFInfo
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Abstract
The invention discloses a kind of Forecasting Methodology of machine components table planar residual stress field, comprise the following steps：1) the binary n rank trigonometrical number multinomial models of main residual stress component are established；2) preliminary test analysis determines model order n values；3) according to model order n, number of checkpoints and testing scheme is determined and is tested；4) according to forecast model and test result, matrix form is carried out to model and arranges solving model coefficient；5) according to residual stress component test result, the model coefficient asked and self-balancing condition, other residual stress component forecast model coefficients are solved；6) table planar residual stress field image characterizes.Inventive process avoids the blindness of test, the model solution process of foundation is simply and method has taken into full account the self-equilibrium characteristic of table planar residual stress field, it can conveniently enter in program of finite element, can not know that manufacturing process or prediction using the table planar residual stress field of the machine components of process have a clear superiority in detail.
Description
Technical field
The invention belongs to technical field of nondestructive testing, more particularly, to a kind of machine components table planar residual stress field
Forecasting Methodology.
Background technology
Proved from theory and practice in a variety of subjects, the destructive process of machine components, put down from table
What face started, and residual stress is an important indicator of the machined table plane quality of part, it is with machine components manufacturer
Formula is different and different, to the static mechanical property of material, stress corrosion performance, anti-fatigue performance, dimensional stability and use
There is significant impact in life-span, and the quality of parts list planar residual stress directly influences the service behaviour of part or machine
And working life.For a long time, parts list planar residual stress is studied, always as many researchers and engineering technology
The hot subject of personnel.
The research work to residual stress mostly concentrates on the formation mechenism of residual stress, influence factor etc. at present
Research, it is less to carry out model sign research for workpiece table planar residual stress field.It is to understand and grasp work that residual stress, which characterizes,
The residual stress distribution characteristic of part processing table plane, improve product manufacture, improve the important of fatigue life prediction precision etc.
Basis, quick sign table planar residual stress field accurate Characterization are always the problem of engineering circles.
The research method of only table planar residual stress field distribution includes experimental method, finite element method, stress function method
And inherent strain method.However, experimental method is big to concentrate generally use least square method, to the number of the limited several discrete points provided
According to being fitted to obtain complete residual stress field, this method due to test point position and the randomness of number, prediction should
The field of force may not meet self-balancing condition, cause prediction accuracy relatively low, and be not easy to input program of finite element；Finite element side
Method knows part manufacturing process due to needing, but part manufacturing process is complicated, and description is got up, and difficulty is very big, and sign is got up bothersome
And accuracy is not easy to ensure, and residual stress field prediction can not be carried out when part manufactures or can not know in detail using process；Stress
Function method and inherent strain method are that one kind assumes that residual stress component or the components of strain are some form of function, pass through perimeter strip
Part solve to obtain method, and the method function that needs to make repeated attempts is debugged, and is not used widely.However, in big portion
In the case of point, such as being commented when influencing each other, before and after product remanufacturing between research Crack Extension or welding residual stress relaxation
When estimating, it is necessary to which rapidly construction meets the residual stress field of self-balancing condition, can be easily by the residual stress field of identification
It is input in program of finite element, consequently facilitating influence of the evaluation residual stress to structural behaviour.
As can be seen here, there is an urgent need to research and develop one kind, can quickly to characterize suitable blank, manufacturing process complicated or not now
The method for the various machine components table planar residual stress fields that Yi Xiang knows, meanwhile, the result of the method identification should conveniently enter
Into program of finite element, for machine components structure design and manufacture, remanufacture, assess, using provide important technology branch
Hold, so that machine components preferably meet actual demands of engineering.
The content of the invention
For the disadvantages described above or Improvement requirement of prior art, the present invention provides a kind of machine components table planar residual stress
The Forecasting Methodology of field, is predicted with giving machine components table planar residual stress field, is the structure design and system of machine components
Make, remanufacture, assessing, using important technology support is provided, so that machine components preferably meet actual demands of engineering.
To achieve the above object, it is proposed, according to the invention, provide a kind of prediction side of machine components table planar residual stress field
Method, it is characterised in that comprise the following steps：
1) forecast model of main residual stress component is established：X-y coordinate system is established in prediction table plane, then selects three
Individual residual stress component σx、σyAnd τxyIn a residual stress component as main residual stress component, wherein, σxTo act on
Direct stress on the face of x-axis and along the x-axis direction, σyTo act on the face of y-axis and along the y-axis direction
Direct stress, τxyTo act on shearing stress on the face of x-axis and along the y-axis direction, this main residual stress component is resettled
Forecast model, and characterize with binary n rank trigonometrical number multinomials the forecast model of this main residual stress component；
2) the polynomial model order n of binary n rank trigonometrical numbers in step 1) is determined：Respectively choose more than two parallel to
The straight line of x-axis and more than two straight lines parallel to y-axis, multiple estimation points are chosen on every straight line respectively and to each estimation
The main residual stress component of point is tested, wherein the spacing of adjacent two estimations point is equal on every straight line, and every straight line
Unitary n' rank trigonometrical numbers multinomial sign is respectively adopted in the appraising model of the main residual stress component of upper estimation point, then will measure
Every straight line on estimate that the data of main residual stress component a little substitute into corresponding unitary n' ranks trigonometrical number multinomial respectively,
And the polynomial exponent number of each unitary n' rank trigonometrical numbers is obtained respectively, and then obtain the polynomial model of binary n rank trigonometrical numbers
Exponent number n；
3) the binary n rank triangles that the binary n rank trigonometrical number multinomials and step 2) established according to step 1) are obtained
The polynomial model order n of series, chosen in prediction table plane and be more than 2n2+ 2n+1 test point, wherein, selection these
Test point prediction table plane on into ranks arrange, and often go in adjacent 2 points spacing it is equal, adjacent 2 points in each column
Spacing is equal, and the main residual stress component of these above-mentioned test points is then measured in prediction table plane, in addition, choose three with
On test point measure as full measuring point and respectively three residual stress components of each full measuring point；
4) master of the test point obtained by the binary n rank trigonometrical number multinomials and step 3) established according to step 1) is residual
Residue stress component measurement data, matrix form arrangement is carried out to binary n rank trigonometrical numbers multinomial, solved using least square method
The polynomial model coefficient of binary n rank trigonometrical numbers described in step 1)；
5) two residual stress components beyond the main residual stress component of the full measuring point obtained according to step 3) test number
According to the polynomial coefficient of binary n rank trigonometrical numbers obtained with step 4), established according to self-balancing condition except main residual stress point
The forecast model of two other residual stress component beyond amount simultaneously solves the forecast model of the two residual stress components；
6) forecast model for the main residual stress component established according to step 1) and step 4), step 5) obtained its
Remaining two component residual stress component forecast models, carry out residual stress field image and characterize, to obtain each point in prediction table plane
Residual stress distribution situation.
Preferably, in step 1), components of stress σ is selectedxAs main residual stress component, its forecast model：
Wherein, x " being coordinate x function and x "=π (x-xmin)/(xmax-xmin), y " being coordinate y function and y "
=π (y-ymin)/(ymax-ymin), xmin、xmaxThe minimum value and maximum of coordinate on the x coordinate axle on surface are respectively predicted,
ymin、ymaxRespectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, a00,aj0,bj0,a0j,b0j,cjk,djk,
fjk,hjkIt is the polynomial model coefficient of binary n rank trigonometrical numbers, wherein j=1,2 ... n, k=1,2 ... j-1, n bis-
First polynomial model order of n ranks trigonometrical number.
Preferably, in step 2), components of stress σ is selectedxAs main residual stress component, parallel to the straight line peace of x-axis
Row is in estimating main residual stress component σ a little on the straight line of y-axisxIt is as follows：
Wherein, x " being coordinate x function and x "=π (x-xmin)/(xmax-xmin), y " being coordinate y function and y "
=π (y-ymin)/(ymax-ymin), xmin、xmaxThe minimum value and maximum of coordinate on the x coordinate axle on surface are respectively predicted,
ymin、ymaxRespectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, a0, aj', bj'And a'0, a'j', b'j'
For the polynomial model coefficient of unitary n' rank trigonometrical numbers, wherein j'=1,2 ... n', n' are that unitary n' rank trigonometrical numbers are multinomial
The model order of formula.
Preferably, in step 2), the polynomial model order n of binary n rank trigonometrical numbers process is as follows：
According to linear regression method, the polynomial mould of unitary n' rank trigonometrical numbers estimated on each straight line a little is solved respectively
Type exponent number n', the maximum for then choosing above-mentioned all model order n' are more as the binary n rank trigonometrical numbers that step 1) is established
The model order n of item formula.
Preferably, step 2) also includes following sub-step：
(2-1) is tested the main residual stress component of each estimation point on all straight lines parallel to x-axis, obtains m
Group test dataWith m group estimated datasWherein xiExist for i-th of estimation point
Coordinate in x-axis,It is x for x coordinate on straight lineiEstimation point main residual stress component test value,It is x for x coordinate on straight lineiEstimation point main residual stress component estimate, obtained according to linear regression analysisWherein,It is vectorial for unitary n' rank trigonometrical number multinomial models estimation coefficient, andFor unitary n' rank trigonometrical number polynomial modules
Type estimation coefficient, σxResidual stress test for the test value composition of the main residual stress components of stress of m estimation point is vectorial, and
AndX is unitary n' rank trigonometrical number multinomial model design matrixes, andxi" it is that coordinate is xiEstimation point calculated value,
And xi"=π (xi-xmin)/xmax-xmin, xmin、xmaxThe minimum value of coordinate on the x coordinate axle on prediction surface respectively measured
And maximum；
(2-2) is increased n' successively since 1 with step-length 1 carries out value, using coefficient of determinationOr adjustment coefficient of determination Ra 2=1- (1-R2) (m-1)/(m-2n-2) weigh estimation equation to sample
The fitting degree of test value, whereinFor it is all estimation point main residual stress component test datas averages andWork as R2WithFitting degree meets pre-provisioning request when being all higher than 0.9, and then obtains described flat
Row is in the polynomial model order n ' of unitary n' rank trigonometrical numbers of each straight line of x-axis.
Preferably, step 4) comprises the following steps that：
(4-1) matrix form arranges：Obtain the m' group test datas of m' test point
Wherein i'=1,2 ... m', xi'For coordinate of i-th ' the individual test point in x-axis,For the main residual stress of the i-th ' individual test point
The test value of component, and
Matrix σ is setx'=A γ, wherein σx' it is the residual stress test that the main residual stress component of m' test point forms
Vector, andA is prediction table planar design matrix, andγ is that binary n rank trigonometrical numbers are more
The model coefficient vector of item formula, and
γ=[a00 a10 a10' b01 b01' … an0 an0' b0n b0n' c(n-1)1 d(n-1)1 f(n-1)1 h(n-1)1 …
c1(n-1) d1(n-1) f1(n-1) h1(n-1)]′’
Wherein,For the test value of the main residual stress component of the i-th ' individual test point, x "i′For the x " of the i-th ' individual test point
Calculated value and xi′"=π (xi'-xmin)/(xmax-xmin), y "i′For the y " calculated values and y of the i-th ' individual test pointi′"=π
(yi'-ymin)/(ymax-ymin), in addition, xmin、xmaxRespectively predict the minimum value and maximum of coordinate on the x coordinate axle on surface
Value, ymin、ymaxRespectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, a00,aj0,bj0,a0j,b0j,cjk,
djk,fjk,hjkIt is binary n rank trigonometrical number multinomial model coefficients, wherein j=1,2 ... n, k=1,2 ... j-1, n are
The polynomial model order of binary n rank trigonometrical numbers；In addition, Ai'For the design matrix of the i-th ' individual measuring point, and
Ai=[1 cos (xi′″) sin(xi′″) cos(yi′″) sin(yi′″) cos(2xi′″) sin(2xi′″) cos
(2yi′″)；
sin(2yi′″) cos(xi′″)cos(yi′″) cos(xi′″)sin(yi′″) sin(xi′″)cos(yi′″)
sin(xi′″)sin(yi′″) … cos(nxi′″) sin(nxi′″) cos(nyi′″) sin(nyi′″)
cos[(n-1)xi′″]cos(yi′″) sin[(n-1)xi′″]cosyi′″ cos[(n-1)xi′″]sin(yi′″)
sin[(n-1)xi″]sin(yi″) … cos(xi′″)cos[(n-1)yi′″] cos(xi′″)sin[(n-1)yi′″]
sin(xi′″)cos[(n-1)yi′″] sin(xi′″)sin[(n-1)yi′″]]
(4-2) least square method coefficient is estimated：Obtain m' estimated data of m' test pointIf
PutFor the average of the components of stress of m' test point along the x-axis direction, and
For m' test point σxEstimate, andEstimate for binary n rank trigonometrical number multinomial models
Coefficient vector is counted, and
Then binary n rank trigonometrical number multinomial models estimation coefficient vector is obtained according to linear regression analysisAnd then obtain the polynomial model coefficient of binary n rank trigonometrical numbers.
Preferably, step 5) detailed process is as follows：
(5-1) obtains σyAnd τxyResidual stress component forecast model：
Predict in table plane, according to stress boundary condition and balance differential equation, obtain residual stress component σx, σyAnd τxy
Meet following relation：
Wherein fx,fyThe respectively direction muscle power component of x coordinate axle and y-coordinate axle；
Then residual stress component σ is obtained according to residual stress equilibrium equationyAnd τxyForecast model；
(5-2) solves σyAnd τxyResidual stress component forecast model newly-increased coefficient：By the residual stress of step 4) point
Amount test value substitutes into the σ of acquisition respectivelyyAnd τxyResidual stress component forecast model, solve σyAnd τxyResidual stress component it is pre-
That surveys model inserts coefficient, then takes the average value for inserting coefficient, that is, obtains σyAnd τxyResidual stress component forecast model it is new
Increase coefficient.
Preferably, in the step (5-1), the σ of foundationyAnd τxyResidual stress component forecast model be：
Wherein, x " being coordinate x function and x "=π (x-xmin)/(xmax-xmin), y " being coordinate y function and y "
=π (y-ymin)/(ymax-ymin), xmin、xmaxThe minimum value and maximum of coordinate on the x coordinate axle on surface are respectively predicted,
ymin、ymaxRespectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, x0=π/(xmax-xmin), y0=π/
(ymax-ymin), b00For residual stress component σyForecast model insert coefficient, c00For residual stress component τxyPrediction mould
Type inserts coefficient, aj0,bj0,a0j,b0j,cjk,djk,fjk,hjkIt is the polynomial model coefficient of binary n rank trigonometrical numbers, its
Middle j=1,2 ... n, k=1,2 ... j-1, n are the polynomial model order of binary n rank trigonometrical numbers.
Preferably, in the step (5-2), m " is set to organize test dataIts
In," the components of stress σ of individual test point for i-thy、τxyTest value, Bi″、Ci″Respectively
(xi″,yi″) pointCorrelation matrix, then
Wherein,
In general, by the contemplated above technical scheme of the present invention compared with prior art, have below beneficial to effect
Fruit：
The present invention proposes that a kind of residual stress test value using limited table planar point is remaining to machine components table plane
The method that stress field is predicted.This method is residual as table planar residual stress field one using binary n rank trigonometrical number multinomials
The universal model of residue stress component；And propose by table plane and straight line parallel to reference axis Points on Straight Line corresponding remnants
The components of stress are tested, and the polynomial model order of binary n rank trigonometrical numbers is determined by unitary trigonometrical number multinomial model
Number n；And then determine table planar residual stress site test number and distribution；Should to remnants according to least square method using test value
Force component forecast model coefficient is solved, and then draws the model of the residual stress component in a direction；According to self-balancing bar
Part and a small number of test points and then the other components of solution；Finally, parts list plane remnants are obtained so as to image using software programming
Stress field.This method avoid the blindness of test, the model solution process of foundation is simple and has taken into full account table plane remnants
The self-equilibrium characteristic of stress field, obtained regression model is solved, can not only solve any point residual stress, and can utilize
Programming software is programmed image and characterizes residual stress field distribution, can conveniently enter in program of finite element, be machine
The structure design of tool part and manufacture, remanufacture, assess, using important technology support is provided, especially can not know manufacture work in detail
Skill, using the machine components and the prediction of table planar residual stress field that remanufactures part of process there is obvious unique advantage.
Brief description of the drawings
Fig. 1 is the flow chart of the Forecasting Methodology of machine components table planar residual stress field of the present invention；
Fig. 2 is the structural representation of the embodiment of the present invention；
Fig. 3 is the distribution schematic diagram of test point on the straight line of the embodiment of the present invention；
Fig. 4 is the distribution schematic diagram of the table planar residual stress field prediction model measuring point of the embodiment of the present invention；
Embodiment
In order to make the purpose , technical scheme and advantage of the present invention be clearer, it is right below in conjunction with drawings and Examples
The present invention is further elaborated.It should be appreciated that specific embodiment described herein is only to explain the present invention, not
For limiting the present invention.As long as in addition, technical characteristic involved in each embodiment of invention described below that
Conflict can is not formed between this to be mutually combined.
The present invention is in view of machine components table plane discrete point residual stress measurability, and table planar point residual stress balance
Theory, using the residual stress test value of limited table planar point, machine components table planar residual stress field is predicted.
Fig. 1 show the flow chart of the Forecasting Methodology of machine components table planar residual stress field of the present invention, this Forecasting Methodology
Specifically include following steps：
Step 1 establishes the forecast model of main residual stress component
X-y coordinate system is established in prediction table plane, then selects three residual stress component σx、σyAnd τxyIn one
Residual stress component is as main residual stress component, wherein σxTo act on the face of x-axis and along the x-axis direction just
Stress, σyTo act on direct stress on the face of y-axis and along the y-axis direction, τxyTo act on the face of x-axis
And shearing stress along the y-axis direction, the forecast model of this main residual stress component is resettled, and it is more with binary n rank trigonometrical numbers
Item formula characterizes the forecast model of this main residual stress component, for example selects σxFor main residual stress component, i.e.,
Wherein, x " being coordinate x function and x "=π (x-xmin)/(xmax-xmin), y " being coordinate y function and y "
=π (y-ymin)/(ymax-ymin), xmin、xmaxThe minimum value and maximum of coordinate on the x coordinate axle on surface are respectively predicted,
ymin、ymaxRespectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, a00,aj0,bj0,a0j,b0j,cjk,djk,
fjk,hjkIt is the polynomial model coefficient of binary n rank trigonometrical numbers, wherein j=1,2 ... n, k=1,2 ... j-1, n bis-
First polynomial model order of n ranks trigonometrical number.
The analysis of step 2 preliminary test determines model order n
Choose the main residual stress component that straight line is carried out Points on Straight Line needed for step 1 to test, the master for establishing Points on Straight Line
Residual stress component forecast model, the main residual stress component test result of the point on each straight line is substituted into residual stress minute respectively
Measure forecast model, the optimal model order n of solving model.Specifically include following sub-step：
(2-1) chooses rectilinear point and carries out residual stress test
More than two are chosen respectively parallel to the straight line of x-axis and more than two straight lines parallel to y-axis, every straight line
It is upper to choose multiple estimation points respectively and each main residual stress component for estimating point is tested, wherein adjacent two on every straight line
Estimate that the spacing of point is equal, record test point coordinate and test value.
(2-2) establishes Points on Straight Line residual stress forecast model
Establish corresponding unitary n ranks trigonometrical number polynomial function conduct respectively to the point on the straight line parallel to two reference axis
Forecast model.Parallel to x-axis, the remaining residual stress component σ of y-axis Points on Straight LinexRegression model be respectively：
Wherein, a0, aj', bj'And a'0, a'j', b'j'It is the polynomial model coefficient of unitary n' rank trigonometrical numbers, wherein
J'=1,2 ... n', n' are the polynomial model order of unitary n' rank trigonometrical numbers.
(2-3) determines optimal models exponent number n
For a practical problem, the point residual stress component on the straight line parallel to x-axis is tested, if obtained
M group test datas (x1, x2…xm；), wherein i=1,2 ... m.Then have：[a0 a1 b1 … an' bn'] ' i.e.：
For convenience of estimation coefficient is carried out, then with matrix form equation
σx=X β (5)
Wherein, σxFor the residual stress test vector of the test value composition of the main residual stress components of stress of m estimation point；
X is unitary n' rank trigonometrical number multinomial model design matrixes；β be unitary n' ranks trigonometrical number multinomial model regression coefficient to
Amount；Have：
β=[a0 a1 b1 … an' bn']T (1+2n)×1 (5c)
NoteFor it is all estimation point main residual stress component test datas averages andFor estimating for all main residual stress components for estimating point
Evaluation,To estimate regression coefficient vector, then have
Using Least Square Method coefficient, to m measuring point, e is madei(i=1,2...m) is i-th of test point residual error, its
Residual sum of squares (RSS) Q is
To make Q reach minimum, according to extremum principle, condition should be met： I.e.
1+2n equation of the above is referred to as normal equation, and solution normal equation group can obtain the least-squares estimation value of parameter.For
Facilitate providing for parameter Estimation formula, then above-mentioned normal equation group is expressed in matrix as
By regression modelBoth sides are same respectively to multiply XT
Formula (9) is substituted into (10) to obtain,
To obtain regression coefficient, number of checkpoints should be greater than independent variable number, i.e. m ＞ 1+2n', then n'＜ 0.5 (m-1).
With n' increase in the range of satisfaction, function will preferably approach reality, but the weight for increasing term coefficient can be less and less, due to
The point that the bigger needs of n' values are tested is more, uneconomical therefore theoretical according to linear regression analysis, increases 1 successively since 1 to n'
Value is carried out, needs to use coefficient of determinationOr adjustment coefficient of determination Ra 2=1- (1-R2)(m-1)/(m-
2n-2) weigh fitting degree of the estimation equation to test sample value, it is believed that work as R2WithIt is all higher than being fitted when 0.9 more satisfactory.
It is straight parallel to x-axis direction parallel to the point model of fit exponent number on the straight line of y-axis, the similar above-mentioned solution of method for solving
Line point, difference hypothetical model should be y " functions.The preferably minimum n' values of fitting effect are solved to the point on every straight line,
Choose the value for the model order n that the n' maximums needed for all straight lines are established as step 1.
Step 3 determines number of checkpoints and distribution and tested
The binary n rank trigonometrical numbers that the binary n rank trigonometrical number multinomials and step 2 established according to step 1 are obtained
Polynomial model order n, chosen in prediction table plane and be more than 2n2+ 2n+1 test point, wherein, these tests of selection
Point prediction table plane on into ranks arrange, and often go in adjacent 2 points spacing it is equal, adjacent 2 points of spacing in each column
It is equal, the main residual stress component of these above-mentioned test points is then measured in prediction table plane, in addition, choosing more than three
Test point is as full measuring point and measures three residual stress component σ of each full measuring point respectivelyx,σy,τxy。
Step 4 solving model coefficient
The main remnants of test point obtained by the binary n rank trigonometrical number multinomials and step 3 established according to step 1 should
Force component measurement data, matrix form arrangement is carried out to binary n rank trigonometrical numbers multinomial, utilizes least square method solution procedure
The polynomial model coefficient of binary n rank trigonometrical numbers described in 1；
Specifically include following sub-step：
(4-1) model matrix form arranges
For a practical problem, if obtaining m' group test datasWherein i'
=1,2 ... m'.Then have：
So as to have
Wherein Ai'For the design matrix of the i-th ' individual measuring point, γ is the polynomial model coefficient vector of binary n rank trigonometrical numbers.
A is made as prediction table planar design matrix and is hadσ'xFor the main residual stress stress of m' estimation point
The residual stress test vector of the test value composition of component, andThen have
σx'=A γ (15)
(4-2) least square method coefficient is estimated
Obtain m' estimated data of m' test pointSetIt is m' test point along x
The average of the components of stress of direction of principal axis, and For m' test point σxEstimate,
And It is vectorial for binary n rank trigonometrical number multinomial models estimation coefficient, and
Then have
I.e.
Using Least Square Method coefficient, to m' measuring point, the quadratic sum of its error is Q, thenReach minimum, according to extremum principle, that is, meet condition：γjRepresent j-th of unknown number in γ vectors.Order is Ai'kRepresent Ai'K-th in vector
Element.Then have
Above 2n2+ 2n+1 equations are referred to as normal equation, and solution normal equation group can obtain the least-squares estimation of parameter
Value.For convenience of providing for parameter Estimation formula, then above-mentioned normal equation group is expressed in matrix as
By regression modelBoth sides are same respectively to multiply AT
Formula (19) is substituted into (20), so as to
Substitute into and stress vector test value is responded in step 3, you can solution obtains estimation coefficient vector
Step 5 solves remaining residual stress component
According to step 3 obtain full measuring point main residual stress component beyond two residual stress component test datas and
The polynomial coefficient of binary n rank trigonometrical numbers that step 4 obtains, established according to self-balancing condition in addition to main residual stress component
Two other residual stress component forecast model and solve the forecast model of the two residual stress components, specifically include as
Lower sub-step：
(5-1) obtains σyAnd τxyResidual stress component forecast model
Parallel in xy table plane, only existing three residual stress component σx, σyAnd τxy.Surface point is according to stress boundary
The self-balancing condition that condition and balance differential equation are formed, by calculating, each surface point residual stress meets equation below：
Wherein, fx,fyThe respectively direction muscle power component of x coordinate axle and y-coordinate axle.
According to equation (22), residual stress component σ is establishedy、τxyForecast model is as follows：
Wherein x0=π/(xmax-xmin), y0=π/(ymax-ymin), b00For residual stress component σyForecast model insert
Coefficient, c00For residual stress component τxyForecast model insert coefficient.
(5-2) solves σyAnd τxyResidual stress component forecast model newly-increased coefficient b00、c00
For convenience of estimation coefficient is carried out, to equation (23a) and (23b), matrix form equation can be obtained：
Wherein, B, C are respectively the σ of (x, y) pointy, τxyCorrelation matrix, and
By remaining corresponding residual stress component test value of the step 4, the forecast model of foundation is substituted into respectively, solves phase
Answer remaining unknowm coefficient of model, the estimate then averaged as unknown-model coefficient.Remember m " group test datas (x1,y1,
x2,y2,…xm″,ym″；σyi″,τxyi″), Bi″、Ci″Respectively (xi″,yi″) point σyi″、τxyi″Correlation matrix, so as to have
Step 6 residual stress field image characterizes
The forecast model and step 4, step 5 for the main residual stress component established according to step 1 obtained remaining two
Component residual stress component forecast model, carry out residual stress field image and characterize, to obtain the remnants of each point in prediction table plane
Stress distribution situation.
With reference to the coupon that embodiment specification is 600mm × 400mm × 100mm, to 600mm × 400mm upper table
Above-mentioned steps are specifically described the residual stress Forecasting Methodology of plane：
Step 1 establishes the forecast model of main residual stress component
Fig. 2 is the structural representation of the embodiment of the present invention, and residual stress component σ is chosen in upper table planexFor the two of coordinate
First n ranks trigonometrical number multinomial.Wherein xmin=-300mm, xmax=300mm, ymin=-200mm, ymax=200mm, x "=π
(x-xmin)/(xmax-xmin), y "=π (y-ymin)/(ymax-ymin), shown in forecast model such as formula (1).
The analysis of step 2 preliminary test determines model order n
Rectilinear direction residual stress component forecast model is established, to table plane and straight line point stress test, is sat parallel to two
2 straight lines, uniform intensive test model residual stress component direction residual stress are respectively chosen on the straight line of parameter, and records test
Point coordinates and test value, substitute into the optimal n values of model solution.Specifically include following sub-step：
(2-1) establishes Points on Straight Line residual stress forecast model
It is shown parallel to the forecast model such as formula (3) of x-axis, the Points on Straight Line in y-axis direction.
(2-2) chooses Points on Straight Line and carries out residual stress test
Test point distribution schematic diagram on straight line as shown in figure 3, be respectively separated on the straight line parallel to x coordinate axle 30mm,
20mm is tested, and parallel to 2 straight lines of each selection are tested on the straight line of x-axis, is tested, boundary point is not easy to measure at its
5mm is moved in face and carries out test σxTest, 21 points of record test point coordinates and test are recorded on every straight line parallel to reference axis
Value.
(2-3) determines optimal models exponent number n
The 21 groups of test datas that will be obtained on every straight lineWherein i=1,2 ... 21.Substitution formula
(11), according to R2Solution formula, solve the model order minimum values of four straight lines is respectively 5,5,3,3, therefore take n=5 as σx
The exponent number of component 2.3 yuan series multinomial model.
Step 3 determines number of checkpoints and distribution and tested
According to the model order 5 of solution, determine that number of checkpoints is more than at 61 points, can be distributed according to measuring point as shown in Figure 4
Schematic diagram tests the σ of 64 black real pointsx, also can be suitably some more again, and record test result, at the same random 3 points measure it is all residual
Residue stress component σx,σy,τxy, can be as shown in Fig. 4 hollow dots.
Step 4 solving model coefficient
The test result described in forecast model and the step 3 according to the step 1, carry out model coefficient and ask
Solution, specifically includes following sub-step：
(4-1) model matrix form arranges
M'=64 is obtained, shown in matrix form such as formula (15).
(4-2) least square method coefficient is estimated
Test number is substituted into formula (21) solution and obtains model estimation coefficient
Step 5 solves remaining residual stress component
(5-1) establishes two outer two residual stress component forecast models
Remaining residual stress component value of test is substituted into (26a) and (26b), wherein m "=3, can solve to obtain b00,
c00。
Step 6 residual stress field image characterizes
The forecast model and step 4, step 5 for the main residual stress component established according to step 1 obtained remaining two
Component residual stress component forecast model, carry out residual stress field image using programming software and characterize, so as to which prediction table can be obtained
The residual stress distribution situation of each point in plane.
By adopting the above-described technical solution, the present invention proposes general one direction residual stress regression model, utilize
Point on straight line, which makes a preliminary test, determines model order, according to model order size, further determines that experiment test scheme, keeps away
Exempt from the blindness of test, while according to balance differential equation and a small number of residual stress component test value solving model coefficients, can
Table plane any point residual stress is obtained, and image can be programmed using programming software and characterizes residual stress field distribution,
It can conveniently enter in program of finite element, structure design for machine components and manufacture, remanufacture, assess, using
Important technology is provided to support, especially can not know in detail manufacturing process, using process machine components and remanufacture the table of part and put down
The prediction of face residual stress field has obvious unique advantage.
As it will be easily appreciated by one skilled in the art that the foregoing is merely illustrative of the preferred embodiments of the present invention, not to
The limitation present invention, all any modification, equivalent and improvement made within the spirit and principles of the invention etc., all should be included
Within protection scope of the present invention.
Claims (9)
1. a kind of Forecasting Methodology of machine components table planar residual stress field, it is characterised in that comprise the following steps：
1) forecast model of main residual stress component is established：X-y coordinate system is established in prediction table plane, then selection three is residual
Residue stress component σx、σyAnd τxyIn a residual stress component as main residual stress component, wherein, σxIt is vertical to act on
In direct stress on the face of x-axis and along the x-axis direction, σyFor act on the face of y-axis and along the y-axis direction just should
Power, τxyTo act on shearing stress on the face of x-axis and along the y-axis direction, the pre- of this main residual stress component is resettled
Model is surveyed, and the forecast model of this main residual stress component is characterized with binary n rank trigonometrical number multinomials；
2) the polynomial model order n of binary n rank trigonometrical numbers in step 1) is determined：More than two are chosen respectively parallel to x-axis
Straight line and more than two straight lines parallel to y-axis, multiple estimation points are chosen on every straight line respectively and to each estimation point
Main residual stress component tested, wherein the spacing of adjacent two estimations point is equal on every straight line, and on every straight line
Estimate that unitary n ' rank trigonometrical numbers multinomial sign is respectively adopted in the appraising model of the main residual stress component of point, then will measure
Estimate that the data of main residual stress component a little substitute into corresponding unitary n ' ranks trigonometrical number multinomial respectively on every straight line, and
The polynomial exponent number of each unitary n ' rank trigonometrical numbers is obtained respectively, and then obtains the polynomial model order of binary n rank trigonometrical numbers
Number n；
3) the binary n rank trigonometrical numbers that the binary n rank trigonometrical number multinomials and step 2) established according to step 1) are obtained
Polynomial model order n, chosen in prediction table plane and be more than 2n2+ 2n+1 test point, wherein, these tests of selection
Point prediction table plane on into ranks arrange, and often go in adjacent 2 points spacing it is equal, adjacent 2 points of spacing in each column
It is equal, the main residual stress component of these above-mentioned test points is then measured in prediction table plane, in addition, choosing more than three
Test point is as full measuring point and measures three residual stress components of each full measuring point respectively；
4) the main remnants of the test point obtained by the binary n rank trigonometrical number multinomials and step 3) established according to step 1) should
Force component measurement data, matrix form arrangement is carried out to binary n rank trigonometrical numbers multinomial, utilizes least square method solution procedure
1) the polynomial model coefficient of binary n rank trigonometrical numbers described in；
5) according to step 3) obtain full measuring point main residual stress component beyond two residual stress component test datas and
Step 4) obtain the polynomial coefficient of binary n rank trigonometrical numbers, according to self-balancing condition establish except main residual stress component with
The forecast model of two other outer residual stress component simultaneously solves the forecast model of the two residual stress components；
6) forecast model for the main residual stress component established according to step 1) and step 4), step 5) obtained remaining two
Component residual stress component forecast model, carry out residual stress field image and characterize, to obtain the remnants of each point in prediction table plane
Stress distribution situation.
2. Forecasting Methodology as claimed in claim 1, it is characterised in that in step 1), select components of stress σxShould as main remnants
Force component, its forecast model：
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Wherein, x " being coordinate x function and x "=π (x-xmin)/(xmax-xmin), y " being coordinate y function and y "=π
(y-ymin)/(ymax-ymin), xmin、xmaxRespectively predict the minimum value and maximum of coordinate on the x coordinate axle on surface, ymin、
ymaxRespectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, a00,aj0,bj0,a0j,b0j,cjk,djk,fjk,
hjkIt is the polynomial model coefficient of binary n rank trigonometrical numbers, wherein j=1,2 ... n, k=1,2 ... j-1, n are binary n
The polynomial model order of rank trigonometrical number.
3. Forecasting Methodology as claimed in claim 1, it is characterised in that in step 2), select components of stress σxShould as main remnants
Force component, straight line parallel to x-axis and parallel to the main residual stress component σ estimated on the straight line of y-axis a littlexIt is as follows：
Wherein, x " being coordinate x function and x "=π (x-xmin)/(xmax-xmin), y " being coordinate y function and y "=π
(y-ymin)/(ymax-ymin), xmin、xmaxRespectively predict the minimum value and maximum of coordinate on the x coordinate axle on surface, ymin、
ymaxRespectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, a0, aj′, bj′With a '0, a 'j′, b 'j′It is one
First polynomial model coefficient of n ' ranks trigonometrical number, wherein j '=1,2 ... n ', n ' are that unitary n ' rank trigonometrical numbers are polynomial
Model order.
4. Forecasting Methodology as claimed in claim 1, it is characterised in that in step 2), the polynomial mould of binary n rank trigonometrical numbers
Type exponent number n process is as follows：
According to linear regression method, the polynomial model order of unitary n ' rank trigonometrical numbers estimated on each straight line a little is solved respectively
Number n ', then chooses the binary n rank trigonometrical number multinomials that above-mentioned all model order n ' maximum is established as step 1)
Model order n.
5. Forecasting Methodology as claimed in claim 1, it is characterised in that step 2) also includes following sub-step：
(2-1) is tested the main residual stress component of each estimation point on all straight lines parallel to x-axis, is obtained m groups and is surveyed
Try dataWith m group estimated datasWherein xiIt is that i-th of estimation is put in x-axis
On coordinate, i=1,2 ... m,It is x for x coordinate on straight lineiEstimation point main residual stress component test value,
It is x for x coordinate on straight lineiEstimation point main residual stress component estimate, obtained according to linear regression analysisWherein,It is vectorial for unitary n ' rank trigonometrical number multinomial models estimation coefficient, and For unitary n ' rank trigonometrical number multinomial models
Estimation coefficient, σxResidual stress test for the test value composition of the main residual stress components of stress of m estimation point is vectorial, andX is unitary n ' rank trigonometrical number multinomial model design matrixes, andx″iIt is x for coordinateiEstimation point calculated value,
And x "i=π (xi-xmin)/xmax-xmin, xmin、xmaxThe minimum value of coordinate on the x coordinate axle on prediction surface respectively measured
And maximum；
(2-2) is increased n ' successively since 1 with step-length 1 carries out value, using coefficient of determinationOr
Adjust coefficient of determinationFitting degree of the estimation equation to test sample value is weighed, whereinIt is all
Estimate point main residual stress component test data average andWork as R2WithIt is all higher than
Fitting degree meets pre-provisioning request when 0.9, and then it is multinomial to obtain the unitary n ' rank trigonometrical numbers parallel to each straight line of x-axis
The model order n ' of formula.
6. Forecasting Methodology as claimed in claim 1, it is characterised in that step 4) comprises the following steps that：
(4-1) matrix form arranges：Obtain m ' groups of test datas of the individual test points of m 'Its
Middle i '=1,2 ... m ', xi′For coordinate of the i-th ' individual test point in x-axis,For the main residual stress point of the i-th ' individual test point
The test value of amount, and
Matrix σ is setx'=A γ, wherein σx' for the individual test points of m ' main residual stress component form residual stress test to
Amount, andA is prediction table planar design matrix, andγ is that binary n rank trigonometrical numbers are multinomial
The model coefficient vector of formula, and
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Wherein,For the test value of the main residual stress component of the i-th ' individual test point, x "i′For the x " calculated values of the i-th ' individual test point
And xi′"=π (xi′-xmin)/(xmax-xmin), y "i′For the y " calculated values and y of the i-th ' individual test pointi′"=π (yi′-ymin)/
(ymax-ymin), in addition, xmin、xmaxRespectively predict the minimum value and maximum of coordinate on the x coordinate axle on surface, ymin、ymax
Respectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, a00,aj0,bj0,a0j,b0j,cjk,djk,fjk,hjk
For binary n rank trigonometrical number multinomial model coefficients, wherein j=1,2 ... n, k=1,2 ... j-1, n are binary n rank triangles
The polynomial model order of series；In addition, Ai′For the design matrix of the i-th ' individual measuring point, and
Ai=[1cos (xi′″)sin(xi′″)cos(yi′″)sin(yi′″)cos(2xi′″)sin(2xi′″)cos(2yi′″)；
sin(2yi′″)cos(xi′″)cos(yi′″)cos(xi′″)sin(yi′″)sin(xi′″)cos(yi′″)
sin(xi′″)sin(yi′″)...cos(nxi′″)sin(nxi′″)cos(nyi′″)sin(nyi′″)
cos[(n-1)xi′″]cos(yi′″)sin[(n-1)xi′″]cosyi′″cos[(n-1)xi′″]sin(yi′″)
sin[(n-1)xi″]sin(yi″)…cos(xi′″)cos[(n-1)yi′″]cos(xi′″)sin[(n-1)yi′″]
sin(xi′″)cos[(n-1)yi′″]sin(xi′″)sin[(n-1)yi′″]]
(4-2) least square method coefficient is estimated：Obtain the individual estimated datas of m ' of the individual test points of m 'SetFor the average of the components of stress of the individual test points of m ' along the x-axis direction, and
For the individual test point σ of m 'xEstimate, and Estimate for binary n rank trigonometrical number multinomial models
Coefficient vector is counted, and
Then binary n rank trigonometrical number multinomial models estimation coefficient vector is obtained according to linear regression analysisAnd then obtain the polynomial model coefficient of binary n rank trigonometrical numbers.
7. Forecasting Methodology as claimed in claim 6, it is characterised in that step 5) detailed process is as follows：
(5-1) obtains σyAnd τxyResidual stress component forecast model：
Predict in table plane, according to stress boundary condition and balance differential equation, obtain residual stress component σx, σyAnd τxyMeet
Following relation：
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</mfrac>
<mo>+</mo>
<mfrac>
<mrow>
<mo>&part;</mo>
<msub>
<mi>&tau;</mi>
<mrow>
<mi>x</mi>
<mi>y</mi>
</mrow>
</msub>
</mrow>
<mrow>
<mo>&part;</mo>
<mi>x</mi>
</mrow>
</mfrac>
<mo>+</mo>
<msub>
<mi>f</mi>
<mi>y</mi>
</msub>
<mo>=</mo>
<mn>0</mn>
</mtd>
</mtr>
</mtable>
</mfenced>
Wherein fx,fyThe respectively direction muscle power component of x coordinate axle and y-coordinate axle；
Then residual stress component σ is obtained according to residual stress equilibrium equationyAnd τxyForecast model；
(5-2) solves σyAnd τxyResidual stress component forecast model newly-increased coefficient：The residual stress component of step 4) is surveyed
Examination value substitutes into the σ of acquisition respectivelyyAnd τxyResidual stress component forecast model, solve σyAnd τxyResidual stress component prediction mould
Type inserts coefficient, then takes the average value for inserting coefficient, that is, obtains σyAnd τxyThe newly-increased system of residual stress component forecast model
Number.
8. Forecasting Methodology as claimed in claim 7, it is characterised in that in the step (5-1), the σ of foundationyAnd τxyRemnants
Components of stress forecast model is：
<mfenced open = "" close = "">
<mtable>
<mtr>
<mtd>
<mrow>
<msub>
<mi>&sigma;</mi>
<mi>y</mi>
</msub>
<mo>=</mo>
<msub>
<mi>b</mi>
<mn>00</mn>
</msub>
<mo>-</mo>
<msub>
<mi>f</mi>
<mi>y</mi>
</msub>
<mi>y</mi>
<mo>+</mo>
<mo>{</mo>
<mfrac>
<mrow>
<msup>
<mi>j</mi>
<mn>2</mn>
</msup>
<msup>
<mi>y</mi>
<mn>2</mn>
</msup>
<msup>
<msub>
<mi>y</mi>
<mn>0</mn>
</msub>
<mn>2</mn>
</msup>
</mrow>
<mn>2</mn>
</mfrac>
<munderover>
<mi>&Sigma;</mi>
<mrow>
<mi>j</mi>
<mo>=</mo>
<mn>1</mn>
</mrow>
<mi>n</mi>
</munderover>
<mo>&lsqb;</mo>
<mo>-</mo>
<msub>
<mi>a</mi>
<mrow>
<mi>j</mi>
<mn>0</mn>
</mrow>
</msub>
<mi>cos</mi>
<mrow>
<mo>(</mo>
<msup>
<mi>jx</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mo>)</mo>
</mrow>
<mo>-</mo>
<msub>
<mi>b</mi>
<mrow>
<mi>j</mi>
<mn>0</mn>
</mrow>
</msub>
<mi>sin</mi>
<mrow>
<mo>(</mo>
<msup>
<mi>jx</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mo>)</mo>
</mrow>
<mo>&rsqb;</mo>
</mrow>
</mtd>
</mtr>
<mtr>
<mtd>
<mrow>
<mo>+</mo>
<munderover>
<mi>&Sigma;</mi>
<mrow>
<mi>j</mi>
<mo>=</mo>
<mn>1</mn>
</mrow>
<mi>n</mi>
</munderover>
<munderover>
<mi>&Sigma;</mi>
<mrow>
<mi>k</mi>
<mo>=</mo>
<mn>1</mn>
</mrow>
<mrow>
<mi>j</mi>
<mo>-</mo>
<mn>1</mn>
</mrow>
</munderover>
<mfrac>
<msup>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<mn>2</mn>
</msup>
<msup>
<mi>k</mi>
<mn>2</mn>
</msup>
</mfrac>
<mo>{</mo>
<msub>
<mi>c</mi>
<mrow>
<mi>j</mi>
<mi>k</mi>
</mrow>
</msub>
<mi>cos</mi>
<mo>&lsqb;</mo>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<msup>
<mi>x</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msup>
<mi>&rsqb;cos(ky</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mi>)</mi>
<mo>+</mo>
<msub>
<mi>d</mi>
<mrow>
<mi>j</mi>
<mi>k</mi>
</mrow>
</msub>
<mi>cos</mi>
<mo>&lsqb;</mo>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<msup>
<mi>x</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msup>
<mi>&rsqb;sin(ky</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mi>)</mi>
</mrow>
</mtd>
</mtr>
<mtr>
<mtd>
<mrow>
<mo>+</mo>
<msub>
<mi>f</mi>
<mrow>
<mi>j</mi>
<mi>k</mi>
</mrow>
</msub>
<mi>sin</mi>
<mo>&lsqb;</mo>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<msup>
<mi>x</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msup>
<mi>&rsqb;cos(ky</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msub>
<mi>)+h</mi>
<mrow>
<mi>j</mi>
<mi>k</mi>
</mrow>
</msub>
<mi>sin</mi>
<mo>&lsqb;</mo>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<msup>
<mi>x</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msup>
<mi>&rsqb;sin(ky</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mi>)}}</mi>
<mfrac>
<mrow>
<msup>
<msub>
<mi>x</mi>
<mn>0</mn>
</msub>
<mn>2</mn>
</msup>
</mrow>
<mrow>
<msup>
<msub>
<mi>y</mi>
<mn>0</mn>
</msub>
<mn>2</mn>
</msup>
</mrow>
</mfrac>
</mrow>
</mtd>
</mtr>
</mtable>
</mfenced>
<mfenced open = "" close = "">
<mtable>
<mtr>
<mtd>
<mrow>
<msub>
<mi>&tau;</mi>
<mrow>
<mi>x</mi>
<mi>y</mi>
</mrow>
</msub>
<mo>=</mo>
<msub>
<mi>c</mi>
<mn>00</mn>
</msub>
<mo>-</mo>
<msub>
<mi>f</mi>
<mi>x</mi>
</msub>
<mi>y</mi>
<mo>+</mo>
<mo>{</mo>
<msub>
<mi>yy</mi>
<mn>0</mn>
</msub>
<munderover>
<mi>&Sigma;</mi>
<mrow>
<mi>j</mi>
<mo>=</mo>
<mn>1</mn>
</mrow>
<mi>n</mi>
</munderover>
<mo>&lsqb;</mo>
<msub>
<mi>a</mi>
<mrow>
<mi>j</mi>
<mn>0</mn>
</mrow>
</msub>
<mi>j</mi>
<mi> </mi>
<mi>sin</mi>
<mrow>
<mo>(</mo>
<msup>
<mi>jx</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mo>)</mo>
</mrow>
<mo>-</mo>
<msub>
<mi>b</mi>
<mrow>
<mi>j</mi>
<mn>0</mn>
</mrow>
</msub>
<mi>j</mi>
<mi> </mi>
<mi>cos</mi>
<mrow>
<mo>(</mo>
<msup>
<mi>jx</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mo>)</mo>
</mrow>
<mo>&rsqb;</mo>
</mrow>
</mtd>
</mtr>
<mtr>
<mtd>
<mrow>
<mo>+</mo>
<munderover>
<mi>&Sigma;</mi>
<mrow>
<mi>j</mi>
<mo>=</mo>
<mn>1</mn>
</mrow>
<mi>n</mi>
</munderover>
<munderover>
<mi>&Sigma;</mi>
<mrow>
<mi>k</mi>
<mo>=</mo>
<mn>1</mn>
</mrow>
<mrow>
<mi>j</mi>
<mo>-</mo>
<mn>1</mn>
</mrow>
</munderover>
<mfrac>
<mrow>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
</mrow>
<mi>k</mi>
</mfrac>
<mo>{</mo>
<msub>
<mi>c</mi>
<mrow>
<mi>j</mi>
<mi>k</mi>
</mrow>
</msub>
<mi>sin</mi>
<mo>&lsqb;</mo>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<msup>
<mi>x</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msup>
<mi>&rsqb;sin(ky</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mi>)</mi>
<mo>-</mo>
<msub>
<mi>d</mi>
<mrow>
<mi>j</mi>
<mi>k</mi>
</mrow>
</msub>
<mi>sin</mi>
<mo>&lsqb;</mo>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<msup>
<mi>x</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msup>
<mi>&rsqb;cos(ky</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mi>)</mi>
</mrow>
</mtd>
</mtr>
<mtr>
<mtd>
<mrow>
<mo>-</mo>
<msub>
<mi>f</mi>
<mrow>
<mi>j</mi>
<mi>k</mi>
</mrow>
</msub>
<mi>cos</mi>
<mo>&lsqb;</mo>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<msup>
<mi>x</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msup>
<mi>&rsqb;sin(ky</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msub>
<mi>)+h</mi>
<mrow>
<mi>j</mi>
<mi>k</mi>
</mrow>
</msub>
<mi>cos</mi>
<mo>&lsqb;</mo>
<mrow>
<mo>(</mo>
<mi>j</mi>
<mo>-</mo>
<mi>k</mi>
<mo>)</mo>
</mrow>
<msup>
<mi>x</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<msup>
<mi>&rsqb;cos(ky</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mi>)}}</mi>
<mfrac>
<msub>
<mi>x</mi>
<mn>0</mn>
</msub>
<msub>
<mi>y</mi>
<mn>0</mn>
</msub>
</mfrac>
</mrow>
</mtd>
</mtr>
</mtable>
</mfenced>
Wherein, x " being coordinate x function and x "=π (x-xmin)/(xmax-xmin), y " being coordinate y function and y "=π
(y-ymin)/(ymax-ymin), xmin、xmaxRespectively predict the minimum value and maximum of coordinate on the x coordinate axle on surface, ymin、
ymaxRespectively predict the minimum value and maximum of coordinate on y-coordinate axle on surface, x0=π/(xmax-xmin), y0=π/(ymax-
ymin), b00For residual stress component σyForecast model insert coefficient, c00For residual stress component τxyForecast model put
Enter coefficient, aj0,bj0,a0j,b0j,cjk,djk,fjk,hjkIt is the polynomial model coefficient of binary n rank trigonometrical numbers, wherein j=
1,2 ... n, k=1,2 ... j-1, n are the polynomial model order of binary n rank trigonometrical numbers.
9. Forecasting Methodology as claimed in claim 7, it is characterised in that in the step (5-2), set m " to organize test dataWherein, i "=1,2 ..., m ", For i-th, " individual test point is answered
Force component σy、τxyTest value, Bi″、Ci″Respectively (xi″,yi″) pointCorrelation matrix, then
<mrow>
<msub>
<mi>b</mi>
<mn>00</mn>
</msub>
<mo>=</mo>
<munderover>
<mo>&Sigma;</mo>
<mrow>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mo>=</mo>
<mn>1</mn>
</mrow>
<msup>
<mi>m</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</munderover>
<mrow>
<mo>(</mo>
<msub>
<mi>&sigma;</mi>
<msub>
<mi>y</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</msub>
</msub>
<mo>-</mo>
<msub>
<mi>B</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</msub>
<mover>
<mi>&gamma;</mi>
<mo>^</mo>
</mover>
<mo>+</mo>
<msub>
<mi>f</mi>
<mi>y</mi>
</msub>
<msub>
<mi>y</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</msub>
<mo>)</mo>
</mrow>
<mo>/</mo>
<msup>
<mi>m</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</mrow>
<mrow>
<msub>
<mi>c</mi>
<mn>00</mn>
</msub>
<mo>=</mo>
<munderover>
<mo>&Sigma;</mo>
<mrow>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mo>=</mo>
<mn>1</mn>
</mrow>
<msup>
<mi>m</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</munderover>
<mrow>
<mo>(</mo>
<msub>
<mi>&tau;</mi>
<mrow>
<msub>
<mi>xy</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</msub>
</mrow>
</msub>
<mo>-</mo>
<msub>
<mi>C</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</msub>
<mover>
<mi>&gamma;</mi>
<mo>^</mo>
</mover>
<mo>-</mo>
<msub>
<mi>f</mi>
<mi>x</mi>
</msub>
<msub>
<mi>y</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</msub>
<mo>)</mo>
</mrow>
<mo>/</mo>
<msup>
<mi>m</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</mrow>
Wherein,
<mfenced open = "" close = "">
<mtable>
<mtr>
<mtd>
<mtable>
<mtr>
<mtd>
<mrow>
<msub>
<mi>B</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
</msub>
<mo>=</mo>
<msubsup>
<mi>x</mi>
<mn>0</mn>
<mn>2</mn>
</msubsup>
<mo>/</mo>
<msubsup>
<mi>y</mi>
<mn>0</mn>
<mn>2</mn>
</msubsup>
<mo>&lsqb;</mo>
<mn>0</mn>
</mrow>
</mtd>
<mtd>
<mrow>
<mo>-</mo>
<msubsup>
<mi>y</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mn>2</mn>
</msubsup>
<msubsup>
<mi>y</mi>
<mn>0</mn>
<mn>2</mn>
</msubsup>
<mo>/</mo>
<mn>2</mn>
<mi>cos</mi>
<mrow>
<mo>(</mo>
<msubsup>
<mi>x</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msubsup>
<mo>)</mo>
</mrow>
</mrow>
</mtd>
<mtd>
<mrow>
<mo>-</mo>
<msubsup>
<mi>y</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mn>2</mn>
</msubsup>
<msubsup>
<mi>y</mi>
<mn>0</mn>
<mn>2</mn>
</msubsup>
<mo>/</mo>
<mn>2</mn>
<mi>sin</mi>
<mrow>
<mo>(</mo>
<msubsup>
<mi>x</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msubsup>
<mo>)</mo>
</mrow>
</mrow>
</mtd>
<mtd>
<mn>0</mn>
</mtd>
<mtd>
<mn>0</mn>
</mtd>
<mtd>
<mrow>
<mo>-</mo>
<mn>2</mn>
<msubsup>
<mi>y</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mn>2</mn>
</msubsup>
<msubsup>
<mi>y</mi>
<mn>0</mn>
<mn>2</mn>
</msubsup>
<mi>cos</mi>
<mrow>
<mo>(</mo>
<mn>2</mn>
<msubsup>
<mi>x</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msubsup>
<mo>)</mo>
</mrow>
</mrow>
</mtd>
</mtr>
<mtr>
<mtd>
<mrow>
<mo>-</mo>
<mn>2</mn>
<msubsup>
<mi>y</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mn>2</mn>
</msubsup>
<msubsup>
<mi>y</mi>
<mn>0</mn>
<mn>2</mn>
</msubsup>
<mi>sin</mi>
<mrow>
<mo>(</mo>
<mn>2</mn>
<msubsup>
<mi>x</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msubsup>
<mo>)</mo>
</mrow>
</mrow>
</mtd>
<mtd>
<mn>0</mn>
</mtd>
<mtd>
<mn>0</mn>
</mtd>
<mtd>
<mrow>
<mi>cos</mi>
<mrow>
<mo>(</mo>
<msubsup>
<mi>x</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msubsup>
<mo>)</mo>
</mrow>
<mi>cos</mi>
<mrow>
<mo>(</mo>
<msubsup>
<mi>y</mi>
<msup>
<mi>i</mi>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msup>
<mrow>
<mo>&prime;</mo>
<mo>&prime;</mo>
</mrow>
</msubsup>
<mo>)</mo>
</mrow>
</mrow>
</mtd>
<mtd>
<mrow>
<mi>cos</mi>
<mrow>
<mo>(</mo>
<msubsup>
<mi>x</mi>
<msup>
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