CN107335848A - Three-dimensional milling residual stress Forecasting Methodology - Google Patents

Three-dimensional milling residual stress Forecasting Methodology Download PDF

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CN107335848A
CN107335848A CN201710491661.2A CN201710491661A CN107335848A CN 107335848 A CN107335848 A CN 107335848A CN 201710491661 A CN201710491661 A CN 201710491661A CN 107335848 A CN107335848 A CN 107335848A
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CN107335848B (en
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万敏
叶翔宇
张卫红
杨昀
马颖超
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Northwestern Polytechnical University
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    • B—PERFORMING OPERATIONS; TRANSPORTING
    • B23—MACHINE TOOLS; METAL-WORKING NOT OTHERWISE PROVIDED FOR
    • B23C—MILLING
    • B23C3/00—Milling particular work; Special milling operations; Machines therefor
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    • B23Q—DETAILS, COMPONENTS, OR ACCESSORIES FOR MACHINE TOOLS, e.g. ARRANGEMENTS FOR COPYING OR CONTROLLING; MACHINE TOOLS IN GENERAL CHARACTERISED BY THE CONSTRUCTION OF PARTICULAR DETAILS OR COMPONENTS; COMBINATIONS OR ASSOCIATIONS OF METAL-WORKING MACHINES, NOT DIRECTED TO A PARTICULAR RESULT
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Abstract

The invention discloses a kind of three-dimensional milling residual stress Forecasting Methodology, for solving the technical problem of existing milling residual stress Forecasting Methodology poor practicability.Technical scheme is that milling process is divided into several small three-dimensional inclined cutting infinitesimals first, and then these infinitesimals are analyzed.The angular relationship of cutting force and cutting speed is calculated again, is next based on J C constitutive models to calculate the shear flow stress of shear surface, so as to release the normal stress on shear surface.Then the behavior of cutting of point of a knife plough is analyzed, obtains ploughing shear force and plough cuts the length in region.The stress contact process of inclined cutting is modeled again, using stress tensor three-dimensional coordinate transformation method will shear and plough cut caused by stress distribution be overlapped, obtain the ess-strain course of inside workpiece in milling process.Constitutive behavior of the workpiece in the case of D elastic-plastic CYCLIC LOADING finally is predicted with the plastoelastic method of increment, obtains the residual stress of milling workpiece surface, practicality is good.

Description

Three-dimensional milling residual stress Forecasting Methodology
Technical field
The present invention relates to a kind of milling residual stress Forecasting Methodology, more particularly to a kind of three-dimensional milling residual stress prediction side Method.
Background technology
When being predicted to cutting residual stress, conventional method has two kinds of theoretical modeling and finite element simulation.It is existing The theoretical modeling method of prediction machining residual stress is all based on the theoretical model of two dimension, there is very strong limitation, it is impossible to It is effectively used for the Milling Processes of three-dimensional.And when using finite element simulation prediction residue stress, due to the three-dimensional of milling Structure too complex, generally require to simplify model, the result so obtained is just necessarily deviated.
" S.Agrawal, S.S.Joshi, Analytical modelling of the residual stresses of document 1 in orthogonal machining of AISI4340steel,Journal of Manufacturing Processes 15 (1) (2013) 167-179. " discloses a kind of residual stress forecast model suitable for two-dimentional orthogonal cutting process, the model The integral that contact mechanics establishes inside workpiece contact stress is primarily based on, is then built using plastoelasticity method The residual stress forecast model of vertical orthogonal cutting process, this method may be only available for two-dimentional right angle turning situation, it is impossible to be used in Increasingly complex three-dimensional milling process.
" Nejah, Tounsi, Tahany, EI-Wardany, Finite element analysis of chip of document 2 formation and residual stresses induced by sequential cutting in side milling with microns to submicron uncut chip thickness and finite cutting edge Radius, Advances in Manufacturing 3 (4) (2015) 309-322. " disclose a kind of suitable for prediction milling The finite element modeling method of process residues stress, this method assume that milling cutter's helix angle is 0 °, then a section of milling cutter are entered Row finite element analysis, and then establish the residual stress calculation model for considering that chip formation and tool arc influence.But this method Actually it is still a kind of two-dimentional method, does not consider situation when helical angle is not 0 °, does not also account for milling three-dimensional Structure.
The typical feature of document above is：In workpiece cutting residual stress modeling, increasingly complex milling feelings are not accounted for Condition, or the method model for the consideration milling established have ignored due to milling cutter three-dimensional structure caused by helical angle.
The content of the invention
In order to overcome the shortcomings of existing milling residual stress Forecasting Methodology poor practicability, the present invention provides a kind of three-dimensional milling Residual stress Forecasting Methodology.Milling process is divided into several small three-dimensional inclined cutting infinitesimals by this method first, then These infinitesimals are analyzed according to three-dimensional inclined cutting theory.The angular relationship of cutting force and cutting speed is calculated again, then The shear flow stress of shear surface is calculated based on J-C constitutive models, so as to release the normal stress on shear surface.Then with cunning Lineation opinion is moved to analyze the behavior of cutting of point of a knife plough, obtains ploughing shear force and plough cuts the length in region.Inclined cutting is answered again Power contact process is modeled, using stress tensor three-dimensional coordinate transformation method will shear and plough cut caused by stress distribution enter Row superposition, obtains the stress-strain course of inside workpiece in milling process.Finally workpiece is predicted with the plastoelastic method of increment Constitutive behavior in the case of three-dimensional elastic-plastic CYCLIC LOADING, obtains the residual stress of milling workpiece surface, and practicality is good.
The technical solution adopted for the present invention to solve the technical problems：A kind of three-dimensional milling residual stress Forecasting Methodology, its Feature is to comprise the following steps：
Step 1: milling process is divided into several three-dimensional inclined cutting infinitesimals.
Step 2: calculated using below equation in three-dimensional Oblique Cutting Process, the angular relationship of cutting force and cutting speed：
sinθi=sin βa sinηflow
tan(θn+αn)=tan βa cosηflow
In formula, β is milling cutter's helix angle, αnIt is tool orthogonal rake, βaIt is angle of friction, ηflowIt is chip flow angle, θiAnd θnIt is fixed Adopted cutting force deflection, φiAnd φnIt is shear rate deflection.
Step 3: the shear flow stress on shear surface is calculated using following formula：
In formula, τsIt is shear flow stress, ε is plastic strain,It is plastic strain rate, ε0It is to refer to strain rate, T is shearing Face temperature, T0It is room temperature, TmIt is material melting point, A, B, C, m and n are the J-C constitutive parameters of material.
Step 4: Cutting Force Coefficient is calculated using following formula：
In formula, KtcIt is the Cutting Force Coefficient in cutting speed direction, KfcIt is perpendicular to the cutting force in workpiece machining surface direction Coefficient.
Step 5: cutting force is calculated using following formula：
Ft=Ktcdaptc+Ktedap
Ff=Kfcdaptc+Kfedap
In formula, dapIt is the axial width of inclined cutting infinitesimal, tcIt is the thickness of cutting of inclined cutting infinitesimal, KteAnd KfeIt is Plough shear force coefficient.FtIt is the cutting force in cutting speed direction, FfIt is perpendicular to the cutting force in workpiece machining surface direction.
Step 6: the normal direction cutting force F on shear surface is calculated using below equationn：
Fn=Ftcosβsinφn+Ffcosφn
In formula, FnIt is the normal direction cutting force on shear surface.
Step 7: the normal stress on shear surface is calculated using below equation
In formula, psIt is the normal stress on shear plane.
Step 8: the length of shear surface is calculated using below equation：
In formula,It is the length of shear surface.
Step 9: the stress distribution that tool arc plough cuts region is considered uniform, calculated using below equation：
In formula, peIt is to plough the normal stress for cutting region, qeIt is to plough the tangential stress for cutting region, PthrustIt is normal direction plough shear force, PcutIt is tangential plough shear force,It is the length that plough cuts region, μ is coefficient of friction.
Step 10: using following formula come calculate respectively shearing and plough cut caused by stress distribution.
In formula, σxxIt is stress along the x-axis direction, σzzIt is stress along the z-axis direction, τxzIt is the shear stress in xz directions, a It is the half of the length of contact area, coordinate value x and z represent the position of calculation point of stress, and p (s) and q (s) are positive and tangential Stress value.For share zone, based on step 3 and step 7 result, by p (s)=ps, q (s)=qs,In substitution Formula；Region is cut for plough, based on step 8 result, by p (s)=pe, q (s)=qe,Substitute into above formula.
Step 11: using following formula will shear and plough cut caused by stress distribution be overlapped.
[σ1']=Q1 T[σ1]Q1
[σ2']=Q2 T[σ2]Q2
[σ3]=[σ1']+[σ2']
In formula, [σ1] represent value of the stress distribution under the coordinate system of shear zone caused by shearing, [σ2] represent that plough cuts caused by Value of the stress distribution under Li Qie areas coordinate system, [σ1'] value of the stress distribution under workpiece coordinate system caused by shearing is represented, [σ2'] represent plough cut caused by value of the stress distribution under workpiece coordinate system, [σ3] represent stress point actual under workpiece coordinate system Implantation.Q1And Q2It is the three-dimensional coordinate transformation matrix of stress tensor.
Step 12: the stress distribution value under inclined cutting unit is transformed into milling coordinate system using following formula.
In formula, φwIt is that milling cutter immerses angle, [σ] is the stress distribution value under milling coordinate system.
Step 13: the stress increment in a tiny time dt is calculated using following formula：
[d σ]=[σ (t+dt)]-[σ (t)]
In formula, [d σ] is the stress increment in a tiny time, and [σ (t)] represents what is obtained based on step 12 result Milling process stress distribution and the relation of time.
Step 14: judge whether workpiece enters mecystasis at some moment using following formula：
In formula, J2It is the invariant of deviatoric tensor of stress second, F is yield surface, τsyIt is the shear yield strength of material, SijRepresent Deviatoric stress, αijRepresent back stress.
Step 15: using the stress increment under following formula computational plasticity state：
In formula, E is elasticity modulus of materials, and v is Poisson's ratio, d σijIt is the stress increment obtained based on step 13 result,WithIt is the stress increment under mecystasis, d εxxWith d εyyIt is plastic strain increment, nijIt is on plastic strain rate direction Unit normal vector.
Step 16: the boundary condition of stress relaxation is judged using following formula：
In formula,It is the residual stress before relaxation,It is overstrain, f1,2,...,6(z) represent one has with coordinate value z The non-zero amount closed.
Step 17: represent often to walk relaxed length using following formula：
In formula, M is loose total step number, Δ σijExpression often walks stress relaxation number, Δ εijRepresent often to walk strain relaxation amount.
Step 18: the stress increment of elastic release process is calculated using following formula：
Δσxy=2G Δs εxy
In formula, G represents elastic properties of materials modulus of shearing.
Step 19: the stress increment of process is discharged with following formula computational plasticity：
Step 20: the final residual stress after relaxation step terminates, obtained is calculated with following formula：
In formula,WithIt is the final residual stress in workpiece coordinate system x directions and y directions respectively.
The beneficial effects of the invention are as follows：Milling process is divided into several small three-dimensional inclined cuttings by this method first Infinitesimal, then these infinitesimals are analyzed according to three-dimensional inclined cutting theory.The angle of cutting force and cutting speed is calculated again Relation, J-C constitutive models are next based on to calculate the shear flow stress of shear surface, should so as to release the normal direction on shear surface Power.Then the behavior of cutting of point of a knife plough is analyzed with Slip Line Theory, obtains ploughing shear force and plough cuts the length in region.Again to oblique The stress contact process of angle cutting is modeled, and will be sheared using the method for stress tensor three-dimensional coordinate transformation and caused by plough cuts Stress distribution is overlapped, and obtains the stress-strain course of inside workpiece in milling process.Finally use the plastoelastic method of increment To predict constitutive behavior of the workpiece in the case of three-dimensional elastic-plastic CYCLIC LOADING, the residual stress of milling workpiece surface is obtained, it is real It is good with property.
The present invention is elaborated with reference to the accompanying drawings and detailed description.
Brief description of the drawings
Fig. 1 is the contact stress change schematic diagram in the three-dimensional milling residual stress Forecasting Methodology embodiment of the present invention.
Fig. 2 is the residual stress prediction result in the three-dimensional milling residual stress Forecasting Methodology embodiment of the present invention.
Fig. 3 is three-dimensional milling process geometric representation in the three-dimensional milling residual stress Forecasting Methodology of the present invention.
Embodiment
Reference picture 1-3.The three-dimensional milling residual stress Forecasting Methodology of the present invention comprises the following steps that：
Step 1: milling process is divided into several small three-dimensional inclined cutting infinitesimals.
Step 2: measure for determining from the cutting force in three-dimensional Oblique Cutting Process and the angular relationship of cutting speed Milling cutter's helix angle β is 20 °, tool orthogonal rake αnFor 5 °, angle of friction βaFor 17 °.Substitute them in following formula and calculate cutting force deflection θi And θn, shear rate deflection φiAnd φn, and chip flow angle ηflow。
sinθi=sin βa sinηflow
tan(θn+αn)=tan βa cosηflow
Step 3: the shear flow stress on shear surface is calculated using following formula：
Reference literature " J.C.Su, S.Y.Liang, Residual stress modeling in machining Database disclosed in processes, Georgia Institute of Technology. " determines material melting point Tm, plasticity should Become ε, plastic strain rate ε, with reference to strain rate ε0And material J-C constitutive parameters A, B, C, m and n.
Step 4: Cutting Force Coefficient is calculated using following formula：
In formula, KtcIt is the Cutting Force Coefficient in cutting speed direction, KfcIt is perpendicular to the cutting force in workpiece machining surface direction Coefficient.
Step 5: cutting force is calculated using following formula：
Ft=Ktcdaptc+Ktedap
Ff=Kfcdaptc+Kfedap
In formula, dapBe value be 0.01mm inclined cutting infinitesimal axial width, tcIt is the cutting of inclined cutting infinitesimal Thickness, KteAnd KfeIt is plough shear force coefficient.FtIt is the cutting force in cutting speed direction, FfIt is perpendicular to workpiece machining surface direction Cutting force.
Step 6: the normal direction cutting force on shear surface is calculated using below equation：
Fn=Ftcosβsinφn+Ff cosφn
In formula, FnIt is the normal direction cutting force on shear surface.
Step 7: the normal stress on shear surface is calculated using below equation：
In formula, psIt is the normal stress on shear plane.
Step 8: the length of shear surface is calculated using below equation：
In formula,It is the length of shear surface.
Step 9: the stress distribution in region is cut using below equation calculating tool arc plough：
In formula, PthrustIt is that normal direction ploughs shear force, PcutIt is tangential plough shear force, obtained plough cuts the normal stress p in regioneFor 1324.6MPa, plough cut the tangential stress q in regioneFor 2317.4MPa, plough cuts the length in regionFor 0.087mm, coefficient of friction μ is 0.3.
Step 10: using following formula come calculate respectively shearing and plough cut caused by stress distribution.
In formula, σxxIt is stress along the x-axis direction, σzzIt is stress along the z-axis direction, τxzIt is the shear stress in xz directions, just To with tangential stress value.For share zone, the result based on step 3 and step 8, by p (s)=ps, q (s)=qs,Substitute into above formula；Region is cut for plough, the result based on step 9, by p (s)=pe, q (s)=qe,Generation Enter above formula.
Step 11: using following formula will shear and plough cut caused by stress distribution be overlapped：
[σ1']=Q1 T[σ1]Q1
[σ2']=Q2 T[σ2]Q2
[σ3]=[σ1']+[σ2']
In formula, [σ1] represent value of the stress distribution under the coordinate system of shear zone caused by shearing, [σ2] represent that plough cuts caused by Value of the stress distribution under Li Qie areas coordinate system, [σ1'] value of the stress distribution under workpiece coordinate system caused by shearing is represented, [σ2'] represent plough cut caused by value of the stress distribution under workpiece coordinate system, [σ3] represent stress point actual under workpiece coordinate system Implantation.Q1And Q2It is the three-dimensional coordinate transformation matrix of stress tensor.
Step 12: the stress distribution value under inclined cutting unit is transformed into milling coordinate system using following formula.
In formula, φwIt is that milling cutter immerses angle, [σ] is the stress distribution value under milling coordinate system.
Step 13: the stress increment in a tiny time dt is calculated using following formula：
[d σ]=[σ (t+dt)]-[σ (t)]
In formula, [d σ] is the stress increment in a tiny time, and [σ (t)] represents what is obtained based on step 12 result Milling process stress distribution and the relation of time.
Step 14: stress variation schematic diagram in the milling process represented in reference picture 3.Judge workpiece at certain using following formula Whether one moment enters mecystasis：
In formula, J2It is the invariant of deviatoric tensor of stress second, the shear yield strength τ of materialsyFor 497MPa, F is yield surface, SijRepresent deviatoric stress, αijRepresent back stress.
Step 15: using the stress increment under following formula computational plasticity state：
In formula, elasticity modulus of materials E is 114GPa, and Poisson's ratio v is 0.33, d σijObtained based on step 13 result Stress increment,WithIt is the stress increment under mecystasis, d εxxWith d εyyIt is plastic strain increment, nijIt is that plasticity should Unit normal vector on variability direction.
Step 16: the boundary condition of stress relaxation is judged using following formula：
In formulaIt is the residual stress before relaxation,It is overstrain, f1,2,...,6(z) expression one is relevant with coordinate value z A non-zero amount.
Step 17: represent often to walk relaxed length using following formula：
In formula, relaxation total step number M value takes 1000, Δ σijExpression often walks stress relaxation number, Δ εijRepresent that often step should fluff Relaxation amount.
Step 18: the stress increment of elastic release process is calculated using following formula：
Δσxy=2G Δs εxy
In formula, elastic properties of materials shear modulus G is 42.8GPa.
Step 19: the stress increment of process is discharged using following formula computational plasticity：
Step 20: the final residual stress after relaxation step terminates, obtained is calculated with following formula：
In formula,WithIt is the final residual stress in workpiece coordinate system x directions and y directions respectively, by its value and experiment The value surveyed is contrasted.
The milling residual stress that the present embodiment prediction is can be seen that from the prediction result of Fig. 2 residual stress is residual with surveying Residue stress coincide preferably, illustrates that the residual stress Forecasting Methodology has reliability.

Claims (1)

1. a kind of three-dimensional milling residual stress Forecasting Methodology, it is characterised in that comprise the following steps：
Step 1: milling process is divided into several three-dimensional inclined cutting infinitesimals；
Step 2: calculated using below equation in three-dimensional Oblique Cutting Process, the angular relationship of cutting force and cutting speed：
sinθi=sin βasinηflow
tan(θn+αn)=tan βacosηflow
<mrow> <msub> <mi>tan&amp;eta;</mi> <mrow> <mi>f</mi> <mi>l</mi> <mi>o</mi> <mi>w</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>&amp;beta;</mi> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;phi;</mi> <mi>n</mi> </msub> <mo>-</mo> <msub> <mi>&amp;alpha;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <msub> <mi>cos&amp;alpha;</mi> <mi>n</mi> </msub> <msub> <mi>tan&amp;phi;</mi> <mi>i</mi> </msub> </mrow> <mrow> <msub> <mi>sin&amp;phi;</mi> <mi>n</mi> </msub> </mrow> </mfrac> </mrow>
<mrow> <msub> <mi>sin&amp;phi;</mi> <mi>i</mi> </msub> <mo>=</mo> <msqrt> <mn>2</mn> </msqrt> <msub> <mi>sin&amp;theta;</mi> <mi>i</mi> </msub> </mrow>
<mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;phi;</mi> <mi>n</mi> </msub> <mo>+</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>tan&amp;theta;</mi> <mi>i</mi> </msub> </mrow> <mrow> <msub> <mi>tan&amp;phi;</mi> <mi>i</mi> </msub> </mrow> </mfrac> </mrow>
In formula, β is milling cutter's helix angle, αnIt is tool orthogonal rake, βaIt is angle of friction, ηflowIt is chip flow angle, θiAnd θnIt is that definition is cut Cut force direction angle, φiAnd φnIt is shear rate deflection；
Step 3: the shear flow stress on shear surface is calculated using following formula：
<mrow> <msub> <mi>&amp;tau;</mi> <mi>s</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <msqrt> <mn>3</mn> </msqrt> </mfrac> <mrow> <mo>(</mo> <mi>A</mi> <mo>+</mo> <msup> <mi>B&amp;epsiv;</mi> <mi>n</mi> </msup> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>C</mi> <mi>l</mi> <mi>n</mi> <mfrac> <mover> <mi>&amp;epsiv;</mi> <mo>&amp;CenterDot;</mo> </mover> <msub> <mi>&amp;epsiv;</mi> <mn>0</mn> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&amp;lsqb;</mo> <mn>1</mn> <mo>-</mo> <msup> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>T</mi> <mo>-</mo> <msub> <mi>T</mi> <mn>0</mn> </msub> </mrow> <mrow> <msub> <mi>T</mi> <mi>m</mi> </msub> <mo>-</mo> <msub> <mi>T</mi> <mn>0</mn> </msub> </mrow> </mfrac> <mo>)</mo> </mrow> <mi>m</mi> </msup> <mo>&amp;rsqb;</mo> </mrow>
In formula, τsIt is shear flow stress, ε is plastic strain,It is plastic strain rate, ε0It is to refer to strain rate, T is shear surface temperature Degree, T0It is room temperature, TmIt is material melting point, A, B, C, m and n are the J-C constitutive parameters of material；
Step 4: Cutting Force Coefficient is calculated using following formula：
<mrow> <msub> <mi>K</mi> <mrow> <mi>t</mi> <mi>c</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>&amp;tau;</mi> <mi>s</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>cos&amp;theta;</mi> <mi>n</mi> </msub> <mo>+</mo> <msub> <mi>tan&amp;theta;</mi> <mi>i</mi> </msub> <mi>tan</mi> <mi>&amp;beta;</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mo>&amp;lsqb;</mo> <mi>cos</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>+</mo> <msub> <mi>&amp;phi;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <msub> <mi>cos&amp;phi;</mi> <mi>i</mi> </msub> <mo>+</mo> <msub> <mi>tan&amp;theta;</mi> <mi>i</mi> </msub> <msub> <mi>sin&amp;phi;</mi> <mi>i</mi> </msub> <mo>&amp;rsqb;</mo> <msub> <mi>sin&amp;phi;</mi> <mi>n</mi> </msub> </mrow> </mfrac> </mrow>
<mrow> <msub> <mi>K</mi> <mrow> <mi>f</mi> <mi>c</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>&amp;tau;</mi> <mi>s</mi> </msub> <msub> <mi>sin&amp;theta;</mi> <mi>n</mi> </msub> </mrow> <mrow> <mo>&amp;lsqb;</mo> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;theta;</mi> <mi>n</mi> </msub> <mo>+</mo> <msub> <mi>&amp;phi;</mi> <mi>n</mi> </msub> <mo>)</mo> </mrow> <msub> <mi>cos&amp;phi;</mi> <mi>i</mi> </msub> <mo>+</mo> <msub> <mi>tan&amp;theta;</mi> <mi>i</mi> </msub> <msub> <mi>sin&amp;phi;</mi> <mi>i</mi> </msub> <mo>&amp;rsqb;</mo> <msub> <mi>cos&amp;beta;sin&amp;phi;</mi> <mi>n</mi> </msub> </mrow> </mfrac> </mrow>
In formula, KtcIt is the Cutting Force Coefficient in cutting speed direction, KfcIt is perpendicular to the Cutting Force Coefficient in workpiece machining surface direction；
Step 5: cutting force is calculated using following formula：
Ft=Ktcdaptc+Ktedap
Ff=Kfcdaptc+Kfedap
In formula, dapIt is the axial width of inclined cutting infinitesimal, tcIt is the thickness of cutting of inclined cutting infinitesimal, KteAnd KfeIt is that plough is cut Force coefficient；FtIt is the cutting force in cutting speed direction, FfIt is perpendicular to the cutting force in workpiece machining surface direction；
Step 6: the normal direction cutting force F on shear surface is calculated using below equationn：
Fn=Ftcosβsinφn+Ffcosφn
In formula, FnIt is the normal direction cutting force on shear surface；
Step 7: the normal stress on shear surface is calculated using below equation
<mrow> <msub> <mi>p</mi> <mi>s</mi> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>F</mi> <mi>n</mi> </msub> <msub> <mi>cos&amp;beta;sin&amp;phi;</mi> <mi>n</mi> </msub> </mrow> <mrow> <msub> <mi>da</mi> <mi>p</mi> </msub> <msub> <mi>t</mi> <mi>c</mi> </msub> </mrow> </mfrac> </mrow>
In formula, psIt is the normal stress on shear plane；
Step 8: the length of shear surface is calculated using below equation：
<mrow> <mover> <mrow> <mi>D</mi> <mi>F</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mo>=</mo> <mfrac> <msub> <mi>t</mi> <mi>c</mi> </msub> <mrow> <msub> <mi>sin&amp;phi;</mi> <mi>n</mi> </msub> <msub> <mi>cos&amp;phi;</mi> <mi>i</mi> </msub> </mrow> </mfrac> </mrow> 1
In formula,It is the length of shear surface；
Step 9: the stress distribution that tool arc plough cuts region is considered uniform, calculated using below equation：
<mrow> <msub> <mi>p</mi> <mi>e</mi> </msub> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <msub> <mi>P</mi> <mrow> <mi>t</mi> <mi>h</mi> <mi>r</mi> <mi>u</mi> <mi>s</mi> <mi>t</mi> </mrow> </msub> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>&amp;beta;</mi> </mrow> <mrow> <msub> <mi>&amp;pi;da</mi> <mi>p</mi> </msub> <mo>&amp;CenterDot;</mo> <mover> <mrow> <mi>G</mi> <mi>E</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> </mrow> </mfrac> </mrow>
<mrow> <msub> <mi>q</mi> <mi>e</mi> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>&amp;mu;P</mi> <mrow> <mi>c</mi> <mi>u</mi> <mi>t</mi> </mrow> </msub> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>&amp;beta;</mi> </mrow> <mrow> <msub> <mi>da</mi> <mi>p</mi> </msub> <mo>&amp;CenterDot;</mo> <mover> <mrow> <mi>G</mi> <mi>E</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> </mrow> </mfrac> </mrow>
In formula, peIt is to plough the normal stress for cutting region, qeIt is to plough the tangential stress for cutting region, PthrustIt is that normal direction ploughs shear force, PcutIt is Tangential plough shear force,It is the length that plough cuts region, μ is coefficient of friction；
Step 10: using following formula come calculate respectively shearing and plough cut caused by stress distribution；
<mrow> <msub> <mi>&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mn>2</mn> <mi>z</mi> </mrow> <mi>&amp;pi;</mi> </mfrac> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mo>-</mo> <mi>a</mi> </mrow> <mi>a</mi> </msubsup> <mfrac> <mrow> <mi>p</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> <msup> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>z</mi> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mi>d</mi> <mi>s</mi> <mo>-</mo> <mfrac> <mn>2</mn> <mi>&amp;pi;</mi> </mfrac> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mo>-</mo> <mi>a</mi> </mrow> <mi>a</mi> </msubsup> <mfrac> <mrow> <mi>q</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>3</mn> </msup> </mrow> <msup> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>z</mi> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mi>d</mi> <mi>s</mi> </mrow>
<mrow> <msub> <mi>&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mn>2</mn> <msup> <mi>z</mi> <mn>3</mn> </msup> </mrow> <mi>&amp;pi;</mi> </mfrac> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mo>-</mo> <mi>a</mi> </mrow> <mi>a</mi> </msubsup> <mfrac> <mrow> <mi>p</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>z</mi> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mi>d</mi> <mi>s</mi> <mo>-</mo> <mfrac> <mrow> <mn>2</mn> <msup> <mi>z</mi> <mn>2</mn> </msup> </mrow> <mi>&amp;pi;</mi> </mfrac> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mo>-</mo> <mi>a</mi> </mrow> <mi>a</mi> </msubsup> <mfrac> <mrow> <mi>q</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>z</mi> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mi>d</mi> <mi>s</mi> </mrow>
<mrow> <msub> <mi>&amp;tau;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mn>2</mn> <msup> <mi>z</mi> <mn>2</mn> </msup> </mrow> <mi>&amp;pi;</mi> </mfrac> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mo>-</mo> <mi>a</mi> </mrow> <mi>a</mi> </msubsup> <mfrac> <mrow> <mi>p</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> </mrow> <msup> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>z</mi> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mi>d</mi> <mi>s</mi> <mo>-</mo> <mfrac> <mrow> <mn>2</mn> <mi>z</mi> </mrow> <mi>&amp;pi;</mi> </mfrac> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mo>-</mo> <mi>a</mi> </mrow> <mi>a</mi> </msubsup> <mfrac> <mrow> <mi>q</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> <msup> <mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mi>s</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <mi>z</mi> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mi>d</mi> <mi>s</mi> </mrow>
In formula, σxxIt is stress along the x-axis direction, σzzIt is stress along the z-axis direction, τxzIt is the shear stress in xz directions, a is to connect The half of the length in region is touched, coordinate value x and z represent the position of calculation point of stress, and p (s) and q (s) are positive and tangential stresses Value；For share zone, based on step 3 and step 7 result, by p (s)=ps, q (s)=qs,Substitute into above formula；It is right Region is cut in plough, based on step 8 result, by p (s)=pe, q (s)=qe,Substitute into above formula；
Step 11: using following formula will shear and plough cut caused by stress distribution be overlapped；
[σ1']=Q1 T[σ1]Q1
[σ2']=Q2 T[σ2]Q2
[σ3]=[σ1']+[σ2']
In formula, [σ1] represent value of the stress distribution under the coordinate system of shear zone caused by shearing, [σ2] represent plough cut caused by stress The value being distributed under Li Qie areas coordinate system, [σ1'] represent value of the stress distribution under workpiece coordinate system caused by shearing, [σ2'] table Show plough cut caused by value of the stress distribution under workpiece coordinate system, [σ3] represent stress distribution value actual under workpiece coordinate system；Q1 And Q2It is the three-dimensional coordinate transformation matrix of stress tensor；
Step 12: the stress distribution value under inclined cutting unit is transformed into milling coordinate system using following formula；
<mrow> <mo>&amp;lsqb;</mo> <mi>&amp;sigma;</mi> <mo>&amp;rsqb;</mo> <mo>=</mo> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>cos&amp;phi;</mi> <mi>w</mi> </msub> </mrow> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mrow> <msub> <mi>sin&amp;phi;</mi> <mi>w</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>-</mo> <msub> <mi>sin&amp;phi;</mi> <mi>w</mi> </msub> </mrow> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mrow> <msub> <mi>cos&amp;phi;</mi> <mi>w</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mo>&amp;lsqb;</mo> <msub> <mi>&amp;sigma;</mi> <mn>3</mn> </msub> <mo>&amp;rsqb;</mo> <msup> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msub> <mi>cos&amp;phi;</mi> <mi>w</mi> </msub> </mrow> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mrow> <msub> <mi>sin&amp;phi;</mi> <mi>w</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>-</mo> <msub> <mi>sin&amp;phi;</mi> <mi>w</mi> </msub> </mrow> </mtd> <mtd> <mn>0</mn> </mtd> <mtd> <mrow> <msub> <mi>cos&amp;phi;</mi> <mi>w</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> <mi>T</mi> </msup> </mrow>
In formula, φwIt is that milling cutter immerses angle, [σ] is the stress distribution value under milling coordinate system；
Step 13: the stress increment in a tiny time dt is calculated using following formula：
[d σ]=[σ (t+dt)]-[σ (t)]
In formula, [d σ] is the stress increment in a tiny time, and [σ (t)] represents the milling obtained based on step 12 result Process stress is distributed the relation with the time；
Step 14: judge whether workpiece enters mecystasis at some moment using following formula：
<mrow> <msub> <mi>J</mi> <mn>2</mn> </msub> <mo>=</mo> <msqrt> <mrow> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mrow> <mo>(</mo> <msub> <mi>S</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>S</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </msqrt> </mrow>
<mrow> <mi>F</mi> <mo>=</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mrow> <mo>(</mo> <msub> <mi>S</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msub> <mi>S</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>i</mi> <mi>j</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <msubsup> <mi>&amp;tau;</mi> <mrow> <mi>s</mi> <mi>y</mi> </mrow> <mn>2</mn> </msubsup> <mo>=</mo> <mn>0</mn> </mrow>
In formula, J2It is the invariant of deviatoric tensor of stress second, F is yield surface, τsyIt is the shear yield strength of material, SijPartially should it represent Power, αijRepresent back stress；
Step 15: using the stress increment under following formula computational plasticity state：
<mfenced open = "" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>d&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mi>E</mi> </mfrac> <mrow> <mo>(</mo> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>-</mo> <mi>v</mi> <mo>(</mo> <mrow> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>+</mo> <msub> <mi>&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> </mrow> <mo>)</mo> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mi>h</mi> </mfrac> <mrow> <mo>(</mo> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mi>&amp;psi;</mi> <mo>&amp;lsqb;</mo> <mfrac> <mn>1</mn> <mi>E</mi> </mfrac> <mrow> <mo>(</mo> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <mi>v</mi> <mo>(</mo> <mrow> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>+</mo> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> </mrow> <mo>)</mo> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mi>h</mi> </mfrac> <mrow> <mo>(</mo> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> </mtable> </mfenced>
<mfenced open = "" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>d&amp;epsiv;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mi>E</mi> </mfrac> <mrow> <mo>(</mo> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>-</mo> <mi>v</mi> <mo>(</mo> <mrow> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>+</mo> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> </mrow> <mo>)</mo> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mi>h</mi> </mfrac> <mrow> <mo>(</mo> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msubsup> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mrow> <mi>p</mi> <mi>l</mi> </mrow> </msubsup> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>d&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> </mtable> </mfenced>
In formula, E is elasticity modulus of materials, and v is Poisson's ratio, d σijIt is the stress increment obtained based on step 13 result,WithIt is the stress increment under mecystasis, d εxxWith d εyyIt is plastic strain increment, nijIt is the unit on plastic strain rate direction Normal vector；
Step 16: the boundary condition of stress relaxation is judged using following formula：
<mrow> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <mn>0</mn> <mo>,</mo> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <mn>0</mn> <mo>,</mo> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <mn>0</mn> <mo>,</mo> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <msub> <mi>f</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>,</mo> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <msub> <mi>f</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>,</mo> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <msub> <mi>f</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> </mrow>
<mrow> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <mn>0</mn> <mo>,</mo> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <mn>0</mn> <mo>,</mo> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <mn>0</mn> <mo>,</mo> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <msub> <mi>f</mi> <mn>4</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>,</mo> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <msub> <mi>f</mi> <mn>5</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>,</mo> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> <mi>r</mi> </msubsup> <mo>=</mo> <msub> <mi>f</mi> <mn>6</mn> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> </mrow>
In formula,It is the residual stress before relaxation,It is overstrain, f1,2,...,6(z) expression one is relevant with coordinate value z One non-zero amount；
Step 17: represent often to walk relaxed length using following formula：
<mrow> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mi>M</mi> </mfrac> <mo>,</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mi>M</mi> </mfrac> <mo>,</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> <mi>r</mi> </msubsup> <mi>M</mi> </mfrac> </mrow>
<mrow> <msub> <mi>&amp;Delta;&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mi>r</mi> </msubsup> <mi>M</mi> </mfrac> <mo>,</mo> <msub> <mi>&amp;Delta;&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> <mi>r</mi> </msubsup> <mi>M</mi> </mfrac> <mo>,</mo> <msub> <mi>&amp;Delta;&amp;epsiv;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <msubsup> <mi>&amp;epsiv;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mi>r</mi> </msubsup> <mi>M</mi> </mfrac> </mrow>
In formula, M is loose total step number, Δ σijExpression often walks stress relaxation number, Δ εijRepresent often to walk strain relaxation amount；
Step 18: the stress increment of elastic release process is calculated using following formula：
<mrow> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>E&amp;Delta;&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>+</mo> <mi>v</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>v</mi> <mo>)</mo> </mrow> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mi>v</mi> <mn>2</mn> </msup> </mrow> </mfrac> </mrow> 3
<mrow> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>vE&amp;Delta;&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>+</mo> <mi>v</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>v</mi> <mo>)</mo> </mrow> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> </mrow> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mi>v</mi> <mn>2</mn> </msup> </mrow> </mfrac> </mrow>
Δσxy=2G Δs εxy
In formula, G represents elastic properties of materials modulus of shearing；
Step 19: the stress increment of process is discharged with following formula computational plasticity：
<mfenced open = "" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;&amp;epsiv;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mi>E</mi> </mfrac> <mo>&amp;lsqb;</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>-</mo> <mi>v</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mi>h</mi> </mfrac> <mrow> <mo>(</mo> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mn>0</mn> </mrow> </mtd> </mtr> </mtable> </mfenced>
<mfenced open = "" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mi>E</mi> </mfrac> <mo>&amp;lsqb;</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>-</mo> <mi>v</mi> <mrow> <mo>(</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mi>h</mi> </mfrac> <mrow> <mo>(</mo> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced>
<mfenced open = "" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>&amp;Delta;&amp;epsiv;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> </mrow> <mrow> <mn>2</mn> <mi>G</mi> </mrow> </mfrac> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mi>h</mi> </mfrac> <mrow> <mo>(</mo> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>n</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>z</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>z</mi> </mrow> </msub> <mo>+</mo> <mn>2</mn> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>n</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>y</mi> </mrow> </msub> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> </mfenced>
Step 20: the final residual stress after relaxation step terminates, obtained is calculated with following formula：
<mrow> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mrow> <mi>r</mi> <mi>e</mi> </mrow> </msubsup> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>M</mi> </munderover> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> <mi>r</mi> </msubsup> <mo>-</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>x</mi> <mi>x</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>t</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>t</mi> </mrow>
<mrow> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mrow> <mi>r</mi> <mi>e</mi> </mrow> </msubsup> <mo>=</mo> <munderover> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>M</mi> </munderover> <msubsup> <mi>&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> <mi>r</mi> </msubsup> <mo>-</mo> <msub> <mi>&amp;Delta;&amp;sigma;</mi> <mrow> <mi>y</mi> <mi>y</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>t</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>t</mi> </mrow>
In formula,WithIt is the final residual stress in workpiece coordinate system x directions and y directions respectively.
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