CN104680018B - A kind of hydrostatic slideway suppresses the analysis and optimization method of inertia force impact - Google Patents

A kind of hydrostatic slideway suppresses the analysis and optimization method of inertia force impact Download PDF

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CN104680018B
CN104680018B CN201510102727.5A CN201510102727A CN104680018B CN 104680018 B CN104680018 B CN 104680018B CN 201510102727 A CN201510102727 A CN 201510102727A CN 104680018 B CN104680018 B CN 104680018B
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蔡力钢
王语莫
刘志峰
程强
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Beijing University of Technology
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Beijing University of Technology
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Abstract

A kind of hydrostatic slideway suppresses the analysis and optimization method of inertia force impact, and this method belongs to static pressure support and Lubrication Design technical field.Hydrostatic slideway is in heavy machine tool using very extensive.Gantry of heavy machine tool, because the influence of inertia force can rock, not only influences machining accuracy in startup or braking, while dangerous.Hydrostatic slideway will not only meet bearing capacity demand, it is also contemplated that the influence of start/stop impact in design.The present invention introduces the Reynolds equation that oil film thickness changes according to heavy machine tool hydrostatic slideway model.Flexural deformation is introduced simultaneously in the change of oil film thickness with linear deflection to simulate actual condition.With finite difference method, the numerical solution for asking for pressure is iterated by successive overrelaxation method.The change of load-carrying properties in the case of being toppled according to the solution of pressure from fuel delivery parameters analysis caused by different inertia force.Prioritization scheme need to be looked for by attempting the analysis result under different oil pad structures, improve the resistance inertia force impact capacity of hydrostatic slideway.

Description

A kind of hydrostatic slideway suppresses the analysis and optimization method of inertia force impact
Technical field
The present invention is a kind of analysis and optimization method that hydrostatic slideway suppresses inertia force impact, belongs to static pressure support and lubrication Design field.
Background technology
For static pressure support system using widely in heavy machine tool, hydrostatic slideway is that a kind of typical case therein should at present With.External oil feed pump in the lubricating pad installed under pipeline direction guiding rail slide by providing pressure oil, pressure of the fluid in oily pocket It is the major part of hydrostatic slideway bearing capacity, hydrostatic effects of the oil when flowing through lubricating pad sealing oil edge are to maintain pressure in oily pocket It is crucial.Generally, gantry of heavy machine tool is highly some rice, and own wt is more than hundred tons, so starting or making When dynamic, because the influence of inertia force will necessarily rock.This rock can not only influence machining accuracy, at the same exist damage cutter or The potential danger of workpiece.So hydrostatic slideway not only needs meet the needs of static bearing capacity, while also to examine in design Consider the influence of start/stop impact.It can ensure safety if it can improve the inertia force impact resistant ability of heavy gantry machine tool Acceleration of motion with improving gantry on the basis of machining accuracy, so as to improve operating efficiency.The inertia force punching of gantry machine tool Hitting resistivity can be judged in terms of two:The inclined degree of guide rail or same when being influenceed by equal inertia force impact Deng torque size caused by inclined degree lower guideway.
In the problem of pressure distribution in sealing oil edge in solving hydrostatic slideway, Reynolds equation is main analysis method.Thunder The solution of promise equation is to analyze the basis of hydrostatic slideway load-carrying properties, but because its equation is partial differential equation of second order in itself, Asking for for analytic solutions is relatively difficult, so in the current analysis method of static pressure support system again based on numerical method.Finite difference Point method is a kind of very practical numerical method, and the differential equation can be changed into algebraic equation by finite difference method approximation, Solved again by the method for value solving of algebraic equation, finally give the approximate solution that pressure is distributed in sealing oil edge.
The content of the invention
The present invention is according to the hydrostatic slideway model of heavy machine tool, the Reynolds equation changed using a kind of introducing oil film thickness. Flexural deformation is introduced simultaneously in the change of oil film thickness with linear deflection to simulate actual condition.With finite difference method, The numerical solution for asking for pressure is iterated by successive overrelaxation method.Analyzed according to the solution of pressure from fuel delivery parameters different used The change of load-carrying properties in the case of being toppled caused by property power.Wherein fuel system selects common quantitative oil feed pump.Attempt not Prioritization scheme need to be looked for the analysis result under oil pad structure, improves the resistance inertia force impact capacity of hydrostatic slideway.
Bending that is a kind of while considering guide rail slide is established according to the structure of gantry machine tool and inertia force situation first to become The oblique model of shape and linear deflection, to simulate actual condition.Substituted into again according to this oblique model as oil clearance condition Reynolds equation, solve, obtain pressure distribution.Bearing capacity and moment of resistance are solved according to pressure distribution and fuel system afterwards. Finally by different oil pad structures is attempted, the size of moment of resistance is contrasted, draws the optimum results of oil pad structure.
The hydrostatic slideway load-bearing capacity analysis method provided by the invention for considering guide rail facial disfigurement comprises the following steps:
S1. the oil film thickness being firstly introduced under inertia force percussion:
Wherein:dzFor the offset of oil film thickness;S is bending segment length;δ is maximum offset.Each different δ values A kind of corresponding drift condition.
S2. the nondimensionalization of variable is carried out to the parameter in hydrostatic slideway again;
Wherein:P is pressure;p0For pressure in oily pocket;L is hydrostatic slideway lubricating pad length;B is hydrostatic slideway lubricating pad width;W For bearing capacity;MyTilting moment is resisted for y directions;Q is flow;UxFor guide moving velocity;H is oil film thickness;η is fluid Viscosity.For dimensionless pressure;For non-dimensional length;For dimensionless width;For dimensionless thickness;Carried for dimensionless Power;Tilting moment is resisted for y directions dimensionless;Flow is held for dimensionless;For dimensionless guide moving velocity;For nothing Dimension oil film thickness.
S3. according to model Reynolds equation is simplified and by the Reynolds equation after simplification by finite difference method from Dissipate;Generally, the translational speed of hydrostatic slideway is less demanding, so heat problem and unobvious, that is, support the viscous of liquid Degree change can be ignored with variable density, and the Reynolds equation after simplifying is:
It is by the Reynolds equation after finite difference method Approximation Discrete:
Wherein:For x directions discrete steps;For y directions discrete steps;I counts for x directions infinitesimal;J is y directions Infinitesimal counts.
S4. bearing capacity and moment of resistance are solved after obtaining the numerical solution of pressure:
Wherein:QeFor the fuel supply flow rate of each lubricating pad;It is i for coordinate, the distance at j nodal point separation slides center.
S5. according to the above results, the analysis that a variety of different lubricating pad sizes carry out bearing capacity is attempted, finds same tilt The maximum optimal solution of moment of resistance under degree.
Consider that the hydrostatic slideway load-bearing capacity analysis method of guide rail facial disfigurement has the following advantages that.
1st, flexural deformation and linear deflection are introduced to simulate reality simultaneously when analyzing guide rail slide caused by inertia force and tilting Border condition of work, the oil film oblique form to be tallied with the actual situation.
2nd, solved using finite difference method and hardly result in the Reynolds equations of analytic solutions, and according to the pressure drawn be distributed into Row load-carrying properties are analyzed.
3rd, hydrostatic slideway caused moment of resistance under equal inclined degree is improved by way of changing lubricating pad size, So as to improve its inertia force impact resistant ability, the reliability of guide rail work is improved.
Brief description of the drawings
Fig. 1 is the inertia force impact resistant capability analysis and optimization method flow chart of hydrostatic slideway.
Fig. 2 is the structural representation of hydrostatic slideway.
Fig. 3 is the structural representation of hydrostatic slideway lubricating pad.
Fig. 4 a are situation of change of the hydrostatic slideway moment of resistance with offset.
Fig. 4 b are situation of change of the oil film thickness of hydrostatic guide way with offset.
Embodiment
With guide rail slide offset delta=2 × 10-5M, Qe=3.3 × 10-6m3/ s, single slide are when 5 lubricating pads are housed Example.As shown in Fig. 2 set lubricating pad long L, wide B;The wide b of the oily long l of pocket;Design oil film thickness is H0
When solving pressure, Reynolds equation boundary condition is introduced:
The flow chart according to Fig. 1, pressure distribution and guide rail surface deformation in each lubricating pad is calculated and is distributed, and according to Result calculates the bearing capacity and moment of resistance of guide rail accordingly.The bearing capacity of wherein hydrostatic slideway act as supporting the weight of gantry frame Amount, and the deadweight of gantry frame hardly changes in the course of the work, so the actual carrying capacity of hydrostatic slideway is constant, but due to The change of dimensionless bearing capacity, oil film will be caused thinning, this is the embodiment that a kind of bearing capacity reduces.Result of calculation such as Fig. 4 a- Shown in 4b.
The contribution that the guide rail of each lubricating pad supports resistance to capsizing is different, if only changing the length L of each lubricating pad without changing Become oily pocket proportion l/L and slide overall length, static load-carrying properties of the hydrostatic slideway under without inclination conditions be it is constant, this Sample can be with the ability of the improvement resistance inertia force impact on the premise of static supporting effect is not influenceed.Adding in view of inertia force Symmetry when speed or deceleration, so when changing L, it is ensured that 1stWith 5thIt is equal, 2ndWith 4thEqual, 5 lubricating pads length and The size of constant while each lubricating pad is kept to be more than zero.Optimizing constraints is:
By attempting the length dimension of a variety of lubricating pads, oil film thickness and moment of resistance is calculated, finds wherein optimal Solution.Under this constraints, work as L1st=0.9, L2nd=0.15, L3rdOptimal solution is obtained when=0.9, only have lost less than Resistance tilting moment is just improved about 100% by 10% oil film thickness.

Claims (1)

1. a kind of hydrostatic slideway suppresses the analysis and optimization method of inertia force impact, it is characterised in that:The implementation process of this method It is as follows,
S1. the oil film thickness being firstly introduced under inertia force percussion:
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Wherein:dzFor the offset of oil film thickness;S is bending segment length;δ is maximum offset;Each different δ values are corresponding A kind of drift condition;
S2. the nondimensionalization of variable is carried out to the parameter in hydrostatic slideway again;
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Wherein:P is pressure;p0For pressure in oily pocket;L is hydrostatic slideway lubricating pad length;B is hydrostatic slideway lubricating pad width;W is to hold Loading capability;MyTilting moment is resisted for y directions;Q is flow;UxFor guide moving velocity;H is oil film thickness;η is oil viscosity;For dimensionless pressure;For non-dimensional length;For dimensionless width;For dimensionless thickness;For dimensionless bearing capacity; Tilting moment is resisted for y directions dimensionless;Flow is held for dimensionless;For dimensionless guide moving velocity;For dimensionless oil Film thickness;
S3. according to model Reynolds equation is simplified and the Reynolds equation after simplification is discrete by finite difference method;It is quiet Press the translational speed of guide rail less demanding, so the viscosity B coefficent and variable density of heat problem and unobvious, i.e. support liquid It can ignore, the Reynolds equation after simplifying is:
<mrow> <mfrac> <mo>&amp;part;</mo> <mrow> <mo>&amp;part;</mo> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> </mrow> </mfrac> <mrow> <mo>(</mo> <msup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mn>3</mn> </msup> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <mo>&amp;part;</mo> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> </mrow> <mrow> <mo>&amp;part;</mo> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mfrac> <mi>L</mi> <mi>B</mi> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mfrac> <mo>&amp;part;</mo> <mrow> <mo>&amp;part;</mo> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> </mrow> </mfrac> <mrow> <mo>(</mo> <msup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mn>3</mn> </msup> <mo>&amp;CenterDot;</mo> <mfrac> <mrow> <mo>&amp;part;</mo> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> </mrow> <mrow> <mo>&amp;part;</mo> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>=</mo> <mn>6</mn> <mfrac> <mo>&amp;part;</mo> <mrow> <mo>&amp;part;</mo> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> </mrow> </mfrac> <mrow> <mo>(</mo> <mover> <msub> <mi>U</mi> <mi>x</mi> </msub> <mo>&amp;OverBar;</mo> </mover> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mo>)</mo> </mrow> </mrow>
It is by the Reynolds equation after finite difference method Approximation Discrete:
<mrow> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msubsup> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> <msubsup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msubsup> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> <msubsup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mfrac> <mi>L</mi> <mi>B</mi> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msubsup> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> <msubsup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mfrac> <mi>L</mi> <mi>B</mi> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msubsup> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> <msubsup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>-</mo> <mn>1</mn> </mrow> <mn>3</mn> </msubsup> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>-</mo> <mn>1</mn> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mn>6</mn> <mrow> <mo>(</mo> <mrow> <msub> <mover> <mrow> <mi>U</mi> <mi>x</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> </msub> <msub> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mo>-</mo> <msub> <mover> <mrow> <mi>U</mi> <mi>x</mi> </mrow> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> </msub> <msub> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <msub> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> </msub> <msubsup> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> </mrow> </mtd> </mtr> </mtable> </mfenced> <mfenced open = "[" close = "]"> <mtable> <mtr> <mtd> <mrow> <msubsup> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> <msubsup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> <mo>+</mo> <msubsup> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> <msubsup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>-</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mfrac> <mi>L</mi> <mi>B</mi> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msubsup> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> <msubsup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> <mn>3</mn> </msubsup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mfrac> <mi>L</mi> <mi>B</mi> </mfrac> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msubsup> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> <mn>2</mn> </msubsup> <msubsup> <mover> <mi>h</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>-</mo> <mn>1</mn> </mrow> <mn>3</mn> </msubsup> </mrow> </mtd> </mtr> </mtable> </mfenced> </mfrac> </mrow>
Wherein:For x directions discrete steps;For y directions discrete steps;I counts for x directions infinitesimal;J is y directions infinitesimal Count;
S4. bearing capacity and moment of resistance are solved after obtaining the numerical solution of pressure:
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <mover> <mi>W</mi> <mo>&amp;OverBar;</mo> </mover> <mo>=</mo> <mi>&amp;Sigma;</mi> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> <mrow> <mo>(</mo> <mrow> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <msub> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> </msub> <mo>&amp;CenterDot;</mo> <msub> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>W</mi> <mo>=</mo> <mfrac> <mrow> <mover> <mi>W</mi> <mo>&amp;OverBar;</mo> </mover> <mi>L</mi> <mi>B</mi> <mi>&amp;eta;</mi> </mrow> <mrow> <mover> <mi>q</mi> <mo>&amp;OverBar;</mo> </mover> <msubsup> <mi>H</mi> <mn>0</mn> <mn>3</mn> </msubsup> </mrow> </mfrac> <mo>&amp;CenterDot;</mo> <msub> <mi>Q</mi> <mi>e</mi> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mover> <mi>M</mi> <mo>&amp;OverBar;</mo> </mover> <mi>y</mi> </msub> <mo>=</mo> <mi>&amp;Sigma;</mi> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> <mrow> <mo>(</mo> <mrow> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>p</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mrow> <mo>(</mo> <mrow> <msub> <mover> <mi>r</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>,</mo> <mi>j</mi> </mrow> </msub> <mo>+</mo> <msub> <mover> <mi>r</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>,</mo> <mi>j</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>&amp;CenterDot;</mo> <msub> <mover> <mi>x</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> </msub> <mo>&amp;CenterDot;</mo> <msub> <mover> <mi>y</mi> <mo>&amp;OverBar;</mo> </mover> <mrow> <mi>s</mi> <mi>t</mi> <mi>e</mi> <mi>p</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>M</mi> <mi>y</mi> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mover> <mi>M</mi> <mo>&amp;OverBar;</mo> </mover> <mi>y</mi> </msub> <msup> <mi>L</mi> <mn>2</mn> </msup> <mi>B</mi> <mi>&amp;eta;</mi> </mrow> <mrow> <mover> <mi>q</mi> <mo>&amp;OverBar;</mo> </mover> <msubsup> <mi>H</mi> <mn>0</mn> <mn>3</mn> </msubsup> </mrow> </mfrac> <mo>&amp;CenterDot;</mo> <msub> <mi>Q</mi> <mi>e</mi> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced>
Wherein:QeFor the fuel supply flow rate of each lubricating pad;It is i for coordinate, the distance at j nodal point separation slides center;
S5. according to the above results, the analysis that a variety of different lubricating pad sizes carry out bearing capacity is attempted, finds same tilt degree The maximum optimal solution of lower moment of resistance;Hydrostatic slideway model has 5 lubricating pads, in optimization process when changing L, it is ensured that 1stWith 5thIt is equal, 2ndWith 4thEqual, 5 lubricating pads length and the size of constant while each lubricating pad is kept to be more than zero;Optimization constraint bar Part is:
<mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>L</mi> <mrow> <mn>1</mn> <mi>s</mi> <mi>t</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>L</mi> <mrow> <mn>5</mn> <mi>t</mi> <mi>h</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>L</mi> <mrow> <mn>2</mn> <mi>n</mi> <mi>d</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>L</mi> <mrow> <mn>4</mn> <mi>t</mi> <mi>h</mi> </mrow> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>&amp;Sigma;L</mi> <mi>k</mi> </msub> <mo>=</mo> <mn>3</mn> <mi>m</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>L</mi> <mi>k</mi> </msub> <mo>&gt;</mo> <mn>0</mn> </mrow> </mtd> </mtr> </mtable> </mfenced>
By attempting the length dimension of a variety of lubricating pads, oil film thickness and moment of resistance is calculated, finds optimal solution wherein; Under this constraints, work as L1st=0.9, L2nd=0.15, L3rdOptimal solution is obtained when=0.9, only have lost the oil less than 10% Resistance tilting moment is just improved about 100% by film thickness.
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