CN104175831B - The design method of the inner circle sleeve thickness of suspension stabiliser bar rubber bushing - Google Patents

The design method of the inner circle sleeve thickness of suspension stabiliser bar rubber bushing Download PDF

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CN104175831B
CN104175831B CN201410476274.8A CN201410476274A CN104175831B CN 104175831 B CN104175831 B CN 104175831B CN 201410476274 A CN201410476274 A CN 201410476274A CN 104175831 B CN104175831 B CN 104175831B
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msub
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delta
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CN104175831A (en
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周长城
宋群
于曰伟
张云山
潘礼军
曹海琳
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Shandong University of Technology
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Abstract

The present invention relates to the design method of the inner circle sleeve thickness of suspension stabiliser bar rubber bushing, belongs to vehicle suspension technical field, and previous home and abroad fails to provide reliable resolution design method always.Present invention be characterized in that:Within round buss thickness be parameter to be designed, according to the roll angular rigidity design requirement value of stabilizer bar system, the wheelspan of vehicle, the structure and material characteristic parameter of stabiliser bar and rubber bushing, the mathematical model of optimizing design of the inner circle sleeve thickness of rubber bushing is established, can obtain inner circle sleeve thickness using Matlab programs optimizes design load.The optimization design value of accurate, reliable inner circle sleeve thickness is can obtain using this method, the design level of stabiliser bar is improved, i.e., only by the optimization design of internal round buss Thickness ness stabilizer bar system can be made to reach the design requirement of roll angular rigidity;Meanwhile design and testing expenses can be reduced using this method, improve vehicle ride performance and handling safety.

Description

The design method of the inner circle sleeve thickness of suspension stabiliser bar rubber bushing
Technical field
The present invention relates to the design of the inner circle sleeve thickness of vehicle suspension stabiliser bar, particularly suspension stabiliser bar rubber bushing Method.
Background technology
Not only the structure by stabiliser bar, diameter are influenceed the roll angular rigidity of suspension system, while are also served as a contrast by rubber The influence of the factors such as length, inner circle radius, exradius, material property and the installation displacement of set, wherein, the inner circle of rubber bushing Radius, determined by the inner circle sleeve thickness for stablizing shank diameter and rubber bushing.However, due to radially being become by rubber bushing Shape and end part of stabilizer rod vertical deviation deformation analytical Calculation theory and the restriction to intercouple key issues of influence, given steady In the case of fixed pole structure and material characterisitic parameter and rubber bushing exradius, axial length and installation site, to stabiliser bar rubber The design of the inner circle sleeve thickness of glue bushing, at present home and abroad fail to provide reliable resolution design method always.State at present Inside and outside scholar is mostly to utilize simulation analysis software, QS system variant under given structure and load condition and Rigidity carry out Numerical Simulation Analysis, still, this method can only to give structure and load condition under stabilizer bar system deformation and Rigidity carries out simulating, verifying, no analytical formula, it is impossible to meets stabilizer bar system analytical design method and modernizes wanting for CAD design Ask.
With the fast development of Vehicle Industry and the raising of travel speed, the design to suspension stabilizer bar system proposes more High design requirement.How given stabiliser bar structure, diameter and material property, it is and the material property of rubber bushing, cylindrical In the case that radius and two rubber bushing installation sites are constant, i.e., on the premise of design and production cost is not increased, only pass through To the optimization design of the inner circle sleeve thickness of rubber bushing, stabilizer bar system can be made to reach the design requirement of roll angular rigidity, It is current enterprise technical problem in the urgent need to address.Therefore, it is necessary to establish a kind of accurate, reliable suspension stabiliser bar rubber lining The design method of the inner circle sleeve thickness of set, vehicle suspension design level is improved, it is only logical on the premise of production cost is not increased Cross the optimization design of the inner circle sleeve thickness of stabiliser bar rubber bushing so that the design that stabilizer bar system reaches roll angular rigidity will Ask, improve vehicle ride performance and handling safety.
The content of the invention
For defect present in above-mentioned prior art, the technical problems to be solved by the invention be to provide it is a kind of easy, The design method of the inner circle sleeve thickness of reliable suspension stabiliser bar rubber bushing, its design flow diagram is as shown in figure 1, stabiliser bar Structural representation is as shown in Figure 2.
In order to solve the above technical problems, the inner circle sleeve thickness of suspension stabiliser bar rubber bushing provided by the present invention is set Meter method, it is characterised in that using following steps:
(1) the vertical deviation deformation coefficient G of end part of stabilizer rod is calculatedw
According to the total length l of suspension stabiliser barc, brachium l1, the mounting distance l between two rubber bushings0, transition arc Radius R, the central angle θ of transition arc, the elasticity modulus of materials E and Poisson's ratio μ of stabiliser bar, to the vertical deviation of end part of stabilizer rod Deformation coefficient GwCalculated, i.e.,:
In formula,
G6=-32 (μ+1) [R (cos θ -1)-l1sinθ]2[2l1cosθ-lc+2Rsinθ];
(2) rubber bushing RADIAL stiffness K is establishedxExpression formula:
According to stablizing shank diameter d, the axial length L of rubber bushing, the elasticity modulus of materials E of rubber bushingx, Poisson's ratio μx, Exradius rb, inner circle radius ra=d/2+ δ, wherein, δ is the design parameter of the inner circle sleeve thickness of rubber bushing, with δ As parameter to be designed, rubber bushing RADIAL stiffness K is establishedxCalculation expression, i.e.,
Wherein, Kx(δ) is the expression formula on inner circle sleeve thickness δ;
b1(δ)=(d/2+ δ) ((d/2+ δ)2+3rb 2)[I(1,αa)K(0,αa)+K(1,αa)I(0,αa)],
b2(δ)=- rb(rb 2+3(d/2+δ)2)[I(1,αhb)K(0,αa)+K(1,αhb)I(0,αa)],
Bessel correction functions:I(0,αhb), K (0, αhb);I(1,αhb), K (1, αhb);
I(1,αa), K (1, αa);I(0,αa), K (0, αa);
(3) foundation and design calculating of the inner circle sleeve thickness design mathematic model of rubber bushing:
The wheelspan B, the diameter d of stabiliser bar, total length l of suspension according to where stabiliser barc, stabilizer bar system roll angular rigidity Design requirement valueThe vertical deviation deformation coefficient G of end part of stabilizer rod obtained by being calculated in step (1)w, and step (2) Middle the rubber bushing radial direction Line stiffness expression formula K establishedx(δ), establish the inner circle sleeve thickness δ of rubber bushing design number Model is learned, i.e.,:
Using Matlab calculation procedures, above-mentioned mathematical modeling is solved, can be obtained in stabilizator rod structure and rubber bushing peace Under holding position permanence condition, meet the inner circle thickness δ of the rubber bushing of stabilizer bar system roll angular rigidity design requirement design Value.
The present invention has the advantage that than prior art:
Due to being deformed analytical Calculation theory by rubber bushing radial deformation and end part of stabilizer rod vertical deviation and intercoupled The restriction of key issues of influence, at present home and abroad fail to provide always for the inner circle sleeve thickness of stabiliser bar rubber bushing Reliable resolution design method, mostly it is to utilize finite simulation element analysis software, line number is entered to stabilizer bar system deformation and rigidity It is worth simulation analysis, still, this method can only be carried out to the deformation and rigidity for giving the stabilizer bar system under structure and load condition Simulating, verifying, no analytical formula, it is impossible to meet stabilizer bar system analytical design method and modernize the requirement of CAD design.
The present invention can be according to the wheelspan of vehicle, the structure and material characteristic parameter of stabiliser bar, the exradius of rubber bushing, Axial length, and the roll angular rigidity design requirement value of stabilizer bar system, served as a contrast by the deformation coefficient and rubber of end part of stabilizer rod The radial rigidity expression formula of set, the mathematical model of optimizing design of the inner circle sleeve thickness of stabiliser bar rubber bushing is established, utilized Matlab programs can obtain the optimization design value of the inner circle sleeve thickness of rubber bushing.Using this method can obtain it is accurate, can The optimization design value of the inner circle sleeve thickness of the rubber bushing leaned on, on the premise of design and production cost is not increased, only pass through To the optimization design of the inner circle sleeve thickness of rubber bushing, stabilizer bar system can be made to reach the design requirement of roll angular rigidity, It is horizontal to improve vehicle suspension design, improves vehicle ride performance and handling safety;Meanwhile it can also be reduced and set using this method Meter and testing expenses, improve the economic benefit of enterprise.
In order to more fully understand that invention is described further below in conjunction with the accompanying drawings:
Fig. 1 be suspension stabiliser bar rubber bushing clipping room away from design flow diagram;
Fig. 2 is the structural representation of lateral stability lever system;
Fig. 3 is the structural representation of rubber bushing;
Fig. 4 be embodiment one stabilizer bar system roll angular rigidity with rubber bushing clipping room away from change curve;
Fig. 5 be embodiment three stabilizer bar system roll angular rigidity with rubber bushing clipping room away from change curve.
Specific embodiment
The present invention is described in further detail below by embodiment.
Embodiment one:The wheelspan B=1600mm of certain automobile front-axle, the structure of stabiliser bar is used, as shown in Fig. 2 wherein, lcFor the total length of stabiliser bar, lc=800mm;l1For brachium, l1=150mm;l0Mounting distance between rubber bushing, l0= 400mm;R is transition arc radius, R=50mm;θ is transition arc central angle, θ=60 °;The elastic modulus E of stable bar material =210GPa, Poisson's ratio μ=0.3.The structure of rubber bushing is as shown in figure 3, wherein, stabiliser bar 1, interior round buss 2, rubber bushing 3, outer round buss 4, the diameter d=20mm of stabiliser bar 1, the elastic modulus E of rubber bushing 3x=7.84MPa, Poisson's ratio μx= 0.47, axial length L=25mm, exradius rb=30mm, inner circle radius ra=(10+ δ) mm, wherein, δ is to be designed Rubber bushing wall thickness of internal cylindrical sleeve δ parameter.The design requirement value of the roll angular rigidity of the vehicle front suspension stabilizer bar systemIn the case of given stabiliser bar and rubber bushing installation site, to rubber bushing inner circle sleeve wall Thick δ is designed.
The design method of the inner circle sleeve thickness for the suspension stabiliser bar rubber bushing that present example is provided, it designs stream Journey is as shown in figure 1, comprise the following steps that:
(1) the vertical deviation deformation coefficient G of end part of stabilizer rod is calculatedw
According to the total length l of stabiliser barc=800mm, brachium l1=150mm, transition arc radius R=50mm, transition arc Central angle θ=60 °, the elasticity modulus of materials E=210GPa of stabiliser bar, between Poisson's ratio μ=0.3, and two rubber bushings Mounting distance l0=400mm, to the vertical deviation deformation coefficient G of end part of stabilizer rodwCalculated, i.e.,:
In formula,
G6=-32 (μ+1) [R (cos θ -1)-l1sinθ]2[2l1cosθ-lc+ 2Rsin θ]=- 0.5624m3
(2) rubber bushing RADIAL stiffness K is establishedxExpression formula:
According to stablizing shank diameter d=20mm, axial length L=25mm of rubber bushing, the elastic properties of materials mould of rubber bushing Measure Ex=7.84Mpa, Poisson's ratio μx=0.47, exradius rb, inner circle radius ra=d/2+ δ, wherein, δ is rubber bushing The design parameter of inner circle sleeve thickness, using δ as parameter to be designed, establish rubber bushing RADIAL stiffness KxComputational chart Up to formula, i.e.,
Wherein, Kx(δ) is the expression formula on inner circle sleeve thickness δ;
b1(δ)=(d/2+ δ) ((d/2+ δ)2+3rb 2)[I(1,αa)K(0,αa)+K(1,αa)I(0,αa)],
b2(δ)=- rb(rb 2+3(d/2+δ)2)[I(1,αhb)K(0,αa)+K(1,αhb)I(0,αa)],
Bessel correction functions:I(1,αa), K (1, αa);I(0,αa), K (0, αa);
I(0,αhb)=1444.8, K (0, αhb)=3.7285 × 10-5;I(1,αhb)=1364.7, K (1, αhb)= 3.9242×10-5
(3) foundation and design calculating of the inner circle sleeve thickness design mathematic model of rubber bushing:
The wheelspan B=1600mm, the diameter d=20mm of stabiliser bar, total length l of suspension according to where stabiliser barc= 800mm, the design requirement value of stabilizer bar system roll angular rigidityObtained by being calculated in step (1) The vertical deviation deformation coefficient G of end part of stabilizer rodw=1.5935 × 10-12m5The rubber bushing footpath established in/N, and step (2) To Line stiffness expression formula Kx(δ), the inner circle sleeve thickness δ of rubber bushing design mathematic model is established, i.e.,:
Using Matlab calculation procedures, above-mentioned mathematical modeling is solved, can be obtained in stabilizator rod structure and rubber bushing peace Under holding position permanence condition, meet that the inner circle sleeve thickness of the rubber bushing of stabilizer bar system roll angular rigidity design requirement is set Evaluation, δ=4mm.
Wherein, in the case of other structures and parameter constant, the stabilizer bar system roll angular rigidity design requirement value, with The inner circle sleeve thickness δ of rubber bushing change curve, as shown in Figure 4.
Embodiment two:The structural parameters of certain vehicle front suspension, the structural parameters of stabiliser bar and suspension bushes inner circle radius and Material characteristic parameter, all identical with embodiment one, simply the roll angular rigidity required by front suspension stabilizer bar system is set It is different to count required value, i.e.,To under this roll angular rigidity design requirement, to the inner circle set of rubber bushing Barrel thickness δ is designed.
Using the design procedure of embodiment one, the inner circle sleeve thickness δ of the Chinese herbaceous peony suspension stabiliser bar rubber bushing is carried out Design.Due to the structural parameters of the vehicle front suspension, the structural parameters of stabiliser bar and rubber bushing exradius and material property Parameter, all identical with embodiment one, simply roll angular rigidity design requirement value is different.Therefore, being somebody's turn to do obtained by designing Inner circle sleeve thickness δ=8mm of vehicle front suspension rubber bushing.
Understood compared with embodiment one, due to roll angular rigidity design requirement value10kN.m/rad is added, then only The inner circle sleeve thickness δ of rubber bushing is increased into 4mm, i.e., 8mm is increased to by previous 4mm, it is possible to do not changing other In the case of structural parameters, the stabilizer bar system of roll angular rigidity design requirement is met.
Embodiment three:The wheelspan B=1600mm of certain automobile front-axle, use the structure and material and embodiment one of stabiliser bar Identical, the diameter d=21mm of stabiliser bar;Clipping room between two rubber bushings is away from l0=400mm, rubber bushing it is outer Radius of circle rb=30.5mm, axial length L=25mm;The elastic modulus E of rubber bushingx=7.84MPa, Poisson's ratio μx=0.47. The design requirement value of the roll angular rigidity of the vehicle front suspension stabilizer bar systemIn given stabiliser bar and In the case of rubber bushing installation site, the inner circle sleeve thickness δ of rubber bushing is designed.
Using the design procedure of embodiment one, the inner circle sleeve thickness δ of the Chinese herbaceous peony suspension stabiliser bar rubber bushing is carried out Design:(1) the vertical deviation deformation coefficient G of end part of stabilizer rod is calculatedw
Due to the mounting distance l between stabilizator rod structure parameter, material characteristic parameter and two rubber bushings0, all with reality The identical of example one is applied, therefore, the vertical deviation deformation coefficient G of end part of stabilizer rodwIt is also identical with embodiment one, I.e.:
(2) rubber bushing RADIAL stiffness K is establishedxExpression formula:
According to the diameter d=21mm of stabiliser bar, axial length L=25mm of rubber bushing, elastic modulus Ex= 7.84Mpa, Poisson's ratio μx=0.47, exradius rb=30.5mm, inner circle radius ra=d/2+ δ=(10.5+ δ) mm, wherein, δ The as inner circle sleeve thickness of rubber bushing, using δ as parameter to be designed, establish rubber bushing RADIAL stiffness KxCalculating Expression formula, i.e.,
Wherein, Kx(δ) is the expression formula on inner circle sleeve thickness δ;
b1(δ)=(d/2+ δ) ((d/2+ δ)2+3rb 2)[I(1,αa)K(0,αa)+K(1,αa)I(0,αa)],
b2(δ)=- rb(rb 2+3(d/2+δ)2)[I(1,αhb)K(0,αa)+K(1,αhb)I(0,αa)],
Bessel correction functions:I(1,αa), K (1, αa);I(0,αa), K (0, αa);
I(0,αhb)=1444.8, K (0, αhb)=3.7285 × 10-5;I(1,αhb)=1364.7, K (1, αhb)= 3.9242×10-5
(3) foundation and design calculating of the inner circle sleeve thickness design mathematic model of rubber bushing:
The wheelspan B=1600mm, the diameter d=21mm of stabiliser bar, total length l of suspension according to where stabiliser barc= 800mm, the design requirement value of stabilizer bar system roll angular rigidityObtained by being calculated in step (1) The vertical deviation deformation coefficient G of end part of stabilizer rodw=1.5935 × 10-12m5The rubber bushing footpath established in/N, and step (2) To Line stiffness expression formula Kx(δ), the inner circle sleeve thickness δ of rubber bushing design mathematic model is established, i.e.,:
Using Matlab calculation procedures, above-mentioned mathematical modeling is solved, can be obtained in stabilizator rod structure and rubber bushing peace Under holding position permanence condition, meet that the inner circle sleeve thickness of the rubber bushing of stabilizer bar system roll angular rigidity design requirement is set Evaluation, δ=3mm.
Wherein, in the case of other structures and parameter constant, the roll angular rigidity design of the vehicle stabilization lever system will Evaluation, with the inner circle sleeve thickness δ of rubber bushing change curve, as shown in Figure 5.

Claims (1)

1. the design method of the inner circle sleeve thickness of suspension stabiliser bar rubber bushing, it is comprised the following steps that:
(1) the vertical deviation deformation coefficient G of end part of stabilizer rod is calculatedw
According to the total length l of suspension stabiliser barc, brachium l1, the mounting distance l between two rubber bushings0, transition arc radius R, the central angle θ of transition arc, the elasticity modulus of materials E and Poisson's ratio μ of stabiliser bar, the vertical deviation of end part of stabilizer rod is deformed Coefficient GwCalculated, i.e.,:
<mrow> <msub> <mi>G</mi> <mi>w</mi> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>G</mi> <mn>1</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>2</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>3</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>4</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>5</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>6</mn> </msub> </mrow> <mrow> <mi>&amp;pi;</mi> <mi>E</mi> </mrow> </mfrac> <mo>;</mo> </mrow>
In formula,
<mrow> <msub> <mi>G</mi> <mn>3</mn> </msub> <mo>=</mo> <mn>64</mn> <mi>R</mi> <mo>&amp;lsqb;</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msubsup> <mi>l</mi> <mn>1</mn> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <mi>&amp;theta;</mi> <mo>+</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <mi>&amp;theta;</mi> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msup> <mi>R</mi> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mi>&amp;theta;</mi> <mo>-</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <mi>&amp;theta;</mi> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>l</mi> <mn>1</mn> </msub> <mi>R</mi> <mi> </mi> <msup> <mi>sin</mi> <mn>2</mn> </msup> <mi>&amp;theta;</mi> <mo>&amp;rsqb;</mo> <mo>,</mo> <msub> <mi>G</mi> <mn>4</mn> </msub> <mo>=</mo> <mfrac> <mrow> <mn>8</mn> <msub> <mi>l</mi> <mn>0</mn> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>l</mi> <mi>c</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> <mn>3</mn> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>G</mi> <mn>5</mn> </msub> <mo>=</mo> <mn>64</mn> <mi>R</mi> <mrow> <mo>(</mo> <mi>&amp;mu;</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&amp;lsqb;</mo> <msup> <mi>R</mi> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mfrac> <mrow> <mn>3</mn> <mi>&amp;theta;</mi> </mrow> <mn>2</mn> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <mi>&amp;theta;</mi> </mrow> <mn>4</mn> </mfrac> <mo>-</mo> <mn>2</mn> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>&amp;theta;</mi> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msubsup> <mi>l</mi> <mn>1</mn> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <mi>&amp;theta;</mi> <mo>-</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <mi>&amp;theta;</mi> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mn>4</mn> <msub> <mi>l</mi> <mn>1</mn> </msub> <mi>R</mi> <mi> </mi> <msup> <mi>sin</mi> <mn>4</mn> </msup> <mfrac> <mi>&amp;theta;</mi> <mn>2</mn> </mfrac> <mo>&amp;rsqb;</mo> <mo>,</mo> </mrow>
G6=-32 (μ+1) [R (cos θ -1)-l1sinθ]2[2l1cosθ-lc+2Rsinθ];
(2) rubber bushing RADIAL stiffness K is establishedxExpression formula:
According to stablizing shank diameter d, the axial length L of rubber bushing, the elasticity modulus of materials E of rubber bushingx, Poisson's ratio μx, it is cylindrical Radius rb, inner circle radius ra=d/2+ δ, wherein, δ is the design parameter of the inner circle sleeve thickness of rubber bushing, using δ as Parameter to be designed, establish rubber bushing RADIAL stiffness KxCalculation expression, i.e.,
<mrow> <msub> <mi>K</mi> <mi>x</mi> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <mi>u</mi> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow>
Wherein, Kx(δ) is the expression formula on inner circle sleeve thickness δ;
<mrow> <mi>u</mi> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> <mrow> <mn>2</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <mi>L</mi> </mrow> </mfrac> <mo>&amp;lsqb;</mo> <mi>l</mi> <mi>n</mi> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>-</mo> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> <mrow> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> </mrow> </mfrac> <mo>&amp;rsqb;</mo> <mo>,</mo> </mrow>
<mrow> <mi>y</mi> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <mi>L</mi> </mrow> </mfrac> <mrow> <mo>(</mo> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>+</mo> <mfrac> <mrow> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> </mrow> <mrow> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>,</mo> </mrow>
αhb=α rba=α (d/2+ δ),
<mrow> <msub> <mi>a</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>)</mo> <mo>&amp;lsqb;</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mn>3</mn> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>)</mo> <mo>-</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>(</mo> <mn>3</mn> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>)</mo> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>&amp;lsqb;</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>&amp;rsqb;</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> <mo>-</mo> <mi>I</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>(</mo> <mn>3</mn> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>)</mo> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <mi>L</mi> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>)</mo> <mo>&amp;lsqb;</mo> <msub> <mi>b</mi> <mn>1</mn> </msub> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> <mo>+</mo> <msub> <mi>b</mi> <mn>2</mn> </msub> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> <mo>+</mo> <msub> <mi>b</mi> <mn>3</mn> </msub> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <mi>L</mi> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mfrac> <mo>,</mo> </mrow>
b1(δ)=(d/2+ δ) ((d/2+ δ)2+3rb 2)[I(1,αa)K(0,αa)+K(1,αa)I(0,αa)],
b2(δ)=- rb(rb 2+3(d/2+δ)2)[I(1,αhb)K(0,αa)+K(1,αhb)I(0,αa)],
<mrow> <msub> <mi>b</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mrow> <mo>(</mo> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msup> <mrow> <mo>(</mo> <mrow> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mi>ln</mi> <mrow> <mo>(</mo> <mi>d</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>h</mi> <mi>b</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mo>;</mo> </mrow>
Bessel correction functions:I(0,αhb), K (0, αhb);I(1,αhb), K (1, αhb);
I(1,αa), K (1, αa);I(0,αa), K (0, αa);
(3) foundation and design calculating of the inner circle sleeve thickness design mathematic model of rubber bushing:
The wheelspan B, the diameter d of stabiliser bar, total length l of suspension according to where stabiliser barc, stabilizer bar system roll angular rigidity sets Count required valueThe vertical deviation deformation coefficient G of end part of stabilizer rod obtained by being calculated in step (1)w, and in step (2) The rubber bushing radial direction Line stiffness expression formula K establishedx(δ), establish the inner circle sleeve thickness δ of rubber bushing design mathematic Model, i.e.,:
Using Matlab calculation procedures, above-mentioned mathematical modeling is solved, can be obtained in stabilizator rod structure and rubber bushing installation position Put under permanence condition, meet the inner circle thickness δ of the rubber bushing of stabilizer bar system roll angular rigidity design requirement design load.
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