CN104331578B - The design method of outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length - Google Patents

The design method of outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length Download PDF

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CN104331578B
CN104331578B CN201410665583.XA CN201410665583A CN104331578B CN 104331578 B CN104331578 B CN 104331578B CN 201410665583 A CN201410665583 A CN 201410665583A CN 104331578 B CN104331578 B CN 104331578B
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cabin
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CN104331578A (en
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周长城
周超
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Shandong University of Technology
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Abstract

The present invention relates to the design method of outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length, belong to cab mounting technical field.Structure and material characteristic parameter of the invention according to driver's cabin stabilizer bar system, utilize roll angular rigidity design requirement value and the pendulum arm length of stabiliser bar, the equivalent Line stiffness of torsion tube, the loading coefficient of reversed rubber bushing, relation between radial rigidity and equivalent combinations Line stiffness, the design mathematic model of outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length is established, and solution design is carried out to it using Matlab programs.By designing example and simulating, verifying, the available accurately and reliably rubber sleeve Design of length value of this method, reliable technical foundation has been established in the exploitation of design and CAD software for driver's cabin stabilizer bar system.Using this method, the design level and quality of driver's cabin stabilizer bar system can be not only improved, improves vehicle ride performance and security;Meanwhile it can also reduce design and testing expenses.

Description

The design method of outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length
Technical field
The present invention relates to vehicle cab suspension, particularly outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length Design method.
Background technology
In driver's cabin actual design, often in the case where keeping stabiliser bar other structures parameter constant, using passing through Rubber sleeve length is adjusted, driver's cabin stabilizer bar system is reached the design requirement of roll angular rigidity.For outer biasing non-coaxial Driver's cabin stabilizer bar system, although being only made up of swing arm, torsion tube and rubber bushing, but one by rigid body, elastomer and The coupling body of flexible body three composition, and because being biased outside torsion tube also in the presence of torsion and the coupling of bending, simultaneously as rubber serves as a contrast The Rigidity Calculation of set is extremely complex, therefore, for it is outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length adjusted design, Fail to provide reliable resolution design method always.At present, the design for driver's cabin stabilizer bar system both at home and abroad, mostly it is profit With ANSYS simulation softwares, simulating, verifying is carried out to the characteristic for giving the driver's cabin stabilizer bar system of structure by solid modelling, to the greatest extent Pipe this method can obtain reliable simulation numerical, however, because ANSYS simulation analysis can only be to the stabiliser bar of given parameters Characteristic carries out simulating, verifying, can not provide accurate analytical design method formula, it is impossible to meets analytical design method and CAD design software development It is required that.Therefore, it is necessary to establish a kind of design side of accurate, reliable outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length Method, on the premise of product cost is not increased, only by the adjusted design to rubber sleeve length, make stabilizer bar system angle of heel firm Degree meets the design requirement of cab mounting, improves product design level and quality, improves vehicle ride performance and security; Meanwhile design and testing expenses are reduced, accelerate product development speed.
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 reliable outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length, its design flow diagram are as shown in Figure 1;Outside The structural representation for biasing non-coaxial driver's cabin stabilizer bar system is as shown in Figure 2;The structural representation of stabiliser bar rubber bushing As shown in Figure 3;Stabilizer bar system deforms and the geometrical relationship figure of swing arm displacement is as shown in Figure 4.
In order to solve the above technical problems, outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length provided by the present invention Design method, it is characterised in that use following design procedure:
(1) the inclination line stiffness K of driver's cabin stabilizer bar systemwsThe calculating of design requirement value:
According to the design requirement value of driver's cabin stabilizer bar system roll angular rigidityThe suspension distance L of stabiliser barc, to driving Sail the inclination line stiffness K of room stabilizer bar systemwsDesign requirement value is calculated, i.e.,
(2) the equivalent Line stiffness expression formula K of non-coaxial driver's cabin torsion tube is biased outsideTFoundation:
According to torsion tube length Lw, internal diameter d, outer diameter D, outer amount of bias T, elastic modulus E and Poisson's ratio μ, and pendulum arm length l1, To the torsion tube of stabiliser bar cab mounting installed position equivalent line stiffness KTCalculated, i.e.,
(3) the equivalent combinations Line stiffness expression formula K of non-coaxial stabiliser bar rubber bushing is biased outsidex(Lx) foundation:
1. establish the radial rigidity expression formula k of rubber bushingx(Lx)
According to the inner circle radius r of rubber sleevea, exradius rb, elastic modulus ExWith Poisson's ratio μx, with rubber sleeve length LxFor Parameter to be designed, establish the radial rigidity expression formula k of rubber bushingx(Lx), i.e.,
Wherein,
Bessel correction functions I (0, α rb), K (0, α rb), I (1, α rb), K (1, α rb),
I(1,αra), K (1, α ra), I (0, α ra), K (0, α ra);
2. calculate the loading coefficient η of the reversed rubber bushing of outer biasing non-coaxial stabilizer bar systemF
According to torsion tube length LW, Poisson's ratio μ, outer amount of bias T, and pendulum arm length l1, to the loading coefficient of reversed rubber bushing ηFCalculated, i.e.,
3. the equivalent combinations Line stiffness expression formula K of outer biasing non-coaxial stabiliser bar rubber bushingx(Lx) foundation
According to the radial rigidity expression formula k for the rubber bushing established in 1. stepx(Lx), and be 2. calculated in step Reversed rubber bushing loading coefficient ηF, establish the outer equivalent combinations Line stiffness table for biasing non-coaxial stabiliser bar rubber bushing Up to formula Kx(Lx), i.e.,
(4) the rubber sleeve length L of non-coaxial driver's cabin stabiliser bar is biased outsidexThe foundation and design of design mathematic model:
According to the design requirement value K of the inclination line stiffness for the driver's cabin stabilizer bar system being calculated in step (1)ws, step Suddenly the equivalent line stiffness K for the torsion tube being calculated in (2)T, and the rubber bushing established of 3. step in step (3) is equivalent Combine Line stiffness expression formula Kx(Lx), establish the outer rubber sleeve length L for biasing non-coaxial driver's cabin stabiliser barxDesign mathematic Model, i.e.,
KTKX(Lx)-KwsKX(Lx)-KwsKT=0;
Using Matlab programs, solve above-mentioned on LxEquation, it is stable that outer biasing non-coaxial driver's cabin can be obtained Bar rubber sleeve length LxDesign load;
(5) the ANSYS simulating, verifyings of non-coaxial driver's cabin stabilizer bar system roll angular rigidity are biased outside:
I utilizes ANSYS finite element emulation softwares, according to rubber sleeve length LxDesign load, and it is outer biasing non-coaxial drive The other structures parameter and material characteristic parameter of room stabilizer bar system are sailed, establishes ANSYS simulation models, grid division, and putting The suspension installed position of arm applies load F, and the deformation to stabilizer bar system carries out ANSYS emulation, and it is non-coaxial to obtain outer biasing The deformation displacement amount f of formula stabilizer bar system swing arm outermost endA
II is according to designed rubber sleeve length Lx, the other structures and material characteristic parameter of rubber bushing, utilize step (3) the rubber bushing radial rigidity expression formula k that the 1. step in is establishedx(Lx), try to achieve the radial rigidity of designed rubber bushing kx
III emulates the deformation displacement amount f of resulting swing arm outermost end according to ANSYSA, pendulum arm length l1, the suspension of swing arm Distance, delta l of the installation site to outermost end1, the suspension distance L of stabiliser barc, arm suspension installed position apply load F, And the radial rigidity k for the rubber bushing being calculated in II stepsx, closed using the geometry of stabilizer bar system deformation and swing arm displacement System, to the ANSYS simulating, verifying values of designed outer biasing non-coaxial driver's cabin stabilizer bar system roll angular rigidityEnter Row calculates, i.e.,
By the ANSYS simulating, verifying values of the non-coaxial driver's cabin stabilizer bar system roll angular rigidityWith stablizing leverage The design requirement value of system roll angular rigidityIt is compared, so as to outer biasing non-coaxial driver's cabin provided by the present invention The design method and parameter design value of stabiliser bar rubber sleeve length are verified.
The present invention has the advantage that than prior art
It it is one by rigid body, elastomer and flexible body three's group due to outer biasing non-coaxial driver's cabin stabilizer bar system Into coupling body, and reversed and the coupling of bending because biasing outside torsion tube also to exist, simultaneously as the Rigidity Calculation of rubber bushing It is extremely complex, therefore, the design for outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length, fail to provide reliably always Resolution design method.At present, the design for driver's cabin stabilizer bar system both at home and abroad, mostly it is to utilize ANSYS simulation softwares, Simulating, verifying is carried out to the characteristic for giving the driver's cabin stabilizer bar system of structure by solid modelling, although this method can be compared Relatively reliable simulation numerical, however, being tested because ANSYS simulation analysis can only carry out emulation to the stabiliser bar characteristic of given parameters Card, can not provide accurate analytical design method formula, it is impossible to meet the requirement of analytical design method and CAD design software development.
The present invention utilizes roll angular rigidity design requirement according to the structure and material characteristic parameter of driver's cabin stabilizer bar system Value and the pendulum arm length of stabiliser bar, the equivalent Line stiffness of torsion tube, the loading coefficient of reversed rubber bushing, radial rigidity and equivalent group Relation between zygonema rigidity, the design mathematic model of outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length is established, And solution design is carried out to it using Matlab programs.It can obtain by designing example and ANSYS simulating, verifyings, this method Accurately and reliably rubber sleeve length LxDesign load, reliable design side is provided for the design of cab mounting and stabilizer bar system Method, and established reliable technical foundation for the exploitation of driver's cabin stabilizer bar system CAD software.Using this method, can not increase Add product cost, only by the adjusted design of rubber sleeve length, improve design level, the matter of cab mounting and stabilizer bar system Amount and performance, meet design requirement of the cab mounting to stabiliser bar roll angular rigidity, further improve the traveling smooth-going of vehicle Property and security;Meanwhile design and testing expenses can be also reduced, accelerate product development speed.
For a better understanding of the present invention, it is described further below in conjunction with the accompanying drawings.
Fig. 1 is the design flow diagram of outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length;
Fig. 2 is the structural representation of outer biasing non-coaxial driver's cabin stabilizer bar system;
Fig. 3 is the structural representation of rubber bushing;
Fig. 4 is the geometrical relationship figure of the outer deformation of biasing non-coaxial stabilizer bar system and swing arm displacement;
Fig. 5 is the radial rigidity k of the rubber bushing of embodiment onexWith rubber sleeve length LxChange curve;
Fig. 6 is the equivalent combinations Line stiffness K of the stabilizer bar system of embodiment onexWith rubber sleeve length LxChange curve;
Fig. 7 is the outer biasing non-coaxial stabilizer bar system roll angular rigidity of embodiment oneWith rubber sleeve length LxChange Change curve;
Fig. 8 is the deformation simulation cloud atlas of the outer biasing non-coaxial driver's cabin stabilizer bar system of embodiment one;
Fig. 9 is the outer biasing non-coaxial stabilizer bar system roll angular rigidity of embodiment twoWith rubber sleeve length LxChange Change curve;
Figure 10 is the deformation simulation cloud atlas of the outer biasing non-coaxial driver's cabin stabilizer bar system of embodiment two.
Specific embodiment
The present invention is described in further detail below by embodiment.
Embodiment one:The structure of certain outer biasing non-coaxial driver's cabin stabilizer bar system is symmetrical, as shown in Fig. 2 bag Include:Swing arm 1, suspended rubber bushing 2, reversed rubber bushing 3, torsion tube 4;Wherein, torsion tube 4 and reversed rubber bushing 3 be not coaxial, turns round The outer amount of bias T=30mm of pipe 4;The distance between the swing arm 1 of left and right two distance Lc=1550mm, as stabiliser bar suspension away from From;The distance between suspended rubber bushing 2 and reversed rubber bushing 3 l1=380mm, as pendulum arm length;The suspension peace of swing arm Holding position C to outermost end A distance, delta l1=47.5mm;The length L of torsion tube 4w=1500mm, internal diameter d=35mm, outer diameter D= 50mm, elastic modulus E=200GPa, Poisson's ratio μ=0.3;The structure and material characteristic of the rubber bushing of left and right four is identical, As shown in figure 3, including:Interior round buss 5, rubber sleeve 6, outer round buss 7, wherein, the interior circular diameter d of interior round buss 5x=35mm, Wall thickness δ=2mm;The inner circle radius r of rubber sleeve 6a=19.5mm, exradius rb=34.5mm, elastic modulus Ex=7.84MPa, Poisson's ratio μx=0.47, the length L of rubber sleevexFor parameter to be designed.The roll angular rigidity design of the driver's cabin stabilizer bar system will EvaluationTo the rubber sleeve length L of the outer biasing non-coaxial driver's cabin stabiliser barxSet Meter, and ANSYS simulating, verifyings are carried out to stabilizer bar system roll angular rigidity in the case of load F=5000N.
The design method for the outer biasing non-coaxial driver's cabin stabiliser bar rubber sleeve length that present example is provided, it sets Flow is counted as shown in figure 1, comprising the following steps that:
(1) the inclination line stiffness K of driver's cabin stabilizer bar systemwsThe calculating of design requirement value:
According to the design requirement value of stabilizer bar system roll angular rigidityStabiliser bar hangs Put distance Lc=1550mm, to the inclination line stiffness K of driver's cabin stabilizer bar systemwsDesign requirement value calculated, i.e.,
(2) the equivalent Line stiffness expression formula K of non-coaxial driver's cabin torsion tube is biased outsideTFoundation:
According to torsion tube length Lw=1500mm, internal diameter d=35mm, outer diameter D=50mm, outer amount of bias T=30mm, springform Measure E=200GPa and Poisson's ratio μ=0.3, and pendulum arm length l1=380mm, to the torsion tube of the outer biasing non-coaxial stabiliser bar Equivalent line stiffness K at cab mounting installation site CTCalculated, i.e.,
(3) the equivalent combinations Line stiffness expression formula K of non-coaxial stabiliser bar rubber bushing is biased outsidex(Lx) foundation:
1. establish the radial rigidity expression formula k of rubber bushingx(Lx)
According to the inner circle radius r of rubber sleevea=19.5mm, exradius rb=34.5mm, elastic modulus Ex=7.84MPa With Poisson's ratio μx=0.47, with rubber sleeve length LxFor parameter to be designed, the radial rigidity table of the stabiliser bar rubber bushing is established Up to formula kx(Lx), i.e.,
Wherein,
Bessel correction functions I (0, α rb), K (0, α rb), I (1, α rb), K (1, α rb),
I(1,αra), K (1, α ra), I (0, α ra), K (0, α ra);
Wherein, the radial rigidity k of rubber bushingxWith rubber sleeve length LxChange curve, as shown in Figure 5;
2. calculate the loading coefficient η of the reversed rubber bushing of outer biasing non-coaxial stabilizer bar systemF
According to torsion tube length LW=1500mm, Poisson's ratio μ=0.3, outer amount of bias T=30mm, and pendulum arm length l1= 380mm, to the loading coefficient η of reversed rubber bushingFCalculated, i.e.,
3. the equivalent combinations Line stiffness expression formula K of outer biasing non-coaxial stabiliser bar rubber bushingx(Lx) foundation
According to the k established in 1. stepx(Lx), and the η being 2. calculated in stepF=0.15808, establish outer bias The equivalent combinations Line stiffness expression formula K of non-coaxial stabiliser bar rubber bushingx(Lx), i.e.,
Wherein, the equivalent combinations Line stiffness K of stabiliser bar rubber bushingxWith rubber sleeve length LxChange curve, such as Fig. 6 institutes Show;
(4) the rubber sleeve length L of non-coaxial driver's cabin stabiliser bar is biased outsidexThe foundation and design of design mathematic model:
According to the K being calculated in step (1)ws=2.461 × 105N/m, the K being calculated in step (2)T= 2.84488×105What the 3. step in N/m, and step (3) was establishedOuter biasing non-coaxial is established to drive Sail room stabiliser bar rubber sleeve length LxDesign mathematic model, i.e.,
KTKX(Lx)-KX(Lx)+KwsKw=0;
Using Matlab programs, solve above-mentioned on LxEquation, the non-coaxial driver's cabin stabiliser bar rubber can be obtained Cover length LxDesign load, i.e.,
Lx=25mm;
Wherein, the roll angular rigidity of the stabilizer bar systemWith rubber sleeve length LxChange curve, as shown in Figure 7;
(5) the ANSYS simulating, verifyings of non-coaxial driver's cabin stabilizer bar system roll angular rigidity are biased outside:
I utilizes ANSYS finite element emulation softwares, according to the rubber sleeve length L obtained by designx=25mm, and outer biasing The other structures parameter and material characteristic parameter of non-coaxial driver's cabin stabilizer bar system, ANSYS simulation models are established, divide net Lattice, and apply load F=5000N at the suspension installation site C of swing arm, the deformation to stabilizer bar system carries out ANSYS emulation, Resulting deformation simulation cloud atlas, as shown in figure 8, wherein, deformation displacement amount f of the stabilizer bar system at swing arm outermost end AAFor
fA=19.984mm;
II is according to designed rubber sleeve length Lx=25mm, the other structures and material characteristic parameter of rubber bushing, profit The rubber bushing radial rigidity expression formula k established with the 1. step in step (3)x(Lx), try to achieve the footpath of designed rubber bushing To rigidity kx=2.1113 × 106N/m;
III emulates the deformation displacement amount f at resulting swing arm outermost end A according to ANSYSA=19.984mm, swing arm length Spend l1=380mm, the suspension installation site C to outermost end A of swing arm distance, delta l1=47.5mm, the suspension distance L of stabiliser barc =1550mm, the load F=5000N applied at the suspension installation site C of swing arm, and the k being calculated in II stepsx= 2.1113×106N/m, using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, as shown in figure 4, non-to the outer biasing The ANSYS simulating, verifying values of coaxial driver's cabin stabilizer bar system roll angular rigidityCalculated, i.e.,
fws=fC+F/kx=20.1317mm;
Understand, the ANSYS simulating, verifying values of the outer biasing non-coaxial driver's cabin stabilizer bar system roll angular rigidityWith the design requirement value of stabilizer bar system roll angular rigidityPhase It coincide, relative deviation is only 0.916%;As a result the outer biasing non-coaxial driver's cabin stabiliser bar rubber that the invention is provided is shown The design method for covering length is correct, and parameter design value is accurately and reliably.
Embodiment two:The structure type of certain outer biasing non-coaxial driver's cabin stabilizer bar system is identical with embodiment one, As shown in Fig. 2 wherein, torsion tube 4 is not coaxial with reversed rubber bushing 3, the outer amount of bias T=50mm of torsion tube 4;The swing arm of left and right two The distance between 1 Lc=1400mm, i.e. stabiliser bar suspension distance;Between suspended rubber bushing 2 and reversed rubber bushing 3 away from From l1=350mm, i.e. pendulum arm length;The suspension installation site C of swing arm to outermost end A distance, delta l1=52.5mm;Torsion tube 4 Length Lw=1000mm, internal diameter d=42mm, outer diameter D=50mm, elasticity modulus of materials E=200GPa, Poisson's ratio μ=0.3;It is left The structure of right four rubber bushings is all identical, as shown in figure 3, wherein, the interior circular diameter d of interior round buss 5x=35mm, wall Thick δ=5mm;The inner circle radius r of rubber sleeve 6a=22.5mm, exradius rb=37.5mm, elastic modulus Ex=7.84MPa, pool Pine compares μx=0.47, the length L of rubber sleevexFor parameter to be designed.The design requirement of the driver's cabin stabilizer bar system roll angular rigidity ValueTo the rubber sleeve length L of the outer biasing non-coaxial driver's cabin stabiliser barxIt is designed, And ANSYS simulating, verifyings are carried out to stabilizer bar system roll angular rigidity in the case of load F=5000N.
Using the step identical with embodiment one, to the rubber sleeve length L of the outer biasing non-coaxial driver's cabin stabiliser barx It is designed, i.e.,:
(1) the inclination line stiffness K of driver's cabin stabilizer bar systemwsThe calculating of design requirement value:
According to the design requirement value of stabilizer bar system roll angular rigidityStabiliser bar hangs Put distance Lc=1400mm, to the inclination line stiffness K of the driver's cabin stabilizer bar systemwsDesign requirement value is calculated, i.e.,
(2) the equivalent Line stiffness expression formula K of non-coaxial driver's cabin torsion tube is biased outsideTFoundation:
According to torsion tube length Lw=1000mm, internal diameter d=42mm, outer diameter D=50mm, outer amount of bias T=30mm, springform Measure E=200GPa and Poisson's ratio μ=0.3, and pendulum arm length l1=350mm, to the torsion tube of the outer biasing non-coaxial stabiliser bar Equivalent line stiffness K at cab mounting installation site CTCalculated, i.e.,
(3) the equivalent combinations Line stiffness expression formula K of non-coaxial stabiliser bar rubber bushing is biased outsidex(Lx) foundation:
1. establish the radial rigidity expression formula k of rubber bushingx(Lx)
According to the inner circle radius r of rubber sleevea=22.5mm, exradius rb=37.5mm, elastic modulus Ex=7.84MPa With Poisson's ratio μx=0.47, with rubber sleeve length LxFor parameter to be designed, the radial rigidity for establishing stabiliser bar rubber bushing is expressed Formula kx(Lx), i.e.,
Wherein,
Bessel correction functions I (0, α rb), K (0, α rb), I (1, α rb), K (1, α rb),
I(1,αra), K (1, α ra), I (0, α ra), K (0, α ra);
2. calculate the loading coefficient η of the reversed rubber bushing of outer biasing non-coaxial stabilizer bar systemF
According to torsion tube length LW=1000mm, Poisson's ratio μ=0.3, outer amount of bias T=30mm, and pendulum arm length l1= 350mm, to the loading coefficient η of reversed rubber bushingFCalculated, i.e.,
3. the equivalent combinations Line stiffness expression formula K of outer biasing non-coaxial stabiliser bar rubber bushingx(Lx) foundation
According to the k established in 1. stepx(Lx), and the η being 2. calculated in stepF=0.3276, establish stabiliser bar rubber The equivalent combinations Line stiffness expression formula K of glue bushingx(Lx), i.e.,
(4) the rubber sleeve length L of non-coaxial driver's cabin stabiliser bar is biased outsidexThe foundation and design of design mathematic model:
According to the K being calculated in step (1)ws=2.97455 × 105N/m, the K being calculated in step (2)T= 3.28257×105What the 3. step in N/m, and step (3) was establishedOuter biasing non-coaxial is established to drive Sail room stabiliser bar rubber sleeve length LxDesign mathematic model, i.e.,
KTKX(Lx)-KX(Lx)+KwsKw=0;
Using Matlab programs, solve above-mentioned on LxEquation, the non-coaxial driver's cabin stabiliser bar rubber can be obtained Cover length LxDesign load, i.e.,
Lx=40mm;
Wherein, the driver's cabin stabilizer bar system roll angular rigidityWith rubber sleeve length LxChange curve, such as Fig. 9 institutes Show;
(5) the ANSYS simulating, verifyings of non-coaxial driver's cabin stabilizer bar system roll angular rigidity are biased outside:
I utilizes ANSYS finite element emulation softwares, according to the rubber sleeve length L obtained by designx=40mm, and outer biasing The other structures parameter and material characteristic parameter of non-coaxial driver's cabin stabilizer bar system, ANSYS simulation models are established, divide net Lattice, apply load F=5000N at the suspension installation site C of swing arm, the deformation to stabilizer bar system carries out ANSYS emulation, institute Obtained deformation simulation cloud atlas, as shown in Figure 10, wherein, deformation displacement amount f of the stabilizer bar system at swing arm outermost end AAFor
fA=17.881mm;
II is according to designed rubber sleeve length Lx=40mm, the other structures and material characteristic parameter of rubber bushing, profit The rubber bushing radial rigidity expression formula k established with the 1. step in step (3)x(Lx), try to achieve the footpath of designed rubber bushing To rigidity kx=4.2085 × 106N/m;
III emulates the deformation displacement amount f at resulting swing arm outermost end A according to ANSYSA=17.881mm, swing arm length Spend l1=350mm, the suspension installation site C to outermost end A of swing arm distance, delta l1=52.5mm, the suspension distance L of stabiliser barc =1400mm, the load F=5000N applied at swing arm suspension installation site C, and the k being calculated in II stepsx= 4.2085×106N/m, using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, as shown in figure 4, non-to the outer biasing The ANSYS simulating, verifying values of coaxial-type driver's cabin stabilizer bar system roll angular rigidityCalculated, i.e.,
fws=fC+F/kx=16.7367mm;
Understand, the ANSYS simulating, verifying values of the outer biasing non-coaxial driver's cabin stabilizer bar system roll angular rigidityWith the design requirement value of stabilizer bar system roll angular rigidity Match, relative deviation is only 0.431%;As a result the outer biasing non-coaxial driver's cabin stabiliser bar rubber that the invention is provided is shown The design method of gum cover length is correct, and parameter design value is accurately and reliably.

Claims (1)

1. the design method of non-coaxial driver's cabin stabiliser bar rubber sleeve length is biased outside, its specific design step is as follows:
(1) the inclination line stiffness K of driver's cabin stabilizer bar systemwsThe calculating of design requirement value:
According to the design requirement value of driver's cabin stabilizer bar system roll angular rigidityThe suspension distance L of stabiliser barc, to driver's cabin The inclination line stiffness K of stabilizer bar systemwsDesign requirement value is calculated, i.e.,
(2) the equivalent Line stiffness expression formula K of non-coaxial driver's cabin torsion tube is biased outsideTFoundation:
According to torsion tube length Lw, internal diameter d, outer diameter D, outer amount of bias T, elastic modulus E and Poisson's ratio μ, and pendulum arm length l1, to steady Equivalent line stiffness K of the torsion tube of fixed pole in cab mounting installed positionTCalculated, i.e.,
<mrow> <msub> <mi>K</mi> <mi>T</mi> </msub> <mo>=</mo> <mfrac> <mrow> <mi>&amp;pi;</mi> <mi>E</mi> <mrow> <mo>(</mo> <msup> <mi>D</mi> <mn>4</mn> </msup> <mo>-</mo> <msup> <mi>d</mi> <mn>4</mn> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mn>32</mn> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>&amp;mu;</mi> <mo>)</mo> </mrow> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mn>1</mn> </msub> <mo>+</mo> <mi>T</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msub> <mi>L</mi> <mi>W</mi> </msub> </mrow> </mfrac> <mo>;</mo> </mrow>
(3) the equivalent combinations Line stiffness expression formula K of non-coaxial stabiliser bar rubber bushing is biased outsidex(Lx) foundation:
1. establish the radial rigidity expression formula k of rubber bushingx(Lx)
According to the inner circle radius r of rubber sleevea, exradius rb, elastic modulus ExWith Poisson's ratio μx, with rubber sleeve length LxTo wait to set Parameter is counted, establishes the radial rigidity expression formula k of rubber bushingx(Lx), i.e.,
<mrow> <msub> <mi>k</mi> <mi>x</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <mi>u</mi> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow>
Wherein,
<mrow> <mi>y</mi> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mn>3</mn> </msub> <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> <msub> <mi>L</mi> <mi>x</mi> </msub> </mrow> </mfrac> <mrow> <mo>(</mo> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>+</mo> <mfrac> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>1</mn> </msub> <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;r</mi> <mi>a</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>a</mi> </msub> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> <mo>-</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>x</mi> </msub> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>2</mn> </msub> <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;r</mi> <mi>a</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>a</mi> </msub> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <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;r</mi> <mi>b</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>x</mi> </msub> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>3</mn> </msub> <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>(</mo> <msub> <mi>b</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>b</mi> <mn>2</mn> </msub> <mo>+</mo> <msub> <mi>b</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>x</mi> </msub> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>1</mn> </msub> <mo>=</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <msub> <mi>r</mi> <mi>a</mi> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>2</mn> </msub> <mo>=</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>3</mn> </msub> <mo>=</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mo>&amp;lsqb;</mo> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mrow> <mo>(</mo> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> <mo>)</mo> </mrow> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mi>a</mi> </msub> </mrow> <mo>&amp;rsqb;</mo> <mo>,</mo> </mrow>
<mrow> <mi>&amp;alpha;</mi> <mo>=</mo> <mn>2</mn> <msqrt> <mn>15</mn> </msqrt> <mo>/</mo> <msub> <mi>L</mi> <mi>x</mi> </msub> <mo>,</mo> </mrow>
Bessel correction functions I (0, α rb), K (0, α rb), I (1, α rb), K (1, α rb),
I(1,αra), K (1, α ra), I (0, α ra), K (0, α ra);
2. calculate the loading coefficient η of the reversed rubber bushing of outer biasing non-coaxial stabilizer bar systemF
According to torsion tube length LW, Poisson's ratio μ, outer amount of bias T, and pendulum arm length l1, to the loading coefficient η of reversed rubber bushingFEnter Row calculates, i.e.,
<mrow> <msub> <mi>&amp;eta;</mi> <mi>F</mi> </msub> <mo>=</mo> <mfrac> <mrow> <mn>24</mn> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>&amp;mu;</mi> <mo>)</mo> </mrow> <msub> <mi>l</mi> <mn>1</mn> </msub> <mi>T</mi> </mrow> <msubsup> <mi>L</mi> <mi>W</mi> <mn>2</mn> </msubsup> </mfrac> <mo>;</mo> </mrow>
3. the equivalent combinations Line stiffness expression formula K of outer biasing non-coaxial stabiliser bar rubber bushingx(Lx) foundation
According to the radial rigidity expression formula k for the rubber bushing established in 1. stepx(Lx), and the torsion being 2. calculated in step Turn the loading coefficient η of rubber bushingF, establish the outer equivalent combinations Line stiffness expression formula for biasing non-coaxial stabiliser bar rubber bushing Kx(Lx), i.e.,
<mrow> <msub> <mi>K</mi> <mi>X</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>k</mi> <mi>x</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;eta;</mi> <mi>F</mi> </msub> </mrow> </mfrac> <mo>;</mo> </mrow>
(4) the rubber sleeve length L of non-coaxial driver's cabin stabiliser bar is biased outsidexThe foundation and design of design mathematic model:
According to the design requirement value K of the inclination line stiffness for the driver's cabin stabilizer bar system being calculated in step (1)ws, step (2) In the equivalent line stiffness K of torsion tube that is calculatedT, and the equivalent combinations of rubber bushing that 3. step in step (3) is established Line stiffness expression formula Kx(Lx), establish the outer rubber sleeve length L for biasing non-coaxial driver's cabin stabiliser barxDesign mathematic model, I.e.
KTKX(Lx)-KwsKX(Lx)-KwsKT=0;
Using Matlab programs, solve above-mentioned on LxEquation, can obtain it is outer biasing non-coaxial driver's cabin stabiliser bar rubber Cover length LxDesign load;
(5) the ANSYS simulating, verifyings of non-coaxial driver's cabin stabilizer bar system roll angular rigidity are biased outside:
I utilizes ANSYS finite element emulation softwares, according to rubber sleeve length LxDesign load, and it is outer biasing non-coaxial driver's cabin it is steady The other structures parameter and material characteristic parameter of fixed pole system, establish ANSYS simulation models, grid division, and hanging in swing arm Put installed position and apply load F, the deformation to stabilizer bar system carries out ANSYS emulation, and it is stable to obtain outer biasing non-coaxial The deformation displacement amount f of lever system swing arm outermost endA
II is according to designed rubber sleeve length Lx, the other structures and material characteristic parameter of rubber bushing, using in step (3) The rubber bushing radial rigidity expression formula k that is established of 1. stepx(Lx), try to achieve the radial rigidity k of designed rubber bushingx
III emulates the deformation displacement amount f of resulting swing arm outermost end according to ANSYSA, pendulum arm length l1, the suspension installation of swing arm Distance, delta l of the position to outermost end1, the suspension distance L of stabiliser barc, in the load F that the suspension installed position of arm applies, and II The radial rigidity k for the rubber bushing being calculated in stepx, using stabilizer bar system deformation and swing arm displacement geometrical relationship, To the ANSYS simulating, verifying values of designed outer biasing non-coaxial driver's cabin stabilizer bar system roll angular rigidityCounted Calculate, i.e.,
<mrow> <msub> <mi>f</mi> <mi>C</mi> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>l</mi> <mn>1</mn> </msub> <msub> <mi>f</mi> <mi>A</mi> </msub> </mrow> <mrow> <msub> <mi>l</mi> <mn>1</mn> </msub> <mo>+</mo> <msub> <mi>&amp;Delta;l</mi> <mn>1</mn> </msub> </mrow> </mfrac> <mo>;</mo> </mrow>
<mrow> <msub> <mi>f</mi> <mrow> <mi>w</mi> <mi>s</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>f</mi> <mi>C</mi> </msub> <mo>+</mo> <mfrac> <mi>F</mi> <msub> <mi>k</mi> <mi>x</mi> </msub> </mfrac> <mo>;</mo> </mrow>
By the ANSYS simulating, verifying values of the non-coaxial driver's cabin stabilizer bar system roll angular rigidityWith stabilizer bar system side The design requirement value of inclination angle rigidityIt is compared, so as to stable to outer biasing non-coaxial driver's cabin provided by the present invention The design method and parameter design value of bar rubber sleeve length are verified.
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