CN104361176B - The design method of outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve - Google Patents

The design method of outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve Download PDF

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CN104361176B
CN104361176B CN201410665508.3A CN201410665508A CN104361176B CN 104361176 B CN104361176 B CN 104361176B CN 201410665508 A CN201410665508 A CN 201410665508A CN 104361176 B CN104361176 B CN 104361176B
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周长城
周超
孙重锋
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Shandong University of Technology
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Abstract

The present invention relates to the design method of outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve, belong to vehicle cab suspension technical field.Structure and material characterisitic parameter of the invention according to stabilizer bar system, using wall thickness of internal cylindrical sleeve as parameter, pass through the equivalent Line stiffness of torsion tube, relation between the loading coefficient and equivalent combinations Line stiffness expression formula of rubber bushing, the design mathematic model of wall thickness of internal cylindrical sleeve is established, and solution design is carried out to it using Matlab programs.Pass through example design and ANSYS simulating, verifyings, the design load of the available accurately and reliably wall thickness of internal cylindrical sleeve of this method, reliable technical foundation is provided for stabilizer bar system design and CAD software exploitation, can be in the case where not increasing product cost, only pass through the adjusted design of wall thickness of internal cylindrical sleeve, the performance of driver's cabin stabilizer bar system is improved, improves vehicle ride performance and security;Meanwhile also reduce design and testing expenses.

Description

The design method of outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve
Technical field
The present invention relates to vehicle cab suspension, particularly outer offset cab stabiliser bar rubber bushing inner circle sleeve wall Thick design method.
Background technology
Rubber bushing is made up of interior round buss, rubber sleeve and outer round buss.In the case where rubber sleeve exradius is given, The inner circle radius and thickness of rubber sleeve are relevant with the wall thickness of interior round buss., can be in driver's cabin stabilizer bar system actual design In the case that stabilizator rod structure is constant, only by the adjusted design to wall thickness of internal cylindrical sleeve, reach rubber sleeve inner circle radius and The adjusted design of its thickness, the adjusted design to driver's cabin stabilizer bar system roll angular rigidity is realized, so as to reach driver's cabin side The design requirement for rigidity of inclining.However, because outer offset cab stabilizer bar system is one by rigid body, elastomer and flexible body The coupling body of three's composition, and torsion tube also has bending and the coupling reversed, so its analysis calculating is extremely complex, therefore, Design for outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve, home and abroad fail to provide reliably always Resolution design method.At present, mostly it is to utilize ANSYS simulation softwares, passes through entity for driver's cabin stabilizer bar system both at home and abroad Model to give structure driver's cabin stabilizer bar system characteristic carry out simulating, verifying, although this method can obtain it is reliable Simulation numerical, however, because ANSYS simulation analysis can only carry out simulating, verifying to the stabiliser bar characteristic of given parameters, it is impossible to carry For accurate analytical design method formula, so analytical design method can not be realized, it can not meet that driver's cabin stabilizer bar system CAD software is developed Requirement.Therefore, it is necessary to establish a kind of accurate, reliable outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve Design method, meet the actual adjusted design requirement of cab mounting and stabilizer bar system roll angular rigidity, do not increasing product On the premise of cost, design level, quality and the performance of stabilizer bar system are improved, 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 offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve, its design flow diagram such as Fig. 1 institutes Show;The structural representation of outer offset cab stabilizer bar system is as shown in Figure 2;The structural representation of stabiliser bar rubber bushing is such as Shown in Fig. 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, round buss in outer offset cab stabiliser bar rubber bushing provided by the present invention The design method of wall thickness, 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 roll angular rigidity design requirement value of stabilizer bar systemThe suspension distance L of stabiliser barc, it is steady to driver's cabin The inclination line stiffness K of fixed pole systemwsDesign requirement value calculated, i.e.,
(2) the equivalent line stiffness K of the torsion tube of outer offset stabiliser barTAnd reversed rubber bushing loading coefficient ηFCalculating:
1. the equivalent line stiffness K of the torsion tube of stabiliser barTCalculating
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, Equivalent line stiffness K of the torsion tube of external offset stabiliser bar in cab mounting installed positionTCalculated, i.e.,
2. the loading coefficient η of reversed rubber bushingFCalculating
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 offset cab stabiliser bar rubber bushingxThe foundation of (δ):
I establishes the radial rigidity expression formula k of rubber bushingx(δ)
According to the length L of rubber sleevex, exradius rb, elastic modulus ExWith Poisson's ratio μx, the interior circular diameter of interior round buss dx, using wall thickness of internal cylindrical sleeve δ as parameter, then the inner circle radius r of rubber sleeveaIt is represented byThe footpath of rubber bushing To rigidity kx(δ) can be expressed as, i.e.,
Wherein,
Bessel correction functions I (0, αb), K (0, αb), I (1, αb), K (1, αb),
I(1,αa), K (1, αa), I (0, αa), K (0, αa);
αb=α rb,
II establishes the equivalent combinations Line stiffness expression formula K of outer offset stabiliser bar rubber bushingx(δ):
According to the radial rigidity expression formula k for the rubber bushing established in I stepsx2. step in (δ), and step (2) The loading coefficient η for the reversed rubber bushing being calculatedF, establish the equivalent combinations Line stiffness expression formula of stabiliser bar rubber bushing Kx(δ), i.e.,
(4) foundation and design of the wall thickness of internal cylindrical sleeve δ design mathematic models of stabiliser bar rubber bushing:
According to the design requirement value K for the driver's cabin stabilizer bar system inclination line stiffness being calculated in step (1)ws, step (2) the equivalent line stiffness K for the torsion tube that the 1. step in is calculatedT, and the stabiliser bar rubber that II steps in step (3) are established The equivalent combinations Line stiffness expression formula K of glue bushingx(δ), establish the wall thickness of internal cylindrical sleeve δ of stabiliser bar rubber bushing design mathematic Model, i.e.,
KTKX(δ)-KwsKX(δ)-KTKws=0;
Using Matlab programs, the above-mentioned equation on δ is solved, outer offset cab stabiliser bar rubber lining can be obtained Cover wall thickness of internal cylindrical sleeve δ design load, and the inner circle radius of rubber sleeve
(5) the ANSYS simulating, verifyings of outer offset cab stabilizer bar system roll angular rigidity:
A) ANSYS finite element emulation softwares are utilized, the wall thickness of internal cylindrical sleeve δ and rubber of the rubber bushing obtained according to design The inner circle radius r of seta, and the other structures parameter and material characteristic parameter of the outer offset cab stabilizer bar system, establish ANSYS simulation models, grid division, and apply load F in the suspension installed position of swing arm, the deformation to stabilizer bar system is entered Row ANSYS is emulated, and obtains deformation displacement amount f of the stabilizer bar system in swing arm outermost endA
B) according to the inner circle radius r that obtained wall thickness of internal cylindrical sleeve δ and rubber sleeve are designed in step (4)a, and rubber bushing Other structures parameter and material characteristic parameter, the radial rigidity expression formula k established using the I steps in step (3)x(rb), The radial rigidity k of designed rubber bushing is calculatedx
C) the deformation displacement amount f of the swing arm outermost end obtained by being emulated 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, the load that is applied at the suspension installation site of swing arm Lotus F, and the radial rigidity k for rubber bushing b) being calculated in stepx, using stabilizer bar system deformation and swing arm displacement it is several What relation, to the outer offset cab stabilizer bar system roll angular rigidityANSYS simulating, verifying values, calculated, i.e.,
By the ANSYS simulating, verifying values of the outer offset cab stabilizer bar system roll angular rigidityWith design requirement ValueIt is compared, so as to the design side of the outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve provided Method and parameter design value are verified.
The present invention has the advantage that than prior art
Because outer offset cab stabilizer bar system not only has the coupling of rigid body, elastomer and flexible body, Er Qieyin Torsion tube biases outside, also has bending and the coupling reversed, and its analysis calculating is extremely complex, therefore, for outer offset cab The design of stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve, home and abroad fail to provide reliable resolution design method always.At present, Mostly it is to utilize ANSYS simulation softwares both at home and abroad for driver's cabin stabilizer bar system, given structure is driven by solid modelling The characteristic for sailing room stabilizer bar system carries out simulating, verifying, although this method can obtain reliable simulation numerical, however, due to ANSYS simulation analysis can only carry out simulating, verifying to the stabiliser bar characteristic of given parameters, it is impossible to accurate analytical design method formula is provided, So the requirement of driver's cabin stabilizer bar system analytical design method and CAD software exploitation can not be met.
The present invention is according to the structure of outer offset cab stabilizer bar system and outer the amount of bias T, torsion tube of torsion tube bending Deformation and the relation between torsional deflection and load, have obtained the equivalent line stiffness K of torsion tubeTWith the system of load of reversed rubber bushing Number βF, and using the wall thickness of internal cylindrical sleeve δ of rubber bushing as parameter to be designed, establish the radial rigidity expression formula of rubber bushing kx(δ) and equivalent combinations Line stiffness expression formula Kx(δ);According to the structural parameters and material of outer offset cab stabilizer bar system Characterisitic parameter, using the roll angular rigidity of stabilizer bar system, the equivalent combinations line with the equivalent Line stiffness and rubber bushing of torsion tube Relation between rigidity, outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve δ design mathematic model is established, And Matlab calculation procedures are utilized, solution design is carried out to it.By designing example and ANSYS simulating, verifyings, this method Available accurately and reliably wall thickness of internal cylindrical sleeve δ design load, for the design of cab mounting and stabilizer bar system, provide can The design method leaned on, and established reliable technical foundation for the exploitation of driver's cabin stabilizer bar system CAD software.Utilize the party Method, only by the adjusted design of wall thickness of internal cylindrical sleeve, it can meet that stabilizer bar system rolls in the case where not increasing product cost The design requirement of angular rigidity, the design level and performance of stabilizer bar system are improved, improve the ride performance and security of vehicle; Meanwhile design and testing expenses are 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 offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve;
Fig. 2 is the structural representation of outer offset cab stabilizer bar system;
Fig. 3 is the structural representation of rubber bushing;
Fig. 4 is the geometrical relationship figure of the outer deformation of offset stabilizer bar system and swing arm displacement;
Fig. 5 is the rubber bushing radial rigidity k of embodiment onexWith wall thickness of internal cylindrical sleeve δ change curve;
Fig. 6 is the equivalent combinations Line stiffness K of the rubber bushing of embodiment onexWith wall thickness of internal cylindrical sleeve δ change curve;
Fig. 7 is the stabilizer bar system roll angular rigidity of embodiment oneWith wall thickness of internal cylindrical sleeve δ change curve;
Fig. 8 is the deformation simulation cloud atlas of the outer offset cab stabilizer bar system of embodiment one;
Fig. 9 is the stabilizer bar system roll angular rigidity of embodiment twoWith wall thickness of internal cylindrical sleeve δ change curve;
Figure 10 is the deformation simulation cloud atlas of the outer offset cab 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 offset cab stabilizer bar system is symmetrical, as shown in Fig. 2 including: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, outside torsion tube 4 Amount of bias T=30mm;The distance between the swing arm 1 of left and right two Lc=1550mm, i.e. stabiliser bar suspension distance;Suspended rubber serves as a contrast Cover the distance between 2 and reversed rubber bushing 3, i.e. pendulum arm length l1=380mm, the suspension installation site C to outermost end A of swing arm 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 are ginseng to be designed Number;The exradius r of rubber sleeve 6b=34.5mm, length Lx=25mm, elastic modulus Ex=7.84MPa, Poisson's ratio μx= 0.47.The required roll angular rigidity of driver's cabin stabilizer bar system designTo the outer offset The wall thickness of internal cylindrical sleeve δ of driver's cabin stabiliser bar rubber bushing is designed, and to stablizing leverage in the case of load F=5000N The roll angular rigidity of system carries out ANSYS simulating, verifyings.
The design method for the outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve that present example is provided, Its design cycle is as shown in figure 1, comprise 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 rigidityThe suspension of stabiliser bar Distance Lc=1550mm, 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 K of the torsion tube of outer offset stabiliser barTAnd reversed rubber bushing loading coefficient ηFCalculating:
1. the equivalent line stiffness K of the torsion tube of stabiliser barTCalculating
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, the torsion tube of the outer offset stabiliser bar is being driven Equivalent line stiffness K at the suspension installation site C of roomTCalculated, i.e.,
2. the loading coefficient η of reversed rubber bushingFCalculating:
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 offset cab stabiliser bar rubber bushingxThe foundation of (δ):
I establishes the radial rigidity expression formula k of rubber bushingx(δ)
According to the length L of rubber sleevex=25mm, exradius rb=34.5mm, elastic modulus Ex=7.84MPa and Poisson Compare μx=0.47, the interior circular diameter d of interior round bussx=35mm, using wall thickness of internal cylindrical sleeve δ as parameter, then the inner circle of rubber sleeve half Footpath raIt is represented byThe radial rigidity of rubber bushing can be expressed as kx(δ), i.e.,
Wherein,
Bessel correction functions I (0, αb), K (0, αb), I (1, αb), K (1, αb),
I(1,αa), K (1, αa), I (0, αa), K (0, αa);
αb=α rb,
Wherein, in exradius rbAnd length LxIn the case of given, the radial rigidity k of rubber bushingxWith wall thickness of internal cylindrical sleeve δ change curve, as shown in Figure 5;
II establishes the equivalent combinations Line stiffness expression formula K of outer offset stabiliser bar rubber bushingx(δ):
According to the k established in I stepsxThe η being calculated in 2. step in (δ), and step (2)F=0.15808, build The equivalent combinations Line stiffness expression formula K of vertical stabiliser bar rubber bushingx(δ), i.e.,
Wherein, the equivalent combinations Line stiffness K of stabiliser bar rubber bushingxWith wall thickness of internal cylindrical sleeve δ change curve, such as Fig. 6 It is shown;
(4) foundation and design of the wall thickness of internal cylindrical sleeve δ design mathematic models of stabiliser bar rubber bushing:
According to the K being calculated in step (1)ws=2.4608 × 105N/m, what the 1. step in step (2) was calculated KT=2.84488 × 105What the II steps in N/m, and step (3) were establishedEstablish stabiliser bar rubber lining Wall thickness of internal cylindrical sleeve δ design mathematic model is covered, i.e.,
KTKX(δ)-KwsKX(δ)-KTKws=0;
Using Matlab programs, the above-mentioned equation on δ is solved, the outer offset cab stabiliser bar rubber lining can be obtained Wall thickness of internal cylindrical sleeve δ is covered, i.e.,
δ=2.0mm;
Wherein, the inner circle radius r of rubber sleevea=dx/ 2+ δ=19.5mm;Stabilizer bar system roll angular rigidityWith inner circle Sleeve wall thickness δ change curve, as shown in Figure 7;
(5) the ANSYS simulating, verifyings of outer offset cab stabilizer bar system roll angular rigidity:
A) ANSYS finite element emulation softwares are utilized, according to design obtain rubber bushing wall thickness of internal cylindrical sleeve δ= 2.0mm and rubber sleeve inner circle radius ra=19.5mm, and the other structures parameter of the outer offset cab stabilizer bar system And material characteristic parameter, ANSYS simulation models, grid division are established, and apply load F at the suspension installation site C of swing arm =5000N, the deformation progress ANSYS emulation to stabilizer bar system, resulting deformation simulation cloud atlas, as shown in figure 8, wherein, Maximum distortion displacement f of the stabilizer bar system in swing arm outermost end AAFor
fA=19.984mm;
B) according to the inner circle radius r that obtained wall thickness of internal cylindrical sleeve δ=2.0mm and rubber sleeve are designed in step (4)a= 19.5mm, and the other structures parameter and material characteristic parameter of rubber bushing, the footpath established using the I steps in step (3) To rigidity expression formula kx(rb), the radial rigidity k of designed rubber bushing is calculatedx=2.1113 × 106N/m;
C) the deformation displacement amount f at swing arm outermost end A obtained by being emulated according to ANSYSA=19.984mm, pendulum arm length 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 k b) being calculated in stepx= 2.1113×106N/m, using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, as shown in figure 4, to the outer offset Driver's cabin stabilizer bar system roll angular rigidityANSYS simulating, verifying values, calculated, i.e.,
Understand:The ANSYS simulating, verifying values of the outer offset cab stabiliser bar roll angular rigidity With design requirement valueMatch, relative deviation is only 0.916%;As a result show to be provided is outer The design method of offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve is correct, and parameter design value is accurately and reliably 's.
Embodiment two:The structure type of certain outer offset cab stabilizer bar system is identical with embodiment one, such as Fig. 2 institutes Show, torsion tube 4 is not coaxial with reversed rubber bushing 3, the outer amount of bias T=30mm of torsion tube 4;The distance between swing arm 1 of left and right two Lc=1400mm, i.e. stabiliser bar suspension distance;The distance between suspended rubber bushing 2 and reversed rubber bushing 3, i.e. swing arm are grown Spend l1=350mm, the suspension installation site C to outermost end A of swing arm distance, delta l1=52.5mm;The length L of torsion tube 4w= 1000mm, internal diameter d=42mm, outer diameter D=50mm;The structure of four rubber bushings in left and right is all identical, as shown in figure 3, its In, the interior circular diameter d of interior round buss 5x=35mm, wall thickness δ are parameter to be designed;The length L of rubber sleeve 6x=40mm, cylindrical half Footpath rb=37.5mm.The material property of stabiliser bar and the material property of rubber bushing, identical with embodiment one, the i.e. bullet of torsion tube Property modulus E=200GPa, Poisson's ratio μ=0.3;The elastic modulus E of rubber sleevex=7.84MPa, Poisson's ratio μx=0.47.The driving The required roll angular rigidity of room stabiliser bar designTo the outer offset cab stabiliser bar rubber The wall thickness of internal cylindrical sleeve δ of glue bushing is designed, and to the roll angular rigidity of stabilizer bar system in the case of load F=5000N Carry out ANSYS simulating, verifyings.
Using the step identical with embodiment one, to the inner circle sleeve wall of the outer offset cab stabiliser bar rubber bushing Thick δ 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 driver's cabin stabilizer bar system inclination line stiffness KwsDesign requirement value calculated, i.e.,
(2) the equivalent line stiffness K of the torsion tube of outer offset stabiliser barTAnd reversed rubber bushing loading coefficient ηFCalculating:
1. the equivalent line stiffness K of the torsion tube of stabiliser barTCalculating
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, the torsion tube of external offset stabiliser bar is in driver's cabin Suspend the equivalent line stiffness K at installation site CTCalculated, i.e.,
2. the loading coefficient η of reversed rubber bushingFCalculating
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 offset cab stabiliser bar rubber bushingxThe foundation of (δ):
I establishes the radial rigidity expression formula k of rubber bushingx(δ)
According to the length L of rubber sleevex=40mm, exradius rb=37.5mm, elastic modulus Ex=7.84MPa and Poisson Compare μx=0.47, the interior circular diameter d of interior round bussx=35mm, using wall thickness of internal cylindrical sleeve δ as parameter, then the inner circle of rubber sleeve half Footpath raIt is represented byThe radial rigidity of rubber bushing can be expressed as kx(δ), i.e.,
Wherein,
Bessel correction functions I (0, αb), K (0, αb), I (1, αb), K (1, αb),
I(1,αa), K (1, αa), I (0, αa), K (0, αa);
αb=α rb,
II establishes the equivalent combinations Line stiffness expression formula K of outer offset stabiliser bar rubber bushingx(δ):
According to the k established in I stepsxThe η being calculated in 2. step in (δ), and step (2)F=0.3276, build The equivalent combinations Line stiffness expression formula K of vertical stabiliser bar rubber bushingx(δ), i.e.,
(4) foundation and design of the wall thickness of internal cylindrical sleeve δ design mathematic models of stabiliser bar rubber bushing:
According to the K being calculated in step (1)ws=2.97455 × 105N/m, the 1. step in step (2) are calculated KT=3.28257 × 105What the II steps in N/m, and step (3) were establishedEstablish wall thickness of internal cylindrical sleeve δ design mathematic model, i.e.,
KTKX(δ)-KwsKX(δ)-KTKws=0;
Using Matlab programs, the above-mentioned equation on δ is solved, the outer offset cab stabiliser bar rubber lining can be obtained Wall thickness of internal cylindrical sleeve δ is covered, i.e.,
δ=5.0mm;
Wherein, the inner circle radius r of rubber sleevea=dx/ 2+ δ=22.5mm;Stabilizer bar system roll angular rigidityWith inner circle Sleeve wall thickness δ change curve, as shown in Figure 9;
(5) the ANSYS simulating, verifyings of outer offset cab stabilizer bar system roll angular rigidity:
A) ANSYS finite element emulation softwares are utilized, according to design obtain rubber bushing wall thickness of internal cylindrical sleeve δ= 5.0mm and rubber sleeve inner circle radius ra=22.5mm, and the other structures parameter of the outer offset cab stabilizer bar system And material characteristic parameter, ANSYS simulation models, grid division are established, and apply load F at the suspension installation site C of swing arm =5000N, the deformation progress ANSYS emulation to the stabilizer bar system, resulting deformation simulation cloud atlas, as shown in Figure 10, its In, deformation displacement amount f of the stabilizer bar system at swing arm outermost end AAFor
fA=17.881mm;
B) according to the inner circle radius r that obtained wall thickness of internal cylindrical sleeve δ=5.0mm and rubber sleeve are designed in step (4)a= 22.5mm, and the other structures parameter and material characteristic parameter of rubber bushing, the footpath established using the I steps in step (3) To rigidity expression formula kx(rb), the radial rigidity k of designed rubber bushing is calculatedx=4.2085 × 106N/m;
C) the deformation displacement amount f at swing arm outermost end A obtained by being emulated according to ANSYSA=17.881mm, pendulum arm length 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 the suspension installation site C of swing arm, and k b) being calculated in stepx= 4.2085×106N/m, using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, as shown in figure 4, to the outer offset Driver's cabin stabilizer bar system roll angular rigidityANSYS simulating, verifying values, calculated, i.e.,
Understand, the ANSYS simulating, verifying values of the outer offset cab stabilizer bar system roll angular rigidityWith design requirement valueMatch, relative deviation is only 0.431%;The design method for the outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve for showing to be provided is correct , parameter design value is accurately and reliably.

Claims (1)

1. the design method of outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve, its specific design step are as follows:
(1) the inclination line stiffness K of driver's cabin stabilizer bar systemwsThe calculating of design requirement value:
According to the roll angular rigidity design requirement value of stabilizer bar systemThe suspension distance L of stabiliser barc, to driver's cabin stabiliser bar The inclination line stiffness K of systemwsDesign requirement value calculated, i.e.,
(2) the equivalent line stiffness K of the torsion tube of outer offset stabiliser barTAnd reversed rubber bushing loading coefficient ηFCalculating:
1. the equivalent line stiffness K of the torsion tube of stabiliser barTCalculating
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, externally Equivalent line stiffness K of the torsion tube of offset stabiliser bar in cab mounting installed positionTCalculated, i.e.,
<mrow> <msub> <mi>K</mi> <mi>T</mi> </msub> <mo>=</mo> <mfrac> <mrow> <mi>&amp;pi;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>
2. the loading coefficient η of reversed rubber bushingFCalculating
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 offset cab stabiliser bar rubber bushingxThe foundation of (δ):
I establishes the radial rigidity expression formula k of rubber bushingx(δ)
According to the length L of rubber sleevex, exradius rb, elastic modulus ExWith Poisson's ratio μx, the interior circular diameter d of interior round bussx, with Wall thickness of internal cylindrical sleeve δ is the inner circle radius r of parameter, then rubber sleeveaIt is represented byThe radial rigidity of rubber bushing kx(δ) can be expressed as, 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,
<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> <mi>b</mi> </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> <mi>b</mi> </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> <msub> <mi>L</mi> <mi>x</mi> </msub> </mrow> </mfrac> <mo>&amp;lsqb;</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> <msup> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>x</mi> </msub> <mo>/</mo> <mn>2</mn> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> </mfrac> <mo>&amp;rsqb;</mo> <mo>,</mo> </mrow>
<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> <mi>&amp;alpha;</mi> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> <mo>)</mo> <mo>(</mo> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <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>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>(</mo> <mn>3</mn> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <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;</mi> <mrow> <mi>a</mi> <mi>b</mi> </mrow> </msub> <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;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;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <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> <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> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <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> <mi>b</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>(</mo> <mn>3</mn> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <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;</mi> <mrow> <mi>a</mi> <mi>b</mi> </mrow> </msub> <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;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;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <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> <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>(</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>)</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;</mi> <mrow> <mi>a</mi> <mi>b</mi> </mrow> </msub> <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> <mi>b</mi> </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> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <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> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</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>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</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;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <msub> <mi>d</mi> <mi>x</mi> </msub> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <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> </mrow> <mo>,</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</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;</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</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> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <mi>d</mi> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>
<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;</mi> <mrow> <mi>a</mi> <mi>b</mi> </mrow> </msub> <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> <mi>b</mi> </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> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mo>&amp;lsqb;</mo> <msup> <mrow> <mo>(</mo> <mfrac> <mi>d</mi> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <mrow> <mo>(</mo> <msup> <mrow> <mo>(</mo> <mrow> <mfrac> <mi>d</mi> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mi>ln</mi> <mrow> <mo>(</mo> <mfrac> <mi>d</mi> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mo>,</mo> </mrow>
Bessel correction functions I (0, αb), K (0, αb), I (1, αb), K (1, αb),
I(1,αa), K (1, αa), I (0, αa), K (0, αa);
<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> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>=</mo> <mi>&amp;alpha;</mi> <mrow> <mo>(</mo> <mfrac> <mi>d</mi> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <mo>=</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mrow> <mi>a</mi> <mi>b</mi> </mrow> </msub> <mo>=</mo> <mi>&amp;alpha;</mi> <mrow> <mo>(</mo> <mfrac> <mi>d</mi> <mn>2</mn> </mfrac> <mo>+</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>;</mo> </mrow>
II establishes the equivalent combinations Line stiffness expression formula K of outer offset stabiliser bar rubber bushingx(δ):
According to the radial rigidity expression formula k for the rubber bushing established in I stepsx2. step in (δ), and step (2) is calculated The loading coefficient η of obtained reversed rubber bushingF, establish the equivalent combinations Line stiffness expression formula K of stabiliser bar rubber bushingx(δ), I.e.
<mrow> <msub> <mi>K</mi> <mi>X</mi> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>k</mi> <mi>x</mi> </msub> <mrow> <mo>(</mo> <mi>&amp;delta;</mi> <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) foundation and design of the wall thickness of internal cylindrical sleeve δ design mathematic models of stabiliser bar rubber bushing:
According to the design requirement value K for the driver's cabin stabilizer bar system inclination line stiffness being calculated in step (1)ws, in step (2) The equivalent line stiffness K of torsion tube that is calculated of 1. stepT, and the stabiliser bar rubber lining that II steps in step (3) are established The equivalent combinations Line stiffness expression formula K of setx(δ), establish the wall thickness of internal cylindrical sleeve δ of stabiliser bar rubber bushing design mathematic mould Type, i.e.,
KTKX(δ)-KwsKX(δ)-KTKws=0;
Using Matlab programs, the above-mentioned equation on δ is solved, can be obtained in outer offset cab stabiliser bar rubber bushing Round buss wall thickness δ design load, and the inner circle radius of rubber sleeve
(5) the ANSYS simulating, verifyings of outer offset cab stabilizer bar system roll angular rigidity:
A) ANSYS finite element emulation softwares are utilized, according to the wall thickness of internal cylindrical sleeve δ's and rubber sleeve for designing obtained rubber bushing Inner circle radius ra, and the other structures parameter and material characteristic parameter of the outer offset cab stabilizer bar system, establish ANSYS Simulation model, grid division, and apply load F in the suspension installed position of swing arm, the deformation to stabilizer bar system is carried out ANSYS is emulated, and obtains deformation displacement amount f of the stabilizer bar system in swing arm outermost endA
B) according to the inner circle radius r that obtained wall thickness of internal cylindrical sleeve δ and rubber sleeve are designed in step (4)a, and rubber bushing its His structural parameters and material characteristic parameter, the radial rigidity expression formula k established using the I steps in step (3)x(rb), calculate Obtain the radial rigidity k of designed rubber bushingx
C) the deformation displacement amount f of the swing arm outermost end obtained by being emulated 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, the load F that is applied at the suspension installation site of swing arm, And the radial rigidity k for rubber bushing b) being calculated in stepx, closed using the geometry of stabilizer bar system deformation and swing arm displacement System, to the outer offset cab stabilizer bar system roll angular rigidityANSYS simulating, verifying values, calculated, 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 outer offset cab stabilizer bar system roll angular rigidityWith design requirement value It is compared, so as to the design method and ginseng of the outer offset cab stabiliser bar rubber bushing wall thickness of internal cylindrical sleeve provided Number design load is verified.
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