CN104200043B - The design method of suspension stabiliser bar rubber bushing length - Google Patents

The design method of suspension stabiliser bar rubber bushing length Download PDF

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CN104200043B
CN104200043B CN201410476381.0A CN201410476381A CN104200043B CN 104200043 B CN104200043 B CN 104200043B CN 201410476381 A CN201410476381 A CN 201410476381A CN 104200043 B CN104200043 B CN 104200043B
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CN104200043A (en
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周长城
提艳
张云山
宋群
潘礼军
程正午
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Shandong University of Technology
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Abstract

The present invention relates to the design method of suspension stabiliser bar rubber bushing length, belong to vehicle suspension technical field.Previous home and abroad fails to provide reliable resolution design method always to rubber bushing length.Present invention be characterized in that:According to the wheelspan of vehicle, the roll angular rigidity design requirement value of stabilizer bar system, end part of stabilizer rod deformation coefficient Gw, the radial rigidity expression formula K of rubber bushingx, the mathematical model of optimizing design of suspension stabiliser bar rubber bushing length is established, the optimization design value of rubber bushing length can be obtained using Matlab programs.Accurate, reliable rubber bushing length optimization design load is can obtain using this method, i.e., on the premise of design and production cost is not increased, only by the optimization design to rubber bushing length, stabilizer bar system can be made to reach the design requirement of roll angular rigidity;Meanwhile design and testing expenses can be reduced using this method, vehicle suspension design level is improved, improves vehicle ride performance and handling safety.

Description

The design method of suspension stabiliser bar rubber bushing length
Technical field
The present invention relates to the design method of vehicle suspension stabiliser bar, particularly suspension stabiliser bar rubber bushing length.
Background technology
QS is one of important composition part of vehicle suspension system, when Vehicular turn travels to prevent vehicle body Excessive inclination occurs and rolls angular oscillation.QS is added in vehicle suspension system can reduce this cross side Incline, if the forward and backward suspension roll angular rigidity distribution design of vehicle is improper, it will influence the steering characteristic of vehicle, general front suspension system The roll angular rigidity of system is more than the roll angular rigidity of rear-suspension system.The roll angular rigidity of suspension system is not only tied by stabiliser bar The influence of structure, diameter, while also by the length of rubber bushing, inner circle radius, exradius, material property and installation displacement Etc. the influence of factor.It is however, theoretical due to being deformed analytical Calculation by rubber bushing radial deformation and end part of stabilizer rod vertical deviation And the restriction to intercouple key issues of influence, the design for stablizing shank diameter and rubber bushing length, home and abroad at present Fail to provide reliable resolution design method always.Home and abroad scholar is mostly to utilize simulation analysis software at present, to laterally steady Fixed pole system variant and rigidity carry out Numerical Simulation Analysis, still, can only be to given structure and load using simulation analysis software In the case of stabilizer bar system deformation and rigidity carry out simulating, verifying, no analytical formula, it is impossible to meet stabilizer bar system parse Design and the requirement of modernization CAD design.
With the fast development of Vehicle Industry and the raising of travel speed, the design to suspension stabilizer bar system proposes more High design requirement.How structure, diameter and the material property of stabiliser bar, and the material property of rubber bushing, inner circle given Radius, exradius and two rubber bushing installation sites it is constant in the case of, i.e., do not increasing design and production cost premise Under, only by the optimization design to rubber bushing length, stabilizer bar system can be made to reach the design requirement of roll angular rigidity, be Enterprise's technical problem in the urgent need to address at present.Therefore, it is necessary to establish a kind of accurate, reliable suspension stabiliser bar rubber bushing The design method of length, vehicle suspension design level is improved, on the premise of production cost is not increased, is served as a contrast by stabiliser bar rubber Cover the optimization design of length so that stabilizer bar system reaches the design requirement of roll angular rigidity, improve vehicle ride performance and Security.
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 suspension stabiliser bar rubber bushing length, its design flow diagram is as shown in figure 1, suspension stabilizer bar system Structural representation, as shown in Figure 2.
In order to solve the above technical problems, the design method of suspension stabiliser bar rubber bushing length provided by the present invention, its It is characterised by using following steps.
(1) the vertical deviation deformation coefficient G of end part of stabilizer rod is calculatedw
According to the total length l of QSc, brachium l1, transition arc radius R, the central angle θ of transition arc, material bullet Mounting distance l between property modulus E and Poisson's ratio μ, and two rubber bushings0, to the vertical deviation deformation coefficient G of end part of stabilizer rodw Calculated, i.e.,:
In formula,
G6=-32 (μ+1) [R (cos θ -1)-l1sinθ]2[2l1cosθ-lc+2Rsinθ];
(2) rubber bushing RADIAL stiffness K is establishedxExpression formula:
Using rubber bushing length L as parameter to be designed, according to the diameter d of stabiliser bar, the inner circle radius of rubber bushing ra, exradius rb, elastic modulus Ex, Poisson's ratio μx, establish rubber bushing RADIAL stiffness KxCalculation expression, i.e.,
Wherein, Kx(L) it is expression formula on rubber bushing length L;
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);
(3) foundation and design calculating of rubber bushing length L design mathematic models:
According to vehicle propons or the wheelspan B of back axle, the diameter d of stabiliser bar, total length lc, the installation between two rubber bushings Distance l0, the design requirement value of stabilizer bar system roll angular rigidityResulting end part of stabilizer rod is calculated in step (1) Vertical deviation deformation coefficient Gw, and the expression formula K for the rubber bushing radial direction Line stiffness established in step (2)x(L) rubber, is established Liner sleeve length L mathematical model of optimizing design, i.e.,:
Using Matlab calculation procedures, above-mentioned mathematical modeling is solved, can be obtained in the structure of stabiliser bar, rubber bushing Inner circle radius ra, exradius rbAnd in the case that installation site is all constant, meet stabilizer bar system roll angular rigidity design requirement Rubber bushing length L optimization design value.
The present invention has the advantage that than prior art:
Due to being deformed analytical Calculation theory by rubber bushing radial deformation and end part of stabilizer rod vertical deviation and intercoupled The restriction of key issues of influence, design of the home and abroad for stabiliser bar rubber bushing length at present, fail to provide reliably always Resolution design method.Home and abroad scholar is mostly to utilize simulation analysis software at present, to QS system variant and just Degree carries out Numerical Simulation Analysis, still, can only be to giving the stable leverage under structure and load condition using simulation analysis software System deformation and rigidity carry out simulating, verifying, no analytical formula, it is impossible to meet that stabilizer bar system analytical design method and modernization CAD are set The requirement of meter.The present invention is according to the wheelspan of vehicle, the roll angular rigidity design requirement value of stabilizer bar system, the change of end part of stabilizer rod Shape coefficient, using rubber bushing length as parameter to be designed, the optimization for establishing suspension stabiliser bar rubber bushing length is set Mathematical modeling is counted, the optimization design value of rubber bushing length can be obtained using Matlab programs.This method can exist to stabiliser bar Given structure, diameter and material property, and the material property of rubber bushing, inner circle radius, exradius and two rubber bushings peace In the case that holding position is constant, i.e., on the premise of design and production cost is not increased, only by the excellent of rubber bushing length Change design, stabilizer bar system can be made to reach the design requirement of roll angular rigidity.Therefore, it is available accurately and reliably using this method Rubber bushing length optimization design value, it is horizontal to improve vehicle suspension design, only logical on the premise of production cost is not increased The optimization design of stabiliser bar rubber bushing length is crossed, stabilizer bar system can be made to reach the design requirement of roll angular rigidity;Meanwhile Vehicle ride performance and handling safety can be improved.Therefore, the present invention designs for the Optimized Matching of suspension stabilizer bar system, carries Reliable design method and technology are supplied.
Brief description of the drawings
In order to more fully understand that invention is described further below in conjunction with the accompanying drawings.
Fig. 1 is the design flow diagram of suspension stabiliser bar rubber bushing length;
Fig. 2 is the structural representation of lateral stability lever system;
Fig. 3 is the structural representation of rubber bushing;
Fig. 4 is the stabilizer bar system roll angular rigidity of embodiment one with the change curve of rubber bushing length;
Fig. 5 is the stabilizer bar system roll angular rigidity of embodiment three with the change curve of rubber bushing length.
Embodiment
The present invention is described in further detail below by embodiment.
Embodiment one:The wheelspan B=1600mm of certain vehicle propons, the structure of stabiliser bar is used, as shown in Fig. 2 wherein, lcFor the total length of stabiliser bar, lc=800mm;l1For brachium, l1=150mm;l0Mounting distance between rubber bushing, l0= 400mm;R is transition arc radius, R=50mm;θ is transition arc central angle, θ=60 °;The elastic modulus E of stable bar material =210GPa, Poisson's ratio μ=0.3.The structure of rubber bushing is as shown in figure 3, wherein, stabiliser bar 1, interior round buss cylinder 2, rubber lining Set 3, outer round buss 4, the diameter d=20mm of stabiliser bar 1, the inner circle radius r of rubber bushing 3a=13mm, exradius rb= 30mm, elastic modulus Ex=7.84MPa, Poisson's ratio μx=0.47, the length L of rubber bushing is parameter to be designed.Before the vehicle The design requirement value of the roll angular rigidity of suspension stabilizer bar systemStructure, rubber lining in stabiliser bar The inner circle radius r of seta, exradius rbAnd in the case that installation site is constant, design is optimized to the length L of rubber bushing.
The design method for the suspension stabiliser bar rubber bushing length that present example is provided, its design cycle such as Fig. 1 institutes Show, comprise the following steps that:
(1) the vertical deviation deformation coefficient G of end part of stabilizer rod is calculatedw
According to the total length l of QSc=800mm, brachium l1=150mm, transition arc radius R=50mm, transition Central angle θ=60 ° of circular arc;Elastic modulus E=210GPa, Poisson's ratio μ=0.3;And the locating distance between two rubber bushings From l0=400mm, to the vertical deviation deformation coefficient G of end part of stabilizer rodwCalculated, i.e.,:
In formula,
G6=-32 (μ+1) [R (cos θ -1)-l1sinθ]2[2l1cosθ-lc+ 2Rsin θ]=- 0.5624m3
(2) rubber bushing RADIAL stiffness K is establishedxExpression formula:
Using rubber bushing length L as parameter to be designed, according to the diameter d=20mm of stabiliser bar, rubber bushing it is interior Radius of circle ra=13mm, exradius rb=30mm, elastic modulus Ex=7.84MPa, Poisson's ratio μx=0.47, establish rubber lining Cover RADIAL stiffness KxCalculation expression, i.e.,:
Wherein, Kx(L) it is expression formula on rubber bushing length L;
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);
(3) foundation and design calculating of rubber bushing Design of length mathematical modeling:
According to the wheelspan B=1600mm of the vehicle propons, the diameter d=20mm of stabiliser bar, total length lc=800mm, two Mounting distance l between rubber bushing0=400mm, the design requirement value of stabilizer bar system roll angular rigidityThe vertical deviation deformation coefficient G of end part of stabilizer rod obtained by being calculated in step (1)w=1.5935 × 10-12m5The expression formula K for the rubber bushing radial direction Line stiffness established in/N, and step (2)x(L) rubber bushing length L, is established Mathematical model of optimizing design, i.e.,:
Using Matlab calculation procedures, above-mentioned mathematical modeling is solved, can be obtained in the structure of stabiliser bar, rubber bushing Inner circle radius ra, exradius rbAnd in the case that installation site is constant, meet stabilizer bar system roll angular rigidity design requirement The optimization design value L=27mm of rubber bushing length.
Wherein, the structure in stabiliser bar, the inner circle radius r of rubber bushinga, exradius rbAnd the bar that installation site is constant Under part, the roll angular rigidity design requirement value of the vehicle front suspension stabilizer bar system, with rubber bushing length L change curve, As shown in Figure 4.
Embodiment two:It is the structural parameters of certain vehicle front suspension, the structural parameters of stabiliser bar, rubber bushing inner circle radius, outer Radius of circle and material characteristic parameter, all identical with embodiment one, the simply inclination of the vehicle front suspension stabilizer bar system Angular rigidityDesign requirement value, the difference with embodiment one, wherein,In other structures parameter not In the case of change, design is optimized to rubber bushing length L, to meet the design requirement of stabilizer bar system roll angular rigidity.
Using the design procedure of embodiment one, the length L of the vehicle front suspension stabiliser bar rubber bushing is designed.By The structural parameters of structural parameters, stabiliser bar in the vehicle front suspension and inner circle radius, exradius and the material of rubber bushing Characterisitic parameter, all identical with embodiment one, the simply roll angular rigidity design requirement value and embodiment of stabilizer bar system One differs.Therefore, in roll angular rigidity design requirement valueIn the case of, design the resulting car The length L=42mm of front suspension rubber bushing.
Understood compared with embodiment one, due to roll angular rigidity design requirement value10kN.m/rad is added, then only The length of rubber bushing is increased into 42mm by previous 27mm, you can in the case where not changing other structures parameter, obtain To the stabilizer bar system for meeting the roll angular rigidity design requirement.
Embodiment three:The wheelspan B=1600mm of certain vehicle propons, stabiliser bar is used in addition to diameter d=21mm, its Its structural parameters and mounting structure parameter and material characteristic parameter, it is all identical with embodiment one;The inner circle of rubber bushing Radius ra=13.5mm, exradius rb=30mm;The elastic modulus E of rubber bushingx=7.84MPa, Poisson's ratio μx=0.47; Length L is amount to be designed.The design requirement value of the roll angular rigidity of the vehicle front suspension stabilizer bar system Under conditions of the given structure of stabiliser bar, the inner circle radius of rubber bushing, exradius and installation site, to rubber bushing Length L is designed.
Using the design procedure of embodiment one, the length L of the vehicle front suspension stabiliser bar rubber bushing is designed.
(1) the vertical deviation deformation coefficient G of end part of stabilizer rod is calculatedw
Due to the mounting distance l between horizontal stabilizer bar structure parameter, material characteristic parameter and two rubber bushings0, all With applying the identical of example one, therefore, the vertical deviation deformation coefficient G of end part of stabilizer rodwIt is also identical with embodiment one, I.e.:
(2) rubber bushing RADIAL stiffness K is establishedxExpression formula:
Using rubber bushing length L as parameter to be designed, according to the diameter d=21mm of stabiliser bar, rubber bushing it is interior Radius of circle ra=13.5mm, exradius rb=30mm, elastic modulus Ex=7.84MPa, Poisson's ratio μx=0.47, establish rubber Bushing RADIAL stiffness KxCalculation expression, i.e.,:
Wherein, Kx(L) it is expression formula on rubber bushing length L;
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);
(3) foundation and design calculating of rubber bushing Design of length mathematical modeling:
According to the wheelspan B=1600mm of propons, the diameter d=21mm of stabiliser bar, total length lc=800mm, two rubber lining Mounting distance l between set0=400mm, the design requirement value of stabilizer bar system roll angular rigidityStep Suddenly the vertical deviation deformation coefficient G of the end part of stabilizer rod obtained by being calculated in (1)w=1.5935 × 10-12m5/ N, and step (2) Middle established rubber bushing radial direction Line stiffness expression formula Kx(L) rubber bushing length L design mathematic model, is established, i.e.,:
Using Matlab calculation procedures, above-mentioned mathematical modeling is solved, can be obtained in the structure of stabiliser bar, rubber bushing Inner circle radius ra, exradius rbAnd in the case that installation site is all constant, meet stabilizer bar system roll angular rigidity design requirement Rubber bushing length optimization design value L=25mm.
Wherein, the structure in stabiliser bar, the inner circle radius r of rubber bushinga, exradius rbAnd the bar that installation site is constant Under part, the roll angular rigidity design requirement value of the vehicle front suspension stabilizer bar system, with rubber bushing length L change curve, As shown in Figure 5.

Claims (1)

1. the design method of suspension stabiliser bar rubber bushing length, it is comprised the following steps that:
(1) the vertical deviation deformation coefficient G of end part of stabilizer rod is calculatedw
According to the total length l of QSc, brachium l1, transition arc radius R, the central angle θ of transition arc, elastic properties of materials mould Measure the mounting distance l between E and Poisson's ratio μ, and two rubber bushings0, to the vertical deviation deformation coefficient G of end part of stabilizer rodwCarry out Calculate, i.e.,:
<mrow> <msub> <mi>G</mi> <mi>w</mi> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mi>G</mi> <mn>1</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>2</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>3</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>4</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>5</mn> </msub> <mo>+</mo> <msub> <mi>G</mi> <mn>6</mn> </msub> </mrow> <mi>&amp;pi;E</mi> </mfrac> <mo>;</mo> </mrow>
In formula, <mrow> <msub> <mi>G</mi> <mn>1</mn> </msub> <mo>=</mo> <mfrac> <msubsup> <mrow> <mn>64</mn> <mi>l</mi> </mrow> <mn>1</mn> <mn>3</mn> </msubsup> <mn>3</mn> </mfrac> <mo>,</mo> <msub> <mi>G</mi> <mn>2</mn> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mn>64</mn> <mo>[</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mn>1</mn> </msub> <mi>cos</mi> <mi>&amp;theta;</mi> <mo>+</mo> <mi>R</mi> <mi>sin</mi> <mi>&amp;theta;</mi> <mo>)</mo> </mrow> <mn>3</mn> </msup> <mo>+</mo> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>l</mi> <mi>c</mi> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> <mo>]</mo> </mrow> <mn>3</mn> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>G</mi> <mn>3</mn> </msub> <mo>=</mo> <mn>64</mn> <mi>R</mi> <mo>[</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msubsup> <mi>l</mi> <mn>1</mn> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <mi>&amp;theta;</mi> <mo>+</mo> <mfrac> <mrow> <mi>sin</mi> <mn>2</mn> <mi>&amp;theta;</mi> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msup> <mi>R</mi> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mi>&amp;theta;</mi> <mo>-</mo> <mfrac> <mrow> <mi>sin</mi> <mn>2</mn> <mi>&amp;theta;</mi> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>l</mi> <mn>1</mn> </msub> <mi>R</mi> <msup> <mi>sin</mi> <mn>2</mn> </msup> <mi>&amp;theta;</mi> <mo>]</mo> <mo>,</mo> <msub> <mi>G</mi> <mn>4</mn> </msub> <mo>=</mo> <mfrac> <mrow> <msub> <mrow> <mn>8</mn> <mi>l</mi> </mrow> <mn>0</mn> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>l</mi> <mi>c</mi> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> <mn>3</mn> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>G</mi> <mn>5</mn> </msub> <mo>=</mo> <mn>64</mn> <mi>R</mi> <mrow> <mo>(</mo> <mi>&amp;mu;</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>[</mo> <msup> <mi>R</mi> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mfrac> <mrow> <mn>3</mn> <mi>&amp;theta;</mi> </mrow> <mn>2</mn> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>sin</mi> <mn>2</mn> <mi>&amp;theta;</mi> </mrow> <mn>4</mn> </mfrac> <mo>-</mo> <mn>2</mn> <mi>sin</mi> <mi>&amp;theta;</mi> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msubsup> <mi>l</mi> <mn>1</mn> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <mi>&amp;theta;</mi> <mo>-</mo> <mfrac> <mrow> <mi>sin</mi> <mn>2</mn> <mi>&amp;theta;</mi> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mrow> <mn>4</mn> <mi>l</mi> </mrow> <mn>1</mn> </msub> <mi>R</mi> <msup> <mi>sin</mi> <mn>4</mn> </msup> <mfrac> <mi>&amp;theta;</mi> <mn>2</mn> </mfrac> <mo>]</mo> <mo>,</mo> </mrow>
G6=-32 (μ+1) [R (cos θ -1)-l1sinθ]2[2l1cosθ-lc+2Rsinθ];
(2) rubber bushing RADIAL stiffness K is establishedxExpression formula:
Using rubber bushing length L as parameter to be designed, according to the diameter d of stabiliser bar, the inner circle radius r of rubber bushinga, outside Radius of circle rb, elastic modulus Ex, Poisson's ratio μx, establish rubber bushing RADIAL stiffness KxCalculation expression, i.e.,
<mrow> <msub> <mi>K</mi> <mi>x</mi> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <mi>u</mi> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow>
Wherein, Kx(L) it is expression formula on rubber bushing length L;
<mrow> <mi>u</mi> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> <mrow> <mn>2</mn> <mi>&amp;pi;</mi> <msub> <mi>E</mi> <mi>x</mi> </msub> <mi>L</mi> </mrow> </mfrac> <mo>[</mo> <mi>ln</mi> <mfrac> <msub> <mi>r</mi> <mi>b</mi> </msub> <msub> <mi>r</mi> <mi>a</mi> </msub> </mfrac> <mo>-</mo> <mfrac> <mrow> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>-</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> </mrow> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> </mrow> </mfrac> <mo>]</mo> <mo>,</mo> </mrow>
<mrow> <mi>y</mi> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>L</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>L</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>L</mi> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> <mrow> <mn>5</mn> <mi>&amp;pi;L</mi> <msub> <mi>E</mi> <mi>x</mi> </msub> </mrow> </mfrac> <mo>[</mo> <mi>ln</mi> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>+</mo> <mfrac> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mrow> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> </mrow> </mfrac> <mo>]</mo> <mo>,</mo> </mrow>
αb=α rba=α ra, <mrow> <mi>&amp;alpha;</mi> <mo>=</mo> <mn>2</mn> <msqrt> <mn>15</mn> </msqrt> <mo>/</mo> <mi>L</mi> <mo>;</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</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> <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> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <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> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mrow> <mn>5</mn> <mi>&amp;pi;L</mi> <msub> <mi>E</mi> <mi>x</mi> </msub> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <msub> <mi>r</mi> <mi>a</mi> </msub> <mo>[</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>]</mo> <mo>[</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>]</mo> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mrow> <mo>(</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>[</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <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> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mrow> <mn>5</mn> <mi>&amp;pi;</mi> <msub> <mi>E</mi> <mi>x</mi> </msub> <mi>L</mi> <msub> <mi>r</mi> <mi>a</mi> </msub> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>[</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>]</mo> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>)</mo> </mrow> <mo>[</mo> <msub> <mi>b</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>b</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>b</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>]</mo> </mrow> <mrow> <mn>5</mn> <mi>&amp;pi;L</mi> <msub> <mi>E</mi> <mi>x</mi> </msub> <msub> <mi>r</mi> <mi>a</mi> </msub> <msub> <mi>&amp;alpha;</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>[</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>]</mo> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</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> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>)</mo> </mrow> <mo>[</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>]</mo> <mo>,</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>-</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msup> <msub> <mi>r</mi> <mi>b</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>[</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>]</mo> <mo>,</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>3</mn> </msub> <mrow> <mo>(</mo> <mi>L</mi> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <msub> <mi>r</mi> <mi>a</mi> </msub> <mo>[</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mi>ln</mi> <msub> <mi>r</mi> <mi>a</mi> </msub> <mo>]</mo> <mo>[</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>]</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);
(3) foundation and design calculating of rubber bushing length L design mathematic models:
According to vehicle propons or the wheelspan B of back axle, the diameter d of stabiliser bar, total length lc, the mounting distance between two rubber bushings l0, the design requirement value of stabilizer bar system roll angular rigidityThe vertical of resulting end part of stabilizer rod is calculated in step (1) Displacement deformation coefficient Gw, and the expression formula K for the rubber bushing radial direction Line stiffness established in step (2)x(L) rubber bushing, is established Length L mathematical model of optimizing design, i.e.,:
Using Matlab calculation procedures, above-mentioned mathematical modeling is solved, structure, the inner circle of rubber bushing in stabiliser bar can be obtained Radius ra, exradius rbAnd in the case that installation site is all constant, meet the rubber of stabilizer bar system roll angular rigidity design requirement Glue liner sleeve length L optimization design value.
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Citations (4)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
JP2002019437A (en) * 2000-07-04 2002-01-23 Daihatsu Motor Co Ltd Rear suspension
CN200958539Y (en) * 2006-07-07 2007-10-10 中国第一汽车集团公司 Composite rubber lining
CN101462476A (en) * 2009-01-12 2009-06-24 奇瑞汽车股份有限公司 Torsion beam type suspension fork
CN102758871A (en) * 2012-07-17 2012-10-31 山东理工大学 Radial deformation superposition analysis calculation method for car stabilizer bar rubber bushing

Patent Citations (4)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
JP2002019437A (en) * 2000-07-04 2002-01-23 Daihatsu Motor Co Ltd Rear suspension
CN200958539Y (en) * 2006-07-07 2007-10-10 中国第一汽车集团公司 Composite rubber lining
CN101462476A (en) * 2009-01-12 2009-06-24 奇瑞汽车股份有限公司 Torsion beam type suspension fork
CN102758871A (en) * 2012-07-17 2012-10-31 山东理工大学 Radial deformation superposition analysis calculation method for car stabilizer bar rubber bushing

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