CN104346497B - The design method of interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter - Google Patents
The design method of interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter Download PDFInfo
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Abstract
The present invention relates to the design method of interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter, belong to cab mounting technical field.The present invention can be according to the design requirement value of roll angular rigidity, the structural parameters and material characteristic parameter of interior biasing non-coaxial driver's cabin stabilizer bar system, by the loading coefficient of the radial rigidity of rubber bushing, equivalent combinations Line stiffness and reversed rubber bushing, analytical design method is carried out to the torsion tube internal diameter of stabilizer bar system.By designing example and simulating, verifying, the torsion tube internal diameter design load of the available accurately and reliably interior biasing non-coaxial driver's cabin stabilizer bar system of this method, reliable design method is provided for stabilizer bar system design, and reliable technical foundation has been established for CAD software exploitation.Using this method, cab mounting and the design level of stabilizer bar system can be not only improved, improves vehicle ride performance and security;Meanwhile design and testing expenses can be also reduced, accelerate product development speed.
Description
Technical field
The present invention relates to vehicle cab suspension, particularly interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter
Design method.
Background technology
Interior biasing non-coaxial driver's cabin stabilizer bar system, torsion tube internal diameter not only influence the torsional deflection of torsion tube, but also
The load of flexural deformation, reversed rubber bushing to torsion tube and deformation, and the performance of whole stabilizer bar system have material impact.
Interior biasing non-coaxial driver's cabin stabilizer bar system is made up of swing arm, torsion tube and rubber bushing, but one by firm
The coupling body of body, elastomer and flexible body three composition, it is extremely complex that it analyzes calculating.In stabiliser bar actual design, often
The internal diameter and wall thickness of torsion tube can be designed in the case of given torsion tube external diameter.However, parsed due to being deformed by rubber bushing
Calculating, the torsional deflection of torsion tube and flexural deformation intercouple, and key issues of reversed rubber bushing load incrementss
Restrict, the design for the torsion tube internal diameter of interior biasing non-coaxial driver's cabin stabilizer bar system, fail to provide reliable solution always
Analyse 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, passes through
Solid modelling carries out simulating, verifying to the characteristic for giving the driver's cabin stabilizer bar system of structure, although can obtain comparing can for this method
The simulation numerical leaned on, however, because ANSYS simulation analysis can only carry out simulating, verifying, nothing to the stabiliser bar characteristic of given parameters
Method provides accurate analytical design method formula, it is impossible to realizes analytical design method, 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 interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter set
Meter method, meet the requirement of cab mounting and stabilizer bar system design, improve product design level and quality, improve vehicle row
Sail ride comfort 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 interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter, its design flow diagram are as shown in Figure 1;
The structural representation of interior biasing non-coaxial driver's cabin stabilizer bar system is as shown in Figure 2;The structural representation of rubber bushing such as Fig. 3
It is shown;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, in interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube provided by the present invention
The design method in footpath, it is characterised in that use following design procedure:
(1) driver's cabin stabilizer bar system inclination line stiffness KwsThe calculating of design requirement value:
According to the roll angular rigidity design requirement value of driver's cabin stabilizer bar systemSuspension distance Lc, it is steady to the driver's cabin
The inclination line stiffness K of fixed pole systemwsDesign requirement value calculated, i.e.,
(2) driver's cabin stabiliser bar rubber bushing radial rigidity kxCalculating:
According to the inner circle radius r of rubber sleevea, exradius rb, length Lx, elastic modulus ExWith Poisson's ratio μx, to driver's cabin
The radial rigidity k of stabiliser bar rubber bushingxCalculated, i.e.,
Wherein,
Bessel correction functions I (0, α rb), K (0, α rb), I (1, α rb), K (1, α rb), I (1, α ra), K (1, α ra), I
(0,αra), K (0, α ra);
(3) the loading coefficient β of reversed rubber bushingFCalculating:
According to the length L of torsion tubeW, Poisson's ratio μ, interior biasing amount T, and pendulum arm length l1, to the system of load of reversed rubber bushing
Number βFCalculated, i.e.,
(4) the equivalent combinations Line stiffness K of interior biasing non-coaxial stabiliser bar rubber bushingxCalculating:
According to pendulum arm length l1, the interior biasing amount T of torsion tube, the radial direction of the rubber bushing obtained by calculating in step (2) is firm
Spend kx, and the loading coefficient β for the reversed rubber bushing being calculated in step (3)F, to the equivalent combinations of stabiliser bar rubber bushing
Line stiffness KxCalculated, i.e.,
(5) interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter d design:
According to the length L of torsion tubew, outer diameter D, elastic model E and Poisson's ratio μ, interior biasing amount T, pendulum arm length l1, step (1)
The design requirement value K of the inclination line stiffness of driver's cabin stabilizer bar system obtained by middle calculatingws, and be calculated in step (4)
Stabiliser bar rubber bushing equivalent combinations Line stiffness Kx, torsion tube internal diameter d design mathematic models are established, and it is designed,
I.e.
(6) the ANSYS simulating, verifyings of interior biasing non-coaxial driver's cabin stabilizer bar system roll angular rigidity:
Using ANSYS finite element emulation softwares, according to torsion tube internal diameter d design loads and the other structures parameter of stabilizer bar system
And material characteristic parameter, ANSYS simulation models, grid division are established, and apply load F at the suspended position of swing arm, to steady
The deformation of fixed pole system carries out ANSYS emulation, obtains deformation displacement amount f of the stabilizer bar system in swing arm outermost endA;
According to the maximum distortion displacement f of the swing arm outermost end obtained by ANSYS emulationA, pendulum arm length l1, swing arm it is outstanding
Distance, delta l of the seated position to outermost end1, the suspension distance L of stabiliser barc, the load F that is applied at the suspended position of swing arm, and
The rubber bushing radial rigidity k being calculated in step (2)x, using stabilizer bar system deformation and swing arm displacement geometrical relationship,
To the ANSYS simulating, verifying values of the roll angular rigidity of interior biasing non-coaxial driver's cabin stabilizer bar systemCalculated, i.e.,
By the ANSYS of non-coaxial driver's cabin stabilizer bar system roll stiffness simulating, verifying valueWith design requirement
ValueIt is compared, so as to the design method of the interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter provided
And parameter design value is verified.
The present invention has the advantage that than prior art
Due to being deformed analytical Calculation, the torsional deflection of torsion tube and intercoupling for flexural deformation by rubber bushing, and reverse
The restriction of key issues of rubber bushing load incrementss, in the torsion tube of interior biasing non-coaxial driver's cabin stabilizer bar system
Footpath is designed, and fails 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 to utilize ANSYS simulation softwares, the characteristic for giving the driver's cabin stabilizer bar system of structure is imitated by solid modelling
True checking, although this method can obtain reliable simulation numerical, however, because ANSYS simulation analysis can only be to given ginseng
Several stabiliser bar characteristics carries out simulating, verifying, can not provide accurate analytical design method formula, it is impossible to realize analytical design method, can not expire
The requirement of sufficient driver's cabin stabilizer bar system CAD software exploitation.
The present invention according to the flexural deformation of the torsion tube of interior biasing non-coaxial driver's cabin stabilizer bar system and torsional deflection and
Relation between load, establish the equivalent combinations Line stiffness of reversed rubber bushing loading coefficient, rubber bushing;Utilize driver's cabin
Between the roll angular rigidity and linear rigidity of stabilizer bar system, with the structure of stabiliser bar and the equivalent combinations Line stiffness of rubber bushing
Relation, establish the design mathematic model of the torsion tube internal diameter of stabilizer bar system;Stabilizer bar system can be rolled according to driver's cabin
The design requirement of angular rigidity, the structural parameters and material characteristic parameter of stabilizer bar system are steady to interior biasing non-coaxial driver's cabin
Fixed pole system torsion tube internal diameter carries out analytical design method.By designing example and ANSYS simulating, verifyings, this method can obtain accurately
Reliable torsion tube internal diameter design load, reliable design method is provided for the design of cab mounting and stabilizer bar system, and
Reliable technical foundation has been established for the exploitation of driver's cabin stabilizer bar system CAD software.Using this method, driver's cabin can be not only improved
Suspension and the design level and quality of stabilizer bar system, meet design requirement of the cab mounting to stabiliser bar roll angular rigidity,
Further improve the ride performance and security of vehicle;Meanwhile design and testing expenses can be also reduced, accelerate product development speed
Degree.
Brief description of the drawings
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 interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter;
Fig. 2 is the structural representation of interior 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 deformation of interior biasing non-coaxial stabilizer bar system and swing arm displacement;
Fig. 5 is the stabiliser bar roll angular rigidity of embodiment oneWith torsion tube internal diameter d change curve;
Fig. 6 is the deformation simulation cloud atlas of the interior biasing non-coaxial driver's cabin stabilizer bar system of embodiment one;
Fig. 7 is the stabiliser bar roll angular rigidity of embodiment twoWith torsion tube internal diameter d change curve;
Fig. 8 is the deformation simulation cloud atlas of the interior biasing non-coaxial driver's cabin stabilizer bar system of embodiment two.
Embodiment
The present invention is described in further detail below by embodiment.
Embodiment one:The structure of certain interior 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 interior biasing amount T=30mm of pipe 4;The distance between the swing arm 1 of left and right two Lc=1550mm, i.e. stabiliser bar suspension distance;It is outstanding
Put the distance between rubber bushing 2 and reversed rubber bushing 3, i.e. pendulum arm length l1=380mm;The suspended position C of swing arm is to most
Outer end A distance, delta l1=47.5mm;The length L of torsion tube 4w=1500mm, outer diameter D=50mm, internal diameter d are parameter to be designed, elasticity
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;Rubber
The length L of gum cover 6x=25mm, inner circle radius ra=19.5mm, exradius rb=34.5mm, elastic modulus Ex=7.84MPa, pool
Pine compares μx=0.47.The required roll angular rigidity of driver's cabin stabiliser bar designIt is interior partially to this
The torsion tube internal diameter d for putting non-coaxial driver's cabin stabilizer bar system is designed, and firm to the angle of heel under load F=5000N
Degree carries out ANSYS simulating, verifyings.
The design method for the interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter that present example is provided, its
Design cycle is as shown in figure 1, comprise the following steps that:
(1) driver's cabin stabilizer bar system inclination line stiffness KwsThe calculating of design requirement value:
According to the design requirement value of stabilizer bar system roll angular rigiditySuspension distance Lc=
1550mm, to the inclination line stiffness K of the driver's cabin stabilizer bar systemwsDesign requirement value calculated, i.e.,
(2) driver's cabin stabiliser bar rubber bushing radial rigidity kxCalculating:
According to the inner circle radius r of rubber sleevea=19.5mm, exradius rb=34.5mm, length Lx=25mm, springform
Measure Ex=7.84MPa and Poisson's ratio μx=0.47, to the radial rigidity K of driver's cabin stabiliser bar rubber bushingxCalculated, i.e.,
Wherein,
Bessel correction functions I (0, α rb)=5.4217 × 10-3, K (0, α rb)=8.6369 × 10-6;
I(1,αrb)=5.1615 × 103, K (1, α rb)=9.0322 × 10-6;
I(1,αra)=63.7756, K (1, α ra)=0.0013,
I(0,αra)=69.8524, K (0, α ra)=0.0012;
(3) the loading coefficient β of reversed rubber bushingFCalculating:
According to the length L of torsion tubeW=1500mm, Poisson's ratio μ=0.3, interior biasing amount T=30mm, and pendulum arm length l1=
380mm, to the loading coefficient β of reversed rubber bushingFCalculated, i.e.,
(4) the equivalent combinations Line stiffness K of interior biasing non-coaxial stabiliser bar rubber bushingxCalculating:
According to pendulum arm length l1K obtained by being calculated in=380mm, the interior biasing amount T=30mm of torsion tube, and step (2)x
=2.1113 × 106N/m, the β being calculated in step (3)F=0.1456, it is firm to the equivalent combinations line of stabiliser bar rubber bushing
Spend KxCalculated, i.e.,
(5) interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter d design:
According to the length L of torsion tubew=1500mm, outer diameter D=50mm, elastic modulus E=200GPa, Poisson's ratio μ=0.3,
Interior biasing amount T=30mm, pendulum arm length l1=380mm, step (1) is middle to calculate resulting Kws=2.514 × 105N/m, and step
Suddenly the K being calculated in (4)x=7.061428 × 105N/m, torsion tube internal diameter d design mathematic models are established, it is non-coaxial to interior biasing
The torsion tube internal diameter d of formula driver's cabin stabilizer bar system is designed, i.e.,
Wherein, the inclination line stiffness K of the stabilizer bar systemws, with torsion tube internal diameter d change curve, as shown in Figure 5.
(6) the ANSYS simulating, verifyings of interior biasing non-coaxial driver's cabin stabilizer bar system roll angular rigidity:
Using ANSYS finite element emulation softwares, according to the torsion tube internal diameter d=35mm obtained by design and stabilizer bar system
Other structures parameter and material characteristic parameter, simulation model, grid division are established, and apply load F at swing arm suspended position C
=5000N, the deformation progress ANSYS emulation to stabilizer bar system, resulting deformation simulation cloud atlas, as shown in fig. 6, wherein,
Deformation displacement amount f of the stabilizer bar system at swing arm outermost end AAFor
fA=19.811mm;
According to the deformation displacement amount f of the swing arm outermost end A obtained by ANSYS emulationA=19.811mm, pendulum arm length l1=
380mm, the suspended position C to outermost end A of swing arm distance, delta l1=47.5mm, the suspension distance L of stabiliser barc=1500mm,
The k being calculated in the load F=5000N applied at the suspended position C of swing arm, and step (2)x=2.1113 × 106N/m,
Using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, as shown in figure 4, stable to the interior biasing non-coaxial driver's cabin
The ANSYS simulating, verifying values of lever system roll angular rigidityCalculated, i.e.,
Understand, the ANSYS of the roll angular rigidity of driver's cabin stabiliser bar simulating, verifying value
With design requirement valueMatch, relative deviation is only 0.386%;Show to be provided it is interior partially
The design method for putting non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter is correct, and parameter design value is accurately and reliably.
Embodiment two:The structure type of certain interior biasing non-coaxial driver's cabin stabilizer bar system and the system of embodiment one,
As shown in Fig. 2 wherein, torsion tube 4 is not coaxial with reversed rubber bushing 3, the interior biasing amount T=30mm 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 i.e. pendulum arm length l1=350mm;The suspended position C of swing arm to outermost end A distance, delta l1=52.5mm;The length of torsion tube 4
Lw=1000mm, outer diameter D=50mm, internal diameter d are parameter to be designed;The structure of four rubber bushings in left and right is all identical, such as
Shown in Fig. 3, wherein, the interior circular diameter d of interior round buss 5x=35mm, wall thickness δ=5mm;The length L of rubber sleeve 6x=40mm, inner circle
Radius ra=22.5mm, exradius rb=37.5mm.The material property of stabiliser bar and the material property of rubber bushing, with implementation
Identical, i.e. elastic modulus E=200GPa of torsion tube of example one, Poisson's ratio μ=0.3;The elastic modulus E of rubber sleevex=7.84MPa,
Poisson's ratio μx=0.47.The required roll angular rigidity of driver's cabin stabiliser bar designIt is interior to this
The torsion tube internal diameter d of biasing non-coaxial driver's cabin stabilizer bar system is designed, and to the angle of heel under load F=5000N
Rigidity carries out ANSYS simulating, verifyings.
Using the step identical with embodiment one, to the torsion tube internal diameter of the interior biasing non-coaxial driver's cabin stabilizer bar system
D is designed, i.e.,:
(1) driver's cabin stabilizer bar system inclination line stiffness KwsThe calculating of design requirement value:
According to the design requirement value of stabilizer bar system roll angular rigiditySuspension distance Lc
=1400mm, to driver's cabin stabilizer bar system inclination line stiffness KwsDesign requirement value calculated, i.e.,
(2) driver's cabin stabiliser bar rubber bushing radial rigidity kxCalculating:
According to the inner circle radius r of rubber sleevea=22.5mm, exradius rb=37.5mm, length Lx=40mm, springform
Measure Ex=7.84MPa, Poisson's ratio μx=0.47, to the radial rigidity k of the driver's cabin stabiliser bar rubber bushingxCalculated, i.e.,
Wherein,
Bessel correction functions I (0, α rb)=214.9082, K (0, α rb)=3.2117 × 10-4;
I(1,αrb)=199.5091, K (1, α rb)=3.4261 × 10-4;
I(1,αra)=13.5072, K (1, α ra)=0.0083,
I(0,αra)=15.4196, K (0, α ra)=0.0075;
(3) the loading coefficient β of reversed rubber bushingFCalculating:
According to the length L of torsion tubeW=1000mm, Poisson's ratio μ=0.3, interior biasing amount T=30mm, and pendulum arm length l1=
350mm, to the loading coefficient β of reversed rubber bushingFCalculated, i.e.,
(4) the equivalent combinations Line stiffness K of interior biasing non-coaxial stabiliser bar rubber bushingxCalculating:
According to the pendulum arm length l of stabiliser bar1=350mm, the interior biasing amount T=30mm of torsion tube, and calculated in step (2)
The k arrivedx=4.2085 × 106N/m, the β being calculated in step (3)F=0.29952, to equivalent group of stabiliser bar rubber bushing
Zygonema stiffness KxCalculated, i.e.,
(5) interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter d design:
According to the length L of torsion tubew=1000mm, outer diameter D=50mm, elastic modulus E=200GPa, Poisson's ratio μ=0.3,
Interior biasing amount T=30mm, pendulum arm length l1=350mm, step (1) is middle to calculate resulting Kws=3.0308 × 105N/m, and
The K being calculated in step (4)x=8.7787 × 105N/m, torsion tube internal diameter d design mathematic models are established, it is non-to the interior biasing same
The torsion tube internal diameter d of shaft type driver's cabin stabilizer bar system is designed, i.e.,
Wherein, the inclination line stiffness K of the stabilizer bar systemws, with torsion tube internal diameter d change curve, as shown in Figure 7.
(6) the ANSYS simulating, verifyings of interior biasing non-coaxial driver's cabin stabilizer bar system roll angular rigidity:
Using ANSYS finite element emulation softwares, according to the torsion tube internal diameter d=42mm obtained by design and stabilizer bar system
Other structures parameter and material characteristic parameter, simulation model is established, grid division, applies load F=at swing arm suspended position C
5000N, the deformation progress ANSYS emulation to stabilizer bar system, resulting deformation simulation cloud atlas, as shown in figure 8, wherein, surely
Deformation displacement amount f of the fixed pole system in swing arm outermost end AAFor
fA=17.637mm;
According to the deformation displacement amount f at the swing arm outermost end A obtained by ANSYS emulationA=17.637mm, pendulum arm length l1
The distance, delta l of=350mm, swing arm suspended position C to outermost end A1=52.5mm, stabiliser bar suspension distance Lc=1400mm, putting
The k being calculated in the load F=5000N applied at the suspended position C of arm, and step (2)x=4.2085 × 106N/m, profit
With stabilizer bar system deformation and swing arm displacement geometrical relationship, as shown in figure 4, to the interior biasing non-coaxial driver's cabin stabiliser bar
The ANSYS simulating, verifying values of system roll angular rigidityCalculated, i.e.,
Understand, the ANSYS of the roll angular rigidity of driver's cabin stabiliser bar simulating, verifying value
With design requirement valueMatch, relative deviation is only 0.166%;Show to be provided it is interior partially
The design method for putting non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter is correct, and parameter design value is accurately and reliably.
Claims (1)
1. the design method of interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter, its specific design step are as follows:
(1) driver's cabin stabilizer bar system inclination line stiffness KwsThe calculating of design requirement value:
According to the roll angular rigidity design requirement value of driver's cabin stabilizer bar systemSuspension distance Lc, to the driver's cabin stabiliser bar
The inclination line stiffness K of systemwsDesign requirement value calculated, i.e.,
(2) driver's cabin stabiliser bar rubber bushing radial rigidity kxCalculating:
According to the inner circle radius r of rubber sleevea, exradius rb, length Lx, elastic modulus ExWith Poisson's ratio μx, it is stable to driver's cabin
The radial rigidity k of bar rubber bushingxCalculated, i.e.,
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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);
(3) the loading coefficient β of reversed rubber bushingFCalculating:
According to the length L of torsion tubeW, Poisson's ratio μ, interior biasing amount T, and pendulum arm length l1, to the loading coefficient β of reversed rubber bushingF
Calculated, i.e.,
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<mi>F</mi>
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</msub>
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</mrow>
<mi>T</mi>
</mrow>
<msubsup>
<mi>L</mi>
<mi>W</mi>
<mn>2</mn>
</msubsup>
</mfrac>
<mo>;</mo>
</mrow>
(4) the equivalent combinations Line stiffness K of interior biasing non-coaxial stabiliser bar rubber bushingxCalculating:
According to pendulum arm length l1, the interior biasing amount T of torsion tube, the middle radial rigidity k for calculating resulting rubber bushing of step (2)x,
And the loading coefficient β for the reversed rubber bushing being calculated in step (3)F, it is firm to the equivalent combinations line of stabiliser bar rubber bushing
Spend KxCalculated, i.e.,
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<msub>
<mi>K</mi>
<mi>X</mi>
</msub>
<mo>=</mo>
<mfrac>
<mrow>
<msub>
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<mi>X</mi>
</msub>
<mi>T</mi>
</mrow>
<mrow>
<msub>
<mi>&beta;</mi>
<mi>F</mi>
</msub>
<msub>
<mi>l</mi>
<mn>1</mn>
</msub>
<mo>+</mo>
<mrow>
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<msub>
<mi>&beta;</mi>
<mi>F</mi>
</msub>
<mo>)</mo>
</mrow>
<mi>T</mi>
</mrow>
</mfrac>
<mo>;</mo>
</mrow>
(5) interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter d design:
According to the length L of torsion tubew, outer diameter D, elastic model E and Poisson's ratio μ, interior biasing amount T, pendulum arm length l1, step (1) falls into a trap
The design requirement value K of the inclination line stiffness of driver's cabin stabilizer bar system obtained by calculatingws, and be calculated in step (4) steady
The equivalent combinations Line stiffness K of fixed pole rubber bushingx, torsion tube internal diameter d design mathematic models are established, and it is designed, i.e.,
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<mi>d</mi>
<mo>=</mo>
<mroot>
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<mrow>
<msup>
<mi>D</mi>
<mn>4</mn>
</msup>
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<msub>
<mi>K</mi>
<mrow>
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<mi>s</mi>
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</mrow>
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<msub>
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<mi>s</mi>
</mrow>
</msub>
<msub>
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<mi>X</mi>
</msub>
<mn>32</mn>
<mrow>
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<mi>&mu;</mi>
<mo>)</mo>
</mrow>
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<msub>
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</msub>
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<mi>T</mi>
<mo>)</mo>
</mrow>
<mn>2</mn>
</msup>
<msub>
<mi>L</mi>
<mi>W</mi>
</msub>
</mrow>
<mrow>
<mo>(</mo>
<msub>
<mi>K</mi>
<mrow>
<mi>w</mi>
<mi>s</mi>
</mrow>
</msub>
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<msub>
<mi>K</mi>
<mi>X</mi>
</msub>
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<mi>&pi;</mi>
<mi>E</mi>
</mrow>
</mfrac>
<mn>4</mn>
</mroot>
<mo>;</mo>
</mrow>
(6) the ANSYS simulating, verifyings of interior biasing non-coaxial driver's cabin stabilizer bar system roll angular rigidity:
Using ANSYS finite element emulation softwares, according to torsion tube internal diameter d design loads and the other structures parameter and material of stabilizer bar system
Expect characterisitic parameter, establish ANSYS simulation models, grid division, and apply load F at the suspended position of swing arm, to stabiliser bar
The deformation of system carries out ANSYS emulation, obtains deformation displacement amount f of the stabilizer bar system in swing arm outermost endA;
According to the maximum distortion displacement f of the swing arm outermost end obtained by ANSYS emulationA, pendulum arm length l1, the suspension position of swing arm
Put the distance, delta l of outermost end1, the suspension distance L of stabiliser barc, the load F that is applied at the suspended position of swing arm, and step
(2) the rubber bushing radial rigidity k being calculated inx, using stabilizer bar system deformation and swing arm displacement geometrical relationship, internally
Bias the ANSYS simulating, verifying values of the roll angular rigidity of non-coaxial driver's cabin stabilizer bar systemCalculated, 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>&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>
<mrow>
<msub>
<mi>K</mi>
<mrow>
<mi>w</mi>
<mi>s</mi>
</mrow>
</msub>
<mo>=</mo>
<mfrac>
<mi>F</mi>
<msub>
<mi>f</mi>
<mrow>
<mi>w</mi>
<mi>s</mi>
</mrow>
</msub>
</mfrac>
<mo>;</mo>
</mrow>
By the ANSYS of non-coaxial driver's cabin stabilizer bar system roll stiffness simulating, verifying valueWith design requirement valueBe compared, so as to the design method of interior biasing non-coaxial driver's cabin stabilizer bar system torsion tube internal diameter provided and
Parameter design value is verified.
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Citations (2)
Publication number | Priority date | Publication date | Assignee | Title |
---|---|---|---|---|
CN102320337A (en) * | 2011-06-30 | 2012-01-18 | 三一重工股份有限公司 | A kind of automobile cab front hung holder and heavy motor vehicle |
CN102923201A (en) * | 2012-11-27 | 2013-02-13 | 东风柳州汽车有限公司 | Front suspension device for heavy-duty truck cab |
-
2014
- 2014-11-03 CN CN201410610661.6A patent/CN104346497B/en not_active Expired - Fee Related
Patent Citations (2)
Publication number | Priority date | Publication date | Assignee | Title |
---|---|---|---|---|
CN102320337A (en) * | 2011-06-30 | 2012-01-18 | 三一重工股份有限公司 | A kind of automobile cab front hung holder and heavy motor vehicle |
CN102923201A (en) * | 2012-11-27 | 2013-02-13 | 东风柳州汽车有限公司 | Front suspension device for heavy-duty truck cab |
Non-Patent Citations (3)
Title |
---|
Optimal Power System Stabilizer Tuning in Multi-machine System via an Improved Differential Evolution;G. Y. Yang 等;《Proceedings of the 17th World Congress The International Federation of Automatic Control》;20080711;全文 * |
前双横臂独立悬架前双横臂独立悬架的建模仿真与改进设计的建模仿真与改进设计;乐升彬;《中国优秀博硕士学位论文全文数据库 (硕士) 工程科技Ⅱ辑》;20041215(第04期);C035-38 * |
基于ADAMS的越野车独立悬架仿真研究;程康;《中国优秀硕士学位论文全文数据库 工程科技Ⅱ辑》;20110915(第09期);C035-87 * |
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