CN107506560A - The method of calibration of driver's cabin stabilizer bar system roll angular rigidity - Google Patents

The method of calibration of driver's cabin stabilizer bar system roll angular rigidity Download PDF

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CN107506560A
CN107506560A CN201710821289.7A CN201710821289A CN107506560A CN 107506560 A CN107506560 A CN 107506560A CN 201710821289 A CN201710821289 A CN 201710821289A CN 107506560 A CN107506560 A CN 107506560A
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李栋
李胜
曹旭光
胡金蕊
黄德惠
张凯
向建东
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FAW Group Corp
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Abstract

The method of calibration of driver's cabin stabilizer bar system roll angular rigidity of the present invention, belong to driver's cabin stabiliser bar technical field, including following calculation procedures:(1) deformation coefficient G of the driver's cabin stabiliser bar at suspended positionwCalculating;(2) driver's cabin stabiliser bar rubber bushing radial rigidity KxAnalytical Calculation;(3) the equivalent line stiffness K of coaxial-type driver's cabin stabilizer bar systemwsCalculating;(4) coaxial-type driver's cabin stabilizer bar system roll angular rigidityCalculating;(5) checking computations of coaxial-type driver's cabin stabilizer bar system rigidity and ANSYS simulating, verifyings.The calculated value of the available accurately and reliably coaxial-type stabilizer bar system roll angular rigidity of the present invention, can not only improve cab mounting and the design level of stabilizer bar system, improve the ride comfort and security of vehicle traveling;Meanwhile can also reduce design and testing expenses, the exploitation for driver's cabin stabilizer bar system CAD software provide reliable technical support.

Description

The method of calibration of driver's cabin stabilizer bar system roll angular rigidity
Technical field
The present invention relates to the method for calibration of vehicle cab suspension, particularly driver's cabin stabilizer bar system roll angular rigidity.
Background technology
Full-float cab is suspended and stabilizer bar system design must is fulfilled for when turn inside diameter travels to roll angular rigidity Design requirement, wherein, roll angular rigidity is not only relevant with the rigidity of suspension spring, but also has with the rigidity of stabilizer bar system Close, and stabilizer bar system mainly includes swing arm, rubber bushing and torsion tube.However, due to by rubber bushing deform analytical Calculation and The restriction for key issues of being intercoupled with stabilizer bar system rigidity, for coaxial-type driver's cabin stabilizer bar system roll angular rigidity Calculation and check, fail to provide reliable Analytic Calculation Method always, shadow that can only be by rubber bushing to stabilizer bar system rigidity Ring, with a conversion factor, approximate estimation is carried out to stabilizer bar system rigidity.At present, it is steady for coaxial-type driver's cabin both at home and abroad Fixed pole system angle of heel Stiffness evaluation, mostly it is to utilize ANSYS simulation softwares, roll angular rigidity is imitated by solid modelling True analysis and checking, this method is although can obtain reliable simulation numerical, however, due to that can not provide accurate parsing meter Formula, so the requirement of coaxial-type driver's cabin stabilizer bar system CAD software exploitation can not be met.As Vehicle Industry is fast-developing And the continuous improvement of Vehicle Speed, higher requirement is proposed to coaxial-type cab mounting and stabilizer bar system design, There is an urgent need to coaxial-type driver's cabin stabilizer bar system CAD software for vehicle manufacture producer.Therefore, it is necessary to establish a kind of accurate, reliable Coaxial-type driver's cabin stabilizer bar system roll angular rigidity computational methods, improve that product design is horizontal and quality, improve 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 method of calibration of reliable driver's cabin stabilizer bar system roll angular rigidity, its calculation flow chart is as shown in figure 1, coaxial-type driver's cabin The structural representation of stabilizer bar system is as shown in Fig. 2 the structural representation of rubber bushing is as shown in Figure 3.
In order to solve the above technical problems, the verification side of driver's cabin stabilizer bar system roll angular rigidity provided by the present invention Method, using following calculation procedure:
(1) deformation coefficient G of the driver's cabin stabiliser bar at suspended positionwCalculating:
According to pendulum arm length l1;The half length l of torsion tube2, internal diameter d, outer diameter D, and the ratio between inside and outside footpath kd=d/D, material Elastic modulus E and Poisson's ratio μ, to deformation coefficient G of the stabiliser bar at suspended positionwCalculated, i.e.,
(2) driver's cabin stabiliser bar rubber bushing radial rigidity KxAnalytical Calculation:
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 stiffness 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,αra), K (1, α ra), I (0, α ra), K (0, α ra);
(3) the equivalent line stiffness K of coaxial-type driver's cabin stabilizer bar systemwsCalculating:
According to the outer diameter D of torsion tube, the G that is calculated in step (1)w, the K that is calculated in step (2)x, it is steady to coaxial-type Stiffness K of the fixed pole system in cab mounting installed positionwsCalculated, i.e.,
(4) coaxial-type driver's cabin stabilizer bar system roll angular rigidityCalculating:
According to the suspension distance L of stabiliser barc, and the K being calculated in step (3)ws, to the stable leverage of coaxial-type driver's cabin The roll angular rigidity of systemCalculated, i.e.,
(5) checking computations of coaxial-type driver's cabin stabilizer bar system rigidity and ANSYS simulating, verifyings:
I applies a certain load F at the suspended position of swing arm, and in the situation for not considering cab mounting spring rate Under, utilize the K obtained by being calculated in step (3)ws, to deformation f of the swing arm at suspended positionwsCCalculated, i.e. fwsC=F/ Kws;According to the suspension distance L of stabiliser barc, pendulum arm length l1, and swing arm at suspended position between outermost end apart from Δ l1, Using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, to deformation displacement amount of the swing arm at outermost endThe side tilt angle of driver's cabinAnd the roll angular rigidity of driver's cabin stabilizer bar systemCalculated;
II utilizes ANSYS finite element emulation softwares, according to the structure and material characteristic parameter of stabilizer bar system, establishes emulation Model, grid division, and apply to enter with I step identical load F, the deformation to stabilizer bar system at the suspended position of swing arm Row ANSYS is emulated, and obtains maximum distortion f of the swing arm at outermost endwsA
According to the maximum distortion f at the swing arm outermost end obtained by ANSYS emulationwsA, pendulum arm length l1, swing arm suspending Distance, delta l of the opening position at outermost end1, and the suspension distance L of stabiliser barc, utilize stabilizer bar system deformation and swing arm displacement Geometrical relationship, to deformation of the swing arm at suspended positionThe side tilt angle of driver's cabinAnd drive The roll angular rigidity of room stabilizer bar systemCalculated;
III calculates step (4) the roll angular rigidity value of resulting stabilizer bar systemRespectively with being calculated in I steps Obtained roll angular rigidity valueRoll angular rigidity value obtained by II step simulation calculationsIt is compared, so as to right The computational methods and Rigidity Calculation value of coaxial-type driver's cabin stabilizer bar system roll angular rigidity provided by the present invention are verified.
Technique effect
Home and abroad is mostly to utilize for the calculation and check of coaxial-type driver's cabin stabilizer bar system roll angular rigidity at present Simulation software, analysis calculating is carried out to stabilizer bar system roll angular rigidity by modeling and simulating, but this method can not provide solution Calculating formula is analysed, it is thus impossible to meet the requirement of coaxial-type driver's cabin stabilizer bar system CAD software exploitation.Although also there is expert will Rubber bushing to stabilizer bar system rigidity influence with 0.75~0.85 a certain conversion factor, to coaxial-type driver's cabin stabiliser bar The rigidity of system carries out approximate estimation, but as Vehicle Industry is fast-developing and Vehicle Speed improves constantly, to coaxial Formula cab mounting and stabilizer bar system design propose higher design requirement, therefore, traditional calculating evaluation method, it is impossible to Meet the requirement of Cab Mounting System design.
The present invention can suspended according to stabiliser bar and the structural parameters and material characteristic parameter of rubber bushing using stabiliser bar The deformation coefficient G of opening positionWAnd rubber bushing radial rigidity KxAnalytical formula, to the vertical line of coaxial-type stabilizer bar system Rigidity and angle of heel carry out analytical Calculation.It is available accurately and reliably by example calculation and ANSYS simulating, verifyings, this method Coaxial-type stabilizer bar system roll angular rigidity calculated value, be coaxial-type cab mounting and stabilizer bar system design, carry Reliable roll angular rigidity computational methods have been supplied, and technical foundation has been established for the exploitation of coaxial-type stabilizer bar system CAD software. Using this method, coaxial-type cab mounting and the design level and quality of stabilizer bar system can be not only improved, improves vehicle Ride performance and security;Meanwhile design and testing expenses can be reduced using this method, accelerate product development speed.
Brief description of the drawings
Fig. 1 is the calculation flow chart of coaxial-type driver's cabin stabilizer bar system roll angular rigidity;
Fig. 2 is the structural representation of coaxial-type stabilizer bar system;
Fig. 3 is the structural representation of rubber bushing;
Fig. 4 is the geometrical relationship figure of stabilizer bar system deformation and swing arm displacement;
Fig. 5 is the deformation simulation checking cloud atlas of the coaxial-type driver's cabin stabilizer bar system of embodiment one;
Fig. 6 is the deformation simulation checking cloud atlas of the coaxial-type driver's cabin stabilizer bar system of embodiment two.
Specific embodiment
The present invention is described in further detail below by embodiment.
Embodiment one:The structure of certain coaxial-type driver's cabin stabilizer bar system is symmetrical, as shown in Fig. 2 comprising:Swing arm 1, Suspended rubber bushing 2, reversed rubber bushing 3, torsion tube 4;Wherein, torsion tube 4, reversed rubber bushing 3 are coaxial;The swing arm 1 of left and right two The distance between Lc=1550mm, i.e. stabiliser bar suspension distance;Between suspended rubber bushing 2 and reversed rubber bushing 3 away from From l1=380mm, i.e. pendulum arm length;The suspended position C of swing arm to outermost end A distance is Δ l1=47.5mm;The length of torsion tube 4 Spend Lw=1500mm, i.e. torsion tube half length l2=Lw/ 2=750mm;The internal diameter d=35mm of torsion tube, outer diameter D=50mm;Turn round The elasticity modulus of materials E=200GPa of pipe, Poisson's ratio μ=0.3;The complete phase of structure and material characteristic of the rubber bushing of left and right four Together, as shown in figure 3, including:Interior round buss 5, rubber sleeve 6, outer round buss 7, wherein, the internal diameter d of interior round buss 5x=35mm, wall Thick δ=2mm;The length L of rubber sleeve 6x=25mm, inner circle radius ra=19.5mm, exradius rb=34.5mm, modulus of elasticity Ex=7.84MPa, Poisson's ratio μx=0.47.According to the structure and material characterisitic parameter of stabiliser bar and rubber bushing given herein above, The roll angular rigidity of the coaxial-type driver's cabin stabilizer bar system is calculated, and to the deformation in the case of load F=5000N Carry out checking computations and ANSYS simulating, verifyings.
The method of calibration for the driver's cabin stabilizer bar system roll angular rigidity that present example is provided, its calculation process is as schemed Shown in 1, comprise the following steps that:
(1) deformation coefficient G of the driver's cabin stabiliser bar at suspended positionwCalculating:
According to pendulum arm length l1=380mm;The half length l of torsion tube2=750mm, internal diameter d=35mm, outer diameter D=50mm, And the ratio between inside and outside footpath kd=d/D=0.7, elastic modulus E=200GPa of torsion tube material and Poisson's ratio μ=0.3, it is stable to this Deformation coefficient G of the bar at suspended positionwCalculated, i.e.,
(2) the radial rigidity K of driver's cabin stabiliser bar rubber bushingxAnalytical Calculation:
According to the inner circle radius r of rubber sleevea=19.5mm, exradius rb=34.5mm, length Lx=25mm, and elasticity Modulus Ex=7.84MPa and 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)=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 equivalent line stiffness K of coaxial-type driver's cabin stabilizer bar systemwsCalculating:
According to outer diameter D=50mm of torsion tube, the G obtained by being calculated in step (1)w=1.8872 × 10-11m5/ N, step (2) K obtained by being calculated inx=2.1113 × 106N/m, to the coaxial-type stabilizer bar system in cab mounting installed position Stiffness KwsCalculated, i.e.,
(4) coaxial-type driver's cabin stabilizer bar system roll angular rigidityCalculating:
According to the suspension distance L of stabiliser barcThe K being calculated in=1550mm, and step (3)ws=2.8627 × 105N/ M, to the roll angular rigidity of the coaxial-type driver's cabin stabilizer bar systemCalculated, i.e.,
(5) checking computations of coaxial-type driver's cabin stabilizer bar system rigidity and ANSYS simulating, verifyings:
I applies load F=5000N at the suspended position C of swing arm, and in the feelings for not considering cab mounting spring rate Under condition, the K obtained by being calculated in step (3) is utilizedws=2.8627 × 105N/m, to deformation f of the swing arm at suspended position CwsC Calculated, i.e.,
fwsC=F/Kws=17.5mm;
According to the suspension distance L of stabiliser barc=1550mm, pendulum arm length l1=380mm, and swing arm are arrived at suspended position C Distance, delta l between outermost end A1=47.5mm, utilize the geometrical relationship of stabilizer bar system deformation and swing arm displacement, such as Fig. 4 It is shown, it is calculated respectively:
Deformation displacement amount of the swing arm at outermost end A
The side tilt angle of driver's cabin
The roll angular rigidity of driver's cabin stabilizer bar system
II utilizes ANSYS finite element emulation softwares, according to the structure and material characteristic parameter of stabilizer bar system, establishes emulation Model, grid division, and application and I step identical load F=5000N at the suspended position C of swing arm, to stabilizer bar system Deformation carry out ANSYS emulation, obtained deformation simulation cloud atlas is as shown in figure 5, wherein, the maximum distortion at swing arm outermost end A Displacement fwsASimulation value be
fwsA=19.738mm;
According to the deformation displacement amount f at the swing arm outermost end A obtained by emulationwsA=19.738mm, pendulum arm length l1= 380mm, swing arm is at suspended position C to the distance, delta l at outermost end A1=47.5mm, and the suspension distance L of stabiliser barc= 1550mm, using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, as shown in figure 4, being calculated respectively:
Deformation displacement amount of the swing arm at suspended position C
The side tilt angle of driver's cabin
The roll angular rigidity of driver's cabin stabilizer bar system
The III roll angular rigidity value that will be calculated in step (4)Respectively with I steps Checking computations valueAnd the ANSYS simulating, verifying values in II steps It is compared, it is known that:The calculated value of the roll angular rigidity of driver's cabin stabilizer bar system, with checking computations value and ANSYS simulating, verifying values Match, relative deviation is respectively 0.017%, 0.44%, shows coaxial-type driver's cabin stabilizer bar system provided by the present invention The computational methods of roll angular rigidity are correct.
Embodiment two:The structure of certain coaxial-type driver's cabin stabilizer bar system is symmetrical, as shown in Fig. 2 two swing arms it Between distance Lc=1400mm, i.e. stabiliser bar suspension distance;The distance between suspended rubber bushing 2 and reversed rubber bushing 3 l1 =350mm, i.e. pendulum arm length;The suspended position C of swing arm to outermost end A distance is Δ l1=52.5mm;The total length of torsion tube 4 Lw=1000mm, i.e. torsion tube half length l2=Lw/ 2=500mm, internal diameter d=42mm, outer diameter D=50mm;The rubber of left and right four The structure and material characteristic of glue bushing is identical, as shown in figure 3, wherein, the internal 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 of stabiliser bar is special The material property of property and rubber bushing, identical with embodiment one, the i.e. elasticity modulus of materials E=200GPa of torsion tube, Poisson's ratio μ =0.3;The elastic modulus E of rubber sleevex=7.84MPa, Poisson's ratio μx=0.47.Served as a contrast according to stabiliser bar given herein above and rubber The structure and material characterisitic parameter of set, the roll angular rigidity of the coaxial-type driver's cabin stabilizer bar system is calculated, and to Deformation in the case of load F=5000N carries out checking computations and ANSYS simulating, verifyings.
Using the step identical with embodiment one, the roll angular rigidity of the coaxial-type driver's cabin stabilizer bar system is counted Calculate, i.e.,:
(1) deformation coefficient G of the driver's cabin stabiliser bar at suspended positionwCalculating:
According to pendulum arm length l1=350mm, the half length l of torsion tube2=500mm, internal diameter d=42mm, outer diameter D=50mm, And the ratio between inside and outside footpath kd=d/D=0.8, elastic modulus E=200GPa of torsion tube material and Poisson's ratio μ=0.3, to stabiliser bar Deformation coefficient G at suspended positionwCalculated, i.e.,
(2) driver's cabin stabiliser bar rubber bushing RADIAL stiffness KxAnalytical Calculation:
According to the inner circle radius r of rubber sleevea=22.5mm, exradius rb=37.5mm, length Lx=40mm, and elasticity Modulus Ex=7.84MPa and Poisson's ratio μx=0.47, to the RADIAL stiffness K of the driver's cabin stabiliser bar rubber bushingxCounted Calculate, 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 equivalent line stiffness K of coaxial-type driver's cabin stabilizer bar systemwsCalculating:
According to outer diameter D=50mm of torsion tube, the G obtained by being calculated in step (1)w=1.3737 × 10-11m5/ N, and step (2) K obtained by being calculated inx=4.2085 × 106N/m, to the coaxial-type driver's cabin stabilizer bar system in suspension installed position Stiffness KwsCalculated, i.e.,
(4) coaxial-type driver's cabin stabilizer bar system roll angular rigidityCalculating:
According to the suspension distance L of stabiliser barcK obtained by being calculated in=1400mm, and step (3)ws=4.1058 × 105N/m, to the roll angular rigidity of the coaxial-type driver's cabin stabilizer bar systemCalculated, i.e.,
(5) checking computations of coaxial-type driver's cabin stabilizer bar system rigidity and ANSYS simulating, verifyings:
I applies a load F=5000N at swing arm suspended position C, and in the feelings for not considering cab mounting spring rate Under condition, the K obtained by being calculated in step (3) is utilizedws=3.7408 × 105N/m, to deformation f of the swing arm at suspended position CwsC Calculated, i.e.,
According to the suspension distance L of stabiliser barc=1400mm, pendulum arm length l1=350mm, and swing arm are arrived at suspended position C Distance, delta l between outermost end A1=52.5mm, utilize the geometrical relationship of stabilizer bar system deformation and swing arm displacement, such as Fig. 4 It is shown, it is calculated respectively:
Deformation displacement amount of the swing arm at outermost end A
The side tilt angle of driver's cabin
The roll angular rigidity of driver's cabin stabilizer bar system
II utilizes ANSYS finite element emulation softwares, according to the structure and material characteristic parameter of stabilizer bar system, establishes emulation Model, grid division, and application and I step identical load F=5000N at the suspended position C of swing arm, to stabilizer bar system Deformation carry out ANSYS emulation, obtained deformation simulation cloud atlas is as shown in fig. 6, wherein, the deformation displacement at swing arm outermost end A Measure fwsAFor:
fwsA=13.915mm;
According to the maximum distortion f at the swing arm outermost end A obtained by emulationwsA=13.995mm, pendulum arm length l1= 350mm, swing arm is at suspended position C to the distance, delta l at outermost end A1=52.5mm, and the suspension distance L of stabiliser barc= 1400mm, using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, as shown in figure 4, being calculated respectively:
Deformation of the swing arm at suspended position C
The side tilt angle of driver's cabin
The roll angular rigidity of driver's cabin stabilizer bar system
The III roll angular rigidity value that will be calculated in step (4)Respectively with I steps Checking computations valueAnd the ANSYS simulating, verifying values in II steps It is compared, it is known that:The roll angular rigidity calculated value of driver's cabin stabilizer bar system, with checking computations value and ANSYS simulating, verifying value phases It coincide, relative deviation is respectively 0%, 0.212%, and the result of calculation under two kinds of different structure vehicles shows provided by the present invention The computational methods of coaxial-type driver's cabin stabilizer bar system roll angular rigidity are correct.
Although the present invention is disclosed above with preferred embodiment, but is not limited to the present invention.It is any to be familiar with ability The technical staff in domain, in the case where not departing from technical solution of the present invention scope, coaxial-type driver's cabin stabilizer bar system can be existed Corresponding variation, modification or equivalent are done in vehicle, but these corresponding variations should all fall the model in technical solution of the present invention protection In enclosing.

Claims (6)

1. the method for calibration of driver's cabin stabilizer bar system roll angular rigidity, it is characterised in that
Including following calculation procedures:
(1) deformation coefficient G of the driver's cabin stabiliser bar at suspended positionwCalculating;
(2) driver's cabin stabiliser bar rubber bushing radial rigidity KxAnalytical Calculation;
(3) the equivalent line stiffness K of coaxial-type driver's cabin stabilizer bar systemwsCalculating;
(4) coaxial-type driver's cabin stabilizer bar system roll angular rigidityCalculating;
(5) checking computations of coaxial-type driver's cabin stabilizer bar system rigidity and ANSYS simulating, verifyings.
2. the method for calibration of driver's cabin stabilizer bar system roll angular rigidity according to claim 1, it is characterised in that
Deformation coefficient G of (1) the driver's cabin stabiliser bar at suspended positionwCalculating:
According to pendulum arm length l1;The half length l of torsion tube2, internal diameter d, outer diameter D, and the ratio between inside and outside footpath kd=d/D, the bullet of material Property modulus E and Poisson's ratio μ, to deformation coefficient G of the stabiliser bar at suspended positionwCalculated, i.e.,
<mrow> <msub> <mi>G</mi> <mi>w</mi> </msub> <mo>=</mo> <mfrac> <mrow> <mn>64</mn> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>&amp;mu;</mi> <mo>)</mo> </mrow> <msubsup> <mi>l</mi> <mn>1</mn> <mn>2</mn> </msubsup> <msub> <mi>l</mi> <mn>2</mn> </msub> </mrow> <mrow> <mi>&amp;pi;</mi> <mi>E</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>k</mi> <mi>d</mi> <mn>4</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>.</mo> </mrow>
3. the method for calibration of driver's cabin stabilizer bar system roll angular rigidity according to claim 1, it is characterised in that
(2) the driver's cabin stabiliser bar rubber bushing radial rigidity KxAnalytical Calculation:
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 stiffness K of bar rubber bushingxCalculated, i.e.,
<mrow> <msub> <mi>K</mi> <mi>x</mi> </msub> <mo>=</mo> <mfrac> <mn>1</mn> <mrow> <mi>u</mi> <mrow> <mo>(</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>y</mi> <mrow> <mo>(</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow>
Wherein,
<mrow> <mi>y</mi> <mrow> <mo>(</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>a</mi> <mn>1</mn> </msub> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mn>2</mn> </msub> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mn>3</mn> </msub> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>x</mi> </msub> </mrow> </mfrac> <mrow> <mo>(</mo> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>+</mo> <mfrac> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mrow> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>1</mn> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>)</mo> <mo>&amp;lsqb;</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>a</mi> </msub> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> <mo>-</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>x</mi> </msub> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>2</mn> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>a</mi> </msub> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> <mo>-</mo> <mi>I</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>x</mi> </msub> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>,</mo> </mrow>
<mrow> <msub> <mi>a</mi> <mn>3</mn> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&amp;mu;</mi> <mi>x</mi> </msub> <mo>)</mo> <mo>(</mo> <msub> <mi>b</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>b</mi> <mn>2</mn> </msub> <mo>+</mo> <msub> <mi>b</mi> <mn>3</mn> </msub> <mo>)</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&amp;pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>x</mi> </msub> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>1</mn> </msub> <mo>=</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <msub> <mi>r</mi> <mi>a</mi> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>2</mn> </msub> <mo>=</mo> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <msub> <mi>r</mi> <mi>b</mi> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>
<mrow> <msub> <mi>b</mi> <mn>3</mn> </msub> <mo>=</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <msub> <mi>r</mi> <mi>b</mi> </msub> <mo>&amp;lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>a</mi> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&amp;alpha;r</mi> <mi>b</mi> </msub> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> <mo>&amp;lsqb;</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mi>a</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mi>b</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mi>a</mi> </msub> <mo>&amp;rsqb;</mo> <mo>,</mo> </mrow>
<mrow> <mi>&amp;alpha;</mi> <mo>=</mo> <mn>2</mn> <msqrt> <mn>15</mn> </msqrt> <mo>/</mo> <msub> <mi>L</mi> <mi>x</mi> </msub> <mo>,</mo> </mrow>
Bessel correction functions I (0, α rb), K (0, α rb), I (1, α rb), K (1, α rb),
I(1,αra), K (1, α ra), I (0, α ra), K (0, α ra);
Bessel correction functions are the solutions obtained when solving rubber bushing deformation equation herein, are deformed by bushing further Obtain bushing radial rigidity.
4. the method for calibration of driver's cabin stabilizer bar system roll angular rigidity according to claim 1, it is characterised in that
The equivalent line stiffness K of (3) the coaxial-type driver's cabin stabilizer bar systemwsCalculating:
According to the outer diameter D of torsion tube, the G that is calculated in step (1)w, the K that is calculated in step (2)x, to coaxial-type stabiliser bar Stiffness K of the system in cab mounting installed positionwsCalculated, i.e.,
<mrow> <msub> <mi>K</mi> <mrow> <mi>w</mi> <mi>s</mi> </mrow> </msub> <mo>=</mo> <mfrac> <msub> <mi>K</mi> <mi>x</mi> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>K</mi> <mi>x</mi> </msub> <msub> <mi>G</mi> <mi>w</mi> </msub> <mo>/</mo> <msup> <mi>D</mi> <mn>4</mn> </msup> <mo>)</mo> </mrow> </mfrac> <mo>.</mo> </mrow>
5. the method for calibration of driver's cabin stabilizer bar system roll angular rigidity according to claim 1, it is characterised in that
(4) the coaxial-type driver's cabin stabilizer bar system roll angular rigidityCalculating:
According to the suspension distance L of stabiliser barc, and the K being calculated in step (3)ws, to coaxial-type driver's cabin stabilizer bar system Roll angular rigidityCalculated, i.e.,
6. the method for calibration of driver's cabin stabilizer bar system roll angular rigidity according to claim 1, it is characterised in that
The checking computations of (5) the coaxial-type driver's cabin stabilizer bar system rigidity and ANSYS simulating, verifyings:
I applies a certain load F at the suspended position of swing arm, and in the case where not considering cab mounting spring rate, profit With the K obtained by calculating in step (3)ws, to deformation f of the swing arm at suspended positionwsCCalculated, i.e. fwsC=F/Kws;Root According to the suspension distance L of stabiliser barc, pendulum arm length l1, and swing arm at suspended position between outermost end apart from Δ l1, utilize Stabilizer bar system deforms and the geometrical relationship of swing arm displacement, to deformation displacement amount of the swing arm at outermost end The side tilt angle of driver's cabinAnd the roll angular rigidity of driver's cabin stabilizer bar systemCalculated;
II utilizes ANSYS finite element emulation softwares, according to the structure and material characteristic parameter of stabilizer bar system, establishes emulation mould Type, grid division, and apply at the suspended position of swing arm and carried out with I step identical load F, the deformation to stabilizer bar system ANSYS is emulated, and obtains maximum distortion f of the swing arm at outermost endwsA;At the swing arm outermost end obtained by ANSYS emulation Maximum distortion fwsA, pendulum arm length l1, swing arm is at suspended position to the distance, delta l at outermost end1, and the suspension distance of stabiliser bar Lc, using the geometrical relationship of stabilizer bar system deformation and swing arm displacement, to deformation of the swing arm at suspended position The side tilt angle of driver's cabinAnd the roll angular rigidity of driver's cabin stabilizer bar systemCalculated;
III calculates step (4) the roll angular rigidity value of resulting stabilizer bar systemRespectively with being calculated in I steps Roll angular rigidity valueRoll angular rigidity value obtained by II step simulation calculationsIt is compared, so as to this hair The computational methods and Rigidity Calculation value of bright provided coaxial-type driver's cabin stabilizer bar system roll angular rigidity carry out testing III card.
CN201710821289.7A 2017-09-13 2017-09-13 The method of calibration of driver's cabin stabilizer bar system roll angular rigidity Pending CN107506560A (en)

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