CN104182597B

CN104182597B - The check method of vehicle suspension roll angular rigidity

Info

Publication number: CN104182597B
Application number: CN201410476073.8A
Authority: CN
Inventors: 周长城; 郭剑; 提艳; 安艳; 高炳凯; 于曰伟
Original assignee: Shandong University of Technology
Current assignee: Shandong University of Technology
Priority date: 2014-09-18
Filing date: 2014-09-18
Publication date: 2018-04-10
Anticipated expiration: 2034-09-18
Also published as: CN104182597A

Abstract

The present invention relates to the check method of vehicle suspension roll angular rigidity, belongs to vehicle suspension technical field.Previously fail to provide the parsing calculation and check method of roll angular rigidity always.The present invention utilizes the deformation coefficient G of end part of stabilizer rod according to vehicle and suspension parameter, and the structure and material characterisitic parameter of designed stabiliser bar and rubber bushing_WAnd rubber bushing radial rigidity K_xAnalytical formula, calculation and check is carried out to forward and backward suspension roll angular rigidity and the total roll angular rigidity of vehicle.Using the calculation and check value of the available accurately and reliably vehicle roll angular rigidity of this method, vehicle suspension and the design level and quality of stabilizer bar system can be not only improved, improves the ride performance and security of vehicle；Meanwhile design and testing expenses can be reduced using this method, accelerate product development speed, and reliable technical support is improved for the exploitation of stabiliser bar CAD software.

Description

The check method of vehicle suspension roll angular rigidity

Technical field

The present invention relates to the check method of vehicle suspension, particularly vehicle suspension roll angular rigidity.

Background technology

Vehicle suspension and stabiliser bar design must are fulfilled for when turn inside diameter travels to the design requirement of roll angular rigidity.But Due to being deformed analytical Calculation by stabiliser bar, rubber bushing deforms analytical Calculation and the key issue that intercouples is restricted, for car The calculation and check of roll angular rigidity, fails to provide reliable Analytic Calculation Method always.At present, both at home and abroad for vehicle roll Angular rigidity is checked, and is mostly to utilize ANSYS simulation softwares, is carried out simulation analysis to roll angular rigidity by solid modelling and is tested Card, this method is although can obtain reliable simulation numerical, yet with accurate analytical formula can not be provided, therefore, The requirement of suspension stabilizer bar system CAD software exploitation can not be met.As Vehicle Industry is fast-developing and Vehicle Speed Improve constantly, higher requirement is proposed to vehicle suspension system and stabiliser bar design, there is an urgent need to stable for vehicle manufacture producer Lever system CAD software.Therefore, it is necessary to establish a kind of check method of accurate, reliable vehicle suspension roll angular rigidity, production is improved Product design level and quality, improve vehicle ride performance and security；Meanwhile design and testing expenses are reduced, accelerate product Development rate.

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 check method of reliable vehicle suspension roll angular rigidity, its calculation and check flow chart is as shown in figure 1, vehicle roll motion model Figure is as shown in Fig. 2 the structural representation of stabiliser bar is as shown in Figure 3.

In order to solve the above technical problems, the check method of vehicle suspension roll angular rigidity provided by the present invention, its feature It is to use following calculation procedure.

(1) total roll angular rigidity required for vehicle suspensionCalculating：

According to automobile body quality m_s, the distance h of vehicle body barycenter and inclination between centers_s, radius of wheel r, side acceleration a_y, And the vehicle body max. roll required by Car designTotal roll angular rigidity required for vehicle suspension is calculated, i.e.,：

Wherein, g is acceleration of gravity；

(2) roll angular rigidity of forward and backward bearing springWithCalculating：

According to the front tread B of vehicle_fWith rear tread B_r, forward swing arm lengths T_1f, rear-swing arm length T_1r, forward and backward bearing spring Installation site is to the distance between swing arm hinge point T_2fAnd T_2r, and the Line stiffness k of forward and backward bearing spring_sfAnd k_sr, to forward and backward outstanding The roll angular rigidity of frame spring is respectively calculated, i.e.,：

(3) the deformation coefficient G of forward and backward suspension end part of stabilizer rod vertical deviation_wfAnd G_wrCalculating：

According to the total length l of forward and backward suspension stabiliser bar_cfAnd l_cr, the mounting distance l of middle two rubber bushings_0fAnd l_0r, The brachium l of forward and backward stabiliser bar_1fAnd l_1r, the transition arc radius R of forward and backward stabiliser bar_fAnd R_r, the central angle θ of transition arc_fWith θ_r, and elastic properties of materials model E and Poisson's ratio μ, to the deformation coefficient G of forward and backward suspension end part of stabilizer rod vertical deviation_wfAnd G_wrCarry out Calculate, be respectively：

In formula,

Q_6f=32 (μ+1) [R_f(cosθ_f-1)-l_1fsinθ_f]²[2l_1fcosθ_f-l_cf+2R_fsinθ_f]；

Q_6r=32 (u+1) [R_r(cosθ_r-1)-l_1rsinθ_r]²[2l_1rcosθ_r-l_cr+2R_rsinθ_r]；

(4) forward and backward stabiliser bar rubber bushing RADIAL stiffness K_xfAnd K_xrAnalytical Calculation：

According to the inner circle radius r of forward and backward rubber bushing_afAnd r_ar, exradius r_bfAnd r_br, axial length L_fAnd L_r, and rubber The elastic modulus E of glue bushing_x, Poisson's ratio μ_x, the radial direction Line stiffness of forward and backward rubber bushing of hanger bracket is respectively calculated, i.e.,：

Wherein,

Bessel correction functions I (0, α_fr_bf), K (0, α_fr_bf), I (1, α_fr_bf), K (1, α_fr_bf)；

I(1,α_fr_af), K (1, α_fr_af), I (0, α_fr_af), K (0, α_fr_af)；

Wherein,

Bessel correction functions I (0, α_rr_br), K (0, α_rr_br), I (1, α_rr_br), K (1, α_rr_br)；

I(1,α_rr_ar), K (1, α_rr_ar), I (0, α_rr_ar), K (0, α_rr_ar)；

(5) roll angular rigidity of forward and backward stabilizer bar systemWithCalculation and check：

According to the wheelspan B of the forward and backward bridge of vehicle_fAnd B_r, the diameter d of forward and backward suspension stabiliser bar_fAnd d_r, length l_cfAnd l_cr, preceding, Rear stabilizer bar rubber bushing mounting distance length l_0fAnd l_0r, at the end points of the forward and backward stabiliser bar obtained by calculating in step (3) Deformation coefficient G_wfAnd G_wr, the middle radial rigidity K for calculating resulting forward and backward rubber bushing of step (4)_xfAnd K_xr, to forward and backward The roll angular rigidity of stabilizer bar systemWithCalculation and check is carried out respectively, i.e.,：

(6) the total roll angular rigidity of vehicleCalculation and check：

According to the roll angular rigidity of forward and backward bearing spring resulting in step (2)WithGained in step (5) The roll angular rigidity of the forward and backward stabilizer bar system arrivedWithCalculation and check is carried out to the total roll angular rigidity of vehicle, i.e.,

If the calculation and check value of the total roll angular rigidity of vehicleResulting more than or equal to calculating in step (1) Design load required by vehicleI.e.Then vehicle roll angular rigidity meets Car design requirement；Otherwise, ifThen vehicle roll angular rigidity is unsatisfactory for Car design requirement, it is necessary to be adjusted to fore suspension and rear suspension stabilizer bar system Design.

The present invention has the advantage that than prior art：

Home and abroad is checked for vehicle suspension roll angular rigidity at present, is mostly to utilize simulation software, is passed through modeling and simulating Check analysis is carried out to vehicle roll angular rigidity, but this method can not provide analytical formula, it is thus impossible to meet stabiliser bar The requirement of the CAD software exploitation of system.The present invention can be according to vehicle parameter and suspension parameter, and designed stabiliser bar and rubber The structural parameters and material characteristic parameter of bushing, utilize the deformation system of the roll angular rigidity of forward and backward bearing spring, end part of stabilizer rod Number G_wAnd rubber bushing radial rigidity K_xAnalytical formula, forward and backward suspension roll angular rigidity and the total roll angular rigidity of vehicle are entered Row calculation and check.The calculation and check value of the available accurately and reliably vehicle roll angular rigidity of this method, for vehicle suspension and stably The design of bar provides reliable roll angular rigidity calculation and check method, and has established technology for the exploitation of stabiliser bar CAD software Basis.Using this method, vehicle suspension and the design level and quality of stabilizer bar system can be not only improved, improves the traveling of vehicle Ride comfort and security；Meanwhile design and testing expenses can be reduced using this method, accelerate product development speed.

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 flow chart of vehicle suspension roll angular rigidity calculation and check；

Fig. 2 is vehicle roll motion illustraton of model；

Fig. 3 is the structural representation of stabilizer bar system；

Fig. 4 is the structural representation of rubber bushing.

Embodiment

The present invention is described in further detail below by embodiment.

Embodiment one：The body quality m of certain vehicle_s=4690kg, side acceleration a_y=0.4g, vehicle body barycenter is with rolling The distance h of between centers_s=1069mm, the design requirement value of vehicle roll angleFront suspension pendulum arm length T_1f=675mm, bullet Spring Line stiffness k_sf=102.45N/mm, spring center to distance T between swing arm hinge point_2f=430mm；Rear suspension pendulum arm length T_1r=650mm, spring wire stiffness K_sr=261N/mm, spring center to distance T between swing arm hinge point_2r=400mm；The vehicle Front axle wheel is away from B_f=1650mm, rear axle wheel is away from B_r=1485mm；The vehicle only sets stabilizer bar system in front suspension, and its structure is shown It is intended to as shown in figure 3, wherein, the diameter d of stabiliser bar_f=20mm, total length l_cf=800mm, brachium l_1f=150mm, transition circle Arc radius R_f=50mm, transition arc central angle θ_f=60 °, the mounting distance l between two rubber bushings_0f=400mm, stabiliser bar Elastic modulus E=210GPa of material, Poisson's ratio μ=0.3.The structure of front stabilizer rubber bushing as shown in figure 4, stabiliser bar 1, Interior round buss 2, rubber bushing 3, outer round buss 4, outer round buss 4 and interior round buss 2 and rubber bushing 3 are as one, by interior Round buss 2 matches with stabiliser bar 1 and merged on stabiliser bar 1, wherein, the axial length L of rubber bushing 1_f=25mm, thickness h_f=10mm, the wall thickness Δ l of interior round buss 2_f=2.0mm, i.e. rubber bushing inner circle radius r_af=12mm, exradius r_bf= 22mm, elastic modulus E_x=7.84MPa, Poisson's ratio μ_x=0.47.Check meter is carried out to the roll angular rigidity of the vehicle suspension system Calculate.

The check method for the vehicle suspension roll angular rigidity that present example is provided, its calculation and check flow such as Fig. 1 institutes Show, comprise the following steps that：

(1) total roll angular rigidity required for vehicle suspensionCalculating：

According to automobile body quality m_s=4690kg, side acceleration a_y=0.4g, vehicle body barycenter and the distance for rolling between centers h_s=1069mm, vehicle roll angleIgnoring unsprung mass m_uIn the case of, to total angle of heel required for vehicle suspension Rigidity is calculated, i.e.,：

(2) roll angular rigidity of the forward and backward bearing spring of vehicleWithCalculating：

According to the front tread B of vehicle_f=1650mm and rear tread B_r=1485mm, forward swing arm lengths T_1f=675mm, rear pendulum Arm lengths T_1r=650mm, forward and backward bearing spring installation site to the distance between swing arm hinge point T_2f=430mm and T_2r= 400mm, and the Line stiffness k of forward and backward bearing spring_sf=102.45N/mm and k_sr=261N/mm, to the side of forward and backward bearing spring Inclination angle rigidity is respectively calculated, i.e.,：

(3) the deformation coefficient G of front suspension end part of stabilizer rod_wfCalculating：

According to the total length l of front suspension stabiliser bar_cf=800mm, the mounting distance l of middle two rubber bushings_0f= 400mm, the brachium l of front stabilizer_1f=150mm, the transition arc radius R of front stabilizer_f=50mm, the central angle of transition arc θ_f=60 °, elastic properties of materials model E=210GPa and Poisson's ratio μ=0.3, the deformation coefficient of front suspension end part of stabilizer rod is carried out Calculate, i.e.,：

In formula,

Q_6f=32 (u+1) [R_f(cosθ_f-1)-l_1fsinθ_f]²[2l_1fcosθ_f-l_cf+2R_fsinθ_f]=- 0.5624m³；

(4) front suspension stabiliser bar rubber bushing RADIAL stiffness K_xfAnalytical Calculation：

According to the inner circle radius r of front suspension stabiliser bar rubber bushing_af=12mm, exradius r_bf=22mm, axial length L_f=25mm, and the elastic modulus E of rubber bushing_x=7.84MPa, Poisson's ratio μ_x=0.47, to the radial direction of front suspension rubber bushing Line stiffness K_xfCalculated, i.e.,：

Wherein,

Bessel correction functions：

I(0,α_fr_bf)=25.0434, K (0, α_fr_bf)=0.0041,

I(1,α_fr_bf)=22.3175, K (1, α_fr_bf)=0.0045,

I(1,α_fr_af)=2.1439, K (1, α_fr_af)=0.0922,

I(0,α_fr_af)=2.8801, K (0, α_fr_af)=0.0769；

(5) roll angular rigidity of front stabilizer systemCalculation and check：

According to the wheelspan B of vehicle propons_f=1650mm, the diameter d of stabiliser bar_f=20mm, length l_cf=800mm, two rubbers Mounting distance l between glue bushing_0f=400mm, the middle deformation coefficient calculated at resulting front stabilizer end points of step (3) G_wf=1.5935 × 10^-12m⁵The radial rigidity K of preceding rubber bushing obtained by being calculated in/N, and step (4)_xf=2106.8N/ Roll angular rigidities of the mm to front stabilizer systemCalculation and check is carried out, i.e.,：

(6) the total roll angular rigidity of vehicleCalculation and check：

According to the roll angular rigidity for the forward and backward bearing spring being calculated in step (2)With

The roll angular rigidity of front stabilizer system obtained by being calculated in step (5)Calculation and check is carried out to the total roll angular rigidity of vehicle, i.e.,

Understand the calculation and check value of the total roll angular rigidity of the vehicleFallen into a trap more than step (1) The design load required by vehicle obtained by calculatingI.e.The vehicle roll angular rigidity is expired Sufficient Car design requirement.

Embodiment two：Known certain automobile body quality m_s=5000kg, side acceleration a_y=0.4g, vehicle body barycenter and side Incline the distance h of between centers_s=1150mm, vehicle roll angleThe spring Line stiffness k of vehicle front suspension_sf=90.761N/mm, The spring Line stiffness k of rear suspension_sr=176.23N/mm；Front tread B_f=1650mm, rear tread B_r=1600mm；Front suspension swing arm Length T_1f=660mm, front suspension spring center to distance T between homonymy swing arm hinge point_2f=450mm；Rear-swing arm length T_1r= 650mm, rear suspension spring installation center to distance T between homonymy transverse arm pin joint_2r=400mm.Forward and backward suspension stabiliser bar and rubber Glue bushing, in addition to diameter differs, other parameters are all identical with embodiment one, wherein, front stabilizer diameter d_f= 22mm, rear stabilizer bar diameter d_r=19mm；The inside radius r of preceding rubber bushing_af=13mm, outer radius r_bf=23mm, rear rubber lining The inner circle radius r of set_ar=11.5mm.Exradius r_br=21.5mm.School is carried out to the roll angular rigidity of the vehicle suspension system Assess calculation.

The step of using embodiment one, calculation and check is carried out to the roll angular rigidity of the vehicle suspension system, i.e.,:

(1) total roll angular rigidity required for vehicle suspensionCalculating：

According to automobile body quality m_s=5000kg, side acceleration a_y=0.4g, vehicle body barycenter and the distance for rolling between centers h_s=1150mm, the max. roll required by vehicle body designTotal roll angular rigidity required for vehicle suspension is carried out Calculate, i.e.,：

In formula, g acceleration of gravity, g=9.8m/s²；

According to the front tread B of vehicle_f=1650mm and rear tread B_r=1600mm, forward swing arm lengths T_1f=660mm, rear pendulum Arm lengths T_1r=650mm, forward and backward bearing spring installation site to the distance between swing arm hinge point T_2f=450mm and T_2r= 400mm, and the Line stiffness k of forward and backward bearing spring_sf=90.761N/mm and k_sr=176.23N/mm, to forward and backward bearing spring Roll angular rigidity be respectively calculated, i.e.,：

(3) the deformation coefficient G of forward and backward suspension end part of stabilizer rod_wfAnd G_wrCalculating：

Because the Chinese herbaceous peony, rear suspension stabiliser bar are in addition to diameter, other parameters are all identical with embodiment one, because This, the deformation coefficient of the Chinese herbaceous peony rear suspension end part of stabilizer rod is also all identical with embodiment one, i.e.,：

(4) forward and backward suspension stabiliser bar rubber bushing RADIAL stiffness K_xfAnd K_xrAnalytical Calculation：

Because the inner circle radius and exradius of the stable shank diameter of forward and backward suspension and rubber bushing differ, therefore, rubber The RADIAL stiffness K of bushing_xfAnd K_xrAlso differ.According to the inside radius r of preceding rubber bushing_af=13mm, outer radius r_bf= 23mm, the inner circle radius r of rear rubber bushing_ar=11.5mm, exradius r_br=21.5mm, and the length L of rubber bushing_f= 25mm, elastic modulus E_x=7.84MPa, Poisson's ratio μ_x=0.47, it is stable to forward and backward suspension using the computational methods of embodiment one Bar rubber bushing RADIAL stiffness K_xfAnd K_xrIt is respectively calculated, i.e.,：

Wherein,

According to the wheelspan B of the forward and backward bridge of vehicle_f=1650mm and B_r=1600mm, the diameter d of forward and backward stabiliser bar_f=22mm And d_r=19mm, length l_cf=l_cr=800mm, the mounting distance l between two rubber bushings_0f=l_0r=400mm, in step (3) The deformation coefficient G at forward and backward stabiliser bar end points obtained by calculating_wf=G_wr=1.5935 × 10^-12m⁵/ N, and step (4) are fallen into a trap The radial rigidity K of forward and backward rubber bushing obtained by calculating_xf=2267.0N/mm and K_xr=2026.7N/mm, to forward and backward stabilization The roll angular rigidity of lever systemWithCalculation and check is carried out respectively, i.e.,：

(6) the total roll angular rigidity of vehicleCalculation and check：

According to the roll angular rigidity for the forward and backward bearing spring being calculated in step (2)WithThe roll angular rigidity of forward and backward stabilizer bar system obtained by being calculated in step (5)WithCalculation and check is carried out to the total roll angular rigidity of vehicle, i.e.,

Understand the calculation and check value of the total roll angular rigidity of the vehicleFallen into a trap more than step (1) The design load required by vehicle obtained by calculatingI.e.The vehicle roll angular rigidity meets Car design requirement.

Embodiment three：Certain vehicle is except the stable shank diameter d of front suspension_fAnd the inner circle radius r of rubber bushing_afAnd exradius r_bfOutside differing, other all parameters are identical with embodiment two, wherein, the diameter d of front stabilizer_f=21mm, rubber The inner circle radius r of glue bushing_af=12.5mm, and exradius r_bf=22.5mm.To the roll angular rigidity of the vehicle suspension system Carry out calculation and check.

The step of using example one is applied, calculation and check is carried out to the roll angular rigidity of the vehicle suspension system:

(1) total roll angular rigidity required for vehicle suspensionCalculating：

Because vehicle parameter is identical with embodiment two, therefore, total roll angular rigidity required for the vehicle suspension, Also it is identical with embodiment two, i.e.,：

It is identical with embodiment two due to the vehicle parameter and suspension stiffness of the vehicle, therefore, before vehicle, The roll angular rigidity of rear suspension spring, it is also identical with embodiment two respectively, i.e.,：

Because the forward and backward suspension stabiliser bar of the vehicle is in addition to diameter, other parameters are all identical with embodiment two, Therefore, the Chinese herbaceous peony, the deformation coefficient of rear suspension end part of stabilizer rod, it is also all identical with embodiment two, i.e.,：

Due to the stable shank diameter of rear suspension and the inner circle radius and exradius of rubber bushing, the complete phase with embodiment two Together, therefore, rear suspension stabiliser bar rubber bushing RADIAL stiffness K_xrAlso it is all identical with embodiment two, i.e.,：

The inner circle radius and exradius of the stable shank diameter of the vehicle front suspension and rubber bushing, the not phase with embodiment two Together.According to the inside radius r of preceding rubber bushing_af=12.5mm, outer radius r_bf=22.5mm, elastic modulus E_x=7.84MPa, Poisson Compare μ_x=0.47, using the computational methods of embodiment one to front suspension stabiliser bar rubber bushing RADIAL stiffness K_xfCalculated, I.e.：

Wherein,

Because the stable shank diameter of the vehicle rear suspension and rubber bushing are identical with embodiment two, therefore, rear suspension The roll angular rigidity of stabilizer bar systemAlso it is identical with embodiment two, i.e.,：

According to the wheelspan B of vehicle propons_f=1650mm, the diameter d of stabiliser bar_f=21mm, length l_cf=800mm, two rubbers Mounting distance l between glue bushing_0f=400mm, the middle deformation coefficient calculated at resulting front stabilizer end points of step (3) G_wf=1.5935 × 10^-12m⁵The radial rigidity K of preceding rubber bushing obtained by being calculated in/N, and step (4)_xf=2186.9N/ Mm, to the roll angular rigidity of front stabilizer systemCalculation and check is carried out, i.e.,：

(6) the total roll angular rigidity of vehicleCalculation and check：

Understand the calculation and check value of the total roll angular rigidity of the vehicleFallen into a trap less than step (1) The design load required by vehicle obtained by calculatingI.e.The vehicle roll angular rigidity is discontented with Sufficient Car design requirement.

Design is adjusted to the stable shank diameter of front suspension, i.e., is d by the diameter adjusted design of front suspension stabiliser bar_f= 22mm, makes the roll angular rigidity of vehicle can increase toMeet that the design of vehicle roll angular rigidity will Evaluation；Design can also be adjusted to the mounting distance between the rubber bushing of front suspension stabiliser bar two, by two rubber bushings it Between l₀Mounting distance adjusted design be l₀=420mm, clipping room make the roll angular rigidity of vehicle can increase to away from increase 10mmMeet the design requirement value of vehicle roll angular rigidity.

Claims

1. the check method of vehicle suspension roll angular rigidity, it is comprised the following steps that：

(1) total roll angular rigidity required for vehicle suspensionCalculating：

According to automobile body quality m_s, the distance h of vehicle body barycenter and inclination between centers_s, radius of wheel r, side acceleration a_y, and car The required vehicle body max. roll of designTotal roll angular rigidity required for vehicle suspension is calculated, i.e.,：

Wherein, g is acceleration of gravity；

According to the front tread B of vehicle_fWith rear tread B_r, forward swing arm lengths T_1f, rear-swing arm length T_1r, forward and backward bearing spring installation Position is to the distance between swing arm hinge point T_2fAnd T_2r, and the Line stiffness k of forward and backward bearing spring_sfAnd k_sr, to forward and backward suspension bullet The roll angular rigidity of spring is respectively calculated, i.e.,：

According to the total length l of forward and backward suspension stabiliser bar_cfAnd l_cr, the mounting distance l of middle two rubber bushings_0fAnd l_0r, it is forward and backward The brachium l of stabiliser bar_1fAnd l_1r, the transition arc radius R of forward and backward stabiliser bar_fAnd R_r, the central angle θ of transition arc_fAnd θ_r, and material Elastic model E and Poisson's ratio μ is expected, to the deformation coefficient G of forward and backward suspension end part of stabilizer rod vertical deviation_wfAnd G_wrCalculated, Respectively：

In formula,

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<mrow> <msub> <mi>Q</mi> <mrow> <mn>5</mn> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mn>64</mn> <msub> <mi>R</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <mi>&mu;</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msup> <msub> <mi>R</mi> <mi>f</mi> </msub> <mn>2</mn> </msup> <mrow> <mo>(</mo> <mfrac> <mrow> <mn>3</mn> <msub> <mi>&theta;</mi> <mi>f</mi> </msub> </mrow> <mn>2</mn> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <msub> <mi>&theta;</mi> <mi>f</mi> </msub> </mrow> <mn>4</mn> </mfrac> <mo>-</mo> <mn>2</mn> <msub> <mi>sin&theta;</mi> <mi>f</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msubsup> <mi>l</mi> <mrow> <mn>1</mn> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>&theta;</mi> <mi>f</mi> </msub> <mo>-</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <msub> <mi>&theta;</mi> <mi>f</mi> </msub> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mn>4</mn> <msub> <mi>l</mi> <mrow> <mn>1</mn> <mi>f</mi> </mrow> </msub> <msub> <mi>R</mi> <mi>f</mi> </msub> <msup> <mi>sin</mi> <mn>4</mn> </msup> <mfrac> <msub> <mi>&theta;</mi> <mi>f</mi> </msub> <mn>2</mn> </mfrac> <mo>&rsqb;</mo> <mo>,</mo> </mrow>

Q_6f=32 (μ+1) [R_f(cosθ_f-1)-l_1fsinθ_f]²[2l_1fcosθ_f-l_cf+2R_fsinθ_f]；

<mrow> <msub> <mi>Q</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>64</mn> <msubsup> <mi>l</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> <mn>3</mn> </msubsup> </mrow> <mn>3</mn> </mfrac> <mo>,</mo> <msub> <mi>Q</mi> <mrow> <mn>2</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>64</mn> <mo>&lsqb;</mo> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> </msub> <msub> <mi>cos&theta;</mi> <mi>r</mi> </msub> <mo>+</mo> <msub> <mi>R</mi> <mi>r</mi> </msub> <msub> <mi>sin&theta;</mi> <mi>r</mi> </msub> <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> <mrow> <mn>0</mn> <mi>r</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>l</mi> <mrow> <mi>c</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>3</mn> </msup> <mo>&rsqb;</mo> </mrow> <mn>3</mn> </mfrac> <mo>,</mo> </mrow>

<mrow> <msub> <mi>Q</mi> <mrow> <mn>3</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mn>64</mn> <msub> <mi>R</mi> <mi>r</mi> </msub> <mo>&lsqb;</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msubsup> <mi>l</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> <mo>+</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msubsup> <mi>R</mi> <mi>r</mi> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> <mo>-</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>l</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> </msub> <msub> <mi>R</mi> <mi>r</mi> </msub> <msup> <mi>sin</mi> <mn>2</mn> </msup> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> <mo>&rsqb;</mo> <mo>,</mo> <msub> <mi>Q</mi> <mrow> <mn>4</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>8</mn> <msub> <mi>l</mi> <mrow> <mn>0</mn> <mi>r</mi> </mrow> </msub> <msup> <mrow> <mo>(</mo> <msub> <mi>l</mi> <mrow> <mn>0</mn> <mi>r</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>l</mi> <mrow> <mi>c</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mrow> <mn>3</mn> </mfrac> <mo>,</mo> </mrow>

<mrow> <msub> <mi>Q</mi> <mrow> <mn>5</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mn>64</mn> <msub> <mi>R</mi> <mi>r</mi> </msub> <mrow> <mo>(</mo> <mi>&mu;</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>r</mi> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <mfrac> <mrow> <mn>3</mn> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> </mrow> <mn>2</mn> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> </mrow> <mn>4</mn> </mfrac> <mo>-</mo> <mn>2</mn> <msub> <mi>sin&theta;</mi> <mi>r</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msubsup> <mi>l</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> <mo>-</mo> <mfrac> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> </mrow> <mn>2</mn> </mfrac> <mo>)</mo> </mrow> <mo>+</mo> <mn>4</mn> <msub> <mi>l</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> </msub> <msub> <mi>R</mi> <mi>r</mi> </msub> <msup> <mi>sin</mi> <mn>4</mn> </msup> <mfrac> <msub> <mi>&theta;</mi> <mi>r</mi> </msub> <mn>2</mn> </mfrac> <mo>&rsqb;</mo> <mo>,</mo> </mrow>

Q_6r=32 (u+1) [R_r(cosθ_r-1)-l_1rsinθ_r]²[2l_1rcosθ_r-l_cr+2R_rsinθ_r]；

According to the inner circle radius r of forward and backward rubber bushing_afAnd r_ar, exradius r_bfAnd r_br, axial length L_fAnd L_r, and rubber lining The elastic modulus E of set_x, Poisson's ratio μ_x, the radial direction Line stiffness of forward and backward rubber bushing of hanger bracket is respectively calculated, i.e.,：

Wherein,

<mrow> <mi>y</mi> <mrow> <mo>(</mo> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mi>f</mi> </mrow> </msub> <mi>I</mi> <mrow> <mo>(</mo> <mrow> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mi>f</mi> </mrow> </msub> <mi>K</mi> <mrow> <mo>(</mo> <mrow> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mrow> <mn>3</mn> <mi>f</mi> </mrow> </msub> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&mu;</mi> <mi>x</mi> </msub> </mrow> <mrow> <mn>5</mn> <msub> <mi>&pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>f</mi> </msub> </mrow> </mfrac> <mrow> <mo>(</mo> <mrow> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>+</mo> <mfrac> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mrow> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>,</mo> </mrow>

<mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&mu;</mi> <mi>x</mi> </msub> <mo>)</mo> <mo>&lsqb;</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> <mo>-</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> <mo>&rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>f</mi> </msub> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>,</mo> </mrow>

<mrow> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <msub> <mi>&mu;</mi> <mi>x</mi> </msub> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>&lsqb;</mo> <mi>I</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> <mo>-</mo> <mi>I</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> <mo>&rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>f</mi> </msub> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>,</mo> </mrow>

<mrow> <msub> <mi>a</mi> <mrow> <mn>3</mn> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&mu;</mi> <mi>x</mi> </msub> <mo>)</mo> <mo>(</mo> <msub> <mi>b</mi> <mrow> <mn>1</mn> <mi>f</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>b</mi> <mrow> <mn>2</mn> <mi>f</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>b</mi> <mrow> <mn>3</mn> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>f</mi> </msub> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow>

<mrow> <msub> <mi>b</mi> <mrow> <mn>1</mn> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>

<mrow> <msub> <mi>b</mi> <mrow> <mn>2</mn> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>

<mrow> <msub> <mi>b</mi> <mrow> <mn>3</mn> <mi>f</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> <mrow> <mo>&lsqb;</mo> <mrow> <mi>I</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mo>&rsqb;</mo> </mrow> <mrow> <mo>&lsqb;</mo> <mrow> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mrow> <mo>(</mo> <mrow> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>f</mi> </mrow> <mn>2</mn> </msubsup> </mrow> <mo>)</mo> </mrow> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>f</mi> </mrow> </msub> </mrow> <mo>&rsqb;</mo> </mrow> <mo>,</mo> </mrow>

<mrow> <msub> <mi>&alpha;</mi> <mi>f</mi> </msub> <mo>=</mo> <mn>2</mn> <msqrt> <mn>15</mn> </msqrt> <mo>/</mo> <msub> <mi>L</mi> <mi>f</mi> </msub> <mo>,</mo> </mrow>

I(1,α_fr_af), K (1, α_fr_af), I (0, α_fr_af), K (0, α_fr_af)；

Wherein,

<mrow> <mi>y</mi> <mrow> <mo>(</mo> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> </msub> <mi>I</mi> <mrow> <mo>(</mo> <mrow> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mi>r</mi> </mrow> </msub> <mi>K</mi> <mrow> <mo>(</mo> <mrow> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mi>a</mi> <mrow> <mn>3</mn> <mi>r</mi> </mrow> </msub> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <msub> <mi>&mu;</mi> <mi>x</mi> </msub> </mrow> <mrow> <mn>5</mn> <msub> <mi>&pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>r</mi> </msub> </mrow> </mfrac> <mrow> <mo>(</mo> <mrow> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>+</mo> <mfrac> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mrow> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>,</mo> </mrow>

<mrow> <msub> <mi>a</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&mu;</mi> <mi>x</mi> </msub> <mo>)</mo> <mo>&lsqb;</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> <mo>-</mo> <mi>K</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> <mo>&rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>r</mi> </msub> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>,</mo> </mrow>

<mrow> <msub> <mi>a</mi> <mrow> <mn>2</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mo>(</mo> <msub> <mi>&mu;</mi> <mi>x</mi> </msub> <mo>+</mo> <mn>1</mn> <mo>)</mo> <mo>&lsqb;</mo> <mi>I</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> <mo>-</mo> <mi>I</mi> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>(</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> <mo>&rsqb;</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>r</mi> </msub> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>,</mo> </mrow>

<mrow> <msub> <mi>a</mi> <mrow> <mn>3</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>&mu;</mi> <mi>x</mi> </msub> <mo>)</mo> <mo>(</mo> <msub> <mi>b</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> </msub> <mo>-</mo> <msub> <mi>b</mi> <mrow> <mn>2</mn> <mi>r</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>b</mi> <mrow> <mn>3</mn> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mrow> <mn>5</mn> <msub> <mi>&pi;E</mi> <mi>x</mi> </msub> <msub> <mi>L</mi> <mi>r</mi> </msub> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>;</mo> </mrow>

<mrow> <msub> <mi>b</mi> <mrow> <mn>1</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>

<mrow> <msub> <mi>b</mi> <mrow> <mn>2</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>+</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mn>3</mn> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>,</mo> </mrow>

<mrow> <msub> <mi>b</mi> <mrow> <mn>3</mn> <mi>r</mi> </mrow> </msub> <mo>=</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>&lsqb;</mo> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mi>K</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mi>I</mi> <mrow> <mo>(</mo> <mn>1</mn> <mo>,</mo> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <msub> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>&lsqb;</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <mrow> <mo>(</mo> <msubsup> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>+</mo> <msubsup> <mi>r</mi> <mrow> <mi>b</mi> <mi>r</mi> </mrow> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mi>ln</mi> <mi> </mi> <msub> <mi>r</mi> <mrow> <mi>a</mi> <mi>r</mi> </mrow> </msub> <mo>&rsqb;</mo> <mo>,</mo> </mrow>

<mrow> <msub> <mi>&alpha;</mi> <mi>r</mi> </msub> <mo>=</mo> <mn>2</mn> <msqrt> <mn>15</mn> </msqrt> <mo>/</mo> <msub> <mi>L</mi> <mi>r</mi> </msub> <mo>;</mo> </mrow>

I(1,α_rr_ar), K (1, α_rr_ar), I (0, α_rr_ar), K (0, α_rr_ar)；

According to the wheelspan B of the forward and backward bridge of vehicle_fAnd B_r, the diameter d of forward and backward suspension stabiliser bar_fAnd d_r, length l_cfAnd l_cr, it is forward and backward steady Fixed pole rubber bushing mounting distance length l_0fAnd l_0r, the change at the end points of the forward and backward stabiliser bar obtained by calculating in step (3) Shape coefficient G_wfAnd G_wr, the middle radial rigidity K for calculating resulting forward and backward rubber bushing of step (4)_xfAnd K_xr, to forward and backward stabilization The roll angular rigidity of lever systemWithCalculation and check is carried out respectively, i.e.,：

(6) the total roll angular rigidity of vehicleCalculation and check：

According to the roll angular rigidity of forward and backward bearing spring resulting in step (2)WithObtained by step (5) The roll angular rigidity of forward and backward stabilizer bar systemWithCalculation and check is carried out to the total roll angular rigidity of vehicle, i.e.,

If the calculation and check value of the total roll angular rigidity of vehicleMore than or equal to the vehicle institute obtained by being calculated in step (1) It is required that design loadI.e.Then vehicle roll angular rigidity meets Car design requirement；Otherwise, ifThen vehicle roll angular rigidity is unsatisfactory for Car design requirement, it is necessary to be adjusted to fore suspension and rear suspension stabilizer bar system Design.