CN104182597A - Vehicle suspension roll angle rigidity checking method - Google Patents

Abstract

The invention relates to a vehicle suspension roll angle rigidity checking method and belongs to the technical field of vehicle suspensions. Previously, a roll angle rigidity analysis checking and calculating method is not given. With the method, the roll angle rigidity of a front suspension and a rear suspension and the total roll angle rigidity of a vehicle are checked and calculated according to vehicle and suspension parameters, structure and material characteristic parameters of a designed stabilizer bar and a designed rubber bushing, the deformation coefficient GW of the end of the stabilizer bar, and the analysis calculation formula of the radial rigidity Kx of the rubber bushing. An accurate and reliable vehicle roll angle rigidity checking and calculation value can be obtained with the method, the design level and the design quality of a vehicle suspension and stabilizer bar system and driving smoothness and driving safety of the vehicle can be improved; meanwhile, design and experiment cost can be reduced with the method, product development speed can be increased, and reliable technical support is provided for development of stabilizer bar CAD software.

Description

The check method of vehicle suspension roll angular rigidity

Technical field

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

Background technology

The designing requirement to roll angular rigidity while travelling that the design of vehicle suspension and stabilizer bar must meet turn inside diameter.Yet owing to being subject to, stabilizer bar Deformation analyses calculates, rubber bushing Deformation analyses calculates and the restriction of the key issue that intercouples, and for the check of vehicle roll angular rigidity, calculates, and fails to provide reliable Analytic Calculation Method always.At present, for vehicle roll angular rigidity, check both at home and abroad, mostly to utilize ANSYS simulation software, by solid modelling, roll angular rigidity is carried out to simulation analysis and checking, although the method can obtain reliable simulation numerical, yet due to accurate analytical formula can not be provided, therefore, can not meet the requirement of suspension stabilizer bar system CAD software development.Along with improving constantly of Vehicle Industry fast development and Vehicle Speed, vehicle suspension system and stabilizer bar design are had higher requirement, Railway Car Plant man is in the urgent need to stabilizer bar system CAD software.Therefore, must set up a kind of check method of accurate, reliable vehicle suspension roll angular rigidity, improve product design level and quality, improve Vehicle Driving Cycle ride comfort and security; Meanwhile, reduce design and testing expenses, accelerate product development speed.

Summary of the invention

For the defect existing in above-mentioned prior art, technical matters to be solved by this invention is to provide a kind of check method of easy, reliable vehicle suspension roll angular rigidity, it checks calculation flow chart as shown in Figure 1, as shown in Figure 2, the structural representation of stabilizer bar as shown in Figure 3 for vehicle roll motion model figure.

For solving the problems of the technologies described above, the check method of vehicle suspension roll angular rigidity provided by the present invention, is characterized in that adopting following calculation procedure.

(1) the needed total roll angular rigidity of vehicle suspension calculating:

According to automobile body quality m _s, the distance h between vehicle body barycenter and roll axis _s, radius of wheel r, side acceleration a _y, and vehicle designs desired vehicle body max. roll the needed total roll angular rigidity of vehicle suspension is calculated, that is:

Wherein, g is acceleration of gravity;

(2) roll angular rigidity of forward and backward bearing spring with calculating:

According to the front tread B of vehicle _fwith rear tread B _r, front swing arm length T _1f, rear-swing arm length T _1r, forward and backward bearing spring installation site is to the distance T between swing arm hinge point _2fand T _2r, and the line rigidity k of forward and backward bearing spring _srand k _sr, the roll angular rigidity of forward and backward bearing spring is calculated respectively, that is:

(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 stabilizer bar _cfand l _cr, the mounting distance l of middle two rubber bushings _0fand l _0r, the brachium l of forward and backward stabilizer bar _1fand l _1r, the transition arc radius R of forward and backward stabilizer bar _fand R _r, the central angle θ of transition arc _fand θ _r, and elastic properties of materials model E and Poisson ratio μ, the deformation coefficient G to forward and backward suspension end part of stabilizer rod vertical deviation _wfand G _wrcalculate, be respectively:

$G_{\mathrm{wf}}=\frac{Q_{1f}-Q_{2f}+Q_{3f}+Q_{4f}+Q_{5f}-Q_{6f}}{\mathrm{\πE}},$

$G_{\mathrm{wr}}=\frac{Q_{1r}-Q_{2r}+Q_{3r}+Q_{4r}+Q_{5r}-Q_{6r}}{\mathrm{\πE}},$

In formula, $Q_{1f}=\frac{{64l}_{1f}^3}{3},Q_{2f}=\frac{64[{(l_{1f}\cos {\mathrm{\θ}}_f+R_f{\sin \mathrm{\θ}}_f)}^3+\frac{1}{8}{(l_{0f}-l_{\mathrm{cf}})}^3]}{3},$

$Q_{3f}={64R}_f[\frac{1}{2}{l}_{1f}^2({\mathrm{\θ}}_f+\frac{{\sin 2\mathrm{\θ}}_f}{2})+\frac{1}{2}{R}_f^2({\mathrm{\θ}}_f-\frac{{\sin 2\mathrm{\θ}}_f}{2})+l_{1f}{R}_f{\sin }^2{\mathrm{\θ}}_f],Q_{4f}=\frac{{8l}_{0f}{(l_{0f}-l_{\mathrm{cf}})}^2}{3},$

$Q_{5f}={64R}_f(\mathrm{\μ}+1)[{R_f}^2(\frac{{3\mathrm{\θ}}_f}{2}+\frac{{\sin 2\mathrm{\θ}}_f}{4}-{2\sin \mathrm{\θ}}_f)+\frac{1}{2}{l}_{1f}^2({\mathrm{\θ}}_f-\frac{{\sin 2\mathrm{\θ}}_f}{2})+{4l}_{1f}{R}_f{\sin }^4\frac{{\mathrm{\θ}}_f}{2}],$

Q _6f＝32(μ+1)[R _f(cosθ _f-1)-l _1fsinθ _f] ²[2l _1fcosθ _f-l _cf+2R _fsinθ _f]；

$Q_{1r}=\frac{{64l}_{1r}^3}{3},Q_{2r}=\frac{64[{(l_{1r}\cos {\mathrm{\θ}}_r+R_r{\sin \mathrm{\θ}}_r)}^3+\frac{1}{8}{(l_{0r}-l_{\mathrm{rc}})}^3]}{3},$

$Q_{3r}={64R}_r[\frac{1}{2}{l}_{1r}^2({\mathrm{\θ}}_r+\frac{{\sin 2\mathrm{\θ}}_r}{2})+\frac{1}{2}{R}_r^2({\mathrm{\θ}}_r-\frac{{\sin 2\mathrm{\θ}}_r}{2})+l_{1r}{R}_r{\sin }^2{\mathrm{\θ}}_r],Q_{4r}=\frac{{8l}_{0r}{(l_{0r}-l_{\mathrm{cr}})}^2}{3},$

$Q_{5r}={64R}_r(\mathrm{\μ}+1)[R_r^2(\frac{{3\mathrm{\θ}}_{\text{r}}}{2}+\frac{\sin 2{\mathrm{\θ}}_r}{4}-{2\sin \mathrm{\θ}}_r)+\frac{1}{2}{l}_{1r}^2({\mathrm{\θ}}_r-\frac{{\sin 2\mathrm{\θ}}_r}{2})+{4l}_{1r}{R}_r{\sin }^4\frac{{\mathrm{\θ}}_r}{2}],$

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 stabilizer bar rubber bushing radial line 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 _braxial length L _fand L _r, and the elastic modulus E of rubber bushing _x, Poisson ratio μ _x, the radial line rigidity of forward and backward rubber bushing of hanger bracket is calculated respectively, that is:

$K_{\mathrm{xf}}=\frac{1}{u_r\left(r_{\mathrm{bf}}\right)+y\left(r_{\mathrm{bf}}\right)},K_{\mathrm{xr}}=\frac{1}{u_r\left(r_{\mathrm{br}}\right)+y\left(r_{\mathrm{br}}\right)};$

Wherein, $u_r\left(r_b\right)=\frac{1+{\mathrm{\μ}}_x}{{2\mathrm{\πE}}_{x}L}(\ln \frac{r_b}{r_a}-\frac{r_b^2-r_a^2}{r_a^2+r_b^2}),$

$y\left(r_b\right)=a_{1}I(0,{\mathrm{\αr}}_b)+a_{2}K(0,{\mathrm{\αr}}_b)+a_3+\frac{1+{\mathrm{\μ}}_x}{{5\mathrm{\πE}}_{x}L}(\ln r_b+\frac{r_b^2}{r_a^2+r_b^2}),$

$a_1=\frac{(1+{\mathrm{\μ}}_x)[K(1,{\mathrm{\αr}}_a)r_a(r_a^2+{3r}_b^2)-K(1,{\mathrm{\αr}}_b)r_b({3r}_a^2+r_b^2)]}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_2)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)},$

$a_2=\frac{({\mathrm{\μ}}_x+1)[I(1,{\mathrm{\αr}}_a)r_a(r_a^2+{3r}_b^2)-I(1,{\mathrm{\αr}}_b)r_b({3r}_a^2+r_b^2)}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)},$

$a_3=-\frac{(1+{\mathrm{\μ}}_x)(b_1-b_2+b_3)}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)};$

$b_1=[I(1,{\mathrm{\αr}}_a)K(0,{\mathrm{\αr}}_a)+K(1,{\mathrm{\αr}}_a)I(0,{\mathrm{\αr}}_a)]r_a(r_a^2+{3r}_b^2),$

$b_2=[I(1,{\mathrm{\αr}}_b)K(0,{\mathrm{\αr}}_a)+K(1,{\mathrm{\αr}}_b)I(0,{\mathrm{\αr}}_a)]r_b(r_b^2+{3r}_b^2),$

$b_3={\mathrm{\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)][r_a^2+(r_a^2+r_b^2)\ln r_a],$

$\mathrm{\α}=2\sqrt{15}/L,$

Bessel correction function I (0, α r _b), K (0, α r _b), I (1, α r _b), K (1, α r _b),

I(1,αr _a)，K(1,αr _a)，I(0,αr _a)，K(0,αr _a)；

I(0,α _af)，K(0,α _af)；

Wherein, $u\left(d_r\right)=\frac{1+{\mathrm{\μ}}_x}{{2\mathrm{\πE}}_x{L}_r}(\ln \frac{d_r+{2h}_r+{2\mathrm{\Δl}}_r}{d_r+{2\mathrm{\Δl}}_r}-\frac{(d_r+{2\mathrm{\Δl}}_r+h_r)h_r}{N_r})$

$y\left(d_r\right)=a_{1r}I(0,{\mathrm{\α}}_{\mathrm{br}})+a_{2r}K(0,{\mathrm{\α}}_{\mathrm{br}})+a_{3r}+\frac{1+{\mathrm{\μ}}_x}{{5\mathrm{\πE}}_x{L}_r}(\ln (\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)+\frac{{(d_r+{2h}_r+{2\mathrm{\Δl}}_r)}^2}{{4N}_r}),$

$a_{1r}=\frac{(1+{\mathrm{\μ}}_x)[K(1,{\mathrm{\α}}_{\mathrm{ar}})M_r-K(1,{\mathrm{\α}}_{\mathrm{br}})P_r]}{{5\mathrm{\πE}}_x{L}_r{\mathrm{\α}}_r{R}_{\mathrm{abr}}[I(1,{\mathrm{\α}}_{\mathrm{ar}})K(1,{\mathrm{\α}}_{\mathrm{br}})-I(1,{\mathrm{\α}}_{\mathrm{br}})K(1,{\mathrm{\α}}_{\mathrm{ar}})]N_r},$

$a_{2r}=\frac{({\mathrm{\μ}}_x+1)[I(1,{\mathrm{\α}}_{\mathrm{ar}})M_r-I(1,{\mathrm{\α}}_{\mathrm{br}})P_r]}{{5\mathrm{\πE}}_x{L}_r{\mathrm{\α}}_r{R}_{\mathrm{abr}}[I(1,{\mathrm{\α}}_{\mathrm{ar}})K(1,{\mathrm{\α}}_{\mathrm{br}})-I(1,{\mathrm{\α}}_{\mathrm{br}})K(1,{\mathrm{\α}}_{\mathrm{ar}})N_r},$

$a_{3r}=\frac{(1+{\mathrm{\μ}}_x)(b_{1r}+b_{2r}+b_{3r})}{{5\mathrm{\πE}}_x{L}_r{\mathrm{\α}}_r{R}_{\mathrm{abr}}[I(1,{\mathrm{\α}}_{\mathrm{ar}})K(1,{\mathrm{\α}}_{\mathrm{br}})-I(1,{\mathrm{\α}}_{\mathrm{br}})K(1,{\mathrm{\α}}_{\mathrm{ar}})]N_r},$

$N_r=2{(\frac{d_r}{2}+{\mathrm{\Δl}}_r)}^2+{h_r}^2+2(\frac{d_r}{2}+{\mathrm{\Δl}}_r)h_r;$

$M_r=(\frac{d_r}{2}+{\mathrm{\Δl}}_r)[{(\frac{d_r}{2}+{\mathrm{\Δl}}_r)}^2+3{(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)}^2],$

$P_r=(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)[3{(\frac{d_r}{2}+{\mathrm{\Δl}}_r)}^2+{(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)}^2],$

${\mathrm{\α}}_{\mathrm{ar}}={\mathrm{\α}}_r(\frac{d_r}{2}+{\mathrm{\Δl}}_r),{\mathrm{\α}}_r=2\sqrt{15}/L_r,{\mathrm{\α}}_{\mathrm{br}}={\mathrm{\α}}_r(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r),$

$R_{\mathrm{abr}}=(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)(\frac{d_r}{2}+\mathrm{\Δ}{l}_r);$

b _1r＝[I(1,α _ar)K(0,α _ar)+I(0,α _ar)K(1,α _ar)]M _r，

b _2r＝-[I(1,α _br)K(0,α _ar)+I(0,α _ar)K(1,α _br)]P _r，

$b_{3r}={\mathrm{\α}}_r{R}_{\mathrm{abr}}[I(1,{\mathrm{\α}}_{\mathrm{ar}})K(1,{\mathrm{\α}}_{\mathrm{br}})-I(1,{\mathrm{\α}}_{\mathrm{br}})K(1,{\mathrm{\α}}_{\mathrm{ar}})][{(\frac{d_r}{2}+{\mathrm{\Δl}}_r)}^2+N_r\ln (\frac{d_r}{2}+{\mathrm{\Δl}}_r)];$

Bessel correction function: I (0, α _br), K (0, α _br), I (1, α _br), K (1, α _br);

I(1,α _ar)，K(1,α _ar)，I(0,α _ar)，K(0,α _ar)；

(5) roll angular rigidity of forward and backward stabilizer bar system with check calculate:

According to the wheelspan B of the forward and backward bridge of vehicle _fand B _r, the diameter d of forward and backward suspension stabilizer bar _fand dr, length l _cfand l _cr, forward and backward stabilizer bar rubber bushing mounting distance length l _0fand l _0r, the deformation coefficient G at the end points place of the resulting forward and backward stabilizer bar of calculating in step (3) _wfand G _wr, the radial rigidity K of the resulting forward and backward rubber bushing of calculating in step (4) _xfand K _xr, the roll angular rigidity to forward and backward stabilizer bar system with check respectively calculating, that is:

(6) gross vehicle roll angular rigidity check calculate:

According to the roll angular rigidity of resulting forward and backward bearing spring in step (2) with the roll angular rigidity of resulting forward and backward stabilizer bar system in step (5) with gross vehicle roll angular rigidity is checked to calculating,

If the check calculated value of gross vehicle roll angular rigidity be more than or equal to the desired design load of the resulting vehicle of calculating in step (1) ? vehicle roll angular rigidity meets vehicle designing requirement.Otherwise, if vehicle roll angular rigidity does not meet vehicle designing requirement, need to carry out adjusted design to fore suspension and rear suspension stabilizer bar system.

The present invention has advantages of than prior art:

Check for vehicle suspension roll angular rigidity home and abroad at present, be mostly to utilize simulation software, by modeling and simulating, vehicle roll angular rigidity is carried out to check analysis, but the method can not provide analytical formula, therefore, can not meet the requirement of the CAD software development of stabilizer bar system.The present invention can be according to vehicle parameter and suspension parameter, and the structural parameters of designed stabilizer bar and rubber bushing and material characteristic parameter, utilizes the roll angular rigidity of forward and backward bearing spring, the deformation coefficient G of end part of stabilizer rod _wand rubber bushing radial rigidity K _xanalytical formula, forward and backward suspension roll angular rigidity and gross vehicle roll angular rigidity are checked to calculating.The method can obtain the check calculated value of vehicle roll angular rigidity accurately and reliably, checks computing method, and established technical foundation for stabilizer bar CAD software development for the design of vehicle suspension and stabilizer bar provides reliable roll angular rigidity.Utilize the method, not only can improve design level and the quality of vehicle suspension and stabilizer bar system, improve ride performance and the security of vehicle; Meanwhile, utilize the method can reduce design and testing expenses, accelerate product development speed.

In order to understand better invention, below in conjunction with accompanying drawing, be described further.

Fig. 1 is that vehicle suspension roll angular rigidity is checked the process flow diagram calculating;

Fig. 2 is vehicle roll motion model figure;

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

Fig. 4 is the structural representation of rubber bushing.

Specific embodiments

Below by embodiment, the present invention is described in further detail.

Embodiment mono-: the body quality m of certain vehicle _s=4690kg, side acceleration a _y=0.4g, the distance h between vehicle body barycenter and roll axis _s=1069mm, the designing requirement value of vehicle roll angle front suspension pendulum arm length T _1f=675mm, spring wire rigidity k _sf=102.45N/mm, spring center is to distance T between swing arm hinge point _2f=430mm; Rear suspension pendulum arm length T _1r=650mm, spring wire stiffness K _rs=261N/mm, spring center is to distance T between swing arm hinge point _2r=400mm; This vehicle front axle wheel is apart from B _f=1650mm, rear axle wheel is apart from B _r=1485mm; This vehicle only arranges stabilizer bar system at front suspension, its structural representation as shown in Figure 3, wherein, the straight d of stabilizer bar _f=20mm, total length l _cf=800mm, brachium, l _1f=150mm, transition arc radius R _f=50mm, transition arc central angle θ _f=60o, the mounting distance l between two rubber bushings _0f=400mm, the elastic modulus E=210GPa of stabilizer bar material, Poisson ratio μ=0.3.The structure of front stabilizer rubber bushing as shown in Figure 4, stabilizer 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, and by interior round buss 2 and stabilizer bar 1, matching to merge is arranged on stabilizer 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. the inner circle radius r of rubber bushing _a=12mm, exradius r _b=22mm, elastic modulus E _x=7.84MPa, Poisson ratio μ _x=0.47.The roll angular rigidity of this vehicle suspension system is checked to calculating.

The check method of the vehicle suspension roll angular rigidity that example of the present invention provides, it checks calculation process as shown in Figure 1, and concrete steps are as follows:

(1) the needed total roll angular rigidity of vehicle suspension calculating:

According to automobile body quality m _s=4690kg, side acceleration a _y=0.4g, the distance h between vehicle body barycenter and roll axis _s=1069mm, vehicle roll angle ignoring in unsprung mass mu situation, the needed total roll angular rigidity of vehicle suspension is being calculated, that is:

(2) roll angular rigidity of the forward and backward bearing spring of vehicle with calculating:

According to the front tread B of vehicle _f=1650mm and rear tread B _r=1485mm, front swing arm length T _1f=675mm, rear-swing arm length T _1r=650mm, forward and backward bearing spring installation site is to the distance T between swing arm hinge point _2f=430mm and T _2r=400mm, and the line rigidity k of forward and backward bearing spring _sf=102.45N/mm and k _sr=261N/mm, calculates respectively the roll angular rigidity of forward and backward bearing spring, that is:

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

According to the total length l of front suspension stabilizer 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 θ of transition arc _f=60., elastic properties of materials model E=210GPa and Poisson ratio μ=0.3, calculate respectively the deformation coefficient of front suspension end part of stabilizer rod, that is:

$G_{\mathrm{wf}}=\frac{Q_{1f}-Q_{2f}+Q_{3f}+Q_{4f}+Q_{5f}-Q_{6f}}{\mathrm{\πE}}=1.5935\×{10}^{-12}{m}^5/N;$

In formula, $Q_{1f}=\frac{{64l}_{1f}^3}{3}={0.0720m}^3,$

$Q_{2f}=\frac{64[{(l_{1f}\cos {\mathrm{\θ}}_f+R_f{\sin \mathrm{\θ}}_f)}^3+\frac{1}{8}{(l_{0f}-l_{\mathrm{cf}})}^3]}{3}={-0.1353m}^3,$

$Q_{3f}={64R}_f[\frac{1}{2}{l}_{1f}^2({\mathrm{\θ}}_f+\frac{{\sin 2\mathrm{\θ}}_f}{2})+\frac{1}{2}{R}_f^2({\mathrm{\θ}}_f-\frac{{\sin 2\mathrm{\θ}}_f}{2})+l_{1f}{R}_f{\sin }^2{\mathrm{\θ}}_f]={0.0737m}^3,$

$Q_{4f}=\frac{{8l}_{0f}{(l_{0f}-l_{\mathrm{cf}})}^2}{3}={0.1707m}^3,$

$Q_{5f}={64R}_f(\mathrm{\μ}+1)[{R_f}^2(\frac{{3\mathrm{\θ}}_f}{2}+\frac{{\sin 2\mathrm{\θ}}_f}{4}-{2\sin \mathrm{\θ}}_f)+\frac{1}{2}{l}_{1f}^2({\mathrm{\θ}}_f-\frac{{\sin 2\mathrm{\θ}}_f}{2})+{4l}_{1f}{R}_f{\sin }^4\frac{{\mathrm{\θ}}_f}{2}]={0.0371m}^3,$

Q _6f＝32(u+1)[R _f(cosθ _f-1)-l _1fsinθ _f] ²[2l _1fcosθ _f-l _cf+2R _fsinθ _f]＝-0.5624m ³；

(4) front suspension stabilizer bar rubber bushing radial line stiffness K _xfanalytical Calculation:

According to the inner circle radius r of front suspension stabilizer 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 ratio _{μ x}=0.47, the radial line stiffness K to front suspension rubber bushing _xfcalculate, that is:

$K_{\mathrm{xf}}=\frac{1}{u\left(r_{\mathrm{bf}}\right)+y\left(r_{\mathrm{bf}}\right)}=2106.8N/\mathrm{mm};$

Wherein, $u\left(r_b\right)=\frac{1+{\mathrm{\μ}}_x}{{2\mathrm{\πE}}_{x}L}(\ln \frac{r_b}{r_a}-\frac{r_b^2-r_a^2}{r_a^2+r_b^2})={7.7271\×10}^{-5}\mathrm{mm}/N;$

$y\left(r_b\right)=a_{1}I(0,{\mathrm{\αr}}_b)+a_{2}K(0,{\mathrm{\αr}}_b)+a_3+\frac{1+{\mathrm{\μ}}_x}{{5\mathrm{\πE}}_{x}L}(\ln r_b+\frac{r_b^2}{r_a^2+r_b^2})={3.9738\×10}^{-4}\mathrm{mm}/N,$

$a_1=\frac{(1+{\mathrm{\μ}}_x)[K(1,{\mathrm{\αr}}_a)r_a(r_a^2+{3r}_b^2)-K(1,{\mathrm{\αr}}_b)r_b({3r}_a^2+r_b^2)]}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_2)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)}={-3.3642\×10}^{-11},$

$a_2=\frac{({\mathrm{\μ}}_x+1)[I(1,{\mathrm{\αr}}_a)r_a(r_a^2+{3r}_b^2)-I(1,{\mathrm{\αr}}_b)r_b({3r}_a^2+r_b^2)}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)}={3.3376\×10}^{-12},$

$a_3=-\frac{(1+{\mathrm{\μ}}_x)(b_1-b_2+b_3)}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)}={1.8565\×10}^{-6},$

$b_1=[I(1,{\mathrm{\αr}}_a)K(0,{\mathrm{\αr}}_a)+K(1,{\mathrm{\αr}}_a)I(0,{\mathrm{\αr}}_a)]r_a(r_a^2+{3r}_b^2)={5.1511\×10}^{-6},$

$b_2=-[I(1,{\mathrm{\αr}}_b)K(0,{\mathrm{\αr}}_a)+K(1,{\mathrm{\αr}}_b)I(0,{\mathrm{\αr}}_a)]r_b(r_b^2+{3r}_b^2){=-4.065\×10}^{-5},$

$b_3={\mathrm{\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)][r_a^2+(r_a^2+r_b^2)\ln r_a]={4.877\×10}^{-4};$

Bessel correction function:

I(0,αr _b)＝25.0434，K(0,αr _b)＝0.0041，

I(1,αr _b)＝22.3175，K(1,αr _b)＝0.0045，

I(1,αr _a)＝2.1439，K(1,αr _a)＝0.0922，

I(0,αr _a)＝2.8801，K(0,αr _a)＝0.0769，

$\mathrm{\α}=2\sqrt{15}/L=309.8387;$

(5) roll angular rigidity of front stabilizer system check calculate:

According to the wheelspan B of vehicle propons _f=1650mm, the diameter d of stabilizer bar _f=20mm, length l _cf=800mm, the mounting distance l between two rubber bushings _0f=400mm, the deformation coefficient G at calculating resulting front stabilizer end points place in step (3) _wf=1.5935 * 10 ^-12m ⁵/ N, and the middle radial rigidity K that calculates resulting front rubber bushing of step (4) _xfthe roll angular rigidity of=2106.8N/mm to front stabilizer system check calculating, that is:

(6) gross vehicle roll angular rigidity check calculate:

According to the roll angular rigidity of the forward and backward bearing spring calculating in step (2) with in step (5), calculate the roll angular rigidity of resulting front stabilizer system gross vehicle roll angular rigidity is checked to calculating,

The check calculated value of known this gross vehicle roll angular rigidity be greater than the desired design load of the resulting vehicle of calculating in step (1) ? this vehicle roll angular rigidity meets vehicle designing requirement.

Embodiment bis-: known certain automobile body quality m _s=5000kg, side acceleration a _y=0.4g, the distance h between vehicle body barycenter and roll axis _s=1150mm, vehicle roll angle the spring wire rigidity k of vehicle front suspension _sf=90.761N/mm, the spring wire rigidity k of rear suspension _sr=176.23N/mm; Front tread B _f=1650mm, rear tread B _r=1600mm; Front suspension pendulum arm length T _1f=660mm, front suspension spring center is to distance T between homonymy swing arm hinge point _2f=450mm; Rear-swing arm length T _1r=650mm, rear suspension spring mounting center is to distance T between homonymy transverse arm pin joint _2r=400mm.Forward and backward suspension stabilizer bar and rubber bushing, except diameter is not identical, other parameter all with execute identical in example one, wherein, front stabilizer diameter d _f=22mm, rear stabilizer bar diameter d _r=19mm; The inside radius r of front rubber bushing _af=13mm, external radius r _bf=23mm, the inner circle radius r of rear rubber bushing _ar=11.5mm.Exradius r _br=21.5mm.The roll angular rigidity of this vehicle suspension system is checked to calculating.

The step of example one is executed in employing, the roll angular rigidity of this vehicle suspension system is checked to calculating, that is:

(1) the needed total roll angular rigidity of vehicle suspension calculating:

According to automobile body quality m _s=5000kg, side acceleration a _y=0.4g, the distance h between vehicle body barycenter and roll axis _s=1150mm, the desired max. roll of vehicle body design the needed total roll angular rigidity of vehicle suspension is calculated, that is:

In formula, g acceleration of gravity, g=9.8m/s ²;

(2) roll angular rigidity of the forward and backward bearing spring of vehicle with calculating:

According to the front tread B of vehicle _f=1650mm and rear tread B _r=1600mm, front swing arm length T _1f=660mm, rear-swing arm length T _1r=650mm, forward and backward bearing spring installation site is to the distance T between swing arm hinge point _2f=450mm and T _2r=400mm, and the line rigidity k of forward and backward bearing spring _sf=90.761N/mm and k _sr=176.23N/mm, calculates respectively the roll angular rigidity of forward and backward bearing spring, that is:

(3) the deformation coefficient G of forward and backward suspension end part of stabilizer rod _wfand G _wrcalculating:

Because the forward and backward suspension stabilizer bar of this car is except diameter, other parameter is identical with embodiment mono-all, therefore, the deformation coefficient of this car fore suspension and rear suspension end part of stabilizer rod, also identical with embodiment mono-all, that is:

$G_{\mathrm{wr}}=G_{\mathrm{wf}}=\frac{Q_{1f}-Q_{2f}+Q_{3f}+Q_{4f}+Q_{5f}-Q_{6f}}{\mathrm{\πE}}=1.5935\×{10}^{-12}{m}^5/N;$

(4) forward and backward suspension stabilizer bar rubber bushing radial line stiffness K _xfand K _xranalytical Calculation:

Because interior radius of circle and the exradius of forward and backward suspension stabilizer bar diameter and rubber bushing are not identical, therefore, the radial line stiffness K of rubber bushing _xfand K _xrnot identical yet.According to the inside radius r of front rubber bushing _af=13mm, external 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 ratio μ _x=0.47, adopt the computing method of embodiment mono-to forward and backward suspension stabilizer bar rubber bushing radial line stiffness K _xfand K _xrcalculate respectively, that is:

$X_{\mathrm{xf}}=\frac{1}{u\left(r_{\mathrm{bf}}\right)+y\left(r_{\mathrm{bf}}\right)}=2267.0N/\mathrm{mm};$

$X_{\mathrm{xr}}=\frac{1}{u\left(r_{\mathrm{br}}\right)+y\left(r_{\mathrm{br}}\right)}=2026.7N/\mathrm{mm};$

Wherein, $u\left(r_{\mathrm{bf}}\right)=\frac{(1+{\mathrm{\μ}}_x)}{{2\mathrm{\πE}}_{x}L}(\ln \frac{r_{\mathrm{bf}}}{r_{\mathrm{af}}}-\frac{r_{\mathrm{bf}}^2-r_{\mathrm{af}}^2}{r_{\mathrm{af}}^2+r_{\mathrm{bf}}^2})={6.54\×10}^{-5}\mathrm{mm}/N;$

$y\left(r_{\mathrm{bf}}\right)=a_{1}I(0,{\mathrm{\αr}}_{\mathrm{bf}})+a_{2}K(0,{\mathrm{\αr}}_{\mathrm{bf}})+a_3+\frac{(1+{\mathrm{\μ}}_x)}{{5\mathrm{\πE}}_{x}L}(\ln r_{\mathrm{bf}}+\frac{r_{\mathrm{bf}}^2}{r_{\mathrm{af}}^2+r_{\mathrm{bf}}^2})={3.757\×10}^{-4}\mathrm{mm}/N,$

$u\left(r_{\mathrm{br}}\right)=\frac{(1+{\mathrm{\μ}}_x)}{{2\mathrm{\πE}}_{x}L}(\ln \frac{r_{\mathrm{br}}}{r_{\mathrm{ar}}}-\frac{r_{\mathrm{br}}^2-r_{\mathrm{ar}}^2}{r_{\mathrm{ar}}^2+r_{\mathrm{br}}^2})={8.429\×10}^{-5}\mathrm{mm}/N;$

$y\left(r_{\mathrm{br}}\right)=a_{1}I(0,{\mathrm{\αr}}_{\mathrm{br}})+a_{2}K(0,{\mathrm{\αr}}_{\mathrm{br}})+a_3+\frac{(1+{\mathrm{\μ}}_x)}{{5\mathrm{\πE}}_{x}L}(\ln r_{\mathrm{br}}+\frac{r_{\mathrm{br}}^2}{r_{\mathrm{ar}}^2+r_{\mathrm{br}}^2})={4.091\×10}^{-4}\mathrm{mm}/N,$

(5) roll angular rigidity of forward and backward stabilizer bar system with check calculate:

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 stabilizer bar _f=22mm and d _r=19mm, length l _cf=l _ct=800mm, the mounting distance l between two rubber bushings _0f=l _0r=400mm, the deformation coefficient G at calculating resulting forward and backward stabilizer bar end points place in step (3) _wf=G _wr=1.5935 * 10 ^-12m ⁵/ N, and the middle radial rigidity K that calculates resulting forward and backward rubber bushing of step (4) _xf=2267.0N/mm and K _xr=2026.7N/mm, the roll angular rigidity to forward and backward stabilizer bar system with check respectively calculating, that is:

(6) gross vehicle roll angular rigidity check calculate:

According to the roll angular rigidity of the forward and backward bearing spring calculating in step (2) with in step (5), calculate the roll angular rigidity of resulting forward and backward stabilizer bar system with gross vehicle roll angular rigidity is checked to calculating,

The check calculated value of known this gross vehicle roll angular rigidity be greater than the desired design load of the resulting vehicle of calculating in step (1) ? this vehicle roll angular rigidity meets vehicle designing requirement.

Embodiment tri-: certain vehicle is except front suspension stabilizer bar diameter d _fand the inner circle radius r of rubber bushing _afand exradius r _bfoutside not identical, other all parameters are identical with embodiment's bis-, wherein, and the diameter d of front stabilizer _f=21mm, the inner circle radius r of rubber bushing _af=12.5mm, and exradius r _bf=22.5mm.The roll angular rigidity of this vehicle suspension system is checked to calculating.

The step of example one is executed in employing, and the roll angular rigidity of this vehicle suspension system is checked to calculating:

(1) the needed total roll angular rigidity of vehicle suspension calculating:

Identical due to vehicle parameter and embodiment bis-, therefore, the needed total roll angular rigidity of this vehicle suspension, also identical with embodiment bis-, that is:

(2) roll angular rigidity of the forward and backward bearing spring of vehicle with calculating:

Due to vehicle parameter and the suspension stiffness of this vehicle, identical with embodiment bis-, therefore, and the roll angular rigidity of the forward and backward bearing spring of vehicle, also identical with embodiment bis-respectively, that is:

(3) the deformation coefficient G of forward and backward suspension end part of stabilizer rod _wfand G _wrcalculating:

Because the forward and backward suspension stabilizer bar of this car is except diameter, other parameter is identical with embodiment bis-all, therefore, the deformation coefficient of the forward and backward suspension end part of stabilizer rod of this car, also identical with embodiment bis-all, that is:

$G_{\mathrm{wr}}=G_{\mathrm{wf}}=\frac{Q_{1f}-Q_{2f}+Q_{3f}+Q_{4f}+Q_{5f}-Q_{6f}}{\mathrm{\πE}}=1.5935\×{10}^{-12}{m}^5/N;$

(4) forward and backward suspension stabilizer bar rubber bushing radial line stiffness K _xfand K _xranalytical Calculation:

Due to interior radius of circle and the exradius of rear suspension stabilizer bar diameter and rubber bushing, identical with embodiment bis-, therefore, rear suspension stabilizer bar rubber bushing radial line stiffness K _xralso identical with embodiment bis-all, that is:

$X_{\mathrm{xr}}=\frac{1}{u\left(r_{\mathrm{br}}\right)+y\left(r_{\mathrm{br}}\right)}=2026.7N/\mathrm{mm};$

Interior radius of circle and the exradius of this vehicle front suspension stabilizer bar diameter and rubber bushing, not identical with embodiment bis-.According to the inside radius r of front rubber bushing _af=12.5mm, external radius r _bf=22.5mm, elastic modulus E _x=7.84MPa, Poisson ratio μ _x=0.47, adopt the computing method of embodiment mono-to front suspension stabilizer bar rubber bushing radial line stiffness K _xfcalculate, that is:

$K_{\mathrm{xf}}=\frac{1}{u\left(r_{\mathrm{bf}}\right)+y\left(r_{\mathrm{bf}}\right)}=2186.9N/\mathrm{mm};$

Wherein, $u\left(r_{\mathrm{bf}}\right)=\frac{(1+{\mathrm{\μ}}_x)}{{2\mathrm{\πE}}_{x}L}(\ln \frac{r_{\mathrm{bf}}}{r_{\mathrm{af}}}-\frac{r_{\mathrm{bf}}^2-r_{\mathrm{af}}^2}{r_{\mathrm{af}}^2+r_{\mathrm{bf}}^2})={7.1\×10}^{-5}\mathrm{mm}/N;$

$y\left(r_{\mathrm{bf}}\right)=a_{1}I(0,{\mathrm{\αr}}_{\mathrm{bf}})+a_{2}K(0,{\mathrm{\αr}}_{\mathrm{bf}})+a_3+\frac{(1+{\mathrm{\μ}}_x)}{{5\mathrm{\πE}}_{x}L}(\ln r_{\mathrm{bf}}+\frac{r_{\mathrm{bf}}^2}{r_{\mathrm{af}}^2+r_{\mathrm{bf}}^2})={3.863\×10}^{-4}\mathrm{mm}/N,$

(5) roll angular rigidity of forward and backward stabilizer bar system with check calculate:

Identical due to this vehicle rear suspension stabilizer bar diameter and rubber bushing and embodiment bis-, therefore, the roll angular rigidity of rear suspension stabilizer bar system also identical with embodiment bis-, that is:

According to the wheelspan B of vehicle propons _f=1650mm, the diameter d of stabilizer bar _f=21mm, length l _cf=800mm, the mounting distance l between two rubber bushings _0f=400mm, the deformation coefficient G at calculating resulting front stabilizer end points place in step (3) _wf=1.5935 * 10 ^-12m ⁵/ N, and the middle radial rigidity Kxf=2186.9N/mm that calculates resulting forward and backward rubber bushing of step (4), the roll angular rigidity to front stabilizer system check calculating, that is:

(6) gross vehicle roll angular rigidity check calculate:

According to the roll angular rigidity of the forward and backward bearing spring calculating in step (2) with in step (5), calculate the roll angular rigidity of resulting forward and backward stabilizer bar system with gross vehicle roll angular rigidity is checked to calculating,

The check calculated value of known this gross vehicle roll angular rigidity be less than the desired design load of the resulting vehicle of calculating in step (1) ? this vehicle roll angular rigidity does not meet vehicle designing requirement.

Front suspension stabilizer bar diameter is carried out to adjusted design, and the diameter adjusted design that is about to front suspension stabilizer bar is d _f=22mm, can be increased to the roll angular rigidity of vehicle meet the designing requirement value of vehicle roll angular rigidity; Also can carry out adjusted design to the mounting distance between front suspension stabilizer bar two rubber bushings, by l between two rubber bushings ₀mounting distance adjusted design be l ₀=420mm, adds installing space and increases 10mm, and the roll angular rigidity of vehicle can be increased to meet the designing requirement value of vehicle roll angular rigidity.

Claims (1)

1. the check method of vehicle suspension roll angular rigidity, its concrete steps are as follows:

(1) the needed total roll angular rigidity of vehicle suspension calculating:

According to automobile body quality m _s, the distance h between vehicle body barycenter and roll axis _s, radius of wheel r, side acceleration a _y, and vehicle designs desired vehicle body max. roll the needed total roll angular rigidity of vehicle suspension is calculated, that is:

Wherein, g is acceleration of gravity;

(2) roll angular rigidity of forward and backward bearing spring with calculating:

According to the front tread B of vehicle _fwith rear tread B _r, front swing arm length T _1f, rear-swing arm length T _1r, forward and backward bearing spring installation site is to the distance T between swing arm hinge point _2fand T _2r, and the line rigidity k of forward and backward bearing spring _srand k _sr, the roll angular rigidity of forward and backward bearing spring is calculated respectively, that is:

(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 stabilizer bar _cfand l _cr, the mounting distance l of middle two rubber bushings _0fand l _0r, the brachium l of forward and backward stabilizer bar _1fand l _1r, the transition arc radius R of forward and backward stabilizer bar _fand R _r, the central angle θ of transition arc _fand θ _r, and elastic properties of materials model E and Poisson ratio μ, the deformation coefficient G to forward and backward suspension end part of stabilizer rod vertical deviation _wfand G _wrcalculate, be respectively:

$G_{\mathrm{wf}}=\frac{Q_{1f}-Q_{2f}+Q_{3f}+Q_{4f}+Q_{5f}-Q_{6f}}{\mathrm{\πE}},$

$G_{\mathrm{wr}}=\frac{Q_{1r}-Q_{2r}+Q_{3r}+Q_{4r}+Q_{5r}-Q_{6r}}{\mathrm{\πE}},$

In formula, $Q_{1f}=\frac{{64l}_{1f}^3}{3},Q_{2f}=\frac{64[{(l_{1f}\cos {\mathrm{\θ}}_f+R_f{\sin \mathrm{\θ}}_f)}^3+\frac{1}{8}{(l_{0f}-l_{\mathrm{cf}})}^3]}{3},$

$Q_{3f}={64R}_f[\frac{1}{2}{l}_{1f}^2({\mathrm{\θ}}_f+\frac{{\sin 2\mathrm{\θ}}_f}{2})+\frac{1}{2}{R}_f^2({\mathrm{\θ}}_f-\frac{{\sin 2\mathrm{\θ}}_f}{2})+l_{1f}{R}_f{\sin }^2{\mathrm{\θ}}_f],Q_{4f}=\frac{{8l}_{0f}{(l_{0f}-l_{\mathrm{cf}})}^2}{3},$

$Q_{5f}={64R}_f(\mathrm{\μ}+1)[{R_f}^2(\frac{{3\mathrm{\θ}}_f}{2}+\frac{{\sin 2\mathrm{\θ}}_f}{4}-{2\sin \mathrm{\θ}}_f)+\frac{1}{2}{l}_{1f}^2({\mathrm{\θ}}_f-\frac{{\sin 2\mathrm{\θ}}_f}{2})+{4l}_{1f}{R}_f{\sin }^4\frac{{\mathrm{\θ}}_f}{2}],$

Q _6f＝32(μ+1)[R _f(cosθ _f-1)-l _1fsinθ _f] ²[2l _1fcosθ _f-l _cf+2R _fsinθ _f]；

$Q_{1r}=\frac{{64l}_{1r}^3}{3},Q_{2r}=\frac{64[{(l_{1r}\cos {\mathrm{\θ}}_r+R_r{\sin \mathrm{\θ}}_r)}^3+\frac{1}{8}{(l_{0r}-l_{\mathrm{rc}})}^3]}{3},$

$Q_{3r}={64R}_r[\frac{1}{2}{l}_{1r}^2({\mathrm{\θ}}_r+\frac{{\sin 2\mathrm{\θ}}_r}{2})+\frac{1}{2}{R}_r^2({\mathrm{\θ}}_r-\frac{{\sin 2\mathrm{\θ}}_r}{2})+l_{1r}{R}_r{\sin }^2{\mathrm{\θ}}_r],Q_{4r}=\frac{{8l}_{0r}{(l_{0r}-l_{\mathrm{cr}})}^2}{3},$

$Q_{5r}={64R}_r(\mathrm{\μ}+1)[R_r^2(\frac{{3\mathrm{\θ}}_{\text{r}}}{2}+\frac{\sin 2{\mathrm{\θ}}_r}{4}-{2\sin \mathrm{\θ}}_r)+\frac{1}{2}{l}_{1r}^2({\mathrm{\θ}}_r-\frac{{\sin 2\mathrm{\θ}}_r}{2})+{4l}_{1r}{R}_r{\sin }^4\frac{{\mathrm{\θ}}_r}{2}],$

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 stabilizer bar rubber bushing radial line 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 _braxial length L _fand L _r, and the elastic modulus E of rubber bushing _x, Poisson ratio μ _x, the radial line rigidity of forward and backward rubber bushing of hanger bracket is calculated respectively, that is:

$K_{\mathrm{xf}}=\frac{1}{u_r\left(r_{\mathrm{bf}}\right)+y\left(r_{\mathrm{bf}}\right)},K_{\mathrm{xr}}=\frac{1}{u_r\left(r_{\mathrm{br}}\right)+y\left(r_{\mathrm{br}}\right)};$

Wherein, $u_r\left(r_b\right)=\frac{1+{\mathrm{\μ}}_x}{{2\mathrm{\πE}}_{x}L}(\ln \frac{r_b}{r_a}-\frac{r_b^2-r_a^2}{r_a^2+r_b^2}),$

$y\left(r_b\right)=a_{1}I(0,{\mathrm{\αr}}_b)+a_{2}K(0,{\mathrm{\αr}}_b)+a_3+\frac{1+{\mathrm{\μ}}_x}{{5\mathrm{\πE}}_{x}L}(\ln r_b+\frac{r_b^2}{r_a^2+r_b^2}),$

$a_1=\frac{(1+{\mathrm{\μ}}_x)[K(1,{\mathrm{\αr}}_a)r_a(r_a^2+{3r}_b^2)-K(1,{\mathrm{\αr}}_b)r_b({3r}_a^2+r_b^2)]}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_2)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)},$

$a_2=\frac{({\mathrm{\μ}}_x+1)[I(1,{\mathrm{\αr}}_a)r_a(r_a^2+{3r}_b^2)-I(1,{\mathrm{\αr}}_b)r_b({3r}_a^2+r_b^2)}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)},$

$a_3=-\frac{(1+{\mathrm{\μ}}_x)(b_1-b_2+b_3)}{{5\mathrm{\πE}}_x{\mathrm{L\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)](r_a^2+r_b^2)};$

$b_1=[I(1,{\mathrm{\αr}}_a)K(0,{\mathrm{\αr}}_a)+K(1,{\mathrm{\αr}}_a)I(0,{\mathrm{\αr}}_a)]r_a(r_a^2+{3r}_b^2),$

$b_2=[I(1,{\mathrm{\αr}}_b)K(0,{\mathrm{\αr}}_a)+K(1,{\mathrm{\αr}}_b)I(0,{\mathrm{\αr}}_a)]r_b(r_b^2+{3r}_b^2),$

$b_3={\mathrm{\αr}}_a{r}_b[I(1,{\mathrm{\αr}}_a)K(1,{\mathrm{\αr}}_b)-K(1,{\mathrm{\αr}}_a)I(1,{\mathrm{\αr}}_b)][r_a^2+(r_a^2+r_b^2)\ln r_a],$

$\mathrm{\α}=2\sqrt{15}/L,$

Bessel correction function I (0, α r _b), K (0, α r _b), I (1, α r _b), K (1, α r _b),

I(1,αr _a)，K(1,αr _a)，I(0,αr _a)，K(0,αr _a)；

I(0,α _af)，K(0,α _af)；

Wherein, $u\left(d_r\right)=\frac{1+{\mathrm{\μ}}_x}{{2\mathrm{\πE}}_x{L}_r}(\ln \frac{d_r+{2h}_r+{2\mathrm{\Δl}}_r}{d_r+{2\mathrm{\Δl}}_r}-\frac{(d_r+{2\mathrm{\Δl}}_r+h_r)h_r}{N_r})$

$y\left(d_r\right)=a_{1r}I(0,{\mathrm{\α}}_{\mathrm{br}})+a_{2r}K(0,{\mathrm{\α}}_{\mathrm{br}})+a_{3r}+\frac{1+{\mathrm{\μ}}_x}{{5\mathrm{\πE}}_x{L}_r}(\ln (\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)+\frac{{(d_r+{2h}_r+{2\mathrm{\Δl}}_r)}^2}{{4N}_r}),$

$a_{1r}=\frac{(1+{\mathrm{\μ}}_x)[K(1,{\mathrm{\α}}_{\mathrm{ar}})M_r-K(1,{\mathrm{\α}}_{\mathrm{br}})P_r]}{{5\mathrm{\πE}}_x{L}_r{\mathrm{\α}}_r{R}_{\mathrm{abr}}[I(1,{\mathrm{\α}}_{\mathrm{ar}})K(1,{\mathrm{\α}}_{\mathrm{br}})-I(1,{\mathrm{\α}}_{\mathrm{br}})K(1,{\mathrm{\α}}_{\mathrm{ar}})]N_r},$

$a_{2r}=\frac{({\mathrm{\μ}}_x+1)[I(1,{\mathrm{\α}}_{\mathrm{ar}})M_r-I(1,{\mathrm{\α}}_{\mathrm{br}})P_r]}{{5\mathrm{\πE}}_x{L}_r{\mathrm{\α}}_r{R}_{\mathrm{abr}}[I(1,{\mathrm{\α}}_{\mathrm{ar}})K(1,{\mathrm{\α}}_{\mathrm{br}})-I(1,{\mathrm{\α}}_{\mathrm{br}})K(1,{\mathrm{\α}}_{\mathrm{ar}})N_r},$

$a_{3r}=\frac{(1+{\mathrm{\μ}}_x)(b_{1r}+b_{2r}+b_{3r})}{{5\mathrm{\πE}}_x{L}_r{\mathrm{\α}}_r{R}_{\mathrm{abr}}[I(1,{\mathrm{\α}}_{\mathrm{ar}})K(1,{\mathrm{\α}}_{\mathrm{br}})-I(1,{\mathrm{\α}}_{\mathrm{br}})K(1,{\mathrm{\α}}_{\mathrm{ar}})]N_r},$

$N_r=2{(\frac{d_r}{2}+{\mathrm{\Δl}}_r)}^2+{h_r}^2+2(\frac{d_r}{2}+{\mathrm{\Δl}}_r)h_r;$

$M_r=(\frac{d_r}{2}+{\mathrm{\Δl}}_r)[{(\frac{d_r}{2}+{\mathrm{\Δl}}_r)}^2+3{(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)}^2],$

$P_r=(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)[3{(\frac{d_r}{2}+{\mathrm{\Δl}}_r)}^2+{(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)}^2],$

${\mathrm{\α}}_{\mathrm{ar}}={\mathrm{\α}}_r(\frac{d_r}{2}+{\mathrm{\Δl}}_r),{\mathrm{\α}}_r=2\sqrt{15}/L_r,{\mathrm{\α}}_{\mathrm{br}}={\mathrm{\α}}_r(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r),$

$R_{\mathrm{abr}}=(\frac{d_r}{2}+h_r+{\mathrm{\Δl}}_r)(\frac{d_r}{2}+\mathrm{\Δ}{l}_r);$

b _1r＝[I(1,α _ar)K(0,α _ar)+I(0,α _ar)K(1,α _ar)]M _r，

b _2r＝-[I(1,α _br)K(0,α _ar)+I(0,α _ar)K(1,α _br)]P _r，

$b_{3r}={\mathrm{\α}}_r{R}_{\mathrm{abr}}[I(1,{\mathrm{\α}}_{\mathrm{ar}})K(1,{\mathrm{\α}}_{\mathrm{br}})-I(1,{\mathrm{\α}}_{\mathrm{br}})K(1,{\mathrm{\α}}_{\mathrm{ar}})][{(\frac{d_r}{2}+{\mathrm{\Δl}}_r)}^2+N_r\ln (\frac{d_r}{2}+{\mathrm{\Δl}}_r)];$

Bessel correction function: I (0, α _br), K (0, α _br), I (1, α _br), K (1, α _br);

I(1,α _ar)，K(1,α _ar)，I(0,α _ar)，K(0,α _ar)；

(5) roll angular rigidity of forward and backward stabilizer bar system with check calculate:

According to the wheelspan B of the forward and backward bridge of vehicle _fand B _r, the diameter d of forward and backward suspension stabilizer bar _fand dr, length l _cfand l _cr, forward and backward stabilizer bar rubber bushing mounting distance length l _0fand l _0r, the deformation coefficient G at the end points place of the resulting forward and backward stabilizer bar of calculating in step (3) _wfand G _wr, the radial rigidity K of the resulting forward and backward rubber bushing of calculating in step (4) _xfand K _xr, the roll angular rigidity to forward and backward stabilizer bar system with check respectively calculating, that is:

(6) gross vehicle roll angular rigidity check calculate:

According to calculating the roll angular rigidity of resulting forward and backward bearing spring in step (2) with the roll angular rigidity of resulting forward and backward stabilizer bar system in step (5) with gross vehicle roll angular rigidity is checked to calculating, that is:

If the check calculated value of gross vehicle roll angular rigidity be more than or equal to the desired design load of the resulting vehicle of calculating in step (1) ? vehicle roll angular rigidity meets vehicle designing requirement; Otherwise, if vehicle roll angular rigidity does not meet vehicle designing requirement, need to carry out adjusted design to fore suspension and rear suspension stabilizer bar system.