Design and calculation of double arm suspension of a car

3.1. Major parameters to understand

1) Caster angle. Consider a wheel from the side view of a car. Inclination of the upper point of the steering axis towards vehicle frame is considered as a positive caster and away from frame is Negative caster [2].

Fig. 1Side view of a front left side wheel to define caster

2) Camber. Consider a car from the front view. Difference in distance between top end and bottom end of a pair of wheels coins the term camber.[2]

Fig. 2Positive camber (front view)

Fig. 3Negative camber (front view)

3) Toe-in and toe-out. Consider a pair of wheels from the top view. Difference in distance between front end and rear end of a pair of wheels coins the term toe in and toe out [2].

Fig. 4Toe-in (top view)

4) Kingpin inclination. Considering from the front view, the angular displacement of the steering axis coins the term King pin Inclination (KPI). It is also known as Steering Axis Inclination (SAI) [2].

Fig. 5Toe-out (top view)

3.2. Force consideration

– Static-force due to gravity, reaction force in wheels.

– Dynamic – longitudinal acceleration and hard braking resulting in dive, lateral forces resulting in rolling of vehicle, vertical gravitational force under dynamic condition of vehicle [3, 4].

3.3. Design consideration

– The shape and dimension of the control arm or wishbone depends on the available space and the stress withstanding capability (moment of inertia as a factor).

– The front arm can take “A” equivalent shape, the rear can take “A” or “H”-equivalent shape.

– The desired ground clearance was fixed GC = 10” (As per requirement).

– Track width: 1143 mm.

– Wheel base: 1524 mm.

– The roll centre axis and pitch centre axis are imaginary axis, they can be determined from the inclination of the control arms with respect to the roll cage and wheel assembly.

1) Roll Center. The axis about which the car tends to roll. This axis intersects the point at which the lines extended from the patch contact point of tyre to instantaneous points of both wheels meet.

Fig. 6Roll center fixing (front or rear view). Note: it should be near to centre of gravity for good cornering performance

2) Pitch center. The axis about which the car tends to dive or squat.

– Required jounce: 2”; required rebounce: 2” (As per requirements).

– Fixed roll center: 260 mm (from ground).

– Obtained center of gravity: 480 mm (from ground).

– Methods to find the center of gravity height:

a) Practical: either front or rear side of the vehicle have to be lifted to approximate height. Weighing machines have to be placed on both the front wheels and rear wheels of the vehicle. Weight before lifting and after lifting have to be noted and applied in the formula to get approx CG.

Formula = (level wheelbase×raised wheel base×added weight on front wheels)/(distance raised×total vehicle weight).

To find CG distance from rear axle = (weight at front×wheel base)/total weight.

b) CAD Method: the components must be designed assuming their mass and the position of Centre of gravity must be found in the complete assembled design of vehicle.

4.1. Stiffness calculation

Considering the vehicle to enter a corner at $V=$41.01 ft/s. The centripetal force due to friction:

The centrifugal force:

From Eq. (1) and Eq. (2) we get: $r=$ 74 ft ≅ 82 ft.

Considering longitudinal acceleration and drive torque to be zero, $T_D=$ 0 lbs-ft, $A_x=$ 0 g’s, and:

where $g$-in terms of ft/s²:

Condition: If the roads are banked. Longitudinal weight transfer is much smaller than lateral weight transfer so we lateral weight transfer.

The component of lateral acceleration along the axis of the car: $\alpha =$ 5, $A_{\gamma }=A_{\alpha }\mathrm{c}\mathrm{o}\mathrm{s}\alpha -\mathrm{s}\mathrm{i}\mathrm{n}\alpha =0.6369\mathrm{c}\mathrm{o}\mathrm{s}5°-\mathrm{s}\mathrm{i}\mathrm{n}5°$, $A_{\gamma }=$–0.5498 g’s (negative sign indicates deceleration), Note: $A_{\gamma }=A_{\alpha }$ (if road is level).

Position of Centre of Gravity. The cad model reference was taken for finding the position of C.G in side view, $a=$ 990 mm = 3.25 ft and $b=$ 533.4 mm = 1.75 ft.

Resulting in: $W_1=W_2=$ 214.825 lbs, $W_R=429.65\mathrm{}\left(W_1+W_2\right)$, $W_3=W_4=$ 115.675 lbs, $W_F=231.35\mathrm{}\left(W_3+W_4\right)$.

Effective weight transfer due to banking: $W^{\text{'}}=W\left(A_{\alpha }\mathrm{c}\mathrm{o}\mathrm{s}\alpha -\mathrm{s}\mathrm{i}\mathrm{n}\alpha \right)=661\left(0.6369\mathrm{c}\mathrm{o}\mathrm{s}5+\mathrm{s}\mathrm{i}\mathrm{n}5\right)$, $W\text{'}=$ 695.13 lbs, $W{\mathrm{\text{'}}}_R=$ 451.83 lbs, $W{\mathrm{\text{'}}}_F=$ 243.29 lbs.

Roll gradient: Assuming roll rate $K_{\varphi R}+K_{\varphi F}=$ 200 lbs-ft/°, $\varphi /A_{\gamma }=-W\mathrm{*}H/\left(K_{\varphi F}+K_{\varphi R}\right)=-661\mathrm{*}0.74/200\text{,}$$\varphi /A_{\gamma }=2.44/g$ (desired for high performance in corners), $K_{\varphi R}=$ 65 % of 200 = 130 lbs ft/° = 7448.45 lbs ft/rad, $K_{\varphi F}=$ 35 % of 200 = 70 lbs ft/° = 4010 lbs ft/rad.

Lateral load transfer due to lateral acceleration:

Load on individual wheel after the effect of lateral acceleration for inner and outer: $W_{Fo}=$ 243.29/2 + 53.24 = 174.885 lbs, $W_{Fi}=$ 24.29/2 – 53.24 = 68.40 lbs, $W_{Ro}=$ 451.83/2 + 98.89 = 324.805 lbs, $W_{Ri}=$ 451.83/2 – 98.89 = 127.025 lbs.

Change from static load to dynamic load: $W_{Fo}=$ 174.885 – 115.65 = 59.235 lbs, $W_{Fi}=$ 68.40 – 115.65 = – 47.25 lbs, $W_{Ro}=$ 324.80 – 214.825 = 109.0975 lbs, $W_{Ri}=$ 127.025 – 214.825 = –87.8 lbs.

Total braking force, consideration (varies vehicle to vehicle):

1) Mass $M=$ (sprung mass + unsprung mass) = 400 kg.

2) Speed = 45 km/hr.

3) Displacement $d=$ 3 m.

$K.E=$ 1/2$M{V}^2$= 1/2·400·12.5² = 3.12·10⁴ J, $F_B=$ workdone / $d$ = (3.12·10⁴) / 3 = 10400 N.

Effective load transfer due to braking: $W_{BF}=$ Static Load + Load Transfer effect = $W_F·a/l+F_B·H_{CG}/l=$ 231.35·990/1524 + 10400·480 / 1524 = 150.29+3275.5, $W_{BF}=$ 3425.8 N = 770.15 lbs, $W_{BF3}=W_{BF4}=$ 770.15/2 = 385 lbs.

The ride rate of the vehicle, considering ride travel 2 inches in front and 1.5 inches in the rear: $K_R=W$ – change/ride travel, $K_{RF}=\mathrm{}$59.325/2" lbs/inch, $K_{RF}=\mathrm{}$29.66 lbs/inch, $K_{RR}=$ 109.975/1.5" lbs/inch, $K_{RR}=$ 73.3 lbs/inch.

Considering longitudinal load transfer $K_{RF}^{\mathrm{\text{'}}}=$ (385 – 115.65) / 4" = 269.65 / 4" = 67.4125 lbs/inch.

Ride frequency:

To verify assumed roll rate:

Hence the assumed roll rate is proved.

The new roll rate $K_{\varphi }=$ 11822.92 lbs ft/rad.

For independent suspension the wheel center rate is, $K_W=\left(K_R\mathrm{*}{K}_T\right)/\left(K_T-K_R\right)$.

Considering, $K_T=\mathrm{}$2015 lbs/inch or 36 kg/mm (Standard value) [1-6], $K_{WF}=\mathrm{}$(67.41·2015) / (2015 – 29.66) = 69.74 lbs/inch, $K_{WR}=$ (73.3·2015) / (2015 – 73.3) = 76.06 lbs/inch.

Installation ratio. Consider the shock absorber to be mounted in the lower arm and the lower arm is parallel to the ground level.

Fig. 7Installation ratio (IR) and mounting angle determination, where L – load on the spring, θ –mounting angle of shock, R – normal reaction force on wheels, X – distance of mounting point from inner end of arm, Y – length of mounting control arm, A – point of control arm on roll cage, B – wheel centre point

Taking moment about “$A$” as zero:

After several iterations, considering the fact increasing “$IR$” makes the ride stiffer and decreasing “$IR$” increases spring force: $\theta =$ 60° and $X/Y=$ 0.68 is chosen, $L=$ 1252.62 N, $L\mathrm{c}\mathrm{o}\mathrm{s}\theta =$ 542.4 N, $L\mathrm{s}\mathrm{i}\mathrm{n}\theta =$ 1138.38 N.

Lateral force considering zero banking $F_L=A_a\mathrm{*}{W}_{ro}$, $F_L=$ 654.93 N.

Here the horizontal component “$L\mathrm{c}\mathrm{o}\mathrm{s}\theta$”and “$F_L$” constitutes two couples on the wheel, these forces produce two equal and opposite forces to them respectively on the upper half of the wheel pressing the upper arm inwards i.e. towards the chassis as shown in Fig. 9.

Considering vertical forces due to uneven roads “3g force is considered”, thus in case of coilover shock absorber, spring chosen or manufactured must be progressive spring or dual rate springs.

Fig. 8Skeleton sketch of Wishbones (WB) mounting

Thus the second rate of spring be, $K_{SF}^{\mathrm{\text{'}}}=K_{WF}/I{R}^2=$ 3·150.82 lbs/inch, $K_{SR}^{\mathrm{\text{'}}}=\mathrm{}{K}_{WR}/I{R}^2=$ 3·164.48 lbs/inch.

Critical damping $C_{Cr}=2\sqrt{\left[\left(KRF·W·12\right)/32.2\right]}$, $C_{Cr}=$ 107.8 lbs-s/ft.

For passenger cars: damping ratio $\zeta$ = 0.3, $C/C_{Cr}=\zeta$, 0.3 = $C$ / 107.8, coefficient of damping $C=\mathrm{}$32.34 lbs-s/ft.

4.2. In case of manufacturing a coilover shock absorber

Shear stress:

${\tau }_s=$ 410 N/mm² (considering F.O.S = 1.5) and $P=$1687.32 N, also:

We get $d^2-5.23d-10.25C=0$.

If $C=$5 then $d=$10 mm and $D=$ 50 mm. From the above equation assuming $C$. We can calculate $d$ and $D$.

Also, from:

From above equations we can determine the parameters of required spring parameters.

4.3. To choose Wishbone material

where $Q$ – first moment of area, $V$ – load, $t$ – thickness of pipe, $I=\left(\pi /4\right)\left(r_o^4-r_i^4\right)$.

Based on market availability and CAD analysis $r_o$ and $r_i$ values can be determined:

Fig. 9Centroid Point from the central axis of a pipe

Thickness of pipe $t=$ 1.5 mm, ${\tau }_s=\left(Q{V}_{max}\right)/It=$ 89.23 N/mm².

A prototype material “Chromoly” is taken for consideration. Based upon the $G$ value of chromoly, $G=$ 80 GPa (AISI 4130 ) = 80·10³ N/mm², shear stress/shear strain = $G$, shear strain = 89.23 / (80·10³) = 0.0011154 (Too small).

Hence the material selected undergoes very less deformation with a given load. The material chosen is perfectly suitable for Control arms.

5.1. Knuckle design

In order to achieve the required riding condition, the knuckle design plays a major role [13]. The fixing of the parameters such as Caster, Camber, Toe angles and King Pin Inclination majorly depends on the design of a knuckle. The dimension and shape of a knuckle is completely dependent on those parameters in order to obtain the efficient ride quality.

Fig. 10Knuckle design-front

5.1.2. Castor angle

Positive castor induces a aligning torque whereas negative castor reduces steering effort. Aligning torque is a favourable factor, so many designs go with a positive castor.

The angle depends on wheel diameter, i.e $t_A\alpha N$.

Caster angle of 4.43° is considered for the prototyping vehicle.

Fig. 11Positive caster

5.2. Selection and design of wishbones and its material

– There are plenty of materials available in many cross-sectional shapes. Each material has its own Young’s modulus value.

– The required material and shape are decided based on the strength of the material.

– The dimension of wishbones and its design is carried out based on the space available along with other important parameters such as Caster, Camber, KPI, roll center and pitch center.

– Wishbone mounting points: Designing the mounting points for the wishbones is an important part. Fixing must be carried out by considering the roll center, pitch center and RC member’s strength.

Fig. 12A-equivalent front lower wishbone design

Fig. 13H-equivalent rear lower wishbone design

The maximum shear stress was calculated and the material for wishbone was selected as Chromoly AISI 4130 seamless tubes (Not limited to this). The Diameters and minimal thickness can be determined with the help of software such as ANSYS.

5.2.1. Part analysis

To determine Diameter and Thickness required, Part analysis is carried out with the software called “ANSYS”. By fixing the maximum load and gradually decreasing the diameter and thickness of the pipe, a safer dimension could be found.

As a result of analysis with factor of safety:

– Diameter of the pipe: 25 mm.

– Thickness $t=$ 1.5mm.

Fig. 14 and Fig. 15 shows the analysis of designed wishbones via ANSYS Workbench software. That depicts that the calculated values to design wishbones are prominently best suited for the considered prototype vehicle.

Fig. 14a) Equivalent strain, b) maximum principal stress, c) total deformation

a)

b)

c)

Fig. 15a) Equivalent strain, b) maximum principal stress, c) total deformation

a)

b)

c)

5.3. Design and selection of shock absorbers

Based upon the required stiffness value of the vehicle, shock absorbers could be selected or manufactured as required among various types of shock absorbers.

To design a coilover shock absorber, the calculated Diameters ratio and material stiffness of the spring coil have to be considered as primary parameters.

Fig. 16Coilover shock absorber design

Based on the calculations a coilover shock absorber with damper is chosen.

– Mean coil diameter $D=$ 74 mm.

– Coil diameter $d=$ 11 mm.

– Number of turns of coil $N_t=N+3=$ 6 + 3.

– Solid length $L_s=$ 99 mm ($L_s=Nt*d$).

– Free length $L_f=$170 mm.