Suspension automotive

Spring Rates, Wheel Rates, Motion Ratios and Roll Stiffness

Appendix 1

ME 470 Vehicle Structural DesignME 470 Vehicle Structural DesignDr. Richard Hathaway, P.E., ProfessorDr. Richard Hathaway, P.E., Professor

Mechanical and Aeronautical EngineeringMechanical and Aeronautical Engineering

Spring Rate Calculations


■ Coil Spring Calculations:Coil Spring Calculations:

K = Spring Rate in lbs/in K = Spring Rate in lbs/in G = Modulus of rigidityG = Modulus of rigidity

d = Spring Wire Diameterd = Spring Wire Diameter R = Mean Radius of the SpringR = Mean Radius of the SpringN = Number of Active CoilsN = Number of Active Coils

Squared and Ground EndsSquared and Ground Ends -1.75 turns-1.75 turnsSquared or Closed EndsSquared or Closed Ends ---- ----Plain EndsPlain Ends -0.5 turns-0.5 turnsPlain ends GroundPlain ends Ground -1.0 turns-1.0 turns

N R 64dG

= K3

4

) + (1 2

E =G

µ


■ Coil Spring Calculations:Coil Spring Calculations:

◆ If Steel is used:If Steel is used: E = 30,000,000 E = 30,000,000 psipsi

N D

d 1,500,000 K

3

4

≅


■ Torsion Bar Rates:Torsion Bar Rates:

L = Bar Length L = Bar Length d = Bar d = Bar Diameter Diameter r = lever arm r = lever arm lengthlength

L

r

d

θ

JG

TL=θ

32

4dJ

π=

L

JGT =θ

L

GdT

32

4πθ

=

Let the deflection at the end = δ r

δθ =


■ Torsion Bar Rates:Torsion Bar Rates: L

r

d

θL

GdT

32

4πθ

=

Then the deflection rate at the free end is found

r

δθ =Since T = F x r &

L

Gd

r

rxF

32

4πδ =

krL

GdF ==2

4

32

πδ


■ The deflection rate at the free end is

L

r

d

θ

The deflection rate at the wheel can now be found through analysis of the motion ratio

krL

GdF ==2

4

32

πδ


■ Torsion Bar Calculations:Torsion Bar Calculations:

◆ If Steel is used:If Steel is used: E = 30,000,000 E = 30,000,000 psipsi

2

000,200,2

rLd K

4

≅

L = Bar Length L = Bar Length d = Bar Diameter d = Bar Diameter r = lever arm lengthr = lever arm length

Typical Leaf Spring

Typical Leaf Spring

Typical deflection behavior:

Typical Leaf Spring

Typical Path behavior on deflection

Motion Ratio Analysis



■ Spring PositionSpring Position

■ The displacement relationship between the spring and the The displacement relationship between the spring and the wheel determines the actual rate the wheel works against wheel determines the actual rate the wheel works against for any spring rate. This displacement relationship may be for any spring rate. This displacement relationship may be defined as a motion ratio. The rate at the wheel is defined defined as a motion ratio. The rate at the wheel is defined as the wheel rate (Kas the wheel rate (K

ww). The rate of the spring itself is ). The rate of the spring itself is called the spring rate (Kcalled the spring rate (K

ss). The displacement relationship ). The displacement relationship is a function of both spring position on the load carrying is a function of both spring position on the load carrying member and the angular orientation of the spring to that member and the angular orientation of the spring to that member.member.

■ Wheel Rate - Location Dependent.Wheel Rate - Location Dependent.

◆ The spring position is important as it defines the mechanical The spring position is important as it defines the mechanical advantage which exists between the wheel and the spring. Figure 1 advantage which exists between the wheel and the spring. Figure 1 depicts a spring acting on a simple lever.depicts a spring acting on a simple lever.


Figure 1

■ From the simple lever system a number of relationships From the simple lever system a number of relationships can be drawn.can be drawn.

b

a F = F AB

a

b d = d AB


2

=

=b

ak

ab

d

ba

F k=

dF

A

A

A

BB

B

■ Motion Ratio in the Road Vehicle.Motion Ratio in the Road Vehicle.◆ The motion ratio describes the displacement ratio between the The motion ratio describes the displacement ratio between the

spring and the centerline of the wheel. The motion ratio spring and the centerline of the wheel. The motion ratio squared times the spring rate gives the wheel rate.squared times the spring rate gives the wheel rate.


Figure 2

■ Using the previous analysis and Figure 2, the following Using the previous analysis and Figure 2, the following

apply.apply.

◆ The above analysis assumes minimal camber change at the wheel.The above analysis assumes minimal camber change at the wheel.

◆ The motion ratio can be determined experimentally and the The motion ratio can be determined experimentally and the measured distance ratio squared for an accurate value.measured distance ratio squared for an accurate value.

αCos2

2

sw b

a K = K

centerline wheelof travel vertical

axis springalong travel K = K

2

sw


Suspension Roll Stiffness

Suspension Roll Stiffness■ ROLL STIFFNESS due to wheel Rates:ROLL STIFFNESS due to wheel Rates:

◆ The roll stiffness (KThe roll stiffness (Kφφ) can be determined using elementary ) can be determined using elementary

analysis techniques. If the wheel rates (K) are determined analysis techniques. If the wheel rates (K) are determined and the spring spacing (t) is known then the roll stiffness and the spring spacing (t) is known then the roll stiffness relationship to spring stiffness follows.relationship to spring stiffness follows.

Note: t is equal Note: t is equal to the wheel track to the wheel track if the wheel rates if the wheel rates are usedare used

■ The torque to rotate the chassis about the roll axis is The torque to rotate the chassis about the roll axis is shown in the following equation.shown in the following equation.

■ For equal spring rates, left and right the above For equal spring rates, left and right the above equation reduces to the following.equation reduces to the following.

θ 2

t K

2

t +

2

t K

2

t = T RL

θ )K + K(

4t = T RL

2

θ(K) 2t = T

2


■ The roll stiffness is then as shown below.The roll stiffness is then as shown below.

■ For roll stiffness in N-m/DegFor roll stiffness in N-m/Deg

(K) 2t =

T = K

2

θφ

( )3.57•2

Kt = T

= K2

θφ

K = Individual wheel rate (N/m) t = track width (m)


■ In English units this can be reduced to Lb-Ft/DegIn English units this can be reduced to Lb-Ft/Deg

K

ftin

12 raddeg

57.3 2

t = (K) 2t = K

22

••φ


T = track width (in) K = Individual Wheel Rate (lb/in)

K 1375

t = K2

φ

■ The total roll stiffness KThe total roll stiffness Kφφ is equal to is equal to

K + K + K = KdevicesRsFst φφφφ

K φ F = Front Roll Stiffness K φR = Rear Roll Stiffness

K φ(devices) = Stabilizer etc contributions


Lateral Spring Center Position

■ The Spring Center to Cg distance (x) at either end of The Spring Center to Cg distance (x) at either end of

the vehicle is important.the vehicle is important.

)K + K(l)K+K( - t K = x

rl

lrlrcg-sc

l - t)K+K(

K = x lrl

rcg-sc

Which reduces to


◆ Then from Then from

■ The spring center to cg distance (x) is positive (to right of The spring center to cg distance (x) is positive (to right of cg) ifcg) if

K + K

l)K+K( - )tK( = x

rl

lrlrcg-sc

l)K+K(> tK if 0 > x lrlrcg-sc∴

lK < lK rrll


■ The location of the Cg from the inside wheel centerline, The location of the Cg from the inside wheel centerline, distance ldistance l

ll, at each axle can be found from the scale , at each axle can be found from the scale

weights at each wheel location.weights at each wheel location.

■ Then by substitution into equation 1 yields equation 6 Then by substitution into equation 1 yields equation 6 indicating the distance between the spring center (sc) and indicating the distance between the spring center (sc) and the center of gravity (cg).the center of gravity (cg).

t)W+W(

W = lrl

rl

t )W + W(

W - )K+K(

K = xrl

r

rl

rcg-sc


Roll Stiffness (Asymmetric Chassis)

■ Roll stiffness should be calculated using the distance Roll stiffness should be calculated using the distance from the instantaneous spring center to each of the from the instantaneous spring center to each of the wheel locations.wheel locations. ◆ The spring center location from the left tire The spring center location from the left tire

centerline is as shown.centerline is as shown.

◆ Therefore the roll stiffness for asymmetric Therefore the roll stiffness for asymmetric springing is,springing is,

t)K+K(

K = )x + l(rl

rcg-scl

)x-l(K + )x+l(K = k

2rr

2ll

3.57φ

Roll Stiffness (Asymmetric Chassis)

■ Recall, for equal spring rates,Recall, for equal spring rates,

( )3.57•2

Kt = T

= K2

θφ

K+K

K )K + K( t

= klr

r

2

rl2

3.57

φ

Then by substitutionThen by substitution becomes,

Example:

Symmetric Setup:

LRw = 175 lb/in

RRw = 175 lb/in

Track = 68 inches

deg

lbs-ft588 = K175

137568 = K

2

φφ

Roll Stiffness

Example:

Roll Stiffness

Asymmetric Setup:

LRw = 200 lb/in RRw = 175 lb/in

deg

lbs-ft 550 = K

688

200+175175

175) + (200 68

= k

2

2

φφ

Asymmetric Setup:

LRw = 200 lb/in RRw = 150 lb/in

Note: Avg = 175 lb/in Track = 68 inches

deg

lbs-ft 432 = K

688

200+150150

150) + (200 68

= k

2

2

φφ

◆ The rotational stiffness of the rear axle (kThe rotational stiffness of the rear axle (kφφ ax ax) due to the tire ) due to the tire

stiffness isstiffness is

◆ The rotational stiffness of the rear springs and rear stabilizer bar areThe rotational stiffness of the rear springs and rear stabilizer bar are

( )T

rtax K

x

txkK ==

3.572

2

φ

kt = tire stiffness (N/m)

tr = rear track width

kφ ax = Rotational stiffness (N-m/deg)

( )S

ssbsuspr K

x

txkkK =+=

3.572

2

φφ

ks = spring stiffness (N/m)

ts = rear spring spacing

kφb = Rear stabilizer bar (N-m/deg)

kφr susp = Rotational stiffness (N-m/deg)


■ The moment produced on the rear axle due to the tire stiffness isThe moment produced on the rear axle due to the tire stiffness is

■ The moment produced on the rear axle due to the springs and anti-The moment produced on the rear axle due to the springs and anti-roll bar isroll bar is

( )a

rtaaxt x

txkKM θθφ 3.572

2

−=−=

( ) ( ) ( )acss

bacsusprs x

txkkKM θθθθ φφ −

+=−=

3.572

2

θa = Axle roll angle θc = Chassis roll angle


■ If no stabilizer bar is present the front suspension springs and the If no stabilizer bar is present the front suspension springs and the tire stiffness can be combined as a series system of springs to tire stiffness can be combined as a series system of springs to determine an equivalent ride rate.determine an equivalent ride rate.

■ The rotational stiffness of the rear axle due to the tire stiffness isThe rotational stiffness of the rear axle due to the tire stiffness is

'

'

spt

sptride KK

KxKK

+=

( ) ( )

+=

3.572

22

x

txmrxkkK ffsp

bcsuspf φφ

2' mrxKK spsp =

■ If a stabilizer bar is present, the front springs and the stabilizer bar act together (parallel) to contribute to the stiffness, this is then translated to the tires.

mr = motion ratio


■ Combining chassis roll rate with the tire Combining chassis roll rate with the tire contributioncontribution

( ) ( )

+=

3.572

22

x

txmrxkkK ffsp

bcsuspf φφ

( )3.572

2

x

txkK rt

ax =φ

axfsusp

axfsuspF KK

KxKK

φφ

ϕϕ

ϕ+

=


Anti-Roll (Stabilizer) Bar Analysis

Anti Roll Bar Analysis

■ The deflection rate at the free end of a torsion bar.

L

r

d

θ

The deflection rate at the wheel can now be found through analysis of the motion ratio previously defined.

krL

GdF ==2

4

32

πδ

2

=

=b

ak

ab

d

ba

F k=

dF

A

A

A

BB

B


■ The deflection rate at the wheel is based on the motion ratio. 2

1

2

=

r

rkk barwh

r1 = length of the attachment arm r2 = the pivot to attachment length

2

1

22

4

32

=

r

r

rL

Gdkwh

π

The Roll stiffness has previously been defined as

( )3.57•2

Kt = T

= K2

θφ


The Roll stiffness has previously been defined as

( )3.57•2

Kt = T

= K2

θφ

=

3.57232

22

1

22

4

x

t

r

r

rL

Gdk

bar

πφ

The stabilizer bar contribution to roll stiffness is now

The end!

Thank YouThank You

Documents

Suspension automotive