View
35
Download
2
Category
Preview:
DESCRIPTION
automotive suspension
Citation preview
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
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
Spring Rate Calculations
■ 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
≅
Spring Rate Calculations
■ 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 =
Spring Rate Calculations
■ 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
Spring Rate Calculations
■ 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
Spring Rate Calculations
■ 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
Motion Ratio Analysis
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.
Motion Ratio Analysis
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
Motion Ratio Analysis
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.
Motion Ratio Analysis
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
Motion Ratio Analysis
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
Suspension Roll Stiffness
■ 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)
Suspension Roll Stiffness
■ 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
Suspension Roll Stiffness
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
Suspension Roll Stiffness
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
Lateral Spring Center Position
◆ 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
Lateral Spring Center Position
■ 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
Lateral Spring Center Position
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)
Suspension Roll Stiffness
■ 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
Suspension Roll Stiffness
■ 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
Suspension Roll Stiffness
■ 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+
=
Suspension Roll Stiffness
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
Anti Roll Bar Analysis
■ 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
Anti Roll Bar Analysis
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
Recommended