1. Spring Rates, Wheel Rates, Motion
Ratios and Roll Stiffness
Appendix 1
ME 470 Vehicle Structural Design
ME 470 Vehicle Structural Design
Dr. Richard Hathaway, P.E., Professor
Dr. Richard Hathaway, P.E., Professor
Mechanical and Aeronautical Engineering
Mechanical and Aeronautical Engineering
3. Spring Rate Calculations
■ Coil Spring Calculations:
Coil Spring Calculations:
K = Spring Rate in lbs/in
K = Spring Rate in lbs/in G = Modulus of rigidity
G = Modulus of rigidity
d = Spring Wire Diameter
d = Spring Wire Diameter R = Mean Radius of the Spring
R = Mean Radius of the Spring
N = Number of Active Coils
N = Number of Active Coils
Squared and Ground Ends
Squared and Ground Ends -1.75 turns
-1.75 turns
Squared or Closed Ends
Squared or Closed Ends ----
----
Plain Ends
Plain Ends -0.5 turns
-0.5 turns
Plain ends Ground
Plain ends Ground -1.0 turns
-1.0 turns
N
R
64
d
G
=
K 3
4
)
+
(1
2
E
=
G
4. Spring Rate Calculations
■ Coil Spring Calculations:
Coil Spring Calculations:
◆ If Steel is used:
If Steel is used: E = 30,000,000 psi
E = 30,000,000 psi
N
D
d
1,500,000
K 3
4
5. Spring Rate Calculations
■ Torsion Bar Rates:
Torsion Bar Rates:
L = Bar Length
L = Bar Length
d = Bar Diameter
d = Bar Diameter
r = lever arm
r = lever arm
length
length
L
r
d
JG
TL
32
4
d
J
L
JG
T
L
G
d
T
32
4
Let the deflection at the end = r
6. Spring Rate Calculations
■ Torsion Bar Rates:
Torsion Bar Rates:
L
r
d
L
G
d
T
32
4
Then the deflection rate at the free end is found
r
Since T = F x r &
L
G
d
r
r
x
F
32
4
k
r
L
G
d
F
2
4
32
7. Spring Rate Calculations
■ The deflection rate at the free end is
L
r
d
The deflection rate at the wheel can now be
found through analysis of the motion ratio
k
r
L
G
d
F
2
4
32
8. Spring Rate Calculations
■ Torsion Bar Calculations:
Torsion Bar Calculations:
◆ If Steel is used:
If Steel is used: E = 30,000,000 psi
E = 30,000,000 psi
2
000
,
200
,
2
r
L
d
K
4
L = Bar Length
L = Bar Length d = Bar Diameter
d = Bar Diameter
r = lever arm length
r = lever arm length
14. Motion Ratio Analysis
■ Spring Position
Spring 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 for
wheel determines the actual rate the wheel works against for
any spring rate. This displacement relationship may be
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 (K
as the wheel rate (Kw
w). The rate of the spring itself is called
). The rate of the spring itself is called
the spring rate (K
the spring rate (Ks
s). The displacement relationship is a
). The displacement relationship is a
function of both spring position on the load carrying
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.
15. ■ 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
16. ■ 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 A
B
a
b
d
=
d A
B
Motion Ratio Analysis
2
b
a
k
a
b
d
b
a
F
k
=
d
F
A
A
A
B
B
B
17. ■ 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
18. ■ 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.
Cos
2
2
s
w
b
a
K
=
K
centerline
wheel
of
travel
vertical
axis
spring
along
travel
K
=
K
2
s
w
Motion Ratio Analysis
20. Suspension Roll Stiffness
■ ROLL STIFFNESS due to wheel Rates:
ROLL STIFFNESS due to wheel Rates:
◆ The roll stiffness (K
The roll stiffness (Kφ
φ) 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 used
are used
21. ■ 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 equation
For equal spring rates, left and right the above equation
reduces to the following.
reduces to the following.
2
t
K
2
t
+
2
t
K
2
t
=
T R
L
)
K
+
K
(
4
t
=
T R
L
2
(K)
2
t
=
T
2
Suspension Roll Stiffness
22. ■ The roll stiffness is then as shown below.
The roll stiffness is then as shown below.
■ For roll stiffness in N-m/Deg
For roll stiffness in N-m/Deg
(K)
2
t
=
T
=
K
2
3
.
57
2
K
t
=
T
=
K
2
K = Individual wheel rate (N/m) t = track width (m)
Suspension Roll Stiffness
23. ■ In English units this can be reduced to Lb-Ft/Deg
In English units this can be reduced to Lb-Ft/Deg
K
ft
in
12
rad
deg
57.3
2
t
=
(K)
2
t
=
K
2
2
Suspension Roll Stiffness
T = track width (in) K = Individual Wheel Rate (lb/in)
K
1375
t
=
K
2
24. ■ The total roll stiffness K
The total roll stiffness K
is equal to
is equal to
K
+
K
+
K
=
K devices
Rs
Fs
t
K F
= Front Roll Stiffness K R
= Rear Roll Stiffness
K (devices)
= Stabilizer etc contributions
Suspension Roll Stiffness
26. ■ The Spring Center to Cg distance (x) at either end of the
The Spring Center to Cg distance (x) at either end of the
vehicle is important.
vehicle is important.
)
K
+
K
(
l
)
K
+
K
(
-
t
K
=
x
r
l
l
r
l
r
cg
-
sc
l
-
t
)
K
+
K
(
K
=
x l
r
l
r
cg
-
sc
Which reduces to
Lateral Spring Center Position
27. ◆ 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) if
cg) if
K
+
K
l
)
K
+
K
(
-
)t
K
(
=
x
r
l
l
r
l
r
cg
-
sc
l
)
K
+
K
(
>
t
K
if
0
>
x l
r
l
r
cg
-
sc
l
K
<
l
K r
r
l
l
Lateral Spring Center Position
28. ■ The location of the Cg from the inside wheel centerline,
The location of the Cg from the inside wheel centerline,
distance l
distance ll
l, at each axle can be found from the scale weights
, at each axle can be found from the scale weights
at each wheel location.
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
=
l
r
l
r
l
t
)
W
+
W
(
W
-
)
K
+
K
(
K
=
x
r
l
r
r
l
r
cg
-
sc
Lateral Spring Center Position
29. 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 centerline is
The spring center location from the left tire centerline is
as shown.
as shown.
◆ Therefore the roll stiffness for asymmetric springing is,
Therefore the roll stiffness for asymmetric springing is,
t
)
K
+
K
(
K
=
)
x
+
l
(
r
l
r
cg
-
sc
l
)
x
-
l
(
K
+
)
x
+
l
(
K
=
k
2
r
r
2
l
l
3
.
57
30. Roll Stiffness (Asymmetric Chassis)
■ Recall, for equal spring rates,
Recall, for equal spring rates,
3
.
57
2
K
t
=
T
=
K
2
K
+
K
K
)
K
+
K
(
t
=
k
l
r
r
2
r
l
2
3
.
57
Then by substitution
Then by substitution becomes,
33. ◆ The rotational stiffness of the rear axle (k
The rotational stiffness of the rear axle (k
ax
ax) due to the tire
) due to the tire
stiffness is
stiffness is
◆ The rotational stiffness of the rear springs and rear stabilizer bar are
The rotational stiffness of the rear springs and rear stabilizer bar are
T
r
t
ax K
x
t
x
k
K
3
.
57
2
2
kt = tire stiffness (N/m)
tr = rear track width
k ax = Rotational stiffness (N-m/deg)
S
s
s
b
susp
r K
x
t
x
k
k
K
3
.
57
2
2
ks = spring stiffness (N/m)
ts = rear spring spacing
kb = Rear stabilizer bar (N-m/deg)
kr susp = Rotational stiffness (N-m/deg)
Suspension Roll Stiffness
34. ■ The moment produced on the rear axle due to the tire stiffness is
The 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 is
roll bar is
a
r
t
a
ax
t
x
t
x
k
K
M
3
.
57
2
2
a
c
s
s
b
a
c
susp
r
s
x
t
x
k
k
K
M
3
.
57
2
2
a = Axle roll angle c = Chassis roll angle
Suspension Roll Stiffness
35. ■ 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.
'
'
sp
t
sp
t
ride
K
K
K
x
K
K
3
.
57
2
2
2
x
t
x
mr
x
k
k
K
f
f
sp
b
c
susp
f
2
'
mr
x
K
K sp
sp
■ 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
36. ■ Combining chassis roll rate with the tire contribution
Combining chassis roll rate with the tire contribution
3
.
57
2
2
2
x
t
x
mr
x
k
k
K
f
f
sp
b
c
susp
f
3
.
57
2
2
x
t
x
k
K r
t
ax
ax
fsusp
ax
fsusp
F
K
K
K
x
K
K
Suspension Roll Stiffness
38. Anti Roll Bar Analysis
■ The deflection rate at the free end of a torsion bar.
L
r
d
The deflection rate at the wheel can now be
found through analysis of the motion ratio
previously defined.
k
r
L
G
d
F
2
4
32
2
b
a
k
a
b
d
b
a
F
k
=
d
F
A
A
A
B
B
B
39. Anti Roll Bar Analysis
■ The deflection rate at the wheel is based on the motion ratio.
2
1
2
r
r
k
k bar
wh
r1 = length of the attachment arm r2 = the pivot to attachment length
2
1
2
2
4
32
r
r
r
L
G
d
kwh
The Roll stiffness has previously been defined as
3
.
57
2
K
t
=
T
=
K
2
40. Anti Roll Bar Analysis
The Roll stiffness has previously been defined as
3
.
57
2
K
t
=
T
=
K
2
3
.
57
2
32
2
2
1
2
2
4
x
t
r
r
r
L
G
d
k bar
The stabilizer bar contribution to roll stiffness is now
