CE 573: Structural Dynamics
Solutions to HW#3
Problem #1
R
y
(x)
x
kc
g
m
v = constant
Frame
Center of Mass
h
y(t)
Y(t)
Wheel
y
o
Given
:
A car can be (crudely) modeled as a single degree of freedom system having the entire car’s mass
m
lumped at the center of mass and connected to the frame (modeled as a rigid bar) by a linear spring
(stiffness =
k
, unstretched length =
l
o
) and a linear damper (damping constant =
c
). The frame, in turn, is a
fixed height
h
above the wheel, which rides over a rough road whose vertical height above the fixed
x

axis is given by a known function
y
R
(x)
. The car starts at the location
x=0
at the initial time
t=0
(assume
that
y
R
(x=0) = 0)
and it moves to the right with a constant horizontal speed
v
. Let
Y(t)
be the height of the
center of mass above the fixed
x
axis, let
y
o
be the static equilibrium height of the center of mass at time
t=0
, and let
y(t)
be the height of the center of mass above this static equilibrium position. Gravity acts
downward as shown.
Required
:
(a) Determine the equation of motion for the car’s center of mass in terms of
y(t)
and the
system and road parameters given above. Hint
: you may find it helpful to express the equation of motion
first in terms of
Y(t)
, then determine the relationship that exists between
y
o
, l
o
,
and
h
in order to simplify
the result and apply it to
y(t)
.
(b) Assume the following values for the parameters: mass
1430
m
=
kg, stiffness
kN/m, and
damping factor
140
k
=
0.40
ξ
=
. Also, assume that the car travels over an idealized harmonic surface
( )
()
s
i
n2
/
R
yx Y
xL
π
=
, with amplitude
80
Y
=
mm and wavelength
30
L
=
meters. Finally, assume the
constant speed of the car is
m/s. Determine the car’s steadystate vibration amplitude and the
steadystate force it transmits to the road.
30
o
v
=
(c) Suppose that the speed of the car is increased to a value above 30 m/s. What effect this would have on
the answers to part (b)? Explain qualitatively why increasing the speed has this effect.
Solution
:
(a) Start with a free body diagram of the center of mass –
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F
d
Y(t)
F
s
mg
Notice that the displacement
is
not
measured from static equilibrium, so the effect of the car’s
weight must be included. Newton’s second law then tells us:
()
Yt
, or
0.
sd
mF
F
m
g
m
Y
m
Y
F
F
m
g
=⇒
−
−−=
+
++=
∑
Fa
We need to evaluate
s
F
and
d
F
, which depend upon the deformation of the center of mass
relative
to the
frame. Starting with the stiffness force, we note that the force is proportional to
spring
current
o
ll
∆= −
l
x
,
where
( )
.
current
R
lY
t
h
y
=
−−
Since the force must be a function of time, we need to replace
with an appropriate time
dependent function. To do this, we note that the constant horizontal velocity of the car implies that
R
yx
o
x
tv
tx
=+
, with
since the car is at the origin when
0
o
x
=
0
t
=
. Thus, we get
{ }
.
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 Fall '05
 Whalen
 Force, fo, DLF

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