MEEG 5113
Set 1
1
2
3
4
Vibration Analysis
A simple model of a single degree-of-freedom
system is a pendulum. The mass is concentrated at a
single point, the cg, and the length of the pendulum
is taken as the distance from the pivot point to the
mass center (the length l in the figure to the right).
When the link is then disturbed from equilibrium, it
will oscillate back and forth forever if there is no air
drag or friction at the pivot.
Lacking these losses or energy dissipation methods,
this system is said to conserve energy which means
that the maximum kinetic energy and maximum
potential energy are equal.
5
Vibration Analysis
For the pendulum shown below to be in dynamic equilibrium, the
following must be true:
Mb  Ib
where Ib = mass moment of inertia of the body
about point b
O
b = any arbitrary point on the body
 = angular acceleration
As point b represents an arbitrary location, two
unknowns can be eliminated by summing
moments about the pivot point, O. Also note
that the angle  is measured from the reference
position (vertically down in this case) and the
positive direction for  defines the positive
direction for summing moments.
y
x

L
m
6
W
Vibration Analysis
2
2
d

d
MO  WL sin  IO  IO 2  2   WL sin
dt
dt
IO
In order to proceed further two issues must
Ox
O
be addressed. First, IO needs to be defined
using the parallel axis theorem. This yields
IO  Icg  md2
Oy
Since the point mass is a sphere, Icg is a
function of the radius of the sphere. However,
for a point mass the radius is so small it can
be taken as zero. Therefore, Icg = 0 and IO is
simply md2 and the equation above becomes
d    WL sin
dt 2
mL2
2

L
m
L sin
7
W
Vibration Analysis
Second, the presence of the sine term makes the above equation a
nonlinear, second order differential equation. However, if  is limited
to small values ( < 7o), sin  can be replaced with  to obtain
d 2   WL    mg    g 
dt 2
mL2
mL
L
In this equation both acceleration and displacement are functions of
time and (t) must be a function which repeats itself every second
derivative with a negative coefficient. Harmonic and complex
exponential functions meet this requirement. For simplicity the
harmonic form will be used so that the angular displacement and
acceleration can be written as
 (t)  A cost  Bsin t
and
d 2   2(A cost  Bsin t)   2 (t)
8
2
dt
Vibration Analysis
By equating the two equations for acceleration we obtain
 2 (t)   g  (t)
L
OR 
   g rad s
L
This value of frequency is called the natural (characteristic)
frequency of oscillation of the pendulum and depends only on the
length of the pendulum and the local acceleration of gravity. If the
pendulum is disturbed from equilibrium, it will oscillate with the
frequency .
The values of A and B from the equation for (t) are obtained from
the initial conditions that were applied to the pendulum to begin
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the oscillation process.
Vibration Analysis
There are four possibilities for those initial conditions which are
1. Initial displacement and initial velocity are both zero. While these
initial conditions satisfy the equation, they produce a trivial
solution as no motion will result.
2. Initial displacement is non-zero and the initial velocity is zero.
These conditions are analogous to an archer preparing to take a
shot at a stationary target. The bow is drawn back and held until
aiming is complete. Then the arrow is released toward the target.
3. Initial displacement is zero and the initial velocity is non-zero.
These conditions are analogous to a golfer hitting a tee shot. For a
very small fraction of a second, the ball deflects without moving as
the golf club impacts it.
4. Initial displacement and initial velocity are non-zero. These
conditions are analogous to plucking a string on a guitar.
10
Vibration Analysis
A more commonly encounter model of a vibrating system is shown
below. In this model, the mass undergoes translation in either the
horizontal or vertical plane. In either case, the reference displacement
location is the EQUILIBRIUM position. For the horizontal translation
model, this represents the free length of the spring (no elongation of
the spring). For the vertical translation model, this represents the
location where the weight of the body is balanced by the force in the
spring. If the free length of the spring is used for the
vertical case, one must account for the energy needed to
extend the spring to the distance where the spring force
balances the weight of the
body. This approach is prone
to error as this energy term
is typically not handled
correctly and this approach
11
is not recommended.
Vibration Analysis
For the translational system shown below, the first step is to draw a
free body diagram of the mass. To do this, assume the body is moved
slightly in the positive coordinate direction which will cause the spring
to extend thus generating a restoring force on the body as shown in (a)
below. As vibration for this model does not take place in the vertical
direction, it is not necessary to include
the weight of the body in this diagram.
The inertia (effective) force is shown in
(b). For dynamic equilibrium, this
system must satisfy d’Alembert’s
principle which states that
Fx   kx(t)  max  mx(t)  x(t)   k x(t)
m
Harmonic and complex
exponential functions
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satisfy this expression.
Vibration Analysis
The harmonic form is simplest to use for the undamped case which
means that x(t) and ax(t) have the form
x(t)  Acost  B sin t
 AND 
ax (t)  x(t)   2  Acost  B sin t 
Substituting these definitions into the previous equation yields the
result
 2 x(t)   k x(t)   2  k     k
m
m
m
This result shows that if the mass is
disturbed from either its equilibrium or
steady-state condition it will oscillate with
a natural or characteristic frequency that is
solely dependent on k and m. And, in the
absence of friction and air drag, the
motion will continue forever.
13
Vibration Analysis
As shown below, if the displacement is a cosine function, the velocity
will be 90o out of phase with it and the acceleration will be 180o out of
phase.
x(t) = A cos t + B sin t
v(t) = (-A sin t + B cos t)
Amplitude
a(t) = -2(A cos t + B sin t)
For a given displacement,
Normalized Displacement, Velocity, and Acceleration
the velocity increases by
1
the angular velocity and the
acceleration by the square
0.5
Disp.
of the angular velocity. This
0
Vel.
means even a slight change
Accel.
in displacement can result
-0.5
in a significant velocity and
-1
even greater acceleration.
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Time
Vibration Analysis
The third type of single degree-of-freedom system is a torsional
system as shown below. In this case, the disk rotates in a horizontal
plane about the shaft that acts like a torsional spring with a torsional
spring constant of kT = GJ/L where G is the shear modulus of the
material, J is the polar moment of the cross section, and L is the length
of the shaft. For dynamic equilibrium, this system must satisfy
d’Alembert’s principle which for this problem is
M   k  (t)  I   I  (t)   (t)   kT  (t)
T
cg
cg
which again yields
 2 (t)   kT  (t)   2  kT    
Icg
Icg
(t) = A cos t + B sin t
I cg
kT
I cg
d/dt = (-A sin t + B cos t)
d2/dt2 = -2(A cos t + B sin t)
15
Vibration Analysis
At times one encounters situations where more than one spring in used
in an application. The physical arrangement of the springs relative to
one another determines the effective or equivalent stiffness of the
system. Shown below are three springs connected in series.
To determine the equivalent stiffness of this set of springs, a free body
diagram is drawn of each spring. When this is done, it can be seen that
each spring must support the same load F. For an extensional spring
the load-deflection equation is F = k(Dx). For the above springs the
equations are
Dx1  F
Dx2  F
Dx3  F
k1
k2
k3
x  Dx1  Dx2
 AND 

 Dx3  F  1 
 k1
1  1  F
k2 k3  ke
16
Vibration Analysis
As springs connected in series add as reciprocals, the magnitude of the
equivalent stiffness is always controlled by the smallest value of k in
the set and will be less than that smallest value. For the above spring
set, let k1 = 10 lb/ft, k2 = 20 lb/ft, and k3 = 100 lb/ft. The equivalent
spring stiffness is
1  1  1  1  0.1  0.05  0.01  0.151ft/lb
ke 10 20 100
- OR-
ke  6.62252 lb/ft
When designing any system that has a series stiffness arrangement,
one must be very careful to attempt to match the stiffnesses as closely
as possible to avoid exceeding the maximum allowable deflection.
This concern explains why the cylinder head gasket on most newer
cars is now a stamped steel component. This allows the stiffness of the
gasket to match that of the cylinder head and block as closely as
17
possible.
Vibration Analysis
Shown below are three springs connected in parallel.
To determine the equivalent stiffness of this set of springs, a free body
diagram is drawn of each spring. When this is done, it can be seen that
each spring must support the same deflection, Dx. For an extensional
spring the load-deflection equation is F = k(Dx). For the above springs
the equations are
F1  k1Dx
F2  k2Dx
F3  k3Dx
 AND 
F  F1  F2  F3   k1  k2  k3 Dx  keDx
18
Vibration Analysis
As springs connected in parallel simply add together, the magnitude of
the equivalent stiffness is always controlled by the largest value of k in
the set and will be greater than that largest value. For the above spring
set, let k1 = 10 lb/ft, k2 = 20 lb/ft, and k3 = 100 lb/ft. The equivalent
spring stiffness is
ke  10  20  100  130 lb/ft
When designing any system that has a parallel stiffness arrangement,
one must be very careful to limit the equivalent stiffness to only the
amount needed to carry the expected maximum load. This concern
partially explains why makers of SUV’s are now claiming their
vehicle rides like a regular passenger car. One benefit to using parallel
springs is space saving. On automotive engines it is sometimes
necessary to stiffen the valve springs without increasing the diameter
of the spring or the diameter of the wire. This can be accomplished
using two springs nested one inside the other. The overall stiffness
19
increases, but the outside dimension does not.
Vibration Analysis
In the preceding discussion, each element was an “ideal” element. The
mass did not contribute stiffness to the system only mass. The spring
did not contribute mass to the system only stiffness. A damper is
considered to have no mass or stiffness and only dissipates energy.
Several common items can be modeled as an “ideal” spring (i.e. they
contribute little to the overall mass of the system).
1. Helical spring – k = (d4G)/(8D3N) where d = wire diameter
D = mean coil diameter
G = shear modulus
N = number of active
coils
2. Rod (axial stiffness) – k = (AE)/L where A = cross sectional area
E = Young’s modulus
L = length
20
Vibration Analysis
3. Rod (torsional stiffness) – k = (GJ)/L where J = polar moment of
the cross section
4. Cantilever beam (tip load) – k = (3EI)/L3 where I = second
moment of the
area about the
neutral axis
5. Simply supported beam (center load) – k = (48EI)/L3
6. Clamped-clamped beam (center load) – k = (192EI)/L3
7. Clamped-pinned beam (center load) – k = (768EI)/7L3
21
Vibration Analysis
When the spring does contribute mass to the system, one must
determine what the dynamic mass of the spring is, i.e. how much of
the spring’s total mass contributes to the motion. In the figure below, it
is easy to see that since one end of the spring is attached to ground that
mass does not move at all. The other end is attached to the moving
mass and it moves a great deal. Therefore, a portion of the spring’s
mass will add to the dynamics. To determine how much mass to add to
M, we first estimate the deflected shape of the spring and then use that
same shape to compute the velocity of a small segment of the spring.
This expression is used to determine the kinetic energy
of the spring. From this result
the dynamic mass can be
found. An example for the
figure to the right follows.
22
Vibration Analysis
For the system shown below, the mass of the spring is considered
large enough to contribute to the motion of the cart, i.e. it a “heavy”
spring. The cart has a mass M, and the spring has a mass m, a free
length l, and a stiffness k. For manner in which the spring is attached
in the system, the static deflection shape of the spring is given by the
equation Dy = (y/l) x. If the dynamic deflection shape of the spring is
the same as the static deflection, the mass and velocity of the spring
element dy are
Dy   y  x  AND  dm   m  dy
 l 
l 
The kinetic energy of the
spring is then equal to
T
T
1
2
2
 m  y 
   x  dy
0 l  l 

l
l
2
3
1 mx  y 

 
2  l 3   3 
0
mdyn = m/3
 1  m  x2
2 3 
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Vibration Analysis
Therefore,
mTotal  M   m 3 
 AND 

k
 3
M m
rad / s
This result indicates that if the mass of the spring cannot be ignored the
natural frequency will be lowered. The amount it is reduced can be
significant if m approaches M as it does in the valve train of an engine.
In this case, the valve spring
can fail due excessive engine
speed if a careful analysis of
the valve train has not been
done. Such a failure results in
considerable damage to the
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engine.
Vibration Analysis
At times it is desirable to alter the vibrational frequency of a system or
design a system that has a specific fundamental frequency. One
example of this situation is the design of a vibrating reed tachometer. A
vibrating reed tachometer is a small handheld instrument which
contains 5 to 10 cantilever beams each tuned to a specific frequency.
Each reed is a cantilever beam with a weight added to the free end
which serves to both tune the beam and act as an indicator.
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Vibration Analysis
To design one reed, the dynamic mass of the cantilever beam must be
determined. The first step is to establish the dynamic deflection of the
beam. A good approximation is the static deflection curve for either a
tip load or due to its own weight. For this example we will use
2
3
 x  x
1
y  y0 [3     ] where y0 = maximum deflection
2
l  l 
x = location along beam from left end
l = beam length
This yields the following velocity and kinetic energy expressions
2
3
 x  x
1
y  y0 [3     ]
2
l l

l
1
T  m y 2 dx
2 0
4
5
6
l
2
 x
 x
 x
1
m
T
y0 [9    6      ]dx
24
l
0
l
l

26
Vibration Analysis
- OR 



T  1 ml y02 [ 9  1  1 ]  1  33 ml  y02  1  33 mbeam  y02  1 meff y02
2 4
5
7 2  140 
2  140 
2
What this result means is that a little less than ¼ of the mass of the
beam is moving enough to contribute to vibration.
Now determine the amount of mass, M, which must be added to the tip
of the cantilever beam to produce a fundamental frequency of 20 Hz if
b = 0.635 cm; h = 0.1016 cm; l = 8.89 cm;
g = 0.07655 N/cm3; E = 200 x 109 N/m2
27
Vibration Analysis
From these values the following can be determined:
ml = 0.00475 kg; I = 0.0000555 x 10-8 m4; k = 473.96 N/m;
which allows the effective mass and then the added mass as follows:
f  1 mk
2 eff
meff  M  33 ml  3 3EI2 2  473.96
140 l 4 f
(4 )400
M  0.0289 kg
28
Vibration Analysis
For the horizontal pendulum shown below, link AB is rigid, has a
uniform cross section, but is not massless and it is desired to
determine the frequency of vibration for the following conditions:
l = 0.3 m; a = 0.15 m; mrod = 10 kg; m = 7 kg; k = 2 kN/m
To proceed, the relationship between , a, and l must be established
for small angle motion. For small angles, the chord length and arc
length approach one another and the chord length represents the
straight line motion at a and l keeping in
mind that our reference
position is the equilibrium
or horizontal position. This
yields the following
equation of motion:
 ka 2 
  
  0
29
I
 A 
Vibration Analysis
For the resulting harmonic motion of the pendulum, the frequency of
vibration is given by
2
ka
1
f 
Hz
2 I A
For this system, the mass moment of inertia is given by
IA = [7 + (1/3)(10)] (0.3)2 = 0.93 kg m2
This gives a frequency value of
 2000 0.15 Hz
0.93
2
f  1
2
f  1.1 Hz
30
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