Normal Modes of Vibration
Shigley does not cover this material directly. However, I believe it will help you
understand the next topic.
Consider the following figure, in which the masses are resting on a horizontal,
frictionless surface. The springs are initially un-stretched, and the displacements of the masses
are measured from that initial configuration.
It is straightforward to develop the equations of motion for the two masses.
   k X  k ( X  X )
m1 X
1
1 1
2
2
1
(25.1)
   k X  k ( X  X )
m2 X
2
3 2
2
2
1
(25.2)
Re-write equations (25.1) and (25.2).
   (k  k ) X  k X
m1 X
1
1
2
1
2 2
(25.3)
  k X  (k  k ) X
m2 X
2
2 1
2
3
2
(25.4)
We make the assumption that there is a normal mode of vibration, a frequency at
which both masses oscillate in phase with each other.
X1 (t )  X10 cos t
(25.5)
X 2 (t )  X 20 cos t
(25.6)
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Substitute equations (25.5) and (25.6) into equations (25.3) and (25.4). Cancel the
common factor of cos t from the resulting equations.
 m1  2 X10   (k 1  k 2 ) X10  k 2 X 20
(25.7)
 m 2  2 X 20  k 2 X10  (k 2  k 3 ) X 20
(25.8)
Re-write equations (25.7) and (25.8).


0  m1  2  (k 1  k 2 ) X10  k 2 X 20

(25.9)

0  k 2 X10  m 2  2  (k 2  k 3 ) X 20
(25.10)
Equations (25.9) and (25.10) are two simultaneous, linear equations involving two unknown
variables, X10 and X 20 . The equations are homogeneous; there are no terms that do not involve
the unknown variables. Therefore, if the two equations are graphed, the result is two straight
lines through the origin.
X 20
X10
The only simultaneous solution to these two equations is the origin, referred to as the
trivial solution. This an uninteresting solution, representing static equilibrium of the system.
The only way to obtain an alternate solution to the system of equations is to choose  2 to make
the slopes of the two lines equal. There will then be an infinite number of solutions, all with the
same ratio of X 20 to X10 .
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m
1
2

 (k 1  k 2 )
k2

2
k2
m 2   (k 2  k 3 )

0  m1  2  (k 1  k 2

) m
2

(25.11)

 2  (k 2  k 3 )  k 2 2
(25.12)
Equation (25.12) can also be obtained by setting the determinant of the coefficient matrix
in equations (25.9) and (25.10) equal to zero.
0
m
1
2
 (k 1  k 2 )
k2


k2
m 2   (k 2  k 3 )
2

(25.13)
There is nothing difficult, in principle, in determining the solution to equation (25.12). It
is a quadratic equation in the variable  2 . There will either be two real solutions or two
complex-conjugate solutions. However, it is a theorem of linear algebra that the eigenvalues of
a determinant such as in equation (25.13), a symmetric matrix with real coefficients, will
always be real.
However, it would not be straightforward to understand the solutions. Therefore, we will
analyze the following special case.
m1  m 2  m
(25.14)
k 3  k1
(25.15)
Re-write equation (25.12) for this special case.


0  m  2  (k 1  k 2 )
2
 k 22
m  2  (k 1  k 2 )   k 2
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(25.16)
(25.17)
Case I (plus sign)
m  2  (k 1  k 2 )  k 2
(25.18)
m 2  k 1  2 k 2
(25.19)
k1  2k 2
(25.20)
m
Substitute equation (25.20) into equation into either equation (25.9) or (25.10). It does
2 
not matter which, because with this choice of  2 , the two equations are identical. Choose
equation (25.9) and set m1  m .
0   k 1  2k 2  (k 1  k 2 ) X10  k 2 X 20
(25.21)
0  k 2 X10  k 2 X 20
(25.22)
X 20   X10
(25.23)
Case II (minus sign)
m  2  (k 1  k 2 )   k 2
(25.24)
m 2  k 1
(25.25)
2 
k1
m
(25.26)
Substitute equation (25.26) into equation into either equation (25.9) or (25.10). It does
not matter which, because with this choice of  2 , the two equations are identical. Choose
equation (25.9) and set m1  m .
0   k 1  (k 1  k 2 ) X10  k 2 X 20
(25.27)
0   k 2 X10  k 2 X 20
(25.28)
X 20  X10
(25.29)
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Examine the two cases.
In case I, the two masses move in opposite directions with equal amplitudes. The center
of the center spring remains stationary. The effective spring constant of the center spring, as
seen by either mass, is equal to 2 k 2 . The total spring constant, as seen by either mass, is equal
to k 1  2 k 2 .
In case II, the two masses move in the same direction with equal amplitudes. The center
spring remains un-stretched. The effective spring constant of the center spring, as seen by either
mass, is equal to zero. The total spring constant, as seen by either mass, is equal to k 1 .
For a system with N degrees of freedom, two in our system, there will always be exactly
N natural frequencies, corresponding to N normal modes of vibration. Case II represents the
first (smaller) natural frequency, and case I represents the second (larger) natural frequency.
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Calculation of Natural Frequencies
An important problem in the analysis of shafts is the calculation of the natural
frequencies of a shaft supporting some number, N, of concentrated masses. There are N degrees
of freedom of such a system and N natural frequencies. The following example will consider
three concentrated masses. The figures appear to be symmetric left-to-right- but this is not
necessary.
The shapes of the three modes should be thought of as schematic. E.g., the amplitude of
the central mass in the second mode will be zero only if the shaft is symmetric about its center.
This symmetry must include both the magnitude of the masses and their locations.
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We again assume that, at each natural frequency, the three masses oscillate in phase with
each other.
Y1 (t )  Y10 cos t
(25.30)
Y2 (t )  Y20 cos t
(25.31)
Y3 (t )  Y30 cos t
(25.31)
When each mass is in its position of maximum displacement, cos t  1 . Omit numerical
subscripts for notational clarity.
    2 Y cos t    2 Y
Y
0
0
(25.32)
Therefore, the shaft must exert a restoring force (inward) on each mass equal to, again
omit numerical subscripts,
FS   m  2 Y0
(25.33)
This requires each mass to exert a force, equal in magnitude and opposite in direction
(outward) on the shaft.
FM  m  2 Y0
(25.34)
These outward forces cause the deflection of the shaft. Make use of flexibility factors,
introduced in lesson 15. f IJ denotes the deflection of the shaft at the location of mass I caused
by a unit force at the location of mass J. Note that it does not matter which direction is chosen as
positive, as long as the same direction is chosen for both the force and deflection.
The following three simultaneous equations result.
Y10  m1  2 Y10 f11  m 2  2 Y20 f12  m 3  2 Y30 f13
(25.35)
Y20  m1  2 Y10 f 21  m 2  2 Y20 f 22  m 3  2 Y30 f 23
(25.36)
Y30  m1  2 Y10 f 31  m 2  2 Y20 f 32  m 3  2 Y30 f 33
(25.37)
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Re-write equations (25.35) through (25.37).
0  (m1  2 f11  1) Y10  (m 2  2 f12 ) Y20  (m 3  2 f13 ) Y30
(25.38)
0  (m1  2 f 21 ) Y10  (m 2  2 f 22  1) Y20  (m 3  2 f 23 ) Y30
(25.39)
0  (m1  2 f 31 ) Y10  (m 2  2 f 32 ) Y20  (m 3  2 f 33  1) Y30
(25.40)
Equations (25.38) through (25.40) are simultaneous, homogeneous linear equations in the
variables Y10 , Y20 , and Y30 . A non-trivial solution exists only if the determinant of the
coefficients is equal to zero.
(m 1  2 f11  1)
0  (m 1  2 f 21 )
(m 1  f 31 )
2
(m 2  2 f12 )
(m 2  2 f 22  1)
(m 2  f 32 )
2
(m 3  2 f13 )
(m 3  2 f 23 )
(m 3  f 33  1)
2
Expansion of equation (25.41) yields a cubic equation in the variable  2 .
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(25.41)
Energy Method
Consider the case of a horizontal shaft with multiple concentrated masses. In the
following figure, the shaft is shown in its un-deflected position and the position of maximum
deflection when vibrating at its first (lowest) natural frequency.
The displacement of the I-th mass is described by the following relationship.
YI (t )  YI 0 cos 1t
(25.41)
YI 0 denotes the amplitude of the I-th mass, and 1 denotes the first angular frequency.
Note that in the lower, deflected position, 1 t  0 , and in the upper, un-deflected position,

1 t  .
2
   Y  sin  t
(25.42)
Y
I
I0 1
1
Determine the kinetic energy of the system in the upper position.
m I YI 0 2 1 2
K .E.  
 1 2
2
I
m I YI 0 2
 2
I
(25.43)
In the lower position, there is no kinetic energy. All the energy in the system is
elastic strain energy in the shaft. In general, the calculation of the elastic strain energy in a
deflected shaft requires the second derivative as a function of position on the shaft. However, it
turns out that it is possible to obtain an accurate estimate of the first natural frequency by
approximating the actual deflected amplitudes with the static-deflection displacements. The
elastic strain energy in the shaft in the static-deflection configuration is the work done in slowly
applying the weights to the un-deflected configuration.
U
I
m I g YI 0
m Y
 g  I I0
2
2
I
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(25.44)
Equate the two energies.
1
2
1 2
m I YI 2
m Y
 2  g  I2 I
I
I
(25.45)
  m I YI 


I


g
  m I YI 2 


 I

(25.46)
Note that equation (25.46) can be manipulated into the following form.
N C  188
 WI YI
I
(25.47)
 WI YI 2
I
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Locations of Keyways
Consider the following alignment of keyways on a shaft. This is generally the least
expensive configuration to manufacture.
The X - axis is the axis of minimum second-area moment (the so-called moment of
inertia), and the Y - axis is the axis of maximum second-area moment. As the shaft makes one
complete revolution, the second-area moment completes two cycles. This will tend to set up a
vertical vibration in the shaft at twice the run speed of the shaft. This can cause problems if the
run speed is close to one-half a natural frequency.
One solution is to align the keyways as shown in the following figure.
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