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3.3 Mechanical Impedance (and Receptance) Method
By this method behavior of complete system is obtained from the behavior of individual
components of the system. It is particularly convenient to use when the characteristics of shaft and
those of bearings are determined from independent experimental or theoretical investigations
carried over a range of frequencies. The simple addition of shaft impedance and bearing
impedance gives the impedance of complete system, which may be used to find system critical
speeds and forced response. General principles are discussed first with reference to a simple
spring-mass system as shown in Figure 3.18(a). Mechanical impedance Z: is defined as the force
required to produce unit displacement. It is generally a complex quantity because of the phase lag
of displacement behind force due to damping.
Figure 3.18 A simple spring-mass system
The impedance of spring alone is given as (Figure 3.18(b))
ss
f kxZ k
x x= = = (49)
where fs is the force on the spring, x is the corresponding displacement and k is the stiffness of the
spring. The impedance of the mass alone is given as (Figure 3.18(c))
22m
m
f m xmxZ m
x x x
ωω
−= = = = −
�� (50)
m
m
f
fm
x
(a) A simple spring-mass
system
m
f
(d) Equivalent system for analysis
purpose (since mass & the spring
have same amount of displacement
they can be thought of connected
in parallel)
k x
fs
(b) Impedance of spring
alone
(c) Impedance of spring
alone
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
160
where fm is the force on the mass, x is the corresponding displacement, m is the mass and ω is the
simple harmonic forcing frequency to the system. The spring and mass are effectively connected in
parallel (Figure 3.18(d)) as far as force is concerned. For subsystems connected in parallel the net
system impedance is given by the sum of individual subsystem impedances (for example the
equivalent stiffness of two spring connected in parallel is given by the sum of the individual spring
stiffness). So for the system shown in Figure 3.18(d) the impedance at the forcing point is
2
s mZ Z Z k mω= + = − (51)
At the resonance (i.e. forcing frequency is equal to the system natural frequency) system impedance
will be zero (since any force produces an infinite amplitude, exception being when the forcing point is
at a node whereupon impedance tends towards infinitely) so that
2 0nk mω− = that is n
k
mω = (52)
where nω is the natural frequency of the system. It is noteworthy that for subsystems connected in
series, the net system ‘receptance’ is sum of the individual subsystem receptances (receptance being
the inverse of impedance). The above approach may be applied to machines whose shafts carry many
rotor inertias when the shaft free-free impedance has been determined independently of those of the
bearing, pedestals and foundations. Shaft free-free impedance means the impedance when the shaft is
not constrained at either support point. This can be determined experimentally by suspending the shaft
so that it is supported only in the vertical direction as shown in Figure 3.19, then determining the
impedance horizontal response to a known horizontal forcing. The individual impedances component
impedances may be combined according to the rules described above to determine the impedance of
the complete system.
Figure 3.19 Suspended shaft in vertical direction: Free-free end conditions
F(t)
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Figure 3.20 A flexible rotor under forced excitation
The impedance of a light flexible shaft carrying a number of disc masses may be determined
theoretically as follows. The shaft is considered to be forced, in the first instance, at locations A and B
(see Figure 3.20). The forcing causes the reaction forces and moments …2211 ,,, MFMF etc. to be
set up as a consequence of disc mass inertias and these deform the shaft according to the relationship
11 12 13 14 1,21
21 22 23 24 2,22
1 2 3 4 ,2
1,1 1,2 1,3 1,4 1,21
2,1 2,2 2,3 2,4 2,22
2 ,1 2 ,2 2 ,3 2 ,4 2
nl
nl
n n n n n nl n
n n n n n na
n n n n n na
n n n na n
α α α α αδα α α α αδ
α α α α αδα α α α αδα α α α αδ
α α α α αδ
+ + + + +
+ + + + +
=
…
…
� � � � � ��
…
…
…
� � � � � ��
…
1
2
1
2
,2
n
n n n
F
F
F
M
M
M
�
�
(53)
where ijα will be determined from the beam theory and back subscripts : l refers to the linear and a
refers to the angular shaft displacement. In general the loading applied to any rotor mass i to cause its
acceleration is
( ) ( )xmxmF liliim +−=+= δωδ 2���� (54)
and
( ) ( )2
i im i d a d aM I Iδ φ ω δ φ= + = − +�� �� (55)
Light flexible shaft
Zero displacement, “datum position”
x A
B δl φ
δα
Tangent to shaft Line parallel to AB
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162
where mi and id
I are the disc mass and the disc diametral mass moment of inertia respectively, x and
φ are shaft linear and angular displacements respectively, caused by movements at shaft locations A
and B and not caused by the inertia forces. The reaction loading on the shaft due to inertia will be
equal and opposite and is given by
( )2
s i i lF m xω δ= + (56)
and
( )2
is i d aM Iω δ φ= + (57)
where ω is the excitation frequency. Substituting equations (56) and (57), written for each disc, into
equation (53), we get
11 12 13 14 1,21
21 22 23 24 2,22
1 2 3 4 ,2
21,1 1,2 1,3 1,4 1,21
2,1 2,2 2,3 2,4 2,22
2 ,1 2 ,2 2 ,3 2 ,4
1
nl
nl
n n n n n nl n
n n n n n na
n n n n n na
n n n na n
α α α α αδα α α α αδ
α α α α αδα α α α αδωα α α α αδ
α α α αδ
+ + + + +
+ + + + +
=
…
…
� � � � � ��
…
…
…
� � � � � ��
( )( )
( )( )( )
( )
1
2
1 1 1
2 2 2
1 1
2 2
2 ,2n
l
l
n l n n
d a
d a
n n d a n n
m x
m x
m x
I
I
I
δ
δ
δ
δ φ
δ φ
α δ φ
+ + + + + +
�
�
…
(58)
which can be rearranged as
}]{[}]{[ xRA =δ (59)
with
[ ]
( )( )
( )
2
1 11 2 12 1,2
2
1 21 2 22 2,2
2
1 2 ,1 2 2 ,2 2 ,2
1/
1/
1/
n
n
n
d n
d n
n n d n n
m a m a I a
m a m a I aA
m a m a I a
ω
ω
ω
− − − − − −
=
− − −
…
…
� � � �
…
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163
[ ] { } { }
11 1 12 2 1,2 1 1
21 1 22 2 2,2 2 2
2 ,1 1 2 ,2 2 2 ,2
; ;
n
n
n
n d l
n d l
a n nn n n n d
a m a m a I x
a m a m a I xR x
a m a m a I
δδ
δ
δ φ
= = =
…
…
� �� � � �
…
(60)
which gives
{ } [ ]{ }C xδ = with 1[ ] [ ] [ ]C A R−= (61)
In general the application of inertia loads iF and iM to the shaft at some point causes proportional
reaction forces AF and BF at points A and B, where
=
i
i
BB
AA
B
A
M
F
bb
bb
F
F
21
21 (62)
the generalized form of equation (62), allowing for all inertias, is
1 2 ,2
1 2 1 2
1 2 ,2
TA A A nA
n n
B B B nB
b b bFF F F M M M
b b bF
=
�� �
� (63)
where ijb may be determined from static equilibrium considerations. Substituting for disc inertia
forces and moments from equations (56) and (57) into equation (63), we get
( )( )
( )( )
( )
1
2
1 1 1
2
2 2 2
21 2 ,2
21 2 ,2
1 1
2
n
l
l
A A A nAn l n n
B B B nBd a
d a n n
m x
m x
b b bFm x
b b bFI
I
ω δω δ
ω δω δ φ
ω δ φ
+
+ +=
+
+
�…
…
�
(64)
On substituting for s'δ in equation (64) from equation (61) (i.e. 1 11 1 12 2 1,2l n nc x c x cδ φ= + + +… etc.)
then gives
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati, ([email protected])
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[ ]{ }A
B
FD x
F
=
(65)
with
[ ]
( )( )
( )
{ } { }
1 11 1 12 1 13 1 1,2
1 2 ,2 2 21 2 22 2 23 2 2,22
1 2 ,2
2 ,1 2 ,2 2 ,3 2 ,2
1 2
1
1
1n n n n
n
A A A n n
B B B n
d n d n d n d n n
T
n
m c m c m c m c
b b b m c m c m c m cD
b b b
I c I c I c I c
x x x
ω
φ
+ + =
+
=
…
… …
… � � � � �
…
�
Figure 3.21 Rigid body linear and angular displacements of the shaft
However, displacements 2211 ,,, φφ xx etc. are related to displacements at forcing points, Ax and
Bx ,
by a relationship of the form
=B
A
x
xGx ][}{ (66)
with
xB
xA a
l
x
al
xxxx BAA
−+=
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165
{ } [ ]
1
11 12
2
1 2
1,1 1,2
1
2 ,1 2 ,2
; n n
n
n n
n n
n
xg g
x
g gx x G
g g
g g
φ
φ
+ +
= =
� ��
� ��
Element of [G] can be obtained by simple consideration of geometry. Substituting equation (66) into
equation (65), it gives
[ ]A A AA AB A
B B BA BB B
F x Z Z xZ
F x Z Z x
= =
(67)
with
[ ] [ ][ ]Z D G=
where [Z] is the impedance matrix for the shaft and rotor assembly, relating forcing and
displacements at points A and B. In the matrix [Z] now all quantities, one can known by theoretical
analysis. The more general form of equation (67) that allows also for the motion in y-direction is
0 0
0 0
0 0
0 0
xA AA AB A
xB BA BB B
yA AA AB A
yB BA BB B
F Z Z x
F Z Z x
F Z Z y
F Z Z y
=
(68)
The response at any other location along the shaft can be obtained by pre-multiplying inverse of
impedance matrix in equation (68) to get displacements at A and B, then substituting these in
equations (66) and (61). If BF is chosen to be zero, the point A can be chosen as any convenient
location along the shaft and corresponding (direct) impedance so evaluated. Equation (68) can be
expanded for any number of forcing points, for example for third forcing point C
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0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
xA AA AB AC A
xB BA BB BC B
xC CA CB CC C
yA AA AB AC A
yB BA BB BC B
yC CA CB CC C
F Z Z Z x
F Z Z Z x
F Z Z Z x
F Z Z Z y
F Z Z Z y
F Z Z Z y
=
�
�
�
… … … … … … … … …
�
�
�
(69)
If forcing is caused by imbalance and AF and BF are reaction forces at pinned bearing supports, the
forced response of the system can be determined by assigning zero values to , , and A B A Bx x y y . From
equation (69) the third and sixth equations gives
xc cc cF Z x= and yc cc cF Z y=
which gives displacements at C as
/ and /c xc cc c yc ccx F Z y F Z= = (70)
From equation (69), 1st, 2nd, 4th, and 5th equations gives
; ; ; xA AC c xB BC c yA AC c yB BC cF Z x F Z x F Z y F Z y= = = = (71)
Noting equation (70), equation (71) can be combined as
0 0
/0 0=
/0 0
0 0
xA AC AC
xB xc ccBC c BC
yA yc ccAC c AC
yB BC BC
F Z Z
F F ZZ x Z
F F ZZ y Z
F Z Z
=
(72)
Equation (72) gives reaction forces at A and B due to forcing at C. These reaction forces may be
substituted back into equation (69), which may be written more generally, to give response at any
other locations as
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0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
xAAA AB ACA
xBBA BB BCB
xDDA DB DCD
yAAA AB ACA
yBBA BB BCB
yDDA DB DCD
FR R Rx
FR R Rx
FR R Rx
FR R Ry
FR R Ry
FR R Ry
=
�
�
�
…… … … … … … ……
�
�
�
(73)
with [ ] 1[ ]R Z −=
where R are component of receptance matrix. In equation (73), [R] is already known to us
corresponding to new sets of chosen points, {F} matrix is now completely known. So the new Dx and
Dy can be obtained as
D DA xA DB xB DD xC DA xA DB xB DD xDx Z F Z F Z F R F R F R F′ ′ ′= + + = + + (74)
and
D DA yA DB yB DD yC DA yA DB yB DD yDy Z F Z F Z F R F R F R F′ ′ ′= + + = + + (75)
Example 3.5 Obtain transverse synchronous critical speeds of a rotor system as shown in Figure 3.3.
Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia, Id = 0.02 kg-m2, the polar
mass moment of inertia, Ip = 0.04 kg-m2. The disc is placed at 0.25 m from the right support. The
shaft is having diameter of 10 mm and total span length of 1 m. The shaft is assumed to be massless.
Take shaft Young’s modulus E = 2.1 × 1011 N/m2. Neglect gyroscopic effects. Take one plane motion
only.
a b
l = a + b
Figure 3.22 A rotor system
Influence coefficients are defined as:
11 12
21 22
y F
M
α α
α αθ
=
with
( )( )
2 2 2 3 2
11 12
2 2
21 22
/ 3 ; 3 2 / 3
( ) / 3 ; 3 3 / 3
a b EIl a l a al EIl
ab b a EIl al a l EIl
α α
α α
= = − − −
= − = − − −
Solution: Figure 3.23 shows a schematic of the rotor system.
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Figure 3.23 Schematic diagram of a rotor system
From equation (53), we have
1 11 12 1
1 21 22 1
l
a
F
M
δ α αδ α α
=
with
4 4 10 4(0.01) 4.909 10 m64 64
I d xπ π −= = = ; 0.75a = m; 0.25b = m; 1.0l = m.
( ) { }
2 2 2 24
11 11 10
2 3 2 2 3 2
4
12 11 10
21
(0.75) (0.25)1.137 10 m/N
3 3 2.1 10 4.909 10 1
3 2 3 (0.75) 1 2 0.75 0.75 13.3031 10 m/N
3 3 2.1 10 4.909 10 1
( ) 0.75 0.25 (0.75 0.25)
3 3 2.1 1
a b
EIl
a l a al
EIl
ab b a
EIl
α
α
α
−−
−−
×= = = ×
× × × × ×
− − × × − × − ×= − = − = − ×
× × × × ×
− × × −= =
× ×
( )
4
1211 10
2 22 23
22 11 10
3.0314 10 m/N0 4.909 10 1
3 0.75 1 3 0.75 1(3 3 )1.4146 10 m/N
3 3 2.1 10 4.909 10 1
al a l
EIl
α
α
−
−−
= − × =× × ×
× × − × −− −= − = − = ×
× × × × ×
From equation (59), we have
1
1
2 2
1 11 12
2 2
1 21 22
1[ ]
1
d
d
m IA
m I
ω α ω α
ω α ω α
− −=
− − and [ ] 1
1
11 1 122
21 1 22
d
d
m IR
m I
α αω
α α
=
and
[ ] 1 1
12 21 22 12
2 2
1 21 1 11
11 1since
1
d dI I a b d bA
c d c aA ad bcm m
ω α ω α
ω α ω α
−− − − = = −−−
with
0.75m 0.25m
A B
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1 1
4 2
11 22 12 21 1 1 11 22( ) ( ) 1d dA m I m Iω α α α α ω α α= − − + +
From equation (61), we have
11 1
1
1 1 1 1 1 1
1
2 211 1 1222 121
2 221 1 221 12 1 11
2 2 2 2
22 11 12 21 1 22 12 12 22
2 2 2
1 12 11 1 1 11 21 1 1 12 12
(1 )1[ ] [ ] [ ]
(1 )
(1 ) (1 )1
(1 )
dd d
d
d d d d d d
d
m II IC A R
m IA m m
I m I m I I I I
A m m m m m I
α αω α ω α
α αω α ω α
ω α α ω α α ω α α ω α α
ω α α ω α α ω α α
− −
= = −
− + − +=
+ − +1
2
1 11 22(1 ) dm Iω α α
−
(A)
Figure 3.24 Displacement of the shaft
From free body diagram of shaft, we have
( )
1
1 1 1 1
1 1 1
0
10 0
so that
11
A B
A B B
A B
F F F F
aM F l M Fa F F Ml l
aF F F F Ml l
= ⇒ = +
= ⇒ − − = ⇒ = +
= − + = − −
∑
∑
Combining equations (B) & (C), we get
1
1
( / ) (1/ )
( / ) (1/ )
A
B
F Fb l l
F Ma l l
− =
(D)
(B)
(C)
F1
FA
FB
M1
aδ
l
a
Tangent to shaft
Parallel
Rigid body displacements
of the shaft
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Noting equation (D), from equation (65), we have
1
1
[ ]A
B
F xD
F φ
=
(E)
with
[ ]
( ) ( )
1 1
1 1
1 1
1 11 1 12
21 22
1 11 21 1 12 22
1 11 21 1 12 22
(1 )( / ) (1/ )
(1 )( / ) (1/ )
/ (1 ) 1/ ( / ) (1/ ) (1 )
( / ) (1 ) (1/ ) ( / ) (1/ ) (1 )
d d
d d
d d
m c m cb l lD
I c I ca l l
b l m c l I c b l m c l I c
a l m c l I c a l m c l I c
+ − = +
+ − − +=
+ + + +
Figure 2.25 Rigid body displacement of the shaft
From Figure 2.25, we have
( )1 1
b A
A A B A B
x x a a b ax x a x x x x
l l l l l
− = + = − + = +
(F)
and
( ) ( ) BAAB xlxl
l
xx/1/11 +−=
−=φ (G)
On combining equations (F) and (G), we get
1
1
[ ]A
B
x xG
xφ
=
with ( / ) ( / )
[ ](1/ ) (1/ )
b l a lG
l l
= −
(H)
Hence, from equations (E) and (H), we have
xA xB x1
1φ
a b
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171
( ) ( )
−
++++
+−−+=
B
A
dd
dd
B
A
x
x
ll
lalb
cIlcmlacIlcmla
cIlcmlbcIlcmlb
F
F
)/1()/1(
)/()/(
)1()/1()/()/1()1()/(
)1()/1()/(/1)1(/
2212121111
2212121111
11
11
which can be simplified as
A AA AB A
B BA BB B
F Z Z x
F Z Z x
=
(I)
with
( ) ( ){ }{ }( )
( ) ( ){ }( )
{ }( )
{ }( )
{ }( )
1 1
1 1
1
1
1 11 21 1 11 21
1 12 22 1 12 22
1 11 21 1
1 12 22
/ (1 ) 1/ ( 1/ ) / (1 ) 1/ /
( / ) (1/ ) (1 ) 1/ ( / ) (1/ ) (1 ) 1/
( / ) (1 ) (1/ ) / ( / )
( / ) (1/ ) (1 ) 1/
d d
d d
AA AB
BA BB
d
d
b l m c l I c l b l m c l I c a l
b l m c l I c l b l m c l I c lZ Z
Z Za l m c l I c b l a l m
a l m c l I c l
+ − − + + − +
− + − − +
=
+ + +
+ + −
{ }( )
{ }( )1
1
11 21
1 12 22
(1 ) (1/ ) /
( / ) (1/ ) (1 ) 1/
d
d
c l I c a l
a l m c l I c l
+ + + + +
Since A & B are pinned support 0A Bx x= = even at critical speeds hence AA ABZ Z= = ∞ i.e.
AAB BA
B
FZ Z
x= = ∞ = . Hence denominator of any of impedance can be put equal to zero to get the
frequency equation. Noting the equation (A), the common denominator of the [Z] matrix (or its
components) are determinant of matrix [A] i.e.
1 1
4 2
11 22 12 21 1 1 11 22( ) ( ) 1 0n d n dm I m Iω α α α α ω α α− − + + =
or
1
1 1
1 11 224 2
11 22 12 21 1 1 11 22 12 21
( ) 10
( ) ( )
d
n n
d d
m I
m I m I
α αω ω
α α α α α α α α
+− + =
− −
From the present problem data, we have the frequency equation of the following form
42 2
2 8 2 8
(10 1.137 0.02 14.146) 10 10
(10 0.02) (1.137 14.146 3.03 ) 10 (10 0.02) (1.137 14.146 3.03 ) 10n nω ω
−
− −
× + × ×− + =
× × × − × × × × − ×
or
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172
4 2 4 8(8.44 10 ) 0.7243 10 0n nω ω− × + × =
which can be solved as
4 4 2 4
2 48.44 10 (8.44 10 ) 4 0.7243 10 8.44 8.26710
2 2nω
+ × ± × − × × ±= = ×
Natural frequencies of the system is given as
129.45nω = rad/sec and
2289.23nω = rad/sec.
3.4 Dynamic Stiffness Matrix Method
It is similar to the transfer matrix method, in that it involves a division of the shaft into a number of
smaller elements for the purpose of analysis. The difference is there in system equation formulation
and the response at all stations in the system is determined simultaneously. The method has the
advantage that once the system matrix has been assembled it can be used to calculate system stability
thresholds as well as response and critical speeds. The major disadvantage is that it requires the
storing and manipulation of large matrices and so is more demanding of computer power than is the
transfer matrix method.
(a) A disc (b) A shaft segment
Figure 3.26 Free body diagram
If a beam element alone as shown in Figure 3.26(b), without concentrated mass, is considered, the
relationships between applied forces and moments and resulting deflections and slopes, are given as
yA yB
θA
θB
QA QB
MB MA MA
MC
QA
QC
Uv
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173
=
b
B
A
A
B
B
A
A
y
y
kkkk
kkkk
kkkk
kkkk
Q
M
Q
M
θ
θ
44434241
34333231
24232221
14131211
(76)
Elements of equation (76) can be obtained by rearranging equations (24-26) of the transfer matrix
method. From the transfer matrix method, a field matrix relates state vectors at two stations as follows
BAQ
M
y
lEI
lEI
lEI
lEI
ll
Q
M
y
−
−=
−
θθ
1000
1002
10
621
2
32
(77)
which can be expanded as
( ) ( )
( ) ( )
2 3
2
/ 2 / 6
/ / 2
A B B B B
A b B B
A B B
A B
y y l l EI M l EI Q
l EI M l EI Q
M M lQ
Q Q
θ
θ θ
− = − + + +
= + +
= −
=
(78)
Equation (78) can be regrouped as
2 3
2
1 2 6
0 12
A B B
l ly l y MEI EI
Ql lEI EI
θ θ
− − = +
(79)
and
0 0 1
0 0 0 1A B B
M y l M
Q Qθ− −
= +
(80)
On rearranging equations (79-80), we have
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2 3
2
1 0 1 0 0 / 2 / 6
0 1 0 1 0 0 / / 2A B A B
y l y M Ml EI l EI
Q Ql EI l EIθ θ− −
− = +
(81)
and
0 0 0 0 1 0 1
0 0 0 0 0 1 0 1A B A B
y y M l M
Q Qθ θ− − −
+ = − +
(82)
On combining equations (81-82), we get
2 3
2
1 0 1 0 0 / 2 / 6
0 1 0 1 0 0 / / 2
0 0 0 0 1 0 1
0 0 0 0 0 1 0 1
A A
B B
y Ml l EI l EI
Ql EI l EI
ly M
Q
θ
θ
− − − = − −− −
(83)
which takes the form of equation (76), with
12 3
11 12 13 14
221 22 23 24
31 32 33 34
41 42 43 44
1 0 10 0 / 2 / 6
0 1 0 10 0 / / 2
0 0 0 01 0 1
0 0 0 00 1 0 1
k k k k ll EI l EI
k k k k l EI l EI
k k k k l
k k k k
−−
− = − −
−
(84)
For the most common type of forcing (i.e. the unbalance) the applied moments and shear forces takes
the form
j j j j; ; ; t t t t
A A A A B B B BM M e Q Q e M M e Q Q eω ω ω ω= = = = (85)
where AM , … are complex in general. Deflections and slopes, similarly can be written as
j j j j; ; ; t t t t
A A A A B B B By Y e e y Y e eω ω ω ωθ θ= = Θ = = Θ (86)
where AY , … are complex in general. Substituting equations (85) and (86) into equation (76), it gives
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175
11 12 13 14
21 22 23 24
31 32 33 34
41 42 43 44
AA
AA
BB
bB
k k k k YM
k k k kQ
k k k k YM
k k k kQ
Θ = Θ
(87)
which can be written in more compact form as
=
}{
}{
][][
][][
}{
}{
2221
1211
B
A
B
A
d
d
uu
uu
F
F (88)
with
{ }M
FQ
=
; { }Y
d
= Θ
; [ ] 13 14
12
23 24
k ku
k k
=
; …
Equation (88) can be expanded to allow for the shear forces, moments, slopes and displacements in
the horizontal direction, as
11 12
11 12
21 22
21 22
{ } { }[ ] 0 [ ] 0
{ } { }0 [ ] 0 [ ]
{ } { }[ ] 0 [ ] 0
{ } { }0 [ ] 0 [ ]
A A
Ah Ah
B B
Bh Bh
F du u
F du u
F du u
F du u
ν ν
ν ν
=
(89)
where subscripts: v and h refer to the vertical and horizontal directions. Equation (89) relates to the
forces & moments at ends A and B to displacements at ends A and B. Now considering the forces and
moments acting on the concentrated mass (i.e. disc as shown in Figure 3.26(a)) at the end of the
element, equations of motion for disc mass are
;
;
A v cv v Ah h ch h
c A P Ah d A ch Ah P A d Ah
Q U Q B my Q U Q B mx
M M I I M M I I
ν
ν ν ν νωθ θ ωθ θ
+ − − = + − − =
− − = − + =
�� ��
� �� � ��
(90)
where p AI ωθ� is the gyroscopic moment, where B is the bearing force (equal to zero if the station
considered is not a baring location), U is a known imbalance, M is the magnitude of concentrated
mass, PI is the polar mass moment of inertia ( 0PI = for gyroscopic effects are to ignored) and Id is
the diametral moment of inertia (Id is related to the rotary inertia). Note that the slopes and
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176
displacements on each side of the concentrated mass are the same. The bearing reaction force may be
expressed in the form
j j; and t t
v v h hB B e B B eω ω= = (91)
where vB and
hB are complex in general. On substituting equation (91) into equation (90) and it
gives: (shear force & bending moment at end A are related with at end C).
2
2
2 2
2 2
;
j ;
j
A c
Ah h ch h
A c p Ah d Av
Ah ch p A d Ah
Q B m Y Q U
Q B m X Q U
M M I I
M M I I
ν ν ν ν
ν ν
ν
ω
ω
ω ω
ω ω
= − + −
= − + −
= − Θ + Θ
= + Θ + Θ
(92)
On substituting equation (92) into equation (89), it gives
2 2
11 12 13 14
2
21 22 23 24
2 2
11 12 13 14
2
21 22 23 24
31 32 33 34
41 42 43 44
31 32
0 j 0 0
0 0 0 0
j 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
v
v
h
h
v
v
h
h
C
d p
C
Cp d
C h h
B
B
B
B
Mk k I I k k
Q U Bk m k k k
M I k k I k k
Q U B k m k k k
M k k k k
k k k kQ
k kM
Q
ν ν
ω ωωω ω
ω
−
− + +
− − − + +
=
33 34
41 42 43 440 0 0 0
C
C
C
C
B
B
B
B
Y
Z
Y
Zk k
k k k k
Θ
Φ Θ Φ
(93)
which can be written as
{ }{ }
0 00 01 0
10 11 11
[ ] [ ] { }
[ ] [ ] { }
F M M d
M M dF
=
(94)
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where subscripts refer to node numbers and the matrix ][ 0M is the dynamic stiffness matrix for
element 0. Similarly for element 1, it will take the form:
{ }{ }
[ ][ ] [ ]
0 11 12 1
21 21 22
{ }
{ }
F M M d
dF M M
=
(95)
where 1{ }F is similar to }{ 1F but also contains imbalance forcing terms and bearing force terms (as
does 0{ }F ). Equations (94) and (95) may be combined to eliminate the internal force and moment
terms of the matrix }{ 1F and it will give overall equation for element 0 and 1
{ }{ }{ }
000 01 0
*
1 10 11 11 12 1
2 21 22 2
[ ] [ ] [0] { }
[ ] [ ] [ ] [ ] { }
[0] [ ] [ ] { }
FM M d
F M M M M d
F M M d
= +
(96)
where { }*1F contains only the imbalance and bearing force terms of matrix { }1F . Equation (96) may
be extended for any number of system elements to give an overall system matrix equation of the form
=
−
}{
}{
}{
}{
}{
*1
*2
*1
0
n
n
F
F
F
F
F
…
}]{[}{
}{
}{
}{
}{
}{
*
1
2
1
0
sZPor
d
d
d
d
d
n
n
=
−
…
(97)
The left hand side of equation (97) contains only known imbalance forcing terms, known applied
forces and moments at the shaft ends (usually zero), and unknown bearing reaction forces. Bearing
reaction forces can be found in the following steps
}{][}{ 1 PZs −= (98)
Equation (98) will give an expression for displacement at each bearing location. In the case of
bearings, which behave as pinned supports, these expressions can be equated to zero and solved
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178
simultaneously to give values for the bearing reaction forces. The back substitution of these forces
into equation (98) then enables the shaft displacements at all other locations to be evaluated. It is
noteworthy that the system dynamic stiffness matrix [Z] is banded about the leading diagonal; this can
be made use of when storing the matrix in the computer, since only non-zero elements values need to
be stored.
Example 3.6 Obtain transverse synchronous critical speeds of a rotor system as shown in Figure 3.27.
Take the mass of the disc, m = 10 kg, the diametral mass moment of inertia, Id = 0.02 kg-m2, the polar
mass moment of inertia, Ip = 0.04 kg-m2. The disc is placed at 0.25 m from the right support. The
shaft is having diameter of 10 mm and total span length of 1 m. The shaft is assumed to be massless.
Take shaft Young’s modulus E = 2.1 × 1011 N/m2. Neglect gyroscopic effects. Take one plane motion
only.
a b
l = a + b
Figure 3.27 A rotor system
Solution: Figure 3.28 shows the free body diagram of shaft elements and disc without gyroscopic
effects.
(a) (b)
(c) (d) (e)
Figure 3.28 A shaft-disc element free-body diagram
QA, yA
A
MA,
θA
MB,
θB
B
QB, yB
A B
C MC
QC
MA
QA
C A
C A
y y
θ θ
=
=
D C A B D C
MD QD QC MC
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For the shaft element DC as shown in Figure 3.28(b), we have
2 2
1
2
1
12 6 12 6
4 6 2
12 6
sym 4
D D
D D
C C
C C
Q yl l
M l l lk
Q yl
M l
θ
θ
− − = − −− − −
(A)
with
1 3
1
EIk
l=
For the beam element AB as shown in Figure 3.28(c), we have
2 2
2
2
2
12 6 12 6
4 6 2
12 6
sym 4
A A
A A
B B
B B
Q yl l
M l l lk
Q yl
M l
θ
θ
− − = − −− − −
(B)
with
2 3
2
EIk
l=
For a disc as shown in Figure 2.38(e), we have the following relationship
andA C C C A d CQ Q my M M I θ− = − = ����
which can be rearranged as
andA C C A C d CQ Q my M M I θ= + = − ���� (C)
Following conditions hold at disc
CA yy = and AC θθ = (D)
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Substituting equations (C) and (D) into equation (B), we get
2 2
2
2
2
12 6 12 6
4 6 2
12 6
sym 4
C C C
C d C C
B B
B B
Q my yl l
M I l l lk
Q yl
M l
θ θ
θ
+ − − − = − −− − −
��
��
(E)
For SHM, we have 2
C n Cy yω= −�� and 2
C n Cθ ω θ= −�� , from equation (E), we have
( )( )
2
2
2 2 2
22
2
2
12 / 6 12 6
4 / 6 2
12 6
sym 4
nC C
C Cd n
B B
B B
m k l lQ y
M l I k l lk
Q yl
Ml
ω
θω
θ
+ − − −= − − − − −
(F)
On combining equations (A) and (F), we get
1 1 1 1 1 1
2 2
1 1 1 1 1 1 1
1 2
1 1 1 1 1 2 2 2 2 22
2 2
1 1 2 22
1 1 1 1 1 1 2 2 2 22
12 6 12 6 0 0
6 4 6 2 0 0
12 1212 6 6 6 12 6
4 46 2 6 6 6 2
D
D
nC C
C C
Bd n
B
k l k k l k
l k l k l k l kQ
k kMk l k l k l k k l k
mQ Q
M M l k l kl k l k l k l k l k
Q I
M
ω
ω
−
− − + − − + −
+ − + =
− + − + + − − − −
2
2 2
2 2 2 2 2 2
2 2
2 2 2 2 2 2 2 2
0 0 12 6 12 6
0 0 6 2 6 4
D
D
C
C
B
B
y
y
l ky
k l k k l k
l k l k l k l k
θ
θ
θ
− − −
−
(G)
Boundary conditions are
0,0 ==== BDBD MMyy (H)
On application of boundary conditions, equation (G) can be written as
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181
( )
( )
2 2
1 1 1 1 1 1
1 2
1 1 1 1 2 2 2 22
2 2
1 1 2 22 2
1 1 1 1 2 2 2 22
2 2
2 2 2 2 2 2
4 6 2 0
12 1206 6 6 6
0
0 4 42 6 6 2
0
0 6 2 4
D
n C
C
Bd n
l k l k l k
k kl k l k l k l k
m y
l k l kl k l k l k l k
I
l k l k l k
θω
θθω
−
− + − + − + = − + + − −
−
(I)
Equation (I) represents an eigen value problem. The non-trivial solution can be obtained by equating
determinant equal to zero, which will give natural frequencies of the system.
( ) ( ){ } ( )24 2 2 2 2 2 2 2 2 2 2 2
1 2 1 2 1 2 1 2 1 1 2 2 2 1 1 2 1 2 1 2(16 ) 48 144 0n d n dmI k k l l k k l l m k l k l I k k k k l l l lω ω + − + − + + =
After substituting values k1 = 2.443×102 N/m , k2 = 6.597×10
2 N/m, l1 = 0.75 m, l2 = 0.25 m, m = 10
kg and 0.02dI = kgm2, we get
5 4 9 2 131.814 10 7.132 10 1.316 10 0n nω ω× − × + × =
Natural frequencies of the system is given as
144.05nω = rad/sec and
2193.36nω = rad/sec
Example 3.7 Obtain the bending critical speed of the rotor system as shown in Figure 3.29. Take the
mass of the disc, m = 5 kg and the diametral mass moment of inertia, Id = 0.02 kg-m2. Take shaft
length a = 0.3 m and b = 0.7 m. The diameter of the shaft is 10 mm. Use the dynamic stiffness matrix
method. Neglect the gyroscopic effects.
a b
A B
Figure 3.29 An overhang rotor system
For a beam element as shown in Figure 3.30, following relations holds
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1 1
2 2
1 1
3
2 2
2
2 2
12 6 12 6
4 6 2
12 6
sym 4
F yl l
M l l lEI
F yll
M l
θ
θ
− − = −
M1, θ1 F1, y1 F2, y2 M2, θ2
Figure 3.30 Positive sense of beam displacements and forces
Solution: Figure 3.31 shows the free body diagram of the rotor elements with neglecting the
gyroscopic effects.
(b) (c)
lCM
QC
2Q DM
D
C D
Figure 3.31 Free body diagram of rotor elements
For shaft segment AC as shown in Figure 3.31(b), we have following relations
1 1
2
1 1 1
3
11
2
1
12 6 12 6
4 6 2
12 6
sym 4
A A
A A
C C
C C
Q Ql l
M Ml l lEI
Q Qll
M Ml
− − = − −− − −
(A)
For the disc as shown in Figure 3.31 (c) for SHM, we have
ba
C DB A
(l )1 (l )2
AQ
lMA QCMC
1 MB
QB
QAMA
Shaft
Disc
(a)
(d)
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2
2
A b A B n
B A d B A B d n
Q Q my Q Q y
M M I M M I
ω
θ θω
− = ⇒ = −
− = ⇒ = +
��
��
(B)
On substituting and rearranging equation (B) into equation (A), we get
21 2 1
221 1 1
3
11
2
1
12 6 6
4 6 2
12 6
sym 4
BB n
Bb d n
CC
CC
yl l lQ m y
l l lM I EI
yllQ
lM
ωθω θ
θ
− − −+ = −−− −−
(C)
On rearranging equation (C), we get
2
1 1 1 1 1 1
2 2 2
1 1 1 1 1 2
1 1 1
2
1 1
(12 ) 6 12 6
(4 ) 6 2
12 6
sym 4
B Bn
B Bd n
C C
C C
Q yk m l k k l k
M l k I l k l k
Q yk l k
M l k
ωθω
θ
+ − − − = − −− − −
(D)
where 311
EIkl
=
For shaft C-D as shown in Figure 3.31(d), we have
−
−
−
−
−
=
−
−
D
D
C
C
D
D
C
C
y
y
kl
klk
klklkl
klkklk
M
Q
M
Q
θ
θ
2
2
2
222
2
2
2222
2
2
222222
4
612
1264
612612
(E)
where 3
22 / lEIk =
By assembling equations (D) and (E), we get
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184
−
−−−
−++−
−+−+−−
−−
−+
=
−
−
+−
+−
D
D
C
C
b
B
d
D
D
CC
CC
B
B
y
y
y
klklklkl
klkklk
klklklklklklklkl
klkklklkkklk
klklIklkl
klkklmk
M
Q
MM
M
Q
θ
θ
θωω
222222
2222
222222
22
22222
221
2122111
2111
222221121111
12111
21
2111
111112
1
4612600
61261200
126)44()66(26
612)66()1212(612
0026)4(6
006126)12(
The boundary conditions are 0B B DQ M M= = = and 0C Dy y= = . On right hand side already third
and fourth columns elements are zero. So fifth row and fifth column has eliminated from equation (3)
and it is given as
( ) ( )( )
2
1 1 1 1 1 1
2 2
1 1 1 1 1 1 1 1
1 1 1 1 2 1 1 2 2 2 2
2 2 2 2
1 1 1 1 1 1 2 2 1 1 2 2 2 2
2 2
2 2 2 2 2 2
(12 ) 6 12 6 00
6 (4 ) 6 2 00
12 6 12 12 6 6 6
6 2 6 6 (4 4 ) 12
0 0 6 12 4
B
d
B
c
D
k m l k k l ky
l k l k I l k l k
k l k k k l k l k l k
l k l k l k l k l k l k l k
l k l k l k
ωω
θ
θθ
+ −
− − =− − + − + − − + +
0
0
[ ]{ } {0}K d =
The non-trivial solution can be obtained by equating determinant equal to zero, which will give
natural frequencies of the system, as
( ) ( ){ } ( )4 2 2 2 2 2 4 2 2 2 2 2 2 4 4
2 2 2 2 1 1 1 2 2 1 1 2 1 2 2 2 1 1 1 2 1 24 3 4 48 3 144 0n d n dmI k l k l k l k k l m k l k l l I k l k l k k l lω ω − + − + − − =
Given data: mass of the disc, m = 5 kg, diametral mass moment inertia 02.0=dI kg-m2, shaft
lengths 0.3 m and 0.7 m. Diameter of the shaft d = 0.01 m. After substituting values k1 = 3.818×103
N/m, k2 = 3.005×102 N/m, l1 = 0.3, l2 = 0.7, m = 5 kg and Id = 0.02kgm
2, we get
4 4 9 2 125.494 10 2.4375 10 4.0967 10 0n nω ω− × + × − × =
Natural frequencies of the system is given as
141.83nω = rad/sec and
2206.44nω = rad/sec
Exercise 3.6. Obtain the bending critical speeds of an overhang rotor system as shown in Figure E3.6.
The end B1 of the shaft is having fixed end conditions. Length of the shaft is 0.4 m and diameter is 0.1
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185
m. The disc is thin and has 1 kg of mass, 0.04 kg-m2 of polar mass moment of inertia and 0.02 kg-m2
of diametral mass moment of inertia. Neglect the mass of the shaft and consider the gyroscopic
effects. Take the shaft speed of 10,000 rpm. Use the dynamic stiffness method.
3.5 Dunkerley’s Formula
It can be used to calculate the machine natural frequencies without recourse to the numerical
methods. This method gives very crude estimation of natural frequency. From influence coefficient
method the natural frequency of the system is obtained by the following conditions
0=′A (99)
For three degrees of freedom system; above equation will give
…+
+++
−2
2333222111
3
2
1)(
1
ωωmamama (100)
but for a polynomial whose first coefficient is unity, the second coefficient is equal to minus of the
sum of the roots of the equation
( ) ( ) ( )2 2 2
1 2 3 11 1 22 2 33 31/ 1/ 1/ a m a m a mω ω ω+ + = + + (101)
2 2 2
1 1 2 2 3 3 11 22 33/ / / 1/ 1/ 1/m k m k m k ω ω ω= + + = + + (102)
where 2211 ,ωω and 33ω are the natural frequencies of the system (i.e. roots of the polynomial in
equation (100)) when only mass 1 2 3, or m m m is present. In most cases the fundamental frequency
1ω will be much lower than the other natural frequencies, so equation (4) may be approximated, in
general, to
B1
D1
Figure E3.6
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186
2 2 2 2
1 11 22 33
1 1 1 1
ω ω ω ω= + + (103)
Dunkerley first suggested this. Equation (103) always gives a value for fundamental frequency, which
is slightly lower than the true value, by virtue of the approximation involved.
Example 3.8 Find the fundamental critical speed of the rotor system shown in Figure3.32. Take EI =
2 MN.m2 for the shaft and mass moment of inertia of disc is negligible.
3 m
1.5 m
Fixed end 80 kg 100 kg
Figure 3.32 An overhang rotor system
Solution:
F1 F2
l l/2 l/2
(a) Case 1 (b) Case 2
Figure 3.33 Overhang rotor systems with a single disc
For Figure 3.33, we have
11 223 3
1 2
3 3;
EI EIk k
l l= = with 1l l= and 2 / 2l l=
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187
2 2
11 223 3
1 1 2 2
3 3;
EI EIω ω
m l m l= =
The system critical speed is given as
3 3
1 2
2 2 2
11 22
41 21 1 14.1625 10
3 3
m l m l
EI EIω ω ω−= + = + = ×
or 49.014ω = rad/sec (Fundamental frequency)
Exercise 3.7 Find the fundamental bending critical speed of the rotor system shown in Figure E3.7.
B1 and B2 are simply supported bearings and D1 and D2 are rigid discs. The shaft is made of steel with
modulus of rigidity E = 2.1 (10)11 N/m2 and uniform diameter d = 10 mm. The various shaft lengths
are as follows: B1D1 = 50 mm, D1D2 = 75 mm, and D2B2 = 50 mm. The mass of discs are: md1 = 4 kg
and md2 = 6 kg. Consider the shaft as massless and neglect the diametral mass moment of inertia of
both discs.
Finite element method is now very popular among all other methods for continuous and complex
rotor-bearing systems. In subsequent chapters we will discuss this method in detail.
B1 B2
D1 D2
Figure E3.7