DYNAMICS OF ROTATING
MACHINES
Michael I Friswell, John E T Penny, Seamus D Garvey and
Arthur W Lees
SOLUTION MANUAL
Version 1. July 2011
The authors welcome any comments and corrections.
Supporting MATLAB scripts and functions have been written with emphasis on
clarity, not necessarily on efficiency or compactness.
1
Chapter 2
Problem 2.1
From Equation (2.24), the response is given
by x ( t ) e −ζωnt ( a0 cos ωd t + b0 sin ωd t ) .
=
Thus x =−ζωn eζωnt ( a0 cos ωd t + b0 sin ωd t ) + ωd eζωnt ( −a0 sin ωd t + b0 cos ωd t ) .
When t = 0 , x ( 0=
) x=
0 a0 and x ( 0 ) = x0 = −ζωn a0 + ωd b0 .
Hence =
b0
( x0 + ζωn x0 )
ωd .
Now m = 1kg , k = 9 N m , x0 = 10−3 m , x0 = 0 .
Thus ζ =0 and b0 = 0 , ωn =ωd = k m = 9 =3rad s . So x ( t ) = 10−3 ( cos 3t ) m .
If c = 1Ns m , 2ζω
ζ c ( 2mω=
3) 0.1666 and
=
m 1 . Then=
n ) 1 ( 2 ×=
n c=
0.169 x0 .
ωd = ωn 1 − ζ 2 = 3 1 − 0.16662 = 2.958 rad s . b0 =
( 0.1666 × 3x0 ) 2.958 =
Hence x ( t ) 10 −3 {cos ( 2.958t ) + 0.169sin ( 2.958t )} m
=
Problem 2.2
For the shaft, d =
0.01m, L =
0.40m, J =
9.818 10−10 m 4 .
πd 4 32 =×
ksh =GJ L =80 ×109 × 9.818 ×10−10 0.15 =523.60N m .
For the disk, I = ρπhD 4 32 = 0.1960kg m 2 . Thus
=
ω
=
ksh I
523.60 0.1960
= 51.68 rad s . ω=
( 2π ) 8.23Hz .
( 2π ) 51.68=
Problem 2.3
ω2 =k m . Thus k = (1× 2π ) 700 = 27635 N m = 27.635 kN m .
2
Problem 2.4
. x0 1mm
= 10−3 m .
ω= 50 × 2π= 314.159 rad s =
v0 =
ωx0 =
314.159 ×10−3 =
0.3142 m s , a0 =
ω2 x0 =
314.1592 ×10−3 =
98.7 m s 2 .
Problem 2.5
Critical damping ζ =1 . Now 2ζωn = c m so that
cc = 2ωn m = 2 × (1× 2π ) × 700 = 8796.5 Ns m .
Problem 2.6
−ζωn ( t +T )
c0 cos ( ωd ( t + T ) − φ ) . Now
x ( t + T ) e −ζωn ( t +T )
cos ( ωd ( t + T ) −=
φ ) cos ( ωd t − φ ) so that= =
e−ζωnT .
−ζω
t
n
x (t )
e
=
x ( t ) e −ζωnt c0 cos ( ωd t − φ=
) and x ( t + T ) e
Hence
 x (t ) 
x (t )
2π
=
= eζωnT . Thus δ = log e 
 = ζωnT = ζωn
ωd
x (t + T )
 x (t + T ) 
2
2πζ
1 − ζ2
.
Problem 2.7
Let xs=
( t ) As cos ωt + Bs sin ωt . Substitute into equation (2.29) we have
−ω2 ( As cos ωt + Bs sin ωt ) + 2ζω ( − As sin ωt + Bs cos ωt )
 + ω2n ( As cos ωt + B
=
s sin ωt )
( f0
m ) cos ωt
Collecting together terms in cos ωt and sin ωt and letting f 0 = 1 gives
{( ω
{( ω
(1 m ) cos ωt
)
}
2
2
n − ω ) Bs − 2ζωn ωAs } sin ωt = 0
2
n
− ω2 As + 2ζωn ωBs=
cos ωt
Dividing by ω2n and noting that generally, cos ωt ≠ 0 and sin ωt ≠ 0 we have
(1 − r 2 ) As + 2ζrB=s 1 ( mωn2 )
(1 − r 2 ) Bs − 2ζrAs =0
Solving this pair of equations gives expressions for As and Bs identical to Equation
(2.33).
Problem 2.8
In Equation (2.35) let r 2 = λ .
Then,
Cs
=
(
)
1 mω2n
1 
2
2  −1 2
.
+
λ
ζ
1
2
=
−
λ
(
)
(
)
2 

2
2
ω
m
n
(1 − λ ) + λ ( 2ζ )
Differentiating this expression gives
d ( Cs )
1 
2
2  −3 2 
2
1
2
2 (1 − λ )( −1) + ( 2ζ )  .
=−
−
λ
+
λ
ζ
(
)
(
)
2




dλ
2mωn
Setting this expression to zero to determine the value of λ for the maximum response,
2
we have  2 (1 − λ )( −1) + ( 2ζ )  = −2 (1 − λ ) + 4ζ 2 = 0 . Thus at maximum response


(resonance) λ = 1 − 2ζ 2 and hence r =
1 − 2ζ 2 . Thus the resonant frequency ωr is
given by ωr =ωn 1 − 2ζ 2 . Note that ωr is not equal to ωd .
Problem 2.9
5 Hz= 10π s. r= ω ωn= 10π 11π= 1.1 .
(a) From Equation (2.34) it is clear that the response is proportional to the input
force. Doubling input force doubles response. No change in phase.
(b) If the phase of the input force changes, the phase of the response changes by
the same amount. Magnitude of response unchanged.
(c) If the forcing frequency is halved, r is halved. Using Equation (2.35) with
ζ =0 ,
3
(
then
=
( Cs )r =1.1
Thus
)
)
(
)
1 mω2n
1 mω2n
, ( Cs )r =0.55
=
=
2
0.21
1 − 1.12
(
( Cs )r =0.55
=
( Cs )r =1.1
(
)
(
)
1 mω2n
1 mω2n
.
=
2
0.6975
1 − 0.552
(
)
0.21
= 0.3011 - a reduction to 30% of the original response.
0.6975
Because the forcing frequency is now below the natural frequency, the phase
changes by 180° to zero.
(d) Initially,
=
( Cs )ζ=0
(
(
( Cs )ζ=0.1
=
( Cs )ζ=0.0
(
)
(
)
(
)
1 mω2n
1 mω2n
,
=
0.3041
2
2 2
1 − 1.1 + ( 2 × 0.1×1.1)
With damping, ( Cs )ζ=0.1
=
Thus
)
)
1 mω2n
1 mω2n
.
=
0.21
2 2
1 − 1.1
(
)
0.21
= 0.6905 - a reduction to 69% of the original response.
0.3041
2ζr 
 2 × 0.1× 1.1 
tan −1 
=
−46° . i.e. −46 + 180 = 134° .
=
2
 1− r 
 1 − 1.12 
( φs )ζ=0.1 =tan −1 
Problem 2.10
In this case f ( t ) = t T . Thus from Equation (2.43) b0 = 0 and
T
T
T
1
1 t
1 t2
1
=
a0
=
f
t
dt
=
dt
=
(
)
∫
∫
T0
T 0T
T 2T
2
0
T
T
2
2
 2πnt 
From Equation (4.42)=
bn
f ( t ) sin ω0 nt
dt
f ( t ) sin 
=
 dt .
∫
∫
T0
T0
 T 
Let 2πnt T =
dt
x so that=
2
Hence bn = 2
( 2πn )
Similarly, a=
n
2 πn
∫
0
(T
2πn ) dx . When
=
t 0,=
x 0 and when t = T , x = 2πn .
2
2 πn
x sin x dx = 2 sin x − x cos x 0 =
−1 ( nπ ) .
( 2πn )
2 πn
2
∫
( 2πn ) 0
2
x cos x dx=
2
( 2πn )
2 πn
2
cos x + x sin x 0 = 0 .
t 1 ∞ 1
 2πnt 
Thus the Fourier series for the function is =
− ∑ sin 
.
T 2 n=1 nπ  T 
In this example, T = 2 . The MATLAB script, Problem_02_10.m, uses this series to
determine the Fourier approximation of the function using 10, 50 and 100 terms in the
series and provides the following graphical output.
4
10 term Fourier series
50 term Fourier series
1
Force (N)
Force (N)
1
0.5
0.5
0
0
-2
-2
0
2
4
Time (s)
100 term Fourier series
0
2
Time (s)
4
Force (N)
1
0.5
0
-2
0
2
Time (s)
4
Problem 2.11
 5 −4 
1 0 
For this system M = 
kg and K = 
 N m . The equations of motion are

 −4 5 
0 2 
 + Kq =
Mq
0 . Solving these equations for free vibrations leads to an eigenvalue
problem of the form Ku = λMu where ωn =
λ . This equation can be solved by
finding roots of the quadratic 2λ 2 − 15λ + 9 = 0 . Alternatively the eigenvalue problem
can be solved directly, see the MATLAB script Problem_02_11.m. Running this
script gives the following output:
1st mode u2/u1 = 1.0856
2nd mode u2/u1 = -0.46058
1st natural frequency = 0.81097 rad/s
2nd natural frequency = 2.6158 rad/s
1st natural frequency = 0.12907 Hz
2nd natural frequency = 0.41632 Hz
Problem 2.12
The radial force due to each out of balance mass is =
fu mu r Ω 2 , Thus when the
masses have rotated from the vertical by an angle θ , then in the vertical direction the
sum of the forces is f vert =mu r Ω 2 cos θ + mu r Ω 2 cos θ =2mu r Ω 2 cos Ωt , since
θ = Ωt . (Refer to Figure 2.36). In the horizontal direction the sum of the forces is
f hz = mu r Ω 2 sin θ − mu r Ω 2 sin θ = 0 . Thus there is no horizontal force and, so, in the
x direction the equation of motion becomes mx + kx= 2mu r Ω 2 cos Ωt .
5
If a vibration absorber is added, and the displacement of this mass is xa then the
equation of motion for free vibrations are
− ka   xa  0 
xa   ka
 ma 0   
+ 
 .
 =
 0 m   

  x   − k a k a + k   x  0 
Thus, assuming a harmonic solution, we have
 ka − ma ω2
  xa  0 
− ka
0.

   =   . Thus ka − ma ω2 ka + k − mω2 − ka2 =
ka + k − mω2   x  0 
 −ka
(
)(
)
 k + k ka  2 ka k
Multiplying out, and dividing by ma m gives ω4 −  a
+
=0 .
ω +
ma 
ma m
 m
The forced vibration equations are
 ka − ma ω2
  xa   0 
− ka
=

  
 cos Ωt
ka + k − mω2   x   f vert 
 −ka
Inverting the 2 × 2 coefficient matrix, we have
2
ka f vert

 0 
ka
 xa  1  ka + k − mω
1 
=
=
cos
Ω
t


 



 cos Ωt .
D  ka − ma ω2 f vert 
 x  D 
ka
ka − ma ω2   f vert 


(
)
where D is the determinant of coefficient matrix. The response x is zero if k=
ma ω2 .
a
The response of the absorber is ( ka D ) f vert cos ωt where
D=
( ka − maω2 )( ka + k − mω2 ) − ka2 . When k=
Thus xa =− (1 ka )
ma ω2 , D = −ka2 .
a
−2mu r Ω 2
f vert cos Ωt =
cos Ωt
ka
Problem 2.13
 + Kq = 0 and assuming a harmonic solution,
The equation of motion are Mq
Ku = λMu where ωn =
λ . The system mass and stiffness matrices are
10 0 0 
 k2 + k3


 −k
M =  0 10 0  kg , K =
3

 0 0 40 
 −k2
−k3
k3 + k4
− k4
− k2
  20 −10 −10 
=

 kN
−k4
  −10 20 −10  m
k1 + k2 + k4   −10 −10 40 
−1.1861
1


 
T
T
u2 =
−1.1861 , u3 =
−1 . The products u 2 M u3 and u 2 K u3 are both zero.
 1.0000 
0


 
Now, the triple product u3T M u3 = 20 .
Thus uTN 3 =
u3T 20 =
{1 −1 0} 4.47214= {0.22361 −0.22361 0} .
6
Problem 2.14
(a) q ( t )
=
3
∑ u Ni ci cos ( ωit + φi )
and since the masses are initially at rest,
p =1
φ1 =φ2 =φ3 =0 . When t =
= 0 , q0
the coefficients ci . Thus C =
0.17132
Now U N = 0.17132
 0.10161
 1.71324
 −1.43703
and U −N1 =

 2.23603
3
=
u Ni ci U N C , where C is a column vector of
∑
p =1
U −N1q 0 .
−0.14369 0.22361 
1
−0.14369 −0.22361
kg

0.12114
0
1.71324 4.06432 
1

−1.43703 4.84584  kg . Given q 0 = 5 ×10−3 1 m , then
1
0 
−2.23603

0.03745


−1
C U=
Since q ( t )
=
N q 0 0.00986  kg m .=
 0 


3
∑ u Ni ci cos ( ωit ) then
p =1
 0.17132 
−0.14369 




q ( t )  0.17132  0.03745cos ( ω1t ) + −0.14369  0.00986 cos ( ω2t ) . Hence
=
0.101611
 0.12114 




0.006417 
−0.001417 




q ( t ) 0.006417  cos ( ω1t ) m + −0.001417  cos ( ω2t ) m ,
=
0.003806 
 0.001194 




6.42 
−1.42 




or q ( t ) 6.42  cos ( ω1t ) mm + −1.42  cos ( ω2t ) mm .
=
 3.81
 1.19 




 + Kq =
(b) We have Mq
Q ( t ) . Letting q = U N p and pre-multiplying by U TN we
 +
 + U TN K U N p =
have U TN M U N p
U TN Q ( t ) or pΛp
P=
(t ) .
This set of uncoupled equations can then be solved individually for pi .
P (t )
=U TN Q
 0.17132 0.17132 0.10161  0 
10.16 
 


( t ) = −0.14369 −0.14369 0.12114  0  cos ( ωt ) =12.11 cos ( ωt )
 0 
0  100 
 0.22361 −0.22361


2
2
2
, ω22 42.9310
=
ω12 20.1725
=
406.9297 rad 2 s=
=
1843.0707 rad 2 s 2 ,
2
=
ω32 54.7723
=
3000.0048 rad 2 s 2 so that
7
0
0
 406.9297


 rad 2 s 2 . Hence the uncoupled equations of
Λ=
0
1843.0707
0


0
0
3000.0048
p2 + 1843.0707 p2 =
12.11cos ( ωt ) and
p1 + 406.9297 p1 =
10.16 cos ( ωt ) , 
motion are 

p3 + 3000.0048 p3 =
0 . Note ω = 6 Hz = 6 × 2π = 37.699 rad s . Solving these three
second order differential equations gives the following steady state solutions:
p1 = −0.0100 , p2 = 0.0287 and p3 = 0 . Thus
0.17132 −0.14369 0.22361  −0.010 


q ( t ) =U N p ( t ) =0.17132 −0.14369 −0.22361  0.0287  cos ( ωt ) m
 0.10161 0.12114
  0 
0
−0.00584 


=
−0.00584  cos ( ωt ) m
 0.00246 


Fk1 =
k1q3 =
30000 × 0.00246 =
73.8N
Fk 2 =k2 ( q1 − q3 ) =10000 × ( −0.00584 − 0.00246 ) =−83N .
=
Fk 3 k3 ( q2 −=
q1 ) 10000 × ( −0.00584 + 0.00584
=
) 0.
Fk 4= k4 ( q3 − q2 )= 10000 × ( 0.00246 + 0.00584 )= 83N .
(c) Response at coordinate 1 due to a harmonic force of 100 N at coordinate 3 is
−0.00584 m . Therefore the response at coordinate 1 due to a harmonic force of 1 N is
−5.84 ×10−5 m . Thus the receptance α13 =
−5.84 ×10−5 m N .
(
)
Mobility Y13 = jωα13 = j 37.6991× −5.84 ×10−5 = −2.2016 ×10−3 j m Ns .
(
)
Inertance, A13 = −ω2α13 = −37.69912 × −5.84 × 10−5 = −8.3039 × 10−2 m Ns 2
The MATLAB script Program_02_14.m repeats these calculations and gives the
following output:
q_1(t) = {6.4167*cos(omega1*t)}+ {-1.4167*cos(omega2*t)} (mm)
q_2(t) = {6.4167*cos(omega1*t)}+ {-1.4167*cos(omega2*t)} (mm)
q_3(t) = {3.8056*cos(omega1*t)}+ {1.1944*cos(omega2*t)} (mm)
Response at q1 = -5.8428 mm
Response at q2 = -5.8428 mm
Response at q3 = 2.4611 mm
Force
Force
Force
Force
in
in
in
in
spring
spring
spring
spring
1
2
3
4
=
=
=
=
73.8337 N
-83.0392 N
-0.0000 N
83.0392 N
Receptance = -5.8428e-005 m/N
Mobility
= -2.2027e-003 j m/Ns
Inertance = 8.3039e-002 m/Ns^2
8
Problem 2.15
 + Kq = 0 and assuming a harmonic solution,
The equations of motion are Mq
Ku = λMu where ωn =
λ . The system mass and stiffness matrices are
 q1 
1 0 0 0 
q 
0 1 0 0
 2
 kg ,
q =
M 
=
,


q
0
0
1
0
 3


q4 
0 0 0 1 
− k2
0
0 
 k1 + k2
 −k
k2 + k3
−k3
0 
2

K =
 0
k3 + k4 − k4 
− k3


0
k4 
− k4
 0
0 
 20 −10 0
 −10 14
−4
0  3 N

10
.
 0
−4 14 −10 
m


0 −10 10 
 0
(a) Remove q3 and q4 : Partition the matrices thus
0 
 20 −10 0
1 0 0 0 
 −10 14
0 1 0 0
−4
0  3 N



.
kg , K =
10
M=
0 0 1 0
 0
−4 14 −10 
m




0 −10 10 
0 0 0 1 
 0
With reference to Equation (2.114)
 14 −10  3
0 −4  3
−1  0.25 0.25 −3
10 , K ss
10 , K sm 
=
=
K ss =



 10
 −10 10 
0.25 0.35
0 0 
1 0 
I

 0 1 
.
Thus, using Equation =
(2.115), Ts =
 
−1


0
1
K
K
−
 ss sm 


0 1 
1 0 
 20 −10  3 N
T
T
From Equation (2.113)=
kg,
=
M r T=
MT 
K r T=
KT 

 10 m .
0 3
 −10 10 
The eigenvalue problem of this reduced system can be solved to give two natural
frequencies.
(b) Stage 1: The ratio of kii to mii is 20000, 14000, 14000 and 10000. The coordinate
with the highest ratio (i.e. coordinate 1) is the first coordinate to be eliminated from
the systems as follows. Reorder the rows and columns so that the new order is
[2 3 4 1] . Hence
1 0 0 0 
0 1 0 0
 kg K
=
M r 0 =
r0
0 0 1 0


0 0 0 1 
Thus K ss = 20000, K sm =
and
−4
0 −10 
 14
 −4 14 −10 0 

 103 N q
=
r0
 0 −10 10
0 
m


0
20 
 −10 0
[ −10
q2 
q 
 3
q  .
 4
 
 q1 
0 0]103 . Hence, P =
−K −ss1K sm =
[0.5 0 0 ]
9
1

I  0
=
T =

P   0

0.5
0 0
1.25 0 0 
1 0 
T
. Hence,=
M r1 T=
MT  0 1 0  kg and
0 1
 0 0 1 

0 0 
0 
 9 −4
q2 
 


T
3 N
K r1 ==
T KT  −4 14 −10  10
and q r1 =
 q3  (from Equation (2.115) ).
m
q 
 0 −10 10 
 4
Stage 2: The ratio of kii to mii is 7200, 14000 and 10000 for coordinates 2, 3 and 4.
Thus we remove coordinate with the highest ratio, coordinate 3. Reordering the rows
and columns into the sequence [ 2 4 3] we have
1.25 0 0 
9
0
−4 



 3 N
M r1 =
 0 1 0  kg K r1 =
 0 10 −10  10
m
 0 0 1
 −4 −10 14 




q2 
 
q r1 =
q4  .
q 
 3
Thus K ss =
14000, K sm =
−K −ss1K sm =
[ −4 −10]103 , and so P =
[0.2857 0.7143]
 1
0 

I 
and=
T =
0
1  . Hence, using Equation (2.115) gives


P  

0.2857 0.7143
q 
1.3316 0.2041
 7.8571 −2.8571 3 N
M r 2 =
=
qr 2  2  .
kg K r 2 
10


m
0.2041 1.5102 
 −2.8571 2.8571 
q4 
The eigenvalue problem of this reduced system can be solved to give two natural
frequencies.
(c) Adding the masses at coordinates 1 and 2 together, and also at 3 and 4 together
gives the following mass and stiffness matrices.
 k1 + k4 −k4  14 −4  3 N
2 0
10
.
M ==
 0 2  kg , K =
k4   −4 4 
m


 − k4
The eigenvalue problem of this reduced system can be solved to give two natural
frequencies.
 q1  1 0 
q  1 0  q
 
  1  .
(d) If q1 = q2 and q3 = q4 then  2  = 
 q3  0 1  q3 
q4  0 1 
Applying Equation (2.113) to the original mass and stiffness matrices gives
14 −4  3 N
2 0
kg and K = 
M=
 10 m . These matrices are identical to those of

 −4 4 
0 2
part (c).
The MATLAB script Problem_02_15.m solves this problem and gives the following
output:
10
Full system
1st natural
2nd natural
3rd natural
4th natural
model
frequency
frequency
frequency
frequency
=
=
=
=
4.8538 Hz
13.2727 Hz
23.5933 Hz
26.6981 Hz
Retaining q1 and q2
1st natural frequency = 6.2228 Hz
2nd natural frequency = 23.5014 Hz
Reducing system by elimination coord with highest k/m term
thus Retaining q2 and q4
1st natural frequency = 4.9189 Hz
2nd natural frequency = 13.8693 Hz
Constraining q1 = q2 and q3 = q4
1st natural frequency = 5.735 Hz
2nd natural frequency = 13.9672 Hz
Note that the reduction in stages, eliminating the coordinate with the highest kii to
mii ratio gives the most accurate estimate for the two lowest natural frequencies.
Problem 2.16
If the amplitude of motion is x0 then the energy dissipated over a quarter of a cycle
by the force f dry is f dry x0 . Hence over one complete cycle the energy dissipated is
W = 4 f dry x0 . The energy dissipated over one cycle by viscous friction is
W =∫
t = 2π ω
t =0
Let x x0 sin ( ωt ) , x =x0 ω cos ( ωt ) and so x 2 =
x02 ω2 cos2 ( ωt ) .
cx 2 dt . =
Hence letting θ = ωt , W = cx02 ω∫
2π
0
cos2 θ d θ = cωπx02 . The equivalent viscous
damping can be determined by equation energies, thus W = cωπx02 = 4 f dry x0 . Hence
( πωx0 ) . The equation of motion is
mωn ) 4 f dry ( πωx0 × 2mωn ) . Hence
( 2=
the equivalent viscous damping=
is ceq 4 f dry
=
ζ ceq
mx + ceq x + kx
= f 0 cos ωt . Thus
=
2ζr 4 f dry
( πx0 k )
system is x0 =
where r =ω ωn . The amplitude of the response of a forced
f0 k
(1 − r )
2 2
. Thus, substituting for ζ for a system with a dry
+ ( 2ζr )
friction damper, we have x02 =
2
f 02 k 2
(1 − r )
2 2
x02
(1 − r )
2 2
 4 f dry 
+

 πx0 k 
2
2
. Rearranging we have
 4 f dry 
f 02
. Further rearrangement gives
+
=

k2
 πk 
11
x02
(1 − r )
2 2
2
4 f dry
f0 1 − δ2
f 02   4 f dry  

x
=
.
Thus
where δ =
. To ensure
= 2 1 − 

0
2
πf 0
k   πf 0  
k
1
−
r


(
)
that x0 is real, 1 − δ2 > 0 and hence f0 f dry > 4 π .
We can also determine the phase of the response because for a single degree of
4 f dry ( πx0 k )
2ζr
.
Thus
, but
freedom system with damping tan φ =
tan
φ
=
1− r2
1− r2
4 f dry
f 1 − δ2
δ
x0 k = 0
and hence
. Thus we see that the phase
=
=
tan φ
1− r2
πf 0 1 − δ2
1 − δ2
(
)
angle is not a function of frequency for this system. For the above analysis to be valid,
δ < 1 , and this implies that f dry < πf 0 4 . In the example, m = 1kg ,=
k 100 ×103 N m ,
ωn
ω = 25 Hz , f dry = 10 N and f 0 = 50 N . Thus =
k=
m 316.23rad/s ,
.δ
=
ω= 25 × 2π= 157.08rad/s and hence r =ω ωn = 0.4967
4 f dry 4 ×10
= = 0.2546 .
πf 0
π× 50
f 1 − δ2
50 1 − 0.25462
=
=
6.42 ×10−4 m ≡ 0.642 mm .
Thus x0 =0
2
5
2
k 1− r
10 1 − 0.4967
(
)
(
)
 180  −1  δ 
=
φ 
=
 14.75° . The MATLAB script Problem_02_16.m repeats
 tan 
2 
 π 
1
−
δ


these calculations and gives the following output and graphical output.
Natural frequency = 50.3292 Hz
delta = 0.2546
Excitation frequency =
25
Response = 0.6419 mm at 14.7527 degree
Excitation frequency =
55
Response = 2.4895 mm at -14.7527 degree
Phase (degrees)
Amplitude
Coulomb damping. F(dry) = 10 N
10
0
0
0.5
1
Frequency ratio r
1.5
2
0
0.5
1
Frequency ratio r
1.5
2
150
100
50
0
12
Problem 2.17
3
Equation of motion is mx + kx + hx=
f 0 cos ωt
If we let
=
x3 x13 cos3 ωt . Expanding this
=
x x1 cos ωt , 
x = −ω2 x1 cos ωt and
(a)
expression in terms of multiple angles gives
=
x3 x13 ( cos 3ωt + 3cos ωt ) 4 .
Considering only the cos ωt terms, we have
=
x3
3 x3 cos ωt
4 1
. Substituting in the
equation of motion and cancelling the cos ωt factor we obtain
−ω2 mx1 + kx1 + 34 hx13 = f 0 . Thus,
3 hx3
4 1
(
)
+ k − ω2 m x1 − f 0 = 0 .
Given ω= 50 × 2π= 314.1593/s k = 106 N m and =
h 40 ×106 N m3 we can solve
this cubic equation for various values of f 0 .(see MATLAB function
Problem_02_17.m) to obtain
=
f 0 30
=
N
x1 2.27 mm
=
f 0 60
=
N
x1 4.40 mm
=
f 0 120
=
N
x1 8.02 mm
=
f 0 240
=
N
x1 13.2 mm
If ω =55 Hz , f 0 = 240=
N , x1 81.1 and − 1.24 mm
When f 0 = 0 ,
3 hx3
4 1
(
)
+ k − ω2 m x1 = 0 . Thus
3 αx 2
1
4
through by m we have
(
)
3 hx 2
4 1
(
)
+ k − ω2 m = 0 and, dividing
+ ω02 − ω2 = 0 where α =h m and ω02 =k m .
Therefore ω2 =ω02 + 34 αx12 . Note that if α > 0 ,
3 αx 2
1
4
6
> 0 and hence ω > ω0 .
When α = 40 ×106 10 = 4 ×106 / m 2s 2 ,=
ω02 10 =
10 105 rad 2 / s 2 x = 0.010 m .
(
)
Thus ω2= 105 + 34 4 ×106 × 0.012 = 100300rad 2 / s 2 and so
=
ω
100300
=
( 2π ) 50.4046 Hz .
(b) Letting=
x = −ω2 x1 cos ωt − 9ω2 x3 cos 3ωt
x x1 cos ωt + x3 cos 3ωt , 
and
=
x3 x13 cos3 ωt + 3 x12 x2 cos 2 ωt cos3ωt + 3 x1x22 cos ωt cos 2 3ωt + x33 cos3 3ωt .
Expanding the powers of the trigonometric functions in terms of multiple angles, and
neglecting terms in cos pωt , where p = 5, 7 and 9 we have
=
x3
1
4
(3x13 + 3x12 x3 + 6 x1x32 ) cos ωt + 14 ( x13 + 6 x12 x3 + 3x33 ) cos 3ωt
Substituting in the equation of motion we have
{( k − ω m ) x + h (3x
2
3
1
1
4
1
)}
+ 3x12 x3 + 6 x1x32 cos ωt +
{( k − 9ω m ) x + h ( x
2
3
3
1
1
4
)}
+ 6 x12 x3 + 3x33 cos 3=
ωt f 0 cos ωt
Thus, equating coefficients of cos ωt and cos 3ωt gives a pair of equations thus:
{( k − ω m ) x + h (3x + 3x x + 6x x )} − f=
f = {( k − 9ω m ) x + h ( x + 6 x x + 3 x )}= 0
=
f1
2
2
2
3
1
1
4
1
3
1
4
3
1
2
1 3
2
1 3
2
1 3
3
3
13
0
0
Let=
ε
f12 + f 22 . Clearly when the values of x1 and x3 are the roots, f=
f=
1
2 0
and ε =0 . The MATLAB function fminsearch.m iterates from initial values to
minimise ε and hence solve the equations. This function is used in the MATLAB
function Problem_02_17.m. From this function we have the following output
Single term
force = 30
force = 60
force = 120
force = 240
solution. Forcing frequency = 50Hz
N, x1 = 2.2736 mm
N, x1 = 4.4048 mm
N, x1 = 8.0172 mm
N, x1 = 13.1609 mm
Single term solution: Forcing frequency = 55Hz
force = 240N. x1 = 81.0725 mm and -1.2360 mm
Magnitude of vibn, 10 mm. Frequency of unforced response = 50.4046 Hz
Two term solution,
omega = 50 Hz
x1
omega = 55 Hz (1st
omega = 55 Hz (2nd
force = 240N
= 13.1604 mm, x3 = 2.8954e-003 mm
solution)
x1 = 80.7894 mm, x3 = 0.5636 mm
solution)
x1 = -1.2360 mm, x3 = -1.9390e-006 mm
Note that to obtain the second solution, when ω =55 Hz , the initial values were zero.
14
Chapter 3
Equations of motion for a rigid body, pinned at one end.
The equations of motion for the system shown in Fig 3.26 can be developed by using
Newton’s 2nd law. A free body diagram for the system is shown below.
Moments acting in the θ direction: f y L= I d 
θ + I p Ωψ where I d is the moment of
inertia of the rotor about the left end.
 − I p Ωθ . In general, f x is the
Moments acting in the ψ direction: − f x L= I d ψ
force to extend a spring or give velocity to a viscous damper. Similarly for f y . Now
(
v =− Lθ and u= Lψ . Hence f y= k y v + c y v= k y ( − Lθ ) + c y − Lθ
)
and
 k x ( Lψ ) + c x ( Lψ ) . Hence the equations of motion are
f x= k x u + c x u=
I d 
θ + I p Ωψ + c y L2 θ + k y L2 θ = 0
 − I p Ωθ + c x L2 ψ + k x L2 ψ =0
Id ψ
Problem 3.1
In this problem there is no damping in the bearing and the stiffness is the same in the
x and y directions. Hence
I d 
θ + I p Ωψ + kL2 θ = 0
or in matrix notation,
 − I p Ωθ + kL2 ψ = 0
Id ψ
I p   θ   kL2 0   θ  0 
   =  .
 + 
0  ψ   0 kL2  ψ  0 
Seeking solutions of the form θ = θ0e jωt and ψ = ψ 0e jωt , results in the equations
 Id
0

 0
0   
θ
+
Ω



I d  ψ
 
 − I p
 −ω2 I d + L2 k

 − jωΩI p

  θ  0 
  0  =   . This matrix must be singular and hence its
2
2  ψ
−ω I d + L k   0  0 
jωΩI p
determinant is zero. Thus ω is the solution of
15
( −ω2 Id + L2k ) + ( ωΩI p )2 = 0
2
or
( −ω2 Id + L2k ) = ( ωΩI p )2 . Hence ( −ω2 Id + L2k ) = ± ( ωΩI p )
2
and so
ω2 I d ± ωΩI p − L2 k = 0 . For example, solving this pair of quadratics for
 10000 × 2π 
2
6
Ω =10, 000 rev min we have 10ω2 + 0.6 
 ω − 0.5 × 10 = 0 . And
60


 10000 × 2π 
2
6
10ω2 − 0.6 
 ω − 0.5 ×10 = 0 . Solving these equations give
60


=
ω 129.7888 and − 192.6206 and ω = −129.7888 and 192.6206 . Thus, taking the
positive roots, ω =129.7888 rad s and 192.6206 rad s or, converting to Hz by
dividing by 2π gives ω =20.6565 Hz and 30.6565 Hz .
Alternatively, we can solve the eigenvalue problem as described by equations (3.48),
(3.50), (3.51) and (3.52). The MATLAB script Problem_03_01.m gives the user the
choice of solving either the characteristic equation or the eigenvalue problem to
determine the system natural frequencies. Of course, both methods give the same
numeric values for the frequencies which are as follows
Solution of the characteristic equation
Rotor speed = 0 rev/min
Natural frequency = 25.1646 Hz
Natural frequency = 25.1646 Hz
Rotor speed = 3000 rev/min
Natural frequency = 23.7093 Hz
Natural frequency = 26.7093 Hz
Rotor speed = 10000 rev/min
Natural frequency = 20.6565 Hz
Natural frequency = 30.6565 Hz
Once ω is calculated the relative displacements of the rotor are determined from the
matrix equation above as
− jωΩI p
θ0
L2 k − ω2 I d
. Note that for Ω =0 the modes are not unique.
= =
ψ 0 L2 k − ω2 I d
jωΩI p
At 3000 rev min , ω =148.9699 rad s and 167.8195 rad s . θ0 ψ 0 =− j and j
respectively. Since v =− Lθ and u= Lψ , then =
v0 u0 j and − j .
 cos θ 
u ( t )  1  jθ  1  − jθ
For the first mode, 
2
=
=
 e +  e
 where θ = ωt .
− j 
− sin θ 
 v ( t )   j 
cos θ 
u ( t )   1  jθ 1  − jθ
For the second mode, 
= 2
 =  e +  e
.
 j
 sin θ 
 v ( t )  − j 
The above is identical to Equations (3.34) and (3.35) and the text with these equations
explains why the first mode (23.71 Hz) is a backward rotating mode and the second
mode (26.71 Hz) is a forward rotating mode. The orbits are circular.
16
Problem 3.2.
In this problem there is no damping in the bearing. Hence
 + k y L2 θ = 0
I d 
θ + I p Ωψ
or in matrix notation,
 − I p Ωθ + k x L2 ψ = 0
Id ψ
 Id

0
 0
0   
θ
 +Ω

I d  ψ
 
 − I p
I p   θ   k y L2
 + 
0  ψ   0

0   θ  0 
   =  .
2  ψ  0 
kx L 
Seeking solutions of the form θ = θ0e jωt and ψ = ψ 0e jωt , results in the equations
 −ω2 I d + L2 k y

 − jωΩI p

  θ  0 
  0  =   . This matrix must be singular and hence
−ω2 I d + L2 k x  ψ 0  0 
its determinant is zero. Thus ω is the solution of
jωΩI p
( −ω2 Id + L2k y )( −ω2 Id + L2kx ) − ( ωΩI p )2 = 0 . Letting λ = ω2 this leads to a
quadratic in λ thus: λ 2 I d2 − ( I 2p Ω 2 + I d L2 ( k x + k y ) ) λ + L2 k x k y = 0 which is easily
solved for given parameters. For example, when
=
Ω 3000 rev min
= 3000 × 2 π =
60 314.159 rad s we have
100λ 2 − 6.1448 ×106 λ + 81.25 ×109 = 0 . Solving this equation gives
λ = −23997 and − 33858 and hence s = λ = ±138.77 j and ± 205.40 j Taking the
positive roots we have
=
ω 138.77 ( 2 π ) and 205.40
=
( 2π ) 24.6548Hz and 29.2854Hz
Alternatively, we can solve the eigenvalue problem as described by equations (3.48),
(3.50), (3.51) and (3.52). The MATLAB script Problem_03_02.m gives the user the
choice of solving either the characteristic equation or the eigenvalue problem to
determine the system natural frequencies. Of course, both methods give the same
numeric values for the frequencies which are as follows
Solution of the characteristic equation
Rotor speed = 0 rev/min
Natural frequency = 25.1646 Hz
Natural frequency = 28.6921 Hz
Rotor speed = 3000 rev/min
Natural frequency = 24.6548 Hz
Natural frequency = 29.2854 Hz
Rotor speed = 10000 rev/min
Natural frequency = 22.0866 Hz
Natural frequency = 32.6906 Hz
Problem 3.3.
In this problem there is damping and the stiffness in the bearing and the properties are
the same in the x and y directions. Hence
I d 
θ + I p Ωψ + cL2 θ + kL2 θ = 0
or in matrix notation,
 − I p Ωθ + cL2 ψ + kL2 ψ =0
Id ψ
17
 Id
0

2
0   
θ   cL
 + 
I d  ψ
   −ΩI
p

ΩI p   θ   kL2
 + 
cL2  ψ   0
0   θ  0 
 =
 
kL2  ψ  0 
Since damping is included we now look for solutions of the form θ = θ0e st and
ψ = ψ 0e st , where s is complex, results in the equations
 s 2 I d + sL2c + L2 k
  θ  0 
sΩI p

 0 =  

s 2 I d + sL2c + L2 k  ψ 0  0 
− sΩI p

Setting the determinant of the above array to zero we have
( s2 Id + sL2c + L2k ) + ( sΩI p )2 =0 . Hence
2
s 2 I d + sL2c + L2 k =± jsΩI p . This gives the following quadratic equation (with one
(
)
0 . For example, when
complex coefficient) I d s 2 + L2c  jΩI p s + L2 k =
=
Ω 3000 rev min
= 3000 × 2 π =
60 314.159 rad s we have
10 s 2 + (125.00 ± 188.50 j ) s + 25 ×104 =
0
Each equation has two roots so the two equations have together four roots (forming
complex conjugate pairs) thus s =
−6.62 ± 167.70 j and − 5.88 ± 148.85 j . From these
roots we can obtain the damped natural frequency by taking the imaginary part of s,
the natural frequency by taking the absolute values of s, and the damping, ζ , by
changing the sign of the real value divided by the absolute value of s. For example,
, ωd 167.7=
when s =
Hz , ζ 6.62
=
=
167.7 0.395
−6.62 ± 167.70 j=
( 2π ) 26.69
and=
ωn
6.622 + 167.72 =
( 2π ) 26.71Hz .
Alternatively, we can solve the eigenvalue problem as described by equations (3.48),
(3.50), (3.51) and (3.52). The MATLAB script Problem_03_03.m gives the user the
choice of solving either the characteristic equation or the eigenvalue problem to
determine the system damped and natural frequencies. Of course, both methods give
the same numeric values for the frequencies which are as follows
Solution of the characteristic equation
Rotor speed = 0 rev/min
Damped natural freq (Hz)
Natural freq (Hz)
25.1449
25.1646
25.1449
25.1646
Rotor speed = 3000 rev/min
Damped natural freq (Hz)
23.6897
26.6897
Zeta
0.0395
0.0395
Natural freq (Hz)
Zeta
23.7082
0.0395
26.7105
0.0395
Rotor speed = 10000 rev/min
Damped natural freq (Hz)
Natural freq (Hz)
20.6380
20.6535
30.6380
30.6610
18
Zeta
0.0388
0.0388
Problem 3.4.
In this case the force applied to the support in the x and y directions is
f x= ku + kc v= kLψ − kc Lθ and f=
kcu + kv
= kc Lψ − kLθ . The moment acting on
y
the rotor in the θ direction is Lf=
kc L2ψ − kL2θ . Similarly the moment acting on
y
the rotor in the ψ direction is − Lf x = −kL2ψ + kc L2θ and the equations of motion
become
θ + I p Ωψ + L2 k θ − L2 kc ψ = 0
I d 
or, in matrix notation,
 − I p Ωθ − L2 kc θ + L2 k ψ = 0
Id ψ
 Id
0

 0
0   
θ
 +Ω

I d  ψ
 
 − I p
I p   θ   kL2
 + 
0  ψ   − kc L2

−kc L2   θ  0 
   = 
kL2  ψ  0 
Seeking solutions of the form θ = θ0e st and ψ = ψ 0e st , gives the following equation
for s
 s 2 I d + L2 k
sΩI p − L2 kc 

 = 0 or
det
 − sΩI p − L2 kc s 2 I d + L2 k 


( s2 Id + L2k ) − ( −sΩI p − L2kc )( sΩI p − L2kc ) =0 and hence
I d2 s 4 + ( I 2p Ω 2 + 2 I d L2 k ) s 2 + L4 ( k 2 − kc2 ) =0 , a quadratic in s 2 .
2
For example, when Ω =3, 000 rev min and letting λ =s 2 , we have
100λ 2 + 5.0355 ×106 λ + 6 × 1010 = 0 . Solving this quadratic gives
λ = −1.935 × 10 4 and − 3.100 × 10 4 . Hence s = 139.12 j and 176.07 j and so
=
ωn 139.12
=
( 2π ) 22.14 Hz and 176.07=
( 2π ) 28.02 Hz .
Alternatively, we can solve the eigenvalue problem as described by equations (3.48),
(3.50), (3.51) and (3.52). The MATLAB script Problem_03_04.m gives the user the
choice of solving either the characteristic equation or the eigenvalue problem to
determine the system natural frequencies. Of course, both methods give the same
numeric values for the frequencies which are as follows
Solution of the characteristic equation
Rotor speed = 0 rev/min
Natural frequency = 22.5079 Hz
Natural frequency = 27.5664 Hz
Rotor speed = 3000 rev/min
Natural frequency = 22.1415 Hz
Natural frequency = 28.0227 Hz
Problem 3.5.
This machine has the equations of motion given by Equation (3.9), with kT = 2k ,
kC = 0 and k R = 2a 2 k , where L is the distance between the bearings (hence
a= b= L 2 ). Thus the equations of motion are
19
mu + 2ku =
0
mv + 2kv =
0
or in matrix notation,
θ + I p Ωψ + 2a 2 k θ = 0
I d 
 − I p Ωθ + 2a 2 k ψ = 0
Id ψ
0
0   u  0 
0   u   2k 0
0 0 0
0   u 
 0 2k
0 0 0






0
0   v  0 
0   v  
0   v 

+
  + Ω 0 0 0
  = 
I p   θ   0 0 2a 2 k
0   
θ
0   θ  0 






 
 
I d  ψ
 0 0 − I p 0  ψ   0 0
0
2a 2 k  ψ  0
The translational and rotational equations of motion decouple. Thus, from
translational equations of motion, natural frequencies are 2k m (twice).
m 0 0
0 m 0

 0 0 Id

 0 0 0
We now consider the 3rd and 4th equations and seek solutions of the form θ = θ0e st
and ψ = ψ 0e st . Thus
 I d s 2 + 2 a2 k
det 
 − I Ωs
p


 = 0 gives the following equation for s
2
2 
I d s + 2a k 
I p Ωs
0 or
( Id s2 + 2a2k ) + ( I pΩs )2 =
2
I d s 2 + 2a 2 k =
± jI p Ωs giving the pair of
quadratics I d s 2 ± jI p Ωs + 2a 2 k =0 . At Ω =3, 000 rev min ,
50 s 2 ± j 9424.8s + 2 ×106 =
0 and solving this equation gives
s=
±315.34 j and ± 126.85 j . Taking the positive roots,
=
ω 315.34
=
=
( 2π ) 50.19Hz and 126.85
( 2π ) 20.19Hz .
Alternatively, we can solve the eigenvalue problem as described by equations (3.48),
(3.50), (3.51) and (3.52). The MATLAB script Problem_03_05.m gives the user the
choice of solving either the characteristic equation or the eigenvalue problem to
determine the system natural frequencies. Of course, both methods give the same
numeric values for the frequencies which are as follows
Solution of the characteristic equation,
Natural frequency = 14.2353 Hz (twice)
Natural frequency = 20.1882 Hz
Natural frequency = 50.1882 Hz
Problem 3.6.
This system is described by Equation (3.84). In matrix notation, we have
0   u   kT
0 0 0
kC   u  0 
0
0
 m 0 0 0   u 


 0 m 0 0   v 
0 0 0
0   v   0
kT −kC 0   v  0 

   + Ω 

+
  = 
0 0 0
I p   θ   0 −kC k R
 0 0 I d 0   
0   θ  0 
θ





 


 
0
0
k R  ψ  0 
 0 0 0 I d  ψ
0 0 − I p 0  ψ   kC
Using the stiffness formulae gives in Appendix 2, Table A2.1, System 5, we have
20
k=
k=
T
uu k=
vv
12 EI ( a + 3b )
b ( 4 a + 3b )
3
kψψ
= k=
, k=
R
θθ
12 EI ( a + b )
b ( 4 a + 3b )
and
6 EI ( 2 a + 3b )
. Letting u = u0e st etc. we can solve these
−kθv =
− 2
kC =
kψ u =
b ( 4 a + 3b )
equations either by forming the characteristic equation or by solving the eigenvalue
problem. The characteristic equation for this system is
Aλ 4 + Bλ3 + C λ 2 + Dλ + E = 0 where λ =s 2 and A = m 2 I d2 ,
=
B
( mI Ω
2
p
2
)
+ 2 mI d k R + 2 I d2 kT m , E =
kC4 + kT2 k R2 − 2 kT k R kC2
=
C m 2 k R2 + 4 mI d kT k R + 2 mI 2p kT Ω2 − 2 mI d kC2 + I d2 kT2 and
=
D 2 mkT k R2 − 2 mkC2 k R + I 2p kT2 Ω2 − 2 I d kT kC2 + 2 I d kT2 k R . Clearly, these expressions are
complicated and deriving them requires a fair amount of tedious algebra. Finally a
quartic equation must be solved and that requires a numerical procedure. In this
problem it is probably easier to solve the eigenvalue problem directly as described by
equations (3.48), (3.50), (3.51) and (3.52). To determine the values of u0 ψ 0 we
can either (i) extract the information from the eigenvalues, or (ii) substitute the values
of s into one of the system equations. Thus, from the first system equation, since
 k

u
ms 2u0 + kT u0 + kC ψ 0 =,
0 then 0 = −  2 C
 ms + k 
ψ0

T 
The MATLAB script Problem_03_06.m gives the user the choice of solving either
the characteristic equation or the eigenvalue problem to determine the system natural
frequencies and u0 ψ 0 . Of course, both methods give the same numeric values
which are as follows
Solution of the EVP
Rotor speed = 0 rev/min
Natural frequency = 15.239 Hz. u/psi = 0.30996
Natural frequency = 15.239 Hz. u/psi = 0.30996
Natural frequency = 67.1535 Hz. u/psi = -0.21508
Natural frequency = 67.1535 Hz. u/psi = -0.21508
Rotor speed = 1000 rev/min
Natural frequency = 9.9593 Hz. u/psi = 0.28808
Natural frequency = 22.1203 Hz. u/psi = 0.36325
Natural frequency = 60.0598 Hz. u/psi = -0.3347
Natural frequency = 79.1488 Hz. u/psi = -0.12689
Using the values of u0 ψ 0 we can make sketches of the mode shapes as shown
below. The left column diagrams are for the stationary rotor, the right column
diagrams are for the spinning rotor.
21
0.5
0.5
0
0
-0.5
0
0.5
1
-0.5
0.5
0.5
0
0
-0.5
0
0.5
1
-0.5
0
0.5
1
0
0.5
1
Problem 3.7.
The equations of motion are the same as Equation (3.4), except that here a force f x1
is required to enforce the constraint u1 = u − aψ = 0 . The equations of motion are
then (noting that a= b= L 2 )
mu + k0 ( u + aψ ) =− f x1
mv + k1 ( v + aθ ) + k2 ( v − aθ ) =0
I d 
θ + I p Ωψ + ak1 ( v + aθ ) − ak2 ( v − aθ ) = 0
 − I p Ωθ + ak0 ( u + aψ ) = af x1
Id ψ
Adding a times the first equation to the last equation, gives
mv + k1 ( v + aθ ) + k2 ( v − aθ ) =0
I 
θ + I Ωψ + ak ( v + aθ ) − ak ( v − aθ ) = 0
d
p
1
2
 − I p Ωθ + 2ak0 ( u + aψ ) = 0
mau + I d ψ
Now applying the constraint u = aψ to eliminate u gives
mv + ( k1 + k2 ) v + a ( k1 − k2 ) θ =0
I d 
θ + I p Ωψ + a ( k1 − k2 ) v + a 2 ( k1 + k2 ) θ = 0
( Id + ma2 ) ψ − I pΩθ + 4a2k0ψ = 0
There is an alternative way of deriving these equations. In the y-z plane the system
identical to the system shown in Fig 3.6 and described by Equation (3.6). In the x-z
pane the system is identical to the system described in Problem 3.1.
Thus in the y-z plane (from Equations (3.6) and noting that a= b= L 2 ) we have
mv + ( k1 + k2 ) v + a ( k1 − k2 ) θ =0
I d 
θ + I p Ωψ + a ( k1 − k2 ) v + a2 ( k1 + k2 ) θ = 0
In the x-z plane (from Problem 3.1) we have
 − I p Ωθ + k0 L2 ψ = 0 where I d1 is the moment of inertia bearing 1 and
I d1ψ
2
I d=
1 I d + ma so that
22
mv + ( k1 + k2 ) v + a ( k1 − k2 ) θ =0
θ + I p Ωψ + a ( k1 − k2 ) v + a2 ( k1 + k2 ) θ = 0 , as before.
I d 
(I
d
)
 − I p Ωθ + 4 a2 k0 ψ = 0
+ ma2 ψ
Letting θ = θ0e st , etc, we obtain
 2

a ( k1 − k2 )
0
 ms + k1 + k2

2
2

=
det a ( k1 − k2 )
I d s + a ( k1 + k2 )
I p Ωs
0




− I p Ωs
0
I d + ma2 s 2 + 4 a2 k0 


From this determinant we can obtain the characteristic equation thus
(
)
Part 1 When Ω =0 . Here the 3rd equation is uncoupled from the other two and so we
obtain
(I
d
)
(
+ ma2 s 2 + 4 k0 a2 =
0 and mI d s 4 + I d + ma2
)(k + k ) s
1
2
2
+ 4 a2 k1k2 =
0.
The linear equation in s 2 and the quadratic equation in s 2 can easily be solved.
st
Part 2. When k=
1 k=
2 k . Here the 1 equation uncouples from the other two and so
(
) {
(
)
}
we obtain I d I d + ma2 s 4 + 2 ka2 I d + ma2 + 4 k0 a2 I d + Ω2 I 2p s 2 + 8a 4 k0 k = 0
and ms 2 + 2 k =
0 . The linear equation in s 2 and the quadratic equation in s 2 can
easily be solved.
Alternatively, in matrix notation we have
0 0
m 0
  
0
0   v 
v

 

   

0
 0 Id
  θ  + Ω 0 0 I p   θ 

 

2  ψ
 
0 − I p 0  ψ 
 0 0 I d + ma   
 k1 + k2
0    v  0 
a ( k1 − k2 )

   
0 θ =
+  a ( k1 − k2 ) a2 ( k1 + k2 )
0 





 
0
0
4 a2 k0   ψ  0 

These equations can be rearranged into the form Ax + Bx=0 , and the eigenvalue
problem can be solved as described by equations (3.48), (3.50), (3.51) and (3.52). The
MATLAB script Problem_03_07.m gives the user the choice of solving either the
characteristic equation or the eigenvalue problem to determine the system natural
frequencies. Of course, both methods give the same numeric values for the
frequencies which are as follows
Solution of the characteristic equation
Rotor speed = 0 rev/min
Stiffness k1 = 1800000 N/m
Stiffness k2 = 2200000 N/m
Natural frequency = 28.8598 Hz
Natural frequency = 36.5126 Hz
Natural frequency = 56.3715 Hz
23
Rotor speed = 9550 rev/min
Stiffness k1 = 2000000 N/m
Stiffness k2 = 2000000 N/m
Natural frequency = 29.0576 Hz
Natural frequency = 33.8302 Hz
Natural frequency = 60.7316 Hz
Problem 3.8.
Using the stiffness formulae gives in Appendix 2, Table A2.4, System 6 with
D6
a= b= L 2 and k1 = k2 , we have =
(
(3EI + a k ) ,
3
)
2
kC =
kψ u =
−kθv =
0,
(
)
6EI
6 EI
EIa2 k + a5 k 2 . The
3EIk + a3k 2 and kψψ
= k=
θθ
D6
D6
equations of motion for the system are
mu + kT u =
0
mv + kT v =
0
I 
θ + k θ =0
k=
k=
T
uu k=
vv
d
R
 + k R ψ =0
Id ψ
Note that because kC = 0 and gyroscopic effects are ignored the four equations are
uncoupled from each other and can be solved independently of each other. Thus,
letting u = u0e jωt etc. we have we have ωn = kT m (twice) and ωn = k R I d .
Simply supported beam with central disk: From Appendix 2, Table A2.1, System 1
with a= b= L 2 we have kTss = 6 EI a3 , k Rss = 6 EI a and kCss = 0 . Thus we can
compute the natural frequencies as above, i.e. ωn = kTss m (twice) and
ωn = k Rss I d .
Rigid rotor: The stiffness coefficients for this system are given (3.7) so that
kTrgd = 2 k , k Rrgd = 2 a2 k and kCrgd = 0 . Again, the equations uncouple so that the
frequency equations are as above. Although these calculations are simple we have
written a MATLAB script to carry them out. Thus The MATLAB script
Problem_03_08.m determines the system natural frequencies for the three models, as
follows.
Bearing stiffness = 50000N/m
Full model (Hz) Simply supported beam (Hz)
1st Mode: 13.2231
72.1384
2nd mode: 34.9851
190.8604
Rigid rotor (Hz)
13.4510
35.5881
Bearing stiffness = 1000000N/m
Full model (Hz) Simply supported beam (Hz)
1st Mode: 46.1998
72.1384
2nd mode: 122.2332
190.8604
Rigid rotor (Hz)
60.1549
159.1549
Bearing stiffness = 100000000N/m
Full model (Hz) Simply supported beam (Hz)
1st Mode: 71.6252
72.1384
2nd mode: 189.5026
190.8604
Rigid rotor (Hz)
601.5491
1591.5494
These results show that when the bearing stiffness is low the system behaves like a
24
rigid rotor on flexible supports. When the bearing stiffness is high, the system behaves
like a simply supported flexible beam. At the intermediate stiffness (1MN/m) only the
full model adequately describes the system.
Problem 3.9.
The data of Table 3.4 gives the eigenvalues and eigenvectors for a rotor supported by
(a) isotropic bearings and (b) anisotropic bearings. The natural frequencies of the
system are determined from the eigenvalues,=
i.e. ωn imag ( s ) ( 2 π ) . The method to
determine the shape of an orbit and direction of whirl is described in Section 3.6.1.
Equation (3.55) requires amplitude and phase of the displacements in the x and y
directions. From this the shape of the orbit can be determined (κ, given in Equation
(3.60)) and the direction of whirl given by the sign of κ.
The displacements in the x and y directions at the two ends of the rotor are given by
Equation (3.61). The MATLAB script Problem_03_09.m analyses the isotropic and
the anisotropic bearing case. The script begins by regenerates the data of Table 3.4 and
then determines T (Equation (3.56)), H = T TT , and then the eigenvalues of H .
From these eigenvalues the script determines κ and the direction if rotation. The
MATLAB script Problem_03_09.m determines the system natural frequencies and
values of κ , thereby giving the shape and direction of the orbit at various locations
along the rotor.
[-ve kappa: backward whirl, +ve kappa: forward whirl]
Isotropic bearings
omega_n (Hz)
x-y orbit at rtr centre
Rotation at rtr centre
x-y orbit at bearing 1
x-y orbit at bearing 2
47.4044
kappa: -1.0000
kappa: -1.0000
kappa: -1.0000
kappa: -1.0000
47.4233
1.0000
1.0000
1.0000
1.0000
88.8962
-1.0000
-1.0000
-1.0000
-1.0000
154.9324
1.0000
1.0000
1.0000
1.0000
Isotropic bearings
omega_n (Hz)
x-y orbit at rtr centre
Rotation at rtr centre
x-y orbit at bearing 1
x-y orbit at bearing 2
47.4140
kappa: -0.0067
kappa: -0.2558
kappa:
0.0012
kappa: -0.0212
50.3203
0.0076
-0.2926
0.0297
-0.0286
91.6220
-0.4441
-0.8889
-0.9120
-0.8770
157.8805
0.3827
0.9348
0.9286
0.9380
As expected, for the isotropic bearings, the orbits are circular. For the anisotropic
bearings, the orbits are elliptic and some of the modes are forward, some are backward
and some are a mixture. For example the mode at frequency 47.414 Hz has a forward
whirl at bearing 1, but backward at the centre of the rotor and at bearing 2.
Problem 3.10.
The equations of motion of this system are given by Equation (3.9) (or Equation
(3.26)) with kT = 2k , kC = 0 and k R = 0.5 L2 k (since a= b= L 2 ). Thus
mu + 2ku =
0
mv + 2kv =
0
.
θ + I p Ωψ + 0.5 L2 k θ = 0
I d 
 − I p Ωθ + 0.5 L2 k ψ = 0
Id ψ
We know that m = 30 kg . The first and second equations are uncoupled from the
25
third and fourth equations and are independent of speed. Thus one natural frequency
is independent of speed and from the 1st or 2nd equation is given by ω1 = 2k m .
2
2
Hence =
k 0.5mω=
0.5 × 30 × ( 2π× 7.1)= 29.8516 kN m
1
The other natural frequency, when Ω =0 , is obtained from the 3rd or 4th equation is
given by ω2 = 0.5 L2 k I d Thus
=
Id
0.5 L2 k 0.5 × 0.82 × 29.8516 ×103
=
= 0.53838 kg m 2
2
2
ω2
( 2π× 21.2 )
When the rotor spins the second pair of natural frequencies are given by the solutions
to equation (3.29), that is
(
−ω2 I d + k R
) (
2
− I p Ωω
)
2
= 0 . Thus I p =
−ω2 I d + k R
Ωω
where the natural frequency is ω = 2π× 46.7 rad s and the rotor spin speed is
=
Ω 2000 × 2π =
60 209.44 rad s . k R =0.5 L2 k =0.5 × 0.82 × 29851.6 =9552.5 Nm
and hence I p = 0.59883kg m 2 .
Problem 3.11.
The development in Section 3.6.1 is reworked with the eigenvalue si = jωi and the
corresponding mode. Consider first the calculation of the lengths of the semi-major
and semi-minor axes - the mode considered here will be the complex conjugate of the
mode used in Section 3.6.1, and hence the matrix T will be post multiplied by
1 0 
0 −1 . Thus H will be as given in Section 3.6.1 and thus so will be the lengths of


the semi-major and semi-minor axes. The only question that remains is to determine
the direction of whirl. If ηu and ηv are the angle corresponding to the new mode
with eigenvalue si =− jωi then the equivalent equations to Equation (3.59) is
ru cos ( ωi t )
ru cos ( −ωi t )
u ( t )  
 

=

 =
 

 v ( t )  ru cos ( ηv − ηu − ωi t )   ru cos ( − ( ηv + ηu ) + ωi t ) 
and hence the same conclusions can be draw about the direction of whirl by making
( ηv − ηu ) negative. Thus in this case 0 < ( ηv − ηu ) < π implies forward whirl and
−π < ( ηv − ηu ) < 0 implies backward whirl.
Problem 3.12
Since the rotor is stationary and the bearings are also isotropic we only need to
consider motion in, say, the x direction. Using the stiffness formulae gives in
Appendix 2, Table A2.4, System 6, and letting k1 → ∞ gives
3EI 
=
D6 a3 k1 3EI + b3 k2 =
, kuu
3EIk1 + a3 + b3 k1k2  ,


D6
3EI 
=
kψ u
3EI ( −ak1 ) + ab a2 − b2 k1k2  ,

D6 
(
)
(
(
)
26
)
3EI 
3EIa2 k1 + a2 b2 ( a + b ) k1k2  .


D6
Since k1 is a factor in both the numerators and the denominator of the above
equations, these simplify to
3EI 
=
D6 a3 3EI + b3 k2 , kT = kuu =
3EI + a3 + b3 k2 

D6 
kψψ
=
(
)
(
(
)
)
3EI 
3EI
−3EIa + ab a2 − b2 k2  , k R =
kψψ = 3EIa2 + a2 b2 ( a + b ) k2 


D6 
D6 
Thus the equations of motion are
 m 0   u   kT kC   u  0 
 .
 0 I     +  k
 =
d  ψ   C k R  ψ  0 

 + Kq =
From this equation we can obtain Mq
0 and hence the eigenvalue
kC =kψu =
s 2 Mq 0 + Kq 0 =
0 . This can be solved to obtain the two system natural frequencies.
Using the reduction formulae of Appendix 2, kred= kT − kC2 k R and
mred= m + I D kC2 k R2 . Hence the frequency of the reduced model is
ωn _ red =
kred mred . The MATLAB script Problem_03_12.m solves the eigenvalue
problem to determine the system natural frequencies and the natural frequency of the
reduced model, which are as follows
Disk diameter/thickness
0.650/0.065
1st natural frequency (Hz)
17.4505
2nd natural frequency (Hz)
68.5833
Reduced model, nat freq (Hz)
17.4785
1.200/0.120
6.4971
15.8084
6.6074
Problem 3.13
(a) Using the formulae gives in Appendix 1:
For the cylinder, I p =
MD 2 8 =
100 × 0.42 8 =
2kg m 2 ,
I d = I p 2 + Mh 2 12 = 2 2 + 100 × 0.62 12 = 4 kg m 2 .
For the disk, I p =
MD 2 8 =
100 × 1.42 8 =
24.5 kg m 2
I d = I p 2 + Mh 2 12 = 24.5 2 + 100 × 0.12 12 = 12.3333 kg m 2
The center of gravity of the system, relative to bearing number 1 is given by
∑ Mz =z ∑ M and hence 100 × 0.8 + 100 × 1.6 =
200z . Thus z = 1.2m .
For the total system, I p =
I p( cyl ) + I p( dsk ) =
2 + 24.5 =
26.5 kg m 2
(
) + ( I + Mh )
=( 4 + 100 × 0.4 ) + (12.3333 + 100 × 0.4 ) =48.3333 kg m
I d = I dC + Mh 2
cyl
2
dC
dsk
2
2
(b) M x = I d α = 48.3333 × ( −5 ) = −241.66 N m
 7000 × 2 π 
M y= I p Ωω= 26.5 × 
 × 2= 38,851N m
60


(c) Acceleration of the centre of gravity is =
ac 0.4 g − z α + g . Thus
27
2
ac= 0.4 × 9.81 + 1.2 × ( −5 ) + 9.81= 19.7340 m s2 . Denoting a vertical force by V,
Vc =
Mac =
200 × 19.7340 =
3946.8 N . The distance from bearing 1 to bearing 2 is
zb 2 = 1.3m . z2 = zb 2 − z = 1.3 − 1.2 = 0.1m .
Vb1 =
117, 7 N
( M x + z2Vc ) zb2 =( −241.66 + 0.1 × 3946.8) 1.3 =
Vb 2 =Vc − Vb1 =3946.8 − 117.7 =3829.1N
Denoting a vertical force by =
H, Hb M
=
=
1.3 29,885 N The
y zb 2 38851
horizontal force acting n the bearings are equal and opposite.
The MATLAB script Problem_03_13.m repeats these calculations thus:
CofG. Distance from brg 1 = 1.2000 m
Combined system, Id = 48.3333 kg m^2
Combined system, Ip = 26.5000 kg m^2
Moment_y = 38851.0291 Nm
Moment_x = -241.6667 Nm
Vertical load brg 1 = 117.7026 N
Vertical load brg 2 = 3829.0974 N
Horizontal loads = 29.8854 kN and -29.8854 kN
28
Chapter 4
In this chapter almost all the solutions require either the element matrices for the axial
deflection of a bar or the lateral deflection of a beam. These are given in text and are
repeated here for convenience.
For a bar in axial vibration (see Equation (4.12) for details)
ρe Aele  2 1 
Ee Ae  1 −1
.
=
, Ke
Me =


le  −1 1 
6 1 2
For a beam in bending (see Equations (4.24) and (4.25) for details)
54
−13le 
6le
 156 22le
 12



2
4le2
13le
−3le2 
ρe Aele  22le
Ee I e  6le 4le
Ms =
K
s
13le
156 −22le 
420  54
le3  −12 −6le



2 
 −13le −3le2 −22le
 6le 2le2
4le 
−12
−6le
12
−6le
6le 

2le2 
.
−6le 

2 
4le 
Problem 4.1
Assembling two axial deflection elements of equal length ( le = L 2 ) gives:
2 1 0
0 0 0 
2 1 0
ρA  L  
ρA  L  
ρA  L  



M=
  1 2 0 +
  0 2 1  =
  1 4 1  .
6 2
6 2
6 2
 0 0 0 
0 1 2 
 0 1 2 
 1 −1 0 
0 0 0 
 1 −1 0 
 2 
 2 
 2 


=
K EA    −1 1 0  + EA   0 1 =
−1 EA    −1 2 −1 .
L
L
L
 0 0 0 
0 −1 1 
 0 −1 1 
With free-free boundary conditions there are no constraints to be applied. Forming
and solving the eigenvalue problem Ku = λMu , where λ = ω2 , gives three natural
frequencies. The lowest one is zero, because the unconstrained bar can move as a
rigid body.
With fixed-free boundary conditions we must eliminate the first row and column from
the system matrices. Thus
ρA  L   4 1 
 2   2 −1
=
M =
  1 2  , K EA    −1 1  .
6  2 
 L 


Forming and solving this eigenvalue problem gives two natural frequencies.
The MATLAB script Problem_04_01.m solves the above eigenvalue problems to
give
In model, rho = A = E = L = 1
Free-free system - 1st elastic freq = 3.4641 rad/s
Fixed-free system - 1st freq = 1.6114 rad/s
 0 
 + Kq =
Adding a force f ( t ) to the free end gives Mq

.
 f ( t )
29
Problem 4.2
Assembling two axial deflection elements of equal length gives:
2 1 0
0 0 0 
2 1 0
ρA  L  
ρA  L  
ρA  L  



M=
  1 2 0 +
  0 2 1  =
  1 4 1  .
6 2
6 2
6 2
 0 0 0 
0 1 2 
 0 1 2 
 1 −1 0 
0 0 0 
 1 −1 0 
 2 
 2 
 2 


=
K EA    −1 1 0  + EA   0 1 =
−1 EA    −1 2 −1 .
L
L
 
 
L
 0 0 0 
0 −1 1 
 0 −1 1 
Note that le = L 2 .
With fixed-fixed ends we must eliminate the first and third rows and columns to give
ρAL
2 EA
=
M =
[ 2] . These are, of course, scalar quantities and so
[ 4] , K
12
L
=
ωn
12 E=
ρL2 3.4641 E ρL2
Assembling three axial deflection elements of equal length gives
2 1 0 0 
0 0 0 0 
0 0 0 0 
1 2 0 0 
0 2 1 0 


ρA  L  
ρA  L  
ρA  L  0 0 0 0 


’
M=
+
+
6  3  0 0 0 0  6  3  0 1 2 0  6  3  0 0 2 1 






0 0 0 0 
0 0 0 0 
0 0 1 2 
 1 −1 0 0 
0 0 0 0 
0 0 0 0 
 −1 1 0 0 
0 1 −1 0 


 3 
 3 
 3  0 0 0 0 


.
K = EA  
+ EA  
+ EA  
 L   0 0 0 0
 L  0 −1 1 0 
 L  0 0 1 −1






 0 0 0 0
0 0 0 0 
0 0 −1 1 
Hence
2 1 0 0
 1 −1 0 0 
1 4 1 0 


ρA  L  
 3   −1 2 −1 0 

.
, K EA  
M =
 
6  3  0 1 4 1 
 L   0 −1 2 −1




0 0 1 2
 0 0 −1 1 
Eliminating the first and fourth rows and columns gives
ρAL  4 1 
3EA  2 −1
. The eigenvalue problem can be formed and
=
, K
M =


18 1 4 
L  −1 2 
solved to give two natural frequencies
Assembling four axial deflection elements of equal length gives
2 1 0 0 0
 1 −1 0 0 0 
1 4 1 0 0 
 −1 2 −1 0 0 



ρA  L 
 4 




,
M =
K
EA
0
1
4
1
0
0
−
1
2
−
1
0
 
 
6  4 
 L 


0 0 1 4 1 
 0 0 −1 2 −1
 0 0 0 1 2 
 0 0 0 −1 1 
Eliminating the first and fifth rows and columns gives
30
4 1 0
 2 −1 0 
ρAL 
4 EA 

1 4 1 , K s =
Ms =
−1 2 −1 . The eigenvalue problem can be


L
24
 0 1 4 
 0 −1 2 
formed and solved to give three natural frequencies.
The MATLAB script Problem _04_02.m solves the above formulations to give
In model, rho = A = E = L =
clamped-clamped, 2 elementclamped-clamped, 3 elementclamped-clamped, 4 element-
1
1st nat freq = 3.4641 rad/s
1st nat freq = 3.2863 rad/s
1st nat freq = 3.2228 rad/s
Problem 4.3
Modelling the system with two elements
 1 −1 0 
0 0 0  0 0 0 
 2 
 2 

=
K EA    −1 1 0  + EA   0 1 −1 + 0 0 0 
L
L
 0 0 0 
0 −1 1  0 0 k 
2 1 0
0 0 0 
ρA  L  
ρA  L  


=
M
  1 2 0  +
  0 2 1  . Thus
6 2
6 2
 0 0 0 
0 1 2 
 1 −1
0 
2 1 0

ρA  L  
 2 

M=
−1  where k * = kL 2 EA .
  1 4 1  , K = EA    −1 2
6 2
 L 
*
 0 1 2 
 0 −1 1 + k 
The bar is clamped at the left hand end, so we must eliminate the first row and
column to give
−1 
2 EA  2
ρAL  4 1 
*
, K
=
M =

 . Since k = EA 2 L , k = 0.25 and hence


*
1
2
12 
L  −1 1 + k 

−1 
ρAL  4 1 
2 EA  2
. Solving the eigenvalue problem (see
=
M =
,
K
12 1 2 
L  −1 1.25
MATLAB script Problem_04_03.m gives
In model, rho = A = E = L = 1, k0 = E*A/(2*L)
Clamped-free system with spring - 1st freq = 1.9027 rad/s
Problem 4.4
Modelling the system with two elements
 1 −1 0 
0 0 0 
 2 
 2 

K =EA    −1 1 0  + EA   0 1 −1
L
L
 0 0 0 
0 −1 1 
2 1 0
0 0 0  0 0 0 
ρA  L  
ρA  L  

 

M=
  1 2 0 +
  0 2 1  + 0 0 0  . Thus
6  2 
6  2 
 0 0 0 
0 1 2  0 0 m 
31
2 1
0 
 1 −1 0 

ρA  L  
 2 
M=
1  , K = EA    −1 2 −1 where =
m* 12m ρAL .
  1 4
6  2 
L
*
+
0
1
2
m
 0 −1 1 


The bar is clamped at the left hand end, so we must eliminate the first row and
column to give
1 
ρAL  4
2 EA  2 −1
. Since m = 3ρAL , m* = 36 and hence
M =

, K


*
−
1
1
12 1 2 + m 
L 

ρAL  4 1 
2 EA  2 −1
. Solving the eigenvalue problem must be
=
,
K
12 1 38
L  −1 1 
solved, (see MATLAB script Problem_04_04.m) gives
=
M
In model, rho = A = E = L = 1, m0 = 3*rho*A*L
Clamped-free system with mass - 1st freq = 0.54733 rad/s
Problem 4.5
Assembling an axial deflection element of length 3L 4 and an axial deflection
element of length L 4 gives
2 1 0
0 0 0 
6 3 0 
ρ A  3L  
ρA  L  
ρA  L  



M=
  1 2 0 +
  0 2 1  =
  3 8 1 
6  4 
6 4
6 4
 0 0 0 
0 1 2 
0 1 2 
 1 −1 0 
0 0 0 
 1 −1 0 
 4 
 4 
 4 


=
K EA    −1 1 0  + EA   0 1 =
−1 EA    −1 4 −3
 3L 
L
 3L 
 0 0 0 
0 −1 1 
 0 −3 3 
Since the left hand end is clamped, we must eliminate the first row and column, thus
ρAL 8 1 
4 EA  4 −3
. The eigenvalue problem can be formulated
=
M =
, K


24 1 2 
3L  −3 3 
and solved, see MATLAB script Problem_04_05.m.
Assembling two axial deflection elements, each of length L 2 , and applying the
boundary conditions has been carried out in Problem 4.1, part (b) and the system
matrices are
ρAL  4 1 
2 EA  2 −1
. Again the eigenvalue problem can be
=
, K
M =


12 1 2 
L  −1 1 
formulated and solved. Problem_04_05.m gives
In model, rho = A = E = L = 1
Clamped-free with equal length elements - 1st freq = 1.6114 rad/s
Clamped-free with unequal length elements - 1st freq = 1.6157 rad/s
Problem 4.6
The arrangement of rows and columns for a lateral displacement element is
[u1 ψ1 u2 ψ 2 ] . Thus using a single element to model a pinned-pinned beam, we
must eliminate the first and third rows and columns. Hence
32
2
2
2
−3L2 
ρAL  4 L
EI  4 L 2 L 
M =

 K


420  −3L2 4 L2 
L3  2 L2 4 L2 
We can formulate and solve the eigenvalue problem to determine the system natural
frequencies, see MATLAB script Problem_04_06.m. Running this scripts gives
In model, rho = A = E = I = L = 1
Pinned-Pinned system with one element - 1st freq = 10.9545 rad/s
Problem 4.7
Assembling two lateral deflection elements of equal length gives:
0
0
  156 22le
54
−13le 0 0  0 0



0
0
4le2
13le
−3le2 0 0  0 0
  22le

13le
156 −22le 0 0  0 0 156 22le
ρAle   54

+
M
2
4le2
420   −13l −3l 2 −22l
 0 0 22le
l
4
0
0
e
e
e
e

 
54
13le
0
0
0
0 0  0 0
 0
 0
0
0
0
0 0  0 0 −13le −3le2

K
  12
6le

2
  6le 4le

EI   −12 −6le

le3   6le 2le2

0
 0

0
 0
−12
6le
−6le
12
2le2
−6le
−6le
0
0
4le2
0
0
0 0 0
 
0 0 0
0 0 0
+
0 0 0
 
0 0 0
0 0  0
0
0
0
0
0
6le
0
0
−12
0 6le 4le2
0 −12 −6le
−6le
12
2le2
−6le
0
0
0
12
6le
0
0
54
13le
156
−22le
0 

0  
−13le  

−3le2  

−22le  

4le2  





2le2  

−6le  

4le2  
0
0
6le
where le = L 2 . Note that pairs of rows and columns are overlapped because we are
making both the slope and deflection at the right hand end of the first element to be
equal to the corresponding slope and column of the left hand end of the second
element. For a clamped-clamped beam the first, second, fifth and sixth rows and
columns must be eliminated to give
ρAle 312 0 
EI  24 0 
,
=
M =
K



.
420  0 8le2 
le3  0 8le2 
k11
EI × 24 × 420
EI
Thus=
ω1
=
= 22.7359
4
m11
ρAL4
ρA ( L 2 ) 312
The MATLAB script Problem_04_07.m) repeats this calculation thus:
In model, rho = A = E = I = L = 1
Clamped-clamped system with two element - 1st freq = 22.7359 rad/s
Problem 4.8
The assembly of two lateral deflection elements of equal length is shown in Problem
4.7 (above) where le = L 2 . For a clamped-pinned beam the first, second and fifth
rows and columns must be eliminated to give
33
0
−13le 
 24 0 6le 
 312



ρAle 
EI
M
=
8le2
−3le2  K = 3  0 8le2 2le2  . Solving the resultant
 0
420 
le 

2
2
2
4le2 
6le 2le 4le 
 −13le −3le
eigenvalue problem (see MATLAB script Problem_04_08.m) gives
In model, rho = A = E = I = L = 1
Clamped-pinned system with two element - 1st freq = 15.5608 rad/s
Problem 4.9
Using a single element, le = L . Because the beam is clamped at the left hand end, the
first and second rows and columns of the element matrices must be eliminated.
Introducing an extra coordinate ( q3 in Figure 4.21) and adding the effect of a
concentrated mass m at q3 , we have
 156 −22 L 0  0 0 0 
 156 −22 L 0 



ρAL 
ρ
AL
0  where
M =  −22 L 4 L2 0  + 0 0 0  =  −22 L 4 L2
420 
420 

0
0
0  0 0 m 
0
m* 


 0
=
m* 420m ρAL . Adding a spring of stiffness k between q1 and q3 , gives
 12 −6 L 0   k

 
EI
2
K=
−
6
L
4
L
0
+ 0


L3 
0
0
0   −k


12 + k * −6 L −k * 
0 −k 
 q1 


EI
 

2
 −6 L
0 0 =
4L
0  q=
q2 
3
L 

q 
*
0 k 
k* 
0
 3
 −k

 + Kq =
where k * = kL3 EI . Thus the equation of motion is Mq
0 . Note that there is
an error in the stiffness matrix given in the solution for this problem in the text book.
Problem 4.10
Consider Equation (4.37). Since the cross sectional area is now a function of ξ we
must rewrite the integral in the first equation of (4.37) as
T
 N e′1 ( ξ )   N e′1 ( ξ ) 
A
N
ξ
+
A
N
ξ
(
)
(
)
(
)


 dξ
1
1
2
2
e
e
∫0
 N e′ 2 ( ξ )   N e′ 2 ( ξ ) 
Hence
2
( A1Ne1 + A2 Ne 2 ) Ne′1Ne′ 2 
le  ( A1 N e1 + A2 N e 2 ) N e′1

N e1 N e1 ( ξ ) , etc.
∫0 ( A N + A N ) N ′ N ′ ( A N + A N ) N ′2  d ξ where =
e1 e 2
e2 
2 e2
1 e1
2 e2
 1 e1
1
1
Now N e′1 ( ξ ) =− and N e′1 ( ξ ) = .
le
le
Consider (for example) the element in row 1, column 2.
le
l
e
 ξ
ξ  1  1 
ξ2 
ξ2 
1 

 + A2
 A1 1 −  + A2   −    d ξ = − 2  A1  ξ −
2le 
2le 
le   le   le 
le  
  le 
0
le 
∫0
l 
l 
1 
1A A 
=
− 2  A1  le − e  + A2 e  =
−  1+ 2
le  2
2
2
2 
le  
34
Completing all the integrations gives
E  A + A2   1 −1
Ke =  1

le  2   −1 1 
Rewriting the integral in the first equation of Equation (4.39) to account for the
variation of cross section area,
T
 N e1 ( ξ )   N e1 ( ξ ) 
∫0 ρ ( A1Ne1 ( ξ ) + A2 Ne2 ( ξ ) )  Ne2 ( ξ )  Ne2 ( ξ ) d ξ
Hence
2
( A1Ne1 + A2 Ne2 ) Ne1Ne2 
le  ( A1 N e1 + A2 N e 2 ) N e1

 ρ dξ
∫0 ( A N + A N ) N N
2
A
N
+
A
N
N
(
)
e1 e 2
e2 
2 e2
1 e1
2 e2
 1 e1

Consider (for example) the element in row 1, column 2.
le
) ρ d ξ ∫0
∫0 ( A1Ne1Ne2 + A2 Ne1Ne2 =
le
2
2
le
( A (1 − 2
1
ξ
le
2
+ ξl 2
e
)
ξ
le
( ) )ρ dξ .
+ A2 1 − lξ
e
ξ2
le2
Competing the integration process gives ρle ( A1 + A2 ) 12 . Completing all the
integrations, we have
ρl 3 A + A2 A1 + A2 
Me = e  1
12  A1 + A2 A1 + 3 A2 
(a) Using tapered elements
Let A1 , A2 , A3 be the cross section area at the clamped end, at the mid point and at the
free end of the bar. Then the assembled matrices are
 1 −1 0 
0 0 0 
E  A1 + A2  
E  A2 + A3  

K
=
0 1 −1 and

  −1 1 0  + 


le  2 
l  2 
 0 0 0  e
0 −1 1 
3 A1 + A2
ρle 
M=
A1 + A2
12 
 0
A1 + A2
A1 + 3 A2
0
0
ρl
0  + e
12
0 
0
0
0 3 A + A
2
3

0 A2 + A3
0


A2 + A3  where le = L 2 .
A2 + 3 A3 
(a) Using uniform elements
Let A1 , A2 , A3 be the cross section area at the clamped end, at the mid point and at the
free end of the bar. The mean cross sectional area of the first element is ( A1 + A2 ) 2
and for the second element is ( A2 + A3 ) 2 . Then the assembled matrices are
1
E  A1 + A2  
K
=

 −1
le  2  
 0
2
ρle  A1 + A2  
M

 1
6  2 
 0
−1 0 
0
A
+
A
E


3 
1 0  +  2
0
le  2  
0
0 0 
1 0
0
ρ
l
A
+
A


3 
2 0  + e  2
0
6  2  
0
0 0 
0 0
1 −1 and
−1 1 
0 0
2 1  where le = L 2
1 2 
Note that in both models the assembled stiffness matrices are identical, In each case
we apply the boundary condition we must eliminate the first row and column of the
35
assembled matrices. The MATLAB script Problem_04_10.m develop both models
for the linearly tapered bar and solves the resulting eigenvalues as follows:
In model, rho = E = I = L = 1. A varies linearly from 0.2 to 0.1
Clamped-free system: two tapered element - 1st freq = 1.8279 rad/s
Clamped-free system: two uniform element - 1st freq = 1.7933 rad/s
Problem 4.11
(a) Modelling the bar with three elements gives
2 1 0 0
0 0 0 0 
0 0 0 0 
1 2 0 0 
0 2 1 0 


ρA  L  
 + ρA  L  
 + ρA  L  0 0 0 0 
M=
 
 
 
6  3   0 0 0 0  6  3  0 1 2 0  6  3  0 0 2 1 






0 0 0 0
0 0 0 0 
0 0 1 2 
 1 −1 0 0 
0 0 0 0
0 0 0 0 
 −1 1 0 0 
0 1 −1 0 


3
 + EA  3  
 + EA  3  0 0 0 0 
K = EA   
  0 −1 1 0 
  0 0 1 −1
 L   0 0 0 0
L
L






 0 0 0 0
0 0 0 0 
0 0 −1 1 
Applying the boundary condition by eliminating the first row and column gives
 2 −1 0 
4 1 0
ρ
A
L
 3 




K = EA    −1 2 −1 and M =
  1 4 1 
L
6
3
 
 
 0 −1 1 
 0 1 2 
The eigenvalue problem can be formulated and solved, see the MATLAB script
Problem_04_11.m.
(b) Using a single element
ρAL  2 1 
EA  1 −1
=
, K
M =


6 1 2 
L  −1 1 
Applying the boundary condition by eliminating the first row and column gives
=
ω
m=
2ρAL 6, k =
EA L and thus
=
k m
3 E ρL2
(c) To reduce the model
 2 −1 0 

 3   2 −1
 3 
K = EA    −1 2 −1 . Thus K ss = EA   
 and hence
−
1
2
L


 L 



 0 −1 1 
 3  0 
 L  1   2 1 
K −ss1 = 
  1 2  . K sm = EA   −1 and thus
 L  
 3EA  3  

−K −ss1K sm
1 3 
 
1  2 1   0  1 3 
=
− 
 =
  . Thus we have T = 2 3

3 1 2  −1 2 3
 1 
 
T
=
K r T=
KT EA L and M r = TT MT = ρAL 3 .
Now
=
ω =
k m
3 E ρL2 . This is identical to the frequency given by a single
degree of freedom system.
36
The MATLAB script Problem_04_11.m solves .the three cases of this problems and
gives the following results
In model, rho
Clamped-free:
Clamped-free:
Clamped-free:
= A = E = L = 1
3 elements - 1st freq = 1.5888 rad/s
1 element - 1st freq = 1.7321 rad/s
Reduced model - 1st freq = 1.7321 rad/s
Problem 4.12
Assembling a large number of elements can only be done realistically using a
computer. The MATLAB script Problem_04_12.m assembles a number of axial
deflection elements and applies the boundary conditions, The first three system
natural frequencies are computed using 3, 4, 6 and 8 elements, thus:
In model, rho = A = E = I = L = 1
No of elements
1st nat freq
2nd nat freq
3
9.8776
39.9451
4
9.8722
39.6342
6
9.8701
39.5104
8
9.8698
39.4887
Exact
9.8696
39.4784
3rd nat freq (rad/s)
98.5901
90.4495
89.1770
88.9407
88.8264
% Error
The also script computes the first three natural frequencies using up to 64 elements
and plots the percentage error against the number of elements,
10
2
10
0
10
-2
10
-4
10
-6
10
-8
0
10
20
30
40
50
Number of Elements
37
60
70
Chapter 5
Note. Solving Problems 5.1, 5.2, 5.3, 5.8, 5.9 and 5.11 requires a finite element
analysis that allows shafts to be modelled and include gyroscopic effects, etc. Here
we use the rotordynamics software developed to accompany this book, but other
appropriate software can be used.
Problem 5.1
Modelling this problem requires finite element analysis (FEA) and can only be solved
using appropriate FEA software. Here MATLAB script Problem_05_01.m makes
use the Rotordynamics Software package to model and analyse the system. The
diagrams below, generated by the script, shows the solid and hollow rotors modelled
with 16 elements.
Node 17
Node 16
Node 15
Node 14
Node 13
Node 12
Node 11
Node 10
Node 9
Node 8
Node 7
Node 6
Node 5
Node 4
Brg Type 1
Node 3
Node 2
Node 1
Brg Type 1
Solid shaft modelled with 16 elements.
Hollow shaft modelled with 16 elements
The output of the script is as follows:
38
Node 17
Node 16
Node 15
Node 14
Node 13
Node 12
Node 11
Node 10
Node 9
Node 8
Node 7
Node 6
Node 5
Node 4
Brg Type 1
Node 3
Node 2
Node 1
Brg Type 1
Solid shaft, 16 Timoshenko elements
Natural Frequency 1 = 16.1328 Hz
Natural Frequency 2 = 17.1523 Hz
Natural Frequency 3 = 64.0008 Hz
Natural Frequency 4 = 68.8696 Hz
Natural Frequency 5 = 144.2202 Hz
Solid shaft, 16 Euler-Bernoulli elements
Natural Frequency 1 = 16.1527 Hz
Natural Frequency 2 = 17.1776 Hz
Natural Frequency 3 = 64.3551 Hz
Natural Frequency 4 = 69.3417 Hz
Natural Frequency 5 = 144.601 Hz
Hollow shaft, 16 Timoshenko elements
Natural Frequency 1 = 16.6622 Hz
Natural Frequency 2 = 17.7754 Hz
Natural Frequency 3 = 64.7167 Hz
Natural Frequency 4 = 69.5014 Hz
Natural Frequency 5 = 145.6799 Hz
Hollow shaft, 16 Euler-Bernoulli elements
Natural Frequency 1 = 16.7702 Hz
Natural Frequency 2 = 17.9158 Hz
Natural Frequency 3 = 66.6536 Hz
Natural Frequency 4 = 72.1014 Hz
Natural Frequency 5 = 147.602 Hz
Solid
Solid
Solid
Solid
Solid
Hollow
Hollow
Hollow
Hollow
Hollow
shaft:
shaft:
shaft:
shaft:
shaft:
shaft:
shaft:
shaft:
shaft:
shaft:
difference
difference
difference
difference
difference
difference
difference
difference
difference
difference
in
in
in
in
in
in
in
in
in
in
nat
nat
nat
nat
nat
nat
nat
nat
nat
nat
freq
freq
freq
freq
freq
freq
freq
freq
freq
freq
1
2
3
4
5
1
2
3
4
5
=
=
=
=
=
=
=
=
=
=
0.1234%
0.1479%
0.55356%
0.68546%
0.26401%
0.64794%
0.78969%
2.993%
3.7408%
1.3194%
The inside and outside diameters of the hollow shaft are chosen to make the natural
frequencies of the hollow shaft system very close to those of the solid shaft. However,
the differences between the natural frequencies obtained using Euler-Bernoulli and
the Timoshenko elements are greater for the hollow shaft than for the solid shaft.
39
Problem 5.2
Modelling this system requires finite element analysis (FEA) and can only be solved
using appropriate FEA software. Here MATLAB script Problem_05_02.m makes
use the Rotordynamics Software package to model and analyse the system. The
diagram below, generated by the script, shows the rotor, modelled with 16 elements.
Node 17
Node 16
Node 15
Node 14
Node 13
Node 12
Node 11
Node 10
Node 9
Node 8
Brg Type 1
Node 7
Node 6
Node 5
Node 4
Node 3
Node 2
Node 1
Brg Type 1
Rotor modelled with 16 elements
The output of the script is as follows:
First five natural Frequencies (Hz)
4 elements
8 elements 16 elements
25.0929
25.0928
25.0928
30.1380
30.1378
30.1378
59.4160
59.4145
59.4144
65.9957
65.9937
65.9935
162.5512
162.5205
162.5182
Note that as the number of elements is increased the change in a particular estimated
frequency decreases.
Problem 5.3
Modelling this problem involves finite element analysis (FEA) and can only be
solved using appropriate FEA software. Here MATLAB script Problem_05_03.m
makes use the Rotordynamics Software package to model and analyse the system.
The diagrams below, generated by the script, shows the rotor models. In model 1,
four tapered elements and three uniform elements are used, in model 2, seven uniform
elements of different diameters are used and in model 3, eleven uniform elements if
different diameters are used. In models 2 and 3 the diameters of the uniform elements
are chosen to be the average of the shaft diameters at each end of the element.
40
Node 8
Node 7
Node 5
Node 4
Node 3
Node 2
Node 1
Node 6
Brg Type 1
Brg Type 1
Model 1. Model with 4 tapered elements and 3 uniform elements.
Node 8
Node 7
Node 6
Node 5
Brg Type 1
Node 4
Node 3
Node 2
Node 1
Brg Type 1
Model 2. Model with 7 uniform elements of three different diameters.
Node 12
Node 11
Node 10
Node 9
Node 8
Node 7
Brg Type 1
Node 6
Node 5
Node 4
Node 3
Node 2
Node 1
Brg Type 1
Model 3. Model with 11 uniform elements of different diameters.
41
The output of script is as follows:
Model 1. Seven tapered and uniform elements
Natural frequency 1 = 27.7103 Hz, Error =
Natural frequency 2 = 32.5623 Hz, Error =
Natural frequency 3 = 40.8942 Hz, Error =
Natural frequency 4 = 50.9622 Hz, Error =
Natural frequency 5 = 93.1906 Hz, Error =
Model 2. Seven uniform elements
Natural frequency 1 = 27.4093
Natural frequency 2 = 32.1892
Natural frequency 3 = 39.6128
Natural frequency 4 = 48.3880
Natural frequency 5 = 109.0774
Hz,
Hz,
Hz,
Hz,
Hz,
Error
Error
Error
Error
Error
0.0183
0.0343
0.1192
0.2002
2.2136
= 0.2827
= 0.3388
= 1.1622
= 2.3740
= 18.1004
Model 3. Eleven uniform elements
Natural frequency 1 = 27.6201 Hz, Error =
Natural frequency 2 = 32.4496 Hz, Error =
Natural frequency 3 = 40.5212 Hz, Error =
Natural frequency 4 = 50.2079 Hz, Error =
Natural frequency 5 = 95.7711 Hz, Error =
0.0719
0.0784
0.2538
0.5541
4.7941
Problem 5.4
The equations of motion for this system are
I d 
θ + I p Ωψ + cuu L2 θ + cuv L2 ψ + kuu L2 θ + kuv L2 ψ = 0
 − I p Ωθ + cvu L2 θ + cvv L2 ψ + kvu L2 θ + kvv L2 ψ = 0
Id ψ
These equation are similar to those given in the solution of Problem 3.3, except that
the stiffness and damping properties are different to the x and y directions and there is
cross coupling between the stiffness and damping in the x and y directions. Thus we
have in matrix notation,
ΩI p   θ  2 cuu cuv   θ  2  kuu kuv   θ  0 
 I d 0   
θ  0
+

 + L 
 
 0 I     −ΩI
   + L k
 =
0  ψ 
d  ψ  

 cvu cvv  ψ 
 vu kvv  ψ  0 

p
The stiffness and damping coefficients are based on the hydrodynamic bearing
properties. To determine the stiffness and damping of the bearing at 3000 rev/min,
from Equation (5.84)
3
DΩηL3b 0.1 ( 3000 × 2 π 60 ) 0.03 × 0.020
0.0393 where.
Ss =
=
=
2
−3
8 fc 2
8 × (1200 2 ) × 0.2 × 10
(
)
2
 Ss   D 
From Equation (5.85), the Somerfeld=
number, S =
   0.3125 .
 π   Lb 
To determine the eccentricity it is necessary to solve the quartic equation given by
Equation (5.83).
(
(
))
(
)
β4 − 4β3 + 6 − 0.0400 2 16 − π2 β2 − 4 + π2 0.0400 2 β + 1 =0 where β = ε2 .
The equation must be solved numerically, and the smallest root is β =0.6557 . Thus
ε =0.8098 . Using Equation (5.61)
42
DΩηL3b ε2
πDΩηL3b ε
and
521.47N
=
−
=
296.76N
fr =
−
=
f
t
2
32
2
2
2
2
2c 1 − ε
8c 1 − ε
(
)
(
)
π 1 − ε2
= 0.5691 . Hence
4ε
−1
=
γ tan ( 0.5726
=
or γ 0.5201 × 180
=
π 29.7974°
) 0.5174rad=
To determine the stiffness and coefficients for the bearings we must evaluate
Equation (5.87). This is a tedious process and the details are not shown. The results
are cuu =
10348 Ns m , cuv =
cvu =
−18184 Ns m , cvv =
79640 Ns m
From Equation (5.86)
=
tan γ
kuu =
5.508 MN m , kuv =
−2.188 MN m , kvu =
−16.323 MN m , kvv =
28.684 MN m .
Using the data provided and computed, the equations of motion in matrix form when
the rotor spins at 3000 rev/min are
0
188.496 
10 0 

θ
q = , M = 
Nms2 , G = 
Nms ,

0 
 0 10 
 −188.496
ψ 
 0.2587 −0.4546  4
 1.3770 −0.5470  6
C=
10 Nms , K = 

 10 N m
 −0.4546 1.9910 
 −4.0809 7.1711 
 + ( G + C ) q + Kq = 0 . This equation can be transformed into an eigenvalue
where Mq
problem, see section 5.8. The MATLAB script Problem_05_04.m determines the
bearing stiffness and damping matrix elements and solves the resultant eigenvalue
problem. The output of the script is as follows:
Rotor speed = 3000 rev/min: --Somerfeld number = 0.3125
Eccentricity = 0.80978
Radial force = 521.4747N
Tangential force = 296.756N
Gamma = 29.6429degrees
Roots = (-1605.9467)rad/s and (-563.3804)rad/s
Roots = (-40.18492+287.8334i)rad/s and (-40.18492-287.8334i)rad/s
Natural freq (Hz) Damped nat freq (Hz) Zeta
46.2544
45.8101
0.13827
46.2544
45.8101
0.13827
Rotor speed = 6000 rev/min: --Somerfeld number = 0.625
Eccentricity = 0.73606
Radial force = 486.3917N
Tangential force = 351.3163N
Gamma = 35.8402degrees
Roots = (-504.1876+573.1688i)rad/s and (-504.1876-573.1688i)rad/s
Roots = (-22.53692+325.6684i)rad/s and (-22.53692-325.6684i)rad/s
Natural freq (Hz) Damped nat freq (Hz) Zeta
121.4935
91.2226
0.66048
121.4935
91.2226
0.66048
51.9557
51.8317
0.06904
51.9557
51.8317
0.06904
43
Problem 5.5
Using Equation (5.97), the equations of motion for this system are
mu + cu + ku + k s v =
0
ms 2 + cs + k
st
. Let u = u0e , etc to give
mv + cv − k s u + kv =
0
−ks
ks
ms 2 + cs + k
=0.
Hence (taking square roots) ms 2 + cs + k =± jk s . Let s = jω (i.e. assuming the real
part of the solution is zero) gives k − mω2 + jcω ± jk s = 0 . Considering the real and
imaginary parts we have k − ω2 m = 0 and ± ks + cω =0 . Thus ω = k m = ωn and
c= ks ωn= ccrit . Since ks = βτ ( Dm Lb ) and τ= P Ω then at the critical conditions,
ks =βP ( Dm Lb Ω ) =ccrit ωn .
Thus Pmax = ccrit ωn Dm Lb Ω β = ( Ωccrit Dm Lb β )
(
k m
)
Given that the power P = 30000 W at a rotor speed of Ω =9600 rev min , then
=
Ω 9600 × 2π =
60 1005.3rad s and τ= P Ω= 29.8416 N m . For this system, β =3 ,
and the blades are of length Lb = 0.05 m with a mean diameter of Dm = 0.15 m .
Thus ks = βτ ( Dm Lb ) = 3 × 29.8416 ( 0.15 × 0.05 ) = 11937 N m .
The rotor is of length L = 0.3m (not 300m as stated in question!) with a diameter
d = 0.015 m carries a central mass ( m ) of 3kg . Thus, for the shaft,
I = πd 4 64 = π× 0.0154 64 = 2.4850 ×10−9 m 4 .
Since E= 2 ×1011 Pa , at the mid-span of the shaft,
k = 48 EI L3 = 48 × 2 × 1011 × 2.4850 × 10−9 0.33 = 883570 N m . Thus we have
=
ωn
k=
m
883570=
3 542.7 rad s and hence ccrit = ks ωn= 21.9948 Ns m
These calculations are repeated in the MATLAB script Problem_05_05.m.
In order to determine the system properties for various values of damping, it is
necessary to solve an eigenvalue problem derived from the system equations. The
eigenvalue problem is solved in the MATLAB script Problem_05_05.m. From the
roots s, we have
s 2π and ζ = −real ( s ) s . To determine
=
ωd imag ( s ) 2π , ω=
n
the direction of the orbit we must compute κ This is done in Problem_05_05.m
using the MATLAB function whirl.m which is provided in the Rotordynamics
software package. The output from the script is
k_sw = 11936.6207 N/m
Critical damping = 21.9948 Ns/m
Damping (Ns/m)
0
0
Damped freq (Hz)
86.3755
86.3755
Nat freq (Hz)
86.3775
86.3775
zeta
0.006754
-0.006754
kappa
-1
1
20
20
86.3739
86.3739
86.3739
86.3811
-0.000613
0.012896
1
-1
40
40
86.3690
86.3690
86.3703
86.3846
0.005529
0.019037
1
-1
44
Note that (i) the value of κ = ±1 implying that the forward or backward whirl orbit is
circular, (2) that when c = 0 and c = 20 Ns m one value of ζ is negative, i.e. the
system is unstable, as we expect, since the critical damping is 21.995 Ns m .
Problem 5.6
We now introduce an error into Equation (5.97), the equations of motion for this
system by changing the sign of ks . Thus we have
mu + cu + ku − k s v =
0
ms 2 + cs + k
st
. Let u = u0e , etc to give
mv + cv + k s u + kv =
0
ks
−ks
ms 2 + cs + k
=0.
Hence (taking square roots) ms 2 + cs + k =± jk s . Let s = jω (i.e. assuming the real
part of the solution is zero) gives k − mω2 + jcω ± jk s = 0 . Considering the real and
imaginary parts we have k − ω2 m = 0 and ± ks + cω =0 . Note that these equations are
identical to those of Problem 5.5, so the frequencies computed will be identical to
u
ms 2 + cs + k
in the correctly
those of Problem 5.5. However, where as 0 =
v0
ks
u0 ms 2 + cs + k
formulated analysis,
in the incorrectly formulate analysis. Thus the
=
v0
− ks
sign of u0 v0 will be reversed in the incorrect formulation. So if u0 v0 is positive
the incorrect formulation will be negative and vice-versa. The direction of whirl will
be reversed and in error.
Problem 5.7
From Equation (5.96)
0  u c0 + cd
 m0 + md
 +

m0 + md   v  − md Ω
 0
md Ω  u 
 
c0 + cd   v 
 k0 + kd − 1 md Ω 2
4
+

− 12 cd Ω

 u  0 
 =
 
2 v
1
k0 + kd − 4 md Ω    0 
1c Ω
2 d
Let =
k k0 + k d , =
m m0 + md , =
c c0 + cd and u = u0e st , etc. Hence,
( 12 cd + md s ) Ω
ms 2 + cs + k − 14 md Ω 2
−
( 12 cd + md s ) Ω
ms 2 + cs + k − 14 md Ω 2
=0
To determine the stability boundary, the real part of the solution is zero, and hence we
let s = jω . Under these conditions, Ω → Ωc and ω → ωc (where subscript c implies
the critical value at the boundary of stability. Thus
( 12 cd + jωc md ) Ωc
−mωc2 + jωc c + k − 14 md Ωc2
−
( 12 cd + jωc md ) Ωc
−mωc2 + jωc c + k − 14 md Ωc2
( −mωc2 + jωcc + k − 14 md Ωc2 ) + ( 12 cd + jωc md )2 Ωc2 =0
2
45
= 0 . Thus
−mωc2 + jωcc + k − 14 md Ωc2 =± j
( 12 cd + jωc md ) Ωc . Hence
−mωc2 + k − 14 md Ωc2 =± ( −ωc md Ωc ) and cωc =± 12 cd Ωc
Taking the positive sign in the imaginary equation, the limit of stability is when
ωc =Ωc cd ( 2c ) =Ωc r 2 where r = cd c . Substituting this value of ωc into the real
part of the equation (and taking the positive sign) gives
k
2
2
.
−m ( Ωc r 2 ) + k − 14 md ( Ωc r 2 ) =− rmd Ωc2 2 and hence Ωc =2
mr 2 + md − 2md r
Note that when c0 = 0 , =
r cd
( c0 + cd=)
1.
k
k
and Ωc 2 = 2
Thus ωc =Ωc 2 =
m − md
m0
Note also that as c0 → ∞ , r → 0 and then Ωc =2
k
and ωc =0 .
md
For the shaft, I =
πd 4 64 =×
3.068 10 −7 m 4 . The shaft stiffness at the centre is (see
Appendix 2, Table A2.1, system 1) =
is k0 48EI
=
L3 1.3635 × 10 7 N m .
7
=
k 1.3635 × 10 7 + 2 × 10
=
3.3635 × 10 7 N m . m0 = 600kg , md = 120kg ,
cd = 200 Ns m and kd = 20 kN m ,
k
3.3635 ×107
Ωc 2 = 2
= 473.53rad s ≡ 4521.9 rev min and
For c0 = 0 , =
m0
600
ωc =Ωc 2 =473.53 2 =236.76 rad s ≡ 37.68Hz .
The critical speed and frequency can be easily calculated for other system damping.
However, MATLAB script Problem_05_07.m calculates these speeds and
frequencies (see output below) and also plots various parameters against rotor speed
for a system damping of c0 = 160 Ns m .
System damping c0 = 0. Conditions at limit of stability
Shaft speed
= 4521.9312 rev/min
Whirl frequency = 37.6828 Hz
System damping c0 = 80 Ns/m. Conditions at limit of stability
Shaft speed
= 6231.7804 rev/min
Whirl frequency = 37.0939 Hz
System damping c0 = 160 Ns/m. Conditions at limit of stability
Shaft speed
= 7663.7602 rev/min
Whirl frequency = 35.4804 Hz
System damping c0 tend to infinity. Conditions at limit of stability
Shaft speed
= 10111.3456 rev/min
Whirl frequency = 0 Hz
46
Problem 5.8
This problem involves finite element analysis (FEA) and can only be solved using
appropriate FEA software.
Case 1. When the bearings are rigid, the bearings act in their nominal position, i.e.
0.1m from end of the shaft. There is no axial tension in the shaft.
Case 2. When contact angle is 20° the point of contact of the bearing on the shaft is
shifted by
=
δ 0.035
=
tan ( 20° ) 0.0127m so that the bearings are now 0.0873m from the ends of
the shaft.. A tensile force of 500N acts on the shaft.
Case 3. If δ =0.0127m a tensile force on the shaft can be applied to the shaft which
will negate the effect of this shift in bearing position. To do this it is necessary to
calculate the first natural frequency for a range of tensile forces and interpolate to find
the force that gives a first natural frequency identical to that of Case 1 (above).
Here MATLAB script Problem_05_08.m makes use the Rotordynamics Software
package to model and analyse the system. The diagrams below, generated by the
script, shows the rotor modelled with 12 elements.
Node 13
Node 12
Node 11
Node 10
Node 9
Node 8
Node 7
Node 6
Node 5
Brg Type 1
Node 4
Node 3
Node 2
Node 1
Brg Type 1
The output of the script is as follows:
Tensile force applied = 0 N
Displacement of the bearings outwards = 0mm
Frequency 1 = 23.2513 Hz
Frequency 2 = 24.3349 Hz
Frequency 3 = 79.8648 Hz
Frequency 4 = 127.6658 Hz
Frequency 5 = 168.3636 Hz
Tensile force applied = 500 N
Displacement of the bearings outwards = 12.739mm
Frequency 1 = 22.3558 Hz
Frequency 2 = 23.3372 Hz
Frequency 3 = 77.8035 Hz
Frequency 4 = 123.1271 Hz
Frequency 5 = 163.4651 Hz
Force to negate the effect of a contact angle of 20 deg: 14905.5202 N
47
Problem 5.9
This problem involves finite element analysis (FEA) and can only be solved using
appropriate FEA software. In order to compute the bearing properties we require the
static load acting on each bearing. From the data of Table 5.11, we can compute the
volume and hence the mass of the shaft. Similarly, the question gives the details of
the disks and the volume and hence the mass can be determined. Thus we have
msft = 101.914 kg , mdsk = 34.184 kg and adding these together the total mass of the
Rotor spd = 0, Isotropic brgs
Natural frequency 1 = 19.0763
Natural frequency 2 = 19.0763
Natural frequency 3 = 38.7344
Natural frequency 4 = 38.7344
Natural frequency 5 = 67.2388
Natural frequency 6 = 67.2388
Hz
Hz
Hz
Hz
Hz
Hz
Rotor spd = 3000rev/min, Isotropic brgs
Natural frequency 1 = 18.9058 Hz, kappa
Natural frequency 2 = 19.2469 Hz, kappa
Natural frequency 3 = 38.4893 Hz, kappa
Natural frequency 4 = 38.9753 Hz, kappa
Natural frequency 5 = 65.8526 Hz, kappa
Natural frequency 6 = 68.5804 Hz, kappa
Rotor spd = 0, Anisotropic brgs
Natural frequency 1 = 18.8252 Hz
Natural frequency 2 = 19.0763 Hz
Natural frequency 3 = 36.9948 Hz
Natural frequency 4 = 38.7344 Hz
Natural frequency 5 = 63.3190 Hz
Natural frequency 6 = 67.2388 Hz
48
= -1.0000
= 1.0000
= -1.0000
= 1.0000
= -1.0000
= 1.0000
Node 29
Node 28
Node 27
Node 26
Node 25
Node 24
Node 23
Node 22
Node 21
Node 20
Node 19
Node 18
Node 17
Node 16
Brg Type 3
Node 15
Node 14
Node 13
Node 12
Node 11
Node 9
Node 10
Node 8
Node 7
Node 6
Node 5
Brg Type 3
Node 4
Node 3
Node 2
Node 1
rotor is 136.097 kg . Thus total force on the bearings
is F 136.097
=
=
g 13.35kN .
This can then be divided in ratios 1:1:1, 1:3:1 and 2:1:2, i.e 445N: 445N: 445N,
267N:801N:267N and 534N:267N:534N.
The MATLAB script Problem_05_09.m makes use the Rotordynamics Software
package to model and analyse the system. The diagrams below, generated by the
script, shows the rotor modelled with 28 Timoshenko elements. Note that all the
hydrodynamic bearing configurations give a stable system.
Brg Type 3
Rotor spd = 3000rev/min, Anisotropic brgs
Natural frequency 1 = 18.7426 Hz, kappa = -0.4948
Natural frequency 2 = 19.1590 Hz, kappa = 0.5010
Natural frequency 3 = 36.9661 Hz, kappa = -0.1596
Natural frequency 4 = 38.7591 Hz, kappa = 0.0978
Natural frequency 5 = 62.9348 Hz, kappa = -0.2357
Natural frequency 6 = 67.5727 Hz, kappa = 0.2983
Hydrodynamic brg, force = 1:1:1
Damped frequency 1 = 18.8745 Hz,
Damped frequency 2 = 19.6652 Hz,
Damped frequency 3 = 34.9988 Hz,
Damped frequency 4 = 45.7624 Hz,
Damped frequency 5 = 52.7040 Hz,
Damped frequency 6 = 53.0997 Hz,
Zeta
Zeta
Zeta
Zeta
Zeta
Zeta
=
=
=
=
=
=
0.0024,
0.0045,
0.0628,
0.0387,
0.3367,
0.4313,
Hydrodynamic brg, force = 1:3:1
Damped frequency 1 = 18.9386 Hz,
Damped frequency 2 = 19.6256 Hz,
Damped frequency 3 = 35.6859 Hz,
Damped frequency 4 = 45.9458 Hz,
Damped frequency 5 = 44.2690 Hz,
Damped frequency 6 = 49.1626 Hz,
Zeta
Zeta
Zeta
Zeta
Zeta
Zeta
=
=
=
=
=
=
-0.0016, kappa = 0.2121
0.0072, kappa = -0.0632
0.0747, kappa = 0.4200
0.0359, kappa = -0.1442
0.3387, kappa = 0.5882
0.3617, kappa = 0.1765
Hydrodynamic brg, force = 2:1:2
Damped frequency 1 = 18.6005 Hz,
Damped frequency 2 = 19.5302 Hz,
Damped frequency 3 = 32.4314 Hz,
Damped frequency 4 = 45.5308 Hz,
Damped frequency 5 = 56.5182 Hz,
Damped frequency 6 = 55.7581 Hz,
Zeta
Zeta
Zeta
Zeta
Zeta
Zeta
=
=
=
=
=
=
-0.0093, kappa = 0.3250
0.0148, kappa = -0.1593
0.0560, kappa = 0.2676
0.0650, kappa = -0.2367
0.2383, kappa = 0.5638
0.4457, kappa = 0.4325
kappa
kappa
kappa
kappa
kappa
kappa
= 0.1463
= -0.0047
= 0.2970
= -0.1739
= 0.4842
= 0.4302
Problem 5.10
Part a: Solve problem 5.1(b).
π 4
For the shaft, I=
. EI s 1.2332 × 10 5 Nm 2
d0 − di4 = 6.1662 × 10 −7 m 4 =
s
64
π
For the shaft, ρA =ρ d02 − di2 =6.5611kg m
4
π 2
For each disk, =
Md
(dd − d02 )=
td ρ 75.2776kg
4
Assume the shape of the first mode of vibration can be approximated by
u1 (=
z ) sin ( πz L ) .
(
)
(
)
Thus u1′ ( z ) =− ( π L ) cos ( πz L ) and u1′′ ( z ) =− ( π L ) sin ( πz L ) . Note that when
2
z = 0 and z = L , both u1 ( z ) and u1′′ ( z ) are zero. Thus the boundary conditions are
satisfied.
U=
max
1 L EI
s
2 0
∫
{
}
4
 EI  π   L 
5
dz  s    =
− ( π L ) sin ( πz L ) =
 7.3321× 10 and
2
2
L

   
2
2
49
Tmax = 12 ω12 ∫ ρA{sin ( πz L )} dz + 12 ω12 M d {sin ( πz D1 L )} + 12 ω12 M d {sin ( πzd 2 L )}
L
2
2
2
0
{
}
 L

= 12 ω12 ρA   + M d sin 2 ( π0.5 L ) + sin 2 ( π L ) 
 2

ω12
=
2


 1.6 
2
6.5611 2  + 75.2776 {0.6913 + 0.8536} =60.7724ω1




Tmax = U max . Thus 60.7724=
ω12 7.3321 × 10 5 so that
=
ω1
7.3321 × 10 5 60.7724
= 109.64 s .
= 109.64
=
Thus the approximate natural frequency
( 2π ) 17.4816Hz .
Part b: Solve Problem 5.2
π 4
For the shaft,
. EI s 3.1063 × 10 5 Nm 2
=
Is =
ds 1.5532 × 10 −6 m 4=
64
π
For the shaft, ρA =ρ ds2 =34.4593 kg m
4
π 2
For each disk, =
(dd − ds2 )=
Md
td ρ 75.6574kg
4
Assume that the shape of first mode of vibration is approximated by
u1 ( z ) =0.2148 − 0.8815 z + 1.6667 z 4 − z 5 .
(
Thus u1′ ( z ) =
−0.8815 + 6.6667 z 3 − 5 z 4
)
L and u=
1′′ ( z )
( 20 z 2 − 20 z3 )
L2
When z = 0.4 then z = 0.25 and when z = 1.6 then z = 1 . Substituting these values
into u1 ( z ) , we have u1=
( 0.4 ) u=
1 (1.6 ) 0 so that there is no deflection at the
bearings. When z = 0 then z = 0 and when z = 1.6 then z = 1 . Substituting these
u1′′ ( 0 ) u=
values into u1′′ ( z ) , we have =
1′′ (1.6 ) 0 . Thus there is no bending moment at
the ends of the shaft.
2
 EI s  400  1 
1 L EI 202 z 2 L4 − z 3 L5 dz =
1.4445 ×105 and
U max =
s
 2   3  105  =
2 ∫0


  L 
L
2
2
2
2
Tmax = 12 ω12 ∫ ρA{u1 ( z )} dz + 12 ω12 M d {u1 ( zd 1 )} + {u1 ( zd 2 )} + {u1 ( zd 3 )} 
0


{
}
{
}
2
=12 ω12 34.4593 ×1.6 × 0.0158 + 75.6574 ( −0.2148 ) + 0.15302 +0.15632  =3.9896ω12


Tmax = U max . Thus 3.9896=
ω12 1.4445 × 10 5 so that
=
ω1
1.4445 × 10 5 3.9896
= 190.28 s .
Thus the approximate natural frequency
= 190.28
=
( 2π ) 30.2844Hz .
50
Problem 5.11
Part (a): For the 2 degree of freedom system the mass matrix is
0

 mdsk 0  312.9kg
=
M dsk =



Id   0
12.71kg m 2 
 0
From Appendix 2, Table A2.1, system 1,
(
=
k =
3EI ( a
1.60 × 10 N m
) ab =
−b ) a b =
2.75 × 10 N
kT =
kuu =
3EI a3 + b3
kC
ψu
2
2
3 3
2 2
6
5
2.75N  5
16.0 N m
10 .
kR =
kψψ =
3EI ( a + b ) ab =
3.30 × 10 5 N m and so K = 
3.30N m 
 2.75N
The eigenvalue problem can then be solved.
Part (b). For the 2 degree of freedom system mass coefficients are given in Appendix
17ρA ( a + b ) 35 =4.76kg ,
2, Table A2.2, system 1. muu =
(
mψu =
13ρA b2 − a2
−0.168kg m , m
) 35 =
ψψ
(
)
=
2ρA a3 + b3 105 =
0.052kg m 2 .
−0.168kg m 
 4.76kg
Hence M shft = 
. Thus=
M M shft + M dsk , giving
2
−
0.168kg
m
0.052kg
m


−0.168kg m 
 317.63kg
. Using K from (a), the eigenvalue can then be
M=
2
−
0.168kg
m
12.765kg
m


solved.
Part (c) The MATLAB script Problem_05_11.m uses the rotordynamics package to
solve the 28 degree of freedom mode. The script also solves the two, 2 degree of
freedom models and also computes the errors in the 2 degree of freedom models,
compared with the 20 element model. The output of Problem_05_11.m is given
below:
Brg Type 1
Node 1
Node 2
Node 3
Node 4
Node 5
Node 6
Node 7
Node 8
Node 9
Node 10
Node 11
Node 12
Node 13
Node 14
Node 15
Node 16
Node 17
Node 18
Node 19
Node 20
Node 21
Brg Type 1
20 Element FE model
Disk
diam/thickness
1
0.3000/0.0300
2
0.4000/0.0400
3
0.5000/0.0500
4
0.6000/0.0600
5
0.7000/0.0700
6
0.8000/0.0800
1st Nat freq (Hz)
40.3017
28.0620
20.5934
15.8173
12.5873
10.2940
51
2nd Nat freq (Hz)
226.6899
136.5622
81.8612
52.6797
36.1118
26.0200
2dof model: Massless shaft,
Disk 1st Nat freq (Hz) 2nd
1
45.7859
2
29.6902
3
21.1981
4
16.0807
5
12.7161
6
10.3627
no central hole in disk
Nat freq (Hz) 1st % error
298.3024
13.6077
145.5528
5.8020
83.5017
2.9363
53.0839
1.6657
36.2349
1.0235
26.0641
0.6667
2dof model: Including shaft
Disk 1st Nat freq (Hz) 2nd
1
40.3289
2
28.0665
3
20.5944
4
15.8175
5
12.5874
6
10.2941
mass, disk with a central hole
Nat freq (Hz) 1st % error 2nd % error
242.1845
0.0676
6.8351
137.4105
0.0158
0.6212
81.9180
0.0046
0.0694
52.6853
0.0016
0.0106
36.1125
0.0006
0.0021
26.0201
0.0003
0.0005
2dof model: Including shaft
Disk 1st Nat freq (Hz) 2nd
1
40.0549
2
27.9428
3
20.5332
4
15.7842
5
12.5678
6
10.2819
mass, no central hole in disk
Nat freq (Hz) 1st % error 2nd % error
242.0939
-0.6123
6.7952
137.3867
-0.4247
0.6037
81.9091
-0.2924
0.0585
52.6809
-0.2087
0.0024
36.1099
-0.1546
-0.0051
26.0184
-0.1182
-0.0062
2nd % error
31.5905
6.5835
2.0041
0.7673
0.3409
0.1693
This output show that the results are substantially improved by accounting for the
mass of the shaft, particularly where the disk is small.
The question asks the reader to (a) model the system assuming a mass-less shaft and a
disk without a central hole, and (b) model the system accounting for the shaft mass
and modelling the disk with a central hole. By mistake the solutions in the text for
part (b) gives the results of modelling the system accounting for the shaft mass and
modelling the disk without a central hole. Clearly the mass at the centre of the disk is
accounted for twice. However, the above results show that this over-estimation of
mass makes very little difference to the results.
52
Chapter 6
Note. Solving Problems 6.10, 6.11 and 6.12 requires a finite element analysis that
allows shafts to be modeled and include gyroscopic effects, etc. Here we use the
rotordynamics software developed to accompany this book, but other appropriate
software can be used.
Problem 6.1
The equations of motion for free vibration of this rotor are developed in Problem 3.1
and are as follows:
 + L2 k θ = 0
I d 
θ + I p Ωψ
 − I p Ωθ + L2 k ψ = 0
Id ψ
To obtain the critical speed we can solve these equations to determine the roots
(natural frequencies) and then equate the speed of rotation to the natural frequencies.
A slightly different but equivalent approach is to combine the pair of equations above
by letting ϕ = ψ − jθ . Then subtracting j × the second equation from the first we
 − jΩI p ϕ + kL2 ϕ = 0 . We can now solve this equation by letting ϕ = ϕ0 e jΩt
have I d ϕ
(
)
etc. to obtain −Ω2 I d − j 2 Ω2 I p + kL2 ϕ0 = 0 and hence
=
Ω
(
)
L2 k I d − I p . This is
a forward whirling critical speed. If we let ϕ = ϕ0 e− jΩt in the previous equation we
obtain
=
Ω
(
)
L2 k I d + I p . Given that L = 0.5m , I p = 0.6kg m 2 , I d = 10kg m 2 and
L2 k
0.52 × 10 6
k = 10 N m . Then =
Ω
=
= 2.3585 × 10 4 and 2.6596 × 10 4 .
Id ± I p
10 ± 0.6
6
2
Hence the critical speeds are 153.57 and 163.08 rad s or 1467 rev min (backwards)
and 1557 rev min (forwards).
To determine the response of the rotor to an unbalance we must add the out of balance
forcing term to the equations of motion. The out of balance force at the right end of
the rotor due to an out of balance mass m0 at a radius ε is fub= m0 εΩ2 . This
rotating force can be resolved into two components in the x and y directions, i.e
2
f=
m0 εΩ2 cos Ωt and f=
x
y m0 εΩ sin Ωt . (Here it is assumed that the out of
balance is in the x direction when t = 0 ). A force in the x direction causes a moment
Mψ = Lf x = m0 εLΩ2 cos Ωt to act about the left end of the rotor in the direction ψ . A
force in the y direction causes a moment Mθ =− Lf y =−m0 εLΩ2 sin Ωt to act about
the left end of the rotor in the direction θ . Thus the forced equations of motion are
 + L2 k θ = − m0 εLΩ2 sin Ωt
I d 
θ + I p Ωψ
 − I p Ωθ + L2 k ψ
Id ψ
= m0 εLΩ2 cos Ωt
Again letting ϕ = ψ − jθ etc. we obtain
 − jΩI p ϕ + kL2 ϕ= m0 εLΩ2 ( cos Ωt + j sin Ωt ) and hence
Id ϕ
 − jΩI p ϕ + kL2 ϕ= m0 εLΩ2 e jΩt
Id ϕ
53
(
)
t
Letting ϕ = ϕ0 e jΩt we have −Ω2 I d + Ω2 I p + kL2 ϕ0 e jΩ=
m0 εLΩ2 e jΩt and hence
m0 εLΩ2
ϕ0 =
kL2 − Ω2 I d − I p
{
(
)}
. The orbit at the left end of the rotor r0 is given by
r0 = Lϕ0 . Now Ω = 1500 × 2 × π / 60 rad/s and thus
=
θ0
0.1 × 0.1 × 0.5 × Ω2
= 6.8295 × 10 −3 rad and hence the radius of the orbit
0.5 × 10 − Ω × (10 − 0.6 )
2
6
2
at the right end is Lθ0 =
3.41mm . The MATLAB script Problem_06_01.m repeats
these calculations and gives the following output.
Critical speeds are 1466.52 and 1557.32 rev/min
Radius of orbit = 3.4147 mm
Problem 6.2.
The equations of motion for free vibrations are
 + L2 k0 + k1Ω + k2 Ω2 θ = 0
I d 
θ + I p Ωψ
(
 − I Ωθ + L ( k
I ψ
2
d
p
)
+ k Ω + k Ω )ψ = 0
2
0
1
2
To obtain the critical speed we can solve these equations to determine the roots
(natural frequencies) and then equate the speed of rotation to the natural frequencies.
A slightly different but equivalent approach is to combine this pair of equations by
letting ϕ = ψ − jθ . Then subtracting j × the second equation from the first we have
(
)
 − jΩI p ϕ + kL2 k0 + k1Ω + k2 Ω2 ϕ = 0 .
Id ϕ
We can now solve this equation by letting ϕ = ϕ0 e jΩt etc. to obtain
{−Ω I − j Ω I + ( k + k Ω + k Ω ) L } ϕ = 0 and hence
{( k L − ( I + I )) Ω + k L Ω + k L } ϕ =0 .
Thus {( k L − ( I + I ) ) Ω + k L Ω + k L } = 0 . Here Ω is a backward critical
2
2
2
2
d
p
0
2
1
2
d
2
2
1
2
0
2
d
2
0
2
p
2
2
0
2
p
2
1
0
speed. Solving this quadratic equation using the system data gives Ω =179.48 and
−141.41 Similarly, if we let ϕ = ϕ0 e− jΩt , then we have
{( k L − ( I
2
2
d
}
))
− I p Ω2 + k1L2 Ω + k0 L2 = 0 Here Ω is a forward critical speed.
Solving this quadratic equation using the system data=
gives Ω 193.06 and − 149.70 .
Taking the positive roots we have Ω =179.48 and 193.06 rad s or 1714 and 1844
rev min .
To determine the response we must add the out of balance moments to the system
thus:
 + L2 k0 + k1Ω + k2 Ω2 θ = − M Ω2 sin Ωt
I d 
θ + I p Ωψ
(
 − I Ωθ + L ( k
I ψ
2
d
p
)
+ k Ω + k Ω ) ψ=
2
0
1
2
M Ω2 cos Ωt
54
To determine the response, let ϕ = ψ − jθ . Then subtracting j × the second equation
from the first we have
(
)
 − jΩI p ϕ + k0 + k1Ω + k2 Ω2 L2 ϕ= M Ω2 ( cos Ωt + j sin Ωt )
Id ϕ
(
)
 − jΩI p ϕ + k0 + k1Ω + k2 Ω2 L2 ϕ= M Ω2 e jΩt .
and hence I d ϕ
{ (
) (
) }
Letting ϕ = ϕ0 e jΩt we have −Ω2 I d − I p + k0 + k1Ω + k2 Ω2 L2 ϕ0= M Ω2 and
M Ω2
hence ϕ0 =
. The radius of the orbit at the
−Ω2 I d − I p + k0 + k1Ω + k2 Ω2 L2
{ (
) }
) (
right end of the rotor is r0 = Lθ0 . At 1500 rev min the bearing stiffness is
1.3096 MN m . Thus
=
θ0
(
5 × 10 −3 Ω2
5 × 10 −3 × Ω2
=
=
1.2922 × 10 −3 rad
2
2
6
2
2
2
−Ω × (10 − 0.6 ) + 0.5 × 1.3096 × 10
−Ω I d1 + L k + Ω I p
)
and hence the radius of the orbit at the right end is 0.646 mm.
The MATLAB script Problem_06_02.m repeats these calculations and gives the
following output. It also plots the Campbell diagram.
Backward critical speed = 1713.93 rev/min
Forward critical speed = 1843.57 rev/min
Radius of orbit = 0.6461 mm
Natural frequency (Hz)
30
25
20
15
10
5
0
0
500
1000
Rotor speed rev/min
1500
2000
Problem 6.3.
From the solution to Problem 3.2 the natural frequencies are given by
(
(
))
ω4 I d2 − I 2p Ω2 + I d L2 k x + k y ω2 + L4 k x k y =0 . The critical speeds are obtained by
(
)
(
)
setting ω = Ω so that Ω 4 I d2 − I 2p − I d L2 k x + k y Ω2 + L4 k x k y =0 which is a
55
quadratic in Ω2 . Hence the solutions for Ω2 are 24717.4 and 32990.4, thus the
critical speeds are 157.22 and 181.63 rad s or 1501 and 1734 rev min .
From Problem 3.2 we have the natural frequencies at 0, 3000 and 10,000 rev min .
This data provides 3 points for each frequency line in the Campbell diagram.
Furthermore, we have one extra point for each frequency line because we know that at
a critical speed ω = Ω . The MATLAB script Problem_06_03.m computes the critical
speeds and plots the Campbell diagram, using 4 points to define each frequency line.
The points are joined by straight lines. The dotted line shows the exact frequency
lines. The difference between the exact and approximate lines is very small. The
output is as follows:
Critical speeds are 1501.3181 and 1734.4623 rev/min
34
Natural frequencies, Hz
32
30
28
26
24
22
0
2000
4000
6000
Rotor speed rev/min
8000
10000
Problem 6.4.
The equations of motion for free vibration of this system are
mu + kuuu + kuψ ψ =0
mv + kvv v + kvθθ =0
 + kθθθ + kθv v = 0
I d 
θ + I p Ωψ
 − I p Ωθ + kψψ ψ + kψuu =0
Id ψ
where the stiffnesses are obtained from Appendix 2, Table A2.1, System 4, and are
12 EI
6 EI
4 EI
kT =
kuu =
kvv = 3 , kC =
kuψ =
kψu =
kθθ =
kψψ =
−kvθ =
−kθv =
− 2 , kR =
a
a
a
 + ΩGq + Kq = 0 where
Writing the equations in matrix notation we have Mq
M = diag ( m, m, I d , I d ) and
56
0
0
G=
0

0
0
0
0
0
 kT

0
0
, K=
Ip
0


0 
 kC
0
0
0
0 −I p
0
0
kT
−kC
−kC
0
kR
0
kC 
0 
. To solve the eigenvalue
0

k R 
problem we must let q = q 0 e jΩt and thus: Ω2 ( M − jC ) q 0 =
Kq 0 . Solving this
eigenvalue problem with the problem data gives the following critical speed, The
resulting values of critical speed are 150.47, 175.32, 669.93 and 1002.1j rad/s - note
that the last critical speed is imaginary and hence there are only 3 critical speeds for
this 4 dof system. These are 1436.9, 1674.2, and 6397.4 rev/min. Since the supports
are isotropic only the critical speed at 1674.2 rev/min will be excited by unbalance.
mu + kuuu + kuψ ψ= mεΩ2 cos Ωt
mv + kvv v + kvθθ= mεΩ2 sin Ωt
 + kθθθ + kθv v = 0
θ + I p Ωψ
I d 
 − I p Ωθ + kψψ ψ + kψuu =0
Id ψ
Alternatively to obtain the forward whirling critical speeds excited by unbalance, we
write the equations of motion in terms of complex co-ordinates r = u + jv and
ϕ = ψ − jθ , then
mr + kuur + kuψ ϕ= mεΩ2 e jΩt
 − I p jΩϕ + kψψ ϕ + kuψ r = 0
Id ϕ
 −mΩ2 + kuu
The critical speeds are then given by det 

kuψ

which gives the quadratic in Ω2 thus
(
)
((

=0
2
Ω + kψψ 

kuψ
(
− Id − I p
)
)
)
m I d − I p Ω 4 − I d − I p kuu + mkψψ + kuu kψψ − ku2ψ =0 with solutions
Ω2 = 3.0738 × 10 4 or − 1.0041 × 10 6 . Note the second solution is not real - that is
this critical speed cannot be attained. The response to the unbalance is given by
q = q0 e jΩt
 −mΩ2 + kuu
where q 0 = 

kuψ



Ω2 + kψψ 

kuψ
(
− Id − I p
)
−1
mεΩ2 

 which gives
 0 
 −3.6058 × 10 −4 , − 1.2997 × 10 −3  , and hence the maximum orbit has radius
q 0T =


0.361mm.
From Figure 6.57, for r = 3 / 0.1 = 30 we have α
= 2.4 × 10 −3 and hence
α = αω2n = 73.77rad/s2 . The torque required is I p α =11.44Nm .
The steady state response for an angular misalignment of the disk is
 −mΩ2 + kuu
q0 = 

kuψ



2
Ω + kψψ 

kuψ
(
− Id − I p
)
−1
57
0



2
 I d − I p βΩ 
(
)
If 
β = 1o, then q0 = 1.2264 × 10 −3 , 3.8798 × 10 −3  , and hence the maximum β in
degrees is 0.361/1.2264=0.285º.
The MATLAB script Problem_06_04.m repeats these calculations and gives the
following output:
Critical speeds = 1437, 1674 and 6397 rev/min
Radius of orbit due to o/b = 0.3606 mm
Equivqlent angular misalignment = 0.2852 degree
r_bar = 30.00
For this value of r_bar obtain alpha_bar = 2.4e-3 from Fig 6.57
Acceleration thro critical speed = 73.7715 rad/s^2
Torque to accelerate thro critical speed = 11.4395 Nm
Problem 6.5.
Consider Equation (6.32). The stiffness coefficients for this system are given in
Appendix 2, Table A2.1, System 1. When the disk is at the mid-span, a = b and
hence kC = 0 . Furthermore, there is no damping and thus c=
c=
c=
T
C
R 0 . Thus, for
pure angular misalignment of disk in complex coordinates
mr + kT r =
0
 − jI p Ωϕ + k R=
Id ϕ
ϕ
( I d − I p ) βΩ2e jΩt
The first equation gives no response. Letting ϕ = ϕ0 e jΩt we obtain
I d − I p ) βΩ2
(
. Now =
ϕ0 =
kR
k R − Ω2 ( I d − I p )
k=
ψψ
12 EI
= 126kNm/rad . Thus, when β =1 ,
L
ϕ0 =−3.3923 × 10 −3 . Thus the maximum value of θ0 or ψ 0 is
θ0 = ψ 0 = 3.3923 × 10 −3 × 180 π = 0.1944° .
The MATLAB script Problem_06_05.m repeats these calculations and gives the
following output
Response of a Jeffcott rotor at 6000 rev/min = 0.1944 degrees
Problem 6.6.
Let r = u + jv . Then we have the response, from Equation (6.47), for the bend and the
48EI
j Ωt +φ0 )
unbalance mr +=
where k = 3 = 381.70kN/m ,
kr krb e jΩt + m0 aΩ2 e (
L
m = 6.9360kg and φ0 is the phase of the unbalance relative to the bend. With just the
bend the response at speed Ω =1000 rev min , or 104.72 rad s , is
krb
= 0.6244mm
r=
−mΩ2 + k
To balance the bend at speed Ω, the right side (i.e. excitation) of the equation must be
zero, and hence we must have φ0 =
180 and m0 =
krb
=
k × 0.5 × 10 −3
Ω2 × 0.075
Ω2 a
kr − m0 aΩ2
At 1500 rev/min the response is then r = b
= −1.1330mm
−mΩ2 + k
58
= 34.81g
The MATLAB script Problem_06_06.m repeats these calculations and gives the
following output:
Mass of shaft = 5.55 kg
Mass of disk = 6.94 kg
Response to bent shaft at 1000 rev/min = 0.6244 mm
Required balance mass = 0.2320 kg
Response at 1500 rev/min = 1.1330 mm
Note that mass of the shaft is similar to the mass of the disk. Thus, neglecting the
mass of the shaft is likely to lead to large errors in computed natural frequency.
Note also that in the text book the solution contains an error, the balance mass is
incorrect.
Problem 6.7.
Using the stiffness formulae gives in Appendix 2, Table A2.4, System 6 with
48EIkb
. For this rotor,
=
I 4.3216 × 10 −8 m 4 and
a= b= L 2 , gives kuu =
3
24 EI + L kb
(
)
hence
=
EI 8.6431 × 103 Nm 2 . Hence
=
kuu 3.9834 × 10 5 N m . If the mass of the shaft
kuu m where m is the disk mass.
is neglected then the first critical speed is Ωcrit =
Evaluating, we have Ωcrit =
240.41rad s or 2296 rev/min. When an out of balance
(
)
m0 εΩ2 .
acts, mu + kuu=
u m0 εΩ2 cos Ωt Letting
=
u u0 cos Ωt , −mΩ2 + kuu u=
0
Thus the unbalance response is given by u0 =
(
m0 εΩ2
−mΩ2 + kuu
)
. At a rotor speed of
1500 rev min or 157.08 rad s , u0 = 1.62mm .
The force required to produce this deflection, f is given by kuuu0 . This force is
transmitted to the ground via the bearings, so the deflection at each bearing is f 2 .
(
)
Thus
=
ubrg f=
2 kbrg kuuu0
Thus ubrg
=
( 2kbrg ) .
3.9834 × 10 5
× 1.62mm
= 0.065mm
2 × 5 × 10 6
The stiffness of the system with short bearings, kuu , can also be derived by
recognising that the shaft stiffness ( kuu )ss is effectively in series with the stiffnesses
at the bearings. Using Equation (2.9) we have 1 k=
uu 1 ( kuu )ss + 1 kb + 1 kb . From
Appendix 2, Table A2.1, System 1, with a= b= L 2 , we have ( kuu )ss = 48EI L3 .
Using this expression in Equation (2.9) gives the same expression for kuu as above.
The same line of reasoning can be applied to a shaft with long bearings. (i.e. clampedclamped supports). From Appendix 2, Table A2.1, System 2, with a= b= L 2 , we
192 EIkb
have ( kuu )cc = 192 EI L3 . Substituting in Equation (2.9), ( kuu )cl =
.
96 EI + L3 kb
(
59
)
Evaluating this system stiffness, ( kuu )cl = 1.4233×10 6 N m . This has raised the
stiffness by a factor of 3.573 and hence the critical speed of 3.573 = 1.89 . The new
critical speed is 4340 rev/min. The MATLAB script Problem_06_07.m gives the
following output:
Critical speed = 2295.7921 rev/min
Deflection at the disk = 1.6212 mm
Deflection at the bearing = 0.064579 mm
Stiffness increased by a factor of 3.573
First critical raised to 4339.6013 rev/min
Note that the solution to this problem given in the text book contains errors.
Problem 6.8.
We will begin by plotting a Campbell diagram. This system is described by Equation
(3.84). In matrix notation, the equations of motion for the system are
0   u   kT
0 0 0
kC   u  0 
0
0
 m 0 0 0   u 


 0 m 0 0   v 



0 0 0
0   v 
kT −kC 0   v  0 
0

   + Ω 

+
  = 
0 0 0
I p   θ   0 −kC k R
 0 0 I d 0   

0
θ
 θ  0 

  
 


 
0
0
k R  ψ  0 
 0 0 0 I d  ψ
0 0 − I p 0  ψ   kC
 + ΩGq + Kq = 0 . Using the stiffness formulae gives in Appendix 2, Table
or Mq
A2.1, System 5, we have
12 EI ( a + b )
12 EI ( a + 3b )
k
=
k
=
k
=
,
and
k=
k=
ψψ
θθ
R
T
uu k=
vv
b ( 4 a + 3b )
b3 ( 4 a + 3b )
6 EI ( 2 a + 3b )
. Letting q = q0e st , then
kC =
kψ u =
−kθv =
− 2
b ( 4 a + 3b )
s 2 Mq 0 + sΩGq 0 + Kq 0 =0 . This leads to an eigenvalue problem of the form
ΩG M   q 
K 0   q 
s
  = −

   . This eigenvalue problem can be solved for a
 M 0   sq 
 0 − M   sq 
particular value of Ω to give the eigenvalues s. From s we can obtain the system
natural frequencies and these frequencies can be plotted against rotor speed to create a
Campbell diagram, see below. Examining this diagram shows immediately why there
are two backward critical speeds but only one forward critical speed. The highest
natural frequency (a forward whirl) is increasing with rotor speed (due to the
gyroscopic effect) faster than the increase in the excitation frequency with rotor
speed, so that the two can never be equal, however high the system rotation speed.
 + ΩGq + Kq = 0 , we let
We now compute the critical speeds directly. From Mq
q = q 0 e jΩt to give −Ω2 Mq 0 + jΩ2 Gq 0 + Kq 0 = 0 . This leads to the eigenvalue
problem Ω2 ( M − jG ) q 0 =
Kq 0 where Ω is a critical speed. Solving this equation
numerically gives the 4 critical speeds, 3 real and one imaginary, see below.
Alternatively, we can go back to the equations of motion and let r= u + jv and
ϕ = ψ − jθ . Adding j × the second equation to the first and subtracting j × the third
60
equation from the fourth, we have
Letting r = r0 e
jΩt
(k
etc gives
(k
mr + kT r + kC ϕ =0
 − jΩI p ϕ + kC r + k R ϕ = 0
Id ϕ
)
T
− Ω2 m r0 + kC ϕ0 = 0
R
− Ω I d + Ω I p ϕ0 + kC r0 = 0
2
2
equations we must determine the roots of
(
{ (
)
)
)
(
}
.
To solve this pair of
)
m I d − I p Ω 4 − kT I d − I p + mk R Ω2 + kT k R − kC2 =0
The roots of this equation are the forward whirling critical speeds. To determine the
backward whirling critical speeds we let r = r0 e − jΩt and this leads to
(
{ (
)
)
(
}
)
m I d + I p Ω 4 − kT I d + I p + mk R Ω2 + kT k R − kC2 =0
Both of these equation are quadratics in Ω2 and if the value of Ω2 is positive, then
Ω is a real value
The MATLAB script Problem_06_08.m computes the critical speeds and plots the
Campbell diagram. The output is shown below.
Solving eigenvalue problem
Critical speeds rev/min
683+
0j
Critical speeds rev/min
1537+
0j
Critical speeds rev/min
0+2562j
Critical speeds rev/min
3183+
0j
Forward critical speeds (rev/min)
1537
Backward critical speeds (rev/min)
3183
683
Note that one of the four critical speeds is imaginary so the solution exists
mathematically but not in reality. Note also that in the text book solutions the forward
critical speed is incorrect.
Nat frequency (Hz)
150
100
50
0
0
1000
2000
3000
4000
Rotor speed (rev/min)
61
5000
6000
Problem 6.9
On Figure 6.62 we must draw the 1x line, i.e. from the origin to the point (12000
rev/min, 200 Hz), a 3x line, i.e. from the origin to the point (12000 rev/min, 600 Hz)
and a line at a constant speed of 10000 rev/min. Where these lines cuts the natural
frequency lines we can obtain the following data (approximately).
(1) From the left hand diagram of Figure 6.62 (which we can identify as the rotor
on isotropic bearings because when the rotor is stationary all frequencies are
repeated because the frequency in the x and y directions is the same) we have
(a) 2,200 and 5,000 rev/min due to synchronous unbalance. (b) 720 and 1,200
rev/min due to the 3x force. (c) At 10,000 rev/min, 22, 38, 119, 244 and 573
Hz.
(2) From the right hand diagram of Figure 6.62 (which we can identify as the
rotor on anisotropic bearings because when the rotor is stationary the
frequencies are distinct because the frequencies in the x and y directions are
different due to the bearing stiffnesses) we have (a) 2,200, 2,600, 2,900 and
5,600 rev/min due to synchronous unbalance. (b) 740, 900, 1,000, 1,400,
5,800 rev/min due to the 3x force. (c) At 10,000 rev/min, 26, 38, 46, 124, 244,
and 573Hz.
Note that in the case of the rotor supported by isotropic bearings only the forward
critical speeds are excited. In the case of the rotor supported by anisotropic bearings
forward and backward critical speeds can be excited.
Problem 6.10.
Modeling this problem requires finite element analysis (FEA) and can only be solved
using appropriate FEA software. Here MATLAB script Problem_06_10.m makes use
the Rotordynamics Software package to model and analyze the system. The diagrams
below, generated by the script, shows mode shapes, the Campbell diagram and
response plots for the three sets of out of balance.
First four critical speeds
1577
1746
3531
4027 rev/min
Only the forward whirling critical speeds will be excited (i.e. 1,746 and 4,027
rev/min).
Undamped natural frequencies (Hz)
90
80
70
60
50
40
30
20
10
0
0
1000
2000
3000
4000
Rotor spin speed (rev/min)
62
5000
Nat Freq = 26.7733Hz
Nat Freq = 61.5787Hz
Response magnitude (m)
Mode shapes
-5
10
Node 1, x
Node 5, x
Node 7, x
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
o/b case (a)
4000
4500
Phase (degrees)
200
100
0
Node 1, x
Node 5, x
Node 7, x
-100
-200
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
4000
4500
Response magnitude (m)
Note this mass distribution only weakly excites the first critical speed (above).
-5
10
Node 1, x
Node 5, x
Node 7, x
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
o/b case (b)
4000
4500
Phase (degrees)
200
100
0
Node 1, x
Node 5, x
Node 7, x
-100
-200
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
4000
4500
Note this mass distribution only weakly excites the second critical speed (above).
63
Response magnitude (m)
-5
10
Node 1, x
Node 5, x
Node 7, x
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
o/b case (c)
4000
4500
Phase (degrees)
200
100
0
Node 1, x
Node 5, x
Node 7, x
-100
-200
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
4000
4500
Note this mass distribution excites both the first and second critical speed (above).
Problem 6.11
This problem is based on Problem 5.9 and since it requires a finite element analysis
(FEA) it can only be solved using appropriate FEA software. In order to compute the
bearing properties we require the static load acting on each bearing. From the data of
Table 5.11, we can compute the volume and hence the mass of the shaft. Similarly,
the question gives the details of the disks and the volume and hence the mass can be
determined. Thus we have msft = 101.914 kg , mdsk = 34.184 kg and adding these
together the total mass of the rotor is 136.097 kg . Thus total force on the bearings is
=
F 136.097
=
g 13.35kN . This can then be divided in ratios 1:1:1. The MATLAB
script Problem_06_11.m makes use the Rotordynamics Software package to model
the system with 28 Timoshenko elements and analyze the system.
The system is initially modeled assuming that the outer bearings are placed at the
extremities of the rotor. (This is the model used to generate Figures 6.63 and 6.64).
The MATLAB script gives the following output:
Isotropic brgs. Critical speeds (rev/min)
1015.60
1021.15
2037.86
2053.11
Anisotropic brgs. Critical speeds (rev/min)
1006.85
1019.01
1966.53
2046.06
Hydrodynamic brg, force = 1:1:1. Critical speeds (rev/min)
Initial values
600.0
700.0 1000.0 1050.0 1800.0 2210.0 2350.0 2720.0
Final values
600.4
703.0
986.8 1055.7 1825.1 2201.6 2362.7 2716.8
In the above computation, it is difficult to accurately locate the critical speeds at 2,201
and 2,716 rev/min, even when the initial values of critical speed used in the iteration
are close to these values. The MATLAB script Problem_06_11e.m models the same
64
system, except that the bearings are now moved to the same position as in Problem
5.9 and gives the following output:
Isotropic brgs. Critical speeds (rev/min)
1140.69
1148.49
2312.75
2335.34
Anisotropic brgs. Critical speeds (rev/min)
1128.63
1145.48
2218.74
2324.96
Hydrodynamic brg, force = 1:1:1. Critical speeds (rev/min)
Initial values
820.0 1020.0 1105.0 1195.0 2025.0 2755.0
Final values
819.5 1018.8 1107.7 1196.0 2023.0 2754.5
The relatively small change in the position of the outer bearings make some quite
significant change to the critical speeds, particularly in the case of the hydrodynamic
bearings.
Problem 6.12
Modeling this problem requires finite element analysis (FEA) and can only be solved
using appropriate FEA software. Here MATLAB script Problem_06_12.m makes use
the Rotordynamics Software package to model and analyze the system. The diagrams
below, generated by the script, show the Campbell diagram and response plots for the
of out of balance and the bent shaft.
First four critical speeds
989
1010
3795
4180 rev/min
Response magnitude (m)
The graphs below show the response due to out of balance and bend. In this problem
the responses at the first critical speed is similar.
-5
10
Node 6, x
Node 11, x
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
out of balance
4000
4500
Phase (degrees)
200
100
0
-100
-200
Node 6, x
Node 11, x
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
65
4000
4500
Response magnitude (m)
-2
10
-4
10
Node 6, x
Node 11, x
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
bent shaft
4000
4500
Phase (degrees)
200
100
0
-100
-200
Node 6, x
Node 11, x
0
500
1000
1500 2000 2500 3000 3500
Rotor spin speed (rev/min)
4000
4500
Problem 6.13.
We begin with the proof of the expression for the solution of pi ( t ) , starting with
Duhamel’s integral given in the question. Thus
t
t
s t −τ )
s t −τ
=
p t
f u τ e i (=
dτ
f u e −βτe i ( ) d τ .
i
( ) ∫0
i g
( )
∫0
i 0
t −βτ+ s ( t −τ )
sit t −βτ− si τ
i
Thus pi ( t ) fi u0=
e
d
f
u
e
dτ .
=
τ
0
i
∫
∫e
0
0
t
 1
fi u0 e sit  ( −β−si )t 
−β− si )τ 
(
Hence pi ( t ) fi u0 e  =
=
− 1
e

e

 −β − si
 0 − ( si + β ) 
fi u0
Finally
=
pi ( t )
esit − e−βt
( si + β )
From the equations of motion developed for Problem 3.1 we have to add the forcing
term. The forced equations of motion are (with k x = k y = k )
sit
(
)
 + cθ + L2 k θ = 0
I d 
θ + I p Ωψ
 − I p Ωθ + cψ
 + L k ψ = kLug ( t )
Id ψ
2
Or in matrix notation
 c
 I d 0   
θ
+
Ω





 
 0 I d  ψ
 − I p
. where c ∝ kL2 .
I p   θ   L2 k
 + 
c  ψ   0
0   θ   0 
   =
,
L2 k  ψ  kLug ( t ) 
0
 + ΩGq + Kq = Qug ( t ) where Q =   .
i.e. Mq
kL 
66
0 
0 
10 0 
0.25
 20 188.5
Here M = 
, K = 10 6 
, C=
, Q = 10 6   .



0.25
0.5
 0 10 
 0
188.5 20 
Writing these equations in state space form gives
ΩG M 
ΩG M  d q  K 0  q  Q 
  ug ( t ) . Let A = 
 M 0  dt q  +  0 −M  q  =
,

   
  0 
M 0
K 0 
q 
and u =   . Setting the right side of the above equations to zero,
B=

 0 −M 
q 
and letting u = u R est leads to the eigenvalue problem sAu R = −Bu R . Solving this
eigenvalue problem (with 4 degrees of freedom) provides the eigenvalues or roots si
and the eigenvectors u Ri , i = 1, , 4 and the subscript R denotes that this is the right
eigenvector. Alternatively, we have su L A = − u L B or sA T u L = −BT u L . Solving this
eigenvalue problem provides identical eigenvalues, but the left eigenvectors, u Li ,
i = 1, , 4 . (See Section 5.8). Note that B = BT but A ≠ A T . Consider the equation
Q 
for the forced system, Au + Bu =
gug ( t ) where g =   . In this example
0 
=
gT 0 0.5 × 10 6 0 0  .


Let U R = [ u R1 u R 2 u R 3 u R 4 ] , U L = [ u L1
uL2
u L3
u L 4 ] and note that U R
and U L are square matrices. In this example
 0.0002 − 0.0057 j 0.0002 + 0.0057 j − 0.0062 − 0.0004 j − 0.0062 + 0.0004 j 
 −0.0057 − 0.0002 j − 0.0057 + 0.0002 j 0.0004 − 0.0062 j
0.0004 + 0.0062 j 

UR =
 0.9534 + 0.0466 j
0.9534 − 0.0466 j 0.0728 − 0.9272 j
0.0728 + 0.9272 j 


0.0466 + 0.9534 j 0.9272 + 0.0728 j
0.9272 − 0.0728 j 
 0.0466 − 0.9534 j
0.0001 + 0.0058 j
0.0061 + 0.0006 j
0.0061 − 0.0006 j 
 0.0001 − 0.0058 j
 0.0058 + 0.0001 j
0.0058 − 0.0001 j
0.0006 − 0.0061 j
0.0006 + 0.0061 j 
UL = 
 0.9773 + 0.0227 j
0.9773 − 0.0227 j − 0.0913 + 0.9087 j − 0.0913 − 0.9087 j 


0.9087 + 0.0913 j
0.9087 − 0.0913 j 
 −0.0227 + 0.9773 j − 0.0227 − 0.9773 j
Note that the modes shapes are not unique, they can be scaled in amplitude and a
phase angle can be added to each element of the vector.
Let u ( t ) = U R p ( t ) and pre-multiply the equation of motion by U TL . Thus we have
U TL AU R p + U TL BU R p =
U TL gug ( t ) . Letting U TL AU R = A* and U TL BU R = B* . A*
0.0126 − 0.2095 j 
0.0126 + 0.2095 j 
*
*
.
and B are diagonal matrices. In this system, diag A = 
0.0399 − 0.2392 j 


0.0399 + 0.2392 j 
Furthermore B* A* = −s (where s is a diagonal matrix of the roots) so that
( )
=
p − sp
U gu ( t )
( A )=
*
−1
T
L
g
( )
fug ( t ) and hence f = A*
67
−1
U TL g . In this example,
2.9121 + 0.0493 j 
0.0595 + 1.3862 j 
2.9121 − 0.0493 j 


 and f = 10 4 0.0595 − 1.3862 j  .
UTL g = 103 
0.2870 − 3.0520 j 
1.2608 − 0.0905 j 




0.2870 + 3.0520 j 
1.2608 + 0.0905 j 
This set of differential equations in p comprise four uncoupled equations of the form
p i − si pi =
fi ug ( t ) where i = 1, , 4 and fi =
{( U g ) A } . Each equation can be
T
L
*
i
i
solved using for the function ug ( t ) using Duhamel’s integral to obtain pi ( t ) . Given
p ( t ) we can determine u ( t ) from u ( t ) = U R p ( t ) . From u ( t ) we can derive θ ( t )
and ψ ( t ) . Hence we can compute the displacements at the flexible bearing since
(
)
fi u0
esit − e−βt to
( si + β )
obtain at large number of values in the time series requires a computer.
The MATLAB script Problem_06_13.m solves this problem and gives the following
numeric and graphical output:
u ( t )= Lψ ( t ) and v ( t ) =− Lθ ( t ) . Clearly, to solve
=
pi ( t )
Response orthogonal to ground disturbance mm
Response in direction of ground disturbance mm
Root = -1.0595+/-j167.8162
Root = -0.9405+/-j148.9666
2
1.5
1
0.5
0
-0.5
0
2
1
3
1
0.5
0
-0.5
-1
Orbit during the time 2 to 3 sec
Orbit during the time 0 to 1 sec
2
0.15
1.5
displacement mm
displacement mm
3
Time, s
0.2
0.1
0.05
0
-0.05
1
0.5
0
-0.1
-0.15
-0.2
2
1
0
Time, s
-0.1
0
0.1
displacement mm
0.2
-0.5
-1
-0.5
0
0.5
displacement mm
1
Removing the damping in the system obviously removes the decay in the time series.
68
Chapter 7
Note. Solving Problems 7.10 and 7.11 requires a finite element analysis that allows
shafts to be modeled and include gyroscopic effects, etc. Here we use the
rotordynamics software developed to accompany this book, but other appropriate
software can be used.
Problem 7.1
For a Jeffcott rotor, ignoring the mass of the shaft, the only mass is the disc and the
stiffness is the stiffness of the rotor at mid-span. Thus in the x and y-directions,
mu + kuu u =
0 and mv + kvv v =
0 . In rotation about the x and y-directions,
 − I Ωθ + k ψ = 0 . These equations are based on
I 
θ + I Ωψ + k θ = 0 and I ψ
d
θθ
p
d
ψψ
p
those given in Equation (3.83). Letting, k=
k R . Thus in
=
k=
θθ k=
ψψ
uu
vv kT and k
 + ΩGq + Kq = 0 where in this case
coordinates fixed in space we have Mq
m 0 0
0 m 0
M=
 0 0 Id

 0 0 0
0
0
0
0 
, G=
0

0


I d 
0
0
0
0
0
 kT

0
0
, K=
Ip
0


0 
 0
0
0
0
0 −I p
0
kT
0
u 

v 
0
 
, q =  .

0
θ

ψ 
k R 
0
0
kR
0
0
0
Here,
=
kT 48 EI
=
L3 1.1685 ×106 N m and
=
k R 12=
EI L 1.4314 ×105 Nm
(
)
 = s 2q0e st ) gives Ms 2 + ΩGs + K q 0 =
Letting q = q0e st , (and q = sq 0e st , q
0 . We
can either solve the resulting eigenvalue problem with 4 degrees of freedom or, obtain
 I d s 2 + kR
ΩI p s   θ0  st 0 
2
2
st
st
  e =   .
ms + kT u0i e =0, ms + kT v0i e =0 , 
2
ψ
 −ΩI p s

0 
I
s
k
+
d
R 0

From the first pair of equations, s= −kT m (twice) and from the second set of
equations
(
)
(
I d s 2 + kR
ΩI p s
−ΩI p s
I d s 2 + kR
)
(
= 0 . Thus I d s 2 + k R
) + ( ΩI p s )2 =0 . Rearranging this
2
equation gives I d s 2 + k R =± jΩI p s or I d s 2  jΩI p s + k R =
0 . We can solve this
quadratic equation to find the roots. The MATLAB script Problem_07_01.m carries
out these calculations. It also solves the eigenvalue for the complete system.
Converting the equations of motion from the fixed coordinates to coordinates rotating
with the shaft using the transformation given in Equations (7.5) – (7.10) we have
 q + Ω 2 M
 +K
 =G, K
  + Ω M
 +G
 +G
 q = 0 where M
 =M, G
 =K,
Mq
(
1
) { (
2
 0 −2m 0

0
0
 =  2m
 = −M , M
M
1
2
 0
0
0

0
2Id
 0
1
)
}
0
0 
0
0 

and G1 = 
0
−2 I d 


0 
0
69
0 0
0 0
0 Ip
0
0





I p 
0
0
0
 = s 2q e st we can either solve the
Letting q = q 0e st , we have q = sq 0e st and q
0
resulting eigenvalue problem with 4 degrees of freedom or, obtain a pair of 2 degree
eigenvalue problems thus:
ms 2 − mΩ 2 + kT
−2mΩs
2mΩs
ms 2 − mΩ 2 + kT
(
)
(
= 0 and
)
I d s 2 + Ω2 I p − I d + kR
− 2 I d − I p Ωs
( 2I d − I p ) Ωs
I d s 2 + Ω2 I p − I d + kR
(
Thus we have
)
(
=0
)
(
)
ms 2 ± j 2mΩs + kT − mΩ 2 =0 and I d s 2 ± j 2 I d − I p Ωs + k R + Ω 2 I p − I d = 0 . We
can solve these quadratic equations to find the roots. The MATLAB script
Problem_07_01.m gives the user the choice of solving either the characteristic
equations or the eigenvalue problem to determine the system natural frequencies. Of
course, both methods give the same numeric values for the frequencies which are as
follows
Solution of characteristic equations
In fixed coordinates at 3000 rev/min
Natural frequency 1 = 16.4507 Hz
Natural frequency 2 = 31.4102 Hz
Natural frequency 3 = 31.4102 Hz
Natural frequency 4 = 110.2007 Hz
In rotating coordinates at 3000 rev/min
Natural frequency 1 = 18.5898 Hz
Natural frequency 2 = 60.2007 Hz
Natural frequency 3 = 66.4507 Hz
Natural frequency 4 = 81.4102 Hz
Problem 7.2
) { (
(
}
)
 q + Ω 2 M
 +K
  + Ω M
 +G
 +G
 q = 0
In rotating coordinates we have Mq
1
2
1
where from Equations (7.80), (7.83) and (7.84)
0
 0 −2m
0 
m 0 0

0 m 0
0
0
0 
 2m



, M1 =  0
M=
0
0
 0 0 I dx 0 



 0
 0 0 0 I dy 
I dx + I dy
0

(
0
0
0 
 −m 0
0
 0 −m
0
0 



, From (7.85) G1 = 
M2 =
0
 0
0 − I dy
0 



− I dx 
0
0
0
 0
0
0 0 0
0
 kT
0 0 0

0
0
kT
 =
= TT KT
.K
= K= 
G
0 0 0
Ip
 0 −kC



0
 kC
0 0 − I p 0 
70
0
0
(
− I dx + I dy
)
0
0 0
0 0
0 Ip
0
0
−kC
kR
0
0
)



 and





 and


I p 
0
0
0
kC 
0 
0

k R 
For a simply supported shaft, from Appendix 2, Table A2.1, system 1 we have
k=
k=
T
uu k=
vv
k=
kψψ
= k=
R
θθ
(
3EI a3 + b3
3 3
), k
C
a b
3EI ( a + b )
=
kψ u =
−kθv
(
)
3EI a2 − b2
and
= 2 2
a b
, where a = 0.2m and b = 0.5m . To solve the equations
ab
of motion we must rearrange them as a set of state space equation thus

 +K
 +G
  d q  Ω 2 M
 +G

Ω M
M
0  q 
1
2
1
 

  = 

  q 
M
0  dt q  
0
−M

(
)
(
)
Letting q = q 0e st , (and q = sq 0e st etc.) leads to an 8 × 8 eigenvalue problem in s. The
MATLAB script Problem_07_02.m formulates and solves this eigenvalue problem
to determine the eigenvalues. The script also checks that the real part of the
eigenvalues are positive implying an unstable system. The output of the script is as
follows. Note that the system is unstable under certain conditions.
Roots at a rotor speed of 2400 rev/min
Real part
Imag part
Real part
-0.0000
14.4102
-0.0000
-0.0000
365.3374
-0.0000
0.0000
484.6082
0.0000
0.0000
662.7558
0.0000
Roots at a rotor speed of 2600 rev/min
Real part
Imag part
Real part
-7.5741
0.0000
7.5741
-0.0000
380.7460
-0.0000
0.0000
492.6906
0.0000
-0.0000
678.9032
-0.0000
Roots at a rotor speed of 2800 rev/min
Real part
Imag part
Real part
0.0000
17.7441
0.0000
-0.0000
396.4589
-0.0000
0.0000
501.3517
0.0000
0.0000
695.5383
0.0000
Problem 7.3
(
Imag part
-14.4102
-365.3374
-484.6082
-662.7558
Unstable = 1
0
0
0
0
Imag part
0.0000
-380.7460
-492.6906
-678.9032
Unstable = 1
1
0
0
0
Imag part
-17.7441
-396.4589
-501.3517
-695.5383
Unstable = 1
0
0
0
0
) { (
)
}
 q + Ω 2 M
 +K
  + Ω M
 +G
 +G
 q = 0 where from
In rotating coordinates, Mq
1
2
1
Equations (7.80), (7.83) and (7.84)
m 0 0 0 
 0 −2m 0
0 m 0 0 
 2m
0
0




M=
, M1 = 
 0 0 Id 0 
 0
0
0



0
2Id
 0 0 0 I d 
 0
0
 −m 0
 0 −m 0
 =
M
2
 0
0 −Id

0
0
 0
0 
0 
and
−2 I d 

0 
0
0 
0
0 

. From (7.85) G1 = 
0
0 


− I d 
0
71
0 0
0 0
0 Ip
0
0


 and


I p 
0
0
0
0
0 0 0
0
0 
 kuu 0
0 0 0


0
0 kvv 0
0 
T



= K=
and K= T KT
G=
0 0 0
Ip
 0
0 kθθ
0 




0
0 kψψ 
 0
0 0 − I p 0 
For a simply supported shaft, from Appendix 2, Table A2.1, system 1 we have
48EI yy
12 EI yy
48EI xx
12 EI xx
,
,
and
where, for an
kuu =
k
=
k
=
=
k
vv
θθ
ψψ
L
L
L3
L3
elliptical cross section I xx = πd y d x3 64 etc. Since the first pair of equations is
uncoupled from the second pair we can solve them separately. Thus, letting u = u0 e st
 ms 2 + kuu − mΩ2
 u  st 0 
−2 mΩs
etc., we have 
  e =   .

0 
2 mΩs
ms 2 + kvv − mΩ2   v 
Setting the determinant of the coefficients to zero gives
{
} (
)(
)
m 2 s 4 + m ( kuu + kvv ) + 2 m 2 Ω2 s 2 + kuu − mΩ2 kvv − mΩ2 =
0 . This is a quadratic
in s 2 and can readily be solved. From the second pair of equations and letting
ψ = ψ 0 est etc, we have
(
)
(
)
 I d s 2 + kθθ − I d − I p Ω2

− 2 I d − I p Ωs

 u  est = 0  .

0 
2 I d − I p Ωs
I d s 2 + kψψ − I d − I p Ω2   v 


Setting the determinant of the coefficients to zero gives
(
{ (
)
(
) (
)
) }
) Ω )(k
I d2 s 4 + I d kθθ + kψψ + 2 I d2 − 2 I d I p + I 2p Ω2 s 2 +
(k − (I
θθ
d
− Ip
2
ψψ
(
) )
− I d − I p Ω2 =
0
This is a quadratic in s 2 and can readily be solved. Alternatively the 4 equations of
motion can be solved by forming the state equations and solving the 8 × 8 eigenvalue
problem in s (see Problem 7.2). The MATLAB script Problem_07_03.m gives the
user the choice of solving either the characteristic equations or the eigenvalue
problem to determine the system natural frequencies. The script also checks that the
real part of the roots or eigenvalues are positive implying an unstable system. The
output of the script is as follows. Note that the system is unstable under certain
conditions.
Solution of the characteristic equations
Roots at a rotor speed of 1900 rev/min
Real part
Imag part
Real part
0.0000
9.4646
0.0000
0.0000
326.4443
0.0000
0.0000
355.7124
0.0000
0.0000
409.7592
0.0000
Roots at a rotor speed of 2000 rev/min
Real part
Imag part
Real part
6.6870
0.0000
-6.6870
0.0000
331.4147
0.0000
0.0000
361.6804
0.0000
0.0000
420.2268
0.0000
72
Imag part
-9.4646
-326.4443
-355.7124
-409.7592
Unstable = 1
0
0
0
0
Imag part
0.0000
-331.4147
-361.6804
-420.2268
Unstable = 1
1
0
0
0
Roots at a rotor speed of 2100 rev/min
Real part
Imag part
Real part
0.0000
6.3311
0.0000
0.0000
336.5616
0.0000
0.0000
367.8495
0.0000
0.0000
430.6949
0.0000
Imag part
-6.3311
-336.5616
-367.8495
-430.6949
Unstable = 1
0
0
0
0
Problem 7.4
In rotating coordinates, the internal shaft viscous damping is given by
0
0
ciT 0
0 c
0
0 
iT

=



f d Cq
=
. Now in rotating coordinates,
0
0 ciR 0 


0
0 ciR 
 0
(
)
 TC

=
 TT q + T
 TT C
 and
 T q . Thus
=
f d Tf
=
TC
=
Ci TC
=
d
iq
i
i
i
 0
 −c
T



 iT
=
=
Ci1 TC
iT
 0

 0
ciT
0
0
0
0
0 
. Thus, in fixed coordinates, the equations of
ciR 

0 
0
0
0
−ciR
 + ( Ce + Ci + ΩG ) q + ( ΩCi1 + K ) q = 0 where
motion are Mq
m 0 0
0 m 0
M=
 0 0 Id

 0 0 0
ce
0
Ce = 
0

0
0
0
0

0
, G=
0
0


I d 
0
0
0
0
0
0
0
0 −I p
0
 kT

0
0
, K=
Ip
0


0 
 0
0
kT
0
0
0
0
kR
0
0
0 
and
0

k R 
0
0 
0
0

0
0
 + ( Ce + Ci ) q + Kq =
When the system is at rest, Mq
0 . Considering the first pair of
equations we have
mu + ( ce + ciT ) u + kT u =
0
. Hence, for these separate single degree of freedom
mv + ( ce + ciT ) v + kT v =
0
0
ce
0
0
0
0
equations, we have ζ =
ce + ciT
2 mkT
(twice). Considering the second pair of equations,
I d 
θ + ciR θ + k R θ = 0
. Hence, , for these separate single degree of freedom equations,
 + ciR ψ + k R ψ =0
Id ψ
we have ζ =
ciR
2 I d kR
(twice). Consider now the set of system equation. Letting
q = q0e st , we have Ms 2 + ( Ce + Ci + ΩG ) s + ( ΩCi1 + K )  q0e st = 0 .Because the


first pair and the second pair of equations are uncouple from each other we can
proceed as follows:
73
ms 2 + ( ce + ciT ) s + kT
ciT Ω
−ciT Ω
ms 2 + ( ce + ciT ) s + kT
{ms2 + ( ce + ciT ) s + kT }
2
Similarly
(
)
− ciR + I p Ω
+ ciT 2Ω 2 =
0 . Hence ms 2 + ( ce + ciT ) s + kT ± jciT Ω =0 .
( ciR + I p ) Ω
I d s 2 + ciR s + k R
= 0 and thus
I d s 2 + ciR s + k R
= 0 and thus
( Id s2 + ciR s + kR ) + ( ciR + I p )2 Ω2 =0 . Hence Id s2 + ciR s + kR ± j ( ciR + I p ) Ω =0 .
2
We can solve these two quadratic equations to obtain the 4 system roots.
Alternatively we can the 4 equations together, (see Problem 7.2). The MATLAB
script Problem_07_04.m gives the user the choice of solving either the characteristic
equations or the eigenvalue problem to determine the system natural frequencies. The
script also checks that the real part of the roots or eigenvalues are positive implying
an unstable system. The output of the script is as follows. Note that the system is
unstable under certain conditions.
zeta = 0.005067 (twice) and 0.0074759 (twice)
Solution of the characteristic equations
Roots at a rotor speed of 2200 rev/min
Real part
Imag part
Real part
Imag part
-2.0838
127.8392
-2.0838
-127.8392
-0.0272
197.3557
-0.0272
-197.3557
-1.9728
197.3557
-1.9728
-197.3557
-1.9162
559.8082
-1.9162
-559.8082
Roots at a rotor speed of 2300 rev/min
Real part
Imag part
Real part
Imag part
-2.0860
124.2722
-2.0860
-124.2722
0.0170
197.3560
0.0170
-197.3560
-2.0170
197.3560
-2.0170
-197.3560
-1.9140
575.8762
-1.9140
-575.8762
Problem 7.5
{
Unstable = 1
0
0
0
0
Unstable = 1
0
1
0
0
}
 . An out of
  + Ω ( M + G ) q + Ω 2 ( M + G ) + K q =
In rotating coordinates Mq
Q
1
2
1
balance force rotates with the shaft so that in rotating coordinates the out of balance
 is constant, there will be no velocities or
force is in a fixed direction. Thus, since Q
.
Q
accelerations in the rotating coordinate system, so that Ω 2 ( M + G ) + K q =
{
2
1
}
Separating the translation and rotation coordinate gives
 
 2  m 0  0 0   k T 0   q T  Q
T
+
=
 + 
Ω  








0 I d  0 I p    0 k R   q R   0 
  

Since there is no excitation or coupling to the rotation coordinates, we have
 kTx − mΩ 2
 u  Q 
0

 =  x
0
kTy − mΩ 2  v  Q y 

74
If the out of balance force is in the x-direction,
 kTx − mΩ 2
 u 
0
m0εΩ 2
2 1 


 =
and
hence
. Similarly
u
=
εΩ
m

 
0
kTx − mΩ 2

0
kTy − mΩ 2  v 
0 
when the out of balance force is in the y direction v =
m0εΩ 2
kTy − mΩ 2
. Finally, when the
out of balance force is at 45° to the x direction is
 kTx − mΩ 2
 u 


0
m0εΩ 2
2 1 2 


 =
and
 m0εΩ 
 , Hence u =

0
kTy − mΩ 2   v 
2 kTx − mΩ 2
1 2 
(
v =
(
m0εΩ 2
2 kTy − mΩ
r =
u 2 + v 2 =
)
2
)
. Thus the response is
m0 εΩ 2
1
kTx − mΩ
2
2
+
1
kTy − mΩ 2
The MATLAB script Problem_07_05.m gives the following output:
Rotor speed = 1900 rev/min
Response due to o/b in direction of major axis = 85.8894 mu_m
Response due to o/b in direction of minor axis = 336.9978 mu_m
Response due to o/b at 45 degree to major axis = 245.9111 mu_m
Problem 7.6
kuu =
48EI xx
L3
, kvv =
48EI yy
L3
the gravity critical speed.
ω2x ω2y
. From Equation (7.42), Ω =
where Ω is
2 ω2x + ω2y
2
(
)
(
(
)(
)
)
From Equation (7.22), s 4 + ω2x + ω2y + 2Ω 2 s 2 + ω2x − Ω 2 ω2y − Ω 2 = 0
Let s = jΩ , then s 2 = −Ω 2 and s 4 = Ω 4 . Substituting in Equation (7.22) gives
(
)
(
)(
)
Ω 4 − ω2x + ω2y + 2Ω 2 Ω 2 + ω2x − Ω 2 ω2y − Ω 2 = 0
−2
(
ω2x
+ ω2y
)Ω + (
2
ω2x ω2y
) =0 and thus Ω
2
ω2x ω2y
. Now, for this rotor,
=
2 ω2x + ω2y
(
)
ω2y kTy=
m 4.1546 × 10 4 . Thus
=
ω2x kTx=
m 4.7270 × 10 4 and=
ω2x ω2y
4.7270 × 4.1546 ×108
=
Ω
=
=
1.1056 ×104 . Hence
4
2
2
2 ( 4.7270 + 4.1546 )10
2 ωx + ω y
2
(
)
=
Ω
1.1056 × 10 4 × 60=
( 2π ) 1004.1rev min
The MATLAB script Problem_07_06.m calculates the gravity critical thus:
Gravity critical = 1004.0798 rev/min
75
Problem 7.7
For a Jeffcott rotor, ignoring the mass of the shaft, the only mass is the disc and the
system stiffness is the stiffness of the shaft at mid-span. Let k=
k=
uu
vv kT and
k=
kR .
θθ k=
ψψ
 + ΩGq + Kq = 0
Fixed Coordinates: In coordinates fixed in space we have Mq
where, in this case,
0
0 0 0
0
m 0 0 0 
 kT 0 0
u 
0 0 0

0 m 0 0 


v 
0
0 kT 0
0
 
, G = 
M=
, K=
, q =  .
0 0 0
Ip
 0 0 Id 0 
 0 0 kR 0 
θ






ψ 
 0 0 0 k R 
 0 0 0 I d 
0 0 − I p 0 
(
)
 = s 2q0e st ) gives Ms 2 + ΩGs + K q 0 =
Letting q = q0e st , (and q = sq 0e st , q
0.
Noting that the first two equations are uncoupled from each other and from the third
and fourth equations, the first two equations give
( ms2 + kT ) u0est =0, ( ms2 + kT ) v0est =0 and hence s=
1
s=
j kT m ,
2
s5 = s6 = − j kT m . Since s =± jωn then ωn1 =
ωn 2 = kT m . The normal
convention is to order the natural frequencies using the amplitude; here we have
assumed that the lower natural frequencies arise from the translational modes, which
is often the case for realistic systems but is not guaranteed.
From the third and fourth equations,
 I d s 2 + kR
I d s 2 + kR
ΩI p s
ΩI p s   θ0  st 0 

   e =   . Hence
= 0 . Thus
2
2
ψ0 
0
 −ΩI p s



−Ω
I
s
I
s
+
k
I
s
k
+
p
d
R
d
R

( Id s2 + kR ) + ( ΩI p s )2 =0 . Rearranging, Id s2 ± jΩI p s + kR =0 . Dividing by I
2
we
d
have s 2 ± jΩαs + ω02 = 0 where α =I p I d and ω02 =k R I d . Solving the quadratic
equations gives s =
(
( − jΩα ±
−Ω 2α 2 − 4ω02
) 2 = ± j 
) 2 and s = ( jΩα ±
−Ω 2α 2 − 4ω02
)2

 . Since the natural

2

frequency is defined to be positive we have, ωn3 =  − 12 Ωα + ω02 + 12 Ωα  and


2
1
2
1
ωn=
4  2 Ωα + ω0 + 2 Ωα  , where ωn3 ≤ ωn 4 . The eigenvalues may be written


as s3 = jωn3 , s4 = jωn 4 , s7 =− jωn3 , s8 =− jωn 4 .
or s = ± jΩα ± −Ω 2α 2 − 4ω02
1 Ωα ±
2
ω02 +
( 12 Ωα )
2
(
(
)
)
Rotating Coordinates: Converting the equations of motion from the fixed
coordinates to coordinates rotating with the shaft using the transformation given in
Equations (7.5) – (7.10) we have
 q + Ω 2 M
 +K
 =G, K
  + Ω M
 +G
 +G
 q = 0 where M
 =M, G
 =K,
Mq
(
1
) { (
2
1
)
}
76
 0 −2m 0
 2m
0
0


M 2 = −M , M1 = 
 0
0
0

0
2Id
 0
0
0 
0

0 

and G1 = 
0
−2 I d 


0 
0
0 0
0 0
0 Ip
0
0


.


I p 
0
0
0
Letting q = q 0e st etc. we obtain four equations. (Note we are writing the root as s to
distinguish it from the roots in the fixed coordinates, s). The first pair of equations are
uncoupled from the second pair of equations and each pair can be solve separately
from the other pair. Thus, from the first pair of equations,
ms 2 − mΩ 2 + kT
−2mΩs
2mΩs
ms − mΩ + kT
2
2
= 0 , and hence ms 2 ± j 2mΩs + kT − mΩ 2 =0 .
(
)
Dividing by m gives s 2 ± j 2Ωs + ω2n1 − Ω 2 = 0 since ω2n1 =
kT m . Solving these
two quadratic equations, and simplifying we have s = ± j ( Ω ± ωn1 ) = s ± jΩ . Now
 n1 = ωn1 − Ω and ω
 n 2 = ωn1 + Ω = ωn 2 + Ω . Note that taking the
 n and so ω
s =± jω
absolute value ensures that the natural frequencies are positive. Furthermore the
ordering is arbitrary as the relative amplitudes of the natural frequencies will vary
with rotor spin speed. From the second pair of equations, we have
( I p − 2I d ) Ωs
I d s 2 − I d Ω 2 + I p Ω 2 + k R
(
− I p − 2Id
(
)
Ωs
I d s 2 − I d Ω 2 + I p Ω 2 + k R
)
(
= 0 . Hence,
)
I d s 2 ± j I p − 2 I d Ωs + k R − Ω 2 I d − I p = 0 . Dividing by I d and noting that
α =I p I d and ω02 =k R I d we have s 2 ± j ( α − 2 ) Ωs + Ω 2 ( α − 1) + ω02 = 0 . Solving
2

these two quadratic equations gives s = ± j  12 α − 1 Ω ± ω02 + 12 Ωα  = s ± jΩ


from the definition of s above. Also, from the definition of ωn3 and ωn 4 above we
(
 n3 = −
have ω
 n4 =
ω
( 12 α − 1) Ω +
( 12 α − 1) Ω +
ω02 +
ω02 +
( 12 Ωα )
( 12 Ωα )
2
2
)
(
)
= ωn3 + Ω and
= ωn 4 − Ω . Since ωn3 ≤ ωn 4 we know that ωn3
corresponds to a backward whirling mode (see Section 3.6.1), and we have to add Ω
to obtain the natural frequency in the rotating frame. Conversely ωn 4 is a forward
whirling mode and we have to subtract Ω to obtain the natural frequency in the
rotating frame. This is consistent with the discussion at the end of Section 7.3.1,
although note that this discussion considers the transformation from the rotating to the
stationary coordinates, whereas here we consider the transformation from the
stationary to rotating coordinates. In Problem 7.1 the natural frequencies equal to
16.4507, 31.4102 (twice) and 110.2007 Hz. Applying these results to these natural
frequencies in fixed coordinate computed to convert to rotating coordinates gives
18.5898Hz ,
31.4102 + 50 =
81.4102Hz , 31.4102 − 50 =
110.2007 − 50 =
60.2007Hz and 16.4507 + 50 =
66.4507Hz (as in Problem 7.1).
77
Problem 7.8
In rotating coordinate the equations of motion are of the form
 q + Ω 2 M
 +K
  + Ω M
 +G
 +G
 q = 0 where for this system
Mq
(
) { (
1
 2
 =  kL
K
 0
2
1
 0
0 

M
=
1 
I dy 
 I dx + I dy
0    I dx
 M=
kL2 
 0
 0
0 
 =
, G
− I dx 
 − I p
− I
 =  dy
M
2
 0
}
)
Ip

0 
I
 = p
G
1
 0
(
)
− I dx + I dy 


0
0
θ
 and q =  
I p 
ψ 
For this system, L = 0.5 m , k = 1× 106 N m , I dx = 10.6 kg m 2 , I dy = 10.2 kg m 2 and
I p = 0.8 kg m 2 and hence
(
)

0
I p − I dx + I dy   0 −20 
 =

,
 I +I −I
  20 0 
0
p
 dx dy

0   −9.4
 I p − I dy
0 

 +G
,
=
M
=

 
2
1
I p − I dx   0
−9.8
 0

 +G
M
1
(
)
 kL2 0  0.25 ×106

0
 =
=
K
 
 . Letting q = q 0 e st we have
2
6
 0 kL  
0
0.25 ×10 
 s + Ω2 M
 +K
 s2 + Ω M
 +G
 +G
  q =
M
0 . The system roots can be
1
2
1

 0
determined from the determinant of the coefficient matrix thus
 s + Ω2 M
 +K
 s2 + Ω M
 +G
 +G
 =
M
0
) { (
(
(
) { (
1
2
1
)
}
)
}
This simplifies to
10.6 s 2 − 9.4Ω 2 + 0.25 ×106
−20Ωs
20Ωs
10.2 s 2 − 9.8Ω 2 + 0.25 × 106
=0
This leads to a quadratic in s 2 . Alternatively we can solve an eigenvalue problem.
The gravity critical speed can be derived following the approach of Equations (7.37)
and (7.38). Thus we have
 q + Ω 2 M
 +K
  + Ω M
 +G
 +G
 q = −mg  sin Ωt  . Thus
Mq
1
2
1
cos Ωt 
(
) { (
)
(
)
I p − I dx + I dy 
 q +

0

2
2
 kL + Ω I p − I dy

0
sin Ωt 

 q = −mg 


cos Ωt 
0
kL2 + Ω 2 I p − I dx 


Let q1 = θ = θ 0 sin Ωt and q2 = ψ = ψ 0 cos Ωt . Then
 I dx
 0


0
0 
 + Ω 
q
I dy 
 I +I −I
p
 dx dy
}
(
)
(
)
(
78
)
(
)
(
)
 kL2 + Ω 2 I p − I dy − I dx
−Ω 2 I p − I dx − I dy   θ 
1

  0  = −mg  
 −Ω 2 I − I − I
1
kL2 + Ω 2 I p − I dx − I dy  ψ 0 
p
dx
dy


The maximum response occurs when the determinant of the coefficient matrix is zero.
(
{
)
(
Now D = kL2 + Ω 2 I p − I dy − I dx
(
)
(
(
)}
2
(
)
− Ω 4 I p − I dx − I dy
)
2
and hence, if D = 0 ,
)
kL2 + Ω 2 I p − I dy − I dx =
±Ω 2 I p − I dx − I dy . This leads to gravity critical speed is
kL2
given by Ω gr =
.
2 I dx + I dy − I p
(
)
Out of balance

 q + Ω 2 M
 +K
  + Ω M
 +G
 +G
 q = m r Ω 2 L 1
Mq
1
2
1
0
2 1
) { (
(
(
)
)
}
2 
 . Thus
2 
 kL2 + Ω 2 I p − I dy

0
1 2 

 q = m0 r Ω 2 L 
 . Hence we can

2 1 2 
0
kL2 + Ω 2 I p − I dx 


determine the response q.
The MATLAB script Problem_07_08.m carries out these calculations and gives the
following output
(
)
Solution of the characteristic equations
Roots at a rotor speed of 1500 rev/min
Real part
Imag part
Real part
0.0000
3.8210
0.0000
0.0000
306.2299
0.0000
Roots at a rotor speed of 1540 rev/min
Real part
Imag part
Real part
-1.6091
0.0000
1.6091
0.0000
310.2638
0.0000
Imag part
-3.8210
-306.2299
Unstable = 1
0
0
Imag part
0.0000
-310.2638
Unstable = 1
1
0
Gravity critical = 754.9382 rev/min
At 1500 rev/min, responses are 0.96584 mm and 2.1291 mm
Orbit radius = 2.3379 mm
Problem 7.9
In this case the force applied to the support in the x and y directions is
f x= ku + kc v= kLψ − kc Lθ and f y = −kcu + kv = −kc Lψ − kLθ . The moment acting
on the rotor in the θ direction is −kc L2ψ − kL2θ . Similarly the moment acting on the
rotor in the ψ direction is −kL2ψ + kc L2θ and, in fixed coordinates the equations of
motion become
θ + I p Ωψ + L2 k θ + L2 kc ψ = 0
I d 
or, in matrix notation,
 − I p Ωθ − L2 kc θ + L2 k ψ = 0
Id ψ
 Id
0

 0
0   
θ
 +Ω

I d  ψ
 
 − I p
I p   θ   kL2
 + 
0  ψ   − kc L2

79
kc L2   θ  0 
   = 
2

kL  ψ  0 
Seeking solutions of the form θ = θ0e st and ψ = ψ 0e st , gives the following equation
for s
 s 2 I d + L2 k
sΩI p + L2 kc 
 = 0 or
det 
 − sΩI p − L2 kc s 2 I d + L2 k 


( s2 Id + L2k ) + ( sΩI p + L2kc )
2
2
=
0 and hence
I d s 2 ± jI p Ωs + L2 ( k ± jkc ) =0 ..
Alternatively, we can solve the eigenvalue problem as described by equations (3.48),
(3.50), (3.51) and (3.52). The MATLAB script Problem_07_09.m gives the user the
choice of solving either the characteristic equation or the eigenvalue problem to
determine the system natural frequencies. Of course, both methods give the same
numeric values for the frequencies which are as follows
Solution of the characteristic equations
Roots at a rotor speed of 0 rev/min
Real part
Imag part
Real part
Imag part Unstable = 1
-15.7337
158.8948
-15.7337
-158.8948
0
15.7337
158.8948
15.7337
-158.8948
1
Natural frequencies at a rotor speed of 0 rev/min
25.2889 Hz
25.2889 Hz
Roots at a rotor speed of 3000 rev/min
Real part
Imag part
Real part
Imag part Unstable = 1
-15.7063
149.7466
-15.7063
-149.7466
0
15.7063
168.5961
15.7063
-168.5961
1
Natural frequencies at a rotor speed of 3000 rev/min
23.8329 Hz
26.8329 Hz
Problem 7.10
Modelling this system requires finite element analysis (FEA) and can only be solved
using appropriate FEA software. The rotor is the same as that of Problem 5.2 except
that now the rotor has internal damping. This can lead to instability at high speeds.
The MATLAB script Problem_07_10.m makes use of the Rotordynamics Software
package to model and analyse the system. The model has 8 elements and the stability
is determining by checking that all the eigenvalues have negative real parts. This is
done every 0.5 rev/min, until any one eigenvalue has a positive real part.. The output
of the script is as follows:
Rotor is unstable above 4521.5 rev/min
80
Node 9
Node 8
Node 7
Node 6
Node 5
Brg Type 3
Node 4
Node 3
Node 2
Node 1
Brg Type 3
Real(eigenvalues)
0.1
0
-0.1
-0.2
-0.3
-0.4
-0.5
0
1000
2000
3000
Rotor Speed (rev/min)
4000
5000
Problem 7.11
This is the same system as in Problem 5.2 and 7.10 except that now the rotor is
asymmetric. There shaft stiffness in one direction is 10% higher than in the other
direction. Modelling this system requires finite element analysis (FEA) and must be
solved using appropriate FEA software. Here the MATLAB script Problem_07_11.m
uses the Rotordynamics Software package to model and analyse the system in
rotating coordinates and plots the real part of the eigenvalues. When the real part of
the eigenvalues are positive the system is unstable. It can be seen there are three
speed ranges where that is the case. The MATLAB script gives the following output:
Unstable
Unstable
Unstable
Unstable
Unstable
Unstable
above
below
above
below
above
below
1746.5 rev/min
1831.5 rev/min
2944.5 rev/min
2949.5 rev/min
4030 rev/min
4226.5 rev/min
At 3000 rev/min
Natural freq 1 = 17.4679
Natural freq 2 = 19.1392
Natural freq 3 = 75.8073
Pesudo-nat freq from nat
Pesudo-nat freq from nat
Pesudo-nat freq from nat
Pesudo-nat freq from nat
Pesudo-nat freq from nat
Pesudo-nat freq from nat
Hz
Hz
Hz
freq
freq
freq
freq
freq
freq
1
1
2
2
3
3
=
=
=
=
=
=
32.5321 Hz
67.4679 Hz
30.8608 Hz
69.1392 Hz
25.8073 Hz
125.8073 Hz
81
10
Real(eigenvalues)
5
0
-5
-10
0
1000
2000
3000
Rotor Speed (rev/min)
82
4000
5000
Chapter 8
Problem 8.1
From Equation (8.1) we have
4
fub =
Ω 2 ∑ mi εi
i =1
(
=
Ω 2 0.4 × 0.2e j 0 + 0.2 × 0.2e j 76 π 180 + 0.7 × 0.1e j132 π 180 + 0.2 × 0.3e j 212 π 180
)
=Ω 2 {( 0.080 ) + ( 0.0097 + 0.0388 j ) + ( 0.0468 + 0.0520 j ) + ( −0.0509 + 0.0318 j )}
=
Ω 2 {−0.0080 + 0.0590 j} kg m
4
M ub =
Ω ∑ mi εi zi
2
i =1
2
=Ω
{( 0.080 ) × 0 + ( 0.0097 + 0.0388 j ) × 0.5 +
( 0.0468 + 0.0520 j ) ×1.0 + ( −0.0509 + 0.0318 j ) ×1.5}
=
Ω 2 {−0.1183 + 0.0237 j} kg m 2
Since for balance ( bB + bD + fub ) Ω 2 =
0 and ( bB z B + bD z D + M ub ) Ω 2 =
0.
Cancelling the Ω 2 terms from the above equations, then from Equation (8.4) we have
Ab = v where
1 1
1
 fub  0.0080 − 0.0590 j 
1
=
=
A =
and v =
 
.



 z B z D  0.5 1.5
 M ub  0.1183 − 0.0237 j 
0.1245
−0.1063 − 0.0648 j 
−1
Thus
=
b A=
v 
 kg m . Hence b = 
 kg m and
0.1144 
 0.1143 + 0.0058 j 
−146.6149 
θ = ∠b ×180 π = 
 degree . Thus the unbalance correction masses are
 2.8974 
0.1245 1 1.2447 
m
= b =
ε 
 = 
 kg
0.1144  0.1 1.1445 
The MATLAB script Problem_08_01.m repeats these calculations to give.
Required balance mass at B = 1.2447 kg at -148.6149 degree
Required balance mass at D = 1.1445 kg at 2.8974 degree
Problem 8.2
From Equation (8.1) we have
3
fub =
Ω 2 ∑ mi εi
i =1
(
=
Ω 2 10 × 0.15e j120 π 180 + 50 × 0.10e j15π 180 + 20 × 0.20e− j 45π 180
)
=
Ω 2 {( −0.7500 + 1.2990 j ) + ( 4.8296 + 1.2941 j ) + ( 2.6264 − 2.6264 j )}
=
Ω 2 {6.9081 + 0.2353 j} kg mm=Ω 210−3 {6.9081 + 0.2353 j} kg m
83
3
M ub =
Ω 2 ∑ mi εi zi
i =1
2
=
Ω 10−3 {( −0.7500 + 2.2990 j ) × 0.4 +
( 4.8296 + 1.2941 j ) × 0.8 + ( 2.6264 − 2.6264 j ) ×1.6}
=
Ω 210−3 {8.0892 − 2.9706 j} kg m 2
Part (a) Since for balance ( bb1 + bb 2 + fub ) Ω 2 =
0 and
( bb1zb1 + bb 2 zb 2 + M ub ) Ω2 =0 . From Equation (8.4) we have Ab = v where
1  1 1 
1
 f 
 −6.9081 − 0.2353 j 
=
A =
and v =  ub  = Ω 210−3 
.



−8.0892 + 2.9706 j 
 zb1 zb 2  0 1.2 
 M ub 
 −0.1671 − 2.2402 j 
Thus fbrg =
Ω 2b =
A −1v =
Ω 210−3 
 kg m .
−6.7140 + 2.4755 j 
At 800 rev min , the forces at the bearings are
 −0.1671 − 2.2402 j   −1.1725 − 15.7226 j 
2
fbrg
= ( 800 × 2π 60 ) 10−3 
=
 
N.
−6.7140 + 2.4755 j  −47.3109 + 17.3740 j 
15.7662 
−94.2651
Thus fbrg = 
 N and θ = ∠b ×180 π = 
 degree
50.4001
159.8353 
Part (b). Since for balance ( bD1 + bD3 + fub ) Ω 2 =
0 and
( bD1zD1 + bD3 zD3 + M ub ) Ω2 =0 . Cancelling the Ω2
terms from the above equations,
then from Equation (8.4) we have Ab = v where.
1  1
1
 1
 fub 
−3  −6.9081 − 0.2353 j 
=
A =
=
and v =
 10 
.



−8.0892 + 2.9706 j 
 z D1 z D3  0.4 1.6 
 M ub 
−2.4698 − 2.1618 j 
−2.4698 − 2.1618 j 
−1
Thus
=
b A=
v 10−3 
 kg m= 
g m .
 −4.4383 + 2.3971 j 
 −4.4383 + 2.3971 j 
Hence the unbalance corrections are
3.2822 
−138.8044 
b =
 g m and θ = ∠b ×180 π = 
 degree
5.0442 
 151.6272 
The MATLAB script Problem_08_02.m repeats these calculations to give
Force at brg 1 = 15.7662 N at -94.2651 degree
Force at brg 2 = 50.4001 N at 159.8353 degree
Required balance at dsk 1 = 3.2822 g m at -138.8044 degree
Required balance at dsk 3 = 5.0442 g m at 151.6272 degree
Problem 8.3
At 3000 rev/min, the initial rotor response is:
=
r0 0.02
=
e j 30 π 180
( 0.0173 + 0.0100j ) mm
Adding the trial balance mass b1 = 10e j180 π 180 = −10 g m gives the response
60 π 180
=
r1 0.03e − j=
Thus
( 0.0150 − 0.0260 j ) mm
84
rd =r1 − r0 =0.0173 + 0.0100 j − ( 0.0150 − 0.0260 j ) =−
( 0.0023 − 0.0360 j ) mm
From Equation (8.14)
bc =− r0b1 rd =− ( 0.0173 + 0.0100j ) × ( −10 ) ( −0.0023 − 0.0360 j )
=
( −3.0769 + 4.6154 j ) g m
Thus the product of the balance mass times the radius at which it acts is
bc = 5.5470 g m and θb = ∠bc ×180 π = 123.69° . The MATLAB script
Problem_08_03.m repeats these calculations and give the following output
Required balance = 5.547 g m at 123.6901 degree
Problem 8.4
The initial rotor response at 1500 rev/min in planes 1 and 2 is
0.35e− j 64 π 180  0.1534 − 0.3146 j 
r0 =
=
 
 mm The first trial balance mass added in
− j 89 π 180
0.0063
0.3599
j
−


0.36
e


0.5e
−3 
j0 

−3 0.5
plane 1 is b1 10
=
=

 10   kg m . The rotor response is then
0 
 0 
 0.35e− j 94 π 180  −0.0305 − 0.3487 j 
r1 =
=
 
 mm
− j122 π 180
0.45e
  −0.2385 − 0.3816 j 
 −0.1839 − 0.0341 j 
Then rd 1 = r1 − r0 = 
 mm
−0.2447 − 0.0217 j 
0

 0 

−3 
The second trial mass added
in plane 2 is b 2 10
10−3 
=
=

 kg m
j 90 π 180 
0.5e

0.5 j 
0.59e− j 73π 180  0.1725 − 0.5642 j 
The rotor response
is then r2 =
=
 
m
− j 86 π 180
j
0.0384
0.5487
−


e
0.55


0.0191 − 0.2496 j 
Then rd 2 = r2 − r0 = 
 mm
0.0321 − 0.1887 j 
 −0.1839 − 0.0341 j 0.0191 − 0.2496 j 
 −0.2447 − 0.0217 j 0.0321 − 0.1887 j  mm then


 6.8256 + 1.7542 j −9.1008 − 1.4630 j  1
rd−1 = 

 −1.5437 + 8.9132 j 1.8132 − 6.6441 j  mm
0 
−3  0.5
Defining,
=
b [b
=
1 b 2 ] 10 
 kg m , then from Equation (8.26)
 0 0.5 j 
 −0.5076 − 0.6943 j 
10−3 
bc =
−b rd−1r0 =
 kg m
 0.5794 − 0.0935 j 
 0.8601
 −126.1743
Thus b c = 
g m and θc = ∠bc ×180 π = 

 degree
0.5869 
 −9.1636 
The MATLAB script Problem_08_04.m repeats these calculations and gives the
following results
Defining
=
rd
[r=
d 1 rd 2 ]
85
Required balance, plane 1 = 0.86007 g m at -126.1743 degree
Required balance, plane 2 = 0.58688 g m at -9.1636 degree
Problem 8.5
The initial rotor response at 3000 rev/min at bearings 1 and 2 is
j 49 π 180 
0.18e
  0.1181 + 0.1358 j 
r0 =
=
 
 mm . The first trial balance mass added at
j 52 π 180
0.62e
 0.3817 + 0.4886 j 
0.5e j 0 
−3 
−3 0.5
disk 1 is b1 10
=
=

 10   kg m . The rotor response is then
0 
 0 
0.17e j 54 π 180  0.0999 + 0.1375 j 
r1 =
=
 
 mm .
j 40 π 180
0.76e
 0.5822 + 0.4885 j 
−0.0182 + 0.0017 j 
Then rd 1 = r1 − r0 = 
 mm . The second trial mass is added at disk
0.2005


2 (and the first trial mass at disk 1 is left in place). Thus
j0


 0.5 
−3  0.5e
=
b 2 10
=
10−3 


 kg m
j 90 π 180
0.5 j 
0.5e

0.10e− j 22 π 180  0.0927 − 0.0375 j 
The rotor response
is then r2 =
=
 
 mm
j 24 π 180
j
0.5755
0.2562
+


 0.63e

−0.0254 − 0.1733 j 
Thus rd 2 = r2 − r0 = 
 mm
 0.1938 − 0.2323 j 
 −0.0182 + 0.0017 j −0.0254 − 0.1733 j 
Defining
mm then
=
rd [r=
d 1 rd 2 ] 
0.2005
0.1938 − 0.2323 j 

 −5.6519 − 5.2157 j 4.4320 − 0.4241 j 
mm −1
rd−1 = 

 −0.2534 + 5.0898 j 0.0197 + 0.4634 j 
0.5 
−3  0.5
Defining
=
b [b
=
1 b 2 ] 10 
 kg m , then from Example (8.26)
 0 0.5 j 
 −0.4999 − 0.6864 j 
10−3 
bc =
−b rd−1r0 =
 kg m
 0.3766 + 0.4701 j 
0.8492 
 −126.0642 
Thus b c = 
g m and θc = ∠bc ×180 π = 

 degree .
 0.6023
 51.3043 
The MATLAB script Problem_08_05.m repeats these calculations and gives the
following output:
Required balance, plane 1 = 0.84919 g m at -126.0642 degree
Required balance, plane 2 = 0.60233 g m at 51.3043 degree
This machine is flexible and we have performed a 2 plane balance. Therefore above
3000rev/min the residual unbalance will excite the higher flexible modes and hence
the machine will not be balanced at higher speeds. Indeed the machine will not be
exactly fully balanced at any speeds, and is only balanced at the sensor location for
3000rev/min.
86
Problem 8.6
From Equation (6.55), the response of a complex rotor to an out of balance is
−1
K − Ω 2M + jΩ ( ΩG + C )  Ω 2b 0 where Ω 2b0 is a vector of unknown
q0=
u


unbalance forces. From Equation (6.60), the response of a bent rotor is
−1
K − Ω 2M + jΩ ( ΩG + C )  Kqb 0 where qb 0 is a vector defining the shape of
q 0=
b


the bent rotor. Let us assume that a bent rotor spins at Ω0 , but fact that the rotor is
bent is unknown and the response is assumed to be due to unbalance. Then this
unbalance can be corrected, and the response reduced to zero, by introducing
appropriate balance masses thus:
−1
) (
(
)
=
0 K − Ω02M + jΩ0 ( Ω0G + C )  Kqb 0 − Ω02b 0 or Kqb 0 − Ω02b0 = 0 . Thus the


rotor is successfully balanced at speed Ω0 . When the speed changes to Ω1 , for
(
)
example, then Kqb 0 − Ω12b0 and clearly this is not equal to a vector of zeros. Thus
when the rotor speed changes the response is no longer zero and the rotor is not
balanced.
Problem 8.7
At 1500 rev/min the initial response at bearings 1 and 2 is
0.20e j150 π 180  −0.1732 + 0.1000 j 
r0 =
=
 
 mm
j120 π 180
0.30e
  −0.1500 + 0.2598 j 
0.1×150e j 0 
−3 
−3 15
The first trial mass added
to disk 1 is b1 10
=
=

 10   kg m
0


0
The rotor response at bearings 1 and 2 is then
0.10e j140 π 180   −0.0766 + 0.0643 j 
r1 =
=
 
 mm
j150 π 180
0.20e
 −0.1732 + 0.1.000 j 
 0.0966 − 0.0357 j 
Then rd 1 = r1 − r0 = 
 mm
−0.0232 − 0.1598 j 
The second trial mass added to disk 2 is
j0

 15 

−3  0.1× 150e
b 2 10
=
10−3   kg m


j 90 π 180
15 j 
0.1× 150e

The rotor response at bearings 1 and 2 is then
 0.45e j 90 π 180  
0.4500 j

r2 =
=


 mm
j 300 π 180
0.40e
 0.2000 − 0.3464 j 
0.1732 + 0.3500 j 
Then rd 2 = r2 − r0 = 
 mm
0.3500 − 0.6062 j 
 0.0966 − 0.0357 j 0.1732 + 0.3500 j 
Defining
=
rd [r=
d 1 rd 2 ] 
 mm then
 −0.0232 − 0.1598 j 0.3500 − 0.6062 j 
87
 2.6420 + 12.9045 j 6.8085 + 2.7647 j 
mm −1
rd−1 = 

 −2.3220 − 1.9600 j −0.9139 + 1.7090 j 
−3 15 15 
Defining
=
b [b
=
1 b 2 ] 10 
 kg m , then from Equation (8.26)
 0 15 j 
 0.4795 + 0.1505 j 
10−1 
bc =
−b rd−1r0 =
 kg m
 −0.0580 − 0.0437 j 
0.3350 
 17.4255 
Thus m
kg and θc = ∠bc ×180 π = 
=
bc =
ε 
c

 degree .
0.0484 
 −143.0003
At 3000 rev/min the initial response at bearings 1 and 2 is
 0.10e j 70 π 180   0.0342 + 0.0940 j 
r0 =
=
 
 mm
j110 π 180
0.20e
 −0.0684 + 0.1879 j 
Adding the first trial mass described by b1 (above), the rotor response at bearings 1
and 2 is then
 0.25e j 90 π 180  
0.2500 j

r1 =
=


 mm
j150 π 180
0.2598
0.1500
j
−
+


0.30
e


−0.0342 + 0.1560 j 
Then rd 1 = r1 − r0 = 
 mm
 −0.1914 − 0.0379 j 
Adding the second trial mass described by b 2 (above), the rotor response at bearings
1 and 2 is then
 0.30e j130 π 180  −0.1928 + 0.2298 j 
r2 =
=
 
 mm
j 300 π 180
0.40e
  0.2000 − 0.3464 j 
−0.2270 + 0.1358 j 
Then rd 2 = r2 − r0 = 
 mm
 0.2684 − 0.5343 j 
 −0.0342 + 0.1560 j −0.2270 + 0.1358 j 
Defining
=
rd [r=
d 1 rd 2 ] 
 mm then
 −0.1914 − 0.0379 j 0.2684 − 0.5343 j 
 −5.1845 − 5.1720 j −0.7086 − 3.1617 j 
mm −1
rd−1 = 

1.6834 + 0.9965 j 
 1.1757 − 2.0804 j
 −1.3270 + 0.4402 j 
Thus from Equation (8.26), bc =
10−2 
−b rd−1r0 =
 kg m
 0.4313 + 0.1001 j 
0.0932 
161.6485
Thus m
and
kg
180
=
bc =
ε 
θ
=
∠
b
×
π
=
c
c
c

 13.0647  degree .
 0.0295


The MATLAB script Problem_08_07.m repeats these calculations and gives the
following results
Rotor speed 1500 rev/min
Required balance, plane 1 = 0.33501 kg at 17.4255 degree
Required balance, plane 2 = 0.048397 kg at -143.0003 degree
Rotor speed 3000 rev/min
Required balance, plane 1 = 0.093205 kg at 161.6485 degree
Required balance, plane 2 = 0.029517 kg at 13.0647 degree
88
Problem 8.8
The initial response at bearings 1 and 2 in the x and y directions is
0.0420

 0.0420
 
mm
=
r0 x =
j10 π 180  0.0443 + 0.0078 j 
0.045e
 

0.065e− j 40 π 180  0.0498 − 0.0418 j 
r0 y =
=
 
 mm
− j 30 π 180
0.0580
0.0335
j
−


0.067
e


0.03 × 250e
−3 
j0 

−3 7.5
The first trial mass added
to disk 1 is b1 10
=
=

 10   kg m
0
0 


The rotor response at bearings 1 and 2 in the x and y directions is then
0.076e− j 20 π 180  0.0714 − 0.0260 j 
r1x =
=
 
 mm
− j10 π 180
0.0788
0.0139
j
−


e
0.080


−
j
60
π
180
0.120e
 0.0600 − 0.1039 j 
r1 y =
=
 
 mm
− j 60 π 180
0.120e
 0.0600 − 0.1039 j 
0.0294 − 0.0260 j 
Hence rd 1x = r1x − r0 x = 
 mm
0.0345 − 0.0217 j 
 0.0102 − 0.0621 j 
rd 1 y = r1 y − r0 y = 
 mm
0.0020 − 0.0704 j 
0

0 

−3 
The second trial mass added
to disk 2 is b 2 10
10−3   kg m
=
=

j0 
0.03 × 250e 
7.5
The rotor response at bearings 1 and 2 in the x and y directions is then
0.048e− j10 π 180  0.0473 − 0.0083 j 
=
r2 x =
 
 mm
0.0500
0.0500

 

0.074e− j 40 π 180  0.0567 − 0.0476 j 
r2 y =
=
 
 mm
− j 40 π 180
0.075e
 0.0575 − 0.0482 j 
 0.0053 − 0.0083 j 
Hence rd 2 x = r2 x − r0 x = 
 mm
0.0057 − 0.0078 j 
 0.0069 − 0.0058 j 
rd 2 y = r2 y − r0 y = 
 mm
−0.0006 − 0.0147 j 
0.0294 − 0.0260 j 0.0053 − 0.0083 j 
 rd 1x rd 2 x  0.0345 − 0.0217 j 0.0057 − 0.0078 j 
 mm
Thus rd =
=
 
r
r


0.0102
0.0621
j
0.0069
0.0058
j
−
−
 d1 y d 2 y 


0.0020 − 0.0704 j −0.0006 − 0.0147 j 
The 4 × 2 matrix rd cannot be inverted, but we can obtain the pseudoinverse (see
Equation (8.37) and the following text) to give
4.359 − 6.403 j −9.258 + 16.119 j 2.701 + 1.890 j 
 2.930 − 5.503 j
rd† = 100 
−13.53 + 21.86 j 
 −1.897 + 42.16 j −7.414 + 45.28 j 54.35 − 61.71 j
89
7.5 0 
10−3 
 kg m then we have
 0 7.5
 −0.0060 − 0.0055 j 
bc =
−b rd†r0 =
 0.0052 − 0.0007 j  kg m


0.0324 
 −137.6634 
Thus m
kg and θc = ∠bc ×180 π = 
=
bc =
ε 
c

 degree .
 0.0208
 −8.0893 
Using x direction data only
0.0294 − 0.0260 j 0.0053 − 0.0083 j 
=
rd [r=
d 1x rd 2 x ] 
 mm
0.0345 − 0.0217 j 0.0057 − 0.0078 j 
 −2.0602 + 0.7769 j 2.0467 − 0.9277 j 
mm −1
rd−1 = 100 

 9.2800 + 0.1796 j −8.8545 + 1.2691 j 
 −0.0086 − 0.0056 j 
Thus, using Equation (8.26), bc =
−b rd−1r0 =
 0.0094 + 0.0041 j  kg m


0.0410 
 −146.6913
Thus m
kg and θc = ∠bc ×180 π = 
=
bc =
ε 
c

 degree .
0.0410 
 23.3087 
The MATLAB script Problem_08_08.m repeats these calculations and gives the
following results and also gives a plot of the rotor orbit at the bearings, shown below.
Defining
=
b
[b=
1 b2 ]
Using x direction data only
Required balance, plane 1 = 0.041016 kg at -146.6913 degree
Required balance, plane 2 = 0.041016 kg at 23.3087 degree
Using x and y direction data
Required balance, plane 1 = 0.032418 kg at -137.6634 degree
Required balance, plane 2 = 0.020814 kg at -8.0893 degree
Initial orbit
1st trial orbit
2nd trial orbit
0.1
y direction (mm)
0.05
0
-0.05
-0.1
-0.15
-0.1
-0.05
0
0.05
x direction (mm)
90
0.1
0.15
Problem 8.9
The initial rotor response at bearings 1 and 2 in the x and y directions is
0.0420

 0.0420  
r0 x =
mm
=
j 9 π 180  0.0444 + 0.0070 j 
0.045e
 

0.065e− j 39 π 180  0.0505 − 0.0409 j 
r0 y =
=
 
 mm
− j 34 π 180
0.0555
0.0375
j
−


0.067
e


0.03 × 250e
−3 
j0 

−3 7.5
The first trial mass added
to disk 1 is b1 10
=
=

 10   kg m
0
0 


The rotor response at bearings 1 and 2 in the x and y directions is then
0.076e− j 22 π 180  0.0705 − 0.0285 j 
r1x =
=
 
 mm
− j13π 180
0.0779
0.0180
j
−


e
0.080


−
j
61
π
180
 0.120e
 0.0582 − 0.1050 j 
r1 y =
=
 
 mm
− j 56 π 180
0.120e
  0.0671 − 0.0995 j 
Hence
0.0285 − 0.0285 j 
rd 1x = r1x − r0 x = 
 mm
0.0335 − 0.0250 j 
0.0077 − 0.0640 j 
rd 1 y = r1 y − r0 y = 
 mm
0.0116 − 0.0620 j 
0


−3 
−3  0 
The second trial mass added
to disk 2 is b 2 10
10
=
=


  kg m
j0
7.5
0.03 × 250e 
The rotor response at bearings 1 and 2 in the x and y directions is then
0.048e− j 7 π 180  0.0476 − 0.0058 j 
r2 x =
=
 
 mm
j 3π 180
0.0549
0.0026
j
+


0.0500
e


0.074e− j 45π 180   0.0523 − 0.0523 j 
r2 y =
=
 
 mm
− j 40 π 180
0.075e
 0.0575 − 0.0482 j 
Hence
0.0056 − 0.0058 j 
rd 2 x = r2 x − r0 x = 
 mm
0.0055 − 0.0044 j 
0.0018 − 0.0114 j 
rd 2 y = r2 y − r0 y = 
 mm
0.0019 − 0.0107 j 
 0.0285 − 0.0285 j 0.0056 − 0.0058 j 
 rd 1x rd 2 x   0.0335 − 0.0250 j 0.0055 − 0.0044 j 
 mm
Thus rd =
=
 
r
r


0.0077
0.0640
j
0.0018
0.0114
j
−
−
d
1
y
d
2
y




 0.0116 − 0.0620 j 0.0019 − 0.0107 j 
The 4 × 2 matrix rd cannot be inverted, but we can obtain the pseudoinverse (see
Equation (8.37) and the following text) to give
 −0.632 − 0.922 j 0.645 + 0.060 j −0.490 + 0.0687 j 0.236 + 0.447 j 
rd† = 100 

 3.691 + 5.290 j −3.452 − 0.204 j 2.778 − 0.0812 j −1.272 − 2.198 j 
91
7.5 0 
10−3 
 kg m then we have
 0 7.5
 −0.0073 − 0.0060 j 
bc =
−b rd†r0 =
 0.0097 + 0.0025 j  kg m


0.0376 
 −140.5128
Thus m
kg and θc = ∠bc ×180 π = 
=
bc =
ε 
c

 degree .
 0.0401
 14.5691 
Using x direction data only,
 0.0285 − 0.0285 j 0.0056 − 0.0058 j 
=
rd [r=
d 1x rd 2 x ] 
 mm
0.0335 − 0.0250 j 0.0055 − 0.0044 j 
 −0.9753 − 0.7759 j 1.2276 + 0.7479 j 
mm −1
rd−1 = 100 

 5.6155 + 4.8156 j −6.0137 − 3.8134 j 
 −0.0063 − 0.0070 j 
Thus, using Equation (8.26), bc =
−b rd−1r0 =
 0.0034 + 0.0072 j  kg m


0.0375
 −131.8990 
Thus m
kg and θc = ∠bc ×180 π = 
=
bc =
ε 
c

 degree .
0.0318
 64.3837 
The MATLAB script Problem_08_09.m repeats these calculations and gives the
following results and also gives a plot of the rotor orbit at the bearings, shown below.
Defining
=
b
[b=
1 b2 ]
Using x direction data only
Required balance, plane 1 = 0.037452 kg at -131.899 degree
Required balance, plane 2 = 0.031837 kg at 64.3837 degree
Using x and y direction data
Required balance, plane 1 = 0.03761 kg at -140.5128 degree
Required balance, plane 2 = 0.04009 kg at 14.5691 degree
Initial orbit
1st trial orbit
2nd trial orbit
0.1
y direction (mm)
0.05
0
-0.05
-0.1
-0.15
-0.1
-0.05
0
0.05
x direction (mm)
92
0.1
0.15
Problem 8.10
The initial rotor response at bearings 1 and 2 in the x and y directions is
0.0310
0.0310


 
mm
=
r0 x =
j 20 π 180  0.0479 + 0.0174 j 
0.051e
 

0.050e− j 30 π 180  0.0433 − 0.0250 j 
r0 y =
=
 
 mm
− j 30 π 180
0.0624
0.0360
j
−


0.072
e


0.03 × 250e
−3 
j0 

−3 7.5
The first trial mass added
to disk 1 is b1 10
=
=

 10   kg m
0
0 


The rotor response at bearings 1 and 2 in the x and y directions is then
0.034
0.0340


 
mm
=
r1x =
− j 20 π 180  0.0498 + 0.0181 j 
0.053e
 

0.055e− j 40 π 180   0.0421 − 0.0354 j 
r1 y =
=
 
 mm
− j 30 π 180
0.0658
0.0380
j
−


0.076
e


Hence
0.3000


rd 1x = r1x − r0 x =10−2 
 mm
0.1879 + 0.0684 j 
 0.1169 − 1.0353 j 
rd 1 y = r1 y − r0 y =10−2 
 mm
0.3464 − 0.2000 j 
0

0 

−3 
The second trial mass added
to disk 2 is b 2 10
10−3   kg m
=
=

j0 
7.5
0.03 × 250e 
The rotor response at bearings 1 and 2 in the x and y directions is then
0.032

 
0.0320

=
r2 x =


 mm
j 20 π 180
0.0510e
 0.0479 + 0.0174 j 
0.052e− j 30 π 180  0.0450 − 0.0260 j 
r2 y =
=
 
 mm
− j 30 π 180
0.072e
 0.0624 − 0.0360 j 
Hence
0.1000 
rd 2 x = r2 x − r0 x = 10−2 
 mm
 0 
0.1732 − 0.1000 j 
rd 2 y = r2 y − r0 y = 10−2 
 mm
0


0.3000
0.1000




 rd 1x rd 2 x 
0.1879 + 0.0684 j
0

 mm
Thus rd =
=

r
r


−
0.1169
−
1.0353
j
0.1732
−
0.1000
j
d
1
y
d
2
y




0
 0.3464 − 0.2000 j

The 4 × 2 matrix rd cannot be inverted, but we can obtain the pseudoinverse (see
Equation (8.37) and the following text) to give
0.189 − 0.982 j 0.483 − 0.176 j −0.327 + 0.378 j 2.701 + 1.890 j 
rd† = 100 

5.323 + 2.945 j −0.423 + 2.243 j 2.762 − 0.106 j −3.980 + 2.236 j 
93
7.5 0 
10−3 
 kg m then we have
 0 7.5
 −0.0760 + 0.0044 j 
bc =
−b rd†r0 =
 −0.0412 − 0.3004 j  kg m


0.3046 
 −176.6739 
Thus m
kg and θc = ∠bc ×180 π = 
=
bc =
ε 
c

 degree .
1.2128 
 −97.8016 
Using x direction data only
0.3000
0.1000 

mm
=
rd [r=
d 1x rd 2 x ] 
0 
0.1879 + 0.0684 j
0.4698 − 0.1710 j 
 0
mm −1
rd−1 = 103 

1.0000 −1.4095 + 0.5130 j 
 −0.1913
Using Equation (8.26), bc =
−b rd−1r0 =
 0.3413  kg m


0.7650 
180.0000 
Thus m
kg and θc = ∠bc ×180 π = 
=
bc =
ε 
c

 degree
1.3650 
 0.0000 
The MATLAB script Problem_08_10.m repeats these calculations and gives the
following results and also gives a plot of the rotor orbit at the bearings, shown below.
Defining
=
b
[b=
1 b2 ]
Using x direction data only
Required balance, plane 1 = 0.765 kg at 180 degree
Required balance, plane 2 = 1.365 kg at -7.4562e-014 degree
Using x and y direction data
Required balance, plane 1 = 0.3046 kg at 176.7 degree
Required balance, plane 2 = 1.213 kg at -97.8 degree
y direction (mm)
0.05
Initial orbit
1st trial orbit
2nd trial orbit
0
-0.05
-0.06
-0.04
-0.02
0
0.02
x direction (mm)
94
0.04
0.06
Problem 8.11
The rotor system is described in EXAMPLE 8.5.5, except that the diameter of the left
hand disk is increase to 0.35 m. The MATLAB script Problem_08_11.m models and
simulates the rotor, including the inherent unbalance, using the Rotordynamics
Software Package. From this script Problem_08_11.m we obtain the mode shapes
and the response due to the inherent out of balance. For th first mode of the stationary
rotor, the ratio of displacements between the right hand disk to the left hand disk ( a1 )
1
is −2.5241 . For the second mode, this ratio ( a2 ) is 2.1588 . Thus u1 =   and
a1 
1
u 2 =   . Since we wish to balance the effect of the first two modes, we must to
a2 
define the vectors e1 and e2 according to Equation (8.43).
−a 
Fir the first mode we must define e1 so that u1T e1 = 1 . If e1 =  2  then
 1 
1 −a2 
T
u1T e=

 then u1 e1 = 1 . There is in fact a
1 a1 − a2 . Thus if we make e1 =
a1 − a2  1 
wide range of possible forms of e1 .
−a 
For the second mode we define e2 so that u T2 e 2 = 1 and u1T e 2 = 0 . If e 2 =  1  ,
 1 
1 −a1 
then u1T e 2 = 0 and uT2 e=
  then
2 a2 − a1 . Thus, if we make e 2 =
a2 − a1  1 
u T2 e 2 = 1 . (Note that the trial forces b1 and b 2 are determined by b1 = δ1e1 and
b 2 = δ2e 2 . The parameters δ1 and δ2 are chosen to make the trail forces a suitable
size, i.e. not so small that it is difficult to measure the response accurately, but not so
large as to risk damaging the rotor due to an excessive level of vibration. It does not
matter if e1 is chosen such that u1T e1 is not equal to unity, because the scale factor in
e1 is multiplied by δ1 . The parameter δ1 is still chosen to make b1 have a suitable
magnitude. The same argument applies to e2 ,)
Initial Response. From the rotor system model, we find that the response at 1110
rev/min at nodes 5 and 13 due to the inherent unbalance is
 33.9991 − 8.5685 j 
r01 = 
 mm
−85.0928 + 21.4472 j 
−2.1588  0.4610 
1 −a2 
1
Balance the first
mode. e1 =
=
=



 
.
a1 − a2  1  −4.6829  1  −0.2135
Let δ1 =0.001 . Then b1 = δ1 e1 .
0.4610 
 0 
Thus b1 = 
 g m and ∠b1 ×180 π =  degrees
0.2135
180 
With the trial balance b1 added to the rotor, the simulation gives the following
response at 1110 re/min at nodes 5 and 13:
95
 1.8516 − 8.5685 j 
−32.1475
r1 = 
 mm and hence rd 1 =r1 − r01 =
 mm
−4.6511 + 21.4472 j 
 80.4417 
 0.4876 − 0.1229 j 
rH r
α1 = −δ1 dH1 01 = 10−3 (1.0578 − 0.2666 j ) and bc1 =
α1e1 =

g m .
rd 1 rd 1
−0.2259 + 0.0569 j 
0.5029 
345.85
Hence, bc1 = 
 g m and ∠bc1 ×180 π =

0.2329 
165.85 
With the trial balance forces b1 removed and replaced by the balance forces b c1 , the
first mode is balanced and the simulation gives the following response at 2,315
rev/min at nodes 5 an 13:
0.4532 − 0.1620 j 
r02 = 100 
 mm
1.0134 − 0.3623 j 
Balance the second mode.
1 −a1 
1 2.5241 0.5390 
=
e2 =
=
0.001 . Then b 2 = δ2e 2 .
 

 
 . Let δ2 =
a2 − a1  1  4.6829  1  0.2135 
0 
0.5390 
Thus b 2 = 
 g m and ∠b1 ×180 π =  degrees
0 
0.2135
With the trial balance forces b 2 added to the rotor (in addition to the balance forces
b c1 ), the simulation gives the following response at xxx re/min at nodes 5 and 13:
−0.6799 − 0.1620 j 
 −113.3145
r2 = 100 
 mm and thus rd 2 =r2 − r02 =
 mm
 −1.5205 − 0.3623 j 
−253.3900 
 0.2156 − 0.0771 j 
rH r
α 2 = −δ2 dH2 02 = 10−3 ( −0.3999 − 0.1430 j ) . b c 2 =
α 2e 2 =

g m .
rd 2 rd 2
0.0854 − 0.0305 j 
0.2289 
340.33
Hence, bc 2 = 
 g m and ∠b c 2 ×180 π =

0.0907 
340.33
 0.7032 − 0.2000 j 
Combining b c1 and b c 2 gives b c = b c1 + b c 2 = 
g m .
−0.1405 + 0.0264 j 
344.13
 0.7311
Hence, bc = 
 degrees
 g m and ∠b c ×180 π =
169.36 
0.1429 
The MATLAB script Problem_08_11.m simulates the rotor system and repeats these
calculations and gives the following results:
Required balance, mode 1 at disk 1: 0.50288 g m at -14.1463 degrees
Required balance, mode 1 at disk 2: 0.23295 g m at 165.8537 degrees
Required balance, mode 2 at disk 1: 0.22893 g m at -19.6714 degrees
Required balance, mode 2 at disk 2: 0.090699 g m at -19.6714 degrees
Total balance, modes 1 & 2 at disk 1: 0.73109 g m at -15.874 degrees
Total balance, modes 1 & 2 at disk 2: 0.14294 g m at 169.3564 degrees
96
Problem 8.12
The rotor system is described in EXAMPLE 8.5.4. The MATLAB script
Problem_08_12.m models and simulates the system, including the inherent
unbalance, using the Rotordynamics Software Package. From this script we obtain the
response due to the inherent out of balance. However, the mode shapes are assumed to
0.95
1 
be have been measured inaccurately and are u1 =   and u 2 = 
 . If we make
 −1 
0.9 
0.5391
 1 
T
T
T
=
e1 e=
e1t = 

 , then e1 u1 = 1 . Of
 then e1t u 2 = 0 . Letting
1t e1t u1
0.5121
0.95




(
)
course, e1T u 2 = 0 .
(Note that the trial forces b1 and b 2 are determined by b1 = δ1e1 and b 2 = δ2e 2 . The
parameters δ1 and δ2 are chosen to make the trail forces a suitable size, i.e. not so
small that it is difficult to measure the response accurately, but not so large as to risk
damaging the rotor due to an excessive level of vibration. It does not matter if e1 is
chosen such that u1T e1 is not equal to unity, because the scale factor in e1 is
multiplied by δ1 . The parameter δ1 is still chosen to make b1 have a suitable
magnitude. The same argument applies to e2 .)
Initial Response. From the rotor system model, we find that the response at 820
rev/min at nodes 5 due to the inherent unbalance is
=
r01 (1.0100 − 1.8433 j ) mm
0.5291
−3
Balance the first mode. e1 = 
 . Let δ1= 0.4 ×10 . Then
0.5121
0.2156 
0.2156 
0 
b1 =
δ1e1 =
 g m and ∠b1 ×180 π =  degrees

 g m . Thus b1 = 
0.2049 
0.2049 
0 
With the trial balance forces b1 added to the rotor, the simulation gives the following
response at 820 rev/min at nodes 5:
=
r1 ( 0.0844 − 2.1567 j ) mm and hence rd 1 =
r1 − r01 =
( −0.9255 − 0.3133 j ) mm
rd 1 r01
= 10−4 (1.4965 − 8.4731 j ) .
rd 1 rd 1
0.4638
0.0807 − 0.4568 j 
b c1 =
α1e1 =
 g m and

 g m . Hence, bc1 = 
0.4406 
0.0766 − 0.4339 j 
−79.9841
∠bc1 ×180 π =
 degrees. With the trial balance forces b1 removed and
−79.9841
replaced by the balance forces b c1 , the first mode is balanced and the simulation
gives the following response at 3075 rev/min at nodes 5:
=
r02 ( 0.6355 + 0.1399 j ) mm
Using the data from node 5 only, α1 = −δ1
97
Balance the second mode.
−0.9 
1 
T
u1 =   . If we make e 2t = 
 then e2t u1 = 0 . Letting
 1 
0.9 
 0.4852 
T
T
T
=
e2 e=

 , then e2 u 2 = 1 . Of course, e2 u1 = 0 .
2t e 2t u 2
−0.5391
(
)
 0.1941 
 0.1941
Let δ2 = 0.4 ×10−3 . Then b 2 =
δ 2e 2 =

 g m . Thus b 2 = 
 g m and
−0.2156 
0.2156 
 0 
∠b 2 ×180 π =  degrees
180 
With the trial balance forces b 2 added to the rotor (in addition to the balance forces
b c1 ), the simulation gives the following response at 3075 rev/min at nodes 5:
=
r2
( 0.6425 − 0.2428 j ) mm
and thus rd 2 =r2 − r02 =( 0.0070 − 0.3827 j ) mm
rd 2 r02
= 10−4 (1.3405 − 6.6663 j ) .
rd 2 rd 2
0.3299 
 0.0650 − 0.3234 j 
bc 2 =
α 2e 2 =
 g m and

 g m . Hence, bc 2 = 
0.3666 
 −0.0723 + 0.3594 j 
−78.6300 
∠bc 2 ×180 π =
 degrees
101.3700 
Using that data from node 5 only, α 2 = −δ2
0.1457 − 0.7802 j 
Combining b c1 and b c 2 gives b c = b c1 + b c 2 = 
g m .
0.0044 − 0.0746 j 
−79.4213
0.7937 
Hence bc = 
 degrees .
 g m and ∠bc ×180 π =
−86.6443
0.0747 
The MATLAB script Problem_08_12.m simulates the rotor system and repeats these
calculations and gives the following results:
Using approximate mode
e1 = [ 0.5391
0.5121
Required balance, mode
Required balance, mode
shapes
]
1 at disk 1: 0.4638 g m at 280.0159 degrees
1 at disk 2: 0.4406 g m at 280.0159 degrees
e2 = [ 0.4852 -0.5391 ]
Required balance, mode 2 at disk 1: 0.3299 g m at 281.3700 degrees
Required balance, mode 2 at disk 2: 0.3666 g m at 101.3700 degrees
Total required balance at disk 1: 0.7937 g m at 280.5787 degrees
Total required balance at disk 2: 0.0747 g m at 273.3557 degrees
Response at disk 1 at 820 rev/min = 87.6393 mu_m
Response at disk 1 at 3075 rev/min = 0.0000 mu_m
98
Using the exact mode shapes
e1 = [ 0.5000
0.5000 ]
Required balance, mode 1 at disk 1: 0.4522 g m at 280.0167 degrees
Required balance, mode 1 at disk 2: 0.4522 g m at 280.0167 degrees
e2 = [ 0.5000 -0.5000 ]
Required balance, mode 2 at disk 1: 0.3598 g m at 281.4885 degrees
Required balance, mode 2 at disk 2: 0.3598 g m at 101.4885 degrees
Total required balance at disk 1: 0.8120 g m at 280.6689 degrees
Total required balance at disk 2: 0.0930 g m at 274.3107 degrees
Response at disk 1 at 820 rev/min =
Response at disk 1 at 3075 rev/min =
2.6806 mu_m
0.0000 mu_m
Problem 8.13
Using four measurements. (This is a restatement of the solution of EXAMPLE 8.6.1).
Initial estimate for the required balance. The largest difference between the original
response magnitude and the modified response occurs when the trial balance is fitted
at 240° . i.e r3 −=
r0 2.6759 − 2.0
= 0.6759 mm s . Therefore, to balance the as found
response of 2.0 mm s , the initial estimate of the corrective balance is
bc 0 =
20 × 2.0 0.6759 =
59.18g m . Because the trial unbalance at 240° increases the
unbalance response, to reduce the unbalance response the initial estimate for the
position of the corrective balance must be 240 − 180 =60° .
The initial estimate for R0 ( Ω
=
bc 0 2.0 59.18
= 0.0338 kg −1s −1 at 60° .
) r0 =
We must now determine the values of the three unknown parameters, i.e. R ( Ω ) and
the real and imaginary parts of bc to achieve balance. Letting R ( Ω ) =σ1 ,
real ( bc ) = σ2 and imag ( bc ) = σ3 , we must adjust these parameters to make
ri − σ1 × bi − ( σ2=
+ jσ3 ) 0,=
where i 1,  , 4 .
To estimate the unknown parameters, we minimise the sum of the squares of the
, σ3 )
errors, i.e. minimize J ( σ1 , σ2=
3
∑ ( ri − σ1 bi − ( σ2 + jσ3 ) )
2
i =0
To minimize this function requires an iterative scheme. Thi sis done in MATLAB
script Problem_08_13.m, together with the MATLAB user defined function bnpr.m.
Using 4 measurements with an initial balance of 59.18 g m at 60 degrees gives a final
balance of 57.139g m at 42.4897 degrees.
Using three measurements. Case 1. Starting with the same initial balance as in
Example 8.6.1 (above), i.e. 59.18g m at 60° and minimising the function
J ( σ1 , σ2 , σ3 ) gives a balance of 57.1353g m at 42.491 degrees.
Using three measurements. Case 2. Starting with an initial balance of 25g m at 40°
and minimising the function J ( σ1 , σ2 , σ3 ) gives a balance of 29.654g m at 51.0166
degrees.
Thus to obtain a unique solution this non-linear problem of three unknowns requires
four equations, based on the as found conditions and the response to three trial
balance masses. Three equations, (as found and two trial balance masses) do not give
a unique (correct) solution. The apparent solution depends on the initial conditioned.
Using script Problem_08_13.m gives
99
Using 3 trial masses. Initial balance = 59.18 g m at 60 degrees
Required balance = 57.139g m at 42.4897 degrees
Final value of |R(Omega)| = 0.035002 1/(kg s)
Using only 2 trial masses. Initial balance = 59.18 g m at 60 degrees
Required balance = 57.1353g m at 42.491 degrees
Final value of |R(Omega)| = 0.035005 1/(kg s)
Using only 2 trial masses. Initial balance = 25 g m at 40 degrees
Required balance = 29.654g m at 51.0166 degrees
Final value of |R(Omega)| = 0.067444 1/(kg s)
Problem 8.14
The rotor system is described in EXAMPLE 8.5.4. The MATLAB script
Problem_08_14.m models and simulates the rotor, including the inherent unbalance,
using the Rotordynamics Software Package.
Initial Response
From the rotor system model, we find that the response at 1000 rev/min at nodes 5
and 13 due to the inherent unbalance is
1.2173 − 2.0176 j 
r01 = 
 mm
1.2184 − 2.0227 j 
Balance the first mode.
Since we wish to balance the effect of the first mode by adding a mass to the LH disk
1
1 
only, we must to define the vector e1t as e1t =   . The first mode shape is u1 =   .
1
0 
1 
T
T
Hence e1 e=
=
  and then e1 u1 = 1 . The parameters δ1 is chosen to make
1t e1t u1
0
 
the trial forces a suitable size, i.e. not so small that it is difficult to measure the
response accurately, but not so large as to risk damaging the rotor due to an excessive
level of vibration. It does not matter if e1 is chosen such that u1T e1 is not equal to
(
)
unity, because the scale factor in e1 is multiplied by δ1 . The parameter δ1 is still
chosen to make b1 have a suitable magnitude. (The same argument applies to e2 ,)
0.4 
0.4 
Let δ1= 0.4 ×10−3 . Then b1 =
10−3   and hence b1 =   g m and
δ1e1 =
0 
0 
0 
∠b1 ×180 π =  degrees . With the trial balance forces b1 added to the rotor, the
0 
simulation gives the following response at 1000 re/min at nodes 5 and 9:
0.2466 − 2.3925 j 
−0.9707 − 0.3749 j 
r1 = 
 mm and hence rd 1 =r1 − r01 =
 mm
0.2447 − 2.3976 j 
−0.9737 − 0.3749 j 
α1 = −δ1
0.1572 − 0.8918 j 
= 10−4 (1.5717 − 8.9177 j ) . bc1 =
α1e1 =

g m .
0


0.9055
−80.0045
=
 g m and ∠bc1 ×180 π =
 degrees
0
 0 


rdH1 r01
rdH1 rd 1
Hence, bc1
100
With the trial balance b1 removed and replaced by the balance b c1 , the first mode is
balanced and the simulation gives the following response at 3075 rev/min at nodes 5
an 9:
−0.1738 − 0.0177 j 
r02 = 
 mm
 0.1731 + 0.0172 j 
1
1
−1
Balance the second mode. u1 =   and u 2 =   If we make e2t =   then
1
1
−1
(
)
 0.5000 
T
T
T
Letting e2 e=
eT2t u1 = 0 . =

 , then e2 u 2 = 1 . Of course, e2 u1 = 0 .
2t e 2t u 2
−
0.5000


 0.2 
0.2 
Let δ1= 0.4 ×10−3 . Then b 2 =
δ 2e 2 =

 g m and hence b 2 =   g m and
−0.2 
0.2 
 0 
∠b 2 ×180 π =  degrees . With the trial balance forces b 2 added to the rotor (in
180 
addition to the balance forces b c1 ), the simulation gives the following response at
−0.1633 − 0.3911 j 
3000 rev/min at nodes 5 and 9: r2 = 
 mm and thus
 0.1626 + 0.3906 j 
 0.0104 − 0.3734 j 
rd 2 =r2 − r02 =
 mm
−0.0104 + 0.3734 j 
−0.0068 + 0.0931 j 
= 10−4 ( −0.1353 + 1.8614 j ) . bc 2 =
α 2e 2 =

g m .
 0.0068 − 0.0931 j 
0.9331
 94.1568 
Hence, bc 2 = 
 g m and ∠b c 2 ×180 π =
 degrees
0.9331
−85.8432 
0.1504 − 0.7987 j 
Combining b c1 and b c 2 gives bc = b c1 + b c 2 = 
 g m . Hence,
 0.0068 − 0.0931 j 
0.8127 
280.6647 
bc = 
 g m and ∠bc ×180 π =
 degrees
 0.0933
 274.1568 
The MATLAB script Problem_08_14.m simulates the rotor system and repeats these
calculations and gives the following results (very similar to EXAMPLE 8.5.4):
α 2 = −δ2
rdH2 r02
rdH2 rd 2
Required balance, mode 1 at disk 1: 0.90552 g m at -80.0045 degrees
Required balance, mode 1 at disk 2: 0 g m at 0 degrees
Required balance, mode 2 at disk 1: 0.093314 g m at 94.1568 degrees
Required balance, mode 2 at disk 2: 0.093314 g m at -85.8432 degrees
Total balance, modes 1 & 2 at disk 1: 0.81274 g m at 280.6647 degrees
Total balance, modes 1 & 2 at disk 2: 0.093314 g m at 274.157 degrees
101
Problem 8.15
Balance mode 1 at 830 rev/min. Converting the as found response at 830 rev/min and
response to the trial balance mass to complex numbers gives
 1.616e j 81π 180 
0.2528 + 1.5961 j 

 

− j 85π 180  
=
r01 0.633e =
  0.0552 − 0.6306 j  mm

j 98π 180   −0.5861 + 4.1700 j 

 4.211e
 
 1.726e j 72 π 180 
0.5334 + 1.6415 j 

 

− j 94 π 180  
r1 =
0.651e
=
−0.0454 − 0.6494 j  mm

j 87 π 180   0.2246 + 4.2861 j 

 4.292e
 
Thus
 0.2806 + 0.0454 j 


rd 1 =
r1 − r01 =−
 0.1006 − 0.0188 j  mm
 0.8107 + 0.1161 j 


0 
 
Now b1 has been chosen to be proportional to e1 . Since b1 = 5  g m then e1 must
0 
 
0
 
be of the form  p  . The element p must be chosen to make u1T e1 = 1 .
0
 
0
 0.3780 

  0 



 

Since u1 = −0.1526  , e1 =
1 −0.1526  =
−6.5531
 1 

  0 
0



 

δ1 =b1 e1 =( b1 )2
α1 =δ1 rdH1 r01
(
( e1 )2 =5 −6.5531 =−0.7630 .
rdH1 rd 1
)
0




=0.1590 + j 3.9549 . bc1 =
α1e1 =
−1.0419 − j 25.9168 g m


0


 0 
 0 




and hence b c1 = 25.94  g m and ∠bc1 =−
 92  degrees
 0 
 0 




Balance mode 2 at 1080 rev/min. Converting the as found response at 1080 rev/min
and response to the trial balance mass to complex numbers gives
9.142e − j156 π 180 
−8.3516 − j 3.7184 

 

j15π 180  
=
r02  0.447e =
  0.4318 + j 0.1157  mm

 

2.299e j 25π 180   2.0836 + j 0.9716 


102
13.664e− j160 π 180 
−12.8400 − 4.6734 j 

 

j15π 180  
=
r2  0.682e =
  0.6588 + 0.1765 j  mm . Thus

j 21π 180   3.2068 + 1.2310 j 

 3.435e
 
−4.4883 − 0.9550 j 


rd 2 =r2 − r02 = 0.2270 + 0.0608 j  mm . Now b 2 has been chosen to be
 1.1232 + 0.2594 j 


 p1 
 5 
 


proportional to e2 . Since b 2 =  0  g m , e2 must be of the form  0  . Also, we
p 
−1.89 


 3
p 
 1 
u1T   1  0 


T
T
must make u1 e2 = 0 and u 2 e2 = 1 where u 2 = −0.0514  . Thus    0  =   or
uT2    1 
 −0.2513


 p3 
 ( u1 )1

( u 2 )1
U
=
( u1 )3   p1  0
 =  
( u 2 )3   p3  1 
 ( u1 )1 ( u1 )3  0.3780
=
 
( u 2 )1 ( u 2 )3   1
1 
 p1 
−1 0   0.9132 
. Hence=
U=


  


−0.2513
1  −0.3452 
 p3 
 0.9132 


Thus e 2 =  0  , Check on orthogonality:
−0.3452 


=
δ2 b 2 =
e2
( b 2 )1 ( e=
2 )1
Thus α 2 =δ2 rdH2 r02
bc 2
u1T 
0
 1
 T  [e1 e2 ] = 
’
0.3368
1


u
 2 
5 0.9132
= 5.4750 .
(rdH2 rd 2 ) =−10.6695 − j 2.2603 .
−9.7440 − 2.0642 j 


=
α 2e 2 =
0

g m
 3.6832 + 0.7803 j 


−168
9.9602 




or bc 2 =  0  g m and ∠b c 2 =
 0  degrees
 12 
3.7650 




Balance mode 3 at 2900 rev/min. Converting the as found response at 2900 rev/min
and response to the trial balance mass to complex numbers gives
 0.791e j156 π 180 
−0.7226 + 0.3217 j 

 

j152 π 180  
r03 =
10.290e
=
 −9.0855 + 4.8309 j  mm

 

0.891e j152 π 180   −0.7867 + 0.4183 j 


103
 0.823e j169 π 180 
−0.8079 + 0.1570 j 

 

j164 π 180  
r3 =
10.731e
=
−10.3153 + 2.9579 j  mm

j165π 180   −0.8973 + 0.2404 j 

 0.929e
 
Thus
 −0.0853 − 0.1647 j 


rd 3 =
r3 − r03 =−
 1.2298 − 1.8730 j  mm
−0.1106 − 0.1779 j 


We must make u1Te3 = 0 , uT2 e3 = 0 and u3Te3 = 1 . Thus,
u1T 
0 
 
 
T
=
U
u 2  e3 = 0  . Letting
 T
1 
u
 
 3 
0 
−1  
e3 U=
=
0 
1 
 
u1T  0.3780 −0.1526
1 
  

T
−0.0514 −0.2513 . then
u 2   1
 T   0.0763
1
0.0866 
u
 3  
0.0806 


0.9835 .
0.1196 


u1T 
 
Check on orthogonality: uT2  [e1 e 2
 T
u
 3 
0
0
 1

e 3 ] = 0.3368
1
0 
 6.5531 0.0398 1 
0.410 


Now b3 should have been chosen to be proportional to e3 . b3 =  5  g m and this
0.608


implies that δ3 =0.4100 0.0806 or 5 0.9835 or 0.608 0.1196 . Thus
δ3 =5.0862 or 5.0839 or 5.0832 . This means that b3 proportional to e3 to an
accuracy of 3 significant figures. The mean value is for δ3 is 5.0844.
α3 =δ3 rdH3 r03
(rdH3 rd 3 ) =−2.1402 + 23.2255 j .
 −0.1725 + 1.8722 j 


bc3 =
α3e3 =
−2.1048 + 22.8421 j  g m
 −0.2560 + 2.7780 j 


or b c3
 1.881 
95


 
= 22.939  g m and ∠bc3 =
95 degrees
 2.790 
95


 
The MATLAB script Problem_08_15.m repeats these calculations and gives the
following results:
104
Mode 1 at 830 rev/min
Vector e1 = [ 0.0000 -6.5531 0.0000]
delta = -0.763
alpha = 3.9581 at 87.6979 degree
Amplitude
of bc1 =
0.0000 25.9378
Phase (deg)
of bc1 =
0.0000 -92.3021
Mode 2 at 1080 rev/min
Vector e2 = [ 0.9132 0.0000 -0.3452]
delta = 5.475
alpha = 10.9063 at -168.0391 degree
Amplitude
of bc2 =
9.9602
0.0000
Phase (deg)
of bc2 = -168.0391
0.0000
Orthogonality test
[ 1.0000 0.0000 ]
[ 0.3368 1.0000 ]
Mode 3 at 2900 rev/min
Vector e3 = [ 0.0806 0.9835 0.1196]
delta = 5.0844
alpha = 23.3239 at 95.2648 degree
Amplitude
of bc3 =
1.8801 22.9388
Phase (deg)
of bc3 = 95.2648 95.2648
Orthogonality test
[ 1.0000 0.0000 -0.0000 ]
[ 0.3368 1.0000 -0.0000 ]
[ -6.5531 0.0398 1.0000 ]
105
0.0000
0.0000
3.7650
11.9609
2.7898
95.2648
Chapter 9
Problem 9.1
(a) Axial vibration. In this case, A =
πd 2 4 =
0.0020m 2 ,
3
, k2 EA=
=
k1 EA
=
L1 9.8175 ×108 N m =
L2 6.545 ×108 N m , k=
c 50 × 10 N m
3
2
and k=
b 30 × 10 N m . M D = πρhD 4 = 22.054 kg . The equations of motion (with
 + Kq = 0 where
4 degrees of freedom) for both axial vibrations are Mq
K
−k1
0
0 
 kb + k1
 −k
k1 + kc
− kc
0 
1
=
M
 0
kc + k 2 − k 2 
− kc


0
k2 
− k2
 0
M D
 0

 0

 0
0
0
MD
0
0
0
MD
0
0 
0 
0 

M D 
This leads to the eigenvalue problem λMq 0 =
Kq 0 where λ = ω2n . This is solved in
the MATLAB script Problem_09_01.m and the natural frequencies are shown below.
If we divide the system into to parts then the system matrices are shown below.
0 
0 
 k1 −k1 
M D
 k2 − k2 
M
, K r 2 =
=
K r1 =
M r1 =
Mr 2  D



MD 
M D 
 −k1 k1 
 0
 − k2 k2 
 0
Thus we have two systems with 2 degrees of freedom and each resultant eigenvalue
problem can be solved separately. See the output of the MATLAB script below.
If we join the central inertias together then
0 
 kb + kc −kc 
 2M D
=
K rb =
M
. This is a 2 degree of freedom system
rb
 0
kc 
2 M D 

 − kc
and the resultant eigenvalue can be solved, see the output of the MATLAB script
Problem_09_01.m as follows:
Axial vibrations
Natural frequency
Natural frequency
Natural frequency
Natural frequency
1
2
3
4
=
=
=
=
2.711 Hz
8.204 Hz
1226.1684 Hz
1501.7444 Hz
Treating each rotor separately
System A, natural frequency 1 =
System A, natural frequency 2 =
System B, natural frequency 1 =
System B, natural frequency 2 =
0 Hz
1501.7291 Hz
0 Hz
1226.1566 Hz
Treating rotor as two rigid bodies coupled together
System A & B joined, natural frequency 1 = 2.7111 Hz
System A & B joined, natural frequency 2 = 8.204 Hz
The coupling stiffness is very low compared to the shaft axial stiffnesses. For the two
highest frequencies, the coupling has little effect so treating the two rotors separately
gives accurate results. Conversely, joining the two rotors together predict the two
lowest frequencies very accurately.
106
(b) Torsional vibration. In this case J =
πd 4 32 =
6.136 ×10−7 m 4 ,
, k2 GJ=
=
k1 GJ=
L1 1.227 ×105 N m =
L2 8.181×104 N m , kc = 6 ×105 N m and
2
2
=
ID M
=
D D 8 0.248 kg m . The equations of motion (with 4 degrees of freedom)
 + Kq = 0 where
for both axial vibrations are Mq
K
−k1
0
0 
 k1
 −k k + k
− kc
0 
1
1
c
=
M
 0
kc + k 2 − k 2 
− kc


0
k2 
− k2
 0
ID
0

0

 0
0
0
ID
0
0
0
ID
0
0
0 
0

I D 
This leads to the eigenvalue problem λMq 0 =
Kq 0 where λ = ω2n . This is solved in
the MATLAB script Problem_09_01.m and the natural frequencies are shown below.
If we divide the system into to parts then the system matrices are shown below.
0
 k1 −k1 
ID 0 
 k2 − k2 
I
=
K r1 =
M r1 =
Mr 2  D
, K r 2 =





 −k1 k1 
 0 ID 
 − k2 k2 
 0 ID 
Thus we have two systems with 2 degrees of freedom and each resultant eigenvalue
problem can be solved separately. See the output of the MATLAB script below.
If we join the central inertias together then
0 
 kc − kc 
2I
. This is a 2 degree of freedom system and
=
K rb =
M rb  D

2 I D 
 0
 − kc kc 
the resultant eigenvalue can be solved, see the output of the MATLAB script
Problem_09_01.m as follows:
Torsional vibrations
Natural frequency 1 =
Natural frequency 2 =
Natural frequency 3 =
Natural frequency 4 =
0 Hz
94.5794 Hz
146.2986 Hz
365.9606 Hz
Treating each rotor separately
System A, natural frequency 1 =
System A, natural frequency 2 =
System B, natural frequency 1 =
System B, natural frequency 2 =
0 Hz
158.2961 Hz
0 Hz
129.2483 Hz
Treating rotor as two rigid bodies coupled together
System A & B joined, natural frequency 1 = 0 Hz
System A & B joined, natural frequency 2 = 247.5005 Hz
Note that because the torsional stiffness of the coupling is similar to the shaft
stiffnesses, analysing the rotors separately, or joining then with a rigid connection
give poor prediction of the natural frequencies of the 4 degree of freedom system.
107
Problem 9.2
 + Kq = 0 where the
For this relatively simple system the equations of motion are Mq
stiffness and inertia matrices are
0
0
0 
 I cr 0
0
0
0 
 k1 −k1

 − k 2k
0
0 
−k1
0
0 
 0 I cg 0
1
 1
0 I te 0
0 
K =
0 −k1 k1 + k2 −k2
0  M 0




0
0
0
I
0
−
−
0
0
k
2
k
k


cg
2
2
2



 0
−k2 k2 
0
0
0
0
0 I gr 
 0
Substituting numerical values give
 1 −1 0 0 0 
100 0 0 0 0 
 −1 2 −1 0 0 
 0 10 0 0 0 




 0
25 ×106  0 −1 2 −1 0  M =
K=
0 50 0 0 




0 0 10 0 
 0 0 −1 2 −1
 0
0 0 0 80 
 0 0 0 −1 1 
 0
Solving these equations to determine the natural frequencies leads to an eigenvalue
problem. This eigenvalue problem is solved in the MATLAB script
Problem_09_02.m. The output is as follows:
Natural
Natural
Natural
Natural
Natural
frequency
frequency
frequency
frequency
frequency
1
2
3
4
5
=
=
=
=
=
0 Hz
58.7123 Hz
124.8542 Hz
360.9772 Hz
378.8684 Hz
Not that the first natural frequency is zero, because the system is unconstrained.
Problem 9.3
φ2
φ1
ID1
ID1
k1
k2
ID2
ID1
gφ2
φ3
Suppose the rotation of the small gear is ϕ . The kinetic energy of gear is
=
T
But ϕ = gφ2 and hence the kinetic energy of the gear
is T
=
1 I g2 φ
2 .
2
2 D2
1I ϕ
2.
2 D2
Thus, referred
to φ2 , the moment of inertia of the small gear is g2 I D 2 . Similarly, considering the
strain energy stored in the right side shaft,=
V
108
1k
2 2
2
( ϕ − φ=
3)
1k
2 2
( gφ2 − φ3 )2 .
Hence
=
V
1k
2 2
(g φ
2 2
2
)
− 2 gφ2 φ3 + φ32 . This implies that there are terms in g2 k2 and
− gk2 appear in the equation of motion – see the stiffness matrix below. Thus, for this
 + Kq = 0 where the stiffness
relatively simple system the equations of motion are Mq
and inertia matrices are
−k1
0 
0
0 
 k1
 I D1




2
2
K=
 −k1 k1 + g k2 − gk2  M =
 0 I D1 + g I D 2 0  .
 0
 0
− gk2
k2 
0
I D1 




Using the data given, the second moment of area for the shaft is 6.136 × 10 −7 m 4 6
and hence=
k1
1.227 × 10 5 Nm and
=
k2
8.181 × 10 4 Nm , I D1 = 0.248 kg m 2 ,
I D 2 = 0.0155 kg m 2 and g = 2 . The MATLAB script Problem_09_03.m solves
the eigenvalue problem for this problem and gives the following output:
Natural frequency 1 = 0 Hz
Natural frequency 2 = 106.2585 Hz
Natural frequency 3 = 215.2714 Hz
Note that since this torsional system is unconstrained, the first natural frequency is
zero.
Problem 9.4
Consider the system with all the constraints removed, i.e. there are no gear
U + KU qU =
interactions. Thus (see Equation (9.10)), MU q
QU + QC where QC are
the internal forces that enforce the constraints between the degrees of freedom. The
unconstrained mass and stiffness matrices are
 I g1

 0
 0

MU =  0

 0

 0
 0
 km1
 0

 0

KU =  0
 −km1

 0
 0

0
0
0
0
0
Ig2
0
0
0
0
0
I g3
0
0
0
0
0
Ig4
0
0
0
0
0
0
0
0
0
0
0
I m1
0
Ie
0 0
0 0
0 ke
0 0
0 0
0 − ke
0 0
0
0
0
km 2
0
0
− km 2
0
0
0
0 

0 
0 

0 

0 

0 
I m 2 
−km1 0
0
0
0
− ke
0
0
km1
0
0
ke
0
0
109
0 
0 
0 

− km 2 
0 

0 
km 2 
Let us refer the unconstrained coordinates φ1 , φ2 and φ4 to φ3 . From Figure 9.16 we
can see that φ2 =γ 23φ3 where γ 23 =
−r3 r4 .
−r3 r2 and φ4 =γ 43φ3 where γ 43 =
φ1 =γ12 φ2 where γ12 =
−r2 r1 . In matrix notation
1
−γ 23 0 0
0
1 −γ
0
0 0
12

0
−γ 43 1 0
0
1
−γ 23
0

T
=
E
0
1 −γ12
0
0
−γ 43
0 0   φ1  0 
   
0 0     = 0  or E T qU = 0 . (Equation (9.12). Thus
0 0  φ7  0 
0 0 0 0
0 0 0 0 
1 0 0 0 
Using φ1 =γ12 φ2 and φ2 =γ 23φ3 we have φ1 =γ12 ( γ 23φ3 ) . Let qU = Tq R where
 φ1   γ12 γ 23 0 0 0 
φ   γ
0 0 0 
 2   23
 φ3 
 φ3   1
0 0 0  
  
 φ5 
0 0 0   .
φ4  =  γ 43
φ6
φ   0
1 0 0  
5
  
  φ7 
0 1 0  
 φ6   0
  
0 0 1 
 φ7   0
Thus when we apply constraints the system has 4 degrees of freedom. Note that
γ 43 =
−2.8571 and γ 23 =
ET T = 0 . Note also that in this system γ12 =
−1 . Since
U + KU qU =
MU q
QU + QC , (Equation (9.12)) and qU = Tq R , TT QU = Q R ,
 R + K R q R =
TT QC = 0 and M R = TT MU T etc, we have M R q
Q R This gives
0 0
 5.469 − 0.257 − 4.000 0.257 
0.063 0

 −0.257 0.090



0
0
0
0.180 0 0
5




10 M R
KR =
0

 −4.000 0

4.000 0
0
20 0




0
0.090 
0
0 0.180 
0
 0.257 0
We can determine the system natural frequencies by solving the eigenvalue problem
ω2M R q R 0 =
K R q R 0 . (For the system natural frequencies, see the output from
MATLAB script Problem_09_04.m, below).
Gear error: The gear error is φ2 + φ3 = 1 × 10 −5 sin ( 3000t ) . This can be combined
with the existing constraints by E T qU = e ( t ) (Equation (9.30)) where in this case
1×10−5 


=
e ( t )  0  sin Ωt . (Recall that γ 23 =
−1 ). Let =
qU Tq R + q ref (Equation
 0 


(9.31)). Here q ref is chosen so that E T q ref = e ( t ) . Thus E E T q ref = Ee ( t ) and
(
hence q ref = E ET E
)
−1
e ( t ) (Equation (9.32)).
110
T
In this problem q ref
=
10 4 {−0.143 0.050 0.050 − 0.143 0 0 0} sin Ωt .
U + KU qU =
Substitute =
QU + QC (Equation (9.12)) and
qU Tq R + q ref into MU q
premultiplying by TT , noting that and TT QC = 0 and M R = TT MU T etc. gives
(
)
 R + K R q R =
 ref + KU q ref , Equation (9.33). This can be written
M Rq
Q R − TT MU q
(
)
 ref + KU q ref . In this
 R + K R q R =
as M R q
−TT MU q
Q R + Q Rref where Q Rref =
problem Q R = 0 . Now since q ref is a harmonic,
(
)
ref = −Ω2 q ref and hence Q Rref =
−TT KU − Ω 2 MU q ref . In this problem
q
−2.000 
−0.129 


=
Q Rref 
 sin Ωt where Ω =3000 rad s
 2.000 
−0.129 
qR
Solving Equation (9.33), we have =
(K
2
R − Ω MR
)
−1
Q Rref . In this problem
 0.1316 


−3  −0.0020 
=
q R 10 
 sinΩt
−0.0003 
 0.0022 
Now =
qU Tq R + q ref .and in this problem
=
qUT 10 −3 {0.3618 − 0.1266 0.1366 − 0.3904 − 0.0020 − 0.0003 0.0022} sin Ωt
We can determine τke
= ke ( φ6 − φ3 ) where φ6 etc. is a component of qU . From
Equation (9.12), assuming a harmonic solution and QU = 0 , then the torques due to
the constraints and gear error are =
QC
(K
U
)
− Ω 2 MU qU . This gives, for this
problem QCT =
{−3.238 17.096 36.331 3.494 0 0 0} sin Ωt
Hence f12 = QC1 r1 , f34 = QC 4 r4=
, f23 QC 2 r2 − f12 or
=
f23 QC 3 r3 − f34 .
The MATLAB script Problem_09_04.m solves this problem and gives the following
output.
Natural
Natural
Natural
Natural
frequency
frequency
frequency
frequency
1
2
3
4
=
0.0000
= 32.5860
= 35.5881
= 470.9944
Hz
Hz
Hz
Hz
Max torque on shaft ke = -54.7769 Nm
Max tangential force = 263.4817 N
Problem 9.5
From Equation (9.31) =
qU Tq R + q ref , we know that ET qU = e (Equation (9.30))
and hence ET qU = ET Tq R + ET q ref = ET q ref (since ET T = 0 ).
Thus ET q ref = e . Equation (SM.1)
111
Suppose we have two choices for q ref , denoted by q ref 1 and q ref 0 . Thus from
(
)
Equation (SM.1) ET q ref 1 − q ref 0 = e − e = 0 . Equation (SM.2)
Since ET T = 0 and [ E T] is full rank, we can write q ref 1 − q ref 0 in terms of the
columns of [ E T] . Thus q ref 1 − q ref 0 =
Th + E g Equation (SM.3)
for some vectors h and g . From Equation (SM.2)
(
)
ET q ref 1 − q ref 0 = ET Th + ET E g = 0 Equation (SM.4)
Since ET T = 0 , this implies that ET E g = 0 Equation (SM.5)
Since E is full rank we must have g = 0 and hence q ref 1 − q ref 0 =
Th Equ’n (SM.6).
The two possible q ref vectors will lead to different reduced states q R , denoted by
q R 0 and q R1 . From Equation (9.33) these are given by
(
)
 R1 + K R q R1 =
 ref 1 + KU q ref 1 )
M Rq
Q R − TT ( MU q
 R 0 + K R q R 0 =
 ref 0 + KU q ref 0
M Rq
Q R − TT MU q
Subtracting, and using Equation (SM.6), we have
 + K Th )
 R1 − q
 R 0 ) + K R ( q R1 − q R 0 ) =
M R (q
−TT ( MU Th
U
 − K h
=
−M R h
R
 ) + K ( q − q + h ) =
 R1 − q
 R 0 + h
and hence M R ( q
0 Equation (SM.7)
R
R1
R0
Initial conditions: From Equations (9.31) and (SM.6),
qu1 − qu=
=
0 at t = 0 . (Similarly for
) 0 and thus q R1 − q R 0 + h =
0 T ( q R1 − q R 0 + h
the initial velocities). Then from Equation (9.31)
qu 0= Tq R 0 + q ref 0= Tq R1 + Th + q ref 0
= Tq R1 + q ref 1 − q ref 0 + q ref 0 = qu1
Problem 9.6
The equations of motion can be derived from energy principles or by applying
Newton’s 2nd Law.
q2
f3
m2
q4
q3
f4
m3
m4
For example, considering mass m3 in the free body diagram shown above, then
m3q=
f3 − f 4 where f3 and f 4 are spring forces and are given by
3
f3 =
−k3 ( q3 − q2 ) and
=
f 4 k4 ( q3 − q4 ) . Form these three relationships, and
0 . Applying Newton’s 2nd
rearranging we obtain m3q3 − k3q2 + ( k3 + k4 ) q3 − k4 q4 =
Law to each mass gives the following equation, expressed here in matrix notation
 + Kq =
thus Mq
0 where
112
M = diag [ m1
K=
m2
m3
m4
m5
m6
m7
m8
m9 ] and
− k2
0
0
0
0
0
0
0 
 k1 + k2
 −k
k2 + k3
−k3
0
0
0
0
0
0 
2

 0
−k3
k3 + k4
− k4
0
0
0
0
0 


0
− k4
k 4 + k5
− k5
0
0
0
0 
 0
 0
0
0
− k5
k5 + k6
− k6
0
0
0 


0
0
0
− k6
k6 + k7 + k10
−k7
0
0 
 0
 0
−k7
−k8
0
0
0
0
k7 + k8
0 


−k8
0
0
0
0
0
k8 + k9 −k9 
 0
 0
− k9
k9 
0
0
0
0
0
0

To obtain the first three axial; natural frequencies we must solve the eigenvalue
problem Kq 0 = λMq 0 where λ = ω2n . The MATLAB script Problem_09_06.m
formulates the mass and stiffness matrices from the given data, solves the eigenvalue
problem and gives the following output:
Natural frequency 1 = 1.5479 Hz
Natural frequency 2 = 11.9264 Hz
Natural frequency 3 = 17.9239 Hz
Mode 1
0.01
0.005
0
1
2
3
4
5
6
Axial position
Mode 2
7
8
9
1
2
3
4
5
6
Axial position
Mode 3
7
8
9
1
2
3
4
5
6
Axial position
7
8
9
0.05
0
0.05
0
-0.05
Problem 9.7
Although the constraint is posed in the question in terms of angular velocity the same
constrain must exist in terms of angular position. Since the torques are specified in
terms of shafts 1 and 2, then it is sensible to write the position of shaft 3 in terms of
shafts 1 and 2. Let φ1 =q1 , φ2 =q2 . The constraint is φ1 − φ2 5 + φ3 = 0 . Thus
113
 1 0


φ3 =−q1 + q2 5 . In matrix for this becomes φ = Tq where T =  0 1  . Now
 −1 1 
5

M R = TT M T . Here M = diag [ J1
J2
J3 ] . Expressing the torques applied to shafts
τ 
 = Γ . Thus q
 = M −R1Γ . This
1 and 2 as a vector we have Γ =  1  and hence M R q
 τ2 
gives the accelerations of shafts 1 and 2 only. Hence to obtain the accelerations of all
 . These relatively simple calculations can be done by
three shafts we have 
φ = Tq
hand but the MATLAB script Problem_09_07.m implements these calculations and
gives the following output:
Angular acceleration of shaft 1 = 64.5361 rad/s^2
Angular acceleration of shaft 2 = 371.134 rad/s^2
Angular acceleration of shaft 3 = 9.6907 rad/s^2
Problem 9.8
To refer a inertia from one shat speed to another we consider the kinetic energy of the
inertia. Let I m be the polar moment of inertia of the motor, I g be the polar moment
of inertia of the small gear and I c be the polar moment of inertia of the large gear and
cone combined. Let ωs be the angular velocity of the shaft and ωc be the angular
velocity of the cone. Thus the kinetic energy of the cone and large gear, is =
T
1 I ω2 .
2 c c
Now, ωc = gωs where g is the gear ratio and in this case is equal to 1 6 . Thus the
kinetic energy of the cone and large gear can be expressed =
as T
1 I g 2 ω2 .
s
2 c
Hence
the polar moment of inertia of the cone and large gear is I c g2 when it is referred to
the shaft, i.e. it is assumed to be rotating at the same speed as the shaft.
If the torque acting on the crusher cone is τc , then the work done by this torque is
τc φc where φc is the angular displacement of the cone. Since φc = gφs the work
done by the torque is τc gφs when related to the shaft. The stiffness of the shaft can be
derived from =
1 k 1 / k1 + 1 k2 . The polar moment of inertia of the cone and large
gear combined is the sum of the polar moments of inertia of the individual
components. Thus, relating the cone inertia and torque to the shaft, and applying
Newton’s 2nd Law it is easy to show that M
φ + Kφ =
Γ where
0
 Im

 0 
 k −k 
and
Γ
=
K
=
M=
,

 .


2
 0 I m + g I c 
 −k k 
 τc g 
The MATLAB script Problem_09_08.m implements these calculations and gives the
following output:
Referred inertia of cone and large gear = 50kg m^2
equivalent shaft stiffness = 1.2MN/rad
Natural frequency = 28.4705 Hz
Mode shape = [-0.040825 0.12247]
114
Peak
Peak
Peak
Peak
Peak
shaft
shaft
shaft
shaft
shaft
torque
torque
torque
torque
torque
at
at
at
at
at
10
20
30
40
50
Hz
Hz
Hz
Hz
Hz
=
=
=
=
=
427.7746
740.3462
3398.879
385.0415
179.9207
Nm
Nm
Nm
Nm
Nm
Problem 9.9
Let ne is the number of elements and the coordinates are numbered 1 to ne + 1 . The
driving machine is located at coordinate 1 and the drill bit is at coordinate ne + 1 . We
can model the drill pipe using the stiffness and inertia elements given in Equations
(4.45) and (4.48) respectively. Using these matrices we can assemble the inertia
matrix (M) and the stiffness matrix (K) for the drill string. The inertia of the driving
machine must be added to the inertia matrix at coordinate 1 and the inertia of the drill
bit is added to the inertia matrix at coordinate ne + 1 . This analysis has been
implemented in the MATLAB script Problem_09_09.m and gives the following
output:
First five natural Frequencies (Hz)
5 elements
8 elements 50 elements
0.0000
0.0000
0.0000
1.9417
1.9225
1.9106
4.0719
3.9187
3.8232
6.5437
6.0625
5.7394
9.1547
8.4185
7.6613
Problem 9.10
 + Kq =
The equations of motion for this system are Mq
Q ( t ) where M and K are
given in Problem 9.6. We now need to determine the vector of forces acting on the
system. The axial force due to the cylinders firing once per cycle is
=
f ( t ) 6, 000 cos (10 πt ) + 1, 500 cos ( 30 πt ) − 500 sin ( 30 πt )
Only force terms at a frequency of 15 Hz (i.e. 30π ) will excite the system at 15 Hz,
f ( t ) a cos ( 30 πt ) + b sin ( 30 πt ) where
i.e. at this frequency the force is effectively =
a = 1, 500 and b = −500 . Due to the firing timing of each cylinder the axial force
separating the masses either side of the cylinder are
=
f ( t ) a cos ( 30 πt − nα ) + b sin ( 30 πt − nα ) , where α = 2 π 5 (i.e. 72° ). and n
depends on the cylinder number, see below. Expanding this force function gives
=
f ( t ) {a cos ( nα ) + b sin ( nα )} cos ( 30 πt ) + {a sin ( nα ) − b cos ( nα )} sin ( 30 πt ) .
When the system runs at 5 Hz (i.e. 10π ) then only the terms at this frequency excite
=
f ( t ) a cos (10 πt ) where a = 6, 000 . Due to
the system, i.e. the force is effectively
f ( t ) a cos (10 πt − nα ) where n
the firing timing of each cylinder the force will be=
depends on the cylinder number. Thus
=
f ( t ) {a cos ( nα ) + b sin ( nα )} cos (10 πt ) + {a sin ( nα ) − b cos ( nα )} sin (10 πt ) where
in this case, b = 0 .
Since the firing order is 1-3-5-2-4 then n = 0, 1, , 4 . Thus, for cylinders 1, 2, , 5 the
values of n are 0, 3, 1, 4 and 2 in that order.
Each coordinate (corresponding to a mass) has a force acting on each side.
115
q2
q4
q3
f2
f3
m2
m3
m4
For example, the axial force acting at coordinate 3 is Q=
f3 ( t ) − f 4 ( t ) and hence
3 (t )
Q
=
3 (t )
{a cos ( 3α ) + b sin ( 3α )} cos (10 πt ) + {a sin ( 3α ) − b cos ( 3α )} sin (10 πt )
Si
− {a cos ( α ) + b sin ( α )} cos (10 πt ) − {a sin ( α ) − b cos ( α )} sin (10 πt )
mplifying this expression
=
Q3 ( t ) a ( cos ( 3α ) − cos ( α ) ) + b ( sin ( 3α ) − sin ( α ) ) cos ( Ωt )
{
}
{
}
+ a ( sin ( 3α ) − sin ( α ) ) − b ( cos ( 3α ) − cos ( α ) ) sin ( Ωt )
=
or Q
3 ( t ) Q3C cos ( Ωt ) + Q3S sin ( Ωt ) .
No forces act at coordinates 7, 8 and 9. The response is determined from
q=
C
(K − Ω M )
2
−1
QC and q=
C
(K − Ω M )
2
The response at the rim (coordinate 9) is=
q9
the rim and hub (coordinate 8) is=
F89
−1
QC
qC2 9 + qS29 . The axial force between
FC 89 k9 ( qC 9 − qC 8 )
FC289 + FS289 where=
and=
FS 89 k9 ( qS 9 − qS 8 ) .
The MATLAB script Problem_09_10.m determines the system response at the rim
and the axial force between the hub and rim. The output is as follows:
At 5 Hz, Axial reponse at gen rim = 5.9878 mu_m
Axial force between gen hub and rim = 130.0146 N
At 15 Hz, Axial reponse at gen rim = 5.0966 mu_m
Axial force between gen hub and rim = 995.9699 N
116