rtiwari rd book 01a
TRANSCRIPT
-
8/10/2019 Rtiwari Rd Book 01a
1/21
-
8/10/2019 Rtiwari Rd Book 01a
2/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
2
When the rotor is not eccentric, however, a small imbalance mass, mi is attached at a relatively larger
radius of rias shown in Figure 1.2(c), the imbalance force can be written as
trmtF ii sin)(2= (1.2)
For the case when the rotor is eccentric and a small imbalance mass is attached as shown in Figure 2(d),
the imbalance force will be
)sin(sin)( 22 ++= trmtemtF ii (1.3)
No imbalance
Fig 1.2(a)Rotor geometrical centre and
centre of gravity coincident
Fig 1.2(b)Rotor geometrical centre and centre of
gravity not coincident
Imbalance force =mass of rotor eccentricity
square of spin speed
Imbalance force =
mass of rotor eccentricity
square of spin speed
Fig 1.2(c)Rotor geometrical centre, centre of gravity
and an additional imbalance mass
Fig 1.2(d)Rotor geometrical center, centre of gravity
and imbalance mass
Imbalance force is the vectoraddition of forces due to the rotor
and imbalance forces
Fig 1.1(a)
A rigid rotor mounted on flexible bearings
bkbk
2effk k=
Fig 1.1(b)
A flexible rotor mounted on rigid bearings
348 /effk EI L=
F(t)
Fig 1.1(c)An equivalent single degree of freedom
spring-mass system
effk
F(t)
Fig 1.1(d)Free body diagram of disc mass
effk y
y
-
8/10/2019 Rtiwari Rd Book 01a
3/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
3
where is the phase difference between the vectors of imbalance forces due to the rotor eccentricity and
the imbalance mass. Figure 1.3 shows the unbalance location on a rotor system. For a constant angular
velocity of the rotor, , the location of the unbalance is given as t = .
On application of the Newtons law on the free body of the rotor mass as shown in Figure 1.1(d), i.e.
equating sum of external forces to the mass of the rotor multiplied by the acceleration of the center of
gravity of the rotor mass, we have
ymtemykeff =+ sin2
(1.4)
where effk is the effective stiffness of the rotor system (see Figure 1.1). Equation (1.4) is a standard
equation of motion of a single DOF spring-mass system and can be written as
temykym eff sin2=+ (1.5)
From the free vibration of the when the external imbalance force is absent, the rotor mass will be having
oscillation and that will be given by
)sin()( tYty n= (1.6)
where n is the frequency of oscillation during the free vibration and that is called the natural frequency
of the system. On substituting equation (1.6) into the homogeneous part of equation of motion (1.5), it
gives
0)sin()( 2 =+ tYkm neffn (1.7)
For the non-trivial solution of equation (1.7), the natural frequency of the system can be written as
Figure 1.3 Unbalance location on a rotor
o
At time, t= 0
C G
o
At time, t
G
C
-
8/10/2019 Rtiwari Rd Book 01a
4/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
4
mkeffnf /= (1.8)
The steady state forced response can be modeled as
)sin()( = tYty (1.9)
where Yis the amplitude of displacement and is the phase lag of the displacement with respect to the
imbalance force. On substituting equation (1.9) into equation (1.5), the steady state forced response
amplitude can be written as
[ ]
2 2
2 2
( ) sin( ) sin( )
or
( ) sin cos cos sin sin( )
eff
eff
m k Y t me t
m k Y t t me t
+ =
+ =
On separating the sine and cosine terms, we have
2 2 2( ) cos and ( ) ( sin ) 0eff eff m k Y me m k Y + = + =
Since the term2( )effm k + in general may not be zero, from the second expression we have 0=
and the first expression can be simplified as
2 2 2
2 2 2 2
/1eff nf
mY Y e
k m
= = = =
with / nf = (1.10)
where Y is the non-dimensional unbalance response (ratio of the unbalance response to the eccentricity)
and is the frequency ratio (ratio of the spin speed of the rotor to the natural frequency of the rotor
system). The non-dimensional unbalance response (absolute value) is plotted with respect to the
frequency ratio as shown in Figure 1.4. The response changes its sign after critical speed, which means
that shaft deflects in the opposite direction to the unbalance eccentricity. Both linear and semi-log plot is
shown to have clarity of response variation. It can be seen that as the frequency ratio increases the non-
dimensional response asymptotically approaches to unity. It means unbalance response approaches to theeccentricity of the rotor. Physically it implies that the rotor rotates about its center of gravity at high
frequency ratio. From Figure 3 and equation (1.10) it should be noted that we have unbounded unbalance
response when the denominator2(1 ) becomes zero i.e when the spin speed is
1cr = or /cr eff nf k m = = (1.11)
-
8/10/2019 Rtiwari Rd Book 01a
5/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
5
This is a resonance conditionand the spin speed corresponding to the resonance is defined as critical
speed. The subscript: crrepresent the critical. For the present case the critical speed is equal to the natural
frequency of the system as given in equation (8), the undamped steady state forced response amplitude
tends to infinity. The natural frequency concept has come from free vibrations and critical speed fromforced vibrations. It should be noted that in rotor dynamics in general the critical speed and the natural
frequency might not be same (e.g. when gyroscopic couple is considered in the analysis). The sign
represent that the rotor will have critical speed while rotating in either clockwise or counter clockwise.
Since damping is not considered in the analysis phase angle, , becomes zero (or 1800after the critical
speed). The analysis presented in this section can be applied to the transverse, torsional and axial
vibrations of rotors and accordingly critical speed can be termed by prefixing respective names of
vibrations. For torsional vibration care should be taken that mass will be replaced by the polar mass
moment of inertia of rotor and stiffness will be torsional stiffness. Similarly, for axial vibration mass willremain same as transverse vibration, however, the stiffness will be the axial stiffness.
(a) Linear plot (b) Semi-log plotFigure 1.4 Non-dimensional unbalance response versus frequency ratio
% Non-dimensional unbalance response wrt to freq ratio "Figure_1_4.m"
% Copywriters: Dr R Tiwari, Dept of Mechanical Engg., IIT Guwahati.% 13-01-2005clear all;deta_freq=0.005;
freq_ratio(1)=deta_freq;N_pt=1000;
for ii = 1:1:N_pty_resp(ii)=freq_ratio(ii)^2/(1-freq_ratio(ii)^2);if(ii
-
8/10/2019 Rtiwari Rd Book 01a
6/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
6
freq_ratio(ii+1)=freq_ratio(ii)+deta_freq;end
endfigure(1)
plot(freq_ratio,abs(y_resp), 'k-', [0 5], [1 1], 'k--');xlabel('w/wn ---->');
ylabel('y/e ---->');figure(2)semilogy(freq_ratio,abs(y_resp), 'k-', [0 5], [1 1], 'k--');
xlabel('w/wn ---->');ylabel('y/e ---->');
Unbalance will always be present in a rotor. However, the unbalance response can be reduced by the
following methods.
(i) Correction at source i.e. balancing the rotor: Balancing the rotor is the most direct approach, since it
attacks the problem at source. However, in practice a rotor cannot be balanced perfectly and that the best
achievable state of balance tends to degrade during operation of a rotor (e.g. turbomachinery). There are
two type of unbalances (a) static unbalance: The principal axis of the polar mass moment of inertia of the
rotor is parallel to the centerline of the shaft as shown in Figure 1.5a. The rotor can be balanced by a
single plane balancing and (b) dynamic unbalance: The principal axis of the polar mass moment of inertia
of the rotor is inclined to the centerline of the shaft as shown in Figure 1.5b. For balancing such rotors
minimum of two planes are required.
(a) Perfectly balance (No force and moment) (b) Static unbalance (pure radial force)
(c) Dynamic unbalance (pure moment) (d) Dynamic unbalance (both force and moment)
Figure 1.5 Classification of unbalances for a short rigid rotor
G
F
G
GG
-
8/10/2019 Rtiwari Rd Book 01a
7/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
7
(ii) Operate rotor away from the critical speed: i.e. during design itself or during operation by providing
temporary auxiliary support: Moving the machine operating speed farther away from the critical speed
can be achieved by changing the rotor operational speed or by changing the critical speed itself. The
critical speed can be changed either at the design stage or during operation. At design stage changingrotor mass or its distributions and dimensions of the rotor and its support lengths can alter the critical
speed. During operation auxiliary support can be provided to increase the effective stiffness of the rotor,
which in turn increases the critical speed. By this arrangement the actual rotor critical speed can be safely
traversed and then the auxiliary support can be withdrawn which brings the critical speed of the rotor
below the operation speed. In general, changing the critical speed is useful for constant operational speed
machines or for machines with a narrow range of operational speed.
(iii)Add damping to the system or active control of the rotor: If a critical speed must be traversed slowly
or repeatedly, or if machine operation near a critical speed can not be avoided, then the most effective
way to reduce the amplitude of the synchronous whirl is to add damping. On the other hand the damping
(e.g. the shaft material or hysterisis or internal damping) may lead to rotor instability (self excited
vibration). The squeeze film and magnetic bearings are often used to control the dynamics of such
systems. Squeeze-film bearings (SFB) are, in effect, fluid-film bearings in which both the journal and
bearing are non-rotating. The ability to provide damping is retained but there is no capacity to provide
stiffness as the latter is related to journal rotation. They are used extensively in applications where it is
necessary to eliminate instabilities and to limit rotor vibration and its effect on the supporting structures of
rotor-bearing systems, especially in aeroengines. In recent years, advanced development of
electromagnetic bearing technology has enabled the active control of rotor bearing systems through active
magnetic bearings (AMB). In particular the electromagnetic suspension of a rotating shaft without
mechanical contact has allowed the development of supercritical shafts in conjunction with modern
digital control strategies. With the development of smart fluids (for example electro and magneto-
rheological fluids) now new controllable bearings are in the primitive development stage. The basic
premise of such smart fluids that their dynamic properties (i.e. damping and stiffness) can be controlled
by changing current or magnetic flux in micro-second time. Schematics of typical passive and active (i.e.
smart or controllable) squeeze film dampers and active magnetic bearings are shown in Figure 1.6.
-
8/10/2019 Rtiwari Rd Book 01a
8/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
8
Figure 1.6 (a) Schematic diagram of squeeze film dampers
Figure 1.6 (b) Smart (active) fluid-film dampers
Figure 1.6 (c) Basic principle of active magnetic bearings
Bearing bush
Outer raceway of rolling
bearing (can displace
radially and constraint
not to rotate.Squeeze film
Rotor
Oil feed groove
Rotor
Rolling bearing
Electrodes
Teflon
Sensor
Rotor
Power Amplifier Electromagnet
Controller
-
8/10/2019 Rtiwari Rd Book 01a
9/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
9
Example 1.1: A rotor has a mass of 10 kg. The operational speed of the rotor system is at (100 1) Hz.
What should be the bounds of rotor effective stiffness so that the critical speed should not fall within 5%
of operating speeds? Assume there is no damping in the rotor system.
Solution:Operation speed range: 99 to 101 Hz
5% of lower operational speed: 99-990.05=99(1-0.05)=990.95=99.05 Hz
5% of upper operational speed: 101+1010.05=101(1+0.05)=1011.05= 106.05 Hz
The effective stiffness corresponding to lower operating speed = 99.05210 = 98.1 kN/m
The effective stiffness corresponding to upper operating speed = 106. 05210 = 112.5 kN/m
So the effective stiffness should not fall in he range of 98.1 to 112.5 kN/m.
Example 1.2: A rotor has 100 Hz as critical speed and its operating speed is 120 Hz. If we want to avoid
altogether crossing of the critical speed then what should be the increase in the support stiffness by
auxiliary support system. We should have at least 5 Hz of gap between the operating speed and the
critical speed. The rotor has a mass of 10 kg.
Solution: Initial stiffness of the support = 100210 = 100 kN/m
First we will reach 95 Hz of rotor speed, which makes 5 Hz of gap with original critical speed of the
rotor. Now since we cannot increase the rotor speed further we need to increase the critical speed of the
rotor to al least 125 Hz.
Hence the corresponding effective of the support stiffness should be = 125210 = 156.25 kN/m.
Hence the auxiliary support system should increase the effective stiffness by 56.25 kN/m.
1.2 Rankine Rotor Model
The single DOF rotor model has limitations that it cannot represent the orbital motion of the rotor in two
transverse directions. Rankine (1869) used a two DOF model to describe the motion of the rotor in two
transverse directions as shown in Figure 1.7(a). The shape of orbit produced depends upon the relative
amplitude and phase of the motions in two transverse directions and the orbit could of circular, elliptical
or straight line, inclined tox andyaxis, as shown in Figure 1.8. However, as shown in Figure 1.7(b) the
free body diagram of the rotor, for the constant spin speed the radius of the whirling of the rotor center
-
8/10/2019 Rtiwari Rd Book 01a
10/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
10
will increase parabolically and will be given as kFr c /= whereFcis the centrifugal force and it can be
physically also visualized as there will not be any resonance condition, as found in the single DOF model,
when the spin speed is increased gradually. This is a serious limitation of the Rankine model. Moreover,
this model does not represent the realistic rotating imbalance force.
1.3 Jeffcott Rotor Model
Figure 1.9 shows a typical Jeffcott rotor. It consists of a simply supported flexible massless shaft with a
rigid disc mounted at the mid-span. The disc center of rotation, C, and its center of gravity, G, is offset by
a distance, e. The shaft spin speed is, and the shaft whirls about the bearing axis with whirl frequency,
. For the present case synchronous whirl is assumed (i.e. = ). The synchronous motion occurs
between the earth and the moon. Because of the synchronous motion the moon orbital speed and its own
spin speed are equal as shown in Figure 1.10(a). The unbalance force in general leads to synchronous
whirl conditions. Other kinds of whirl condition, which can occur in real system, are: anti-synchronous
(i.e. = ) and asynchronous (i.e. ) as shown in Figure 1.10(b). The antisynchronous whirl
occurs when there are rubs between the rotor and stator, however, very rarely it occurs. Asynchronous
condition occurs when the speed conditions are high (e.g. when gyroscopic effects are predominates) or
Fig 1.7(b) Free body diagram of the
model
Fr
Fig 1.7(a) Rankine rotor model
(Two degree of freedom spring-
mass rotor model)
Fig 1.8(a)
Circular motion
efk
Fig 1.8(c)
Straight line motion
efk
Fig 1.8(b)
Elliptical motion
efk
-
8/10/2019 Rtiwari Rd Book 01a
11/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
11
when rotor is asymmetric or when the bearing is anisotropic in dynamic properties. The stiffness of the
shaft is expressed as
3load/deflection 48 /k EI L= = (1.12)
Coordinates to define the position of the center of rotation of the rotor are xu and yu . The location of the
imbalance is given by . Thus, three dofs are needed to define the position of the Jeffcott rotor.
Figure 1.10(a) Synchronous whirl Figure 1.10(b) Anti-synchronous whirl
From Figure 1.9(c) the force balance in xu , yu and directions can be written as
( )cos2
2
eudt
dmucku xxx += (1.13)
Fig 1.9(b)A Jeffcott rotor model iny-zplane
Fig 1.9(a)
A Jeffcott rotor model
v
CG e=
Fig 1.9(c)Free body diagram of the
disc inx-yplane
Shaft spin
direction
Shaft whirling
direction
Shaft
Shaft spin
direction
Shaft whirling
direction
Shaft
-
8/10/2019 Rtiwari Rd Book 01a
12/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
12
( )sin2
2
eudt
dmmgucku yyy += (1.14)
and
dImge = cos (1.15)
Apart from restoring force contribution from the shaft, the damping force is also considered. The damping
force is idealized as viscous damperand it is mainly coming from the support and aerodynamic forces at
disc. The material damping of the shaft will not contribute viscous damping and it may leads to instability
in the rotor and that will be considered in detail latter.
For the case = ti.e. when the disc is rotating at constant spin speed, the Jeffcott rotor model is reduces
to two DOF rotor model. Neglecting the effect of gravity force, equations of motion in thexandycan be
written as
( )teudt
dmucku xxx cos2
2
+= (1.16)
and
( )teudt
dmucku yyy sin2
2
+= (1.17)
Equations of motion can be written in the standard from as
temkuucum xxx cos2=++ (1.18)
temkuucum yyy sin2=++ (1.19)
It should be noted that equations of motion are uncoupled and motion can be analysed independently in
two transverse planes. Noting equation (1.8), from the undamped free vibration analyses it can be seen
that since the rotor is symmetric rotor will be having two natural frequencies that are equal and given as
mknf /2,1 = (1.20)
The damping does not affect the natural frequency of the system appreciably. However, their effect is
more predominate for suppressing the resonance amplitude. By extending the previous method the steady
state forced response can be written as
-
8/10/2019 Rtiwari Rd Book 01a
13/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
13
[ ]
cos( )
cos ( ( / 2 ) sin( )
x x
y y y
u U t
u U t U t
=
= + = (1.21)
where Uxand Uyare the steady state forced response amplitudes in the xandydirections, respectively.
is the phase lag of the displacement with respect to the imbalance force. The phase difference between thetwo direction responses will be of 900as two directions are orthogonal to each other. For the direction of
whirling shown in Figure 1.9 i.e. counter clockwise (ccw) for the present axis system the response in y
direction will lag behind xdirection response by /2 radians. Hence the lag of the ydirection response
with respect to the force will be (/2 + ). On taking the first and second derivatives of the response with
respect to time, t, we get
)cos(
)sin(
=
=
tUu
tUu
yy
xx
and
)sin(
)cos(
2
2
=
=
tUu
tUu
yy
xx
(1.22, 1.23)
On substituting equations (1.21) to (1.23) into equation (1.18) and separating the in-phase (i.e. cost) and
quadrature (i.e. sint) terms, we get
emkUcUUm xxx22
cossincos =++ (1.24)
0sincossin2 =+ xxx kUcUUm (1.25)
Equation (1.25) gives
2tan
mk
c
= (1.26)
which gives
( ) ( )222sin
cmk
c
+= and
( ) ( )222
2
cos
cmk
mk
+
= (1.27, 1.28)
Substituting equations (1.27) and (1.28) into equation (1.24), we get
( ) ( )222
2
cmk
emUx
+= (1.29)
Similarly, we can obtain response amplitude inydirection from equation (1.19) as
-
8/10/2019 Rtiwari Rd Book 01a
14/21
-
8/10/2019 Rtiwari Rd Book 01a
15/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
15
2tan
mk
c
= (1.39)
On substitution of phase from equations (1.39) to (1.37) the whirl amplitude can be written as
( ) ( )222
2
cmk
emUr+
= (1.40)
Equations (1.39) and (1.40) are similar to previous results i.e. equations (1.26) to (1.30). The non-
dimensional form of equations (1.39) and (1.40) can be written as
21
2tan
= (1.41)
( ) ( )222
2
21/
+== eUU rr (1.42)
with
kmcccmk ccnn 2;/;/;/ ==== (1.43)
where is the frequency ratio, n is the natural frequency of non-rotating rotor, is the damping ratio
and ccis the critical damping of the system for which the damping ratio is equal to unity. Figure 1.11(a)
shows that the maximum amplitude occurs at slightly higher frequency than the n when damping is
present in the system, however maximum amplitude occurs at n for the undamped case. The increase in
the damping results in increase in the critical speed, however damping is the most important parameter for
reducing the whirl amplitude at critical speed. Since the measurement of amplitude of vibration at critical
speed is difficult, hence the determination of precise critical speed is difficult. To overcome this problem
the measurement of the phase is advantageous at least to determine the undamped natural frequency of the
system.
-
8/10/2019 Rtiwari Rd Book 01a
16/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
16
% Computer code for generating Jeffcott Rotor Response (Figure 1.11)% Copywrites: Dr R Tiwari, Dept of Mechanical Engg., IIT Guwahati, 16-1-2005clear all;
jj=0;zeta=[0.0 0.01 0.05 0.1 0.4 1.0 2.0 10.0 100.0 10000.0];
for jj = 1:1:length(zeta);freq_ratio(1)=0;ii=1;
for ii = 2:1:300freq_ratio(ii)=freq_ratio(ii-1)+0.01;
phase(ii,jj)=(180/pi)*atan2(2*zeta(jj)*freq_ratio(ii),(1-freq_ratio(ii)^2));resp_ratio(ii,jj)=freq_ratio(ii)^2/sqrt((1-freq_ratio(ii)^2)^2+(2*zeta(jj)*freq_ratio(ii)));ii=ii+1;
endend
figure(1)plot(freq_ratio, phase);%title('Phase verus frequency ratio plot');xlabel('Frequnecy ratio');ylabel('Phase');
figure(2)%semilogy(freq_ratio, resp_ratio);
plot(freq_ratio, resp_ratio, [1 1], [0 10], 'b-');%title('Non-dimensional response verus frequency ratio plot');axis([0 freq_ratio(end) 0 10]);
xlabel('Frequency ratio');ylabel('Non-dimensional Response');
Fig 1.11(a)Variation of the non-dimensional
response versus frequency ratio for
different damping ratios
Fig 1.11(b)Variation of the phase versus frequency
ratio for different damping ratios
-
8/10/2019 Rtiwari Rd Book 01a
17/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
17
As it can be seen from Figure 1.11(b) the phase angle is 900at frequency n even for the case of damped
system. For lightly underdamped system the phase angle changes from 0 0 to 900 as the spin speed is
increased and becomes 1800 as the spin speed is increased to higher frequency ratio. For very high-
overdamped system the phase angle always remain at 900before and after the resonance, which may be a
physically unrealistic case. As the spin speed crosses the critical speed the center of the mass of the disc
of Jeffcott rotor comes inside of the whirl orbit and rotor tries to rotate about the center of gravity. As can
be seen from the graph at the spin speed approaches infinity the displacement of the shaft tends to the
equal to the disc eccentricity. The change in phase between the force and the response is also shown in
Figure 7 for three difference spin speeds i.e. below critical speed, at critical speed and above the critical
speed. Since for the present analysis the synchronous whirl condition is assumed, at a particular speed
shaft will not be having any flexural vibration and it (in a particular bend configuration) will be whirling
(orbiting) about its bearing axis as shown in Figure 1.10(a). It can be seen that the black point in the shaft
will be having compression during the whirling. However, it can be seen from Figure 1.10(b) for
antisynchronous whirl that the shaft (the black point in shaft) will be having reversal of the bending
stresses twice per whirling (orbiting) of the shaft. For asynchronous whirl the black point inside the shaft
will take all the time different positions during whirling of the shaft.
Fig 1.12(b)Phase angle between the force and
response vectors ( )n =
Fig 1.12(c)Phase angle between the force and
response vectors ( )n >
Fig 1.12(a)Phase angle between the force and
response vectors ( )n <
CG e=/ 2
imbF imbF
imbF
-
8/10/2019 Rtiwari Rd Book 01a
18/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
18
Matrix Method
With the development in software, which can handle complex matrices, the following procedure may be
very helpful for numerical simulation of even the very complicated rotor systems also. Equations (1.18)
and (1.19) can be combined in the matrix form as
=
+
+
tem
tem
u
u
k
k
u
u
c
c
u
u
m
m
y
x
y
x
y
x
sin
cos
0
0
0
0
0
02
2
(1.44)
The force vector in equation (1.44) is expressed as
( )( )
( )
( )
222
22 2
cos sincosRe Re Re
sin cossin
x
y
j t
imb j t
j timb
me e Fm e t j t m e te
Fm e t j t m e t me je
+ = = =
(1.45)
xy imbimb jFF = (1.46)
where the Re(.) represents the real part of the quantity inside the parenthesis andximb
F andyimb
F are the
imbalance force components in x and y directions respectively. On substituting equation (1.45) into
equation (1.44) and henceforth for brevity the symbol Re(.) will be removed and it can be written as
tj
imb
imb
y
x
y
x
y
x
eF
F
u
u
k
k
u
u
c
c
u
u
m
m
y
x
=
+
+
0
0
0
0
0
0
(1.47)
The relationship (1.46) is true for the present axis system and the direction of whirling of the imbalance
force vector chosen as shown in Figure 1.13(a). For this case theyimb
F lag behind theximb
F by 900. Let us
derive this relationship: Ifximb
F Fe = , then
[ ]( / 2) ( / 2) cos( / 2) sin( / 2)y x
j j j j j
imb imbF Fe Fe e Fe j jFe jF = = = + = =
For the direction of whirl opposite to the axis system, as shown in Figure 1.13(b), the following
relationship will hold
[ ]( / 2) ( / 2) cos( / 2) sin( / 2)y x
j j j j j
imb imbF Fe Fe e Fe j jFe jF += = = + = =
so that,
-
8/10/2019 Rtiwari Rd Book 01a
19/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
19
xy imbimb jFF = (1.48)
in which case theyimb
F lead theximb
F by 900. It should be noted in equation (1.49) that the right hand
side force vector elements have significance of real parts only, which is quite clear from the expandedform of the force vector in equation (1.45). Equation (1.47) can be written in more compact form as
[ ]{ } [ ]{ } [ ]{ } { } tjimb eFuKuCuM =++ (1.49)
The solution can be chosen as
{ } { } tjeUu = (1.50)
where the vector {U}elements are, in general, complex.
The above equation gives
{ } { } { } { }2 andj t j tu j U e u U e = = (1.51)
On substituting equations (1.50) and (1.51) into equation (1.49), we get
[ ] [ ] [ ]( ){ } { }imbFUCjKM =++ 2 (1.52)
The above equation can be written as
[ ]{ } { }imbFUZ = (1.53)
with
[ ] [ ] [ ] [ ]( )CjKMZ ++= 2 (1.54)
Fig 1.13(a)The direction of whirl same as the
positive axis direction
Fig 1.13(b)
The direction of whirl opposite to the
positive axis direction
t
v
t
v
imbxFimbxF
imbyFimbyFimbFimbF
y xF jF= y xF jF=
-
8/10/2019 Rtiwari Rd Book 01a
20/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
20
where [Z] is the dynamic stiffness matrix. The response can be obtained as
{ } [ ] { }imbFZU1
= (1.55)
The above method is quite general in nature and it can be applied to multi-dof systems once equations of
motion in the standard form are available.
Example 1.3: Obtain the response of a rotor system with the following equations of motion.
temkuum xx cos2=+ and temkuum yy sin
2=+
Solution: The first equation can be written as
tj
xxx eFkuum =+ with 2meFx =
in which the real part of the right hand side term has meaning. The solution can be assumed as
tj
xx eUu =
where in general Uxis a complex quantity. The above equation gives
tj
xx eUu 2=
On substituting in equation of motion, we get
( ) 22 mekUUm xx =+
which gives
2
2
mk
meUx
=
Hence the solution becomes
2 2 2
2 2 2Re (cos sin ) cosj tx
me me meu e t j t t
k m k m k m
= = + =
Similarly for the second equation of motion can be written as
tj
yyy eFkuum =+ with 2jmeFy =
in which the real part of the right hand side term only has meaning. The solution can be assumed as
-
8/10/2019 Rtiwari Rd Book 01a
21/21
Dr R Tiwari, Associate Professor, Dept. of Mechanical Engg., IIT Guwahati ([email protected])
tj
yy eUu =
where in general Uyis a complex quantity. The above equation gives
tj
yy eUu 2=
On substituting in equation of motion, we get
( ) yyy FkUUm =+ 2 which gives 2mkF
U y
y
=
Hence the solution becomes
2 2
2 2 2
2 2
2 2
(cos sin )
Re ( cos sin ) sin
y j t j t
y
F jme jmeu e e t j t
k m k m k m
me mej t t t
k m k m
= = = +
= + =
(Answer)