Dynamics of Machines - Forced Vibration - Problems.pdf

nines Forced VIbralion7
4.39
9.12 FORCETRANSMISSIBILITY AND VIBRATIONISOLATION
situations. They shoul
SOLATIOON
9.12 FORCE
Sin ot
FT
C.oX
Except in a few cases, vibrations are not desirable in many situations T
be eliminated oratleast reduced. In some applications, the rotating/ reciprocad
smission to adjacen
ocatingpe
should be isolated from the foundation to prevent vibration
transmission to aT
isolated from the
support
90°
structure. Some precise or delicate instruments need to be isolated from the s
SX
orts
which may be subjected to certain vibrations.
9.12.1 Types of Isolation
777mnm
sof isolation
lfthe transmissionofforce is prevented/reduced then the effectivenessoficnl a) Vector diagram of the forces
b) Force Transmissibility
is known as force isolation.
Fig.9.11
Similarly, the effectiveness of isolation measured in terms
ofmotiontransmitted:.
The exciting force is F sin ot. Under steady state conditions, we can represent
called as motion isolation.
the forces acting on the mass as a vector diagram.
Thelesserthe force/motion transmitted, the greater will be the isolation.
The forces acting are:
9.12.2 Isolating Materials
Springforce,sX
Dampingforce, C.o.X
1.
One ofthe methods ofreducing the transmitted vibrations is to place proper isolation 2.
material in between the vibrating body and the support. Afew isolating materials are:
3. Exciting force, F, sinor and,
)Rubberpads or slabs
Centrifugal force, m Xo
4.
(i) Felt
The force transmitted to the foundation is the vector sum ofspring force and damping
i) Cork
force. These forces are 90 out of phase with each other and their vector sum
(iv) Metallic springs
Fy =ysx)3 + (CoX? = X ys*+(Co)*
(v) Spring washers
Substituting the value of X obtained in earlier analysis;
They all are elastic in nature with good damping properties.
9.12.3 Force Transmissibility
E s+(Co)
F
s mo) + (Co)*
Transmissibility is the ratio of the force transmitted to the foundation io hat
impressed (acting ) uon the system
Rewriting the equation,
V1+(2
ie, Force Trans nissibility=orCetransmitted to the foundation
Force acting on the system Transmissibility, Tr =
-+(2*|
Let us consider a mass m
supported on a
foundation by means ofan isolato
an equivalent stiffness ofsand
damping coefficient C as shown.
having

es
ist-a)vhe The ioure 9.12, shows the plot oftransmissibilityVs frequency curve, for diferent
4.41
Iheangleoflagbetveen the transnitteri fone to the impressed force is
Value's
sofdamping ratio, =
tan a = 2 r
(2)
(1-) Thes
salient points are:
All curves start from unity value
oftransmissibility(Tr= 1) and pass through the
and. tan
impressed force
( a ) -Angleoflagbetween the transmitted force and the impree
(applied fore) unit transmissibility at frequency ratio,= y2
),,
-Phaseangleofapplied foree with respect to the displacement
a-Phaseangleoftransmitted foree with respect to the displacement. As approaches infinity, the transmissibility tends to be approaching zero.
300
is large, the isolation is effective. Generally, larger mass gives low natural
frequency and consequently higher value of Damping in this region reduces
the effectiveness ofisolation.
Transmission
I f is small, the transmissibility increases.Larger stiffnessgives high value of
200 natural frequency and consequentlylowervalues of
Isolationiseffectiveif lie, o,<<0.
9.12.4 Motion Transmissibility or Amplitude Transmissibility
Isolation
Motion (or) amplitudetransmissibility (or) Absolute amplitudeofthemass
Displacement transmissibility Amplitude of base excitation
100
If, Xis the steady state absolute amplitude ofthemass
andYisthe amplitude of
the base excitation,
Damoine Ratio
C =1.0
I+(2) and
Motion transmissibility,
- (2 C
25r
tano=
-
1.0 2.0 4.0 tan a 2 r
3.0
FREQUENCY RATIO
Fig.9.12 Transmissibility Curve
(-a)phase angle orangleoflag.

4.42 Dynamics of Machin
s of Machines
EorcedVibration
4.43
Example :9.19 Example: 9.20
having a
total
stiffness
Amachine ofmass 100kg is supportedon a structure having a tos..
chine ofmass 75kg is mounted on
springs ofstiffness1200KN/m and with
assumed danmping factor of 0.2. A piston within the machine of mass 2kg
kas a reciprocating motion with a stroke of80mm and a
speed of 3000 cycles/min.
ssuming the motion to be simple harmonic. Find () the amplitude of motio
of the machine(i) its phase angle with respect to the excitingforce (üi) The
force transmitted to the
foundation, and, (iv) the phase angle of transmited
of 800KNm and has a rotating unbalanced element which
an
in a
damp
disturbing force of 40ON at a speed of 3000 rp.m. Assuming a,
lue to unbalance
ratio of 0.25, determine () the amplitude of vibrations due to unt Assu
and, (i) the transmittedforce.
Solution:
force with respect to the
excitingforce IAU, April/may 2010
Given
Solution:
m =
100kg; s =
800 kN/m =
800 x 10° N/m;
F400N at N 3000 r.p.m; = 0.25.
2TN
2n(3000) 314rad/s
Given:
m=75kg; s =1200 KN/m= 1200 x10'N/m; 5 =
0.2;m= 2kg; stroke,
60 60
(i) Amplitude ofvibration due to unbalance,x
L 80mm and e
= =
40mm= 0.04m; N =
3000 cycles/min
max
F/s)
ya- +(2Er) 2 7TN
2T(3000)314rad Is
max
60 60
Natural circular frequency ofvibration,
o m =126.49radls
Natural circular frequency of vibrations,
75
E 800x10 = 89.4427 rad/s
Vm 9 Amplitude of motion ofthemachine (due to inbalance), xx
100
314
= 3.5106
Maximum unbalanced force due to reciprocating parts,
89.4427
F =
m.eo =
2x0.04x314 7887.68N
,/s)
(400/800 x 10')
5x10
-3.5106) +
(2x0.25x3.5106)? 128.24+3.0811
mar
Amplitude of motion, max
y-+(2r)*
4.363x 10 m (or) 0.04363mm .(Ans) 314
= 2.4824
Frequency Ratio, r =
126.49
(ii) Transmitted Force, F,
7887.68
F Vi+(25P)3
yd-+(2* V-3.5106) + (2x0.5x3.5106)
(400),1+(2x0.25x3.5106)J
6.5731x10
(26.6494+0.986)
1200x10
max
-2.4824)+ (2x0.2x2.4824)
808.067
.(Ans)
.(Ans = 1.2503 x10 m or 1.2503 mm
V128.24+3.08N 0.5IN

Dynamics oJ Machin.
of Machines
4.44
ced Vibration
For 4.45
i) Phase angle with respecet to exeiting force (applied force), ¢ Solution:
Given:
tan = (2,r)
m 90kg: m= 2kg: stroke L=100mm =
0.Im, e==0.05m
(1-)
(2x0.2x2.4824) = -0.1923
(1-2.4824)
tan (-0.1923)=-10.89or-10.89 +180 =
169.11 A
F
N 1000 rp.m.
30
No. of springs, n =
4; Fr, =
Combined stiffness of thesprings,s
(Ans)
. tan tan (180+ 9)1 2TN 2T(1000) 104.72rad s
Circular frequency,
60 60
ii) Force transmitted to the foundation, FT No damping ( = 0);
F VI+(2)?
-ry+(2 -2.4824* + (2x0.2x2.4824)2
Force Transmitted,FT _1
Force applied, F,
(7887.68)1+(2x0.2x2.4824)2
30
- should
1+(2
a-+(2* 30
11115.66826
2114.48N
(26.6494+0.986) Ans) Fo be s0-
(iv) Phase angleoftransmittedforee with respect to the exciting force,(-d
Since 0,1- 30
tana= 2r =(2x0.2x2.4824) +(1-r)=30, Sincer cannotbenegative;
-(1 -)=30,-1=30, P=31,r=5.5678.
= 0.99296
a =
tan (0.99296 = 5.5678
44.8
104.72 18.81 radls
( ot) = 169.11-44.8 124.31
.(Ans) 5.5678
Example: 9.21 n m
18.81
A machine has a total mass of 90kg and unbalanced reciprocatingparis o
mass 2kg which moves
through a vertical stroke of 100mm with simpte
harmonic motion. The machine is mounted on four springs. The machine
is having only one degree offreedom and can undergo verticaldisplacemen
only:. Calculate(i) the combined stiffness ofthespringiftheforce transm
itted
to the foundation is one-thirtieth of the applied force. Neglect danp
and take the speedofrotationofthemachine crank-shaft as 1000rpm."
the machine is actually supportedon the spring, itisfoundthat tne amping
Combined stiffness, s =m x18.81 90x18.81=31843.45N/m Ans)
()Force transmitted to the foundation at 900 rpm, FT
Damping reduces the sucessive amplitudes by 30% (given)
= 1-0.3 =
0.7
211
-2
Logirathmicdecrementlogex
reduces the amplitude of the successive free vibrations by 30% a
2 T = 0.3567
Find
(ii)theforce transmitted to
thefoundation at 900rpm. [AU, Nov/Dec
20

hes
4.46
Solution
2 T =
0.3567 y
-
)., Squaring both sides:
Given:
(2mg) = 0.12722(1
-
5*), 39.4784
=
0.12722 -0.127222a
m= 80kg: n,=2.2kg; Stroke L =
100mm =
0.Im,e--0.05m:
39.6056 = 0.12722.
= 0.0567
Fr N =800rpm
2T(900) = 94.248 rads 20
At 900 rpm, o =
60
Combined
stiffness of the spring, s
Maximum unbalanced force due to reciprocating parts, F, =
m, e.o
Circular frequency, o = 2TN 2(800) -
83.78 rad/s
60 60
F = 2x0.05 x (94.248<) = 888.27N
94.248 5.0105
No damping ie, 0
FTL
Fo
Frequency Ratio,r =
18.81
20
V+(2
ya-(2¢
Froce Transmitted, Fr
V+(2
a-+(2
=
Also, F 20
888.271+(2x0.0567x5.0105)2
va-5.0105)+(2x0.0567x5.0105)
0-rd-*
1
0 20
Since
Force Transmitted, FT
1021.64 Sincer can not be negative;
= 42.37N
581.056+0.323 .Ans - (1-r)=20 -1=20 =
21r =
4.5826
Example: 9.22 = r = 4.5826
A machine supported symmetrically on four springs has mass of80kg. 7he
mass
ofthereciprocating parts is 2.2kg which move through a vertical
stroke of 100mm with simple harmonic motion. Neglecting danping
determine the combined stiffness ofthesprings so thattheforcetransm
to the foundation is (1/20 of the impressedforce. The machine cruna
shaft rotates at 800rpm. If under actual working conditions, the dampiuns
reduces the amplitudes ofsuccessive vibrations by 30% find;
83.78 18.28 radS
Hence, n 4.5826
= 18.28 radls
Combined stiffness,s =m x18.28- =
80 x
18.28 =26732.67N/m ...(Ans)
Force transmitted to the foundation at 800rpm, F7,
(i) The force transmitted to the
foundation at 800 rpm
Damping reduces the sucessive amplitudes by 30% (given)
(ii) The force transmitted to the foundation at rosonance, an"
amplitude of vibrations at
resonance.
he
= 1-0.3 =
0.7,= 0.7
IAU, Now.Dec 2012

o r c e d V b r a t i o n
4.49
2 7T
Logarithmicdecrement, - FarceTransmited, FT, = y(2 -V+ (22/?
v2 2
og, 27
2T = 0.3567
0.7 - - 36.76 1+ (2x0.0567)
(2 x0.0567)
(2m)=
0.3567 (Vi -, squaring both sides;
(2T)=0.12722 (1- 5), 39.4784 4 =0.1272 -0.127222
37
0.1134 = 326.28N
...(Ans)
39.60562=0.12722, Amplitude of vibrations at resonance
0.0567.
Force Transmitted,FTr
At800rpm; o =
27(800) 83.78rad/s We know, Combined spring stiffness, s =
Amplitude,x
60
Force
Maximum unbalanced force, F =
m, eo =
2.2x0.05x83.786 =
772.1N
. Amplitude, x =
Deflection
83.78
Frequency ratio, 326.28
18.28 =0.01221 m or 12.21lmm . (Ans)
26732.67
r = 4.5832
Example: 9.23
E+(2
a-3+(2P*
Force Transmitted, FTr =
A centrifugal compressor ofmass 100kg is supported on isolators having
a damping factor of 0.20. It runs at a constant speed of 1500rpm and
772.11+(2x0.0567x4.5832) has a rotating unbalance of 0.1kg-m. What should be the stiffness of
the isolator if the force transmitted to the foundations is to be less
than 10% oftheunbalancedforce?
Solution1
V-4.5832*+(2x0.0567x4.5832)2
870.16
400.229+0.2701 Given:
F43.48 N
Force transmitted to the foundation at resonance, FT.
..(Ans) m=100kg: =
0.2
N 1500 rpm,
At
resonance,o, =
, and hence, r
=1| 2T(1500)=157.08rad's
60
, 18.28 rad/s
(m.e) =0.1kg -m;
m, eo, =
2.2x0.05x 18.284
T = 10% =0.1.
F
F 36.76N

DynanNICS o
Machines
4.50 Forced Vibration
4.51
F EV+(2 E.
F
+(2)
a-3+(2
Example: 9.24
Compressor unil of mass
200kg is mounted rigidly on a concrete bed
haying a mass oj SU0kg. The
disturbingforce whose frequency is the same
as the compressor speed and which is sinusoidal, has maximum value oj
294N. Ifthe compressor speed is 1000 rpm, determine the stiffnessof rubber
nads to be used beneath the concrete bed such that the force transmitted is
0.5% of the disturbing force. Neglect damping.
-r
-+(2)
+(2x0.2
0-+ (2x0.2r)*
= 0.1
V+(0.4 = 0.1 d-2+(0.4r)? . .
squaring both sides Solution:
(1 +(0.4))=0.1 [(1-3+(0.4r)*] Given:
100(1 +0.16r) = 1 -2r++0.162 m, 200kg; m m,
=
500kg: F, 294N: =0;
compressor
2T(100) 0.5
100+ 16r=r- 1.84r+ 11 N= 1000rpm; o =-
=104.72rad/s;F,=xF =0.005F,
60
- 17.84r -99 0
FT0.005F.=0.005 (294) =
1.47N
( - 17.84 (P) -99 =0
Tr= 0.005=y (2
yd-(2
Solving: 22.2829,r =4.7205
=0]
o
Compressor
r = = 4.7205 = 0.005
a-
157.08
= 33.276 rad/s Concrete bed
4.7205 Forrto be positive, -0005
-1 200 Rubber pads
, m 33.276 r 14.1774 Fig.9.13
S m.o, r = 14.1774
= 100 (33.276)
104.72
4.1774
= 7.3864 rad/s
=
100 (33.276) =
110729 N/m [Note: m = m,tm
= 7.3864
m =500 +200 700kg]
Forto be less than 10%,Stiffness ofthe isolator,s=
110729 N/m Maxm
Also,
s= mx
7.3864*= (500+ 200) x7.38644
.. (Ans)
.(Ans)
Stiffness ofrubberpads, s= 38191 Nm

4.52 Dynamic.mics of Machines
4.53
Forced Vibration
Example: 9.25 transmitted to the
supports, FT,
A machine ofmass100 kg has a 20 kg rotor with 0.5 mm
eccentrici
F, =ov+(25,)?
mounting springs have s =
85 x 1°'N/m. The damping ratio is a
he
constrained to move
vertically
ratio is 0.02. The
Frr
ya-+(27)3
Find: (9 dynamic amplitudeofthe machine and, (i)theforcetrane.
AU, Nov/Dec 2004
operating speed is 600 rpm and the unit is constrained to move ver
rce transmitted
39.479 x
1+(2x 0.02x 2.155)
to the supports.
V-2.155 +(2x 0.02 x2.155)
Solution :
39.6254
Given:
3.645
m= 100 kg; m =
20 kg;e= 0.5 mm =
0.5 x
10 m; s =
85 x 10'N
0.02;N =600 rp.m. Im, FT10.871N .(Ans)
Example: 9,26
An industrial machine weighing 445 kg is supported on a spring with a
Angular velocity, 2TN
2n(600) =62.832 rad Is
static deflection of 0.5 cm. Ifthe machine has rotating imbalanceof
25 kg-cm, determine the force transmitted at 1200 rpm and the dynamic
[AU, Nov/Dec2004
60 60
85x10= 29.155 rad Is
Natural circular
frequency,a, 1
100
100
amplitude at that speed.
(i) Dynamic amplitude of the machine,x Solution:
Given:
62.832
Frequency ratio, r =-
= 2.155
29.155
m 445 kg ; 8=0.5 em =0.005 m; Rotating imbalance, (m, e) =25
kg-cm= 0.25 kg-m; N = 1200 rpm.
CF/s)
ya-+(237)2
Amax
27(1200)=125.664 rad| s
Angular velocity, o =
60 60
Maximum unbalanced force due to the reciprocating parts, F% =
mo eo
9.81 - 44.294 rad is
Vo.005
F 20x 0.5 x 10% x62.8322 Naturalcircularfrequency,
39.479 N
=44.294=. Ym
(39.479/85 x10)
Amax
y1-2.155) + (2 x 0.02 x
2.155) S =
m(44.294) = 445 (44.294) = 873072 N/m
4.6445x 104
Force Transmitted at 1200 rpm, F7r
3.6450
FI+(2)
FTr No
. :
Nodamping, =
0]
=
1.2742 x 10 m
(or)
max0.112742 mm d-+(24 ya-
.(Ans)
max

Machines
4.54 ,
Forced
Pibration o
455
Max. unbalanced force, F, = m,e o
F -0.11
0.25 x (l125.664)
= 3947.86N
...:Nodamping, =
0
125.664 2.837 0--10
Frequency ratio, r =
44.294
(W,
(1-)= t10
(3947.86)
Forr to be positive;
1-2.837
(1-r)= -10
560.09 N
Dynamic amplitude at 1200 rpm,xma
.(Ans)
(i) r= 11 and r =3.3166
(Fos) r=-and , =
max
-+(2)*
43.982
3.3166
a- .no damping, =0] =
13.261 rad / s
(3947.86/ 873072) Also = 13.261
a-2.837
m
=
6.4152 x 10 m (or) s = mx (13.261) =30x (13.261)
0.64152 mm . (Ans) Stiffness, s= 5275.68 N/m for 3 springs
mar
Example:9.27
Find the stiffness ofeach spring when a
refrigerator unit having a mass of
30 kg is to be supported by 3 springs. Theforce transmitted to the supporting
structure is only 10% of theimpressedforce.The refrigerator unit operates
5275.68
. Stiffness ofone spring
3
= 1758.56 N/m .. (Ans)
Example: 9.28
at 420 rp.m.
IAU,Nov/Dec2005. A single cylinder reciprocating engine oftotal mass 250 kg is to be installed on
Solution: an elastic support which permits vibratory motior only in vertical direction.
The mass ofpiston is 3.75 kg and it reciprocates vertically with SHM with a
stroke of 150 mm. The maximumvibratoryforcetransmitted through the elastic
Given:
m=30 kg: no.ofsprings =3;FT.=10%of F =(0.1) Fa;N=420rpm.
Support to the foundation must be limited to 500 N when the engine runs at
750rp.m andless than 500 Nat all higherspeeds. ) Findthe necessarystiflness
ofthe elastic support and the amplitude ofvibration at 800 rp.m (i) Ifthe engine
speed is reduced below 750 rp.m, at what speed will the transmittedforce again
AU, Nov/Dec 2006
Angularvelocity,o = 2TN 2(420) 43.982 rad l
s
60 60
Fovl+(2 r)
F,-
0-+(25* becomes 500 N.

Dynamics ofMachin
4.56 Forced ibration
|4.57
Solution
At 800 rpm
Given: 2TN
27(800) = 83.776 radls
60 60
m
=250kg;m,=3.75kg;L
=
1S0 mm
=
0.15 m;
(Fo/s)
y-+(2
0.1= 0.075m; F = 500 N;N = 750 rpm. Amplitude ofvibration, max .
|r
At 750 rpm
Angular velocity, o = N= £"O= 78.54 radls
60
Fs)
d- No damping, =0]
60
83.776
New frequency ratio, r=
Maximum unbalanced force, Fo =
mo e
o 37.149
r 2.25513
= 3.75 x 0.075 x (78.54)2
F 1784.89N (1784.89 /345013.3)
max
Sinceno damping; = 0 a-2.25513*)*
5.1734x 10-3
Fov+
(2 Fo
=1.2663 x10 m (or)
FT = 500 4.0856
ya-+(2* vd-2 1.2663 mm
max
. (Ans)
1784.89
ie,
Speed at which the transmitted force again becomes 500 N, N,
500
Va- There will bea speed below the resonance at which the transmitted force will again
become equal to 500 N.
ie, o <
O,n
In other words, r, < 1 [In earliercase;r> 1]
Now, maximum unbalanced force, F% = (m) e) oj
a- 1784.89 3.46978
500
1-)= 3.46978
Forrto be positive; (1 -)=-3.46978
Fo =
(3.25 x 0.075) (o) =
0.244 of
Fo_
= 4.46978
r=2.114185 500
Fr
va-
0.244 =500
Frequency ratio, r=-
78.54 (or)
va-
=
=
37.149 rad I s
2.114185
0.244o - 4.88 x10 oj
(or)
Also 500
On 37.149
(or) (1-)=+4.88
x10 oi
s m x
37.149 =
250x37.1492
Stiffness, Only positivevalue to be taken as r,
s 345013 N/m
.. (Ans)

Dynamics ofMachina.
4.58
Forced Vibration
4.59
p litude of steady statem otion
, ,(no damping, =0)
A
m plitude ofsteady.
1- + 4.88 x
10 of New = |
(Fo/s)
va- (2
0,n . max
37.149 oj =
(37.149) (4.88x10 of)
(F%/s)
1380.048 o = + 0.67346 of
a- ... = 0
of = 824.6657
(500/30000) 0.016667
= 28.717 rad/s =
=4.2378x 10 m
3.932841 (or)
V1-2.221
601-274.23 rpm
N =
2T
... (Ans)
xm 4.2378 mm . (Ans)
|Example: 9.29 Maximum force transmitted to the foundation, FT G =
0]
A singlecylinderengine has an out-of-balanceforce of500N at an engine
speed of300 rp.m. The mass ofthe engine is 150 kg and it is carried on a Fov+(2
FTr
Fo
set ofsprings oftotal stiffness of300 N/cm. a-+(27)3 a-22
Find the amplitude of the steady state motion ef the mass and the
500
maximum oscillatingJforce transnitted to thefoundation. = 127.13 N
(Ans)
V1-2.221)
i) Ifa viscous damping is interposed between the mass and the
foundation, thedamping force being 1000Nat I ms ofvelocity,fnd
the amplitude of the forced damped oscillation of the mass and is
angle of lag with disturbing force.
i) Amplitude offorced damped oscillation,* (with damping
Dampingcoefficient, C =
Dampingforce
Velocity
1000 1000 N/(m/s)
IAU, Nov/Dec 200)
ntical dampingcoefticient,Ce =
2mo, = 2 x 150 x 14.142 4242.6
Criti
Solution:
Given:
10000.2357
F500N;N=300rp.m ; m = 150 kg; s =300 N/cm = 300x10
Dampingfactor. 4242.6
N/m = 3x 104 N/m.
(500/30000) 0.016667
Angular velocity, 27tN 2T(300) 4.06982
max
=31.416 rad/s
60 1-2.221 ) + (2 x 0.2357 x 2.221)
6
Circular natural frequency, o, =
.
(or)
= 4.0952 m
10 = 14.142 radls .(Ans)
150 = 4.0952 mm
X.
Frequency ratio, r= 31.416
O 14.142 2.221

4.60 Machines
Dynamics of Ma Forced Vibration.
4.61
stiffness of each spring
Angle oflag between the amplitude and disturbing foree, ¢
(i)
2
tan o =-
(1-r)
Fov1+(2
Vd-*+(25, Ja-2 : 0
Fr, Fo
2x 0.2357x 2.221
(1-2.2214)
-
=-0.2662
a-F
tan(-0.2662)
.
tan tan (180+9))
(1-)=t11
= -
14.91° (or)
Forrtobe positive;
1-)=
-11
P= 12
-14.91+180 =165.09°
.Ans)
Example: 9.30
The mass of an electric motor is 120 kg and it runs at 1500 rp.m. The
armature mass is 35 kg and its CG lies 0.5 mm from 1he axis ofrotation,
The motor is mounted on five springs of negligible damping so that the
r= 3.4611
Also; r= 3.4611
force transmitted is one-eleventh of theimpressedforce. Assume that the
mass ofthe motor is equally distributed amongthefive springs. Determine
(i) the stiffness of each spring (ii) the dynamic force transmitted to the
base at the operating speed and, (i) the
naturalfrequencyofthe system
157.0 45.345 radIs
3.4611
45.345
AU, Nov/Dec 2007
Solution
S =m x 45.345
= 120x45.345
246741.264 N/m
Given:
m=120 kg; N= 1500r.p.m ; m,= 35 kg; e =
0.5 mm =
0.0005 m;no.o Equivalent stiffness of 5 springs = 246741.261N/m
.(Ans)
springs=5; Fr, =:=0 Stiffness of each spring
=
49348.25N/m
11
Angular velocity, o =
60
2tN
2t(1500) -157.08 radIs
Dynamic force transmitted to the base, FT,
60 Fr
a-*
Maximum unbalanced force, Fo moe o 431.8
35x0.0005x (157.08
F 431.8 N
V-3.4611
(Ans)
39.33 N

Dynamics ofMachi forced Vibration
4.65
nines
4.64 The stripofthe instrument is pressed over the vibratingbody tofind its naturalfrequency
strip is changed till the amplitude ofvibration becomes maxin
ximum.
The
lengthofthes
Screw
m
Strip
Ypas
n Y Yobject
7mm TTTTYm
Vibrating object Fig.9.16 Fullarton Tachometer
Atthe instant, the excitation frequency cquals the natural frequency ofthecantilever
strip that can be seen directly from the strip it self. The strip has different frequencies that
Fig.9.15 Accelerometer
ional
Ifr<<1 then; Z =r2 and Z Yr Now, the recorded amplitude Z. is proporti
to the acceleration ofthe vibrating body. The natural frequency ofthe accelerometershou
be at least rwice as high as the highest frequency ofthe acceleration to be recorded.
varies with its length.
The natural frequency, Jn 2T m
Ifthe input acceleration is not sinusoidal, there may be a phase shift dependingon
the amount ofdamping. However,ifthedamping ratio is around 0.7, the phaseshiftis
Where,
E-Modulus ofelasticityofthestrip material,, N/m*.
1-Area moment ofinertiaofthe strip section,m*
m- Attached mass, kg
1-Effectivelengthofstrip, m
almost linear.
9.16 FREQUENCY MEASURING DEVICES
The working ofa frequency measuring instrument is based on the principleofresonance
It may be recalled that at resonance, the excitation frequency is equal to the natural frequency
ofthemeasuring instrument and the amplitudeofvibration is maximum.
9.16.2 Fruhm's Reed Tachometer
This instrument is also known as multi reed Tachometer. It consists ofseveral reeds
as shown with different known natural frequencies. Small difference in the frequencies of
Successive reeds can provide more accurate results.
Two typesoffrequency measuring devices are
popular
Fullarton Tachometer
The instrument is brought into contact with the vibrating object whose frequency is
to be measured. One of the reeds will have the maximum amplitude of vibration and it
i) Fruhm's Reed Tachometer.
9.16.1 Fullarton Tachometer
Can be concluded that node gives frequency of the vibrating object.
This instrument is called as
single Reed Fullarton Tachometer. It consists of a
u
Let, m- mass attached to the end ofeach reed, kg
np
caryinga small mass atached at one of itsfreeends.Bydesign, thestripresembles a
cau
untilever lengthofindividual reed,m
E-modulus ofelasticityofthe reed material,N/m
beam, the length of which can be changed by means ofa screw mechanism as
Sho
.

Dynamics of.Machines
4.66
ForcedVibration
4.67
Here also, the reed can be compared
with a Cantilever beamfixes.
the static deflection of the free end is given by
ixed at one end
and Highlights (Forced Vibrations)
There is an impressed force on the system to keep it vibrating.
mg)"
3EI 8 =9.81m/s2 Amplitude (maximum displacement;
static
Fo (Fo/s)
max
-mo +(Coy ya- +(25/*
Tarea moment ofinertiaofthereed section, m".
F=impressedforce,N; s stiffness, N/m
.b= base; d=depth
For a rectangular section, 12
= frequencyratio C=dampingcoefficient.N/(m/s)
.S-stiffness of reed,N/m
Natural fraquencyofthereed, f, =
damping factor Ccriticaldampingcoefficient,N/(m/s)
E
fn2 m tan 2
a-))
and Angle oflag
From the formula it can be seen that by changing the length of the reed or
mass at the end, we can have a set of known frequencies. The reed that has a
frequency equal to the natural frequency of the vibrating object vibrates with a large
is termed as "zero frequency deflection"
amplitude. Magnification factor
-2+(25
Thus the frequency of the vibrating object can be found out easily. The accuracy of
the result depends upon the difference between the value ofthe natural frequency of Amplitude at resonance, x, =Fo/S
successive reeds which should be as small as possible.
Disturbance caused by unbalance
Cantilever Reeds
) Rotatingparts:
AARE Amplitude (maximumdisplacement);
e m
max
-+25)*
77777777777777777IITI777777777 eccentricity (radius) ofunbalance mass, m
Fig.9.17 Fruhm's Reed Tachometer
munbalance mass, kg
m=total mass of the system, kg

Dynam
amics of Mach
4.68 orcedVibration
|4.69
TWO MARKS QUESTIONS AND
ANSsWERS
orced vibration', give some example and write the differential
Reciprocating parts
Amplitude (maximum displacement)
Define 'For
governing equation.
When a system vibrates under the influence ofan external applied force which continues
ta be a part of the system, the resultant vibration is called as a Forced vibration.
mav
vd-+(25 to
Examples: a music instrument like drum, guitar, abell,vibratory conveyors; air
Forced vibrations due to support excitation
compressors; machine tools etc.
X y+(2 Governing cquation: m(dx/dr') +
C(dr/dt) + sx =
FSin of.
Mention any four areas where vibration is desirable. (AU, May/June 2014)
Displacement transmissibilIny' y 1-23 + (2G r)
Music instruments
)
Mechanical and Electrical Bell
Also; x X.
max
an 6-a) =
tan tan(2g,) Vibration Conveyors
(iv) Mobile instruments
tan(2t-); (¢ -
a) =angleoflagbetweentransmittedforce andimpressedfore
(v) Loud Speakers
a
Define resonance. (AU, April/May 2003)
3.
Relativeamplitude: When the natural frequency ofa vibrating body is equal to the operating frequency
-+253 oftheexternal applied forcethen it is called Resonance. (o=o,).
Define dynamic magnification factor. (AU, April/May 2014)
4.
phase angle ofimpressed(applied) force F with respect to displacement.
a phase angle oftransmitedforce F, with respect to
displacement.
Dynamic magnification factor is the ratio of the steady state amplitude to the zero
frequency deflection.
It is a function offrequency ratio, r and damping factor, .
Force Transmissibility and Vibration isolation
X
Transmissibility, Tr =
, - 1+(2)
Fo d-? +(2r*)
Dynamic magnification factor
Fo s) V-+(2)
Define Transmissibility and Transmissibility ratio and what is its roie in
5.
(AU, Nov/Dec 2014)
vibration analysis?
A fraction oftheapplied force on any machine will be transmitted to the foundation
which is called as the Transmissibility.
FTForcetransmitted =-
a- 25*) Transmissibility ratio is the ratio of the force transmitted to the foundation to the
force acting on the system
Transmissibility ratio or isolation factor is an indication ofthe effectiveness ofisolation
of the system.

Dynamics of Machines - Forced Vibration - Problems.pdf

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