Mechanical Vibrations
Forced Vibration of a Single Degree of Freedom System
Philadelphia University
Faculty of Engineering
Mechanical Engineering Department
Dr. Adnan Dawood Mohammed
(Professor of Mechanical Engineering)
Harmonically Excited Vibration
 Physical system
..
.
m x  c x  kx  F t 
Equat ionof mot ion
F t  is a harmonicforce.It may t akesone of t hefollowingforms:
F t   Fo sin t
F t   Fo cost
F t   Fo e jt
Fo is t heamplit udeof force
 is t hefrequencyof force
Harmonically Excited Vibration
Consider the form:
.. .
m x  c x  kx  Fo e jt
Equation of motion (1)
T his2nd order differential equation is non - homogenuos. Itssolution has
two parts
1. Complementary function(T ransientresponse): It is thesolutionfor
the homogeneous equation (described in previouschapter).for   1
T hesolution was writtenas :
xc (t )  e  nt C1 cosd t  C2 sin d t 
2. P articularintegral(steadystateresponse: It is thesolutionfor the
non homogenuosequation.Because theforce(excitation) is harmaonic
with frequency we can expect tha
t theresponse(solution)is harmonic
with thesame frequency. T hesolutioncan be writtenas
x p (t)  Xe jt
,
X is called the' steadtystateresponseamplitude'
Harmonically Excited Vibration
Subst it ut e t hepart icularint egralint oequat ion (1) and solvefor X :
X 
Fo
k   2 m  jc




m   c  e
By writ ing k   2 m  jc
k  
2
2
2
j
as
, where  t an-1
cw
is t he" phase
2
k  m
angle"bet ween Forceand Response.
 X  X e  j
where X 
Fo
k   m  c 
2
2
2
T herefor t hepart icularint egralcan now be writ t enas
x p  X e j t  
, X denot est he" magnit ude"of t he
st eady st at eresponseamplit ude.
Harmonically Excited Vibration
X and  may be writ t enas :
X 
 Fo 


 k 
1  r   2r 
2 2
2
,
where r 

is t hefrequencyrat io.
n
 2r 
2 
1

r


T he t ot alsolut ion(response)for t he equat ion of mot ion(1) is :
  t an1 
x(t )  xc (t )  x p (t )
x(t )  e  nt C1 cosd t  C2 sin d t   X e j t  
Not e t hatC1 and C2 are const ant st o be derminedfrom
knowing t heinat ialcondit ionsof t hemot ion.
Harmonically Excited Vibration
NOT ES:
1. For theforceform F(t)  Fo sin t  x p (t )  X sin t   
2. For theforceform F(t)  Fo cost  x p (t )  X cost   
3. T ransientresponse( xc (t )) representsa motion t ha
t decays with
timeand can be neglectedaftera certain time.
4.St eady stateresponse( x p (t )) is a harmonicmotion with constant
amplitudeand frequency.
5. T he term
X
 st
Fo
is usually called the" staticdeflection( st ). T heratio
k
is usually called the" Magnification Factor(M)".
M
1
1  r   2r 
2 2
2
Damped Forced Vibration System
 Graphical representation for Magnification factor M and ϕ.
Damped Forced Vibration System
 Notes on the graphical representation of X.
 For ζ = 0 , the system is reduced and becomes un-damped.
 for any amount of ζ > 0 , the amplitude of vibration decreases
(i.e. reduction in the magnification factor M). This is correct for
any value of r.
 For the case of r = 0, the magnification factor equals 1.
 The amplitude of the forced vibration approaches zero when the
frequency ratio ‘r’ approaches the infinity (i.e. M→0 when r →
∞)
Damped Forced Vibration System
 Notes on the graphical representation for ϕ.
 For ζ = 0 , the phase angle is zero for 0<r<1 and 180o for r>1.
 For any amount of ζ > 0 and 0<r<1 , 0o<ϕ<90o.
 For ζ > 0 and r>1 , 90o<ϕ<180o.
 For ζ > 0 and r=1 , ϕ= 90o.
 For ζ > 0 and r>>1 , ϕ approaches 180o.
Harmonically Excited Vibration
Variationof displacement withfrequency:
We have:
X
Fo
k
1  r   2r 
2 2
is thes.s responseamplitude
2
1. At low frequency(r  0) then X 
Fo
which means thedisplacement
k
is Stiffness control.
2. At high frequency(r  1) then X 
Fo
which means thedisplacement
2
m
is Mass control.
Fo
which means thedisplacement
2k
is Dampingcontrol.(for small )
3. At resonance(r  1) then X 
Harmonically Excited Vibration
Frequency of Maximumamplitude:
T hes.s responseamplitudecan be writtenas :
-1/2
Fo
2
2 2
X 
1  r  2r 
k
d
X is maximumwhen ( X )  0
dr
T hisconditiongives :


X is maximumwhen r 


 1  2 2 which is known as a
n
Resonance Frequency, i.e
res  n 1  2 2
Forced Vibration due to Rotating Unbalance
Unbalance in rotating machines is a
common source of vibration
excitation. If Mt is the total mass of
the system, m is the eccentric mass
and  is the speed of rotation, the
centrifugal force due to unbalanced
mass is meω2 where e is the
eccentricity.
 The
vertical
component
(meω2
sin(ωt) is the effective one because
it is in the direction of motion of
the system. The equation of motion
is:
.. .
M t x c x kx  m e 2 sin t 
Forced Vibration due to Rotating Unbalance
X 
X 
 m e 2

 k

1  r 
2 2




 2 r 
,
or
2
 m e 2

M 
r
t 

1  r 
2 2
 2 r 
2
and
 2 r 
  t an 
2 
1 r 
1
T heplot sof X and  are shown in t hefigure below
Transmissibility of Force
If t hedeflect ionof foundat ionis neglugible,
t hen t heforce t ransmited
t t o t hefoundat ionis
Ftr e jt  kx  cx
Assume harmonicmot ion
(k  jc) Fo
Ftr 
(k   2 m)  jc 
i.e,
x  Xe jt
define T R F as t heForceT ransmissibilit y, T R F 
TRF 
1  (2r ) 2
(1  r 2 )  (2r ) 2
,
2r
  t an
(1- r 2 )
1
Ftr
Fo
Transmissibility of displacement (support motion)
Physical system:
The forcing function for the base excitation
..  . . 
Mathematical model: m x  c x  y   k x  y   0


Transmissibility of displacement (support motion)
Substitute the forcing function into the math. Model:
.. .
m x  c x  kx  kY sin t   cY cost   A sin t   
Where:
AY k
2
2


 c
 c 
  tan   
 k 
Assume x(t) Xsint
1
Displacement T ransmissibility(T Rd ) 
(Harmonic motion)
X
Y
k 2  c 
2

2
k  m   c 
2 2
 2r 
2 
1

r


  tan1 
1  2r 
2

1  r   2r 
2 2
2
Transmissibility of displacement (support motion)
Graphical representation of Force or
Transmissibility ((TR) and the Phase angle (
Displacement
 Example 3.1: Plate Supporting a Pump:
A reciprocating pump, weighing 68 kg, is mounted at the middle of a steel
plate of thickness 1 cm, width 50 cm, and length 250 cm. clamped along
two edges as shown in Fig. During operation of the pump, the plate is
subjected to a harmonic force, F(t) = 220 cos (62.832t) N. if E=200
Gpa, Find the amplitude of vibration of the plate.
 Example 3.1: solution
 The plate can be modeled as fixed – fixed beam has the
following stiffness:
192EI
l3
3
1
1
But I  bh3 
50x10 2 1x10 2  41.667x109 m 4
12
12
192 200x109 41.667x109
So, k 
 102,400.82 N / m
2 3
250x10
k








 The maximum amplitude (X) is found as:
-ve means that the
Fo
220
X

 1.32487m m response is out of
2
k  m
102,400.82  6862.832
phase with
excitation
Example 3.2:
Find the total response of a single-degree-of-freedom system
. with m =
10 kg, c = 20 N-s/m, k=4000 N/m, xo = 0.01m and x=o0 when
an external force F(t) = Fo cos(ωt) acts on the system with Fo = 100 N and
ω = 10 rad/sec .
Solution
a. From the given data
n 
 
k

m
4000
 20rad / s
10
c
20

 0.05
2mn 21020
d  1   2 n
d  1  0.052 20
d  19.975rad / s
r
 10

 0.5
n 20
Example 3.2:
Solution
F
100
 st  o 
 0.025 m
k
4000
X 
 st
1  r   2r 
2 2
 0.3326 m
2
 2 r 
o

3
.
814

2
1 r 
  tan1 
Total solution:
X(t) = X c (t) + X p(t)
x (t)  Ae
x (t)  Ae
- n t
cos( n t -  )  X cos( t - 0.066)
- 0.05*20t
cos(19.97t -  )  0.3326cos(20t - 0.066)
at t  0, xt   0.01m
0  Acos  0.3326* cos0.066
 Acos  0.33187
(1)
-t
x (t)  Ae cos(19.97t-  )  0.3326cos(20t - 0.066)
-t
-t
x (t )  Ae * -19.97sin(19.97t-  )  A cos(19.97t-  ) * (-t)e
 0.3326* 20sin(20t - 0.066)
at t  0, x t   0
 0  19.97 A sin  0.438,
A sin  0.0219 (2)
From (1) and (2)
  0.066rad and A  -0.3325
-t
 x(t)  -6.64e cos(19.97t- 0.066) - 0.3326cos(20t - 0.066)