Introuction To numerical methods
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Transcript Introuction To numerical methods
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 cost
F t Fo e jt
Fo is t heamplit udeof force
is t hefrequencyof force
Harmonically Excited Vibration
Consider the form:
.. .
m x c x kx Fo e jt
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 cosd 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 jt
,
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 2r
2 2
2
,
where r
is t hefrequencyrat io.
n
2r
2
1
r
T he t ot alsolut ion(response)for t he equat ion of mot ion(1) is :
t an1
x(t ) xc (t ) x p (t )
x(t ) e nt C1 cosd 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 cost x p (t ) X cost
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 2r
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 2r
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
2k
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 2r
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 jt kx cx
Assume harmonicmot ion
(k jc) Fo
Ftr
(k 2 m) jc
i.e,
x Xe jt
define T R F as t heForceT ransmissibilit y, T R F
TRF
1 (2r ) 2
(1 r 2 ) (2r ) 2
,
2r
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 cY cost A sin t
Where:
AY k
2
2
c
c
tan
k
Assume x(t) Xsint
1
Displacement T ransmissibility(T Rd )
(Harmonic motion)
X
Y
k 2 c
2
2
k m c
2 2
2r
2
1
r
tan1
1 2r
2
1 r 2r
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 41.667x109 m 4
12
12
192 200x109 41.667x109
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 6862.832
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
2mn 21020
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 2r
2 2
0.3326 m
2
2 r
o
3
.
814
2
1 r
tan1
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, xt 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)