1 free vibration damping for class
TRANSCRIPT
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SDOF Definitions
lumped mass
stiffness proportionalto displacement
damping proportional to
velocity linear time invariant
2nd order differentialequations
Assumptions
m
k c
x(t)
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Equation of Motion - Natural Frequency
Newtons Second Law:
(written about static equilibrium position)
Circularfrequencygiven as:
=+= )t(fkxxmmaf && (2.2.2)
mk
n = (2.2.3)
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Equation of Motion - Natural Frequency
Equation of Motion written in standard form
has the general solution
where A and B are two necessary constantsdetermined from the initial conditions of
displacement and velocity
(2.2.4)
(2.2.5)tcosBtsinAx nn +=
0xx2n =+&&
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Equation of Motion - Natural Frequency
This results in
and the frequency can be shown in relation tomass and stiffness or the static displacement ofthe system as
(2.2.6)
(2.2.8)
tcos)0(xtsin)0(x
x nnn
+
=&
=
= g
2
1m
k2
11f
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Energy Method
In a conservative system, the total energy is
constant. Kinetic energy is stored in the mass interms of velocity and potential energy is stored asstrain energy in the spring
From conservation of energy, an equilibrium onstate 1 and state 2 exists - and at the extremesthe maximums result in
(2.3.1)ttanconsUT =+ ( ) 0UTdt
d=+
2211 UTUT +=+ MAXMAX UT =
(2.3.2)
(2.3.3) (2.3.5)
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Rayleigh Method: Effective Mass
The kinetic energy can be written in terms of
velocity for a specific point in the system as
where the effective mass is at some point
If the stiffness is known at that point then thenatural frequency can be calculated as
(2.4.1)
(2.4.2)
2eff2
1 xmT &=
effn m
k=
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Virtual Work - Equilibrium of Bodies
If a system is in equilibrium under the action of a
set of forces is given a virtual displacement, thevirtual work done by the foces will be zero
(1) a virtual displacement is a small variationof the coordinate (must be compatiblewith constraints of the system)
(2) virtual work is work done by all activeforces in avirtual displacement (nogeometry change which implies forces on
system remain unchanged)
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Viscously Damped Free Vibration
The equation of motion for the SDOF with viscous
damping included is
both a homogeneous and particular solution exist
(2.6.2) =++= )t(fkxxcxmmaf &&&
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Homogenous Solution
Homogenous equation is
and assuming an exponential solution form gives
which results in
which yields the characteristic equation
(2.6.3)
(2.6.4)
0kxxcxm =++ &&&
(2.6.4)
(2. 6.5)
0ekcsms st2 =++
0mks
mcs2 =++
m
k
m2
c
m2
cs
2
2,1
=
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Homogenous Solution
This results in a general solution given as
Three distinct solutions result from this general
solution.
(2.6.7)tsts
21 BeAex +=
(1) underdamped
(2) overdamped(3) critically damped
11=
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Homogenous Solution
Some general definitions are:
Critical Damping
% Critical Damping
Roots
(2.6.9)
(2.6.10)
(2.6.11)
km2m2
m
km2c nc ===
cc
c=
1s 2nn2,1 =
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The general solution can be rewritten as
and with adjustingsome terms
the poles can be
observed as seenin the figure uponchan in dam in
(2.6.12)
(2.6.13)
Homogenous Solution
)t(fm
1xx2x 2nn =++ &&&
2
n
2,11i
s=
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Oscillatory Solution (3 different forms)
and the general homogeneous solution becomes
(2.6.14)
(2.6.16)
(2.6.17
(2. 6.18
Homogenous Solution - Underdamped 1
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The damped exponential response
( ) ( )
+
+= t1cos)0(xt1sin
1
)0(x)0(xex n
2n
2
2n
ntn &
Homogenous Solution - Underdamped 1
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Overdamped solution is
with coefficients A and B as
(2.6.19)
(2.6.20)
(2.6.21)
Homogenous Solution - Overdamped 1>
t1t1n
2
n
2
BeAex+ +=
12
)0(x1)0(xA
2
n
n2
++=
&
12
)0(x1)0(xB
2n
n2
=
&
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The critically damped solution is
and the solution becomes
(2.6.22)
(2.6.23)
Homogenous Solution - Critically Damped 1=
( ) tneBtAx +=
( )t
n
t
n
n
et)0(x)0(x
e)0(xx
+
+=
&
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Logarithmic Decrement
From the damped vibration solution, the amount of
damping in the system can be expressed as
and the log decrement becomes
and for small damping
(2.7.2)
(2.7.3)
dn)t(
tdn
d1n
1n
eln
e
eln ===
+
21
2
=
(2.7.4) 2
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Logarithmic Decrement
Plot of Damping Estimates
Log Decrement
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Coulomb Damping
Sliding or dry friction produces a force which is
opposite to the velocity and its direction changeswith each half cycle
or
the decay in amplitude
per cycle is therefore
( ) 0)XX(FXXk21
11d
2
1
2
1 =+
( )d11
FXXk2
1=
d21 F4XXk = (2.8.1)
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Review - SDOF Definitions
lumped mass
stiffness proportionalto displacement
damping proportional to
velocity linear time invariant
2nd order differentialequations
Assumptions
m
k c
x(t)
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Review - SDOF Equations
Equation of Motion
)t(fxkdt
dxc
dt
xdm
2
2
=++ )t(fxkxcxm =++ &&&
Characteristic Equation
Roots or poles of the characteristic equation
0kscsm 2 =++
m
k
m2
c
m2
cs
2
2,1 +
=
or
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Review - SDOF Definitions
Poles expressed as
Damping Factor
Natural Frequency
% Critical Damping
Critical Damping
Damped NaturalFre uenc
POLE
CONJUGATE
j
n
d( ) d
2
n
2
nn2,1 js ==
n=
mk
n=
ccc=
nc
m2c =2
nd 1 =