4.SDOF Systems in Free Vibration

8/2/2019 4.SDOF Systems in Free Vibration

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Free vibration response due to initial conditions

J. Enrique Martinez-Rueda Dynamics of Structures 1

Solution to the equation of motion of a SDOF systemunder free vibration

If there is no excitation force then the equation of motion of a SDOF system becomes:

0)()()( *** =++ t Y K t Y C t Y M &&& (fv.1)

The above equation of motion is a 2 nd degree differential equation with constantcoefficients; the solution of the equation is the displacement response ) (t Y , if this isassumed to be:

t Aet Y =)( (fv.2)Then the velocity and the acceleration responses will be given as:

t e Adt

t dY t Y == )()(& (fv.3)

t e Adt

t Y d t Y 2)()( ==

&&& (fv.4)

Substituting the above functions ) (t Y , )(t Y & and )(t Y && in the equation of motion (fv.1)

0**2* =++ t t t AeK e AC e A M (fv.5)

Dividing everything by t Ae

0**2* =++ K C M characteristic equation (fv.6)

The solution to this equation is

2*

**2

*

*

*

*

*

**2**

4

4

222

4

M

K M

M

C

M

C

M

K M C C

== (fv.7)

or

*

*2

*

*

*

*

22 M

K

M

C

M

C

= (fv.8)

But if we change the variables of eqn (fv.8) using:

*

*

M

K = [rad/sec] = undamped circular frequency (fv.9)

*

*

2 M

C = = damping ratio (fv.10)


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Then eqn. (fv.8) can be rewritten as:

12222 == (fv.11)

Now, there are three possibilities for

case 1 . =1.0 i.e., critically damped system

= 1 = 2 = - (i.e., the two values of are equal) (fv.12)

The Theory of Differential Equations (TED) shows that, in this case, the solution is:

t t teC eC t Y 21)( += (fv.13)

or t t teC eC t Y +=21

)( (fv.14)

or ( )t C eC t Y t 21 1)( += (fv.15)

case 2. > 1.0 i.e ., overdamped system

1 2 and TED shows the solution is:t t eC eC t Y 21 21)(

+= (fv.16)or

+

+=1

2

1

1

22

)( t t

eC eC t Y (fv.17)

case 3. < 1.0 i.e. , underdamped system

Now the term 12 is imaginary and hence 1 and 2 are complex numbers

222 1111 == i (fv.18)

2

1 = i (fv.19) or d i = (fv.20)

where 21 =d [rad/sec] damped circular frequency ( d < ) (fv.21)

Finally, TED shows that the solution in this case is:

( )t C t C et Y d d t sincos)( 21 += (fv.22)


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Physical interpretation of the above solutions

1. systems with critical damping ( = 1)

0.0 0.1 0.2 0.3 0.4 0.50.0

0.2

0.4

0.6

0.8

1.0

=1

t

Y(t)

2. overdamped systems ( > 1)

0.0 0.1 0.2 0.3 0.4 0.50.0

0.5

1.0

1.5

2.0

> 1

t

Y(t)

In both of the above cases:

the displacement response Y(t) is characterised by exponential decay with nooscillation

in general the theory of systems with 1 is used for the design of shock absorvers or damping mechanisms of doors.


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3. underdamped systems ( < 1)

Td

Y(t)

t

infinite number of oscillations (in theory) amplitude of oscillations decay in an exponential fashion

in conventional structures is typically assumed to be 2-5% but in specialstructures it can be as high as 20%

the system vibrates at the damped frequency d with a damped period T d :

d

d T 2= [sec] (fv.23)

For systems with no damping, we define:

2=T [sec] = natural period or period of vibration (fv.24)

2

1 ==T

f [Hertz = cycles/sec ] = natural frequency (fv.25)

The natural period can be seen as the time required by the system to complete a fullcycle of vibration .


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Response of damped SDOF system in free vibration dueto initial conditions

The displacement response of this system is given by:

( )t C t C et Y d d t sincos)( 21 += (fv.26)

where C 1 & C 2 are integration constants that can be found invoking the initial i.e . boundaryconditions of the problem.

Deriving the above eqn. with respect to time we get

( ) ( )t C t C et C t C et Y d d d t

d d d d t

sincoscossin)( 2121 ++=

& (fv.27)

Now, let us assume we know the initial conditions (initial displacement Y o and initialvelocity oY

&) of the response Y(t):

Y o = Y (0) at t = 0 (fv.28)

)0(Y Y o&& = at t = 0 (fv.29)

Substituting the above initial conditions into eqns (fv.26) and (fv.27) we get:

Y o = 1 ( C 1 + 0) Y o = C 1 (fv.30)

=oY & 1(-0 + C 2 d ) - C 1

d

oo Y Y C +=

&2 (fv.31)

Hence the displacement response of the system will be:

++= t Y Y t Y et Y d

d

ood ot sincos)(

&(fv.32)

It can be shown that if the initial conditions are invoked for time 0 thenthe response of the system is given by:

( ) ( ) ( ) ( ) ( ) ( )

++=

t Y Y t Y et Y d d

d t sincos)(

&(fv.33)

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4.SDOF Systems in Free Vibration