STRUCTURAL DYAMICS- critical damping, under-damping and over damping in damped but free vibration.

 DAMPED FREE VIBRATIONS: The vibration of the structural members or materials in the absence of external forced but possessing a resisting force like frictions or dampers is called damped free vibrations.
note that certain  external force is required initially for vibration of body but after the body is set into vibration the forced  is removed out so that body vibrates in it's one or more natural frequency (called free vibration).
and if friction force or damping is considered it called as damped and free vibration.




The equation of dynamics for structural response to vibration is given by:

inertial force+ damping force + spring force = external force.

mẍ + cẋ + kx = p(t)
where ẍ = dynamic acceleration with respect to time.
          ẋ = velocity with respect to time.
          x=displacement with respect to time.
          p(t)= dynamic external load.
          m= mass of system.
          k=stiffness of system.
          c= damping co-efficient.


when external load is absent it resembles to damped but free  vibration case.
i.e. p(t)= 0,then

mẍ + cẋ + kx = 0.

it's a homogenous equations of order 2 and should have two roots.
The  general solution is given by x= A e^st......................................(1)
Now, differentiating equation (1) with repect to time to find ẋ and ẍ
here ẋ =A s e^st.
also, ẍ =A s *s e^st.
Then, the gereral equation  of dyanamics for damped but free vibration becomes,

     mA s² e^st. + c A s e^st.+ kA e^st = 0.

or, A e^st(ms² + c s + k)=0,
   if A e^st  =0 ,it becomes trivial solution so it cannot be zero because it's general solution as represented in equation (1) and if general solution equals to zero we cannot futher proceed.

  the possible roots are:
 s= -c/2m土 √((-c/2m)² - 4mK/4m ²).

also, w² = k/m. where w = natural vibration frequency.
 s= -c/2m土 √((c/2m)² +w²)..................................................... (2).
 this equation (2) is controlled by the square root portion.

the portion in square root can be zero, real and imaginary depending on the condition given.

case 1: when the portion √((-c/2m)² -w²) =0
it's a case of critical damping.
on solving,
      c/2m = w.
or,   c =2mw.

the damping co-efficient in this case is called as critical damping co-efficient.

generally represented by Ccr = 2*m*w.

Note: critical damping is that condition where the damping present in the body is sufficient to cease it's free vibration.

the general solution to this case is given by:
       x = (A1 +A2 t) e^-wt....................................................(3)
also,
diffrentiating equation  ( 3 ) with respect to time, we get.
       ẋ =  -w*(A1 +A2 t) e^-wt .+ A2  e^-wt.............................(4)
     
     ^{applying critical condition: at t=o, x= x(o). and}   ẋ =  ẋ(0).
^{then from equation (3), we get,}
    x(0)  = (A1 +A2 *0) e^-w*0.

 or, x(0)= A1.

Also from equation (4),

      ẋ(0) =  -w*(A1 +A2 *0) e^-w*0 .+ A2  e^-w*0.

or,  ẋ(0)=  - w * A1 + A2.

or, ẋ(0) + w * A1= A2.

i.e. A2 = ẋ(0) + w *x(0).

Therefore the particualar solution becomes

x = ( x(0) + ẋ(0) + w *x(0) *t ) e^-wt  .............................................(required for critical case ).

 case 2: when the portion √((-c/2m)² - w²) <0
it's a case of under- damping.

before solving it let us introduce a factor called damping ratio represented by ε.

ε = given damping co-efficient /critical damping co-efficient.

or, ε=  C/ Ccr .

or, ε =  C/ 2mw .

 i.e. c= 2 m w ε.............................................................................(5).

on solving the square root portion we get,

or   c < 2mw.

or, c < Ccr

 for convinience: let us take square root section of case 2 and solve it:

√((-c/2m)² - w²) <0

squaring both sides,

((-c/2m)² - w²) <0
or, c/2m <w
or c < 2mw.

thus from the definition, the damping ratio for under damped condition becomes less than 1.

or c = 2 m w ε.

we know, from equation 2,

 s= -c/2m土 √((c/2m)² -w²)

or,  s= - 2 m w ε/2m土 √(( 2 m w ε/2m)² -w²)

 or,  s= -  w ε土 √(( w ε)² -w²)

 or,  s= -  w ε土 w√( ε)² -1)

since ε is less than 1 so the root portion yeilds imaginary solutions.


or,  s= -  w ε土 i w√1- ( ε)² )

let,  w√1- ( ε)² ) = wd

then the two possible roots are : s1 = -  w ε +i wd.
  and s2 = -  w ε  -i wd.

The general solution for this underdamped case is given by,

 x = A1 e^s1t  + A2 e^{st .}

^or,  x = A1 e ^(-wε+iwd)t + A2 e^(-wε-iwd)t .

or, x =  e^-wεt   ( A1  e^iwdt +  A2 e^-iwdt ).


we know,  sinØ = ( e^iØ   -e^-iØ   ) /2i.


        also,  cosØ = ( e^{iØ   +}e^-iØ   ) /2  .

finally the equation becomes,

or, x =  e^-wεt   ( A  coswd*t +B sinwd*t.).......................(6).

using underdamped condition and substituting in equation (6) we get,

  when t=0 , x= x(0) and ẋ =ẋ (0).

 x =  e^-wε*0 ( A  coswd*0+B sinwd*0).

or, A= x(0).

also on differentiating with respect to time, we get,

ẋ = -wε e^-wεt   ( A  cos(wd*t) +B sin(wd*t).) +   e^-wεt   ( A* wd* (- sinw*t) +  B wd coswd*t).

^{putting initials conditions, at t=0 ,we get}
 ẋ(t) =ẋ (0)
then

  B =(ẋ (0)+εw A)/wd

therefore the particualar solution to this  case becomes,

x =   e^-wεt   (x(0)coswdt+ (ẋ (0)+εw A)/wd*sinwdt)................................... required solution for this under damped case.

 case 3: when the portion √((-c/2m)² - w²) >0
it's a case of over- damping.


ε = given damping co-efficient /critical damping co-efficient.

or, ε=  C/ Ccr .

or, ε =  C/ 2mw .

 i.e. c= 2 m w ε.............................................................................(5).

on solving the square root portion of case 3 we get,

or   c >2mw.

or, c > Ccr

thus from the definition of damping ratio,  the damping ratio for under damped condition becomes greater than 1.

or c = 2 m w ε

we know, from equation 2,

 s= -c/2m土 √((c/2m)² -w²)

or,  s= - 2 m w ε/2m土 √(( 2 m w ε/2m)² -w²)

 or,  s= -  w ε土 √(( w ε)² -w²)

 or,  s= -  w ε土 w√( ε)² -1)

let,  w√ ( ε)² -1) = wd

then the two possible roots are : s1 = -  w ε +wd.
  and s2 = -  w ε  - wd.

The general solution for this  overdamped case is given by,

 x(t) = A1 e^s1t  + A2 e^{st .}

^or,  x(t) = A1 e ^(-wε+wd)t + A2 e^(-wε-wd)t .

or, x(t) =  e^-wεt   ( A1  e^wdt +  A2 e^-wdt ).

using initial condition ,at time t=0,we get.

x (0)=  e^-wε*0   ( A1  e^wd*0 +  A2 e^-wd*0 ).

so, x (0) = A1+ A2...................................................... (6).

again, differentiating above displacement equation w.r.t time we get,

ẋ(t) = -wε e^-wεt   ( A1  e^wdt +  A2 e^-wdt ). +   e^-wεt   ( A1* wd*  e^wdt +  A2 -wd*e^-wdt).

using initial conditions,

ẋ(0) = -wε (A1 +A2) +  (wd A1 -Wd A2).

  ẋ(0) =(wd-wε )A1 -A2(wε+wd).....................................................(7).


on solving equation (6) and (7) , we get,

x(0) *(wε+wd) + ẋ(0) = 2* wd * A1

A1 =x(0) (wε+wd) + ẋ(0) /(2*wd)


similarly, we get,

 A2 =  x(0)(wd-wε )  -ẋ(0)/(2*wd)


thus on substituting values of A1 and A2 in general solution  x(t) =  e^-wεt   ( A1  e^wdt +  A2 e^-wdt ). we obtain particular solution.

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