Prof. A. Meher Prasad
   Department of Civil Engineering
Indian Institute of Technology Madras
    email: prasadam@iitm.ac.in
Outline

Degrees of Freedom

Idealisation of SDOF System

Formulation of Equation of motion

Free vibration of undamped/da...
What is Dynamics ?
Basic difference between static and dynamic loading




             P                              P(t)




  Resistance ...
Characteristics and sources of Typical Dynamic Loadings
  Periodic Loading:

               (a)                           ...
Dynamic Degrees of Freedom
The number of independent displacement
components that must be considered to
represent the effects of all significant
iner...
Examples
Dynamic Degrees of Freedom




   1.                                    2.
              Massless          Spring         ...
Dynamic Degrees of Freedom

3.                          Flexible and       Finite             Flexible and   Point
       ...
Dynamic Degrees of Freedom



                             Rigid deck

         5.




                                   ...
Idealisation of Structure as SDOF
Mathematical model - SDOF System

                                                      x
                            k

 ...
Newton’s second law of motion

Force = P(t) = Rate of change of momentum of any mass

                d ₩ dx
             ...
The force P(t) includes ,                                      mg

1) Elastic constraints which opposes displacement
     ...
Equations of motion:
                                  fs
Spring force           - fs  x            k
                   ...
Examples
FBD for mass
1.
                            x                fs = kx
       k                                           .....
(3a)       Rigid ,massless       P(t)       m                                      P(t)

                                 ...
(3b)


                   a                Rigid
               b
                                    massless            ...
(3c)
            m                                                               ..
                                      ...
(4a)   μ (distributed mass)
                              P(t)   m                            mL         P(t)
            ...
(4b)

                   a
                           μ
                                                    ₩ a
          ...
Special cases:
(4c)                (4d)              (4e)



              L                                  k
          ...
x
     ke

    ce           me              Pe(t)




 me && + ce x + ke x = Pe (t )
    x       &                        ...
Internal
                          m/2    hinge P(t)       Rigid with uniform mass μL/2 = m/2
    Rigid ,massless

N      ...
7             1               1 ← 16 N                    3
me =      m      ce = c          ke = k ↑1 +            Pe (t ...
Free Vibration

Undamped SDOF System


Damped SDOF systems
Free Vibration of Undamped System

                      && + p 2 x = 0
                      x                           ...
Amplitude of motion
                                                       2
                                             ...
Natural frequencies of other SDF systems

      p–   square root of the coefficient of displacement
           term divide...
Condition of instability

                      1
              N =-      kL = N cr         (22)
                     16

...
Natural frequencies of single mass systems


                               p= k /m                                     (1...
Relationship between Simple oscillator and Simple pendulum



                                               L




       ...
Effective stiffness ke and static deflection δst




                                  ke       g
                        ...
Determination of Force - Displacement relation, F-∆




    1. Apply the static force ,F on the mass in the direction of m...
Examples
Rigid ,massless               m

                                                                      2
(a)              ...
(b)     Rigid bar                    m          (c)                                     m
                                ...
(d)                                      (f)   Flexible but mass less
        k1             kn

                         ...
(g)               Rigid deck; columns mass less &   (h)
                  axially inextensible
                           ...
(i)
         EI
                                                    1 FL3 5 RL3
                                          ...
(j)                               (l)




      L/2                   L/2         L/2                L/2

                ...
(n)                                             (p)
                                                                   L
 ...
Natural frequencies of simple MDF systems treated as SDF


            m

            a

         A, E, I, L


           ...
• For pitching or rocking mode
      1 .. a 2a AE
         my       +     ya = 0
      2      2 3     L                   ...
Free Vibration of Damped SDOF
Free Vibration of damped SDOF systems



   mx + cx + kx = 0
    && &                              k

   && +
        c   ...
Solution of Eq.(A) may be obtained by a function in the form x = ert
where r is a constant to be determined. Substituting ...
rt           rt
Thus e 1 and e 2 are solutions and, provided r1 and r2 are different
from one another, the complete soluti...
For z  1 (Light Damping) :


              x ( t ) = e -z pt [ A cos pd t + B sin pd t ]             (39)

      where, p...
g
                   Td = 2π / pd        Extremum point ( x(t ) = 0 )
                                     Point of tangen...
Motion known as Damped harmonic motion


A system behaving in this manner (i.e., a system for whichz  1 ) is said
to be U...
Rate of Decay of Peaks


          xn +1      -z p
                             2p
                                       ...
Logarithmic decrement


                           xn
    Defined as     d = ln                                           ...
For z  1 (Heavy Damping)

 Such system is said to be over damped or super critically damped.

           x(t ) = C1e( - )...
For z = 1

 Such system is said to be critically damped.

          x (t ) = C1e - pt + C2te - pt

  With initial conditio...
Response to Impulsive Forces

   Response to simple Force Pulses

   Response to a Step Pulse

   Response to a Rectangula...
Response to Impulsive Forces
                                                                         Po
Let the duration ...
For an undamped system, the maximum response is determined
from as ,
                a P0 t1 = a P0 kt1 = a ( x ) pt
     ...
•Damping has much less importance in controlling maximum
   response of a structure to impulsive load.

           The max...
Response to simple Force Pulses


                                                        P (t )
 P(t)                    ...
Response to a Step Pulse


                        For undamped system, x + p2 x = p2 (xst)o
P(t)
                        ...
Response to a Step Pulse….

                                    z=0



                  x (t ) 2
                 ( xst )...
Response to a Rectangular Pulse


 For t  t1, solution is the same as before,

                     x(t ) = ( xst ) ( 1 -...
Response to a Rectangular Pulse


                                  pt
            1-cos pt1      2sin 2 1            pt1
...
Response to a Rectangular Pulse…
x(t)/(xst)0

                            2
                                              ...
Dynamic response of undamped SDF system to rectangular
pulse force. Static solution is shown by dotted lines
(a)                                     Forced response
                                          Free response




 (b)  ...
Response to a Rectangular Pulse…
          3
                Impulsive solution, 2π f t1
     xmax 2
   ( xst ) 0
        ...
Response to Half-Sine Pulse
P(t) = Po sin ωt, where ω = π / t1
                                                           ...
•   Note that in these solutions, t1 and T enter as a ratio and that
    similarly, t appears as the ratio t /T. In other ...
Dynamic response of undamped SDF system to half cycle
sine pulse force; static solution is shown by dashed lines
Response to half cycle sine pulse force (a) response maxima during
forced vibration phase; (b) maximum responses during ea...
2pft1 4pft
               1
              pft1

                                                  t1



                  ...
Response to Half-cycle Force Pulses
 •   For low values of ft1 (say < 0.2), the maximum value of xmax or AF is
     depend...
Conditions under which response is static:

On the basis of the spectrum for the ‘ramp pulse’ presented next, it is
conclu...
Response to Step force

For t  tr                                                    P(t)
                              ←...
Substituting these quantities into x(t), the peak amplitude is found as:

              xmax          1                   ...
For Rectangular Pulse:

x(t ) = ( xst )0 [1 - cos pt ]                      for t    t1       P(t)
              ₩      p...
Design Spectrum for Half-Cycle Force Pulses

                      2
                          A
                         ...
Response to Multi-Cycle Force Pulses

 Effect of Full-Cycle Sine Pulse

 •   The high-frequency, right hand limit is defin...
Effect of n Half-Sine Pulses

 The absolute maximum value of the spectrum in this case occurs
 at a value of, ft1=0.5     ...
Effect of a sequence of Impulses


  I
                                Suppose that t1 = T/2
       t1
            I
     ...
Effect of a sequence of Impulses

•   For n equal impulses, of successively opposite signs, spaced at
    intervals t1 = T...
Response of Damped systems to Sinusoidal Force
Response of Damped systems to Sinusoidal Force

   P(t) = P0 sinωt
   where ω =     p/ t1= Circular frequency          P(t...
This leads to
             (p2-ω2)M - 2 ζpωN = p2(xst)0
                                                                  ...
Steady State Response


         P(t)




                                                          a                  t
 ...
AF          α




                     w
                =
                     p
        w
     =
        p
Effect of damping


   • Reduces the response, and the greater the amount of
     damping, the greater the reduction.


  ...
Resonant Frequency and Amplitude


                              wres = p 1- 2z 2          (72)
                          ...
Transmissibility of system

 The dynamic force transmitted to the base of the SDOF system is
                             ...
Transmissibility of system

 The ratio of the amplitudes of the transmitted force and the applied
 force is defined as the...
Transmissibility of system




        TR




                                     ω
                                =
  ...
•   Irrespective of the amount of damping involved, TR<1 only for
    values of (ω/p)2 >2. In other words, in order for th...
• In the frequency range where TR<1, damping increases the
  transmissibility. In spite of this it is desirable to have so...
Application
   Consider a reciprocating or rotating machine which, due to unbalance
   of its moving parts, is acted upon ...
•   If the support flexibility is such that is less than the value defined
    by Eq.(79), the transmitted force will be g...
Application to Ground-Excited systems

                             k
                                        m
          ...
Rotating Unbalance


                                     Total mass of machine = M
       M          m                  u...
Reciprocating unbalance




        m
                                           e         
          L               F ...
Determination of Natural frequency and Damping

Steady State Response Curves

  Structure subjected to a sinusoidally vary...
Determination of Natural frequency and Damping


           Resonant Amplification Method
           Half-Power or Bandwid...
Determination of Natural frequency and Damping


(a) Resonant Amplification Method



Determine maximum amplification (A.F...
(b) Half-Power or Bandwidth Method:


In this method ζ is determined from the part of the spectrum near the
     peak step...
(xo)max
xo




                           1
                            2   (xo)max
     (xo)st


                  ∆f    ...
Determination of Natural frequency and Damping…


 1.For small amounts of damping, it can be shown that ζ is related
 to t...
Derivation
                                       1
         1 1            1                2
              =         ...
Other Methods for Evaluating response of SDF Systems


(c) Duhamel’s Integral
    In this approach the forcing function is...
I=P (τ )dτ




The strip of loading shown shaded represents and impulse,
                  I = P() d
For an undamped SDF...
Implicit in this derivation is the assumption that the system is initially (at
t=0) at rest. For arbitrary initial conditi...
The effect of the initial motion in this case is defined by Eqn. 41
Eqns. 84 and 87 are referred to in the literature with...
Generalised SDOF System
Generalised SDOF System



         m*q( t ) + c*q( t ) + k *q( t ) = p * ( t )
           &&         &


         q( t ) ...
y( x, t ) =  ( x ) q( t )
(a)




(b)      m(x)           m1,j1                                m2, j2




          x1

 ...
c1                     c2
                         a1(x)
(c)             c(x)




                                       L...
(d)                                      k1                            k2




                          k(x)
 (e)
        ...
P(x,t)



                       Pi(t)




                 l
           p* =  p( x, t ) ( x )dx + pi (t ) i ( x )
   ...
Effect of damping



   Viscous damping

   Coulomb damping

   Hysteretic damping
Effect of damping




   Energy dissipated into heat or radiated away


   •   The loss of energy from the oscillatory sys...
Effect of damping


  Energy dissipated mechanism may emanate from
         (i)   Friction at supports & joints
         (...
Simplified damping models have been proposed .These models
are found to be adequate in evaluating the system response.
Dep...
(a) Viscous damping

             Fd(t) = c    x
                          &
             c        -       coefficient of ...
(b) Equivalent viscous damping:

  p Ceqw X 2 = Wd
                                               Fd+kx       ellipse
    ...
(b) Coulomb damping:                                              Fd
                            -
                       ...
1
             k ( X 12 - X -1 ) - Fd ( X 1 + X -1 ) = 0
                          2

           2
           1
          ...
(c) Hysteretic damping (material damping or structural damping):

    - Inelastic deformation of the material composing th...
Equivalent viscous Coefficient



              e) Coulomb Ceq=        4F
                                    p WX
       ...
Reference

Dynamics of Structures: Theory and Application to
Earthquake Engineering – Anil K. Chopra, Prentice Hall
India
...
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  1. 1. Prof. A. Meher Prasad Department of Civil Engineering Indian Institute of Technology Madras email: prasadam@iitm.ac.in
  2. 2. Outline Degrees of Freedom Idealisation of SDOF System Formulation of Equation of motion Free vibration of undamped/damped systems Forced vibration of systems Steady state response to harmonic forces Determination of natural frequency Duhamel’s Integral and other methods of solution Damping in structures
  3. 3. What is Dynamics ?
  4. 4. Basic difference between static and dynamic loading P P(t) Resistance due to internal Accelerations producing inertia elastic forces of structure forces (inertia forces form a significant portion of load equilibrated by the internal elastic forces of the structure) Static Dynamic
  5. 5. Characteristics and sources of Typical Dynamic Loadings Periodic Loading: (a) Unbalanced rotating machine in building (b) Rotating propeller at stem of ship Non Periodic Loading: (c) Bomb blast pressure on building (d) Earthquake on water tank (a) Simple harmonic (b) Complex (c) Impulsive (d) Long duration
  6. 6. Dynamic Degrees of Freedom
  7. 7. The number of independent displacement components that must be considered to represent the effects of all significant inertia forces of a structure.
  8. 8. Examples
  9. 9. Dynamic Degrees of Freedom 1. 2. Massless Spring Inextensible spring with Spring θ mass (a) (b) (c)
  10. 10. Dynamic Degrees of Freedom 3. Flexible and Finite Flexible and Point Rigid bar with massless mass massless mass distributed mass Massless spring (a) (b) (c) Flexible beam 4. Flexible and Flexible beam with distributed massless with distributed mass mass Point mass y(x) = c11(x)+ c22(x)+…… (a) (b) (c)
  11. 11. Dynamic Degrees of Freedom Rigid deck 5. Massless columns
  12. 12. Idealisation of Structure as SDOF
  13. 13. Mathematical model - SDOF System x k c m P(t) Mass element ,m - representing the mass and inertial characteristic of the structure Spring element ,k - representing the elastic restoring force and potential energy capacity of the structure. Dashpot, c - representing the frictional characteristics and energy losses of the structure Excitation force, P(t) - represents the external force acting on structure.
  14. 14. Newton’s second law of motion Force = P(t) = Rate of change of momentum of any mass d ₩ dx = m dt │ dt .. x, x When mass is not varying with time, P(t) .. mx .. P(t) = m x(t) = mass x acceleration Inertia force D’Alembert’s Principle: This Principle states that “mass develops an inertia force proportional to its acceleration and opposing it”.
  15. 15. The force P(t) includes , mg 1) Elastic constraints which opposes displacement .. 2) Viscous forces which resist velocities kx mx 3) External forces which are independently defined 4) Inertia forces which resist accelerations N mx + k x = 0 &&
  16. 16. Equations of motion: fs Spring force - fs  x k . 1 Viscous damping force - fd  x x .. Inertia Force - fI  x fD External Forces - P(t) c 1 . x
  17. 17. Examples
  18. 18. FBD for mass 1. x fs = kx k .. fI= m . x P(t) c m P(t) fd = cx mx + cx + kx = P (t ) && & (1) dx d 2x x= & ; && = x 2. dt dt 2 .. . mx 0+ cx Kx + w P(t) w δst = w/k x(t) = displacement measured from position of static equilibrium P(t) mx + cx + kx = P(t ) && & (2)
  19. 19. (3a) Rigid ,massless P(t) m P(t) .. mx k c x ₩ a ₩ b a k x c x & │ L │ L b d x – vertical displacement of the mass L measured from the position of static equilibrium 2 2 ₩b ₩a d mx + c && x+k & x = P (t ) (3) │L │L L
  20. 20. (3b) a Rigid b massless ₩a k x d k │L L ₩b c x & │L c P(t) P(t) x .. mx m Stiffness term W=mg 2 2 ₩b ← ₩a W d mx + c && & + ↑k x + x = P (t ) (4) │L ↑ │L → L L Note: The stiffness is larger in this case
  21. 21. (3c) m .. mx x P(t) P(t) c ₩ . b c x & │ L L k d ₩ a k x b │ L Rigid massless a Stiffness term 2 2 ₩b ← ₩a W d mx + c && x + ↑k & - x = P (t ) (5) │L ↑ │L → L L Note: The stiffness is decreased in this case. The stiffness term goes to zero - Effective stiffness is zero – unstable - Buckling load
  22. 22. (4a) μ (distributed mass) P(t) m mL P(t) x && (2/3)L 2 k c .. x mx ₩ a ₩ . b a fs = k x fd = c x & │ L │ L b d L 2 2 ← 1₩ ₩b a d ↑m + 3 m L → && + c x │L x+k & │L x = L P (t )
  23. 23. (4b) a μ ₩ a b fs = k x (2/3)L d k │ L L ₩ b mL fd = c x & x && c │ L 2 P(t) P(t) .. mx x m 2 2 ← 1₩ ₩ b ← a ₩ W 1 d ↑m + 3 m L x && + c & + ↑k x ᄆ + mg x= P (t ) → │L ↑ │L → │L 2 L (Negative sign for the bar supported at bottom)
  24. 24. Special cases: (4c) (4d) (4e) L k L J θ x g 3 g && + x = 0 x && + x x=0 && J q + kq = 0 L 2 L
  25. 25. x ke ce me Pe(t) me && + ce x + ke x = Pe (t ) x & (6) me - equivalent or effective mass Ce - equivalent or effective damping coefficient Ke - equivalent or effective stiffness Pe - equivalent or effective force
  26. 26. Internal m/2 hinge P(t) Rigid with uniform mass μL/2 = m/2 Rigid ,massless N N x(t) k c L/4 L/4 L/8 L/8 L/4 m && x m x ₩ L && m ᅲ P(t) ᅲ = && x 2 2 │ 2 2 4 N N o RL ₩x ₩x & k c │2 │2 7 1 1 ← 16 N 3 (5) m && + c x + k ↑ + x & 1 x= P (t ) 24 4 4 → kL 4
  27. 27. 7 1 1 ← 16 N 3 me = m ce = c ke = k ↑1 + Pe (t ) = P (t ) (7) 24 4 4 → kL 4 For N = - (1/16) k L ke = 0 This value of N corresponds to critical buckling load
  28. 28. Free Vibration Undamped SDOF System Damped SDOF systems
  29. 29. Free Vibration of Undamped System && + p 2 x = 0 x (9) ₩k (10) p =2 │m General solution is, x(t) = A cos pt + B sin pt (or) (11) x(t) = C sin (pt + α) (12) where, (13) C = A +B 2 2 2p m (14) T= = 2p = natural period p k p 1 (15) f = = = natural frequency 2p T p - circular natural frequency of undamped system in Hz.
  30. 30. Amplitude of motion 2 ₩v x x0 + 0 2 │p vo x0 at t 2p T= p 2 v0 ₩0 v x(t ) = x0 cos pt + sin pt or x(t ) = x0 + 2 sin ( pt + a ) (16) p │p x0 X 0 =initial displacement (17) where, tan a = v0 p V0 =initial velocity
  31. 31. Natural frequencies of other SDF systems p– square root of the coefficient of displacement term divided by coefficient of acceleration (18) For Simple Pendulum, p= g L 2 k₩ a g (19) For system considered in (3b) , p = + m │L L For system considered in (5) , p = 6 k ₩ 16 N (20) 1+ 7 m│ kL 6 k For N=0 , p = po = (21) 7m 1 and for N = - kL , p = 0 16
  32. 32. Condition of instability 1 N =- kL = N cr (22) 16 N p = po 1 - (23) N cr p2 Ncr N
  33. 33. Natural frequencies of single mass systems p= k /m (10) Letting m = W/g and noting that W/k = δst (24) δst is the static deflection of the mass due to a force equal to its weight (the force applied in the direction of motion). g (25) p= d st 1 g (26) f = 2p d st (27) δst is expressed in m, T = 2 d st
  34. 34. Relationship between Simple oscillator and Simple pendulum L g g p= p= dst L Hence, δst = L = 0.025 m f ≈ 3.1 cps δst = L = 0.25 m f ≈ 1.0 cps δst = L = 2.50 m f ≈ 0.3 cps
  35. 35. Effective stiffness ke and static deflection δst ke g p= = (28) m d st ke - the static force which when applied to the mass will deflect the mass by a unit amount. δst - the static deflection of the mass due to its own weight the force (weight) being applied in the direction of motion.
  36. 36. Determination of Force - Displacement relation, F-∆ 1. Apply the static force ,F on the mass in the direction of motion 3. Compute or measure the resulting deflection of the mass ,∆ Then , ke = F / ∆ δst = ∆ due to F = W
  37. 37. Examples
  38. 38. Rigid ,massless m 2 (a) L F =  k a k 2 F a a Therefore, ke = =   k L  L F or From Equilibrium, 2 L W (29) d st =   a k L  F F a From Compatibility, 2 L F =  LF a k   ak
  39. 39. (b) Rigid bar m (c) m Rigid bar k3 k1 k2 k1 k2 a a L L F F = F1 + F2 ∆ k3 k1 k2 k1 k2 2 F F a ∆ = ∆1 + ∆2 = + F =   k1 + k 2  k 3  a 2 L   k1 + k 2 L 2 F a  1 1 1 k e = =   k1 + k 2 (30) = = + (31)  L F k e  a 2 k3   k1 + k 2 L
  40. 40. (d) (f) Flexible but mass less k1 kn 3EI ke = ke = k1 +k2 + ……+ kn L3 (32) (e) k1 k2 12EI . . ke = . L3 kn ∆e = ∆1 + ∆2 + ……+ ∆n F F F 3EI (34) = + + ...... + ke = k1 k 2 kn L3  1 1 1 1 (33) = = + ...... + F k e k1 k 2 kn
  41. 41. (g) Rigid deck; columns mass less & (h) axially inextensible EI k2 k1 L1 E2I2 L2 E1I1 L F k2 k2 Lateral Stiffness : kb k1 kb+k1 E1 I1 E2 I 2 ke = 12 3 + 3 3 L L 1 1 1 1 1 = + = + ke k2 kb + k1 k2 3EI + k 1 L3
  42. 42. (i) EI 1 FL3 5 RL3 = - 3 EI 48 EI k Eliminating R, kL3 L/2 L/2 768 + 7 F EI ᅲ1 FL3 = kL3 3 EI ∆ 768 + 32 EI R where, 3 3 5 FL 1 RL R kL3 a = - = 768 + 32 48 EI 24 EI k F EI ᅲ3 EI 5 ke = = where, R = F  kL3 L3 EI 768 + 7 2 + 48 3 EI L
  43. 43. (j) (l) L/2 L/2 L/2 L/2 48 EI 768EI ke = ke = L3 7 L3 (k) (m) L/2 L/2 a b 192 EI 3EIL ke = ke = 2 2 L3 ab
  44. 44. (n) (p) L 2R d Gd 4 AE k = k = 64nR 3 L n – number of turns A – Cross sectional area (0) (q) L EI GJ k = k = L L I - moment of inertia of cross sectional area J – Torsional constant of cross L - Total length section
  45. 45. Natural frequencies of simple MDF systems treated as SDF m a A, E, I, L (i) (ii) (iii) (iv) Columns are massless and can move only in the plane of paper • Vertical mode of vibration 2 AE 2 AE (35) ke = pv = L mL
  46. 46. • For pitching or rocking mode 1 .. a 2a AE my + ya = 0 2 2 3 L AE/L AE/L 1 .. AE m ay + y=0 6 L 1 .. AE my + y=0 6 L (AE/L)y (AE/L)y 6 AE pp = = 3 pv (36) mL • For Lateral mode 2 12 EI EI AE ₩ r ke = 2 3 = 24 3 = 24 L L L │L r is the radius of gyration of cross section of each column 2 12 EI EI AE ₩r ke = 2 3 = 24 3 = 24 L L L │L (37) plateral < paxial < ppitching
  47. 47. Free Vibration of Damped SDOF
  48. 48. Free Vibration of damped SDOF systems mx + cx + kx = 0 && & k && + c k c m x x+ x =0 & m m && + 2 p +x 2 = 0 xζpx & (A) x k where, p= m c c ζ= = (Dimensionless parameter) (38) 2mp 2 km
  49. 49. Solution of Eq.(A) may be obtained by a function in the form x = ert where r is a constant to be determined. Substituting this into (A) we obtain, ert ( r 2 + 2 ζpr p+ 2 ) =0 In order for this equation to be valid for all values of t, r2 + 2 ζpr p+ 2 =0 or ( r1,2 = p -z ᄆ z 2 - 1 )
  50. 50. rt rt Thus e 1 and e 2 are solutions and, provided r1 and r2 are different from one another, the complete solution is x = c1e + c2 e r1t r2t The constants of integration c1 and c2 must be evaluated from the initial conditions of the motion. Note that for ζ >1, r1 and r2 are real and negative for ζ <1, r1 and r2 are imaginary and for ζ =1, r1= r2= -p Solution depends on whether ζ is smaller than, greater than, or equal to one.
  51. 51. For z  1 (Light Damping) : x ( t ) = e -z pt [ A cos pd t + B sin pd t ] (39) where, pd = p 1 - z 2 (40) ‘A’ and ‘B’ are related to the initial conditions as follows A = x0 v0 z B= + x0 pd 1-z 2 In other words, Eqn. 39 can also be written as,  v z   x ( t) = e -z pt  xo cos pd t +  o + x  sin pd t  (41)   pd 1-z 2 o     
  52. 52. g Td = 2π / pd Extremum point ( x(t ) = 0 ) Point of tangency ( cos( pd t - a ) = 1) 2p x T = = Damped natural period pd xn Xn+1 t pd = p 1 - z = Damped circular natural frequency 2 2p Td = = Damped natural period pd pd = p 1 - z 2 = Damped circular natural frequency
  53. 53. Motion known as Damped harmonic motion A system behaving in this manner (i.e., a system for whichz  1 ) is said to be Underdamped or Subcritically damped The behaviour of structure is generally of this type, as the practical range of z is normally < 0.2 The equation shows that damping lowers the natural frequency of the system, but for values of z < 0.2 the reduction is for all practical purpose negligible. Unless otherwise indicated the term natural frequency will refer to the frequency of the undamped system
  54. 54. Rate of Decay of Peaks xn +1 -z p 2p  z  =e pd = exp  -2p  (42) xn   1-z  2  xn +1 0.8 0.7 xn 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 z
  55. 55. Logarithmic decrement xn Defined as d = ln (43) xn +1 It is an alternative measure of damping and is related to z by the equation z d = 2p ; 2pz (44) 1-z 2 For small values of damping, x n (45) d  = 2pz xn When damping is quite small, 1 xn (46) d = ln N xn + N
  56. 56. For z  1 (Heavy Damping) Such system is said to be over damped or super critically damped. x(t ) = C1e( - ) t + C2 e( - ) t i.e., the response equation will be sum of two exponentially decaying curve In this case r1 and r2 are real negative roots. x xo o t
  57. 57. For z = 1 Such system is said to be critically damped. x (t ) = C1e - pt + C2te - pt With initial conditions, x (t ) = ←x0 ( 1 + pt ) + v0t e - pt → The value of ‘c’ for which z =1 is known as the critical coefficient of damping Ccr = 2mp = 2 km (47) C Therefore, z = (48) Ccr
  58. 58. Response to Impulsive Forces Response to simple Force Pulses Response to a Step Pulse Response to a Rectangular Pulse Response to Half-Sine Pulse Response to Half-cycle Force Pulses Response to Step force Response to Multi-Cycle Force Pulses
  59. 59. Response to Impulsive Forces Po Let the duration of force,t1 be small compared to P(t) the natural period of the system The effect of the force in this case is equivalent to an instantaneous velocity change without corresponding change in displacement t t1 << T The velocity,V0 ,imparted to the system is obtained from the impulse-momentum relationship (49) mV0 = I = Area under forcing function = α P0 t1 1 for a rectangular pulse where , α 2 / π for a half-sine wave 1 / 2 for a triangular pulse a Pt0 1 Therefore, V0 = (50) m
  60. 60. For an undamped system, the maximum response is determined from as , a P0 t1 = a P0 kt1 = a ( x ) pt xmax = V = 0 p m p k mp st 0 1 Therefore, xmax = 2pa ft1( xst )0 xmax t1 or = 2pa ft1 = 2pa (51) ( xst ) 0 T
  61. 61. •Damping has much less importance in controlling maximum response of a structure to impulsive load. The maximum will be reached in a very short time, before the damping forces can absorb much energy from the structure. For this reason only undamped response to Impulsive loading is considered. • Important: in design of Vehicles such as trucks, automobiles or traveling cranes
  62. 62. Response to simple Force Pulses P (t ) P(t) && + 2V px + p x = x & 2 (52) m or && + 2V px + p 2 x = p 2 xst (t ) x & (53) P (t ) t where, xst (t ) = k P (t ) = Static displacement induced by k exciting force at time, t General Form of solution: (54) x(t) = xhomogeneous + xparticular
  63. 63. Response to a Step Pulse For undamped system, x + p2 x = p2 (xst)o P(t) Po Po where (xst)o = k x(t) = A cos pt + B sin pt + (xst)o 0 t At t = 0 , x = 0 and v = 0 A = - (xst)o and B = 0 t (55) x(t) = (xst)o [1 – cos pt] = [ 1 – cos 2π T ]
  64. 64. Response to a Step Pulse…. z=0 x (t ) 2 ( xst )o 1 0 (t / T) For damped systems it can be shown that: ₩ ₩ z -z pt x(t ) = ( xst )0 1 - e cos pd t + sin pd t │ │ 1- z 2 zp - xmax 1-z 2 = 1+ e (56) ( xst )0
  65. 65. Response to a Rectangular Pulse For t  t1, solution is the same as before, x(t ) = ( xst ) ( 1 - cos pt ) 0 (55) For t  t1, we have a condition of free vibration, and the solution can be obtained by application of Eq.17a as follows: 2 2 ₩ Vi x(t ) = xi + sin ( p(t -t1) +a ) │p P(t) where, tan a = x i Po Vi / p xi = ( xst ) ( 1 - cos pt ) 0 t1 t Vi = p sin pt1
  66. 66. Response to a Rectangular Pulse pt 1-cos pt1 2sin 2 1 pt1 tan a = = 2 sin pt1 pt1 pt1 = tan 2 2sin cos 2 2 pt hence, a= 1 2 2 2 ₩ t1 So, x(t ) = (1-cos pt1) +sin pt1 ( xst )0 sin p(t -t1) + p (57(a)) │ 2 ₩ t pt ₩ t x(t ) = ( xst )0 2(1 - cos pt1 ) sin p t - 1 = 2 ( xst ) 0 sin 1 sin p t - 1 (57(b)) │ 2 2 │ 2 (Amplitude of motion)
  67. 67. Response to a Rectangular Pulse… x(t)/(xst)0 2 2 t1/T=2 t1/T=1.5 0 1 2 0 1 2 t/T t/T t1/T=1/p x(t)/(xst)0 1.68 2 t1/T=1 1/6 0 1 2 t/T 0 1 t/T In the plots, we have implicitly assumed that T constant and t1 varies; Results also applicable when t1 = fixed and T varies
  68. 68. Dynamic response of undamped SDF system to rectangular pulse force. Static solution is shown by dotted lines
  69. 69. (a) Forced response Free response (b) Overall maximum Response to rectangular pulse force: (a) maximum response during each of forced vibration and free vibration phases; (b) shock spectrum
  70. 70. Response to a Rectangular Pulse… 3 Impulsive solution, 2π f t1 xmax 2 ( xst ) 0 1 0 1 2 3 f t1 = t 1/T This diagram Is known as the response spectrum of the system for the particular forcing function considered. Note that with xmax determined, the maximum spring force Fmax = k xmax Fmax kxmax xmax (58) In fact, = = ( Fst ) 0 P0 ( xst ) 0
  71. 71. Response to Half-Sine Pulse P(t) = Po sin ωt, where ω = π / t1 P(t) POsin ωt x + p2 x = p2 (xst)o sin ωt for t t 1   = 0x(t ) = ( xst )0 [sin wt - w sin pt ] for t t1 t1 t p 1- w 2 ₩ for t t1, │p or x(t ) = ( xst ) 0 ₩ sin pt 1T - sin 2p t (59) 2 t1 2 t1 T 1- 1 T ₩ │ 4 │ t1  w ₩ for t t1 2 p x(t ) = │ 2 cos pt ( xst )0 sin pt - 1 t ₩ 1 1 ₩w 2 │ 2T -1 │ p t cos p t 1 1 (60) or x(t ) = T T ( x ) sin 2p ₩t - 1 t 1 st 0 T 2T 0.25 - t 2 ₩1 │ │T
  72. 72. • Note that in these solutions, t1 and T enter as a ratio and that similarly, t appears as the ratio t /T. In other words, f t1 = t1 / T may be interpreted either as a duration or as a frequency parameter • In the following response histories, t1 will be presumed to be the same but the results in a given case are applicable to any combination of t1 and T for which t1/T has the indicated value • In the derivation of response to a half-sine pulse and in the response histories, the system is presumed to be initially at rest
  73. 73. Dynamic response of undamped SDF system to half cycle sine pulse force; static solution is shown by dashed lines
  74. 74. Response to half cycle sine pulse force (a) response maxima during forced vibration phase; (b) maximum responses during each of forced vibration and free vibration phases; (c) shock spectrum
  75. 75. 2pft1 4pft 1 pft1 t1 t1 t1 ft1 Shock spectra for three force pulses of equal magnitude
  76. 76. Response to Half-cycle Force Pulses • For low values of ft1 (say < 0.2), the maximum value of xmax or AF is dependent on the area under the force pulse i.e, Impulsive-sensitive. Limiting value is governed by Impulse Force Response. • At high values of ft1, rate of application of load controls the AF. The rise time for the rectangular pulse, tr, is zero, whereas for the half-sine pulse it is finite. For all continuous inputs, the high-frequency limit of AF is unity. • The absolute maximum value of the spectrum is relatively insensitive to the detailed shape of the pulse(2 Vs 1.7), but it is generally larger for pulses with small rise times (i.e, when the peak value of the force is attained rapidly). • The frequency value ft1 corresponding to the peak spectral ordinate is also relatively insensitive to the detailed shape of the pulse. For the particular inputs investigated, it may be considered to range between ft1 = 0.5 and 0.8. * AF=Amplification factor
  77. 77. Conditions under which response is static: On the basis of the spectrum for the ‘ramp pulse’ presented next, it is concluded that the AF may be taken as unity when: ftr = 2 (61) For the pulse of arbitrary shape, tr should be interpreted as the horizontal projection of a straight line extending from the beginning of the pulse to its peak ordinate with a slope approximately equal to the maximum slope of the pulse. This can normally be done by inspection. For a discontinuous pulse, tr = 0 and the frequency value satisfying ftr = 2 is, as it should be, infinite. In other words, the high-frequency value of the AF is always greater than one in this case
  78. 78. Response to Step force For t  tr P(t) ←₩ t ₩ pt sin x(t ) = ( xst )0 ↑ - Po →│ tr │ ptr ←₩ t 1 T ₩p t 2 = ( xst )0 ↑ - sin tr t →│ tr 2p tr │ T For t  tr = ₩ t ← sin pt 1 P0 x(t ) = ( xst )0 ↑ - 1 + sin p (t - tr ) │ tr → ptr ptr ← 1 T t 1 T (t - tr ) ( xst )0 ↑1 - sin 2p + sin 2p → 2p tr T 2p tr T + Differentiating and equating to zero, the peak time is (tr - t ) P0 obtained as: tr 1 - cos ptr tan ptr = sin ptr tr
  79. 79. Substituting these quantities into x(t), the peak amplitude is found as: xmax 1 2 pt = 1+ 2(1 - cos ptr ) = 1 + sin r ( xst ) 0 ptr ptr 2 1T pt = [1 + sin r p tr T xmax ( xst )0 f tr
  80. 80. For Rectangular Pulse: x(t ) = ( xst )0 [1 - cos pt ] for t  t1 P(t) ₩ pt1 ₩ t1 Po = ( xst )0 2 sin sin p t - for t  t1 │ 2 │ 2 t Half-Sine Pulse: ( xst )0 ₩ pt 1 T t x(t ) = 2 sin - sin 2p for t  t1 1₩ T │ t1 2 t1 T 1- P(t) 4 │ t1 POsinωt ( xst )0 ( ft1 ) cos p ft1 ( f sin ( 2p ft - p ft1 ) ) 0.25 - ( ft1 ) 2 t1 t
  81. 81. Design Spectrum for Half-Cycle Force Pulses 2 A B xmax 1 C D (x xt )0 ft1=0.6 ft1= 2 o ft1 x • Line OA defined by equation.51 (i.e max = 2 π α f t1 = 2 π α t1 ) ( xst )0 T • Ordinate of point B taken as 1.6 and abscissa as shown • The frequency beyond which AF=1 is defined by equation. 61 • The transition curve BC is tangent at B and has a cusp at C Spectrum applicable to undamped systems.
  82. 82. Response to Multi-Cycle Force Pulses Effect of Full-Cycle Sine Pulse • The high-frequency, right hand limit is defined by the rules given before • The peak value of the spectrum in this case is twice as large as for the half-sine pulse, indicating that this peak is controlled by the ‘periodicity’ of the forcing function. In this case, the peak values of the responses induced by the individual half-cycle pulses are additive • The peak value of the spectrum occurs, as before, for a value ft1=0.6 • The characteristics of the spectrum in the left-handed, low- frequency limit cannot be determined in this case by application of the impulse-momentum relationship. However, the concepts may be used, which will be discussed later.
  83. 83. Effect of n Half-Sine Pulses The absolute maximum value of the spectrum in this case occurs at a value of, ft1=0.5 (62) Where t1 is the duration of each pulse and the value of the peak is (63) approximately equal to: xmax = n (p/2) (xst)o
  84. 84. Effect of a sequence of Impulses I Suppose that t1 = T/2 t1 I t I/mp x(t) Effect of first pulse x(t) I/mp Effect of second pulse x(t) 2I/mp Combined effect of two pulses t
  85. 85. Effect of a sequence of Impulses • For n equal impulses, of successively opposite signs, spaced at intervals t1 = T / 2 and xmax = n I/(mp) (64) • For n equal impulses of the same sign, the above equation holds when the pulses are spaced at interval t1 = T • For n unequal impulses spaced at the critical spacings noted (65) above, xmax = Σ Ij /(mp) (summation over j for 1 to n). Where Ij is the magnitude of the jth impulse • If spacing of impulses are different, the effects are combined vectorially
  86. 86. Response of Damped systems to Sinusoidal Force
  87. 87. Response of Damped systems to Sinusoidal Force P(t) = P0 sinωt where ω = p/ t1= Circular frequency P(t) of the exciting force t1 t Solution: The Particular solution in this case may be taken as x(t) = M sinωt + N cosωt (a) Substituting Eq.(a) into Eq.52, and combining all terms involving sinωt and cos ωt, we obtain [-w 2 M - 2V pw N + p 2 M ]sin wt + [-w 2 N + 2 V pw M + p 2 N ]cos wt = p 2 ( X st )0 sin wt
  88. 88. This leads to (p2-ω2)M - 2 ζpωN = p2(xst)0 (c) 2 ζpωM+ (p2-ω2)N = 0 Where ₩ 2 ₩ω ω 1- 2ζ │p N=- │p (x st )0 M= 2 (x st )0 2 2 2 ← ₩ ω 2 ₩ω 2 ← ₩ ω ₩ω ↑1- +4ζ 2 ↑1- +4ζ 2 ↑ │p → │p ↑ │p → │p The solution in this case is (x st )0 z a x(t) = e- pt(A cos pD t +B sin pD t) + (1- ) +4ζ  2 2 2 2 sin (ωt – ) (66) where w 1 = = p 2ft1 (67) 2z tan a = (68) (1- 2 )
  89. 89. Steady State Response P(t) a t w x(t) x(t) 1 = sin(wt - a ) (69) (x st ) 0 (1-  ) + 4z  2 2 2 2 x max 1 AF = = (70) (x st ) 0 (1- 2 ) 2 + 4z 2 2 w 1 p Note that at = = 1, AF = = (71) p 2z d
  90. 90. AF α w = p w = p
  91. 91. Effect of damping • Reduces the response, and the greater the amount of damping, the greater the reduction. • The effect is different in different regions of the spectrum. • The greatest reduction is obtained where most needed (i.e., at and near resonance). • Near resonance, response is very sensitive to variation in ζ (see Eq.71). Accordingly, the effect of damping must be considered and the value of ζ must be known accurately in this case.
  92. 92. Resonant Frequency and Amplitude wres = p 1- 2z 2 (72) 1 (AF)max = (73) 2z 1- 2z 2 1 These equations are valid only for z  2 1 For values of < z1 2 ωres = 0 (74) (A.F.)max = 1
  93. 93. Transmissibility of system The dynamic force transmitted to the base of the SDOF system is ← cx& (75) F = kx + cx = k ↑x + & → k Substituting x from Eq.(69), we obtain P0 1 cw F= k [sin(wt - a ) + cos(wt - a )] k (1- 2 ) 2 + 4z 2 2 k cw cw w Noting that = 2 = 2z = 2z and combining the sine and k mp p cosine terms into a single sine term, we obtain F(t) 1 + 4z 2 2 = sin(wt - a + g ) (76) P0 (1- 2 ) 2 + 4z 2 2 where g is the phase angle defined by tan g = 2ζ  (77)
  94. 94. Transmissibility of system The ratio of the amplitudes of the transmitted force and the applied force is defined as the transmissibility of the system, TR, and is given by F 1 + 4z 2 2 TR = 0 = (78) P0 (1-  ) + 4z  2 2 2 2 The variation of TR with and ζ is shown in the following figure. For the special case of ζ =0, Eq.78 reduces to 1 TR = (1- 2 ) which is the same expression as for the amplification factor xmax/(xst)0
  95. 95. Transmissibility of system TR ω = p Transmissibility for harmonic excitation
  96. 96. • Irrespective of the amount of damping involved, TR<1 only for values of (ω/p)2 >2. In other words, in order for the transmitted force to be less than the applied force, the support system must be flexible. in cms w f e f e dst Noting that = =  2 p f 4.98 50 (79) The static deflection of the system, dst must be dst  fe 2 0.5 dst in cms 0.1 5 10 40 fe fe is the frequency of the exciting force, in cps
  97. 97. • In the frequency range where TR<1, damping increases the transmissibility. In spite of this it is desirable to have some amount of damping to minimize the undesirable effect of the nearly resonant condition which will develop during starting and stopping operations as the exciting frequency passes through the natural frequency of the system. • When ζ is negligibly small, the flexibility of the supports needed to ensure a prescribed value of TR may be determined from 1 TR = ₩w ₩ 2 (80a) -1 p ││ Proceeding as before, we find that the value of dst corresponding to Eq.(80a) is ← 1 p2 ← 1 25 dst ; ↑1 + 2 (in) or dst ; ↑1 + 2 (cm) (80b) → TR f e → TR f e
  98. 98. Application Consider a reciprocating or rotating machine which, due to unbalance of its moving parts, is acted upon by a force P0 sinωt. If the machine were attached rigidly to a supporting structure as shown in Fig.(a), the amplitude of the force transmitted to the structure would be P0 (i.e., TR=1). If P0 is large, it may induce undesirable vibrations in the structure, and it may be necessary to reduce the magnitude of the transmitted force. This can be done by the use of an approximately designed spring- dashpot support system, as shown in Fig (b) and (c). P0 sinωt P0 sinωt m P0 sinωt m mb m k c k c (a) (b) (c)
  99. 99. • If the support flexibility is such that is less than the value defined by Eq.(79), the transmitted force will be greater than applied, and the insertion of the flexible support will have an adverse effect. • The required flexibility is defined by Eq.(80b), where TR is the desired transmissibility. • The value of may be increased either by decreasing the spring stiffness, k, or increasing the weight of the moving mass, as shown in Fig.(c).
  100. 100. Application to Ground-Excited systems k m c y(t) x(t) For systems subjected to a sinusoidal base displacement, y(t) = y0 sinωt it can be shown that the ratio of the steady state displacement amplitude, xmax, to the maximum displacement of the base motion, yo, is defined in Eq.(78). Thus TR has a double meaning, and Eq.(78) can also be used to proportion the support systems of sensitive instruments or equipment items that may be mounted on a vibrating structure. For systems for which ζ ,may be considered negligible, the value d of st required to limit the transmissibility TR = xmax/y0 to a specified value may be determined from Eq.(80b).
  101. 101. Rotating Unbalance Total mass of machine = M M m unbalanced mass =m e ωt eccentricity =e angular velocity =ω x d2 ( M - m) && + m 2 ( x + e sin wt ) + cx + kx = 0 x & k/2 c k/2 dt Mx + cx + kx = mew 2 sin wt && &
  102. 102. Reciprocating unbalance m  e  L F = mew sin w t + sin 2w t  2 e  L  ωt M e - radius of crank shaft L - length of the connectivity rod e/L - is small quantity second term can be neglected
  103. 103. Determination of Natural frequency and Damping Steady State Response Curves Structure subjected to a sinusoidally varying force of fixed amplitude for a series of frequencies. The exciting force may be generated by two masses rotating about the same axis in opposite direction For each frequency, determine the amplitude of the resulting steady-state displacement ( or a quantity which is proportional to x, such as strain in a member) and plot a frequency response curve (response spectrum) For negligibly small damping, the natural frequency is the value of fe for which the response is maximum. When damping is not negligible, determine p =2πf from Eq.72. The damping factor , ζ may be determined as follows:
  104. 104. Determination of Natural frequency and Damping Resonant Amplification Method Half-Power or Bandwidth Method Duhamel’s Integral
  105. 105. Determination of Natural frequency and Damping (a) Resonant Amplification Method Determine maximum amplification (A.F)max=(x0)max/ (xst)0 Evaluate z from Eq.73 or its simpler version, Eq.71, when z is small Limitations: It may not be possible to apply a sufficiently large P0 to measure (xst)0 reliably, and it may not be possible to evaluate (xst)0 reliably by analytical means.
  106. 106. (b) Half-Power or Bandwidth Method: In this method ζ is determined from the part of the spectrum near the peak steps involved are as follows, 5. Determine Peak of curve, (x0)max ( ) 2. Draw a horizontal line at a response level of 1/ 2 ( x0 ) max and , determine the intersection points with the response spectrum. These points are known as the half-power points of the spectrum f 3. Evaluate the bandwidth, defined as f
  107. 107. (xo)max xo 1 2 (xo)max (xo)st ∆f fe f
  108. 108. Determination of Natural frequency and Damping… 1.For small amounts of damping, it can be shown that ζ is related to the bandwidth by the equation 1 f z = (81) 2 f Limitations: Unless the peaked portion of the spectrum is determine accurately, it would be impossible to evaluate reliably the damping factor. As an indication of the frequency control capability required for the exciter , note that for f = 5cps, and ζ = 0.01, the frequency difference f = 2(0.01)5 = 0.1cps with the Cal Tech vibrator it is possible to change the frequency to a value that differs by one tenth of a percent from its previous value.
  109. 109. Derivation 1 1 1  1  2 = 2 2 2z  (1 -  ) + (2z )  2 1 1 = 8z ( 1 -  ) + ( 4z 2 2 ) 2 2 2  2 = 1 - 2z 2 ᄆ 2z 1 + z 2 12 = 1 - 2z 2 - 2z 1 + z 2 1 = 1 - z - z 2 22 = 1 - 2z 2 + 2z 1 + z 2 2 = 1 + z - z 2 1 1 ( f1 - f 2 ) ( f1 + f 2 ) z = (1 - 2 ) = f ᄏ 2 2 f 2 ( f 2 - f1 ) z ᄏ ( f1 + f 2 )
  110. 110. Other Methods for Evaluating response of SDF Systems (c) Duhamel’s Integral In this approach the forcing function is conceived as being made up of a series of vertical strips, as shown in the figure, the effect of each strip is then computed by application of the solution for free vibration, and the total effect is determined by superposition of the component effects P(t) P() x(t) d o t t  
  111. 111. I=P (τ )dτ The strip of loading shown shaded represents and impulse, I = P() d For an undamped SDF system, this induces a displacement P ( )d x = sin [ p (t -  ) ] (82) mp The displacement at time t induced by integration as 1  =t x(t ) = P ( ) sin ← p ( t -  ) d → (83) mp  = 0 or t x (t ) = p  xst ( ) sin  p ( t -  )  d   (84) 0
  112. 112. Implicit in this derivation is the assumption that the system is initially (at t=0) at rest. For arbitrary initial conditions, Eqn 84 should be augmented by the free vibration terms as follows t V0 x ( t ) = x0 cos pt + sin pt + p xst (  ) sin ← p ( t -  ) d → p 0 For viscously damped system with z  1 ,becomes P (  ) d -z p( t - ) x(t ) = e sin ← pd ( t -  ) → mpd Leading to the following counterpart of Eqn.84 t p x( t) =  xst (  ) e -z p( t - ) sin  pd ( t -  )  d   1-z 2 0
  113. 113. The effect of the initial motion in this case is defined by Eqn. 41 Eqns. 84 and 87 are referred to in the literature with different names. They are most commonly known as Duhamel’s Integrals, but are also identified as the superposition integrals, convolution integrals, or Dorel’s integrals. Example1: Evaluate response to rectangular pulse, take z = 0 and For t ᆪ t1 x0 = v0 = 0 P(t) t x ( t ) = p ( xst ) o sin ← p ( t -  ) d = ( xst ) o cos p ( t -  )  =t Po →  =0 0 t1 t t ᄈ t1 = ( xst ) o [ 1 - cos pt ] For t1 x ( t ) = p ( xst ) 0 sin ← p ( t -  ) d + 0 = ( xst ) 0 ←cos p ( t - t1 ) - cos pt → → 0
  114. 114. Generalised SDOF System
  115. 115. Generalised SDOF System m*q( t ) + c*q( t ) + k *q( t ) = p * ( t ) && & q( t ) = Single generalised coordinate expressing the motion of the system m* = generalised mass c* = generalised damping coefficient k* = generalised stiffness p* = generalised force
  116. 116. y( x, t ) =  ( x ) q( t ) (a) (b) m(x) m1,j1 m2, j2 x1 l m* =  m( x ) ( x ) 2 dx + mi i2 + ji i2 0
  117. 117. c1 c2 a1(x) (c) c(x) L l l c* =  c( x ) ( x ) 2 dx +  EI ( x ) quot; ( x )a1 ( x )dx + ci i2 0 0
  118. 118. (d) k1 k2 k(x) (e) N l l l k * =  k ( x ) 2 ( x )dx +  EI ( x ) quot; ( x ) 2 dx + ki i2 -  N ( x ) ' ( x )2 dx 0 0 0
  119. 119. P(x,t) Pi(t) l p* =  p( x, t ) ( x )dx + pi (t ) i ( x ) 0 Note: Force direction and displacement direction is same (+ve)
  120. 120. Effect of damping Viscous damping Coulomb damping Hysteretic damping
  121. 121. Effect of damping Energy dissipated into heat or radiated away • The loss of energy from the oscillatory system results in the decay of amplitude of free vibration. • In steady-state forced vibration ,the loss of energy is balanced by the energy which is supplied by the excitation.
  122. 122. Effect of damping Energy dissipated mechanism may emanate from (i) Friction at supports & joints (ii) Hysteresis in material ,internal molecular friction, sliding friction (iii) Propagation of elastic waves into foundation ,radiation effect (iv) Air-resistance,fluid resistance (v) Cracks in concrete-may dependent on past load – history etc.., Exact mathematical description is quite complicated &not suitable for vibration analysis.
  123. 123. Simplified damping models have been proposed .These models are found to be adequate in evaluating the system response. Depending on the type of damping present ,the force displacement relationship when plotted may differ greatly. Force - displacement curve enclose an area ,referred to as the hysteresis loop,that is proportional to the energy lost per cycle. Wd = Fd dx In general Wd depends on temperature,frequency,amplitude. For viscous type Wd = Fd dx Fd = cx & 2p w Wd = p cw X = cxdx = cx dt = cw X 2 & & 2 2 2 cos 2 (wt - F )dt = p cw X 2 0
  124. 124. (a) Viscous damping Fd(t) = c x & c - coefficient of damping Wviscous - work done for one full cycle = cpw X 2 Fd cωX X(t) -X
  125. 125. (b) Equivalent viscous damping: p Ceqw X 2 = Wd Fd+kx ellipse Wd Ceq = pw X 2 2k 2Ws Cc = where k = 2 , Ws = strain energy w X x C Wd z = = Cc 4p Ws
  126. 126. (b) Coulomb damping: Fd - - Wcoulomb = 4 F X F It results from sliding of two dry surfaces -X X(t) The damping force=product of the normal force & the coefficient of friction (independent of the velocity once the motion starts. Linear decay 4Fd/k x1 x2 Frequency of oscillation ∆ x-1 k pm = m
  127. 127. 1 k ( X 12 - X -1 ) - Fd ( X 1 + X -1 ) = 0 2 2 1 k ( X 1 - X -1 ) = Fd 2 4 Fd X1 - X 2 = Decay in amplitude per cycle k The motion will cease ,however when the amplitude becomes less than ∆, at which the spring force is insufficient to overcome the static friction.
  128. 128. (c) Hysteretic damping (material damping or structural damping): - Inelastic deformation of the material composing the device Fd ←m - 1 W = 4F X hysteretic y ↑ m Fy → -x Fy is the yield force xy x Kh=elastic damper Xy Displacement at which material first yields stiffness x m= xy (d) Structural damping x& fD = z k x x& WD = 2z kX 2 = a X 2 Energy dissipated is frequency independent.
  129. 129. Equivalent viscous Coefficient e) Coulomb Ceq= 4F p WX 4 Fy  m - 1  h) Hysteretic Ceq= pw X  m    2k s c) Structural Ceq= pw
  130. 130. Reference Dynamics of Structures: Theory and Application to Earthquake Engineering – Anil K. Chopra, Prentice Hall India Reading Assignment Course notes & Reading material

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