This document provides an overview of structural dynamics and single degree of freedom (SDOF) systems. It defines dynamic degrees of freedom and idealizes structures as SDOF systems with mass, spring, and damper elements. The document derives equations of motion for various SDOF systems using Newton's second law and D'Alembert's principle. It also discusses free vibration of undamped systems and determination of natural frequencies. Methods to determine the effective stiffness and natural frequency of SDOF systems are presented through examples.

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

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. What is Dynamics ?

4. Basic difference between static and dynamic loading
P P(t)
Resistance due to internal
elastic forces of structure
Static
Accelerations producing inertia
forces (inertia forces form a
significant portion of load
equilibrated by the internal
elastic forces of the structure)
Dynamic

5. Characteristics and sources of Typical Dynamic Loadings
(a)
(b)
Periodic Loading:
Non Periodic Loading:
Unbalanced rotating
machine in building
Rotating propeller
at stem of ship
(c)
(d)
Bomb blast
pressure on
building
Earthquake on
water tank
(a) Simple harmonic (b) Complex (c) Impulsive (d) Long duration

6. Dynamic Degrees of Freedom

7. The number of independent displacement
components that must be considered to
represent the effects of all significant
inertia forces of a structure.

8. Examples

9. Massless
spring
Spring
with
mass
(a) (b) (c)
Inextensible
Spring
θ
1. 2.
Dynamic Degrees of Freedom

10. 3.
Rigid bar with
distributed mass
Massless
spring
Flexible and
massless
Flexible and
massless
Point
mass
Finite
mass
(a) (b) (c)
4. Flexible and
massless
Point
mass
Flexible beam
with distributed
mass
Flexible beam
with distributed
mass
(a) (b)
Dynamic Degrees of Freedom
y(x) = c11(x)+ c22(x)+……
(c)

11. 5.
Rigid deck
Massless
columns
Dynamic Degrees of Freedom

12. Idealisation of Structure as SDOF

13. Mathematical model - SDOF System
Mass element ,m
Spring element ,k
- representing the mass and inertial
characteristic of the structure
- 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.
P(t)
x
m
k
c

14. Newton’s second law of motion
Force = P(t) = Rate of change of momentum of any mass
d ￦
m
dx
dt ￨ dt
=
When mass is not varying with time,
..
P(t) = m x(t) = mass x acceleration
P(t)
x, x
m x
..
Inertia force
D’Alembert’s Principle: This Principle states that “mass develops
an inertia force proportional to its acceleration and opposing it”.
..

15. mg
kx mx
..
The force P(t) includes ,
1) Elastic constraints which opposes displacement
2) Viscous forces which resist velocities
3) External forces which are independently defined
4) Inertia forces which resist accelerations N
m&x&k x 
0

16. P(t)
Equations of motion:
Spring force - fs  x
.
Viscous damping force - fd  x
..
Inertia Force - fI  x
External Forces -
k
1
fs
x
c
1
fD
x
.

17. Examples

18. P(t)
x
m
k
c
FBD for mass
δst = w/k
P(t)
1.
2.
fd = cx
P(t)
f I = m .
x
fs = kx
..
dx d 2
x
dt dt 2
x
·  ; ·x· 
m·x·cx
·  kx  P(t) (1)
x(t) = displacement measured from
position of static equilibrium
m·x·cx
·  kx  P(t) (2)
P(t) w
Kx + w
0+ cx
mx
..
.

19. Rigid ,massless P(t) m
a
b
d
L
(3a) P(t)
m x
..
x
k
￦
a
￨L
x – vertical displacement of the mass
measured from the position of
static equilibrium
c
￦
b
￨L
x
k c
2 2
d
P ( t )
L
x 
￦b
￨ L
￦a
￨ L
m x  c x  k (3)
x

20. x
P(t)
m x
..
k ￦
a
￨L
c￦
b
￨L
x
Rigid
massless
m
a
b
d
L
k
c
2 2
W d
P(t)
L
￩
 x 
￦b
￨L
￦a
￨L L
mx  c x  ￪k
￪
￫
P(t)
Stiffness term
(3b)
W=mg
(4)
Note: The stiffness is larger in this case
x

21. 2 2
W d
k
L L
￩
￦b ￦a
 x  P(t)
L
￪
￨
￪
￫
m·x· c x 
￨ L
·
Rigid
massless
m
a
b
d
L
c
P(t)
x
k ￦
a
￨L
P(t)
m x
..
Stiffness term
(3c)
c x
·
￦
b .
￨L
(5)
Note: The stiffness is decreased in this case. The stiffness term
goes to zero - Effective stiffness is zero – unstable - Buckling load
x

22. 2 2
3
a d
L L
￩ 1 ￦ ￦
b
x  P ( t )
L
￪
￫ ￨ ￨
m   L x  c x  k
(4a) m
a
b
d
L
k c
μ (distributed mass)
P(t) P(t)
m x
..
￦
a ￦
b
fs  k x fd c x
￨L ￨L
.
2
L
x
(2/3)L
x

23. (4b)
2
3
d
L
￩ 1￦ b
￦
2
L
￩￦ a
ﾱ
W

1
 g x  P(t)
L 2 L
￪
￫ ￨ ￨ ￨
m   L ·x·c x
·  ￪k
￪
￫
m
a
b
d
L
k
c
P(t)
μ
P(t)
m x
..
x
L
￦
a
fs  k
￨
d
￨L
f c￦
b
x
·
2
L
·x·
(Negative sign for the bar
supported at bottom)
(2/3)L
x

24. Special cases:
x
L
L
k
J
θ
J·· k   0
L
·x·
g
x  0
2 L
·x·
3 g
x  0
(4c) (4d) (4e)

25. Pe(t)
x
me
ke
ce
me ·x·ce x
·  ke x  Pe(t)
me - equivalent or effective mass
Ce - equivalent or effective damping coefficient
Ke - equivalent or effective stiffness
Pe - equivalent or effective force
(6)

26. Rigid ,massless
Rigid with uniform mass μL/2 = m/2
L/4 L/4 L/8 L/8 L/4
k c
m/2
N
x(t)
P(t)
k ￦
x
￨2
c￦
x
·
￨2
P(t) ￦
L
ￗ
·x· 
m
·x·
￨ 2 2 4
2 2
m
ￗ
·x·
N
RL
24 4 kL
￩ 
1 6 N x 
3
P(t)
4
￫
7
m ·x·
1
c x
· 
1
k ￪1
4
(5)
o
N
N
Internal
hinge

27. 24 4
e e e
m 
7
m c 
1
c k 
1 ￩ 
16N
k ￪1
4 kL
￫ 4
e
P (t) 
3
P(t)
For N = - (1/16) k L ke = 0
This value of N corresponds to critical buckling load
(7)

28. Free Vibration
Undamped SDOF System
Damped SDOF systems

29. Free Vibration of Undamped System
A cos pt + B sin pt (or)
C sin (pt + α)
A2
x(t) =
x(t) =
where,
C   B2
x  p2
x  0
p2
 ￦
k
￨m
General solution is,
p k
T 
2 2 m
 natural period
f 
p

1
 natural frequency
2 T
p - circular natural frequency of undamped system in Hz.
(9)
(10)
(11)
(12)
(13)
(14)
(15)

30. 0
v
sin pt
0
p
x(t)  x cos pt 
Amplitude of motion
t
x
2
0
0
x2 ￦
v

￨p
T 
2
p
or
2
0
0
x2 ￦
v
x(t)   sin(pt )
￨p
where,
x0
v0 p
tan  
(16)
(17)
vo
x0
X0=initial displacement
V0 =initial velocity
t

31. Natural frequencies of other SDF systems
p – square root of the coefficient of displacement
term divided by coefficient of acceleration
For Simple Pendulum, p  g L
p
6 k ￦1
16N
7 m ￨ kL
2

k ￦
a g
m ￨L L
For system considered in (3b) , p
For system considered in (5) ,
For N=0 , o
6 k
7 m
p  p 
16
and for N  
1
kL , p  0
(18)
(19)
(20)
(21)

32. Condition of instability
16
cr
N  
1
kL  N
o
cr
N
N
p  p 1
p2
Ncr
N
(22)
(23)

33. Natural frequencies of single mass systems
p  k / m
Letting m = W/g
and noting that W/k = δst
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
st
p 
2
g
st
f 
1
st
δst is expressed in m, T  2
(10)
(24)
(25)
(26)
(27)

34. Relationship between Simple oscillator and Simple pendulum
L
g
st
p  g
L
p 
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. Effective stiffness ke and static deflection δst
ke
m
g
st
p  
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.
(28)

36. 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
Determination of Force - Displacement relation, F-∆

37. Examples

38. (a)
Rigid ,massless m
a
L
k

a
F
 L 
F
 L  F
 a  k
 
F
k
 L 
2
F
   a 
 
k
 L 
2
F
   a 
 
Therefore,
 a 
2
ke     k
F
  L 
or
k
 L 
2
W
st   a 
 
From Equilibrium,
 
From Compatibility,
(29)

39. Rigid bar m
a
L
k1
F = F1 + F2
k2
k1 k2
2
1
1 2
L
L
e k  k
 
 
 a 
2

F

k 
 
 a 
2
F    k   k 
(b) (c)
Rigid bar
a
L
k1
F
k1
m
k3
k2
∆
k3
k2
1 2
2
1
F
k
 k  k

2
3  a 
F
∆ = ∆ + ∆ = 
3
2
1
1
 L 
1
k

 k  k

 L 
 
F k 2
e  a 
 1
(30) (31)

40. k1 kn
ke = k1 +k2 + ……+ kn
(d)
(e)
k1
kn
.
k2
.
∆e = ∆1 + ∆2 + ……+ ∆n
k k k
 
F

F
 ...... 
F
1 2 n
F ke k1 k2 kn


1

1

1
...... 
1
(f) Flexible but mass less
L3
ke

3EI
L3
ke

12EI
L3
ke

3EI
(32)
(34)
(33)

41. (g) Rigid deck; columns mass less &
axially inextensible
E I
1 1
E2I2
L1 L2
Lateral Stiffness :
e
L3
L3
k  12
E1I1  3
E2 I2
(h)
EI
k1
k2
L
k2
k1
kb
F
k2
kb+k1
1
1
L3
 
1

1 1

1
ke k2 kb  k1 k2
3EI
 k

42. (i)
EI
L/2
k
L/2
R
F
∆
a
48 EI 24 EI k
3 3
 
5 FL

1 RL

R
F
EI
where, R 
5
2  48
L3
3 3
 
1 FL

5 RL
3 EI 48 EI
Eliminating R,
kL3
kL3
3 EI
EI
768 7
  EI
3
ￗ
1 FL
768 32
where,
kL3
F EI
kL3
EI
768 32
EI ￗ3
L3
ke 


768 7

43. (j)
L/2 L/2
e
L3
k 
48EI
L/2 L/2
e
L3
k 
192EI
L/2 L/2
e
7L3
k 
768EI
a b
e
a2
b2
k 
3EIL
(k)
(l)
(m)

44. 2R
d
G d 4
k 
6 4 n R 3
(n)
n – number of turns
(0)
(p)
(q)
L
k 
AE
L
k 
EI
L
I - moment of inertia of cross sectional area
L - Total length
L
k 
GJ
L
J – Torsional constant of cross
section
A – Cross sectional area

45. m
a
A, E, I, L
Natural frequencies of simple MDF systems treated as SDF
(i) (ii) (iii) (iv)
Columns are massless and can move only in the plane of paper
• Vertical mode of vibration
e v
2AE
L mL
k 
2AE
p  (35)

46. • For pitching or rocking mode
(36)
• For Lateral mode
2
12EI EI
L3
ke  2  24  24
L3
AE ￦
r
L ￨L
r is the radius of gyration of cross section of each column
2
L3
12EI EI AE ￦
r
ke  2  24  24
L3
L ￨L
(37)
6
1
6
p v
2 2 3 L
L
AE
L
3 p
mL
y  ya  0
ay  y  0
my  y  0
 p 
6AE

..
1
 .. AE
1

.. a 2a AE
AE/L AE/L
(AE/L)y (AE/L)y
lateral < axial < pitching
p
p p

47. Free Vibration of Damped SDOF

48. Free Vibration of damped SDOF systems
(A)
2
m·x·cx
·  kx  0
·x·
c
x
· 
k
x  0
m m
·
x
·
ζ

p
x
2 p
· x  0
where, p 
k
m
x
ζ 
c
2mp

2
c
km
(Dimensionless parameter) (38)
m
k
c

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
r2

ζ
p
2
r p 2
 0
In order for this equation to be valid for all values of t,
1,2
r2

ζ
p
2
r p 2
 0
r  p ﾱ 2
1
or

50. Thus are solutions and, provided r1 and r2 are different
er1t
and er2t
1 2
from one another, the complete solution is
x  c er1t
c er2t
1 2
The constants of integration c1 and c2 must be evaluated from the
initial conditions of the motion.
Note that forζ >1, r and r are real and negative
1 2
forζ <1, r and r are imaginary and
1 2
for
ζ =1, r = r = -p
is smaller than, greater than, or
Solution depends on whether ζ
equal to one.

51. For  1 (Light Damping):
0
d
p
B 
v0 
 x
12
x t  e pt
A cos p t  B sin p t
d d
 p 1  2
‘A’ and ‘B’ are related to the initial conditions as follows
A  x0
(39)
(40)
(41)
o 

 pt

 v 
xt e 
xo cos pdt  xo sin pdt
 pd

 12

 
 
In other words, Eqn. 39 can also be written as,
where, pd

52. T 
2  Damped natural period
pd
p  p 12
 Damped circular natural frequency
d
g
Extremum point ( x(t)  0 )
Point of tangency ( cos(pdt ) 1)
Td = 2π / pd
xn Xn+1
t
x
d
d
T
p

2
 Damped natural period
pd  p 1 Damped circular natural frequency
2

53. Motion known as Damped harmonic motion
A system behaving in this manner (i.e., a system for which 1 ) is said
to be Underdamped or Subcritically damped
The behaviour of structure is generally of this type, as the practical range
of  is normally < 0.2
The equation shows that damping lowers the natural frequency of the
system, but for values of  < 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. Rate of Decay of Peaks
x 
 p
2
pd

n1
 e
xn


 exp2

 12


(42)
xn1
xn
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4


55. Defined as
xn
  ln
xn  1
It is an alternative measure of damping and is related to  by
the equation
12
  2  ;2
xn
N xn N
 
1
ln
Logarithmic decrement
(43)
(44)
(46)
xn
When damping is quite small,
For small values of damping,
 
xn
 2
 (45)

57. Such system is said to be over damped or super critically damped.
i.e., the response equation will be sum of two exponentially
decaying curve
In this case r1 and r2 are real negative roots.
1 2
x(t)  C e()t
 C e()t
For   1 (Heavy Damping)
x
xo
o t

58. 0 0
e pt
x(t)  ￩x 1  pt  v t
￫
The value of ‘c’ for which1 is known as the critical
coefficient of damping
Ccr  2mp  2 km
Therefore,
Ccr
 
C
For  1
Such system is said to be critically damped.
x(t)  C e pt
 C te pt
1 2
With initial conditions,
(47)
(48)

59. 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

60. Let the duration of force,t1 be small compared to
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
The velocity,V0 ,imparted to the system is
obtained from the impulse-momentum relationship
mV0 = I = Area under forcing function = α P0 t1
where ,
1 for a rectangular pulse
α 2 / π for a half-sine wave
1 / 2 for a triangular pulse
0
Therefore, V =
0 1
m
Pt
Response to Impulsive Forces
t
P(t)
Po
t1 << T
(50)
(49)

61. For an undamped system, the maximum response is determined
from as ,
Therefore,
or
1
max st 0
x
p m p k mp
 V 0
  P0t1  P0 kt1
 (x ) pt
max
1
st 0
t
x
(x )
1
T
 2 ft  2 (51)
xmax 2 ft1(xst )0

62. •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

63. = Static displacement induced by
exciting force at time, t
st
m
k
·
x·  2 px
·  p2
x 
P(t)
·
x·  2 px
·  p2
x  p2
x (t)
st
or
where, x (t) 
P(t)
P(t)
k
Response to simple Force Pulses
(52)
(53)
(54)
P(t)
t
General Form of solution:
x(t) = xhomogeneous + xparticular

64. Response to a Step Pulse
where (xst)o =
x(t) = A cos pt + B sin pt + (xst)o
At t = 0 , x = 0 and v = 0
A = - (xst)o and B = 0
k
For undamped system, x + p2 x = p2 (xst)o
Po
t
x(t) = (xst)o [1 – cos pt] = [ 1 – cos 2π T ]
t
Po
0
P(t)
(55)

65. For damped systems it can be shown that:
d d
st 0 sin p t

 pt
￦ ￦
x(t)  (x ) 1 e cos p t 
1 2
￨
￨
xmax
(xst )0


1 2
 1 e
Response to a Step Pulse….
(56)
=0
x(t)
( xst )o
(t / T)

66. 2
1
i
￦
Vi
x(t)  x2
 sin p(tt ) 
￨ p
xi
Vi / p
where, tan 
P(t)
t
Po
Response to a Rectangular Pulse
(55)
t1
   
0
i st
1  cos pt
x  x
For t  t1, solution is the same as before,
x(t)  xst  1  c o s p t 
0
For t t1, we have a condition of free vibration,
and the solution can be obtained by application of Eq.17a as follows:
Vi  psin pt1

67. 1
2
2sin 1 cos
2
1
2
2
2sin2 pt1
1-cos pt pt
pt
sin pt1
tan  1  2
pt  tan
  pt1
hence,
1
1 1 1
2
st 0
t
￦
x(t) (1cos pt )2
sin2
pt (x ) sin p(tt ) p
￨
Response to a Rectangular Pulse
  1
1 0
sin
2 2 2
st 0 st
pt
x(t)  (x ) 2(1cos pt ) sin p￦
t 
t1
 2 x sin p￦
t 
t1
￨ ￨
(57(a))
(57(b))
(Amplitude of motion)
So,

68. 1
t /T=1.5
t1/T=1 1/6
1
t /T=2
In the plots, we have implicitly assumed that T constant and t1 varies;
Results also applicable when t1 = fixed and T varies
1
1 1
Response to a Rectangular Pulse…
t1/T=1/
t/T
t/T
t/T
t/T
1.68
2
2
2
x(t)/(x
st
)
0
x(t)/(x
st
)
0

69. Dynamic response of undamped SDF system to rectangular
pulse force. Static solution is shown by dotted lines

70. Forced response
Free response
Overall maximum
Response to rectangular pulse force: (a) maximum
response during each of forced vibration and free
vibration phases; (b) shock spectrum
(a)
(b)

71. Note that with xmax determined, the maximum spring force
Fmax = k xmax
In fact,
0 0
P0
Fst  xst 
Fmax 
kxmax 
xmax
xmax
xst 0
2
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.
Response to a Rectangular Pulse…
3
Impulsive solution, 2π f t1
(58)

72. P(t) = Po sin ωt, where ω = π / t1
st o
x + p2 x = p2 (x ) sin ωt 1
for t t1,
or
1
or
for t  t

 = 0

for t t
(xst )0
p
for t  t1
￦

￨
x(t)  [sint  p sin pt]
2
1
2
1 1
xst 0
t 2 t T
￦ ￨
4￨t1
￦
sin  t  1 T sin2 t
1 1 T
x(t) 
1
2
2
2 st 0
cos pt1
2 T
￦
￨ p
￨
￦
￨ p
x(t)  (x ) sin ￦
pt  1 t
1
1
1
1 t
T
2 T
t1
cos  t1
(xst )0 sin 2
2
￦t
￨T
￨T
x(t)  T 
0.25 ￦t
Response to Half-Sine Pulse
P(t)
t
O
P sin ωt
t1
(59)
(60)

73. • 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

74. Dynamic response of undamped SDF system to half cycle
sine pulse force; static solution is shown by dashed lines

75. 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

76. Shock spectra for three force pulses of equal magnitude
t1
t1
t1
2ft1 4ft1
ft1
ft1

77. • 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
Response to Half-cycle Force Pulses

78. 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
Conditions under which response is static:

79. 0
0
r
t pt
￦
sin pt
x(t)  (xst ) 
￩￦
t
￪
￫￨r ￨ r
￩￦
t
 (xst ) 
1 T
sin￦2 t
￪
t 2 t ￨ T
￫￨r
r
r
For tt
For t t
r r
r r
sin pt 1
pt
1 T
T
sin 2
2 t
￩  sin p(t  t )
pt r
x(t)  (xst )0 ￪1
￫
￩ t 1 T (t  tr )
T 2 t
(xst ) 1  sin 2
0 ￪
￫
r
r
sin pt
tan pt 
1 cos ptr
Response to Step force
P(t)
Po
t
tr
=
+
0
r
t
P
(tr t )
r
P
￦t
0
￨t
tr
Differentiating and equating to zero, the peak time is
obtained as:

80. Substituting these quantities into x(t), the peak amplitude is found as:
0
sin r
2
r
r r
x 1 2 pt
(xst ) pt pt
T
 tr
max
 1 2(1 cos pt )  1
 [1
1 T
sin
 tr
f tr
xmax
(xst)0

81. For Rectangular Pulse:
Half-Sine Pulse:
2 2
x(t)  (xst )0[1 cos pt]
 (x ) ￦
2 sin
pt1
sin p ￦
t 
t1
st 0
￨ ￨
2
1 1
(xst )0 t 1 T t
￦
x(t)  sin  sin 2
t 2 t T
￦
T ￨
1
1
4 ￨t1
1
(xst )0
( ft1) cos  ft1 f sin 2 ft - ft1 
0.25  ( ft )2
for t  t1
for t  t1
for t  t1
P(t)
Po
t
P(t)
t
POsinωt
t1

82. •
• 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.
Design Spectrum for Half-Cycle Force Pulses
C D
ft1
ft1=0.6 ft1= 2
xt 0
xmax
(x )
1
2
A
B
o
• Line OA defined by equation.51 (i.e max
= 2 π α f t1 = 2 π α )
x
(x )
st 0
Ordinate of point B taken as 1.6 and abscissa as shown
T
t1

83. •
•
•
• 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.
Response to Multi-Cycle Force Pulses
Effect of Full-Cycle Sine Pulse

84. The absolute maximum value of the spectrum in this case occurs
at a value of, ft1=0.5
Where t1 is the duration of each pulse and the value of the peak is
approximately equal to: xmax = n (/2) (xst)o
Effect of n Half-Sine Pulses
(63)
(62)

85. x(t)
I/mp
x(t) I/mp
x(t) 2I/mp
I
I
t1 t
Suppose that t1 = T/2
Effect of first pulse
Effect of second pulse
Combined effect of two pulses
t
Effect of a sequence of Impulses

86. • For n equal impulses, of successively opposite signs, spaced at
intervals t1 = T / 2 and xmax = n I/(mp)
• 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
max j
above, x = Σ I /(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
Effect of a sequence of Impulses
(64)
(65)

87. Response of Damped systems to Sinusoidal Force

88. Response of Damped systems to Sinusoidal Force
P(t)
t
t1
Solution:
(a)
The Particular solution in this case may be taken as
x(t) = M sinωt + N cosωt
Substituting Eq.(a) into Eq.52, and combining all terms involving sinωt
and cos ωt, we obtain
P(t) = P0 sinωt
where ω = / t = Circular frequency
1
of the exciting force
[-2
M -2pN  p2
M]sint[-2
N 2 pM  p2
N]cost  p2
(X ) sint
st 0

89. This leads to
Where
The solution in this case is
x(t) = e- pt(A cos pD t +B sin pD t) + sin (ωt – ) (66)
where
(c)
2
2
st 0
2 2
1-
p
+4ζ2
￦
ω
M= ￨ (x )
￦
ω
￨p
￦
ω
￨p
￩
￪1-
￪
￫
(p2-ω2)M - 2ζpωN = p2(xst)0
2ζpωM+ (p2-ω2)N = 0
2
2 2
N=- (xst )0
+4ζ2
￦
ω
2ζ
￨p
￦
ω
￨p
￦
ω
￨p
￩
￪1-
￪
￫
 
(x )
st 0
(1-2
)2
+4ζ2
2
1
p 2ft1
 


tan  
2
(1- 2
)
(67)
(68)

90. Steady State Response
x( t ) 1
(xs t )0
 sin(t - )
(1- 2
)2
 42
2
(69)
xm a x
(xs t )0
A F  
1
(1- 2
)2
 42
2
(70)
P(t)
t
x(t)


Note that at  

1, AF 
1


p 2 
(71)

91. p
 

AF
p
 

α

92. 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.

93. Resonant Frequency and Amplitude
max
1
res  p 1- 22
(AF) 
2 1- 22
(72)
(73)
2
These equations are valid only for 
1
For values of
1
< 1
2
ωres = 0
(A.F.)max = 1
(74)

94. Transmissibility of system
The dynamic force transmitted to the base of the SDOF system is
1
k k
F  k
P0
[sin(t - ) 
c
cos(t - )]
(1- 2
)2
 42
2
k mp2
p
c
 2

 2
Noting that c
 and combining the sine and
cosine terms into a single sine term, we obtain
P0
F ( t )

1  42
2
sin(t -   )
(1- 2
)2
 42
2
where is the phase angle defined by tan = 2ζ 
(76)
(77)
(75)
k
￩
￫
Substituting x from Eq.(69), we obtain

cx
·
F  kx  cx
·  k ￪x

95. 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  42
2
T R  0

P0 (1- 2
)2
 42
2
(78)
1
T R 
(1- 2
)
which is the same expression as for the amplification factor xmax/(xst)0
Transmissibility of system
The variation of TR with and ζ is shown in the following figure. For
the special case of ζ =0, Eq.78 reduces to

96. p
Transmissibility for harmonic excitation
 
ω
TR
Transmissibility of system

97. • 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.
st

fe
Noting that


fe
 2
p f 4.98
in cms
The static deflection of the system, st must be
50
st 
fe2
10 40
fe
0.15
fe is the frequency of the exciting force, in cps
0.5
st in
cms
(79)

98. • 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
2
1
- 1
T R 
￦￦
￨ ￨ p
(80a)
Proceeding as before, we find that the value of st corresponding to
Eq.(80a) is
st 2 2
e e
1 1 25
(cm)
TR f TR f
2
￩ ￩
st
￪ ￪
￫ ￫
 ; 1 (in) or ; 1 (80b)

99. 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
m
0
P sinωt
k c
m
k c
0
P sinωt
m
mb
(a)
(c)
(b)

100. • 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).

101. Application to Ground-Excited systems
x(t)
m
k
c
y(t)
of required to limit the transmissibility TR = xmax/y0 to a
specified value may be determined from Eq.(80b).
st
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

102. Rotating Unbalance
Total mass of machine = M
unbalanced mass
eccentricity
angular velocity
= m
= e
= ω
d2
dt2
(M  m)·x·m (x  esint) cx
·  kx  0
M·x·cx
·  kx  me2
sint
e ωt
M m
k/2 k/2
c
x

103. Reciprocating unbalance
e
L
 
F me2
sint sin2t
 
e - radius of crank shaft
L - length of the connectivity rod
e/L - is small quantity second term
can be neglected
ωt
e
m
M
L

104. 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:
Determination of Natural frequency and Damping
Steady State Response Curves

105. Determination of Natural frequency and Damping
Resonant Amplification Method
Half-Power or Bandwidth Method
Duhamel’s Integral

106. Determine maximum amplification (A.F)max=(x0)max/ (xst)0
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.
Evaluate  from Eq.73 or its simpler version, Eq.71, when
(a) Resonant Amplification Method
Determination of Natural frequency and Damping

107. (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 ma,x and
determine the intersection points with the response spectrum.
These points are known as the half-power points of the spectrum
3. Evaluate the bandwidth, defined as f
f

108. (xo)max
(x )
o max
1
2
(xo)st
f
∆f fe
xo

109. 1.For small amounts of damping, it can be shown that ζ is related
to the bandwidth by the equation
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.
2 f
 
1 f
(81)
Determination of Natural frequency and Damping…

110. 1
2
2
2 2

1 1


 2 
(1
1
)  (2)
 
1 1
2 2
1
1 2
2 2 2
f
f
1
82
2
2
2


1
12
2
 42
2

1 22
ﾱ 2
1 22
 2
1 22
 2
12
12
12
 1 2
 1 2
1 ( f  f2 )  f2 )
 
1
(  )  ﾻ
( f1
ﾻ
( f2  f1 )
( f1  f2 )
Derivation

111. Other Methods for Evaluating response of SDF Systems
t

t
(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

112. The displacement at time t induced by integration as
I = P ( τ ) d τ
mp
sin p(t - )
The strip of loading shown shaded represents and impulse,
I = P() d
For an undamped SDF system, this induces a displacement
x 
P( )d
1  t
mp  0
x(t)  P( ) sin ￩
￫p t   d
or
0
t
x(t)  p  xst ( ) sin p t   d
(82)
(83)
(84)

113. 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
For viscously damped system with  1 ,becomes
Leading to the following counterpart of Eqn.84
     
0
0
t
st
V0
p
d
x t  x cos pt sin pt p x  sin￩p t 
￫
   
 
d
d
P  d
mp
  p t
x(t)  t 
e sin ￩p
￫
2 0
t
p pt
xt  xst e sin
pd t 
d
1

114. For
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   0 and
x0  v0  0
For t ﾣ t1
t ﾳ t1
0
t1
xt pxst  sin￩pt - d 0xst  ￩cos pt -t1-cos pt
0 ￫ 0 ￫
 
0
t
st st
o o
st o
 t
 0
xt px  sin ￩pt   d  x  cos p t 
￫
 x  1cos pt
P(t)
Po
t1 t

115. Generalised SDOF System

116. Generalised SDOF System
q(t) 
m*
q
""(t)c*
q
"(t)k*
q(t)  p*
(t)
Single generalised coordinate expressing the
motion of the system
generalised mass
m*

c*
 generalised damping coefficient
k*

p*

generalised stiffness
generalised force

117. (a)
x1
m1,j1 m2, j2
m(x)
(b)
yx,t  xqt
* 2 2 2
0
i i
l
m  m(x)(x) dx m jii


118. c1 c2
a1(x)
L
(c) c(x)
* 2 " 2
1
0 0
i i
 (x)a
l l
c  c(x)(x) dx  EI(x) (x)dx c
 

119. k1 k2
(d)
* 2 " 2 2 ' 2
0 0 0
l
i i
l l
k  k(x) (x)dx  EI(x)
   (x) dx k N(x) (x) dx
(e)
k(x)
N

120. P(x,t)
Note: Force direction and displacement direction is same (+ve)
*
0
i i
 (x)
l
p  p(x,t)(x)dx p (t)

Pi(t)

121. Effect of damping
Viscous damping
Coulomb damping
Hysteretic damping

122. 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.

123. 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.
Effect of damping

124. 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  Fddx
In general Wd depends on temperature,frequency,amplitude.
For viscous type
Wd  Fddx
Fd cx
·
W cX2
 cx
·dx 
d
2

cx
·2
dt c2
X2
cos2
(t)dt cX2
0

125. c -
Wviscous -
Fd(t) = c x
·
cωX
coefficient of damping
work done for one full cycle = cX 2
Fd
-X
X(t)
(a) Viscous damping

126. (b) Equivalent viscous damping:
2
eq d
eq
c s
W
C
C where
X 2
 X 2

Wd
Cc 4Ws
 C X  W
 d

2k
k 
2Ws
, W  strain energy
 
C

x
ellipse
Fd+kx

127. x1
x2 ∆
Linear decay
4Fd/k
Frequency of oscillation
k
m

p 
x-1
It results from sliding of two dry surfaces
The damping force=product of the normal
force & the coefficient of friction (independent
of the velocity once the motion starts.
(b) Coulomb damping:
-X X(t)
Fd
F

coulomb

W  4F X

128. 1
2
2
d
1 1 d 1 1
1
1
k(X 2
 X 2
)  F (X  X )  0
1
k(X  X )  F
The motion will cease ,however when the amplitude becomes
less than ∆, at which the spring force is insufficient to
overcome the static friction.
1 2
k
X  X 
4Fd
Decay in amplitude per cycle

129. D
f  k x
W  2 kX 2
  X 2
x
x
D
Energy dissipated is frequency independent.
(c) Hysteretic damping (material damping or structural damping):
- Inelastic deformation of the material composing the device

￩ 1
Whysteretic  4Fy X ￪
￫
y
F is the yield force
h
K =elastic
damper
stiffness
xy
- x
x
Fy
Fd
(d) Structural damping
xy
Xy Displacement at which material first yields
 
x

130. eq
e) Coulomb C =
eq
h) Hysteretic C =
c) Structural Ceq=
4F
WX
4Fy
X
2ks


  1
 
 
Equivalent viscous Coefficient

131. Reference
Dynamics of Structures: Theory and Application to
Earthquake Engineering – Anil K. Chopra, Prentice Hall
India
Reading Assignment
Course notes & Reading material