Sdof

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/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

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

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

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

Dynamic Degrees of Freedom

1. 2.
Massless Spring Inextensible
spring with Spring
θ
mass

(a) (b) (c)


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)


Rigid deck

5.

Massless
columns

Idealisation of Structure as SDOF

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.

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”.

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
&&

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

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)

(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

(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

(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

(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 )

(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)

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

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

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

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

Free Vibration

Undamped SDOF System

Damped SDOF systems

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.

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

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

Condition of instability

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

N
p = po 1 - (23)
N cr

p2

Ncr

N

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

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

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.

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

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

(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

(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

(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

(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

(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

(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

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

• 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

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

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 )

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.

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

   

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

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

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

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

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

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

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

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

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

•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

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

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 ]

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


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


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)

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

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

(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

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

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

• 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

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 each of
forced vibration and free vibration phases; (c) shock spectrum

2pft1 4pft
1
pft1

t1

t1

t1
ft1

Shock spectra for three force pulses of equal magnitude

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

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

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

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

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

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.

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.

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

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

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

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)

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

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 )

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

AF α

w
=
p
w
=
p

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.

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

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)


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


TR

ω
=
p
Transmissibility for harmonic excitation

• 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

• 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

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)

• 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).

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).

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
&& &

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

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:


Resonant Amplification Method
Half-Power or Bandwidth Method
Duhamel’s Integral


(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.

(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

(xo)max
xo

1
2 (xo)max
(xo)st

∆f fe

f

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.

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 )

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
 

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

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

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

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

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

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

(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

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)

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 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.

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.

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

(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

(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

(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

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.

(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.

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

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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