AcknowledgementMost of the numerical data and plots
presented in this lecture are taken
from the Text Book
“Dynamics of Structures—Theory and
Application to Earthquake Engineering”
by Prof. Anil K. Chopra
Single Degree of Freedom (SDOF) System• One variable is required to describe the deformation• For example: Mass-Spring-Dashpot System
Newton’s Law of Motion• Resultant force
= Mass X Acceleration
• Resultant Force Applied force, Force offered by spring, Force resisted by dashpot,
( )f tkx
cx&
Equation of Motion
• Initial Conditions:• Standard procedure is available to solve this
differential equation
( )mx cx kx f t+ + =&& &
( ) ( )0 00 ; 0x t x x t v= = = =&
Special Cases• Free vibration---• Undamped---
( ) 0f t =
0c =
Equation of Motion
• Initial Conditions:
( ) ( )0 00 ; and /or 0x t x x t v= = = =&
0mx kx+ =&&
Displacement Response x(t): Natural Frequency
rad/secn k m =
Time Period
2 secn nT =
( )22
0 0 0
Amplitude
nA x v = +
1 00
0
Phase tann
v
x
-æ ö÷ç ÷= ç ÷ç ÷çè ø
Equation of Motion
• Initial Conditions:
( ) ( )0 00 ; and /or 0x t x x t v= = = =&
0mx cx kx+ + =&& &
Displacement Response x(t):
Provided system parameters satisfy certain condition:
2 critical damping
1 damping ratio
cr
cr
c km
c c
= =
= < ®
0
0 0
tan D
n
x
v x
=
+
( )sinnt
Dx e t -
= +
1/22
2 0 00
n
D
v xx
é ùæ ö+ê ú÷ç ÷= + çê ú÷ç ÷çè øê úë û
Effect of Different Damping Ratios
( ) ( )0 00 ; 0 0x t x x t v= = = = =&
Initial Conditions
More the damping more rapidly the vibration dies out
Logarithmic Decay and Estimation
of Damping
Amplitude ratio in two successive cycles
21
2exp
1
n DTi
i
xe
x
+
æ ö÷ç ÷ç= = ÷ç ÷ç ÷÷ç -è ø 1
1ln
2
i
i
x
x
+
æ ö÷ç ÷» ç ÷ç ÷çè ø
Damping can be estimated from free vibration test
Consider a Sinusoidal Force
0( ) sin is forcing frequencyf t p t = ®
Governing Equation
0 sinmx kx p t+ =&& ( ) ( )0 00 ; 0x t x x t v= = = =&Initial Conditions
General Solution
( ) ( ) ( )c px t x t x t= +
( ) Complementary Function
Homogeneous Solution
cx t =
=
( ) Particular Integralpx t =
Transient
Steady State
Special case
0 0 00; ; 0.2n nx v p k = = =
( ) 00Displacement due to amplitude of load if
applied statically (maximum static response)
stx p k= =
( )( )
0
2
1sin
1p
n
px t t
k
é ùê ú= ê ú
-ê úë û
Consider only steady
state motion
Maximum steady
state response
( )
00 2
1
1p
n
px
k
é ùê ú= ê ú
-ê úë û
Deformation Response Factor
( ) ( )
0
2
0
Absolute Maximum Steady State Response 1
Maximum Static Response 1
p
d
st n
xR
x = = =
-
( ) ( ) ( )0
sinp st dx t x R t = -
Consider only steady state response
0
180
n
o
n
í <ïï® = ìï >ïî
Resonance: Salient Points
Forcing frequency = Natural frequency n =
Apparently• Deformation Response Factor is Infinite
• Unbounded response
In reality• Form of Steady State Response is not valid at
• Actual Steady State Response isn =
( ) 0 cos2
p n n
px t t t
k = -
Amplitude
grows to infinity
ONLY at
INFINITE time
Consider Once again the same Sinusoidal Force
0( ) sin is forcing frequencyf t p t = ®
Governing Equation is now
0 sinmx cx kx p t+ + =&& & ( ) ( )0 00 ; 0x t x x t v= = = =&Initial Conditions
General Solution is, as before,
( ) ( ) ( )c px t x t x t= +
( ) Complementary Function
Homogeneous Solution
cx t =
=
( ) Particular Integralpx t =
Transient
Steady State
Typical Response in Damped
Harmonic Forced Vibration
Special case
0 0 00; ; 0.2, 0.05n nx v p k = = = =
Deformation Response Factor
( )
0
0
Absolute Maximum Steady State Response
Maximum Static Response
p
d
st
xR
x= =
( ) ( ) ( )0
sinp st dx t x R t = -
Consider only steady state response
( ){ } ( ){ }
( )
( )22 22
21; tan
11 2
n
d
nn n
R
= =-
- +
1
1
1n
í< <ïïïï »ìïï> >ïïî
Unit Impulse and Its Effect
Unit Impulse
Newton’s 2nd Law
( ) ( )d
mx p tdt
=&
( ) ( )2
1
2 11
t
t
p t dt m x x m x= = - =ò & & &
Assuming AT REST condition just before the impulse
( ) 1x m =&
Therefore, unit impulse impart a velocity to the system and hence,
vibration due to initial velocity
0 00; 1x v m= =
Solution of damped free vibration due to initial velocity lead to
( )1
sinnt
D
D
x t e tm
-= ‘t’ is measured from just after the
impulse, i.e., from the instant
Unit Impulse Response Function
( ) ( ) ( ) ( )1
sinn t
D
D
x t e t h t tm
- - é ù= - = - ® ³ë û
If ‘t’ is measured from the instant when is measured
Unit Impulse Response Function, i.e., Response due to unit impulse
Arbitrary Loading Decomposed
into Impulses
( ) ( ) ( )( ) ( ) ( )sinn t
D
D
pdx t p h t e t
m
- - é ù= - = -ë û
( ) ( ) ( ) ( )0
1sinn
t
t
D
D
x t p e t dm
- - é ù= -ë ûò
Response due to one impulse
Overall Response
Note: This is the response with ZERO Initial Conditions,
Otherwise, Add the result of Free Vibration
The Procedure can be used in Numerically Evaluating the
response of a variety of loadings including that due to
earthquake
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