Vibration of Single Degree

of Freedom SystemDhiman Basu

IIT Gandhinagar

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

Introduction

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 =

Undamped Free

Vibration

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

Some Examples (Assume No

Damping)

33 /k EI L=32 6 /k EI L=

Damped Free Vibration

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

Comparison of Damped and

Undamped Systems

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

Undamped Forced Vibration

due to Harmonic Loading

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

Damped Forced Vibration

due to Harmonic Loading

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 = = = =

Resonance in Damped System

n =

For a Special Case of Zero Initial Conditions and 5% Damping

Effect of Damping at Resonance

For a Special Case of Zero Initial Conditions

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

Response to Arbitrary Force

by Duhamel’s Integral

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

Thank You