SDOF review 061904 - Faculty Server Contact | UMass...

1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Review SDOF Theory

Review SDOF Theory

Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell

[ K ]n

[ M ]n [ M ]a[ K ]a [ E ]a

[ ω ]2

Structural Dynamic Modeling Techniques & Modal Analysis Methods


SDOF Definitions

• lumped mass• stiffness proportional to displacement

• damping proportional to velocity

• linear time invariant• 2nd order differential equations

Assumptions

m

k c

x(t) f(t)


Homogenous Solution

Homogenous equation is

and assuming an exponential solution form gives

which results in

which yields the characteristic equation

0kxxcxm =++ &&&

( ) 0ekcsms st2 =++

0mks

mcs2 =++

mk

m2c

m2cs

2

2,1 −

±−=


Homogenous Solution

This results in a general solution given as

Three distinct solutions result from this general solution.

tsts 21 BeAex +=

(1) underdamped(2) overdamped(3) critically damped

1<ζ

1>ζ

1=ζ


Homogenous Solution

Some general definitions are:

Critical Damping

% Critical Damping

Roots

km2m2mkm2c nc =ω==

ccc

=ζ

1s 2nn2,1 −ζω±ζω−=


The general solution can be rewritten as

and with adjustingsome terms

the poles can be observed as seen in the figure uponchanging damping

(2.6.13)

Homogenous Solution

)t(fm1xx2x 2

nn =ω+ζω+ &&&

2

n

2,1 1is

ζ−±ζ−=ω


Oscillatory Solution (3 different forms)

and the general homogeneous solution becomes

Homogenous Solution - Underdamped 1<ζ

( )t1it1it n2

n2

n BeAeex ωζ−−ωζ−ζω− +=

( )φ+ωζ−= ζω− t1sinXex n2tn

( ) ( )( )t1cosCt1sinCex n2

2n2

1tn ωζ−+ωζ−= ζω−

( ) ( )

ωζ−+ωζ−

ζ−ω

ζω+= ζω− t1cos)0(xt1sin

1)0(x)0(xex n

2n

22

n

ntn &

2nd 1 ζ−ω=ω


The damped exponential response

( ) ( )

ωζ−+ωζ−

ζ−ω

ζω+= ζω− t1cos)0(xt1sin

1)0(x)0(xex n

2n

22

n

ntn &

Homogenous Solution - Underdamped 1<ζ


Overdamped solution is

with coefficients A and B as

Homogenous Solution - Overdamped 1>ζ

( ) ( ) t1t1 n2

n2

BeAex ω−ζ−ζ−ω−ζ+ζ− +=

( )12

)0(x1)0(xA2

n

n2

−ζω

ω−ζ+ζ+= &

( )12

)0(x1)0(xB2

n

n2

−ζω

ω−ζ−ζ−−= &


The critically damped solution is

and the solution becomes

Homogenous Solution - Critically Damped 1=ζ

( ) tneBtAx ω+=

( ) tn

t

n

n

et)0(x)0(x

e)0(xxω−

ω−

ω+

+=

&


Logarithmic Decrement

From the damped vibration solution, the amount of damping in the system can be expressed as

and the log decrement becomes

and for small damping

dn)t(

tdn

d1n

1n

elneeln τζω−===δ τζω−

τ+ζω−

ζω−

212

ζ−

πζ=δ

πζ≅δ 2


Logarithmic Decrement

Plot of Damping Estimates Log Decrement


Laplace Approach

TIME DOMAIN

FREQUENCY

DOMAIN

LAPLACE

DOMAIN

TRANSFORMATION TRANSFORMATION

PARAMETER ESTIMATION SUBSET


Review - SDOF Definitions

• lumped mass• stiffness proportional to displacement

• damping proportional to velocity

• linear time invariant• 2nd order differential equations

Assumptions

m

k c

x(t) f(t)


Review - SDOF Equations

Equation of Motion

)t(fxkdtdxc

dtxdm 2

2

=++ )t(fxkxcxm =++ &&&

Characteristic Equation

Roots or poles of the characteristic equation

0kscsm 2 =++

mk

m2c

m2cs

2

2,1 +

±−=

or


Review - SDOF Definitions

Poles expressed as

Damping Factor

Natural Frequency

% Critical Damping

Critical Damping

Damped NaturalFrequency

POLE

CONJUGATE

jω

σζωn

ωd

( ) d2n

2nn2,1 js ω±σ−=ω−ζω±ζω−=

nζω=σ

mk

n=ω

ccc=ζ

nc m2c ω=2

nd 1 ζ−ω=ω


SDOF - Laplace Domain

Equation of Motion

Laplace Transform

)t(fxkxcxm =++ &&&

002 xxs)s(xs)x( &&& −−=ℑ

0x)s(xs)x( −=ℑ &

)s(x)x( =ℑ

)s(f))t(f( =ℑ



Laplace Domain Equation of Motion

Assuming that initial conditions are zero

)s(f)s(x)kscsm( 2 =++

002 xmx)cms()s(f)s(x)kscsm( &+++=++

CharacteristicPortion

AppliedForce

InitialDisplacement

InitialVelocity



System Characteristic Equation is

The solution is then

( ) )kscsm(sb 2 ++=where)s(f)s(x)s(b =

)s(f)s(h)s(f)s(b)s(x 1 == −

The System Transfer Function is defined as

kcsms1)s(h 2 ++

=


SDOF - Transfer Function

System Transfer Function

kcsms1)s(h 2 ++

=

Complex valued function defines the surface shown

Source: Vibrant Technology


SDOF - Transfer Function

Polynomial Form

Pole-Zero Form

Partial Fraction Form

Exponential Form

kcsms1)s(h 2 ++

=

)ps)(ps(m/1)s(h *

11 −−=

)ps(a

)ps(a)s(h *

1

*1

1

1

−+

−=

tsinem1)t(h d

t

d

ωω

= ζω−


SDOF - Damped Exponential Response

tsinem1)t(h d

t

d

ωω

= ζω−

Amplitud

e

Damping Decay

Period


SDOF - Transfer Function & Residues

Residue

d

ps1

1

jm21

)ps)(s(ha

1

ω=

−

=

→

related to mode shapes

Source: Vibrant Technology


SDOF - Frequency Response Function

The Frequency Response Function (FRF) is the Transfer Function evaluated at s=jω

Partial Fraction Form

)pj(a

)pj(a)s(h)j(h *

1

*1

1

1js −ω

+−ω

==ω ω=

kcj)j(m1)s(h)j(h 2js +ω+ω

==ω ω=

Polynomial Form

( ) ( )ω+ω−=ω

cjmk1)j(h 2

Complex Form


Complex Frequency Response Function

The Complex FRF - real and imaginary parts

2

n

22

n

n

2

n

22

n

2

n

21

2j

21

1)j(h

ωω

ζ+

ωω

−

ωω

ζ−

ωω

ζ+

ωω

−

ωω

−=ω


Complex Frequency Response Function

The Complex FRF at resonance reduces to a simple magnitude and phase as

2

n

n

1

2tan;

21j)j(h

ωω−

ωωζ

=φζ

−=ω


Forced Harmonic Vibration

Consider the SDOF system with a sinusoidally varying forcing function applied to the mass as shown

F=F0sinωt

From the Newton’s Second Law,

∑ ω=++⇒= tsinFkxxcxmmaf 0&&&



The solution consists of the complementary solution (homogeneous solution) and the particular solution. The complementary part of the solution has already been discussed.The particular solution in the one of interest here. Since the oscillation of the response is at the same frequency as the excitation, the particular solution will be of the form

( )φ−ω= tsinXx



Substituting this into the differential equation, the solution is of the form

Note that this is also seen graphically as

(recall that the velocity and acceleration are 90 and 180 degrees ahead of the displacement)

( ) ( )2220

cmk

FXω+ω−

=

ω−ω

=φ −2

1

mkctan



This is expressed in nondimensional form as

and can be further reduced recalling the following expressions for a SDOF

( )222

0

kc

km1

kF

Xω+

ω−

=

ω−

ω=φ −

km1k

ctan 2

1

mk

n =ω nc m2c ω=ccc

=ζ



The nondimensional expression is

2

n

22

n

021

1FXk

ωωζ+

ωω−

=

2

n

n1

1

2tan

ωω

−

ωω

ζ=φ −



This yields the popular plot of forced response



The complex force vector also yields useful information for interpretation of the results



The differential equation describing the system

and the complete solution of this problem is given as

tsinmFxx2x 02

nn ω=ω+ζω+ &&&

( )1n2t

1

2

n

22

n

0

t1sineX

21

)tsin(kF)t(x

n φ+ωζ−+

ωω

ζ+

ωω

−

φ−ω=

ζω−

SDOF review 061904 - Faculty Server Contact | UMass...

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