july 28 - introduction to structural dynamics dr. hw huang

8/13/2019 July 28 - Introduction to Structural Dynamics Dr. HW Huang

http://slidepdf.com/reader/full/july-28-introduction-to-structural-dynamics-dr-hw-huang 1/52

Structural Dynamics: Introduction

Dr. Hongwei Huang

Department of Bridge Engineering

Tongji University

We make our living in dynamics, structural health monitoring

and vibration control



DYNAMICS OF STRUCTURES

∆ Theory and methods for analyzing response of structures

under dynamic loads

∆ Structures: beam, frame building, bridge, etc.

∆ Response:∆ Deformation

∆ Stress

∆ Acceleration



∆ WHY/ HOW

if

M

ck

u

p(t)

Input force

Outputdisplacement

M

k u

c

0)( pt p

t)( 0 sinω pt p k

pu 0

)t1(0 cosωk

pu

)(t pkuucum

Math modeling



1. Essential characteristics of a dynamic problem

1.1 Typical dynamical problems:

Structural vibration under seismic excitation;

Vibration of large span bridge and high-rise building under turbulent windloads;

Vehicle vibration due to road roughness;

Bomb blast pressure on civil engineering structures.



1.2 Differences between static and dynamic loads:

Static loads: magnitude, direction and position remains constant or variesslowly with time, such as structural self weight, snow loads and etc..

Dynamic loads: magnitude, direction and position varies rapidly with time or

the loads are suddenly applied to or removed from the structure.

1.3 Essential characteristics

∆ Time varying nature: a succession of solutions has to be established

corresponding to all times of interest in the response history.

∆ Inertial forces: represent a significant component of the total loads, thedynamic characteristics of the problem are reflected in its solution.



2. Dynamic loads and analyzing methods

2.1 Prescribed dynamic loading:

The time varying characteristics of the loading is fully prescribed, even

though it may be highly oscillatory or irregular in form.

Deterministic analysis: a direct structural time history responses computation

corresponding to the prescribed loading history.

P (t )

t



2.2 Random dynamic loading:

The time varying characteristics is not completely prescribed but can be

defined in a statistical sense.

Nondeterministic analysis: provides only statistical information about the

structural response from the statistically defined loading.

2.3 Prescribed dynamic loading can be categorized as periodic loadings and

nonperiodic loadings:

Periodic loading

(1) harmonic loading

loading in the form of sinusoidal variation

e.g. loading due to rotating machinery

(2) Other periodic loadingsComplex periodic loading which can not be represented by a harmonic

function but by the sum of a series of simple harmonic components

e.g. hydrodynamic pressures generated by a propeller at the stern of a ship



∆ Nonperiodic loading

(1) Impulsive loading

short-duration high intensity loading which is generally described by

sine-wave, triangular or rectangular function

e.g. blast or explosion loadings

(2) General forms of loadingRepresented by any function or just some data sets

e.g. earthquake excitations



(3) Damping Force

Damping force: the force which induces the energy loss of the system. It

equals to the product of the damping coefficient and velocity of mass.

D — Damping

c — Damping coefficient

ù — Velocity of mass

∆ Determination of the damping coefficient c

It cannot be obtained directly from the geometry of the structure. It isgenerally obtained via structural vibration tests.

∆ Viscous (velocity proportional) damping is the most generally used

damping mechanism.

∆ Other general damping mechanism:

∆ Friction damping: the damping due to friction, generally is a constant;

∆ Hysteretic damping: a damping mechanism proportional to the displacementamplitude but in phase with the velocity;

∆ Liquid damping: the damping force is proportional to the square of

velocity of the mass.

uc f D



3.2 Formulation of EOM

(1) The Newton’s second law of motion

s D f f t p F )(

)(t p f f ma s D

ma

ua

ku f s

uc f D

The EOM of a SDOF system

)(t pkuucum



(2) Direct Equilibrium using d’Alembert Principle

D’Alembert’s principle: a mass develops an inertial force proportional to its

acceleration and opposing it. It permits the equation of motion to be

expressed as equations of dynamic equilibrium.

ku f s

uc f D

um f I

)(t pkuucum


0)( s D I f f f t p



(4) Variational approach (Hamilton’s principle)

Hamilton‘s principle is an“integral principle”, which means that it

considers the entire motion of a system between time t1 and t2. In each

time duration [t1, t2], the variation of system kinetic energy, potential

energy and work done by nonconservation force equals to 0.

where,

T —— Kinetic energy of the system；

V —— Potential energy of a system；

Wnc —— Work done by the system due to all nonconservation force

δ —— variation with time.

0)(2

1

2

1

dt W dt V T t

t

t

t nc

j

j

ncjnc u P W



Kinetic energy： Potential energy：

Variation of energy

Variation of the work done by nonconservation force

Substitute them into Hamilton Pinciple:

2

2

1umT 2

2

1kuV

ukuuumV T )(

uucut pW nc )(

0)(

2

1

2

1 dt W dt V T

t

t

t

t nc

0])([2

1

dt ut pukuuucuumt

t

2 2 2

1 1 1

2 2 22

11 1 1

( ) ( )

( )

t t t

t t t

t t t t

t t t t

d d mu udt mu u dt mu u dt

dt dt

mud u mu u mu udt mu udt

2

1

0)]([t

t udt t pkuucum )(t pkuucum



(5) Lagrange equation of motion

Kinetic energy：

Potential Energy：

Work done by non-conservation force：

So，

Substitute them into Lagrange equation：

)(t pkuucum


N jt P u

V

u

T

u

T

dt

d ncj

j j j

,,2,1,)()(

2

21 umT

2

2

1kuV

)(t puc P nc

umumdt

d

u

T

dt

d

)()( 0

u

T ku

u

V

)()( t P u

V

u

T

u

T

dt

d nc



4.2 Free vibration of undamped SDOF system

Free vibration: motions caused by initial disturbance and with no applied forces.

Undamped: the energy loss mechanism is not considered.

No damping：c = 0

Free vibration： p(t) = 0

EOM：

Initial disturbance：

)0(),0( 00 uuuu t t

0 kuum



Assuming the solution to the ODE is of the following form

where s is a coefficient; A is an arbitrary complex constant

s can be determined according to：

The solutions of the above equation are：

st Aet u )(

0 kuum

0)( 2 st Aek ms

2 0ms k

nn i si s w w 21 ,

m

k i n w ，1



Therefore, the total response includes two terms as follows：

Considering the following relationship：

The solution to the EOM is：

where A, B are constant determined by initial conditions.

Velocity can be obtained by taking one step derivatives on u(t)

Concerning the initial conditions

t it it st s nn e Ae Ae Ae At u w w 212121)(

cos sin ; cos sinix ixe x i x e x i x

t Bt At u nn w w sincos)(

t Bt At u nnnn w w w w cossin)(

Buu

Auu

nt

t

w

)0(

)0(

0

0

(0),

(0)

n

A u

u B

w



Dynamic response of free vibration of an undamped SDOF system is

where

is called the natural frequency or resonant frequency depending on the material

and the structure

Response is a simple harmonic motion..

Period:

Maximum response:

t u

t ut u n

n

n w w

w sin)0(

cos)0()(

mk

n w

n

nT w

p 2

22 ])0(

[)]0([n

mω

uuu



4.3 Free vibration of damped SDOF system

Free vibration： p(t) = 0

EOM：

Initial disturbance：

Assuming that the solution is in the following form

where s is a coefficient; A is an arbitrary complex constant.

s can be determined according to：

0 kuucum

)0(),0( 00 uuuu t t

st Aet u )(

2 0ms cs k 22

2,1 )2

(2

nm

c

m

c s w



If , the SDOF system will not vibrate.

If , the system will vibrate.

The c satisfying is called the critical damping, denoted as

0)2

(22 n

m

cw

0)2

(22 n

m

cw

0)2

(22 n

m

cw

kmmc ncr 22 w



Damping ratio: the ratio between the damping coefficient c and the critical

damping coefficient ccr , denoted as ζ:

(1) if ζ＜1, under damped;

imaginary oscillation

(2) if ζ＝1, critically damped;

no vibration

(3) if ζ＞1, over damped.no vibration

For steel structure,

For RC structure,

ncr m

c

c

c

w

z

2

0.01z

0.03

0.05

lowmagnitudevibration

strong vibration

z

z



Underdamped Systems: most of civil engineering structures are underdamped

system.

For underdamped system, substituting into

we obtain

Subsequently, the free vibration response under initial disturbance can be

obtained as

where ω D is the damped natural frequency of the system.

nmc zw 222

2,1 )

2

(

2

n

m

c

m

c s w

22,1 1 z w zw nn i s

]sin))0()0((cos)0([)( t uut uet u D

D

n D

t n w w zw w zw

21 z w w n D

D

DT

w

p 2

22 11

2

z z w

p

n

n

D

T T



Measurement of damping ratio:

The damping ratio of an underdamped system affects greatly the structure

free vibration response, and therefore, has to be evaluated either theoretically or

experimentally.

Observing the free vibration response curve of a SDOF system,

Logarithmic decrement of damping is

and thus the damping ratio can be determined as

If the damping level is relatively low,

it can also be approximated as

)1

2exp()exp(

)(

)(

21 z

pz zw

Dn

Di

i

i

i T T t u

t u

u

u

21 1

2ln

z

pz

i

i

u

u

2)2(1

2

p

p z

p

z

2



4.5 Forced vibration under periodic excitations

Any periodical loading p(t ) can be expanded using Fourier series as follows:

Tp — period of the loading

11

0 sincos)( j

j j

j

j j t bt aat p w w

p

jT

j j p

w w 2

1

p

p

p

T

j

p

j

T

j

p

j

T

p

ndt t t pT

b

ndt t t pT

a

dt t pT

a

0

0

00

,3,2,1)sin()(2

,3,2,1)cos()(2

)(1

w

w



Substitute into the equations of motion

Using matrix representation

Solving for As and Bs, and substitute the values of As and Bs into u p:

2 2

s n s n s 0

2 2

s n s n s

2 2

n s n s 0

2 2

n s n s

( A 2 B A p ) cos t

B 2 A B sin t 0

( )A (2 )B p

( 2 )A ( )B 0

for all time.

w zw w w w

w zw w w w

w w zw w

zw w w w

2 2

sn n 02 2

sn n

A( ) 2 pB 02 ( )

w w zw w zw w w w

10 n p 2 22 2 2 2

nn n

X

p 2u (t) cos( t tan )

( ) (2 )

zw w w w ww w zw w



Things to notice about damped forced response

If z = 0, undamped equations result

Steady state solution prevails for large t

Often we ignore the transient term (how large is z, how long is t ?)

Coefficients of transient terms (constants of integration) are effected by the initial

conditions AND the forcing function



21

12

r

r z tan

Magnitude:

Dynamic magnification

Factor (DMF):

Phase:

Frequency ratio:

Characterization of the response

02 2 2 2

n n

pU( ) (2 )

w w zw w

2

n

2 2 20 0

UU 1

p / k p (1 r ) (2 r)

w

z

n

r w

w



Magnitude plot

0 0.5 1 1.5 2-20

-10

0

10

20

30

40

r

X

( d B )

z =0.01

z =0.1

z =0.3

z =0.5

z =1

d2 2 2

1R

(1 r ) (2 r)

z

Resonance is close to r = 1

For z = 0, r =1 defines

resonance

As z grows resonance movesr <1

The exact value of r , can be

found from differentiating

the magnitude



Phase plot

Resonance occurs at =

p2

The phase changes more

rapidly when the dampingis small

From low to high values

of r the phase always

changes by 1800 or p

radians0 0.5 1 1.5 2

0

0.5

1

1.5

2

2.5

3

3.5

r

P h a s e ( r a d

)

z =0.01

z =0.1

z =0.3

z =0.5

z =1

2

1

1

2r

r z tan



d2 2 2

2

d max 2

d d 1R 0

dr dr (1 r ) (2 r)

r 1 2 1 1/ 2

1R

2 1

peak

z

z z

z z

Compute max peak by differentiating:



d

2 2 2 2

2 2 2

R 1 1

2 2 2 1 (1 r ) (2 r)r (1 2 ) 2 1

z z z z z z

Experimental evaluation of damping using half-power bandwidth:

for small z, z2 = 0

2

r 1 2 r 1 2 z zExpand using Taylor expansionr 1 2 z

a b b a

r 1 higher terms

r 1 r 1 r r 2

, ,

z

z z z

4 7 Response to general dynamic loading



4.7 Response to general dynamic loading

EOM：

(1) Response to impulse Newton’s second law

at ：

Introduce initial velocity, then response is free vibration with initial velocity.

mu cu ku P

( ) 0, ( ) 0, ( )

u ut d u d du

1( ) ( )

du P mu m du P d

d m

t

0

( )( ) ( )

d

P u d u d m

(2) Response to general dynamic loading



(2) Response to general dynamic loading

For a linear undamped system：

Duhamel’s integral (only valid for linear system).

For a linear damped system：

( )sin ( )n

n

P d du t

m

w

w

0

( ) 1sin ( ) ( ) sin ( )

t

n n

n n

P d u du t d P t d

m m

w w

w w

( )( ) cos sin sin ( ) for >n n n

n

uu t A t B t t

w w w

w

( )( ) [ cos ( ) sin ( )]n t D Du t e A t B t zw w w ( )

0

1( ) sin ( )n

t t D

D

u P e t d m

zw w

w

5 Multi-Degree of Freedom System



x 1 x 2

k 1

m1 m2

k 2

5. Multi-Degree of Freedom System

Two Degrees of Freedom

0)()()(

0)()()()(

:gRearrangin

)()()(

)()()()(

221222

2212111

12222

1221111

t xk t xk t xm

t xk t xk k t xm

t xt xk t xm

t xt xk t xk t xm

Initial conditions:

Two coupled, second -order, ordinary differential equations with

constant coefficients

Needs 4 constants of integration to solve

Thus 4 initial conditions on positions and velocities



Solution by Matrix Methods

1 1 1

2 2 2

1 1 2 2

2 2 2

x x x

0

0

0

x (t) x (t) x (t)(t) , (t) , (t)

x (t) x (t) x (t)

m k k k

,m k k

M K

Mx Kx

0

0

221222

2212111

)()()(

)()()()(

t x k t x k t x m

t x k t x k k t x m



Initial Conditions: 10 10

20 20

x x(0) , (0)

x x

x x

2

2

Let ( )

1, , unknown

-

-

j t

j t

t e

j

e

w

w

w

w

w

x u

u 0

M K u 0

M K u 0

2

1

2

-

two algebraic equation in 3 uknowns

= , andu

u

w

w

M K u 0

u

Solution:

Changes ode into algebraicequation:

Condition for Solution:

2

1

2

inv - exists

Require does not exist

or det -

w

w

M K u 0

u 0

M K 0

One equation in one unknown w

B k t ifi t th h t i ti ti



Back to our specific system: the characteristic equation

2

2 1 1 2 2

22 2 2

4 21 2 1 2 2 1 2 2 1 2

det - 0

det 0

( ) 0

m k k k

k m k

m m m k m k m k k k

w

w

w

w w

M K

21 1

22 2

( )

( )

w

w

M K u 0

M K u 0

A vector equation for each square frequency

Calculating the corresponding vectors u1 and u2

and:



Physical interpretation of all the math

Each of the TWO masses is oscillating at TWO natural frequencies w1 and w2

The relative magnitude of each sine term, and hence of the magnitude of

oscillation of m1 and m2 is determined by the value of A1 and A2

The vectors u1 and u2 are called mode shapes

First note that A1, A2, 1 and 2 are determined by the initial conditions

Choose them so that A2 = 1 = 2 =0

Then:

Thus each mass oscillates at (one) frequency w1 with magnitudes

proportional to u1 the 1st mode shape

What is a mode shape?

1 1 1 1

(t) A sin( t ) w x u



Mode shapes:

Mode 1:

k 1

m1

x 1

m2

x 2 k 2

Mode 2:

k 1

m1

x 1

m2

x 2 k 2

x 2 =A

x 2 =A

x 1=A/3

x 1=-A/3

1

31

1u

1

31

2u

Solution as a sum of modes



Solution as a sum of modes

x(t ) u1 cos 1t u2 cos 2t

Determines how the first

frequency contributes to theresponse

Determines how the second

frequency contributes to the

response

Things to note

Two degrees of freedom implies two natural frequencies

Each mass oscillates at with these two frequencies present inthe response

Orthogonality property of mode shapes



Orthogonality property of mode shapes

1 2

1

2

1

2

. .

.

.

.

.

n

T

n

T

n

,

m

m

m

k

k

k

U u u u

U MU

U KU

Normalization of mode shapes



Normalization of mode shapes

ijij

T j j

u

u Mu

When M is diagonal matrix (shear building)

1

ijij

n

k kjk

u

m u

The orthogonal property becomes

21

22

2

..

T

T

n

w

w

w

Φ MΦ I

Φ KΦ



Thank you!

Questions?

july 28 - introduction to structural dynamics dr. hw huang

Documents