Dynsmics of Aerospace Structures Lecture

7/29/2019 Dynsmics of Aerospace Structures Lecture

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ASEN 5022 - Spring 2006

Dynamics of Aerospace Structures

Lecture 01: 17 January 2006

Introduction


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Dynamics according to

Galilei Galileo (Two New Sciences, 1636):A subject of never-ending interest.

Isaac Newton (The Principia, 1687):

We offer this work as the mathematical prin

ophy; for all the difficulty of philosophy see

this from phenomena of motions to investig

nature, and then from these forces to demon

phenomena; and to this end the general pro

first and second book are directed.


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What we will learn/acquire in this course

1. Kinematics which are essential for the descriof masses.

2. Ability to derive the equations of motion for c

ical systems.

3. Understanding of vibration phenomena forments.

4. Ability to model complex vibration problems b

ber of equations.

5. Applications of dynamical principles for eng

applications.

Above All, Become Excellent Dynamist


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Transition from Statics to Dynamics

Stati cs

Force equilibrium :

f or a f r ee body i

j fi j = 0

j f

Moment eq uili bri um :

around pointP

j MP j = 0

MP j

Observation: The crucial aspect of dynamics is t

pute the acceleration vector for every mass in the s


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Landaus Uniqueness Theorem

If all the position vectors {x1, x2, ..., xn} and t

tors {x1, x2, ..., xn} for n particles are given the accelerations {x1, x2, ..., xn } at that insta

defined.

The relations between the accelerations, velo

tion vectors are called the equations of motio

L. D. Landau and E. M. Lifshits, Mechanics (3rd

Press, 1959.


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Reference Frame and Position Vector

Consider two frames, K and K

, where V is cmagnitude and direction, and in which the pro

and time are the same.

r

r '

V t

O

O '

F r a m e K

F r a m e K '

r = V t + r '

E x a m p l e o f T w o I n e r t i a l F r a m


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Reference Frame and Position Vector-Contd

The kinematic relations between frames K and K

r = Vt + r

r = V + r v = V + v

r = r a = a

Conclusion: The acceleration vectors are the same in a

Hence, Galileos relativity principle holds.

In all inertial frames, the laws of mechanics are the sa

ferred to Galileos relativity principle, one of the mos

ciples of mechanics.


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Newtons Eight Definitions

DEFINITION I: The quantity of matter[mass] is the measure

from its density and bulk conjointly.

DEFINITION II: The quantity of motion[linear momentum] is

same, arising from the velocity and quantity of matter conjointly

DEFINITION III: The vis insita, or the innate force of matter[inerof resisting, by which every body, as much as in it lies, continue

whether it be rest, or of moving uniformly forwards in a right[str

DEFINITION IV: An impressed force is an action exerted upo

change its state, either of rest, or of uniform motion in a right lin


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Newtons Eight Definitions - contd

DEFINITION V: A centripetal force is that by which bodies ar

or any way tend, towards a point as to a centre.

DEFINITION VI: An absolute centripetal force is the measure

tional to the efficacy of the cause that propagates it from the centr

round about.

DEFINITION VII: The accelerative quantity of a centripetal for

the same, proportional to the velocity which it generates in a give

DEFINITION VIII: The motive[motion-causing] quantity of a ce

measure of the same, proportional to the motion which it generat


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Newtons Three Laws

Law I: Every body continues in its state of remotion in a right line, unless it is compelled to c

by forces impressed upon it.

Law II: The change of motion is proportional to

impressed; and is made in the direction of the ri

the force is impressed.

Law III: To every action there is always oppo

action; or, the mutual actions of two bodies upo

always equal, and directed to contrary parts.


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Test of Galileos Relativity Principle

m

k

B

F r e e - f r e e

s t a t e

F r a m e K

o

( m a s s c e n t e r )

F r a m e K '

S t a t i c

d i s p l a c e m e n t

A s s u m e d

d y n a m i c s t a t e

X

X '

d

x '

x

h

m g

k ( x

= k

m


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Test of Galileos Relativity Principle-contd

Position vector in frame K : X = h

Position vector in frame K : X = x

Relation between the two frames : x = x

Equation of motion in frame K : mg k(x Equation of motion in frame K : mg kx


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EOM in frame K : mx + kx = (mg + kd)(st

EOM in frame K : mx + kx = mg

Remark 1: The mass-point based coordinate system

include the dead weight mg in the equations of mot

Remark 2: The ground based coordinate system (fra

require to account for the static equilibrium.

Remark 3: When the ground itself moves, i.e., buil

earthquakes and automobile riding on wavy roads,

ify h to X = xg + h d + x where xg is the motion osimilar modification must be made for X.


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What happens to the mass point-based coordinate sground moves?


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Lecture 02: 19 January 2006

Vibration of Two-DOF Systems


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1. Equations of Motion

Consider a two-DOF model as shown in Fig. 1. The free-body

M1 and m2 are also shown in Fig. 1. Note that the inertia forare associated with minus sign, for the inertia forces can be con

forces (cf.,

f ma = 0). Summing the forces acting on each

of motion for the coupled two-mass-spring-damper system can b

For M1 :

f = f1(t) M1 x1 + {k2(x2 x1) + c2( x2 x1)

For m2 :

f = f2(t) m2 x2 {k2(x2 x1) + c2( x2 x1)

M1 x1 = f1(t) K1 x1 + k2 (x2 x1) + c2 ( x2 x

m2 x2 = k2 (x2 x1) c2 ( x2 x1)


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T

w

o

D

O

F

S

p

r

i

n

g

-

M

a

s

s

-

D

a

m

p

e

r

M

o

d

e

l

k

B

C

x

m

x

M

K

F

r

e

e

B

o

d

y

D

i

a

g

r

a

m

f

o

r

T

w

o

D

O

F

S

p

r

i

n

g

-

M

a

s

s

-

D

a

m

p

e

r

M

o

d

e

l

M

f

1

(J

)

K

1

x

1

m

M

1

-(

)

m

2

x

2

-(

)

c2

(

x2

x1

)

-

k

2

c2

(

x2

x1

)

-

k

2

(

x

2x

1

)

-

(

x

2x

1

)

-

B

B

x

1

Fig. 1 Two DOF Simplified Vertical Motion Mo


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As the preceding equation involves two displacements, x1 and x2involves complex matrix differential algebra. For design cons

important insight can be gained by considering the special case

given by

f1(t) = F1ej t, f2(t) = F2e

j t

so that the solution assumes the form of

x1(t) = X1ej t

x2(t) = X2ej t

Substituting (2) and (3) into (1), we obtain

2M1 X1 = F1 K1 X1 + k2 (X2 X1) + j c2

2m2 X2 = F2 k2 (X2 X1) j c2 (X2 X1)

In order to solve for X1 and X2, lets rearrange the preceding equ

(2M1 + j c2 + K1 + k2) X1 = F1 + (k2 + j

(2m2 + j c2 + k2) X2 = F2 + (k2 + j


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Solving for X1 and X2 we obtain

X1 =

H2()F1 + H12()F2

H1() H2() H212()

X2 =H1()F2 + H12()F1

H1()H2() H2

12()

H1() = (2M1 + j c2 + K1 + k2)

H2() = (2m2 + j c2 + k2)

H12() = (k2 + j c2)

Although (6) appears to be very complex, several simplification

vibration designers. This is studied below.


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2. What Are Vibration Modes and Mode Shape?

It turns out that the motions of X1 and X2 given by (6) are not ra

as the solution may suggest. They are interlinked by the propertyUnderstanding the physical properties of mode shapes is importa

design structures subject to vibrations.

As a motivation, let us consider the following special 2-DOF cas

m1

= m2

= 1, k1

= k2

= 2.618 2, c2

= 0, f1

(t

For this model, we apply two different excitations:

f2(t) =

0.2 sin (0.98 t)

sin (0.98 2.618 t)


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Figure 2 illustrates the two responses subject to the two excitati

Observe that, for the case of f2(t) = 0.2 sin (0.98 t), both m

are moving in phase, that is, they move in the same direction in ti

the system is subjected to f2(t) = sin (0.98 2.618 t), mass m

moving out phase, that is, they move in the opposite directions in t

depending on the excitation frequency, the motions of the two m

different. To understand this strange phenomena, one has to und

modes and mode shapes. To this end, let us recast (1) in a matrix

m1 00 m2

x1x2

+

k1 + k2 k2k2 k2

x1x2

=

The characteristic equation of the above coupled 2-dof differentia

obtained as follows. First, we assume the solution of their homog

the form

x1x2

=

x1x2

ej t


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0 0 . 2 0 . 4 0 . 6 0 . 8 1 1 . 2 1 . 4 1 . 6 1 . 8 2

0

0 . 5

1

1 . 5

2

2 . 5

M

a

s

s

P

o

s

i

t

i

o

n

s

f

o

r

M

o

d

e

1

T i m e

2 - d o f s y s t e m r e s p o n s e d u e t o h a r m o n i c i n p u t

a t m a s s 2 w i t h o m e g a = 0 . 9 8 * o m e g a

1

P o s i t i o n o f

M a s s 1

m 1

m 2

P o s i t i o n o f

M a s s 2

0 0 . 2 0 . 4 0 . 6 0 . 8

0

0 . 5

1

1 . 5

2

2 . 5

M

a

s

s

P

o

s

i

t

i

o

n

s

f

o

r

M

o

d

e

2

2 - d o f s y s t e m r e s p o n s e

a t m a s s 2 w i t h o m e g a =

m 1

m 2

k 1

k 2

Figure 2 Motions of two masses under two different


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Second, substituting this into (9) with f1 = f2 = 0 yields:

[2 m1 0

0 m2

+ k1 + k2 k2

k2 k2

] x1

x2

=

Hence, the characteristic equation is obtained by requiring that th

a nontrivial solution:

det k1 + k2 2m1 k2k2 k2

2

m2 = 0 4( k1 + k2

m1

Note that, with m1 = m2 = 1, k1 = k2 = 2.6182, the two

characteristic equation are given by

n1 = , n2 = 2.618

These two values are called characteristic values whose square

natural frequencies or vibration modes of the system.


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The corresponding eigenvectors can be computed from the secon

2 x2 k2 x1 + k2 x2 = 0 x1

x2=

k2 2

k2=

Note that for the two modes computed in (13), we have two diffe

For = ,x1

x2= 1

2

2.618 2= 0.618

For = 2.618,x1

x2= 1

2.6182 2

2.618 2= 1.

The ratios of these eigenvector sets are plotted in Figure 3.

M o d a l A m p l i t u d e f o r M o d e 1

M

a

s

s

P

o

i

n

t

F i x e d

e n d

1

2

0 . 6

1 . 0

0 . 0

M o d e S h a p e f o r M o d e 1

0 1 . 0

- 1 . 6

M o d a l A m p l i t u d e f o r M o d e 2

M

a

s

s

P

o

i

n

t

1

2

M o d e S h a p e f o r M o d e 2

F i x e d

e n d

Figure 3 Mode Shapes of 2-DOF Example Prob


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Observe from Figure 3 that for the case of the first mode when

same direction with its amplitude of 0.616 while mass 2 moves

other words, the two masses move in phase as illustrated in Figure

mode, mass 1 moves in the opposite direction by -1.618 while m

amplitude, which is also illustraed in Figure 2. From these obserthat mode shapes indicate how the system will deform when sub

excitation whose frequency is close to one of the natural frequ

modes) of the system. Thus, mode-shape information is useful in

subjected to harmonic excitations. Examples of such systems

airplanes, ships, motor vehicles, and many other machinery equip


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3. Vibration analysis of 2-DOF Spring-Mass Problem by MA

In Matlab we invoke the following routine:

[X, D] = eig(K, M);where X is the eigenvector, D is the ei

For example, the eigenproblem of the 2-DOF system studied in th

can now be analyzed by MATLAB with

K = 2 2.618 2 2.618 22.618

2

2.6182 , M = 1

0

Upon using the following MATLAB code, we find the following

%% 2 dof eigenvalue analysis

%

%stiffness matrixK = [2*2.618*pi^2 -2.618*pi^2;-2.618*pi^2 2.618*pi^2];


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% mass matrixM = [1 0;0 1];

% call eigenvalue routine

[X, D] = eig(K, M);

% write eigenvector and eigenvalueslambda = diag(D);

disp([Eigenvalues of 2-dof system num2str

disp( );disp([Eigenvectors of 2-dof system num2strdisp([ num2str

% compute the frequencies

freq = diag(D); % extract the two diagonal entri

disp( );freq = sqrt(freq);disp([Frequencies of 2-dof system num2st


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we find the following results:

Eigenvalues of 2-dof system 67.6464 9.869

Eigenvectors of 2-dof system 0.85065 0.5257-0.52573 0.850

Frequencies of 2-dof system 8.2247 3.1416

Note that MATLAB prints out the highest mode first. Hence, t

given by

For the second mode with 2 = 2.618 = 8.2247 : X2 =

x2x2

For the first mode with 1 = = 3.14157 : X1 =

x1x1

The mode shapes computed by MATLAB is normalized so that t

is unity. If the modal amplitude at mass 2, x22, for the second m

to be unity from MATLAB computed value x22 = 0.52573,


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T w o D O F S p r i n g - M a s s - D a m p e r S u s p e n s i o

m

2

x

1

C

2

x

2

1

M

1

B

2

B

k

2

/ 2

1

K

/ 2

Figure 4: 2-DOF Suspension Model


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xst(1) = F1/K2

such that from (7) we have

H1 =X1()xst(1)

= K1 H22()

H11() H22() H2

12()

Dividing both the nominator and the denominator by M1 m2 a

rameters introduced in (20), the frequency response function H1

expressed as

H1 =H22()

H11() H22() H2

12()

H11() = 2 + j 2 2 +

21 +

22

H22() =

2

+ j 2 2 + 2

2

H12() = (j 2 2 + 22)


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1 0

1

1 0

- 1

1 0

0

1 0

1

F r e q u e n c y ( H z )

A

m

p

l

i

t

u

d

e

I n v a r i a n t p o i n t

F r e q u e n c y R e s p o n s e F u n c t i o n s f o r D i f f e r e n t D a m p i n g

o f a 2 - D O F S u s p e n s i o n M o d e l

Figure 5: Frequency Response Function at Ma


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Figure 5 illustrates the frequency response function (FRF) for diff

with

1 = 2 = 10 H z, = 0.1

Note that there are two points that all FRF curves pass through t

It is these invariant points that were first discovered by den Hartin 1928 who subsequently utilized their properties for improved d

shock attenuation.

In their design study by using Figure 1, den Hartog and Ormon

if the magnitudes of the two invariant points are made to be the

mass ratio and the damping ratio, a near optimum design goal ca

For the design of accelerometers using Figure 4, the objective

amplitude of the frequency response of mass 2 in order to decrea

ratio while realizing a flat plateau of frequency interval of interes

For the design of resonators using Figure 4, in addition to maxim

of FRF at mass 2, the gap between the two invariant points shoul

MATLAB code that produced Figure 5 islisted below.


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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% hartog_oscillator.m

% by k c park 15 january 2002

clear

% Define some useful numbers:

dtr=pi/180;

Hz2rps=2*pi; rps2Hz=1/2/pi; % Conversions to/from Hz and rad

% Natural frequencies of individual masses: 10Hzw1=10*Hz2rps;

w2=w1;

mu = 0.1;

for zeta =0:0.05:0.8 %zeta is the damping ratio

% Define a system in transfer function forma0=1; a1 =2*zeta*w2; a2 = w2^2;

b0=1;

b1 =2*zeta*(1+mu)*w2;

b2 = w1^2+w2^2 + mu*(1+4*zeta^2)*w2^2 - 4*mu*zeta^2*w2^2;


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b3 = 2*zeta*(mu*w2^3 + w2*(w1^2+mu*w2^2)) - 4*zeta*mu*w2^3;

b4 = w2^2*(w1^2+mu*w2^2)- mu*w2^4;

sys = tf ( [a0 a1 a2], [b0 b1 b2 b3 b4]);

% Pick a range of frequencies to look at from 0 Hz to 100 Hz

wf= logspace(0.5,1.5,400)*Hz2rps;

% Find the frequency response

H=freqresp(sys,wf); % This ends up being a 3 dimensional arr

H=reshape(H,size(wf)); % this makes it the same size as wf

% Plot the amplitude as a function of frequency in Hz

loglog(wf*rps2Hz,w1^2*abs(H)), grid

xlabel(Frequency (Hz))

ylabel(Amplitude)

hold on;

endaxis([5 30 0.1 30]);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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References

1. R. E. D. Bishop, Vibration, Cambridge University Press, 1979.

2. J. P. Den Hartog, Mechanical Vibrations, Dover Pub., 1984.

3. Maurice Roseau, Vibrations in Mechanical Systems, Springer-Verlag, 1984


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Dynamics of Aerospace Structur

Lecture 03:

Solution of 1 and 2-DOF Vibrating M


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Governing equation: m x + c x + kx = f(t

p = m x = f(t) kx (c/m)p

x = (1/m)p

x

p

=

0 1/m

k c/m

x

p

+

f

Canonical equation: x = Ax + Bu

Output equation: y = Cx + Du


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Solution of the canonical equation:

x = eAtx(0) +

t0

eA(t )Bu( ) d

eAtx(0)-term:

Response due to the initial conditio

t0

eA(t )Bu( ) d-term:

Response due to the external (appli

function.


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Discrete approximation:

x(kT) = eAkTx(0) + kT

0

eA(kT )Bu

x(kT+ T) = eA(kT+T)x(0)

+

kT+T0

eA(kT+T )Bu( )


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Discrete approximation-contd

x(kT+ T) = eAT[eAkTx(0)]

+

kT

0

eA(kT+T )Bu( ) d

+

kT+TkT

eA(kT+T )Bu(


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Discrete approximation -contd

x(kT+ T) = eAT[eAkTx(0)]

+ eAT[

kT

0

eA(kT )Bu(

+

kT+TkT

eA(kT+T )Bu(

x(kT+ T) = eATx(kT)

+ kT+T

kT

eA(kT+T )Bu(


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kT+T

kT

eA(kT+T )Bu( ) d

(via change of variable: = kT+ T

0T

eABu(kT) (d ), with u( )

=

T

0

eABu(kT) (d )

= A1(eAT I) B u(kT) = u(k

where = A1

(eAT

I) B


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Discrete approximation -conclud

x(kT+ T) = eATx(kT) + u(kT

x (k+ 1) = x(k) + u(k)

= A1(eAT I) B


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Useful Matlab routines:

sys= ss(a,b,c,d): Specify a state space mode

H =tf (num, den): Convert a state space mcorresponding transfer function.

[y,t,x]=step(sys): obtain step response.

[y,t,x]=lsim(sys,u,t,x0): obtain response due

input u.[H] =freqresp(sys,w): compute frequency re

a grid of frequencies.

[Hresh]=reshape(H, size(w)): make H the s

w.


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[y,t,x]=impulse(sys,t): obtain impulse respon

[y,t,x]=initial(sys,xo): free response due to

dition.


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% vibex.m

% Program to compute 1DOF computations

% January 16, 2004 (KCP)

clear

% Define some useful numbers:

Hz2rps=2*pi; rps2Hz=1/2/pi; % Conversions to/from Hz and rad/s

% Natural frequency of 1 Hz

omega_n=1*Hz2rps;T_n = 2*pi/omega_n; % period of the response

% critical damping ratio

zeta= 0.05; %5 percent

% state space model

a=[0 1; -omega_n^2 -2*zeta*omega_n];


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b=[0; omega_n^2]; % a force is appled, equivalent to unit static d

c=[1 0; 0 1]; % both momentum and displacement are available

% specify state space model for this problem

sys = ss (a,b,c,0);

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Time response computation

%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% obtain the response, y1 =displ; y2=mementum% step(sys); % simplest step response computation

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

% for specified computations;

%

% the total time of simulation


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Tfinal = 6*T_n; % 6 periods

%specify the output time points:

dt = (1/40)*(T_n); %40 samples per period

%time interval

t = 0:dt:Tfinal;

%obtain response

[y,t,x]=step(sys,t);

%plot the response

figure(2);

plot(t, y(:,1), k);

xlabel(Time(sec));

ylabel(Displacement);title(Step Response computed by specifying the time step);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% end of response computation

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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%

% frequency response calculation

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Pick a range of frequencies to look at from 1 Hz to 10 Hz

wf= logspace(-1,1,200)*Hz2rps;

% Find the frequency response

% generte transfer function

TransF = tf(sys);

H=freqresp(TransF,wf); % This ends up being a 3 dimensional array!

Hreshape=reshape(H(1,:,:),size(wf)); % this makes it the same size

% Plot the amplitude as a function of frequency in Hz

figure(3)

loglog(wf*rps2Hz,abs(Hreshape));

grid on;

xlabel(Frequency (Hz))

ylabel(Amplitude)


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title(Frequency response of1DOF damped system);

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% end the frequency response

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

%harmonic response of 1dofsystem

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Pick a range of times to look at the time response

t=0:0.005:5; % for 5 period of T_n

% sinusoidal forcing function

u=sin(Hz2rps*2*t); % omega_forcing = 2 Hertz!

[ys,ts]=lsim(sys,u,t);


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figure(4)

plot(ts,ys(:,1)),grid

xlabel(Time (sec))

ylabel(Displacement)

title(Time response due to harmonic excitation);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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0 1 2 3 4 5

0

0 . 2

0 . 4

0 . 6

0 . 8

1

1 . 2

1 . 4

1 . 6

1 . 8

2

T i m e ( s e c )

D

i

s

p

l

a

c

e

m

e

n

t

S t e p R e s p o n s e c o m p u t e d b y s p e c i f y i n g t h e t i m e s t e p


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0 0 . 5 1 1 . 5 2 2 . 5 3 3 . 5 4

- 0 . 8

- 0 . 6

- 0 . 4

- 0 . 2

0

0 . 2

0 . 4

0 . 6

0 . 8

1

T i m e ( s e c )

D

i

s

p

l

a

c

e

m

e

n

t

T i m e r e s p o n s e d u e t o h a r m o n i c e x c i t a t i o n


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1 0

- 1

1 0

0

1 0

- 2

1 0

- 1

1 0

0

1 0

1

F r e q u e n c y ( H z )

A

m

p

l

i

t

u

d

e

F r e q u e n c y r e s p o n s e o f 1 D O F d a m p e d s y s t e m


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Lecture 04: January 26

Problem Solution via Newtons 2nd


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A Spring-Mass-Bar System:

x

mK

o

K

o

m

x, i

L

Mg

F

F

A

C

y, j

B

B

C

A

N

o

NANA

kx

mg

A

ACY

Mg

ML2/12)(

C

AACX

ACY


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Position Vectors

Position vector for mass m:

rA = x i

Position vector for bar M:

rC = rA + rAC

rAC =L

2(sin i cos j)


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Velocity Vectors

Velocity vector for mass m:

vA = rA = x i

Velocity vector for bar M:

vC = vA + vAC

vAC =L

2(cos i + sin j)


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Acceleration Vectors

Acceleration vector for mass m:

aA = vA = x i

Acceleration vector for bar M:

aC = aA + aAC

aAC =L

2(cos i + sin j)

L 2

2(sin i cos j)


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Equilibrium equation for mass m:

kx i m x i mgj + NAj

+ XACi + YACj = 0


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Equilibrium equation for bar M:

Fi MaC Mgj XACi YACj

MC = M L2

12 k + rCA (XACi

+rCB

Fi = 0


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Equilibrium equation for mass m - contd

Equation(7) yields:

m x + k x XAC = 0

NA mg + YAC = 0

XAC = m x + kx


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Equation(8) [sum of forces] yields:

F M( x +L

2cos

L 2

2sin )

XAC =

Mg + M(L

2sin +

L 2

2cos ) + YAC


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Equation(9) [sum of moments] yields:

M L2

12

L2

( sin i + cos j) (XACi + YAC

+L

2(sin i cos j) Fi =


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Equilibrium equation for bar M - conclu

M L2

12

+ [L

2cos XAC +

L

2sin YAC]

+L

2cos F = 0


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Equations for

1. Obtain (L2

cos XAC + L2

sin YAC) i

(11) and (12):

(

L

2 cos XAC +

L

2 sin YAC) =

L

2cos (F Mx)

L

2sin Mg

M


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Equations for - concluded

2. Substitute (16) into (14) to obtain:

M L2

3

+M L

2

cos x +Mg L

2

sin

= L cos F


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Summary: Coupled equations for x and

(M + m) x + kx +

M L

2 cos

M L

2sin 2 = F

M L2

3x +

M L

2cos x +

Mg L

2sin

= L cos F


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What ifMA were adopted instead ofMC ?

MA =

M L2

12 k

+rAC (Mgj MaC ) + rAB (Fi) = 0

which leads to (17)!

Observation: There may be a preferrable ch

ocations where one can sum the moment!!!


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Appendix

wo independent variables in EOM:

Position vector (or displacement): r

Linear momentum vector:

p = m v, v =dr

dt

Newtons second law:

F =dp

dt=

d(mv)

dt

Moment of a force and angular momentum

ngular momentum:

Ho = r p = r (mr)


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ate of angular momentum:

Ho = r p + r p = r p

= r F = Mo

M = Ho

Work and Energy

ncrement of work:

d W = F dr = mr dr

= mdr

dt rdt = d( 1

2mr r) = d T

ntegrating both sides, we obtain:

r2r1

F dr =

r2r1

d( 12

mr r)

(V2 V1) = T2 T1


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efinition of potential energy:

V(r) = rr e f

r

F dr

d W = d V(r)

xample of potential energy for a spring undergoing displacement:

V() =

0

Fd x =

0

kx dx = 12

k2

ifting gravity force from y1 to y2:y2

y1

F dy =

y2

y1

(mgj) d y(j)

= -mg (y2 y1) = (V2 V1)


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V2 = mgy2, V1 = mgy1

ystems of discrete bodies

Mass center: For a system consisting of N masses whose position vectors are

2, 3, ...N} , the mass center is defined as

Ni =1

mi ri = 0

Momentum of N masses: ptotal = m vC

m =Ni =1

mi

Kinetic energy of N masses


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ri = rc + ri , vi = vc + v

i

T = 12

N

i =1

mi (vC + vi ) (vC + v

i )

= 12

m vC vC + vC 1

2

Ni =1

mi vi

+ 12

N

i =1mi v

i v

i

ince we have from (8),N

i =1mi ri = 0, which leads to

Ni =1

mi vi = 0

o that (27) reduces to

T = 12

m vC vC +1

2

Ni =1

mi vi v

i


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otational Moment of Inertia

HO = m

r (dmv), v = r

=

m

r ( r) dm

et r and be expressed as

r = x i + yj + zk

= x i + yj + z k

tep 1: Compute r

r = det

i j k

x y z

x y z

= (zy yz )i + (xz zx )j + (yx xy )k


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tep 2: Compute r ( r)

r ( r) = [(y2 + z2)x x yy x zz ]i

+ [yxx

+ (x 2 + z2)y

yzz

]j

+ [zxx zyy (x2 + y2)z ]k

tep 3: Compute m r ( r)dmH = Hx i + Hyj + Hz k

Hx = Ix x x + Ix y y + Ix z z

Hy = Iyx x + Iyy y + Iyz z

Hz = Izx x + Izy y + Izz z

Ix x = m

(y2 + z2) dm, Ix y = m

x y dm, et c.

ther quantities may be computed by changing the subscripts from Ix x and

tep 4: Compute MO = HO


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HO = ( Hx i + Hyj + Hz k)

+ (Hx i + Hy j + Hz k)

ince we havei = i, j = j, k = k

HO becomes

MO = HO = ( Hx i + Hyj + Hz k)

+ (Hx i + Hyj + Hz k)

Kinetic energy of general rigid-body systems

Kinematics:r = rC + r

v = vC + r


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Kinetic energy:

T = 12

m vC vC

+ 12

m

( ri ) ( ri ) dm

or a general point P that is different from point C, we have

T = 12

m vP vP + vP ( rPC)

+ 12

m

( r)2 dm


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here (x, y,z)-components of 12

m

( r)2 dm are given by

1

2 m

( r)2 dm = IP {}

=

0 z y

z 0 xy x 0

{} =

xyz

IP =

Ix x Ix y Ix zIx y Iyy IyzIx z Iyz Ix z

P

here subscript P denotes that the rotatioanl moments of inertia are com

espect to point P .


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Lecture 05: January 31

Energy Principles for Dynamics


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Principle of Virtual Work for Statics

For statics a la Meirovitch:

N

i =1

Fi ri = 0

where

{Fi } are impressed forces, and

{} designates the virtual character of

taneous variations, as opposed to the dsymbol {d} designating actual differen

sitions {ri } taking place in the time inte


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Principle of Virtual Work for Dynamics

For dynamics a la dAlembert:

N

i =1

(Fi mi ri ) ri = 0

where

{ri } is the acceleration of particle mi .


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Generalized Coordinates

Virtual displacements ri may be ex

terms of the generalized virtual dispqk(k = 1, 2, 3, ..., n):

ri

=

n

k=1

ri

qkq

k, i = 1, 2, 3, ..

An example:

r = x i + yj + L(cos i + sin


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Generalized Coordinates-contd

Observe that there are three independent

(x , y, ).

r =r

xx +

r

yy +

r

= x i + yj + L( sin i + cos j)

Note that the dimensional unit of (x , y) i

whereas that of is radian, suggesting tha

of generalized coordinates can be differen


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Interpretation of Equation (2)

Lets revisit the formulation of the equati

tion for the spring-mass-bar problem stuprevious lecture. To this end, first, we must

virtual displacements.


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A Spring-Mass-Bar System:

x

mK

o

K

o

m

x,i

L

Mg

F

F

A

C

y, j

B

B

C

A

N

o

NA

kx

mg

A

ACY

Mg

ML2/12)(

C

AACX

ACY


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Virtual displacements:

Virtual displacement for mass m:

rA = x i

Virtual displacement for bar M:

rC = rA + rAC

rAC =L

2(cos i + sin j)


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Virtual displacements-contd

Virtual rotation for bar M:

Here, one utilizes the angular velocity

obtain the virtual rotation

= k

In general one has

= x i + yj + zk


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Acceleration Vectors:

Acceleration vector for mass m:

aA = vA = x i

Acceleration vector for bar M:

aC = aA + aAC

aAC =L

2(cos i + sin j)

L 2

2(sin i cos j)


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Equilibrium equation for mass m:

o m x

NA

k x

mg

AACX

ACY

fA = kx i m x i mgj + NAj

+ XACi + YACj = 0


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MgF

M rCML

2/12)(

B

C

AACX

ACY

fC = Fi MaC Mgj XACi YACj

MC = M L2

12k + rCA (XACi

+rCB Fi = 0


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An Application of DAlemberts Princi

spring-mass-bar problem

(Fi mi ri ) ri =

fA rA + f

C r

C + M

C

C =

Note that fA rA = 0 and so other two te


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{kx i m x i mgj + NAj

+ XACi + YACj} rA

+ {Fi MaC Mgj

XACi YACj} rC

{M L2

12k + rCB Fi

+ rCA (XACi YACj)}


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The first term in (16) becomes

fA rA ={k x i m x i mgj + NAj

+ XACi + YACj} (

fA rA ={k x m x + XAC} x

Remark: Observe that the reactions forces Y

have played no role in the resulting virtual w


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The second term in (16) becomes

fC rC =

{Fi MaC Mgj XACi YACj}

= {Fi M[ x i +L

2(cos i + sin j)

L 2

2 (sin i cos j)]

Mgj XACi YACj}

{x i +L

2(cos i + sin j) }


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fC rC = {F M[ x +

L

2cos

L 2

2

XAC} x

+ {F M[ x +L

2cos

L 2

2sin ]

XAC}L

2cos

+ {M[L

2sin

L 2

2cos ] Mg

YAC}L2sin


99/391

fC rC = {F M[ x +

L

2cos

L 2

2sin

XAC} x

+ {L cos

2F

M L

2cos x

M L2

4

L

2cos XAC

M Lg

2sin

L

2sin YAC}


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In evaluating the third term, first, carry

rCB Fi =

L2

(sin i cos j) Fi

=F L

2cos k

rCA (XACi YACj) =

{L

2cos XAC +

L

2sin YAC}k


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Finally, we obtain:

MC C =

{M L2

12+

F L

2cos

+ [L

2cos XAC +

L

2sin YAC]}


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Combine (17), (20) and (23) to obtain

{(M + m) x + kx +

M L

2 cos

M L

2sin 2 F} x

+ {M L2

3x +

M L

2cos x

+Mg L

2sin L cos F} = 0


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Since x and are arbitrary, we obtain:

(M + m) x + kx +M L

2 cos

M L

2sin 2 = F

M L2

3 +

M L

2cos x +

Mg L

2sin

= L cos F


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Observations

1. When using Newtons second law, Eqs.(25

are obtained by eliminating the reaction fo

and YAC.

2. On the other hand, one does not need to

reaction forces in applying dAlemberts p

Only apparent forces including inertia forc

be considered. This is a major advantage of

energy principles over Newtons method.


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Lecture 06: 02 February 2006

Hamiltons Principle and

Euler-Lagranges Equations


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Lets begin with dAlemberts principle:

Ni =1

(Fi mi ri ) ri = 0

T + W

Ni =1

d

dt(mi ri ri ) =

W =

Ni =1

Fi ri , Ti = d(12

mi ri


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Lets integrate (1) in time:

t2t1

{T + W

Ni =1

ddt

(mi ri ri )} d

t2t1

{T + W} dt =

t2t1

Ni =1

d

dt(mi ri r


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Noting that we have

t2

t1

N

i =1

d

dt

(mi ri ri ) dt

=

Ni =1

(mi ri ri )|t2t1

so that, if ri (t1) and ri (t2) are specified, we h

ri (t1) = ri (t2) = 0


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Hence, equation (2) becomes

t2

t1

{T + W} dt = 0

which is known as extended Hamiltons princ


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Hamiltons Principle for Conservative System

In general the work done, W, consists of tw

W = Wcons + Wnoncons

where subscripts cons and noncons designate

and nonconservative systems, respectively.


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Hamiltons Principle for Conservative System

For conservative systems, we have have fr

Wnoncons = 0:

t2t1

{T + Wcons } dt = 0

which is known as Hamiltons principle for

systems.


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Action Integral for Conservative Systems

Observe that the work done on the conserva

can be expressed in terms of the correspondienergy:

Wcons = V, V =

Ni =1

(rr e f

rFi


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Substituting (9) into (8), one obtains:

t2t1

{T V} dt = 0

t2

t1

{T V} dt = 0

I =

t2t1

{L} dt = 0, L = T


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Euler-Lagrange Equations of Motion

Lets substitute the general work expression

extended Hamiltons principle (6) to obtaint2t1

{T + Wcons + Wnoncons} dt =

which, with (9) together with L = T V, be

t2t1

{L + Wnoncons} dt = 0


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Since L can be expressed in terms of the

coordinates as

L =n

k=1

{L

qkqk +

L

qkqk}

one obtains

t2t1

L dt =

t2t1

nk=1

{L

qkqk +

L

qkqk


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Integrating by part the first term of the pre

gral, we obtain

t2t1

nk=1

{L

qkqk} dt =

nk=1

{L

qkqk

t2

t1

n

k=1

{d

dt(

L

qk)qk} dt


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Since qi (t1) = q(t2) = 0, we have

t2t1

nk=1

{L

qkqk} dt

=

t2t1

nk=1

{d

dt(

L

qk)qk} dt


118/391

Substituting (16) into (14), then introducing

expression into (12), we finally obtain:

t2t1

nk=1

{ ddt

( Lqk

) + Lqk

+ Qk}qk d

Wnoncons =

n

k=1

Qkqk

Since qk are arbitrary, we obtain:

d

dt(

L

qk)

L

qk=

Qk

which is called Euler-Lagranges equations


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A Spring-Mass-Bar System for Euler-Lagrange

x

K

o

m

x, i

L

Mg F

A

C

y, j

B

B

C

A

xB

yC

Notice how simple the problem descripti

now!


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Kinematics:

Position vectors of mass M, bar at C, and B

nonconservative force F is applied:

rA = x i

rC = x i +L

2

(sin i cos j)

rB = x i + L(sin i cos j)

rC = L

2j


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Kinetic energy:

T = 12

m(rA rA ) + 12M(rC rC ) +12

IC

= 12[(M + m) x2 + M L x cos +

1

3M L


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Potential energy:

Vspring = 0

x

(kx i) drA =1

2

kx

Vgr avi t y =

rCrC

(Mgj) drC

= Mg

L

2 (1 cos )

Total potential energy:

V = Vspring + Vgr avi t y


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Nonconservative work Wnoncons:

Wnoncons = FB rB

= Fi [x i + L(cos i + sin j

= F(x + L cos )

Qx = F, Q = F L cos

Note that there are only two state variables: x

system kinetic energy, potential energy and th

vative work.


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d

dt

(L

x

) L

x

= Qx = F

d

dt(

L

)

L

= Q = F L cos


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d

dt[(M + m) x + 1

2

M L cos ] + kx =

M L

6

d

dt(2L + 3 x cos )

+ 12M L( x + g) sin = F L cos


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Lecture 07: 07 February

The Method of Lagranges Multipliers


127/391

The Euler-Lagranges equations of motion are obtaine

tem energy that consists of the kinetic energy, poten

virtual work due to nonconservative forces. In othe

enabling the derivation process simpler, the Euler-La

ism eliminated an important information: the reaction

and constraints (such as boundaries). Often, both the

analysts need to know joint force levels so that necessa

lations can be carried without jeopardizing the systemin point is the human joints which, repeadedly over loa

bone deformations and arthritis.

The question is: How do we re-introduce the constrai

tion forces) within the Euler-Lagrange formalism? T

oped by Lagrange and described in his book, M ecaniq

published in 1788. This is accomplished as follows.

Lets consider holonomic cases, viz., the constraint con

be expicitly stated in terms of position vectors, and con


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spring-mass system shown below. Suppose we would l

reaction force between mass m1 and spring k2, and the

between spring k1 and the attached boundary. We now

the procedure that leads to the determaination of the re


129/391

k

f

x1

m2

1 2

x2

1m

k

1 f2

k

x1

1 21mk

f1

x3

(a) Assembled system

(b) Partitioned system

x0x

g

Subsystem 1 Subsystem


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A procedure for the method of Lagranges multipliers

Step 1: Partition the system into completely free subsystems

Step 2: Identify the conditions of constraints between the cotems.

Step 3: Construct the energy of each of the completely free

Step 4: Obtain the total energy by summing the energy of each

free subsystems.

Step 5: Append the conditions of constraints by multiply

unknown coefficient, (multiplier), to the total system energ

potential).


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Reference solution based on the assembled s

L = T V

T = 12 m1 x21 +

12 m2 x

22

V = 12 k1x21 +

12 k2(x2 x1)

2

W = f1(t)x1 + f1(t)x2

Hence, we obtain via the Euler-Lagrange fo

m1 x1 + (k1 + k2)x1 k2x2 = f1(t

m2 x2 + k2x2 k2x1 = f2(t


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Application of the method of Lagranges multiplier

Step 1: Done in the preceding figure.

Step 2: Conditions of constraints:

Between k1 and the fixed end : 0g = x0 x

Between m1 and k2 : 13 = x1 x3 = 0

Step 3: Energy of two completely free subsystemsFor subsystem 1:

T1 =12

m1 x21

V1 =12

k1(x1 x0)2

W1 = f1(t)x1


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For subsystem 2:

T2 =12

m2 x22

V2 =1

2

k2(x2 x3)2

W2 = f2(t)x2

Step 4: Sum the total system energy

T = T1 + T2

V = V1 + V2

W = W1 + W2


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Step 5: Construct the constraint functional

= 0g0g + 13 13

= (x0 xg) 0g + (x1 x3) 13

Total Lagrangian of the partitioned system:

L = (T1 + T2) (V1 + V2)

(xg,x0,x1,x3, 0g, 13)


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Partitioned equations of motion:

x1 term: m1 x1 + k1x1 k1x0 + 13 =

x2 term: m2 x2 + k2x2 k2x3 = f2(t)

x3 term: k2x2 + k2x3 13 = 0

x0 term: k1x1 + k1x0 + 0g = 0

0g term: (x0 xg) = 0

13 term: (x1 x3) = 0


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Matrix form of the partitioned equations of motion

m1D2 + k1 0 k1 |0 1

0 m2D2 + k2 k2 0 |0 0

0 k2 k2 0 |0 1

k1 0 0 k1 |1 0

|

0 0 0 1 |0 0

1 0 1 0 |0 0

x1x2x3x0

og13

where D2 = d2dt2

. Symbolically, the above equation can bA C

CT 0

x

=

f

xb


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Hence, solution of (11) provides both the displacem

reaction forces .

Question: Can one obtain the equations of motion for

system (2)?

The answer is yes! To this end, we observe the fol

between the partitioned degrees of freedom and the a

as

x1x2x3x0

= 1 0 0

0 1 0

1 0 0

0 0 1

x1x2x0

x =

Substituting the above assembling operator L into (11


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LT A L LTC

CT L 0

xg

=

LT f

xb

Simple matrix multiplications show

LT C =

0 0

0 0

1 0

K = LTAL =

m1D

2 + k1 + k2 k2k2 m2D

2 + k2k1 0

Therefore, (13) reduces to:


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m1D2 + k1 + k2 k2 k1 |0

k2 m2D2 + k2 0 |0

k1 0 k1 |1

|

0 0 1 |0

x1x2x0

og

=

Finally, invoking the bounday condition

x0 = xg = 0

we arrive at

m1D

2 + k1 + k2 k2k2 m2D

2 + k2

x1x2

=

f1f2

Note that solution of (16) provides the reaction force at the left-en

(17) does not!


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Application of the method of Lagranges multipliers

x

K

o

m

x, i

L

Mg F

A

C

y, j

B

B

C

A

xC

yC

x

K

o

m

x, iA

y, j

A

o

L

A

C

y, j

B

A

xC

mg

Subsystem 1 Subsystem 2


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Kinematics:

Kinematics of Subsystem 1:

rA = x i + yj

vA = x i + yj

Kinematics of Subsystem 2:

rC = xC i + yCj

rA = rC L

2(sin i cos j)

rB

= rC

+L

2(sin i cos j)


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Step 1; Partitioning is done in the preceding figure!

Step 2: Conditions of constraints:

(1) Between mass m and the ground:

(y yg) j = 0

(2) Joint at A:

rA rA = 0

(x xC +L

2sin ) i

+ (y yC L

2 cos ) j = 0


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Step 3: Construct the energy of completely free two su

Energy of Subsystem 1:

T1 = 12 m( x2 + y2)

V1 = mgy +12

kx 2

Energy of Subsystem 2:

T2 =12

M( x 2C + y2C ) +

12

IC2

V2 = Mg yC


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Nonconservative work Wnoncons:

Wnoncons = FB rB

= Fi [xC i + yCj +L

2(cos i + sin

= F(xC +L

2cos )

Qx B

= F, Q = F L cos


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Step 4: The total energy of the system:

T = 12

m( x 2 + y2)

+1

2M( x2

C + y2

C ) +1

2IC

2

V = mgy + 12

kx 2 + Mg yC

Step 5: Construct the constraint functional:

= (y yg) 0g + (x xC +L

2sin )

+ (y yC L

2cos ) Ay


146/391

Partitioned equations of motion via the method of Lagpliers

x : m x + kx + Ax = 0

y : m y + mg + 0g + Ay = 0

xC : MxC Ax = F

yC : MyC + Mg Ay = 0

: IC +L

2cos Ax +

L

2sin Ay = F L cos

0g : (y yg ) = 0

Ax : (x xC +L

2

sin ) = 0

Ay : (y yC L

2cos ) = 0

Note that (0g , Ax , Ay ) corrrespond to the reaction forces (NA, X

obtained via Newtons second law, viz., in equation (10) of Lecture 04 S


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Lecture 8: 21 February

Vibration of String, Bar and Shaft


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w(x,t)

wz

f(x,t)

x dx

x+ dxx

w(x,t)

x

w(x,t)

x+

2w(x,t)

x2dx

T(x )

T(x)+T(x)

xdxf(x, t)dx

(x, t) =

(x + dx, t) =

ds

dx

String in transverse vibration


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Formulation via Hamiltons principle:

t2t1

[L + Wnoncons]dt = 0, L = T

T =

L0

12 (x )w

2 d x

V =L

0

T(x)(ds d x ) + 12 Kw2(L

From the preceding figure, we find

ds = [1 + w2x (x , t)]12 d x [1 + 12 w

2x (x


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Hence, substituting (2) into (1),we obtain

V = L0

1

2

T(x ) w2

x

(x , t) d x + 1

2

Kw2

Transverse displacement: w(x , t) (m)

Time derivative: w = wdt

(m/s)

Spatial derivtive: wx =

w

d x (m/m)String tension: T(x ) (N)

Mass per unit string length: (x ) (kg/m

String length: ds (m)

End spring : K (N/m)


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Since there are impotant differences betwe

tions of discrete model vs. continuum model

L below.

L = T V

T = {

L0

12 (x)w

2(x, t) d x}

V = {

L

0

12 T(x) w

2x (x, t) d x

+ { 12 Kw2(L , t)}


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Evaluation oft2

t1T dt

T = {L

0

12 (x )w

2

(x, t) d x }

=

L0

12 (x){w

2(x, t)} d x }

=L

0(x){w(x , t) w(x , t)}


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Evaluation oft2

t1T dt - contd

With T, lets perform the definite time inte

t2t1

T dt =

t2t1

L0

(x ){w(x , t) w(x , t


154/391

Evaluation oft2

t1T dt - contd

t2t1

T dt = t2t1L

0

(x ){w(x , t) w(x , t

=

L0

{

t2t1

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

=L

0{ [(x )w(x, t) w(x , t)]

t2t1

[

t2t1

(x)w(x , t) w(x, t) dt]


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Evaluation oft2

t1T dt - concluded

Since w(x , t1) and w(x, t1) are considered

where over the spatial domain 0 x L , w

w(x , t1) = w(x, t1) = 0

so that (t2t1

T dt) reduces to

t2t1

T dt =

t2t1

L0

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


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Evaluation oft2

t1V dt

Since the potential energy V does not involv

tive ofw(x , t), it is adequate to carry out sim

V =

L0

12 T(x ) w

2x (x, t) d x +

12 Kw

2

=L

0

T(x ) wx (x, t)wx (x , t) d x

+ Kw(L , t)w(L , t)


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Evaluation oft2

t1V dt - contd

The first term in the previous equation beco

L0

T(x) wx (x, t)wx (x, t) d x

= [T(x ) wx (x , t)w(x , t)]|w(L ,tw(0,t

L

0

T(x ) wx x (x , t)w(x , t) d


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Evaluation oft2

t1V dt - concluded

L0

T(x) wx

(x, t)wx

(x, t) d x

= [T(x ) wx (x , t)w(x , t)]x =L

[T(x ) wx (x , t)w(x , t)]x=0

L

0T(x ) wx x (x , t)w(x , t) d


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Virtual work due to nonconservative force

Wnoncons = L

0

f(x, t)w(x, t)d

Substituting (13), (12), (10) and (9) into (1

Hamiltons principle:t2t1

[(T V) + Wnoncons]dt =


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Hamiltons principle for string

t2

t1

L

0

{[(x )w(x, t)

+ T(x) wx x (x , t) + f(x, t)] w(x , t)}

[(T(x ) wx (x , t) + Kw(x , t))w(x ,

+ [T(x) wx (x , t)w(x, t)]x=0 dt =

In order for the above variational expression

the term inside the brace {.} must vanish, whgoverning equation of motion:

(x )w(x , t) = T(x ) wx x (x, t) + f(


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Similarly, the two terms in the bracket mus

[(T(x ) wx (x , t) + Kw(x , t))w(x , t)]x

[T(x ) wx (x , t)w(x , t)]x=0


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Boundary conditions for continuum string

Two types of boundary conditions:

Essential (or geometric) boundary condi

displacements are specified.

Natural (or force) boundary conditions:

boundary conditions come from force (ance considerations.


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Boundary conditions at x = L:

From (16) we find

T(L) wx (L , t) + Kw(L , t) = 0

or w(L , t)] = 0

Since the right end is left to move, (19) is not

Hence, the correct boundary conditon is g

which is a natural boundary condition.


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Boundary conditions at x = 0:

From (17) we find

T(0) wx (0, t) = 0

or w(0, t) = 0

Since the right end is fixed, (20) is not approp

the correct boundary conditon is given by

an essential or geometric boundary conditon.


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Boundary conditions - concluded

Observe that the variational equation, viz., Ha

ciple (14)provides the information to determinconditions, depending on the end configuratio

condition.

This is a distinct advantage of the variational f

continuum models as opposed to Newtons app


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Analogy: Torsional shaft and axial Rod

String Shaft Ax

Variable w(x , t) (m) (x, t) (r ad) u(xStiffness T(x) (N) G J(x)(N m2) E A

Hence, the equations of motion for bar and

derived by appropriate parameter changes.


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Strings in transverse vibration

Equation (15) with f(x , t) = 0 offers the con

value problem for strings together the bou

tions (18) and (21). To this end, we seek the s

homogeneous equation of(15) in the form o

w(x , t) = W(x ) F(t)

by employing the separation of variable us

lution of partial differential equations. Sub


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into (15), (18) and (21), we obtain

d

d x[T(x )

d W(x )

d x]F(t)

= (x )W(x)d2 F(t)

dt2, 0 < x 1, one root branches out to the infinite negative

axis while the other approaches the origin (see Figure 18.6). This invariance property, that is, the magnitude

of the complex root remains the same for all damping ratios, plays an important role both for controller

synthesis and computational algorithms.

14.3 VARIOUS DAMPING MODELS

Modeling of damping remains a challenge in structural dynamics. Over the years various damping models

have been proposed. Below lists a sample of parameterized damping models.

14.3.1 Mass-Proportional Damping Model

If the damping can be made proportional to the mass such as dynamic friction cases, the damping matrix can

be parameterized according to

D = M (14.23)as already discussed in the previous section. The root loci are a special case of (14.2) governed by

s2

+

s

+ 2

=0 (14.24)

which is plotted in Fig. 18.7 as method A with its parameter = 1. Note that the root loci form a straightvertical line when ( > /2).

14.3.2 Viscous Damping Model

One of the most widely used damping models is the viscous damping characterization. This has been adopted

for modeling of coated damping layers, of lubricants in rotating machines, and of a plethora of unknown

sources. Mathematically, the viscous damping model can be expressed as

D = 2T [p] TT, [p] = diag (1 p1 , 2 p2 ,..., N pN )TTMT = I, TTKT =

(14.25)

Therefore, the characteristic equation of a viscous damped system is given by

s2j + 2j pj sj + 2j = 0 (14.26)

Note that the two cases {p = 0, 1 = 2 = . . . = N = } and {p = 2, 1 = 2 = . . . = N = } havebeen studied in the preceding section.

148


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149 14.3 VARIOUS DAMPING MODELS

-2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-25

-20

-15

-10

-5

0

5

10

15

20

25

Root loci of various proportional damping models

Real part of normalized frequency by sampling rate h

Imaginarypartofnormalizedfrequencybysa

mplingrateh

model Amodel B

model Cmodel D1

model D2

model

D3

Fig. 14.7 Root Loci of Proportionally Damped Structural Dynamics System

The root loci of a viscous damped case, when combined with a mass-proportional damping, are obtained by

{p = 1, 1 = 2 = . . . = N = } s2 + ( + 2 ) s + 2 = 0 (14.27)

which is plotted in Fig. 14.7 labeled as model C with its parameters

{

=1.0,

=0.025

}. Note

that in this model the modal damping ratio is the same ( = const) for all the frequencies greater than{ > /2(1 ), < 1.0 }.

14.3.3 Intermediate-frequency damped case

In practice such as acoustically treated structural systems, both the low and high frequencies are associ-

ated with very low damping while intermediate frequencies are damped depending on damping treatment

employed. Such damping models may be parameterized according to

D = M + T [ (1 tanh h) p] TT (14.28)

which results in the following root loci equation:

s2 + ( + (1 tanh h) p) s + 2 = 0 (14.29)

The root loci of this model are plotted in Fig. 14.7 as method D1, D2, D3 with its parameters

p = 2, = 1.0, h = 1.0, = 0.025, = {0.025, 0.05, 0.1} (14.30)

149


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Lecture 14: MODELING FOR STRUCTURAL VIBRATIONS: FEM MODELS, DAMPING, SIMILARITY L1410

Note that as becomes large, the real part of the root loci approaches

s as (14.31)Hence, if = 0 the root loci will lie on the imaginary axis, indicating no damping for those frequencies.

14.4 USE OF SIMILARITY LAWS IN MODELING

Dimensional analysis has been widely used in experiment design and scale-model construction. The task

in dimensional analysis is to identify physical quantities that influence the physical phenomena on hand,

then deduce a independent set of dimensionless products. These dimensionless products are then utilized for

experiment design and scale-model construction. The underlying principle is Buckinghams theorem that

enables a systematic construction of linearly independent dimensionless products. It should be pointed out

that dimensional analysis is applicable both to statics and dynamics. In dynamics, a useful dimensional

analysis may be carried out, which can provide insight into dynamic behavior. It is referred to mechanical

similarity after Landau and Lifshits. The basic idea is as follow.

The starting point of mechanical similarity is that multiplication of kinetic and potential energy expression

or the Lagrangian by any constant does not affect the equations of motion. It is this observation that can

yield the dynamic behavior of a system without actually solving the equations of motion.

For example, we may pose the question: Does the amplitude of string vibration affect the frequency, assuming

its potential energy is a quadratic function of its amplitudes? To answer this question, let us consider the

simplest case, i.e., a single mass and a single spring case given by

T = 12m x2, U = 1

2kx2 L = T Y

m x + kx = 0, x(0) = x0, x(0) = x0

(14.32)

where m and k are the mass and spring constant, x is the vibration amplitude, and x0 and x0 are the initialamplitude and velocity of the mass.

Of course, we all know that the frequency of the above system is given by

=k/m (14.33)

which clearly shows that the frequency is independent of the amplitude x(t).

Let us scale the new amplitude x by

= x

x(14.34)

so that the potential energy U changes to

U(x ) = 12k(x )2 = 1

2 2x2 (14.35)

Now we introduce the time scale

= t

t(14.36)

where characterizes the ratio of periods of motion or time durations of the two systems.

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The Race between a Sliding Block and a Rolling Cylinder on an Inclined Slope Consider a cube sliding

on an inclined plane without friction, whose Lagrangian can be expressed as

Lcube = 12ms2 mgs sin (14.45)

where m is the mass of the cube, s is the distance along the slope measured from the bottom of the slope, g

is the gravity, is the inclined angle of the slope, and s is the speed of the cube along the slope.

If a cylinder of the same mass m is to roll along the slope without slip, its Lagrangian can be obtained as

Lcyl = 12m(s)2 + 12 ( 12mr2)2 mgs sin (14.46)

where r is the radius of the cylinder, and is the angular velocity of the cylinder during rolling without slip.

Since the angular velocity can be expressed in terms of s as

= s

r, s = s

t(14.47)

(14.46) can be simplified to

Lcyl = 123m

2 (s)2 mgs sin (14.48)

Transforming the above by the scaling

t/t= , s/s = (14.49)

yields

Lcyl = 123m

2

2

2(s)2 mgs sin = 3

2

2

2[ 1

2ms2 2

2

3 mgs sin ] (14.50)

Comparing the above with (14.45), we find for the resulting equation of motion from (14.50) to be the same

as obtained by (14.45) we must have

22

3= 1 = t

t= 3

2(14.51)

Now if the distance is the same, i.e., s = s, we have = 1. Therefore, the cylinder will arrive at the bottomof the inclined slope by a factor of

32

longer over the arrival time of the cube.

In other words, the cube will arrive at the bottom faster than the cylinder, and the arrival time ratio of the

two bodies will be proportional to the ratio of the square root of the kinetic energies. Physically, the reason

for the slower average speed of the cylinder is because for the case of cylinder, the same potential energy

change is translated into both the translational and rotational kinetic energy while the entire change energy

gets into the translational kinetic energy for the cube.

14.4.1 Mechanical Similarity in String, Rod and Shaft Vibrations

The Lagrangian of strings, rods and shafts can be expressed as

L = T U

T = 12

0

m(x) [w(x, t)

t]2 dx

U = 12

0

k(x) [w(x, t)

x]2 dx

(14.52)

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Observe that the similarity laws summarized in Table 14.1 can be derived if one solves for each of the three

vibration problems. The emphasis is to utilize the respective energy expressions rather than the solutions

of the governing differential equations. This is because it is often considerably easier to obtain the energy

expressions of a system than obtain the vibration characteristics or quasi-static deformations.

Finally, let us consider the role of the energy ratio

CE = ( 2 ) (14.58)

which plays a pivotal role in sizing up the experiment design as it offers the experiment energy requirements

relative to the actual system. In practice it is often the energy requirement that dictates the scaling than the

geometrical considerations. Notice that since we have only one constraint for four parameters, the remaining

three parameters can be used in meeting practical considerations in experiment design.

14.4.2 Mechanical Similarity in Euler-Bernoulli Beam

The Lagrangian of a continuous Euler-Bernoulli beam can be expressed as

L

=T

(Um

+Ug)

T = 12

0

A(x) [w(x, t)

t]2 dx

Um = 12

0

E I(x) [2w(x, t)

x2]2dx

Ug = 12

0

P(x) [w(x, t)

x]2 dx

(14.59)

where w(x, t) is the transverse displacement of the neutral axis of the beam, m(x) is the mass per unit beam

length, E I(x) denotes the bending rigidity of the beam, and P(x) denotes the pre-stressed axial force.

Introducing the scaling transformations given by (14.53) as modified to the case of beam, the Lagrangian of

the new system can be expressed as

L = (

2) T

3Um

f

Ug, =

EI

E I, f = P

P

= ( 2

) [T ( 2

4) [Um + (

f 2

) Ug]

L = (

2) L, if (

2

4) = 1 and ( f

2

) = 1

(14.60)

Therefore, the similarity law for a beam is expressed as

p

p=

E I/A

EI/A (

)2 provided

P (x)P(x)

= EI

E I

2

2, ifP(x) = 0 (14.61)

Notice that the appendage arms for a spinning satellite, helicopter rotor blades and turbine blades all experi-

ence axial tensions. Hence, the scaling of beam sizes must consider the axial force scaling as well.

One important application of the preceding scaling law is for the experiment design of rotating members for

micro-machine members as the rotating speed is very high and experiment design often necessitates a scaling

up rather than scaling down.

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.

15Solution ofVibration and

Transient Problems

151


310/391

Chapter 15: SOLUTION OF VIBRATION AND TRANSIENT PROBLEMS 152

15.1 MODAL APPROACH TO TRANSIENT ANALYSIS

Consider the following large-order finite element model equations of motion for linear structures:

Mu(t) + Du(t) + Ku(t) = f(t), D = (M + K), (,) are constant. (15.1)

where the size of the displacement vector, n, ranges from several thousands to several millions.

Now suppose we would like to obtain the displacement response, u(t), for expected applied force, f(t).

There are two approaches: direct time integration and modal superposition. We will differ direct time

integration techniques for transient response analysis to the latter part of the course, and concentrate on

modal superposition techniques. To this end, we first decompose the displacement vector, u(t), in terms of

its modal components by

u(t) = q(t) (15.2)

where is the mode shapes of the free-vibration modes, and q(t) is the generalized modal displacement.

The mode shape matrix has the property of simultaneously diagonalizing both the mass and stiffness

matrices. That is, it is obtained from the following eigenvalue problem:

K = M

= diag(21,...,2N)

(15.3)

where j is the j -th undamped frequency component of (15.1).

In structural dynamics, one often employs the following special form of mode shapes (eigenvector):

T K = , T M = I = diag(1, 1, ..., 1) (15.4)

Substituting (15.2) into (15.1) and pre-multiplying the resulting equation by T results in the following

uncoupled modal equation:

qi (t) + ( + 2i ) qi (t) +

2i qi (t) = pi (t), i = 1, 2, 3, ..., n.

pi (t) = (i, :)T f(t)

(15.5)

The above equation can be cast into a canonical form

xi (t) = Ai xi (t) + b pi (t), xi = [ qi (t) qi (t) ]T

Ai =

0 1

2i ( + 2i )

, b =

0

1

(15.6)

whose solution is given by

xi (t) = eAi t xi (0) +

t0

eAi (t ) b pi ( ) d (15.7)

It should be noted that theabove solution provides only for one of the n-vector generalized modal coordinates,

q(t) (n 1).

Carrying out for the entire n-modal vector, the physical displacement, u(t) (n 1), can be obtained from

(15.2) by

u(t) = (1 : n, 1 : n) q(t) (1 : n, 1), q = [ q1(t), q2(t), ..., qn (t) ]T (15.8)

152


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153 15.2 SOLUTION BY DIRECT INTEGRATION METHODS

While the solution method described in (15.2) - (15.8) appears to be straightfowrad, its practical implemen-

tation needs to overcome several computational challenges, which include:

(a) When the size of discrete finite element model increases, the task for obtaining a large number of modes

(m), if not all, m


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Ttotal = hn max

for n = 1 : nmax

u(n) = M1 (f(n) K u(n))

u(n + 12

) = u(n 12

) + h u(n)

u(n + 1) = u(n) + h u(n + 12

)

end

(15.12)

15.2.2 The Trapezoidal Rule for Undamped Case (D = 0)

This method is also referred to Newmarks implicit rule with itsfreeparameter chosen to be ( = 12

, = 1/4).

Among several ways of implementing the trapezoidal integration rule, we will employ a summed form or

half-interval rule given as follows.

u(n + 12

) = u(n) + 12

h u(n + 12

)

u(n +12 ) = u(n) +

12 h u(n +

12 )

u(n + 1) = 2u(n + 12

) u(n)

u(n + 1) = 2u(n + 12

) u(n)

(15.13)

In using the preceding formula, one multiply the first of (15.13) by M to yield

M u(n + 12

) = M u(n) + 12

h M u(n + 12

) (15.14)

The term M u(n + 12

) in the above equation is obtained from (15.1) as

Mu(n + 12

) = f(n + 12

) Du(n + 12

) K u(n + 12

) (15.15)

which, when substituted into (15.14), results in

M u(n + 12

) = M u(n) + 12

h {f(n + 12

) Du(n + 12

) K u(n + 12

)}

[M + 12

hD] u(n + 12

) = M u(n) + 12

h {f(n + 12

) K u(n + 12

)}

(15.16)

Now multiply the second of (15.13) by [M + 12

hD] to obtain

[M + 12

hD] u(n + 12

) = [M + 12

hD] u(n) + 12

h [M + 12

hD] u(n + 12

) (15.17)

third, substitute the second term in the righthand side of (15.17) by (15.16), one obtains

[M + 12

hD] u(n + 12

) = [M + 12

hD] u(n) + 12

h {M u(n) + 12

h {f(n + 12

) K u(n + 12

)}

[M + 12

hD + ( 12

h)2K] u(n + 12

) = M {u(n) + 12

h u(n)} + 12

h Du(n) + ( 12

h)2 f(n + 12

)

(15.18)

154


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155 15.4 ILLUSTRATIVE PROBLEMS

Implicit integration steps

Assemble: A = [M + 12

hD + ( 12

h)2K]

Factor: A = LU

for n = 0 : nmax

b(n) = M {u(n) + 12

h u(n)} + 12

h Du(n) + ( 12

h)2 f(n + 12

)

u(n + 12

) = A1 b(n), where A1 = U1 L1

u(n + 12

) = {u(n + 12

) u(n)}/( 12

h)

u(n + 1) = 2 u(n + 12

) u(n)

u(n + 1) = 2 u(n + 12

) u(n)

end

(15.19)

15.3 DISCRETE APPROXIMATION OF MODAL SOLUTION

The modal-form solution of the equations of motion for linear structures given by (15.7) and (15.8) involves

the convolution integral of the applied force. For general applied forces an exact evaluation of the convolution

integral can involve a considerable effort. To this end, an approximate solution is utilized in practice. To this

end, (15.7) is expressed in discrete form at time t = nh :

xi (nh ) = eAi nh xi (0) +

nh0

eAi (nh ) b pi ( ) d (15.20)

Likewise, at time t = nh + h, xi (nh + h) is given by

xi (nh + h) = eAi nh+h xi (0) +

nh+h

0

eAi (nh+h ) b pi ( ) d

= eAi h [eAi nh xi (0) +

nh0

eAi (nh ) b pi ( ) d]

+

nh+hnh

eAi (nh+h ) b pi ( ) d

(15.21)

The bracketed term in the above equation is xi (nh ) in view of (15.20) and the second term is approximated

as nh+hnh

eAi (nh+h ) b pi ( ) d [

nh+hnh

eAi (nh+h ) d] b pi (nh )

A1i (eAi h I) b pi (nh )

(15.22)

Substiutting this together the bracketed term by xi (nh ) into (15.21), xi (nh + h) is approximated as

xi (nh + h) = eAi h xi (nh ) + A

1i (e

Ai h I) b pi (nh ), xi = [ qi (nh + h), qi (nh + h) ]T (15.23)

Once xi (nh + h) is computed, the physical displacement u(nh + h) is obtained via the modal summation

expression (15.8).

155


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15.4 ILLUSTRATIVE PROBLEMS

Consider a beam with boundary constraints as shown in Figure 15.1. For illustrative purposes, two beams

will be considered: a fixed-simply supported beam and a beam with boundary constraints with the following

specific boundary conditions:

Fixed and simply supported ends:

w(0, t) =w(0, t)

x= 0, w(L , t) = 0 (15.24)

Beam with boundary springs:

w(0, t) = w(L , t) = 0, k1 =(1.e + 5) E I

L, k2 =

E I

L(15.25)

kw1 kw2

EI, m(x)

x

z

w

L

Beam with boundary constraints

k 1k 2

Fig. 15.1 Beam with boundary constraints

It should be noted that the end condition, (w(0, t) = w(L , t) = 0), is equivalent to (kw1 , kw2 ).

However, in computer implementation it is impractical to use (kw1 , kw2 ) due to limited floating

point precision.

The applied force chosen are

Step load: f(L/2, t) = 1.00, 0 t

Sinusoidal load: f(L/2, t) = sin(2ff t)(15.26)

where the forcing frequency is set to ff = 1.5(1/2/ ), with 1 being the fundamental frequency of the

model problems.

Figures 15.2-3 illustrate time responses of a beam with boundary springs subject to unit mid-span step load.

The responses by solid red lines are those obtained by using the central difference method, the ones with +

are by the trapezoidal rule, and the blue lines by the canonical formula(15.23).

The step increments used for the three methods are

h =

1.8928E 7, for the central difference method

1.5861E 5, for the canonical formula

1.6.3445E 5, for the trapezoidal rule

(15.27)

It should be noted that the stepsie for the central difference method is dictated by the computational stability

whereas taht of the canonical formula and teh trapezoidal rule by accurcy considerations. We hope to revisit

this issue later in the course.

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157 15.4 ILLUSTRATIVE PROBLEMS

0 1 2 3 4 5 6 7

x 10-3

0

1

2

3

4

5

6

7x 10

-6

time (sec.)

Verticaldisplacementatthebeamc

enter

Beam under midspan step load with 4 beam elements

Fig. 15.2 Beam Midspan Vertical Time Response

0 1 2 3 4 5 6 7

x 10-3

-1

-0.5

0

0.5

1

1.5

2

2.5x 10

-5

time (sec.)

Rotationaldisplacementatthebeamc

enter

Beam under midspan step load with 4 beam elements

Fig. 15.3 Beam Midspan Rotational Time Response

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.

16Methods forVibration Analysis

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Chapter 16: METHODS FOR VIBRATION ANALYSIS 162

16.1 PROBLEM CLASSIFICATION

According to S. H. Krandall (1956), engineering problems can be classified into

three categories:

equilibrium problems

eigenvalue problems

propagation problems

Equilibrium problems are characterized by the structural or mechanical deforma-

tions due to quasi-static or repetitive loadings. In other words, in structural and

mechanical systems the solution of equilibrium problems is a stress or deforma-

tion state under a given load. The modeling and analysis tasks are thus to obtainthe system stiffness or flexibility so that the stresses or displacements computed

accurately match the observed ones.

Eigenvalue problems can be considered as extentions of equilibrium problems

in that their solutions are dictated by the same equilibrium states. There is an

additional distinct feature in eigenvalue problems: their solutions are characterized

by a unique set of system configurations such as resonance and buckling.

Propagation problems are to predict the subsequent stresses or deformation states ofa system under the time-varying loading and deformation states. It is called initial-

value problems in mathematics or disturbance transmissions in wave propagation.

Modal testing is perhaps the most widely accepted words for activities involving

the characterization of mechanical and structural vibrations through testing and

measurements. It is primarily concerned with the determination of mode shapes

(eigenvevtors) and modes (eigenvalues), and to the extent possible the damping

ratios of a vibrating system. Therefore, modal testingcan be viewed as experimental

Dynsmics of Aerospace Structures Lecture

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