Dynamics Generalized coordinates

7/29/2019 Dynamics Generalized coordinates

1/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Generalized Single Degree of FreedomSystems

Giacomo Boffi

Dipartimento di Ingegneria Strutturale, Politecnico di Milano

April 13, 2011

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes

Refinement ofRayleighs

Estimates

Outline

Introductory Remarks

Assemblage of Rigid Bodies

Continuous Systems

Vibration Analysis by Rayleighs Method

Selection of Mode Shapes

Refinement of Rayleighs Estimates


2/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Introductory Remarks

Until now our SDOFs were described as composed by asingle mass connected to a fixed reference by means of aspring and a damper.

While the mass-spring is a useful representation, manydifferent, more complex systems can be studied as SDOFsystems, either exactly or under some simplifyingassumption.

1. SDOF rigid body assemblages, where flexibility isconcentrated in a number of springs and dampers, canbe studied, e.g., using the Principle of Virtual

Displacements and the DAlembert Principle.2. simple structural systems can be studied, in an

approximate manner, assuming a fixed pattern ofdisplacements, whose amplitude (the single degree offreedom) varies with time.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Further Remarks on Rigid Assemblages

Today we restrict our consideration to plane, 2-D systems.In rigid body assemblages the limitation to a single shape ofdisplacement is a consequence of the configuration of the

system, i.e., the disposition of supports and internal hinges.When the equation of motion is written in terms of a singleparameter and its time derivatives, the terms that figure ascoefficients in the equation of motion can be regarded as thegeneralised properties of the assemblage: generalised mass,damping and stiffness on left hand, generalised loading onright hand.

mx + cx + kx = p(t)


3/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Further Remarks on Continuous Systems

Continuous systems have an infinite variety of deformation

patterns.By restricting the deformation to a single shape of varyingamplitude, we introduce an infinity of internal contstraintsthat limit the infinite variety of deformation patterns, butunder this assumption the system configuration ismathematically described by a single parameter, so that

our model can be analysed in exactly the same way as astrict SDOF system,

we can compute the generalised mass, damping,stiffness properties of the SDOF system.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Final Remarks on Generalised SDOF Systems

From the previous comments, it should be apparent that

everything we have seen regarding the behaviour and theintegration of the equation of motion of proper SDOFsystems applies to rigid body assemblages and to SDOFmodels of flexible systems, provided that we have the meansfor determining the generalised properties of the dynamicalsystems under investigation.


4/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Assemblages of Rigid Bodies

planar, or bidimensional, rigid bodies, constrained tomove in a plane,

the flexibility is concentrated in discrete elements,

springs and dampers, rigid bodies are connected to a fixed reference and to

each other by means of springs, dampers and smooth,bilateral constraints (read hinges, double pendulumsand rollers),

inertial forces are distributed forces, acting on eachmaterial point of each rigid body, their resultant can bedescribed by

a force applied to the centre of mass of the body,proportional to acceleration vector and total massM =

dm

a couple, proportional to angular acceleration and themoment of inertia J of the rigid body,J =

(x2 +y2)dm.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Rigid Bar

x

G

L

Unit mass m= constant,

Length L,

Centre of Mass xG = L/2,

Total Mass m= mL,

Moment of Inertia J = mL2

12

= mL3

12


5/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Rigid Rectangle

G

y

a

b

Unit mass = constant,

Sides a, b

Centre of Mass xG = a/2, yG = b/2

Total Mass m= ab,

Moment of Inertia J = ma2 + b2

12=

a3b + ab3

12

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Rigid Triangle

For a right triangle.

y

G

a

b


Sides a, b

Centre of Mass xG = a/3, yG = b/3

Total Mass m= ab/2,

Moment of Inertia J = ma2

+ b2

18= a

3

b + ab3

36


6/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Rigid Oval

When a = b = D = 2R the oval is a circle.

x

y

a

b


Axes a, b

Centre of Mass xG = yG = 0

Total Mass m= ab

4,

Moment of Inertia J = mL2

12= m

L3

12

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

trabacolo1

c k c k 2 211 N

m , J2 2

p(x,t) = P x/a f(t)

a 2 a a a a a

The mass of the left bar is m1 = m4a and its moment of

inertia is J1 = m1(4a)2

12= 4a2m1/3.

The maximum value of the external load isPmax = P 4a/a = 4P and the resultant of triangular load is

R = 4P 4a/2 = 8Pa


7/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Forces and Virtual Displacements

c1Z4

m1Z2

3k1Z4

c2Z2m2Z

3kZ

3

NZ(t)

J2Z3a

8Pa f(t)J1Z4a

Z4

Z2

3Z4

Z 2Z3

Z3

u

2 = Z/(3a)1 = Z/(4a)

u= 7a4a cos 13a cos 2, u= 4a sin 11+3a sin 22

1 = Z/(4a), 2 = Z/(3a)

sin 1 Z/(4a), sin 2 Z/(3a)

u=

14a +

13a

Z Z = 7

12aZ Z

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Principle of Virtual Displacements

c1Z4

m1Z2

3k1Z4

c2Z2m2Z

3kZ

3

NZ(t)

J2Z3a

8Pa f(t)J1Z4a

Z4

Z2

3Z4

Z 2Z3

Z3

u

2 = Z/(3a)1 = Z/(4a)

The virtual work of the InertialDampingElasticExternal

forces:

WI = m1Z

2

Z

2 J1

Z

4a

Z

4a m2

2Z

3

2Z

3 J2

Z

3a

Z

3a

=

m1

4+ 4

m2

9+

J1

16a2+

J2

9a2

Z Z

WD = c1Z

4

Z

4c2Z Z = (c2 + c1/16) Z Z

WS = k13Z

4

3Z

4 k2

Z

3

Z

3=

9k116

+k2

9

Z Z

WExt = 8Pa f(t)2Z

3+ N

7

12aZ Z


8/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates


For a rigid body in condition of equilibrium the total virtualwork must be equal to zero

WI + WD + WS + WExt = 0

Substituting our expressions of the virtual work contributionsand simplifying Z, the equation of equilibrium is

m1

4+ 4

m2

9+

J1

16a2+

J2

9a2

Z+

+ (c2 + c1/16) Z +

9k116

+ k29

Z =

8Pa f(t)2

3+ N

7

12aZ

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates


Collecting Z and its time derivatives give us

mZ + cZ + kZ = pf(t)

introducing the so called generalised properties, in ourexample it is

m = 14

m1 + 49

9m2 + 116a2

J1 + 19a2

J2,

c =1

16c1 + c2,

k =9

16k1 +

1

9k2

7

12aN,

p =16

3Pa.

It is worth writing downthe expression of k: k =

9k116

+k2

9

7

12aN

Geometrical stiffness


9/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Lets start with an example...

Consider a cantilever, with varying properties m and EJ,subjected to a load that is function of both time t andposition x,

p = p(x, t).

The transverse displacements v will be function of time andposition,

v = v(x, t)

H

x m= m(x)

N

EJ = EJ(x)v(x, t)

p(x, t)

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

... and an hypothesis

To study the previous problem, we introduce an approximatemodel by the following hypothesis,

v(x, t) = (x)Z(t),

that is, the hypothesis of separation of variablesNote that (x), the shape function, is adimensional, whileZ(t) is dimensionally a generalised displacement, usuallychosen to characterise the structural behaviour.In our example we can use the displacement of the tip of thechimney, thus implying that (H) = 1 because

Z(t) = v(H, t) and

v(H, t) = (H)Z(t)


10/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates


For a flexible system, the PoVD states that, at equilibrium,

WE = WI.

The virtual work of external forces can be easily computed,the virtual work of internal forces is usually approximated bythe virtual work done by bending moments, that is

WI

M

where is the curvature and the virtual increment ofcurvature.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

WE

The external forces are p(x, t), N and the forces of inertiafI; we have, by separation of variables, that v = (x)Zand we can write

Wp =H

0 p(x, t)v dx =H

0 p(x, t)(x) dx

Z = p

(t) Z

WInertia =

H0

m(x)vv dx =

H0

m(x)(x)Z(x) dx Z

=

H0

m(x)2(x) dx

Z(t) Z = mZ Z.

The virtual work done by the axial force deserves a separatetreatment...


11/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

WN

The virtual work of N is WN = Nu where u is thevariation of the vertical displacement of the top of thechimney.We start computing the vertical displacement of the top ofthe chimney in terms of the rotation of the axis line, (x)Z(t),

u(t) = H

H0

cos dx =

H0

(1 cos ) dx,

substituting the well known approximation cos 1 2

2in

the above equation we have

u(t) =

H

0

2

2dx =

H

0

2(x)Z2(t)

2dx

hence

u=

H0

2(x)Z(t)Z dx =

H0

2(x) dx ZZ

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

WInt

Approximating the internal work with the work done bybending moments, for an infinitesimal slice of beam we write

dWInt =1

2Mv(x, t) dx =

1

2M(x)Z(t) dx

with M = EJ(x)v(x)

(dWInt) = EJ(x)2(x)Z(t)Z dx

integrating

WInt = H0

EJ(x)2(x) dx ZZ = k Z Z


12/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Remarks

the shape function must respect the geometricalboundary conditions of the problem, i.e., both

1 = x2 and 2 = 1 cos x2H

are accettable shape functions for our example, as1(0) = 2(0) = 0 and

1(0) =

2(0) = 0

better results are obtained when the second derivativeof the shape function at least resembles the typicaldistribution of bending moments in our problem, so that

between

1 = constant and 2 =2

4H2cos

x

2H

the second choice is preferable.


13/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Example

Using (x) = 1 cos x2H , with m= constant and

EJ = constant, with a load characteristic of seismic

excitation, p(t) = mvg(t),

m = m

H0

(1 cosx

2H)2 dx = m(

3

2

4

)H

k = EJ4

16H4

H0

cos2x

2Hdx =

4

32

EJ

H3

k

G

= N2

4H2H

0

sin2x

2Hdx =

2

8HN

pg = mvg(t)

H0

1 cosx

2Hdx =

1

2

mH vg(t)

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Vibration Analysis

The process of estimating the vibration characteristicsof a complex system is known as vibration analysis.

We can use our previous results for flexible systems,

based on the SDOF model, to give an estimate of thenatural frequency 2 = k/m

A different approach, proposed by Lord Rayleigh, startsfrom different premises to give the same results but theRayleighs Quotient method is important because itoffers a better understanding of the vibrationalbehaviour, eventually leading to successive refinements

of the first estimate of 2

.


14/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Rayleighs Quotient Method

Our focus will be on the free vibration of a flexible,undamped system.

inspired by the free vibrations of a proper SDOF we

write

Z(t) = Z0 sin t and v(x, t) = Z0(x) sin t,

the displacement and the velocity are in quadrature:when v is at its maximum v = 0 (hence V= Vmax,T= 0) and when v = 0 v is at its maximum (henceV= 0, T= Tmax,

disregarding damping, the energy of the system isconstant during free vibrations,

Vmax + 0 = 0 + Tmax

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Rayleigh s Quotient Method

Now we write the expressions for Vmax and Tmax,

Vmax =1

2Z20

S

EJ(x)2(x) dx,

Tmax = 12

2Z20S

m(x)2(x) dx,

equating the two expressions and solving for 2 we have

2 =

S EJ(x)

2(x) dxS m(x)

2(x) dx.

Recognizing the expressions we found for k

and m

wecould question the utility of Rayleighs Quotient...


15/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Rayleighs Quotient Method

in Rayleighs method we know the specific timedependency of the inertial forces

fI = m(x)v = m(x)2Z0(x) sin t

fI has the same shape we use for displacements. if were the real shape assumed by the structure in

free vibrations, the displacements v due to a loadingfI =

2m(x)(x)Z0 should be proportional to (x)through a constant factor, with equilibrium respected inevery point of the structure during free vibrations.

starting from a shape function 0(x), a new shape

function 1 can be determined normalizing thedisplacements due to the inertial forces associated with

0(x), fI = m(x)0(x), we are going to demonstrate that the new shape

function is a better approximation of the true modeshape

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Selection of mode shapes

Given different shape functions i and considering the trueshape of free vibration , in the former cases equilibrium isnot respected by the structure itself.To keep inertia induced deformation proportional to i we

must consider the presence of additional elastic constraints.This leads to the following considerations

the frequency of vibration of a structure with additionalconstraints is higher than the true natural frequency,

the criterium to discriminate between different shapefunctions is: better shape functions give lower estimatesof the natural frequency, the true natural frequency

being a lower bound of all estimates.


16/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Selection of mode shapes 2

In general the selection of trial shapes goes through two

steps,

1. the analyst considers the flexibilities of different parts ofthe structure and the presence of symmetries to devisean approximate shape,

2. the structure is loaded with constant loads directed asthe assumed displacements, the displacements arecomputed and used as the shape function,

of course a little practice helps a lot in the the choice of aproper pattern of loading...

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Selection of mode shapes 3

p = m(x)

P = M

p = m(x)

p=

m(x)

p = m(x)

(a)

(b) (c)

(d)


17/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Refinement R00

Choose a trial function (0)(x) and write

v(0) = (0)(x)Z(0) sin t

Vmax =1

2Z(0)2

EJ(0)2 dx

Tmax =1

22Z(0)2

m(0)2 dx

our first estimate R00 of 2 is

2 =

EJ(0)2 dxm(0)2 dx

.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage of

Rigid Bodies

Continuous

Systems

Vibration Analysis

by RayleighsMethod

Selection of Mode

Shapes


Estimates

Refinement R01

We try to give a better estimate of Vmax computing theexternal work done by the inertial forces,

p(0) = 2m(x)v(0) = Z(0)2(0)(x)

the deflections due to p(0) are

v(1) = 2v(1)

2= 2(1)Z

(1)

2= 2(1)Z(1),

where we write Z(1) because we need to keep the unknown2 in evidence. The maximum strain energy is

Vmax =1

2

p(0)v(1) dx =

1

24Z(0)Z(1)

m(x)(0)(1) dx

Equating to our previus estimate of Tmax we find the R01estimate

2 =Z(0)

Z(1)

m(x)(0)(0) dxm(x)(0)(1) dx


18/18

Generalized

SDOFs

Giacomo Boffi

Introductory

Remarks

Assemblage of

Rigid Bodies

ContinuousSystems

Vibration Analysis

by Rayleighs

Method

Selection of Mode

Shapes

Refinement of

Rayleighs

Estimates

Refinement R11

With little additional effort it is possible to compute Tmaxfrom v(1):

Tmax =

1

2 2

m(x)v(1)2

dx =

1

2 6

Z(1)2

m(x)(1)2

dx

equating to our last approximation for Vmax we have the R11approximation to the frequency of vibration,

2 =Z(0)

Z(1)

m(x)(0)(1) dxm(x)(1)(1) dx

.

Of course the procedure can be extended to compute betterand better estimates of 2 but usually the refinements arenot extended beyond R11, because it would be contradictorywith the quick estimate nature of the Rayleighs Quotientmethod and also because R11 estimates are usually verygood ones.

Refinement Example

m

1.5m

2m

k

2k

3k

(0)

1

11/15

6/15

1

1

1

1

1.5

2

(1)p(0)

2m

T=1

22 4.5 mZ0

V=

1

2 1 3kZ0

2 =3

9/2

k

m=

2

3

k

m

v(1) =15

4

m

k2(1)

Z(1) =15

4

m

k

V(1) =1

2m

15

4

m

k4 (1 + 33/30 + 4/5)

=1

2m

15

4

m

k4

87

30

2 =92m

m 878mk

=12

29

k

m= 0.4138

k

m

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Dynamics Generalized coordinates