dynamic of structures

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks

Assemblage ofRigid Bodies

ContinuousSystems

Vibration Analysisby RayleighsMethod

Selection of ModeShapes

Refinement ofRayleighsEstimates

Generalized Single Degree of FreedomSystems

Giacomo Boffi

Dipartimento di Ingegneria Civile e Ambientale, Politecnico di Milano

April 14, 2015

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Outline

Introductory Remarks

Assemblage of Rigid Bodies

Continuous Systems

Vibration Analysis by Rayleighs Method

Selection of Mode Shapes

Refinement of Rayleighs Estimates

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




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 VirtualDisplacements and the DAlembert Principle.

2. simple structural systems can be studied, in anapproximate manner, assuming a fixed pattern ofdisplacements, whose amplitude (the single degree offreedom) varies with time.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




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 thesystem, 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 asthe generalised properties of the assemblage: generalisedmass, damping and stiffness on left hand, generalisedloading on right hand.

m?x + c?x + k?x = p?(t)

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Further Remarks on Continuous Systems

Continuous systems have an infinite variety of deformationpatterns.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

I our model can be analysed in exactly the same way asa strict SDOF system,

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

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Final Remarks on Generalised SDOF Systems

From the previous comments, it should be apparent thateverything 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 themeans for determining the generalised properties of thedynamical systems under investigation.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Assemblages of Rigid Bodies

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

I the flexibility is concentrated in discrete elements,springs and dampers,

I rigid bodies are connected to a fixed reference and toeach other by means of springs, dampers and smooth,bilateral constraints (read hinges, double pendulumsand rollers),

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

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

dm

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

(x2 + y2)dm.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




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= m

L3

12

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Rigid Rectangle

Gy

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


ContinuousSystems




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,


18=

a3b + ab3

36

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Rigid Oval

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

x

y

ab


Axes a, b

Centre of Mass xG = yG = 0

Total Mass m = piab

4,


16

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




trabacolo1

c k c k2 211N

m , J2 2p(x,t) = P x/a f(t)

a 2 a a a a a

The mass of the left bar is m1 = m 4a and its moment ofinertia 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 isR = 4P 4a/2 = 8Pa

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Forces and Virtual Displacements

c1Z4

m1Z2

3k1Z4

c2Z2m2Z

3kZ3

NZ(t)

J2Z3a

8Pa f (t)J1Z4a

Z4

Z2

3 Z4

Z 2 Z3

Z3

u

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

u = 7a4a cos 13a cos 2, u = 4a sin 11+3a sin 221 = Z/(4a), 2 = Z/(3a)

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

(1

4a +1

3a

)Z Z = 712aZ Z

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Principle of Virtual Displacements

c1Z4

m1Z2

3k1Z4

c2Z2m2Z

3kZ3

NZ(t)

J2Z3a

8Pa f (t)J1Z4a

Z4

Z2

3 Z4

Z 2 Z3

Z3

u

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

The virtual work of the Inertial forces:

WI = m1 Z2

Z

2 J1 Z

4a

Z

4am2 2Z

3

2Z

3 J2 Z

3a

Z

3a

= (m14

+ 4m29

+J1

16a2+J2

9a2

)Z Z

WD = c1 Z4

Z

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

WS = k1 3Z4

3Z

4 k2Z

3

Z

3=

(9k116

+k29

)Z Z

WExt = 8Pa f (t)2Z

3+ N

7

12aZ Z

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





c1Z4

m1Z2

3k1Z4

c2Z2m2Z

3kZ3

NZ(t)

J2Z3a

8Pa f (t)J1Z4a

Z4

Z2

3 Z4

Z 2 Z3

Z3

u

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

The virtual work of the Damping forces:

WI = m1 Z2

Z

2 J1 Z

4a

Z

4am2 2Z

3

2Z

3 J2 Z

3a

Z

3a

= (m14

+ 4m29

+J1

16a2+J2

9a2

)Z Z

WD = c1 Z4

Z

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

WS = k1 3Z4

3Z

4 k2Z

3

Z

3=

(9k116

+k29

)Z Z

WExt = 8Pa f (t)2Z

3+ N

7

12aZ Z

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





c1Z4

m1Z2

3k1Z4

c2Z2m2Z

3kZ3

NZ(t)

J2Z3a

8Pa f (t)J1Z4a

Z4

Z2

3 Z4

Z 2 Z3

Z3

u

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

The virtual work of the Elastic forces:

WI = m1 Z2

Z

2 J1 Z

4a

Z

4am2 2Z

3

2Z

3 J2 Z

3a

Z

3a

= (m14

+ 4m29

+J1

16a2+J2

9a2

)Z Z

WD = c1 Z4

Z

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

WS = k1 3Z4

3Z

4 k2Z

3

Z

3=

(9k116

+k29

)Z Z

WExt = 8Pa f (t)2Z

3+ N

7

12aZ Z

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





c1Z4

m1Z2

3k1Z4

c2Z2m2Z

3kZ3

NZ(t)

J2Z3a

8Pa f (t)J1Z4a

Z4

Z2

3 Z4

Z 2 Z3

Z3

u

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

The virtual work of the External forces:

WI = m1 Z2

Z

2 J1 Z

4a

Z

4am2 2Z

3

2Z

3 J2 Z

3a

Z

3a

= (m14

+ 4m29

+J1

16a2+J2

9a2

)Z Z

WD = c1 Z4

Z

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

WS = k1 3Z4

3Z

4 k2Z

3

Z

3=

(9k116

+k29

)Z Z

WExt = 8Pa f (t)2Z

3+ N

7

12aZ Z

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





c1Z4

m1Z2

3k1Z4

c2Z2m2Z

3kZ3

NZ(t)

J2Z3a

8Pa f (t)J1Z4a

Z4

Z2

3 Z4

Z 2 Z3

Z3

u

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

WI = m1 Z2

Z

2 J1 Z

4a

Z

4am2 2Z

3

2Z

3 J2 Z

3a

Z

3a

= (m14

+ 4m29

+J1

16a2+J2

9a2

)Z Z

WD = c1 Z4

Z

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

WS = k1 3Z4

3Z

4 k2Z

3

Z

3=

(9k116

+k29

)Z Z

WExt = 8Pa f (t)2Z

3+ N

7

12aZ Z

_

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





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 workcontributions and simplifying Z, the equation ofequilibrium is(

m14

+ 4m29

+J1

16a2+J2

9a2

)Z+

+ (c2 + c1/16) Z +

(9k116

+k29

)Z =

8Pa f (t)2

3+ N

7

12aZ

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





Collecting Z and its time derivatives give us

m?Z + c?Z + k?Z = p?f (t)

introducing the so called generalised properties, in ourexample it is

m? =1

4m1 +

4

99m2 +

1

16a2J1 +

1

9a2J2,

c? =1

16c1 + c2,

k? =9

16k1 +

1

9k2 7

12aN,

p? =16

3Pa.

It is worth writing downthe expression of k?: k? =

9k116

+k29 7

12aN

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





Collecting Z and its time derivatives give us

m?Z + c?Z + k?Z = p?f (t)

introducing the so called generalised properties, in ourexample it is

m? =1

4m1 +

4

99m2 +

1

16a2J1 +

1

9a2J2,

c? =1

16c1 + c2,

k? =9

16k1 +

1

9k2 7

12aN,

p? =16

3Pa.

It is worth writing downthe expression of k?: k? =

9k116

+k29 7

12aN

Geometrical stiffness

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




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


ContinuousSystems




... and an hypothesis

To study the previous problem, we introduce anapproximate model by the following hypothesis,

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

that is, the hypothesis of separation of variables

Note 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 ofthe chimney, thus implying that (H) = 1 because

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

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

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems






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 ofthe chimney, thus implying that (H) = 1 because

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

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

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems






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 ofthe chimney, thus implying that (H) = 1 because

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

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

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





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


ContinuousSystems




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 =

H0

p(x, t)v dx =

[ H0

p(x, t)(x) dx

]Z = p?(t) Z

WInertia =

H0

m(x)v v dx = H

0

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

=

[ H0

m(x)2(x) dx]Z(t) Z = m?Z Z.

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

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




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 H

0

cos dx =

H0

(1 cos) dx,

substituting the well known approximation cos 1 22in the above equation we have

u(t) =

H0

2

2dx =

H0

2(x)Z2(t)2

dx

hence

u =

H0

2(x)Z(t)Z dx = H

0

2(x) dx ZZ

and

WN =

[ H0

2(x) dx N]Z Z = k?G Z Z

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




WInt

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

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

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Remarks

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

1 = x2 and 2 = 1 cos pix

2H

are accettable shape functions for our example, as1(0) = 2(0) = 0 and 1(0) =

2(0) = 0

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

1 = constant and 2 =pi2

4H2cos

pix

2H

the second choice is preferable.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Remarks

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1 0

0.2

0.4

0.6

0.8

1

vi"/

Z(t)

v/Z(

t)

x/H

f1=1-cos(pi*x/2)f2=x2



2H


2(0) = 0



4H2cos

pix

2H


GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Remarks

0

0.5

1

1.5

2

2.5

0 0.2 0.4 0.6 0.8 1 0

0.2

0.4

0.6

0.8

1

vi"/

Z(t)

v/Z(

t)

x/H

f1=1-cos(pi*x/2)f2=x2f1"

f2"



2H


2(0) = 0



4H2cos

pix

2H


GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Remarks



2H


2(0) = 0



4H2cos

pix

2H


GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Example

Using (x) = 1 cos pix2H , with m = constant andEJ = constant, with a load characteristic of seismicexcitation, p(t) = mvg(t),

m? = m

H0

(1 cos pix2H

)2 dx = m(3

2 4pi

)H

k? = EJpi4

16H4

H0

cos2pix

2Hdx =

pi4

32

EJ

H3

k?G = Npi2

4H2

H0

sin2pix

2Hdx =

pi2

8HN

p?g = mvg(t) H

0

1 cos pix2H

dx = (

1 2pi

)mH vg(t)

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Vibration Analysis

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

I We can use our previous results for flexible systems,based on the SDOF model, to give an estimate of thenatural frequency 2 = k?/m?

I A different approach, proposed by Lord Rayleigh, startsfrom different premises to give the same results butthe Rayleighs Quotient method is important becauseit offers a better understanding of the vibrationalbehaviour, eventually leading to successive refinementsof the first estimate of 2.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Rayleighs Quotient Method

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

I inspired by the free vibrations of a proper SDOF wewrite

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

I 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,

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

Vmax + 0 = 0 + Tmax

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





Our focus will be on the free vibration of a flexible,undamped system.I inspired by the free vibrations of a proper SDOF we

write

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

I 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,

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

Vmax + 0 = 0 + Tmax

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Rayleigh s Quotient Method

Now we write the expressions for Vmax and Tmax,

Vmax =1

2Z20

S

EJ(x)2(x) dx,

Tmax =1

22Z20

S

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...

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





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

fI = m(x)v = m(x)2Z0(x) sintfI has the same shape we use for displacements.

I if were the real shape assumed by the structure infree vibrations, the displacements v due to a loadingfI =

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

I starting from a shape function 0(x), a new shapefunction 1 can be determined normalizing thedisplacements due to the inertial forces associated with0(x), fI = m(x)0(x),

I we are going to demonstrate that the new shapefunction is a better approximation of the true modeshape

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




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

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

I the criterium to discriminate between different shapefunctions is: better shape functions give lowerestimates of the natural frequency, the true naturalfrequency being a lower bound of all estimates.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





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

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


GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





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 wemust consider the presence of additional elastic constraints.This leads to the following considerationsI the frequency of vibration of a structure with additional

constraints is higher than the true natural frequency,


GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems





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 wemust consider the presence of additional elastic constraints.This leads to the following considerationsI the frequency of vibration of a structure with additional

constraints is higher than the true natural frequency,I the criterium to discriminate between different shape

functions is: better shape functions give lowerestimates of the natural frequency, the true naturalfrequency being a lower bound of all estimates.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Selection of mode shapes 2

In general the selection of trial shapes goes through twosteps,

1. the analyst considers the flexibilities of different partsof the structure and the presence of symmetries todevise an 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


ContinuousSystems




Selection of mode shapes 3

p = m(x)

P = M

p = m(x)

p=m

(x)

p = m(x)

(a)

(b) (c)

(d)

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Refinement R00

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

v (0) = (0)(x)Z(0) sint

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


ContinuousSystems




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

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Refinement R11

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

Tmax =1

22m(x)v (1)2 dx =

1

26Z(1)2

m(x)(1)2 dx

equating to our last approximation for Vmax we have theR11 approximation 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 computebetter and better estimates of 2 but usually therefinements are not extended beyond R11, because it wouldbe contradictory with the quick estimate nature of theRayleighs Quotient method and also because R11estimates are usually very good ones.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Refinement R11


Tmax =1

22m(x)v (1)2 dx =

1

26Z(1)2

m(x)(1)2 dx


2 =Z(0)

Z(1)

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

.

Of course the procedure can be extended to computebetter and better estimates of 2 but usually therefinements are not extended beyond R11, because it wouldbe contradictory with the quick estimate nature of theRayleighs Quotient method

and also because R11estimates are usually very good ones.

GeneralizedSDOFs

Giacomo Boffi

IntroductoryRemarks


ContinuousSystems




Refinement R11


Tmax =1

22m(x)v (1)2 dx =

1

26Z(1)2

m(x)(1)2 dx


2 =Z(0)

Z(1)

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

.

Of course the procedure can be extended to computebetter and better estimates of 2 but usually therefinements are not extended beyond R11, because it wouldbe contradictory with the quick estimate nature of theRayleighs Quotient method and also because R11estimates are usually very good 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 4.5mZ20

V =1

2 1 3k Z20

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

Introductory RemarksAssemblage of Rigid BodiesContinuous SystemsVibration Analysis by Rayleigh's MethodSelection of Mode ShapesRefinement of Rayleigh's Estimates

dynamic of structures

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