Dynamics and Control of Flexible Structures

Dynamics and Control of Flexible Structures Mohammad Tawfik Academy of Knowledge http://academyofknowledge.org

DOI: 10.13140/RG.2.2.29036.26242

Draft printed on: 20. September 2018

Videos explaining these lecture notes are available on http://AcademyOfKnowledge.org under “Dynamics of Structures”)

Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

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Table of Contents 1 Single Degree of Freedom Systems................................................................................................4 1.1 Equations of Motion................................................................................................................4 1.2 Vibration Due to Initial Excitation...........................................................................................7 1.2.1 Illustration of an over-damped system.............................................................................8 1.2.2 Illustration of a critically-damped system......................................................................10 1.2.3 Illustration of an under-damped system.........................................................................11 1.3 Harmonic Excitation..............................................................................................................12 1.4 Periodic Excitation.................................................................................................................16 1.5 General Excitation.................................................................................................................16 1.6 Approximate Evaluation of Time Responses.........................................................................16 2 Multiple Degree of Freedom Systems...........................................................................................17 2.1 2-DOF Equations of Motion..................................................................................................17 2.2 Solution for Motion without Force........................................................................................19 2.3 Forced Response to Harmonic Excitation.............................................................................20 2.4 Vibration Absorbers – First Passive Attenuation Technique..................................................21 3 Variational Methods and Hamilton’s Principle..............................................................................23 3.1 Simplified Concepts of Functional and Variation..................................................................23 3.2 Hamilton’s Principle..............................................................................................................24 3.3 Obtaining the Equations of Motion from Hamilton’s Principle............................................24 3.3.1 Single Degree of Freedom.............................................................................................24 3.3.2 2-DOF Systems..............................................................................................................26 4 Dynamics of Cables, Bars, and Shafts...........................................................................................28 4.1 Cables’ Equations of Motion.................................................................................................28 4.2 Natural Frequencies and Mode Shapes..................................................................................28 4.3 Bars’ Equations of Motion.....................................................................................................28 4.4 Different Boundary Conditions..............................................................................................29 4.4.1 Fixed-Free case..............................................................................................................29 4.4.2 Fixed-Fixed case............................................................................................................30 4.4.3 Fixed-Spring case...........................................................................................................31 4.5 Using Hamilton’s Principle....................................................................................................31 5 Dynamics of Beams.......................................................................................................................34 5.1 Equations of Motion..............................................................................................................34 5.2 Boundary Conditions.............................................................................................................36 5.2.1 Vibration of a Simply-Supported beam..........................................................................37 5.2.2 Vibration of a cantilever beam.......................................................................................38 5.3 Finite Element Model for Beam Vibration............................................................................39 5.3.1 Interpolation function.....................................................................................................40 5.3.2 Element Equations.........................................................................................................42 5.3.3 Assembling the Structure Equations..............................................................................44 5.3.4 Applying the boundary conditions.................................................................................46 5.4 Natural Frequencies and Time Response...............................................................................48 5.5 Newmark Method for Time Domain Solution.......................................................................51 5.6 The Effect of Axial Loads......................................................................................................51 5.7 The Fire-Hose Problem..........................................................................................................53 Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

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6 7

8

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5.8 Beams under Moving Loads..................................................................................................55 Dynamics of Membranes and Plates.............................................................................................57 Periodic Structures for Vibration Attenuation...............................................................................58 7.1 What is a periodic structure?..................................................................................................58 7.2 Periodic Analysis...................................................................................................................59 7.3 Discrete Periodic System.......................................................................................................62 7.3.1 Case of identical masses.................................................................................................62 7.3.2 Case of alternating masses.............................................................................................65 7.4 The Reverse Approach in Periodic Analysis..........................................................................67 7.4.1 Reverse approach for a discrete system.........................................................................67 7.4.2 Generalization of the Inverse Periodic approach...........................................................68 7.5 Periodic Bars..........................................................................................................................71 7.6 Periodic Beams......................................................................................................................77 7.6.1 Numerical Analysis........................................................................................................77 7.6.2 Experimental Analysis...................................................................................................80 7.7 Further Reading.....................................................................................................................88 Attenuation of Structure Dynamics...............................................................................................89 8.1 Viscoelastic Materials............................................................................................................89 8.2 Passive Control Using Piezoelectric Materials......................................................................89 8.3 Active Control Using Piezoelectric Materials.......................................................................89 8.4 Magnetorheological Fluids and Damping..............................................................................89 Aeroelasticity.................................................................................................................................90 9.1 What is Aeroelasticity?..........................................................................................................90 9.2 Static Aeroelastic Problems...................................................................................................90 9.2.1 Divergence.....................................................................................................................90 9.2.2 Control Reversal............................................................................................................90 9.3 Dynamic Aeroelasticity – Wing Flutter.................................................................................90 9.4 Dynamic Aeroelasticity – Panel Flutter.................................................................................90 9.4.1 Introduction....................................................................................................................90 9.4.2 Panel-Flutter Analysis....................................................................................................92 9.4.3 References and Further Readings...................................................................................94


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Single Degree of Freedom Systems

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The single degree of freedom (SDF) is the simplest dynamic system we could study. In-spite of its simplicity, the SDOF may model several engineering problems and provide the necessary insight to its performance. Also, more complex systems may be described as a collection of SDF systems which make the study of such simple systems of great importance to anyone who wants to study the dynamics of structures or any dynamic system.

1.1 Equations of Motion For the system shown in Figure 1.1.1, the mass m (kg) if connected to a rigid wall with a spring with stiffness k (N/m) and a damper c (N.s/m).

Figure 1.1.1.Mass-Spring-Damper System Sketch To write down the equation of motion of this mass, we may draw the free body diagram. (Figure 1.1.2). IN this figure, we may write down the expressions for the forces created by the spring and the damper respectively to be: f k =kx f c =c x˙

Figure 1.1.2.Mass-Spring-Damper System Sketch Summing up the forces in the direction of motion, we may write Σ F x =m x¨ =f −f k −f c =f −k x−c x˙ Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

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Rearranging the terms of the equation, we get: m x¨ +c x˙ + k x=f Which is the differential equation that describes the mass-spring-damper system we have above. The solution of this differential equation may be presented as two parts; namely, the homogeneous solution and the particular integral. x (t)=x homogeneous + x particular integral =x h+ x p The homogeneous solution may be written in the general form as: x h (t)=a eλ t Substituting into the differential equation, we get: m λ2 a e λt +c λ a e λ t +k a e λ t =0 Rearranging, (m λ2 +c λ+ k )a eλ t =0 Since e λ t=0 can not be equal to zero, and if a=0 we get a trivial solution, then we can only say that: m λ2 +c λ+ k=0 Which is called the characteristic equation. Solving for the characteristic value, we get: −c±√ c 2−4 mk λ= 2m According to the above solution of the characteristic equation, we may get one of three casses.

Case #1 c 2−4 mk> 0 - Over Damped In this case, the value of the damping is large enough to preserve the value under the square root to be positive. This is called over damped case. In this case, we get two distinct real negative values for the characteristic equation. −c−√ c 2−4 mk 2m −c+ √ c2 −4 mk λ 2= 2m

λ 1=

Using those two values in the general solution, we get:


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Single Degree of Freedom Systems x h (t)=a 1 e λ t +a 2 e λ t 1

2

In this case, both, the exponential terms decay with time with different rates. We will need to apply the initial conditions to be able to get the solution (see section 1.2).

Case #2 c 2−4 mk=0 - Critically Damped In some cases, we may get the the condition where c 2−4 mk=0 . In this case, the value under the root becomes zero, which leads to the condition where the characteristic value is repeated. λ 1=λ 2=

−c 2m

The solution of such a case may be written as: x h (t)=(a1 +a2 t)e λt Note that the linear term will not lead to instability in the solution since it is multiplied by a decaying exponential. Thus, again, this solution will give a decaying of x(t). This case is called critically damped. In practice, we do not concern ourselves with this case because of the fact that setting the values in such a manner is almost impossible to be adjusted without errors that lead to case one or case three.

Case #3 c 2−4 mk< 0 - Under Damped The third case happens when the damping is low enough such that the value under the square root becomes negative. This case is the most important case when analyzing structure dynamics. In this case, the solutions of the characteristic equation become a pair of complex conjugate numbers. The real part will still be equal to the result obtained from case #2, but the imaginary part is what creates all the fun. −c−i √ 4 mk−c 2 2m −c+ i √ 4 mk−c 2 λ 2= 2m

λ 1=

Substituting the values into the general solution we get to write: x h (t)=e

−c t 2m

−i

(a 1 e

√

2

k c −( )t m 2m

+a e √ i

2

2

k c −( )t m 2m

)

We usually use some symbols to present the above quantities.


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√

k is called the natural frequency of the system. This is the frequency at which the massm spring system will oscillate if excited with initial conditions only (and without damping). ωn =

ζ=

c is called the damping ratio. Note that if it is equal to one, we get the critical damping 2 √ mk

case: 1=

c →2 √ mk=c →4 mk=c2 2 √mk

Hence, whenever it has a value less than 1, we get the case of under damping. If we use the abovedefined quantities into the general solution, we get: x h (t)=e− ζ ω t (a1 e−i √ ω n

n

2

2

−(ζ ω n) t

+a 2 e i √ ω

n

2

2

−(ζ ω n) t

)

Using the complex exponential identity e ia =cos(a)+i sin(a) , we can write: x h (t)=e− ζ ω t ( A 1 cos( √ω n2 −( ζ ωn )2 t)+ A2 sin( √ ω n2 −( ζ ωn )2 t)) n

Which is another form of the same solution where

A 1=a 1+ a2, A 2=i(a 1−a2 ) . Another form of

the general solution is x h (t)=e− ζ ω t A cos ( √ ωn2−(ζ ωn)2 t+ ϕ) n

Where

A= √ A21 + A 22 , ϕ=−arctan ( A1 / A2 )

1.2 Vibration Due to Initial Excitation The homogeneous solution of the differential equation that we got in the previous section requires two initial conditions to obtain a unique solution. Usually, the initial conditions are given in the form of the displacement and velocity at the beginning of the time we are concerned with (strictly speaking, any two values of either velocities or displacements at any point in time will create a unique solution). m x¨ +c x˙ + k x=0 x (0)=x o x˙ (0)=v o If we strict our focus in this section on the response to initial excitation, the general solution reduces to be the homogeneous solution (zero excitation force). Thus, we get: x (t )=x h


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Single Degree of Freedom Systems Case #1 - Over Damped In this case, you may recall, we have two distinct real roots of the characteristic equation and the solution has the general form: x (t)=a 1 e λ t +a 2 e λ 1

2

t

Differentiating, we get: x˙ (t)=a 1 λ 1 eλ t + a2 λ2 e λ t 1

2

Now, let’s apply the initial conditions to be able to obtain the time response: a1 +a 2=x o λ 1 a 1+ λ2 a2=v o λ x o −v o a1= λ2 −λ 2 1 v o−λ 1 x o a2= λ −λ 2 1 λ x o−v o λ t v o−λ1 x o λ t x (t)= λ2 −λ e + λ −λ e 2 1 2 1 1

1.2.1

2

Illustration of an over-damped system

Consider the case where m=1 kg, k=1 N/m, and c=3 Ns/m. When checking the numbers, you will find that the damping ratio is 1.5. Thus, the system has two distinct characteristic values of -2.6 and 0.38. If the system is excited with a unit initial value of the displacement or a unit initial velocity, we get two different responses as shown in Figure 1.2.1.


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Single Degree of Freedom Systems 1.2 1

x(t)

0.8 0.6 X(t) (x0=0, v0=1) X(t) (x0=1, v0=0)

0.4 0.2 0 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Time (s)

Figure 1.2.1.Response of an over-damped system to different initial conditions Case #2 - Critically Damped In this case, you may recall, we have one real root of the characteristic equation and the solution has the general form: λ 1=λ 2=

−c =−ζ ωn 2m

x (t)=(a1 +a2 t )e−ζ ω t n

Differentiating, we get: x˙ (t)=a 2 e−ζ ω t −ζ ωn (a 1+ a2 t )e−ζ ω t n

n

Substituting with the initial conditions: a1=x o a2=v o +ζ ωn xo Hence, the time response becomes: x (t)=( x o+(v o+ ζ ω n x o )t) e−ζ ω t n


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1.2.2

Illustration of a critically-damped system

Consider the case where m=1 kg, k=1 N/m, and c=1 Ns/m. When checking the numbers, you will find that the damping ratio is 1. Thus, the system has a repeated characteristic values of -0.5. If the system is excited with a unit initial value of the displacement or a unit initial velocity, we get two different responses as shown in Figure 1.2.2. 1.2 1

x(t)

0.8 0.6 X(t) (x0=0, v0=1) X(t) (x0=1, v0=0)

0.4 0.2 0 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Time (s)

Figure 1.2.2.Response of an critically-damped system to different initial conditions

Case #3 - Under Damped This is the most interesting case for all those who are interested in working in the field of dynamics and vibrations. In this case, the damping ratio is less than one. That leads to a response that, typically, shows oscillatory motion. The characteristic values of the problem are given in the form: −c−i √ 4 mk−c 2 =−ζ ω n−i √ ωn2−(ζ ωn)2 =−ζ ω n−iω d 2m −c +i √ 4 mk−c 2 λ 2= =−ζ ω n+ i √ ωn2−(ζ ωn)2=−ζ ω n+i ωd 2m

λ 1=

2 2 Where ωd =√ ωn−(ζ ωn) is sometimes referred to as the damped natural frequency. It got that

name form the fact that is appears where the frequency of the system should be in the time response. In practice, the difference between ωn and ωd is not significantly large. When substituting into the general solution, we get:


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Single Degree of Freedom Systems x (t)=e− ζ ω t (A 1 cos(ω d t)+ A2 sin(ω d t )) n

Differentiating: x˙ (t)=e− ζ ω t (− A1 (ω d sin(ωd t )+ ζ ω n cos (ωd t ))+ A2 (ω d cos( ωd t)−ζ ωn sin(ω d t))) n

Applying initial conditions: A 1=x o A 2=

v o + x o ζ ωn ωd

x (t)=e− ζ ω t (x o cos(ω d t)+ n

v o + x o ζ ωn sin(ωd t )) ωd

Using the other formula, we get: x (t)=e− ζ ω t A cos (ωd t +ϕ ) n

Where

√

v 2o + x 2o ω2n A=√ A + A = ω 2d v + x ζ ωn ϕ=−arctan ( A2 / A1 )=−arctan ( o o ) xo 2 1

1.2.3

2 2

Illustration of an under-damped system

Consider the case where m=1 kg, k=1 N/m, and c=0.5 Ns/m. When checking the numbers, you will find that the damping ratio is 0.25. Thus, the system has a natural frequency of 1. If the system is excited with a unit initial value of the displacement or a unit initial velocity, we get two different responses as shown in Figure 1.2.3.


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Single Degree of Freedom Systems 1.2 1 0.8 0.6 x(t)

0.4 X(t) (x0=0, v0=1) X(t) (x0=1, v0=0)

0.2 0 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

-0.2 -0.4 -0.6 Time (s)

Figure 1.2.3: Response of an under-damped system to different initial conditions

1.3 Harmonic Excitation The case of harmonic excitations of dynamic systems is particularly important to the study of structure vibrations and dynamics because of many reasons: •

It is the easiest, among most of excitation functions, to work with analytically.

•

It constitutes the foundation for analysis of all periodic excitations through the use of Fourier series

•

Many practical periodic excitations can be approximated by a single harmonic functions

For an excitation function to be called harmonic is has to be described by a single Sin or Cos function. Sometimes, it may be easier to present the harmonic excitation as an exponential complex function as well. Let us consider the excitation to be in the form: f (t)=f o sin(ω t) Here, f o indicates the magnitude of the excitation force and ω indicates the frequency of the excitation. For such an excitation, the particular integral solution of the differential equation becomes in the form: x p (t)=x o sin (ω t) Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

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Single Degree of Freedom Systems Where

x o is the response magnitude. Note that the response follows the frequency of the

excitation regardless of the natural frequency. Substituting into the differential equation (without damping), we will get: −mω 2 x o sin (ω t)+k x o sin (ω t)=f o sin (ω t) which can be rearranged into the form: (−m ω2+ k ) x o sin (ω t)=f o sin(ω t) From this equation, we may readily get a relation between the force and response magnitudes in the form: x o=

fo 2

−mω + k

=

f o /m ω n2−ω2

Note that the above relation shows that the magnitude of the response will change with frequency. When the excitation frequency is equal to the natural frequency of the system, the response will give an infinite value which indicates the presence of resonance. The change of the sign of the response as the excitation frequncy becomes larger that the natural frequncy indicates that the response is acting in the opposite direction of the excitation force. Figure 1.3.1 illustrates a typical frequency response function for an undamped single degree of freedom system.

The expression frequency response is one that is commonly used in dynamic analysis to indicate the graph that presents the variation of the response magnitude with the excitation frequency. An algebraic form of the frequency response is usually called the frequency response function.


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Single Degree of Freedom Systems 5 4.5 4

ABS(xo/(fo/m))

3.5 3 2.5 2 1.5 1 0.5 0 0

0.5

1

1.5

2

2.5

3

3.5

4

w/wn

Figure 1.3.1: Typical change of the response magnitude with excitation frequency

when damping in introduced into the problem, it is easier to present the excitation and the response in comples exponential form. Thus the excitation function will be presented as: f (t)=f o ei ω t The response follows the excitation to be in the form: x p (t )=x o e

iω t

When substituting into the differential equation, we get: −mω 2 x o ei ω t +i ω c x o e i ωt + k xo e i ω t=f o ei ω t Which may be rearranged to become: (−m ω2+ ic ω+k ) x o e i ωt =f o ei ω t From which, we can readily get the relation between the excitation and response magnitude in the fom: x o=

fo 2

−mω +i c ω+ k

=

f o /m 2

ωn −ω2 +2 i ζ ωn

Multiplying by the complex conjugate to get rid of the complex number in the denominator, we get:


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Single Degree of Freedom Systems x o=

f o /m 2

2

2 2

2

(ω n −ω ) +(2 ζ ωn )

2

( ωn −ω −2 i ζ ωn )

From this relation, we may get the magnitude and the phase of the response in the form: |x o|=

f o /m

√(ω

2

−ω2 )2 +(2 ζ ω n)2 2ζω Arg(x o)=arctan( 2 n 2 ) ωn −ω n

Note that the above relation reduces to the case of undamped results, we got earlier, if we substitute the damping ratio with a zero. Figure 1.3.2 illustrates the effect of the damping ratio on the frequency response of a single degree of freedom system.

5 4.5 Zeta=0 Zeta=0.2 Zeta=0.5

4

ABS(xo/(fo/m))

3.5 3 2.5 2 1.5 1 0.5 0 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

w/wn

Figure 1.3.2: Typical response of a single dgree of freedom system with different damping ratios


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1.4 Periodic Excitation 1.5 General Excitation 1.6 Approximate Evaluation of Time Responses Euler Method Rounge-Kuta Method


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Multiple Degree of Freedom Systems

2

Multiple Degree of Freedom Systems

2.1 2-DOF Equations of Motion A simple 2-DOF system is illustrated in Figure 2.1.1. It is strait-forward expansion of the SDOF system we were discussing before. To obtain the equations of motion, we will draw the free-body diagram of both masses then apply the dynamic equilibrium equation for each of them (Figure 2.1.2).

Figure 2.1.1: Illustration of a 2-DOF system

Figure 2.1.2: FBD for the 2-DOF system

For the first mass, the summation of forces may be written as:

∑ F x 1=m1 x¨1=f 1 +f k + f c −f k −f c 2

2

1

1

The forces due to the springs and dampers may written as:


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Multiple Degree of Freedom Systems f k =k 1 x1 f c =c 1 x˙1 f k =k 2 ( x 2−x 1) f c =c 2 ( x˙ 2− x˙1 ) 1

1

2

2

Substituting into the equilibrium equation, we get: m1 x¨1=f 1+ k 2 (x 2−x 1)+c 2 ( x˙ 2− x˙1 )−k 1 x 1−c 1 x˙1 Performing the same steps for the second mass, we get: m2 x¨2=f 2−k 2 ( x 2−x 1)−c 2( x˙2− x˙1 ) Thus, we get two equations of motion that may be written as: m1 x¨1+(c 1 +c 2) x˙1−c 2 x˙2 +(k 1+ k 2) x 1−k 2 x 2=f 1 m2 x¨2 +c 2 x˙2−c 2 x˙1 +k 2 x 2−k 2 x 1=f 2 Rearranging the equations in the form of matrices, we get:

[

m1 0 0 m2

]{ } [

x¨ 1 c + c −c 2 + 1 2 x¨ 2 −c 2 c2

]{ } [

x˙1 k +k −k 2 + 1 2 x˙2 −k 2 k2

]{ } { } x1 f = 1 x2 f2

Which may be written in shorthand as: [M ]{ X¨ }+[ C]{ X˙ }+[ K ]{X }={F} Notice that the above matrix equation looks exactly like the equation of motion of a single degree of freedom system except that is is made up of matrices and vectors. The general solution for the above differential equation may be written as: X (t )=X h (t )+ X p (t) Another way of writing the above equations of motion can be by introducing a velocity vector such that: Y (t)= X˙ (t)→ Y˙ (t)= X¨ (t) Which can be used in the equation of motion to get: M Y˙ +C Y + K X=F Thus the second order differential equation may now be presented in the form of a first order differential equation with twice as much degrees of freedom.

{XY˙˙ }=[−M0

−1

I K −M −1 C

]{ } { } 0 X + −1 Y M F

Z˙ =A Z +U


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Multiple Degree of Freedom Systems This presentation of the equations of motion is typically used in control. It may also be useful in finding the time response and natural frequencies of the system as we will see later. Thus, we will be switching from one form of presentation to the other seamlessly as both capture the full dynamics of the system.

2.2 Solution for Motion without Force As indicated in section 2.1, the general solution for the equations of motion has a homogeneous solution and a particular integral. In this section we will focus on the homogeneous solution. The general solution for the homogeneous equation may be written as: X (t )=X o e i ωt For a 2-DOF system without damping, we may substitute the general solution into the differential equation to get: i ωt

2

(−ω M + K) X o e =0 This equation will give a trivial solution except in the case where the matrix (−ω2 M + K) becomes singular. Since we are not interested in the trivial solution, we will investigate the case in which the matrix is singular, ie. the determinant is equal to zero. 2

‖K−ω M‖=0 For the 2-DOF system illustrated in Figure 2.1.1, the system we are handling becomes:

([

] [

k 1+ k 2 −k 2 m 0 −ω 2 1 −k 2 k2 0 m2

]){ } { } xo 1 = 0 0 xo 2

Thus the matrix of interest becomes:

‖

‖

k 1+ k 2−ω2 m1 −k 2 =0 −k 2 k 2−ω2 m2

Evaluating the determinant and equating it by zero we get: (k 1 +k 2−ω2 m1 )(k 2−ω2 m2 )−k 22=0 Expanding and rearranging terms, we get: m1 m2 ω4−(m2 (k 1 +k 2)+m1 k 2)ω2 +k 1 k 2=0 Which is a second order equation in ω2 whose solution is given as: ω2=ω 21 , ω22 For the sake of completeness, we will write the solution to be:


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Multiple Degree of Freedom Systems ( m2 ( k 1+ k 2)+m1 k 2)±√(m2 (k 1 +k 2)+m 1 k 2)2−4 m 1 m2 k 1 k 2 ω= 2 m1 m2 2

Each of the natural frequencies is the square-root of one of the results. However, we can show that the frequencies, in this case, are all real with two positive and two negative values. The general solution in this cae become in the form: X (t )=X o 1 e iω + X o2 e−i ω + X o 3 ei ω + X o 4 e−i ω 1

Where

1

2

2

X oi is the ith Eigen vector of the problem multiplied by a scale determined from the initial

conditions (we have to have four initial conditions for this problem to obtain a unique solution). Thus the general solution for a MDOF system with no excitation may be written in the form: X (t )=∑ X oi e iω t i

i

If we get to use the other form of the equations of motion of the system, we get the general solution for the system without excitations in the form: Z (t )=Z o e λ t Substituting in the differential equation, we get: λ Z o e λt =A Z o e λ t Rearranging the terms: (λ I − A)Z o e λ t=0 Which give a non-trivial solution if the equation ‖λ I − A‖=0 is satisfied. This gives another Eigenvalue problem that results in the natural frequencies in the form of complex conjugate couples and the general solution becomes in the form: Z (t )=∑ Z oi e λ t i

i

2.3 Forced Response to Harmonic Excitation As for the single degree of freedom systems, the response to harmonic excitations is quite interesting for the MDOF systems. The excitation may presented in the form: F( t)=F o ei ω t Which gives a general particular integral solution in the form: X (t )=X o e i ωt When substituted in the equations of motion, we get: Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

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Multiple Degree of Freedom Systems (−ω2 M + iω C+ K ) X 0 ei ω t =Fo e i ωt To get threlation between the magnitude of excitation and that of the response, we get: (−ω2 M + iω C+ K ) X 0=F o → X 0 =(−ω 2 M +iω C+ K)−1 F o Which is essentially, and algebraic equation relating the excitation magnitude to the response magnitude. The above relation may be rewritten in the form X 0=K −1 D Fo Where K D =−ω 2 M +iω C+ K is called the dynamic stiffness matrix. The concept of the dynamic stiffness matrix is quite useful in many problems, especially, when we are only interested in the frequency response of the system.

2.4 Vibration Absorbers – First Passive Attenuation Technique Let us revisit the example of Figure 2.1.1. If the dampers were removed, the equations will reduce to: m1 x¨1+(k 1+ k 2) x 1−k 2 x2=f 1 m2 x¨2 + k 2 x 2−k 2 x 1=f 2

[

m1 0 0 m2

]{ } [

x¨ 1 k +k −k 2 + 1 2 x¨ 2 −k 2 k2

]{ } { } x1 f = 1 x2 f2

Now let’s write down the relations explicitly for the case harmonic excitation:

( [ −ω2

][

m1 0 k +k −k 2 + 1 2 0 m2 −k 2 k2

]){ } { } x1 o f = 1o x2 o f 2o

Obtaining the dynamic stiffness matrix, we may write down the relations between the force magnitudes and the response magnitudes as:

{ }[

k 1+ k 2−ω 2 m1 −k 2 x1 o = x2 o −k 2 k 2 −ω2 m2

]{ } −1

f 1o f 2o

Performing the matrix inversion, we get:

{ }

[

k 2−ω 2 m2 k2 k2 k 1 + k 2−ω2 m1

]

{ }

x1 o f 1o = 2 2 2 x 2 o (k 1 + k 2−ω m1)(k 2 −ω m2 )−k 2 f 2 o

From the above, we may write: Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

21

Multiple Degree of Freedom Systems k 2−ω2 m2 k2 x 1 o= f + 1 o Δ Δ f 2o k2 k 1 + k 2−ω2 m1 x 2o = Δ f 1o + f 2o Δ

Now, let us examine the above relations. For the first mass, if the second force is zero, you will find that the response of x1 will reach a zero when: k 2=ω2 m2 → x1 o =0 This result is interesting because of two main things: 1. At that frequency, the response of the mass at which the force is applied is zero while the response of the second mass is finite. 2. The frequency at which the response is zero is that of the spring-mass system of the second mass if it was not connected to the first one. This phenomenon is called the vibration absorber. It is a simple way to reduce the vibration of a mass (primary target) at a certain frequency by connecting it to a secondary mass spring system whose natural frequency is the one we are targeting. Vibration absorbers have been used in a wide variety of applications, and are still used, because of their simplicity and effetiveness.


22

Variational Methods and Hamilton’s Principle

3


3.1 Simplified Concepts of Functional and Variation A functional can be defined, simply, as a function of functions that has a scalar value. The functionals we use in our work are in the form of integration of functions on the domain of interest. For example: x2

Π=∫ f ( x) dx x1

This is a functional that is defined as the integral of f(x) over the domain [x1,x2]. S you may notice, the resultant of the integral will be a scalar. The variation of the functional is similar to the differentiation of the integral with respect to the function f(x). Such that: x2

δ Π=∫ δ f (x) dx x1

The above equations says that the variation of Π is equal to the integration of the variation of f(x) over the domain [x1,x2]. The variation of the function f(x) is said to be equal to zero where the values of the function are defined. if f ( x3 )=f 3 →δ f ( x 3 )=0

The rules of differentiation apply directly to variations. For example, if the functional is related to the square of a function, then the variation of the functional is given as: x2

Π=∫ f 2 (x) dx x1

x2

δ Π=∫ 2 f ( x)δ f ( x )dx x1

Also, the product rule applies: x2

Π=∫ f ( x) g(x)dx x1

x2

δ Π=∫ (g ( x)δ f ( x )+f ( x)δ g ( x))dx x1

These relations are what we need to work with in the coming parts.


23


3.2 Hamilton’s Principle Hamilton’s principle is an approach to obtaining the equations of motion for mechanical systems through the manipulation of the energy terms. In mechanical systems, thre are three main types of energy that we are concerned with. 1. The kinetic energy with reflects the motion information of the masses 2. The potential energy which is concerned with the conservative fields of energy such as the elastic and gravitational energy terms 3. The external work which concerned with non-conservative work done by external forces or damping elements Hamilton’s principle defines the total energy of the system as at any given point as: Π=T −U +W ex Then the principle states that the variation of the total energy over an arbitrary period of time should sum up to zero, or in mathematical terms: t2

∫ δ Π dt =0 t1

The above relation may be rewritten in terms of the components of the total energy to read: t2

∫ (δ T −δU + δ W ex )dt=0 t1

Hamilton’s principle contains everything related to the dynamics of the system. When manipulated to obtain the differential equations, the principle expresses the boundary and initial conditions as well.

3.3 Obtaining the Equations of Motion from Hamilton’s Principle 3.3.1

Single Degree of Freedom

For a single degree of freedom system, the energy terms may be expressed as: 1 U= k x 2 2 1 T = m x˙2 2 W ex=f x−f damper x


24

Variational Methods and Hamilton’s Principle Note that the external work done by the damper has a negative sign as it extracts energy from the system during its motion. If we evaluate the variation of the energy terms, we get: δ U=k x δ x δT =m x˙ δ x˙ δ W ex=f δ x−c x˙ δ x Applying the time integral to the variation, we get: t2

t2

t1

t1

t2

t2

t1

t1

∫ δ U dt =∫ k x δ x dt t2

∫ δ T dt=∫ m x˙ δ x˙ dt=m x˙ δ x| −∫ m x¨ δ x dt t2 t1

t1

Notice that we applied integration by part to the integral of the kinetic energy to transfer the time derivative from the variation of the velocity to obtain the acceleration. t2

t2

t2

t2

t1

t1

t1

t1

∫ δ W ex dt=∫ (f δ x−c x˙ δ x) dt=∫ f δ x dt −∫ c x˙ δ x dt If we recast the expressions we obtained into the Hamilton’s principle, we get: t2

t2

t1

t1

∫ (δ T −δU + δ W ex )dt=∫−(k x δ x +m x¨ δ x+ c x˙ δ x−f δ x)dt+ m x˙ δ x|t =0 t2 1

According the principles of variation, each term in the above result should be equal to zero by itself. That gives the boundary terms in the form: t2

m x˙ δ x|t =0→ 1

m x˙ δ x|t =0 and 1

m x˙ δ x|t =0 2

Which indicates that at the given two points in time, you either know the displacement or velocity, ie. initial conditions. Also, the integral will give us: t2

∫ (k x+ m x¨ + c x˙ −f )δ x dt=0 t1

Since the variation of the position x(t) can not be equal to zero all the time (or else we already have the solution), then the expression in the braces should be equal to zero: k x+ m x¨ +c x−f ˙ =0→m x¨ +c x˙ + k x=f (t) Which is the differential equation describing the motion of the spring mass-damper system as obtained earlier. Notice that through the Hamilton’s principle, we started with an expression for the total energy of the system and ended up with the differential equation and the initial conditions.


25


3.3.2

2-DOF Systems

Now let’s apply the principle to the 2-DOF of Figure 2.1.1. The energy terms may be expressed as: 1 1 U = k 1 x 21 + k 2 (x 2−x 1)2 2 2 1 1 T = m1 x˙ 21 + m1 x˙ 22 2 2 W ex=f 1 x1 + f 2 x 2−f c x 1−f c ( x 2 + x 1) 1

2

Applying the variation on each term, we get: δ U =k 1 x 1 δ x1 +k 2( x 2−x 1 )( δ x 2−δ x 1 ) δ T =m1 x˙1 δ x˙1+ m1 x˙2 δ x˙2 δ W ex =f 1 δ x 1 + f 2 δ x 2−f c δ x 1−f c (δ x 2 +δ x1 ) 1

2

Applying the time integration on each term, we get: t2

t2

t1

t1

∫ δU dt=∫ (k 1 x 1 δ x 1+ k 2 (x 2−x1 )(δ x 2−δ x1 )) dt t2

t2

t1

t1

t2

∫ δ T dt=∫ (m1 x˙1 δ x˙1 +m2 x˙2 δ x˙2 )dt=(m1 x˙ 1 δ x 1+ m2 x˙ 2 δ x 2)|t −∫ (m1 x¨1 δ x1 +m2 x¨2 δ x2 )dt t2 1

t2

t2

t1

t1

t1

∫ δW ex dt=∫ f 1 δ x1 + f 2 δ x 2−f c δ x 1−f c ( δ x 2−δ x 1)dt t2

t2

t1

t1

1

2

→∫ δW ex dt =∫ f 1 δ x1 + f 2 δ x 2−c 1 x˙1 δ x1 −c 2( x˙2− x˙1 )(δ x2−δ x 1 )dt Collecting the terms of δ x1, δ x 2 we get: t2

t2

t1

t1

∫ δ U dt=∫ ((k 1 x1 + k 2 x 1−k 2 x 2) δ x1 +k 2 x 2 δ x 2)dt t2

t2

∫ δ T dt=(m1 x˙1 δ x 1 +m2 x˙2 δ x 2)|t −∫(m1 x¨ 1 δ x 1+ m2 x¨ 2 δ x 2)dt t2 1

t1

t2

t2

t1

t1

t1

∫ δ W ex dt=∫ (f 1−(c 1 +c 2) x˙ 1+ c2 x˙ 2) δ x 1 +(f 2−c2 x˙ 2+ c2 x˙ 1)δ x 2 dt Casting into the Hamilton’s principle, we may separate the integrals into two as follows:


26

Variational Methods and Hamilton’s Principle t2

t2

t1

t1

t2

∫ δ Π dt=∫ (δ T −δ U +δ W ex)dt=∫ F1 δ x1 + F 2 δ x 2 dt=0 t2

(

t1

t2

∫ F 1 δ x 1=(m1 x˙ 1 δ x1 )|t − ∫ ((k 1+ k 2) x 1−k 2 x2 +m1 x¨1 +(c 1+ c2 ) x˙1−c2 x˙ 2−f 1 )dt t2 1

t1

t2

t1

t2

(∫

∫ F 2 δ x 2=(m2 x˙ 2 δ x2 )|t − t1

1

t2

)

)

δ x1=0

(k 2 x 2−k 2 x1 +m 2 x¨2 +c 2 x˙2−c2 x˙ 1−f 2 )dt δ x2 =0

t1

Which gives us the differential equations and the initial conditions as in the case of the SDOF. m1 x¨1+(c 1 +c 2) x˙1−c 2 x˙2 +(k 1+ k 2) x 1−k 2 x 2=f 1 m2 x¨2 +c 2 x˙2−c 2 x˙1 +k 2 x 2−k 2 x 1=f 2

Note: Hamilton’s principle may appear as an over killing when we apply them to the simple problems of SDOF and 2-DOF systems. But this is just an exercise so that you may get familiar with the procedure to be able to apply it for more complicated problems for which the differential equations are not readily available.


27

Dynamics of Cables, Bars, and Shafts

4

Dynamics of Cables, Bars, and Shafts

4.1 Cables’ Equations of Motion 4.2 Natural Frequencies and Mode Shapes 4.3 Bars’ Equations of Motion d σ)( A +dA)−σ A + f dx ¨ ∑ F x =ρ A dx u=(σ+ ρ A dx u=σ dA+ A d σ+ f dx=d ( A σ)+ f dx=d (EA ¨

(

)

∂u )+ f dx ∂x

∂ EA ∂u + f →ρ A u= ¨ ∂x ∂x For the case of constant EA the problem becomes: ρ A u=EA ¨

∂2 u +f ∂ x2

The homogeneous equation reduces to the wave equation in the form: ρA ∂2 u 1 ∂2 u u= → u= ¨ ¨ EA ∂ x2 c2 ∂ x2 Using separation of variables, we may write: u(x ,t )=X (x )T (t) Into the PDE, we get: T¨ (t) 2 X ' ' (x ) 1 ¨ X (x) T ( t)=X ' ' ( x)T (t )→ =c =−ω2 2 T (t) X(x) c Solving the temporal equation, we get: T¨ (t)+ω2 T (t )=0→T (t)=a 1 cos (ω t)+a 2 sin(ωt ) For the spacial equation, we get: 2 X ' ' (x )+ ω2 X (x)=0→ X ( x )=b1 cos ( ω x)+b 2 sin( ω x) c c c


28

Dynamics of Cables, Bars, and Shafts Both results, give: u(x ,t )=(a1 cos (ω t)+a 2 sin( ω t))(b1 cos ( ω x )+ b2 sin ( ω x)) c c

4.4 Different Boundary Conditions As studied in the fundamentals of mechanics of materials, bars have two types of boundary conditions. The displacement boundary conditions occur when you have a fixation support at which the displacement is defined. That may be expressed as: u(x 1 ,t)=0 The other type of boundary condition is what was called the strain boundary condition. That may be expressed as: ∂u(x 1 ,t ) P(t) = ∂x EA

4.4.1

Fixed-Free case

In this case, the boundary conditions may be given as: u(0 , t)=0 ∂u( L , t) =0 ∂x u(0 , t)=(a1 cos (ω t)+a2 sin(ω t))(b1 )=0 Which give b1=0 . Getting the derivative of the displacement function, we get: ∂ u(x ,t )= ω ( a cos(ω t )+ a sin (ω t))(−b sin( ω x)+b cos ( ω x)) 2 1 2 ∂x c 1 c c Applying the strain boundary condition, we get: ∂ u(L , t)= ω (a cos(ωt )+ a sin(ω t))(b cos( ω L))=0 2 2 ∂x c 1 c If b2=0 we get a trivial solution. Thus: cos( ω L)=0→ ω L=( 2n−1) π where(n=1,2,. ..) c c 2 Which indicates that we have an infinite number of solutions and the natural frequencies may be expressed as:


29

Dynamics of Cables, Bars, and Shafts c ωn=(2 n−1) π 2L And the spacial solution may be written as: X n ( x )=b2 sin

x) ( (2 n−1)π 2L

which is called the mode shape or the Eigen-function of the solution. Now the solution becomes in the form: ∞

u(x ,t )=∑ sin( n=1

4.4.2

(

)

(2n−1)π (2 n−1) π c (2 n−1)π c x) A 1 cos ( t)+ A2 sin ( t) 2L 2L 2L

Fixed-Fixed case

In this case, the boundary conditions may be given as: u(0 , t)=0 u(L , t)=0 u(0 , t)=(a1 cos (ω t)+a2 sin(ω t))(b1 )=0 Which give b1=0 . Applying the other boundary condition, we get: u(L , t)=( a1 cos(ω t)+ a2 sin (ω t))( b2 sin( ω L))=0 c If b2=0 we get a trivial solution. Thus: sin( ω L)=0 → ω L=n π where(n=1,2,...) c c Which indicates that we have an infinite number of solutions and the natural frequencies may be expressed as: ωn=n π

c L

Note: For the same mode number, the natural frequency of the fixed-fixed bar is twice as high as the fixed-free bar And the spacial solution may be written as: X n ( x )=b2 sin

( nLπ x)

which is called the mode shape or the Eigen-function of the solution. Now the solution becomes in the form: Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

30

Dynamics of Cables, Bars, and Shafts ∞

u(x ,t )=∑ sin( n=1

4.4.3

(

)

nπ nπc nπc x) A 1 cos ( t)+ A 2 sin( t) L L L

Fixed-Spring case

In this case, the boundary conditions may be given as: u(0 , t)=0 ∂ u(L , t)=− k u(L , t) ∂x EA u(0 , t)=(a1 cos (ω t)+a2 sin(ω t))(b1 )=0 Which give b1=0 . Applying the other boundary condition, we get: ∂ u(L , t)= ω (a cos( ωt )+ a sin(ω t ))(b cos( ω L))=− k ( a cos(ω t)+ a sin (ω t))(b sin ( ω L)) 2 2 2 2 ∂x c 1 c EA 1 c k ( ω cos ( ω L)+ sin( ω L)) b2=0 c c EA c If b2=0 we get a trivial solution. Thus: −

EA ω =tan ( ω L) k c c

Which indicates that we have an infinite number of solutions and the natural frequencies, but, will need to use numerical techniques to get the values of the natural frequencies. But we can still present the general solution as an infinite series

∞

u(x ,t )=∑ sin( n=1

ωn x) ( A 1 cos (ωn t)+ A2 sin (ω n t) ) c

Note: For the extreme cases of the values of the spring support, we get: k =0→ fixed-free case k =∞→ fixed-fixed case

4.5 Using Hamilton’s Principle For the potential energy, we have the elastic energy which may be expressed as:


31

Dynamics of Cables, Bars, and Shafts L

L

1 1 U= ∫ A σ ϵ dx= ∫ EA ϵ2 dx 20 2 0 L

L

0

0

δ U=∫ EA ϵ δ ϵdx=∫ EA

( )

∂u ∂u δ dx ∂x ∂ x

Using integration by parts:

|

L

δ U= EA

L

∂u ∂2 u δ u −∫ EA 2 δ u dx ∂x ∂x 0 0

For the kinetic energy, we may write: L

T=

1 ∫ ρ A u˙ 2 dx 2 0 L

δ T =∫ ρ A u˙ δ u˙ dx 0

The external work may be due to distributed and concentrated loads, so we may write: L

W ex = P u|x=x +∫ f u dx 1

0

L

δ W ex= P δ u|x=x +∫ f δ u dx 1

0

Using the above results into Hamilton’s principle, we get: t2

t2

∫ δ Π dt=∫ δ T −δ U +δ W ex dt t2

t2

∫ δ Π dt=∫ t1 t2

t1

∫ t1

t1

(∫ (

t1

L

2

ρ A u˙ δ u+ ˙ EA

0

((

) (

∂ u ∂u δu+ f δ u dx− EA δu−P δu 2 ∂x ∂x

)| ) dt=0 L

0

)| ) dt=0

∂u − EA δ u−P δ u ∂x

L

0

)| =0

(

EA

(

∂u EA −P δ u =0 ∂x 0

∂u δu−P δ u ∂x

| ) | )

( EA ∂u∂ x −P δ u

L

0

L

x=0

and

( EA ∂u∂ x −P )δ u|

x=L


32

Dynamics of Cables, Bars, and Shafts Which indicates that on both sides, you either have δ u=0 which is the fixation boundary condition, or you have: ∂u P = ∂ x EA Which is the strain boundary condition. On the other hand, the remaining part of the integration is: t2

(( L

∫∫ t1

ρ A u˙ δ u+ ˙ EA

0

) )

∂2 u δ u+f δ u dx dt=0 ∂ x2

Using integration by parts on the time integral, we get: L

∫ 0

(

t2

(

ρ A u˙ δ u| +∫ −ρ A u+ ¨ EA t2 t1

t1

) )

∂2 u + f δ u dt dx=0 ∂ x2

t2

ρ A u˙ δu|t =0 1

ρ A u˙ δu|t=t =0 1

ρ A u˙ δu|t=t =0 which indicates that on both limits of the time span, you either 2

know the given displacement shape or that the velocity is zero. The rest of the integral will give: ∂2 u + f =0 ∂ x2 ∂2 u →ρ A u=EA +f ¨ ∂ x2

−ρ A u+ ¨ EA

Which is the same differential equation we got in section 4.3.


33

Dynamics of Beams

5

Dynamics of Beams

5.1 Equations of Motion We will be using Hamilton’s principle to find the equations of motion for an Euler-Bernoulli beam. Assuming that the cross-section is homogeneous (constant modulus of elasticity, we may write the elastic energy as: L

U=

( )

2

1 ∂2 v EI dx ∫ 20 ∂ x2

Applying the variation: L

δ U=∫ EI 0

( )( )

∂2 v ∂2 v δ dx ∂ x2 ∂ x2

Applying integration by parts twice:

( ))|

(

∂2 v ∂v δ U= EI 2 δ ∂x ∂x

L

) )|

( (

L

∂2 v − ∂ EI 2 δ v ∂x ∂x 0

L

( ( ))

2 ∂2 v +∫ ∂ 2 EI ∂ x2 0 ∂x 0

δ v dx

For the kinetic energy: L

T=

1 ρ A v˙ 2 dx ∫ 2 0

Taking the variation of the kinetic energy: L

δ T =∫ ρ A v˙ δ v˙ dx 0

Applying the time integral: t2

t2 L

L

t1

t1 0

0

∫ δ T dt=∫∫ ρ A v˙ δ v˙ dx dt=∫

(

t2

)

(ρ A v˙ δ v )|t −∫ ρ A v¨ δ v dt dx t2 1

t1

For the work done by external forces: L

W ex =∫ f (x , t )v dx 0

Applying the variation and tim integration: t2

t2 L

t1

t1 0

∫ δ W ex dt=∫∫ f ( x , t) δ v dx dt Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

34

Dynamics of Beams Into Hamilton’s principle, and separating integration terms from boundary terms we get:

(

)

2 ∂2 v ρ A v¨ + ∂ 2 EI =f ( x , t) ∂x ∂ x2

With boundary conditions:

(

( ))| =0

∂2 v ∂ v EI δ ∂ x2 ∂ x

L

0

) )|

( (

2 ∂ EI ∂ v δ v ∂x ∂ x2

L

=0

0

And the initial conditions: t2

( ρ A v˙ δ v )|t

1

For the special case of a beam with constant cross-section, we may write the homogeneous form of the PDE as: ρ A v¨ + EI

∂4 v =0 ∂ x4

Assuming the solution in the form: v (x , t)=X ( x)T (t) By substituting into the PDE, we get: ρA

T¨ (t ) d 4 X ( x )/dx 4 + EI =0 T (t ) X ( x)

Now we may write: T¨ EI d 4 X /d x 4 =− =−ω2 T ρA X Solving the temporal equation, we get: T (t)=a 1 sin(ωt )+ a2 cos(ω t) While for the spacial equation, we may write: ρ Aω d4 X 4 4 −β X=0 where β = 4 EI dx

2

The general solution for this differential equation may be given in the form: X (x )=b1 cosh (β x)+b 2 sinh (β x)+b3 cos (β x)+b 4 sin(β x )


35

Dynamics of Beams Hence, the general solution of the PDE may be written as: v (x , t)=(a1 cos(ω t )+ a2 sin (ω t))(b1 cosh(β x)+b2 sinh(β x )+ b3 cos(β x)+b 4 sin (β x))

5.2 Boundary Conditions The partial differential equation above is fourth-order in space that needs four boundary conditions in order to produce a unique solution. First type of boundary conditions would be knowing the displacement of the beam at one or more points. This may be written as: v (x 1 )=δ Where δ is the known value of the deflection at that point. Note here that the usual value is zero, nevertheless, is may have any non-zero value as we will see in some examples. The value of the displacement may be determined in the cases of pinned, roller, or fixed supports.

The second type of boundary condition may be in the form: ∂v =θ ∂x At such boundaries, the slope of the beam is determined at the point of the boundary condition. For small slopes, the angle of the tangent is equal to the value of the slope. As in the case of the displacement boundary condition, the usual value of the slope fixation is zero, but it can definitely take any value according to the fixation condition. Slopes of the beam may be set using fixed supports as well as sliding supports (not mentioned before in this text).

The third type of boundary condition is the bending moment boundary condition which may be written as: EI z

∂2 v =M ∂ x2

For this boundary condition, the bending moment of the beam is determined at a given point. Free ends, rollers, and pinned supports are all examples of where we may be able to determine the bending moment as a boundary condition. It usually takes the value of zero unless there is a concentrated moment located at that point.

The forth, and last, type of boundary condition that we may set to the beam is the shear boundary condition.


36

Dynamics of Beams EI z

∂3 v =S ∂ x3

This kind of boundary conditions occurs when you know the applied lateral force at a given point. This usually happens at free ends and sliding supports.

Figure 5.2.1: Summary of the 2-D beam supports and boundary conditions

5.2.1

Vibration of a Simply-Supported beam

For the simply-supported beam, the boundary conditions may be given as: d 2 v (0 , t) =0 ∂ x2 ∂ 2 v (L, t) at x=L :v ( L , t )=0 , EI =0 2 ∂x at x =0: v (0 , t)=0 , EI

v (x , t)=T (t)(b 1 cosh (β x )+ b2 sinh(β x)+b 3 cos (β x )+ b4 sin(β x )) ∂v (x , t) =β T (t )(b1 sinh(β x )+ b2 cosh (β x)−b 3 sin (β x)+b 4 cos(β x )) ∂x ∂2 v ( x ,t ) 2 =β T (t)( b1 cosh(β x)+b2 sinh (β x )−b3 cos (β x)−b4 sin(β x )) ∂ x2 Applying the boundary conditions at x=0, we get: Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

37

Dynamics of Beams b 1+ b3=0 b1−b3=0 →b1=b 3=0 While at x=L: b 2 sinh(β L)+b 4 sin(β L)=0 b2 sinh(β L)−b 4 sin(β L)=0 →b2=0, b 4 sin(β L)=0→β L=n π From which:

√

2 nπ 4 ω ρ A = L EI 2 2 n π EI ωn = 2 (n=1,2,...) ρA L

βn =

√

Which gives the General solution to be: ∞

v (x , t)=∑ sin ( n=1

5.2.2

nπ x)( A 1 cos (ωn t)+ A2 sin (ω n t)) L

Vibration of a cantilever beam

For the cantilever beam, the boundary conditions may be given as: ∂ v (0 , t) =0 ∂x ∂2 v ( L , t) ∂ 3 v (L , t) at x=L : EI =0, EI =0 ∂ x2 ∂ x3 at x=0: v (0 , t)=0 ,

v (x , t)=T (t)(b 1 cosh (β x )+ b2 sinh(β x)+b 3 cos (β x )+ b4 sin(β x )) ∂v (x , t) =β T (t )(b1 sinh(β x )+ b2 cosh (β x)−b 3 sin (β x)+b 4 cos(β x )) ∂x ∂2 v ( x ,t ) 2 =β T (t)( b1 cosh(β x)+b2 sinh (β x )−b3 cos (β x)−b4 sin(β x )) ∂ x2 ∂3 v ( x , t) 3 =β T (t )(b1 sinh(β x)+ b2 cosh (β x )+b3 sin(β x )−b 4 cos(β x )) ∂ x3 Applying the boundary conditions at x=0, we get: b 1+b 3=0 b2 +b 4=0


38

Dynamics of Beams While at x=L: b1 cosh(β L)+ b2 sinh (β L)−b3 cos(β L)−b 4 sin(β L)=0 b1 sinh(β L)+b2 cosh(β L)+ b3 sin(β L)−b4 cos (β L)=0 The above set of equations will give a trivial solution except in the case where the determinant of the coefficient matrix is zero:

[ ‖

1 0 1 0 0 1 0 1 CH SH −C −S SH CH S −C

]{ } { } ‖ b1 0 b2 = 0 0 b3 0 b4

1 0 1 0 0 1 0 1 =0 CH SH −C −S SH CH S −C

Where CH =cosh(β L), SH=sinh(β L) ,C=cos (β L), S=sin(β L) . Expanding the determinant, we get: ((C 2 +S 2)−( SH . S−CH .C ))+((CH . C−SH . S)+(CH 2−SH 2))=0 1−2(SH . S−CH . C)−1=0 SH . S=CH .C tanh (β L). tan (β L)=1 Which needs a numerical technique to evaluate the different values of β which should result in an infinite number of solutions just like the case of the simply supported beam. The general slotion for this problem will be given in the form: ∞

v (x , t)=∑ T n (t)(b 1 n cosh (βn x )+ b2 n sinh(βn x)+b 3 n cos (βn x)+b 4 n sin(βn x )) n=1

5.3 Finite Element Model for Beam Vibration As you can see from the previous derivations, the exact solution may be obtained for simple structures. To be able to handle the more complex structures and more sophisticated problems, we will need to use a numerical method. One of the candidates for this problem is the finite element method. The finite element model of a structure may be derived from the differential equation using weighted residual methods, or from the variational form of the total energy expression as presented in Hamilton’s principle. We will adopt the variational approach for the derivation of the finite element model in this text. Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

39

Dynamics of Beams

5.3.1

Interpolation function

The interpolation function for the thin beams needs to satisfy the essential boundary conditions. In our case, the essential boundary conditions are the displacements and slopes of the beam at each end of the element. Thus, we need to satisfy four conditions, which leads to the need of, at least, a third order polynomial. The polynomial may, generally, be written as: v (x )=a 0+ a1 x +a 2 x 2+ a3 x 3 Or, in vector form:

v (x )=⌊ 1 x x 2 x 3 ⌋

{}

a0 a1 =⌊ H (x)⌋{a} a2 a3

The boundary conditions that need to be satisfied by this polynomial are the values of the displacements and slopes at each of the end of the beam elements presented in Figure 5.3.1.

Figure 5.3.1. Beam element with end displacements, slopes, point forces, and point moments. The slope function of the beam elements is given as: dv =v ' ( x)=a1 +2 a2 x +3 a 3 x 2 dx Or: dv =⌊ 0 1 2 x 3 x2 ⌋ dx

{}

a0 a1 =⌊H ' (x )⌋{a} a2 a3


40

Dynamics of Beams Using the boundary conditions, we may get: v (0)=v 1=⌊ H (0)⌋{a} v ' (0)=v ' 1=⌊ H '(0)⌋{a} v ( L)=v 2=⌊ H ( L)⌋{a} v ' (L)=v ' 2=⌊ H '(L)⌋{a} In matrix form:

{ }[

v1 1 0 v '1 = 1 v2 0 v'2

0 0 0 1 0 0 2 L L L3 1 2 L 3 L2

]{ } a0 a1 a2 a3

{v }=[T b ]{a} Which may, readily, be solved to get:

[

1 0 −3 {a}= 2 L 2 L3

0 1 −2 L 1 L2

0 0 3 L2 −2 L3

]

0 0 −1 L {v } 1 L2

{a}=[T b ]−1 {v } From which we may write: v (x )=⌊H (x )⌋[T b ]−1 {v }=⌊N (x )⌋{v } where:

{ }

3 x 2 2 x3 + 3 2 L L 2 2 x x3 x− + L L2 T ⌊N ( x)⌋ ={N ( x )}= 3 x2 2 x3 − 3 L2 L 2 x x3 − 2+ 3 L L 1−


41

Dynamics of Beams

5.3.2

Element Equations

To derive the element equations for the finite element method, we will use the reduced form. The reduced form, in simple terms, is the form of the energy equation before applying the integration by parts. From section 5.1, we may recall the reduced form to be: t2

t2 L

t1

t1 0

∫ δ Πdt=∫∫

(

( )

( )( ) |

∂2 v ∂2 v ∂2 v δ v−EI δ ∂t 2 ∂ x2 ∂ x2 dv dv + f ( x , t )δ v + p1 δ v|x=0+ p2 δ v|x=L + m1 δ + m2 δ dx x=0 dx −ρ A

|

x=L

)

dx dt=0

If we use the interpolation functions we derived in the previous section, we get: dv =⌊ H ' (x)⌋[T b ]−1 {v}=⌊ N ' (x)⌋{v } dx d2 v −1 =⌊H ' ' ( x)⌋[T b ] {v}=⌊N ' ' ( x)⌋{v } 2 dx

{ }

{ }

2 d2 v d2 v −1 d v =⌊H ( x)⌋[T ] =⌊ N (x )⌋ b dt 2 dt 2 dt 2

Where: ⌊H '( x )⌋=⌊ 0 1 2 x 3 x 2 ⌋

⌊H ' ' ( x)⌋=⌊ 0 0 2 6 x ⌋

{ }

−6 x 6 x 2 + 3 L2 L 4 x 3 x2 1− + 2 L L ⌊N ' ( x)⌋T ={N '( x)}= 6 x 6 x2 − L2 L 3 2 x 3 x2 − 2+ 3 L L


42

Dynamics of Beams

{}

−6 12 x + 3 L2 L −4 6 x + L L2 T ⌊N ' ' ( x)⌋ ={N ' ' (x)}= 6 12 x − L2 L 3 2 6x − 2+ 3 L L

when applying the vriation, we get:

δ v (x)=⌊ H (x)⌋[T b ]−1 {δ v }=⌊ N ( x)⌋{δ v } δ v ' (x)=⌊ H ' (x)⌋[T b ]−1 {δ v }=⌊ N '(x)⌋{δ v } δ v ' ' (x)=⌊ H ' ' (x)⌋[T b ]−1 {δ v }=⌊ N ' ' ( x )⌋{δ v}

Using the results above into the Hamilton’s principle, we get: t2

t2 L

∫ δ Πdt={δ v } ∫∫ T

t1

t1 0

(

)

{ }

∂2 v −EI {N ' '}⌊ N ' ' ⌋{v} dx dt=0 ∂ t2 + f ( x , t) {N }+ p1 {N }|x=0+ p2 {N }|x=L + m1 {N ' }|x=0 + m2 {N ' }|x= L − ρ A {N }⌊N ⌋

Note that: ⌊ N ⌋x=0 =⌊1 ⌊ N ' ⌋ x=0 =⌊0 ⌊N ⌋ x=L =⌊0 ⌊N ' ⌋x= L=⌊0

0 1 0 0

0 0 1 0

0⌋ 0⌋ 0⌋ 1⌋

Thus, the equation becomes:

{}

p1 ∫ ρ A {N }⌊N ⌋dx dd tv2 + EI {N ' '}⌊ N ' '⌋dx {v}=∫ f (x ){N }+ mp 1 0 0 2 m2 L

{ } 2

L

Or

[M ]{v¨ }+[ K ]{v}={q } Where [M] is called the mass matrix, [K] is called the stiffness matrix, and {q} is called the generalized force vector. Note that the generalized force vector includes the effect of the distributed Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

43

Dynamics of Beams loads as well as the concentrated ones. Also note that we can not write an explicit expression for the generalized force without knowing the force function. However, we may write:

{ } L

∫ f ( x , t)N 1 dx 0 L

{}

f 1 (t) f (t ) = 2 ∫ f ( x ,t) {N }= L0 f 3 (t ) 0 ∫ f ( x , t)N 3 dx f 4 ( t) 0

∫ f ( x , t)N 2 dx

L

L

∫ f ( x ,t) N 4 dx 0

Meanwhile, we may get the mass and stiffness matrices as:

[

156 22 L 54 −13 L 2 ρ A L 22 L 4L 13 L −3 L2 [M ]= 420 54 13 L 156 −22 L 2 −13 L −3 L −22 L 4 L2

[

12 6 L −12 6 L EI 6 L 4 L2 −6 L 2 L2 [K ]= 3 L −12 −6 L 12 −6 L 6 L 2 L2 −6 L 4 L2

5.3.3

]

]

Assembling the Structure Equations

Let’s consider a simple beam structure that is divided into two elements as in Figure 5.3.2. We should consider two things: 1. The displacement and slope at the second node of the first element is exactly the same as those at the first node of the second element. This should apply to ensure the continuity of the displacement and slope from one element to the neighboring one. 2. The concentrated loads maybe divided among the elements at the connecting node, but, the summation of the divided parts should add up to the total externally applied concentrated load.


44

Dynamics of Beams

Figure 5.3.2. Two-Element beam structures. We may write the element equation for each element to be: [M (1 ) ]{v¨ }(1) +[K (1) ]{v }(1)={q}(1) (2 )

(2)

(2)

(2)

(2)

[M ]{v¨ } +[K ]{v } ={q}

Each of the above equations consists of four scalar equations. Now, if we want to model the whole structure, we will need to add the generalized forces to obtain the total external concentrated loads. Keeping in mind that the displacements at the connecting node is the same, we can get the equations for the assembled structure to be:

[

]{ } ]{ } { } { }

v¨1 156 22 L 54 −13 L 0 0 2 2 v¨ ' 1 22 L 4L 13 L −3 L 0 0 ρ A L 54 13 L 312 0 54 −13 L v¨2 2 420 −13 L −3 L2 0 8L 13 L −3 L2 v¨ ' 2 0 0 54 13 L 156 −22 L v¨3 2 0 0 −13 L −3 L −22 L 4 L2 v¨ ' 3

[

12 6 L −12 6 L 0 0 2 2 6 L 4 L −6 L 2 L 0 0 EI −12 −6 L 24 0 −12 6 L + 3 2 2 2 0 8 L −6 L 2 L L 6 L 2L 0 0 −12 −6 L 12 −6 L 0 0 6L 2 L2 −6 L 4 L2

f (1) 1 v1 p1 (1) f v'1 2 m1 (1) (2) f + f v2 p = 3(1) 1(2) + 2 v'2 m2 f 4 +f 2 (2) v3 p3 f3 (2) v'3 m3 f4

Where we assumed that the element lengths, density, cross-section area, modulus of elasticity, and second moment of area are constant for both elements. Also, the superscripts (1) and (2) denote the information from the first and second element matrices respectively.


45

Dynamics of Beams

5.3.4

Applying the boundary conditions

Up to this moment, all the procedure described for the finite element model may apply to any beam with any number of elements. Differences will appear when different beams are subjected to different boundary conditions. For illustration purposes, we will consider two cases here.

Case of a cantilever beam: For a cantilever beam fixed from the left side, we have: v 1=0 and v ' 1=0 → v¨1=0 and v¨ ' 1=0 Thus, the equation we obtained may be written as:

[

]{ } ]{ } { } { }

156 22 L 54 −13 L 0 0 0 2 2 0 22 L 4L 13 L −3 L 0 0 ρ A L 54 13 L 312 0 54 −13 L v¨2 2 2 420 −13 L −3 L 0 8L 13 L −3 L2 v¨ ' 2 0 0 54 13 L 156 −22 L v¨3 2 0 0 −13 L −3 L −22 L 4 L2 v¨ ' 3

[

12 6 L −12 6 L 0 0 2 2 6 L 4 L −6 L 2 L 0 0 EI −12 −6 L 24 0 −12 6 L + 3 2 2 0 8 L −6 L 2 L2 L 6 L 2L 0 0 −12 −6 L 12 −6 L 0 0 6L 2 L2 −6 L 4 L2

f (1) 1 p1 0 (1) f 0 2 m1 (1) (2) v2 f +f p = 3(1) 1(2) + 2 v'2 m2 f 4 +f 2 (2) v3 p3 f3 v'3 m3 f (2) 4

Which may be separated into two distinct equations given as:

[

]{ }

312 0 54 −13 L v¨2 2 ρ AL 0 8L 13 L −3 L2 v¨ ' 2 420 54 13 L 156 −22 L v¨3 2 −13 L −3 L −22 L 4 L2 v¨ ' 3

[

]{ } { } { }

(2) f (1) v2 3 +f 1 p2 24 0 −12 6 L (1) (2) 2 2 f +f EI 0 v'2 m 8 L −6 L 2 L + 3 = 4 (2) 2 + 2 p3 L −12 −6 L 12 −6 L v 3 f3 2 2 (2) 6 L 2 L −6 L 4 L v'3 m3 f4

Which is called the primary equations. The primary equations could be solved for the time response of the unknown displacements and slopes using any exat or approximate techniques (usually numerical). The second equation becomes:


46

Dynamics of Beams

[

ρ A L 54 −13 L 0 0 420 13 L −3 L2 0 0

+

[

EI −12 6 L 0 0 2 L3 −6 L 2 L 0 0

]

]

{} v¨2 v¨ ' 2 v¨3 v¨ ' 3

{ } { }{ } v2 f (1) v '2 p = 1(1) + 1 v3 m1 f2 v '3

Which is called the secondary equations or the auxiliary equations. The secondary equations have the unknown variables as the time history of the concentrated force and moment at the fixed end, thus, they are the support reactions which are found by substituting the solution obtained from the primary equations directly.

Case of a simply-supported beam: For a simply-supported beam we have: v 1=0 and v 3=0 → v¨1=0 and v¨3=0 Thus, the equation we obtained may be written as:

[

]{ } ]{ } { } { }

156 22 L 54 −13 L 0 0 0 2 2 v '1 ¨ 22 L 4L 13 L −3 L 0 0 ρ A L 54 13 L 312 0 54 −13 L v¨2 2 2 420 −13 L −3 L 0 8L 13 L −3 L2 v¨ ' 2 0 0 54 13 L 156 −22 L 0 2 0 0 −13 L −3 L −22 L 4 L2 v¨ ' 3

[

12 6 L −12 6 L 0 0 2 2 6 L 4 L −6 L 2 L 0 0 EI −12 −6 L 24 0 −12 6 L + 3 2 2 0 8 L −6 L 2 L2 L 6 L 2L 0 0 −12 −6 L 12 −6 L 0 0 6L 2 L2 −6 L 4 L2

f (1) 1 p1 0 (1) f v'1 2 m1 (1) (2) f +f v2 p = 3(1) 1(2) + 2 v'2 m2 f 4 +f 2 (2) 0 p3 f3 v'3 m3 f (2) 4

which gives the primary equations as:


47

Dynamics of Beams

[

4 L2 13 L −3 L2 0 ρ A L 13 L 312 0 −13 L 2 2 420 −3 L 0 8L −3 L2 0 −13 L −3 L2 4 L2

[

]{ } v¨ ' 1 v¨2 v¨ ' 2 v¨ ' 3

]{ } { } { }

f (1) v '1 2 m1 4 L2 −6 L 2 L2 0 (1) (2) 0 6 L v2 f 3 +f 1 EI −6 L 24 p + 3 = (1) + 2 2 2 2 (2) 2 L 0 8 L 2 L v '2 m2 L f 4 +f 2 (2) 0 6 L 2 L2 4 L2 v ' 3 m3 f4

And the secondary equations as:

{}

v¨ ' 1 ρ A L 22 L 54 −13 L v¨2 0 420 0 54 13 L −22 L v¨ ' 2 v¨ ' 3

[

+

[

EI 6 L −12 6 L 0 3 L 0 −12 −6 L −6 L

]

]

{ } { }{ } v'1 f (1) v2 p 1 = (2) + 1 v'2 p3 f3 v'3

Which may be used to evaluate the support reactions at both ends.

5.4 Natural Frequencies and Time Response In the previous section we obtained the equations of motion of the beam using the finnite element method in the form: [M ]{v¨ }+[ K ]{v}={q } Which is the form we have for a MDOF system. From this form we may calculate the natural frequencies as well as obtain the response to harmonic excitation using the dynamic stiffness matrix. The code below evaluates the natural frequencies of a beam. The bounday conditions set are for a simply supported beam. %Clearing the memory and display clear all clc %Problem Data NE=5; %number of elements


48

Dynamics of Beams Length=1.0; %beam length Width=0.02; %beam width Thickness=0.001; %beam thickness Modulus=71e9; %Modulus of Elasticity Aluminum (GPa) Rho=2700; %Density (Kg/m^3) %Cross-section area Area=Width*Thickness; %Second moment of area Imoment=Width*Thickness*Thickness*Thickness/12; Le=Length/NE; %Element Length %Element stiffness matrix Ke=Modulus*Imoment*[12

,6*Le

,-12

,6*Le; ...

6*Le

,4*Le*Le ,-6*Le,2*Le*Le; ...

-12

,-6*Le

6*Le

,2*Le*Le ,-6*Le,4*Le*Le]/Le/Le/Le;

,12

,-6*Le; ...

%Element Mass matrix Me=Rho*Area*Le*[156

,22*Le

,54

,-13*Le; ...

22*Le, 4*Le*Le ,13*Le ,-3*Le*Le; ... 54

,13*Le

,156

,-22*Le; ...

-13*Le,-3*Le*Le ,-22*Le,4*Le*Le]/420; %Global matrices assembly %Initializing empty matrices KGlobal=zeros(2*(NE+1),2*(NE+1)); MGlobal=zeros(2*(NE+1),2*(NE+1)); FGlobal=zeros(2*(NE+1),1); %Global Force Vector %Assembling the global matrix for ii=1:NE KGlobal(2*ii-1:2*(ii+1),2*ii-1:2*(ii+1))= ... KGlobal(2*ii-1:2*(ii+1),2*ii-1:2*(ii+1))+Ke; MGlobal(2*ii-1:2*(ii+1),2*ii-1:2*(ii+1))= ... MGlobal(2*ii-1:2*(ii+1),2*ii-1:2*(ii+1))+Me; end %Applying the boundary conditions BCS=[1,2*(NE+1)-1]; %Simply Supported BCS KGlobal(BCS,:)=[]; KGlobal(:,BCS)=[]; MGlobal(BCS,:)=[]; MGlobal(:,BCS)=[]; FGlobal(BCS)=[];


49

Dynamics of Beams %Evaluating the Natural Frequencies vvv=sort(sqrt(eig(inv(MGlobal)*KGlobal)))

Adding a few lines to obtain the frequency response

%Evaluating the frequency response %The excitation will be through % a concentrated harmonic moment at the % left side (DOF# 1) %The measurement will be measured from % the displacement of the nede just before the % right support (DOF # (2*NE - 1)) FGlobal(1)=1; FREQ=[0]; for ii = 0:1000 FREQ(ii+1,1)=0.5*ii; KD=KGlobal - FREQ(ii+1,1)*FREQ(ii+1,1)*MGlobal; RESP=inv(KD)*FGlobal; Res(ii+1,1)=20*log(abs(RESP(2*NE-1))); endfor plot(FREQ,Res) grid

We get the frequency response shown in


50

Dynamics of Beams

Figure 5.4.1: Beam frequency response function (Log scale)

5.5 Newmark Method for Time Domain Solution 5.6 The Effect of Axial Loads For the special case of a beam with constant cross-section, we may write the homogeneous form of the PDE as: ρ A v¨ + EI

∂4 v ∂2 v + P =0 ∂ x4 ∂ x2

If we go through the finite element model derivation, we will get the equation of motion as: [M ]{v¨ }+([K ]−P[ K G ]){v }={q} Where [KG] is called the geometric stiffness matrix and one the element level, it will be:

[

36 3 L −36 3 L 1 3 L 4 L2 −3 L −L2 [K G ]= 30 L −36 −3 L 36 −3 L 3 L −L2 −3 L 4 L2

]


51

Dynamics of Beams We may use the static equation (removing the acceleration term) to evaluate the buckling load for a beam. However, since we are concerned with the dynamic problem, let’s examine the effect of the axial load on the natural frequencies of the beam. To perform this task, we will add the geometric stiffness matrix to the equation of motion and add a loop to change the values of the axial load and evaluate the minimum natural frequency. The loop will have the line listed below %Evaluating the Natural Frequencies for ii=0:100 PP=ii*0.015; PPP(ii+1)=PP; vvv(ii+1)=min(sort(sqrt(eig(inv(MGlobal)*(KGlobal-PP*KGGlobal))))); endfor

The result is plotted in the graph of Figure 5.6.1. Notice that the natural frequency with no load is the first point on the graph. Also, notice that the natural frequency reduces with the increase of the axial load until it reaches zero as the load reached the buckling load.

Figure 5.6.1: Change of minimum natural frequency of a simply-supported beam with axial load


52

Dynamics of Beams

5.7 The Fire-Hose Problem A very interesting problem in structure dynamics is that of a pipe conveying fluid. The dynamics of such a problem describes some important problems such as that of petroleum pipes. As the fluid flows inside the pipe, it creates thrust against the pipe, much like the axial force, this reduces the bending stiffness of the pipe and may cause buckling. Also, another phenomenon may appear in such problems, namely flutter. Flutter is defines as self excited vibration of a structure. This phenomenon can be seen clearly in cases where the fluid is traveling very fast inside a compliant (soft) pipe such as the fire-hose. In such cases, you may notice that the fire-hose starts moving around without any external excitation vibrating as the body of a snake.

To model this problem, we will use Hamilton’s principle to derive the equations of motion, then use the reduced form to create a finite element model to investigate the dynamic performance under different conditions. The potential energy in our problem will involve the elastic energy as well as the gravitational energy such that: L

U=∫ 0

( ( )

)

2

1 ∂2 v EI +(ρ p A p +ρf A f ) gv dx 2 ∂ x2

Where the subscript p denoted the pipe properties while the subscript f denotes the fluid. Obtaining the variation, we gt: L

( ( )( )

δ U=∫ EI 0

)

∂2 v ∂2 v δ +(ρ p A p+ρ f A f )g δ v dx ∂ x2 ∂ x2

Integrating by parts to get the boundary terms:

( ) ( )| ( ( ))

∂2 v ∂v δ U= EI δ 2 ∂x ∂x

L

∂2 v − ∂ EI ∂x ∂ x2 0

| ( L

L

δ v +∫ EI 0

0

)

( )

∂4 v δ v+(ρ p A p+ρf A f )g δ v dx ∂ x4

The kinetic energy will be: L

T=

(

(

)

)

2 1 ∂v 2 2 ρ A v +ρ A V + v ˙ +ρf A f V f dx ∫ p p ˙ f f f 2 0 ∂x

Obtaining the variation, we get: L

(

(

δ T =∫ (ρ p A p+ρf A f ) v˙ δ v˙ +ρf A f V f 0

2

( )

( )

∂v ∂v ∂v ∂v δ +V f v˙ δ +V f δ v˙ ∂x ∂x ∂x ∂x

))

dx

Performing integration by parts over x, we get:


53

Dynamics of Beams

(

∂v δ T =ρf A f V δ v +V f v˙ δ v ∂x L

(

2 f

(

+∫ (ρ p A p +ρf A f ) v˙ δ v˙ +ρf A f −V 2f 0

)|

L

0

∂2 v ∂ v˙ ∂v δ v−V f δ v+ V f δ v˙ 2 ∂x ∂x ∂x

))

dx

If we ignore writing the boundary terms explicitly, we get the total energy after performing the integration by part as:

t2

t2

t1

t1

(( L

∫ δ Π dt=∫ ∫

(

−(ρ p A p+ρf A f ) v¨ −ρf A f V 2f

0

4

∂2 v ∂v +2V f ˙ 2 ∂ x ∂x

∂ v −EI 4 −(ρ p A p +ρf A f )g ∂x

) )

)

δ v dx dt=0

From which we can get the equation of motion to be: (ρp A p +ρf A f ) v¨ + EI

2 ∂4 v ∂ v˙ 2 ∂ v +ρ A V +2ρ f A f V f =−(ρ p A p +ρ f A f )g f f f 4 2 ∂x ∂x ∂x

The first two terms of the above equation are the terms that we may get in case the fluid is not flowing inside the pipe. The third term describes the effect of the thrust created by the fluid flow similar to the term we had for axial loading of columns, while the last term describes the effect of the Coriolis acceleration due to the motion of the fluid inside a pipe that is also moving. On the right-hand side, the effect of the weight of the pipe and fluid appears as a static constant loading.

Obtaining the finite element model for the above problem, the equation of motion will be: [M ]{v¨ }+([K ]− ρ f A f V 2f [ K G ]){v }+2 ρ f A f V f [C ]{v˙ }={f g } In the above equation, the mass matrix will include the mass of the fluid as well as the pipe, while the stiffness and geometric stiffness matrices will not change. The Coriolis matrix will be: L

[C ]=∫ {N }⌊N ' ⌋dx 0

Which will be:

[

−30 6 L 30 −6 L 1 −6 L 0 6 L −L2 [C ]= 60 −30 −6 L −30 6 L 6L L2 −6 L 0

]

While the gravitational load vector will be given by:


54

Dynamics of Beams L

{f g }=−( ρ p A p + ρ f A f ) g∫ {N }dx 0

Which will be:

{}

6 ( ρ A + ρ A ) Lg L {f g }=− p p f f 12 6 −L

A typical plot of the change of the natural frequency with the fluid speed is shown in Figure 5.7.1. As you may notice, the first natural frequency reaches zero at a certain velocity, which indicate buckling occurring in the beam due to the fluid thrust. At a higher speed, two natural frequencies come to be equal which results in the dynamic instability (flutter). The results below are plotted for a simply supported pipe made of aluminum and conveying water. The pipe length is 2 m, internal radius 10 mm, wall thickness 1 mm. Water density is 1000 kg/m3, Aluminum density is 2700 kg/m3, modulusof ealsticity 71 GPa, and 10 elements were used to run the simulation.

Figure 5.7.1: Typical change of natural frequency with flow speed

5.8 Beams under Moving Loads


55

Dynamics of Beams


56

Dynamics of Membranes and Plates

6

Dynamics of Membranes and Plates


57

Periodic Structures for Vibration Attenuation

7


7.1 What is a periodic structure? A periodic structure is an expression we will b using to describe structures that are composed of identical substructures that are bonded in a periodically repeated manner. Periodic structures are seen in many engineering products, examples of periodic structures may include satellite solar panels, railway tracks, aircraft fuselage, multistory buildings, etc … Following the above definition of periodic structure, there must be a distinction between different substructures that defines the individual unit, that distinction or boundary will introduce a sudden change in the properties of the structure. Two main types of discontinuities may be identifies, namely: geometric discontinuity and material discontinuity. Figure 7.1.1 and Figure 7.1.2 show sketches of the two different types of discontinuities.

Figure 7.1.1 Periodic Structure with Material Discontinuity

Figure 7.1.2: Periodic Structure with Geometric Discontinuity

Recall what happens to a wave as it travels through a boundary between two different media (see Figure 7.1.3); part of the light wave refracts inside the water and another part reflects back into the air. Mechanical waves behave in a similar way! Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

58


Figure 7.1.3: Sketch of light wave behavior when incident on water surface

Now, imagine a rod, an example of 1-D structures. As the wave propagates through the rod, it faces a discontinuity in the structure. A part of the wave reflects and another part propagates into the new part. The reflected part of the wave will, definitely, interfere with the incident wave. The interference between the incident and reflected waves will result, in some frequency band, in destructive interference. In the frequency band where destructive interference occurs, there will be reduced vibration level. This band is what we call Stop-Band. Stop bands are the center of interest for the periodic analysis of structures.

7.2 Periodic Analysis Periodic structures can be modeled like any ordinary structure, but in a periodic structure, the study of the behavior of one cell is enough to determine the stop and pass bands of the complete structure independent of the number of cells. Recall that the equations of motion for a general 1-D structure (Figure 7.2.1) may be presented as:

[

]{ } [

]{ } { }

m11 m12 U¨ 1 k 11 k 12 U 1 F + = 1 ¨ m21 m22 U 2 k 21 k 22 U 2 F2


59


Figure 7.2.1: A sketch of a general 1-D structure

Where U is a vector presenting the displacements at a certain point in the structure, F is a general force vector; m and k are general mass and stiffness terms depending on the modeling method. For harmonic excitation, we may write:

[

]{ } { }

k 11−ω 2 m11 k 12 −ω 2 m12 U 1 F = 1 2 2 F2 k 12−ω m21 k 12 −ω m22 U 2

Which presents a dynamic stiffness matrix relation between the amplitudes of the excitation and the response of the structure. In a more compact form, we may write:

[

]{ } { }

D11 D12 U 1 F = 1 D21 D22 U 2 F2

If we expand the equations, we get: D11 U 1 + D12 U 2=F 1 D21 U 1 + D22 U 2 =F 2 Rearranging the terms, we may get: −1 U 2 =D−1 12 F 1 −D 12 D 11 U 1 −1 F2=D 21 U 1 + D 22( D−1 12 F 1 −D 12 D 11 U 1)

Or:

{ }[

]{ }

−D−1 D−1 U2 U1 12 D 11 12 = −1 −1 F2 D12−D22 D12 D11 D22 D12 F 1

The above equation relates the output of a structure unit to its input. In general, if this unit is a part of a periodic structure, the output of a cell will become the input to the following cell. Since we are working with identical cell, we may write: Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

60


{ } { } U2 U1 =eμ F2 −F 1

The negative sign associated with the force is to maintain the internal equilibrium of the structure. Using this relation into the input output relation, we get:

[

]{ } { }

−D−1 D−1 U1 U 12 D 11 12 =e μ 1 −1 −1 F1 −D12 + D22 D12 D11 −D 22 D12 F 1

Which, in linear algebraic expressions, says that when the matrix is multiplied by the vector, the result is a scale of the same vector; ie an eigenvalue problem. The matrix in the above equation is called the transfer matrix and the scalar  is called the propagation factor. Thus, we may write: e μ =Eigenvalues(T )= λ It can be proven that the eigenvalues of the transfer matrix of any structure (derived in fashion used above) will always appear in pairs of reciprocals ( λ and 1 / λ ). Note that the terms of the transfer matrix are functions of the excitation frequency, since we derived it from the dynamic stiffness matrix, thus, the eigenvalues will be functions of the excitation frequencies as well. Finally, we may write:

μ (ω )= ArcCosh

( λ +1/2 λ )

In general, the propagation factor  may be presented as:

μ =α +i β Notes: 1- If one of the eigenvalues is real then its reciprocal is also real and: |λ|≥1

or |1/ λ|≥1

In case they are positive, we directly get

μ (ω )= ArcCosh

( λ +1/2 λ )

Which is always a real number. However, if they are negative, get:

μ (ω )= ArcCosh

(|λ +12 / λ|)+i π

Which is a complex number with a constant imaginary part. If the propagation factor has a non-zero real part, that will mean that there is either a decay in magnitude between the input and output, or there is a magnification, indicating either destructive or


61

Periodic Structures for Vibration Attenuation constructive interference of the incident and reflected waves respectively. This is what we call the stop-band. 2- If the eigenvalues appear in a complex conjugate pair, the imaginary parts cancel one another, but, the real part has to have an absolute value less than one −1≤Real ( λ )≤1 , which results in a pure imaginary value for  that varies from 0 to p. If the propagation factor is an imaginary number, that indicates no change in the magnitude between the input and output. This is what we call the pass-band.

7.3 Discrete Periodic System 7.3.1

Case of identical masses

If identical masses are connected with identical springs, we have a discrete periodic system with a cell that may be presented as illustrated in Figure 7.3.1.

Figure 7.3.1: Discrete Periodic System Now, we may write down the equations of motion of the cell to be:

[m0 m0 ]{xx¨¨ }+[−kk −kk ]{xx }={ff } 1

1

1

2

2

2

If we assume harmonic excitations, and therefore responses, we get:

[

k −ω 2 m −k −k k −ω 2 m

]{ } { } X1 F = 1 X2 F2

Rearranging the terms of the two equations to get the input-output relation, we get:

[

ω 2m 1 − k k 2 2 ( k −ω m ) ω2m −k+ −1+ k k 1−

]

{ }{ } X1 X = 2 F1 F2


62

Periodic Structures for Vibration Attenuation And the transfer matrix for becomes:

[

ω2m 1 − k k 2 2 ( k−ω m ) ω2m k− 1− k k 1−

]{ } { } X1 μ X1 =e F1 F1

Thus, the eigenvalue problem becomes:

|

|

ω2m 1 −λ − k k =0 2 2 2 ( k−ω m ) ω m k− 1− −λ k k 1−

λ 2 +2

(

)

ω2m −1 λ +1=0 k

Which gives the eigenvalues to be:

) √(

(

ω 2m λ =− −1 ± k

)

2

ω2m −1 −1 k

Which gives us one of two cases: Case #1:

(

)

2

ω2m ω2m k −1 ≤1→−1≤ −1≤1→0≤ω 2≤2 k k m

In this case, the excitation frequency is between 0 and the natural frequency of single mass-spring system. This will give a complex conjugate pare of eigenvalues for which the propagation factor is equal to:

(

μ = ArcCosh 1−

)

(

ω2m ω2m =i ArcCos 1− k k

)

2 Which is an imaginary number varying from 0 to ip for the values of 0≤ω ≤2

k m

Case #2:

2

(

)

2

ω2m −1 ≥1→ k

ω m ω2m −1≥ −1( impossible) or ≥2→ k k k ω 2≥2 m Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

63

Periodic Structures for Vibration Attenuation In this case, the excitation frequency is more than the natural frequency of single mass-spring system. This will give a pair of real eigenvalues:

(

λ 1 =−

) √(

ω 2m −1 − k

) √(

(

ω 2m λ 2 =− −1 + k

)

2

ω 2m −1 −1 k

)

2

ω2m −1 −1 k

Check that λ 1∗λ 2 =1 For this case, the propagation factor is equal to:

((

μ = ArcCosh −

))

(

)

ω 2m ω2m −1 = ArcCoss −1 +i π k k 2

Which is a complex number with a constant imaginary part equal to ip for the values of ω ≥2

k m

The above results for a discrete system (Plotted in Figure 7.3.2 and Figure 7.3.3) agree with all the predictions we have introduced in 7.2.

Figure 7.3.2: Variation of the eigenvalues with frequency for a discrete system


64


Figure 7.3.3: Variation of the propagation factor with frequency for a discrete system Below, is a simple Octave program that generates the results above for mass of 1 kg and stiffness of 1 N/m.

m=1; k=1; mc=[m,0;0,m]; kc=[k,-k;-k,k]; for ii=1:1001 freq(ii)=(ii-1)*0.002; KD=kc-freq(ii)*freq(ii)*mc; TT=[-KD(1,1)/KD(1,2) 1/KD(1,2) KD(2,2)*KD(1,1)/KD(1,2)-KD(2,1) -KD(2,2)/KD(1,2)]; Lamda(:,ii)=sort(eig(TT)); Mew(ii)=acosh(0.5*(Lamda(1,ii)+Lamda(2,ii))); end subplot(2,1,1); plot(freq,Lamda(1,:),freq,Lamda(2,:)); grid subplot(2,1,2); plot(freq,real(Mew),freq,imag(Mew)); grid

7.3.2

Case of alternating masses

No let’s consider a case very similar to the one we have just covered in section 7.3.1. But in this case, the masses will be alternating between m kg and 2m kg. In this case, the cell will need to include a mass m in the middle, and it will have three degrees of freedom. Writing the equations of motion, we get:


65


[

]{ } [

m 0 0 x¨1 k −k 0 + 0 m 0 x¨2 −k 2k −k 0 0 m x¨3 0 −k k

]{ } { } x1 f1 = x2 0 f x3 3

Note that when writing the equations of motion, the force on the internal mass was taken to be zero. That is needed to maintain the periodicity as the f1 and f3 are considered as internal forces when connected to the adjacent cells, but if we have f2, that will not vanish. Using the harmonic excitation, we get:

[

k −ω 2 m −k 0 2 −k 2 k−ω m −k 0 −k k−ω 2 m

]{ } { } X1 F1 = X2 0 F X3 3

Now, we can use the second equation to eliminate the internal degree of freedom from the equation. This is called dynamic condensation and it preserves all the information of the system while reducing the size of the problem. From the second equation, we get: X 2=

k ( X 1+ X 3 ) 2 k−ω 2 m

Into the other two equations, we get:

[ [ |

( k−ω 2 m)2−k ω 2 m 2 k−ω 2 m k2 − 2 2 k−ω m

k2 2 k −ω 2 m (k−ω 2 m)2−k ω 2 m 2 2 k−ω m −

]

{ }{ } X1 F = 1 X3 F3

Rearranging the terms, we get the transfer matrix equation: (k−ω 2 m)2−k ω 2 m k2 (( k−ω 2 m)2−k ω 2 m)2 k2 − 2 2 2 2 k−ω m k (2 k−ω m)

2 k−ω 2 m − k2 (k −ω 2 m)2−k ω 2 m 2 k

]{ } { } | X1 X =e μ 1 F1 F1

From which we get the characteristic equation of the eigenvalue problem to be: (k−ω 2 m)2−k ω 2 m −λ k2 (( k−ω 2 m)2 −k ω 2 m)2 k2 − 2 k−ω 2 m k 2 (2 k−ω 2 m)

2 k−ω 2 m k2 =0 (k −ω 2 m)2−k ω 2 m −λ k2 −

After simplification, you get:


66

Periodic Structures for Vibration Attenuation λ 2 +2

k ω 2 m−(k−ω 2 m)2 λ +1=0 k2

Which has the solution: −k ω 2 m−(k −ω 2 m)2 λ= ± k2

√(

)

2

k ω 2 m−(k −ω 2 m)2 −1 k2

If we focus on the case of complex results, to get the pass bands, we get:

(

)

2

k ω 2 m−(k−ω 2 m)2 ≤1→ k2 ω 2 m ω 2m 2 0≤3 −( ) ≤2→ k k ω 2m ω 2m ω 2m 0≤ ≤3 AND ≤1 AND ≥2→ k k k ω 2m ω 2m 0≤ ≤1 OR 2≤ ≤3 k k Which indicates that we have two pass bands.

7.4 The Reverse Approach in Periodic Analysis So far, we have been working with the model of the periodic structure using the dynamic stiffness matrix, which includes the excitation frequency, then obtaining the relations for the propagation factor. This approach starts from known physical quantity, excitation frequency, and ends up with a mathematical quantity, propagation factor, that presents the pass and stop band of the periodic system. We may call this approach forward approach in periodic analysis.

In contrast, we have the reverse approach. In the reverse approach, we start by setting the relations in terms of the propagation factor, then obtain the frequencies that generate the it (which is much like the natural frequencies). Let’s illustrate using the problem of section 7.3.1.

7.4.1

Reverse approach for a discrete system

Recall from the problem of section 7.3.1 that the dynamic stiffness matrix relation was found to be:

[

k −ω 2 m −k −k k −ω 2 m

]{ } { } X1 F = 1 X2 F2

Meanwhile, we had the input-output relation as: Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

67


{} { } X2 X1 =e μ F2 −F1

Thus, the equations of motion may be written as: ( k−ω 2 m) X 1−k e μ X 1=F 1 −k X 1 +( k−ω 2 m) emu X 1=−e mu F 1 Manipulating the second equation: (k −ω 2 m) X 1−k e μ X 1 =F 1 −k e−μ X 1+(k −ω 2 m) X 1=−F 1 Rearranging erms and adding both equations: ((2−e μ−e− μ )k−2 ω 2 m) X 1=0 Which will give a trivial solution unless ((2−e μ−e− μ )k−2 ω 2 m)=0 Which results in: k ω 2= (1−cosh( μ )) m For this relation, we have one of three cases: Case #1: If m is an imaginary number. In this case, the relation will become: k k ω 2= (1−cos( μ ))→0≤ω 2≤2 m m Case #2: If m is a complex number with the real part constant at the value of ip. This will make the relation change to be: k ω 2= (1+cosh (real ( μ ))) m Case #3: Any other real or complex value of m will result in a non-real value of the excitation frequency.

Note: The relations we got in Cases 1 & 2 above are the inverse of the relations we got section 7.3.1.

7.4.2

Generalization of the Inverse Periodic approach

In the dynamic stiffness matrix relation, we can categorize the degrees of freedom into three categories:


68

Periodic Structures for Vibration Attenuation 1. Internal degrees of freedom that are associated with no inputs or outputs of the cell 2. Output degrees of freedom that are responsible for transferring the wave to the adjacent cell 3. Input degrees of freedom that are responsible to the excitation of the cell In the forward approach for the periodic analysis, we needed to get a transfer matrix that relates the input and output degrees of freedom. Thus, if we had any internal degrees of freedom, we needed to condense them to keep the relation with only the input and output degrees of freedom. The resulting relation starting by determining the excitation frequency then we end up evaluating the propagation factor.

Now, let’s reconsider the equations of motion we wrote for the general system of section 7.2.

[

]{ } { }

k 11−ω 2 m11 k 12 −ω 2 m12 U 1 F = 1 2 2 F2 k 12−ω m21 k 12 −ω m22 U 2

If we introduce the relation:

{ } { } U2 U1 =eμ F2 −F 1

Substituting, we get:

[

k 11 −ω 2 m11 k 12 e μ −ω 2 m12 e μ k 12 e−μ −ω 2 m21 e−μ k 12 −ω 2 m22

]{ } { } U1 F1 = U1 −F1

Rearranging the terms and adding both equations:

( k 11 +k 22 +k 12 e μ +k 21 e−μ −ω 2 (m11 +m22 +m12 e μ +m21 e−μ )) U 1 =0 Which may be written in compact form as:

( K ( μ )−ω 2 M ( μ )) U 1=0 Which is an eigenvalue problem in the excitation frequency. Note: Up to this moment, the inverse approach has not introduce any advantage to the periodic analysis. This is true. There is an implicit assumption that we have been using in the forward approach; that is that the number of degrees of freedom on the input side is equal to that on the output side of the cell. However, in the reverse approach, we do not need that assumption since we do not have to invert


69

Periodic Structures for Vibration Attenuation any of the stiffness or mass matrices (or sub-matrices) during the analysis. This will give a great advantage when the periodic structure becomes two or three-dimensional. In the reverse approach, however, you have to consciously select the values of the propagation factor that you use to evaluate the excitation frequency. This may be a price we are gladly willing to pay in order to be able to perform our periodic analysis.


70


7.5 Periodic Bars Consider the cell presented in Figure 7.1.1 or Figure 7.1.2. Each of those cell can represent a cell in a periodic bar. If we use finite element modeling for the dynamics of the cell, each cell will need to be presenter by at least two elements, each presenting a part one part of the cell. Note: Later, we will present an argument for the use of more than one element, or higher order elements, for each of the cell parts.

Figure 7.5.1 presents a free-body diagram for each of the parts of the cell. Using the finite element modeling, we may write the equations of motion as:

[

]{ } [

]{ } { }

2 m1 m1 0 u¨1 k1 −k 1 0 u1 f1 m1 2 m1 +2 m2 m2 u¨2 + −k 1 k 1+ k 2 −k 2 u 2 = 0 f3 0 m2 2 m2 u¨3 0 −k 2 k2 u3

Figure 7.5.1: Free-body diagram for the two elements of a bar cell Where: A 1 ρ 1 L1 6 A 2 ρ 2 L2 m 2= 6 E1 A 1 k 1= L1 E A k 2= 2 2 L2 m 1=

Where E, A, r, & L are the modulus of elasticity, cross-section area, density, and element length for each of the elements respectively. (note that up to this moment we are allowing for material and geometric discontinuities simultaneously). The dynamic stiffness matrix relation will become: Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

71


[

]{ } { }

k 1−2 ω 2 m1 −k 1−ω 2 m1 0 U1 F1 2 2 2 = −k 1−ω m1 k 1 +k 2 −2 ω (m1+ m2 ) −k 2−ω m2 U 2 0 2 2 F U 0 −k 2−ω m2 k 2−2 ω m 21 3 3

Now, we may follow the forward or the reverse approach (since they are both equivalent). Let us start with the forward approach. To proceed with the forward approach analysis, we will performed dynamic condensation on the dynamic stiffness matrix in order to get rid of the internal degree (s) of freedom:

[

(k 1 +ω 2 m1)U 1+(k 2+ ω 2 m2 )U 3 U2= → k 1 +k 2−2 ω 2 (m1+ m2 ) (k 1 + ω 2 m1)2

(k 1−2 ω 2 m1)−

k 1 +k 2−2 ω 2 (m1+ m2 ) −(k 1 + ω 2 m1)( k 2 +ω 2 m 2) 2

k 1 +k 2−2 ω (m1+ m2 )

−(k 1 +ω 2 m1)(k 2+ ω 2 m2 ) k 1+ k 2−2 ω 2 (m1 +m2) (k 2 +ω 2 m2)2 (k 2−2 ω 2 m2 )− k 1 +k 2−2 ω 2 (m1 +m2)

]{ } { } U1 F = 1 U3 F3

Following the procedure presented in section 7.2, we may write: ( k 1+ ω 2 m1 )2 k 1+ k 2−2 ω 2 (m1 +m2) −( k 1 +ω 2 m1)(k 2 + ω 2 m 2 )

D11=(k 1−2 ω 2 m1)− D21 =

k 1+ k 2−2 ω 2 (m1 +m2) D 11 D 12 D 21 D 22

[

−(k 1 + ω 2 m 1)( k 2+ ω 2 m2 ) k 1 +k 2−2 ω 2(m1 +m2 ) (k 2 + ω 2 m2)2 2 D22 =(k 2−2 ω m2)− k 1 +k 2 −2 ω 2 (m1+ m2 ) U1 F = 1 U3 F3 D12 =

]{ } { }

From which we may get the transfer function relation to be:

[

]{ } { }

−D−1 D−1 U1 U 12 D 11 12 =e μ 1 −1 −1 F1 −D21 + D22 D12 D11 −D 22 D12 F1

Which will yield the eigenvalues we need to evaluate the propagation factor. On the other hand, if we decide to proceed with the reverse approach, we will need to remove the output degrees of freedom in terms of the input degrees of freedom which will give:

[

]{ } { }

(k 1 +k 2)−2 ω 2 ( m1 +m2) −( k 1+ k 2 e−μ )−ω 2( m1 +m2 e−μ ) U 1 = 0 μ 2 μ 2 0 U2 −(k 1 +k 2 e )−ω ( m1 +m2 e ) (k 1 +k 2)−2 ω (m1 +m2)

To get a non-trivial solution, we will get:


72


|[

] [

]|

k 1+ k 2 −k 1+ k 2 e−μ (m1 +m 2 e− μ ) 2 2(m1 +m 2 ) −ω =0 μ μ −k 1+ k 2 e k1 + k2 (m1+ m2 e ) 2(m1+ m2)

Which gives a characteristic equation that may be solved for the excitation frequencies given the value of the propagation factor. Note: As you may expect from the relations we got in the forward and the reverse approach, writing an explicit expression for the relations between the excitation frequency and the propagation factor will become extremely complex. Thus, we will depend on the general formulation to prepare a computer code that handles the problem in a much faster way

To run a program, we will need to set some values. The material of the bar is selected to be Aluminum with modulus of elasticity of 71 GPa and density 2700 kg/m3 with the cross sections are circular with the smaller diameter to be 4 cm and the larger 4 √ 2 cm. The length of each part of the cell will be taken as 1 m.

Figure 7.5.2: The eigenvalue variation of the bar cell Figure 7.5.2 shows the variation of the real part of the eigenvalues, obtained from the forward approach, with the frequency, while Figure 7.5.3 shows the variation of the real and imaginary parts of the propagation factor.


73


Figure 7.5.3: The propagation factor variation for the bar cell The octave code used to plot the above graphs is listed below. Rho=2700; A1=0.02*0.02*pi; A2=2*A1; Ee=71e9; Le=1; m1=Rho*A1*Le/6; k1=Ee*A1/Le; m2=Rho*A2*Le/6; k2=Ee*A2/Le; mc=[2*m1,m1,0;m1,2*m1+2*m2,m2;0,m2,2*m2]; kc=[k1,-k1,0;-k1,k1+k2,-k2;0,-k2,k2]; for ii=1:1001 freq(ii)=(ii-1)*20; KD3=kc-freq(ii)*freq(ii)*mc; X21=-KD3(2,1)/KD3(2,2); X23=-KD3(2,3)/KD3(2,2); KD=[KD3(1,1)+KD3(1,2)*X21, KD3(1,3)+KD3(1,2)*X23; ... KD3(3,1)+KD3(3,2)*X21, KD3(3,3)+KD3(3,2)*X23]; TT=[-KD(1,1)/KD(1,2) 1/KD(1,2) KD(2,2)*KD(1,1)/KD(1,2)-KD(2,1) -KD(2,2)/KD(1,2)]; Lamda(:,ii)=(eig(TT)); Mew(ii)=acosh(0.5*(Lamda(1,ii)+Lamda(2,ii))); end figure(1); plot(freq,Lamda(1,:),freq,Lamda(2,:)); grid figure(2); plot(freq,real(Mew),freq,abs(imag(Mew))); grid

If we plot the results of the excitation frequency variation with the propagation factor, we get the plots of Figure 7.5.4 and Figure 7.5.5. Comparing the plots of both approaches we can readily notice that the results are identical to what we expected. Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

74


Note: when using finite element analysis to evaluate the natural frequencies of a structure it is always recommended to use twice, or more, as many degrees of freedom as the number of frequencies you are interested in. As the periodic analysis problem involves eignvalue problems that are related to the excitation frequencies, we will need to use more degrees of freedom than the number of stop bands we are interested in investigating. Since we analyze, only, one cell of the structure, we will need to introduce internal degrees of freedom by using more than one element per part of the cell or by using higher order elements (either cases introduce extra degrees of freedom). Indeed when comparing the results with experimental results, the higher order elements were able to capture the dynamics more accurately.

Figure 7.5.4: Variation of the excitation frequencies with the imaginary part of the propagation factor


75


Figure 7.5.5: Variation of the excitation frequencies with the real part of the propagation factor


76


7.6 Periodic Beams 7.6.1

Numerical Analysis

The periodic beam problem differs from the above problems in that each node in the beam element has two degrees of freedom, namely, the displacement and the slope. Thus all the elements of the matrix will be sum matrices of size 2x2. Similarly, the displacements will be a 2x1 vector. Hence the transfer matrix will be a 4x4 one yielding 4 eigenvalues instead of two. The eigenvalues will still appear in couples as described before, but, two of them will always be real (as observed from the numerical results). According to section 5.3, have a two-element beam as in Figure 7.6.1.

Figure 7.6.1. Two-Element beam structures. And we may write the element equation for each element to be: [M (1 ) ]{v¨ }(1) +[K (1) ] {v }(1)={f }(1) [M (2 ) ]{v¨ }(2) +[K (2) ]{v }(2)={f }(2) And the assembled equations may be written as:

[

]{ } [

]{ } { }

M (1) M (1) 0 K (1) K (1) 0 v¨1 v1 f1 11 12 11 12 (1) (1) (2) (2) (1) (1) (2) (2) M 21 M 22 + M 11 M 12 v¨2 + K 21 K 22 + K 11 K 12 v 2 = f 2 v¨3 v3 f3 0 M (2) M (2) 0 K (221) K(2) 21 22 22

Where each element of the above matrices is a sub-matrix of size 2x2, and all the elements of the vectors are 2x1 vectors. Using the case of harmonic excitation, we may write the above equation as a dynamic stiffness matrix relation:

[

]{ } { }

K D 11 K D 12 0 V1 F1 K D 21 K D 22 K D 23 V 2 = F2 0 K D 32 K D 33 V 3 F3

Now, we need to set F2=0, then condense the matrix by removing V2, to get:


77

Periodic Structures for Vibration Attenuation V 2=−K −1 D 22 (K D 21 V 1+ K D 23 V 3) Which reduces the above equation in become:

[

]{ } { }

K D 11 −K−1 −K −1 V1 F D 22 K D 21 D 22 K D 23 = 1 −1 −1 F3 −K D 22 K D 21 K D 33−K D 22 K D 23 V 3

Or:

[

]{ } { }

D11 D12 V 1 F = 1 D21 D22 V 3 F3

thus, we may proceed as before to get the transfer matrix and evaluate the eigenvalues and propagation factor. Figure 7.6.2 and Figure 7.6.3 show the variation of the eigenvalues and the propagation factor for a beam cell respectively. The beam cell had modulus of elasticity of 71 GPa and density of 2700 kg/m3, width of 5 cm, each element had a length of 5 cm, while the thickness of the first element was 1 mm and the second 3 mm.

Figure 7.6.2. Eigenvalues (lowest three) of a beam cell


78


Figure 7.6.3. Propagation factor (real and imaginary) for a beam cell

The Octave code used to produce the above plots is listed below. Notice that, in the codes, we used the Log function instead of the Acosh function. That is because we can not guarantee the order of the eigenvalues we get, thus we can not know which couple to use in the Acosh function. Also, the forth eigenvalue is quite large, thus, it is not plotted in order to be able to see the other three clearly. Finally, only the second propagation factor is plotted instead of the four obtained, that is because one is the same as the one plotted and the other two will always be real numbers. Rho=2700; Ee=71e9; B1=0.05;T1=0.001;L1=0.05; B2=0.05;T2=0.003;L2=0.05; A1=B1*T1; A2=B2*T2; I1=B1*T1*T1*T1/12; I2=B2*T2*T2*T2/12; %Element stiffness matrix K1=Ee*I1*[12 ,6*L1 ,-12 ,6*L1; ... 6*L1 ,4*L1*L1 ,-6*L1,2*L1*L1; ... -12 ,-6*L1 ,12 ,-6*L1; ... 6*L1 ,2*L1*L1 ,-6*L1,4*L1*L1]/L1/L1/L1; K2=Ee*I2*[12 ,6*L2 ,-12 ,6*L2; ... 6*L2 ,4*L2*L2 ,-6*L2,2*L2*L2; ... -12 ,-6*L2 ,12 ,-6*L2; ... 6*L2 ,2*L2*L2 ,-6*L2,4*L2*L2]/L2/L2/L2; %Element Mass matrix M1=Rho*A1*L1*[156 ,22*L1 ,54 ,-13*L1; ... 22*L1, 4*L1*L1 ,13*L1 ,-3*L1*L1; ...


79

Periodic Structures for Vibration Attenuation 54 ,13*L1 ,156 ,-22*L1; ... -13*L1,-3*L1*L1 ,-22*L1,4*L1*L1]/420; M2=Rho*A2*L2*[156 ,22*L2 ,54 ,-13*L2; ... 22*L2, 4*L2*L2 ,13*L2 ,-3*L2*L2; ... 54 ,13*L2 ,156 ,-22*L2; ... -13*L2,-3*L2*L2 ,-22*L2,4*L2*L2]/420; KG=zeros(6,6); MG=KG; KG(1:4,1:4)=K1; KG(3:6,3:6)=KG(3:6,3:6)+K2; MG(1:4,1:4)=M1; MG(3:6,3:6)=MG(3:6,3:6)+M2; for ii=1:1001 freq(ii)=(ii-1)*20; KD3=KG-freq(ii)*freq(ii)*MG; X21=-inv(KD3(3:4,3:4))*KD3(3:4,1:2); X23=-inv(KD3(3:4,3:4))*KD3(3:4,5:6); KD=[KD3(1:2,1:2)+KD3(1:2,3:4)*X21, KD3(1:2,5:6)+KD3(1:2,3:4)*X23; ... KD3(5:6,1:2)+KD3(5:6,3:4)*X21, KD3(5:6,5:6)+KD3(5:6,3:4)*X23]; TT=[-inv(KD(1:2,3:4))*KD(1:2,1:2) , inv(KD(1:2,3:4)) KD(3:4,3:4)*inv(KD(1:2,3:4))*KD(1:2,1:2)-KD(3:4,1:2), -KD(3:4,3:4)*inv(KD(1:2,3:4))]; Lamda(:,ii)=sort(eig(TT)); Mew(:,ii)=log(Lamda(:,ii)); end figure(1) plot(freq,Lamda(1:3,:))%,freq,Lamda(4,:)) axis([freq(1),freq(1001),-5,15]) grid xlabel("Frequency (rad/sec)") ylabel("Eigenvalues") figure(2) plot(freq,abs(real(Mew(2,:))),freq,abs(imag(Mew(2,:)))) grid xlabel("Frequency (rad/sec)") ylabel("Propagation Factor")

The reverse analysis of a beam is not as easy as that for a discrete system or a bar, thus we will not go into it in this text.

7.6.2

Experimental Analysis

In order to develop more understanding of the of the behaviour of the periodic beams as well as developing a numerical model to study its characteristics, an experiment was set for a periodic beam with free-free boundary conditions (Figure 7.6.4).


80


Figure 7.6.4. A picture tup of the periodic beam experiment. The beam is aluminum beam which is 40 cm long and 5 cm wide with 1 mm thickness. The periodicity was introduced onto the beam by bonding 5 cm by 5 cm pieces of the same material on both surfaces separated by 5 cm (Figure 7.6.5). The beam is then suspended by a thin wire from one of its end to simulate free-free boundary conditions. Thus, the beam is set up with four identical cells each of which has free-free boundary conditions.


81


Figure 7.6.5. A sketch for one cell of the periodic beam The beam is then excited by a piezostack (model AE0505D16 NEC Tokin, Union City, CA, 94587) at one end and the measurement was taken by an accelerometer from the other end (Figure 7.6.6 and Figure 7.6.7).

The experiment described above was set up and measurements were taken from the two ends of the beam. Figure 7.6.8 shows the transfer function frequency response of the beam for the plain and periodic beams. The attenuation factor of the beam, as calculated by the real part of the propagation factor of the periodic model, is plotted below the frequency response for the sake of comparison. The results shown emphasize the accuracy of the periodic model used to predict the behavior of the beam. Figure 7.6.9 presents the frequency response obtained by the finite element model of the described beam. Comparing the results of both figures, we can note clearly the consistency of results obtained by the three models, experimental, periodic and finite element.


82


Figure 7.6.6. The excitation piezostack Figure 7.6.7. The output accelerometer 40

2.5

20

2

1.5 -20 1

Attenuation Factor

Respence Ampl. (dB)

0

Plain Beam

-40

Periodic Beam Attenuation Factor -60

-80 0

1000

2000

3000

4000

5000

0.5

0 6000

Frequency (Hz)

Figure 7.6.8. The frequency response together with the numerical results of the stop bands for the proposed beam.


83


Figure 7.6.9. Frequency response of the beam using finite element model.


84

Periodic Structures for Vibration Attenuation Another experiment was set up for a set of beams with cantilever boundary conditions. The experiments was set up with two accelerometers and excited by a piezoelectric actutator as shown in Figure 7.6.10 and Figure 7.6.11. Different cases with varying the lengths L1 and L2 were constructed to examine the effect of the geometry on the attenuation characteristics (Figure 7.6.12).

Figure 7.6.10. Sketch of the experimental setup for the cantilever beam.

Figure 7.6.11. A picture of the experimental setup.

Figure 7.6.12. A sketch for the cell geometry of the experiment for the cantilever beam.


85

Periodic Structures for Vibration Attenuation 10 9

20

8 10 7 0 0

500

1000

1500

2000

2500

3000

3500

6 4000

-10

5 4

-20

Plain Beam Periodic Beam Attenuation Factor

-30

-40

A t t e n u a t in F a c t o r ( ra d )

T ra n s f e r F u n c t io n A m p lit u d e ( d B )

30

3 2 1

-50

0 Frequency (Hz)

Figure 7.6.13. Experimental results obtained for case #1 compared to plain beam and attenuation curves obtained by numerical model.

20

10

8 0 0

500

1000

1500

2000

2500

3000

3500

7 4000 6

-10

5 -20

4


-30

A tte n u atin F a c to r (ra d )

T ra n s fer F u n c tio n A m p litu d e (d B )

9 10

3 2

-40 1 -50

0 Frequency (Hz)

Figure 7.6.14. Experimental results obtained for case #2 compared to plain beam and attenuation curves obtained by numerical model.


86

Periodic Structures for Vibration Attenuation 10 9 10 8 0 0

500

1000

1500

2000

2500

3000

3500

7 4000 6

-10

5

-20

-30

-40

4

A tte n u a tin F a c to r (ra d )

T ra n s fe r F u n c tio n A m p litu d e (d B )

20

3


2 1

-50

0 Frequency (Hz)

Figure 7.6.15. Experimental results obtained for case #3 compared to plain beam and attenuation curves obtained by numerical model. As you may see from the results of the above two experiments. The periodic analysis of the beams can accurately predict the results of the experiments. Also, you may notice the considerable reduction in the vibration amplitude of the periodic beams compared to the pain ones.


87


7.7 Further Reading For the purpose of this text, we will not discuss further details about periodic structures. Nevertheless, numerous publications have used the concept for vibration attenuation in bars, beams, plates, and shells. The work covered a broad band of theoretical, numerical, and experimental studies. In the following, you may find a useful list of publications for further reading.

[1]

Mead, D. J. (1999). Passive vibration control. John Wiley & Sons Inc.

[2]

Asiri, S., Baz, A., & Pines, D. (2006). Active periodic struts for a gearbox support system. Smart materials and structures, 15(6), 1707.

[3]

Asiri, S., & Al-Bassyiouni, M. (2011). Vibration attenuation using passive periodic struts. International Journal of Vehicle Noise and Vibration, 7(1), 1-15.

[4]

Xu, M. B., & Huang, L. (2002). Control of multi-span beam vibration by a random wave reflector. Journal of sound and vibration, 250(4), 591-608.

[5]

Asiri, S. (2007). Vibration attenuation of automotive vehicle engine using periodic mounts. International Journal of Vehicle Noise and Vibration, 3(3), 302-315.

[6]

Tawfik, M. (2008). Dynamics and stability of stepped gun-barrels with moving bullets. Advances in Acoustics and vibration, 2008.

[7]

AboElSooud, M. T. (2003). Vibration control of plates using periodically distributed shunted piezoelectric patches (Doctoral dissertation, University of Maryland, College Park).

[8]

Talaat, S., Desoki, A., Tawfik, M., & Negm, H. M. (2012). EXPERIMENTAL AND NUMERICAL INVESTIGATION OF THE CAPABILITIES OF PERIODIC PLATES FOR VIBRATION ATTENUATION. AMME-15, Cairo, Egypt.

[9]

Tawfik, M., Chung, J., & Baz, A. (2004, April). Wave attenuation in periodic helicopter blade. In 5th Jordan International Mechanical Engineering Conference, Amman, Jordan (pp. 26-28).

[10] El-Din, M. A., & Tawfik, M. (2006, July). Vibration attenuation in rotating beams with periodically distributed piezoelectric controllers. In Proceedings of the 13th International Congress on Sound and Vibration (ICSV’06). [11] Mead, D. J., & Yaman, Y. (1991). The response of infinite periodic beams to point harmonic forces: a flexural wave analysis. Journal of Sound and Vibration, 144(3), 507-529.


88

Attenuation of Structure Dynamics

8

Attenuation of Structure Dynamics

8.1 Viscoelastic Materials

8.2 Passive Control Using Piezoelectric Materials

8.3 Active Control Using Piezoelectric Materials

8.4 Magnetorheological Fluids and Damping


89

Aeroelasticity

9

Aeroelasticity

9.1 What is Aeroelasticity? 9.2 Static Aeroelastic Problems 9.2.1

Divergence

9.2.2

Control Reversal

9.3 Dynamic Aeroelasticity – Wing Flutter

9.4 Dynamic Aeroelasticity – Panel Flutter 9.4.1

Introduction

Panel-flutter is defined as the self-excited oscillation of the external skin of a flight vehicle when exposed to airflow along its surface [1] . As early as late 1950’s, the problem of panel-flutter has drawn the researchers’ attention, but it has not been of high interest until the evolution of the National Aerospace Plane, the High Speed Civil Transport (HSCT), the Advanced Tactical Fighter projects, and F-22 fighter [2] [3] [4] among other projects of high-speed vehicles.

At high speed maneuvering of flight vehicles, the external skin might undergo self-excited vibration due to the aerodynamic loading. This phenomenon is known as panel-flutter. Panel-flutter is characterized by having higher vibration amplitude in the third quarter length of the panel (Figure 9.4.1). This phenomena causes the skin panels of the flying vehicle to vibrate laterally with high amplitudes that cause high in-plane oscillatory stresses; in turn, those stresses cause the panel to be a subject to fatigue failure.


90

Aeroelasticity

Figure 9.4.1. Sketch of panel-flutter phenomenon It is thus required to determine the panel-flutter boundary (critical dynamic pressure) as well as the amplitude of vibration at which the panel will be oscillating in post-flutter conditions. Linear analysis of the panel structure could be used to predict the critical dynamic pressure of flutter, but is limited to that, as post flutter vibration is characterized by being of high amplitude which needs the application of nonlinear modeling techniques to describe.

The linear analysis though, predicts exponential growth of the amplitude of vibration with the increase of the dynamic pressure in a post flutter condition. However, it is worth noting that under those conditions, the vibration of the panel will be influenced by inplane as well as bending stresses that lead to the limit cycle oscillation. Thus, failure of the panel does not occur at post-flutter dynamic pressures, but extended exposure to the panel-flutter decreases the fatigue life of the panel.

One of the earliest reviews that covered the topic of panel flutter was that presented by Dowell in 1970 [5] . It covered the variety of available literature dealing with the problem. Analytical and numerical methods predicting the flutter boundary as well as the limit cycle amplitude for panels (beam models) were presented as well as the effect of presence of in-plane loading. Later, Bismarck-Nasr [6] [7] presented reviews of the finite element analysis of the aeroelastic problems of plates and shells. The study also included the cases of in-plane loading and its effects on the problem. Lately, Mei et al. [8] presented another review that covered the topics of analytical and numerical analysis of panel flutter together with the topics involving the flutter control and delay.

Different methods were used to predict the post-flutter (limit cycle) attitude, which is a nonlinear phenomenon by nature, of the panel; modal transformation approach with direct numerical integration, harmonic balance, perturbation method, and nonlinear finite element method [1] [9] were used for that purpose.


91

Aeroelasticity The aerodynamic loading on the panel was also predicted using different approaches; unsteady supersonic potential flow [10] , linearized potential flow[1] [4] , and quasi-steady piston theory. The most popular of which is the first order quasi-steady piston theory that was introduced by Ashley and Zartarian[11] . That approximate theory gives high accuracy results at high Mach numbers (M ∞>1.6).

At the flight condition of panel-flutter (usually supersonic flight conditions), the phenomenon is associated with elevated temperatures, produced from the aerodynamic heating through the boundary layer friction and the presence of shock waves. This heating adds to the complexity of the problem by introducing panel stiffness reduction and thermal loading, which might also be associated with post-buckling deflection. In the following, a literature review of panel-flutter analysis and control topics is presented.

9.4.2

Panel-Flutter Analysis

The non-linear finite element formulation, introduced by Mei[1] , was the basis on which Dixon and Mei [12] , Xue and Mei [3] , and Abdel-Motagalay et al. [13] , build their finite element models to analyze the flutter boundary, limit cycle, and the thermal problems, with the extension to random loading and SMA embedding introduced by Zhong [14] .

Different finite element models were developed to analyze the behavior of panels subject to aerodynamic loading. Mei [1] introduced the use of nonlinear finite element methods to predict the behavior of isotropic panels in the limit-cycle oscillations (LOC). In his work, he used the quasisteady first order piston theory to predict the aerodynamic loading for M∞>1.6. He also presented a comparison of the effect of different structural boundary conditions on the critical dynamic pressure and on oscillation amplitude. Dixon and Mei [12] extended the use of finite element nonlinear analysis to composite panels. von Karman strain-displacement relations were used to represent the large deflections and the aerodynamic load was modeled using quasi-steady first-order piston theory. They solved the equations of motion using the linear-updated mode with a nonlinear time function (LUM/NTF) approximation. Results were also presented for different boundary conditions.

Model enhancements were also developed to include the thermal effects as well as the flow direction. Xue and Mei [3] presented a very good study on the combined effect of aerodynamic forces and thermal loads on panel-flutter problems. They studied the effect of the temperature elevation on the critical dynamic pressure as well as the buckling temperature variation under different dynamic pressure conditions. The study also showed the effect of the different boundary conditions on the amplitude of the limit cycle oscillations (LOC). Abdel-Motagalay et al.[13] studied the effect of flow direction on the panel-flutter behavior using first order shear deformation


92

Aeroelasticity theory for laminated composite panels. They formulated the finite element nonlinear equations in structural node degrees of freedom, and then reduced the number of equations using a modal transformation. The resulting reduced equations were then solved using the LUM/NTF approximation.

Sarma and Varadan [9] adopted two methods to solve the nonlinear panel-flutter problem, the starting point of the first solution was calculated using nonlinear vibration mode and in the second they used linear mode as their starting point. They derived the equations from energy considerations using Lagrange’s equations of motion, and then reduced the equations to nonlinear algebraic equations to solve a double eigenvalue problem.

Frampton et al. [4] [15] applied the linearized potential flow aerodynamics to the prediction and control of the flutter boundary using discrete infinite impulse response (IIR) filters with conventional [15] and modern [4] control theories. To get their solutions, they increased the nondimensional dynamic pressure and calculated the coupled system eigenvalues until the point of coalescence at which the first complex eigenvalue appears. They studied the linear panel-flutter problem, thus, only presenting a prediction of the flutter boundary.

Gray et al. [16] introduced the approximation of the third order unsteady piston theory aerodynamics for the flow over a 2-D panel. Both nonlinear aerodynamic and structure terms were considered in their finite element formulation. They also presented results for different support conditions. They concluded that the third order piston theory introduces a destabilizing effect as compared to the first order quasi-steady theory.

Benamar et al. [17] [18] formulated the large amplitude plate vibration problem and developed the numerical model to apply the analysis to fully clamped plates. They claimed that the assumption of the space-time solution w(x,y,t) can be presented in the form w(x,y,t)=q(t)*f(x,y) may be inaccurate for nonlinear deflections. They suggested that for high amplitudes and low aspect ratios, the effect of nonlinear (plastic) material properties must be taken into consideration as well. They also presented a set of experimental results conducted to investigate the dynamic response characteristics of fully clamped plates at large vibration amplitudes 18.

Different aerodynamic models were introduced to the solution of the panel-flutter problem to enhance the results of the finite element model or introduce new ranges of analysis. Yang and Sung [10] introduced the unsteady aerodynamic model in their research on panel-flutter in low supersonic flow fields where the quasi-steady piston theory fails to produce accurate results. Liu et Dynamics and Control of Flexible Structures 20. Sep. 2018 Mohammad Tawfik

93

Aeroelasticity al. [19] introduced a new approach for the aerodynamic modeling of wings and panel in supersonic-hypersonic flight regimes. Their model was a generalization of the piston theory. Their model also accounts for the effects of wing thickness.

Lately, studies of panel flutter were directed to enhancing the models and introducing different realistic parameters into the problem. Lee et al. modeled a panel with sheer deformable model as well as applying boundary conditions in the form a Timoshenko beams. Surace and Udrescu studied the panel flutter problem with higher order finite element model with the effect of external static pressure. While Bismarch-Nasr and Bones studied the effect of the aerodynamic damping on the panel flutter attitude. Most recently, Young and Lee presented a study of the flutter of the plates subject to in-plane random loading. Those studies introduced more practical models for the problem of panel flutter which was and still is very hard to study experimentally.

9.4.3

References and Further Readings

[1]

Mei, C., “A Finite Element Approach for Non-linear panel-flutter,” AIAA Journal, Vol. 15, No. 8, 1977, pp. 1107-1110.

[2]

Zhou, R. C., Xue, D. Y., and Mei, C., “On Analysis of Nonlinear Panel-flutter at Supersonic Speeds,” Proceedings of the First Industry/Academy Symposium On Research For Future Supersonic And Hypersonic Vehicles, Vol. 1, Greensboro, North Carolina, 1994, pp. 343-348.

[3]

Xue, D. Y., and Mei, C., “Finite Element Non-linear Panel-flutter with Arbitrary Temperature in Supersonic Flow,” AIAA Journal, Vol. 31, No. 1, 1993, pp. 154-162.

[4]

Frampton, K. D., Clark, R. L., and Dowell, E. H., “State-Space Modeling For Aeroelastic Panels With Linearized Potential Flow Aerodynamic Loading,” Journal Of Aircraft, Vol. 33, No. 4, 1996, pp. 816-822.

[5]

Dowell, E. H., “Panel Flutter: A Review of The Aeroelastic Stability of Plates and Shells,” AIAA Journal, Vol. 8, No. 3, 1970, pp. 385-399.

[6]

Bismarck-Nasr, M. N., “Finite Element analysis of Aeroelasticity of Plates and Shells,” Applied Mechanics Review, Vol. 45, No. 12, 1992, pp. 461-482.

[7]

Bismarck-Nasr, M. N., “Finite Elements in Aeroelasticity of Plates and Shells,” Applied Mechanics Review, Vol. 49, No. 10, 1996, pp. S17-S24.


94

Aeroelasticity [8]

Mei, C., Abdel-Motagaly, K., and Chen, R., “Review of Nonlinear Panel Flutter at Supersonic and Hypersonic Speeds,” Applied Mechanics Review, Vol. 52, No. 10, 1999, pp. 321-332.

[9]

Sarma, B. S., and Varadan, T. K. “Non-linear Panel-flutter by Finite Element Method,” AIAA Journal, Vol. 26, No. 5, 1988, pp. 566-574.

[10] Yang, T. Y., and Sung, S. H. “Finite Element Panel-flutter in Three-Dimensional Supersonic Unsteady Potential Flow,” AIAA Journal, Vol. 15, No. 12, 1977, pp. 1677-1683. [11] Ashley, H., and Zartarian, G., “Piston Theory – A New Aerodynamic Tool for the Aeroelastician,” Journal of Aeronautical Sciences, Vol. 23, No. 12, 1956, pp. 1109-1118. [12] Dixon, I. R., and Mei, C., “Finite Element Analysis of Large-Amplitude Panel-flutter of Thin Laminates,” AIAA Journal, Vol. 31, No. 4, 1993, pp. 701-707. [13] Abdel-Motagaly, K., Chen, R., and Mei, C. “Nonlinear Flutter of Composite Panels Under Yawed Supersonic Flow Using Finite Elements,” AIAA Journal, Vol. 37, No 9, 1999, pp. 1025-1032. [14] Zhong, Z. “Reduction of Thermal Deflection And Random Response Of Composite Structures With Embedded Shape memory Alloy At Elevated Temperature”, PhD Dissertation, 1998, Old Dominion University, Aerospace Department, Norfolk, Virginia. [15] Frampton, Kenneth D., Clark, Robert L., and Dowell, Earl H. “Active Control Of Panelflutter With Linearized Potential Flow Aerodynamics”, AIAA Paper 95-1079-CP, February 1995. [16] Gray, C. E., Mei, C., and Shore, C. P., “Finite Element Method for Large-Amplitude Two-Dimensional Panel-flutter at Hypersonic Speeds,” AIAA Journal, Vol. 29, No. 2, 1991, pp. 290-298. [17] Benamar, R, Bennouna, M. M. K., and White R. G. “The effect of large vibration amplitudes on the mode shapes and natural frequencies of thin elastic structures PART II: Fully Clamped Rectangular Isotropic Plates”, Journal of Sound and Vibration, Vol. 164, No. 2, 1993, pp. 295-316. [18] Benamar, R, Bennouna, M. M. K., and White R. G. “The effect of large vibration amplitudes on the mode shapes and natural frequencies of thin elastic structures PARTIII: Fully Clamped Rectangular Isotropic Plates – Measurements of The Mode Shape Amplitude Dependence And The Spatial Distribution Of Harmonic Distortion”, Journal of Sound and Vibration, Vol. 175, No. 3, 1994, pp. 377-395. [19] Liu, D. D., Yao, Z. X., Sarhaddi, D., and Chavez, F., “From Piston Theory to Uniform Hypersonic-Supersonic Lifting Surface Method,” Journal of Aircraft, Vol. 34, No. 3, 1997, pp. 304-312.


95

Aeroelasticity [20] Tawfik, M., Ro, J. J., and Mei, C., “Thermal post-buckling and aeroelastic behaviour of shape memory alloy reinforced plates,” Smart Materials and Structures, Vol 11, No. 2, 2002, pp. 297-307. [21] Xue, D.Y., “Finite Element Frequency Domain Solution of Nonlinear Panel-flutter With Temperature Effects And Fatigue Life Analysis”, PhD Dissertation, 1991, Old Dominion University, Mechanical Engineering Department, Norfolk, Virginia. [22] H. Ibrahim, M. Tawfik, and H.M. Negm, “Thermal Buckling and Nonlinear Flutter Behavior of Shape Memory Alloy Hybrid Composite Plates,” Journal of Vibration and Control, March 2011; vol. 17, 3: pp. 321-333, DOI: 10.1177/1077546309353368 [23] H. Ibrahim and M. Tawfik, “Limit-Cycle Oscillations of Functionally Graded Material Plates subject to Aerodynamic and Thermal Loads,” Journal of Vibration and Control, December 2010; vol. 16, 14: pp. 2147-2166, DOI: 10.1177/1077546309353366 [24] Hesham Hamed Ibrahim, Hong Hee Yoo, Mohammad Tawfik, and Kwan-Soo Lee, “Thermo-acoustic random response of temperature-dependent functionally graded material panels,” Computational Mechanics, 2010, DOI 10.1007/s00466-010-0477-1 [25] H. Ibrahim, M. Tawfik, and H.M. Negm, Hong Hee Yoo, “Aero-Thermo-Acoustic Response of Shape Memory Alloy Hybrid Composite Panels,” Journal of Aircraft, vol. 46, No. 5, pp. 1544-1555, 2009. [26] H. Ibrahim, M. Tawfik, and H.M. Negm, “Limit-Cycle Oscillation of Shape Memory Alloy Hybrid Composite Plates at Elevated Temperatures” Mechanics of Advanced Materials and Structures, vol. 16, No. 6., pp. 429-441, August 2009, DOI: 10.1080/15376490902781159 [27] H. Ibrahim, M. Tawfik, and H.M. Negm, “Thermoacoustic Random Response of Shape Memory Alloy Hybrid Composite Plates” Journal of Aircraft, Vol. 45, No. 3, pp. 962-970, 2008, DOI: 10.2514/1.32843 [28] H. Ibrahim, M. Tawfik, and M. Al-Ajmi, “Nonlinear Panel Flutter with Temperature Effects of Functionally Graded Material Panels with Temperature-Dependent Material Properties" Computational Mechanics, Vol. 41, No. 2, pp. 325-334, 2008 [29] H. Ibrahim, M. Tawfik, and M. Al-Ajmi, “Thermal Buckling and Nonlinear Flutter Behavior of Functionally Graded Material Panels" Journal of Aircraft, Vol. 44, No. 5, pp. 1610-1618, 2007.


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Dynamics and Control of Flexible Structures

Dynamics and Control of Flexible Structures

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