Structural Dynamics Coursework 3

8/10/2019 Structural Dynamics Coursework 3

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UCLCivil, Environmental and Geomatic Engineering

Structural Dynamics2014Tutorial 3 Carmine Russo 14103106 page 1 of 16

Structural Dynamics(CEGEM071/CEGEG071)

Tutorial 3 Student: Carmine Russo14103106

1. Introduction

In general, the system has six degrees of

freedom (if the beams CD and EF are not

rigid, then they are eight) but neglectingthe axial deformations of columns

( ) they become two.We choose as lagrangian coordinates the

horizontal displacements of CD and EF.

Collecting these variables in the vector, we

have: 2 3 2 3In order to write the equilibrium equations, we have to find the stiffness of each element for each

displacement. In this case, we can use the direct method by solving the differential equation of the elastic

beam: With the boundary conditions: We get the constants of integration:

Finally:

Initial data:

- - - - - - -


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Since theres no lateral load applied to the columns, the shear force is constant along their length.

It is convenient to indicate with kthe quantity:

- 0 1Then, the generic expression of the shear force become:

Where 3krepresent the stiffness.We can assemble the stiffness matrix, column by column, simply imposing one deformation at time,

while keeping the other one equal to zero, and finding the equilibrium forces.

Displacement :In this case we have the mass (that is theone who moves) subjected to the

displacement .The beam CD is subjected to the horizontal

forces

and

;

the beam EF, in this case, just to the

horizontal force EF generate by the

deformation of the column EC.

The total forces for each mass are:

We can summarize these relations as:

0 1

Displacement :In this case we have the mass (that is the onewho moves) subjected to the displacement .

In this case we first define the shear on the

column FG that is: ./

The total forces for each mass are: We can summarize these relations as:

0 1 Thus the static equilibrium is represented by the

equations:

0 1 23


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The stiffness matrix

- 0 1 0 1 0 1The mass matrix

- 0

1 -

The energetic approach

Instead of following the equilibrium approach, we can find the equation by calculating the total energy of

the system:

Kinetic Energy The mass matrix can be found simply by calculating the Jacobian:

- 6 7 [

]

Potential Energy

In this exercise the potential elastic energy is given by the horizontal displacements of the beams,

therefore by using the stiffnesses already calculated above, we can directly write the expression

of this energy without calculate the integrals. Simply remembering that the potential energy of a

single spring is:

we just have to sum the potential energies of each deformed element: - - - -The stiffness matrix can be found simply by calculating the Jacobian:

- 0 1 0 1

Exactly the same results obtained with the equilibrium method!

The equations of equilibrium

In case of undamped free vibrations, the equations of motion are:

* 0 1 . / . /


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2. Part a - Modal AnalysisStarting from the equations of motion, for the undamped free vibrating system:- -

* 0

1 . / .

/That is a set of linear differential equations with constant coefficients. We may generate a solution byassuming that each generalized coordinate varies exponentially as . In principle, the coefficient could be any constant but, since damping in not taken in account, the system is conservative. If , the total mechanical energy of the system will grow, while leads to adecaying response: both cases violate conservation of energy.

Because of that we anticipate that , corresponding to harmonic motion, so we construct a solutionbased on the trial form: 2 ./ 3Where and are constants.Every term produced by the substitution of the trial solution exhibits the same time dependence, so the

equation of motion will be satisfied at all instants only if the coefficients of the exponential terms match.Furthermore, the constant factor is common to every term, so it cancels. Therefore, we have:- ./ - ./ - - ./ The nontrivial solution of this system can exist only if the value of is such that - -is notinvertible. We must find the value of for which the determinant of this matrix is equal to zero ( generaleigenvalue problem):- - | | where we set . From this condition we get the characteristic equation:

-

Eigenvalues:

- - -

0 1 0 1The natural frequenciesare:

./ ./ 0 1The periods are: (

) + -

The frequencies:

,

+ + -


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Eigenvectors:

Now that we have determined the natural frequencies, we proceed to evaluate the mode shapes.

When or , at least one of the scalar equations described by - - ./ is notindependent of the other. We can retain one of the two conditions (for instance the first one), and add a

condition on the norm of the first eigenvector.

Mode shape 1

+ +With the condition: We have:

{

{

}

The first eigenvector is:

+ +

Mode shape 2

+ +

With the condition:

We have:

{

{

}

The first eigenvector is:

+

+

We can normalize both vectors with the respect of the maximum absolute value of each mode shape:

+

+ +

+

+

+ +

+

According to our analysis, a vibration in the first mode occurs at

0 1with the amplitude of

times that of. Because . / is positive, both beams move in phase in thismodal vibration. The second mode occurs at 0 1, with the amplitude of


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times . In the second mode . / is negative, which means that at each instant the beams aremoving in opposite directions.

We define the modal masses: - Since each element of an eigenvector is scaled by an arbitrary element, it follows that the modal mass

values depend on the choice of that element; moreover, the modal masses occur throughout the

evaluation of both responses, thereby compensating this arbitrariness contained in the eigenvector.

- + + -

- + + -

Now we can calculate the normal modes(i.e. modes normalized with the respect of the mass matrix)

+ +

+

+ + +

The modal shape matrix:

- -

We can check the orthogonality properties:

- - - - - -- 0 1 - - - 0 1

By using the modal transformation: -we can write the equation of motion in modalcoordinates: - - -- -- Pre-multiplying by -and using the orthogonality property, leads to:

--- --- 0 1 * 6 7 ./ ./ * ./ .

/

The equations are now decoupled and each one represent a single degree of freedom motion.
https://www.google.it/search?hl=it&biw=1366&bih=600&q=pre+multiplying&spell=1&sa=X&ei=NJ1-VKDNLYrV7QazzIG4DA&ved=0CBoQBSgAhttps://www.google.it/search?hl=it&biw=1366&bih=600&q=pre+multiplying&spell=1&sa=X&ei=NJ1-VKDNLYrV7QazzIG4DA&ved=0CBoQBSgA


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Therefore the solutions are: ./ *Where the constant,and ,can be found with the initial conditions:

./

45

And we have:

*

Finally, the responses in the lagrangian coordinates are:

./ - * - -

- - Each one is a combination of single degree of freedom in free vibrations that we found in modal

coordinates.

The initial conditions of the motion are:

./ 45

That in modal coordinates become:

* - And

* - [ ] -

Where the inverse of the mode shapes matrix, can be easily evaluated using a consequence property of

the orthogonality:

- --

3. Part bRayleigh approximationIn general the damped equation of motion is:- - - Assuming a simplified Rayleigh damping matrix - -, the equation of motion become:- - - Where can be found considering that the system is damped at 3% of the critical damping whenoscillating at its first natural mode frequency. In order to do that, first we make a modal transformation:

-- -- --


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Then we pre-multiply by -, and we get:--- --- --- Using the orthogonality property of mode shape matrix, the equation become:

6 7 6 7 Usually the form of the equation of a damped motion is:

Simply comparing the system of equations with the common form written above, we get the relationsthat make us are able to evaluate the damping coefficientsthat comply the Rayleigh approximation:

As specified (damping coefficient at the first natural mode frequency), therefore:

0 1 0 1 0 10 1 We can now write the system of equations as:

6 7 In modal coordinates, the equations are completely decoupled, hence the solution can be easily find.

The system is underdamped; we construct a solution based on the trial form vector:

*By simply substituting this form in the equation, we get the characteristic equations for each modalcoordinate:

Where:

+

+ 0 1

Finally the solutions vectorfor the free underdamped motion is:


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[

]The unknown coefficients,,,(or ,,,) are set by the initial conditions:

* -

-

- 45

.

/ -

. / -* (

) * 645 457

* - -

( ) -- -- . / -

The response in lagrangian coordinates

In matrix form:

-

[

]

- - - Explicit form:

- - - -

We can plot the response for each beam (see next page)

in this case, because


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4. Part cDamped system subjected to an impulse-like

excitationIn the previous paragraph, we found the

complementary solution of the motion.

In the case presented in Point C, we have an

impulse-like force applied on the beam CD

(Level_1). To find and plot the equation of

motion, first we have to find the particular

solution associated to this form of excitation:

{

- - - -

With:

0 1

The force applied can be defined also by

translating the time's origin (we will use this

other formulation in one of the methods of

solution proposed).

Phase I

System is

at rest

Phase II

Impulse excitation

Forced vibrations

Phase III

Free

vibrations

Our system of equations, in phase IIwhen- -, is:- -

- + 4

5Where we used as simplified Rayleighs damping matrix: - -Making the modal transformation, we get:

--- --- --- - 4 5Using the orthogonality property of the mode shape matrix, the equations become:

6 7 6 7 4 5And finally, the form in modal coordinates:

6

7

+

Where: 0 1 0 1 0 1 0 1 - -01 - - - -

Method 1: Direct integration (undetermined coefficients)

- -

-

-


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In both equations, the excitation force has the form of a polynomial of first degree. For this, we will need

the following guess for the particular solution:

So, by differentiating and substituting into the differential equation:

- -

- -

Now, we will need the coefficients of the terms on both sides of the equal sign to be the same:

- -

- - - - - - - -Therefore, the solution can be wrote in matrix form:

./ ./ - - - - ./The total solution:

./ ./ * - * * - - - - - . /

[ - - - - ] . / The unknown coefficients,,,are set by the initial conditions: - ./ ./

0 1

0 1 ./ ./And * ./

{

* }

./

Solving numerically finally, we get:


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-

At the end of phase II:

-

* . / - 4 5 ./ -Those written above are the initial conditions of phase III, when the system is in free vibrations. The

equation of motion at this point is:

- - - - -And the constant of integration can be found as we did in part 3:

* - -- *

* - ./ -* (

) (

) -- + --

+

(

)

-

- -

-

-

- -

-+

. / -


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Method 2: Convolution integral (Duhamel)


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The solution inphase IIcan be found as solution of the convolution between the generic response to the

single impulse (Dirac delta) at time, and the affective force applied:

{

- - -

-

Note that the solution is the difference between a ramp forces, minus a constant force:

{ 6 - - - 7 6 - - - 7

Integrating by parts:

- -

- 6 - -7

-

-

[ - -]

- 6 - -7 -

-

The total solution in phase II, in terms of the lagrangian and , in this case is going to be acombination of the two modal solution, scaled by the eigenvectors.

./ - + -Apparently simpler than the previous one, but just more compact. In fact, the convolution integral holds

inside the complete base of the vector space of solutions.

Afterphase II, the solution has the same form we found with the direct integration:

- - - - -With the constant of integration determined by the initial conditions.


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Structural Dynamics 2014 Tutorial 3 Carmine Russo 14103106 page 17 of 16

Method 4: Discrete formulation (Newmark)

To verify the quality of solutions found, we can try a discrete approach, performing a numerical

integration with the Newmark method (constant average acceleration):

Starting from the decoupled system of equations:

-

-

- -With the initial conditions: we can integrate numerically simply following the steps:

1) -2) - 3) *4) 01

The update the initial condition for the new iteration: And repeating the calculation incrementing the time with the step -By executing this method for both the coordinates, and the recombining the values with the coefficient

given by the components of the eigenvectors, we get the diagram of the total solution. For example forthe diagram in phase II has the aspect of:

That is exactly the same result we get with the analytical solution

-10

-5

0

5

10

15

20

25

30

0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4

Structural Dynamics Coursework 3

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