Structural Dynamics Coursework 2

8/10/2019 Structural Dynamics Coursework 2

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UCL Civil, Environmental and Geomatic EngineeringStructural Dynamics 2014 Tutorial 2 Carmine Russo 14103106 page 1 of 15

Structural Dynamics (CEGEM071/CEGEG071)

Tutorial 2 Student: Carmine Russo 14103106

1. IntroductionInitial data:

In order to proceed to the discussion of the solution, we need first to find some quantities that we will

use further on in this exercise and to make some consideration on the mechanical system.

* + (Youngs modulus)

(mass of the tank)

Columns:

Section area:

Moment of inertia:

Braces:

Section area:


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In general, the system has three degrees offreedom (if the beam AB is not rigid thenthey are four). We choose as lagrangiancoordinates the vertical and horizontaldisplacement of G and the counterclockwise rotation of the beam AB aroundG. Collecting these variables in the vector:

we can express the displacements of theother point of the structure in function ofthe chosen coordinates:Node A:

Node B:

In which we have considered that is very small thus, approximated the cosine with McLaurin series andneglected the squared terms. What we get is that, in the horizontal motion, all points move of the samequantity:

Because of the rigidity of the beam AB, the vertical displacements are tied up by the rotation angle:

Now, in order to write the equilibrium equations, we have to findthe stiffness of each element for each displacement. In this case,we can use the direct method:

Column AC and BD: In general there are three kind ofdisplacements that occurs for the columns: 1. Horizontal displacement on top

By solving the differential equation of the elastic beam:

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


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Finally:

2. Rotational displacement on top

By solving the differential equation of the elastic beam:

With the boundary conditions:

We get the constants of integration: Finally:

Therefore, for the columns AC and BD, we have:

Total shear on top

Total moment on top We indicate:

* + translational stiffnessfor shear forces

And

rotational stiffness for

bending moments 3. Vertical displacement

In which:

Braces AD and BC: We can find a useful expression of stiffness for thebraces by using the following relationships:

and

Therefore, we can evaluate the stiffness in the horizontaldirection of each cable as:

*+ axial stiffness of thecolumns


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We do the same for the vertical direction, and we get:

in in in and Therefore, we can evaluate the stiffness in the verticaldirection of each cable as:

in in Numerically:

The equations of motion:

1) Horizontal equilibrium:

The two bracings elements work only in tension, therefore is like considering the presence of just oneof them that works in compression and in tension:

( )

Where is the horizontal force applied in G.2) Vertical equilibrium:


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In this direction, we have to discuss the equation of equilibrium for various cases:

a) If and both braces are in tension (or with zero tension), then: By substituting, we have:

Factorizing and simplifying: b) If and or and , just one brace is in tension, the equationbecomes:

c) If and , none of the braces is working:

Where is the vertical force applied in G.

3) Rotational equilibrium:

n in


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We can to discuss the equation of equilibrium for various cases The energetic approach to the solution

The same results can be found simply calculating the energy of the system. By using the results found for abeam subjected to a rotational displacement and a lateral displacement, we have:

And for the axial displacements, we have found:

For the bracing elements, we decomposed the force in two components (vertical and horizontal), and foundthe stiffness according to each direction:

Element AD, horizontaldirection Element AD, verticaldirection Element BC, horizontaldirection Element BC, verticaldirection

The total strain energy of the system can be calculated simply integrating, but even with this procedure, weshould discuss the tension in the bracing elements.

In case that all bracings are working in tension and compression, the total strain energy is:

( ) ( ) ( ) ( )


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The stiffness matrix of the system is:

( ) ( )

The result obtained is the same that we achieved with the equilibrium approach, for instance if we take thefirst row and multiply it for the vector of lagrangian coordinates, and adding the inertial forces, we have:

If we consider just one of the braces working, then we have just one stiffness :

Which is exactly the same equation we get previously for the horizontal equilibrium!

2. Part A): Simplified 1 d.o.f. system

If the axial stiffness of the column is infinite, thenthe number of degree of freedom reduces one:

and

Hence, we have only the equilibrium in thehorizontal direction:

( )

That can be written in the form:

Where is the natural circular frequency of the system:

( )

* + * + +

The period is:

The frequency:


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3. Part B): Steady state motion of the system subjected to anharmonic periodic excitation

The equation of motion is: With:

}

And: + Complementary solution

Searching a solution in the form: Substituting in the equation, we get the characteristic equation:

Therefre the solution is: In trigonometric form: in

Particular solutionThe particular solution of a sistem corresponding to an harmonic excitation is als harmonic at theexcitation frequency. The generic harmonic response is:

}

To find the amplitude it is sufficient to substitute the above form in the origin equation. Thus we have:

( ) } } This equation must be satisfied for all values of t, so we match the terms contained within the brackets.This leads to: ( ) In which the quantity at the denominator is the dynamic stiffness.

We can express this quantity in a more convenient form, if we substitute:

We obtain:

is the complex frequency response (that in this case is a real number because damping is zero) :

Total solution in


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Using the initial conditions:

and We can find the constant and :

( ) In case the initial conditions are homogeneous, the solution become:and

The system has no damping, therefore the complementary solution does not vanish and theres not a pointwhere the transient motion has ended . The maximum possible displacement occurs when:

or In this conditions:

The shear is constant along the columns, and has the same value in both:

We can draw a diagram of the maximum shear, where is represented the shear forces for both direction ofexcitation/motion:


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The plot of the steady state motion, for the solution with homogeneous initial conditions:

%*** UCL Civil, Environmental and Geomatic Engineering ***; %*********** Structural Dynamics *****************; %***************** (CEGEM071/CEGEG071) *******************; %***** Tutorial 2 Student: Carmine Russo 14103106 *****; clear; clear all ; T=0.09; t=0:0.0001:20*T; % Input data: X=1.9007317*10^(-4); omegaC=15.0796447; omegaN=63.3672668; % Function: u=X.*(cos(omegaC*t)-cos(omegaN*t)); up=X.*cos(omegaC*t); uc=-X.*cos(omegaN*t); hold on plot(t,uc, 'm.' , 'LineWidth' ,0.5) plot(t,up, 'g--' , 'LineWidth' ,2) plot(t,u, 'LineWidth' ,2) xlabel( 'Time [s]' ) ylabel( 'Displacement [m]' ) grid on hold off


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4. Part C): Damped systemThe equation of the system in case of presence of damping is:

With: }

+

Complementary solution

The system is underdamped.We search a solution that has the form:

By simply substituting this form in the equation, we get the characteristic equation: Where: *+

Finally the complementary solution is: Particular solutionUsing the same procedure used in point b) , starting from the equation of motion written in the form: Where:

The generic harmonic response is:

}

Thus we have: ( ) } }


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This equation must be satisfied for all values of t, so we match the terms contained within the brackets.This leads to:

( ) (the quantity at the denominator is the dynamic stiffness)We can express this quantity in a more convenient form, if we substitute:

and

We obtain:

is the complex frequency response :

Since is a complex number, we can represent in modulus and phase by calculating the

modulus of the complex frequency response:

| | Where: | | And:

a Finally, the steady state response is the real part of the imaginary number:

, - , | | - , | | - | | () The total solution:Adding the complementary to the particular solution, we have:

| |( )

Using the initial condition we can find now the constants and :

| |( ) Using the initial conditions:

and


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| |( ) | |in( )

| |( ) | |( )

If we consider :and

| |( ) | |( ) | |( )

We can calculate this two constant and we have:


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We can see how after a certain time the system moves following the excitation force.At resonance:

Then:

| |

Assuming that the resonance occurs when the transient is already vanished, the structure should have adisplcement equal to:

| | 5. Part D)

Applying a static force the deflection is:

Maximum moments:

The maximum moments in case of non damping and damping are

That means that maximum moment is the double in case of absence of damping, and 1,5 times in case ofpresence of damping

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Structural Dynamics Coursework 2