Structural Dynamics Coursework 3
February 2, 2017 | Author: Carmine Russo | Category: N/A
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Structural Dynamics (CEGEM071/CEGEG071) Tutorial 3 β Student: Carmine Russo β 14103106
1. Introduction
Initial data:
, , ,
-
, -
,
, -
,
-
In general, the system has six degrees of freedom (if the beams CD and EF are not rigid, then they are eight) but neglecting the 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: β
2
3
2
3
In 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:
π UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 1 of 16
Since thereβs no lateral load applied to the columns, the shear force is constant along their length. It is convenient to indicate with k the quantity: [ ] ( , -) Then, the generic expression of the shear force become: π»
β 1
0
Where β3kβ represent 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 the one who moves) subjected to the displacement . The beam CD is subjected to the horizontal forces π π and π ; 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: π
π
π
(
)
π We can summarize these relations as: {
}
0
1
Displacement : In this case we have the mass (that is the one who moves) subjected to the displacement . In this case we first define the shear on the column FG that is: π . / The total forces for each mass are: π π
(
π
)
We can summarize these relations as: {
}
0
1
Thus the static equilibrium is represented by the equations: {
}
0
1 2
3
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 2 of 16
The stiffness matrix
, -
0
1
0
1
, -
[
]
[
β 1
]0
The mass matrix
1,
0
-
The energetic approach Instead of following the equilibrium approach, we can find the equation by calculating the total energy of the system: ο
Kinetic Energy Μ
π»
Μ
The mass matrix can be found simply by calculating the Jacobian:
π , -
π
6 Μ
Μ
Μ Μ
7
π [
ο
π
Μ
Μ
[
π Μ
]
Μ ] Μ
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: ,
β
(
-
)
,
,
,
-
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:
[
]( Μ
Μ
( ) ( )
*
0
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
1
.
( ) ( )
/
. /
page 3 of 16
2. Part a - Modal Analysis Starting from the equations of motion, for the undamped free vibrating system:
, - Μ( [
](
Μ Μ
( ) ( )
*
, -
)
( )
0
( )
1.
( )
/
. /
That is a set of linear differential equations with constant coefficients. We may generate a solution by assuming 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 * + ) will grow, while * + , the total mechanical energy of the system (π leads to a decaying response: both cases violate conservation of energy. Because of that we anticipate that , corresponding to harmonic motion, so we construct a solution based on the trial form: 2 . /
( )
3
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: Where
, -
, - . /
. /
*, -
, -+ . /
, -+ is not The nontrivial solution of this system can exist only if the value of is such that *, invertible. We must find the value of for which the determinant of this matrix is equal to zero (general eigenvalue problem): ( ) , -) (, | | ( ) where we set (
. From this condition we get the characteristic equation: )(
,
)
(
)-
Eigenvalues:
,
(
)-
[
(
β,
)-
, (
)-
β (
)
0
β 1
0
β 1
]
The natural frequencies are: β
.
/
/0
.
β 1
The periods are: π
+, -
( (
)
The frequencies:
π
(
,
( +
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
(
+,
-
page 4 of 16
Eigenvectors: Now that we have determined the natural frequencies, we proceed to evaluate the mode shapes. When
, at least one of the scalar equations described by *, -
or
, -+ . /
is not
independent 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. ο
Mode shape 1 ( [
)
]( )
(
+
( +
With the condition: We have: (
(
)
)
{ [ { (
{
] )
}
] )
}
The first eigenvector is: ( ο
+
(
+
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 times that of
. Because .
modal vibration. The second mode occurs at
β
+
(
+ 0
+
(
+
β 1 with the amplitude of
/ is positive, both beams move in phase in this 0
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
β 1, with the amplitude of page 5 of 16
. In the second mode .
times
β
/ is negative, which means that at each instant the beams are
moving 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)
β
β
(
+
(
(
+
(
+
(
+
+
(
+
The modal shape matrix: , -
,
-
[
]
(
) (
(
)
)
*
+ (
)
We can check the orthogonality properties:
, -
,
-
, -
, -
, -
, -
,
, -
β 1
0
-
, -
, -
By using the modal transformation:
, -
( )
( )
0
β 1
we can write the equation of motion in modal
coordinates:
, - Μ(
)
, -
, -, - Μ (
( )
)
, -, -
( )
Pre-multiplying by , - and using the orthogonality property, leads to:
, - , -, - Μ (
)
, - , -, -
0
( )
Μ ( * Μ
[
]. /
Μ 1( * Μ
6
7. /
. /
. /
Μ The equations are now decoupled and each one represent a single degree of freedom motion. UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 6 of 16
Therefore the solutions are:
Where the constant
,
and
(
.
)
( )
.
( )
( )
( (
(
/
) )
( (
) * )
,
can be found with the initial conditions:
(
)
(
)
/
(
Μ(
)
Μ
4
)
Μ
(
)
(
)
5
(
Μ
) Μ
And we have: ( )
(
*
(
(
)
( )
)
(
Μ
[
]
(
(
)
Μ Μ
)
[
]( Μ
Μ
)
) Μ
[ ] Finally, the responses in the lagrangian coordinates are:
β, - , .
()
, -(
/
()
(
)-
*
( ) ( )β, - , ( ) 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: (
.
)
(
)
(
)
/
(
Μ(
)
Μ
4
)
Μ
(
)
(
)
5
( Μ
Μ
)
That in modal coordinates become: ( )
(
*
, -
(
)
And ( )
(
*
[
, -
]
Μ(
, -
)
[
Μ(
)
]
Where the inverse of the mode shapes matrix, can be easily evaluated using a consequence property of the orthogonality: , -
, -π» , -
[
][
]
[
]
[
]
3. Part b β Rayleigh approximation In general the damped equation of motion is:
, - Μ(
)
, - Μ(
Assuming a simplified Rayleigh damping matrix , -
, - Μ(
)
, - Μ(
, -
)
( )
, -, the equation of motion become:
, -
)
( )
Where can be found considering that the system is damped at 3% of the critical damping when oscillating at its first natural mode frequency. In order to do that, first we make a modal transformation:
, -, - Μ (
)
, -, - Μ (
)
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
, -, -
( )
page 7 of 16
Then we pre-multiply by , - , and we get: , - , -, - Μ (
, - , -, - Μ (
)
)
, - , -, -
( )
Using the orthogonality property of mode shape matrix, the equation become:
Μ(
6
)
7 Μ(
6
)
7
( )
Usually the form of the equation of a damped motion is: Μ( ) Μ( ) ( ) Simply comparing the system of equations with the common form written above, we get the relations that make us are able to evaluate the damping coefficients that comply the Rayleigh approximation:
, { (damping coefficient at the first natural mode frequency), therefore:
As specified
{ We can now write the system of equations as: Μ(
0
β1
0
β1
] Μ(
[
)
0β
β1
0
)
6
7
1
( )
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 modal coordinate:
β , β
{ Where:
(
+
(
(
)
(
)
)
(
+0
β 1
Finally the solutions vector for the free underdamped motion is: ( )
[
[
]
[
]
(
)
] [
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
][
]
page 8 of 16
( ( ( (
π»
[
] [
] [
The unknown coefficients (
*
(
)
, -
(
,
,
,
(or
)
, -π» , - 4
,
,
5
[
,
) ) ) ) ]
) are set by the initial conditions: ].
/, /, -
. Μ (
*
( Μ
(
*
64
)
[
Μ Μ
5
]4
[
57
) (
*
[, -
Μ(
Μ
*, -π , - ( (
Μ
)
)
], -
[
)]
(
] , -π , - (
][
in this case, because Μ (
)+
/, -
.
)
The response in lagrangian coordinates ο
In matrix form: π»
( )
, -
( )
[
][
] [
] [
, -(
( )
ο
( )
,
(
,
(
)
(
)
( ( ( (
) ) ) ) ]
)-
(
)-
)
Explicit form:
(
,
(
)
(
)-
,
(
)
(
,
(
)
(
)-
,
(
)
(
)) )-
We can plot the response for each beam (see next page)
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 9 of 16
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 10 of 16
4. Part c β Damped system subjected to an impulse-like excitation In 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:
β1 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). Our system of equations, in βphase IIβ when , 0
Μ , -( Μ
( )
Phase I System is at rest
Μ
, -,
)
, -(
( )
Μ
( )
( )
, -(
)
)
+
)
4
()
5
( )
( )
, - , -, - Μ (
Phase III Free vibrations
is:
Where we used as simplified Rayleighβs damping matrix: , Making the modal transformation, we get:
, - , -, - Μ (
Phase II Impulse excitation Forced vibrations
, -
, - , -, -
( )
, - 4
( )
5
Using the orthogonality property of the mode shape matrix, the equations become:
Μ(
6
)
7 Μ(
6
)
7
[
( )
()
]4
5
And finally, the form in modal coordinates: Μ( )
] Μ( )
[
6
7
( )(
()
+
Where: ;
;
0
; β 1;
0
ο
β 1; ;
()
,
-
(
, -)
β 1
0
β 1
0 0 1(
, -
, -)
(
, -
, -)
Method 1: Direct integration (undetermined coefficients) Μ ,
( )
Μ
( )
( )
(
, -)
Μ
,( )
Μ
( )
( )
(
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
(
)
,-
, -)
page 11 of 16
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: , β (
)
, -
(
, -
)
, -
, -
, β -
{
, -
(
)
, -
(
)
, -
, -
{
Therefore, the solution can be wrote in matrix form: .
/
[
]. /
, β , β -
[
, ]. / , -
The total solution: .
π» π»
/
.
( ) ( )
/
, -*(
[
(
*
(
)*
(
)
, - [(
(
)
(
)
,
(
)-
,
(
)-
,
*
] [
,
(
)
(
)
{
The unknown coefficients , - .
π» π»
0 And ( Μ
Μ
π» π»
*
/
,
,
(
)
(
)
,
*
(
*]
] [
(
)-+
(
)-+ , , , ,
] [
)
].
[
( ( ( (
))))- ]
[
/+
].
/ }
are set by the initial conditions:
. / π
1 {*
+ [
]*
+
1 . /}
0
. /
. / [ [
] [
]*
+
]
. / *
+ [
{
][
]
(
* }
Solving numerically finally, we get:
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 12 of 16
[
]
], -
[
At the end of phase II: , -
π»( )
(
π»( )
*
/, -
.
4
Μ
π»( )
Μ
π»( )
/, β -
.
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: (
*
(
)
, -
(
)
, -π» , - (
π»( ) π»( )
* ](
[ Μ (
*
Μ [, - , - (
Μ (
)
(
*, -π , - ( (
)
)
Μ
+
[
][
], - , -(
][
] , -π , - (
.
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
+] ( )
( )
[
/, -
. ( )
( )
, β ) , β -
*, -
, , -
++
/, -
page 13 of 16
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 14 of 16
ο
Method 2: Convolution integral (Duhamel)
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 15 of 16
The solution in phase II can be found as solution of the convolution between the generic response to the single impulse (Dirac delta) at time , and the affective force applied:
{
( )
β«
( )
β«
(
, -)
(
)
,
(
)-
(
, -)
(
)
,
(
)-
Note that the solution is the difference between a ramp forces, minus a constant force:
{
( )
6β«
(
)
,
(
)-
β«
, -
( )
6β«
(
)
,
(
)-
β«
, -
(
)
,
(
)-
7
(
)
,
(
)-
7
Integrating by parts: ( ) (
(
)
(
)
)
,
)
(
, -
)|
[( (
, -
{
( ( )
,
6
) (
,
) ,
)-
(
(
)-] | )-7
| }
, -
( ) (
(
(
)
)
)
[( (
( )
,
6
) (
.
( )
)
( ,
,
)-
The total solution in phase II, in terms of the lagrangian and combination of the two modal solution, scaled by the eigenvectors. ( )
,
)
, -
)|
, -
{
(
(
(
)-] | )-7
| }
, -
, in this case is going to be a
( )
/
, -(
(
+
)
, -
( )
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. After phase II, the solution has the same form we found with the direct integration: (
)
, -(
* *
, ,
(
)-
,
(
)-
,
( (
)-+
) )-+
With the constant of integration determined by the initial conditions.
UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 16 of 16
ο
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: Μ ,
Μ
( )
( )
( )
(
, -)
( )
(
-)
(
, -
Μ
Μ
( )
( )
,
)
, -
With the initial conditions: ( )
Μ
( )
Μ
( )
( )
we can integrate numerically simply following the steps: 1) Μ ( ) Μ ( ) ( , -) ( ) 2)
(
)
[ (
3)
Μ
(
)
4)
Μ
(
)
0
( (
(
, -)
)
)
Μ
Μ
(
(
)*
)
1
Μ
Μ
(
(
Μ
)
)]
(
)
The update the initial condition for the new iteration:
Μ
(
Μ
)
Μ
()
(
)
Μ ( ) Μ () Μ ( ) , 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 for the diagram in phase II has the aspect of: 30
25
20
15
10
5
0 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
-5
-10
That is exactly the same result we get with the analytical solution UCL β Civil, Environmental and Geomatic Engineering Structural Dynamics β 2014 β Tutorial 3 β Carmine Russo β 14103106
page 17 of 16
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