User:EML4500.f08.JAMAMA/FE

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7.1 2D heat problem and here[edit]

Given: 2D Strong Form and BCs[edit]

biunit square

(identity matrix)

,

Essential boundary condition:

on

Natural boundary condition:

none

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Find: Temperature inside and error at (0,0)[edit]

Solve for until error is less then at center (x,y)=(0,0) for:

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1) Static (steady state): in

1a) 2d LIBF
1b) 2d LLEBF
1b1) Uniform mesh
1b2) Non-uniform mesh


2) Dynamic(transient):

Initial condition

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2a)

2d LIBF

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2b)

2b1) 2d LIBF
2b2) 2d LLEBF
2b2a) Uniform mesh
2b2b) non-uniform mesh

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Solution[edit]

In general by Prof. Vu Quoc's notes


[1]

, where , where represents local node.

For linear 2D case with boundary being square

Bases, are defined to be

where and corresponds to numbering of nodes on element scale.

Rect

also noting that it's a rectangular shape, such that the distance from x: 1-to-2 = 1-to-3 and node 1 = node 4 in 'x' coordiantes, y: 1-to-4 = 1-to-3 and node 1 = node 2 in 'x' coordinates, we substitute for and for and the same for other coordinates






Jacobian Matrix = BN *[mathbf{x} \ mathbf{y}]


</math>

taking the and values from the definition of parent element and substituting into Jacobian matrix we get:

From given Prof. Vu Quoc's Notes

or

With this last information we have a completely defined stiffness element matrix . Also note that since we have, 4 degrees of freedom in 1 element node, we do expect the gram matrix to be 4x4, therefore,

note, to save some space in typing the formula (else it stretched out of the page), the independent variables for were avoided


for the initial square of length of '1', we divided into 4 smaller squares each 0.5 length, the element nodes stiffness matrix came out to be:

K(:,:,1) = 
[  1/24, -1/96, -1/48, -1/96]
[ -1/96,  1/24, -1/96, -1/48]
[ -1/48, -1/96,  1/24, -1/96]
[ -1/96, -1/48, -1/96,  1/24]


K(:,:,2) = 
[  1/24, -1/96, -1/48, -1/96]
[ -1/96,  1/24, -1/96, -1/48]
[ -1/48, -1/96,  1/24, -1/96]
[ -1/96, -1/48, -1/96,  1/24]

this answer shows 2 elements stiffness matrices located along 'x' axes. Noting that because of the symmetry of each elements, the equivalent spacing of each node in elements, and because the material matrix is identity matrix, all stiffness element matrices will be the same. The only differ in element stiffness matrix will occur when we will change the number of stiffness.

,where

force[edit]

From prof. Vu Quoc's notes for elemental force expression

converting to local coordinates

F(:,:,1) = 
[  1/16,  1/16]
[ -1/16,  1/16]
[ -1/16, -1/16]
[  1/16, -1/16]


F(:,:,2) =  
[  1/16,  1/16]
[ -1/16,  1/16]
[ -1/16, -1/16]
[  1/16, -1/16]

Global Matrix creation[edit]

Then we constructed the Global stiffness matrix by the process of superimposing the local stiffness coefficients. Matlab output below shows the process of superimposing the elements to global stiffness matrix.

The matlab code for global matrix construction was written with respect to this element naming system, and then it was extended to encompass other numbers of elements

Global to Local

Original element matrices

K(:,:,1) = 

[  1/24, -1/96, -1/48, -1/96]
[ -1/96,  1/24, -1/96, -1/48]
[ -1/48, -1/96,  1/24, -1/96]
[ -1/96, -1/48, -1/96,  1/24]
K(:,:,2) = 

[  1/24, -1/96, -1/48, -1/96]
[ -1/96,  1/24, -1/96, -1/48]
[ -1/48, -1/96,  1/24, -1/96]
[ -1/96, -1/48, -1/96,  1/24]
     

Superpositions to global matrix

KGlobal =
      1/24          -1/96           0             -1/96          -1/48           0              0              0              0       
     -1/96           1/12          -1/96          -1/48          -1/48          -1/48           0              0              0       
      0             -1/96           1/24           0             -1/48          -1/96           0              0              0       
     -1/96          -1/48           0              1/12          -1/48           0             -1/96          -1/48           0       
     -1/48          -1/48          -1/48          -1/48           1/6           -1/48          -1/48          -1/48          -1/48    
      0             -1/48          -1/96           0             -1/48           1/12           0             -1/48          -1/96    
      0              0              0             -1/96          -1/48           0              1/24          -1/96           0       
      0              0              0             -1/48          -1/48          -1/48          -1/96           1/12          -1/96    
      0              0              0              0             -1/48          -1/96           0             -1/96           1/24    
FGlobal =
      1/8     
      0       
     -1/8     
      1/4     
      0       
     -1/4     
      1/8     
      0       
     -1/8     

Boundary Conditions[edit]

Noting that essential BCs were specified to be on all sides of the 2x2 square. It would mean (viewing from the above global node figure) all nodes around node 5 will be tagged as essential BC.

Global to Local1

Since we can either eliminate the nodes form Global Stiffness matrix to construct another one with Temperature ('u' function) as unknown or just '0' the node values in Global Stiffness matrix and then do inverse and multiply by Force matrix to find other non-constrained dofs.

We wrote a matlab code that automatically constrains the associate nodes and puts their contribution to force function.

Here how code applies BC's

KGlobal =
      1/24          -1/96           0             -1/96          -1/48           0              0              0              0       
     -1/96           1/12          -1/96          -1/48          -1/48          -1/48           0              0              0       
      0             -1/96           1/24           0             -1/48          -1/96           0              0              0       
     -1/96          -1/48           0              1/12          -1/48           0             -1/96          -1/48           0       
     -1/48          -1/48          -1/48          -1/48           1/6           -1/48          -1/48          -1/48          -1/48    
      0             -1/48          -1/96           0             -1/48           1/12           0             -1/48          -1/96    
      0              0              0             -1/96          -1/48           0              1/24          -1/96           0       
      0              0              0             -1/48          -1/48          -1/48          -1/96           1/12          -1/96    
      0              0              0              0             -1/48          -1/96           0             -1/96           1/24    


Kchange =
      0              0              0              0              0              0              0              0              0       
      0              0              0              0              0              0              0              0              0       
      0              0              0              0              0              0              0              0              0       
      0              0              0              0              0              0              0              0              0       
      0              0              0              0              1/6            0              0              0              0       
      0              0              0              0              0              0              0              0              0       
      0              0              0              0              0              0              0              0              0       
      0              0              0              0              0              0              0              0              0       
      0              0              0              0              0              0              0              0              0       


FGlobal =
     -1/16    
      0       
      1/16    
     -1/8     
      0       
      1/8     
     -1/16    
      0       
      1/16        


Fchange =
     -7/48    
     -7/48    
      0       
     -1/6     
      1/3     
      1/12    
     -7/48    
     -7/48    
      0       


Then the matrices were reduced to the ones of the importance ( Global K matrix after the BCs where applied was isolated so no 0s would be present and attributing F function value(s) where picked by code as well)

KchangeF =
      1/6     


FchangeF =
     1/3     


Finally Temperature at unknown nodes was found by using this formula

u =
     2     

Also output for 4x4 element free node values came out to be:

dcalc =
      2       
      2       
      2       
      2       

as you may see, only 1 node, in this example was present, so only 1 numerical value was generated.

FEATeam5HW7p1

Discussion[edit]

Note 1 important aspect of this output. That is, with forcing function changed to 1, the output of free degrees of freedom is still '2' as if no forcing function was present to begin with. I guess it might be explainable that because of constrained boundary conditions at all 4 ends, in order to keep the temperature constant at all 4 ends while heat was generated, the generated heat had to be instantaneously dissipated though the boundaries of the area, so that boundary temperature would stay unchanged.

As in matrix/numeral explanation, it should be noted, that because essential boundary condition is applied over all 4 sides, it has integration limits (working boundary range) as all element boundary. Therefore, with intorduction with heat source, force instead of being modified, was only scaled (remember force equation):

for f = 0 (HW 6.6)


for f= 1 (HW 7.1)

you may see that force term is a scale only term. Moreover, notice the force global term

For 4 element at f = 0:

FGlobal =
      1/8     
      0       
     -1/8     
      1/4 
      0      

     -1/4     
      1/8     
      0       
     -1/8     


Fchange =
      5/24    
      7/48    
     -1/16    
      7/24   
     -1/3     

     -5/24    
      5/24    
      7/48    
     -1/16    

For 4 element at f = 1:

FGlobal =

     -1/16    
      0       
      1/16    
     -1/8     
      0      

      1/8     
     -1/16    
      0       
      1/16    


Fchange =

     -7/48    
     -7/48    
      0       
     -1/6     
     -1/3     

      1/12    
     -7/48    
     -7/48    
      0       


For 9 element at f = 1

FGlobal =

     -1/36    
      0       
      0       
      1/36    
     -1/18    
      0 
      0     

      1/18    
     -1/18    
      0 
      0     

      1/18    
     -1/36    
      0       
      0       
      1/36    


Fchange =

     -7/108   
     -7/108   
     -7/108   
      0       
    -11/108  
      5/54    
      5/54    

      1/108   
     -5/54  
      5/54    
      5/54    

      1/54    
     -7/108   
     -7/108   
     -7/108   
      0       

, where boxed answer shows free nodes.

Note that force matrix DOES differ, but note that because of the similarity of force functions (scalar multiples of each other, namely, caused because forcing function is a constant and essential boundary condition acts on the same area as forcing function), the global force matrix, since it contained '0's on initial force function, will contain '0's on any other force function, as long as forcing function is scalar and essential B.C is the same as whole element B.C. Therefore, with same used stiffness matrix 'K' the adjusted force matrix with initial values of '0' on free nodes, always came out to be the same for specific number of elements with differing scalar value of forcing function.

As a result, for free node displacements, we observed the constant value of '2' .

--Eml5526.s11.team5.JA 20:51, 6 April 2011 (UTC)

2)[edit]

Transient Term[edit]


Overall Equation[edit]

References[edit]

  1. http://en.wikiversity.org/w/index.php?title=File:Fe1.s11.mtg30.djvu&page=5


NOTE THAT FOR LLIB CASE, THERE WAS ONLY 1 ELEMENT PRESENT (THE GLOBAL MATRIX ELEMENT) AND SINCE ALL BCs ON ELEMENT WERE CONSTRAINED BY ESSENTIAL BOUNDARIES, DUE TO THE LACK OF A FREE ELEMENT, A SOLUTION COULD NOT BE FOUND REGARDLESS OF THE NUMBER OF BASIS FUCNTIONS.

Static using 2d LLEBF uniform mesh[edit]

N=4
N=6
N=8
N=10


As you can see for this case the temperature is 2 at every node which is to be expected.

Static using 2d LLEBF non-uniform mesh[edit]

N=4
N=6
N=8
N=10

Here the answer is also 2 at every node, which agrees with the uniform mesh LLEBF. There is a slight difference between nodes which can be attributed to matlab roundoff error.

Dynamic (case 2b) using 2d LLEBF uniform mesh[edit]

N=2
N=3
N=4
N=5

Dynamic (case 2b) using 2d LLEBF non-uniform mesh[edit]


N=3
N=4
N=5


Eml5526.s11.team5.savery

Eml5526.s11.team5.smith 20:37, 21 April 2011 (UTC)

Eml5526.s11.team5.smith 21:11, 21 April 2011 (UTC)

7.2 Finding the static and dynamic solution of a bi-unit square using 2D LIBF[edit]

Given[edit]

P.D.E[edit]

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Essential Boundary Condition[edit]

Find[edit]

Static(steady state)[edit]

1. Find the static solution to accuracy at the center.

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Dynamic(transient)[edit]

1. Fundamental Frequency and Fundamental Period

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2. Plot for symmetric and Non-Symmetric

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3. Produce a movie of the vibrating membrane for symmetric and Non-Symmetric

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Solution[edit]

Substituting the given values in the P.D.E, we get,

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2D LIBF

Where,

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Consider the Weak Form

where,

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Static case[edit]

For static case, the above equation can be written as,

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where,

Using the above equations in MATLAB, we get,

 K =   33.3257  -13.5386  -13.5386    2.8639
      -13.5386   33.3257    2.8639  -13.5386
      -13.5386    2.8639   33.3257  -13.5386
        2.8639  -13.5386  -13.5386   33.3257
 F =  [0.5625  0.5625  0.5625  0.5625]^T
 d =  [0.0617  0.0617  0.0617  0.0617]^T


Matlab Code for Membrane (Static case)[edit]

Membrane (Static case)[edit]

Figure 1: Static Membrane

Dynamic case[edit]

For this case, It is given that the value of mass density

Free Vibration: Solve Eigen Value problem[edit]

The equation given to solve the Eigen value problem is,

Where,

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Using the above equation and the expression for K, F matrix, we can calculate the fundamental frequency and fundamental time period.

Fundamental Frequency = 0.23725

Fundamental Time Period = 4.2149

MATLAB code for finding the fundamental frequency and Time period of the vibrating membrane[edit]

Transient Analysis with symmetric Uo[edit]

Initial conditions



By using the above initial condition, we can produce the movie of the vibrating membrane.

Movie of the vibrating membrane[edit]

The Movie of dynamic membrane has been uploaded in the youtube and it can be seen from the link below

FE Analysis of a Dynamic Membrane - symmetric case

For T = 0,

Figure 2: Dynamic Membrane
MATLAB code for producing the movie of the vibrating membrane[edit]

Transient Analysis with Non-symmetric Uo[edit]

In this case, initial u value is given by,

By using the above initial condition, we can produce the movie of the vibrating membrane.

Movie of the vibrating membrane[edit]

The Movie of dynamic membrane for the non-symmetric case, has been uploaded in the youtube and it can be seen from the link below

FE Analysis of a Dynamic membrane - Non Symmetric case

Figure 3: Dynamic Membrane


--Eml5526.s11.team5.srv 21:06, 21 April 2011 (UTC)

MATLAB code for producing the movie of the vibrating membrane[edit]

--Lokeshdahiya 21:09, 21 April 2011 (UTC)

7.3 Static Solution of a Circle Using Quadrilateral and Triangular Elements[edit]

Given/Find: Static Solution of the Circle[edit]

Let Omega a circle with a one-unit radius. Find the static solution such that: in Omega on

all to 10^-6 accuracy at the center. Quadrilateral and Triangular elements should be used but for only 1/4 of the circle. Plot the deformed shape in 3D.

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Solution: Static Solution of the Circle Using Quadrilateral and Triangular Elements to an Accuracy of 10^-6[edit]

The Partial Differential Equation for a vibrating membrane is the following via transient analysis:

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Where: Transverse Displacement Tension (Force/Length) = Constant Distributed Transverse Force Mass Density (Mass/Area) &

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The Weighted Residual Form is as follows:

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Which can thereby be broken into three separate terms:

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This final equation will be used later in the Weak Form. For now, the first term above (set equal to ), is equivalent to:

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And by using divergence theorem, the first term of the equation above may be converted to yield:

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Now since

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and the following are true:

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Then the K equation becomes:

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Otherwise written as:

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where the last term is conveniently used to write ultimately the weak form of the weighted residual form which is the following:

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From the Discrete Weighted Form,

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But because g=0 from the boundary conditions, as well as M due to the static requirement, the displacements may be derived:

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Now to solve for the displacements:

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Where

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and

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And setting the Shape Function to the following:

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The deformed shape coordinates may further be derived from these following two equations, where the summations are taken across all coordinate points taken from Abaqus output files:

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Using the following Jacobian matrix:

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The following equations may now be utilized:

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Implementing MATLAB at this point, in conjunction with output data files from Abaqus which contain element coordinate points for three separate quadrilateral meshes and three separate triangular meshes (included), the final displacements may be derived, to an accuracy of 10^-6:

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The above MATLAB script for the 6 different meshes generate the following results; the first three show quadrilateral elements in order of increasing elements. The last three show triangular elements in order of increasing elements.

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Fig quad4.png The figure to the left shows a unit-less displacement of: 0.0623258

Fig quad1.png The figure to the left shows a unit-less displacement of: 0.0624613

Fig quad2.png The figure to the left shows a unit-less displacement of: 0.0625011

The figure to the left shows a unit-less displacement of: 0.0624613

Fig tri4.png The figure to the left shows a unit-less displacement of: 0.0639197

Fig tri1.png The figure to the left shows a unit-less displacement of: 0.0619208