University of Florida/Eml4507/s13.team4ever.R7

Problem R7.1a: Verify the dim of the ${\underline {k}}_{6x6}^{(e)}{\underline {d}}_{6x1}^{(e)}$ matrices (fead.f08.mtgs.[37-41] pg. 2)

On our honor, we did this assignment on our own.

Given: The desired dimensions for k and d with index of 6

${\underline {k}}_{6x6}^{(e)}{\underline {d}}_{6x1}^{(e)}$

Find:

1.Use an index of 6 to verify the dimensions of k and d

Obtain:

${\underline {k}}_{6x6}^{(e)}{\underline {d}}_{6x1}^{(e)}={\underline {f}}_{6x1}^{(e)}$

With supporting: ${\underline {k}}_{6x6}^{(e)}={\underline {\tilde {T}}}_{6x6}^{(e)T}{\underline {\tilde {k}}}_{6x6}^{(e)}{\underline {\tilde {T}}}_{6x6}^{(e)}$

From:

${\underline {\tilde {k}}}_{6x6}^{(e)}{\underline {\tilde {d}}}_{6x1}^{(e)}={\underline {\tilde {f}}}_{6x1}^{(e)}$

Solution: Verify the dimensions of k and d

The constructed 6x6 matrix shown below represents ${\underline {\tilde {T}}}$

${\begin{Bmatrix}{\tilde {d}}_{1}\\{\tilde {d}}_{2}\\{\tilde {d}}_{3}\\{\tilde {d}}_{4}\\{\tilde {d}}_{5}\\{\tilde {d}}_{6}\end{Bmatrix}}={\begin{Bmatrix}R_{11}&R_{12}&0&0&0&0\\R_{21}&R_{22}&0&0&0&0\\0&0&1&0&0&0\\0&0&0&R_{11}&R_{12}&0\\0&0&0&R_{21}&R_{22}&0\\0&0&0&0&0&1\end{Bmatrix}}{\begin{Bmatrix}d_{1}\\d_{2}\\d_{3}\\d_{4}\\d_{5}\\d_{6}\end{Bmatrix}}$

The following matrix represents the element stiffness matrix in local coordinates.

${\underline {\tilde {k}}}_{6x6}^{(e)}={\begin{bmatrix}{\frac {EA}{L}}&0&0&{\frac {-EA}{L}}&0&0\\0&{\frac {12EI}{L^{3}}}&{\frac {6EI}{L^{2}}}&0&{\frac {12EI}{L^{3}}}&{\frac {6EI}{L^{2}}}\\0&{\frac {6EI}{L^{2}}}&{\frac {4EI}{L}}&0&{\frac {-6EI}{L^{2}}}&{\frac {2EI}{L}}\\{\frac {-EA}{L}}&0&0&{\frac {EA}{L}}&0&0\\0&{\frac {-12EI}{L^{3}}}&{\frac {-6EI}{L^{2}}}&0&{\frac {12EI}{L^{3}}}&{\frac {-6EI}{L^{2}}}\\0&{\frac {6EI}{L^{2}}}&{\frac {2EI}{L}}&0&{\frac {-6EI}{L^{2}}}&{\frac {4EI}{L}}\end{bmatrix}}$

Using the above two matricies and the transpose of T, you can construct the following equation:

${\underline {k}}_{6x6}^{(e)}={\underline {\tilde {T}}}_{6x6}^{(e)T}{\underline {\tilde {k}}}_{6x6}^{(e)}{\underline {\tilde {T}}}_{6x6}^{(e)}$

Multiplying ${\underline {\tilde {k}}}_{6x6}^{(e)}$ by ${\underline {\tilde {d}}}_{6x1}^{(e)}$ will yield the required ${\underline {\tilde {f}}}_{6x1}^{(e)}$

Problem R7.1b: Solve 2 element frame system (fead.f08.mtgs.[37-41] pg. 3)

On our honor, we did this assignment on our own.

Given: Information on the two element truss system

Assume a square cross section. Also use same data from fea.f08.mtgs.p5-4.

Figure 1: Two member system (fig.JPG)

Element length: (1) =4

Element length: (2) =2

Young's modulus: (1) =3

Young's modulus: (2) =5

Cross section area: (1) =1

Cross section area: (2) =2

Inclination angle: (1) = 30 deg

Inclination angle: (2) = -45 deg

Solution:

1. Plot Undeformed Shape

Undeformed Shape.

Figure 1: Two member system (2bnodef.JPG)

View of element with single force member view.

Figure 1: Two member system (2b.JPG)

2. Plot Deformed Shape 2-bar Truss

Plotting the deformed shape 2-bar truss for the given problem from p5-2 of the fead08 lecture notes. The deformation can be seen the truss using the forces given from p5-2.

Figure 1: Two member system (2bdef2.JPG)

3. Plot Deformed Shape 2-bar Frame

Plotting the deformed shape 2-bar frame for the given problem from p5-2 of the fead08 lecture notes. The constraints are preserved during the analysis in this mode to keep the frame.

Figure 1: Two member system (2bdef.JPG)

ref

for i=1:2:5
    x_1((i+1)/2)=P(1,((i+1)/2))+V(i,5);
end

for i=2:2:6
    y_1((i/2))=P(2,(i/2))+V(i,1);
end
 
%Required deformation plotting
hold on
for i=1:2
    l1=N(1,i);
    l2=N(2,i);
    x1=[x_1(l1),x_1(l2)];
    y1=[y_1(l1),y_1(l2)];
    axis([-1 5 -1 5])
    plot(x1,y1,'b')
    title(['Constrained Shape'])
    hold on
end

%Required deformation plotting
for i=1:2:5
    x_o((i+1)/2)=P(1,((i+1)/2));
end

for i=2:2:6
    y_o((i/2))=P(2,(i/2));
end
 for i=1:2
    l1=N(1,i);
    l2=N(2,i);
    x1=[x_o(l1),x_o(l2)];
    y1=[y_o(l1),y_o(l2)];
    axis([-1 5 -1 5])
    plot(x1,y1,'-.or')
    hold on
 end
end

Problem R7.2

On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

Description

We are to resolve problem 5.7. We are to solve for motion of the truss using modal superposition using the three lowest eigenvalues.

Solution

The following code was used for problem 5.7 to obtain the K and M matrices and to obtain the eigenvalues and the eigenvectors:

function R7P2a

p = 2;
E = 5;
A = 0.5;
L = 1;
Ld = 1*sqrt(2);
 
%degree of freedom for each element
dof = zeros(10,5);
dof(1,:) = [1 1 2 3 4];
dof(2,:) = [2 1 2 5 6];
dof(3,:) = [3 3 4 5 6];
dof(4,:) = [4 5 6 9 10];
dof(5,:) = [5 5 6 7 8];
dof(6,:) = [6 3 4 9 10];
dof(7,:) = [7 3 4 7 8];
dof(8,:) = [8 7 8 9 10];
dof(9,:) = [9 9 10 11 12];
dof(10,:) = [10 7 8 11 12];
 
%position of each node
pos = zeros(2,6);
pos(:,1) = [0;0];
pos(:,2) = [L;0];
pos(:,3) = [L;L];
pos(:,4) = [2*L;0];
pos(:,5) = [2*L;L];
pos(:,6) = [3*L;0];
 
%Connections
conn = zeros(2,10);
conn(:,1) = [1;2];
conn(:,2) = [1;3];
conn(:,3) = [2;3];
conn(:,4) = [3;5];
conn(:,5) = [3;4];
conn(:,6) = [2;5];
conn(:,7) = [2;4];
conn(:,8) = [4;5];
conn(:,9) = [5;6];
conn(:,10) = [4;6];
 
%seperates into x and y coordinates
x = zeros(10,2);
y = zeros(10,2);
for i = 1:10
    x(i,:) = [pos(1,conn(1,i)),pos(1,conn(2,i))];
    y(i,:) = [pos(2,conn(1,i)),pos(2,conn(2,i))];
end
 
%Set up K and M matrix
K = zeros(12);
M = zeros(12);
for i = 1:10
    xm = x(i,2)-x(i,1);
    ym = y(i,2)-y(i,1);
    L = sqrt(xm^2+ym^2);
    l = xm/L;
    m = ym/L;
    k = E*A/L*[l^2 l*m -l^2 -l*m;l*m m^2 -l*m -m^2;-l^2 -l*m l^2 l*m;-l*m -m^2 l*m m^2;];
    m = L*A*p;
    m = [m/2 0 0 0;0 m/2 0 0;0 0 m/2 0;0 0 0 m/2];
    Dof = dof(i,2:5);
    K(Dof,Dof) = K(Dof,Dof)+k;
    M(Dof,Dof) = M(Dof,Dof)+m;
end
bc = [1;2;12];
K
M
F = [0;0;0;0;0;0;0;-5;0;0;0;0];

[L,X] = eigen(K,M,bc)

From this, we obtain M and K matrices which are:

K =

 Columns 1 through 9

   3.3839    0.8839   -2.5000         0   -0.8839   -0.8839         0         0         0
   0.8839    0.8839         0         0   -0.8839   -0.8839         0         0         0
  -2.5000         0    5.8839    0.8839         0         0   -2.5000         0   -0.8839
        0         0    0.8839    3.3839         0   -2.5000         0         0   -0.8839
  -0.8839   -0.8839         0         0    4.2678         0   -0.8839    0.8839   -2.5000
  -0.8839   -0.8839         0   -2.5000         0    4.2678    0.8839   -0.8839         0
        0         0   -2.5000         0   -0.8839    0.8839    5.8839   -0.8839         0
        0         0         0         0    0.8839   -0.8839   -0.8839    3.3839         0
        0         0   -0.8839   -0.8839   -2.5000         0         0         0    4.2678
        0         0   -0.8839   -0.8839         0         0         0   -2.5000         0
        0         0         0         0         0         0   -2.5000         0   -0.8839
        0         0         0         0         0         0         0         0    0.8839

 Columns 10 through 12

        0         0         0
        0         0         0
  -0.8839         0         0
  -0.8839         0         0
        0         0         0
        0         0         0
        0   -2.5000         0
  -2.5000         0         0
        0   -0.8839    0.8839
   4.2678    0.8839   -0.8839
   0.8839    3.3839   -0.8839
  -0.8839   -0.8839    0.8839

M =

 Columns 1 through 9

   1.2071         0         0         0         0         0         0         0         0
        0    1.2071         0         0         0         0         0         0         0
        0         0    2.2071         0         0         0         0         0         0
        0         0         0    2.2071         0         0         0         0         0
        0         0         0         0    2.4142         0         0         0         0
        0         0         0         0         0    2.4142         0         0         0
        0         0         0         0         0         0    2.2071         0         0
        0         0         0         0         0         0         0    2.2071         0
        0         0         0         0         0         0         0         0    2.4142
        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

 Columns 10 through 12

        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
   2.4142         0         0
        0    1.2071         0
        0         0    1.2071

This will also output the lowest eigenpairs:

$\gamma _{1},\phi _{1}$

$\gamma _{2},\phi _{2}$

$\gamma _{3},\phi _{3}$

Then, with these, we will find what the modal equations are according to the equation:

$z^{''}+\gamma _{i}z=\phi _{i}^{T}F(t)$

This will yield 3 unique differential equations. We solve for the complete solution to these differential equation using the boundary conditions from:

$z_{i}(0)={\bar {\phi }}_{i}^{T}Md(0)$ and

$z_{i}^{'}(0)={\bar {\phi }}_{i}^{T}Md^{'}(0)$

With these solutions for z, we can then find and plot the actual displacements from modal superposition using:

$d(t)=\Sigma _{j=1}^{3}z_{j}\phi _{j}$

Problem R7.2

On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

Description

We are to resolve problem 5.7. We are to solve for motion of the truss using modal superposition using the three lowest eigenvalues.

Solution

The following code was used for problem 5.7 to obtain the K and M matrices and to obtain the eigenvalues and the eigenvectors:

function R7P2a

p = 2;
E = 5;
A = 0.5;
L = 1;
Ld = 1*sqrt(2);
 
%degree of freedom for each element
dof = zeros(10,5);
dof(1,:) = [1 1 2 3 4];
dof(2,:) = [2 1 2 5 6];
dof(3,:) = [3 3 4 5 6];
dof(4,:) = [4 5 6 9 10];
dof(5,:) = [5 5 6 7 8];
dof(6,:) = [6 3 4 9 10];
dof(7,:) = [7 3 4 7 8];
dof(8,:) = [8 7 8 9 10];
dof(9,:) = [9 9 10 11 12];
dof(10,:) = [10 7 8 11 12];
 
%position of each node
pos = zeros(2,6);
pos(:,1) = [0;0];
pos(:,2) = [L;0];
pos(:,3) = [L;L];
pos(:,4) = [2*L;0];
pos(:,5) = [2*L;L];
pos(:,6) = [3*L;0];
 
%Connections
conn = zeros(2,10);
conn(:,1) = [1;2];
conn(:,2) = [1;3];
conn(:,3) = [2;3];
conn(:,4) = [3;5];
conn(:,5) = [3;4];
conn(:,6) = [2;5];
conn(:,7) = [2;4];
conn(:,8) = [4;5];
conn(:,9) = [5;6];
conn(:,10) = [4;6];
 
%seperates into x and y coordinates
x = zeros(10,2);
y = zeros(10,2);
for i = 1:10
    x(i,:) = [pos(1,conn(1,i)),pos(1,conn(2,i))];
    y(i,:) = [pos(2,conn(1,i)),pos(2,conn(2,i))];
end
 
%Set up K and M matrix
K = zeros(12);
M = zeros(12);
for i = 1:10
    xm = x(i,2)-x(i,1);
    ym = y(i,2)-y(i,1);
    L = sqrt(xm^2+ym^2);
    l = xm/L;
    m = ym/L;
    k = E*A/L*[l^2 l*m -l^2 -l*m;l*m m^2 -l*m -m^2;-l^2 -l*m l^2 l*m;-l*m -m^2 l*m m^2;];
    m = L*A*p;
    m = [m/2 0 0 0;0 m/2 0 0;0 0 m/2 0;0 0 0 m/2];
    Dof = dof(i,2:5);
    K(Dof,Dof) = K(Dof,Dof)+k;
    M(Dof,Dof) = M(Dof,Dof)+m;
end
bc = [1;2;12];
K
M
F = [0;0;0;0;0;0;0;-5;0;0;0;0];

[L,X] = eigen(K,M,bc)

From this, we obtain M and K matrices which are:

K =

 Columns 1 through 9

   3.3839    0.8839   -2.5000         0   -0.8839   -0.8839         0         0         0
   0.8839    0.8839         0         0   -0.8839   -0.8839         0         0         0
  -2.5000         0    5.8839    0.8839         0         0   -2.5000         0   -0.8839
        0         0    0.8839    3.3839         0   -2.5000         0         0   -0.8839
  -0.8839   -0.8839         0         0    4.2678         0   -0.8839    0.8839   -2.5000
  -0.8839   -0.8839         0   -2.5000         0    4.2678    0.8839   -0.8839         0
        0         0   -2.5000         0   -0.8839    0.8839    5.8839   -0.8839         0
        0         0         0         0    0.8839   -0.8839   -0.8839    3.3839         0
        0         0   -0.8839   -0.8839   -2.5000         0         0         0    4.2678
        0         0   -0.8839   -0.8839         0         0         0   -2.5000         0
        0         0         0         0         0         0   -2.5000         0   -0.8839
        0         0         0         0         0         0         0         0    0.8839

 Columns 10 through 12

        0         0         0
        0         0         0
  -0.8839         0         0
  -0.8839         0         0
        0         0         0
        0         0         0
        0   -2.5000         0
  -2.5000         0         0
        0   -0.8839    0.8839
   4.2678    0.8839   -0.8839
   0.8839    3.3839   -0.8839
  -0.8839   -0.8839    0.8839

M =

 Columns 1 through 9

   1.2071         0         0         0         0         0         0         0         0
        0    1.2071         0         0         0         0         0         0         0
        0         0    2.2071         0         0         0         0         0         0
        0         0         0    2.2071         0         0         0         0         0
        0         0         0         0    2.4142         0         0         0         0
        0         0         0         0         0    2.4142         0         0         0
        0         0         0         0         0         0    2.2071         0         0
        0         0         0         0         0         0         0    2.2071         0
        0         0         0         0         0         0         0         0    2.4142
        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

 Columns 10 through 12

        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
   2.4142         0         0
        0    1.2071         0
        0         0    1.2071

This will also output the lowest eigenpairs:

$\gamma _{1},\phi _{1}$

$\gamma _{2},\phi _{2}$

$\gamma _{3},\phi _{3}$

Then, with these, we will find what the modal equations are according to the equation:

$z^{''}+\gamma _{i}z=\phi _{i}^{T}F(t)$

This will yield 3 unique differential equations. We solve for the complete solution to these differential equation using the boundary conditions from:

$z_{i}(0)={\bar {\phi }}_{i}^{T}Md(0)$ and

$z_{i}^{'}(0)={\bar {\phi }}_{i}^{T}Md^{'}(0)$

With these solutions for z, we can then find and plot the actual displacements from modal superposition using:

$d(t)=\Sigma _{j=1}^{3}z_{j}\phi _{j}$

Table of Assignments R7

Problem Assignments R5
Problem #	Solved&Typed by	Reviewed by
1a	Vernon Babich, Chad Colocar	All
1b	Vernon Babich, Tyler Wulterkens	All
2	David Bonner, Chad Colocar	All