# University of Florida/Eml4507/s13.team4ever.R7

## Problem R7.1a: Verify the dim of the ${\displaystyle {\underline {k}}_{6x6}^{(e)}{\underline {d}}_{6x1}^{(e)}}$ 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

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

### Find:

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

Obtain:

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

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

From:

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

### Solution: Verify the dimensions of k and d

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

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

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

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

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

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

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

## 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:

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

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

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

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

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

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

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

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

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

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

## 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:

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

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

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

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

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

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

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

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

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

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

## 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