User:Eml4500.f08.Lulz.mitchell.bm

BEN IS FREAKING AWESWOME
Just kidding

HW1 Notes
8-25: Introduced website and went over syllabus.

8-27: Set the deadline for submitting the team structure for Wed, 3 Sep, 2008.

Teams should consist of 5 or 6 students. Teams are to submit a paper at the end of class on 3 Sep with the following information:
 * Team name; should be 4-6 characters
 * Member names and UF-ID numbers
 * Wiki usernames

8-29: The class namespace is announced as being Eml4500.fo8.

9-3: Began lecturing on "Trusses, Matrix Method" from chapter 4 of the book.

Notation:
 * Global Node Number - a number inside a circle; designates a node for the whole structure.
 * Local Node Number - a number inside a square; designates a node for a specific element.
 * Element Number - a number inside a triangle; designates a specific element of the entire structure.

Here we see a truss with 2 elastic, deformable bars

Here we see the global free body diagram for the truss system.

The R's represent unknown reaction forces. The P is a known applied force. This FBD has 4 unknowns and 3 equations of equilibrium which makes it statically indeterminate.

2 FBDs of 2 bar elements:





Notation: fi(e) Ex: f2(1) represents the 2nd internal force in the 1st element.
 * i - ith internal force
 * (e) - element number
 * displacements are written in the same manner: di(e)

Since corresponding forces and displacements act along the same line, they can be represented by the same arrow.

Ex: f1(2),d1(2)

The next big step is to come up with Force-Displacement (FD) relationships.

Below we see the FD relationship of a 1 dimensional spring element with one end fixed followed by the FD relationship for a 1 dimensional spring element with both ends free.



Note: An elastic bar inclined in a plane is a 2 dimensional spring.

The FD relationship for the system can be written in matrix form with forces and displacements arranged as column matrices. The k matrix is called the stiffness matrix.

$$\binom {f1}{f2}=\begin{vmatrix} k & -k\\ -k &k \end{vmatrix} \binom{d1}{d2}$$

Some special cases to consider:

Case 1: An observer sits on node 1.

The observer would move with node 1 and the spring would seem to stretch by d2-d1.

f2=k(d1-d2)

Case 2: An observer sits on node 2.

The observer would move with node 2.

f1=-f2=-k(d2-d1)

9-5: Reading assignment: Ch 4 (Trusses, Beams, Frame) Ch 1 (Big Picture): 1.1 Discretization 1.1.1 Plane Truss elements 1.2 Assembly of Element equations (Ex: 1.4 Five-bar trusses) 1.4 Element solution of model ralidity 1.4.1 Plane Truss elemnts

Steps to solve a simple truss:

1. global picture (description); at structure level
 * global dofs (disp. dofs); disp. unknowns
 * global forces; usually known except for reactions

disp. dofs are partitioned into:
 * a known part; e.g. fixed dofs, constraints
 * an unknown part; solved using FEM

similarly for global forces:
 * a known part; applied forces
 * an unknown part; reaction forces

2. element picture
 * element dofs (either local or global coordinates)
 * element forces (either local or global coordinates)

3. global force-displacement
 * element stiffness matrices in global coordinates
 * element force matrices in global coordinates
 * assembly of element stiffness matrices and element force matrices into global force-displacement relationship: kd=F
 * k is NxN;d is Nx1;F is Nx1
 * N = # of known and unknown displacement dofs
 * "free-free" system, unconstrained, singular, determinate is 0

4. elimination of known dops to reduce the global force-displacement relationship
 * stiffness matrix is non-singular; therefore invertible
 * can solve the system of equations - kd=F
 * only known dofs left
 * k is MxM; d is Mx1;F is Mx1
 * M = # of unknown displacement dofs
 * k is non-singular; therefore k-1 exists
 * therefore d=k-1F

5.compute element forces from now known d
 * gives element stresses

6.compute reactions (unknown forces)

Specific example (to see how the recipe works)

(pic)

Element Lenght:
 * L(1)=4
 * L(2)=2

Young's Modulus:
 * E(1)=3
 * E(2)=5

X-sectional Area:
 * A(1)=1
 * A(2)=2

Inclination Angle:
 * o (1)=30 deg
 * o (2)=-45 deg

1)Global Picture:

-global dofs

(pic)

-numbering the disp. dofs and global forces
 * follow order of nodes
 * for each node: label x then y

-it is possible to arrange global forces and disp. dofs in matrix form

(matrix)

HW 2 Notes
From where we left off.

The second step is to draw the element pictures. In these pictures you should label the element dofs and the element forces in either global or local coordinates.

(Pic)

The third step is to develop the global force-displacement relationship at the element level. As we saw earlier, for a specified element the force-displacement relationship is: $$\textbf{k}^{(e)}*\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

Where k(e) is the element stiffness matrix, d(e) is the element displacement matrix and f(e) the element force matrix.

The axial stiffness of a specified element is determined by the equation:

$$k=\frac{E^{(e)}A^{(e)}}{L^{(e)}}$$

Where E is the Young's modulus for the element, A is the cross-sectional area for the element and L is the length if the element.

Now we need to find the director cosines (l(e) and m(e)) for each element using the following equations:

$$l^{(e)}=\textbf{i}^{\sim}\cdot\textbf{i}=cos\theta^{(e)}$$

$$m^{(e)}=\textbf{j}^{\sim}\cdot\textbf{j}=sin\theta^{(e)}$$

(pic)

Using the director cosines, axial stiffness and the following formula we can determine the element stiffness matrix.

$$\textbf{k}^{(e)}=k^{(e)}\left[\begin{array}{c c c c} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2\\ (l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{array}\right ]$$

For element 1:

$$\theta=30$$

$$l=cos\theta=\frac{\sqrt{3}}2$$

$$m=sin\theta=\frac{1}2$$

$$k=\frac{EA}L=\frac{(3)(1)}4=\frac{3}4$$

This gives us the stiffness matrix for element 1:

$$k^{(1)}=\left[\begin{array}{c c c c} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{array}\right]$$

Notice that the stiffness matrix for element 1 is symmetric.

For element 2:

$$\theta=-45$$

$$l=cos(-45)=\frac{\sqrt{2}}2$$

$$m=sim(-45)=\frac{\sqrt{2}}2$$

$$k=\frac{EA}L=\frac{(5)(2)}2=5$$

This gives us the stiffness matrix for element 2:

$$k^{(2)}=\left[\begin{array}{c c c c} \frac{5}2 & -\frac{5}2 & \frac{5}2 & -\frac{5}2\\ -\frac{5}2 & \frac{5}2 & -\frac{5}2 & \frac{5}2\\ -\frac{5}2 & \frac{5}2 & -\frac{5}2 & \frac{5}2\\ \frac{5}2 & -\frac{5}2 & \frac{5}2 & -\frac{5}2 \end{array}\right]$$

Notice that the stiffness matrix for element 2 is also symmetric.

Now we have to assemble the global stiffness matrix. This is illustrated in the picture below where the shaded area represents the addition of overlapping parts of $$\textbf{k}^{(1)}$$ and $$\textbf{k}^{(2)}$$.



$$K_{33}=k_{33}^{(1)} + k_{11}^{(2)}$$

$$K_{34}=k_{34}^{(1)} + k_{12}^{(2)}$$

$$K_{43}=k_{43}^{(1)} + k_{21}^{(2)}$$

$$K_{44}=k_{44}^{(1)} + k_{22}^{(2)}$$

For this example we get:

$$\textbf{K}=\left[\begin{array}{c c c c c c} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{49}{16} & \frac{-40+3\sqrt{3}}{16} & -\frac{5}2 & -\frac{5}2\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{-40+3\sqrt{3}}{16} & \frac{43}{16} & -\frac{5}2 & -\frac{5}2\\ 0 & 0 & -\frac{5}2 & -\frac{5}2 & \frac{5}2 & \frac{5}2\\ 0 & 0 & -\frac{5}2 & -\frac{5}2 & \frac{5}2 & \frac{5}2 \end{array}\right]$$

Now on the fourth step we eliminate known dofs to reduce the global force-displacement relationship.

Due to the applied constraints, $$d_1=d_2=d_5=d_6=0$$. This allows us to eliminate columns 1,2,5 and 6 due to the principal of virtual work. We also note that $$F_1=F_2=F_3=F_5=F_6=0$$. This allows us to eliminate rows 1,2,5 and 6 by the same principal. The resulting, simplified force-displacement relationship follows:

$$\left[\begin{array}{c c} K_{33} & K_{34}\\ K_{34} & K_{44} \end{array}\right] \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix}=\begin{Bmatrix} F_3\\ F_4 \end{Bmatrix}$$

Let $$F_3=0$$ and $$F_4=10$$.Now plugging in numbers we get:

$$\left[\begin{array}{c c} \frac{49}{16} & \frac{-40+3\sqrt{3}}{16}\\ \frac{-40+3\sqrt{3}}{16} & \frac{43}{16} \end{array}\right] \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix}= \begin{Bmatrix} 0\\ 10 \end{Bmatrix}$$

Now we can solve for the unknown displacement dofs by multiplying both sides by:

$$\textbf{K}^{-1}=\frac{1}{det\textbf{K}} \left[\begin{array}{c c} K_{44} & -K_{34}\\ -K_{43} & K_{33} \end{array}\right]$$.

Plugging in numbers we get:

$$\textbf{K}^{-1}=\frac{1}{3.4988} \left[\begin{array}{c c} \frac{49}{16} & \frac{40-3\sqrt{3}}{16}\\ \frac{40-3\sqrt{3}}{16} & \frac{43}{16} \end{array}\right]$$

It is important to note that: $$\textbf{K}^{-1}\neq\frac{1}{det\textbf{K}}\textbf{K}^T$$

Following through with this, we get:

$$\begin{Bmatrix} d_3\\ d_4\end{Bmatrix}= \textbf{K}^{-1}\begin{Bmatrix} F_3\\ F_4\end{Bmatrix}$$

Plugging in the numbers now gives:

$$\begin{Bmatrix} d_3\\ d_4\end{Bmatrix}= \left[\begin{array} {c c} .8753 & .6217\\ .627 & .7681 \end{array}\right] \begin{Bmatrix} 0\\ 10\end{Bmatrix}$$

Solving this system gives us:

$$d_3=6.217$$

$$d_4=7.681$$

Now in step five we can compute reactions by one of two methods.

The first method is to use the element force-displacement relationship.

$$\textbf{k}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

For element 1 we would have:

$$\left[\begin{array}{c c c c} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{array}\right]\begin{Bmatrix} 0\\ 0\\ 6.217\\ 7.681\end{Bmatrix}= \begin{Bmatrix} f_1^{(1)}\\ f_2^{(1)}\\ f_3^{(1)}\\ f_4^{(1)}\end{Bmatrix}$$

Solving this system we get:

$$f_1^{(1)}=5.992$$

$$f_2^{(1)}=3.459$$

$$f_3^{(1)}=-5.992$$

$$f_4^{(1)}=-3.459$$

For element 2 we have:

$$\left[\begin{array}{c c c c} \frac{5}2 & -\frac{5}2 & \frac{5}2 & -\frac{5}2\\ -\frac{5}2 & \frac{5}2 & -\frac{5}2 & \frac{5}2\\ -\frac{5}2 & \frac{5}2 & -\frac{5}2 & \frac{5}2\\ \frac{5}2 & -\frac{5}2 & -\frac{5}2 & \frac{5}2 \end{array}\right]\begin{Bmatrix} 6.217\\ 7.681\\ 0\\ 0\end{Bmatrix}= \begin{Bmatrix} f_1^{(2)}\\ f_2^{(2)}\\ f_3^{(2)}\\ f_4^{(2)}\end{Bmatrix}$$

Solving this system we get:

$$f_1^{(2)}=-3.660$$

$$f_2^{(2)}=3.660$$

$$f_3^{(2)}=3.660$$

$$f_4^{(2)}=-3.660$$

The second method is to solve by statics. Since this system is statically indeterminate we must use the Euler cut method.

(pic)

HW 3 Matlab, Problem, and Latex
Matlab-

% Matlab Script %***************************************************** % % Purpose: %  Sketch the undeformed and deformed %  two-bar truss system from the clas %  example problem. % % Author: Benjamin Mitchell % % Created: 10-07-2008 00:25:38 EST % %***************************************************** clear; close; dof = 6; %dof per node d1 = 0;        %disp of node 1 in the x direction d2 = 0;        %disp of node 1 in the y direction d3 = 4.3520;   %disp of node 2 in the x direction d4 = 6.1271;   %disp of node 2 in the y direction d5 = 0;        %disp of node 3 in the x direction d6 = 0;        %disp of node 3 in the y direction %input nodal coordinates num_nodes = 3;              %total number of nodes num_elems = 2;              %total number of elements tot_dofs = 2 * num_nodes;   %total dofs of the system %element 1 details L1 = 4; E1 = 3; A1 = 1; theta1 = 30; %deg thetar1 = (pi * theta1)/180; %element 2 details L2 = 2; E2 = 5; A2 = 2; theta2 = -45; %deg thetar2 = (pi * theta2)/180; %undeformed positions pos(1,:)=[0,L1*cos(thetar1),L1*cos(thetar1) + L2*cos(thetar2)]; pos(2,:)=[0,L1*sin(thetar1),0]; %endeformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end %element connectivity array %follows: %node_conn(local node number, elemnt number) = global node number node_conn(1,1)=1;    %element 1 node_conn(2,1)=2; node_conn(1,2)=2;    %element 2 node_conn(2,2)=3; %plot the undeformed 2-bar truss system for i=1:num_elems node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; axis([-4 10 -4 10]) plot(xx,yy,'--rs') hold on end text(x(node_conn(1,1)),y(node_conn(1,1))-.5,'Global Node 1','HorizontalAlignment','center') text(x(node_conn(2,1)),y(node_conn(2,1))+.5,'Global Node 2','HorizontalAlignment','center') text(x(node_conn(2,2)),y(node_conn(2,2))-.5,'Global Node 3','HorizontalAlignment','center') text(x(node_conn(2,1))/2,y(node_conn(2,1))/2,'Element 1','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,2))-x(node_conn(2,1)))/2,(y(node_conn(1,2)))/2,'Element 2','HorizontalAlignment','center') %deformed positions pos(1,:)=[0 + d1,L1*cos(thetar1) + d3,L1*cos(thetar1) + L2*cos(thetar2) + d5]; pos(2,:)=[0 + d2,L1*sin(thetar1) + d4,0 + d6]; %deformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end %plot the deformed 2-bar truss system for i=1:num_elems hold on    node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; plot(xx,yy,'-s') end text(x(node_conn(2,1)),y(node_conn(2,1))-.5,'Global Node 2','HorizontalAlignment','center') text(x(node_conn(2,1))/2,y(node_conn(2,1))/2,'Element 1','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,2))-x(node_conn(2,1)))/2,(y(node_conn(1,2)))/2,'Element 2','HorizontalAlignment','center') title('Deformed and Undeformed Two-bar Truss System') xlabel('X') ylabel('Y')



Latex-

--12-1 eq 1--

$$\textbf{k}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

--12-1 eq 2--

$$k^{(e)}\hat{\textbf{k}}^{(e)}\textbf{q}^{(e)}=\textbf{p}^{(e)}$$

$$k^{(e)}\left[\begin{array}{c c} 1 & -1\\ -1 & 1 \end{array}\right] \begin{Bmatrix} q_1^{(e)}\\ q_2^{(e)} \end{Bmatrix}= \begin{Bmatrix} P_1^{(e)}\\ P_2^{(e)} \end{Bmatrix}$$

--12-2 relationship--

$$\textbf{q}^{(e)}=\textbf{T}^{(e)}\textbf{f}^{(e)}$$

--12-3 q director cosines--

$$q_1^{(e)}=\vec{d_1^{(e)}}\cdot\vec{\tilde{i}}$$

$$q_1^{(e)}=(d_1^{(e)}\vec{i}+d_2^{(e)}\vec{j})\cdot\vec{\tilde{i}}$$

$$q_1^{(e)}=d_1^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+d_2^{(e)}(\vec{j}\cdot\vec{\tilde{i}})$$

$$(\vec{i}\cdot\vec{\tilde{i}})$$ and $$(\vec{j}\cdot\vec{\tilde{i}})$$ are director cosines

$$q_1^{(e)}=l^{(e)}d_1^{(e)}+m^{(e)}d_2^{(e)}$$

$$q_1^{(e)}=\lfloor\begin{array}{c c} l^{(e)} & m^{(e)} \end{array}\rfloor \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

similarly for local node 2: $$q_2^{(e)}=\lfloor\begin{array}{c c} l^{(e)} & m^{(e)} \end{array}\rfloor \begin{Bmatrix} d_3^{(e)}\\ d_4^{(e)} \end{Bmatrix}$$

$$\begin{Bmatrix} q_1^{(e)}\\ q_2^{(e)} \end{Bmatrix}= \left[\begin{array}{c c c c} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{array}\right] \left\{\begin{array}{c} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)} \end{array}\right\}$$

--14-1 similarly--

$$\begin{Bmatrix} P_1^{(e)}\\ P_2^{(e)} \end{Bmatrix}=\textbf{T}^{(e)} \left\{\begin{array}{c} f_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)} \end{array}\right\}$$

elem axial FD rel

$$\hat{\textbf{k}}^{(e)}\textbf{q}^{(e)}=\textbf{P}^{(e)}$$

--14-2--

$$\hat{\textbf{k}}^{(e)}(\textbf{T}^{(e)}\textbf{d}^{(e)})=\textbf{T}^{(e)}\textbf{f}^{(e)}$$

want: $$\hat{\textbf{k}}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

so move $$\textbf{T}^{(e)}$$ to left hand side.

$$\textbf{T}^{(e)}$$ is rectangular so we can't invert

Ans: $$[\textbf{T}^{(e)T}\hat{\textbf{k}}^{(e)}\textbf{T}^{(e)}]\text{d}^{(e)}=\textbf{f}^{(e)}$$

$$\textbf{k}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

--14-3--

$$\textbf{k}^{(e)}=\textbf{T}^{(e)T}\hat{\textbf{k}}^{(e)}\textbf{T}^{(e)}$$

--15-2--

$$q_2^{(1)}=\frac{P_1^{(1)}}{k^{(1)}}=\frac{P_2^{(1)}}{k^{(1)}}$$

--15-3--

$$q_1^{(1)}=0$$

$$q_1^{(2)}=\frac{-P_2^{(2)}}{k^{(2)}}$$

$$\textbf{AC}=\frac{P_2^{(1)}}{k^{(1)}}$$

$$-\textbf{AB}=\frac{-P_2^{(2)}}{k^{(2)}}$$

--15-4--

$$u_y=Rsin(\alpha)$$

$$u_x=R[1-cos(\alpha)]$$

--16-1--

$$\textbf{AC}=\frac{|P_2^{(2)}|}{k^{(1)}}$$

Problem 9
$$\left[\begin{array}{c c c c c c c c} k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}&k_{14}^{(1)}&0&0&0&0\\ k_{21}^{(1)}&k_{22}^{(1)}&k_{23}^{(1)}&k_{24}^{(1)}&0&0&0&0\\ k_{31}^{(1)}&k_{32}^{(1)}&(k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)})& (k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)})&k_{13}^{(2)}&k_{14}^{(2)}& k_{13}^{(3)}&k_{14}^{(3)}\\ k_{41}^{(1)}&k_{42}^{(1)}&(k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)})& (k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)})&k_{23}^{(2)}&k_{24}^{(2)}& k_{23}^{(3)}&k_{24}^{(3)}\\ 0&0&k_{31}^{(2)}&k_{32}^{(2)}&k_{33}^{(2)}&k_{34}^{(2)}&0&0\\ 0&0&k_{41}^{(2)}&k_{42}^{(2)}&k_{43}^{(2)}&k_{44}^{(2)}&0&0\\ 0&0&k_{31}^{(3)}&k_{32}^{(3)}&0&0&k_{33}^{(3)}&k_{34}^{(3)}\\ 0&0&k_{41}^{(3)}&k_{42}^{(3)}&0&0&k_{43}^{(3)}&k_{44}^{(3)} \end{array}\right]$$

Prob. 5 Eigenvalues
Now we want to plot the eigenvectors corresponding to the 0 eigenvalues of the 2-bar truss system that we have been looking at in class for the past few weeks. We want to assume no constraints on the 2-bar truss system so we will use the full 6x6 global stiffness matrix.

We know that the global stiffness matrix K is:

K = 0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

If we use the eig function in Matlab we get that the eigenvalues are as follows:

Eval=eig(K) Eval= -0.0001        -0.0000         -0.0000          0.0000          1.4706         10.0294

Next we can use the function [V,D] = EIG(X)which gives us a diagonal matrix D of eigenvalues and a full matrix V whose columns contain the corresponding eigenvectors that corresponds to the equation:

$$\textbf{X}\times\textbf{V}=\textbf{D}\times\textbf{V}$$

For our particular case, we set X equivalent to K and the diagonal matrix D equivalent to $$\lambda$$ to get the following:

$$\textbf{K}\times\textbf{V}=\lambda\times\textbf{V})$$

Therefore, using Matlab we get:

[V,Lam]=eig(K) Lam = -0.0001        0         0         0         0         0         0   -0.0000         0         0         0         0         0         0   -0.0000         0         0         0         0         0         0    0.0000         0         0         0         0         0         0    1.4706         0         0         0         0         0         0   10.0294  V = -0.4494   0.3877    0.5158    0.0160    0.6174   -0.0139    0.6754   -0.4281    0.4825    0.0244    0.3565   -0.0080    0.1682    0.3877    0.5158    0.0160   -0.5409    0.5123   -0.3942   -0.4281    0.4825    0.0244   -0.4330   -0.4904    0.2812    0.4163   -0.0116    0.7023   -0.0765   -0.4984   -0.2812   -0.3995   -0.0449    0.7107    0.0765    0.4984

Where Lam contains the previously found eigenvalues and V contains the corresponding eigenvectors. It is important to note now that while Matlab calculated the eigenvalues to precise decimals, we can assume that the first four values in the diagonal Lam matrix are 0.

Now we would like to plot the eigenvectors that correspond to these four 0 eigenvalues so we may get a look at their mode shapes. Using the following Matlab function allows us to plot the four mode shapes at the same time.

for i=1:4 figure plot(V(:,i)); title(['Mode Shape ',num2str(i)]) end



Now let us consider the following three-bar truss system.



Assume that all three bars have the same material properties:

L=1    E=2      A=3

$$k=\frac{EA}L=\frac{(2)(3)}1=6$$

For element 1 and element 3:

$$l^{(1,3)}=0$$

$$m^{(1,3)}=1$$

For elemnt 2:

$$l^{(2)}=1$$

$$m^{(2)}=0$$

$$\textbf{k}^{(1)}=\textbf{k}^{(3)}= \left [\begin{array}{c c c c} 0 & 0 & 0 & 0\\ 0 & 36 & 0 & -36\\ 0 & 0 & 0 & 0\\ 0 & -36 & 0 &-36 \end{array}\right]$$

$$\textbf{k}^{(2)}= \left [\begin{array}{c c c c} 36 & 0 & -36 & 0\\ 0 & 0 & 0 & 0\\ -36 & 0 & 36 & 0\\ 0 & 0 & 0 & 0 \end{array} \right]$$

$$\textbf{K}= \left [\begin{array}{c c c c c c c c} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 36 & 0 & -36 & 0 & 0 & 0 & 0\\ 0 & 0 & 36 & 0 & -36 & 0 & 0 & 0\\ 0 & -36 & 0 & 36 & 0 & 0 & 0 & 0\\ 0 & 0 & -36 & 0 & 36 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 36 & 0 & -36\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -36 & 0 & 36 \end{array} \right] $$

In this example we are assuming that global nodes 1 and 4 are constrained so that they don't move. This means that $$d_1=d_2=d_7=d_8=0$$. Therefore we only have to consider the following modified global stiffness matrix.

$$\overline{\textbf{K}}= \left[ \begin{array} {c c c c} 36 & 0 & -36 & 0\\ 0 & 36 & 0 & 0\\ -36 & 0 & 36 & 0\\ 0 & 0 & 0 & 36 \end{array} \right]$$

If we follow the same steps as above we get:

Kbar = 36    0   -36     0     0    36     0     0   -36     0    36     0     0     0     0    36 Eval=eig(Kbar) Eval = 0   36    36    72 [V,Lam]=eig(Kbar) V = -0.7071        0         0   -0.7071         0    1.0000         0         0   -0.7071         0         0    0.7071         0         0    1.0000         0 Lam = 0    0     0     0     0    36     0     0     0     0    36     0     0     0     0    72

In this example we have only one zero eigenvalue and it can be plotted as follows:

plot(V(:,1)); title(['Mode Shape',num2str(1)])



HW 5
Choice 3:


 * $$W_1=0\;, W_2=0, W_3=1, W_4=0, W_5=0, W_6=0$$


 * $$\mathbf{W^T}=\begin{bmatrix} 0&0&1&0&0&0 \end{bmatrix}$$


 * $$\sum_{j=1}^{6} k_{3j}d_{j} - F_{3} = 0$$


 * $$\sum_{j=1}^{6} k_{3j}d_{j} = F_{3}$$

Choice 4:


 * $$W_1=0\;, W_2=0, W_3=0, W_4=1, W_5=0, W_6=0$$


 * $$\mathbf{W^T}=\begin{bmatrix} 0&0&0&1&0&0 \end{bmatrix}$$


 * $$\sum_{j=1}^{6} k_{4j}d_{j} - F_{4} = 0$$


 * $$\sum_{j=1}^{6} k_{4j}d_{j} = F_{4}$$

Choice 5:


 * $$W_1=0\;, W_2=0, W_3=0, W_4=0, W_5=1, W_6=0$$


 * $$\mathbf{W^T}=\begin{bmatrix} 0&0&0&0&1&0 \end{bmatrix}$$


 * $$\sum_{j=1}^{6} k_{5j}d_{j} - F_{5} = 0$$


 * $$\sum_{j=1}^{6} k_{5j}d_{j} = F_{5}$$

Choice 6:


 * $$W_1=0\;, W_2=0, W_3=0, W_4=0, W_5=0, W_6=1$$


 * $$\mathbf{W^T}=\begin{bmatrix} 0&0&0&0&0&1 \end{bmatrix}$$


 * $$\sum_{j=1}^{6} k_{6j}d_{j} - F_{6} = 0$$


 * $$\sum_{j=1}^{6} k_{6j}d_{j} = F_{6}$$