University of Florida/Eml4507/s13.team4ever.R6

Honor Pledge:
On my honor I have neither given nor received unauthorized aid in doing this problem.

Problem Description:
We are tasked with completing Pb.53.9 on p.53.19b by transforming the generalized eigenvalue problem into a standard eigenvalue problem.

Given
$$m_1=3, m_2=2$$ $$k_1=10,k_2=20,k_3=15$$

Solution
Solving the FBD of both $$m_1 $$ and $$ m_2 $$ gives us the equations of motion. $$m= \begin{bmatrix} m_1 & 0\\ 0 & m_2\\ \end{bmatrix}= \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} $$ $$k= \begin{bmatrix} k_1-k_2 & k_2\\ k_2 & -(k_2+k_3)\\ \end{bmatrix}=\begin{bmatrix} -10 & 20\\ 20 & -35\\ \end{bmatrix}$$ Equation of Motion in Matrix Form. $$\begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix}\begin{bmatrix} \ddot{x_1}\\ \ddot{x_2}\\ \end{bmatrix}+\begin{bmatrix} -10 & 20\\ 20 & -35\\ \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$ Now transforming the generalized eigenvalue problem into the standard eigenvalue problem.

$$kx={\lambda}mx$$.............................................. (1)  The $$ m, m^{-1/2},$$ and $$ m^{1/2}$$ matrix will be needed so they are solved and provided below. $$m=\begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix},m^{-1/2}=\begin{bmatrix} 1/\sqrt{3} & 0\\ 0 & 1/\sqrt{2}\\ \end{bmatrix},m^{1/2}=\begin{bmatrix} \sqrt{3} & 0\\ 0 & \sqrt{2}\\ \end{bmatrix}$$ Derivation of the new value $$x^*$$.First premultiply by $$ m^{-1/2}$$ $$m^{-1/2}kx={\lambda}m^{-1/2}mx={\lambda}m^{1/2}x$$ Let, $$x^*=m^{1/2}x$$ Then, $$ x=m^{-1/2}x^*$$ Plugging Into Eq. 1 above. $$(m^{-1/2}km^{1/2})x^*=k^*x^*={\lambda}x^*=0$$ Where, $$k^*=m^{-1/2}km^{1/2}=\begin{bmatrix} 3.333 & 8.165\\ 8.165 & -17.5\\ \end{bmatrix}$$ Now, $$(k^*-{\lambda}I)x^*=0$$............................................ (2)  First, Solving for the eigenvalues by setting the determinant of $$(k^*-{\lambda}I)$$equal to zero. $$det(k^*-{\lambda}I)=det(\begin{bmatrix} -3.333-{\lambda} & 8.165\\ 8.165 & -17.5-{\lambda}\\ \end{bmatrix})=0$$ $$(-3.33-{\lambda})(-17.5-{\lambda})=0$$ $${\lambda}^2 +20.83{\lambda}-8.392 = 0$$ Solving for $$ {\lambda}_1 $$ and $${\lambda}_2$$ using the quadratic. $${\lambda}_1=0.395$$ and $${\lambda}_2=-21.225$$ Plugging these values in one at a time into Eq.2 above will yield two respective eigenvectors. Solving for $$ x^*_1$$, $$\begin{bmatrix} 3.333-0.935 & 8.165\\ 8.165 & -17.50.395\\ \end{bmatrix}\begin{bmatrix} x^*_{11}\\ x^*_{12}\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$ $$-3.725x^*_{11}+8.165x^*_{12}=0$$ $$8.165x^*_{11}-17.895x^*_{12}=0$$ So, $$x^*_1=\begin{bmatrix} 2.19\\ 1\\ \end{bmatrix}$$ Now solving for $$x^*_2$$, $$\begin{bmatrix} 3.333+21.225 & 8.165\\ 8.165 & -17.5+21.225\\ \end{bmatrix}\begin{bmatrix} x^*_{21}\\ x^*_{22}\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$ $$17.895x^*_{21}+8.165x^*_{22}=0$$ $$8.165x^*_{21}+3.725x^*_{22}=0$$ So, $$x^*_2=\begin{bmatrix} -0.456\\ 1\\ \end{bmatrix}$$

Final Solution
Using the transformation $$x=M^{-1/2}x^*$$, it is possible to obtain the eigenvectors $$x_1$$ and $$ x_2$$. $$ x_1=M^{-1/2}x^*_1=\begin{bmatrix} 1/\sqrt{3}& 0\\ 0& 1/\sqrt{2}\\ \end{bmatrix}\begin{bmatrix} 2.19\\ 1\\ \end{bmatrix}=\begin{bmatrix} 1.264\\ 0.707\\ \end{bmatrix}$$ $$ x_2=M^{-1/2}x^*_2=\begin{bmatrix} 1/\sqrt{3}& 0\\ 0& 1/\sqrt{2}\\ \end{bmatrix}\begin{bmatrix} -0.456\\ 1\\ \end{bmatrix}=\begin{bmatrix} -0.263\\ 0.707\\ \end{bmatrix}$$ So, $$x_1=\begin{bmatrix} 1.264\\ 0.707\\ \end{bmatrix}$$ $$x_2=\begin{bmatrix} -0.263\\ 0.707\\ \end{bmatrix}$$

Problem R6.2
On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

Description
We are to resolve problem 4.4/5.4 but by using the method of transforming the generalized eigenvalue problem into a standard eigenvalue problem and then proceeding in the calculations.

Solution
This problem is accomplished by using the following simple transformation:

$$ K^{*} = M^{-1/2} * K * M^{-1/2} $$

The problem is carried out exactly the same as from report 5 until right before the eigenvectors are found. Instead, the eigenvalues and eigenvectors are found from the $$ K^{*} $$ matrix. Once this is done, the true eigenvalues for our system are found according to:

$$ x = M^{-1/2} * x^{*} $$

The code to solve this problem are presented below.

References for Matlab code: Team 3 Report 4 Team 7 Report 4 Team 4 Report 5

The resulting displacements were:

X1 =

0  -0.2705    0.4767   -0.6077    0.6621   -0.6638    0.5912         0   -0.1427    0.3262   -0.4835    0.5702   -0.5548    0.4132

Y1 =

0.0217  -0.0687    0.0978   -0.1125    0.1091   -0.1148   -0.2720         0    0.0032    0.0412   -0.0785    0.1042   -0.1062    0.3054

X2 =

0  -0.3958    0.4536   -0.1539   -0.3379    0.7810   -0.8311         0   -0.3426    0.6375   -0.6625    0.3889    0.0108   -0.2698

Y2 =

0.0623  -0.1362    0.1170   -0.0295   -0.0724    0.1912    0.5228         0   -0.0024    0.1013   -0.1388    0.0939   -0.0144   -0.4913

X3 =

0   0.2496   -0.5547    0.6761   -0.4979    0.1250   -0.1723         0   -0.3624    0.4730   -0.3895    0.3085   -0.3383    0.4022

Y3 =

0.0936  -0.0157   -0.0466    0.0581   -0.0157   -0.1955   -1.0802         0   -0.0593    0.0534    0.0069   -0.0549    0.1049    0.7836

Problem Statement
Redo R3.1. Compute the reactions, and compare the reactions in the case of non-zero prescribed disp dofs with the reactions in the case of zero prescribed disp dofs.

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

Given zero displacement DOFs, the following MATLAB program solves for the unknown global displacement DOF's, unknown for components (reactions), and member forces in the system.

function R6p3 F = 20000; %N E = 206000000000; %Pa L = 1; %m A = .0001; %m^2 % displacement in meters d3 = 0; % =u2 d4 = 0; % =v2 d5 = 0; % =u3 d6 = 0; % =v3 % Table 2.1 KS 2008 p.82 % Element  EA/L      i-->j    phi    l=cos(phi)  m=sin(phi) % 1      206x10^5    1-->3   -pi/6    0.866       -0.5 % 2      206x10^5    1-->2   -pi/2      0           1 % 3      206x10^5    1-->4   -5pi/6  -0.866       -0.5 l1 = 0.866; m1 = -0.5; l2 = 0; m2 = 1; l3 = -0.866; m3 = -0.5; % element stiffness matrices % E2.46 % element 1 stiffness matrix 1-->3 k1 = [l1^2 l1*m1 0 0 -l1^2 -l1*m1 0 0; l1*m1 m1^2 0 0 -l1*m1 -m1^2 0 0; 0 0 0 0 0 0 0 0;      0 0 0 0 0 0 0 0;       -l1^2 -l1*m1 0 0 l1^2 l1*m1 0 0; -l1*m1 -m1^2 0 0 l1*m1 m1^2 0 0; 0 0 0 0 0 0 0 0;      0 0 0 0 0 0 0 0]; % element 1 stiffness matrix 1-->2 k2 = [l2^2 l2*m2 -l2^2 -l2*m2 0 0 0 0; l2*m2 m2^2 -l2*m2 -m2^2 0 0 0 0; -l2^2 -l2*m2 l2^2 l2*m2 0 0 0 0; -l2*m2 -m2^2 l2*m2 m2^2 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]; % element 1 stiffness matrix 1-->4 k3 = [l3^2 l3*m3 0 0 0 0 -l3^2 -l3*m3; l3*m3 m3^2 0 0 0 0 -l3*m3 -m3^2; 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;       -l3^2 -l3*m3 0 0 0 0 l3^2 l3*m3; -l3*m3 -m3^2 0 0 0 0 l3*m3 m3^2]; %global stiffness matrix K = E*A/L*(k1 + k2 + k3) % Force matrix Fpost = [F*cos(pi/4); F*sin(pi/4); 0; 0; 0; 0; 0; 0] % global disp dofs dispg = K\Fpost % u2=v2=u3=v3=u4=v4=0, so delete rows/columns 3 through 8 Knew = 1*10^7*[3.0898 0; 0 3.0900]; % displacement values u1, v1 Fcut = [F*cos(pi/4); F*sin(pi/4)]; disp = Knew\Fcut % forces in each element P1 = E*A/L*(0.866*(d5-disp(1))-0.5*(d6-disp(2))) P2 = E*A/L*(0*(d3-disp(1))+1*(d4-disp(2))) P3 = E*A/L*(-0.866*(0-disp(1))-0.5*(dispg(4)-disp(2))) stress1 = P1/A stress2 = P2/A stress3 = P3/A Global Displacements dispg = 1.0e+12 * NaN NaN NaN 3.5927    3.7905     6.5693     2.3727     3.0716

Unknown disp at node 1(m) disp = 1.0e-03 * 0.4577    0.4577

Element forces (N) P1 = -3.4512e+03 P2 = -9.4281e+03 P3 = 1.2879e+04

Element stresses (Pa) stress1 = -3.4512e+07 stress2 = -9.4281e+07 stress3 = 1.2879e+08

Verifying with Calfem and plotting

function R6p3calfem F = 20000; %N E = 206000000000; %Pa L = 1; %m A = .0001; %m^2 % Checking with CALFEM edof = [1 1 2 5 6; 2 1 2 3 4;        3 1 2 7 8]; coord = [cos(pi/6) sin(pi/6);cos(pi/6) 1+sin(pi/6);2*cos(pi/6) 0;0 0]; dof = [1 2;3 4;5 6;7 8]; [ex,ey] = coordxtr(edof,coord,dof,2); ep = [E A]; K = zeros(8); Fo = zeros(8,1); Fo(1) = F*cos(pi/4); Fo(2) = F*sin(pi/4); for i = 1:3 Ke = bar2e(ex(i,:),ey(i,:),ep); K = assem(edof(i,:),K,Ke); end bc = [3 0; 4 0; 5 0; 6 0; 7 0; 8 0]; Q = solveq(K,Fo,bc); ed = extract(edof,Q); for i = 1:3 N(i)=bar2s(ex(i,:),ey(i,:),ep,ed(i,:)); end plotpar=[1 4 0];scale=100; eldraw2(ex,ey); eldisp2(ex,ey,ed,plotpar,scale); grid on N1 = N(1) N2 = N(2) N3 = N(3) stress1c = N(1)/A stress2c = N(2)/A stress3c = N(3)/A

OUTPUT N1 = -3.4509e+03 N2 = -9.4281e+03 N3 = 1.2879e+04

stress1c = -3.4509e+07 stress2c = -9.4281e+07 stress3c = 1.2879e+08



Original values of non-zero displacement degrees of freedom: (R3.1) Global displacement DOF's (m) $$ u_1 = -0.005 $$ $$ v_1 = -0.0103 $$ $$ u_2 = 0.02 $$ $$ v_2 = -0.01 $$ $$ u_3 = -0.03 $$ $$ v_3 = 0.05 $$ $$ u_4 = -0.0257 $$ $$ v_4 = 0.0764 $$

Element Reaction Forces (N) $$ P^1 = -854900 $$ $$ P^2 = -418180 $$ $$ P^3 = 740079 $$

Element Stresses (GPa) $$ \sigma^1 = -8.5490 $$ $$ \sigma^2 = -4.1818 $$ $$ \sigma^3 = 7.4008 $$

Problem 6.4 Deformation in 3D Space
On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Problem Statement
There is a 3D truss system with applied forces. We need to optimize each beam to meet a factor of safety of 1.5.

Conclusion
The final area of each member was able to be reduced to its minimum of 0.1 inches and stay above a factor of safety of 1.5. Final stresses are in psi:
 * 1) 	13.94
 * 2) 	-8487.15
 * 3) 	-8487.15
 * 4) 	8611.63
 * 5) 	8611.63
 * 6) 	-12871.54
 * 7) 	13045.99
 * 8) 	-12871.54
 * 9) 	13045.99
 * 10) 	63.42
 * 11) 	63.42
 * 12) 	788.95
 * 13) 	-778.48
 * 14) 	-5171.02
 * 15) 	5174.49
 * 16) 	-5171.02
 * 17) 	5174.49
 * 18) 	-7787.74
 * 19) 	-7787.74
 * 20) 	7792.39
 * 21) 	7792.39
 * 22) 	14680.46
 * 23) 	-14679.76
 * 24) 	-14679.76
 * 25) 	14680.46

CALFEM Verification
The Matlab calculations were confirmed using the CALFEM toolbox in Matlab.

The following presents the constants used and options for different designs

This Edof matrix uses column 1 as the as the element number. The next three columns are the DOFs for the first node and the next three columns are for the second node since this is a three dimensional problem.

Coord contains the x, y, and z coordinates for each node.

The bar3e is used to make the stiffness matrix for a 3D system. The bar3s is used to find the member forces in a 3D system. One can see that the 3 denotes the dimension of the system being observed. The stresses can be found by dividing the forces by the areas.

Problem 6.5 3d Mode Shapes
On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Problem Statement
Determine the first 3 mode shapes of the 3D truss system from problem 6.4

Solution
The set up of the problem was the same as problem 6.4. The eigenvalues can be seen below.