University of Florida/Eml4507/Fead-s13-team4-R2

Given 2-D Truss Model and Constants
Consider the 2D truss system



Use the following constants to very calculations

$$l^{(1)}=4$$

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

$$E^{(1)}=3$$

$$E^{(2)}=5$$

$$A^{(1)}=1$$

$$A^{(2)}=2$$

$$\O ^{(1)}=30^{\circ}$$

$$\O ^{(2)}=135^{\circ}$$

$$P=7$$

Construct local stiffness matrices
The stiffness matrices will be constructed for individual elements 1 and 2. The rows and columns will be numbered to represent the force and degree of freedom. For element 1

$$ \vec{K^{(1)}}=\begin{bmatrix} &1&2&3&4\\ 1&(l^{(1)} )^{2} & l^{(1)}m^{(1)} & -(l^{(1)} )^{2} & -l^{(1)}m^{(1)}\\ 2&l^{(1)}m^{(1)} & (m^{(1)} )^{2} & -l^{(1)}m^{(1)} & -(m^{(1)} )^{2}\\ 3&-(l^{(1)} )^{2}& -l^{(1)}m^{(1)} & (l^{(1)} )^{2} & l^{(1)}m^{(1)}\\ 4&-l^{(1)}m^{(1)} & -(m^{(1)} )^{2} & l^{(1)}m^{(1)} & (m^{(1)} )^{2}\\ \end{bmatrix}*K^{(1)} $$

For element 2

$$ \vec{K^{(2)}}=\begin{bmatrix} &3&4&5&6\\ 3&(l^{(2)} )^{2} & l^{(2)}m^{(2)} & -(l^{(2)} )^{2} & -l^{(2)}m^{(2)}\\ 4&l^{(2)}m^{(2)} & (m^{(2)} )^{2} & -l^{(2)}m^{(2)} & -(m^{(2)} )^{2}\\ 5&-(l^{(2)} )^{2}& -l^{(2)}m^{(2)} & (l^{(2)} )^{2} & l^{(2)}m^{(2)}\\ 6&-l^{(2)}m^{(2)} & -(m^{(2)} )^{2} & l^{(2)}m^{(2)} & (m^{(2)} )^{2}\\ \end{bmatrix}*K^{(2)} $$

The variable K can be determined by

$$K^{(e)}=\frac{EA}{L}$$

If the variables are plugged in, we get the following matrices:

For element 1

$$ \vec{K^{(1)}}=\begin{bmatrix} &1&2&3&4\\ 1&0.5625 & 0.3248 & -0.5625 & -0.3248\\ 2&0.3248 & 0.1875 & -0.3248 & -0.1875\\ 3&-0.5625& -0.3248 & 0.5625& 0.3248\\ 4&-0.3248 & -0.1875 & 0.3248 & 0.1875\\ \end{bmatrix} $$

For element 2

$$ \vec{K^{(2)}}=\begin{bmatrix} &3&4&5&6\\ 3&2.5&-2.5&-2.5&2.5\\ 4&-2.5&2.5&2.5&-2.5\\ 5&-2.5&2.5&2.5&-2.5\\ 6&2.5&-2.5&-2.5&2.5\\ \end{bmatrix} $$

It is clear that if we wish to combine the matrices, it will place coordinate 3,3 of element 2 on the location of 3,3 of element 1 due to them referencing the same point in the same direction. When combining the matrices, addition will be used since the matrices represent directional forces and forces always combine in the same direction by addition.

Combining local stiffness matrices into a global matrix
The local matrices from element 1 and 2 will be combined as described above to give

$$ \vec{K}=\begin{bmatrix} &1&2&3&4&5&6\\ 1&0.5625 & 0.3248 & -0.5625 & -0.3248&0&0\\ 2&0.3248 & 0.1875 & -0.3248 & -0.1875&0&0\\ 3&-0.5625& -0.3248 & 3.0625& -2.1752&-2.5&2.5\\ 4&-0.3248 & -0.1875 & -2.1752 & 2.6875&2.5&-2.5\\ 5&0&0&-2.5&2.5&2.5&-2.5\\ 6&0&0&2.5&-2.5&-2.5&2.5\\ \end{bmatrix} $$

MatLab Code
EDU>> Edof = [1 1 2 3 4

2 3 4 5 6];

EDU>> K = zeros(6);

EDU>> f = zeros(6,1);

EDU>> f(4) = -7;

EDU>> E1 = 3; E2 = 5;

EDU>> L1 = 4;L2 = 2;

EDU>> A1 = 1;A2 = 2;

EDU>> ep1 = [E1 A1]

ep1 =

3    1

EDU>> ep2 = [E2 A2]

ep2 =

5    2

EDU>> ex1 = [-3.464 0]; ex2 = [0 1.414];

EDU>> ey1 = [-2 0]; ey2 = [0 -1.414];

EDU>> Ke1 = bar2e(ex1,ey1,ep1)

Ke1 =

0.5625   0.3248   -0.5625   -0.3248    0.3248    0.1875   -0.3248   -0.1875   -0.5625   -0.3248    0.5625    0.3248   -0.3248   -0.1875    0.3248    0.1875

EDU>> Ke2 = bar2e(ex2,ey2,ep2)

Ke2 =

2.5004  -2.5004   -2.5004    2.5004   -2.5004    2.5004    2.5004   -2.5004   -2.5004    2.5004    2.5004   -2.5004    2.5004   -2.5004   -2.5004    2.5004

EDU>> K = assem(Edof(1,:),K,Ke1);

EDU>> K = assem(Edof(2,:),K,Ke2)

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    3.0629   -2.1756   -2.5004    2.5004   -0.3248   -0.1875   -2.1756    2.6879    2.5004   -2.5004         0         0   -2.5004    2.5004    2.5004   -2.5004         0         0    2.5004    -2.5004  -2.5004    2.5004

EDU>> bc = [1 0;2 0;5 0;6 0];

EDU>> [a,r]=solveq(K,f,bc)

a =

0        0   -4.3519   -6.1268         0         0

r =

4.4378   2.5622         0   -0.0000   -4.4378    4.4378

Conclusion
By comparing the value from the hand calculation and the value from the CALFEM program, we can see that the two matrices are identical. This verifies the assembly process of using addition when combining local matrices by addition.

Disccription:
We are asked to consider the L2-ODE-CC system referenced in (1) p.53-2 to have the following complex conjugate roots:

$$ \lambda_{1,2} = -0.5 \pm 2i $$

We are asked to find the homogeneous solution in standard form and to find the damping ratio, $$ \zeta $$

We are then to consider the initial conditions

$$ y(0) = 1, y'(0) = 0 $$

We are to plot this solution and then use that plot to find the log decrement and compare it to the log decrement that would be obtained using (6) p.53-5b.

Finally, we're to find the quality factor, Q, and the loss factor, $$ \eta $$

Solution:
From R1.5, we have already obtained that for the above conditions, $$ y(t) = e^{-0.5 t} cos(2t) + (\frac{1}{4}) e^{-0.5 t} sin(2t) $$

with coefficients to the characteristic equation of

$$ a = 1$$ and $$ b = \frac{17}{4}$$

From (1) and (2) p.53-5, we have that

$$ \omega^{2} = b $$

$$ 2 \omega \zeta = a $$

Knowing both a and b we get

$$ \omega = \sqrt{ \frac{17}{4}} = 2.0615 $$

$$ \zeta = \frac{1}{2 \omega} = \frac{1}{2*2.0615} = 0.2425 $$

We then have from equation (1) p.53-5d

$$ \delta = \frac{1}{3}[\delta^{(1)} + \delta^{(2)} + \delta^{(3)}] $$



Using the above graph of the homogeneous solution, accurate estimates were able to be made as to the various values of y at the peaks

Using equation (3) p.53-5c we are able to find

$$ \delta^{(1)} = log(1/0.2075) = 0.68235 $$ $$ \delta^{(2)} = log(0.2075/0.0432) = 0.68216 $$ $$ \delta^{(3)} = log(0.0432/0.009) = 0.68124 $$

Conclusion
We then arrive at

$$ \delta = \frac{1}{3}[0.68235 + 0.68216 + 0.68124] = 0.68192 $$

For comparison, the equation (6) p.53-5b gives

$$ \delta = \frac{2 \pi \zeta}{\sqrt{1 - \zeta^{2}}} = \frac{2 \pi 0.2425}{\sqrt{1 - 0.2425^{2}}} = 1.5705 $$

We now find the quality factor, Q. From (2) p.53-5d, we have

$$ Q = \frac{\delta}{\pi} = \frac{\sqrt{1 - \zeta^{2}}}{2 \zeta} = \frac{\sqrt{1 - 0.2425^{2}}}{2 *0.2425} = 2 $$

From (5) p.53-5d, we know that the loss factor is just the inverse of the quality factor so

$$ \eta = \frac{1}{Q} = \frac{1}{2} = 0.5 $$

Problem Statement
KS 2008 p.103 sec.2.7 pb.21 write a matlab program to assemble the global stiffness matrix, compute the unknown disp dofs, the reaction forces, the member forces; compare the results with exact hand calculation Statics, Mechanics of Materials).

run CALFEM to verify the results, but now do this problem in a different way (as in commercial FE codes).

first, establish 2 arrays: (1) the global node coordinate array “coord”, and (2) the elem connectivity array “conn”. next, write a matlab code to loop over the number of element to call the function “bar2e” of CALFEM to compute the element stiffness matrices. with this method, you let your matlab code construct the arrays ex1, ey1, ex2, ey2, etc. for you, using the info from the arrays “coord” and “conn”.

MATLAB Code
The following is a MATLAB code that assembles the global stiffness matrix, computes the unknown displacement dofs, reaction forces, and member forces function FEA_R2p3 E = 10000; % Young's modulus A = pi*2^2/4; % Area % Lengths of each element L1 = 12; L2 = 12; L3 = 12; L4 = sqrt((12^2)+(9^2)); L5 = 9; L6 = sqrt((12^2)+(9^2)); L7 = 12; L8 = 9; L9 = sqrt((12^2)+(9^2)); % topology matrix % [element# node1x node1y node2x node2y] Edof = [1 1 2  3  4;2  3  4 5 6;3  5  6 7 8; 4 1 2  9 10;5  9 10 3 4;6  9 10 5 6;            7 9 10 11 12;8 11 12 5 6;9 11 12 7 8];    % element stiffness matrix in local coordinates for elements 1 through 9 esmloc1 = (E*A/L1)*[1 0 -1 0;0 0 0 0;-1 0 1 0;0 0 0 0]; esmloc2 = (E*A/L2)*[1 0 -1 0;0 0 0 0;-1 0 1 0;0 0 0 0]; esmloc3 = (E*A/L3)*[1 0 -1 0;0 0 0 0;-1 0 1 0;0 0 0 0]; esmloc4 = (E*A/L4)*[(L1/L4)^2 (L1/L4)*(L5/L4) -(L1/L4)^2 -(L1/L4)*(L5/L4); (L1/L4)*(L5/L4) (L5/L4)^2 -(L1/L4)*(L5/L4) -(L5/L4)^2; -(L1/L4)^2 -(L1/L4)*(L5/L4) (L1/L4)^2 (L1/L4)*(L5/L4); -(L1/L4)*(L5/L4) -(L5/L4)^2 (L1/L4)*(L5/L4) (L5/L4)^2]; % [cos^2 cossin -cos^2 -cossin;cossin sin^2 -cossin -sin^2; % -cos^2 -cossin cos^2 cossin;-cossin -sin^2 cossin sin^2] esmloc5 = (E*A/L5)*[0 0 0 0;0 1 0 -1;0 0 0 0;0 -1 0 1]; esmloc6 = (E*A/L6)*[(L1/L4)^2 -(L1/L4)*(L5/L4) -(L1/L4)^2 (L1/L4)*(L5/L4); -(L1/L4)*(L5/L4) (L5/L4)^2 (L1/L4)*(L5/L4) -(L5/L4)^2; -(L1/L4)^2 (L1/L4)*(L5/L4) (L1/L4)^2 -(L1/L4)*(L5/L4); (L1/L4)*(L5/L4) -(L5/L4)^2 -(L1/L4)*(L5/L4) (L5/L4)^2]; % [cos^2 -cossin -cos^2 cossin;-cossin sin^2 cossin -sin^2; % -cos^2 cossin cos^2 -cossin;cossin -sin^2 -cossin sin^2] esmloc7 = (E*A/L7)*[1 0 -1 0;0 0 0 0;-1 0 1 0;0 0 0 0]; esmloc8 = (E*A/L8)*[0 0 0 0;0 1 0 -1;0 0 0 0;0 -1 0 1]; esmloc9 = (E*A/L9)*[(L1/L4)^2 -(L1/L4)*(L5/L4) -(L1/L4)^2 (L1/L4)*(L5/L4); -(L1/L4)*(L5/L4) (L5/L4)^2 (L1/L4)*(L5/L4) -(L5/L4)^2; -(L1/L4)^2 (L1/L4)*(L5/L4) (L1/L4)^2 -(L1/L4)*(L5/L4); (L1/L4)*(L5/L4) -(L5/L4)^2 -(L1/L4)*(L5/L4) (L5/L4)^2]; % [cos^2 -cossin -cos^2 cossin;-cossin sin^2 cossin -sin^2; % -cos^2 cossin cos^2 -cossin;cossin -sin^2 -cossin sin^2] % element nodal coordinates % ex1 and ey1 are the x and y coordinates to the nodes of element 1 ex1 = [0 12]; ey1 = [0 0]; ex2 = [12 24]; ey2 = [0 0]; ex3 = [24 36]; ey3 = [0 0]; ex4 = [0 12]; ey4 = [0 9]; ex5 = [12 12]; ey5 = [9 0]; ex6 = [12 24]; ey6 = [9 0]; ex7 = [12 24]; ey7 = [9 9]; ex8 = [24 24]; ey8 = [9 0]; ex9 = [24 36]; ey9 = [9 0]; K = zeros(12); F = zeros(12,1); F(6)=-1200; F(11)=400; ep = [E A]; % assembly of the global stiffness matrix K = assem(Edof(1,:),K,esmloc1); K = assem(Edof(2,:),K,esmloc2); K = assem(Edof(3,:),K,esmloc3); K = assem(Edof(4,:),K,esmloc4); K = assem(Edof(5,:),K,esmloc5); K = assem(Edof(6,:),K,esmloc6); K = assem(Edof(7,:),K,esmloc7); K = assem(Edof(8,:),K,esmloc8); K = assem(Edof(9,:),K,esmloc9); % boundary conditions bc = [1 0;2 0;8 0]; % reaction force vector Q = solveq(K,F,bc); % finding element displacements ed1 = extract(Edof(1,:),Q); ed2 = extract(Edof(2,:),Q); ed3 = extract(Edof(3,:),Q); ed4 = extract(Edof(4,:),Q); ed5 = extract(Edof(5,:),Q); ed6 = extract(Edof(6,:),Q); ed7 = extract(Edof(7,:),Q); ed8 = extract(Edof(8,:),Q); ed9 = extract(Edof(9,:),Q); disp = [ed1;ed2;ed3;ed4;ed5;ed6;ed7;ed8;ed9]; % reaction forces at each element; member forces (stress) N1 = bar2s(ex1,ey1,ep,ed1); MF1 = N1/A; N2 = bar2s(ex2,ey2,ep,ed2); MF2 = N2/A; N3 = bar2s(ex3,ey3,ep,ed3); MF3 = N3/A; N4 = bar2s(ex4,ey4,ep,ed4); MF4 = N4/A; N5 = bar2s(ex5,ey5,ep,ed5); MF5 = N5/A; N6 = bar2s(ex6,ey6,ep,ed6); MF6 = N6/A; N7 = bar2s(ex7,ey7,ep,ed7); MF7 = N7/A; N8 = bar2s(ex8,ey8,ep,ed8); MF8 = N8/A; N9 = bar2s(ex9,ey9,ep,ed9); MF9 = N9/A; RF = [N1;N2;N3;N4;N5;N6;N7;N8;N9]; MF = [MF1;MF2;MF3;MF4;MF5;MF6;MF7;MF8;MF9];

With this, the output is as follows: GlobalStiffnessMatrix = K    dispDofs = disp reactionForces = RF   memberForces = MF

GlobalStiffnessMatrix = 1.0e+003 * 3.9584   1.0053   -2.6180         0         0         0         0         0   -1.3404   -1.0053         0         0    1.0053    0.7540         0         0         0         0         0         0   -1.0053   -0.7540         0         0   -2.6180         0    5.2360         0   -2.6180         0         0         0         0         0         0         0         0         0         0    3.4907         0         0         0         0         0   -3.4907         0         0         0         0   -2.6180         0    6.5764   -1.0053   -2.6180         0   -1.3404    1.0053         0         0         0         0         0         0   -1.0053    4.2446         0         0    1.0053   -0.7540         0   -3.4907         0         0         0         0   -2.6180         0    3.9584   -1.0053         0         0   -1.3404    1.0053         0         0         0         0         0         0   -1.0053    0.7540         0         0    1.0053   -0.7540   -1.3404   -1.0053         0         0   -1.3404    1.0053         0         0    5.2988         0   -2.6180         0   -1.0053   -0.7540         0   -3.4907    1.0053   -0.7540         0         0         0    4.9986         0         0         0         0         0         0         0         0   -1.3404    1.0053   -2.6180         0    3.9584   -1.0053         0         0         0         0         0   -3.4907    1.0053   -0.7540         0         0   -1.0053    4.2446

dispDofs = 0        0    0.3056   -1.4992    0.3056   -1.4992    0.6112   -2.1836    0.6112   -2.1836    1.0695         0         0         0    0.8260   -1.4992    0.8260   -1.4992    0.3056   -1.4992    0.8260   -1.4992    0.6112   -2.1836    0.8260   -1.4992    0.5204   -1.9258    0.5204   -1.9258    0.6112   -2.1836    0.5204   -1.9258    1.0695         0

reactionForces = 1.0e+003 * 0.8000   0.8000    1.2000   -0.5000         0    0.5000   -0.8000    0.9000   -1.5000

memberForces = 254.6479 254.6479  381.9719 -159.1549         0  159.1549  254.6479  286.4789 -477.4648

It can be seen here that the problem done by hand corresponds to the force output of the MATLAB program.

CALFEM MATLAB
% CALFEM % global coordinate matrix Coord = [0 0;12 0;24 0;36 0;12 9;24 9]; % element connectivity array conn = Edof; % global nodal dof matrix Dof = [1 2;3 4;5 6;7 8;9 10;11 12]; % contructing element arrays [ex,ey]=coordxtr(conn,Coord,Dof,2); K=zeros(12); EA=[E A]; % Young's modulus, Area % compute element stiffness matrix Ke and global stiffness matrix for i=1:9 Ke = bar2e(ex(i,:),ey(i,:),EA); % element stiffness matrix K = assem(conn(i,:),K,Ke);     % global stiffness matrix end % element displacement vector ed=extract(conn,Q); % reaction forces for i=1:9 N(i)=bar2s(ex(i,:),ey(i,:),ep,ed(i,:)); end % member forces MF1C = N(1)/A; % stress in AC    MF2C = N(2)/A;  % stress in CE    MF3C = N(3)/A;  % stress in EF    MF4C = N(4)/A;  % stress in AB    MF5C = 0;  % stress in BC    MF6C = N(6)/A;  % stress in BE    MF7C = N(7)/A;  % stress in BD    MF8C = N(8)/A;  % stress in DE    MF9C = N(9)/A;  % stress in DF    MFC = [MF1C;MF2C;MF3C;MF4C;MF5C;MF6C;MF7C;MF8C;MF9C];

The output to the CALFEM code is as follows: CGlobalStiffnessMatrix = K    CdispDofs = ed    CreactionForces = N'    CmemberForces = MFC

CGlobalStiffnessMatrix = 1.0e+003 * 3.9584   1.0053   -2.6180         0         0         0         0         0   -1.3404   -1.0053         0         0    1.0053    0.7540         0         0         0         0         0         0   -1.0053   -0.7540         0         0   -2.6180         0    5.2360         0   -2.6180         0         0         0         0         0         0         0         0         0         0    3.4907         0         0         0         0         0   -3.4907         0         0         0         0   -2.6180         0    6.5764   -1.0053   -2.6180         0   -1.3404    1.0053         0         0         0         0         0         0   -1.0053    4.2446         0         0    1.0053   -0.7540         0   -3.4907         0         0         0         0   -2.6180         0    3.9584   -1.0053         0         0   -1.3404    1.0053         0         0         0         0         0         0   -1.0053    0.7540         0         0    1.0053   -0.7540   -1.3404   -1.0053         0         0   -1.3404    1.0053         0         0    5.2988         0   -2.6180         0   -1.0053   -0.7540         0   -3.4907    1.0053   -0.7540         0         0         0    4.9986         0         0         0         0         0         0         0         0   -1.3404    1.0053   -2.6180         0    3.9584   -1.0053         0         0         0         0         0   -3.4907    1.0053   -0.7540         0         0   -1.0053    4.2446

CdispDofs = 0        0    0.3056   -1.4992    0.3056   -1.4992    0.6112   -2.1836    0.6112   -2.1836    1.0695         0         0         0    0.8260   -1.4992    0.8260   -1.4992    0.3056   -1.4992    0.8260   -1.4992    0.6112   -2.1836    0.8260   -1.4992    0.5204   -1.9258    0.5204   -1.9258    0.6112   -2.1836    0.5204   -1.9258    1.0695         0

CreactionForces = 1.0e+003 * 0.8000   0.8000    1.2000   -0.5000         0    0.5000   -0.8000    0.9000   -1.5000

CmemberForces = 254.6479 254.6479  381.9719 -159.1549         0  159.1549  254.6479  286.4789 -477.4648

The results of all methods verifies their accuracy.

Description:
We are to find an expression for the excitation referenced in (1) p.53-6 and then find the actual solution and the amplification factor, A for the following conditions:

$$ f_{0}/m = r_{0} = 1 $$ $$ \bar{w} = 0.9 w $$

We are then to find the complete solution.

Solution:
$$ y'' + y' +(\frac{17}{4})y = \frac{f_{0}}{m} cos(\bar{w} t) $$

Let us guess the solution is:

$$ y_{p}(t) = A sin(\bar{w} t) + B cos(\bar{w} t) $$

then

$$ y_{p}(t) = A sin(\bar{w} t) + B cos(\bar{w} t) $$ $$ y'_{p}(t) = A \bar{w} cos(\bar{w} t) - B \bar{w} sin(\bar{w} t) $$ $$ y''_{p}(t) = - A \bar{w}^{2} cos(\bar{w} t) - B \bar{w}^{2} sin(\bar{w} t) $$

We then write

$$ [ - A \bar{w}^{2} - B \bar{w} + (\frac{17}{4}) A ] sin(\bar{w} t) + [ - B \bar{w}^{2} + A \bar{w} + (\frac{17}{4}) B] cos(\bar{w} t) = \frac{f_{0}}{m} cos(\bar{w} t) $$

It follows that

$$ [ - A \bar{w}^{2} - B \bar{w} + (\frac{17}{4}) A ] = 0 $$ and that $$ [ - B \bar{w}^{2} + A \bar{w} + (\frac{17}{4}) B] = \frac{f_{0}}{m} $$

We solve for A and B to be

$$ A = \frac{f_{0}/m}{[\bar{w} + (\frac{17}{4})^{2}(\frac{1}{\bar{w}}) - \frac{34 \bar{w}}{4} + \bar{w}^{3} ]} $$ $$ B = \frac{ \frac{f_{0}}{m} [ \frac{17}{4} - \bar{w}^{2}] }{[[\bar{w} + (\frac{17}{4})^{2}(\frac{1}{\bar{w}}) - \frac{34 \bar{w}}{4} + \bar{w}^{3} ] (\bar{w})]} $$

Once given the conditions

$$ f_{0}/m = r_{0} = 1 $$ $$ \bar{w} = 0.9 w $$

we can find

$$ \bar{w} = 0.9 w =  \bar{w} = 0.9 * 2.0615 = 1.855 $$

Plugging these values in yields

$$ A = 0.4528 $$ $$ B = 0.1975 $$

The particular solution is then

$$ y_{p}(t) = 0.4528 sin(1.855 t) + 0.1975 cos(1.855 t) $$

Equations (3) p. 53-8 and (2) p. 53-8 give us

$$ \rho = \frac{\bar{w}}{w} = \frac{1.855}{2.0615} = 0.9$$ $$ A = \frac{1}{[(1 - \rho^{2})^{2} + (2 \zeta \rho)^{2}]^{\frac{1}{2}}} = \frac{1}{[(1 - 0.9^{2})^{2} + (2 *0.2425 *0.9)^{2}]^{\frac{1}{2}}} = 2.628 $$

The following is the graph of the homogeneous solution:



The following is the graph of the particular solution:



The following is the graph of the total solution, $$ y(t) = y_{h}(t) + y_{p}(t) $$:



A graph of all three plots is produced below:



Given a Linearly Elastic Spring System
Consider a system of 3 linearly elastic springs

Use the following diagram for the node locations

Construct the topology matrix
This matrix contains the node numbers and degrees of freedom.

$$ \begin{bmatrix} 1 & 1 & 2 \\ 2 & 2 & 3 \\ 3 & 2 & 3 \end{bmatrix} $$

Construct the global stiffness matrix and load vector f
$$ \mathbf{K}= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

The load vector is in position 2 with F=100

$$ \mathbf{f}= \begin{bmatrix} 0 \\ 100 \\ 0 \end{bmatrix}

$$

Generating the element stiffness matrices
The element stiffness matrices are generated using the CALFEM function spring1e. The matrices are generated using k and 2k where k=1500

$$ \mathbf{Ke1}= \begin{bmatrix} 1500 & -1500 \\ -1500 & 1500 \end{bmatrix} $$

$$ \mathbf{Ke2}= \begin{bmatrix} 3000 & -3000 \\ -3000& 3000 \end{bmatrix} $$

Assembling the element stiffness matrices into the global stiffness matrix
The assembly is done using the assem fucntion from CALFEM.

Assembling the second element at the first row:

$$ \mathbf{K}= \begin{bmatrix} 3000 & -3000 & 0 \\ -3000 & 3000 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

Assembling the first element at the second row:

$$ \mathbf{K}= \begin{bmatrix} 3000 & -3000 & 0 \\ -3000 & 4500 & -1500 \\ 0 & -1500 & 1500 \end{bmatrix} $$

Assembling the second element at the third row:

$$ \mathbf{K}= \begin{bmatrix} 3000 & -3000 & 0 \\ -3000 & 7500 & -4500 \\ 0 & -4500 & 4500 \end{bmatrix} $$

Solve the system of equations using boundary conditions
The CALFEM fucntion solveq is used to solve the system of equations subject to certain boundary conditions.

$$ \mathbf{bc}= \begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix} $$

$$ \mathbf{a}= \begin{bmatrix} 0 \\ 0.0133 \\ 0 \end{bmatrix} $$

$$ \mathbf{a}= \begin{bmatrix} -40 \\ 0 \\ -60 \end{bmatrix} $$

Evaluate the element forces from the element displacements
$$ \mathbf{ed1}= \begin{bmatrix} 0 & 0.0133 \end{bmatrix} $$

$$ \mathbf{ed2}= \begin{bmatrix} 0.0133 & 0 \end{bmatrix} $$

$$ \mathbf{ed3}= \begin{bmatrix} 0.0133 & 0 \end{bmatrix} $$

Evaluate the spring forces from function spring1s
$$ \mathbf{es1}= \begin{bmatrix} 40 \end{bmatrix} $$

$$ \mathbf{es2}= \begin{bmatrix} -20 \end{bmatrix} $$

$$ \mathbf{es3}= \begin{bmatrix} -40 \end{bmatrix} $$

Solution: Results of verification
This section verifies the CALFEM code by manual calculations

Nodal equilibrium equations
$$f_1^{(1)}=F_1 = R_1$$

$$f_2^{(1)}+ f_2^{(2)}+ f_2^{(3)}=F = 100 $$

$$f_3^{(2)}+f_3^{(3)}=F_2=R_2$$

Setting up the equation to solve
$$ \begin{bmatrix}K_s\end{bmatrix}\begin{Bmatrix}Q_s\end{Bmatrix}=\begin{Bmatrix}F_s \end{Bmatrix} $$

$$ 1000\begin{bmatrix}3 &-3 &-2  &0 \\  -3&7.5  &-4.5 \\  0&-4.5  &4.5 \end{bmatrix}\begin{Bmatrix}u_1\\ u_2\\ u_3\end{Bmatrix}=\begin{Bmatrix}R_1\\ 100\\ R_2\end{Bmatrix}$$

Implementing the boundary conditions
$$ 1000\begin{bmatrix}7.5 \end{bmatrix}\begin{Bmatrix} u_2 \end{Bmatrix}=\begin{Bmatrix}100 \end{Bmatrix} $$

Inverting the global stiffness matrix to find displacements
$$ u_2=0.01333 $$

Displacements 1 and 3 are zero due to boundary conditions.

Calculate forces in springs from equation
The forces in the springs are found by the following equations. Negative forces display compressions and positive tensions. $$P=k(u_j-u_i)$$

$$ P^{(1)}=3000(0.01333-0)=40$$

$$P^{(2)}=1500(0-0.01333)=-20$$

$$P^{(3)}=3000(0-0.01333)=-40$$

Discuss if the solution is verified
The solution by CALFEM is verified since both the displacements and the forces are the same by both methods.

Problem Statement
Find the displacements of the nodes as well as the forces in the springs(tension/compression). Also determine the reaction forces at the walls.

Nodal Equilibrium Equations
$$f_1^{(1)}+ f_1^{(4)}=F_1 = R_1$$ $$f_2^{(1)}+ f_2^{(2)}+ f_2^{(3)}=F_2 = 0 $$ $$f_3^{(3)}+ f_3^{(4)}+ f_3^{(5)}=F_3 = 1000$$ $$f_4^{(2)}+ f_4^{(5)}+ f_4^{(6)}=F_4 = 0$$ $$f_5^{(6)}=F_5=R_5$$

$$\begin{bmatrix}K_s\end{bmatrix}\begin{Bmatrix}Q_s\end{Bmatrix}=\begin{Bmatrix}F_s \end{Bmatrix}$$

$$ 100\begin{bmatrix}7 &-5 &-2  &0  &0 \\  -5&15  &-6  &-4  &0 \\  -2&-6  &12  &-4  &0 \\  0&-4&-4  &11  &-3 \\ 0&0  &0  &-3  &3 \end{bmatrix}\begin{Bmatrix}u_1\\ u_2\\ u_3\\ u_4\\ u_5\end{Bmatrix}=\begin{Bmatrix}R_1\\ 0\\ 1000\\ 0\\ R_5\end{Bmatrix}$$ $$100\begin{bmatrix}15 &-6  &-4\\ -6  &12  &-4\\-4&-4  &11\\ \end{bmatrix}\begin{Bmatrix} u_2\\ u_3\\ u_4\end{Bmatrix}=\begin{Bmatrix}0\\ 1000\\ 0\end{Bmatrix}$$

Using Matlab Solutions Displacements These are the displacements at $$u_2$$,  $$u_3$$, and    $$u_4$$ respectively. The negative sign shows that the nodes are moving in the direction of the applied force which is against the convention of positive displacement left to right.

Now it is possible to find the forces in the spring. Negative forces denote compression and positive forces denote tension.

Forces in Springs
$$P=k(u_j-u_i)$$ $$ P^{(1)}=500(-0.8542-0)=-427.1N$$ $$P^{(2)}=400(-0.8750+0.8542)=-8.32N$$ $$P^{(3)}=600(-1.5521+0.8542)=-419N$$ $$P^{(4)}=200(-0.8750-0)=-310N$$ $$P^{(5)}=400(-0.8750+1.5521)=270.8N$$ $$P^{(6)}=300(0+0.8750)=262.5N$$ Now,

Reaction Forces
$$ f_1^{(1)}+ f_1^{(4)}=F_1 = R_1$$

$$R_1=-427N-310N=-737N$$

$$ f_5^{(6)}=F_5=R_5$$

$$R_5=262.5N$$

Verification with Calfem
a values are the nodal displacements 1 through 5. r values are the reaction forces at the walls. es1-es6 are the forces in the springs. Negative values denote compression and positive values denote tension. These answers match those previously calculated.

--EML4507.s13.team4ever.Wulterkens (discuss • contribs) 07:21, 6 February 2013 (UTC)