# University of Florida/Eml4507/s13.team3.Guzy
## Problem 2.5: Finite Analysis Of A Spring System

The free body diagram is defined by the matrix
Next, create the load force matrix
Next each element stiffness matrix is created
Now we create the global stiffness matrix
Boundary conditions are now applied
Displacements are found:
Spring forces are evaluated
By hand:
    $\displaystyle f=kd$
    $\displaystyle f=force$
    $\displaystyle k=stiffness$
    $\displaystyle d=displacement$ In matrix form: ${\begin{bmatrix}f_{1x}\\f_{2x}\\\end{bmatrix}}={\begin{bmatrix}k&-k\\-k&k\\\end{bmatrix}}{\begin{bmatrix}d_{1x}\\d_{2x}\\\end{bmatrix}}$ First we create a matrix to show the systems Degree of Freedoms. The first column represents each spring(#1,#2,#3). The second and third column represent the connectivity of each node. $DOF={\begin{bmatrix}1&1&2\\2&2&3\\3&2&3\\\end{bmatrix}}$ The following is a matrix of the Force vector $F={\begin{bmatrix}F_{1x}\\F_{2x}\\F_{3x}\\\end{bmatrix}}$ Next we need each element stiffness matrix for every spring: Element stiffness matrix for Spring #1: $k_{1}={\begin{bmatrix}3000&-3000\\-3000&3000\\\end{bmatrix}}={\begin{bmatrix}3000&-3000&0\\-3000&3000&0\\0&0&0\\\end{bmatrix}}$ Element stiffness matrix for Spring #2: $k_{2}={\begin{bmatrix}1500&-1500\\-1500&1500\\\end{bmatrix}}={\begin{bmatrix}0&0&0\\0&1500&-1500\\0&-1500&1500\\\end{bmatrix}}$ Element stiffness matrix for Spring #3: $k_{3}={\begin{bmatrix}3000&-3000\\-3000&3000\\\end{bmatrix}}={\begin{bmatrix}0&0&0\\0&3000&-3000\\0&-3000&3000\\\end{bmatrix}}$ To get the global stiffness matrix, we add up each element stiffness matrix: $K={\begin{bmatrix}3000&-3000&0\\-3000&3000&0\\0&0&0\\\end{bmatrix}}+{\begin{bmatrix}0&0&0\\0&1500&-1500\\0&-1500&1500\\\end{bmatrix}}+{\begin{bmatrix}0&0&0\\0&3000&-3000\\0&-3000&3000\\\end{bmatrix}}={\begin{bmatrix}3000&-3000&0\\-3000&7500&-4500\\0&-4500&4500\\\end{bmatrix}}$ Plugging each matrix into Hooke's Law we obtain: ${\begin{bmatrix}3000&-3000&0\\-3000&7500&-4500\\0&-4500&4500\\\end{bmatrix}}{\begin{bmatrix}d_{1x}\\d_{2x}\\d_{3x}\\\end{bmatrix}}={\begin{bmatrix}F_{1x}\\F_{2x}\\F_{3x}\\\end{bmatrix}}$ Because node one and three are attached to the wall, their displacements are zero: $\displaystyle d_{1x}=0$ $\displaystyle d_{3x}=0$ Now we solve the follwing equation: ${\begin{bmatrix}3000&-3000&0\\-3000&7500&-4500\\0&-4500&4500\\\end{bmatrix}}{\begin{bmatrix}0\\d_{2x}\\0\\\end{bmatrix}}={\begin{bmatrix}F_{1x}\\F_{2x}\\F_{3x}\\\end{bmatrix}}$ Solving the System we obtain the following equations: $\displaystyle 3000*0-3000*d_{2x}+0*d_{3x}=F_{1x}$ $\displaystyle -3000*0+7500*d_{2x}-4500*d_{3x}=F_{2x}$ $\displaystyle 0*0-4500*d_{2x}+4500*d_{3x}=F_{3x}$ Simplifying the above equations we obtain: $\displaystyle -3000*d_{2x}=F_{1x}$ $\displaystyle 7500*d_{2x}-4500*d_{3x}=F_{2x}$ $\displaystyle -4500*d_{2x}+4500*d_{3x}=F_{3x}$ Solving we obtain: $\displaystyle F_{1x}=40$ Plugging this value in $\displaystyle {\frac {40}{-3000}}=d_{2x}=-0.01333$ Using substitution for the remaining values, we obtain magnitudes of: $d_{1x}=0$ $d_{2x}=0.01333$ $d_{3x}=0$ $F_{1x}=40$ $F_{2x}=20$ $F_{3x}=40$
* ↑ http://upload.wikimedia.org/wikiversity/en/b/b1/Fead.s13.sec53a.djvu Obtained from calfem34: p.53-2b]
* ↑ http://upload.wikimedia.org/wikiversity/en/b/b1/Fead.s13.sec53a.djvu Obtained from calfem34: p.53-2b]