University of Florida/Eml4507/s13.team3.Guzy

Problem 2.5: Finite Analysis Of A Spring System

File:Report2problem.JPG

Spring-mass-dashpot system

File:FEAproblem2.png

Spring-mass-dashpot system

The free body diagram is defined by the matrix

 EDU>> Edof=[1 1 2;2 2 3;3 2 3];

Next, create the load force matrix

 EDU>> f=[0 100 0];

Next each element stiffness matrix is created

 EDU>> K= [0 0 0;0 0 0;0 0 0];
 EDU>> k=1500;
 EDU>> ep1=k;
 EDU>> ep2=2*k;
 EDU>> Ke1=springle(ep1);
 EDU>> Ke2=springle(ep2);

Now we create the global stiffness matrix

 EDU>> K=assem(Edof(1,:),K,Ke2);
 EDU>> K=assem(Edof(2,:),K,Ke1);
 EDU>> K=assem(Edof(3,:),K,Ke2);

Boundary conditions are now applied

 EDU>> bc=[1 0; 3 0];
 EDU>> [a,r]=solveq(K,f,bc);

Displacements are found:

 EDU>> ed1=extract(Edof(1,:),a)
 ed1 =
     0      0.0133
 EDU>> ed2=extract(Edof(2,:),a)
 ed2 =
     0.0133      0 
 EDU>> ed3=extract(Edof(3,:),a)
 ed3 =
     0.0133      0 

Spring forces are evaluated

 EDU>> es1=spring1s(ep2,ed1)
 es1 =
     40
 EDU>> es2=spring1s(ep1,ed2)
 es2 =
    -20
 EDU>> es3=spring1s(ep2,ed3)
 es3 =
    -40

By hand:

$f=kd$

$f=force$

$k=stiffness$

$d=displacement$ In matrix form: $\left[\begin{array}{c}{f}_{1x}\\ f_{2x}\end{array}\right]=\left[\begin{array}{cc}k& -k\\ -k& k\end{array}\right]\left[\begin{array}{c}{d}_{1x}\\ d_{2x}\end{array}\right]$ 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. $DOF=\left[\begin{array}{ccc}1& 1& 2\\ 2& 2& 3\\ 3& 2& 3\end{array}\right]$ The following is a matrix of the Force vector $F=\left[\begin{array}{c}{F}_{1x}\\ F_{2x}\\ F_{3x}\end{array}\right]$ Next we need each element stiffness matrix for every spring: Element stiffness matrix for Spring #1: $k_1=\left[\begin{array}{cc}3000& -3000\\ -3000& 3000\end{array}\right]=\left[\begin{array}{ccc}3000& -3000& 0\\ -3000& 3000& 0\\ 0& 0& 0\end{array}\right]$ Element stiffness matrix for Spring #2: $k_2=\left[\begin{array}{cc}1500& -1500\\ -1500& 1500\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1500& -1500\\ 0& -1500& 1500\end{array}\right]$ Element stiffness matrix for Spring #3: $k_3=\left[\begin{array}{cc}3000& -3000\\ -3000& 3000\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\\ 0& 3000& -3000\\ 0& -3000& 3000\end{array}\right]$ To get the global stiffness matrix, we add up each element stiffness matrix: $K=\left[\begin{array}{ccc}3000& -3000& 0\\ -3000& 3000& 0\\ 0& 0& 0\end{array}\right]+\left[\begin{array}{ccc}0& 0& 0\\ 0& 1500& -1500\\ 0& -1500& 1500\end{array}\right]+\left[\begin{array}{ccc}0& 0& 0\\ 0& 3000& -3000\\ 0& -3000& 3000\end{array}\right]=\left[\begin{array}{ccc}3000& -3000& 0\\ -3000& 7500& -4500\\ 0& -4500& 4500\end{array}\right]$ Plugging each matrix into Hooke's Law we obtain: $\left[\begin{array}{ccc}3000& -3000& 0\\ -3000& 7500& -4500\\ 0& -4500& 4500\end{array}\right]\left[\begin{array}{c}{d}_{1x}\\ d_{2x}\\ d_{3x}\end{array}\right]=\left[\begin{array}{c}{F}_{1x}\\ F_{2x}\\ F_{3x}\end{array}\right]$ Because node one and three are attached to the wall, their displacements are zero: $d_{1x}=0$ $d_{3x}=0$ Now we solve the follwing equation: $\left[\begin{array}{ccc}3000& -3000& 0\\ -3000& 7500& -4500\\ 0& -4500& 4500\end{array}\right]\left[\begin{array}{c}0\\ d_{2x}\\ 0\end{array}\right]=\left[\begin{array}{c}{F}_{1x}\\ F_{2x}\\ F_{3x}\end{array}\right]$ Solving the System we obtain the following equations: $3000*0-3000*d_{2x}+0*d_{3x}=F_{1x}$ $-3000*0+7500*d_{2x}-4500*d_{3x}=F_{2x}$ $0*0-4500*d_{2x}+4500*d_{3x}=F_{3x}$ Simplifying the above equations we obtain: $-3000*d_{2x}=F_{1x}$ $7500*d_{2x}-4500*d_{3x}=F_{2x}$ $-4500*d_{2x}+4500*d_{3x}=F_{3x}$ Solving we obtain: $F_{1x}=40$ Plugging this value in $\frac{40}{-3000}=d_{2x}=-0.01333$ Using substitution for the remaining values, we obtain magnitudes of: $d_{1x}=0$ $d_{2x}=0.01333$ $d_{3x}=0$ $F_{1x}=40$ $F_{2x}=20$ $F_{3x}=40$