University of Florida/Eml4507/s13.team3.Guzy

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Problem 2.5: Finite Analysis Of A Spring System[edit]

Spring-mass-dashpot system[1]
Spring-mass-dashpot system[2]

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 =
 EDU>> es2=spring1s(ep1,ed2)
 es2 =
 EDU>> es3=spring1s(ep2,ed3)
 es3 =

By hand:

In matrix form:

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.

The following is a matrix of the Force vector

Next we need each element stiffness matrix for every spring:

Element stiffness matrix for Spring #1:

Element stiffness matrix for Spring #2:

Element stiffness matrix for Spring #3:

To get the global stiffness matrix, we add up each element stiffness matrix:

Plugging each matrix into Hooke's Law we obtain:

Because node one and three are attached to the wall, their displacements are zero:

Now we solve the follwing equation:

Solving the System we obtain the following equations:

Simplifying the above equations we obtain:

Solving we obtain:

Plugging this value in

Using substitution for the remaining values, we obtain magnitudes of:

  1. Obtained from calfem34: p.53-2b]
  2. Obtained from calfem34: p.53-2b]