University of Florida/Eml4507/s13.team4ever.Wulterkens.R2

1) Explain the Assembly Process of the Above 2D Truss System

2) Verify With CALFEM

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

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

For element 2

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

The variable K can be determined by

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

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

For element 1

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

For element 2

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

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

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

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