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

Given 2-D Truss Model and Constants

Consider the 2D truss system

File:R2.1.jpg

Figure 1: 2D Truss system analyzed in this problem. ([1])

Use the following constants to very calculations

$l^{(1)}=4$

$l^{(2)}=2$

$E^{(1)}=3$

$E^{(2)}=5$

$A^{(1)}=1$

$A^{(2)}=2$

${\mathrm{\varnothing }}^{(1)}=30^{\circ }$

${\mathrm{\varnothing }}^{(2)}=135^{\circ }$

$P=7$

Solution: Assembly Process for 2d Truss System

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

$\overrightarrow{K^{(1)}}=\left[\begin{array}{ccccc}& 1& 2& 3& 4\\ 1& (l^{(1)}{)}^2& l^{(1)}{m}^{(1)}& -(l^{(1)}{)}^2& -l^{(1)}{m}^{(1)}\\ 2& l^{(1)}{m}^{(1)}& (m^{(1)}{)}^2& -l^{(1)}{m}^{(1)}& -(m^{(1)}{)}^2\\ 3& -(l^{(1)}{)}^2& -l^{(1)}{m}^{(1)}& (l^{(1)}{)}^2& l^{(1)}{m}^{(1)}\\ 4& -l^{(1)}{m}^{(1)}& -(m^{(1)}{)}^2& l^{(1)}{m}^{(1)}& (m^{(1)}{)}^2\end{array}\right]*K^{(1)}$

For element 2

$\overrightarrow{K^{(2)}}=\left[\begin{array}{ccccc}& 3& 4& 5& 6\\ 3& (l^{(2)}{)}^2& l^{(2)}{m}^{(2)}& -(l^{(2)}{)}^2& -l^{(2)}{m}^{(2)}\\ 4& l^{(2)}{m}^{(2)}& (m^{(2)}{)}^2& -l^{(2)}{m}^{(2)}& -(m^{(2)}{)}^2\\ 5& -(l^{(2)}{)}^2& -l^{(2)}{m}^{(2)}& (l^{(2)}{)}^2& l^{(2)}{m}^{(2)}\\ 6& -l^{(2)}{m}^{(2)}& -(m^{(2)}{)}^2& l^{(2)}{m}^{(2)}& (m^{(2)}{)}^2\end{array}\right]*K^{(2)}$

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

$\overrightarrow{K^{(1)}}=\left[\begin{array}{ccccc}& 1& 2& 3& 4\\ 1& 0.5625& 0.3248& -0.5625& -0.3248\\ 2& 0.3248& 0.1875& -0.3248& -0.1875\\ 3& -0.5625& -0.3248& 0.5625& 0.3248\\ 4& -0.3248& -0.1875& 0.3248& 0.1875\end{array}\right]$

For element 2

$\overrightarrow{K^{(2)}}=\left[\begin{array}{ccccc}& 3& 4& 5& 6\\ 3& 2.5& -2.5& -2.5& 2.5\\ 4& -2.5& 2.5& 2.5& -2.5\\ 5& -2.5& 2.5& 2.5& -2.5\\ 6& 2.5& -2.5& -2.5& 2.5\end{array}\right]$

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

$\overrightarrow{K}=\left[\begin{array}{ccccccc}& 1& 2& 3& 4& 5& 6\\ 1& 0.5625& 0.3248& -0.5625& -0.3248& 0& 0\\ 2& 0.3248& 0.1875& -0.3248& -0.1875& 0& 0\\ 3& -0.5625& -0.3248& 3.0625& -2.1752& -2.5& 2.5\\ 4& -0.3248& -0.1875& -2.1752& 2.6875& 2.5& -2.5\\ 5& 0& 0& -2.5& 2.5& 2.5& -2.5\\ 6& 0& 0& 2.5& -2.5& -2.5& 2.5\end{array}\right]$

Solution: Verify With CALFEM

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.

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