exs_bar2_l¶

Purpose:

Analysis of a plane truss.

Description:

Consider a plane truss, loaded by a single force \(P=0.5\) MN.

The corresponding finite element model consists of ten elements and twelve degrees of freedom.

Material properties:

Cross-sectional area: \(A=25.0 \times 10^{-4}\) m²
Young’s modulus: \(E=2.10 \times 10^{5}\) MPa

Example:

The topology is defined by the matrix:

>> Edof=[1   1  2  5  6;
         2   3  4  7  8;
         3   5  6  9 10;
         4   7  8 11 12;
         5   7  8  5  6;
         6  11 12  9 10;
         7   3  4  5  6;
         8   7  8  9 10;
         9   1  2  7  8;
        10   5  6 11 12];

A global stiffness matrix K and a load vector f are defined. The load \(P\) is divided into x and y components and inserted in the load vector f:

>> K=zeros(12);
>> f=zeros(12,1);  f(11)=0.5e6*sin(pi/6);  f(12)=-0.5e6*cos(pi/6);

The element matrices Ke are computed by the function bar2e. These matrices are then assembled in the global stiffness matrix using the function assem:

>> A=25.0e-4;    E=2.1e11;   ep=[E A];

>> Ex=[0 2;
       0 2;
       2 4;
       2 4;
       2 2;
       4 4;
       0 2;
       2 4;
       0 2;
       2 4];

>> Ey=[2 2;
       0 0;
       2 2;
       0 0;
       0 2;
       0 2;
       0 2;
       0 2;
       2 0;
       2 0];

All the element matrices are computed and assembled in the loop:

>> for i=1:10
      Ke=bar2e(Ex(i,:),Ey(i,:),ep);
      K=assem(Edof(i,:),K,Ke);
   end;

The displacements in a and the support forces in r are computed by solving the system of equations considering the boundary conditions in bc:

>> bc=[1 0;2 0;3 0;4 0];
>> [a,r]=solveq(K,f,bc)

a =

         0
         0
         0
         0
    0.0024
   -0.0045
   -0.0016
   -0.0042
    0.0030
   -0.0107
   -0.0017
   -0.0113

r =

  1.0e+005 *

   -8.6603
    2.4009
    6.1603
    1.9293
    0.0000
   -0.0000
   -0.0000
   -0.0000
    0.0000
    0.0000
    0.0000
    0.0000

The displacement at the point of loading is \(-1.7 \times 10^{-3}\) m in the x-direction and \(-11.3 \times 10^{-3}\) m in the y-direction. At the upper support the horizontal force is \(-0.866\) MN and the vertical \(0.240\) MN. At the lower support the forces are \(0.616\) MN and \(0.193\) MN, respectively.

Normal forces are evaluated from element displacements. These are obtained from the global displacements a using the function extract_ed. The normal forces are evaluated using the function bar2s:

ed=extract_ed(Edof,a);

>> for i=1:10
      es=bar2s(Ex(i,:),Ey(i,:),ep,ed(i,:));
      N(i,:)=es(1);
   end

The obtained normal forces are:

The largest normal force \(N=0.626\) MN is obtained in element 1 and is equivalent to a normal stress \(\sigma=250\) MPa.

To reduce the quantity of input data, the element coordinate matrices Ex and Ey can alternatively be created from a global coordinate matrix Coord and a global topology matrix Dof using the function coordxtr:

>> Coord=[0 2;
          0 0;
          2 2;
          2 0;
          4 2;
          4 0];

>> Dof=[ 1  2;
         3  4;
         5  6;
         7  8;
         9 10;
        11 12];

>> [ex,ey]=coordxtr(Edof,Coord,Dof,2);