exs_beambar2¶

Purpose:

Analysis of a combined beam and bar structure.

Description:

Consider a structure consisting of a beam with \(A_1=4.0 \times 10^{-3}\) m² and \(I_1=5.4 \times 10^{-5}\) m⁴ supported by two bars with \(A_2=1.0 \times 10^{-3}\) m². The beam as well as the bars have \(E=200\) GPa. The structure is loaded by a distributed load \(q=10\) kN/m. The corresponding finite element model consists of three beam elements and two bar elements and has 14 degrees of freedom.

Example:

The computation is initialised by defining the topology matrix Edof1 for the beam elements and Edof2 for the bar elements. The matrix K (14×14), and vector f (14×1) are created and filled with zeros:

>>  Edof1=[1  1  2  3  4  5  6;
>>         2  4  5  6  7  8  9;
>>         3  7  8  9 10 11 12];
>>  Edof2=[4 13 14  4  5;
>>         5 13 14  7  8];
>>
>>  K=zeros(14);        f=zeros(14,1);

The element property vectors ep1 and ep2 and the element coordinate vectors ex1, ex2, ex3, ex4, ex5, ey1, ey2, ey3, ey4 and ey5 are defined:

>>  E=200e9;   A1=4.0e-3;   A2=1.0e-3;   I1=5.4e-5;
>>
>>  ep1=[E A1 I1];   ep4=[E A2];
>>
>>  eq1=[0 0];       eq2=[0 -10e3];
>>
>>  ex1=[0 2];       ey1=[2 2];
>>  ex2=[2 4];       ey2=[2 2];
>>  ex3=[4 6];       ey3=[2 2];
>>  ex4=[0 2];       ey4=[0 2];
>>  ex5=[0 4];       ey5=[0 2];

The element stiffness matrices Ke1, Ke2 and Ke3 are computed using beam2e and Ke4 and Ke5 are computed using bar2e. Element load vectors fe2 and fe3 are also given by beam2e:

>>  Ke1=beam2e(ex1,ey1,ep1);
>>  [Ke2,fe2]=beam2e(ex2,ey2,ep1,eq2);
>>  [Ke3,fe3]=beam2e(ex3,ey3,ep1,eq2);
>>  Ke4=bar2e(ex4,ey4,ep4);
>>  Ke5=bar2e(ex5,ey5,ep4);

Based on the topology information, the global stiffness matrix K and load vector f are generated by assembling the element matrices using assem:

>>  K=assem(Edof1(1,:),K,Ke1);
>>  [K,f]=assem(Edof1(2,:),K,Ke2,f,fe2);
>>  [K,f]=assem(Edof1(3,:),K,Ke3,f,fe3);
>>  K=assem(Edof2(1,:),K,Ke4);
>>  K=assem(Edof2(2,:),K,Ke5);

Considering the prescribed displacements in bc, the system of equations is solved using the function solveq, yielding displacements a and support forces r. According to the computation the vertical displacement at the end of the beam is \(13.0\) mm:

>>  bc=[1 0; 2 0; 3 0; 13 0; 14 0];
>>  [a,r]=solveq(K,f,bc)

a =                           r =

         0                       1.0e+04 *
         0
         0                       -8.0702
    0.0002                       -0.6604
   -0.0006                       -0.1403
   -0.0010                             0
    0.0004                       -0.0000
   -0.0046                       -0.0000
   -0.0033                             0
    0.0004                       -0.0000
   -0.0130                        0.0000
   -0.0045                             0
         0                             0
         0                       -0.0000
                                  8.0702
                                  4.6604

The section forces es1, es2, es3, es4 and es5 are calculated using bar2s and beam2s from element displacements ed1, ed2, ed3, ed4 and ed5 obtained using extract. This yields the normal forces \(-35.4\) kN, \(-152.5\) kN in the bars and the maximum moment \(10.00\) kNm in the beam:

>>  Ed1=extract_ed(Edof1,a);
>>  Ed2=extract_ed(Edof2,a);
>>
>>  es1=beam2s(ex1,ey1,ep1,Ed1(1,:),eq1,11)
>>  es2=beam2s(ex2,ey2,ep1,Ed1(2,:),eq2,11)
>>  es3=beam2s(ex3,ey3,ep1,Ed1(3,:),eq2,11)
>>  es4=bar2s(ex4,ey4,ep2,Ed2(1,:))
>>  es5=bar2s(ex5,ey5,ep2,Ed2(2,:))

es1 =

   1.0e+04 *

    8.0702    0.6604    0.1403
    8.0702    0.6604    0.0082
      .         .         .
    8.0702    0.6604   -1.1806

es2 =

   1.0e+04 *

    6.8194   -0.5903   -1.1806
    6.8194   -0.3903   -1.0825
      .         .         .
      .         .         .
    6.8194    1.4097   -2.0000

es3 =

   1.0e+04 *

         0   -2.0000   -2.0000
         0   -1.8000   -1.6200
      .         .         .
         0    0.0000   -0.0000

es4 =

   1.0e+04 *

   -3.5376
   -3.5376

es5 =

   1.0e+05 *

   -1.5249
   -1.5249