Civil Engineering Reference
In-Depth Information
C1= LOG( 1/Radius )*C
CALL
Normal_Jac(Vnorm,Jac,xsi,eta,ldim,nodel,Inci,elcor)
CALL
ShapefunctionDisc(Ni,xsi,eta,ldim,nodel,Inci)
Direction1_2: DO
j=1,2
IF
(Isym == 0)
THEN
iD= 2*(i-1) + j
ELSE
iD= Ndest(i,j)
END IF
IF
(iD == 0)
CYCLE
nD= 2*(ns-1) + j
dUe(iD,nD)= dUe(iD,nD)&
+ Ni(n)*C1*Jac*dxdxb* dxbdxp*Wi(m)
END DO Direction1_2
END DO Gauss_points2
END DO Region_Loop_1
END DO Shape_function_1
END DO Element_nodes_1
END SUBROUTINE
SingularIntegration2D
For the discontinuous version of both programs (prog71_discont and
prog111_discont) the input files of the continuous versions can be used. At the
beginning of each program, the coordinates of the collocation nodes are calculated from
the node coordinates of the element and given values of
d
and
d
. From the input file
given incidences are transformed to the discontinuous version, too, from which the
degrees of freedom are specified. There were no other major changes necessary.
12.2.4
Test Example - Single Region
To test the implementation the following example a cantilever beam (same example as
used in chapter 10) shown in Figure 12.4 is considered. The example is designed to
show that the discontinuous elements give good results even for the case where no
corners are present.
The diagram in Figure 12.5 shows the vertical displacements along the cantilever
beam. The results for the discontinuous meshes shown in Figure 12.5 are extrapolated
from the discontinuous points to the element nodes according to the interpolation
functions and are compared with discontinuous elements for two different meshes. The
first mesh consists of 3 quadratic elements along the length and the second has 5
elements along the length. For both one element for the height is used.
As is shown in Figure 12.5 the vertical displacements at the length side of the beam
are the same and agree very well with the analytical solution, except for the coarse mesh
with continuous elements. For the discontinuous mesh with 3 elements and for the
continuous mesh with 5 elements on the length side the same accuracy of results are
obtained. These two meshes are comparable because they have exact the same number
of degrees of freedom.
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