Civil Engineering Reference
In-Depth Information
is then read, and this takes the form of information relating to nodal forces and/or fixed
nodal displacements.
In the case of loads,
loaded nodes
is read first signifying the number of nodes with
forces applied. Then for each of these, the node number and the applied force are read.
In this rod example there is only one freedom at each node, but in later programs in the
chapter where more than one freedom exists at each node, all the “forces” applied at the
loaded node must be included in the correct sense (even if some of them are zero).
In the case of fixed displacements,
fixed freedoms
is read, signifying the number
of fixed freedoms in the mesh. Then for each of these, the node number and the value to
which the freedom is to be fixed is read. In later programs in the chapter where more than
one freedom exists at each node, data must also be read into the vector
sense
,which
gives the sequential number of the freedom at the node that is to be fixed. If a particular
node has more than one fixed freedom, the node must be entered in the data list for each
fixed “sense”. If either
loaded nodes
or
fixed freedoms
equals zero, no further
data relating to that category is required.
In the case shown in Figure 4.2, a rod of uniform stiffness equal to 10
5
is subjected to
a (negative) uniformly distributed axial load of 5/unit length. The force has been “lumped”
at the nodes as was indicated in (2.10). In this example, there are no fixed displacements,
so
fixed freedoms
is read as zero.
In the case shown in Figure 4.3, the rod has a non-uniform stiffness, and since there
is more than one property group (
np types = 2
), the element type vector
etype
must
be read indicating in this case that elements 1 and 2 have an axial stiffness of 1000.0,
and elements 3 and 4 have an axial stiffness of 2000.0. There are no loaded nodes so
loaded nodes
is read as zero but a fixed displacement is applied to the tip of the rod,
so
fixed freedoms
is read as 1, followed by the node number (5) and the magnitude
of the fixed displacement (0.05).
Following equation solution, the nodal displacements (overwritten as
loads
)arecom-
puted and printed. A final “post-processing” phase is then performed, in which the elements
are scanned once more. In this loop, the element nodal displacements (
eld
) are retrieved
from the global displacements vector and the element stiffness matrices (
km
) re-computed.
Multiplication of the element nodal displacements by the element stiffness matrix using
MATMUL
, results in the element “actions” vector called
action
, which holds the internal
end reaction forces for each element.
The computed results for both cases are reproduced in Figure 4.5. In the first case,
the end deflection at node 1 is given as
10
−
4
which is the exact solution. Note,
however, that the element “actions” indicate that the fourth element sustains a mean tensile
force of 4.375 which is the best this element can do to approximate the true solution of a
linearly varying axial load.
In the second case, the nodal displacements indicate the distribution of displacements
along the length of the rod up to the end node 5, which has the expected displacement of
0.05. All the elements in the rod are sustaining a tensile load of 66.67.
In problems such as this, the nodal displacements are always in exact agreement with
the closed form solution achieved by direct integration of the governing equation. Between
the nodes however, the solution may be approximate, due to the limitations of the shape
functions.
−
0
.
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