5.4
BEAM in PLANE WITH BEAMS NO.13
Copy the example file
B4_X.DXF to Z88X.DXF.
B4_X.DXF ---> Z88X.DXF input file for CAD converter Z88X
CAD:
Import Z88X.DXF
into your CAD program and look at it. Usually you would have designed this
example in a CAD system (makes not much sense because this example is extremely
simple) and then exported as Z88X.DXF.
Z88: (in reduced form, more detailed
instructions cf. examples 5.1, 5.2 and 5.3)
Z88X, conversion, "from Z88X.DXF to
Z88I*.TXT"
Z88P, looking at structure, structure
file Z88I1.TXT
Z88F calculates deflections
Z88D calculates stresses
Z88E calculates nodal forces
Z88P, plot FE structure,
now also deflected (FUX, FUY, FUZ per 10.)
This example deals with a
beam, fixed on both sides, and loaded with 1,648 N in the middle in downward
direction. This mechanical problem is covered in every mechanical and civil
engineering handbook. Geometry: Length 1,000 mm, cross-cut 50 x 10 mm. Thus: A
= 500 mm2, Izz = 4,167 mm4, ezz = 5
mm.
The deflection curve has
inflection points, we therefore take 4 beams No.13. Nodes 1 and 5 will be fixed and
node 3 is loaded.
You would calculate
analytically:
f in the middle: F*L3/(192*E*I) = 10 mm
f in the inflection points: fw = f/2 = 5 mm
The bending moments on the left, middle, on the right: F*L/8 = 206,000 Nmm
The slope angle in the inflection points: Phi = atan (3*f/L) = 0.029991 rad
When interpreting the
results of Z88O2.TXT (deflections) and Z88O4.TXT (nodal forces and moments)
refer to the sign definition of chapter 3.13. Especially Z88O4.TXT, node 3: The
force F(2) = force in Y direction is the sum of the forces of elements 2 and 3,
due to extrinsic force. The force F(3) = bending moment is not a summary of
elements 2 and 3, because it is an intrinsic moment, not an extrinsic load ! Also
the signs of the load F(3) at node 1 and F(3) at node 5 are correct, refer to chapter 4.13. Keep in mind that the classical
mechanical science sometimes uses different conventions.
5.4.1
Input:
This example shows
that a FEA basically needs nodes in all locations where you want to get
results. As the beam is fixed left and right, the maximum of displacements
appears in the middle for x = L/2, but the bending curve features two
inflection points for x = L/4 and x = 3L/4. To calculate results for this
locations, the structure must be subdivided with nodes in x = 0, x = L/4, x =
L/2 and x = 3L/4.
Only the file input is
shown here because CAD use is not worth here.
Z88I1.TXT so becomes:
2 5 4 15 1 0 1 0 0 (2-D,5 nodes,4 ele, 5 DOF,1 mat info, KFLAG
0, IBFLAG 1 ,IPFLAG 0, IQFLAG 0)
1 3 0 0 (1.node,
3 DOF, X and Y coordinate)
2 3 250 0
3 3 500 0
4 3 750 0
5 3 1000 0
1 13 (1. element, type beam in plane No.13)
1 2 (coincidence for 1. element)
2 13
2 3
3 13
3 4
4 13
4 5
1 4 206000 0.3 1 500 0 0 4167 5 0 0 (mat
info for ele 1 to 4, Young's, Poisson's, INTORD (any), QPARA = area, Ixx=0,
exx=0, Izz, ezz, It=0, Wt=0)
The node 1 is fixed in all
degrees of freedom at the boundary conditions. It is important to fix
especially the DOF 1 = displacement in X direction so that the structure cannot
move. Node 5 is fixed in DOF 2 = displacement in Y direction and DOF 3 =
rotation around Z axis. You could also fix DOF 1 for node 5, if you wish. But
in reality one of the bearings or supports will allow for thermal expansion. This
was taken into account in Z88I2.TXT.
Here Z88I2.TXT:
6 (6 Boundary conditions)
1 1 2 0 (Node 1, DOF 1 gets a displacment of 0 =
DOF 1 fixed)
1 2 2 0 (Node 1, DOF 2 fixed)
1 3 2 0 (Node 1, DOF 3 fixed (restraining moment)
3 2 1 -1648
(Node 3, DOF 2 gets load of -1,648 N)
5 2 2 0
5 3 2 0
The parameter file for the
stress processor Z88I3.TXT
can have any content (cf. sections 3.5 and 4.13), because Gauss points, radial
and tangential stresses as well as calculation of the von Mises stresses has no
significance for Beams No.13.
5.4.2
Results
The Cholesky solver Z88F
provides the following output files:
Z88O0.TXT stores the processed structure data. For documentation
purposes.
Z88O1.TXT stores the processed boundary conditions: For documentation
purposes.
Z88O2.TXT, the displacements, the main task and solution of the FEA
problem.
The stress processor Z88D internally uses the calculated displacements from
Z88F and stores Z88O3.TXT, the calculated stresses. The results in
Z88O3.TXT do not depend on the header parameters in Z88I3.TXT for Beams No.13.
The nodal force processor Z88E internally uses the calculated
deflections of Z88F and stores Z88O4.TXT, the computed nodal forces.
The following picture of
the plot program shows the deflected structure for FUX, FUY and FUZ = 10 each
(magnifications of the deflections).
Attention to the results of
the nodal force calculation: Node 3: The force F(2) = force in Y direction is
the sum of the forces of elements 2 and 3, due to extrinsic force. The force
F(3) = bending moment is not a summary of elements 2 and 3, because it is an
intrinsic moment, not an extrinsic load ! Also the signs of the load F(3) at
node 1 and F(3) at node 5 are correct, refer to chapter 4.13. Keep in mind that
the classical mechanical science sometimes uses different conventions.
Additional remark: Such
simple examples are well suitable to become aware of the sign omen definitions.
Experiment with this example and calculate other bend cases from good
handbooks. Frameworks with Beams No.2 are calculated accordingly. However, a real
spatial structure then must be available: At least one Z coordinate must not
equal 0.
View of undeflected and
deflected structure
Take into account: The plot program
Z88P connects the
nodes with straight lines, although the deflection curve represents a cubic
parable in the case of a Beam No.13 or No.2. This means: Z88P shows the
deformations correctly for the node, but straight lines are between the nodes. Therefore,
no deflection curve is shown. If you want to plot a real nice deflection curve
with Z88P, then use basically more nodes, e.g. 15 to 20 nodes for this example
(the cubic bending curve is then featured by a couple of straight lines).