5.6
PIPE UNDER INTERNAL PRESSURE, PLAIN STRESS ELEMENT NO.7
Copy the example file
B6_X.DXF to Z88X.DXF.
B6_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 and then exported it 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"
Z88O, looking at structure, structure
file Z88I1.TXT
Z88F calculates deflections
Z88D calculates stresses
Z88E calculates nodal forces
Z88O, plot FE structure, now also
deflected (FUX, FUY, FUZ per 100.)
We deal with a pipe under
internal pressure of 1,000 bar (=100 N/mm2). Inside diameter
of the
pipe is 80 mm, outside diameter of the pipe is 160 mm. The length is 40
mm. If
one chooses the supports cleverly, a quarter of the pipe is enough to
reflect
the problem.
Such structures are best
suited for polar coordinates. The internal pressure of 1,000 bar
corresponds to
a force of 251,327 N which is loaded onto the inside quadrant. The
251,327 N
have to be distributed onto the nodes 1,6,9,14,17,22,25,30 and 33 in
accordance
with the rules for boundary conditions
(cf. chapter 3.4):
"1/6 points": 10,472 N
"2/3 points": 41,888 N
"2/6 points": 20,944 N
Control:
2*10,472+4*41,888+3*20,944 = 251,328 okay
These forces have an
outwardly directed radial effect. Thus, they must be subdivided into X
and Y
components for boundary conditions. E.g. the node 6 as "2/3 point" is
subdivided into X = 41,083 N and into Y = 8,172 N, because node 6 has
an angle
Phi = 11.25 degrees.
When dealing with a
rotationally symmetrical structure, the additional calculation of
radial
stresses and tangential stresses can be interesting: Set KFLAG to 1 in
Z88I3.TXT. As stresses are calculated in the Gauss points, use linear
extrapolations to get the stresses directly in the inside diameter and
the
outside diameter.
This problem is simple to
check analytically. Consult appropriate machine element books for
proper
calculation formulas or see chapter 5.7.
Plot of the undeflected
structure
5.6.1
Input
With
CAD program:
Proceed after the description chapter 2.7.2. Do
not forget to write on the layer Z88EIO
the element descriptions by TEXT function:
FE 1
7 (1st finite element type
7)
FE 2
7 (2nd finite element type
7)
......... (element
3 to 7
dropped here)
FE 8 7 (
8th finite element type 7)
Write the general
information and material information on the layer Z88GEN, like
Z88I1.TXT 2
37 8
74
1 1
0
0 0 (2D,
37 nodes, 8 ele, 74 DOF, 1 mat info,
polar coor., IBFLAG 0, IPFLAG 0, IQFLAG 0)
MAT 1
1 8
206000
0.3 3
40 (1st
mat info: Ele 1 to 8: Young's, Poisson's,INTORD= 3, QPARA = thickness =
40)
Write the boundary
conditions with the TEXT function onto the layer Z88RBD. Here we have
the case of
edge
loads for the
boundary conditions. You should consult chapter 3.4. and take into
account the
explanation and sketches for load distributions.
Z88I2.TXT 26 (26
boundary conditions)
RBD 1 1 1 1 10472 (1st BC: Node 1, DOF 1(= X), a load of 10,472 N)
RBD 2 1 2 2 0 (2nd BC: Node 1, DOF 2 (=Y), a displacement of 0
(=fixed))
RBD 3 2 2 2 0
RBD 4 3 2 2 0
RBD 5 4 2 2 0
RBD 6 5 2 2 0
RBD 7 6 1 1 41083
RBD 8 6 2 1 8172
RBD 9 9 1 1 19350
RBD 10 9 2 1 8015
RBD 11 14 1 1 34829
RBD 12 14 2 1 23272
RBD 13 17 1 1 14810
RBD 14 17 2 1 14810
RBD 15 22 1 1 23272
RBD 16 22 2 1 34829
RBD 17 25 1 1 8015
RBD 18 25 2 1 19350
RBD 19 30 1 1 8172
RBD 20 30 2 1 41083
RBD 21 33 1 2 0
RBD 22 33 2 1 10472
RBD 23 34 1 2 0
RBD 24 35 1 2 0
RBD 25 36 1 2 0
RBD 26 37 1 2 0
Switch to the layer Z88GEN
and write into any free place:
Z88I3.TXT 3
1 1
(3x3 Gauss points for stresses, KFLAG=1
i.e. additional calculation of radial and tangential stresses, von
Mises
stresses)
Export the drawing as DXF
file with the name Z88X.DXF, then start the CAD converter Z88X with the
option
"from Z88X.DXF to Z88I*.TXT" (DXF -> I*). The CAD converter
produces the three Z88 input files Z88I1.TXT, Z88I2.TXT, Z88I3.TXT.
With
editor:
Write the
structure data file Z88I1.TXT (cf.
chapter 3.2) with an editor:
2 37
8 74
1
1 0
0
0 ( 2D, 37 nodes, 8
elements,
74 DOF, 1 mat info, Polar coor., beam & plate & surface loads flag 0, each)
1 2
40 0
(1st node, 2 DOF, R and Phi coordinate)
2 2
48 0
(2nd node, 2 DOF, R and Phi coordinate)
3 2
56 0
4 2
68 0
5 2
80 0
6 2
40 11.25
7 2
56 11.25
8 2
80 11.25
9 2
40 22.5
..... (Nodes
10.. 35 dropped here
)
36 2
68 90
37 2
80 90
1 7
(element 1, Plain Stress Element No.7)
1 3
11 9
2
7 10
6 (coincidence
1st element)
2 7
3 5
13 11
4
8 12
7
..... (elements 3 .. 7 dropped here)
8 7
(element 8, Plain Stress Element No.7)
27 29
37 35
28
32 36
31 (coincidence
8th element)
1 8
206000 0.3
3
40 (Ele 1 to 8: Young's,
Poisson's, INTORD = 3, thickness = 40)
Here we have the case of edge
loads for the
boundary conditions. Consult chapter 3.4. and take into account the
explanation
and sketches for load distributions. Here is Z88I2.TXT:
26
(26 boundary
conditions)
1 1
1 10472
(Node 1, DOF 1(= X), a load of 10,472 N)
1 2
2 0
(Node 1, DOF 2 (=Y), a
displacement of 0 (=fixed))
2 2
2 0
3 2
2 0
4 2
2 0
5 2
2 0
6 1
1 41083
6 2
1 8172
9 1
1 19350
9 2
1 8015
14 1
1 34829
14 2
1 23272
17 1
1 14810
17 2
1 14810
22 1
1 23272
22 2
1 34829
25 1
1 8015
25 2
1 19350
30 1
1 8172
30 2
1 41083
33 1
2 0
33 2
1 10472
34 1
2 0
35 1
2 0
36 1
2 0
37 1
2 0
This example is very nice
for experiments with the boundary conditions: Enter deflections rather
than
forces into X and Y, e.g. 0.01 mm in radial direction to the outside.
At node 1
you can enter the 0.01 mm directly as X displacement and at node 33 you
can
enter directly the Y displacement of 0.01 mm, but for the other nodes
the
radial displacements of 0.01 mm must be subdivided into X and Y
components
respectively (via sine and cosine). Or enter mixed BC: A couple of
nodes with
displacements, the others with forces.. In practice nobody would do so
for such
a task, however, but Z88 can handle it.
A broad experimenting field
also opens up with Z88I3.TXT: You
have 5 possibilities for the first value
and two possibilities each for the second and third value, cf. Chapters
3.5 and 4.7. Now
we can
produce plenty of results:
Here is Z88I3.TXT:
3 1
1 (3x3 Gauss points for
stresses, KFLAG=1 i.e. additional calculation of radial and tangential
stresses, von Mises stresses)
CAD
and editor:
The structure data Z88I1.TXT, the boundary conditions Z88I2.TXT and the
header
file for the stress processor Z88I3.TXT (with any contents) are ready
to go.
Now launch
> Z88F the
Cholesky solver
> Z88D
the stress processor
> Z88E
the nodal force processor
Boundary
conditions by edge loads:
The
data entry by single forces was somewhat
cumbersome because of dividing the force of 251.327 N to several nodal
points
with respect to the actual angle position. It is much more easier to
enter edge
loads by the surface and pressure loads file Z88I5.TXT. The edge load
is:
q= F/l = F/(r*Phi) =
251327/(40*Pi/2) = 4000 N/mm
This
edge load acts onto the elements 1, 3, 5 and 7.
The edge of element 1 is the edge defined by
the corner
nodes 9 and 1 and the middle node 6 etc. The edge load points normally
to the
edge, there are no tangential loads. Thus,
the surface and pressure loads file Z88I5.TXT
is:
4
1 4000. 0.
9 1
6
3 4000. 0.
17 9
14
5 4000. 0.
25 17 22
7 4000. 0.
33 25 30
To
make the solvers reading in the surface and
pressure loads file you are to set the surface and pressure loads
flag to 1
in the first line of the general data file Z88I1.TXT:
2
37
8 74
1
0 0
0 1
| surface and pressure loads flag IQFLAG
Now
edit the boundary conditions file Z88I2.TXT:
Skip all forces:
10
1
2
2 0.
2
2
2 0.
3
2
2 0.
4
2
2 0.
5
2
2 0.
33
1
2 0.
34
1
2 0.
35
1
2 0.
36
1
2 0.
37
1
2 0.
Please
see the sample files B6_Q_1.TXT, B6_Q_2.TXT and
B6_Q_5.TXT on the CD-ROM or the Internet packages. Now compute
deflections,
stresses and nodal forces as usual:
·
Z88F the
Cholesky solver
·
Z88D the
stress processor
·
Z88E the
nodal force processor
5.6.2 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 nodal force processor Z88E internally uses the calculated
deflections of Z88F and stores Z88O4.TXT, the computed nodal
forces.
This example is very
suitable to demonstrate all the possibilities of the stress calculation
with Z88D and Plain Stress Elements No.7 (or Plain Stress Elements No.11). We recall: Z88I3.TXT
was: 3 1
1, i.e. 3 x 3 Gauss points,
additional calculation of radial and tangential stresses (which is very
meaningful for this example) and von Mises stresses calculation. Enter
in
Z88I3.TXT: 3 0 1, so that you will get von Mises stresses but no radial
and
tangential stresses. The results with 2 0 0 become even shorter (only
still 2 x
2 Gauss points, no radial/tangential stresses and no von Mises
stresses). You
will get the stresses with 0 0 0 into the corner nodes instead of into
the
Gauss points. Now
experiment.. you have 5 x 2 x 2 = 20 possibilities.
Now you should plot the
boundary conditions. The problem here is that some nodes have more than
one BC
e.g. node 1 features a force in X direction and is fixed in Y
direction,
too:
1
1 1
+1.04720E+004
1
2
2 +0.00000E+000
Thus, you select the BC you
want to see.
Suppose you would now choose a
hardened steel. This means to compute the principal
stresses not the von Mises
stresses. Thus, the third entry in Z88I3.TXT would now read 2:
3
1 2
Then do a stress calculation
again with Z88D.