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"
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 100.)
We deal with a pipe under internal pressure of 1,000 bar (=100
N/mm*mm). 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 O.k.
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 (2D, 37 nodes, 8 ele, 74 DOF, 1 mat info, polar coor., IBFLAG 0, IPFLAG 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 ( 2D, 37 nodes, 8 elements, 74 DOF, 1 mat info, Polar coor., beam & plate 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 (1st BC: Node 1, DOF 1(= X), a load of 10,472 N)
1 2 2 0 (2nd BC: 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
> 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.
The following picture of the plot program shows the deflected
structure for FUXand FUY = 100 each (magnifications of the deflections):
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. Take into account that stresses display
in Z88P is impossible for corner nodes.
Now experiment.. you have 5 x 2 x 2 = 20 possibilities.
Plot of the undeflected and the deflected structure
Plot of the von Mises stresses