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.

Plot of the von Mises stresses

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.

Selective plot of the BCs with Z88O.

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.