Beam Problem 13 first created
30/08/9 - last modified
30/08/9 Page Author: Ty HarnessUnderstanding the Z88 beam in space element No.2
The Z88 finite element beam in space no. 2 has 6 DOF per node, ux,uy,uz,rotx,roty,rotz
It's quite complicated to make the switch from 2D to 3D analysis so this tutorial is to help you understand how
to input the 3D geometry and loads and also read back the results.
Starting with a simple example of a cantilever, as shown in figure 1, with depth of 250mm and width of 75mm.
The Length of the cantilever is 1000mm and has a Young's Modulus of 2.05E5 N/mm^2 and a shear modulus of
7.9300E+004 N/mm^2. I've made the cross section rectangular to investigate and identify the correct order of
material properties. Of course this is really a plane problem so it's easy to compare the results with an analytical analysis.
Figure 1 - Cantilever with end load
The depth has been orientated along the global z axis and a 5000N force,P applied in the -ve z direction.
From simple beam theory we can calculate the deflection at the free end.
The maximum shear force is simply equal to P, 5000N.
The maximum bending moment is found at x = 0
$M_max = P*L = 5000*1000 = 5*10^6 [N*mm]
and therefore the maximum stress is found x = 0 at the extreme fibres of the depth hence 125mm from the neutral axis.
$ sigma_max = (M_max*0.5*d)/I = (5*10^6*0.5*250)/ (9.766*10^7) = 6.4 N/(mm^2)$
The derivation of the analytical equations are found in most mechanics books out there. For example, Timoshenko and Young, 'Strength
of Materials'.
For the Z88 3D analysis the geometry file Z88i1.txt contains additional information over the plane problems
previously encountered, Breaking down the format:
The header line : 3 5 4 30 1 0 1 0 Z88I1.TXT,typed in by Ty
Analysis Type : 3 for 3D
Number of Nodes : 5
Number of Elements : 4
Number of DOF : 5x6 = 30
Number of Material lines = 1
Coordinate system : 0 for Cartesian and 1 for Polar
Beam Element : 0 for no 1 for yes
Plate Element : 0 for no 1 for yes
Comment line
The nodal coordinates in a Cartesian system, X,Y,Z, comment
The elements section:
1 2 element #1
1 2
2 2 element #2
2 3
3 2 element #3
3 4
4 2 element #4
4 5
Element ID, Element type Beam 2, Comment
Element 1 has 2 nodes and is connected between node1 and node2. This is the local x axis
of the element. There's more information below. It's coincidental this example that the global
and local axes are aligned.
The material line needs some explanation with each property separated with a space or tab character.
Start Element: 1
End Element: 4
Young's Modulus E: +2.05000E+005
Poisson's Ratio: +3.00000E-001
Integration order: 1 {Z88 INTERNAL PARAMETER}
Cross section Area: +1.87500E+004
Second moment of area, Iyy (bending around the local yy axis): +9.765625E+007
Max dist from the local yy axis: +1.25000E+002
Second moment of area, Izz (bending around the z axis): +8.7891E+006
Max distance from the zz axis: +3.7500E+001
Second Polar Moment of Area, J: +1.0644535E+008
Shear Modulus, G: +7.9300E+004
The material properties refer to element local coordinate system
where the element x axis is aligned with global axis. This is of course not always the case and you
need to be careful you get the Second moment of area's the correct way round see figure 2.
i.e. node 1 has all DOF set to 0 displacement and node 5 has force of 5000 applied in the -ve z direction.
Z88 Deflection Results
Looking at the deflection results file Z88O2.txt as expected we can see a max deflection
on node 5 of -8.3252033E-002 in the U(3) or z direction which agrees with the
above analytical result.
output file Z88O2.TXT : displacements, computed by Z88F V10
*************
element #= 4 type =spbeam node 4 node 5
SIGXX TAUXX SIGZZ1 SIGYY1 SIGZZ2 SIGYY2
+0.000E+000 +0.000E+000 +0.000E+000 -1.600E+000 +0.000E+000 -3.442E-014
Z88 Nodal Force Results
From the Z88 force results we can see the shear force in z direction on node 1 is +5000 which is the opposite
sign to the applied force and hence the beam is in equilibrium in the z direction and of course
the restorative moment around the y axis on node 1 must also be equal to the applied transverse force x the lever arm to be in equilibrium.
output file Z88O4.TXT : nodal forces, computed by Z88E V10
************
starting with nodal forces computed for each element
----------------------------------------------------
element # = 1 type = beam in space
node F(1) F(2) F(3) F(4) F(5) F(6)
1 +0.00000E+000 +0.00000E+000 +5.00000E+003 +0.00000E+000 -5.00000E+006 +0.00000E+000
2 +0.00000E+000 +0.00000E+000 -5.00000E+003 +0.00000E+000 +3.75000E+006 +0.00000E+000
element # = 2 type = beam in space
node F(1) F(2) F(3) F(4) F(5) F(6)
2 +0.00000E+000 +0.00000E+000 +5.00000E+003 +0.00000E+000 -3.75000E+006 +0.00000E+000
3 +0.00000E+000 +0.00000E+000 -5.00000E+003 +0.00000E+000 +2.50000E+006 +0.00000E+000
element # = 3 type = beam in space
node F(1) F(2) F(3) F(4) F(5) F(6)
3 +0.00000E+000 +0.00000E+000 +5.00000E+003 +0.00000E+000 -2.50000E+006 +0.00000E+000
4 +0.00000E+000 +0.00000E+000 -5.00000E+003 +0.00000E+000 +1.25000E+006 +0.00000E+000
element # = 4 type = beam in space
node F(1) F(2) F(3) F(4) F(5) F(6)
4 +0.00000E+000 +0.00000E+000 +5.00000E+003 +0.00000E+000 -1.25000E+006 +0.00000E+000
5 +0.00000E+000 +0.00000E+000 -5.00000E+003 +0.00000E+000 +7.45058E-009 +0.00000E+000
now the nodal sums for each node
--------------------------------
node F(1) F(2) F(3) F(4) F(5) F(6)
1 +0.00000E+000 +0.00000E+000 +5.00000E+003 +0.00000E+000 -5.00000E+006 +0.00000E+000
2 +0.00000E+000 +0.00000E+000 -4.36557E-011 +0.00000E+000 +5.58794E-009 +0.00000E+000
3 +0.00000E+000 +0.00000E+000 -2.61934E-010 +0.00000E+000 -7.45058E-009 +0.00000E+000
4 +0.00000E+000 +0.00000E+000 -3.20142E-010 +0.00000E+000 +1.11759E-008 +0.00000E+000
5 +0.00000E+000 +0.00000E+000 -5.00000E+003 +0.00000E+000 +7.45058E-009 +0.00000E+000
Repeat the analysis but applying and end force in +ve y direction
The maximum shear force is simply equal to P, 5000N.
The maximum bending moment is found at x = 0
$M_max = P*L = 5000*1000 = 5*10^6 [N*mm]
and therefore the maximum stress is found x = 0 at the extreme fibres of the depth hence 37.5mm from the neutral axis.
$ sigma_max = (M_max*0.5*d)/I = (5*10^6*0.5*75)/ (8.7891*10^6 ) = 21.3 N/(mm^2)$
Z88 Analysis
All the hard work of setting up the modal pays off now because z88i1.txt is unchanged and all we have to do is edit the load file.
element #= 4 type =spbeam node 4 node 5
SIGXX TAUXX SIGZZ1 SIGYY1 SIGZZ2 SIGYY2
+2.667E-001 +0.000E+000 +0.000E+000 +0.000E+000 +0.000E+000 +0.000E+000
ASCII to MathML used in this page: ASCIItoMathML homepage
