CCX gravity problem?

1

Look at this simple example:
I have a box 10x10x10 mm, density 7.8e-09 g/mm^3, and gravity of 9800 mm/sec
then I would expect a total reaction force of 0.0764 N but CCX gives 0.0637 N
so does CCX have general gravity problem or does a deeper meaning exist?

*node
1, 0., 0., 0.
2, 10., 0., 0.
3, 10., 10., 0.
4, 0., 10., 0.
**
5, 0., 0., 10.
6, 10., 0., 10.
7, 10., 10., 10.
8, 0., 10., 10.
**
9, 5., 0., 0.
10, 10., 5., 0.
11, 5., 10., 0.
12, 0., 5., 0.
**
13, 5., 0., 10.
14, 10., 5., 10.
15, 5., 10., 10.
16, 0., 5., 10.
**
17, 0., 0., 5.
18, 10., 0., 5.
19, 10., 10., 5.
20, 0., 10., 5.
**
*nset, nset=bottom
1, 2, 5, 6, 9, 13, 17, 18
**
*element, type=c3d20, elset=cube
1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
16, 17, 18, 19, 20
**
*material, name=steel
*density
7.8E-09
*elastic
210000, 0.28
**
*solid section, elset=cube, material=steel
**
*step
*static
**
*boundary, op=new
bottom, 1, 3, 0
**
*Dload
cube, Grav, -9800., 0.0, 1.0, 0.0
**
*node print, nset=bottom, totals=Yes
rf
**
*node file
rf, u
*el file
s, e
*end step
2

check out this point in the manual:

image
image600Ă—684 25.6 KB

3

@JuanP74
I can be wrong, but if I understand correctly, then the reaction force for the nodes will correspond in same way as if a surface load had been distributed to the nodes, and for that reason the mass of the cube distributed to the top nodes should correspond to the same reaction force as if the cube has been put under gravity force.

To proof my supposition, I calculate the mass of the cube and distribute the force to the top nodes in according to figure 156 in the manual as a kind of virtual gravity, and in this case, I get this result.

 forces (fx,fy,fz) for set BOTTOM and time  0.1000000E+01

         1 -9.536286E-04  4.217641E-03 -9.536286E-04
         2  9.536286E-04  4.217641E-03 -9.536286E-04
         5 -9.536286E-04  4.217641E-03  9.536286E-04
         6  9.536286E-04  4.217641E-03  9.536286E-04
         9 -2.943876E-18 -2.332764E-02 -3.784692E-03
        13  1.011649E-18 -2.332764E-02  3.784692E-03
        17 -3.784692E-03 -2.332764E-02 -6.062956E-18
        18  3.784692E-03 -2.332764E-02 -7.401797E-18

 total force (fx,fy,fz) for set BOTTOM and time  0.1000000E+01

       -1.257675E-17 -7.644000E-02 -9.561626E-18

which are exactly as expected and correspond to gravity force of the cube.

But when I instead of the virtual gravity load put gravity load on the cube by the *DLOAD card as show in the datafile, I get this result.

 forces (fx,fy,fz) for set BOTTOM and time  0.1000000E+01

         1  8.423300E-04  4.433178E-03  8.423300E-04
         2 -8.423300E-04  4.433178E-03  8.423300E-04
         5  8.423300E-04  4.433178E-03 -8.423300E-04
         6 -8.423300E-04  4.433178E-03 -8.423300E-04
         9  2.305835E-18  1.149182E-02  3.469199E-03
        13 -5.796352E-19  1.149182E-02 -3.469199E-03
        17  3.469199E-03  1.149182E-02  3.899951E-18
        18 -3.469199E-03  1.149182E-02  6.065179E-18

 total force (fx,fy,fz) for set BOTTOM and time  0.1000000E+01

        3.035766E-18  6.370000E-02  4.241517E-19

which in my mind not really make much sense, so please explain for me how the reaction force should be understood, and how it actual correspond to the real gravity force of the cube

4

The origin of the confusion comes from associating the acronym RF to reaction force when a more appropriate name would be EF (External Forces).
Read carefully what Juan has posted. Your result contains a subtracted part of the weight of the cube.

From understanding this difference you have several options to correctly extract the reaction forces.

imagen
imagen988Ă—571 37.6 KB

2 Likes
5

@Disla
Thanks for the explanation. :+1:
I see the point. I was tricked by the name “Reaction Forces”. Gravity load is a special load case since all nodes in the element has to be weighted and then total external forces will be weights from all nodes except those which have been fixed. I agree, the name “External forces” would make much more sense, since this is exactly the outcome for this “RF” option.

1 Like