Is the Kinematics of 3D Space Truss Affected by Closed Loops in the System?

[topcomer]

Most, or perhaps all, of these modes could be theoretically prevented by changing all diagonals in the post 17 truss to cross braces (i.e., an X in each small square, with the crossing diagonals not connected to each other at their intersection point). Using cross braces and inextensional members, no corner node on any small square can deflect out-of-plane, theoretically. Therefore, I think it eliminates the modes mentioned in my preceding paragraph.

I agree, however this is not allowed. What is allowed, though, is to make the squares "irregular" in their shape. Or, more generally, it is allowed to draw arbitrary triangles, with the only requisite that they must have at least one vertex on the external boundary. And of course, there should be no hanging vertex, that is a node lying on another triangle's edge or, in other words, each node must be a vertex for all the incident triangles.

I mention irregular shapes because of the hexagon example. According to the counting, the hexagon has 5 free vertices and 11 non redundant members. This boils down to 5*3-11=4 dofs, but for the irregular hexagon is somehow not possible to reach other admissible configuration without breaking it down into pieces and then reassemble it. The same is true, for example, in the regular case after you bend along one diagonal: you cannot bend along any other one, but this does not mean that the admissible configuration is unique.

Mathematically, I believe I'm looking for the existence of not connected constrained manifolds. For the regular hexagon, more manifolds intersect in the undeformed state. While for the irregular, the manifolds are only a point set of admissible configurations.

Leaving the maths aside, the question is to find a simple calculation or procedure discerning the structures that can rigid-body deformed into their admissible shapes, and the structures that have to be broken and reassembled.

 

[topcomer]

False statement. I currently retract the above statement and most of the statements below it in post 17.

When you say that it is not possible to use a nonlinear analysis, are you thinking about the following? Say that the nonlinear strain is the following 2x2 tensor (one tensor for each triangle):

\epsilon (u) = \nabla u ^T \nabla u

However, for small displacements, the linearization is the stiffness matrix I was trying to build before, which is always singular for a structure connected with ball joints. Is this what you were saying? So let me try to see it by using a test function:

K = \nabla_u^2 \epsilon (u) = \nabla \phi ^T \nabla \phi

where now:

\nabla \phi = \left[ \nabla \phi_x \nabla \phi_y \nabla \phi_z \right]^T

is a 9x2 tensor (and not 3x2 as when there was u). Note that x,y,z are the displacements in the three directions, while the gradient is taken with respect to some parametrization (u,v) of the undeformed domain. Then:

K_{11} = (\phi_{x,u})^2 + (\phi_{y,u})^2 + (\phi_{z,u})^2
K_{22} = (\phi_{x,v})^2 + (\phi_{y,v})^2 + (\phi_{z,v})^2
K_{12} = K_{21} = \phi_{x,u}\phi_{x,v} + \phi_{y,u}\phi_{y,v} + \phi_{z,u}\phi_{z,v}

where at the moment I don't clearly see any singularity. So apply this to the regular hexagon, basis functions adequate to represent rigid trusses with ball joints are piecewise linear. Let's write their gradient for the six triangles, considering for simplicity only one component (say x) since they are equal, and proceeding counterclockwise from the fixed triangle:

\nabla \phi_x = [-1/10, 1/10, 0; -1/5, -1/5, 1/5] (v1,v2,v0)
\nabla \phi_x = [1/10, -1/10, 0; 1/5, 1/5, -1/5] (v3,v0,v2)
\nabla \phi_x = [0, -1/10, 1/10; 1/5, -1/5, -1/5] (v4,v0,v3)
\nabla \phi_x = [-1/10, 0, 1/10; 1/5, -1/5, 1/5] (v5,v0,v4)
\nabla \phi_x = [-1/10, 1/10, 0; -1/5, -1/5, 1/5] (v6,v0,v5)
\nabla \phi_x = [0, 1/10, -1/10; -1/5, 1/5, 1/5] (v1,v0,v6)

where v0 is the center, and the others are numerated counterclockwise. Well, maybe I should continue in Matlab. But so far it does not look like this approach is flawed as the linear analysis. I'll let you know later.

 

[topcomer]

This is the stiffness matrix computed by Matlab:

(1, 1) -> -10.851
(4, 1) -> 2.1701
(5, 1) -> 1.0851
(7, 1) -> 2.1701
(8, 1) -> -1.0851
(10, 1) -> 1.0851
(13, 1) -> 2.1701
(14, 1) -> 1.0851
(16, 1) -> 2.1701
(17, 1) -> -1.0851
(19, 1) -> 1.0851
(2, 2) -> -15.191
(4, 2) -> 1.0851
(5, 2) -> 4.3403
(7, 2) -> -1.0851
(8, 2) -> 4.3403
(11, 2) -> -1.0851
(13, 2) -> 1.0851
(14, 2) -> 4.3403
(16, 2) -> -1.0851
(17, 2) -> 4.3403
(20, 2) -> -1.0851
(1, 4) -> 2.1701
(2, 4) -> 1.0851
(4, 4) -> -3.7977
(5, 4) -> -0.54254
(7, 4) -> 0.54253
(8, 4) -> 0.54254
(19, 4) -> 1.0851
(20, 4) -> -1.0851
(1, 5) -> 1.0851
(2, 5) -> 4.3403
(4, 5) -> -0.54254
(5, 5) -> -5.9679
(7, 5) -> -0.54254
(8, 5) -> -0.54253
(20, 5) -> 2.1701
(1, 7) -> 2.1701
(2, 7) -> -1.0851
(4, 7) -> 0.54253
(5, 7) -> -0.54254
(7, 7) -> -3.7977
(8, 7) -> 0.54254
(10, 7) -> 1.0851
(11, 7) -> 1.0851
(1, 8) -> -1.0851
(2, 8) -> 4.3403
(4, 8) -> 0.54254
(5, 8) -> -0.54253
(7, 8) -> 0.54254
(8, 8) -> -5.9679
(11, 8) -> 2.1701
(1, 10) -> 1.0851
(7, 10) -> 1.0851
(10, 10) -> -3.2552
(13, 10) -> 1.0851
(2, 11) -> -1.0851
(7, 11) -> 1.0851
(8, 11) -> 2.1701
(11, 11) -> -3.2552
(13, 11) -> -1.0851
(14, 11) -> 2.1701
(1, 13) -> 2.1701
(2, 13) -> 1.0851
(10, 13) -> 1.0851
(11, 13) -> -1.0851
(13, 13) -> -3.7977
(14, 13) -> -0.54254
(16, 13) -> 0.54253
(17, 13) -> 0.54254
(1, 14) -> 1.0851
(2, 14) -> 4.3403
(11, 14) -> 2.1701
(13, 14) -> -0.54254
(14, 14) -> -5.9679
(16, 14) -> -0.54254
(17, 14) -> -0.54253
(1, 16) -> 2.1701
(2, 16) -> -1.0851
(13, 16) -> 0.54253
(14, 16) -> -0.54254
(16, 16) -> -3.7977
(17, 16) -> 0.54254
(19, 16) -> 1.0851
(20, 16) -> 1.0851
(1, 17) -> -1.0851
(2, 17) -> 4.3403
(13, 17) -> 0.54254
(14, 17) -> -0.54253
(16, 17) -> 0.54254
(17, 17) -> -5.9679
(20, 17) -> 2.1701
(1, 19) -> 1.0851
(4, 19) -> 1.0851
(16, 19) -> 1.0851
(19, 19) -> -3.2552
(2, 20) -> -1.0851
(4, 20) -> -1.0851
(5, 20) -> 2.1701
(16, 20) -> 1.0851
(17, 20) -> 2.1701
(20, 20) -> -3.2552

whose null space has 7 components. Let's now neglect the elements corresponding to fixed nodes:

(1, 1) -> -10.851
(4, 1) -> 1.0851
(7, 1) -> 2.1701
(8, 1) -> 1.0851
(10, 1) -> 2.1701
(11, 1) -> -1.0851
(13, 1) -> 1.0851
(2, 2) -> -15.191
(5, 2) -> -1.0851
(7, 2) -> 1.0851
(8, 2) -> 4.3403
(10, 2) -> -1.0851
(11, 2) -> 4.3403
(14, 2) -> -1.0851
(1, 4) -> 1.0851
(4, 4) -> -3.2552
(7, 4) -> 1.0851
(2, 5) -> -1.0851
(5, 5) -> -3.2552
(7, 5) -> -1.0851
(8, 5) -> 2.1701
(1, 7) -> 2.1701
(2, 7) -> 1.0851
(4, 7) -> 1.0851
(5, 7) -> -1.0851
(7, 7) -> -3.7977
(8, 7) -> -0.54254
(10, 7) -> 0.54253
(11, 7) -> 0.54254
(1, 8) -> 1.0851
(2, 8) -> 4.3403
(5, 8) -> 2.1701
(7, 8) -> -0.54254
(8, 8) -> -5.9679
(10, 8) -> -0.54254
(11, 8) -> -0.54253
(1, 10) -> 2.1701
(2, 10) -> -1.0851
(7, 10) -> 0.54253
(8, 10) -> -0.54254
(10, 10) -> -3.7977
(11, 10) -> 0.54254
(13, 10) -> 1.0851
(14, 10) -> 1.0851
(1, 11) -> -1.0851
(2, 11) -> 4.3403
(7, 11) -> 0.54254
(8, 11) -> -0.54253
(10, 11) -> 0.54254
(11, 11) -> -5.9679
(14, 11) -> 2.1701
(1, 13) -> 1.0851
(10, 13) -> 1.0851
(13, 13) -> -3.2552
(2, 14) -> -1.0851

and we get the following null space:

0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
-0.12518 0.96664 -0.22349 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.99196 0.12619 -0.00981 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.01871 -0.22292 -0.97466 -0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 -1.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 1.00000

which seems again to indicate that the structure is free to move in the z-direction regardless of the trusses.

 

[nvn]

Just to clarify my statement you quoted in post 32, I'm saying, in the truss shown in the post 17 diagram, it appears it is possible that nodes can move out-of-plane without elongating members, as discussed in post 30. Therefore, I retracted my statement your post 32 quote referenced, because although there might be cases where the referenced statement (a statement about nonlinear analysis) is true, there appear to be exceptions, as discussed in post 30.

 

[chingkui]

didn't try to go through all posts here, but are you looking for the Chebychev–Grübler–Kutzbach's criterion in 3D?

 

[topcomer]

didn't try to go through all posts here, but are you looking for the Chebychev–Grübler–Kutzbach's criterion in 3D?

I must admit to have never heard about this criterion before, but from the definition on wikipedia it seems that the answer to this question is yes. So do you think it is possible to apply it to my case? Would it work with the hexagon trivial example?

From more googling, it seems that the criterion fails when there are closed loops in the system. This might be bypassed using the "dual" representation of my rigid triangles connected by ball joints (shown in post #27), that is by observing that triangles can bend only about the hinges.