# Kinematic of space truss

1. Oct 5, 2009

### topcomer

Hi,

I am a PhD student in math and in my current research I'm faced with the problem of analyzing the rigidity of a structure of rigid beams in 3D, that is if the structure can move. In particular, something like this:

http://img525.imageshack.us/img525/1545/98322092.th.jpg [Broken]

where all the lower and left lines vertices are fixed. It is thus clear that the row and column formed by little squares are rigid, and the same is true for the bottom-right and the upper-left triangles. However, it seems that this structure is not rigid in 3D since is possible to move the upper-right pair of triangles along the diagonal of the big square by "folding" the little square they form.

Is there a way to choose the orientation of the triangles such that the structure cannot move? Is there a theory I can refer to in order to understand this? 2D is very easy, but the counting of degrees of freedom seems different in 3D, if possible at all when the structure has the topology of a surface.

AQ

PS I don't know if the homework area is more appropriate for this question, sorry.

Last edited by a moderator: May 4, 2017
2. Oct 6, 2009

### nvn

Can you draw on your diagram the boundary conditions (constraints)? I.e., show which degrees of freedom are rigidly fixed to the outside world. Secondly, are the joints of all members pinned or welded? If pinned, are they pinned only on certain axes, or on all three axes (which would be a ball joint)? I didn't follow your comment about folding. Maybe if you label some key points, it would be easier to refer to certain points or lines in your description (?). In 3-D, there can be up to six degrees of freedom per node.

3. Oct 6, 2009

### topcomer

Thanks for the reply. Sorry for having been sloppy, but I'm not used with terminology and notations of this field.

The joints leave 2 axes free, that is they do not allow for a rotation having the truss as axis. Here is the upgraded diagram, where red dots are fixed dofs:

http://img17.imageshack.us/img17/8599/55880634.th.jpg [Broken]

My comment about folding refers to the fact that instead of beams you can think of rigid triangles and squares, so if you think about "folding" the upper-right square along its diagonal, then it seems you can freely move the structure.

Last edited by a moderator: May 4, 2017
4. Oct 6, 2009

### nvn

You said, "The joints leave two axes free." State which dofs of the joints are free, and which dofs are not free. The truss is not an axis, so what you said is unclear. And you said, "Red dots are fixed dofs." But you have not described which dofs are fixed by a red dot; all three translations only, all six dofs, or what? If all six dofs are fixed by a red dot, then those members are inconsistent with the other members, which have joints that "leave two [rotation?] axes free." You need to define coordinate system axes on your diagram, and clearly explain which dofs are constrained, and which dofs are free. There are three translations (Tx, Ty, Tz), and three rotations (Rx, Ry, Rz), per node. Rx means rotation about the x axis.

Last edited: Oct 6, 2009
5. Oct 6, 2009

### topcomer

Maybe some examples are more useful than explanations, since I don't understand why I can't choose my coordinate system to have one axis along the truss. Red dots are the same as the joints between trusses, but also fix positional dofs.

This triangle has 6 dofs but is rigid, i.e. that trusses have fixed angles between them:

http://img84.imageshack.us/img84/9256/angles.th.jpg [Broken]

This triangle is completely fixed, i.e. no truss can move in any way:

http://img75.imageshack.us/img75/8348/fixedb.th.jpg [Broken]

This triangle has 1 dof, i.e. it can rotate about the y-axis:

http://img84.imageshack.us/img84/4100/roty.th.jpg [Broken]

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6. Oct 6, 2009

### nvn

So are you saying all of your member-to-member joints shown in post 3 are welded joints, meaning two connected members cannot rotate relative to each other about any axis?

[strike]Notice there are two types of constraints you need to describe: member-to-member (inter-member) dofs (or we could call it member-to-node dofs), and[/strike] node-to-outside-world (external) dofs.

EDIT: I retract the above sentence. Member-to-node fixity is described by element formulation, and is not separate dofs. The only dofs are node-to-world dofs, and there can be up to six dofs per node. Full connectivity (welding) of a member to a node means the member inherits all dofs of the node.

Last edited: Oct 7, 2009
7. Oct 6, 2009

### topcomer

No, I believe I'm saying the opposite, otherwise the structure would be rigid (not infinitesimally but it does not matter) if trusses were welded. I'm starting to think that ball-joints are a good description, if they don't allow rotations in case (2) of my previous post.

Member-to-member and node-to-outside joints allow the same rotations. The difference is that the red dots constrain also the positions. As I understood from your explanations, the following structure:

http://img203.imageshack.us/img203/374/truss.th.jpg [Broken]

should have 9 free dofs, which would be 12 if there were no red dot. These dofs should be rotations around the 3 cartesian axes for each truss. I'm still not fully convinced though, that a infinitesimally thin truss should have 3 rotations and not 2.

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8. Oct 7, 2009

### nvn

Therefore, it sounds like we have clarified that all of your structural members are connected by ball joints, which means they are space truss members. And therefore, all nodes have only three dofs (translations Tx, Ty, Tz), except for nodes having a red dot, which have all three translations constrained. Note that a red dot does not impose rotational fixity to a member, because truss members do not inherit rotational fixity from the node. Rotational dofs do not exist in this model; they are irrelevant. Truss members can rotate only because their nodes translate, not because their nodes rotate.

What do you mean by, "I'm still not fully convinced, though, that an infinitesimally thin truss should have 3 rotations, and not 2"? Could you elaborate on this comment? A truss member has only up to three translations per node.

9. Oct 7, 2009

### topcomer

OK now that you clarified this, it's also clear that I was assigning wrong dofs. The question "I'm still not fully convinced, though, that an infinitesimally thin truss should have 3 rotations, and not 2" was motivated by my attempt to put frames at nodes, having thus 6 dofs, and to say that a single member should not be allowed to rotate along itself, to reduce the total dofs for a single truss. But your approach and explanation have cleared my ideas.

So, coming back to the original question, is it possible to say if the structure is rigid? And if not, is it possible to change the orientation of some members in order to make it rigid? Thank you again for the patience.

10. Oct 7, 2009

### Mech_Engineer

This basically lookis like a numerical FEA problem to me.

If this framework is being modeled in 3-D using spar elements, it is true that the end-points for each spar will only have 3 degrees of freedom, translations in X,Y, and Z. However, if each element is modeled as a beam element rather than a simple spar, them it is possible to incorporate 6 degrees of reedom in each node, translation in X, Y, and Z, and Rotation about X, Y, and Z. Which element you use will depend on the purpose of the model.

You will have to define what you mean by rigid in the first place. If by rigid you mean fully constrained, that can be determined by looking at the number of nodal degrees of freedom and constraints available in your system. If the connections are only modeled as spar connections, that means they have zero bending stiffness, and are only able to take tension or compression, and cannot be constrained in bending.

To determine if the structure is fully constrained, you will need to take all of your degrees of freedom (72 nodes times 3 DOF each), subtract the fixed degrees of freedom (21 nodes times 3 fixed DOF each), and subtract 1 DOF for each side of a spar that is attached to an unfixed node (51 nodes times 3 spars each). The result of this is 216 total nodal degrees of freedom, minus 63 fixed boundary condition DOF, minus 153 degrees of freedom constrained by attached spars. This results in 0 left over degrees of freedom, which seems to indicate to me that the structure is fully constrained, although it's a little difficult to visualize.

Last edited: Oct 7, 2009
11. Oct 7, 2009

### Mech_Engineer

I'll go through these simples examples too, to help things along...

So we have 3 nodes with 3 DOF each, giving us a total of 9 nodal degrees of freedom. Subtract the number of fixed nodes times 3 (0 x 3 = 0) and subtract the number of connections between a spar and an unconstrained node (3 x 2 = 6). So, this triangle has a total number of free nodal degrees of freedom of 9 - 0 - 6 = 3 (keep in mind we aren't looking at rotational degrees of freedom).

This one is obviously trivial. We can see that it is actually OVERconstrained however, because each node is fixed and has a spar attached to it.

The free degree of freedom for this case is actually that the node can traslate in Z (although for large displacements it will orbit the axis defined by the other two as you pointed out).

So we have 3 nodes with 3 DOF each, giving us a total of 9 nodal degrees of freedom. Subtract the number of fixed nodes times 3 (2 x 3 = 6) and subtract the number of connections between a spar and an unconstrained node (1 x 2 = 2). So, this triangle has a total number of available degrees of freedom of 9 - 6 - 2 = 1

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12. Oct 7, 2009

### nvn

topcomer: Are you referring to displacement in the x, y, or z direction(s) in this sentence?

13. Oct 8, 2009

### topcomer

This framework has indeed origin from a FEM model where a square membrane is discretized used some piecewise linear nonconforming elements. After some simplifications, I could show that the rigidity of such discretization is equivalent to the rigidity of this ribbon modeled using FEA.

By "beams" you mean 3 dimensional elements which can have bending, and by "spars" thin rigid lines? Then I'm using "spars".

Yes, I mean constrained. But I also mean that the spars have infinite and not zero bending stiffness. As for the connections, yes they oppose no resistance to bending.

Let me illustrate an example showing why I think this does not suffice. Consider the following hexagon:

http://img269.imageshack.us/img269/8703/freet.th.jpg [Broken]

which, even without counting, it is clear to have 4 dofs. But with the same connectivity, we can also draw this other hexagon:

http://img14.imageshack.us/img14/7802/nonfree.th.jpg [Broken]

which seems to have only 1 dof. So, is it really enough to count? Are my examples wrong?

I'm referring to the fact that I believe it's possible to move the green point along the blue line (in-plane) without deforming the trusses.

http://img88.imageshack.us/img88/9264/movement.th.jpg [Broken]

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14. Oct 8, 2009

### nvn

When someone speaks of a "rigid" truss member, they are usually referring to the material of the truss member, not the external constraints applied to the nodes. topcomer, my current understanding is that your truss members (having ball joints on each end) are made of a theoretically, infinitely rigid material, which means the member has absolutely zero elongation, which is called inextensional or rigid members.

In your problem, there is a big difference between theoretical rigidity versus actual material having a very high stiffness. The theoretical material has absolutely zero elongation, which is physically impossible. But my current understanding is you do want theoretically rigid truss members. Please let us know if I am misinterpreting.

Last edited: Oct 8, 2009
15. Oct 8, 2009

### topcomer

Yes, I have theoretically rigid inextensible truss members, and I want to show if the global structure made of this theoretical members can move.

16. Oct 8, 2009

### Mech_Engineer

You're right, it's an imperfect solution, especially with large numbers of nodes and rigid body rotation degrees of freedom that are difficult to describe using just nodal DOF. I wonder however, if a similar method will work if we take into account all 6 DOF for each node (3 translation, 3 rotation).

It's difficult to visualize how that would occur without lengthening the diagonal spar that is attached to it, but it is possible the structure could somehow "fold up" out of plane to accomodate it.

I think your best bet for definitively determining the constrained nature of the structure is to develop the stiffness matrix for it in 3-D space, either by hand (which could take a while) or using a program like MatLAB (the raw stiffness matrix will be 216x216, and 153x153 after the boundary conditions are applied). If the stiffness matrix for the system is singular after all of the boundary conditions are applied (det(K) = 0), then the system is unstable (i.e. underconstrained). If you're not sure how to derive a stiffness matrix for a 3-dimensional spar, you can look it up in an intro to Finite Element Analysis book. I have two, both of which have examples of 3-dimensional spars and the derivation of 3-dimensional system matrices for space trusses (albeit with only 4 nodes, 3 spar elements).

You're a PhD math student, so I'm assuming you would have to do something like this manually, but the easy route is of course to plug your structure into an FEA package like ANSYS. ANSYS gives an error for a singular stiffness matrix, which would be a dead giveaway that the system is underconstrained.

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17. Oct 8, 2009

### nvn

Because of the inextensional members, it appears the truss shown in your post 3 diagram, and shown again in my attached file below, is completely rigid in the x and y directions, or any direction in the xy plane, including along the blue line you drew in post 13. A finite element (FE) program will not be able to reflect the theoretical nature of the problem for z-direction (out-of-plane) displacement. A linear FE analysis (static or modal) will show the truss highly unstable in the z direction. It would not be stable until it deflects out-of-plane a finite amount in a nonlinear FE analysis. But the out-of-plane deflection is inconsistent with the assumption of inextensional members, because any z-direction displacement means the upper and right sides of the truss must describe an arc out of the page. The arc means the sides must therefore pull inward, which would bend the upper and right sides of the big square. This "bending" would require extension of the members, which cannot occur if the members are inextensional (rigid).

#### Attached Files:

• ###### space-truss-01.png
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Last edited: Oct 9, 2009
18. Oct 8, 2009

### Mech_Engineer

An FEA program can detect an underconstrained condition using the stiffness matrix (as said in my previous post).

Well that's not necessarily true if the truss is able to somehow fold up on itself (how you show the truss folding up is pretty difficult however).

I was noticing that in the OP's hexagonal examples, the only time there are "folding" degrees of freedom are when two spars are co-linear with each other. However, the space truss in the first post does not have any connections like this, which might be a critical part of showing the truss cannot "fold up" without lengthening a spar in the process.

Last edited: Oct 8, 2009
19. Oct 8, 2009

### Mech_Engineer

Perhaps you can use a method of dividing the structure up. If you first look at the top 3 boxes in the upper right corner and detemine if they by themselves are rigid, and then add a box on each side and analyze that structure, and so on...

#### Attached Files:

• ###### Example.JPG
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20. Oct 8, 2009

### topcomer

Actually I'm doing applied maths, so I do plenty of Matlab/Mathematica/C++. Usually I do Finite Elements and not FEA though, which is somehow different I believe, can you please point me to one of these books? Maybe I can implement the stiffness matrix symbolically in Mathematica, so I will get some insights about the rigidity of the framework. Thank you!

21. Oct 8, 2009

### topcomer

I don't think this is possible, since the deformation mode is global and not local. To visualize it, think about lifting and pushing nodes on the outer and inner part of the non-fixed ribbons, the ribbon will turn inward or outward, and the green point will come closer to the fixed boundary. What you have to achieve, is just moving the green point along the blue diagonal.

I still however have the hope to say that apart from a certain special number of structures, for most of the choices of diagonals in the ribbon, the framework is rigid.

22. Oct 8, 2009

### Mech_Engineer

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23. Oct 8, 2009

### topcomer

I thought a little bit more about this and I have some concern. The stiffness matrix is a linearization of the deformation, so it would only tell if the system is linearly under-constrained, which is always the case for the framework I proposed. Just think about lifting infinitesimally a vertex, this would introduce only a second-order elongation of the spars.

24. Oct 8, 2009

### Mech_Engineer

FEA programs are used all the time to analyze 2-D and 3-D space trusses. I see what you're saying about stiffness in the "micro" scale, but there are tricks you can use in the way the problem is solved to avoid them. As it is the problem should be fully defined because any movement of a node in or out of the plane will cause a lengthening of a link somewhere.

You could also do a modal analysis on the structure to find the lowest modes of vibration, which will give you some insight into the structure's stiffness.

25. Oct 9, 2009

### topcomer

So I think as a start it might be wise to try with the hexagon example by hand.

We have (7-2)*3=15 free dofs and 12 members, after having considered the boundary conditions. Let's say that the nodes have coordinates:

EX1: (+/- 5, +/- 5), (+/- 10, 0), (0, 0)
EX2: (+/- 5, +/- 5), (+/- 10, 0), (0, -2)

Let's do EX1 first. The trusses have lengths 10 and sqrt(50), and thus the geometries are
(counterclockwise):

inner members (starting from the origin)
cos(\tetha) = 1, 5/sqrt(50), -5/sqrt(50), -1, -5/sqrt(50), 5/sqrt(50)
sin(\theta) = 0, 5/sqrt(50), 5/sqrt(50), 0, -5/sqrt(50), -5/sqrt(50)

outer members (starting from the previous endpoints)
cos(\tetha) = -5/sqrt(50), -1, -5/sqrt(50), 5/sqrt(50), 1, 5/sqrt(50)
sin(\theta) = 5/sqrt(50), 0, -5/sqrt(50), -5/sqrt(50), 0, 5/sqrt(50)

So, using a notation in accordance to your ANSYS book p.73, we can set:

cos(\theta)_x = cos(\theta) = c
cos(\theta)_y = sin(\theta) = s
cos(\theta)_Z = 0

Meaning that for one member the stiffness matrix is:

{
{c^2, c*s, 0, - c^2, -c*s, 0},
{c*s, s^2, 0, -c*s, -s^2, 0},
{0, 0, 0, 0, 0, 0},
{-c^2, -c*s, 0, c^2, c*s, 0},
{-c*s, -s^2, 0, c*s, s^2, 0},
{0, 0, 0, 0 ,0 ,0},
}

Well no matter how we assemble the global stiffness matrix, there will never be any nonzero entry in the lines and rows corresponding to nodal z-displacements, so the matrix will always be singular. The same is true for EX2, so this approach will never be able to tell if the framework is rigid in the z-direction. Am I wrong?

With the light shed by the above example, I think you are right, a linear analysis will always show an unstable z direction. But clearly the "regular" hexagon is not rigid, so I don't agree with the statement claiming that the structure is completely rigid, because there can be non-trivial deformation modes that do not require any elongation of the members. Note that the blue line is in-plane, so the green point does not have to move in the z direction.

Or, if you are right, I should be able to come up with a proof, or a calculation, which at the moment seems out of reach.

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