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

In summary, the conversation is about a PhD student's research on analyzing the rigidity of a structure of rigid beams in 3D. The student presents a diagram of the structure and discusses the possibility of it being able to move. They also ask for a theory to help understand the problem. The conversation then delves into terminology and notations, specifically regarding joints and degrees of freedom. The student provides examples of different types of triangles and their degrees of freedom, and the conversation concludes with a discussion on the constraints and dofs of the structure. The student is still unsure about the number of rotations a thin truss should have.
  • #1
topcomer
37
0
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.

Thank you in advance,
AQ

PS I don't know if the homework area is more appropriate for this question, sorry.
 
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  • #2
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
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.
 
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  • #4
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.
 
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  • #5
nvn said:
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 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.

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
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.
 
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  • #7
nvn said:
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?

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 node-to-outside-world (external) dofs.

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
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
nvn said:
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.

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
topcomer said:
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.

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.

topcomer said:
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.

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.
 
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  • #11
topcomer said:
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.

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

topcomer said:
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]

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).

topcomer said:
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 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.

topcomer said:
This triangle has 1 dof, i.e. it can rotate about the y-axis:
http://img84.imageshack.us/img84/4100/roty.th.jpg [Broken]

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
topcomer said:
It seems this structure is not rigid in 3D, since it is possible to move the upper-right pair of triangles along the diagonal of the big square by "folding" the little square they form.
topcomer: Are you referring to displacement in the x, y, or z direction(s) in this sentence?
 
  • #13
Mech_Engineer said:
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.

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".

Mech_Engineer said:
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.

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.

Mech_Engineer said:
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.

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?

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

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
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.
 
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  • #15
nvn said:
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 inextentional or rigid members.

In your problem, there is a very 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.

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
topcomer said:
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?

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).

topcomer said:
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]

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 accommodate 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
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).
 

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  • #18
nvn said:
A computer program will not be able to reflect the theoretical nature of the problem for z-direction (out-of-plane) displacement. A linear analysis 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 analysis.

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

nvn said:
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 would require extension of the members, which cannot occur if the members are inextensional (rigid).

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.
 
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  • #19
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...
 

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  • Example.JPG
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  • #20
Mech_Engineer said:
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.

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
Mech_Engineer said:
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...

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.
 
  • #23
Mech_Engineer said:
An FEA program can detect an underconstrained condition using the stiffness matrix (as said in my previous post).

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
topcomer said:
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.

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

topcomer said:
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?

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?

nvn said:
Because of the inextensional members, it appears your truss 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 computer program will not be able to reflect the theoretical nature of the problem for z-direction (out-of-plane) displacement. A linear analysis 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 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).

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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  • #26
topcomer said:
But clearly the "regular" hexagon is not rigid, so I don't agree with the statement claiming that the structure is completely rigid. 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.
My post, which you quoted in post 25, refers to the truss with the blue line in post 13. I wasn't referring to a hexagonal truss. In case this was unclear, I have now clarified in post 17 specifically which truss I am referring to in post 17. Clearly, the regular hexagonal example truss in post 25 has a hinge line, and thus has a rigid-body mode. But I don't know why you refer to a blue line in the above quote; there is no blue line or green dot in a hexagonal truss. The truss shown in post 17 obviously has no hinge lines, and I thus answered for that truss in post 17.

To come up with a proof of in-plane stability, develop the structure stiffness matrix for your truss in post 3, except model it as a 2-D (planar) truss, instead of a space truss. You would therefore use the member stiffness matrix for 2-D truss members in this derivation.
 
  • #27
nvn said:
My post, which you quoted in post 25, refers to the truss with the blue line in post 13. I wasn't referring to a hexagonal truss. In case this was unclear, I have now clarified in post 17 specifically which truss I am referring to in post 17. Clearly, the regular hexagonal example truss in post 25 has a hinge line, and thus has a rigid-body mode. But I don't know why you refer to a blue line in the above quote; there is no blue line or green dot in a hexagonal truss. The truss shown in post 17 obviously has no hinge lines, and I thus answered for that truss in post 17.

To come up with a proof of in-plane stability, develop the structure stiffness matrix for your truss in post 3, except model it as a 2-D (planar) truss, instead of a space truss. You would therefore use the element stiffness matrix for 2-D truss members in this derivation.

I'm referring to the hexagon because it's a simpler case where you can visualize the rigid body mode, while in the truss with a blue line in post 13 this is more difficult, but that does not mean there isn't any possible rigid body mode.

In-plane stability is obvious, the problem is that the green point can move in (and even out of) plane thanks to the fact that the other points can move out-of-plane compensating each other, as in the following:

http://img185.imageshack.us/img185/6645/nonrigid.th.jpg [Broken]

this image was generated in Mathematica where I implemented a simple multibody on a different but equivalent (in terms of allowed movements) truss network, which however did not lead me to any conclusion, because it didn't give me any insight in the problem.
 
Last edited by a moderator:
  • #28
topcomer said:
The green point can move in (and even out of) plane, thanks to the fact that the other points can move out-of-plane, compensating each other.
The truss in post 17 has no (obvious) hinge lines. Therefore, I currently think any out-of-plane movement of the other points would require elongation of members, as alluded to in post 17 and the last sentence of post 18. But we can keep thinking on your above comment.
 
Last edited:
  • #29
nvn said:
The truss in post 17 has no (obvious) hinge lines. Therefore, I currently think any out-of-plane movement of the other points would require elongation of members, as alluded to in post 17 and the last sentence of post 18. But we can keep thinking on your above comment.

Do you agree that if we remove the diagonal of the upper-right square (to which the green point belongs), then such a movement (to be precise many) is possible? Then, just observe that that diagonal is the hinge about which that square can fold, letting the movement of the green point possible. That's why my idea was to flip the orientation of that diagonal in order to lock the framework, except for the corner point.
 
Last edited:
  • #30
nvn said:
A linear FE analysis (static or modal) will show the [post 17] truss highly unstable in the z direction.

True statement. Even intuitively, there is no expectation that a nontrivial planar space truss, with frictionless ball joints, will have any out-of-plane stability. No matter what configuration of members you use for your post 17 truss, and even if the member material is extremely stiff or rigid, the member axial stiffness initially has no mechanical advantage whatsoever to prevent out-of-plane deflection. Therefore, as you correctly stated in post 25, the structure stiffness matrix will always be singular.
nvn said:
[The post 17 truss] would not be stable until it deflects out-of-plane a finite amount in a nonlinear FE analysis.

False statement. I currently retract the above statement and most of the statements below it in post 17.
nvn said:
The truss in post 17 has no (obvious) hinge lines. Therefore, I currently think any out-of-plane movement of the other points would require elongation of members.

I retract the last sentence. I currently agree that out-of-plane displacement of the post 17 truss is possible without elongating members, with or without the upper right-hand diagonal installed, and regardless of whether or not that diagonal is flipped.
topcomer said:
The green point can move in (and even out of) plane, thanks to the fact that the other points can move out-of-plane, compensating each other.

Agreed. There appear to be innumerable combinations of out-of-plane nodal displacements, causing no elongation of members, that can cause the upper right-hand square to have even rigid-body, in-plane translations. Therefore, flipping the diagonal in the upper right-hand square does not make the truss rigid nor stable. The number of non-elongation modes increases if we include folding along the upper right-hand square current diagonal. Therefore, the truss of post 17 is highly unstable, even after nodes deflect out-of-plane.

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.

However, even with cross braces, you are still faced with the situation mentioned in my first paragraph, above, in which the structure stiffness matrix will always be singular.
 
  • #31
nvn said:
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.
 
  • #32
nvn said:
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):

[tex]\epsilon (u) = \nabla u ^T \nabla u[/tex]

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:

[tex]K = \nabla_u^2 \epsilon (u) = \nabla \phi ^T \nabla \phi[/tex]

where now:

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

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:

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

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:

[tex]\nabla \phi_x = [-1/10, 1/10, 0; -1/5, -1/5, 1/5][/tex] (v1,v2,v0)
[tex]\nabla \phi_x = [1/10, -1/10, 0; 1/5, 1/5, -1/5][/tex] (v3,v0,v2)
[tex]\nabla \phi_x = [0, -1/10, 1/10; 1/5, -1/5, -1/5][/tex] (v4,v0,v3)
[tex]\nabla \phi_x = [-1/10, 0, 1/10; 1/5, -1/5, 1/5][/tex] (v5,v0,v4)
[tex]\nabla \phi_x = [-1/10, 1/10, 0; -1/5, -1/5, 1/5][/tex] (v6,v0,v5)
[tex]\nabla \phi_x = [0, 1/10, -1/10; -1/5, 1/5, 1/5][/tex] (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.
 
  • #33
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.
 
  • #34
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.
 
  • #35
didn't try to go through all posts here, but are you looking for the Chebychev–Grübler–Kutzbach's criterion in 3D?
 
<h2>1. How do closed loops affect the kinematics of 3D space truss? </h2><p>Closed loops in a 3D space truss can affect the kinematics of the system by introducing additional forces and moments, which can alter the overall structural behavior. This can lead to changes in the displacement, velocity, and acceleration of the truss elements.</p><h2>2. Can closed loops in a 3D space truss cause instability? </h2><p>Yes, closed loops in a 3D space truss can cause instability if they are not properly accounted for in the design and analysis. The additional forces and moments introduced by the closed loops can lead to unexpected deformations and failure of the truss structure.</p><h2>3. How can closed loops be incorporated into the kinematic analysis of a 3D space truss? </h2><p>Closed loops can be incorporated into the kinematic analysis of a 3D space truss by using methods such as the finite element method or matrix structural analysis. These methods allow for the inclusion of closed loops in the overall structural model and can accurately predict their effects on the kinematics of the truss.</p><h2>4. Are there any benefits to including closed loops in a 3D space truss design? </h2><p>Yes, closed loops can provide additional support and stability to the truss structure. They can also allow for more efficient use of materials and can help distribute forces more evenly throughout the truss.</p><h2>5. How do closed loops in a 3D space truss affect the overall strength of the structure? </h2><p>The presence of closed loops in a 3D space truss can increase the overall strength of the structure, as they can help distribute forces and resist deformation. However, if not properly accounted for, closed loops can also introduce weak points in the truss and decrease its overall strength.</p>

1. How do closed loops affect the kinematics of 3D space truss?

Closed loops in a 3D space truss can affect the kinematics of the system by introducing additional forces and moments, which can alter the overall structural behavior. This can lead to changes in the displacement, velocity, and acceleration of the truss elements.

2. Can closed loops in a 3D space truss cause instability?

Yes, closed loops in a 3D space truss can cause instability if they are not properly accounted for in the design and analysis. The additional forces and moments introduced by the closed loops can lead to unexpected deformations and failure of the truss structure.

3. How can closed loops be incorporated into the kinematic analysis of a 3D space truss?

Closed loops can be incorporated into the kinematic analysis of a 3D space truss by using methods such as the finite element method or matrix structural analysis. These methods allow for the inclusion of closed loops in the overall structural model and can accurately predict their effects on the kinematics of the truss.

4. Are there any benefits to including closed loops in a 3D space truss design?

Yes, closed loops can provide additional support and stability to the truss structure. They can also allow for more efficient use of materials and can help distribute forces more evenly throughout the truss.

5. How do closed loops in a 3D space truss affect the overall strength of the structure?

The presence of closed loops in a 3D space truss can increase the overall strength of the structure, as they can help distribute forces and resist deformation. However, if not properly accounted for, closed loops can also introduce weak points in the truss and decrease its overall strength.

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