2nd Post: How many diagonal braces are needed to make a cube rigid?

Question

A guy welding together a large cube using angle iron for the edges notices that the cube is not very rigid. So he decides to weld on some diagonal braces, going from any one corner to another. How many braces does he need to make the cube rigid? Assume that each member is rigid, but that all joints are flexible. Crossing diagonals may be welded together.

The reason for this 2nd repost is because some may have more to say on this problem. Can it be done with fewer than 6 diagonals?

See original question and answers:

http://answers.yahoo.com/question/index;_ylt=AlYz35VsYz6Wtfwv0bdO3.7sy6IX;_ylv=3?qid=20071205004208AARoBvf

Dr D & Remo Aviron, with both of your answers (and all the proposals), what you seem to be overlooking is that a square with a single diagonal can "butterfly", or fold along the length of the diagonal. Because of this, with all of your proposals, movement is still possible. It might give you an headache to figure out how movement is even possible, but trust me, it is.

Duke, you got off with a good start proposing that the inner diagonals be welded at the center. However, your final proposal using just 4 diagonals doesn't do it, because face ABCD can rotate relative to face A'B'C'D', if you also consider "butterfly" folding, as I have described above.

Picking the best answer is hard here, but I will go with Remo Aviron, because he suggested a way of looking at the problem in terms of struck compression or stretching. In fact, if 4 internal structs are welded at the center, and a 5th one is a face diagonal, in theory the cube is rigid, but in practice it is not. Why is it not in practice? Because when the face opposite to the one with the face diagonal is rotated, it will indeed cause changes to struct lengths. However, the changes are a square function of the angular displacement, which means that even very tiny compression or stretching of the structs can result in an unacceptable angular motion. In other words, it would behave like a spring.

Answer 1

To get below 6 braces, you are going to have to go beyond looking at things from the normal stress-strain degree of freedom relationship, but instead use a methodology which I believe is called infinitesimal rigidity. In other words if you displace a member A by an infinitesimal amount, it would cause other members to either compress or expand. And if you assume rigid members, this becomes an impossiblity.

Lets try 5 braces first:

Let me answer use the following drawing which is a cube stablized by a tetrahedon on the inside: http://www.robertinventor.com/cubeetc/cubetetrahedron.jpg

Now cut the two right hand braces and replace them with an internal diaganol from the top left to the bottom right.

Now, if you look close, you see that you have a stable tetrahedon (one triangle of which is in the base) that stabilizes the base and the left hand corner point.

On top of the cube the two vertexes in the center appear to be free to move just restricted by the cube edge, but are they? If you try to push them together, the crossmember on the top will prevent this movement. And if you try to move both of them in the same direction (with their bottoms fixed to the base), you will force the far right upright edge member into compression. Even though there appears to be some degrees of freedom left open, there are none. In other words, you have a rigid cube.

I'm not sure, but I believe you might even be able to get down to 4 struts. The first way would be the tetrahedon I mentioned above, with the 3 'floating' top vertices stablized by the fourth fixed vertices and the fixed base vertices. The other way would be to have 4 internal struts criss-crossing the cube. I've looked at this and at least under simple torque, there would have to be some compression or stretching of a member which would mean that it, too, might work.

Lastly, using the same principle of incompressible members, you might even be able to get down to three internal members. But you'd have to get someone smarter than me, i.e., http://answers.yahoo.com/my/profile?show=92f97d29e3d55f3219cc25dfabd347c7aa or http://answers.yahoo.com/my/profile;_ylt=Ao7FcstV9gM3C5.stW7mjCUAxgt.;_ylv=3?show=uDRfzKDFaa or http://answers.yahoo.com/my/profile;_ylt=AtYDGlKa0LMEgwBmkSEuJX0Axgt.;_ylv=3?show=1c25f0773926f8cde3f71d5867bbbc1eaa to solve that problem.

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[Some editing] With 4 internal struts, no matter what you do, it will be rigid. This is because any movement of the cube will cause at least one of these struts to either compress or stretch.

I thought of an interesting way to look at the problem. Circumscribe the Cube with a sphere. Now all of the internal struts are circumferences. You can see that any distortion of the cube with all of the members keeping a constant length will force a change in length of a diagonal brace. If you can't move without causing a strut to change length, then the body is rigid so long as the strut does not change length.

Answer 2

If I have understood correctly "from any one corner to another" that allows to use not only face diagonals, but inner also. Then I think 4 will suffice:

1) Every tetrahedron ABCD is obviously a rigid construction. If we remove an edge, say CD, the 2 remaining triangles ABC and ABD can move as being on a hinge AB. Hence every rectangle ABCD with 2 diagonals (degenerated tetrahedron with 4 co-planar vertexes) is also rigid.

2) Let's consider a triangular prism (it's a pentahedron) ABCC'B'A' (AA' || BB' || CC'). How many braces will make it rigid? Taking, say the lateral face ABB'A' and adding the diagonals AB' and BA' (|AB'| = |BA'|) we fix the rectangle ABB'A' according 1) above. Imagine we put the prism on a table, face ABB'A' down. The construction can move as there are hinges along AB and A'B', CC' can move in a vertical plane, being parallel to the table. Let's add now a diagonal AC' (or A'C) with length as ACC'A' to become a rectangle. The tetrahedron AA'B'C' has now all 6 edges, so it's rigid. But fixing the position of vertex C' fixes exactly the triangle ABC', hence the tetrahedron ABCC' /C can not move any more, otherwise C' would move also!/ and the entire prism. So 3 diagonals are enough the prism to be rigid.

3) Let ABCDA'B'C'D' is our cube, ABCD - lower base /A, B, C, D - sequential vertexes/, A'B'C'D' - upper base /i.e. AA' || BB' || CC' || DD'/., it's a union of 2 triangular prisms: ABCDD'C' and ABB'A'D'C' with common rectangle ABC'D'. According 2) we can fix the vertexes A, B, C', D' with (INNER!) diagonals AC' and BD', the next 2 face digonals AC and A'C' take care of the rest.
Here is how it is looking like:
http://i219.photobucket.com/albums/cc101/Andrey_Dyukmedzhiev/Rigid.gif

Note: I edited my answer thoroughly, initially it was without any explanations, there were some erroneous notations also

Answer 3

I think this could be done with 5.

If the braces are rigid, then each square only needs one diagonal to make it rigid (NOT TWO). So if 6 diagonals are used, it would definitely work, but one of them will be redundant.

Consider the cube ABCDEFGH.
Let's start first of all with 3 braces: AD, DG, BH
http://i14.tinypic.com/73kizys.jpg

So the top, right and front faces cannot shear. However it is still possible for EF to move sideways. So we need another brace to prevent that. Let's add EH, so now the back face is rigid.
http://i10.tinypic.com/6z7tmc1.jpg

The only faces that can shear now are the bottom and the left, and that will happen is AE moves back/front. So we need another brace to prevent that. Let's add AG.
http://i17.tinypic.com/6k8zceh.jpg

Now there is no way any other face can shear. The only face that isn't braced is the bottom face (ha ha bottom face). But in order for it to shear, at least one of the other faces must shear as well, and that cannot happen if 5 of them are braced.

***EDIT***
I thought for a moment about this:
http://i2.tinypic.com/8bq23gn.jpg
But it's clear that the front and back faces can slide relative to each other such that the top and bottom faces become rhombi (or whatever the plural of rhombus is). So this arrangement of 4 braces won't work.

*EDIT*
The thing about using hte long diagonals is that if you have 4 long diagonals, then they all intersect in the middle. Mathematically that is not a problem, but physically it is. If they are all welded at the center, then would it still be considered 4 braces or 8?

Answer 4

5 ... Last and final edit. Looking at Alexander's picture of a cube, it is clear that you can clip out the two braces on the right hand side and replace them with an interior diagonal from the upper left to the lower right. This makes 5 braces. [Note: I've also looked at cubes with three and four diagonals and they appear stable -- except they would be very vulnerable to torque so I discarded those.]