# Force in tetrahedron edges

I am looking for a formula that enables me to calculate the force in a tetrahedron edge such that it relates $F_b$ with $F_z$ through the beam thickness and length. I have the following assumptions:

• The beams are circular and hollow. So the have a radius $r$ and a thickness $t$.
• The static force points in the negative $z$-direction and is active in the c.o.g.
• The contribution of the vertices/joints is neglected.
• The tetrahedron is standing on one vertex as shown in the image below.
• The beam length is $a$.

I have drawn the situation in the following image:

Can someone provide me with a formula or some pointers on how to find the beam force?

Edit, using Jaime's hints.

As the c.o.g. is alined with he lower vertex the structure is in equilibrium. When looking at the lower vertex the vertical component of the three beam forces equals the gravitational force.

$$F_{b_{y1}}+F_{b_{y2}}+F_{b_{y3}}=F_{g}\\ F_{b_{y1}} = F_{b_{y2}} = F_{b_{y3}}\\ F_{b_{y}} = \frac{m \cdot g}{3}$$

From the Encyclopedia Polyhedra written by Robert Gray I found that the half-cone angle is 35$^{o}$. This means that:

$$F_{b} = F_{b_{y}} \cdot cos 35\\ F_{b} = \frac{m \cdot g}{3} \cdot cos 35$$

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So now consider one of the top vertices, looking at it from above. You have the force you just calculated pushing it out and up. Consider only the force in the horizontal plane, which will again require some trig. That force is balanced by the two equal unknown forces coming from the top bars, which should be easy with some more trig. – Jaime Nov 6 '12 at 17:05