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I have a friend who cedes that a framework that is a triangle would be stronger than a square framework. But he maintains that a square solid would not be stronger than a triangle solid. Ie a set of steel beams in a triangle is stronger than a set of steel beams in a square, but a solid square out of Steel would be stronger than a solid triangle out of Steel.

Is he correct? Why or why not?

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    $\begingroup$ What do you mean by “A is stronger that B”? $\endgroup$ – pppqqq Feb 1 '14 at 15:30
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A parallelogram defined by its perimeter is unstable to shear. If the top corner is pushed it will scissor shut. A triangle defined by its perimeter is rigid toward shear. Solid planes cannot scissor shut. However, eccentricity of an axial force results in a bending moment acting on beam elements - buckling. Pattern the surface to resist other failure modes.

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A tetrahedron composed of steel beams is very strong to shear forces, where a cube would easily collapse without other supports. On the other hand, when it comes to compression forces A cube will have four beams supporting each face and a tetrahedron would only have three.

Even then, when a compression force is applied to a face, each steel beam in of itself is stronger to forces parallel to the the length of the beam versus perpendicular to its length. A tetrahedron's supporting beams would be diagonal to the pressure and contrast with a cube's 4 support beams in parallel with the direction of force.

Now when dealing with solids, and assuming the same random non crystalline internal structure between a tetrahedron and a cube, a tetrahedron sitting with one face on the ground will have one point facing up. The the tetrahedron highest point will be comparatively weak to oppose a compression force acting on it, and contrast with the same force applied to a much larger area of face of a cube. Both tetrahedron and cube would be about the same if the force is applied to it were applied to only a little point.

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