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I was wondering, what is the strongest : a rectangle or an arc (same width/height) ? screenshot

you have to imagine the screenshot in 3D : you have 4 walls and a paved layer (the rectangle) or a dome

Reason why the rectangle would be stronger : same size as the dome but more raw materials

Reason why the dome would be stronger : the vertical pressure is directed to the walls -> Keystone

Assumptions :

  • the 4 "walls" are infinitely strong
  • the dome / rectangle has a weight negligible relatively to the force applied
  • The dome / rectangle is not infinitely strong, it can bend / break given the vertical pressure

EDIT

I am making a plastic box and people will walk on it (box size is size of a foot). for some reasons walls must be thin and the box must be pretty thin too and we wondered what shape is stronger / is less likely to bend in its center

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    $\begingroup$ Why would the rectangle have more material? $\endgroup$ – mart Mar 24 '15 at 14:46
  • $\begingroup$ @mart "same width/height" a solid shape would have more material $\endgroup$ – ratchet freak Mar 24 '15 at 15:41
  • $\begingroup$ Has the material significant tension strength? $\endgroup$ – mart Mar 24 '15 at 15:43
  • $\begingroup$ When you say "same height" do you perhaps mean that the same amount of material is used to construct each? Or, that they have the same constant or average thickness? Something else? $\endgroup$ – Air Mar 24 '15 at 23:38
  • $\begingroup$ @Air height as in width x height (in pixels here : w: 420, h: 50). nothing about thickness $\endgroup$ – Thomas Mar 25 '15 at 0:10
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The answer to your question is very simple.

If the arch and the rectangle are made of the same material, and the rectangle height and width are equivalent to the arch rise and span, and they are loaded with the same load in the same place, and the weight of the material is neglected, the rectangle will always have at least as as much load capacity as the arch, and could have more depending on the material properties. Always.

This is a result of the principle of superimposition. If you overlay your arch on top of your rectangle, you'll see that the rectangle, in a manner of speaking, contains the arch:

enter image description here

The extra material above and below the arch, given your assumptions (the most important of which, aside from using the same material and dimensions, is ignoring self weight of the material), cannot negate the load carrying capacity of the arch "inside" of it. Therefore, the rectangle will have at least as much capacity as the arch.

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    $\begingroup$ Given that this is apparently the best answer, i.e. it has been accepted by the OP, all I can say is that the original question was shockingly poor! $\endgroup$ – AndyT Apr 2 '15 at 11:55
  • $\begingroup$ @AndyT It was indeed a pretty basic question. $\endgroup$ – Rick Teachey Apr 2 '15 at 12:57
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Your question is very broad, but at its heart is question about how various shapes transfer load.

Rectangular Beam

When a solid beam is loaded, it will experience moments and shears.

The moments are primarily resisted by the material furthest away from the centroid of the section. This leaves the material at the center of the beam under-utilized. This is the reason that steel beams are commonly shaped like an "H" on its side. Solid beams are not efficient with regard to moments.

Shear is transferred through the entire section of a solid beam, so the entire section is utilized.

Arch

Arches primarily transfer load through axial forces. These forces utilize the entire area of the section. It is for this reason that an arch is the more efficient of the two shapes. The entire area of the section is employed to transfer the load.

Only in comparison

The comparison above only works if the supports are not controlling. Arches require their supports to resist large horizontal forces. They basically convert a portion of the vertical forces to horizontal forces. There is no conversion of this type in beams, so the beam supports are simpler.

An Example

One notable example of an efficient arch is the dome of the basketball stadium at the University of Illinois. This structure is a concrete dome that is 3.5 inch thick at its thinnest. The structure has steel strands wrapped around its base that help to resist the horizontal forces. The dome is self-supporting.

enter image description here

There is an interesting article on the efficiency of arches that mentions the dome above.

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  • $\begingroup$ thanks for the answer. I edited the question in case it helps. does you question still holds (see @thepowerofnone comment) $\endgroup$ – Thomas Mar 26 '15 at 16:06
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A inverted catenary arch will not need any tensile strength to support the force provided that it is uniform across the entire top. However all that force will be transfered to to walls and out to the side. It needs a firm anchor to avoid the outward bulge that would destroy it.

If the rectangle has a similar anchor then you can think away all the material that is not part of a catenary arch. and see above. The normal sag would then be counteracted by the force at the anchor points

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  • $\begingroup$ what if the vertical force is not uniform but a bit more concentrated around the center of the rectangle/dome ? $\endgroup$ – Thomas Mar 25 '15 at 0:12
  • $\begingroup$ thanks for the answer. I edited the question in case it helps. does you question still holds (see @thepowerofnone comment) $\endgroup$ – Thomas Mar 26 '15 at 16:06
  • $\begingroup$ Catenary is the form for where loading is constant per unit length of the structure. If the loading is constant per width (which seems to be the case proposed here) the optimum form is a parabola. $\endgroup$ – achrn Mar 30 '15 at 9:45

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