Which of these structures is the strongest?

have at it boys...
strawpoll.me/12000844

strongest at what? Torsion, bending, axial forces?

sorry
should've clarified bending, but go in depth at your hearts contempt.

second planar moment of area about neutral axis
[math]
I_x = \int\int_{crosssection} y^2 dA
[/math]

what are the dimensions, OP you cunt

this was supposed to be a very broad question, whatever you want I guess

Are we to assume cross sectional area is constant across all beams?

I intentionally left no annotations, scales, or specifications. Assume what you please.

The one with the jew start. I know because if you move the square tower towards the jew tower, the square exactly fits as the square that is inside the jew star.

And if you move the other pillars into the jew tower, they are all smaller aswell. The jew start has extra material to make the bendy edgy spikes jews like. Therefore it is the strongest as it is as solid as the square thingy but has extra material which strenghtens it.

t. PhD in Jew Science

>no triangle

>No triangle.
Wtf you on about? There's plenty of Illuminati confirmed.

The octagon has the most broad angles so I say octagon less (weakpoints)

tempted to say star for the area
tempted to say pentagon for asymmetry
tempted to say hexagon for hexagons rule
tempted to say square for bend-ability without breaking

Whichever is largest. In this case, the 8-point star on the right.

The star on the very right not only has more material, but I think it would withstand bending more because every angle is populated with more arcs than the other shapes, and arcs increase strength.

Well then the square beam is the strongest as it's area is 10000000x larger than the next biggest beam.

not a regular here, but I figger there's a reason forts became star-shaped

if each bar has the same cross sectional area, it looks like the square bar is best when subjected to cantilevered bending.

the furthest right

Forts became star shaped to maximize field of fire.

i think the all-around winner surely should be the star, by way of having more faces closer to a parallel angle to the bending force

roll the columns by 45deg and see if i'm completely wrong, plox

>roll the columns by 45deg and see if i'm completely wrong, plox

gimme 50$.

n0

yep, and, keep goin

What program is this?

> only asymmetric cross section
> lowest votes

t. engineer

only the first was free

Solidworks.

Define strongest.
bending? buckling? torsion?

Read the thread faglord. He said bending.

So for bending stresses ill just use normal stress. The equation is σ = My/I where M is the maximum moment in the beam, Y is the maximum distance from the centroid, and I is the moment of inertia. For the square through the octagon ill assume that the largest height is 1.

Now lets assume that the maximum moment through the beams are the same. So for the square through the octagon M and Y are the same and the answer comes down to the I value. Doing the calculations the square wins out with a moment of inertia of .0833 whatevers^4.

For the 8 pointed star I couldn't find a moment of inertia equation that matched that specific shape, so I just used a regular octagram with a side length of 1. The I value comes out to be .021 whatevers^4. And since the heights don't match up between the octagram and the square, the y value comes into effect. Between the two the Octagram would have to be about 4 times taller than the square to have the same maximum normal stress for any given moment.

So the square wins as long as they are nearly the same size for any given moment in the beam.

AND for polygons with large amounts of sides you can model it as reaching the shape of a circle. A circle with a diameter of 1 has an I value of .04 whatevers^4. So a square wins over a circle and polynomials with large amounts of sides as well.

>tl:dr Square wins

Hey guys, I found the retard.

>bending

The one with the largest section modulus, dipshit.

No dimensions on those.