alcazar

iTrader: (2)

Easy one really, but I cannot get Google to find it, all it wants to come up with is concrete calculations.

How does the diameter of a cylinder affect it's strength, ie. what weight it can support, assuming all cylinders to be the same height?

It SEEMS to come out as a curve*, but firstly, is that right, and how are the diameter and crushing strength linked, I assume by some constant?

Thanks for any links or answers.

* looks almost parabolic, with it being upside down, so that higher diameters support more, but in proportionally smaller increments.

Reason for this: question about crumple zones in cars.

chocolate_o_brian

iTrader: (22)

Is the curve in retrospect to the thickness of the material and the length of the cylinder itself, Jeff? I'm wondering if it has any relevance to the 'pie' symbol in regards to the inner and outer diameter. It's monday, I'm slow on the ball today.

Only engineer I know of is Coffin Dodger but I think he's a sparky.

CREWJ

If it's only head on you are talking about it will all be about the CSA (Cross Sectional Area)

Therefore you can have a huge diameter and a tiny wall thickness.

speedking

Failure mode changes. Short solid circular column fails in crushing, so capacity is proportional to cross-sectional area.

But if the load is eccentric then one side gets loaded more than the other and the core does very little. therefore this unused material can be removed and you get a hollow cylinder which is much more effective for the quantity of material.

When the column is a bit longer then you may get shear failure, where the top slides off the bottom at a 45° failure plane. Or buckling where the column bends and fails sideways.

These are all dependent on the top and bottom restraint applied.

A very complex subject.

For crush zones, this will be the failure mode.

RobJenks

Firstly the effective length of the column - This is dependent on how the column is restrained at each end. Le
The radius of gyration has to be established , which is determined by the Moment of Inertia divided by the cross section area of the column . r is the square root of this function.
L/r can now be determined and an allowable compressive stress f a determined from tables in AISC ( Steel members )
The actual compressive stress can be computed from F/a
Providing fa > F/a then the member is adequate
This is the basic approach .

alcazar

iTrader: (2)

Sorry, but I still don't have an answer as to why the larger diameter cylinders SEEM to be able to support proportionally less than smaller ones?

The graph I have produced looks like the left hand side of an upside down parabola and it must be pretty close since ALL results fit almost perfectly until I got to one with the full 100 diameter and then it collapsed very quickly. That's probably to do with the wall thickness more than owt.

I was interestyed if there was a relationship between mass supported and diameter that would produce a parabola?

ScoobyWon't

What happens if you start changing the wall thickness?

speedking

In a perfect analysis, for the same amount of material, a larger diameter cylinder will be able to support the same load as a smaller one. The limiting factor is the compressive strength of the material.

If you bring buckling into the equation then a very narrow cylinder will fail by buckling, as it gets wider then the capacity increases. When it gets very wide with a thin wall and the load is eccentric then you would get a local buckling failure of part of the wall.

What formula is producing the graph?

alcazar

iTrader: (2)

Thanks for the above, it's exactly what I needed in simple terms.

The graph isn't produced by a formula, but by experimental results.

I know that if it produces a straight line, I can say that one variable is directly dependant on the other, but can I say the same for a curve? Maths seems to say that it would be true, but there would be another variable in play?

speedking

Although it is for concrete, look at page 5 of this.

Then you can graduate to producing this type of interaction diagram.

speedking

There could be another variable, hence the interaction diagrams.

Or, instead of x is proportional to y, x is proportional to y².

Or you haven't defined your variables properly, e.g. maybe you need to consider stress instead of load.

etc. etc.

alcazar

iTrader: (2)

Thanks for that. The semi-parabolic shape does indicate that x varies with y squared is a possiblity.

After I reached 100mm, it collapsed at under 2000g applied mass, so I think buckling of a sidewall was responsible.

Thanks for all replies.