# How do I calculate how much weight an iron cylinder can support?

I would prefer an equation. The premise is that I'm trying to calculate how thick an iron cylinder using a screw jack would need to be to support 20kg of weight for engineering class. We haven't learned anything along these lines and I'd really like to know.

Thanks!

• Can you add a sketch of the problem? basically, In questions like these we have to understand what are the loads (in your case the only load is a 20kg weight) and where they act on the item, what are the supports and what is the item geometry. – Yaniv Ben David Oct 31 '18 at 21:02
• I added a quick sketch I just threw together. @YanivBenDavid – F3ztive Nov 1 '18 at 14:58
• It will fail by buckling so the tensile and yield strength are unimportant. – blacksmith37 Nov 1 '18 at 15:39
• @blacksmith37 why? Is there some equation I can use to find this out? – F3ztive Nov 1 '18 at 17:16
• The compression yield is over 50,000 psi ; If that were the controlling factor , the rod would be about 0.001 " ( one thousandth of an inch ) in diameter. Look up buckling on the net/ Wikipedia. A good cardboard tube would probably do the job. – blacksmith37 Nov 2 '18 at 20:46

There are stablished criteria by codes to design columns, but we ignore these for the benefit of clarity here.

Basically we want to design an Iron column strong enough to support a 20 kg load with a length of $$L= 2\times 3ft = 6ft$$

We multiplied by 2 because the column is a cantilever column, it is not restrained laterally on top.

First we check for the slenderness.

Usually the columns with a radius of gyration (radius of gyration is the radius at which you can assume all the mass is located), greater then L/20 are considered slender and need to be checked for buckling load. Radius of gyration of a cylinder is :

$$R_{gyration} = D/4$$

So we know any diameter less than 4.5 cm should be considered a slender column and checked for buckling.

Say for trial and error we pick as a first guess a 1.5 cm diameter post.

$$p_{critical} = \frac {\pi^2EI}{L^2} \\\ and \ I = \frac {pi D^4}{64}$$

This is the Maximum buckling load.

So I = 0.24cm^4 and

$$P = \pi^2(0.24E)/ 183^2 = \text{ approx} \ 76kg$$ Assuming Iron's E at 100 GPa. This strength is reasonable and has a factor of safety of approx. 4, I would stop here.

But through the same steps picking a size smaller we gat closer to the optimal size.

Finally we check the result Against the compression strength with a factor of safety, and check for likelihood of overturning of the base block, which is missing the width.