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TLDR;

Can you please help me understanding the string in the picture and help me finding the yield stress that my metal can handle?

Full story

I am trying to estimate how much load some zinc-coated steel pipes that I bought can handle.

This webpage provides a calculator in which I specify 6 parameters and I get estimated if my tube will fail or not. I know 5 out of 6 parameters for my system; what I lack is what they call the yield stress (measured in PSI, see the table near the center of webpage for some example of yield stress values).

Examining my pipe I found a string (see picture); is this enough to identify the yield stress of my pipe?

The best I got so far is that by googling "EN 10255 steel" I land into this webpage from which I think that one can deduce that (I stress the fact that I am not sure about the following) the "Upper Yield strength $\text{R_eH min.}\ (MPa)$" is 195 (which amounts to $\approx 28282\ \text{PSI})$.

Beside not being sure if the number above is the one I am looking for, in the picture you can read also other numbers, which seem to correspond to other steel types. Maybe my steel is an alloy?

So my question:

Can you please help me understanding the string in the picture and help me finding the yield stress that my metal can handle?

The string

This is the string I read from the picture

JTL-33.7 OD-L2-EN 10255/ EN 10219/10240/A1-W-20-0436-CF

I am not sure if the last letter is a F or a E. 33.7 is the outer diameter of the pipe.

Where I bought the pipe

I bought the steel pipe from Hornbach (Germany), here is the webpage for the steel. The diameter is 33.7 mm, the wall thickness is 2.65 mm and it costs ~ 7€ per meter. Notice that it is written that the pipe is not suitable for load bearing structure/scaffolding, just for building furniture. But still I would like to know exactly how much load it can hold, since also furniture will have to stand some weight.

enter image description here

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  • $\begingroup$ It's probably just mild steel. $\endgroup$
    – DKNguyen
    Sep 20, 2021 at 21:51
  • $\begingroup$ @DKNguyen I have added some extra details (price included). Does this tell you something more? In the case it is mild steel, do you know what's the yield stress? $\endgroup$
    – Nisba
    Sep 20, 2021 at 21:58
  • $\begingroup$ Once you know the material and dimensions you just look up the material properties and do the calculation. $\endgroup$
    – DKNguyen
    Sep 20, 2021 at 22:10
  • $\begingroup$ The dimensions are 33.7 mm outer diameter and 2.65 mm wall thickness, but I don't know the material except for the information given in the question. $\endgroup$
    – Nisba
    Sep 20, 2021 at 22:11
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    $\begingroup$ That's why I told you it's probably mild steel. $\endgroup$
    – DKNguyen
    Sep 20, 2021 at 22:12

2 Answers 2

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According to this site, the pipe is made in accordance with euro standard EN10219 and conforming to EN10240. The minimum upper limit of yield is 235 MPa. EN10255 seems to be the British standard for tube steel. I suggest to get the relevant standard to verify the strength of your pipe.

https://www.acciaitubi.com/product/construction-tubes https://www.acciaitubi.com/files/prodotti/6/construction-tubes-en.pdf

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How much strength do you need ? As ordinary carbon steel the yield strength will be over 30,000 psi. Because it is ERW ( welded ) it will have some cold work and likely higher strength than 30,000 psi. Some numbers on the tube indicate the mill heat number. The steel has a strength , not a stress. Looking at your "tube calculator" page with a calculated stress of 42,644 psi includes a safety factor of near 1.8. So a yield strength of 30,000 psi will tolerate the stress but with a lower safety factor ( Which I calculate as 1.26, relatively low). I think you need a higher safety factor. The most practical stronger stronger tube (higher safety factor) is thicker wall or larger diameter, or both.

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  • $\begingroup$ I don't have a specific number in mind, I am just exploring the various combinations of (length, load displacement, load) to stay safe. To do so I have created a simple Python script. I'll construct things based on what gives me a good safety. Which safety factor is it considered ok? Is 1.8 already good? $\endgroup$
    – Nisba
    Sep 21, 2021 at 22:09

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