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Working out a Safety factor in a steel tube? steel plate

Question:

I am doing a bit of science work i have a steel tube/pipe with a Circumferential(hoop) stress of 63.96 Mpa, if the tube has a 250Mpa tensile strength and a safety factor is 10 would you stand next to the pipe? What is the formula to work this out?

Anwser:

If a structure is required to meed a safety factor of 10, then that means it is required to survive a load that is 10 times higher than the operating load, without failing. Note that there are different ways to fail, and you can write a safety factor for any of them. It's common to have a safety factor for yielding (permanent deformation, but still able to hold the operating load times the safety factor without seriously compromising the structural integrity) which is relatively low, and a safety factor for ultimate failure (inability to carry the operating load times the safety factor; i.e. structural collapse) which is relatively high.

Now, assuming everything behaves linearly, the hoop stress is a linear function of load. So if the load goes up by a factor of 10, the hoop stress goes up by the same factor, to 639.6 MPa. That exceeds the tensile strength of the material, which means the material would rupture. (If we were talking about yielding, we'd refer to the yield strength instead of the tensile strength.) Thus, you don't meet the required safety factor, and it's probably not a good idea to stand next to the pipe.

On the other hand, steel is a ductile material, and it's possible that, as the load increases beyond the yield point, the load redistributes, which means stress is not linearly related to load. This might keep the hoop stress below the material tensile strength. This is a more accurate, but also much more complicated, way to perform the analysis. Given that we don't know nearly enough about the situation from the problem statement to do this analysis, we can't rely on this.

Remember, the basic idea of a safety factor is, it's the factor you need to increase the operating load by, and still survive. If the problem can be treated as linear, the safety factor can be applied directly to stress (and also deformation), to be compared to an allowable (which might be a material strength, a fatigue threshold, etc). I hope that helps!

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