ASME A17.1 Section 3.6.1 Jack-Supporting Structure

[Simon Nguyen]

Hi guys,

I am designing a pit to support a hydraulic elevator in NY. The building department refereed me to ASME A17.1 for the design requirements. The problem is that my pit is "hanging". The pit's slab will be supported by steel beams. If I understand section 3.6.1 correctly, I will need to design the supporting structure for the rated capacity of the elevator with a minimum factor of safety of 5. In other words, the ultimate strength of my steel supporting structure will have to be at least 5 times the rated capacity of the elevator. This seems a little excessive to me, my steel beams will need to be huge. Does anyone know the reasoning behind this factor of safety of 5? And is there anyway that I could use a smaller factor of safety.

Sincerely,

Simon

 

[JAE]

In case the elevator drops suddenly and impacts the pit springs - impact forces are much higher than static loads.

Now having said this - since this is an extraordinary event, you might be able to design for 5 times the capacity but with no other additional "live load" factors on the load (i.e. don't also add 1.6 factor).
...similar to the logic of using 1.0W with the "ultimate" wind forces from ASCE 7-10 or the tornadic forces from ICC 500.



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[Nor Cal SE]

I completely agree with JAE about not applying the load factors, such as 1.6 for live loads, due to the already-conservative 5-times safety factor. To do so would give you a safety factor in reality of closer to 7.5, and I don't believe this is the code's intent.

The other similar issue of interpretation involves the last line of 3.6.1: "based on ultimate strength". Does this mean you apply the phi factors per AISC equations? Or omit them because the ASME verbiage is ultimate strength? It could be argued that "ultimate strength" means exactly that, all the way up to ultimate strength (or in other words, nominal strength). But due to the lack of clarity, I would lean towards applying the phi factor because this won't penalize your design a great deal since phi will equal 0.9 for many steel design checks.

 

[Simon Nguyen]

Thanks JAE,
I think your are right, impact load is probably the main reason. I also had the same thought. I was planning to use D+L (L is the elevator rated capacity) as my controlling load combo and use Pn/5, Mn/5, and Vn/5 as my design capacity. But my supervisor was not entirely satisfied with this. So, I want to see if anyone know some ways that we could push the boundary a little more.

 

[SAIL3]

I would check if it is a FOS 5 to failure/ultimate stress.......it will still leave the problem of how to handle stability/buckling of the members due to an impact/transient load.....