Destructive Testing Steel, no safety factor.

[WoodDesignCat]

Most of my experience is from building design, where there are well defined safety factors and design stresses to use for steel based on ASD and LRFD design forces. After a career change I am now creating in house destructive tests for wood components, which sometimes include metal plates and welded connections. When I do testing I'm just trying to prove our design works correctly before sending our products out for official testing. Depending on what type of system we test, the tested ultimate failure load gets reduced by between 2.1 to 2.5 times for the usable ASD design capacity.

When I design metal components that get used with our wood products I would like to design them without a safety factor, because they will get the 2.1 to 2.5 safety factor added on. Lets say I'm designed a metal plate in bending and shear, should I just be using the ultimate tensile strength and shear modulus without any factors added on, or are there safety factors built into the ultimate tensile strength and shear modulus for steel? I haven't tried designing a weld without safety factors, but is there anything I should be careful of?

 

[JAE]

I guess I would attempt to design the steel plate - your initial design - using AISC criteria (Ω factor for AISC ASD) with their specification and their equations for bending, shear, compression, tension, buckling capacities etc.

Then if you wish - you can bump the design up a bit to match with your 2.1 to 2.5 SF (Ω factor = 0.4 to 0.48). Then have the wood-steel assembly tested and verify that the "ultimate" capacity of the assembly is still within the desired safety factor limit.

I think your higher safety values (2.1 to 2.5) represent more of the variability in wood than steel so bumping up the steel for this higher level may not make sense.

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[WoodDesignCat]

@JAE,

I think you are saying to design my metal parts of the assemblies for the ASD load, using the standard AISC safety factor, Ω. Let me run through an example and see if that still makes sense to you.

Let's say I'm designing a shearwall that has a built in holdown, and the built in holdown has a steel plate in bending. If I want the shearwall to be able to have 700PLF ASD capacity, and design it as 6' long by 10' tall, then my ASD holdown force would be about 7,000# (700*10'). In order to get a wall with a capacity of 700 PLF, I need to actually get the wall to fail at an ultimate load of 1750PLF (700*2.5), so my actual ultimate load on the holdown needs to be 17,500#. If I design the plate for the 7000# force, with a bending Ω=1.67, designing for yield, but actually test to 17,500# then I will be over yield, but still under ultimate.

Maybe I should do what your saying so that I'm designing my steel components for the the ASD load correctly, but then just need to double check that my ultimate load will be less than the ultimate steel strength without safety factors, and if it isn't then use a higher safety factor.

 

[JAE]

Yes generally agree with that (and myself I guess) - but there's also the issue of seismic design - conducting your orchestra of structural elements - to perform in a certain way - i.e. brittle type elements are kicked up higher than ductile elements so that you have an assurance that the right element goes ductile first without a brittle collapse.

With that - your hold down steel plate bending is a ductile type failure perhaps compared with the hold-down bolt through the wood - or the bolt rupture through the hold down vertical plate...or the anchor bolt failure into the concrete below.

So setting up a chain of possible failures along the load path - then orchestrating the level of safety - and the possible Omega factors used in static seismic design - might be necessary to allow engineers to use your product understanding its limitations.

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