Ultimate strength in safety factor

[Vertical-Eng]

If the definition in a code of "factor of safety" is thus:

the ratio of the ultimate strength to the working stress of a member under maximum static loading...

Would I assume this would be using ultimate stress (Fu) in determining the maximum allowable stress? There are many places where a safety factor of 5 is prescribed which is very high for things like steel ladders and work platforms. I'm used to using AISC steel manual which usually uses Fy/omega for ASD safety factors, and am just not used to seeing a code use Fu.

A36 steel with a S.F. of 5 using Fu, puts allowable stress at 11.6 ksi.
AISC simple bending would be Fy/1.67, puts allowable stress at 22.6 ksi.

Based on the code I'm using the former would be using elastic section modulus for section properties, whereas AISC prescribes Z for many members. So that makes quite a large difference too.

 

[driftLimiter]

Be careful with this. You shouldn't just assume the ultimate stress is the capacity of any given member. For example columns with high slenderness have an allowable stress that is much lower than the yield stress. Beams that are subject to lateral torsional buckling also have an allowable stress that is much less than the yield stress for bending.

The AISC formulas for compression, and bending should be used to determine the allowable stress.
Where does the requirement for 5x FOS come from? Is it supposed to be compared with AISC and ASCE7 Loads? Or is the load source something else?

 

[human909]

There are many places where a safety factor of 5 is prescribed which is very high for things like steel ladders and work platforms.

For most of the things where the safety factor is 5 that is often a multiple of the maximum allowable MASS not the ultimate LOAD.

A 100kg mass person will put a greater ultimate load on a ladder or work platform than 100kg of force (980N).

Likewise a lifting shackled rated for 5tonne will almost certainly experience a higher load than 5tonne of force if being used to lift a 5tonne mass item.

 

[JoshPlumSE]

Would I assume this would be using ultimate stress (Fu) in determining the maximum allowable stress?

You definitely shouldn't assume that. Maybe it's true for a pure material failure like pure axial stress. But, the safety factors in the codes are based on many things.

1) Variability in loading. This is the Load Factor side.
2) Type of failure. Is it ductile (like tensile yielding of A36) or is it brittle (like a weld fracture)? Ductile failure has a lower factor of safety than non-ductile failure.
3) There's also just the concept of buckling. Elastic, or in-elastic and how accurate the formulas are for predicting this. There are so many cases where buckling (of vessels or tanks or such) are "fudged" with approximate formulas that need a better factor of safety.

 

[JLNJ]

"Safety Factor" as a machine designer uses the phase is not the way structural engineers think of "Demand to Capacity Ratio". This can be a disconnect, such as when trying to assess OSHA's fall protection rules which incorporate factors of safety.

 

[ThomasH]

Comparing safety factors without context can in my experience lead to the wrong conclusions.

First you have the capacity of a lifting device, concrete floor, steel beam or whatever. It calculated with the applicable code or standard and includes several factors. Those factors differ with material, code, and application.

And then you have the load with certain assumptions regarding the properties for the load and how it is applied on the structure or device. Or sometimes, how sure are you of the accuracy of the load itself?

You can also have the quality of work expected at a building site or factory. That can be included in a safety factor.

So, I don't think a "safety factor" can be considered as defined in any consistent always applicable manner. I have just seen too many ways of doing it.

 

[Vertical-Eng]

ASME A17.1 Elevator Code is where the definition of safety factor is from. It's an older code (slow to update) and based on ASD. When designing working platforms it prescribes point loads and area loads and then simply states to use a safety factor of 5. I've always just used the steel manual for ASD loads: so an omega of 1.67, I check channels and angles for LTB and all the required failure modes. I use Z, 1.5S, or S as required by AISC. For the stress I divide Fy by 5 for the maximum allowed stress. Then when needing to design a vertical ladder, I found ASME A17.1 points to ANSI A14.3 which is a standard for fixed ladders. In this code (which is obviously older as well) it states, "7.2.1 The factor of safety shall be based on the ultimate stress of the material and comply with... Ductile material shall have a safety factor of not less than 5 times the designed static load." It prescribes point loads for the ladder.

So this got me thinking, should I be using Fu since it says ultimate. There is no definition of "ultimate" in either code, but every engineer knows that the ultimate tensile strength is of a material from looking at a stress/strain curve. No one considers yield to be ultimate... right? And 5 is a very high safety factor for steel given ASD loads. So this clicked for me. I've been using the wrong F. I've been over-sizing my channel and angle supports for platforms. I always thought the safety factor was too high. But secondarily, these older codes didn't consider plastic section modulus, so if I use Fu/5 I think I would also be limited to using S and not Z.

 

[MotorCity]

"Factor of Safety" is really a misnomer when it comes to structural design. The allowable stresses and allowable loads are determined by a statistical analysis (bell curve). For example, given a load of 100 psf, what is the likelihood this load will be exceeded in a given return period. While such an evaluation may translate into a factor of safety, its not the same as arbitrarily applying a uniform factor of safety across the board. It comes down to what code writers feel is a reasonable risk based on statistics and historical data.

 

[driftLimiter]

I think if your using AISC you just use LRFD and treat all your loads other than the material weight as live loads.
If you do AISC design method then divide by 5 at the end its very wasteful.

Attempting to conform to this old code is more headache than its worth. FOS dividing by 5 is a typical approach when we have test data that shows a failure (ultimate capacity) we would divide it by 5 then call it allowable. If your using AISC this concept of FOS doesn't apply in the same sense.

 

[Vertical-Eng]

Attempting to conform to this old code is more headache than its worth.

Does that mean I just ignore the safety factor of 5 and use solely AISC steel design? The ANSI code prescribes loads as well. So for the ANSI ladder code, it says use a 250 lb point load on a ladder rung and use a safety factor of 5. Instead do I use a 1.6 live load combination for that and the phi factor of 0.90?

 

[JStephen]

Note that ASD = "allowable stress design" but also "allowable strength design" and those two are not the same.
In general, I would interpret that requirement to use a 5 factor instead of a 1.67 factor in the Allowable Strength Design (ie, current AISC formulation).
But in the example you give for simple building, the allowable strength is based on plastic yielding and will give a calculated bending stress higher than Fy/FOS.
Also see ASME BTH, which has some of the same design considerations but includes the factor of safety in the equations. But there, they'll use something like 1.25Fy/FOS, which is making use of the same ideas.

 

[phamENG]

Vertical-Eng, the ultimate load or ultimate stress is simply the load or stress at failure. So if you have a simply supported beam with a point load in the middle, if you exceed M=F[sub]y[/sub]*Z the beam will fail. You can't get to F[sub]u[/sub]. There may be some region that gets there...but it doesn't have to. If the entire section plastifies it will be unstable. That is ultimate. So you have the ultimate yield strength and the ultimate tensile strength. They are two very different things. And they are germane to the controlling limit state in different situations (yielding for bending, tensile for rupture in the net section, etc.)

As for the factor of safety, it's what you divide the relevant stress by to get your allowable stress. So for a 50ksi channel with a factor of safety of five, your bending stress can't exceed 10ksi.

Don't mix AISC and ASME. If you're designing the platform for the car (I've done a handful of them), it is not part of the building - it's a piece of equipment. And you're right, the beams will be bigger than you might expect. That's fine. Keep in mind that this thing is going to be moving and vibrating up and down that shaft for the foreseeable future, not just sitting still holding up a few square feet of grating. The factor of safety is what is for several good reasons.

 

[Vertical-Eng]

Don't mix AISC and ASME. If you're designing the platform for the car (I've done a handful of them), it is not part of the building

It's not the car platform, it's a working platform either in the pit, or at the top of the hoistway to service the sheaves. It's directly attached to the building and ASME prescribes required loading and safety factor.

 

[phamENG]

What section are you looking at in ASME 17.1?

 

[Vertical-Eng]

What section are you looking at in ASME 17.1?

A17.1-2016 2.7.5.3

 

[phamENG]

Putting the direction on loading aside (450lbf concentrated load or a 225lb uniform load spread over an area of 0.44sf? What?), the factor of safety thing is pretty straightforward, and I disagree with what JLNJ said about it being significantly different from the Demand:Capacity ratio.

In this case, I would design it like this:

For each limit state, determine the capacity of the structure using the appropriate stress as I mentioned above. Then, compare the demand to that capacity. If your Capacity/Demand is 5 or more, then you meet ASME. For instance:

If the beam in consideration has a capacity (M=F[sub]y[/sub]*Z[sub]x[/sub]) of 16.2k-ft, and the demand is 3.1k-ft, you're good because 16.2/3.1=5.22.

So for this check, you're essentially outside of 'ASD' and 'LRFD' as defined by AISC.

I say 'this check' because I think this is a place where you end up mixing a little in the desing. That platform is part of the building, and needs to meet the requirements of ASCE 7 and AISC 360, too. ASCE 7 says that 'elevated platforms (other than exit ways)' should be designed for 60psf. If the shaft is larger than 8'x8', that could quickly exceed the ASME requirements (even when you account for FoS differences). That said, you shouldn't mix AISC and ASME in the same design check. Do them as separate designs and make sure the design you use satisfies both.