Determining Factor of Safety of Existing Structure

[440MoPar]

A client asked us to verify that a structure of theirs (i.e., down to/including its individual components) meets a safety factor of 5:1.

To summarize my calculation for FoS:

I calculate the design strength (?Rn for LRFD (?=0.75), Rn/? for ASD (?=2)) then divide that by the actual force, P being applied to the member to determine the FoS.

For example:

LL = 10 kips
DL = 10 kips
Rn = 120 kips

LRFD
------
P = 1.2(10) + 1.6(10) = 28 kips
?Rn = 90 kips
FoS = 90/28 = 3.21
CHECK: FoS > 5 - NO!

ASD
------
P = D + L = 20 kips
Rn/? = 60 kips
FoS = 60/20 = 3.00
CHECK: FoS > 5 - NO!

Is this calculation correct? Or am I adding a safety factor to a safety factor?

For example, should the calculation be this:

Rn = 120 kips
P = 20 kips
FoS = Rn/P = 6
CHECK: FoS > 5 - YES!

Please help! And provide any support documents (posting .pdfs, .jpgs, etc.) or document citations if possible.

Thanks!

 

[csd72]

The last calculation is the correct one, when calculating the FOS, this is the only factor involved.

What type of structure is this? 5 is very high, it is normally 1.5 to 2!

 

[MaddEngineer]

Oh the dreaded factor of safety issue,

I deal with this from time to time when designing temporary structures. Manufactured scaffolds are typically load tested until FAILURE (i.e., a leg or a brace buckles) and then that load is factored down by 4.0 to get the safe working load (its a little more complicated based on DL and LL F.O.S.).

So when you are asked to calculate a FOS it always gets tricky since by test method it is based on FAILURE load.

I would not include ? = 0.75 in your analysis since this is a reduction factor. Also Mn is typically based on ZFy, which is the elastic limit, you may want to use Fu... its all based on what you consider failure.

Also if you are using code prescribed live loads, etc. they may have intrinsic factors of safety that you may have to rationalize in your analysis. If you are checking against an actual load then it is more direct.

 

[csd72]

Agreed.

FOS = theoretical load capacity/applied load

no bells, whistles e.t.c. just the raw thoretical values.

 

[ChipB]

I'm more in agreement with MaddEngineer.

Yielding is not failing.

You have to use real loads, not factored loads. Otherwise, you aren't comparing apples to apples.

For an A36 flexural member Fu=58ksi
Therefore, allowable stress with a F.O.S. of 5:
58/5 = 11.6 ksi

In order to meet this F.O.S criteria, the stress in any flexural element, can't exceed 11.6ksi which is about half of the stress you are allowed by ASD.

 

[csd72]

agreed as long as the beam is capable of achieving ultimate capacity prior to buckling.

 

[JAE]

If something yields and the yielding induces extreme deflections in the plastic state, the structure might become unstable and/or unusable (per csd72's remark about buckling)

Failure can be defined as any type of limit state. Just need to be clear on what limit state you want as your failure point.

 

[MechEng2005]

I agree that "failure" needs to be defined to determine if yielding qualifies as failure. In some cases I think it certainly should, and in others maybe not.

For example, picture a 2d equilateral triangle. Set one side on "ground" and pushing down on the point opposite. As the material deforms, the "base" side will elogate, and the top point will lower. As this process continues, the stress will continue to increase each cycle even if the force is the same. I would say that in this specific case, yielding would constitute failure (or need for life-cycle analysis).

-- MechEng2005