Steel Capacity Factor/Utilisation Ratio 1

[mrlm]

Hi all,

When using a program like Microstran for steel design, it provides me with a capacity factor or utilisation ratio which is essentially capacity/load. If the capacity factor is greater than 1, it suggests the member has sufficient capacity for the applied loads. However, when the member is subject to combined actions either uniaxial or biaxial, I cannot workout how the software is calculating the capacity factor. Whether it is phi.Nc/N* or phi.Mi/M* or some ratio between the two. Would someone be able to help me out?
Thank you.

 

[human909]

Probably whatever the combined formula is in your local code.
ie N*/(phi.Nc)+Mx*/(phi.Mcx)+My*/(phi.Mcy) < 1

So the inverse of this give you the capacity factor.

 

[Tomfh]

Create a report. It should be in there.

 

[Agent666]

For combined actions there are often many checks that depending on the constraints it could send you down multiple paths in AS4100 or NZS3404 (which I'm assuming you are referring to based on the nomenclature you're using). Depending on whether or not you have full lateral restraint or not quite different checks are required. So start with determining which case is actually governing and simply follow the code provisions. Can't recall if Microstran reports what the governing case was.

 

[mrlm]

Hi all,

Thanks for the replies. I have spoken to a couple of other guys in the office and they have had similar sorts of issues in not being able to calculate the capacity factor.

I have tried what was recommended above and still came up empty handed.

See attached an example of the results I'm using, if anyone has the time to review it would be appreciated. The example is using AS4100 as the design code
Section is a 165.1X3.5CHS. I have verified all the results from the output and agree with all. Although, the program assumes the CHS is stress relieved making alpha.b -1.0 as opposed to -0.5 for non-stress relieved which means the member compression capacity is slightly higher in the output.

 

[Agent666]

Well I agree with all the design capacities at the bottom, but cannot for the life of me see how they got that ratio either! Because of the way the combined actions checks work it's not really appropriate to compare them as a ratio because they are tied to both moment and axial loads, they are simply a pass or a fail. What I mean by this is you cannot imply because a ratio of 0.5 on a combined check, that you'd be able to increase the moment by 50%. Mainly because many of the checks involve power terms and not being linear. You can put yourself in the mindset that you have much more capacity remaining than you think.

I'm just plugging and chugging combinations of ratios at this point, but no dice.

Suggest you contact Bentley, they should be able to explain it to you.

 

[human909]

This is the danger of being too reliant on computer. Though I admit that this basket also includes me. The guys in your office seem to suffer also from the problem. Anyway enough lecturing here is the answer, and my approach in deciphering:

-My first check was ϕNcx/N* which comes close
-Second check was to actually look at the output and follow up the specific inplane capacity reference mentioned.
-I then proceeded to do the calculation. M*<ϕ(Ms(1-N*/ϕNc) as referenced in 8.4.2.2

It turns out that (1.108 x N*) is threshold number where design load exceeds capacity. (A potentially better capacity factor would be the CF x M* as well. But it turns out that this approach isn't used here. Likely because calculation of CF with Bentley's approach is a simple closed formula.)


{At 1.108 N*=229.6, (1-N*/ϕNc)=0.07127, Mi=1.943 and ϕMi=1.749 ~ Mx*=1.75}

 

[Agent666]

But unless I'm mistaken that doesn't work out just factoring N*?

M*<27.27x(1-(1.108x207.22)/247.22))

1.75<1.94,

doesn't pan out? by it should be a factor by which you can increase both M* and N* by to hit the capacity!

i.e.

1.108xM*<=phi_Msx x (1-(1.108 x N*/phiNc))

1.94<1.94... QED

 

[human909]

Yes and yes.

I double counted the ϕs and by happanstance got the right CF. Your calc is correct you are not mistaken.

 

[Agent666]

Funny, wouldn't have got it without your logic on the multiplier... Two wrongs do make a right...

Rearranged for completeness

x = phiMsx ÷ (M* + (phiMsx x N* / phiNc))

x = 1.108