Gantry Beam Analysis

[BSim]

I'm currently trying to learn how to properly analyze a gantry beam. At our office we have a program supplied to us by the jacking gantry manufacturer that if you put in the beam size and span it will spit out what they say in the maximum allowable load to be placed on the beam. For this example I will say that we were analyzing a W24x250 with a 50 foot span. Now the program that was given to us claims that the beam is capable of supporting a load at midspan of 48.6K (24.3 tons). We have never used this program before and have always used an engineering firm we are familiar with and they have always told us that that beam for our purposes was capable of supporting 37 tons (74K) (we hang the load from the bottom flange with a trolley). I'm an engineer but new at analyzing beams for these purposes and I'm trying to understand how the program calculated this load and how the engineering firm came up with their number (we have operated based on the number they gave us for this beam for years and it has always been safe).

Now the program that was provided to us by the gantry manufacturer is quite old, it's copyrighted 1993 and I have to run is on a DOS emulator. When I input my parameters it asks for:
-AISC Shape (W24x250)
-Material yield stress(50KSI)
-Vertical impact factor in percent (is set default to 10% but can be changed) (I don't know what this is used for and i would like some guidance here)
-Lateral force factor in percent (is set default to 15% but can be changed) (Again I don't know what this is for)
-Span Length (50 feet)
It then just spits out what they claim the max allowable load to be as 48.6 kips.

My confusion here stems from the fact that I have never analyzed a load hung from the bottom flanges of a beam and am not quite sure what the right assumptions are.
I am assuming that the beam will not experience Lateral Torsional Buckling since the load is hung from the bottom, is that assumption correct.

When I analyzed the beam for bending stress I did M=Fy/s and got 2683K-ft
so M/Ω=1607K-ft as the total allowable moment.

I then calculated the moment on the beam due to its own self weight at 78.1K-ft

So the Allowable Moment from the load would be 1529K-ft
and since this is a concentrated load at the center and M=PL/4 I found P=122K which is much greater than either the program of the engineer we have used.

Then I checked shear(even though I knew it wouldn't control). Table 3-2 in the AISC manual gives Vn/Ω as 547K, so my allowable P would be 1094K. (Also I've always been taught to calculate M/Ω by M=Fy/S, and in Table 3-2 they give the values for Mpx/Ωb and the values there are higher than what I calculated, can I use these values of M/Ω instead of calculating it myself all the time?)

...Ok so after that I still have a maximum allowable load of 122K and it's governed by the bending moment. I haven't checked lateral torsional buckling and I'm not sure how to analyze hanging the load from the flange. Any input would be greatly appreciated in how to analyze that and that the purpose of the vertical impact factor and lateral force factor is for.

I know I seem like I have a lot of questions but I am a couple of years out of school and had been working for Thornton Tomasetti for a while before coming to this firm and I didn't do much of this work with them, and no one at this company has ever done this before, they've always hired another firm to do it. I'm not in any kind of rush but I want to learn how to properly analyze this case myself and understand where these numbers are coming from.

 

[WARose]

-AISC Shape (W24x250)
-Material yield stress(50KSI)
-Vertical impact factor in percent (is set default to 10% but can be changed) (I don't know what this is used for and i would like some guidance here)
-Lateral force factor in percent (is set default to 15% but can be changed) (Again I don't know what this is for)
-Span Length (50 feet)
It then just spits out what they claim the max allowable load to be as 48.6 kips.

Running this on a spreadsheet I have, I came out with a beam just barley being adequate with this alone (considering only stress; the lateral deflection was something like 3"). However, if you have a trolley hanging on a beam and your lateral criteria is 15%......since it is on the bottom flange, you also need to consider torsion. Doing that pushes the beam over as per my spread sheet.

 

[canwesteng]

To clarify on what warose said, you need to consider the combined effects of strong axis bending, weak axis bending and torsion and it seems like your error is only considering strong axis moment. It may be best to continue using loads provided by your consultant for this, or read up on the crane design guide here

http://www.cisc-icca.ca/solutions-c...ng-steel-structures-design-guide#.VMK28j34BFo

 

[NOLAENG]

Yes you have to consider LTB.
You can use the values from Table 3-2 if flexural yielding is the controlling factor. See Section F2.
The values are higher than your calculated value due to the fact you are using the elastic section modulus and the values in the table use the plastic modulus.
As WARose and canwesteng have mentioned, you have to consider biaxial bending due to the assumed lateral factor.
For recommended vertical and lateral factors, look at ASCE and CMAA. I'm sure there are other guides too.
For your beam, you also have to consider the local bending of the bottom flange when the trolley load is applied to it.
For local bending, look at the CMAA 70/74 codes and AISC also has a few articles for guidance.

Assuming:
Simply supported beam
W24x250
Load = 48.6 kips
Fy = 50 ksi
Span Length = 50 ft
Unbraced Length of Compression Flange = 50 ft
1 Trolley with 2 wheels
Vertical impact factor = 10%
Lateral force factor = 15%

I agree with WARose again. For the assumed conditions, the beam is already close to full utilization for strength alone.

 

[JedClampett]

Is it possible that the beam was made of (or assumed to be made of) 36 ksi steel. That would explain the difference between the engineering firm's numbers and the programs.

 

[SlideRuleEra]

BSim - You have done a good job of describing the problem. I believe I see reasons for some of the discrepancies; you using current methods. If you use old-style ASD (AISC Manual of Steel Construction, 9th Edition - c. 1989) your engineer's answer of 74 kips comes out of the calcs. You mention that the software is from 1993, maybe the engineer's answer is from that time, too.

I want to... understand where these numbers are coming from.

Looking at the problem per AISC 9th Edition:
You take LTB into account (but not by that name). As NOLAENG stated, the compression flange's "unbraced length" is 50 ft. Use 9th Ed. equations, or more likely the grafts based on those equations, to determine that the allowable moment for a W24x250, 50 ksi, with a 50 ft. unbraced length is 1010 K-ft. Subtract the 78 K-ft, that you calculated for the weight of the beam to get 932 K-ft. Solve for the point load at the center and get your engineer's answer of 74 kips (truncated). That is the total allowable load. AISC 9th Ed. also covers the impact load which is listed as 10% for certain traveling cranes. Therefore, 74 kips is 110% of the allowable load that the crane can lift. The actual allowable load that can be lifted is 67 kips.

The lateral force factor is used to "provide for the effect of moving crane trolleys" and is used to design the crane runway. AISC 9th Ed. lists it at not less than 20%, however, not the 15% default in the software.

I took a look at JedClampett's good suggestion (36 ksi vs. 50 ksi) and got a surprising answer. Per AISC 9th Ed., it does not matter. The unbraced length is very long, so long in fact that the equations (and graphs) limit the allowable moment to 1010 K-ft. for both 36 ksi steel and 50 ksi steel.

Why the software answer for the allowable load is 48.6 kips is not clear to me. Perhaps the manufacturer considered other factors that are unique to his product.

Maybe that will help explain some of the differences you are seeing.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[CTW]

See if this document helps you.