Live load distribution factors 1

[timmyo131]

I am designing a simply supported adjacent box beam bridge. The bridge is a replacement for an existing bridge which is only 16 ft wide (one design lane). So the design is for 4 box beams all 48" wide.

The problem I'm running into is in Section 4.6.2.2.2 in AASHTO/DM-4 which provides equations for calculating the live load distribution factors for moment and shear. My bridge falls under the Type of Beam: Concrete Beams used in Mult-Beam Decks and my applicable cross-section is either f or g. The problem is that the equations that I want to use show that the range of applicability is for 5 <= Nb <=20. Nb = number of beams. My Nb = 4, so I am outside this range. What equations should I use instead?

Pertaining to the moment factor for an interior beam (Table 4.6.2.2.2b-1), I was considering the case below the one mentioned above that shows equations for S/D. There is nothing written in the range of applicability box which would prohibit its use. Would this be the best solution? If so, should I just use the suggested k value of 1.0 that is provided for box section beams, or use the k = 2.5*(Nb)^-0.2 >= 1.5 equation shown in the previous case?

 

[MichSt]

Worst case would be one wheel line on a box, no?? Could compare that to the result you would get using Nb = 4 in the distribution equation.

 

[edward1]

From modeling prestressed box beams using finite elements, I recommend in this case use D= 2 Wheel Lines/4 Beams =0.5

 

[graybeach]

Will you have a concrete deck? Since the Pennsylvania collapse, cases are being made that asphalt topped adjacent box beams should be designed for a factor of 1 and should be considered nonredundant.

 

[miecz]

Looks like the DFs are based on research done by Johnston and Mattock, referenced at the end of Section 4. I would get a hold of that research paper and see if any testing was done on bridges with 4 stringers. Probably not, in which case, I would extrapolate the formulas, and add an ignorance factor. What's the distribution factor for 7 stringers, 6, and 5. As an upper bound, the distribution factor for 4 stringers can't be more than 5/4 times the DF for 5 stringers.

 

[timmyo131]

To Graybeach: Yes, the bridge will have a composite deck. The existing bridge, which we are replacing, is a non-composite deck. As you alluded to, non-composite adjacent box beam bridges seem to run into problems in terms of redundancy and drainage issues. We feel the new composite deck will solve many of these problems that the existing bridge is experiencing.

To Edward1: I see what you are saying, but seeing that it will be a composite bridge deck, a DF = 0.5 seems rather conservative.

To Miercz: I think I looked at this issue very similarly to your advice. The interior moment DF equations per AASHTO/DM-4 are based on b, L, I, J, and Nb. So I made a little table of DF vs Nb that keeps all other variables held constant except for Nb. This gave me a good idea of how the DF decreases as Nb increases. What I noticed (when looking at my particular bridge layout) was that for the 5 beam case, the distribution factor was 0.29. So if you had one beam fail, you would still have 4 beams x 0.29 = 1.16, which means 4 of the 5 beams would still be able to resist 116% of the full design live load. If you used the given equations for the 4 beam case (which are said to be invalid for the 4 beam case), you get a DF = 0.303. So if one beam failed, you would only have 3 beams x 0.303 = 0.91. So if one beam fails, the other 3 will not be capable of resisting the full design live load (which may be why the 4 beam case is invalid for the provided equations). So I decided to increase my moment DF = 0.38 so that if I had one beam failure I would still have 3 beams x 0.38 = 1.14 which is approximately the same redundancy as the 5 beam case. Then, looking at the shear DF, I thought about how in general, we want all concrete beams to be stronger in shear by a certain margin compared to the flexural strength. We do this so that if failure occurs, it will be a slow ductile failure rather than an abrupt shear failure. So again, I looked at the 5 beam case and what the provided equations would give me for a moment DF and a shear DF. Then I just took the ratio of the two which ended up being 1.66 (shear/moment). So then I just took my new moment DF = 0.38 and multiplied it by 1.66 to get a shear DF = 0.63. Does my logic seem reasonable?

Thank you to everyone for the excellent replies. I greatly appreciate it!

 

[miecz]

You need to get a hold of the research paper that led to the code provision. Sometimes the reasoning behind the provision is not even close to what you think it is. While it's possible that you have hit upon the rational behind the provision, it's just as likely that you missed it. Also, you get to check whether the limit in the applicability of the code provision is because a.) that't where the testing stopped, or b.) the test data diverged at that point. In any case, reading the test data always results in more confidence when extrapolating a code provision.

 

[wiktor]

"My bridge falls under the Type of Beam: Concrete Beams used in Mult-Beam Decks and my applicable cross-section is either f or g. The problem is that the equations that I want to use show that the range of applicability is for 5 <= Nb <=20. Nb = number of beams. My Nb = 4, so I am outside this range. What equations should I use instead?"

Your type is rather "Concrete deck on Spread Box Beams" or "double T-beams" . You may use only formulas allowing Nb=4
But there is simple solution - make a small 3-d grid model with 4 beams, some kind of transverse connectors, and load it with unit loads, as separate cases. From this you can develop accurate factors for the load distribution. If you like to learn more on the subject, look for "Guyon-Massonnet method".
But for a small bridge, using precast beam, that's overkill. Use factor 0.5, i.e. 50% of combined truck and lane load for each beam, for both moment and shear.
This type of beams, regardless of thickness of the overlaid deck (until really enormous) have the tendency to work independent if not post-tensioned transversely.

AASHTO is not a cookbook, it's a design specification