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AASHTO Load Question

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BadgerPE

Structural
Joined
Jan 27, 2010
Messages
500
Location
US
Hey all,

Finally got back to the box culvert I was working on last week. This is my first go around with any AASHTO type design and I am having issues with what I feel is a relatively small box culvert (double barrel, 6' wide x 3' tall openings, 10" continuous top slab, 8" walls, 1' cover over top of structure).

My question has to do with all of the overstrength factors for truck loading, and if I am applying them correctly. Here is what I have:
mpf (multiple presence factor for one lane loaded, traffic parallel to span)=1.2
Dynamic Impact Factor= 1+.33(1-.125*fill height)=1.29
LL Load factor=1.75
Total overstrength factor=1.2*1.29*1.75=2.71

So for example, a 32k axle load becomes 86.7k, which seems a little ridiculous. When I distribute this out over the effective area, 17.3ft^2, and transform it into a notational load for a 1' strip x 2' tire legnth, I get a value 10k.

I have this set up in RISA to determine moments and shears over a 1' strip width with moving loads and then I am calculating reinforcing by hand. Right now shear is the controlling factor at the interior support (12.2k). With a 10" slab and 4ksi concrete, I only have a shear capacity of 9.9k.

Assuming my adjustment factors are correct, would it be best to increase my slab thickness to 11 or 12" and be done with it? I am having a hard time believing that a 10" slab can not take the truck loading. Or can I assume the layer of steel that is not in tension acts as shear reinforcing and will contribute to the strength of the slab?

Like I said above, this is the first AASHTO structure I have done and would appreciate any guidance. Also, we are being paid to design this as a CIP structure for a contractor who would like to keep his guys at work. So that is why we are not doing a precast structure.
 
I have obtained load factors int hat range using AASHTO as well. I agree it seems a bit much, but it is what it is.
 
Thanks Dcarr,

It just seems ridiculous to factor the load by almost 3. But like you said, it is what it is.
 
The factors are what they are and look fine. Unless you have a box that is really long, making the top slab larger doesn't make it a whole lot more expensive. However, I'd have to say that your wheel load looks larger than than normal for the given configuration. I didn't run any numbers but it seems to be high, in my opinion.
 
Bridge,

I have less then 2' of cover so I figured the effective area=17.3 ft^2 as follows:
Cover=1'
Max Span=6'
Length of Tire=10 in
LLDF=1.15

Weff=96+1.44*S=96+1.44*6=104.4in
Leff=Lt+LLDF*H*12=10in+1.15*1'*12in/ft=23.8in
Aeff=2485in^2 or 17.3ft^2

For the 32 kip axle load the area load= 32k/17.3ft^2=1.85ksf
Factored Area Load=2.71*1.85=5.01ksf

Notational Load = (Leff/12)*1'Width*5.01ksf=9.94k

Any thoughts on the process?

 
I was getting 9.93Klf. We limit Weff to 2*unit width, but short span i am sure that your over 8' lengths
 
Zten,
What unit width are you referring to? Tire width? The span is short, 6' max.
 
The length of the culvert...6 foot span, but how long is it?
 
You're numbers look fine. I was thinking in terms of the unfactored load for some reason...
 
Thanks for all the help. I didn't see an 11" top slab happening when I started the design....
 
Our standards show a 10" slab but is for 2' of cover. Usually cover will control the shear but when you're next to nothing on cover, the loads are treated differently. One thing to watch out for is that below 2', the shear capacity equation is different. You can no longer use the specific member clause towards the end of the concrete chapter.
 
Are you referring to the Vc=0.0316*Beta*f'c^0.5*b*d_v?
 
Seems this could be subject to your interpretation of the code, but you may be able to ignore DLA due to the fact that your entire structure is entirely buried.

The commentary states "To qualify for relief from impact, the entire component must be buried."
 
See section 5.14.5.3, the articles of 5.8 and 5.13.3.6 shall apply to slabs of box culverts under less than 2ft of fill and to sidewalls. The shear equation that has to be used is the one referred to above.
 
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