AASHTO Load Question
AASHTO Load Question
(OP)
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.
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.






RE: AASHTO Load Question
RE: AASHTO Load Question
It just seems ridiculous to factor the load by almost 3. But like you said, it is what it is.
RE: AASHTO Load Question
RE: AASHTO Load Question
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?
RE: AASHTO Load Question
RE: AASHTO Load Question
What unit width are you referring to? Tire width? The span is short, 6' max.
RE: AASHTO Load Question
RE: AASHTO Load Question
RE: AASHTO Load Question
RE: AASHTO Load Question
RE: AASHTO Load Question
RE: AASHTO Load Question
The commentary states "To qualify for relief from impact, the entire component must be buried."
RE: AASHTO Load Question