Soil bearing, Overturning & Sliding check for Concrete Arena Wall Footing

[oengineer]

I have a concrete wall footing that I am designing for a rodeo. I have attached a picture of the wall footing.I have a question regarding impact loads. Since this wall is being built to separate the grandstands form the rodeo arena where the animals will be,Do any if the codes mention designing for an impact load caused by animals in an arena? Should I apply a distributed load along the face of the wall on the other side of the bleachers/grandstands? If I should apply a load, what would be acceptable (maybe 200 psf, 150 psf, or 100 psf). Any suggestions and/or comments are appreciated.

 

[Leftwow]

I'm glad you added the footing. I would work backwards and see how much force your design can withstand, then make a judgement from there, I usually do that when there is an unknown force. Why do the extra code look-up if your structure can withstand a 50 kip lateral load?

 

[oldestguy]

I'd suspect the impact of animals is too slight to be of concern. However, some day that rodeo may be junk automobiles and so that is the impact you might look at such as a junk truck.

On your drain detail, the gravel serves no purpose. Instead just use a perforated pipe with a geotech "sock" covering to keep the sand out of the pipe.

 

[DaveAtkins]

First, calculate the kinetic energy produced by the animal impacting the wall. KE = (W^2)(V)/2g, where W = the weight of the animal (kips) and V = the velocity of the animal (ft/sec).

Second, calculate the stiffness of the wall. k = the inverse of the deflection (inches) due to a 1 kip load applied at the top of the wall. In other words, k = 1/Δ.

Third, calculate the deflection (inches) caused by the kinetic energy calculated above. Δ = the square root of [(KE)(12)/k].

Finally, you can determine the force of impact (kips) caused by the animal by multiplying the deflection calculated in step three by k determined in step 2. In other words, F = (Δ)(k).

DaveAtkins

 

[oengineer]

@ Leftwow- I just worked backwards and determined my foundation can only handle 74 psf as a distributed wall load. Earlier, I checked the wall and the footing for strength and it can handle a 200 psf distributed load due to the thickness and the reinforcing. I am still having issues with my sliding check whether I use 74 psf or 200 psf. For my overturning, my foundation passes with 74 psf but not with 200 psf. Would 70 psf be acceptable for a lateral load applied to my wall due to impact from animals?

 

[oengineer]

@ oldestguy - I am now just seeing your post. What would be the impact for junk automobiles? Is it located in ASCE 7? Would 70 psf be acceptable? Would designing for automobiles be out of the scope of our work since that is not the attended use of our facility?

@ DaveAtkins - I do not know the weight of the animal nor the speed they move at. Do any of the codes provide this information?

 

[KootK]

It's not quite the same topic but you might find some useful information in this thread: Link

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[oengineer]

@ KootK - Thank you

@ everyone/anyone -Since I am having issues with sliding check for my wall, if I dowel in the pavement shown in the picture to my wall footing should that eliminate sliding of the wall? We are building a new pavement and wall, so I can have the 2 attached together. Does anyone foresee any problems/issues?

 

[SteelPE]

I might be inclined to design the wall to include the loads imposed by a vehicle barrier system as indicated in ASCE-7. This is a 6,000 load located 1'-6" above grade. In your instance. I might increase the height a bit and see what that does the the design. Remember, you can distribute this load through the wall to your footing below. Some might take a 30 degree spread for the load, others might push it to a 45 degree spread. The higher the load is placed on the wall the larger the spread.

 

[Leftwow]

Are you taking a 1 foot section of the footing? Because I wouldn't assume it would be equivalent to an active earth pressure load, I would assume it would be more of a point load.

Also, another thing I would add is this point load would be spread across a length of a footing, not a one foot section. So your resisting moment should be weight of concrete per area times the length of your footing.

So you could say

Resisting Moment = Weight of footing*entire length

So if your footing is 20 feet long, you could say, 6 ft tall * 1 foot wide * 20 FT LONG * 150 lbs/ft cu = 18000 lbs
and also! the weight of your foot 5*1.5*20*150=22500

Then for the Resisting moment = (18k+22.5k) lbs*centroidx, which would be half your footing length so

40.5k*2.5 ft =101,250 ft lbs

Overturning Moment = mass of cow*approximate speed of cow*height of footing

Weight of cow =4600 lbs
mass of cow = 4600lbs/32.2 lbs/s^2=143 lbs m

momentum of cow (lets say 15 mph to 22 ft per second) = 143 lbs m * 22 ft/s = 3146 lbs F

Give it, a safety factor for time purposes say 5,

3146*5=15730 lbs F
multiply by entire height of wall so, 15730*6 = 94380 ft lbs

So resisting = 101250 ft-lbs
and overturning = 94380 ft-lbs
which resisting>overturning with a safety factor of 5.

I think impact loads should be a concentrated load.

Unless you have a stampede of cows... then you have a serious issue. =)

 

[Leftwow]

Oh and there is also ASCE 7-10 C5.4.5 Impact Loads

 

[oengineer]

@ Leftwow- I am taking a foot section of the wall footing. I did all my calculations as if I only had a foot length of the wall. I am basing my design on a safety factor of 1.5 for sliding and overturning.If I use 70 psf, with a wall height above grade of 4 ft, with 1 ft section, my point load is 280 lbs.

@ SteelPE - 6000 lb is quite a force. The drawing I have attached is what is existing at the site right now. We also took pictures (please see additional attachment). The timber planks you see are currently cover up the existing concrete wall that we will be replacing.

 

[oengineer]

the wall I am designing will separate the arena from the bleachers.

 

[Leftwow]

So you are assuming,

you are going to take a 1 foot wide footing, and have a bull ram into it. No offense but I just don't see the logic. Then you would have to take your cow point load and distribute that along the entire wall.