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Footings with Off-Center Columns

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lacyma1

Structural
Jun 2, 2021
13
Hello,

So I've recently started working on the design of some unique footings with columns largely off center due to the construction occurring in an existing building (our client wants to continue operations in the adjacent parts of the building, so along walls and in corners columns have large eccentricities). What I've done so far is to provide a large footing that extends from floor level down 4' below grade and sized this footing such that the resultant of the footing weight and column weight occurs inside the kern of the footing. Vertical Loads are quite low and range from 10k to 30k.

I have a few questions regarding this:
1. Are these footings designed properly so far? are there additional concerns I should incorporate into my strategy?
2. Would this be a good location to use plain concrete?

Thank you
 
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That seems a reasonable way to design them. You don't HAVE to have the resultant be within the kern, though it is probably a good idea for gravity loads.

I would not think this would be the ideal scenario for plane concrete.
 

Will you post proposed foundation plan and provide more info ?
 
JoshPlumSE said:
You don't HAVE to have the resultant be within the kern, though it is probably a good idea for gravity loads.
I believe its always better for the resultant to be within the kern of footing. Isn't it necessary to do a non-linear analysis if the resultant happens to fall outside the kern?

lacyma1, You need to give more details to comment as suggested by HTURKAK.
 
Thank you all for your responses. I've attached one of the more extreme cases. As you can see I have a column 9" away from the corner in both directions and have sized the footing such that the centroid falls just within the kern. Any knowledge or advice would be greatly appreciated. I know it says that there are 2 columns on the pad, but there is only one. That's an error from when I was flexing my spreadsheet.
 
 https://files.engineering.com/getfile.aspx?folder=b5889c78-6f2e-402d-8c84-518bcd668f39&file=Off-Center_Column_Footing.pdf
I'm having a hard time wrapping my head around the idea of where the "kern" is on the diagonal of a footing.

If you look at your 10 kip load you are inducing 27.5 '-k moment about each axis. Using My/I that's 480 psf in each of two directions. So the tip stress is 480x2=960 psf. Your P/A is about 40k/49ft^2 = 816 psf. This says to me that you get net uplift and stress redistribution based on a rather complicated shape under the footing.

I'm not saying it's unsafe (you have 30,000 pounds of concrete for a 10,000 column load) I'm saying the methodology might be suspect. Or maybe my logic is.

There are charts out there for eccentrically loaded rectangular footings. I think I'd have a look at those. I've also modeled these in RISA to get an idea of actual soil stresses.
 
The kern is a geometric property and should be independent of the loading. I was under the impression that tipping was not possible if the resultant of all loads lay within the kern, as the whole footing would be engaged in compression.

I see the logic there, but I guess my rationale is essentially that I'm designing a combined footing where the second column load is the footing itself. If the resultant of the footing weight and the column load lie within the footing's kern, then as I understand it tipping cannot occur. Using a pressure calculation p = P/A + + (Mx/I)*x[sub]1[/sub] + (My/I)*y[sub]1[/sub], it's easy to see that I should get uplift on the opposite corner still. I just don't understand why this happens if my resultant loading is within the kern of my footing.
 
My first impression is that such a massive footing is not economical, but there seems to be something wrong with your calculations (or mine).

The column load is 9.87 factored, so maybe 7.3k unfactored? The footing weight is 49*4*150 = 29,400# = 29.4k

Ftg. A = 49ft^2 I = ab[sup]3[/sup]/12 = 200 ft^4
P = 7.3 + 29.4 = 36.7k; M[sub]x[/sub] = M[sub]y[/sub] = 7.3*(3.5 - 9/12) = 20.1'k
P/A + M[sub]x[/sub]*y/I[sub]x[/sub] + M[sub]y[/sub]*x/I[sub]y[/sub]
= 36.7/49 +- 2*20.1*3.5/200 = 0.749 +- 0.704 = 1.45 or 0.045ksf (Edited)

Your value of 1.5 ksf ultimate soil bearing seems extremely low. Is that the allowable soil bearing?

BA
 
Alrighty,

So thank you all for the input.

Celt83: I looked at your spreadsheet (very cool), I moved the load application point to where my resultant is then recalculated my moments based on the full load of my footing + column load at the resultant. This makes more sense to me rather than applying all of my vertical load in the center of my footing. Though I may have been wrong to decrease my moments. Adjusted file is attached.

BAretired:
I believe your I values are very off since bending should be about the 4' thickness (making I equal to 37.33. My bearing pressure was an assumed value taken from the IBC, this may be allowable rather than ultimate (good catch). Also, it looks like your x and y values for the top right corner went missing.
 
 https://files.engineering.com/getfile.aspx?folder=9e6a6744-e68d-47bc-8e0c-40eab572d2e7&file=05_Rigid_Base_Plate_Analysis_MR.xlsm
lacyman1 said:
I believe your I values are very off since bending should be about the 4' thickness
no, I is based off of the plan dimensions of the foundation, P/A +/- M/S for the bearing pressure is looking at the foundation as if it were a cross-section

lacyman1 said:
I moved the load application point to where my resultant is then recalculated my moments based on the full load of my footing
this should give you the same results as ultimately the sheet computes moments about the center of the rigid plate, so if you placed all of the load at the resultant location then the applied moments should be 0.

My Personal Open Source Structural Applications:

Open Source Structural GitHub Group:
 
Celt83,
Yes, I have an awful time typing and thinking simultaneously. I really should get into the habit of writing it all out before typing it. I edited my post several times.

lacyma1 said:
Using a pressure calculation p = P/A + + (Mx/I)*x1 + (My/I)*y1, it's easy to see that I should get uplift on the opposite corner still. I just don't understand why this happens if my resultant loading is within the kern of my footing.

The moment of inertia to which I was referring was in connection with soil bearing, not bending of the footing. So the correct 'I' value to use is 7*7^3/12 = 200ft^4. I assume that is the 'I' value you mentioned above. There is no uplift on the opposite corner by my calculations.

BA
 
Whoops! Thanks for setting me straight on the moment of inertia! Alrighty, Basically if I set that spreadsheet to sum moments about the resultant, my moments applied will be zero (basic statics) and my footing is still entirely in compression. The Moment of Inertia issue explains why I was getting large negative pressures on the corner of my footing rather than very small positive pressures when I used that equation. Thank you both very much!

When it comes to reinforcement detailing, I'm rather confused. Do you all have any recommendations for how I start?
 
Depending on soil properties and building geometry, a pile or strap footing might be considered in lieu of the massive footing.

With the proposed footing, an orthogonal top mat is required to resist the negative bending moment resulting from the offset column. If you wish to add shrinkage reinforcement, you might consider a mat on the bottom and four sides as well, resulting in a steel cage wrapping the entire footing.



BA
 
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