EIT needs a little help from the more experienced on how to design for a combined footing supporting two identically loaded exhaust stacks that aligned in one direction and are 10'-0" o.c. in the other. The loading provided by the stack manufacturer has a 50k dead/live combination and nearly 2000 k-ft of moment at the base. This moment seemed unusually large, but I have verified the design with the manufacturer. I am limited to 22'-0" of length and the width/depth do not have limitation (within reason). I have designed combined footings with axial load before but I'm at a loss on how to design for the moment at the base. Any help would be greatly appreciated.
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Combined Footing With Axial Load & Moment
- Thread starter YoungMind
- Start date
JedClampett
Structural
You have a real problem. The eccentricity is roughly 40 ft. without the foundation weight. "Foundation Engineering" by Peck, Hanson and Thornburn will give you a method to figure out soil pressures, but you'll have to get the eccentricity within the foundation and preferably within the middle third. The solution is to either provide enough bulk in the foundation that its self weight modifies the eccentricity or use piles or piers. "Structural Enginering Handbook" by Gaylord and Gaylord gives some examples of stack designs.
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- #3
Thanks for your help, JedClampett.
By common sense, I seemingly understand the concept of increasing my footing size to pull back the eccentricity. I am at a loss, though, on how to make this relation in my calculations (increase in ftg size vs decrease in eccentricity). Any suggestions?
By common sense, I seemingly understand the concept of increasing my footing size to pull back the eccentricity. I am at a loss, though, on how to make this relation in my calculations (increase in ftg size vs decrease in eccentricity). Any suggestions?
JedClampett
Structural
What you have to do is to increase the weight of the foundation to decrease the eccentricity. If your length and/or width are restricted, you'll have to increase the thickness.
YoungMind.
The eccentricity that you have to worry about is the eccentricity of the total load on your foundation, including the footing selfweight.
Since the applied moment will remain constant, increasing the total vertical load will result in a reduced eccentricity.
As JeddClampett has suggested, it is preferable to keep the resultant load within the 'middle third' of the base. Adhering to that rule would mean that your eccentricity would need to be reduced to 22/6 = 3.7 ft. With an applied moment of 4000 kip.ft, your total load would need to be no less than 1080 kips leading to a footing weight of 1000 kips or so.
Personally, I would not attempt to stay within the middle third, in view of the immense footing that would result.
The other approach is to accept some uplift at the 'rear' of the footing under peak moment conditions, and aim for a larger eccentricity. Just how large is acceptable depends entirely on the permissible bearing pressure under the footing.
As a start point, if you were to assume a final eccentricity of 6 ft, the total load would come down to 4000/6 = 666k, or 566k in the footing itself. Assuming a 22 ft square base, that would work out at 8 ft. thick.
With a linear distribution of bearing pressure and resultant 5 ft from an edge, your bearing pressure diagram would be a wedge, 15 ft long, 22 ft. wide, with maximum pressure = 2*666/(15*22) = 4 kips/sq.ft.
If you can accept more than 4 kips/sq.ft. then you could use a larger eccentricity and a lighter base. A few trials should quickly get you to a reasonable design.
Good luck.
The eccentricity that you have to worry about is the eccentricity of the total load on your foundation, including the footing selfweight.
Since the applied moment will remain constant, increasing the total vertical load will result in a reduced eccentricity.
As JeddClampett has suggested, it is preferable to keep the resultant load within the 'middle third' of the base. Adhering to that rule would mean that your eccentricity would need to be reduced to 22/6 = 3.7 ft. With an applied moment of 4000 kip.ft, your total load would need to be no less than 1080 kips leading to a footing weight of 1000 kips or so.
Personally, I would not attempt to stay within the middle third, in view of the immense footing that would result.
The other approach is to accept some uplift at the 'rear' of the footing under peak moment conditions, and aim for a larger eccentricity. Just how large is acceptable depends entirely on the permissible bearing pressure under the footing.
As a start point, if you were to assume a final eccentricity of 6 ft, the total load would come down to 4000/6 = 666k, or 566k in the footing itself. Assuming a 22 ft square base, that would work out at 8 ft. thick.
With a linear distribution of bearing pressure and resultant 5 ft from an edge, your bearing pressure diagram would be a wedge, 15 ft long, 22 ft. wide, with maximum pressure = 2*666/(15*22) = 4 kips/sq.ft.
If you can accept more than 4 kips/sq.ft. then you could use a larger eccentricity and a lighter base. A few trials should quickly get you to a reasonable design.
Good luck.
- Thread starter
- #6
austim,
My apologies for posting in two locations. I am new to the forum and did not find the foundation engineering forum until I had already posted in this one.
I understand the "middle-third" concept (Kern) to prevent the stress distribution from transfering into tension. But, I noticed that you doubled the applied moment to reach 4000 k-ft. I wasn't really sure if I could assume the moments from each stack (2000 k-ft each) to react as one moment at the center of the footing. Is that what you were doing? If so, can you tell me how that works. Please excuse my ignorance. I'm learning, though.
My apologies for posting in two locations. I am new to the forum and did not find the foundation engineering forum until I had already posted in this one.
I understand the "middle-third" concept (Kern) to prevent the stress distribution from transfering into tension. But, I noticed that you doubled the applied moment to reach 4000 k-ft. I wasn't really sure if I could assume the moments from each stack (2000 k-ft each) to react as one moment at the center of the footing. Is that what you were doing? If so, can you tell me how that works. Please excuse my ignorance. I'm learning, though.
- Thread starter
- #7
JedClampett
Structural
When the eccentricity is within the middle third, you are using P/a + Mc/I. However, when you are not, you have to account for the fact that the soil cannot take tension. That is why it's helpful to go to a textbook that has all that derived for you.
YoungMind,
No apology need for learners. I should warn you that, if all goes well, you'll still be learning when you are 70 or over.
Yes, the total moment on the base is 4000. Assuming that your two stacks are identical, then each has a resultant lateral load F at a height H, where FH = 2000. So the total horizontal load is 2F, also acting at H.
Jedd has said all that is needed about P/A+/- My/I.
No apology need for learners. I should warn you that, if all goes well, you'll still be learning when you are 70 or over.
Yes, the total moment on the base is 4000. Assuming that your two stacks are identical, then each has a resultant lateral load F at a height H, where FH = 2000. So the total horizontal load is 2F, also acting at H.
Jedd has said all that is needed about P/A+/- My/I.
If you are planning for a rectangular foundation for the stacks, your foundation is not subjected to just uniaxial moment. since the moment is due to wind, whose direction may vary, you have to consider a case of the wind force on the stack along the diagonal of the rectangle, which introduces biaxial eccentricity on the footing. so, your soil pressure has to be calculated from the formula
p(max/min.) = P/A +/- Mx/Zx +/- My/Zy
Where, Mx and My are components of the wind moment (4000 k-ft. in your case, applied diagonally)along x and y directions. Each of these moments will be smaller than M but the resultant pressures can be more than the uniaxial case.
However, if you choose a circular footing for the problem, due to the axisymmetric nature, you can design the foundation for uniaxial moment.
p(max/min.) = P/A +/- Mx/Zx +/- My/Zy
Where, Mx and My are components of the wind moment (4000 k-ft. in your case, applied diagonally)along x and y directions. Each of these moments will be smaller than M but the resultant pressures can be more than the uniaxial case.
However, if you choose a circular footing for the problem, due to the axisymmetric nature, you can design the foundation for uniaxial moment.
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