OCTAGON FOOTING BASE PRESSURE

[thanhtang]

I have an 1' thick existing OCTAGON FOOTING down below the grade. On top of the octagon footing is an smaller existing octagon pedestal. The pedestal is 6' thick and 1' above the grade.

The existing foundation is currently supporting an old vessel. The old vessel will be replaced with a new vessel of the same kind. The existing foundation will be reused if possible. My task is to check if the EXISTING FOUNDATION is adequate to support the new vessel under new Codes.

There are vertical load, moment, and shear on top of the pedestal. I can calculate the total moment M and vertical load P at the bottom of the footing.

I need to CALCULATE THE FOOTING BASE PRESSURE to compare against soil bearing capacity provided from Geotechnical Report. I also need to CALCULATE STABILITY RATIO.

I know how to get it if the footing is rectangle, square, or round, but not octagon. My text books did not help.

Please advise me with approach, formulas, etc. for an OCTAGON FOOTING, not round footing.

Footing and pedestal size or details and applied load orientations are not necessary to be posted though.

Thanks

PS: I have posted this in Soil Mechanic Forum, but I think here is more appropriate.

 

[csd72]

modelling it as round will be a good approximation.

csd

 

[Lion06]

I would find the section modulus of the footing (this will require hand calcs, but should be pretty straightforward), then apply your P/A-M/S to verify you have full bearing.
If you have full bearing, use P/A+M/S to get your maximum bearing pressure.
If you don't have full bearing, you will have to draw a free-body of your footing with the eccentric axial load on the top and the soil pressure on the bottom. When figuring out the max pressure by using the total soil load, be sure to account for your footing base not being rectangular. This will require some geometry and algebra skills, but I have done something similar with shearwall footings where the footing went from a wall footing to a column footing at the end and I needed to account for both.
Just hope that you have full bearing, then the exercise is easy!!

 

[thanhtang]

csd72,
Thanks for the comment. I would, but octagon is preferable and very popular in PetroChem Industry.

StructuralEIT,
Thanks for your reply
Yes, if full bearing, then P/A ± M/S is enough.
But I need to analyze it in some load cases such as empty vessel plus full wind of 130 mph that may cause the base is not in full bearing.

I am not quite understand your commnent in the "... don't have full bearing"" case.
The concrete, soil, and vessel dead loads will distribute evently. The moment due to wind on vessel will cause max and min pressure.
First question is when (e=?) the min pressure is zero at one edge of the footing. Second question is if pressure equal zero at somewhere in the base, not at the edge, where is the location of the zero pressure and how to get the max pressure ?
The zero pressure location may be in a range that cause the foundation overturns. So how to calculate the stability ratio ?

 

[Lion06]

calc the "resisting" moment of all the loads you can count on to keep the footing from overturning - This would include the DL of the empty vessel and the self-weight of the ftg and any other loads that will ALWAYS be there - about the toe of the footing. Compare that to the overturning moment from the wind. I always include the base shear*ftg thickness in this value.
If you do not have full bearing, then you will have a geometry and algebra exercise on your hands.
What I have done is this: Draw a free-body of the footing. Show the location of the eccentric vertical load on top of the ftg with the distance from the centerline of the ftg. From statics, your soil reaction has to be at the same location and have the same magnitude. You won't know how far in you have zero pressure until you do the geometry and algebra exercise. I would have two expressions for the soil pressure, one would be for the tip of the ftg back to where it becomes a "square" and one for where it is a "square". You will end up with a quadratic equation that will tell you the distance of the pressure diagram and the maximum value. You may have to adjust this slightly to fit your circumstances, but this should work for you.

 

[SlideRuleEra]

As a first try, suggest that you look at the conditions the same way they would have been addressed in pre-computer days - bound the problem by analyzing the simpler best case and the worst case.

Think of the octagonal footing as being "less than" a square (length of side of square = distance across flats of the octagon) and "more than" a circle (diameter of circle = distance across the flats of the octagon).

Compute results for the square and the circle, if they both pass, then the octagon is ok.

If both the square and circle fail, the octagon is not adequate.

If the square passes, and the circle fails, the octagon may or may not be ok. Then you could consider more refined calculations... but, for an industrial footing, it is probably best to replace one that has marginal capacity for the new load.

http://www.SlideRuleEra.net

 

[csd72]

thanhtang,

I was basically trying to say what Sliderule said. But personally I would just analyse it as a circle of the same width if the circle works then so will the octagon (as the circle will have less area).

I can understand the use of the octagon as formwork will be much cheaper.

csd

 

[thanhtang]

StructuralEIT,
Thanks. I understand your first paragraph, but not the second paragraph.

I am very interested in :
1) knowing "do the geometry and algebra exercise"
2) knowing how to "end up with a quadratic equation that will tell you the distance of the pressure diagram and the maximum value"
3) knowing how "to adjust this slightly to fit your circumstances"

It sounds like I ask you to do my work, but I cannot get anything practically from your comments. I would be appriciate if some formulas or more specific and constructive info. posted.

SlideRuleEra,
Logical thoughts. Thanks for advise.
The foundation was designed for a 30' high x 66" dia. vessel with wind speed of 100 mph around more than 20 years ago.
The Client is gonna replace the old vessel with a new one of the same kind (approximate same size) with current wind speed criteria of 130 mph. With wind pressure formula contains V^2, the new wind speed of 130 mph will impose 69% more load than the old wind speed of 100 mph. So the existing foundation may have marginal capacity under new load (with the hope that people often over design a little bit for foundation). That's why I insist to analyze it as OCTAGON fdn.
Replace the old foundation with a new one is not feasible. It requries at least 2 weeks to do a new fdn.. It costs approximate $1 mil. in profit to shut down the vessel. No one thinks the Client will be happy to lose millions in profit just to replace a vessel.

 

[csd72]

One place to start:

http://www.structuremag.org/archives/2004/june/Engineers Notebook.pdf

 

[WillisV]

Part 2 of that same article actually discusses circular footings (and says to conservatively treat octogonal footings using the circumscribed circular dimension):

http://www.structuremag.org/archives/2005/June 2005/Circular-Footings-by-Allen-June-05.pdf

 

[DWHA]

Thanhtang
The way I see the situation is that you are looking for an exact answer in an unexact from of engineering. Soil mechanics is not an exact science, and looking for an exact answer is not (in my opinion) the best way to approach the situation. I understand that spending time to do this correctly is important to the client to prevent the shutdown of the machine.
You have an existing foundation that either has been adequate or has not been adequate to resist the loads for the past 20 years. That is a better answer to your question that using exact formulas on approximate soil parameters. Cause that is what soil parameters are, approximate. End the end your exact solution for an octagon will be no more accurate than assumtions you made with the soil.

 

[Lion06]

thanhtang-
First, let me apologize and say that I meant a system of equations, not a quadratic equation.
1&2)
you have the eccentric load at the top of the footing and you know exactly where it acts.
draw the triangular soil pressure diagram under the footing. Start with qmax at the right of the diagram. Let's say that the distance from that point to where the footing turns into a "square" is 10'. Let's say the entire length of the pressure diagram is x'. Now we know the slope of the pressure diagram and can say that the pressure at the point where it turns into a "Square" is (qmax-10qmax/x). Currently, we are looking at an elevation of the footing to view these pressure diagrams. Let's say we look at the bottom of the footing and divide it into (4) parts. (2) triangular pieces, and (2) rectangular pieces. Use the areas of these triangles and rectangles along with the pressure diagrams to find the load on each of the (4) pieces. Knowing that this load must equal the eccentric load applied from above will give you the first equation involving "qmax" and "x".
The second equation will come from the location of the resultant of the (4) loads. Form an equation (using moments about a point) knowing that the resultant of all (4) loads must act at the same distance from the centroid as the eccentric load above.
Now you have two equations and two unknowns. Now it is all algebra.

3) You can use the same principles but apply it to any specifice case, maybe you have a footing that is rectangular but gets wider near the end or maybe you have this octagonal footing. The geometry is different, but the idea is the same.

 

[thanhtang]

Thanks to all. All comments help me in certain way.

The analysis with Octagon shape was done last weekend . I came up with the Net Allowable Bearing of 3 ksf. The Geotechnical Report is not submitted to us yet, but it is likely that the Net Allowable Soil Bearing capacity will be 3 ksf which is just in line. (I have 3ksf on other Geo. Report performed in the vincity of the investigated location)
You can see it is extremly marginal.

Again, I appreciate all of your helps.