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Spring Stifness 1

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kieran1

Structural
Feb 27, 2002
178
Sorry if this has been covered before.

Can anyone point out a rule of thumb for estimating the soil spring stiffness from the bearing capacity. It has been suggested to me a value of 40 times the bearing capacity divided by a factor of safety. (KN/m3 units)

This is for modelling a raft foundation

Kieran
 
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I don't know of any general rule of thumb like the kind you're referring to.

Generally speaking we get our spring stiffnesses based on the Modulus of Subgrade Reaction provided by the Geotech in his / her report.

This is an elastic spring stiffness which the geotechs always seems to be a bit reluctant to provide. Possibly because it pains them to approximate the highly non-linear soil behavior with simple elastic stiffness. But, if you beg and plead enough (and write it into their contract) then they should eventually give it to you.

If you do not have a geotech report for the project then you might come up with a rough ballpark value based on the soil type. The Bowles book (Foundation Analysis and Design) has a table which lists different ranges for the subgrade modulus based on whether the soil is sandy, silty, dense, or such.

The ranges are pretty wide, so it really is better to get your numbers from a geotech report. But, at least this book might give you a basis for enveloping the spring stiffnesses.
 
I don't know any correlation between bearing pressure and subgrade modulus. I'm not sure I'd trust one if I found it either. Bearing pressure is dependent on soil properties that extend to depths that are dependent on the size of the footing and the pressure that the footing exerts. Also, bearing pressure is typically derived based on servicability rather than ultimate strength (or a factor of ultimate strength). I mean what value is it to have a bearing pressure that doesn't "fail" but allows too much settlement.

I have often heard the term "raft" foundation, but it warrants some measure of clarification. Are you using a thick slab to hold some floor load (i.e., 500 psf) and at the edges or interior areas, thickening or strengthing the "raft" to allow for point or line loads? Will you really construct a monolithic slab that does it all?

If you are really designing a monolithic slab you will need to know both bearing pressure AND subgrade modulus. You'll have to apply the bearing pressure to the portion of the slab that has limited, but concentrated loads derived from the columns and walls. I'm not enough of a structural engineer to know how you'll do that (computer model, tributary area or such). You may have some chicken-or-egg issues with this - don't know.

What I do when I provide subgrade modulus to structural engieers is I run a CBR (California Bearing Ratio) test and correlate the subgrade modulus to CBR value. This is a meaningful correlation, as the CBR is a point load applied in a small but concentrated area - one that mobilizes the strength of the soil in the near surface. I think this models the behavoir of a subgrade as it relates to slab-on-grade design.

Hope I've helped some. The most widely used value in Central Virginia is 125 pci and I almost never see a value greater than 200 pci. Don't know how to convert this to metric units, 'cause I'm anti-metric - ha.

f-d

¡papá gordo ain’t no madre flaca!
 
Thanks guys

What I’m try to analyze is essentially a combined base but the Point loads can occur anywhere on the base. The loads are not high and the ground is relatively good. I’m trying to minimize the top steel for the hogging moments induced between point loads. I would normally calculate these moments like an inverted flat slab with a distributed load equal to the bearing pressure and spanning between column locations. However in this case the top steel required would be excessive.

What I’ve done is to analyze the base with a number of different spring stiffness to see the effects this has on the slab. I’ve settled on a stiffness which ensures that the spring support reactions did not exceed the allowable bearing capacity.

Is this a fair assumption??


Kieran
 
What spring support stiffness did you end up back-calculating? If it's much more than 200 pci I'd say you're on a flyer and need to reconsider what you're doing.

Then again, I don't know the soil. Do you have a geotechnical engineering study that shows soil layering and index soil tests?

f-d

¡papá gordo ain’t no madre flaca!
 
I ended up with a spring stiffness of 24000KN/m3 (88.32 PCI) The soil is a medium dense silty clay.

Unfortunately the geotechnical guys are reluctant to help out

Kieran
 
I'd say you're still coloring inside the lines then.

f-d

¡papá gordo ain’t no madre flaca!
 
The geotechs in my area always seem to give 100 pci. Something tells me they are doing a rigor analysis.
 
Bowles gives a range of 15000 - 30000 kN/m3 for a stiff clay, 30000 - 60000 for very stiff. (medium dense silty clay is not a description I recognise)

Adjusting the spring stiffness to ensure the bearing capacity is not exceeded is NOT the correct procedure. That is similar to adjusting the bearing capacity so that it isn't exceeded.
I would use a range of stiffnesses based on soil description, and adjust the raft size/thickness to ensure bearing capacity is not exceeded. I ignore any local minor infringement of the bearing capacity.
 
I think apsix is correct on all counts. That said, 88 pci is quite a low value and well within the typical range for most soils.

Just a BIG caution: If somebody gives you a subgrade modulus of 200 pci, NEVER go through a similar calculation to guess at bearing capacity. You may find yourself in court with a building that's lopsided!

f-d

¡papá gordo ain’t no madre flaca!
 
Thanks guys. Apsix the bearing capacity was never going to be exceeded due to the size of base and loads acting on it. I was just using some different spring stiffness to see the impact on the base, But thanks again for your help

Kieran
 
kieran1,

If you look at Bowles Table 9-1, modulus subgrade, kv (kN/m3)
approx equal to 240 Cu (kPa).
Allowable bearing pressure, qa (kPa) = Nc Cu / SF = (5.14 /3) Cu

Combine the two equations we get
kv (kN/m3) = {(240 x 3)/5.14 } qa = 140 qa (kPa)

Conclusion : vertical mudolus subgrade reaction (kN/m3) approx. equal to 140 x allowable bearing pressure
 
Thanks irawanfirmansyah thats exactly what I was looking for.

Kieran
 
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