UU triaxial test interpretation - sample not failed 1

[Baztien]

When a UU result show a curve like the one attached what are the explanation possible?
my suggestion is:
- sample not fully saturated and therefore phi>0

the specimen is low plasticity very silty very sandy CLAY

Is this behaviour on UU test curve frequent for this type of cohesive soil?

 

[fattdad]

one test cannot confirm whether you have phi=0 conditions. Not sure that has anything to do with it.

The sample is clearly mushing under the load as there's no peak strength. this is mostly typical for softer soils.

Failure for these soils should be assigned based on too much strain. I don't have the citation, but I recall that conventional wisdom is to parallel the initial tangent modulus from 2 percent strain and then take failure as the intersection between the stress-strain curve and the parallel line that you draw.

f-d

¡papá gordo ain’t no madre flaca!

 

[dgillette]

Is there a link to a picture? None shows up for me.

 

[Baztien]

Sorry I deleted the first version
here it is with just the graph

 

[Baztien]

Fattdad looking at the stress and the undrained shear strength profile the sample is not soft

 

[dgillette]

I don't know what to make of it (although I have not done many UU tests, and that was a long time ago, back when we trained wooly mammoths to push the Shelby tubes). It's still climbing at 20% strain. The apparent deviator strength of 460 kPa = 67 psi = 9600 psf is rather high, definitely not what I would call soft. For UNCONFINED compression, Terzaghi and Peck show 4000-8000 psf for "very stiff" clay. E at small strain is about 4800 kPa = 700 psi, typical of medium clay.

Is this sample from great depth, >30 m or so, so it would have preconsolidation stress high enough to explain the strength? How did the confining pressure in the test compare with the overburden?

It is possible that the sample is not fully saturated, so application of confining pressure is taken at least in part by effective stress, not all by pore pressure (as with a fully sat'd sample). What happens if you pretend that the test is drained? Assume that the confining pressure is Sigma3' in a drained test and that cohesion is zero, then find the friction angle.

Bon chance!

 

[fattdad]

O.K. I'll retract the "soft" statement. And, I think dgillette makes a good point: Is your confining stress greater, less or equal to the effective vertical stress from the sample depth?

It's definatly not shearing like a "stiff" soil and confining stress can be a factor.

f-d

¡papá gordo ain’t no madre flaca!

 

[Baztien]

Water depth at the site in 130m, the sample is 21m down
so total stress as confining pressure was 1700kPa

The geology is definitely overconsolidated very stiff clay but also interstratified sand & clay

Doing the calculation as drained a find phi=7 for the first curve

on a similar material I have the same curve which reached even higher deviator stress

atterberg is low plasticity clay (100% of the material)
phi calculated as test drained is 14

what do you reckon?

 

[dgillette]

I am puzzled.

Is it the other way around, sample at 130 and water at 21m? If I use 130 m * 2.1 tonne/m^3 * 9.81 m/s^2 - 109 m * 1 tonne/m^3 * 9.81 m/s^2, I get 1600 kPa for effective overburden and 2600 for total overburden. With sample at 21 m and water at 130 m, I get only 430 kPa (effective=total). (Forgive me if I have botched this - I normally work only in ft and lb.)

If you are coming up with such small friction angles assuming that it is drained with c=0, I think that means your first sample is pretty close to saturated. (But if the sample is really 109 m above the water table, it probably would not be saturated. If the water table is 130 m down, you must be working on a hill in a desert. Or, if the sample is 130 m deep, I must wonder why you want to know its strength, tunneling?)

With your first sample, the deviator stress at the maximum strain is only about 1/4 of the effective overburden of 1700. In a CU test, we would expect it to be on the order of 1/2 of the preconsolidation pressure (Su/sigma'c ~ 0.25 or a bit more). The second sample has peak deviator of 2/3 the effective overburden stress, which could fit with it being somewhat OC.

Here is where I just admit that I don't know what's going on with the samples and the very large strains to failure, and I give up.

Regards,
DRG

 

[Baztien]

dgillette

Sorry if I confused you but that's offshore sample which was taken.

Water table is 130m above ground and sample is 21m down

 

[fattdad]

the effective stress at a depth of 21 meters (assuming typical unit density) would be 190 KPa - NOT 1,700 KPa. You have placed the sample in a stress conditon that's much greater than the in-situ overburden stress. In that higher-stress environment, the sample is behaving as "soft" by not showing a peak strength.

Re-run the test at the existing overburden pressure and I bet you see a peak strength.

Just some thoughts.

f-d

¡papá gordo ain’t no madre flaca!

 

[Baztien]

It's a UU test so the cell was loaded to TOTAL stress pressure

I did the calculation that dgilette suggested by pretending the cell pressure (total stress) was an effective stress

Was I meant to ignore the cell pressure applied and do the calculation with a calculated effective stress

FYI the test can't be redone but it's not the problem
I'm just trying to get a better understanding for future reference

 

[dgillette]

I was suggesting that to see if your sample had a low degree of saturation. If you did, much of the applied confining pressure would be carried by effective stress, similar to a drained test, and there would not be big changes in the pore pressure due to shearing (because the air in voids is very compressible compared to water). You would calculate a friction angle that could be as high as the drained value for your clay, perhaps 25. If it was highly saturated, the effective stress would not increase very much with the confining pressure, and you would calculate a very low friction angle, like the 7 degrees that you found. That makes me think that your sample was pretty well saturated.

 

[fattdad]

so you are including the weight of the column of water above the mudline? I just don't get that at all.

Good luck.

f-d

¡papá gordo ain’t no madre flaca!

 

[Baztien]

Dgillette

I have reviewed another test which showed similar curve.
again low plasticity CLAY tested under TOTAL stress.
Water table is 130m above ground so the confining pressure was quite high.
doing the calculation phi=tan-1((sigma1-sigma3)/)sigma1+sigma3))
as if there was no pore pressure built up (the pore pressure is not recorded during the UU test)
I find a friction angle of 25 degree.

My thoughts are that the samples have a dilatant behaviour.

What do you think?

 

[SixDegrees]

Could this be from capillary pressure? The water pressure at 130 m is 1274 KPa, which is on the same order of magnitude was what your measuring from strengths. Perhaps the pore pressure in your test starts out as -1274 KPa (or less depending on how much time has passed to allow drainage). Just a thought.

 

[dgillette]

Baztien - Sorry I didn't respond sooner, but SixDegrees reminded me.

Yes, I think it's probably a sign of dilatant, or at least noncontractive behavior.

On the subject of confining pressure in the test, yes, with 130 m of water the total stress is, strictly speaking,

130 m * 9.81 kPa/m + 21 m * 2.1/m * 9.81 kPa/m = 1710,

but the VERTICAL effective stress is only

1700 - 151 m * 9.81 = 230 kPa

and the HORIZONTAL is something smaller than that, maybe 120 - 140.

If you apply a confining stress (total) of 1700 kPa, can you do that without unintentionally applying horizontal and/or vertical effective stresses greater than what the specimen felt in the ground, thereby consolidating it? Therefore, I'm wondering whether it makes sense to try to match the cell pressure to the total stress at the location of the sample.

I'm not a lab guy and, like I said, it's been a long time since I have done any of this testing myself.

SixDegrees - I don't think you could explain this with capillarity. Capillary rise in fine-grained soils is generally a few m; therefore, I think only a few tens of kPa can be applied by capillarity.

DRG

 

[Baztien]

Hi dgillette and sixdegrees

Many thanks for your replies which lead to further reading (mainly Manual of soil testing - Head 1980, a bit old but that's what I've got to hand)

to answer sixdegrees:
when a sample is taken of the ground it is often that the pore pressure becomes negative due to release of the stresses. But I think that a cell pressure of 1700kPa should recreate a positive pore pressure in the specimen.

However as dgilette mentioned with no back pressure and no pore pressure measurement it is impossible to observe the behaviour and to ensure the full saturation of the sample. Therefore the effective stress state is unknown.
I don't think it could be consolidated by the cell pressure because no drainage was allowed and the time lapse would have been quite small between confinement and compression.

What I'm coming at is that the sample being heavily overconsolidated (OCR>10) and very sandy CLAY probably behaved in a dilatant manner during shearing. With the absence of back pressure it probably created negative pore pressure and increased the effective stress.

Normally with a well saturated sample when applying the cell pressure high enough the pore air becomes dissolved in the pore water.
I understand from that if the sample dilate during shear and pore pressure decrease the air could reappear in the pores and volume change will be made easier.

In conclusion I think the sample behaved in a partially undrained manner during the UU test.

Thanks for your input,
please tell me if you think it isn't correct.
It helps me to put it down to see if it make sense as I write it.