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Long term and short term shear strength from CU test 2

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Ilight

Civil/Environmental
Dec 2, 2018
10
Hey everyone:

For a typical CU triaxial test, I can't seem to get my head around why the effective failure envelope corresponds to the long-term drained effect for a clay sample.

Let's say the confining stress is Sigma 3 and the deviatoric stress is DeltaSigma, hence total stress sigma1 = sigma3 + DeltaSigma and minor principle stress sigma3 stays constant. Then to develop the effective stress Mohr circle, excess pore pressure measured from the shearing stage would need to be subtracted from total stress.

Now from my understanding, excess pore pressure will dissipate completely in the long term, hence u=0, if that's the case, shouldn't the total stress mohr circle represent the long-term drained effect? I.e: effective sigma 1 = total sigma 1 - u, with u =0.

I feel I've missed something trivial yet critical, but can't seem to grasp

Also for the drained shear strength obtained from CD and CU test (T'=c'+a'tan@'), is the a' value used in the equation similar for both tests (assuming same confining stress)? Because for CU test, the effective stress Mohr circle is offset by pore pressure U from the total stress Mohr circle.

Pretty standard stuff, help would be much appreciated xD
 
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If a natural clay deposit is going to fail it will fail undrained. If you want to use effective stress parameters you would need to use the pore pressure during failure and not the steady state pore pressure before failure, otherwise use undrained strengths.

Also Clay material has a very low permeability therefore a true drained failure requires months, years, or decades to drain after construction to reach the long term and the same amount of time to fail. But following construction if the clay didn't fail it gains strength as it drains back to the tau over sigma ratio of 0.22 with the increased stress.

If your dealing with compacted clay that is a different story.

The first stage of site investigation is desktop and it informs the engineer of the anticipated subsurface conditions. By precluding the site investigation the design engineer cannot accept any responsibility for providing a safe and economical design.
 
The total pore-water pressure, u, is calculated as follows:

u = u0 + Δu

Where u0 is the hydrostatic (at rest) pore-water pressure and Δu is the excess pore-water pressure. In a triaxial test, u0 will be equal to the back pressure, which is not always equal to the in situ hydrostatic pore-water pressure (this may need to vary from in situ conditions due to sample disturbance effects, amongst other issues).

Consolidated triaxial specimens are consolidated until Δu = 0 (i.e. full pore-water pressure dissipation), therefore u = u0 = applied back pressure.

Given the above, you will see that your definition is not correct since u is not equal to 0, but instead equal to u0.
 
Thankyou for the reply GeoEnvGuy

So by undrained failure, you're referring to the short term state when the overburden pressure is taken by the pore pressure right? Which is what CU test resembles, since in the shearing stage the drains are closed.

But I don't quite understand where the long terms effect comes into play for the CU test, because the drains are kept closed, how is pore pressure able to dissipate. It occurs to me that the effective stress mohr circle actually resembles the short term effect and not the long term, because the effective stress doesn't equal to total stress, due to excess pore pressure being generated.

However I've seen in various resources the effective Mohr circle is equivalent to the one generated from the CD test without much explanation. At the moment all I could come up with is that because the effective strength parameters c' & phi' can be obtained, so they can resemble the long term drained failure....

CU_test_vbxkmf.png


Thanks in advance
 
Hi LRJ

So for the undrained shearing stage in CU test, it would mean that the pore pressure at failure actually equals to
U=U0 + U(from applied vertical load)?

But that counts as short term effect right? I don't quite understand how it suppose to resemble long term effect (if you could refer to the mohr circle picture above), since the excess pore pressure would have dissipated [ponder]

Thankyou :)
 
Ilight,

In a CU test the pore pressure can be measured; the complete name is a CU test with pore pressure measurement. If the test is ran without measuring the pore pressure, then only the undrained strength can be determined. If the test is ran with pore pressure measurement, then both the undrained and drained strengths can be determined as discussed by LRJ.

GeoEnvGuy - your statement "If a natural clay deposit is going to fail it will fail undrained." is not always true. Many clay slopes fail under drained conditions. When dealing with clay or plastic silt deposits it is important to check both drained and undrained failure conditions.

Mike Lambert
 
I'm just writing to say that I disagree with GeoEnvGuy.

The softening of natural clay slopes (i.e., from temporal variation in moisture, weather, temperature, etc.) can and does reduce the effective shear strength of natural clay slopes from peak strength to fully-softened strength. Those failures are clearly drained failures from the tedious reduction in friction angle.

f-d

ípapß gordo ainÆt no madre flaca!
 
Ilight said:
Hi LRJ

So for the undrained shearing stage in CU test, it would mean that the pore pressure at failure actually equals to
U=U0 + U(from applied vertical load)?
The undrained shear strength in a CU test would be the same as a test without the measurement of pore-water pressure, since the deviator stress is the same in total stress and effective stress space.

The Δu is dependent on the applied load as well as the rate of drainage (permeability). It is therefore time (rate) dependent.

But that counts as short term effect right? I don't quite understand how it suppose to resemble long term effect (if you could refer to the mohr circle picture above), since the excess pore pressure would have dissipated ponder

Thankyou :)

You should avoid thinking that total stress = undrained = short-term loading. Think of it in a critical state framework: constant volume (undrained) or constant stress (drained) or somewhere between the two (partially drained). I'd also recommend looking up real stress paths rather than Mohr circles, as this would help to understand what is actually happening during the test.

Also, a total stress friction angle is pretty much a meaningless parameter, unless you are in the world of partial saturation.
 
Just wanted to comment on Geopavetraffic and fattdad's point regarding drained failures of over consolidated clay and yes they do happen. In my experience which is predominantly on slightly over consolidated clays, construction always loads up clay into the normally consolidated range which exceeds the pre-consolidation pressure and generates positive pore pressures, which is illustrated by the figures below regarding the effective stress path and total stress path.

But that may not be the case here which may consider the case where the construction may be less than the pre-consolidation pressure. Which is why i endorse the previous comments of checking both drained and undrained failure conditions, with drained failures from the tedious reduction in friction angle.

Ilight I believe Abramsons book on slope stability does endorse the long-term condition tested with the CD test or the CU with pore pressure measurements.

clay_OC_NC_ss5wgt.jpg


The first stage of site investigation is desktop and it informs the engineer of the anticipated subsurface conditions. By precluding the site investigation the design engineer cannot accept any responsibility for providing a safe and economical design.
 
I don't want to be confused any more. It's the obligation of the responsible carge engineer to know the local geology, do the appropriate strength testing and consider modes of failure. I don't need stress paths to make my claim that natural clays can fail in either undrained or drained behavior. The earlier claim that it will always be undrained failure needed to be addressed.

Beyond all that I understand all the text book references.

f-d

ípapß gordo ainÆt no madre flaca!
 
I'm with fattdad on this one. Do what you want.

However, if you only do undrained/total stress analysis; you will have a slope fail on you eventually. Best of luck.

Mike Lambert
 
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