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Cohesion c vs c' from Triaxial CU with pore pressure

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detate

Geotechnical
Dec 13, 2002
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Hello, When i plot my total stress Mohr's circles for an overconsilidated fat clay (LL=81, PI=58) I get a decent looking diagram with c=3.21 psi and phi = 8.2°. However, when I plot the effective stress Mohr's circles and plot the c'-phi' line tangent to the circles I get c'=10.28 psi and phi' = 8.2°. I am not sure how effective cohesion can be so much larger (3x +)than the total cohesion, is that usual / possible? Should I consider a curved envelope? If so, what effective cohesion intercept should be used - the total stress value? Perhaps I should plot the c'-phi' line using the total stress cohesion and an average fit tangent line?

Doug Tate
southeast USA
 
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take a look at your stress paths.

To really vet the results, we'd need to see more of the data.

f-d

¡papá gordo ain’t no madre flaca!
 
First thing you need to do is replot the data at a natural scale. The vertical exageration makes it very difficult to glance at the data and make any sense of it. Plot it by hand if you need to.

Initial comments are that the stress paths look odd, I suggest you check all the data. Were these single stage or multi-stage tests?

Second, why did you run the tests at such high normal stresses?

Mike Lambert
 
I'm not sure what you are looking at, but the plots are 1:1 vertical to horizontal on my screen. They are single stage tests. I didn't get to assign the tests, I was asked to review the result, so I don't know why the engineer chose the high normal stress (the samples were taken at 15' deep), perhaps to aid in saturation? Below is a link to the output data for the effective stress curves. The stress strain curves look pretty normal.

In any case, It bothers me that the effective cohesion appears to plot so much higher than the total cohesion, I'm not sure I beleve it. I am strongly tempted to believe that the effective cohesion can't be higher that the total cohesion. I can see a believable envelope line that meets that criteria at c`=3.21 psi and a believable phi` angle of about 16°.

Doug Tate
southeast USA
 
 http://files.engineering.com/getfile.aspx?folder=0cf68e07-57e1-47d2-a062-ee61c56e0917&file=Triaxial_Data.pdf
yes, it appears the aspect on the pdf files is a bit off (at least on my monitor).

I have no idea what the stress strain curves look like. I do know that the effective stress paths do not seem to justify the line that seems to suggest your angle alpha. Alpha is usually coincident with the stress path. Please don't use stress path data to graphically measure phi. You have to actually convert alpha to phi (it's a geometric conversion).

So, I'd start with the stress-strain curves and ask yourself, what defines failure. I mean if one failure is at 5 percent strain and the other is at 15 percent strain, is that a meaningful failure description?

f-d

¡papá gordo ain’t no madre flaca!
 
I'll take a look at the data you posted today.

As for the high normal stresses, they would have been picked high to help with saturation; if anything the high normal stresses hurt your ability to saturate the samples due to cell limitations on the total pressure.

One more line of questions:
are all the samples from the same boring?
are all the samples from the same depth?
are all the samples the same material?

Mike Lambert
 
F-D - all speciments were taken to 15% strain, which defines failure. i.e., there was no peak stress at lower strain. We did indeed compute phi and c based upon alpha and a from the stress path (phi=arcsin(tan(alpha)), c=a/cos(alpha)) but the pore pressure shift seem to make the lower two specimens coincide for the effective case.

Mike- all the specimens are from the same boring and same Shelby tube, but slightly different depths (by a few 1/10 of feet). They appear (visually) to be the same soil type.

I could be happy with a c'=0, phi' = 20°, which is very nearly parallel with the upper part of the stress paths.

Doug Tate
southeast USA
 
. . . just another thing. . .

Any summary table that shows saturation greater than 100 percent is suspect. If we know it can't occur, it shouldn't be calculated. Some other assumption is incorrect.

f-d

¡papá gordo ain’t no madre flaca!
 
I agree with fattdad, you need to figure out what data is correct and what is not since saturation can't be over 100%.

I looked at the effective stress data and using stress paths and have the following comments/observations:
* all of the samples are over consolidated at the tested stresses
* Red data indicates a phi of ~27.8 degrees with an effective cohesion of 2.2 psi,
* Blue data indicates a phi of ~20.1 degrees with an effective cohesion of 3.8 psi, and
* Green data indicates a phi of ~18.2 degrees with a negative effective cohesion.

Either the soil is different, or there was a problem with the tests. The effective stress paths for the Bule and Green data show a loss of normal stress with no or little increase in shear stress at the start of shearing; looks suspect to me.

If I had this data, I would be looking for additional tests to figure out what is really going on. I can draw no further conclusions from this data.

Lastly, Lambe and Whitman (amoung others) would predict somewhere around 24 degrees for this material based on PI.

Mike Lambert
 
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