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Fully Softened Shear Strength 2

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SDUT

Civil/Environmental
Mar 18, 2021
8
Hello

For a slope stability project I am involved in, we ran CU tests with pore pressure. We also ran hydrometer tests for the same sample. The drained angle of friction i am getting from the CU envelope is smaller than the drained fully softened angles that i get using the correlation between Fully softened angle, liquid limit and Clay fraction (Stark et al.). Am I missing something here? Should not the angle obtained from CU envelope (peak angle) be larger than the fully softened angle? Any insight will be appreciated.

Thanks
 
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Have you checked what the standard deviation is for the correlation? Wouldn't surprise me if one SD with a correlation between liquid limit and clay fraction gave you a spread of 5-7 degrees (eg 15-25).
 
Which phi angle are you using from Stark? Depending on the specific correlation you are using, those give you three of four phi angles. Additionally, are you sure that the CU test is a good test? I ask, because about half of the test results I see provided are just wrong. It is a difficult test to run correctly.

Mike Lambert
 
Mike, I am using the correlations given in 2014 ASCE paper by stark. The correlations are between LL and fully softened angles for different clay fractions. The paper is
Gamez, J. and Stark, T.D., “Fully Softened Shear Strength at Low Stresses for Levee and Embankment Design”, ASCE Journal of Geotechnical and Geoenvironmental Engineering, June. 2014, pp. 06014010-1-06014010-6.
Some of the angles from CU lab tests are quite low. I may have to ask the lab to rerun.

Geotechguy1, i have not checked the SD of the correlations yet. will check. I suspect it might be a case of erroneous lab test though.
 
So you have completed consolidated undrained lab tests and you want to determine the fully softened friction angle? This makes no sense to me why you are getting lower values and not doing direct shear or ring shear as Stark does in illinois.

 
Perhaps sample disturbance is affecting the TX CU results? Is your clay medium stiff/stiff or soft? Drained strengths may be more suitable for overconsolidated medium stiff/stiff clays. Do you have a drained cohesion reported in the test (this is typically assumed as zero but perhaps your test has some c' some it may result in a low phi')?


For Stark's FSS calculations, you can download a spreadsheet from Professor's Stark website:


You will notice that the strength envelope is not linear as it depends on the effective normal stress.
 
I ask again, which phi angle from the correlations are you using? Since the fully softened strength is non-linear; there is a different phi angle between zero and each point on the fully softened curve.

Mike Lambert
 
Mike,

I don't fully follow what you are referring by "which phi angle". There are three equations relating fully softened angle with liquid limit and clay fraction. I used the relevant equation based on LL and clay fraction. Am i missing something?

Okiryu,

I downloaded the spreadsheet and it gives me the same results. The sheet essentially incorporates the equations that are in Stark's paper. Samples are medium stiff to very stiff. There are drained cohesions reported. Should you get zero cohesion from the tests? I have assumed zero drained cohesion for my analyses though.

Thanks
 
SDUT,

Stark's equations give you a shear stress at a specific normal stress. Depending on the equation you are using, you get either 3 or 4 pairs of shear and normal stress. If you plot those points, they do not form a straight line. To get a phi angle you must have a straight line. So my question to you is how are you drawing the straight line that gives you a phi angle?

Options to draw a straight line include:

1. Best fit between the 3 or 4 points. If you do this, you will end up with a phi angle and a cohesive intersecpt.
2. Straight line with zero intersecpt through one of the other 3 or 4 points. Each of the points will give you a different phi angle.

Mike Lambert
 
For drained analysis, cohesion intercept is normally assumed as zero (option 2 from Mike's post above).
 
I'm seconding Geopave's post. Fully softened strengths are a non-linear curve with a cohesion of zero. There are ways to develop a formula based on the normal/shear data that can then be used to calculate an equivalent phi angle.

Check out Lade's 2010 Paper I've attached. Castellano's 2014 Thesis is also probably the current state of the art for Fully Softened Strength [URL unfurl="true"]https://vtechworks.lib.vt.edu/bitstream/handle/10919/56631/Castellanos_BA_D_2014.pdf[/url]
 
I also wrote a paper on the use of FSS in slope stability analyses for my Masters. Here's the section on applying different strength selections. I hope this helps
 
 https://files.engineering.com/getfile.aspx?folder=25b01bf7-5f0a-4712-906a-c5eaaa7f79f6&file=Pages_from_Best_Practices_for_Designing_and_Constructing_with_Clays_successtible_to_Fully_Softened_Strengths_-_revised05152020.pdf
just run a normally consolidated DDS with the sample hydrated to the liquid limit. Peak is FSS, residual is residual.

Not to take away from Tim's correlations; however. But, if you have a lab, run the test and record the value. I don't believe you will get there with a CU-bar or a Shelby-tube sample.

So, mush the sample up with DI to the LL and run a DDS. (give those incrementation steps time to consolidate; however!)

f-d

ípapß gordo ainÆt no madre flaca!
 
Agree with f-d. The peak strength of specimens remolded at a W% higher than the LL (which simulates a normal consolidation condition for the same clay composition) is similar to the FSS of an overconsolidated stiff clay.
 
Can you repost the link to the Castellanos paper, Milliontown? It wont open for me
 
Thank you so much to all of you for the replies. Much appreciated and things are much clearer now.
 
geotechguy, try the link again, the VT website was down for maintenance for a little while.
 
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