Cn and Phin or Cu + Phiu


Dear all,

I have been dealing with processes from Poland and most recently from Russia and I have a question that maybe the experience of some of you will be helpfull. 

For cohesive soils they don't use the undrained shear test - Cu, it is not commom at all to have this parameter as a "basic" one for clays. Instead of this they use Cu and Phiu, in Poland, and Cn and Phin, in Russia. My question is if any of you know how to estimate the Cu, from those parameters. I don't think they are drained parameters.

I will give you some examples:

- A silty clay with Cu=8kPa and Phi=9º

- a silty clay/clay with Cu=31kPa and Phi=9º

Furthermore I was wondering how they will do to input parameters in softwares once the majority of softwares assume Mohr Coulomb for a drained state and Tresca for undrained. I don't think it will be correct to assume, for a short term (so undrained) the numbers given above in Mohr Coulomb criterion (once this way the software will assume a drained condition which is not the case).

Does any of you have experience with those type of parameters and geotechnical design softwares?

Thanks in advance.

 

 

 

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1 Answer

Votes: 0

David Nash

Dear Diana

In saturated clays we expect that undrained strength = cu with phi=0.

However phi(u) is commonly found to be >0 in sets of unconsolidated undrained triaxial tests. This arises because the samples are not fully saturated so that the effective stress changes as the cell pressure is increased (contrary to the expectation that phi(u) = 0.)

Sometimes consolidated undrained triaxial tests are carried out on normally consolidated samples of clay at several different cell pressures, and the increase of strength with confining pressure is then described by phi(cu). The use of phi(cu) was discussed in a classic paper by Bishop and Bjerrum (1960) who strongly advised against its use since its value depends on how the mean stress varies during shear; despite this it has remained in some text books. In the UK we advocate interpretation of tests in terms of effective stress and I would not wish to see the use of phi(cu) promoted further.

I recommend that you examine the original test data including stress paths and try to interpret the tests in terms of effective stress. You may be able to draw conclusions about a suitable value of undrained strength cu with phi=0 and this can be input into a total stress analysis of short term stability.

Best regards,

David Nash