×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Undrained Strength Ratio from CU Triaxial

Undrained Strength Ratio from CU Triaxial

Undrained Strength Ratio from CU Triaxial

(OP)
Hi all,

Can someone please enlighten me as to the correct way of calculating the undrained strength ratio (Su/σv0’) from the results of CU (compression) triaxial tests?

Not a single textbook or course notes I have seen actually describes how to do this, yet all the time in the literature you see Su/σv0’ values which have been derived from Tx results.

I know that one is able to get the friction angle (phi) and cohesion in total stress terms from the CU test. In a NC clay this envelope is supposed to pass through the origin. So, if the ΦT is 20°, can I convert this to a Su/σv’ by tan 20° = 0.36?

Or is the more correct thing to do to look at the stress path in q-p’ space (or t-s space) and select the maximum shear stress divided by the consolidation effective stress (σ3 at the end of consolidation/ start of the shearing phase)? This would seem more correct than the first approach since you might have contractive behaviour (as in the attached) in which case the peak shear strength for a given consolidation stress will be lower than the Mohr-Coulomb envelope suggests.

By the way, quite a few textbooks will explain the strength parameters in terms of total stress but will then go on to say that it is seldom used in practice. Also, confusingly a lot of textbooks refer to this as the TOTAL STRESS strength rather than UNDRAINED strength and then call the results of UU tests undrained, as if there is some kind of important distinction to be made (which they never go on to make). Is it incorrect to call the strength from a CU test an undrained strength? After all, the test is most certainly undrained.

RE: Undrained Strength Ratio from CU Triaxial

Generally you cannot use the friction angle to get the undrained shear strength (su). This is because su is equal to the deviator stress at failure (qf) divided by 2 (think of a Mohr circle), which is affected by the pore-water pressure (i.e. contractant versus dilatant behaviour during undrained shearing, as you note).

As for σ'v0 in the test, that is the vertical effective stress at the start of shearing (= end of the final consolidation stage). Using σ'v0 is probably not the best normalising parameter for overconsolidated or anisotropic materials, however (the mean effective stress, p'0 or σ'm, would be better).

Undrained tests can be used to obtain effective stress (and total stress) parameters if you measure the pore-water pressure. This is because you are able to calculate what the effective stresses are, so can plot the effective stress path. As an aside, increasingly engineers appear to be scheduling UU triaxials with pore-water pressure measurement as a 'cheap' alternative to doing a full CU test. This penny-pinching practice is utterly ridiculous and indicative of engineers who haven't got a clue what they are doing with respect to lab testing.

To be honest I never really liked the 'total stress = undrained' and 'effective stress = drained' definitions. It's just confusing. Whenever you think of drained or undrained behaviour, just think of critical state soil mechanics: is it constant volume (undrained) or constant stress (drained)?

RE: Undrained Strength Ratio from CU Triaxial

A CU-bar is typically performed at three confinements. Each confinement is consolidated to some load, the drainage is turned off and the sample is sheared. You get a total stress for failure and you get a pore pressure that accompanies that failure.

Ignoring the pore pressures, you'll have a maximum deviator stress for each consolidation load (i.e., for each P'). That's your Su.

f-d

ípapß gordo ainÆt no madre flaca!

RE: Undrained Strength Ratio from CU Triaxial

There's no need to undertake multiple CU tests as you can define the critical state line with a single CU test which is sheared to sufficiently large strain to reach critical state. Likewise, plotting Mohr circles is anachronistic: the full stress path from a single test is much more useful and relevant, especially if the strain level is noted at certain points on the stress path.

RE: Undrained Strength Ratio from CU Triaxial

(OP)
Thanks for the help everyone. Much appreciated!

Although I do still wonder if one should be very clear about saying the MAXIMUM deviator stress divided by 2, or the final deviator stress divided by 2 when the stress path reaches the critical state (I think this is what LRJ means by "deviator stress at failure"), because in the case of a contractive sample that deviator stress is lower than the deviator stress at the collapse surface (which would be the peak deviator stress), and must surely be the true undrained strength of the sample. Its hard to put that in words without an illustration to go with it, but I imagine you know what I mean.

RE: Undrained Strength Ratio from CU Triaxial

if the point is to get the Su/P ratio, you'd need multiple points. The PvQ drawing can't get you there. Well, maybe I'm missing something?

f-d

ípapß gordo ainÆt no madre flaca!

RE: Undrained Strength Ratio from CU Triaxial

Ah yes, for su/σ'v0 you are right that you'd need multiple tests. But you don't need multiple tests to define the friction angle with an effective stress path triaxial test.

RE: Undrained Strength Ratio from CU Triaxial

I agree that there is a relationship between phi and alpha. I was just reacting to the question about undrained shear strength.

f-d

ípapß gordo ainÆt no madre flaca!

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members! Already a Member? Login


Resources


Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close