TailignsEddie
Geotechnical
- Oct 20, 2018
- 5
Hi all,
Can someone please enlighten me as to the correct way of calculating the undrained strength ratio (Su/σ[sub]v0[/sub]’) 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/σ[sub]v0[/sub]’ 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 Φ[sub]T[/sub] 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.
Can someone please enlighten me as to the correct way of calculating the undrained strength ratio (Su/σ[sub]v0[/sub]’) 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/σ[sub]v0[/sub]’ 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 Φ[sub]T[/sub] 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.