Undrained shear strength 3

[Atros]

Hello there, i was wondering if someone could help me interpreting this triaxial undrained test.

I need to obtain the undrained shear strength (Su) in terms of the effective vertical pressure: Su/Sigma'v.

Also cohesion and friction angle.

Im attaching the spreadsheet with the results. Thanks a lot.

 

[fattdad]

With the drainage turned off, you should have some maximum deviator stress (total stress) that occurred at undrained failure for that level of confinement. To give me drained P' v. q plots doesn't provide the data to obtain Su, unless you want me to do the homework?

Not sure why the lab is not giving the reporting requirements for CU-txc?

f-d

p.s., also if there is back-pressure saturation, you are assuming that the Su/P is informed only by completely saturated soils. That may or may not be the case?


ípapß gordo ainÆt no madre flaca!

 

[LRJ]

The shear strength is q/2 at failure. Usually 'failure' is defined either as the peak value or at a specified strain level, though this strictly depends on the application. If there is no peak before 10% then the value of q at 10% is taken and divided by 2 to give su. It really is important to note that what is defined as failure is dependent on the specific application, however; certain analyses, for instance, might consider the peak principal effective stress ratio as failure instead.

You get c' and φ' by fitting a line to the stress paths from those tests. Usually the peak (q[sub]max[/sub]) or the ultimate values (at the end of the test) are used to fit the line. However, once again, the line fitting may depend on the specific application. Also, for what it's worth, c' and φ' is not a great idealisation, despite its wide application in geotechnical engineering.

 

[fattdad]

You do not get c' and phi' from the P' v. q' graph! There's trig to relate these two drawings!

You do not get Su from the P' v. q' plot. Those are effective stresses and you are looking for undrained strength.

f-d

ípapß gordo ainÆt no madre flaca!

 

[LRJ]

Ah yes, the φ' angle actually comes from putting the fitted angle into a fairly standard equation (sinφ' = 3M/(6+M) for compression; sinφ' = 3M/(6-M) for extension; M = q/p').

s[sub]u[/sub] is half of q at failure. So the three tests would each have an s[sub]u[/sub]. But the OP wants a stress ratio, so they'll have to derive the s[sub]u[/sub] for each test and normalise it by the vertical effective stress. That will give three results, but the s[sub]u[/sub] to use would depend on the stress expected in the ground.

 

[Okiryu]

I think that LRJ is correct. Critical state model. Attached are some extracts from Budhu's book. See section 6.4 at page 6 of the attached.

 

[Atros]

THanks a lot for the help.

Im reattaching the spreadsheet with some results. Please let me know what you think.

Thanks.

 

[LRJ]

I'm not sure why you've assumed an internal friction angle and a K[sub]0[/sub] value in your calculations: you have σ'[sub]v0[/sub] already (the value at the start of the test). Also, you appear to have defined s[sub]u[/sub] as the last value, whereas it is more typically to take the peak deviator stress and divide by 2. Are you certain that your definition of 'failure' is appropriate for your design purposes?

If you want s[sub]u[/sub]/σ'[sub]v0[/sub] then you're going to have three different results: one for each test. Determine s[sub]u[/sub] (typically q[sub]max[/sub]/2) and then normalise by your initial effective vertical stress for each test; this will give you one result per test. You can fit a line through the three results if needed, but the angle of this line won't be your internal friction angle; you do not get the friction angle from a plot of s[sub]u[/sub] versus σ'[sub]v0[/sub] (see the fifth post of this thread).

 

[Atros]

Hi,

Im reataching the spreadsheet, first with phi' and cohesion results (effective). Please let me know if you think they are ok.

Thanks.

 

[fattdad]

it seems your peak phi is less than your residual phi, which can't be, based on the stress strain drawings.

the graphs do look good though!

f-d

ípapß gordo ainÆt no madre flaca!

 

[LRJ]

I'm unsure how you've determined your M values - it's very useful to actually plot this information. Also, to confirm, is your apparent cohesion intercept half of the intercept when plotting q?