Phi>0 for cohesive undrained material

[Baztien]

The reason for this post is that after some soil mechanics reading I am still confused about the phi=0 condition or rather the phi>0 for certain cohesive material in undrained condition.

Beeing in the offshore industry we follow API as recommendation for driven pile capacity. Cohesive material are treated as undrained and we normally perform UUs at different depths in a layer.
We use the phi=0 to assess Su. We often have sandy clays and overconsolidated clays where we also apply phi=0 but soil mechanics books state phi is >0 for these soil.

Could someone explain why phi should be >0 in these types of material in undrained condition?

 

[ItsOnlyDirt]

I have never heard of this is bog standrad soil mechanics.

It is the norm to assume Phi = 0 for undrained conditions

Phi can be, and usually is > 0 in Drained conditions, i.e. (typically long term for clays)

This is the only thing I can think of, do you have the reference for Phi>0 in undrained conditions?? I am interested now....

cheers

 

[dgillette]

Are you perchance looking at a phi-cu analysis? Phi-cu is sometimes used to fake a Su/p' strength ratio. It gives you Su as a fn of Sigma'fc (consolidation stress on the failure plane)? It's a poor second to SHANSEP or similar sort of undrained analysis. For explanation of the difference, see Chuck Ladd's 1986 Terzaghi Lecture, published in the ASCE JGE about 1989 or 1990. That's one of the best general references on undrained strength that there is.

Phi is never really zero anyway; that's just a convenient fiction (sometimes).

 

[fattdad]

when a clay is fully saturated, under short term loading the increase in pore pressure occurs at the same rate as the application of stress. As such, you truly get a phi=0 condition - the only strength coming from the undrained shear strength.

If the clay (in situ) is not fully saturated, you may still have poor drainage, but the application of stress doesn't result in the same increase in pore pressure and you will get some undrained friction angle. The soil will behave like this until the stress increase (and resulting pore pressure) is sufficient to dissolve the air and act as saturated (well at that point is actually saturated). Then the failure envelope flat-lines and the friction angle goes to zero.

An undrained friction angle flattens with confining stress, so to use Su and Phi in a total stress analysis you'd have to confirm unsaturation and you'd have to consider stress dependence.

Dusting off the gray matter. . .

f-d

¡papá gordo ain’t no madre flaca!

 

[Baztien]

fattdad that's exactly what I'm talking about.
So if I understand well, provided a very sandy clay or an overconsolidated clay is fully saturated the phi=0 condition is applicable?

 

[fattdad]

Not sure about your reference to soil type. . .

this is a topic of degrees. Undrained conditions can develop in other soil types too. We use undrained conditions when the application of load is faster than the dissipation of pore pressure. This can happen in clean sands as liquifaction, for example.

When fine-grained soils (i.e., silt, elastic silt, clay, fat clay and their sandy counterparts) are saturated, phi=0 analysis is appropriate. When they're not saturated, it's also appropriate unless you have data to show otherwise. Often you don't.

Bear in mind that in many instances, a clay layer is below the water table and it's still not "saturated." I'd still go into the engineering problem assuming it was though. . .

Good luck on your project. Hope I helped.

f-d

¡papá gordo ain’t no madre flaca!

 

[irawanfirmansyah]

Baztien,
You will get phi >0 if during shesr there is water flow from one part to another part within the soil specimen or from soil specimen to the porous stone.

In other word, you will get phi=0 if the soil specimen is fully saturated and the test apparatus is totally water tight