Hallo Simon,
Yes this is a good discussion.
The mountain of sand is simple, but a nice illustration, I have used it
for that purpose. You were right to use it as a first example.
The critical wedge or taper has its name because the whole wedge was
supposed to be in the critical state (constantly close to the Mohr/Coulomb
failure ctriterion). I have published on that subject (critical taper and
FTB formation) and have done many experiments that show that the
equilibrium in a wedge is in fact a 'punctuated' equilibrium. The wedge
deforms in systematic cycles, a new paper is in preparation.
You are absolutely correct with calling phi a material property instead of
only rock property. I have done many experiments where we determined phi
for experiments in dry sand. Sand is in fact a cohesionless rock. The phi
of sand is in the normal range for sedimentary rocks and is a bit higher
than the average 30 degrees.
Which brings me to the next comment, the range of values for the internal
friction angle. The majority of sedimentary rocks has values of phi of
slightly higher than 30. Very strong rocks go up to about 45 (I have dealt
with a quartz cemented sandstone with a phi of some 42). However, if one
has no measurements and a guess has to be made, then 30 is a very good
guess. The range is wide, 25<phi<45 is what I use, but the bulk of the
sedimentary rocks do not come as high as 45 and vary between 30 and 35.
Assuming an average of 30 (also just because it is an easy number), then
25 is only 5 degrees off and an error of 2.5 is made. When the true phi is
45 (extremely high) the difference is 15, which devided by 2 is 7.5. Given
that field measurements of dips and strikes have an accuracy of about 5
degrees, I think that our rough estimate of 30 is not bad. If a stronger
rock is assumed, one can use for example 33 as a best guess and still do a
very good guessing job.
Bedding as anisotropy is relevant for relativelysmall structures. On a
scale of say basin analysis, anisotropies on bedding scale are not
relevant anymore, faults will cut right across that and the overall
picture will not be influenced by individual beds.
For very strong rocks you will find that phi does not approach zero, this
would mean that a failure plane would have to be at 45 degrees angle to
the maximum principle stress, which is not the case. In fact we observe
that very strong rocks tend to fail in tension, indicating that phi=90.
Tension fractures are parallel to Sigma-1 and perpendicular to Sigma-3,
which is why they are always open when they form!
Many faults in basement rocks have erroneously been interpreted as e.g.
normal faults, but are in fact very large vertical tension fractures (or
joints) along which later displacement took place.
Regards, Dirk Nieuwland
> Thanks Dirk, makes for an interesting discussion.
>
> On 30/09/2011, at 8:41 AM, Dirk Nieuwland wrote:
>
>> Dear All,
>>
>> The mountain of sand experiment is nice, but not perfect, too many
>> variables affect it to get consistent results. The demonstration cannot
>> be used to determine the internal friction angle of various types of
>> sandstones, let alone a limestone or a crystalline rock.
> Agreed, I was just trying to keep things simple for the original question
> seeking the difference between internal and external friction.
>
>> The internal friction angle, as I understand it, explains the
>> difference between the angle of a newly formed shear plane in a rock and
>> the maximum principle stress. This angle is not 45 degrees, which would
>> correspond with the angle between the maximum shear stress and the
>> maximum principle stress, but is 45 degrees minus phi (the internal
>> friction angle)/2. [a=45-phi/2]
> In other words, what you appear to be describing to this geotechnical
> engineer (& definitely not a 'rock mechanic'), is the angle of failure of
> an 'active wedge'. Conversely, the passive failure wedge forms at 45+ø/2.
>
>> Phi is a rock property.
> Just a rock property?! Also a soil property. To broaden the field, I guess
> we could say it is a material property.
>
>> The relation ship a=45-phi/2 is a very powerful tool to determine for
>> example stress orientations from tectonic structures. Phi is not a
>> constant as is often thought. With low stresses Phi becomes larger and
>> increases to 90 degrees in proper tension situations. This is why the
>> angle between a tension fracture and the maximum principle stress is 0
>> (zero): [a= 45 - 90/2 = 0.]
>> The best way to determine Phi is to do a series of tri-axial tests at
>> different confining stresses. The resulting series of Mohr circles at
>> failure, connect to the Mohr/Coulomb failure envelope, the slope of this
>> envelope is the internal friction angle Phi.
> Yes, agreed, but can this be done on intact rock that is not subject to
> say bedding anisotrophies, in which case, orientation of confining
> stresses can play a large part in the outcome? For such intact rock, I
> would imagine that some substantial confining pressures would be required,
> meaning some pretty heavy duty tri-axial gear. If the rock is so strong
> that the confining stresses don't alter the failure load, then aren't you
> going to approach a total stress (ie undrained) situation, for which phi
> will surely approach zero?
>
>> The tensile strength can conveniently be estimated using the Griffith
>> fracture criterium, according to which the tensile strength is half the
>> cohesion. The cohesion was obtained with the triaxial test results.
> Way outside my game
>
>> Surprisingly, phi is close to 30 for most rocks.
> Hmm, I would have thought this was a big call. Literature on my shelves
> suggests a range from <30 to >45°, seemingly dependent on rock strength.
>
> Best regards, Simon Woodward
>
>
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