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
|