Hi,
I'm in need of a clarification regarding the forces that act at an obliquity "phi" to the circular arc (slip surface), when looking at the friction circle method for slope stability.
What are those forces?
Why do they act in the direction they do (at an angle "phi", which is the internal friction angle, from the normal to the slip surface)?
Hi,
Each of those forces represents the resultant of the normal and the tangent forces acting at a given point on the failure surface. In fact, for a soil obeying to the Mohr-Coulomb failure criterion:
t= sigma'(N)*tan(phi)
where sigma'(N) is the normal effective stress and phi is the friction angle of the soil.
As you can see, the ratio of t to sigma'(N) equals tan(phi). By integrating t and sigma'(N) along the base of a generic slice of soil mass you would obtain T and N (respectively tangential force and normal force) whose resultant is represented by those forces displayed on your attachment.
Thank you for the answer! I was missing the Mohr-Coulomb criterion. Since it is not written anywhere in the chapter about the failure circle, it didn't came to my mind even though it is something as easy as that. :)
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RE: Friction circle method
Each of those forces represents the resultant of the normal and the tangent forces acting at a given point on the failure surface. In fact, for a soil obeying to the Mohr-Coulomb failure criterion:
t= sigma'(N)*tan(phi)
where sigma'(N) is the normal effective stress and phi is the friction angle of the soil.
As you can see, the ratio of t to sigma'(N) equals tan(phi). By integrating t and sigma'(N) along the base of a generic slice of soil mass you would obtain T and N (respectively tangential force and normal force) whose resultant is represented by those forces displayed on your attachment.
Cheers
RE: Friction circle method