Friction circle method

Sjotroll · Jan 3, 2018

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)?

I put a scheme of the method in the attachment if someone needs it. (

http://files.engineering.com/getfil...0-8439-f0f2ab317aa3&file=friction_circle.JPG)

Perhaps it's just some kind of assumption for the purpose of the method that I'm missing here...

Any help is appreciated!

lukeng · Jan 4, 2018

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.

Cheers

Sjotroll · Jan 8, 2018

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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Friction circle method

Sjotroll

Geotechnical

lukeng

Geotechnical

Sjotroll

Geotechnical

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