Cohesionless soil 1

[marcpreston]

I am analysing a soil which has a cohesion value of 0Kn/m2. The soil is a granular material. I am trying to analyse the foundation in sliding.

When calculating sliding the formula is

Rv tan angle of friction/Rh

If you take out the angle of friction the formula becomes

Rv/Rh

This results in no internal friction in the soil. The factor of safety will be bigger than soils with a angle of friction say 10 degrees which can not be true. As the 10 degrees soil has internal friction and the so of zero angle hasnt.

What would the formula be for a soil with no internal friction.

 

[Ron]

Why would you assume no angle of internal friction? Most sands are going to have an angle of internal friction somewhere between 30 and 40 degrees, usually about 34-35 degrees for a fine to medium sand. If it is a silt, the angle can vary all over the place.

I think you have something backwards...usually as cohesion increases, the angle of internal friction decreases.

 

[marcpreston]

I written it wrong. If there was cohesion but internal friction.

 

[marcpreston]

If there was cohesion but no internal friction.

 

[mdhshanwil]

Your formula is only a subset of the general formulation.

A more general formulation is:

[(Rv * tan phi') + B*c + Pp] / [Rh]

B = ftg width
Pp = passive pressure from embedment

Note that even this is simplified somewhat.

Your formula has been simplified for a footing on a cohesionless soil (i.e., c=0) and with no embedment (i.e., Pp = 0)