Taylor’s Failure Criterion
Description:
Taylor (1948) used an energy method to derive a simple soil model. He assumed that the shear strength of soil is due to sliding friction from shearing and the interlocking of soil particles. Consider a rectangular soil element that is sheared by a shear stress t under a constant vertical effective stress s9z (Figure ). Let us assume that the increment of shear strain is dg and the increment of vertical strain is dεz. The external energy (force 3 distance moved in the direction of the force or stress 3 compatible strain) is t dg.
The internal energy is the work done by friction, mf s9z dg, where mf is the static, sliding friction coeffi cient and the work done by the movement of the soil against the vertical effective stress, 6s9zdεz. The negative sign indicates the vertical strain is in the opposite direction (expansion) to the direction of the vertical effective stress. The energy 6s9zdεz is the interlocking energy due to the arrangement of the soil particles or soil fabric. For equilibrium,
Dividing by s9zdg, we get
At critical state, mf 5 tanf9cs andTherefore,
At peak shear strength,
where the subscripts, cs and p, denote critical state and peak, respectively. Unlike Coulomb failure criterion, Taylor failure criterion does not require the assumption of any physical mechanism of failure, such as a plane of sliding. It can be applied at every stage of loading for soils that are homogeneous and deform under plane strain conditions similar to simple shear.
This failure criterion would not apply to soils that fail along a joint or an interface between two soils. Taylor failure criterion gives a higher peak dilation angle than Coulomb failure criterion. Equation (10.13) applies to two-dimensional stress systems. An extension of Taylor failure criterion to account for three-dimensional stress is presented in Chapter 11. Neither Taylor nor Coulomb failure criterion explicitly considers the rotation of the soil particles during shearing.