reservoir geomecanics

102 Reservoir geomechanics
Drucker–Prager criterion
The extended von Mises yield criterion, or Drucker–Prager criterion, was originally
developed for soil mechanics (Drucker and Prager 1952). The von Mises criterion may
be written in the following way
J 2 = k 2 (4.25)
where k is an empirical constant. The yield surface of the modified von Mises criterion in
principal stress space is a right circular cone equally inclined to the principal stress axes.
The intersection of the π-plane with this yield surface is a circle. The yield function
used by Drucker and Prager to describe the cone in applying the limit theorems to
perfectly plastic soils has the form:
J 1/2
2
= k + α J 1 (4.26)
where α and k are material constants. The material parameters α and k can be determined
from the slope and the intercept of the failure envelope plotted in the J 1 and (J 2 ) 1/2
space. α is related to the internal friction of the material and k to the cohesion of
the material. In this way, the Drucker–Prager criterion can be compared to the Mohr–
Coulomb criterion. When α is equal to zero, equation (4.26) reduces to the von Mises
criterion.
The Drucker–Prager criteria can be divided into an outer bound criterion (or circumscribed
Drucker–Prager) and an inner bound criterion (or inscribed Drucker–
Prager). These two versions of the Drucker–Prager criterion come from comparing
the Drucker–Prager criterion with the Mohr–Coulomb criterion. In Figure 4.6 the two
Drucker–Prager criteria are plotted in the π-plane. The inner Drucker–Prager circle
only touches the inside of the Mohr–Coulomb criterion and the outer Drucker-Prager
circle coincides with the outer apices of the Mohr–Coulomb hexagon.
The inscribed Drucker–Prager criterion is obtained when (Veeken, Walters et al.
1989; McLean and Addis 1990)
3sin φ
α = √ (4.27)
9 + 3sin 2 φ
and
3C 0 cos φ
k =
2 √ q √ 9 + 3sin 2 φ
(4.28)
where φ is the angle of internal friction, as defined above.
The circumscribed Drucker–Prager criterion (McLean and Addis 1990; Zhou 1994)
is obtained when
6 sin φ
α = √ (4.29)
3 (3 − sin φ)