The Delft Sand, Clay & Rock Cutting Model, 2019a

Clay Cutting.
7.5. The Tear Type.
7.5.1. Introduction.
In the previous chapter, the equations for the cutting forces of the Flow Type cutting mechanism have been
derived. These equation however do not take into consideration that normal forces and thus stresses may become
negative and may exceed the tensile strength of the clay. If the tensile stresses exceed the tensile strength, tensile
failure will occur and the clay will not fail by plastic shear failure, but by tensile failure. The failure mechanism in
this case is named the Tear Type mechanism. Based on the Mohr circle, tensile cracks will occur under and angle
of 45 degrees downwards with respect of the shear angle as is shown in Figure 7-26. When the blade is progressing
with the cutting velocity, after a short while a so called secondary crack will occur under 90 degrees with the first
(primary) crack. The model as derived in this chapter, does not assume that the tensile strength is exceeded at the
moment of tensile crack forming over the full length of the tensile crack. The model assumes that the tensile
strength is exceeded at the start of the tensile crack only. In order to determine whether the tensile strength is
exceeded, the average shear stress in the shear plane is used. Of course there may be a stress distribution in the
shear plane, leading to locally higher and lower shear stresses and thus normal stresses, but these cannot be
determined with the methodology used. Only average stresses can be determined. The methodology applied
however gives reasonable and practical tools to determine whether the Tear Type cutting mechanism will occur
or not.
Figure 7-26: The Tear Type cutting mechanism in clay.
7.5.2. The Normal Force on the Shear Plane.
In order to determine the normal (possibly tensile) stresses on the shear plane, first the normal force on the shear
plane has to be determined.
Ccos( ) A
N1
sin( )
(7-63)
Substituting the equations for the cohesive force C and the adhesive force A gives:
N
1
sch i w s ah b
w
cos( )
sin sin
sin( )
(7-64)
The average normal stress on the shear plane equals the normal force on the shear plane N1, divided by the cross
sectional area of the shear plane, giving:
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