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

The Delft Sand, Clay & Rock Cutting Model.
in the equations is the shear angle β. Assuming that the mechanism will choose a shear angle where the cutting
energy is at a minimum, a shear angle β is found according to:
(7-89)
4 2
If we substitute this solution in the cutting force equations we find:
cos
Fh 2 s c hi
w
1 sin
sin
F
2 s c hi w
1 sin
(7-90)
(7-91)
The horizontal force will increase with an increasing blade angle, the vertical force also, but upwards directed. In
the case of the Curling Type, the ac ratio r is not fully mobilized giving:
sin
rm
2
1 sin
(7-92)
The condition of having a normal force of zero on the blade can never fulfill the condition of having an equilibrium
of moments on the layer cut, since the normal force on the blade is zero and is therefore rejected. Still this condition
gives insight in the behavior of the equations of clay cutting and is therefore mentioned here.
7.6.3. The Equilibrium of Moments.
As mentioned in the previous paragraph, the equilibrium of moments on the layer cut has to be fulfilled. If we take
the equilibrium of moments around the tip of the blade, there are only two forces participating in the equilibrium
of moments, the normal force on the shear plane N1 and the normal force on the blade N2. These forces have acting
points R1 and R2 on the shear plane and on the blade. If the normal stresses are uniformly distributed, both acting
points will be at the center (half way) the corresponding planes. The acting point of the normal force on the shear
plane will be at half the length of the shear plane and the acting point of the normal force on the blade will be at
half the (mobilized) length of the blade. Two factors are introduced to give the exact location of these acting points,
λ1 on the shear plane and λ2 on the blade. When the moment N2·R2 on the blade is greater than the moment N1·R1
on the blade curling will occur in such a way that both moments are equal. The contact length between the clay
and the blade will be reduced to a mobilized contact length hb,m.
The normal force on the shear plane is now equal to the force N1, because the internal friction angle is zero:
Ccos( ) A
N1
sin( )
(7-93)
The normal force on the blade is now equal to the force N2, because the external friction angle is zero:
N
2
C Acos( )
sin( )
(7-94)
This gives for the equilibrium of moments:
N1 R1 N2 R2
(7-95)
For both acting points we can write:
R
h
1 i
1 , R2
sin
h
2 b,m
sin
(7-96)
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