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

3.6.5. The Resulting Cutting Forces.
The General Cutting Process.
The resulting cutting forces in x, y and z direction can be determined once the deviation velocity components are
known. However, it can be seen that the second velocity condition equation (3-67) requires the horizontal and
vertical cutting forces perpendicular to the cutting edge, while these forces can only be determined if the mobilized
internal and external friction angles and the mobilized cohesion and adhesion (equations (3-68), (3-69), (3-70) and
(3-71)) are known. This creates an implicit set of equations that will have to be solved by means of an iteration
process. For the cutting forces on the blade the following equation can be derived:
x2 h d2
F F cos F sin
(3-72)
y2 h d2
F F sin F cos
(3-73)
F
z2
F
(3-74)
v
The problem of the model being implicit can be solved in the following way:
A new variable λ is introduced in such a way that:
vd1
vd
sin
1
1
vd2
vd
sin
1
(3-75)
(3-76)
This satisfies the condition from equations (3-61) and (3-62) for the sum of these 2 velocities:
d1 d2 d
v v v sin
(3-77)
The procedure starts with a starting value for λ=1. Based on the velocities found with equations (3-75), (3-76),
(3-52) and (3-53), the mobilized internal φe and external δe friction angles and the cohesion ce and adhesion ae can
be determined using the equations (3-68), (3-69), (3-70) and (3-71). Once these are known, the horizontal Fh and
vertical Fv cutting forces in the plane perpendicular to the cutting edge can be determined with equations (3-8) and
(3-9). With the equations (3-48), (3-50), (3-64) and (3-65) the friction and deviation forces on the blade and the
shear plane can be determined. Now with equation (3-67) the value of the variable λ can be determined and if the
starting value is correct, this value should be found. In general this will not be the case after one iteration. But
repeating this procedure 3 or 4 times should give enough accuracy.
3.7. Example Program in Visual Basic 6.
Start:
Labda = 1
'In case of deviation angle
If Iota 0 Then
Vr1 = Vd * cos(Iota) * sin(Alpha) / sin(Alpha + Beta)
Vr2 = Vd * cos(Iota) * sin(Beta) / sin(Alpha + Beta)
Vd1 = Vd * sin(Iota) * Labda / (1 + Labda)
Vd2 = Vd * sin(Iota) / (1 + Labda)
'So Vd1+Vd2 = Vd * sin(Iota)
Phi_e = atn(Tan(Phi) * cos(atn(Vd1 / Vr1)))
Delta_e = atn(Tan(Delta) * cos(atn(Vd2 / Vr2)))
Cohesion_e = Cohesion * cos(atn(Vd1 / Vr1))
Adhesion_e = Adhesion * cos(atn(Vd2 / Vr2))
End If
(3-52)
(3-53)
(3-75)
(3-76)
(3-68)
(3-69)
(3-70)
(3-71)
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