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

z2
r2
The Delft Sand, Clay & Rock Cutting Model.
v v sin
(3-60)
Since both approaches will have to give the same resulting velocity components, the following condition for the
transverse velocity components can be derived:
x1 x2 d1 d2 d
v v v v v sin
(3-61)
y1 y2 d1 d2 d
v v v v v sin
(3-62)
v
z1
v
(3-63)
z2
3.6.4. The Deviation Force.
Since a friction force always has a direction matching the direction of the relative velocity between two bodies,
the fact that a deviation velocity exists on the shear surface and on the blade, implies that also deviation forces
must exist. To match the direction of the relative velocities, the following equations can be derived for the deviation
force on the shear surface and on the blade (Figure 3-18):
F
d1
vd1
Ff1
(3-64)
v
r1
vd2
Fd2
Ff2
(3-65)
vr2
Since perpendicular to the cutting edge, an equilibrium of forces exists, the two deviation forces must be equal in
magnitude and have opposite directions.
F
d1
F
(3-66)
d2
By substituting equations (3-64) and (3-65) in equation (3-66) and then substituting equations (3-48) and (3-50)
for the friction forces and equations (3-52) and (3-53) for the relative velocities, the following equation can be
derived, giving a second relation between the two deviation velocities:
v
vd1 Ff2 vr1
Fh cos F sin sin
vd2 Ff1 v
r2 Fh cos Fv
sin sin
(3-67)
To determine Fh and Fv perpendicular to the cutting edge, the angle of internal friction φe and the external friction
angle δe mobilized perpendicular to the cutting edge, have to be determined by using the ratio of the transverse
velocity and the relative velocity, according to:
v
d1
tan e tan cos
atn v r1
v
d2
tan e tan cos
atn
v r2
(3-68)
(3-69)
For the cohesion c and the adhesion a this gives:
v
d1
ce
c cos
atn
v r1
v
d2
ae
a cos
atn
v r2
(3-70)
(3-71)
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