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

A Wedge in Hyperbaric Rock Cutting.
F K cos( ) W cos( ) C sin( )
v 1 1 1
C sin( ) W cos( ) K cos( ) 0
2 2 2
(15-6)
The force K1 on the shear plane is now:
K
1
W2 sin( ) W1 sin( ) C1 cos( ) C2 cos( )
sin( )
(15-7)
The force K2 on the pseudo blade is now:
K
2
W2 sin( ) W1 sin( ) C1 cos( ) C2 cos( )
sin( )
(15-8)
From equation (15-8) the forces on the pseudo blade can be derived. On the pseudo blade a force component in
the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
Fh W2 sin( ) K2sin( ) C2cos( )
(15-9)
F W2 cos( ) K2 cos( ) C2
sin( )
(15-10)
The normal force on the shear plane is now:
W2 sin( ) W1 sin( )
N1
cos( )
sin( )
sin( )
C1 cos( ) C2
cos( ) cos( )
(15-11)
The normal force on the pseudo blade is now:
W2 sin( ) W1 sin( )
N2
cos( )
sin( )
sin( )
C1 cos( ) C2
cos( ) cos( )
(15-12)
Now knowing the forces on the pseudo blade A-C, the equilibrium of forces on the wedge A-C-D can be derived.
The horizontal equilibrium of forces on the wedge is:
F W sin K sin K sin
h 4 4 3
C W sin C cos K sin 0
3 2 2 2
(15-13)
The vertical equilibrium of forces on the wedge is:
F W cos K cos W K cos
v 4 4 3 3
W cos C sin K cos 0
2 2 2
(15-14)
The unknowns in this equation are K3 and K4, since K2 has already been solved. Two other unknowns are, the
external friction angle δ, since the external friction does not have to be fully mobilized, and the wedge angle θ.
These 2 additional unknowns require 2 additional conditions in order to solve the problem. One additional
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