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

Rock Cutting: Hyperbaric Conditions.
Figure 9-7: The forces on the layer cut in rock
(hyperbaric).
Figure 9-8: The forces on the blade in rock
(hyperbaric).
These forces are shown in Figure 9-8. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive
force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By
taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
2 2
2 2 2
K N S
(9-3)
The horizontal equilibrium of forces:
F K sin( ) W sin( ) Ccos( )
h 1 1
A cos( ) W sin( ) K sin( ) 0
2 2
(9-4)
The vertical equilibrium of forces:
F K cos( ) W cos( ) C sin( )
v 1 1
A sin( ) W cos( ) K cos( ) 0
2 2
(9-5)
The force K1 on the shear plane is now:
K
1
W2 sin( ) W1 sin( ) C cos( ) A cos
sin( )
(9-6)
The force K2 on the blade is now:
K
2
W2 sin( ) W1 sin( ) C cos( ) A cos
sin( )
(9-7)
From equation (9-7) the forces on the blade can be derived. On the 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( )
(9-8)
F W2 cos( ) K2
cos( )
(9-9)
The normal force on the shear plane is now:
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