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

The Delft Sand, Clay & Rock Cutting Model.
Figure 8-27: The Flow Type cutting mechanism in ductile rock cutting.
The specific energy Esp is defined as the amount of energy used/required to excavate 1 m 3 of soil/rock. This can
be determined by dividing the cutting power Pc by the production Q and results in the cutting force Fh in the
direction of the cutting velocity vc, divided by the cross section cut hi·w:
E
sp
Pc Fh
vc Fh
Q h w v h w
i c i
(8-97)
The model for rock cutting under atmospheric conditions is based on the Flow Type of cutting mechanism.
Although in general rock will encounter a more brittle failure mechanism and the Flow Type considered represents
the ductile failure mechanism, the Flow Type mechanism forms the basis for all cutting processes. The definitions
of the Flow Type mechanism are shown in Figure 8-27.
Figure 8-28 illustrates the forces on the layer of rock cut. The forces shown are valid in general. The forces acting
on this layer are:
1. A normal force acting on the shear surface N1 resulting from the grain stresses.
2. A shear force S1 as a result of internal friction N1·tan(.
3. A shear force C as a result of the shear strength (cohesion) c or c. This force can be calculated by multiplying
the cohesive shear strength c with the area of the shear plane.
4. A force normal to the blade N2 resulting from the grain stresses.
5. A shear force S2 as a result of the soil/steel friction N2·tan( or external friction.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1.
The forces acting on a straight blade when cutting rock, can be distinguished as:
6. A force normal to the blade N2 resulting from the grain stresses.
7. A shear force S2 as a result of the soil/steel friction N2·tan(or external friction.
These forces are shown in Figure 8-29. If the forces N2 and S2 are combined to a resulting force K2 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.
The horizontal equilibrium of forces:
F h K 1 sin( ) C cos( ) K 2 sin( ) 0
(8-98)
The vertical equilibrium of forces:
F v K 1 cos( ) C sin( ) K 2 cos( ) 0
(8-99)
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