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

The Delft Sand, Clay & Rock Cutting Model.
The normal force on the pseudo blade is now:
W2 sin( ) W1 sin( ) G1sin( )
N2
cos( )
sin( )
sin( )
I cos( ) C1 cos( ) C2
cos( ) cos( )
(10-12)
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 A cos W sin K sin K sin
h 4 4 3
C W sin C cos K sin 0
3 2 2 2
(10-13)
The vertical equilibrium of forces on the wedge is:
F A sin W cos K cos W K cos
v 4 4 3 3
W cos C sin K cos G 0
2 2 2 2
(10-14)
The unknowns in this equation are K3 and K4, since K2 has already been solved. Three other unknowns are the
adhesive force on the blade A, since the adhesion does not have to be mobilized fully if the wedge is static, the
external friction angle δ, since also the external friction does not have to be fully mobilized, and the wedge angle
θ. These 3 additional unknowns require 3 additional conditions in order to solve the problem. One additional
condition is the equilibrium of moments of the wedge, a second condition the principle of minimum required
cutting energy. A third condition is found by assuming that the external shear stress (adhesion) and the external
shear angle (external friction) are mobilized by the same amount. Depending on whether the soil pushes upwards
or downwards against the blade, the mobilization factor is between -1 and +1. Now in practice, sand and rock have
no adhesion while clay has no external friction, so in these cases the third condition is not relevant. However in
mixed soil both the external shear stress and the external friction may be present.
The force K3 on the bottom of the wedge is now:
K
3
sin
W2 sin W3 sin W4 sin
sin
K2 sin G2 sin
(10-15)
3 2
sin
A cos C cos C cos
+
The force K4 on the blade is now:
K
4
sin
W2 sin W3 sin W4 sin K 2 sin G2 sin
3 2
sin
A cos C cos C cos
+
This results in a horizontal force of:
(10-16)
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