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

The Delft Sand, Clay & Rock Cutting Model.
The length of the front of the wedge A-C is:
L
2
h
sin
b
(13-17)
The length of the bottom of the wedge A-D is:
L
1 1
h
tan
3 b
tan
(13-18)
The length of the blade C-D is:
L
4
hb
sin
(13-19)
The distance of the tip of the blade perpendicular to the front of the wedge is:
(13-20)
L L sin
5 3
The distance from point A to the intersection point of the line going from the tip of the blade perpendicular to the
front of the blade is.
(13-21)
L L cos
6 3
The distance of the acting point of the force N2 to the intersection point of the line going from the tip of the blade
perpendicular to the front of the blade is:
L7 L6 R2
(13-22)
R2 follows from the equilibrium of moments on the layer cut, assuming the forces on the shear plane act at half
the length of the shear plane.
N1 R1 N2 R2
(13-23)
Now the equilibrium equation of moments can be derived according to:
N
L
N
L
4 4 3 3
M N2 L7 C2 L5
0
(13-24)
2 2
Both equation (13-11) and equation (13-24) don not contain the mobilized adhesive force Am, giving the
possibility to solve the two unknowns N3 and N4. To solve the normal force N3 first an expression for the normal
force N4 has to be derived based on the equilibrium of moments.
N
4
4
2 N3 L 3 /2 N2L7 C2
L5
(13-25)
L
Equation (13-11) and equation (13-25) should give the same result for the normal force N4, thus:
2 N L /2 N L C L
3 3 2 7 2 5
L
4
C sin N cos C sin N cos
2 2 3 3
(13-26)
This can be written as:
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