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

The Delft Sand, Clay & Rock Cutting Model.
Substituting equation (2-67) gives:
F N tan cos Nsin 0
G Ncos N tan sin 0
(2-70)
Writing the full tangent and multiplying with cos(φ) gives:
F cos Nsin cos N sin cos 0
G cos Ncos cos N sin sin 0
(2-71)
Now the terms with the normal force N can be combined to:
Fcos Nsin 0
G cos Ncos 0
(2-72)
Cross multiplying with sine and cosine to give the normal force the same terms:
F cos cos Nsin cos 0
Gcos sin Ncos sin 0
(2-73)
Adding up the two equations gives:
Fcoscos Gcossin
(2-74)
Solving the first 3 equations with the first 3 unknowns gives for the force on the retaining wall:
F G tan
(2-75)
With the equation for the weight of the sand.
1
2
G s
g h cot
w
(2-76)
2
The equation for the force on the retaining wall is found.
sin
1
2
cos
F s
g h w
2 sin cos
(2-77)
This equation still contains the angle of the shear plane as an unknown. Since we are looking for the maximum
possible force, a value for β has to be found where this force reaches a maximum. The derivative of the force and
the second derivative have to be determined.
dF 0
d
(2-78)
2
dF 0 2
d
(2-79)
Since the equation of the force on the retaining wall contains this angle both in the nominator and the denominator,
determining the derivative may be complicated. It is easier to simplify the equation with the following trick:
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