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

The Delft Sand, Clay & Rock Cutting Model.
5.3. The Equilibrium of Forces.
Figure 5-4: The forces on the layer cut in dry sand.
Figure 5-5: The forces on the blade in dry sand.
Figure 5-4 illustrates the forces on the layer of soil 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 effective grain stresses.
2. A shear force S1 as a result of internal friction, N1·tan(.
3. A gravity force G as a result of the weight of the layer cut.
4. An inertial force I, resulting from acceleration of the soil.
5. A force normal to the blade N2, resulting from the effective grain stresses.
6. A shear force S2 as a result of the soil/steel friction N2·tan(.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1.
2 2
1 1 1
K N S
(5-1)
The forces acting on a straight blade when cutting soil, can be distinguished as:
7. A force normal to the blade N2, resulting from the effective grain stresses.
8. A shear force S2 as a result of the soil/steel friction N2·tan(.
These forces are shown in Figure 5-5. 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
(5-2)
Pure sand is supposed to be cohesion less, meaning it does not have shear strength or the shear strength is zero and
the adhesion is also zero. The shear stresses, internal and external, depend completely on the normal stresses. In
dry sand the pores between the sand grains are filled with air and although dilatation will occur due to shearing,
Miedema (1987 September), there will be hardly any generation of pore under pressures because the permeability
for air flowing through the pores is high. This means that the cutting forces do not depend on pore pressure forces,
nor on adhesion and cohesion, but only on gravity and inertia, resulting in the following set of equations:
The horizontal equilibrium of forces:
F h K 1 sin( ) I cos( ) K 2 sin( ) 0
(5-3)
The vertical equilibrium of forces:
F v K 1 cos( ) I sin( ) G K 2 cos( ) 0
(5-4)
Page 106 of 454 TOC Copyright © Dr.ir. S.A. Miedema