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

The Delft Sand, Clay & Rock Cutting Model.
16.8.2. Calc.: Cutting Forces & Mechanisms.
Consider a rock with a compressive strength of 100 MPa and a tensile strength of -10 MPa. The angle of
internal friction is 20 degrees, the angle of external friction is 13.33 degrees. A blade angle of 60 degrees is
used and a blade height of 0.1 m and blade width w=1 m. The layer thickness is 0.1 m.
The shear angle is:
= 43.33 degrees
2 2
The cohesion or shear strength is:
UCS 1
sin
c =35 MPa
2
cos
The normal stress on the shear plane is:
ccos( ) N1
cos( ) 21.52 M Pa
sin( )
The shear stress on the shear plane is:
S1
N1
c tan =42.84 MPa
The normal stress in the center of the Mohr circle is:
C N1 S1
tan 37.1 MPa
The radius of this Mohr circle is now:
S1
R
cos
45.59 MPa
The minimum principle stress of this Mohr circle is:
R 8.48 MPa
min C
Since -8.48 MPa>-10 MPa there is no tensile failure but shear failure.
The horizontal force is now:
2ch i wcos( )sin( )
Fh HF chi w 1.912 35 0.11 6.691 MN
1
cos( )
The vertical force is now:
2ch i wcos( )cos( )
F
VF
chi
w 0.572 35 0.11 2.0 MN
1
cos( )
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