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

Exercises.
16.8.4. Calc.: Cutting Forces & Mechanisms.
Consider a rock with a compressive strength of 60 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 =21 MPa
2
cos
The normal stress on the shear plane is:
ccos( ) N1
cos( ) 12.9 M Pa
sin( )
The shear stress on the shear plane is:
S1
N1
c tan =25.7 MPa
The normal stress in the center of the Mohr circle is:
C N1 S1
tan 22.27 MPa
The radius of this Mohr circle is now:
S1
R
cos
27.35 MPa
The minimum principle stress of this Mohr circle is:
R 5.09 MPa
min C
Since -5.09 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 21 0.11 4.02 MN
1
cos( )
The vertical force is now:
2ch i wcos( )cos( )
F
VF
chi
w 0.572 21 0.11 1.20 MN
1
cos( )
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