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

sin
Rock Cutting: Atmospheric Conditions.
sin sin
2 0
(8-89)
(8-90)
2 2
Using this, gives for the force F:
F
1 2ch i wcos
n 1 1 cos
(8-91)
This gives for the horizontal force Fh and the vertical force Fv:
1 2ch wcos( )sin( )
1
F c h w
i
h HF i
n 1 1 cos( ) n
1
(8-92)
(8-93)
This solution is the same as the Merchant solution (equations (8-109) and (8-110)) that will be derived in the next
chapter, if the value of the stress distribution factor n=0. In fact the stress distribution factor n is just a factor to
reduce the forces. From tests it appeared that in a type of rock the value of n depends on the rake angle. It should
be mentioned that for this particular case n is about 1 for a large cutting angle. In that case tensile failure may give
way to a process of shear failure, which is observed by other researches as well. For cutting angles smaller than
80 degrees n is more or less constant with a value of n=0.5. Figure 8-31 and Figure 8-32 show the coefficients λHF
and λVF for the horizontal and vertical forces Fh and Fv according to equations (8-109) and (8-110) as a function
of the blade angle α and the internal friction angle φ, where the external friction angle δ is assumed to be 2/3·φ. A
positive coefficient λVF for the vertical force means that the vertical force Fv is downwards directed. Based on
equation (8-97) and (8-109) the specific energy Esp can be determined according to:
E
sp
Pc Fh vc Fh
1
HF
c
Q h w v h w n
1
i c i
(8-94)
1.0
Stress Distribution Nishimatsu
0.9
0.8
0.7
p/p 0
0.6
0.5
0.4
0.3
n=0.00
n=0.25
n=0.50
n=1.00
n=2.00
n=4.00
n=8.00
0.2
0.1
0.0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Distance along the shear plane
© S.A.M.
Figure 8-25: The stress distribution along the shear plane.
Copyright © Dr.ir. S.A. Miedema TOC Page 271 of 454