Shear-Angle Models for Oblique Metal Cutting

​
inclination
i
15.
normal rake
α
n
10.
rake face friction
β
a
42.8
shear angle model
Merchant Minimum Energy
Krystof Maximum Shear Stress
Armarego-Whitfield
chip flow
η = 7.62 deg
normal shear
ϕ
n
= 29.16 deg
oblique shear
ϕ
i
= 10.80 deg
oblique force
θ
i
= 5.17 deg
normal force
θ
n
= 32.55 deg
chip ratio
r
c
= 0.53
This Demonstration illustrates three shear-angle models for oblique metal cutting: Merchant's minimum energy, Krystof's maximum shear stress, and the Armarego–Whitfield model. The equations can be found in[1].

Details

Oblique metal cutting is the basic building block for understanding all metal removal processes. However, it presents a 3D assemblage of velocities and forces that is complicated even for the thin shear plane approximation. This Demonstration attempts to collect all the equations given in Sec. 2.5 of[1]. Three shear angle models are incorporated, the minimum energy principle of Merchant (MME), maximum shear stress (on the shear plane) of Krystof (KMSS), and the Armarego–Whitfield (AW) model. The full nomenclature used and the basic relations are available in[1].
The normal plane
P
n
perpendicular to the cutting edge is shown as a green plane, transparent to let you see various angles. The three input angles are the normal rake
α
n
, inclination
i
, and the rake-face friction
β
a
. The five output angles are chip-flow
η
, normal shear
ϕ
n
, oblique shear
ϕ
i
, oblique force
θ
i
between resultant force
F
c
and
P
n
, and the normal force
θ
n
between the projection of
F
c
on
P
n
and the
x
axis.
However, two extra relations are derived for the purpose of illustration.
First is the formula for chip thickness ratio
r
c
=h/
h
c
. Let


,


s
, and


c
represent the cutting, shear, and chip velocity. Let their magnitudes be
v
,
v
s
, and
v
c
, respectively. Let the uncut chip width be
b
and cut chip width be
b
c
. The thin shear plane requires that


c
=


s
+


. It is easy to see from material conservation that
hvb=
h
c
v
c
b
c
,
so
r
c
=
v
c
b
c
/(vb)
.
b
and
b
c
are related through
b
c
/b=cosη/cosi
. Suppose

n
represents the unit-vector normal to the shear plane; then
(


c
-


)·

n
=0
; the net material accumulating at the shear plane is zero. If the cutting velocity is taken to be of unit magnitude, that is,
v=1
,


=(cosi,sini,0)
,


c
=
v
c
(cosηsin
α
n
,sinη,cosηcos
α
n
)
,

n
=(sin
ϕ
n
,0,cos
ϕ
n
)
,
v
c
=sin
ϕ
n
/cos(
α
n
-
ϕ
n
)
,
r
c
=cosηsin
ϕ
n
/(cosicos(
α
n
-
ϕ
n
))
.
Second, a unit vector that lies in the intersection of the shear plane
P
S
and the plane
P
V
formed by the
z
axis and


is desired. Consider the vector


=(-cos
ϕ
n
,ψ,sin
ϕ
n
)
. This needs to be coplanar with


and

ℯ
z
=(0,0,1)
, which may be written as the scalar vector triple-product equation
(


,


,

ℯ
z
)=0
. Then



1+
2
ψ
is the desired vector. It can be found that



1+
2
ψ
=(11+
2
tan
i
2
cos
ϕ
n
)(-cos
ϕ
n
,-tanicos
ϕ
n
,sin
ϕ
n
)
.
A little more vector algebra can solve for the points on the intersection of the shear plane and the uncut work surface. In this Demonstration,


,


s
, and


c
are shown in red. The lengths of the vectors show relative magnitudes.
With a given friction angle
β
a
of the chip sliding on the tool, the direction of the resultant force
F
c
can be calculated taking only the two components
F
u
along the chip velocity direction and
F
v
normal to the tool surface.
F
c
can also be computed from the components (given in[1]) in the directions of the cutting velocity as
F
tc
, thrust as
F
fc
, and normal as
F
rc
. It can be seen in this Demonstration that the two resultant force vectors line up as expected. From the postulated direction of the resultant force vector, two angles
θ
n
and
θ
i
are obtained. The lengths of all vectors show relative magnitudes.
Thus with the three input angles
i
,
α
n
, and
β
a
and five output angles
η
,
ϕ
n
,
ϕ
i
,
θ
n
, and
θ
i
, the complete force and velocity diagrams (like the Merchant's circle diagram in orthogonal cutting) can be constructed.
The calculation of the two shear angles
ϕ
n
and
ϕ
i
from some hypotheses drives the remaining outputs. Unlike orthogonal cutting, there is no single relation connecting the rake, friction, and shear angles. A set of five equations needs to be solved simultaneously under different hypotheses.
The three hypotheses coded in the Demonstration are MME, KMSS, and AW. However, one common approximation used in oblique cutting is Stabler's rule, which says that the chip flow angle is equal to the inclination angle. It is of interest to see which of the hypotheses agree with Stabler's rule.
The MME model involves minimization of the nondimensionalized power
P
with respect to
ϕ
n
and
ϕ
i
.
P=
cos
θ
n
+tan
θ
i
tani
(cos(
θ
n
+
ϕ
n
)cos
ϕ
i
+tan
θ
i
sin
ϕ
i
)sin
ϕ
n
.
That is,
∂P/∂
ϕ
n
=∂P/∂
ϕ
i
=0
gives rise to the equations
cos
ϕ
i
cos(
θ
n
+2
ϕ
n
)/sin
ϕ
n
+cot
ϕ
n
sin
ϕ
i
tan
θ
i
=0
,
-cos(
θ
n
+
ϕ
n
)/sin
ϕ
i
+cos
ϕ
i
tan
θ
i
=0
.
This Demonstration plots the five output angles as a "web" plot with a point along the
η
axis corresponding to
i
. The baseline is orthogonal cutting shown as the blue curve. When
i
is changed from 0, all the angles change to form the red curve.
For any setting of
i
,
α
n
and
β
a
, Stabler's rule
(η=i)
can be compared to the models' results. When
i=0
, the MME and AW models converge to the well-known Merchant's model, while KMSS converges to the Lee and Shaffer model.

References

[1] Y. Altintas, Manufacturing Automation: Metal Cutting Mechanics, Machine Tool Vibrations, and CNC Design, 2nd ed., Cambridge: Cambridge University Press, 2012.

Permanent Citation

Raja Kountanya
​
​"Shear-Angle Models for Oblique Metal Cutting"​
​http://demonstrations.wolfram.com/ShearAngleModelsForObliqueMetalCutting/​
​Wolfram Demonstrations Project​
​Published: March 28, 2014