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Solution Homework #3:
Metal Cutting
2.008 Design and Manufacturing II
Spring 2004
Problem 1:
(a) Consider the Merchant’s Cutting Force diagram in Figure 1 and the chipworkpiece
interface during orthogonal cutting in Fig. 2.
α
φ
R
F
t
F
c
F
n
F
s
F
N
α
β
β−α
Fig 1. Merchant’s diagram
Fig 2. Cutting FBD
Merchant’s hypothesis is that the shear plane is located to minimize the cutting force, or
where the shear stress is maximum. Derive the Merchant’s relationship between shear
angle, rake angle, and friction angle as below from the diagram above.
o
α
β
φ
=
45
+
−
2
2
(b) Consider the directions of the cutting force and the thrust force. Fc, cutting force, is
always positive, since the material is removed. Is the thrust force, Ft, also positive at all
times? If not, explain why. Also explain how you can make Ft = 0 for a given friction
coefficient between the tool and the work piece.
Solution
(a)
From the Force diagram (we discussed it in class so you know how each of the elements
came in), we know that
Fs = Fc*cos
φ
 Ft*sin
φ
(1)
Fn = Fc*sin
φ
+ Ft*cos
φ
(2)
Where
Fc = net horizontal cutting force
Ft = net vertical thrust force
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View Full DocumentFs = shear force along shear plane
Fn = shear force normal to shear plane
There are two ways to derive Merchant´s Relationship: find the minimum of the cutting
force or the maximum of the shear stress. We will have a look at both.
i) Find the minimum of the cutting force Fc
A
t
0
i
w
(3)
F
=σ
i
A
=σ
s
s
s
s
s
sin
ϕ
sin
ϕ
F
t
=
tan(
β−α
)
(4)
F
c
With
σ
s
as the shear stress and A
s
the shear area; t
0
is the thickness of the uncut chip,
called the depth of cut, w is the width of the work.
(3) and (4) in (2):
tw
i
σ
0
=
Fcos
ϕ−
F
s
in
ϕ
tan
(
)
s
c
c
sin
ϕ
ii
σ
1
⇔
F
=
0
s
i
(5)
c
sin
ϕ
cos
ϕ
−
sin
ϕ
tan(
β
−
α
)
In order for the cutting force to be minimum at angle
ϕ
, the first derivative has to be
zero:
∂
F
c
=
0
∂ϕ
Substituting equation (5) and solving:
!
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 Spring '04
 JungHoonChun

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