BULETINUL INSTITUTULUI POLITEHNIC DIN IAŞI - Universitatea ...

76 Ana-Maria Matei and Marius-Nicolae Milea
2. Theoretical Models of Cutting Force Components in Face Milling
depending on the Cutting Forces developed on a Single-Tooth
The face milling forces components acting on the tool FZ , FX si FY are
depending on the cutting forces components developed on a single tooth level
Fz , Fx si Fy , which are influenced by the cutting tooth geometry (γ, k, λ), the
standard compression yield point of the material being cut, the cross - sectional
area of the chip, the chips’ friction coefficient μ, and the chips´ contraction
coefficient Cd [3,4].
Considering the specific elements of the symmetrical or unsymmetrical
milling, there are several possibilities of modelling the cutting force
components in face milling.
2.1. Theoretical Models of Cutting Force Components
in Symmetrical Face Milling
Fig. 1 shows the geometrical models of complete and incomplete face
milling. The values of specific elements for each milling variant result as
follows: the contact angle Ψ = 180°, t = D, the number of teeth that
simultaneously cut zs = z/2 at complete face milling and Ψ < 180°, t < D,
z t
= arcsin at incomplete face milling [2].
z s
π
D
The theoretical models for the evaluation of cutting force components in
face milling are developed in the light of two simplifying assumptions, as
follows:
a) The cutting force components on a tooth, respectively Fz – the main cutting
force (the tangential force), Fx – the radial force and Fy – the axial force, take
values depending on the cross-sectional area of undeformed chip;
b) The tooth z1 is the one to which the tooth coordinates system xyz
corresponds to the mill coordinates system (FZ = Fz, FX = Fx, FY = Fy), and the
other teeth that simultaneously cut are placed equidistant (φ = 2π/z).
The values of cutting force components in symmetrical face milling depend
on zs and Fz, Fx , Fy , as follows:
a) For symmetrical milling with odd zs the Eq. (1) results.
(1)
F
F
Z
X
( zs
/ 2)
−1
( zs
/ 2)
−1
2π
⎡ 2π
⎤
= Fz
+ 2 ∑ Fz
cos( zi
) = Fz
⎢1
+ 2 ∑ Fz
cos( zi
) ⎥ ,
1 z ⎣ 1 z ⎦
( zs
/ 2)
−1
( zs
/ 2)
−1
2π
⎡ 2π
⎤
= Fx
+ 2 ∑ Fx
cos( zi
) = Fx
⎢1
+ 2 ∑ Fx
cos( zi
) ⎥ ,
1 z ⎣ 1 z ⎦
F = F ⋅ z .
Y
y
s