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Machining with double cutting edge tools—I. Symmetrical triangular cuts
Int. .I. Mach. Tool Des. Res. Vol. 7, pp. 23-37. PergamonPress 1967. Printedin Great Britain
MACHINING WITH DOUBLE CUTTING TOOLS-I. SYMMETRICAL TRIANGULAR
EDGE CUTS
E. J. A. ARMAREGO* (Received
29 August 1966)
Abstract-The plane deformation models are developed for symmetrical triangular cuts. Single-edge orthogonal cuts, full-depth and non-full-depth triangular cuts are tested and compared. It is shown that theoretical and experimental force functions are consistent when the force intercepts are considered as tool edge forces. It is also found that the thin plane model, with the chip critically stressed, correlates with triangular cut results for the conditions tested. INTRODUCTION NEED to study the mechanics of cutting of double-edged tools is obvious when one considers the geometry of practical cutting tools. The analysis for single-edged tools, although not completely solved, has proved very useful to describe the influence of various parameters on the deformation and power in cutting. In general the thin plane (or thin zone) models suggested by Piispanen [l], Merchant [2], Lee and Shaffer [3] and other workers have been more widely used than the thick zone models studied by Palmer and Oxley [4] or Okushima and Hitomi [SJ. This follows since the thick zone models are too complex for practical use and furthermore there is some doubt that such models are representative of the deformation at the usual cutting speeds. It is therefore important to study the full potential of the thin plane models when applied to cases other than those with a single cutting edge. THE
NOMENCLATURE
bc T TC
Workpiece velocity. Chip velocity. Shear velocity. Rake angle. Tool included angle, groove included angle. Normal clearance angle. Reference clearance angle. Length of workpiece cut. Length of chip Maximum width of triangular cut; width of cut for single-edge orthogonal cutting. Maximum chip width for triangular cuts. Maximum undeformed chip thickness for full-depth triangular cuts; maximum groove depth for non-full-depth triangular cuts. Maximum chip thickness for full-depth triangular cuts. Undeformed chip thickness for non-full depth triangular cuts and for single-edge orthogonal cutting. * Department
of Mechanical Engineering,
University of Melbourne, 23
Victoria, Australia.
24
E. J. A. ARMAREGO
CJI,02,
“3
P dcz 2.4s 24 FS F* FP FQ FH
R F N K~P, KIP KZQ, KIQ k2p, klp, kzu,
Shear angle. Friction angle. Shear stress in the shear plane. Norma1 stress on the shear plane. Maximum shear stress. Principal stresses. Hydrostatic stress. Strain increment. Total shear plane area. Projection of total shear plane area in maximum shear stress plane. Shear force in one shear plane. Norma1 force on one shear plane. Force in velocity plane along the work velocity direction. Thrust force in the velocity plane perpendicular to the work velocity direction. Side force on one shear plane perpendicular to the velocity plane. Resultant force in velocity plane. Friction force in the rake face. Force normal to the rake face. Experimental force constants of Fp and FQ components for single-edge orthogonal cutting. Experimental force constants of Fp and FQ components for triangular cuts.
km
MECHANICS
OF
CUTTING
FOR
SYMMETRICAL
TRIANGULAR
CUTS
Cutting with symmetrical triangular tools as shown in Fig. 2 is a special case of machining with two cutting edges. The cuts taken may be “full-depth” or “non-full-depth” as seen in Fig. 2 a and b respectively. It is readily expected that due to symmetry the workpiece and chip velocities will be in planes parallel to the plane of symmetry. Before proceeding with the analysis of the deformation and forces in cutting it is necessary to consider the tool geometry.
FIG. I. The geometry
and specification
of symmetrical
triangular
tools.
Machining with Double Cutting Edge Tools-I.
Symmetrical Triangular Cuts
2.5
In Fig. 1 the relevant shape of the triangular tool is shown. The tool consists of a rake face ABO and two clearance faces GOA, GOB. The cutting edges are represented by A0 and BO. The rake angle a is specified in the plane of symmetry of the tool and cut, and is thus also in the velocity plane. It will be shown that this rake angle is comparable to the rake angle for orthogonal cutting with a single cutting edge. The included angle 26 is peculiar to this type of tools and is therefore unrelated to any of the angles for single-edged tools. The angle 26 is specified in the plane perpendicular to the work velocity VW and is therefore equal to the angle of the machined groove. The clearance angles are considered in planes perpendicular to the cutting edges A0 and BO i.e. normal clearance angles Cl,. These angles represent the actual clearance between the newly-cut surface and the tool clearance faces. The clearance angle Cl in Fig. 1 is a useful reference angle for checking the normal clearance angles ground on the tool. The clearance angles are related by the equations below tan Cl, tan Cl = (1) sm S2/(1 + C0S2 6 tan2 a) + cos2 6 tan a tan Cl, when a = 0 tan Cl, tan Cl = ___ (2) sin 6 There is little difficulty in grinding these tools to the correct rake the specified angles can be directly set on the machine vice axes. It to grind the clearance faces by setting the clearance angles about axes edges if the correct geometry is to be readily achieved. The problems rectly grinding the clearance angles for lathe tools (i.e. double cutting shown by the author [6]. Thin plane deformation
model for full-depth
and included angles; is however important parallel to the cutting introduced by incoredge tools) have been
cuts
The deformation for full-depth cuts may be considered as two shear planes pivoted about the cutting edges as shown in Fig. 2 a. The work, chip and shear velocities will be in planes parallel to the plane of symmetry. It may be assumed that plane strain conditions will prevail so that the chip width b, is equal to the groove width b. From continuity and incompressibility conditions it follows that TJ,bc
= Tlb
(3)
T Te
(4)
thus le r=-_=_ 1 From the deformation
geometry
tan 4 = [(T/Te)
cos
the shear angle is given by a]/[1 -
(T/Tc) sin a] = r
Equation 5 is thus the same as for orthogonal From plane strain conditions the stresses shows the tool rake face and the shear planes, stresses on elements close to the shear plane. a2 = (al + v3)/2 = p and from the equilibrium shear plane, the shear stress in the shear plane
cos
a/(1
-
r sin
a)
(9
cutting with a single cutting edge. on the shear plane can be studied. Figure 3 from a number of views, together with the It is seen that for plane strain (i.e. drz = 0) of element x, an element sectioned by the is given by
s, = 7 sin 6’ = rAr/Ag
(6)
E. J. A. ARMAREGO
26
and the normal
stress on the shear planes
Us is equal to the hydrostatic
stress
i.e.
% = p where T = maximum shear stress, A, = area of one shear plane, A, = projected area of shear plane in the maximum 8 ,y”
Y
(7)
shear stress plane.
6
T> Cr
A'--,
1
‘\ ‘N
’
I 1: I /’
,’ /’
W
v b=b,
B
/
AB\\ *. -----_1
. .
. .
\
\
+
/’ 0
v,
b=b,
FIG. 2. Thin plane deformation
for symmetrical triangular cuts. (a) Full-depth cuts; (b) Non-fuIl-depth cuts.
7
Machining with Double Cutting Edge Tools-I.
Symmetrical Triangular Cuts
FIG. 3. Shear plane stresses for plane strain conditions
27
(full-depth triangular cuts).
Jt is thus apparent that the shear stress in the shear plane is not the maximum shear stress and in fact the shearing process approximates planes originating from the cutting edges. Whether or not 03 equals zero will depend on the deformation model envisaged. Thus if it is considered that the chip is loaded to the yield stress between the shear planes and the rake face, 03 = 0 as for the Lee and Shaffer type solution. Alternatively if we have a thin plane, and the chip is not critically stressed, the Oxley- [7] or Merchant-type solution may be used in which cases us is determined or ignored respectively.
Force analysis (i) Merchant-type solution. As for the Merchant model it may be assumed that the resultant force acting on the chip at the shear planes is balanced by an equal, opposite and colinear force acting on the chip at the tool-chip interface. It is also assumed that the cutting edges are sharp and no forces act on these edges. Figure 4 shows the forces acting on the
End an view of
FIG. 4. Force diagram
shear
plones
for thin plane model (full-depth triangular cuts).
28
E.J.A. ARMAREGO
shear planes and the components of these forces in and perpendicular to the plane of symmetry. It is noted that the shear forces in the shear plane are colinear with the shear velocity. Considering the forces on one shear plane, the shear force in the shear plane is found from the force triangle, thus Fs = 2” COS(+ + X - a) where F8 = R = h = 4 = a =
shear force in one shear plane, resultant force in the plane of symmetry friction angle in the plane of symmetry, shear angle, rake angle.
The shear force can also be expressed
(or velocity
(8)
plane),
as Fs = &A,
(9)
From equations (8) and (9) and the geometry of Fig. 4, the power force F,/2, Fo/2 and side force FR due to one shear plane are given by
FP _ &As cos (A 2
&A, sin (A -
a) a)
FR = &As tan (4 + A -
force
(10)
u)
cos (4 + A -
FQ _
2
a)
cos (#J -t A -
thrust
(11) a) cot 6’.
(12)
Equations (IO), (11) and (12) apply for the second shear plane except that the side force Fn acts in the opposite sense resulting in no total side force. From equations (IO), (11) and (6) the force components can be expressed in terms of the maximum shear stress T and the corresponding shear area, i.e. F
_ 27A, cos (A P
COS(4
+
/I-
a) = a)
7T2 tan 6 cos (h - a) sin #J cos (4 + h - a)
FQ = yT2 tan 6 sin (A - a) sm 95cos (4 + h - a)
(13)
(14)
The above equations are equivalent to those for orthogonal cutting with a single cutting edge and a, 4 and h have the usual meaning (note that T2 tan 6 is the area of cut corresponding to bt for “rectangular” orthogonal cuts). This analysis does not relate A, 4 and CLin the form of a shear angle relation, although the minimum energy principle can be adopted for this case provided one accepts the assumptions applied by Merchant. (ii) Lee and ShafSer-type solution. The previous analysis of forces will be unaltered if the chip is considered to be loaded to the yield stress as suggested by Lee and Shaffer. The resultant force, however, will act along the al direction in the plane of symmetry since the
Machining with Double Cutting Edge Tools-I.
Symmetrical Triangular Cuts
side forces are balanced by the hydrostatic stresses and ~3 = 0. This analysis and CLby the Mohr stress circle and is given below
29
relates
A, +
Equation (15) can be applied in conjunction with equations (13) and (14) to arrive at the FP and FQ forces. The deformation and stress affected region is shown in Fig. 5.
stress
Shear
View of stressed region m 7 directIon
p=T=uz
stress on PIones porollel to oc Direct stress
stress on rake face
A
FIG.
5.
Deformation
Thin plane deformation
2+a)
Y +I!!!
c3
ml
and stresses for thin plane model (chip critically stressed).
model for non-fill-depth
cuts
The analysis for full-depth cuts is the same as for the above cases but equations and (14) are modified to account for the appropriate area of cut, thus F P
(13)
= 9(2T - I) tan 6 cos (A - a) sin 4 cos (4 + X - a)
FQ = 7t(2T - t) tan 6 sin (A - -~a) sin 4 cos (4 + h - a)
(17)
where t = undeformed chip thickness, T = maximum groove depth. EXPERIMENTAL
RESULTS
AND
DISCUSSIONS
were devised to study the mechanics of cutting for full-depth triangular cuts. Orthogonal cutting (with a single cutting edge) tests were initially run to represent “standard” data for comparison with the full-depth triangular tests. Further experiments were run to compare full-depth results to non-full-depth tests. The experiments were run on a universal milling machine. Cutting tools (18-4-1 H.S.S.) were fitted to a stationary three-component Tests
30
E. J. A. ARMAREGO
strain gauge dynamometer mounted on the machine column. The cutting forces were registered on a pen recorder and the workpiece (65ST6 aluminium alloy 4 in. x 6 in. Y 6 in.) was held in a vice on the milling machine table. All the cuts were performed by feeding the workpiece past the stationary tool at 9$ in./min. Orthogonal
cutting with a single cutting edge
Tools with rake angles of lo”, 20” and 30”, and 6” clearance angles were used. Each tool was tested for a range of undeformed chip thicknesses and four widths of cut. The undeformed chip thickness was set with the aid of a dial gauge and was further checked by the chip weight technique. The forces and shear angles for the 20” rake angle tool are shown in Figs. 6 and 5. From Fig. 6 it is seen that the force per unit width is a linear function of the undeformed chip thickness, with a positive intercept at t = 0. It follows that Fpib = Kzpt i
IGp
FQ/b = &jt
KIQ
+
(18)
or Fp = Kz,bt
+ Kl,b (18a)
FQ = K2Qbt + Klob
Undeformed FIG. 6. Influence
of undeformed
chip thickness,
chip thickness orthogonal
t in.
on the force per unit width cutting.
~1 :z 20’, 18-4-l H.S.S. tool. 65ST6 Al alloy workpiece. Dry cutting. V = 92 in/m& 0 h = 0.24 in. c b = 0.20 in. a b : 0.15 in. 0 h 7 0.10 in.
for single-edge
Machining with Double Cutting Edge Tools-I.
0
I
I
I
0,002
0.004
Undeformed
Symmetrica
I
I
0.006
Triangular Cuts
0~008
0~010
31
I 0.012
t in.
chip thickness,
FIG. 7. Influence of undeformed chip thickness on the shear angle for single edge orthogonal cutting. Cutting conditions as in Fig. 6.
The above equations suggest that the cutting forces depend on the area cut as well as the length of the cutting edge in contact with the work. Figure 7 shows that the shear angle, and hence the deformation, is unaffected by the undeformed chip thickness or the width of cut in these tests. The interpretation of the force functions is of some interest. The physical significance of the intercept force has been discussed by various authors. A number of workers [8,9] have assumed that the intercepts represent tool edge forces which do not contribute to the chip formation process while others [lo] claimed that a size effect occurs in metal cutting which leads to higher shear stresses and hence relatively higher forces at the smaller undeformed chip thickness values. It may be qualitatively shown that the former explanation is consistent with the experimental data in this work. Considering the force relations for orthogonal cutting with a single cutting edge it is known that Tbt cos (A - u) Fp= . sm + cos (4 + X - CC) (19) I;Q = . Tbt sin (X - a) sin 4 cos (4 + h - a) ’ The above relations are derived on the assumption that no rubbing or edge forces act at the tool tip. If it is now assumed that the shear stress exhibits a size effect as in equation (20),
r-c+” where C and D are constants, .
(20)
then the force equations
ccoso-
4
sm $ cos (4 + h C sin (X - a) sin + cos (4 + X 3
t
a) 1 a)
bt +
(19) become
D cos (A - a) b sin 4 cos (4 + h a) [ I D sin (h - a) sin + cos (4 + A -
(21)
b a) I
.
32
E. J. A. ARMAREGO
,
Equations (21) and (18a) are of the same form provided $J, a and h are constant, in which case the whole measured force contributes to the deformation process. Experimentally it is seen that 4 is independent of b and t. The experimental friction angle X is found by
(22) which shows that X is dependent of t (since K&Kzp # Kl~/Klp) and is not constant. Thus equation (21) is not consistent with the experimental data. When it is considered that the shear stress is constant and that an edge force if existent is not included in the analysis then equations (19) lead to F
=
. rbt cos (A - a) ~ ~ = constant. sm 4 cos (4 t--X - U)
bt
FQ =
Tbt sin (A - a,) ~ constant. sm + cos (f$ + X - a)
bt
P
(23)
provided 4, a and X are constant. Correlation between equations (18a) and (23) occurs if Klpb and KlQb are considered as edge forces (unaccounted in the analysis of cutting). It is also necessary to show that 4 and h are constant for a given a. Experimentally # is constant and h is found from tan (A -
K~Q U) = -= constant K~P
giving a constant A. Thus the experimental and analytical force relations are consistent when the intercept forces are considered as edge forces not contributing to the chip formation process. and given in From equations (18) (24) and (19) Kep, KZQ, h and 7 are determined Table 1. The shear angle relation shown in Fig. 8 leads to a linear relation but does not fit the Merchant or Lee and Shaffer solutions. Similar discrepancies with these and other theories have been reported by the author [l I].
(X-a)
FIG. 8. Comparison
of experimental and theoretical shear angle relations orthogonal cutting. Conditions as in Fig. 6.
for single-edge
Machining with Double Cutting Edge Tools--T. Symmetrical Triangular Cuts TABLE
I.%NGLE-EDGEORTHOGONALCUTTING
Kzp ( x IO3psi)
KP& (X IO3 psi)
123 110 85.7
35.8 18.5 5.3
Full-depth symmetrical
DATA
1” 24.5 29.7 35.1
IO 20 30
33
T(X 103psi)
26.3 29.5 33.5
40 42.8 38.6
triangular cuts
Three sets of tools with rake angles O”, 10” and 20”, and 6” normal clearance angles were tested. For each rake angle four included angles were used. The undeformed chip thickness was set using a dial gauge and the values checked by measuring the groove width with a Brine11 microscope. Each tool was tested for a range of undeformed chip thickness values. From the previous single cutting edge tests it was expected that the cutting forces would depend on the area of cut and the length of cutting edge in contact with the work. For full-depth triangular cuts the forces could be represented by FP = kzpT2 + kl,T
(25)
FQ = k2QT2 + klQT or F,IT =
K2pT
+
kl,
(254
FQIT = k2QT + klQ.
0
I 0~010
Maximum ,FIG.9. Force
I 0.02
I 0.03
I 0.04
I 0.05
I
0.06
undeformed chip thickness, Tin.
per unit depth variation for full-dkpth sykmetr’ical triangular cuts. 18-4-l H.S.S tool, a = 0”, 6 = 45” V = 9H in,‘min. 65ST6 Al alloy workpiece. Dry cutting.
E. J. A. ARMAREG~
0
I
I
0.01
0.02
Maximum FIG.
IO.
Shear
angle
undeformed
I 0.03 chip
I
I
0.04
I
0.05
0.06
thickness,
T
for full-depth symmetrical Conditions in Fig. 9.
in.
triangular
variation
cuts.
Figures 9 and 10 show the results of one such test. It is seen that the shear angle is unaffected by the maximum undeformed chip thickness and the forces are represented by equation (25a). By considering kl,T and klQT as edge forces, the friction angle and shear stress were determined from modified forms of equations (24) and (13) tan (A -
i.e.
f~) = 4”
(26)
2P
and 7 = kzp +Acos (4 -t A - a) tan 6 cos (A - CI)
.
(27)
The calculated values for the tests run are shown in Table 2. In plotting the shear angle relation in Fig. 11 it is seen that the results correlate fairly well with the Lee and Shaffer shear angle relation. This correlation may appear to be unexpected when one considers that the results for orthogonal cutting with a single cutting edge usually show poor correlation with this theory. However, the plane strain assumption used in cutting analyses is more likely to occur with triangular tools, where the deformation is bounded by the material cut, than with orthogonal cutting with a single cutting edge. It is also noticed that the shear stress values in Table 2 show reasonable correlation with those found from single-edge cuts.
FIG. II. Comparison symmetrical triangular
of experimental and theoretical shear angle relations for full-depth cuts. 18-4-1 H.S.S. tool, V = 9$ in/min. 65ST6 Al alloy workpiece Dry cutting. W 01 = 0”; n a = 10”; n 01 = 20”.
Machining
with
Double Cutting Edge Tools-I.
Symmetrical
Triangular
Cuts
35
TABLE~.FULL-DEPTHSYMMETRICALTRIANGULARCUTHNGDATA kzp(lo4psi) 0 0 0 0 10 10 10 10 20 20 20 20
k,Q
68.05 28-08 14.75 7.96 49.68 23.10 1240 7.30 39.72 19.40 10.64 6.14
15 60 45 30 15 60 45 30 75 60 45 30
(lo4
psi)
rl”
+
14.2 17.1 20.8 24.3 20.3 24.8 27.0 29.1 29.3 30.1 34.6 27.0
34.60 1443 6.94 3.19 20.96 7.90 4.20 2.18 9.12 3.24 1.55 0.64
21.0 27.2 25.2 21-9 32.2 28.9 28.8 26.7 33.0 29.5 28.2 26.0
It thus appears that the slip line field solution for rigid ideally-plastic describe the cutting process for full-depth triangular cuts. Non-full-depth
symmetrical
t (103 psi)
38.0 38.5 40.0 42.3 37.5 42.6 41.4 44.7 40.0 43.9 45.3 47.0
conditions
can
triangular cuts
A number of tests were run with non-full-depth cuts keeping the undeformed chip thickness t constant and varying the groove depth T. Figures 12 and 13 show the force variation and shear angle for the tool with a = 0, 6 = 45” and t = 0.010 in. It is seen that the forces are linearly related to T and yield negative force intercepts. These results can be explained in terms of the full-depth cut results. It is reasonable to assume that the force difference between two full-depth cuts will result in the force for a non-full-depth cut 400-
300
-
2
e
200-
: u.
IOO-
Predicted -+
0
I 0.02
Maximum
FIG.12. Force
0.04
from
Experiment01
I
I
I
0.06
0.06
0.10
groove
depth.
T in.
variation for non-full-depth symmetrical triangular cuts. 18-4-l H.S.S. tool; 01 = 0”; 6 = 45”, I = 0.010 in. V = 95 in/min. 65ST6 Al alloy workpiece. Dry cutting.
36
E. J. A.
Maximum
FIG. 13. Shear angle variation
ARMAREGO
groove
depth,
T
in.
for non-full-depth symmetrical triangular cuts. Conditions as in Fig. 12.
provided due consideration is given to the edge forces and the difference in the shear and friction angles. The force for a non-full depth cut of maximum groove depth T and undeformed chip thickness t = T - Tl is given by the difference in deformation-contributing forces for full-depth cuts T and Tl plus the edge force for a full depth T. Thus from equations (25) Fp = kzp[T’
-
T12] + klpT
i.e. Fp = T[kl,
+ 2tk2,] -
kzpt2
FQ = T[klQ $- 2tkfQ] -
k2gt2
(28)
and similarly (284
For a fixed t the forces for a non-full-depth cut will have the same form as in Fig. 12 but when T < t the cuts become full-depth cuts (equations (25) rather than (28) apply) and the forces drop to zero when T equals nought. When the shear angles for full depth and non-full depth cuts are the same then kzp, kzQ, k~,, klQ from full depth cuts can be directly applied to equations (28). In the above test the shear angle for non-full depth is smaller than that for the corresponding full depth cuts. The shear angle relation in equation (15) was used to find the new h and from equations (27) and (26) the corrected values of kpz, and k2Q were determined. The edge force constants kl, and klQ were considered the same for both tests. The new constants were applied to equations (28) to correlate full-depth results to non-full-depth experimental data. The dashed line in Fig. 12 represents the force function obtained from full-depth data. It is seen that the correlation is good suggesting that the deformation for non-full-depth cuts is similar to the full-depth case and that the edge force interpretation is further verified.
CONCLUSIONS (i) It has been shown that the cutting forces are dependent on the area of cut and the length of the cutting edge in contact with the work. (ii) The force intercepts may be considered as tool edge forces not contributing to the chip formation. (iii) Thin plane deformation models have been developed for symmetrical triangular cuts. (iv) Experimental results showed that the thin plane model, with the chip critically stressed, adequately describes the cutting process for symmetrical triangular cuts.
Machining
with Double
Cutting Edge Tools-I.
Symmetrical Triangular cuts
REFERENCES [I] V. PIISPANEN,Teknillinen Aika Kauslenti 27, 315 (1937). [2] M. E. MERCHANT,J. Appl. Phys. 16,267 and 318 (1945). [3] E. H. LEE and B. W. SHAFFER,J. Appl. Mech. 18,405 (1951). [4] W. B. PALMERand P. L. B. OXLEY, Proc. Znstn mech. Engrs 173, 623 (1959). [5] K. OKUSHIMAand K. HITOMI,J. Engng Znd. 83, 545, (1961). [6] E. J. A. ARMAREGO,Int. J. Mach. Tool. Des. Res. 4, 189 (1965). [7] P. L. B. OXLEY and A. P. HATTON, Znt. J. mech. Sci. 3, 68, (1961). [8] S. KOBAYASHIand E. G. THOMSEN,J. Engng Znd. 81,251 (1959). [9] R. H. BROWN and E. J. A. ARMAREGO,Znt. J. Mach. Tool Des. Res. 4, 9 (1964). IlO] W. R. BACKER, E. R. MARSHALLand M. C. SHAW, Trans. Am. Sot. mech. Engrs 74, 61 (1952). -1I] E. J. A. ARMAREGO,Znt. J. Mach. Tool Des. Res., 6, 139 (1966).
37
MACHINING WITH DOUBLE CUTTING TOOLS-I. SYMMETRICAL TRIANGULAR
EDGE CUTS
E. J. A. ARMAREGO* (Received
29 August 1966)
Abstract-The plane deformation models are developed for symmetrical triangular cuts. Single-edge orthogonal cuts, full-depth and non-full-depth triangular cuts are tested and compared. It is shown that theoretical and experimental force functions are consistent when the force intercepts are considered as tool edge forces. It is also found that the thin plane model, with the chip critically stressed, correlates with triangular cut results for the conditions tested. INTRODUCTION NEED to study the mechanics of cutting of double-edged tools is obvious when one considers the geometry of practical cutting tools. The analysis for single-edged tools, although not completely solved, has proved very useful to describe the influence of various parameters on the deformation and power in cutting. In general the thin plane (or thin zone) models suggested by Piispanen [l], Merchant [2], Lee and Shaffer [3] and other workers have been more widely used than the thick zone models studied by Palmer and Oxley [4] or Okushima and Hitomi [SJ. This follows since the thick zone models are too complex for practical use and furthermore there is some doubt that such models are representative of the deformation at the usual cutting speeds. It is therefore important to study the full potential of the thin plane models when applied to cases other than those with a single cutting edge. THE
NOMENCLATURE
bc T TC
Workpiece velocity. Chip velocity. Shear velocity. Rake angle. Tool included angle, groove included angle. Normal clearance angle. Reference clearance angle. Length of workpiece cut. Length of chip Maximum width of triangular cut; width of cut for single-edge orthogonal cutting. Maximum chip width for triangular cuts. Maximum undeformed chip thickness for full-depth triangular cuts; maximum groove depth for non-full-depth triangular cuts. Maximum chip thickness for full-depth triangular cuts. Undeformed chip thickness for non-full depth triangular cuts and for single-edge orthogonal cutting. * Department
of Mechanical Engineering,
University of Melbourne, 23
Victoria, Australia.
24
E. J. A. ARMAREGO
CJI,02,
“3
P dcz 2.4s 24 FS F* FP FQ FH
R F N K~P, KIP KZQ, KIQ k2p, klp, kzu,
Shear angle. Friction angle. Shear stress in the shear plane. Norma1 stress on the shear plane. Maximum shear stress. Principal stresses. Hydrostatic stress. Strain increment. Total shear plane area. Projection of total shear plane area in maximum shear stress plane. Shear force in one shear plane. Norma1 force on one shear plane. Force in velocity plane along the work velocity direction. Thrust force in the velocity plane perpendicular to the work velocity direction. Side force on one shear plane perpendicular to the velocity plane. Resultant force in velocity plane. Friction force in the rake face. Force normal to the rake face. Experimental force constants of Fp and FQ components for single-edge orthogonal cutting. Experimental force constants of Fp and FQ components for triangular cuts.
km
MECHANICS
OF
CUTTING
FOR
SYMMETRICAL
TRIANGULAR
CUTS
Cutting with symmetrical triangular tools as shown in Fig. 2 is a special case of machining with two cutting edges. The cuts taken may be “full-depth” or “non-full-depth” as seen in Fig. 2 a and b respectively. It is readily expected that due to symmetry the workpiece and chip velocities will be in planes parallel to the plane of symmetry. Before proceeding with the analysis of the deformation and forces in cutting it is necessary to consider the tool geometry.
FIG. I. The geometry
and specification
of symmetrical
triangular
tools.
Machining with Double Cutting Edge Tools-I.
Symmetrical Triangular Cuts
2.5
In Fig. 1 the relevant shape of the triangular tool is shown. The tool consists of a rake face ABO and two clearance faces GOA, GOB. The cutting edges are represented by A0 and BO. The rake angle a is specified in the plane of symmetry of the tool and cut, and is thus also in the velocity plane. It will be shown that this rake angle is comparable to the rake angle for orthogonal cutting with a single cutting edge. The included angle 26 is peculiar to this type of tools and is therefore unrelated to any of the angles for single-edged tools. The angle 26 is specified in the plane perpendicular to the work velocity VW and is therefore equal to the angle of the machined groove. The clearance angles are considered in planes perpendicular to the cutting edges A0 and BO i.e. normal clearance angles Cl,. These angles represent the actual clearance between the newly-cut surface and the tool clearance faces. The clearance angle Cl in Fig. 1 is a useful reference angle for checking the normal clearance angles ground on the tool. The clearance angles are related by the equations below tan Cl, tan Cl = (1) sm S2/(1 + C0S2 6 tan2 a) + cos2 6 tan a tan Cl, when a = 0 tan Cl, tan Cl = ___ (2) sin 6 There is little difficulty in grinding these tools to the correct rake the specified angles can be directly set on the machine vice axes. It to grind the clearance faces by setting the clearance angles about axes edges if the correct geometry is to be readily achieved. The problems rectly grinding the clearance angles for lathe tools (i.e. double cutting shown by the author [6]. Thin plane deformation
model for full-depth
and included angles; is however important parallel to the cutting introduced by incoredge tools) have been
cuts
The deformation for full-depth cuts may be considered as two shear planes pivoted about the cutting edges as shown in Fig. 2 a. The work, chip and shear velocities will be in planes parallel to the plane of symmetry. It may be assumed that plane strain conditions will prevail so that the chip width b, is equal to the groove width b. From continuity and incompressibility conditions it follows that TJ,bc
= Tlb
(3)
T Te
(4)
thus le r=-_=_ 1 From the deformation
geometry
tan 4 = [(T/Te)
cos
the shear angle is given by a]/[1 -
(T/Tc) sin a] = r
Equation 5 is thus the same as for orthogonal From plane strain conditions the stresses shows the tool rake face and the shear planes, stresses on elements close to the shear plane. a2 = (al + v3)/2 = p and from the equilibrium shear plane, the shear stress in the shear plane
cos
a/(1
-
r sin
a)
(9
cutting with a single cutting edge. on the shear plane can be studied. Figure 3 from a number of views, together with the It is seen that for plane strain (i.e. drz = 0) of element x, an element sectioned by the is given by
s, = 7 sin 6’ = rAr/Ag
(6)
E. J. A. ARMAREGO
26
and the normal
stress on the shear planes
Us is equal to the hydrostatic
stress
i.e.
% = p where T = maximum shear stress, A, = area of one shear plane, A, = projected area of shear plane in the maximum 8 ,y”
Y
(7)
shear stress plane.
6
T> Cr
A'--,
1
‘\ ‘N
’
I 1: I /’
,’ /’
W
v b=b,
B
/
AB\\ *. -----_1
. .
. .
\
\
+
/’ 0
v,
b=b,
FIG. 2. Thin plane deformation
for symmetrical triangular cuts. (a) Full-depth cuts; (b) Non-fuIl-depth cuts.
7
Machining with Double Cutting Edge Tools-I.
Symmetrical Triangular Cuts
FIG. 3. Shear plane stresses for plane strain conditions
27
(full-depth triangular cuts).
Jt is thus apparent that the shear stress in the shear plane is not the maximum shear stress and in fact the shearing process approximates planes originating from the cutting edges. Whether or not 03 equals zero will depend on the deformation model envisaged. Thus if it is considered that the chip is loaded to the yield stress between the shear planes and the rake face, 03 = 0 as for the Lee and Shaffer type solution. Alternatively if we have a thin plane, and the chip is not critically stressed, the Oxley- [7] or Merchant-type solution may be used in which cases us is determined or ignored respectively.
Force analysis (i) Merchant-type solution. As for the Merchant model it may be assumed that the resultant force acting on the chip at the shear planes is balanced by an equal, opposite and colinear force acting on the chip at the tool-chip interface. It is also assumed that the cutting edges are sharp and no forces act on these edges. Figure 4 shows the forces acting on the
End an view of
FIG. 4. Force diagram
shear
plones
for thin plane model (full-depth triangular cuts).
28
E.J.A. ARMAREGO
shear planes and the components of these forces in and perpendicular to the plane of symmetry. It is noted that the shear forces in the shear plane are colinear with the shear velocity. Considering the forces on one shear plane, the shear force in the shear plane is found from the force triangle, thus Fs = 2” COS(+ + X - a) where F8 = R = h = 4 = a =
shear force in one shear plane, resultant force in the plane of symmetry friction angle in the plane of symmetry, shear angle, rake angle.
The shear force can also be expressed
(or velocity
(8)
plane),
as Fs = &A,
(9)
From equations (8) and (9) and the geometry of Fig. 4, the power force F,/2, Fo/2 and side force FR due to one shear plane are given by
FP _ &As cos (A 2
&A, sin (A -
a) a)
FR = &As tan (4 + A -
force
(10)
u)
cos (4 + A -
FQ _
2
a)
cos (#J -t A -
thrust
(11) a) cot 6’.
(12)
Equations (IO), (11) and (12) apply for the second shear plane except that the side force Fn acts in the opposite sense resulting in no total side force. From equations (IO), (11) and (6) the force components can be expressed in terms of the maximum shear stress T and the corresponding shear area, i.e. F
_ 27A, cos (A P
COS(4
+
/I-
a) = a)
7T2 tan 6 cos (h - a) sin #J cos (4 + h - a)
FQ = yT2 tan 6 sin (A - a) sm 95cos (4 + h - a)
(13)
(14)
The above equations are equivalent to those for orthogonal cutting with a single cutting edge and a, 4 and h have the usual meaning (note that T2 tan 6 is the area of cut corresponding to bt for “rectangular” orthogonal cuts). This analysis does not relate A, 4 and CLin the form of a shear angle relation, although the minimum energy principle can be adopted for this case provided one accepts the assumptions applied by Merchant. (ii) Lee and ShafSer-type solution. The previous analysis of forces will be unaltered if the chip is considered to be loaded to the yield stress as suggested by Lee and Shaffer. The resultant force, however, will act along the al direction in the plane of symmetry since the
Machining with Double Cutting Edge Tools-I.
Symmetrical Triangular Cuts
side forces are balanced by the hydrostatic stresses and ~3 = 0. This analysis and CLby the Mohr stress circle and is given below
29
relates
A, +
Equation (15) can be applied in conjunction with equations (13) and (14) to arrive at the FP and FQ forces. The deformation and stress affected region is shown in Fig. 5.
stress
Shear
View of stressed region m 7 directIon
p=T=uz
stress on PIones porollel to oc Direct stress
stress on rake face
A
FIG.
5.
Deformation
Thin plane deformation
2+a)
Y +I!!!
c3
ml
and stresses for thin plane model (chip critically stressed).
model for non-fill-depth
cuts
The analysis for full-depth cuts is the same as for the above cases but equations and (14) are modified to account for the appropriate area of cut, thus F P
(13)
= 9(2T - I) tan 6 cos (A - a) sin 4 cos (4 + X - a)
FQ = 7t(2T - t) tan 6 sin (A - -~a) sin 4 cos (4 + h - a)
(17)
where t = undeformed chip thickness, T = maximum groove depth. EXPERIMENTAL
RESULTS
AND
DISCUSSIONS
were devised to study the mechanics of cutting for full-depth triangular cuts. Orthogonal cutting (with a single cutting edge) tests were initially run to represent “standard” data for comparison with the full-depth triangular tests. Further experiments were run to compare full-depth results to non-full-depth tests. The experiments were run on a universal milling machine. Cutting tools (18-4-1 H.S.S.) were fitted to a stationary three-component Tests
30
E. J. A. ARMAREGO
strain gauge dynamometer mounted on the machine column. The cutting forces were registered on a pen recorder and the workpiece (65ST6 aluminium alloy 4 in. x 6 in. Y 6 in.) was held in a vice on the milling machine table. All the cuts were performed by feeding the workpiece past the stationary tool at 9$ in./min. Orthogonal
cutting with a single cutting edge
Tools with rake angles of lo”, 20” and 30”, and 6” clearance angles were used. Each tool was tested for a range of undeformed chip thicknesses and four widths of cut. The undeformed chip thickness was set with the aid of a dial gauge and was further checked by the chip weight technique. The forces and shear angles for the 20” rake angle tool are shown in Figs. 6 and 5. From Fig. 6 it is seen that the force per unit width is a linear function of the undeformed chip thickness, with a positive intercept at t = 0. It follows that Fpib = Kzpt i
IGp
FQ/b = &jt
KIQ
+
(18)
or Fp = Kz,bt
+ Kl,b (18a)
FQ = K2Qbt + Klob
Undeformed FIG. 6. Influence
of undeformed
chip thickness,
chip thickness orthogonal
t in.
on the force per unit width cutting.
~1 :z 20’, 18-4-l H.S.S. tool. 65ST6 Al alloy workpiece. Dry cutting. V = 92 in/m& 0 h = 0.24 in. c b = 0.20 in. a b : 0.15 in. 0 h 7 0.10 in.
for single-edge
Machining with Double Cutting Edge Tools-I.
0
I
I
I
0,002
0.004
Undeformed
Symmetrica
I
I
0.006
Triangular Cuts
0~008
0~010
31
I 0.012
t in.
chip thickness,
FIG. 7. Influence of undeformed chip thickness on the shear angle for single edge orthogonal cutting. Cutting conditions as in Fig. 6.
The above equations suggest that the cutting forces depend on the area cut as well as the length of the cutting edge in contact with the work. Figure 7 shows that the shear angle, and hence the deformation, is unaffected by the undeformed chip thickness or the width of cut in these tests. The interpretation of the force functions is of some interest. The physical significance of the intercept force has been discussed by various authors. A number of workers [8,9] have assumed that the intercepts represent tool edge forces which do not contribute to the chip formation process while others [lo] claimed that a size effect occurs in metal cutting which leads to higher shear stresses and hence relatively higher forces at the smaller undeformed chip thickness values. It may be qualitatively shown that the former explanation is consistent with the experimental data in this work. Considering the force relations for orthogonal cutting with a single cutting edge it is known that Tbt cos (A - u) Fp= . sm + cos (4 + X - CC) (19) I;Q = . Tbt sin (X - a) sin 4 cos (4 + h - a) ’ The above relations are derived on the assumption that no rubbing or edge forces act at the tool tip. If it is now assumed that the shear stress exhibits a size effect as in equation (20),
r-c+” where C and D are constants, .
(20)
then the force equations
ccoso-
4
sm $ cos (4 + h C sin (X - a) sin + cos (4 + X 3
t
a) 1 a)
bt +
(19) become
D cos (A - a) b sin 4 cos (4 + h a) [ I D sin (h - a) sin + cos (4 + A -
(21)
b a) I
.
32
E. J. A. ARMAREGO
,
Equations (21) and (18a) are of the same form provided $J, a and h are constant, in which case the whole measured force contributes to the deformation process. Experimentally it is seen that 4 is independent of b and t. The experimental friction angle X is found by
(22) which shows that X is dependent of t (since K&Kzp # Kl~/Klp) and is not constant. Thus equation (21) is not consistent with the experimental data. When it is considered that the shear stress is constant and that an edge force if existent is not included in the analysis then equations (19) lead to F
=
. rbt cos (A - a) ~ ~ = constant. sm 4 cos (4 t--X - U)
bt
FQ =
Tbt sin (A - a,) ~ constant. sm + cos (f$ + X - a)
bt
P
(23)
provided 4, a and X are constant. Correlation between equations (18a) and (23) occurs if Klpb and KlQb are considered as edge forces (unaccounted in the analysis of cutting). It is also necessary to show that 4 and h are constant for a given a. Experimentally # is constant and h is found from tan (A -
K~Q U) = -= constant K~P
giving a constant A. Thus the experimental and analytical force relations are consistent when the intercept forces are considered as edge forces not contributing to the chip formation process. and given in From equations (18) (24) and (19) Kep, KZQ, h and 7 are determined Table 1. The shear angle relation shown in Fig. 8 leads to a linear relation but does not fit the Merchant or Lee and Shaffer solutions. Similar discrepancies with these and other theories have been reported by the author [l I].
(X-a)
FIG. 8. Comparison
of experimental and theoretical shear angle relations orthogonal cutting. Conditions as in Fig. 6.
for single-edge
Machining with Double Cutting Edge Tools--T. Symmetrical Triangular Cuts TABLE
I.%NGLE-EDGEORTHOGONALCUTTING
Kzp ( x IO3psi)
KP& (X IO3 psi)
123 110 85.7
35.8 18.5 5.3
Full-depth symmetrical
DATA
1” 24.5 29.7 35.1
IO 20 30
33
T(X 103psi)
26.3 29.5 33.5
40 42.8 38.6
triangular cuts
Three sets of tools with rake angles O”, 10” and 20”, and 6” normal clearance angles were tested. For each rake angle four included angles were used. The undeformed chip thickness was set using a dial gauge and the values checked by measuring the groove width with a Brine11 microscope. Each tool was tested for a range of undeformed chip thickness values. From the previous single cutting edge tests it was expected that the cutting forces would depend on the area of cut and the length of cutting edge in contact with the work. For full-depth triangular cuts the forces could be represented by FP = kzpT2 + kl,T
(25)
FQ = k2QT2 + klQT or F,IT =
K2pT
+
kl,
(254
FQIT = k2QT + klQ.
0
I 0~010
Maximum ,FIG.9. Force
I 0.02
I 0.03
I 0.04
I 0.05
I
0.06
undeformed chip thickness, Tin.
per unit depth variation for full-dkpth sykmetr’ical triangular cuts. 18-4-l H.S.S tool, a = 0”, 6 = 45” V = 9H in,‘min. 65ST6 Al alloy workpiece. Dry cutting.
E. J. A. ARMAREG~
0
I
I
0.01
0.02
Maximum FIG.
IO.
Shear
angle
undeformed
I 0.03 chip
I
I
0.04
I
0.05
0.06
thickness,
T
for full-depth symmetrical Conditions in Fig. 9.
in.
triangular
variation
cuts.
Figures 9 and 10 show the results of one such test. It is seen that the shear angle is unaffected by the maximum undeformed chip thickness and the forces are represented by equation (25a). By considering kl,T and klQT as edge forces, the friction angle and shear stress were determined from modified forms of equations (24) and (13) tan (A -
i.e.
f~) = 4”
(26)
2P
and 7 = kzp +Acos (4 -t A - a) tan 6 cos (A - CI)
.
(27)
The calculated values for the tests run are shown in Table 2. In plotting the shear angle relation in Fig. 11 it is seen that the results correlate fairly well with the Lee and Shaffer shear angle relation. This correlation may appear to be unexpected when one considers that the results for orthogonal cutting with a single cutting edge usually show poor correlation with this theory. However, the plane strain assumption used in cutting analyses is more likely to occur with triangular tools, where the deformation is bounded by the material cut, than with orthogonal cutting with a single cutting edge. It is also noticed that the shear stress values in Table 2 show reasonable correlation with those found from single-edge cuts.
FIG. II. Comparison symmetrical triangular
of experimental and theoretical shear angle relations for full-depth cuts. 18-4-1 H.S.S. tool, V = 9$ in/min. 65ST6 Al alloy workpiece Dry cutting. W 01 = 0”; n a = 10”; n 01 = 20”.
Machining
with
Double Cutting Edge Tools-I.
Symmetrical
Triangular
Cuts
35
TABLE~.FULL-DEPTHSYMMETRICALTRIANGULARCUTHNGDATA kzp(lo4psi) 0 0 0 0 10 10 10 10 20 20 20 20
k,Q
68.05 28-08 14.75 7.96 49.68 23.10 1240 7.30 39.72 19.40 10.64 6.14
15 60 45 30 15 60 45 30 75 60 45 30
(lo4
psi)
rl”
+
14.2 17.1 20.8 24.3 20.3 24.8 27.0 29.1 29.3 30.1 34.6 27.0
34.60 1443 6.94 3.19 20.96 7.90 4.20 2.18 9.12 3.24 1.55 0.64
21.0 27.2 25.2 21-9 32.2 28.9 28.8 26.7 33.0 29.5 28.2 26.0
It thus appears that the slip line field solution for rigid ideally-plastic describe the cutting process for full-depth triangular cuts. Non-full-depth
symmetrical
t (103 psi)
38.0 38.5 40.0 42.3 37.5 42.6 41.4 44.7 40.0 43.9 45.3 47.0
conditions
can
triangular cuts
A number of tests were run with non-full-depth cuts keeping the undeformed chip thickness t constant and varying the groove depth T. Figures 12 and 13 show the force variation and shear angle for the tool with a = 0, 6 = 45” and t = 0.010 in. It is seen that the forces are linearly related to T and yield negative force intercepts. These results can be explained in terms of the full-depth cut results. It is reasonable to assume that the force difference between two full-depth cuts will result in the force for a non-full-depth cut 400-
300
-
2
e
200-
: u.
IOO-
Predicted -+
0
I 0.02
Maximum
FIG.12. Force
0.04
from
Experiment01
I
I
I
0.06
0.06
0.10
groove
depth.
T in.
variation for non-full-depth symmetrical triangular cuts. 18-4-l H.S.S. tool; 01 = 0”; 6 = 45”, I = 0.010 in. V = 95 in/min. 65ST6 Al alloy workpiece. Dry cutting.
36
E. J. A.
Maximum
FIG. 13. Shear angle variation
ARMAREGO
groove
depth,
T
in.
for non-full-depth symmetrical triangular cuts. Conditions as in Fig. 12.
provided due consideration is given to the edge forces and the difference in the shear and friction angles. The force for a non-full depth cut of maximum groove depth T and undeformed chip thickness t = T - Tl is given by the difference in deformation-contributing forces for full-depth cuts T and Tl plus the edge force for a full depth T. Thus from equations (25) Fp = kzp[T’
-
T12] + klpT
i.e. Fp = T[kl,
+ 2tk2,] -
kzpt2
FQ = T[klQ $- 2tkfQ] -
k2gt2
(28)
and similarly (284
For a fixed t the forces for a non-full-depth cut will have the same form as in Fig. 12 but when T < t the cuts become full-depth cuts (equations (25) rather than (28) apply) and the forces drop to zero when T equals nought. When the shear angles for full depth and non-full depth cuts are the same then kzp, kzQ, k~,, klQ from full depth cuts can be directly applied to equations (28). In the above test the shear angle for non-full depth is smaller than that for the corresponding full depth cuts. The shear angle relation in equation (15) was used to find the new h and from equations (27) and (26) the corrected values of kpz, and k2Q were determined. The edge force constants kl, and klQ were considered the same for both tests. The new constants were applied to equations (28) to correlate full-depth results to non-full-depth experimental data. The dashed line in Fig. 12 represents the force function obtained from full-depth data. It is seen that the correlation is good suggesting that the deformation for non-full-depth cuts is similar to the full-depth case and that the edge force interpretation is further verified.
CONCLUSIONS (i) It has been shown that the cutting forces are dependent on the area of cut and the length of the cutting edge in contact with the work. (ii) The force intercepts may be considered as tool edge forces not contributing to the chip formation. (iii) Thin plane deformation models have been developed for symmetrical triangular cuts. (iv) Experimental results showed that the thin plane model, with the chip critically stressed, adequately describes the cutting process for symmetrical triangular cuts.
Machining
with Double
Cutting Edge Tools-I.
Symmetrical Triangular cuts
REFERENCES [I] V. PIISPANEN,Teknillinen Aika Kauslenti 27, 315 (1937). [2] M. E. MERCHANT,J. Appl. Phys. 16,267 and 318 (1945). [3] E. H. LEE and B. W. SHAFFER,J. Appl. Mech. 18,405 (1951). [4] W. B. PALMERand P. L. B. OXLEY, Proc. Znstn mech. Engrs 173, 623 (1959). [5] K. OKUSHIMAand K. HITOMI,J. Engng Znd. 83, 545, (1961). [6] E. J. A. ARMAREGO,Int. J. Mach. Tool. Des. Res. 4, 189 (1965). [7] P. L. B. OXLEY and A. P. HATTON, Znt. J. mech. Sci. 3, 68, (1961). [8] S. KOBAYASHIand E. G. THOMSEN,J. Engng Znd. 81,251 (1959). [9] R. H. BROWN and E. J. A. ARMAREGO,Znt. J. Mach. Tool Des. Res. 4, 9 (1964). IlO] W. R. BACKER, E. R. MARSHALLand M. C. SHAW, Trans. Am. Sot. mech. Engrs 74, 61 (1952). -1I] E. J. A. ARMAREGO,Znt. J. Mach. Tool Des. Res., 6, 139 (1966).
37