A note on the shape of the shear-plane in orthogonal cutting
Citation for published version (APA):
Mot, E., & Veenstra, P. C. (1967). A note on the shape of the shear-plane in orthogonal cutting. International
journal of machine tool design and research, 7(2), 169-173. https://doi.org/10.1016/0020-7357(67)90031-5
DOI:
10.1016/0020-7357(67)90031-5
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Published: 01/01/1967
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A NOTE ON THE SHAPE OF THE SHEAR-PLANE IN
O R T H O G O N A L CUTTING
E. MOT* a n d P. C. V E E N S T R A t
(Received 23 November 1966)
Abstract--In the classical model of orthogonal cutting, as developed by M. E. Merchant, the shear-plane is represented by a flat plane. Microphotographs have shown that this assumption can be considered a reasonable approximation of reality.
However, it can be proved that this model is inconsistent with the fact that the chip curls during the cutting-process.
In this paper an effort has been made to produce an extension of the Merchant model, assuming a curved chip and, consequently, a curved shear-plane.
T H E O R Y
The classical m o d e l
T o DESCRIBE this m o d e l the f o l l o w i n g quantities are defined t ~ feed per rev
tl ~ thickness o f chip d ~ d e p t h o f cut dx --- w i d t h o f chip ~ n o r m a l side r a k e angle ~b == shear angle V - - cutting speed
V T ~ speed o f chip a l o n g the rake
A tensile strain parallel to the c u t t i n g edge.
0 V
FIG. 1. The classical model of orthogonal cutting.
* Graduate student, Department of Production Engineering, Technological University Eindhoven, Netherlands.
t Professor of Production Engineering, Technological University Eindhoven, Netherlands. 169
170 E. MOT and P. C. VEENSTRA
T h e tensile strain in the direction parallel to the cutting edge (A) follows f r o m
d l = d ( 1 t A ) . (1) Further, f r o m geometrical considerations, it can be seen (Fig. 1) that
t cos (4 -- ~)
tl -- - - : - . (2) sin 6
The relation between V and VT follows f r o m the continuity equation for the material which passes along the shear-plane
t . d . V = t l . dl . VT. (3)
Using (1) and (2) we find from (3)
V T = V si_n__~ .... (4) (1 + A) cos (4 - - c0"
Thus we find V T as a function o f 4.
W h e n VT does not vary in the X-direction, the chip will c o m e off straight. W e know, however, that this is not correct: the chip actually comes off in a curved shape. This is only possible when a speed distribution V T = VT(X) exists, thus, using (4): V T = V T ( 6 ) ,
leading to 6 - - 6(x) • This implies a curved shear-plane.
T h e curved shear-plane m o d e l
W e will assume a chip contact-length o f lc. The velocity-field is chosen in such a way that all along any line A B C which is pseudo-parallel to the chip-boundary, the speed o f the chip is constant. We shall also assume that the curved part o f the chip is a rigid body. I f the outside radius o f the chip is R, we find the pole M according to Fig. 2. T h e relation between
Vr(O) and VT(X) is then given by
V T ( x ) - VT(0) R - - x cos (4 - - ~) (5) R
qb-a
T h e a s s u m e d speed distribution implies t h a t the p a r t O D E F c a n n o t be entirely rigid. Actually, f r o m experiments it is k n o w n t h a t secondary plastic d e f o r m a t i o n does take place in t h a t region, especially n e a r the r a k e plane o f the tool. T o describe the shape o f the shear plane, we superimpose a variable angle ~b = ¢(x) on the constant angle ¢. Then, localisation o f (4) gives Vsin [¢ + ~b(x)] (6) (1 + A) cos [¢ -- c, + ~b(x)]" VT(x) = Substituting x = 0 in (6) we obtain v ~ ( 0 ) = (6) and (7) in (5) gives sin [¢ + ¢(x)] c o s [ 4 - a + ¢ ( x ) ] V sin [¢ + ~b(O)] (1 + A) cos [¢ - - a + ~b(O)] (7) R - - x c o s ( ¢ - - ~) sin [¢ + ¢ ( 0 ) 1 R " c o s [4 - ~ + ¢ ( 0 ) ] " ( 8 ) T h e angle ~b is small. T h u s we a p p r o x i m a t e cos ~b ~ 1 s i n ~ m ~ b ~ t a n ¢ - - d y _ y , . dx F r o m (8) we can n o w easily derive
sin ¢ + y ' cos ¢ R - - x c o s ( ¢ - =) R
cos (¢ - - ~) - - y ' sin (¢ - - a)
or
in which
sin ¢ + y'(0) cos ¢ cos (¢ - - a) - - y'(O) sin (¢ - - a)
A x + B
y ¢ - - _ _
C x + D A = - - c o s 2 (¢ - - a){sin ¢ + y'(0) cos ¢}
B = y ' ( 0 ) R cos a
C = - - c o s (¢ - - a) sin (4' - - a){sin ¢ + y'(0) cos ¢} D = R cos c~. (9) (10) ( l l ) (12) Integration o f (11) gives A y = ~ . x + with b o u n d a r y conditions BC -- AD C2 In (Cx + D) + C1 (113) y(0) = 0 a n d y = 0. (13a) By substitution o f (13a) in (13) we find two equations, which enable us to solve the u n k n o w n quantities C1 a n d y'(0). T h e n the shape o f the shear-plane is known.
172 E. MOT and P. C. VEENSTRA We will illustrate this consideration with a numerical example
30 ° ~ - - 10 ° R = 5 m m
t :: 0.6 mm/rev.
F r o m these assumed data, which m a y according to (12) A - ~ 0 . 4 4 1 5 - 0.76477'(0) B = 4"9240y'(0) C 0.1607 -- 0"2783y'(0) D 4-9240. Substitution o f (15) in (13) gives 4 " 9 2 4 o y ' ( 0 ) - - 13.5287 y 2.7475x -- 0"2783y'(0)~- 0"1609 The b o u n d a r y conditions y(0) 0; can now be substituted in (16). Then, we find
(14)
be obtained f r o m experiments, we calculate
05)
In [{ 0.1607 - - 0 . 2 7 8 3 y ' ( 0 ) } x + 4.9240] + C1. (16) y(1.200) =: 0 (16a) 0 4"9240y'(0) -- 13.5287 In 4.9240 + C1 (17a) 0.2783y'(0) q- 0.1607 0 ~= 3.297 4.9240y'(0) - 13.5287 In {4.7312 - 0.334oy'(0)} + G . (lVb) 0.27837'(0) + 0.1607Elimination o f C1 from (17a) and (17b) gives
In {4.731 -- 0.3340y'(0)} 8.7669y'(0) -- 21-0363 = 0. (18) 4 : 9 2 4 0 y ' ( 0 ) - 13-5287
(18) can be solved by successive approximation according to Newton. Then (17a) gives (71. We find
y'(0) -- 0.0576 (19)
Cj - --119.4901
F o r the equation o f the shear-plane, we find in this case, by substitution o f (19) in (16) y : 2.7475x 119.4901 ~ 74.9581 In ( 0"1767x + 4"9240). (20) F r o m (11) we find
--0.4856x + 0"2836
T a b l e 1 gives s o m e local values o f the a c t u a l shear angle.
TABLE 1
x w ~ Shear
(ram) (tad) (degrees) angle 0 0-0576 3 ° 19' 33 ° 19' 0-2 0-0385 2 ° 13' 32 ° 13' 0"4 0-0184 1 ° 4' 31 ° 4' 0"6 0-0014 --0 ° 5' 29 ° 55' 0"8 --0'0215 -- 1 ° 14' 28 ° 46' 1'0 --0"0415 --2 ° 22' 27 ° 38' 1'2 --0"0622 3 ° 34' 26 ° 26'
Remark. It seems only correct to m e n t i o n a l i m i t a t i o n o f this m o d e l ; we have, indeed, neglected the fact t h a t a t e m p e r a t u r e d i s t r i b u t i o n T = T(x) exists, which will influence the value o f R as l o n g as the c h i p is in t o u c h with the tool. A f t e r the chip has c o m e off, we m a y m e a s u r e a different o u t s i d e radius, since the t e m p e r a t u r e d i s t r i b u t i o n t h e n has d i s a p p e a r e d .
C O N C L U S I O N A N D P R O S P E C T S
W e have s h o w n t h a t f r o m e x p e r i m e n t a l d a t a the local value o f the shear angle can be c o m p u t e d . Its v a r i a t i o n m a y be in the o r d e r o f :k l0 p e r cent. I n the existing l i t e r a t u r e o n o r t h o g o n a l cutting, b a s e d on the M e r c h a n t m o d e l , in m a n y cases average stresses are calculated, d e p e n d e n t o n the value o f the shear angle.
W i t h the help o f this t h e o r y , these c a l c u l a t e d values m a y be localised, thus giving a stress d i s t r i b u t i o n . I n practice this m e a n s t h a t the m e a s u r e m e n t o f the r a d i u s R in a d d i t i o n to the usual q u a n t i t i e s m e a s u r e d , enables us to find a stress d i s t r i b u t i o n o n the shear p l a n e f r o m a n u m b e r o f e x p e r i m e n t s which each i n d i v i d u a l l y p r o v i d e us with values o f average stresses. T h e s o l u t i o n will, however, be only k i n e m a t i c a l l y admissible. I f ey(X) is the n o r m a l stress d i s t r i b u t i o n o n the s h e a r - p l a n e a n d P T the n o r m a l pressure on the tool, we m a y find the w o r k i n g p o i n t o f P T using the e q u i l i b r i u m o f m o m e n t o f the chip r o u n d O
t/sin 4
f d ~ru(x) x dx = P T 4 b
(22)
0where b is the distance between O a n d the w o r k i n g p o i n t o f PT. In (22) we m a y substitute either d ---- c o n s t a n t o r d = d(x).
