A Textbook of Production Enginwring
I
= 1%min.
Example 2. For a metal m a c h i n i ~ ; ~ & e f i winformation in is available : Tool change time, = 8 min -I ' 1i1iri { V i ~ a ! ,iI < , .IT, . - * i t !tv,tv ~ o ore-grid i timea=3 nr'in' Machine running cost, = @ 5 per hour Tool depreciation per re-grind, = 30 p Ra * : ~ r f i ~
1
a!. r,- ,,. , , I ' . . ,. ., Calculating the optimum cutting speed Solution. Tooling cost C, = Tool,changecost + tool regrind cost + tool depreciation , ' 3 ~ .J 5. 5 t '
= -~8+-x5+030
1..
,
5
i
snidagrr, silt
..
=
0
GO--,*<
Rs,1:38
,
@$?-I x; .lo ,i
7 .
bne
,
Example 3. In an orthogonal Wthg operatiout, t kfol&?wjn~$$$h~e been gbewed : Uncut chip thgkness, . t = 0.127 mm .--. ikji:~,~ I :,.. . Mdthofcut,. .<":.; 3 b = 6.35 mm I ,'..c:, . l~ .:: I Cutring speed V =t m h ' RO& angle, a = 10° F, = 567 N , Cutting farce, 1 Thrust force, E, i.627 N Chip thickness, t, = 0.228 mm I , Determine :SIsesrr fthe@&ti~i@atk he shew plane and €her power in &@ ond shew strain rate. for the cutting operatioa Also find &\@&~vgk#*.i%. 1
,
.
1
'
.
s
11
t
Solution. (i) Shear angle,
1-rsina '
1
r
t =ah-
= -'
4-
Ill,
I
0.127 " 0.557 0228 , < , .
r=,pB = F, sina + F, cosa
.st
*A. -
F, cosa-F, sina
$ = ) tan-' (6.64) = 32.62O
''
(iii) Shear force, I
F8=Fccoscp-F,sincp ,
= 567x0855-227x0519
Cutting power, =
(v) Now
FV 567x2 = loo0 1mo
v Chip velocity, Ve = 2 x 0557 = 1.1 14 mls Shear strain, s = cot p'+ tan (rp - a)
A Textbook of Protjtptjm Am@nq@ig ... (vii)
Shear plane
lens =
0.127 -sln cp - 0519 t
"'"
'
= 0.245 mm. Taking the thickness of deformation zone equal to one-tenth of shear plane length. t$,(eq. 14.1 I ) = 0.0245 mm Shear strain rate, 3 =
vs 1s
v- = cosV (9cos-aa)
Now ,:
*:.
.-..
:
5 r
2 x 0.985 .. . = 2.1 1 m/s ~0~21.25
-.
Example 4. The following equation fir tool life is given for a turning operation :
FE=
:b
0.77 0.37
~ ~ " 1 f3
d
.?a
=
C
A 60 minute tool life was obtained while cutting at V = 30 mlmin, f d = 2.5 mm.
= 0.3 mm/rev. and
Determine the change in f q l life the cutting speed, feed and depth of cut are increased by 20% individually and also taken together:
( i ) Now
..
YW'
-
3
k
, -- o
g*fi*~
".$ $F.-% ' - s..;J,. f = 0.3 x 12 = 0.36 mdrev. -$ ,*p!t -&'gw-=&?
(ii) Now
'
*i@ 1
'&
. ' ,&q (1.48)'" = 20.39 min . ";- " - . . - _ .. ~!.i)d~25xld=3mm. ,
. . -
P''-.
(iii) Now
a
.['.
-
C Z P 28-38 ~ 8 ,L;~ i 38 x 03% x (3)'" I
.
A
6~'"9
ij7.a =
1.591
=
P8&
35.38 hin
h~y$tl(<~4
:.
The maximum tffeet on tool life is of cutting speed, and the least effkct is of depth of cut. (iv) Now V = 36 &in, f = Q.36 mmlrev. d = 3 mm
b ~ (1.154)'" = *oil nin. yk &hi = n mnl T i Example 5A Bring an orthogonal machining (turnihh operation of ~ 4 steel, 0 the - - 7 < : I * I I 2i;i?lI...\ Chip thickness = 0.45 mm Width of cut = 2.5 mm. !(I$: .iP:-frlr r,ur3r;uS31-q1;1.+ Feed = 0.25 mdrev - Tangential cut force = IJ3@ N 111 t:c! 3 i ~ ' r e >ssdZ Feed thrust force = 295 N Cutting speqi = 2.5 4 s ..
' m 3b ~
3 ,w
?
*-
,
Emtfib
w e = + 100
=
of shear at the shear plane. (b) Kinetic co-eflcient ofpiction at the
4k*&43 mm, Q = 2.5 mm, a
lo0, t
=
f
=
0.25 mm,
= F c c o s 9 - F,ein cg, where cp is.sbeararsglw, '1
tap=
l-rsi,,
~ L * , ~ H C ' , \ ~3iTT .>~
ti
I
-
A%.u\+.i<*, J *$%s,r-~
.
,.,I
,* 4
.;(,it
.:>:
p
A Textbook of P&t&n
;:
(ii)
- --
-
-
+ F, (eqn. 14.9) 4,- F, tan a
(1' .. F 0 CL = F, tan a
Example 6. The following data relate to an orthogonal turning process
3
:
Chip thickness = 0.62 mm itli k*,:?*, .SS~I~ n t k : r n l jam % ~ h !O ZF )3&> l~zxtlM+ br,
5P.
.a,
Engrngr&g
"6
(y
i 3 * 3 t t b + - ~ ~ n o ~ \ ~rwl, ~ o + b , 9
(0
Cutting ratio, r
=
,
~
~
,
,-c
4:.
..,.
. -.
,
,<
!, i
1 s
7-
t"
'I
r m a 1-rsjna
,=-,
.a
r t
- = 0.322
Chip-reduction co-efficient = - = 3.1
Shear angle, tan cp =
~ % ;*
t
\ \ ,
1
*
i,,:<
. iis..
, %
= 2.947+ 0.065 = 3.012
Example 7. The Tayloriein tool-lfe equation for machining C-40 steel with a I 8 : 4 : I H.S.S. Euiflrc;gMat!afeCd-~O.B ntnr/Wn d a depth of cut of 2 mm is given by VP = C, where n and C are constants. TkfilIowing V and T observations have bem noted : C: d m i n 25 35 $- - ; ; : , q,' 11' I Z min 90 20 > r. . , C 7 , : t ii - - -Calculate (i) n and C. ,:# (1 (ii) Hence recommend the cutting speed for a desired tool l f e of 60 minutes. Solution. > < %. . ,, . n"=C
- ., V = 2739 mlmin. Example 8. The following data from an orthogonal cutting test is available'? '.' ' ! Rake angle, = lJO b9w. is-\%I Chip-thickness ratio. = 0.383 :iizmrl ~Au:'. t = 0.5 mm Uncut chip thickness, Width of cut, b=3mm Yield stress of material in shear, = 280 Ni'mm' Average co-eflcietlr of friction = 0.7 on the tool face
.
*
'
Determine the nglrpal and t~ngentialforces on the tool face.
-----
Solution. Now,
nii
i 3E 1:
tan cp =
r 00s a 1-rsinq -
4
:,;,I
'~2qf-"< la,:
8
'
+
r = 0.353 and @- p 15" <
..
.:*>:
r,i)-,:+-t
19rw(1., ' 1 , ,C,i,l .,
z,.t;r:tcwP (3*7-+) '
Friction angle, $ = 350-
-(3) jr =
Fc=
Now
,r
;
.vbt
sec [fi
- a)-cos(cp +,$ -a)sin cp
Now
" , ~ k r2& F '
-,.
F l = S l l . 3 N .- h .1>'iL. 3:1Eit2 W P ~ . Yw
>w..&
-
A Textbook of Produdon Enginewing
N = Nonnel force on tool face :. = Fc.cos cr - F,sin a = 1405.7 x 0.966 - 51 1.3 x 0.259
I!
Example 9. Thefollowing observations w e e wade during orthogonal cutting of steel tube a&l&42&'Rs. % . , on a lathe : m*h of ~ ~ k i r y l ~ ~prt\u=, &&f5 T $ r n ~ % , , ,,4,4~<9,.,! Cutting speed. P-z'8.2m/@p , Rake angle, .a = 20° *Q. '
-
4
= 3~
= 784 @)@+$'
NI t
Shear angle, tan cp =
rcosa 1
; i na
From here,
cp = 20'.5O
..
p = 34.5"
reo
Now shear strain
i
',
,hi
I
l i ~ 3 i
I
~~
.
'
\YUL()LNin
mmz
. *, . ,%?,,?&
-.,
.nl.t i.s;@.
5i+~~.~~w l ,4c i L ~ ,
.
--
!GOL151 x sin ZOOeQvlirrilqtjl\r
c
4
- 20°)
Now from the relations of chapter 13, (Eqn. 13.6), fiK simple tension test, + j L . ~ C 3;
r
k
ef
For garnliaos-
G.
,$3t5fx\
'.\
a@+tm(q-a) 921 = & 20.5" + tan (20.5' ?c.sS ='2.474
; . '=
,.,-.,C:
mkw%& attw-,~tp . , ,~qc.r,.:t
Find Fc and F, given tensile properly of material as
a
.
.
t = o.a-@qD,q3.
r = 0.351.-C.& = 9
+ p- a
.Y 2;tl! j m r 2
a= k
A
'
'
t:
(E)~
B = K.(T]*
- -- P
92
and
I
(Considering Von Mise's yield condition)ll i . 5 384i (4,605)o."
.
Yield shear stress, 7, =
S 841.66 =f i \-ld
= 485.95
Now shear plane area, A = & :)?
j:>c
:
N/mm2
P8P
,
&. 5x025 = @;., sin 205 I
lr
3.65 mm2 1
ti;.;
.
-!$: '
Now
-
R' cos (P - a) = 2165.7cos (345- 20) r 2 20914 N F, = R' sin ($ -a) = 2165.7 sin 14.5 and = 541.5 N Example 10. During machining of C-25 steel wit% 0 - . I 0 6 - .6 8 90 - 1 m m (ORS) 7
r i r y c Fo =
1;
-
-..
.-
-
- -
shaped tripple carbide cutting tool, the foll~~cing observations have been made : Depth of cut ;mn\l'. tYl015 2 = Q t r n s zr:rh~3itr;rwZ Feed ; = 0.2 m m / m ~ F V ~ ~ + Li b i ~ u , \ \ > 'k: CS$.F, 2:i.;L.&: btsic ?kt:..:! in6:i y \ r r v w i I . . I f . ; i ; i : b 3 ~ . 1 . 3 = 200 i, 7 \\?I.: t,bn+* y %< , -Cb...-,*,?, .v<,ti ~4ee41\ iiti ,,L,-va 9>tit.w. .:&I .. , . Tangential cutting force = I600 N WJYL im .:. ,:u :, F d t h t f o r c e mm 1 L - '. ."F 850 N 5 = 'Ir 2:s t;;,,b fts.,i;r sif? nm,ui,r*.
d"k & ,.,
Chip t h i h e s s 9 2.5 n4vi9 .*i Cakulate : -. (0shear few- ?- L tr-: at c i!
9 , . 3 ! 1 '
,#
,+,,;,!;h~
Qr.19.amv 01
-
'au ;>niz [+R (;3m (ii) Normal farce at shear plane - . - - ,;w + +-oA+, .-
1)
-
01 I-.:-
i:
+.
L, .-
(iii) Friction force r =:?&I 1-3 flh ~ ~ 3 i l f @ i l ( i ' $?, p l c f W t i 3rtil-3T':l$3 WI t (iv) Kinetic co-eficient of fiction ( v ) Specifc cutting energy. EM.0 - ,% Solution. From tool designation, .m\!'t 5s" a-IO,A=?OO . t Other given data are : d = 2 mm, t = / = O.?,,mn(Sipce*kr 90°, Equ. 1 4 . 1 3 ) , f , '"rr V = 200 m/min, t = 0.39 mm, Fc = 1600 N*El E 85,O N:? ,. , ,.,, , - , ( i ) Shear force, F, = Fc YF, cp, Lg,, x,,<:.,i '. Ti._iaJ. J.
5
a
:;>.;:; -
,
=
1
,
.I
,
0513 x cos 10" = 0.551 1 - 0513 x sin 10" cp = 32" Fs= 1600 x cos 32" - 850 x sin 32"
tan 9 =
-, ,
(ii) Normal force at shear plane,
I I
dfiA
Fcsinp+F, COSP r l@Q,x 0.482 +850 x 0.76 = 1515.8 N2i+nrr. ~ ~ - . % c ~ .** , :d ~lI ; (&,.@F" =
:
';
(iii)
8
., 7s
, ti:ncu=,*
A
k
~ r i c t i 4force, F = F, sin a + F, C P ~ O
A Textbook of WWwAitM ErrgrrrgrmMng
.. (iv)
,
,ii
<;
-
i
4-.
&
..:?(:'I
'=
.r.,,,--f*l-
F, tan a + F, Fc-ctana
' - 1600x tan 10" +850
-
1600 850 x tan lo0
tr
= 0.753
Fc Specific cutting energy = b.t - \ I : Now b = d = 2 mm (See equ. 14.13) Specific cutting energy =
4
1600 = 4000 N/mm2 2x02
:
I
1 J
Example 11. A turning tool with side and end cutting edge of 20" and 30" respectively, operates at a feed of 0.1 mdrev. CaIcuIate the CLA of the sMace produced if the tool nose OE: = WWii A?~:J ! ~ b i i-:~... radius is 3.00 mm. Solution. The given data are : C, = 20°, Ce= 30°, f = 0.1 mmlrev, R = 3.00 mm Refer to Fig. 14.26 (a),the peak to valley roughness is given as, eqn. 14.46,' h
=
(I - cos Ce)R +f sin Ce cos Ce - J ( 2 f ~sin3 Ce -f '.sin4
0.402 + 0.0433 - 0.2727 = 0.1726 mm The centre-line average roughness can be taken roughly as, =
'
', ''"l"
-
43.15 pm. Example 12. In 'ORS', the tool angles are : ~hctimtionangle (i) = 0° Orthogonal rake (a) = lo0 Principal cutting edge angle' (A) = 75O Calculate :( i ) Back rake (ii) Side rake.
c,)
'nw'F'' .
.
*-,.A
,
=
a, = Back rake = tan
Also,
,'
-!
'/
-
(AMIE 1974 W)
tan ab = cosAtana+sinAtani
Solution. We know
-
,-
)I<
-' (0.0456) = 2'37'
tan a , = sinhtana-coshtani
ZvJ- . a J r ' 8 4'
sin 75" tan 100 fh "t?tbi I I * , u' ' = 0.966 x 0.176 = 0.17 (side rake) as= tan-' (0.17) = 9",40' .. Example 13. In a single point cutting tool used for turning, the geometry as per ASA is : Back rake L. g ~ - f l j ? '* nc r:3:1 : :
=
Si&rake=Joz.WaI
Thheory of Metal Cwng c, A m ! 4 . a
a
Side cutting edge angle = I S 0 , ,. , ,i. Find the values of inclination angle and rake angle in ORS of tool nomenclature. (AMIE 1975 S)
Solution. As per ASA system, In ORS of npmenclature, ' ~ i ( / l pi.
cr, = 8O,a, = 4O, Cs = 15O A = approach angle :> = 90" - Cs = 900- 150 = 750
r
tan a
Now we know
:. Orthogonal rake angle,
:. Inclination angle,
a
i a q aiiiiii'
=
4;
r;
-
1,.
tana,sinA+tana,, cosh
a = tan-' 0.i04
tan i
Also,
t).l#J
=
sin A tan a,,- cos A tan a,
i = tan-' (0.1 $ewe=I: = 6.70
176)
Example 14. For a turning operation w!th H.S.S. tool for hot rolled 0.2% C-Steel the following data is given : Cutting speed = 0.3 d s Depth of cut Feed (
cs Determine : Cutting power, motor power, specific cutting r@istance and unit power. id3 Solution. The cutting force is, Fc = 162.4 f Os5 kgf AT
Cutting Power, PC =
5,x v 1000
T:
,IW i-
A Textbook of PraducWmh~~
Motor Power, Pm = P, 1q, , L!L $2 -
I ;
7-
G.'\ft
,<.&
:\,
Let q,, for lathe = 0.85
Area of uncut chip, A= = t x b = f x d = 05 x 3 2 = 1.6 mm2 (Eqb 14.12) Specific cutting resistance =
:?r
Unit power = LO
.- ?{I+
1 (1 ,
Fc = 1727.34 N/mm2 f xd
e! = A,V
0553 x lo00 1.6x02xlOOO
:
=
1.728 W/mmz/s.
agiExample 15. Using Taylor equa r6n an using n = 0.5,C = 400. Calculate the percentage increase in tool life when cutting speed is redreed by 50%. Solution. ruu-a
1 ' 1 1
[:y
F,
*.,
1.
%,!I
- ( I
-.z,,
.5!3r5 59;: :
=?* 2
r;-j-
.
,
,;
> >
.
(I
T, = 4T,
..
;
Percentage increase =
T
*,ikr!x,
T
tloifsrii! 3111
ax 100 = 300% G
, gnbi\l\r\ o 103 .L1 ;iqi~:b*.? Exptapk 1.6, For an 'Orthghmcutting i dmcess : .<<>41i> *, &,)L\'> n , *y>: Uncut chip t h i c k s = 0.127 mm L.! "?? i j ' v,, , ' ' %...,'I' Cutting force = 556.25 N 3 ,I<*+ 21!i#\.' i : . i $ f n , i : ~ i ~ z ! = 222.50 N Calculate the percentage ofiota1 energy that goes into overcomingfi.iction at the tool-chip inte$ace.
r
314
kt,?
I.\$
!,:8,3
TbtyiCe ,{;
Solution.
Frictib d&i$ Total energy
:t
''
- -F.Vcc' Fc .VdrL
. ;
Now
(Equ. 14.9)
-
Friction energy = F.r -.-.i Total energy Fr
It is cbar from Fig. 14.19,
* \,
ti
F = A' sin p
F,
4'.cos ($ - a)
=
and
.. From here,
..
<
7
'
'
L T
-
F
ri.
&~Q&-x sin (320)
.
8
,
'&Fit(& q*?ka* and feed of 0.3 mnv'rev. L
>
, :,,,
:',,&i~e#d fice
= 450
N
1
' . I
.
I
I
'"
*
.
.
@ c e m : ~ : q l > t ,;L;A * *. (iii) Energy conrmed if the total, pjetal,femmed during the hhing operation is , (ii) W
(i)
I
- -
:. Cutting power,
'
-
.-- -
PC
,
s
--.
I*
'
,rF; swig zl
+ +)~&3irn
F d power = @
-..5+ 1 rLf
c
, , E , m 9-C'
-- =.---
i ni* "{a- 4
J
=.*
i[\sW
-4 +
0 3 x 4 5 .i4,jb -~aox.l~
33i8r.1
p n u ,
31 i K T $ / :
m
A Textbook of PCscrw~mEn-
(ii)Now,
V
=dxfxV
Specific cutting energy
!.:.
=
2.545 x 10' mm3 / min.
=
1308 x '60 2545 x 10'
\;
.
'
.
''I
(See Equ. 14.12)
,3.08 W.s/mm3
=
Energy consumed = 3.08 x 25 x lo6, W.s
Example 18. A M.S.bar of 100 mm is being turned with a tool having ASA tool significant - 0.5 m m Determine the various components of the machining force and the power consumption. Take : depth.of cut = 2.5 mm, feed = 0.125 mdrev, turning as: 6"- 10"- 5" - 7" - lo0- 30'
speed of job = 300 rev./min., co-eflcient oyfriction at the tool-work interface = 0.6, ultimate shear stress of the work material = 400 MPa. 3 , S~lutCon. It is ,olarr.&an the tool design* that the side cutting edge angle 4s 30". Therefore, the tool approach sngle, A. = 90" 30" = 60" ;, >:, -rT v33 3~~34 1 0 .A, ' Y : ~ ~rictionangle, p=tan-I p=tan-l 0.6 = 3 0 $ ~a 31" Now, ortkogonal rake angle is given as, C . tan a = tan a, . sin h + tan a, cos X 'I.~~'L >\F U 1 Now, as = 10" and a, = 60°, From here, a = 11.6" Now, from Merchant's relation, the shear angle is, .tl>\i ,% p y x 3 41. ' d,=4S0 + d 2 - p12 .:.w, ' 0 : -' = 45" + 5.8" - 15.5" 7 3 5 f 0 Merchant's theory is more accurate for plastics but agrees w l y &K machining &, With Lee and ShafFer relation, T-
8 %
-
&
-
.$
.
?$.:
#jm Tf
5 -3-*-- ='.I
g
4
2'
X A
- as,
I
45' -+jO + 6.62 >25.6* t:t)ktj
Now, the cutting force is given
Fw:9
,,
Now,
b = width c$ cyt =
and uncut chip thickness,
t =f
Fc=
!
depth of cut
*ill
- =-
d
sill&
- sin X '
'
400x2.'5x0.125 . . = 384 N. (19-49 kp#p 4S0 x sin 25.p I
Now, from equation (14.18a), the thrust component 3s;-
"
.
,.#
,
:
*
F,= FL . tan (p - a ) F. = 386 x tan 19.4" =' 135.932 N The thrust force is normal to the tool-job inte;fa=e, that is, normal to the principal cutting -,',,' I . , . , edge of the tool, see Fig. 14.17. ,.F ..> - ,. :. Feed force (along- the axis of the job), ' ' ' '! .,., > ., ,., - F;= Ft sin A = 135.932 i s i n 60' = 1 17.7 N , 2 ' 1 ' ~ .. ' . Radial force (Normal to the ~ i ofsye job),,(u rn@rat lo; :.'A, t'ii''q'
,
-A"
1
kc
I
'=;f;;~p~ t - ~ . ~ dn*sd u 3vsd
r.
cutting Power -&
NOW,
$8W
'
L.J14f.r
67-96' &qo aantj. b-
W
.
lfi,qa, 1 1 ,
t
'
~~zin -,u
and
.
Power = 386
1.
2. 3. 4. ;r
u."t!i
5.
I,,-J~
-
,
7,
=
I:
606 watd
~IULD!!j '5,3!
.
IJ
. snrmt>rr$l
.I@
'
itsi it^^ ni;ilc I R ~ ~ .t@l3 F. ni r r , b ~ f r Jw11341fi Define Machining Pr-W: * , :, -,* .* . , - J p 3 ~ r t t w c ~ l l7dr o ~ .&!, Expla,in a basic machining operation d t h the,help of a neat dia$pnl Explain the various elements of a single-point cutting tool with the help of a neat diagram. 11 '1 , . i 8 * z*tr 3 51 ~ . r + , f itirr'i A What is meant 'try 'band' of a single point cutting tool? With the hplp of a neat sketch, disc"ss the principal surfaces and planes IS metal cutting. >: ' - ;o! , I ~ I L J J>~ @ 11,b*qt 84 s?t&tm: !IS Name the two systems of tool d e s i ~ t i ~ % ~ ( , ~ , , . ~b;ml>pi ~ ~ ! , : 371. , 041, !,, . W& a ne&atiue,raJceygle ig n~,qnally,employedfar cutting b d and strong materials? Showzha 0$5 uftaol sngieswfth the belpaf a sketchad1 3n *JIJI::,.rr1+ 1 1 -~ ~.
.=
p r ' ~
!b
18.
"'
9. Write the relations betweeri ASA nd ORS systems of to6lAangt&': -- ;RXs-0, 10. Wh? is meant by Orthogonal cutting and Oblique cutting?
t
.
,
''
-d'
>!.. "-.;5q1 ~ $ t v r ft; ai
;,, -,-!., ..:.,, Differentiate between positive and negative rake angl@ttlm3 ; . !cc. :I
13. How is thejaoseradiw of a cutting bol selected?
.
IJ
it1 11,&.I
ll;~~owdow,~ea~le~qtthe@eofthecunin&~!:!.,., 12.
'
1.57
,A
6. *
x
+ I ~ Oqdt) IJ~ , ~ ~ d Wj Jq ~ ? sdri! gntrbi 2
rfisrrj
.s
4,.l+;,,
,>I~I -L~:IJI*
I!.;.)
'I.:
L
-
.
"" 2'"' - ';'
14. Discuss the various types of chips produced during metal cutting. 15. Why are 'discantiriuhs ty$e' chips p$e%rFed over the bntinuou's 16. Explain. why built up edge on a cutting tool is undesirable? 17. Name the factors that contribute to the formation of discontinqpus chi s. A kn6 &,!I 18. Name the factors that contribute to the formation of B.U.E.
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20. Discuss the two metha& of metal cu@kg. ewau&er& & m&al ming.;a $Rl?!o, isnr 2t. o h u ~ a-w~s ,'$l~,i~ !. 22. Explain 'Merchant h e circle'. sas1 *is { i l ! 23. Define Tool Life, I 4
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25. Discuss various types of tool wears. 26. Enumerate the factors pr~nbigh,tool wpar q d tgRI tifa +pmd, ~, 27. Nmpe the,@ctprs tRat cotpiby@ to f@& mar. . .. - t ' , ' , f ,, R'T@fl &j -&XU~ jtiikt11.~>if1 28; kame the facton that contribute to crater wear. 3% ,lt)cl s*,, ,&!A? .. . 29. Which two pressure areas of the cutting tools @e~ k j v. tod w-.q d 3+m7 ,: 30. With of a sketch, show c ~ a t awear d;d ha& wear on a ytting rol. . help \ I ,T , 31. ' Discuss Taylor's re&ion$hip for cutting qki-tool life. .i %r; & $ i ' # ~ ~ W ) ' s h n ' it r ; ~ i ~ i ; 32. Derive an expnssion for optimum val& df itmini ,sgeed: : 33. ,In an orthogond cutting o& the folkwing d& &e been observed : . .. . ' - Cuttingspeed -, d.223 &fs7w9.-. .&,.fa L+ .: ,b* .. - . Uncut chip thickness = aiasmun .>' , . "' =r j-&:&"4, . -bfri, Width of cut '.... L3tip-tkicknc.s~ ra%io = 0.51 -&,,'e~;+,.~~.~~ . . : ..?cuttiigtforce N H ~ , + L - - -- ,. : , .
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Determine : Shear angle, friction and% %R pla&, chip velocity, shear strain in chip, shear strain rate and the power'fir the cutting operatio34. The following equation for tool life h
' yyOi?3sSg6.d0.3 , +w.$f&ib b dSi I A 60 min. tool life was obtained using V = 40 dmin, f = 0.25 mm; d = 2.0 mm >.;,:!I t=: ~ ! ' J I J . 1 1 Calculate the effect tool life if speed, feed and dspth of cut are topher increyd by 25% and also if they are increased individually'by 25%. ' "' ' ''. '@ w'' " ' 35. ' 'bilrihg m&tnhg b f C-20 steel with an orthogohsll tmi havifrg a d e of 1O0 dt d feed of 0.2 mm/rev., the value of the sirear ande h a t ~ k e naeseFwd to be W under a shear angle ##&&&t-iijvalue of cutting ratio or the microscope. If the principal cutting Mge yp%ci$WO, chip reduction co-efficient. 36. In a turning operation, it w& observed that the tool life was 100 minutes and 50 minutes at i the to01 fife at 200 dmin cutting speeds of 25 mlmin. and 180 mlmia r&pe&eIy. Ftnd khr under the same cutting conditions. _ 37. The end of a t@gbh!ing turn& on a +lathe.at200 d d n .md at a feed fate of .rI 25.4 mmlmin. , 1 The tube is J5 cp'h diametsr and 2.5 mp thick. The tube meterial obeys the equation, ,, , r-jc-*7< &i 1,. s p i , F ~ \ ~ . . . & Jf! I nh 9rb.j qu j@%wnBli,x3 .e( I 1 I I ~ r t t3rit r,? 4flKfi71?~-,%~6~1 &la+ - gtit ~ n r e p .-i I &me n = 0.26 and K = 5fU Nlm2 L,
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d f the principal cutting edge angk is40°, how.muoh ae ,. ( i ) back rake (ii) Side &e. 39. During machining of C-20 steel, a double carbide cutting tool of 0 10 - 6 6 - 8 - 75 - 1 mm (ORS) shape has been used. Feed is 0.15 mmlrev,, depth of.cqt of 1.4 mm at a cutting g m of 120.,m/win., p cKip thiqbqcss,of Q.30 mm have been obtaiped. Calcdate : - . i(J) the &p -dw&i011 rm43&nt ;;i b k t (ii) the &ak angle, r~.:z ? r * k~ Ib: 'SkeZCh a &g16-6&t kdng to6t'and sho~-&'itthe various tool and tool angles. 41. Give the function of each tool element. List the various tool angles and discuss their significance. 42. Discuss the two construction of tipped tools. 43. Why indexable inserts are better than k h d Pdbt fibs. 44. Why do carbide tools employ negative rake angles more often than H.S.S.tools. 45. When the use of positive rake angles and negative rak; $i;gles is recommended? Give the sigificape of providing no* +us od todl !ip. , ! C . 47. What do you understand by the tenn. '~oolbesignation' or 'Tool Signature'. 48. Describe thptadrepmmted by.10, 1 4 6 9 6 , , % 8 , .win,ASA$ystem. l 49. What is orthogonal rake angle? S8. Dcfmeeutting nstfr). . ' ' 51. What is the a h x i m a l e fhickness @shear m e in meW cutffng? 52. In orthogonal cutting dperation;'hc fecd isXkl'Omin and thexhip thickness is 0.25 mni. The f a thrust f6pee.is 730*W:hi rake mgfe afthe tool is + lo0. cuttitlg force 2s~1360 ad ,, 1 * Find : (a) The shear angle. (bj The size of the force exerted by the too? on the'chip. {c) The -c@t of @@on on the faoc of $e tool. . (4The Sizes k i ~ )or# wd awldo& m 'me ~ mi face. ,wd W a r n fbw, og &e.&ear ,plane. (6) The i s S ~ 6 ~4 ‘ d~ 53. An orthogonal cut 2.5 mm wide is made at a speed of 0.5 m/s and feed of 0.26 mm with a H.S.S. tool huviag,a XI0 *,*, chip th&@wattio ia Run4 do be 0.58, the cutting force is 1400 N and the feed thrust force is 360 N. Find : ( a ) Chip thickness. (b) Shear plane angle. (c) Resultant force. (d) Co-efficient of ofctio" on the &of the tool. (e) Friction force and normal force on the chip. V) Shearing .force and normal force on the shear plane. (g) Specific energy. 54. In orthogonal cutting, the feed is 0.127 mm and the d@ sf aut m a l tebe plane of the paper is 2.54 mm. The cutting s p e d is 4 mls. The cutting foroe is found tQ ba !l$@,@:pd the feed tbrust force 900 N. The M e angle , ~the f the is $ 8'- Find : * p r.& mfl') If
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57. What is understood by tog! life? #~$,:$~,$e: qignificw~;t;t~ an e;?~' ek,,w@ i>i$etp$ed in productivity? What diRnrnf a h d a are u r e s b identify ifid th;.\ooXds reachgd tts Smiting life? ,, 1 $ 1 , :i.~;,s,i<::r),lT ,J;.:: ry18 , rq,- ,<,!f;; ,! -
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58. Establish BW (todlrii% cq (Taylor's tool-life equation) : , ,,:, ,!a.2 b. :::,+,,{,,, : , - f,i A tool life of 100 min is obtained from a cutting tool at a cut$&& ~ I D ~ ; , S - , W * and , lo at 33.3 ,~~G:FH&W~~MA~,.Y;%%M.#~$~, , : c
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(c) W i v e rtil e x p r d m for the most economic cutting speed .,f. (4 1s the most p r o d ~ t i v e~.v t t j, , .' l@~ ~ $ ~ f ! $ ~ m,qJ$$ e ~ ,e:$;p$y~?mif spe* for vow answer. i ;>+ !'a & % & ~ O ~ ~ & ~ i ~ ,#ri$iai9,tH,1&e ~ (e) Make a rough plot of the d a n
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t - :!:' !i 9 ,,j c, iflty. E $ t m p Ir$wW@l% ~ ~ ~ : 1 , , k k t ] , , '" , (0 Wwk material micro-swture. ,,.<, A , , , : , (ii) aLpe of cut. !;'''8 '" i , (iii) Tool take angle. 7 ' '' 62. Fotlowing data were collected from an orthogonal machine test oit%eef L' -,;I ! o 2 . , > ' = l ~ ( , l j f ~ & t f ~ , Cuttlng speed '* ..' >Ll Rake angle - ~3~ *it .' k If)" ' Clearance angle Width of cut = 3.2mm . ,I.*
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v'7-@5m/mkr>.?!t , :r,rl,: .,-, I rv. I. :: take "$"':fi" r 1" I . " . ' H.S.S. ~&f5SIi i .I f i b I : ' 0 9-u qntJ4113 r, la 311: ;(, .I!? . . , I I : b LL i s 4 1 15rlc~ 10 S ~ I I 1ooj ~ e i .lnw (1 q-15 mdr~v. lo L n g l ~ r i l t i o4 , ~1,. In ' \ t ! t I = '1 Chip thickness ratio 0.35 ,x>irk ~ r b 1r.tu3c @ 1331~a t ! - ~ i & A t ~ f f i a O 2 I,.L-,I 0.60 15 q!~~unn.,u=,~ i>r~cy?T lrn} c'. t i ! t" mm aod ! 16 -,.>r,?
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force 6h t I i e ~ k i i 1 d t f i n ~ w ame&dwHI&~'W#sWh.* s1 and was the nvtg of the m i for&* s f i a angk and the specific 2:: rrne,ts d+q,~&hM ~duittrnL to ? r , l c bnt, nrnrl~ R 16 arrrwvqo nrrdw rb-bnvrt .r ... . - . , .. , ,+, 1n:an o&oF9na1 cuFinp operation, the cutti'ng'speekb~>.5mls, rake angle is 8 p d the width "" '' ' hf'thd ad is lo'thtk "Mie uriderfomied chip thidur'dsf 150.2 nun. 13.36 &'ns'of steel chips with a t&t#'>h of 50 cm are obtained, l%e tool post dynamometer gives cutting and dvust farees i 0 CI %r: ~ ~ M&IdN : a ~ . $ # . -& : &,I Arr~d * I t , 8 m u ?4 ! lit--) dC
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(c) shear an&~y~m the proeatage ef total energy. . H M cjla:~-Ih ) \)T (So* 30.9% 69.3%) 66. Id &$onlrl~rcu%&g of a low carbon steel, the specific cutting energy is 4080 N/mmz. The , uwut c & 1 , € 4 i c ~ is ,Q,F am and pha chip ,width .i~:5 mm. T@ecutting speed b 1.J d 3 , and , . thc rake angle of the tool is 10". Assuming eo-efiokt of fEieffsnat the taol-ohjp interface as ,, 07, *c@~P : (a) the cutting force (b) the aver& shear s t e i n the &ear plane (c) the normal $tress on tfiO shear plhe (d) ihe av&i shear stdin i i ~tiutting ' (e) thd1';rveiageshear stiain Use the relation of Lee and Shaffer to find the shear angle. (4080 N, 861.8'N/1*~, 861.8 N/&, 3.%, 4.44 x 105 5') . Iri is - h i with a tripple carbide cutting tool having 0 - 10 - 6 - 6 - 8 rm.nQ& ABRS6 pb anlllc 3,-'1 mp OqS, $happ,q feqi of 0.2 rndrev and depth of cut of 2 mrn at the cutting ,,speed bu9-t bm ~ f1 ; dmia:kav'e &en. employed. A chip thicJrness of 0.36 mm has been obtained. Calculate the chip reduction co-efficient and shear angle. 68. Select the speed in r e v h b fbr tUreing a m d steel bar of diameter 320 man with a H.S.S. T.I tosl, hying a M E life o$# minutes. A feed of 0.2 mmlrev. and depth of cut 1 mm have been chosen. The cutting speed equation is given in the following Taylorian form.l ,
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fanddareinmm. I). . ,, . (A=. 165.2 rev./min) e t 24 @min. i b r qae bwr. C i c c s arc such that it becomes desirable 69, , A H.S.S. tool to run the tdQl for lW0 M&. Eqitgate the suitable speed. Take n = 0.25 In the Taylor quation. +i. (15.74 mlmin) - (
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A Textbook of
Aimed apprawh aagie iW, of mtation. The mean I -the cutting ratio. (0.32) 7 t 1 71. When turning 19 mm d,imeer bar on an .+wddid^@.^^^^^& i;l ,#lvId)?id< g i v ~ tool life of 6 hours. If a length of iq 0.16 w d r e ~ what is the ?Hween tool ~Rmgces. (122@y/rnin.. 16.5 s. i3W) SI 72, W M cutt11ag steel with a H.5.S. the tool life at a ouning & $39 mpm is 60 rnin. .ad at a cutting speed of bg'&% 30 mia Wha is the m 1 liR of t(n c u m on the ms&rid. ?E.i; fv:2 I + ~ ~ : : J h . 2 ( r = 21 1 15/kln4) 73. A' HSS tool requires regrinding af€#3 hours and 2(r m b f t s w b a @WMng steel at a cutting ,$$jc&{$&$ life if1 increased t?~JOSmlmin. (2.7 min.) &16-~irll1irwl. Thg of 100 M w e e i ( d q m ~ t i ~ e ng a4 W d~falrttin.and a life d a ~ e e d . Z & 4 9 e i ~ $~etennine rd~~~
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@f the Tqloa tost a$&quatiom,2P q8.F C = 54) Wt. The pnnr r w i n d m turn a medium C - S d is a p p r n ~ i m a t ~ , & ~ +Tpe ~~~ power fi., avait* I mwhine spindle is 3.73 kW. ~ e r m i k e:'eyPi"" -r JfiR "6 '1,' I 1.t r : , ",G' , (3 MaximumMm' $$--ti .?+.>f!d @ ,- ?i); rev. &%/ntia. W.I oh^ Fcrtd is O. Y - >; L i e d 31rfq!7a L Ftrbc:(lbcil#ifm 316 kbl, 6.53 mm) e e i i cutting an 3i$ mrn and feed
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~ t y e h 3 83. While turning a &d stel bar on a lathe, the following data were obtained. = 150mm a r k f A . w s , , , ~ - .. Machined len& c, , (B .Jlfi .I Speed * 170 rev.lmin r ~n!txlo~*k nr OUT r wya
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w h i n i n g time and the total amount of heat ( N a p Universi~) (mg+WkW9 1.765.-+, 94&5 1 kl) awehiggies; o f i W-atting,aud &&ace various i force angle relationships. Determine shear angle, fiction angle, shear and normal stress on shear f%i*g plaieI khear strain by using the formula dPea. nS;iRH di% U w dl$ tukknrdss er$l)b 3tL 0.b2gm $r1tJl:1~> $&emtW .fi? , Chip thi~knes+q,~, . ..: 0.250 mm ,: ,:#&'5", , - .. %+ . . - " r Width of cut #j$ asnun &&&,-: . ,, * : .'. . ; Cutting speed :-, Ahg@f41(FO @mi&,: .>-.: .'- . .,J *., - . Rake angle ' lo0 . ._ - Cutting force . , Thrust force osa 23 N *. J@WY 3 & 29.G0, 61.24 N/mm2, 6 hm e-td *UP t b * ~ ~ ~ W W ~ &fMgro->$ly!cCbeen H&F nod 3,
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86. A50mmdimettrbarofstct min. The speed was changed @.p2v w a n d the tooi failed in 60 min. of cutting time. Assuming a straight line reletionship exim what cutting speed should be used to obtain a 30 (Aas: V,, = q939 m/min.) min. tool life? 4&$ag a reke angle of lo0, the uncut chip thicknrn 87. For ortho$oml w t t i q w i 4 ,a mm.-Ddemine (a) the cutting ratio. (6) the shear plane is 0.15 mar and chip t h m ' an& wd (c) the sheat s b. . n k rw- :.,i ~(,+&~;~~;‘+75, " ~ 2 3 5 02.734) , ,operation, the following oar8 have been observed: !-0.25 Ihm , * o, ' kiY.75 mm
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. !,,+a:, !,&=475 , Determine: shear angle, d c i e n t of &on, the fiction angle, ultimate shear stress of the mattrial. .,3. (Aas: 18.4O, 0.5, 26.574 379.4 MPa) In h A m i E h i n i n g ~ ~ - ~ & - ~ ~data i n have g been o b m e d : Uneut chip thickness,?^ ? = i02 : tRm ,% I ~ , ~ ~Il T l of cui,ww ~1~.to % ,, h;i.
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+=450 + (I @ t6 i(etermine, (. 2 2' 90. Solve the above probleh, by psuming the machining cutstant to be 70° in Merchant's second equation for angles. AIM, f& out the results by Uihg relation given by Lee and Shaffer Itlwla WI. (Ans: ~ e r i A Th&ry d :Ec= 468.7 N ;F, = 139.46 N) relation. - 2 ! ,,! :, Yc: bnfi x i a r t ~f:!,-tld.Li*Lee and ~ h a f f kt h e w 1 F, m,W.OAJ; E, = 154.6 N) n!.?. kr I # . ah& hY'&&g during orthogonal machining of mild steel with an uncut chip ~'r#iid and width of cut being 2.5 mm. Take a ,- 09, = 0.5, aad 1, = 4 0 ' MPa. of 0.25 9igr:t r~rmrit wad2 snrmts,.ii! c
Hint: Use eqn. 14.19
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Qctmnine the cutting f m and the thrust h q e W thE fpaing ratio when machining M.S. Take: .= ,.t. * ; Uncut chip Wkness 0.25 mnn * c wjdtfiofcut=2m * $ .
p = 0.5
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relation fbr s b bmbpd. " - '* * ~ k * i W . 6 5 N ; 4803 N:8.333) 93. ( a ) Detefmine the various cornpotfenti df the machining force d e f i machining a C.1. block on a shkqj&'\nifth depth of cut = 4 mm, feed = 0:rS muW!hr61yd;'*tlt i&e angle of the tool .3@', . Principal cutting edge angle = 30°, p = 0.6. 7, 340 MPa.
- ~ ~ 82 O ~~ tJ wW
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D ~ c & ~ the & average power consumition if $hqht~ration€&s.wit% -.1s1i141t I (b)the length of the job is 200 mm. . L en wrnl .. ". @am(aJ,U&g p: Lee and Shaffer relation for shear angle. 1
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odr b nrm\ 13: kRl rs bmuf oaw txatz %I rsjsr;lr qn~li;r>I!> fiim tB bhd jool ?hl D f E IIIPJC';~o f 1I~c's34 bit&e tzr~q.:p:rw rtcnl~t:.~ $6 :*#A). @=tan-' p = tan-1 0.6 = 30.960 = 3 t i h - I T , >. ,-- , t n h ,~r!* a'& ~ 1 dt3 ,%i~ 20 3!q1!?i ~ S biI pT I:\ -":.-.-k745O (31 101 = 2y ,. i :!, 7, - . , ,lt;,r, ,LT.Tf&".,& r * , a s , *-I--fil
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"
n~
Thrust force,
=f
_
,
:;I3,
,
cos C,
sM,IoX c0s4,pXPin240 = ?LW-,Nr t ; i* .r d .,, : jGfq . gn (J3 - a),it acts norm4 to tool-job interface. 111 '1'111 s7q ,,, (Eqn. 14.18~)
?ti.'
'.&?
n
:i
c.i
12
= 1103.8 x tan 21' = 423.7 N F, =F,. sin l (normal t&&p wlociQ,y.*j, sSE4=423.7 x sin QOO = 366.93 N *!,: . ..,e *!tj rr )g)nrral f m component+ t;rn? rerr F,, *F,.eas t=311-85 N mim mud tq the mekined surface) I GCI$I: :rnA) ,
Now fead force component,
i
I
340 x 1
' ~ r ~i s?q-\\ 1
and t
.,
-
.iaf
'
. -
i
i
Theory of Metal Cutling (b) Now
Work done= Fc x length of stroke = 1103.8
2 . ..
x '
Average Power= 220.76
x
200 = 220.76 1000
J
60
- = 220.76 watts 60
D e t Bat.owehind at 200 rev./min, with a feed of 94. tM$u$e$he i f r ~ i ~ m ~a ~ ~ rW pe 0.5 mmlrev. and a de of cut o 4 mm.(Ans. 5.78 x 10' mm3/min) %pb dh$I maohlSik& 8 2 I& &meter bar at 40 rev./min. with depth of cut of 2 mm and a feed of 0.3 ,---mmhw., . the cutting fonx at the tool point was 1800 N and the feed force was 400 N. Calculatethe power consumption. (Ans. 935 W, 0.08 W) Sol: Power consumption = Force ~Telocity
r* P"
; ! r v ! ~ . e .. I
~ I
and Velocity =
.;
x,P,,&'JV. = ' 1080x60
w ds 1000xf@
Power = 1800x n x 2 4 8 x 4 0 =935 1000x60
w>4IQQ
Now velociQ ofpfeed afQrce=, LOOO
,-:
.-.P"M-~
' I ,
-
4W'x 0 3 x @
'= o - i
Which is very n li ble as c m p & to Ppwer consux,nption for cutting force. i . Now MRR = d.f. Dave.n N
9@,.
:. Specific energy eonsumption
Power
=I
4-
rrt.8tub1 psint,is f 200 N and its angle s f inclination to the. horktn$al is 96. Tht wmlt8#t &We 3 P . The approach a g l e to whi& itacts perpendicularly is 26". Determine the three +w&ipal cornponeat forees. . ,., . - t I >a a 54b = 970.8 N LJd,', iYI ff"*.ili %hitian. b
4,5.
I
s
1
8
Ff=F,,.tan 26"
-- F;=
ad
!
'? r-
F . + Ff+
(Fn= 635.5 N; Ff= 310 N).
17n2
97. A W C cutting tool myhning MS gave a life between ~ g r i n d sof 100 min. when operating at 80 fiJtnjn. and 33 m h when operating at 100 mlmin. Determine tfie value of the index and the constant_ -.in tool life equatian. , -... -
Sol. - . -
and
-
'
rc>t,'"
C
VTn = C
-
'
>
2
--A-
.,
1 r ,
I
-
:.Log 80+nlog100,= logC
...(1)
loglOO+nlog33=logC
..-(21,
from these two eqw-
a = 0.2 and L3 = 28 1.
I
