A Textbook of Production Enginwring

I

= 1% min.

Example 2. For a metal m a c h i n i ~ ; ~ & e f i w i n information is available : Tool change time, = 8 min - I * : ~ r f i ~ ' 1 1i1iri { V i ~ a i ! , I< , .IT, . - * i t !tv,tv ~ o o i re-grid timea= 3 nr'in'

Machine running cost, = @ 5 per hour Tool depreciation per re-grind, = 30 p Ra

Calculating the optimum cutting speed a!. r,- ,,. , , I ' . . ,. t ' .,

Solution. Tooling cost C, = Tool,change cost + tool regrind cost + tool depreciation , ' 3 ~ . J

5 . 5 = - ~ 8 + - x 5 + 0 3 0

1.. 5 , 0 GO--,* < , @$?-I x; .lo ,i 7 .

i = Rs, 1:38

snidagrr, silt bne

. . ,

Example 3. In an orthogonal Wthg operatiout, tk fol&?wjn~$$$h~e been gbewed : 1 . 1 '

, . Uncut chip thgkness, . t = 0.127 mm .-- . ikji:~,~ I :,.. . Mdthofcut,. .<":.; 3 b = 6.35 mm I ,'..c:, s . l~ .:: t

I Cutring speed V = t m h 1 1

' RO& angle, a = 10° , Cutting farce, 1

F, = 567 N Thrust force, E, i. 627 N Chip thickness, t, = 0.228 mm I ,

Determine : SIsesrr f t h e @ & t i ~ i @ a t k he shew plane and €her power for the cutting operatioa Also find &\@&~vgk#*.i%. in &@ ond shew strain rate.

Solution. ( i ) Shear angle, 1-rs ina 1 '

t 0.127 r = a h - = '- " 0.557

0228 4 - I , < , . Ill,

F, sina + F, cosa . st r=,pB = *A. - F, cosa-F, sina

$ = ) tan-' (6.64) = 32.62O

(v) Now

FV 567x2 Cutting power, = = -

loo0 1mo

v Chip velocity, Ve = 2 x 0557 = 1 .1 14 mls

Shear strain, s = cot p' + tan (rp - a)

(iii) Shear force, '' F8=Fccoscp-F,sincp I ,

= 567x0855-227x0519

(vii)

A Textbook of Protjtptjm Am@nq@ig ...

t 0.1 27 Shear plane lens = - - -

sln cp - 0519 "'" ' = 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

v s Shear strain rate, 3 = - 1s

Now V cos a v- =

cos (9 - a)

:, *:. 5 2 x 0.985 .-.. r .. . = 2.1 1 m/s :<L:, -. ~0~21.25

Example 4. The following equation fir tool life is given for a turning operation :

FE= :b 0.77 0.37 .?a ~ ~ " 1 3 f d = C

A 60 minute tool life was obtained while cutting at V = 30 mlmin, f = 0.3 mm/rev. and d = 2.5 mm.

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 (ii) Now ". $F.-% ' ' g*fi*~ - ..J,;...s f = 0.3 x 12 = 0.36 mdrev.

$ -$ ,*p! t -&'gw-= &?<?. . 2 6 ..:i,.*~ - . : *i@ , 1 . 28.38

'E" I,\ 9 = 1.48 ,'.i. gp-y .r?iJ*j -- - 4a3(a36)0'77 x 1.403 x 30

qk ": o el;r 3

. , - ' '& , &q (1.48)'" = 20.39 min P ' ' - . . ..- . ";- " - -

(iii) Now . a - _ . ~ ! . i ) d ~ 2 5 x l d = 3 m m . . - .['. -

8,~i P8& I C Z P ~ 28-38

= 1.591 L; 38 x 03% x (3)'"

. A 6 ~'"9 h~y$tl(<~4 ij7.a = 35.38 hin

:. 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

' m 3 ~ b b ~ yk (1.154)'" = *oil nin. . . 3 ,w &hi = n mnl T i Example 5A Bring an orthogonal machining (turnihh operation of ~ 4 0 steel, 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 Feed thrust force = 295 N

111 t:c! 3i~'re >ssdZ

Cutting speqi = 2.5 4 s , Emtfib we = + 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,

=Fccos 9 - F,ein cg, where cp is.sbeararsglw, .;(,it

' 1 ~ L * , ~ H C ' , \ ~ . > ~ 3iTT A%.u\+.i<*, J *$%s, r-~ .:>: p

t a p = l-rsi,, ti I - . , . , I ,* 4

(ii)

A Textbook of P&t&n Engrngr&g

;: - -- - - (1' .. F 0 CL = F, tan a + F, (eqn. 14.9)

4, - F, tan a

Example 6. The following data relate to an orthogonal turning process :

Chip thickness = 0.62 mm <i J 3 t ? - > ~ h !O ZF )3&> l~zxtl M+ br, ~ekii '= 0.2 m d w . itli k*,: ?*, .SS~I~ ntk:rnl jam %<I' .

~ o k e ang@ i= 5P. VYI di 6, "6 .i:li':' 3; - +iofi t t t c l

( i) Calculate cutting, V i q and chip co-acient. (ii) Calculate shear angle,

(iii) Calculate the &n* pkar strain iavalved in the deformation process.

S l r on t = f = 0 2 rn (y 14.13) t = 0.62 mm, a = 15O - . .a, i 3 * 3 t t b + - ~ ~ n o ~ \ ~ ~ o + b rwl, , 9 , ~ ~ , ~ * ;% ,-c r t .a ,=-, .., . . -. , t , < (0 Cutting ratio, r = - = 0.322 4:. !, i 1 *

1 s 7 - i,, :< Chip-reduction co-efficient = - = 3.1

t" 'I

r m a \ \ , . , Shear angle, tan cp =

1 -rs jna 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;g Mat!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: dmin 25 35 :;;--$ , q,' 11' I

Z min 90 20 > r. ., C

Calculate (i) n and C. 7 ,: t ii - - -- ,::# (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:'.

Uncut chip thickness, t = 0.5 mm Width of cut, b = 3 m m Yield stress of material in shear, = 280 Ni'mm' Average co-eflcietlr of friction

on the tool face = 0.7

Determine the nglrpal and t~ngential forces on the tool face. -----

r 00s a 4 :,;,I '~2qf-"< la,: 8 ' nii i 3E 1: Solution. Now, tan cp =

1-rsinq - +

r = 0.353 and @ p 15" < -

.:*>: r,i)-,:+-t z,.t;r:tcwP' (3*7-+) 19rw(1., ' 1 , ,C,i,l ., ,r -,.

. . Friction angle, $ = 350-

Now -(3) jr = ;

Fc = .vbt

sec [fi - a)-cos (cp +,$ -a) sin cp >w..& -

Now

F l = S l l . 3 N .- " , ~ k r 2& F ' - h .1>'iL. 3:1Eit2 W P ~ . Y ~ . w<!4

F = Tangential force on tool Eace

A Textbook of Produdon Enginewing

N = Nonnel force on tool face

I! .: = Fc.cos cr - F, sin a = 1405.7 x 0.966 - 51 1.3 x 0.259

Example 9. The following observations wee wade during orthogonal cutting of steel tube on a lathe : a&l&42&'Rs. % . ,

m*h of ~ ~ k i r y l ~ ~ prt\u=, &&f5 T $ r n ~ % , , ,,4,4~<9,.,! .Y 2;tl! j m r 2 Cutting speed. P-z'8.2 m/@p , . ,.,-.,C:

r = 0.351.-C. & = 9 ' . ,hi . *, . ,%?,, ?&

+ p -a = 3~ l i ~ i 3 I \YUL()LN in -., .nl.t i.s;@.

Find Fc and F, given tensile properly of material as mkw%& attw-,~tp . , ,~qc .r,.:t

a = 784 @)@+'$ N I mmz 5i+~~.~~wci L ~ , l , 4 G.

I t '.\ ,$3t5fx\ rcosa - ~~ Shear angle, tan cp = -

1 ; i n a . !GOL151 x sin ZOOeQvl irrilqtjl\r i

From here, cp = 20'.5O ', c . . p = 34.5"

and

Rake angle, ' .a = 20° *.Q 4 - - t = o.a-@qD,q3. .

reo Now shear strain ;.' = 4 a @ + t m ( q - a ) 921 = & 20.5" + tan (20.5' - 20°)

?c.sS ='2.474

Now from the relations of chapter 13, (Eqn. 13.6), fiK simple tension test, A ' '

r + j L . ~ C 3; t:

a = k ( E ) ~ k - -- P

For garnliaos- ef B = K.(T]* 92

(Considering Von Mise's yield condition)ll i

I . . 5 384i (4,605)o."

S \ - ld 841.66 Yield shear stress, 7, = - = - = 485.95 N/mm2 f i , P8P

&. I 5x025 Now shear plane area, A = - = - lr 3.65 mm2

):&? j:>c : @;., sin 205 1

t i ; . ; ' .

-!$:

Now

and

F = R' cos (P - a) = 2 165.7 cos (345 - 20) r i r y c o 7 1; - .- -.. r 2 - 20914 N

F, = R' sin ($ -a) = 2165.7 sin 14.5

= 541.5 N Example 10. During machining of C-25 steel wit% 0 - . I 0 - 6 - .6 - 8 - 90 - 1 mm (ORS)

shaped tripple carbide cutting tool, the foll~~cing observations have been made : Depth of cut ;mn\l'. tYl015 2 = Qtrns 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 d"k &,., i, 7 .v<,ti ~4ee41\ iiti ,,L,-va 9>tit.w. .:&I . . , . #, \ \?I . : t,bn+* y %< , -Cb...-,*,?,

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*.

Chip th ihess 9 2.5 n4vi9 .*i 9,.3!1' Qr.19.amv 01 ,+, ,;,!; h~ 01 I-.:- i:

Cakulate : -. - (0 shear few- ?- L tr-: - at c i! 'au ;>niz [+R (;3m - 1 ) - +. (ii) Normal farce at shear plane - . - , .- ;w + +-oA+, 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 J. - ,%

Solution. From tool designation, .m\!'t 5s"

a - I O , A = ? O O . 5 t Other given data are : d = 2 mm, t =/= O.?,,mn (Sipce*kr 90°, Equ. 14.13),

f , ' " r r

V = 200 m/min, t = 0.39 mm, Fc = 1600 N* El E 85,O N:? ,. , ,.,, , a - , ( i ) Shear force, F, = Fc YF, cp, Lg,, x,,<:.,i '. Ti._iaJ.

0513 x cos 10" :;>.;:; tan 9 = = 0.551

1 - 0513 x sin 10" - = , 1 , . I cp = 32"

, -, , Fs = 1600 x cos 32" - 850 x sin 32"

(ii) Normal force at shear plane, I dfiA I

(&,.@ F" = Fcsinp+F, COSP r l@Q,x 0.482 +850 x 0.76 = 1515.8 N2i+nrr.

: ' ; 8 ., 7s , ti:ncu=,* ~ ~ - . % c ~ . , ~ I ; **:dl A k

(iii) ~ r i c t i 4 force, F = F, sin a + F, C P ~ O

A Textbook of WWwAitM ErrgrrrgrmMng ..

&

, i i , F, tan a + F, (iv) <; .. : ?(:'I '= F c - c t a n a

i - 4-. . r . , , , - - f * l - ' 1600x tan 10" +850 - - = 0.753 1600 - 850 x tan lo0

Fc Specific cutting energy = - b. t

t r -\I: Now b = d = 2 mm (See equ. 14.13) 4 1

: I J

1600 Specific cutting energy = - = 4000 N/mm2

2 x 0 2

Example 11. operates at a feed radius is 3.00 mm.

A turning tool with side and end cutting edge of 20" and 30" respectively, of 0.1 mdrev. CaIcuIate the CLA of the sMace produced if the tool nose

OE: = W W i i A?~:J ! ~ b i i - : ~ . . .

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 c,) = 0.402 + 0.0433 - 0.2727 = 0.1726 mm ' ', ''"l " 'nw'F' ' .

. , - * - , . A

The centre-line average roughness can be taken roughly as,

= 43.15 pm. Example 12. In 'ORS', the tool angles are :

~hctimtion angle ( i ) = 0° Orthogonal rake (a) = lo0

Principal cutting edge angle' (A) = 75O , -

Calculate : ( i ) Back rake (ii) Side rake. (AMIE 1974 W)

Solution. We know tan ab = cosAtana+sinAtani

a, = Back rake = tan -' (0.0456) = 2'37' - ) I< tan a , = sinhtana-coshtani

,' -! '/ - ZvJ- 4 ' . a J r ' 8 = 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

Also,

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, cr, = 8O, a, = 4O, Cs = 15O

In ORS of npmenclature, A = approach angle :> t ) . l # J a < B r ;

= 90" - Cs

Now we know

= 900- 150 = 750 ' ~ i ( / l pi. r i a q aii i i i i ' 4; - 1,.

tan a = tana,sinA+tana,, cosh

:. Orthogonal rake angle, a = tan-' 0.i04

Also, tan i = sin A tan a,, - cos A tan a,

:. Inclination angle, i = tan-' (0.1 176) $ewe= I: =

6.70 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

5, x v T: Cutting Power, PC =

1000 ,IW i-

A Textbook of PraducWm h~~

Motor Power, Pm = P, 1 q,, , L!L 7- ,<.& : \,

Let q,, for lathe = 0.85 $ 2 - I ; G . ' \ f t

Area of uncut chip, A= = t x b = f x d = 05 x 32 = 1.6 mm2 (Eqb 14.12)

Fc Specific cutting resistance = - = 1727.34 N/mm2 f x d

:?r e! Unit power = - = 0553 x lo00 A,V 1.6x02xlOOO

L O .- ?{I+ 1 ( 1 , : = 1.728 W/mmz/s.

agi- Example 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 .5!3r5 59;: : F, *., 1 .

1 ' 1 1 - ( I %,!I

- . z , , =?* 2 r;- j- , > > . [:y . , ; (I

T, = 4T,

; T T *,ikr!x, tloifsrii! 3111

. . Percentage increase = a x 100 = 300%

314 r G

Exptapk 1.6, For an 'Orthghmi cutting dmcess : , gnbi\l\r\ o 103 .L1 ;iqi~:b*.?

.<<>41i> * , &,)L\'- > n , *y>:

Uncut chip t h i c k s = 0.12 7 mm L.! k t , ? "??<t\\l V = 120 dmin

Rake angle = lo0 , Wiath of cut = 6.35 rpln

.c * t .> i j ! , : 8 , 3 ' v,, , Chip thickness = 0.229 mm I.\$ ' ' %...,'I'

Cutting force = 556.25 N TbtyiCe = 222.50 N ;{, :t 3 ,I<*+ 21!i#\.' i:. i $ f n , i : ~ i ~ z !

Calculate the percentage ofiota1 energy that goes into overcomingfi.iction at the tool-chip inte$ace.

Solution.

Now

Frictib d&i$ '' - - F. Vcc' Total energy Fc . VdrL .;

(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 &~Q&-x sin (320) ri. . 8 ,

'&Fit(& q*?ka* and feed of 0.3 mnv'rev. > .

L , - :,,, : ',,&i~e#d f i c e = 450 N '"

' . I 1 I I

. . (ii) W @ c * e m

: ~ : q l > t , ; L ; A * I * . J c

(iii) Energy conrmed if the total, pjetal,femmed during the hhing operation is ,

( i ) :. Cutting power, - - , ,E,m 9-C' ' PC I* ' ,rF; swig zl 33i8r.1 p n u , 31 i K T $ / :

= ---. *.= -4 - .- - - -- i ni* "{a - 4 + +)~&3irn i [ \ s W +

03x45 .i4,jb F d power =

, - ~ a o x . l ~ 5+ 1 rLf

@ s --. -..

m (ii) Now,

A Textbook of PCscrw~m En-

= d x f x V V . ' . ' ' I (See Equ. 14.12)

= 2.545 x 10' mm3 / min.

1308 x '60 Specific cutting energy = = 3.08 W.s/mm3

!.:. ;\ 2545 x 10' ,

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 as: 6" - 10" - 5" - 7" - lo0 - 30' - 0.5 mm Determine the various components of the machining force and the power consumption. Take : depth. of cut = 2.5 mm, feed = 0.125 mdrev, turning 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" T- 30" = 60" ;, >:, 8 % -rT v33 3~~34 1 0 A,. ' Y : ~

~riction angle, 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 .~~'LU >\F 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

= 45" + 5.8" - 15.5" 7 35f0 .:.w, ' 0 : . - '

Merchant's theory is more accurate for plastics but agrees w l y &K machining &, With Lee and ShafFer relation,

?$..: X A #jm Tf 5 -3-*- - ='.I g I

4 2' - 45' -+jO + 6.62 >25.6* Now, the cutting force is given as, t : t )k t j !

Now,

Fw:9 ,, depth of cut d

b = width c$ cyt = - =- *ill sill&

and uncut chip thickness, t = f - sin X '

' 400x2.'5x0.125 . . Fc = = 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) *

t'ii''q' 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

edge of the tool, see Fig. 14.17. -,',,' I . , . ,

,.F ..> - ,. :. Feed force (along the axis of the job), ' ' ' '! - . ,.., > ., , ., , - 1 -A" F;= Ft sin A = 135.932 i s i n 60' = 1 17.7 N , I 2 ' 1 ' ~ . . ' . -

Radial force (Normal to the ~ i s of ye job),,(u rn @rat kc lo; :.'A, L . J 1 4 f . r

'=;f;;~p~ ' 67-96' &qo aantj. lfi,qa , 1 1 , . ' t r. t - ~ . ~ d u n*sd 3vsd NOW, cutting Power -& $8 W W b- ~~zinu -,

and

. Power = 386 x 1.57 = 606 watd I :

rfisrrj +I~OIJ~ qdt) , ~ ~ d j q W J ~ ? sdri! gntrbi 2 ~IULD! !j '5,3! .. IJ .I@ . snrmt>rr$l Jw11341fi '

itsi it^^ ni;ilc I R ~ ~ F . .t@l3 ni r r , b ~ f r 1. Define Machining Pr-W: * , :, -,* .= .* . , - J p 3 ~ r t t w c ~ l l o ~ 7dr .&!,

2. Expla,in a basic machining operation d t h the,help of a neat dia$pnl

3. Explain the various elements of a single-point cutting tool with the help of a neat diagram. 11 '1 , . i 8 * z*tr 351 ~ . r + , f i t i rr ' i A

4. What is meant 'try 'band' of a single point cutting tool?

5. With the hplp of a neat sketch, disc"ss the principal surfaces and planes IS metal cutting. u."t!i ;r ,A 11 ,b*qt 84 s?t& tm: !IS >: ' - ;o! ,I~ILJJ~@ >

6. Name the two systems of tool de s i~ t i~%~( ,~ , , . ~~~! , : b;ml>pi 371. p r ' ~ , 041, !,, .

* I , , - J ~ 7, W& a ne&atiue,raJce ygle ig n~,qnally,employed far cutting b d and strong materials? - !b 18. Showzha 0$5 uftaol sngieswfth the belpaf a sketchad1 3n *JIJI::, .rr1+ t 1 1 ~ ~ - . "' 9. Write the relations betweeri ASA nd ORS systems of to6lAangt&': '' - d ' IJ .

-- ;RXs-0, it1 11 ,&.I >!.. 10. Wh? is meant by Orthogonal cutting and Oblique cutting? . , :I "-.;5q1 ~ $ t v r f t; ai .s - - , ll;~~owdow,~ea~le~qtthe@eofthecunin&~!:!. ,., ;,, ,-- !., ..:.,, 4 , . l + ; , ,

12. Differentiate between positive and negative rake angl@ttlm3 ; I!.;.) . !cc. ,>I~I - L ~ : I J I *

' 13. How is thejaoseradiw of a cutting bol selected? 'I.: L IV b , - -I : . ' 14. Discuss the various types of chips produced during metal cutting. "" 2'"' - ';' 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 4 % - . ! ,

18. Name the factors that contribute to the formation of B.U.E. I *

20. Discuss the two metha& of metal cu@kg.

2t. o h u ~ a-w~s ewau&er& & m&al ming.;a $Rl?!o, isnr I ~ . f b J 7 ~ 1 nc r ' i 8:. , ' $ l ~ , i ~ !. 4 - - - ' /: .F*

22. Explain 'Merchant h e circle'. l

sas1 *is { i l ! 23. Define Tool Life, I >1?, , - 1

--@@I A Textbook of F&&cti~@~

24. ~ h y t o o f w e a r i s i m p o r t a n t i n l l s ~ ~ ~ t t i e g ? 6 , , ! , , ~ . I ~ I , : , r . , % f . . r

25. Discuss various types of tool wears. . 26. Enumerate the factors pr~ nbigh,tool wpar q d tgRI tifa +pmd, ~,

4 27. Nmpe the,@ctprs tRat cotpiby@ to f@& mar. ' > i , ' , . I,, f 1. ,, . . -

C . . t R'T@fl &j -&XU~ jtiikt11.~ >if1 28; kame the facton that contribute to crater wear. 3% ,lt)cl s*,, a> ,&!A? .. \ . 29. Which two pressure areas of the cutting tools @e ~ k j v d to w q d 3+m7 ,: . i - . 30. With 9 help of a sketch, show c ~ a t a wear d;d ha& wear on a ytting rol. . \ I ,T " - A % , - 31. ' Discuss Taylor's re&ion$hip for cutting qki-tool life. , I

.i %r; & $ i ' # ~ ~ W ) ' s h n ' t i 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* .. - .

, * I . . Uncut chip thickness = aiasmun .>' *

"' =r j-&:& "4, . Width of cut -bfri, '. . < ,

L3tip-tkicknc.s~ ra%io = 0.51 . .. - & , , ' e ~ ; + , . ~ ~ . ~ ~ . . : . -

+ .?-

cuttiigt force :a N H ~ , + L - - - , : - . , .

A':+ a@ .< 5 2- 1 - I I

Determine : Shear angle, friction and% %R & pla&, chip velocity, shear strain in chip, shear strain rate and the power'fir the cutting operatio-

34. The following equation for tool life h

' yyOi?3sSg6.d0.3 , +w.$f&ib b d S i L, I

A 60 min. tool life was obtained using

V = 40 dmin, f = 0.25 mm; d = 2.0 mm >.;,:!I - t 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 bf C-20 steel with an orthogohsll tmi havifrg a d e of 1 O0 dt d feed of 0.2 mm/rev., the value of the sirear ande h a t ~ k e n aeseFwd to be W under a shear angle microscope. If the principal cutting Mge yp%c i$ WO, ##&&&t-iijvalue of cutting ratio or the chip reduction co-efficient.

36. In a turning operation, it w& observed that the tool life was 100 minutes and 50 minutes at cutting speeds of 25 mlmin. and 180 mlmia r&pe&eIy. Ftnd khir the to01 fife at 200 dmin under the same cutting conditions. . I _ I

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. - - . j < ; , 1

The tube is J5 cp'h diametsr and 2.5 mp thick. The tube meterial obeys the equation, , , r , r-jc-*7< a - . . L

&i 1,. s p i , F ~ \ ~ . . . & J I f! nh 9rb.j qu $6 j@%w nBli,x3 .e( I 1 I I b l . . a ~ r t t 3rit r,? 4flKfi71?~-,%~6~1 &la+ gtit ~ n r e p .-i I &me n = 0.26 and K = 5fU Nlm2 - I , s

I 1 - I,) I I , 3!3i GI yigd<stim- ~,i , i l@pt3~3 -943 ~ A I ' .31

f Y/V is - and a = se< The friction 3 'cea2 of the 4o#l onargy consumed. Determine the power in optration. . '

38. In an orthogonal cutting opw@b, ~ ' , o angle is 75O. Calcalate,

(0 b d ,,. . x- - r.3

6 :* . .; *='

-'> $- -%-{'.> *) " -';-, :+: t - -wc_;

:: I f 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 gm of 120.,m/win., p cKip thiqbqcss,of Q.30 mm have been obtaiped. Calcdate : - . i(J) the &p -dw&i011 rm43&nt ;;ib k t (ii) the &ak angle, r~.:z ?

r* k~

Ib: 'SkeZCh a &g16-6&t kdng to6t'and sho~-&'it the 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? Ki r Give the sigificape of providing no* +us od todl !ip.

8 " , ! C . >

47. What do you understand by the tenn. '~ool besignation' or 'Tool Signature'.

48. Describe thptadrepmmted by.10, 14696, ,%8, . l win,ASA$ystem. 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? ' '

V) Shearing .force and normal force on the shear plane.

52. In orthogonal cutting dperation;'hc fecd isXkl'O min and thexhip thickness is 0.25 mni. The cuttitlg force 2s~ 1360 ad f a thrust f6pee.is 730*W:hi rake mgfe afthe tool is + lo0. Find : , , 1 *

. \ \

(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. . I %

(4 The Sizes k i ~ k )or# wd awldo& m ' ~ me mi face. (6) The i s S ~ 6 ~ ‘ d 4 ~ - ,wd W a r n fbw, og &e.&ear ,plane.

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 :

1

(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.

(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 , ~ f the the is $ 8'- Find : * p r m f l ' ) .&

(c) The unit power in W/mm'ls. I

59. A workpiecc is being cut %l?P6rmtk-W b;fed!@ hRd kW- 'k feed is 0.25 mmlrev. and the de@$,clf;q~t is 5 mm. E - . .fia] Cutting farce 4.v: , 3t., ( . . , ,

i . ... , I i r ' I I

cut 6.35 mm, thwlf@m& Pi7 .&Is, and a dyn ma sun^ .&e ,mt&vg @ce to be

.d k djrip gic&e$:'t . ,_ I I , , , ' 4 , , ' 1 J . t .. < (

' ! 4 .

(6) Co-efficient of friction. ..' i > , i ' , , ., . i'.

(c) S b r and normal stresses qn .,! $ e & ~ # w , . . . ', . , , . & (4 *Shw!p& @tr+n,7% , .* , . I , ,:+I :4J# i $ i , r . ? , . ' b : , ;~: . .J , > t , 4 ,

. * 4 % . ?.!911f,i'-'<.,2,1*: s * ' J' .is'. : O j i, ,, 8 4 . :d

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? ..I ;,I; ,, 1 - 1 $ 1 , : i .~; , s , i<::r) , lT ,J;.:: r y 1 8 , r q , - , < , ! f ; ; , > ,! -<t;: T-!\

58. Establish BW (todlrii% cq (Taylor's tool-life equation) : , , ,::,, ,!a.2 b. :::,+,,{,,, : , - c f , i

A tool life of 100 min is obtained from a cutting tool at a cut$&& ~ I D ~ ; , S - , W * , and l o at 33.3 w;l~Yms ,~~G:FH&W~~MA~,.Y;%%M.#~$~, , :

%&%2%fl&%~Jb. IT:..>. ! ' 7 ;* . ia,-~s&a~,rJ,f:-~zsw ctdtSjng. speed for

' +:

(c) Wive rtil e x p r d m for the most economic cutting speed . ,f . (4 1s the most prod~t ive ~v t t j , ' l~ . , . . @ ~ $ ~ f ! $ ~ m e ~ ,qJ$$ ,e:$;p$y~?mif spe* reasons

for vow answer. i ld ;>+ '!a & % & ~ O ~ ~ & ~ i ~ ~ ,#ri$iai9,tH,1&e

(e) Make a rough plot of the d a n spa& &;a#&& 'WOW =HI0 rwmw

t - :!:' !i 9 ,I.* ,,j r , c, 3 :. iflty. E$tmp Ir$wW@l% ~ ~ ~ ~ ~ i ~ ~ ~ t i ~

2, : 1 , , 1 k k t ] , , ' " , (0 Wwk material micro-swture.

,, .<, A , ,i , , : , r

(ii) aLpe of cut. !;'''8 ' " i , ,

(iii) Tool take angle.

62. Fotlowing data were collected from an orthogonal machine test oit%eef 7 ' ' '

L' - , ; I !o 2 . , > ' = l ~ ( , l j f ~ & t f ~ , Cuttlng speed

'* ..' >Ll Rake angle - ~3~

*I

Clearance angle -61 * i t '. 8 ~ k If)" '

Width of cut = 3.2mm . . -, = ' . a r ~ ~ g . { , ~ . : I +I

t ,. ,

ff vel*~tt&tI&6~ '* - ,' ' ' - , .

H. ipraii?irYtnC d&ele diagram of forex and evaluate : shear aaelc, shesr wain, U*.agsiw chip flm, Eriction force an the rake b.

1 2 . I . .

63. S h w that &ring osdwgonal cx@@ with a zen, degm of rake w l e , the ratio of t k shear . , . . . stra&,&,& %,work material t% qptfi4fip p~ltbing, m. e. is, giyen by

' i~b(41*1 >d ITS: mm aod ! 16 -,.>r,? 7- 1 v'7-@5 m/mkr>.?! t , :r,rl,: . ,- , I rv. I. : : ( - I ? e ! . 3 1 .

~&f5SIi H.S.S. take "$"':fi" r 1" I . " . ' -

i .I f i b I : ' 0 9-u qntJ4113 r, la 311: ;(, I ! .I!? t.1 . . , I I : b LL i s 4 1 15rlc~ 10 t . $$ S ~ I I 1ooj ~ e i l.nw (1 f~ q-15 mdr~v. lo Lngl ~ r i l t i o ~ 4 , 1,. In

' \ t ! t I = '1 Chip thickness ratio 0.35 fu:3 ,x>irk ~ r b

3; 1r.tu3 c @ 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"

'"" ' ''~he%'&h force 6h t I i e ~ k i i 1 d t f i n ~ w a s 1 me&dwHI&~'W#sWh.* and was r i ' a ~ . ~ : , wu3A&,r&o N! -1 the nvtg of the m i for&* s f i a angk and the specific "$

2:: rrne,ts d+q,~&h- M ~duittrn L to ? r , l c bnt, nrnrl~ '5 R 16 arrrwvqo nrrdw rb-bnvrt , .r ... L . - . , .. , ,+, 14 1n:an o&oF9na1 cuFinp operation, the cutti'ng'speekb~>.5 mls, 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#'&gth of 50 cm are obtained, l%e tool post dynamometer gives cutting and dvust farees

i 0 CI %r: ~ ~ M I N & d : a ~ . $ # . -& : &,I Arr~d * I t , 8 m u ? 4 ! lit--) dC

(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@e cutting speed b 1 .J d 3 , and

, . thc rake angle of the tool is 10". Assuming eo-efiokt of fEieffsn at 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';rveiage shear stiain &. '

Use the relation of Lee and Shaffer to find the shear angle.

. I r i (4080 N, 861.8'N/1*~, 861.8 N/&, 3.%, 4.44 x 105 5')

rm.nQ& ABRS6 is - h i with a tripple carbide cutting tool having 0 - 10 - 6 - 6 - 8 - 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 ~ f ; 1 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 ME 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 ,

?

fanddareinmm. - ( I ) . . ,, . (A=. 165.2 rev./min) 69, ,A H.S.S. tool et 24 @min. ibr qae bwr. Ci-ccs arc such that it becomes desirable

r 1. +i. to run the tdQl for lW0 M&. Eqitgate the suitable speed. Take n = 0.25 In the Taylor quation. (15.74 mlmin)

%a $Ws A Textbook of

~ t - t b ; l ' & I Aimed apprawh aagie iW, of mtation. The mean I - - the cutting ratio.

7 t 1 (0.32)

71. When turning 19 mm d,imeer bar on an .+wddi;l id^@.^^^^^& ,#lv ?id<

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. ,,, SI (122@y/rnin.. 16.5 s. i3W)

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 m1 liR wh of t(n c u m on the ms&rid. ?E.i; ~ 1 1 : ~ ~ : 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 if 1 increased t?~JOSmlmin. , (2.7 min.) &16-~irll1irwl. T h g 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 ~ $ r d ~ ~ ~ ~etennine

dmin : 4B.74 %~4&%, c

@f the Tqloa tost a$&quatiom,2P ?-gkm$ q8.F C = 54) Wt. The pnnr rwind m turn a medium C - S d is a p p r n ~ i m a t ~ , & ~ ~ ~ ~ f i . , +Tpe power

avait* I mwhine spindle is 3.73 kW. ~ e r m i k e :'eyPi"" (3 Maximum Mm' $$--ti .?+.>f!d @ &Gi% -r JfiR "6 '1,' I 1.t r : , ",G' ,

,- ?i); &%/ntia. W.I oh^ Fcrtd is O. rev. 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

was 6,25 ~ t i a snQ wm -med tk &hip t h k b a was 1.25 mm. The N and the *st was found out to k 810 N.

?3wiwdMetdm- . ~ t y e 3 h

83. While turning a &d stel bar on a lathe, the following data were obtained. - . . Machined len& , ,c % , (B .Jlfi

= 150mm a r k f A . w s , , , ~

[,I .I Speed e , * 170 rev.lmin I r ~n!txlo~*k nr OUT r wya M

whining time and the total amount of heat ( N a p Universi~) - (m g+W kW9 1.765. -+, 94&5 1 kl)

awehiggies ; o f i W -atting,aud &&ace various i 3 , 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,~, . ..: I 0.250 mm - , : - ,:#&'5", , - . .

&a - 7

$ ~ %+ .

< I a > . g - " r Width of cut #j$ asnun &&&,-: . , ,

* : . '. . ; Cutting speed :-, Ahg@f41(FO @mi&,: A:. .>-. : .'- . . , J * . , - . Rake angle ' c lo0 .t'* . ._ - Cutting force

. , Thrust force osa ai 23 N * . J@W Y 3 & 29.G0, 61.24 N/mm2, 6 - hm e-td *UP tb * ~ ~ ~ W W ~ & H & F fMgro->$ly!cCbeen n o d . .

at the cutting edge is 25 mm.

86. A50mmdimettrbarofstct min. The speed was changed @.p2 v w a n d the tooi failed in 60 min. of cutting time.

1

Assuming a straight line reletionship exim what cutting speed should be used to obtain a 30 I

min. tool life? (Aas: V,, = q939 m/min.) 87. For ortho$oml wt t i q wi4 ,a ---: 4 &$ag a reke angle of lo0, the uncut chip thicknrn

is 0.15 mar and chip thm mm.-Ddemine (a) the cutting ratio. (6) the shear plane an& wd (c) the sheat s b. . n k rw- :.,i ~(,+&~;~~;‘+75, " ~ 2 3 5 0 , 2.734) '

,operation, the following oar8 have been observed:

* o , , - !-0.25 Ihm

' kiY.75 mm - -'.-$3.L4. - C' *b,=2.3 mm ' If

. ; , Rake angle, '1 a08 I--? n,u a-- fa? - ,w&. Cutting forceY: '

2

EC* N ~ - f w c e . ' . ;, !,,+a:, !,&=475 v , Determine: shear angle, d c i e n t of &on, the fiction angle, ultimate shear stress of the

I mattrial. ., 3. (Aas: 18.4O, 0.5, 26.574 379.4 MPa) : a. In h A m i E h i n i n g ~ ~ - ~ & - ~ ~ i n g data have been o b m e d : ' Uneut chip thickness,?^

2S+--- ?i=0:2 tRm

of cui,ww - ~1~ .to - 'e %,, h;i. ,%Ill I ~ ~ ~ T ,

1''- - Cutting speed . ;Y=2QO mhnin. Rake angle, 6rI ? $55 fid . &dl@

I C w of fiic#iW2 hi3 '4 I -1 .cP.@~ ,%, , ' , & M ~ W J ww - Ultimate Shear b t k s ef W d , . ! ' t*lsO-MPa, ' ,

i i u s .t- : SPOE* tbe cirm ibd c o ~ m ~ r m & 1 1 9 . :- r d * 1- . . PJ

yuiiik: 36.T. 420 N, 125 N? ? -- ",.'- ; - - . ,. z.. , - 4 .

, * 5 , ~ ~ .ny;z.,;.<%. ;:;* t g=- . - 5;. :-; , I

. . . # F:

m . - : - - .

,,l -. . %

A Textbook of P~3Wctbn ' l ! ? m n g

Hint: Use eqn. 14.19 +=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 relation. Itlwla WI. (Ans: ~ e r i A d Th&ry : Ec = 468.7 N ; F, = 139.46 N)

- 2 ! ,,! :, Yc: bnfi x i a r t ~f:!,-tld.Li*- n!.?. Lee and ~ h a f f k t h e w 1 F, m,W.OAJ; E, = 154.6 N)

kr I # . ah& hY'&&g during orthogonal machining of mild steel with an uncut chip ~ 'r# i id of 0.25 and width of cut being 2.5 mm. Take a ,- 09, = 0.5, aad 1, = 4 0 ' MPa. ,LA$ t , ,?SY& h m a bns 9igr:t r~rmrit wad2 snrmts,.ii! c<Grlz$:~rf&m alg

q%h i1. , - .~ .3 I .(a ~.:u.r-. 4 . 4 hynr~r. c o d lAns: N, m. Qctmnine the cutting f m and the thrust hqe W thE fpaing ratio when machining M.S.

Take: .= , . t . * ;

Uncut chip Wkness - 0.25 mnn * c

wjdtf iofcut=2m -1: . 5 * $ .

. I ? i

Raks angle = 0' ,. . , , ?

(a.?ra~svl.!!.J ~ u ~ b l l , i4 a p = 0.5 ..3,, , , f - *I

((!tt$l 5 <.pp.\U Ad*?.. : if 1A ->*rar TxFm+m

- ~ ~ O ~ t 82 ~ J w W a d 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 def 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.

(b) D ~ c & ~ & the average power consumition if $hq ht~ration €&s.wit% 60 strokedmin. and -. 1s1i141t I L I - 1 r

Len 4 wrnl the length of the job is 200 mm. - 2 . ! ' . .

: - 8 ". @amp: (aJ,U&g Lee and Shaffer relation for shear angle.

fil ~ I I ~ ~ I T W J ? ~ stuflist odr b nrm\ 13: kRl rs bmuf oaw txatz %I rsjsr;lr %!ti qn~li;r> I!> fiim tB bhd jool ?hl $,it&

D f E I I I P J C ' ; ~ o f 1I~c's 34 bit&e tzr~q.: p:rw t rtcnl~t:.~ $6 :*#A). @=tan-' p = tan-1 0.6 = 30.960 =3t ih I . tnh ,~r!* a'& ~ 1 3 ~ dt ,%i 20 3!q1!?i ~ S I bi pT - I T , > . , - - - , ,lt;,r, ,LT.Tf&".,& r * , as, *-I--- * I:\ -": .-.- k745O (31 - 101 = 2y ,. _ i :!, 7, - . , . ~1 1 NOW, fjqblX (it -8 along velocity vector), . , , : ; I 3 , ,

Now, d b= - and t = f cos C,

bl msc,

-. tr '1J. ' . i ,'t? 4: .? f.i ' ~ r ~i s?q-\\ 340 x 1

. . i.2 1 ~ " " n~ ?ti.' '.&? sM,IoX c0s4,pXPin240 = ?LW-,N r t ; i*

n : i . r c..i d 12 .,, : Thrust force, jGfq . gn (J3 - a), it acts norm4 to tool-job interface.

111 '1'111 s7q , ,, (Eqn. 14.18~) = 1103.8 x tan 21' = 423.7 N

Now fead force component, 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= 31 1-85 N mim mud tq the mekined surface) i I . i a f GCI- ' $ I : :rnA) , i . - i

Theory of Metal Cutling

2 . (b) Now 200

Work done= Fc x length of stroke = 1103.8 x - ' 1000

= 220.76 J

. . 60 Average Power= 220.76 x - = 220.76 watts

60

94. tM$u$e$he i f r ~ i ~ m ~ ~ ~ r p a WeDet Bat. owehind at 200 rev./min, with a feed of r* P" 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. Calculate the

power consumption. (Ans. 935 W, 0.08 W)

Sol: Power consumption = Force ~Telocity

; ! r v ! ~ . e .. I ~ I x,P,,&'JV. = wds and Velocity = ' 1080x60 1000xf@

;. Power = 1800x nx248x40 =935

1000x60

w>4 IQQ Now velociQ ofp feed afQrce =, LOOO

' I , - 4W'x 0 3 x @ '= o - i

.-.P"M-~ :-, Which is very n li ble as c m p & to Ppwer consux,nption for cutting force. 9@, i . Now MRR = d.f. Dave. n . N

Power :. Specific energy eonsumption =I - 4-

96. Tht wmlt8#t &We rrt.8 tub1 psint,is f 200 N and its angle s f inclination to the. horktn$al is 3 P . The approach agle to whi& itacts perpendicularly is 26". Determine the three +w&ipal cornponeat forees. . b 4 , 5 . I s ,., . - t I > a

%hitian. a - 54b = 970.8 N LJd,', iYI 8 1 f f " * . i l i

Ff= F,,.tan 26" (Fn = 635.5 N; Ff = 310 N).

a d F;= F . + Ff+ 17n2 -- 97. A W C cutting tool myhning MS gave a life between ~ g r i n d s of 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. , -... 2 > -

'? rc>t,'" ' 1 - - - A - .,

C r ,

r- Sol. VTn = C I

- . - - -

:.Log 80+nlog100,= logC ...( 1)

and ' loglOO+nlog33=logC ..- (21,

from these two eqw- a = 0.2 and L3 = 28 1. I