Theory-of-Metal-Cutting-Solved-Problems.pdf

April 3, 2018 | Author: smrutirekha | Category: Machining, Physics & Mathematics, Physics, Materials, Mechanical Engineering


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A Textbook of Production EnginwringI = 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) 5 r 2 x 0.127 -sln cp . The following equation fir tool life is given for a turning operation : FE= :b 0.1 I ) = 0..1 1 m/s ~0~21.0519 t "'" ' = 0....985 .(eq. :<L:. Example 4. t$.3 mm/rev. 14.?a = C A 60 minute tool life was obtained while cutting at V = 30 mlmin. = 0.0245 mm Shear strain rate.A Textbook of Protjtptjm Am@nq@ig .: *:. 3 = vs 1s v. . .245 mm. = 2. feed and depth of cut are increased by 20% individually and also taken together: ( i ) Now .= cosV (9cos-aa) Now . .77 0. Taking the thickness of deformation zone equal to one-tenth of shear plane length. f d = 2.25 -.5 mm.-.37 ~ ~ " 1 f3 d . and Determine the change in f q l life the cutting speed. (vii) Shear plane lens = 0.. 7 < : I * I I 2i..48)'" = 20.. \ ~3iTT . f = Q. . .~ i 38 x 03% x (3)'" I ..38 hin h~y$tl(<~4 :.&q (1.38 'E" I. - C Z P 28-38 ~ 8 . -$ .sbeararsglw.ur3r.i<*.. .48 . 0 the .['.o g*fi*~ ".YW' - 3 k . -.i)d~25xld=3mm. a lo0.-% ' .*~ . '1 tap= l-rsi.r 3 (ii) Now ' *i@ 1 '& . !(I$: . (b) Kinetic co-eflcient ofpiction at the 4k*&43 mm.I .:i. d = 3 mm b ~ (1. 2 6 . -4a3(a36)0'77 x 1...+ Feed = 0. f = 0... ' . - P''-.: .. t = f = 0.\ 9 = 1. and the least effkct is of depth of cut.39 min . A 6~'"9 ij7.154)'" = *oil nin.. Emtfib w e = + 100 = of shear at the shear plane.>~ ti I - A%.45 mm Width of cut = 2. .s.$ $F. Q = 2. ' m 3b ~ 3 . (iii) Now a .i?lI.a = 1.5 4 s .36 mdrev.F.403 x 30 qk ": o el...uS31-q1.(. 28.36 mmlrev. = F c c o s 9 .5 mm. .L...25 mdrev . (iv) Now V = 36 &in.*p!t -&'gw-=&?<?..r?iJ*j i .591 = P8& 35.\ Chip thickness = 0. ". ..1.it .r-~ .25 mm. J *$%s. yk &hi = n mnl T i Example 5A Bring an orthogonal machining (turnihh operation of ~ 4 steel.iP:-frlr r.* 4 .Tangential cut force = IJ3@ N 111 t:c! 3 i ~ ' r e >ssdZ Feed thrust force = 295 N Cutting speqi = 2. y ." . ~ L * . . where cp is. g ' p .ein cg.5 mm.. The maximum tffeet on tool life is of cutting speed. . ~!. ~ H C ' .J.3 x 12 = 0.u\+.w ? *- .:>: p ._ . : t ii . where n and C are constants.> ~ h !O ZF )3&> l~zxtlM+ br. F 0 CL = F. . The following data relate to an orthogonal turning process <i J 3 : Chip thickness = 0..~ ~ n o ~ \ ~rwl..< !.S. (ii) Calculate shear angle.1 Shear angle. tan cp = ~ % .9) 4. q.' 11' I Z min 90 20 > r.:< .: (ii) . .i:li':' 3.947+ 0. 14.:# (1 (ii) Hence recommend the cutting speed for a desired tool l f e of 60 minutes.. n"=C . i 1 s 7- t" 'I r m a 1-rsjna . % = 2.. a = 15O . . 5P. ~ o + b .B ntnr/Wn d a depth of cut of 2 mm is given by VP = C. S l r on t = f = 0 2 rn 14.. .a r t .. Euiflrc. Solution. (iii) Calculate the &n* pkar strain iavalved in the deformation process. .012 Example 7.* t \ \ .62 mm itli k*.. iis. 1 * i.. VYI di 6. Engrngr&g "6 (y i 3 * 3 t t b + . > < %. .13) t = 0. tan a (1' .:?*.gMat!afeCd-~O. r = .62 mm. . -.-- - - + F. ~ekii'=0.322 Chip-reduction co-efficient = .. . (eqn.SS~I~ n t k : r n l jam %<I' .=-.A Textbook of P&t&n .V i q and chip co-acient. . The Tayloriein tool-lfe equation for machining C-40 steel with a I 8 : 4 : I H. .S. : .065 = 3. . .. t ? . tan a Example 6.= 0.a. TkfilIowing V and T observations have bem noted : C: d m i n 25 35 $.-c 4:. +iofi t t t c l ~ o k eang@ i= (i) Calculate cutting. C 7 . .. .= 3. .2 m d w . ..-Calculate (i) n and C. 9 (0 Cutting ratio. ~ ~ . .F. Yw<!4 ~.i.:+-t 19rw(1. * ' Determine the nglrpal and t~ngentialforces on the tool face.' ' ! Rake angle. . 3 N .$ -a)sin cp Now " .7 on the tool face . .. = lJO b9w..t.. Example 8. Width of cut..C. nii i 3E 1: tan cp = r 00s a 1-rsinq - 4 :.p 15" < . V = 2739 mlmin. 3:1Eit2 W P ~ . The following data from an orthogonal cutting test is available'? '. F = Tangential force on tool Eace >w. ~ k r2& F ' -. is-\%I Chip-thickness ratio.r:tcwP (3*7-+) ' Friction angle..& - .I '~2qf-"< la.. ..l . ' 1 .r . = 0. Now. ----- Solution.a)-cos(cp +.. z.383 :iizmrl ~Au:'.1>'iL.i)-.. = 280 Ni'mm' Average co-eflcietlr of friction = 0. t = 0.. F l = S l l ..353 and @..5 mm Uncut chip thickness.:*>: r. b=3mm Yield stress of material in shear.vbt sec [fi .h .: 8 ' + r = 0. $ = 350- -(3) jr = Fc= Now . nl.~~w l .. 5i+~~.@. cp = 20'.~tp . t = o.5" reo Now shear strain i '. = S 841.%?.-.. .6). A = & :)? j:>c : N/mm2 P8P .A Textbook of Produdon Enginewing N = Nonnel force on tool face :.& = 9 + p. .. ..F...65 mm2 1 ti. i na From here. . = Fc. 7.. . .95 Now shear plane area. . &. 5x025 = @..474 . given tensile properly of material as a .(T]* .-.4~<9.966 . 13.~qc. .a = 20° *Q. ..5' ?c. sin 205 I lr 3. Rake angle. ' - 4 = 3~ = 784 @)@+$' NI t Shear angle.. fiK simple tension test. (Eqn.q3. . on a lathe : m*h of ~ ~ k i r y l ~ ~prt\u=. '= .sin a = 1405..3 x 0.! Cutting speed." .a .51 1.259 I! Example 9.4.2m/@p .5" + tan (20.7 x 0.C: mkw%& attw-. *. .cos cr . -- !GOL151 x sin ZOOeQvlirrilqtjl\r c 4 .sS ='2. . tan cp = rcosa 1 . . % . Thefollowing observations w e e wade during orthogonal cutting of steel tube a&l&42&'Rs.a-@qD. P-z'8..s. ' \YUL()LNin mmz .P 92 and I (Considering Von Mise's yield condition)ll i .4c i L ~ . r = 0..$3t5fx\ '.?& -.:t Find Fc and F. Yield shear stress.. &&f5 T $ r n ~ % . + j L .351.Y 2. -!$: ' . .605)o.-C. ~ C 3.hi I l i ~ 3 i I ~~ .tl! j m r 2 a= k A ' ' t: (E)~ B = K.66 =f i \-ld = 485.5O .r.20°) Now from the relations of chapter 13.t i. . p = 34. 5 384i (4. r k ef For garnliaos- G.\ a@+tm(q-a) 921 = & 20. 7 sin 14.... . .O N:? .5 n4vi9 . - -.bn+* y %< . I I dfiA Fcsinp+F.v<.. \ \ > 'k: CS$.?.+. EM. . F.amv 01 - 'au .% Solution. Tangential cutting force = I600 N WJYL im .. 3 = 200 i. -Cb.mn\l'.w + +-oA+.39 mm.. the foll~~cing observations have been made : Depth of cut .: t. 2:i.. From tool designation. = R' sin ($ -a) = 2165. Fc = 1600 N*El E 85.482 +850 x 0.I .76 = 1515. 7s . ( i ) Shear force.19. . .?. . During machining of C-25 steel wit% 0 . .*. p l c f W t i 3rtil-3T':l$3 WI t (iv) Kinetic co-eficient of fiction ( v ) Specifc cutting energy.-.mn(Sipce*kr 90°.>niz [+R (.. f .* A k ~ r i c t i 4force. '"rr V = 200 m/min. 1 4 .i.6 8 90 .. t = 0.rwZ Feed .. I f . COSP r [email protected] shaped tripple carbide cutting tool..20) r 2 20914 N F. 3 ! 1 ' .** . = 0. t Other given data are : d = 2 mm. (0shear few..:.."F 850 N 5 = 'Ir 2:s t. F d t h t f o r c e mm 1 L .. 7 \\?I.. .:! in6:i y \ r r v w i I .- 1) - 01 I-. Chip t h i h e s s 9 2.5 and = 541. . . (&.r sif? nm. .i '. Equ.850 x sin 32" tan 9 = -.- (iii) Friction force r =:?&I 1-3 flh ~ ~ 3 i l f @ i l ( i ' $?.0513 x sin 10" cp = 32" Fs= 1600 x cos 32" .:.:&I .0 ... 0513 x cos 10" = 0.. 1 . sin a + F. i . ..>. 1 3 ) .?.7cos (345. . d"k & . .!. L.-va 9>tit. Lg.. i : b 3 ~ . (iii) 8 . = 1 .5 N Example 10.. .ti ~4ee41\ iiti . t = / = O. .... . J.<:.._iaJ.r*..*i Cakulate : -.Now - R' cos (P .8 N2i+nrr. C P ~ O ..1 m m (ORS) 7 r i r y c Fo = 1..b fts...L tr-: at c i! 9 . F = F.2 m m / m ~ F V ~ ~ + Li b i ~ u .x 0. I 0 6 ..:- i: +.w.L.&: btsic ?kt:. x.'.3m (ii) Normal farce at shear plane . - .@F" = : '.# . Ti.- - . = Fc YF. ti:ncu=. 5 a :..ui. cp.. tYl015 2 = Q t r n s zr:rh~3itr. :d ~lI .L. (ii) Normal force at shear plane.a) = 2165. .:u :.. .551 1 . .m\!'t 5s" a-IO. % c ~ .A=?OO .h~ Qr..F.. ~ ~ . 46.0456) = 2'37' tan a . ''"l" - 43. radius is 3.A .26 (a)..176 = 0.0.' -! '/ - (AMIE 1974 W) tan ab = cosAtana+sinAtani Solution. 14.A Textbook of WWwAitM ErrgrrrgrmMng .753 Fc Specific cutting energy = b. = sinhtana-coshtani ZvJ.1 mmlrev.. A turning tool with side and end cutting edge of 20" and 30" respectively.\ I : Now b = d = 2 mm (See equ. Ce= 30°. 14. . .402 + 0.13) Specific cutting energy = 4 1600 = 4000 N/mm2 2x02 : I 1 J Example 11. eqn.. R = 3. Solution. (iv) . & .00 mm. tan a + F.40' .0433 .. f = 0.cos Ce)R +f sin Ce cos Ce .. Example 12.t . Example 13.15 pm.J ( 2 f ~sin3 Ce -f '. In a single point cutting tool used for turning. u' ' = 0. a J r ' 8 4' sin 75" tan 100 fh "t?tbi I I * . = 20°..00 mm Refer to Fig. the geometry as per ASA is : Back rake L. 14.- )I< -' (0. We know - . = a.17) = 9".--f*l- F. Fc-ctana ' . - i 4-.1600x tan 10" +850 - 1600 850 x tan lo0 tr = 0. c.966 x 0.' h = (I .) 'nw'F'' . .17 (side rake) as= tan-' (0.WaI .:?(:'I '= .. CaIcuIate the CLA of the sMace produced if the tool nose OE: = WWii A?~:J ! ~ b i i-:~. g ~ . In 'ORS'. operates at a feed of 0. = Back rake = tan Also.2727 = 0. the tool angles are : ~hctimtionangle (i) = 0° Orthogonal rake (a) = lo0 Principal cutting edge angle' (A) = 75O Calculate :( i ) Back rake (ii) Side rake.sin4 0.the peak to valley roughness is given as..ii <. The given data are : C.. = ' '.1726 mm The centre-line average roughness can be taken roughly as.. *-.r.1 mdrev.f l j ? '* nc r:3:1 : : = Si&rake=Joz. a a Side cutting edge angle = I S 0 . a i a q aiiiiii' = <B 4. The cutting force is.. . Fc = 162.S.. Find the values of inclination angle and rake angle in ORS of tool nomenclature.4 f Os5 kgf AT Cutting Power. tana.x v 1000 T: . . specific cutting r@istance and unit power. cr.sinA+tana.i04 tan i Also. PC = 5.IW i- . r... - 1.a.. id3 Solution. t). :. motor power.150 = 750 r tan a Now we know :. = 8O. i = tan-' (0. (AMIE 1975 S) Solution. = 4O.cos A tan a.S.Thheory of Metal Cwng c. Orthogonal rake angle. Inclination angle. In ORS of npmenclature. As per ASA system. tool for hot rolled 0. Cs = 15O A = approach angle :> = 90" .i.70 176) Example 14.1 $ewe=I: = 6. . A m ! 4 . ' ~ i ( / l pi. For a turning operation w!th H.2% C-Steel the following data is given : Cutting speed = 0.3 d s Depth of cut Feed ( cs Determine : Cutting power.l#J = sin A tan a. cosh a = tan-' 0.Cs = 900. V 0553 x lo00 1.9) - Friction energy = F.-j- .50 N Calculate the percentage ofiota1 energy that goes into overcomingfi. &. . 7- G. .I<*+ 21!i#\. ..' i : . Now (Equ. . for lathe = 0.& :\.. Fc = 1727. Solution. ruu-a 1 ' 1 1 [:y F. Using Taylor equa r6n an using n = 0.{.iqi~:b*.. agiExample 15. 14.'\ft .. For an 'Orthghmcutting i dmcess : . > i j ' v.C = 400..25 N 3 .6x02xlOOO : = 1. gnbi\l\r\ o 103 . 1.5.728 W/mmz/s.5!3r5 59.i Total energy Fr .( I -.85 Area of uncut chip. 1q. . r 314 kt.\$ !.L1 . Solution.229 mm .? I. Calculate the percentage increase in tool life when cutting speed is redreed by 50%. Pm = P.z. Frictib d&i$ Total energy :t '' .!I .iction at the tool-chip inte$ace.r -. *.: : =?* 2 r. i : ~ i ~ z ! = 222. Wiath of cut = 6. %. = 4T.-F.? Exptapk 1. A= = t x b = f x d = 05 x 3 2 = 1.c* t .VdrL .6.127 mm L.. i $ f n ..:8.Vcc' Fc .)L\'> n .A Textbook of PraducWmh~~ Motor Power.?{I+ 1 (1 . Percentage increase = T *. .34 N/mm2 f xd e! = A.35 rpln Chip thickness = 0.-. Let q. T tloifsrii! 3111 ax 100 = 300% G .6 mm2 (Eqb 14.12) Specific cutting resistance = :?r Unit power = LO . .! "??<t\\l V = 120 d m i n Rake angle = lo0 ..<. > > . (I T. L!L $2 - I .. *y>: Uncut chip t h i c k s = 0.3 TbtyiCe .'I' Cutting force = 556.ikr!x. ' ' %. ..<<>41i> *. .femmed during the hhing operation is .5+ 1 rLf c .rF.4 J =. I . (ii) W (i) I . 14.-.&i~e#d fice = 450 N 1 ' .jb -~aox. I I '" * . . 8 . :. s --. &~Q&-x sin (320) .- :.l~ 33i8r. .. :'.cos ($ ... '&Fit(& q*?ka* and feed of 0.A * *. swig zl + +)~&3irn F d power = @ -. From here.. ti F = A' sin p F. pjetal. (iii) Energy conrmed if the total. ' - . 31 i K T $ / : .3 mnv'rev.--- i ni* "{a.L.i4. @ c e m : ~ : q l > t . I* ' . m 9-C' -.19.It is cbar from Fig. Cutting power.=. < 7 ' ' L T - F ri.- PC . E . 4'.* i[\sW -4 + 0 3 x 4 5 . * \...a) = and .1 p n u . L > . . >:.12) . the cutting force is given Fw:9 .S.s/mm3 = Energy consumed = 3. -rT v33 3~~34 1 0 .A..='. co-eflcient oyfriction at the tool-work interface = 0. the tool approach sngle.% p y x 3 41.=- d sill& . ' 0 : -' = 45" + 5.:. . Take : depth. turning as: 6". = 60°. 14. Therefore. Now. . ortkogonal rake angle is given as. V =dxfxV Specific cutting energy !. W. b = width c$ cyt = and uncut chip thickness.3.6" Now.08 W.'5x0. p=tan-I p=tan-l 0.6.62 >25. A. 3 . ?$.125 . = 90" 30" = 60" . feed = 0.$ . from Merchant's relation. ' Y : ~ ~rictionangle. the shear angle is.w. ''I (See Equ.=4S0 + d 2 .as.: #jm Tf 5 -3-*-. .6* t:t)ktj Now. = 2. T- 8 % - & - .30' speed of job = 300 rev.m A Textbook of PCscrw~mEn- (ii)Now. S~lutCon.10".p12 .bar of 100 mm is being turned with a tool having ASA tool significant .08 x 25 x lo6.olarr.5 mm..125 mdrev. = 384 N. .sin X ' ' 400x2. It is .~~'L >\F U 1 Now.5 m m Determine the various components of the machining force and the power consumption.15.s Example 18.6 = 3 0 $ ~a 31" Now. ultimate shear stress of the work material = 400 MPa.of cut = 2.0.:.p I . (19-49 kp#p 4S0 x sin 25. a = 11.8" .&an the tool design* that the side cutting edge angle 4s 30".545 x 10' mm3 / min.5" 7 3 5 f 0 Merchant's theory is more accurate for plastics but agrees w l y &K machining &. ' d. I 45' -+jO + 6.5" . C . ' ./min. A M.I g 4 2' X A . cos X 'I. With Lee and ShafFer relation. t =f Fc= ! depth of cut *ill .lo0.. = 1308 x '60 2545 x 10' \. tan a = tan a.7" . as = 10" and a.tl>\i . From here. sin h + tan a. * x + I ~ Oqdt) IJ~ ... Define Tool Life.'.A 6.a ) F.the axis of the job). ~b. .. Wh? is meant by Orthogonal cutting and Oblique cutting? t . . normal to the principal cutting -..17. .>I~I -L~:IJI* I!..3! . Write the relations betweeri ASA nd ORS systems of to6lAangt&': -...: L - ..(u rn@rat lo.' I .U.. .rr1+ 1 1 -~ ~. r + . . r'i l ..57 . b ~ f r Jw11341fi Define Machining Pr-W: * .. = 386 x tan 19.. from equation (14. edge of the tool. W& a ne&atiue.qnally.~~owdow. Expla.= FL . !. Name the factors that contribute to the formation of B.u and . 10.. i 8 * z*tr 3 51 ~ .7 N .~ .Now...5q1 ~ $ t v r ft.'. IJ it1 11.E.- " .. . 15.I ll. !cc. see Fig.. -. .o! . How is thejaoseradiw of a cutting bol selected? ..' /: . ~ ~ d Wj Jq ~ ? sdri! gntrbi 2 rfisrrj ."t!i 5. tan (p . Differentiate between positive and negative rake angl@ttlm3 .F. : * F.= p r ' ~ !b 18. . sas1 *is { i l ! 23. 2 ' 1 ' ~ . lfi. cutting Power -& NOW.= Ft sin A = 135. :.l+. . ewau&er& & m&al ming. -A" 1 kc I '=. Power = 386 1. Explain 'Merchant h e circle'.~p~ t . I 4 4 %-. 22..# .. f itirr'i A What is meant 'try 'band' of a single point cutting tool? With the hplp of a neat sketch. . ..-!. :. . "" 2'"' . . * I ~ . Radial force (Normal to the ~ i ofsye job).> . Name the factors that contribute to the formation of discontinqpus chi s.-J~ - .r u.. "' 9. 7. ' ' ' '! . 041.-1 8:.* ..ml>pi ~ ~ ! .. :. . > . why built up edge on a cutting tool is undesirable? 17.fa=e..ilc I R ~ ~ .. o h u ~ a-w~s . Explain. ~ dn*sd u 3vsd r. t'ii''q' .~ea~le~qtthe@eofthecunin&~!:!. I ~ I L J J>~ @ 11. 3.r 67-96' &qo aantj.I@ ' itsi it^^ ni. 1 1 .employedfar cutting b d and strong materials? Showzha 0$5 uftaol sngieswfth the belpaf a sketchad1 3n *JIJI::. . 11 '1 .help of a neat dia$pnl Explain the various elements of a single-point cutting tool with the help of a neat diagram. 12..932 N The thrust force is normal to the tool-job inte. ~ . 4. -.!I 18. ni r r . Discuss the various types of chips produced during metal cutting.' 14.&. isnr 2t. that is.!.. "-. Feed force (along. the thrust component 3s. . f b J 7 ~nc 1 ..) 'I. >: ' . $8W ' L. ..18a). 14.:.. b .in a basic machining operation d t h the.. = I: 606 watd ~IULD!!j '5. 2..i~ !.raJceygle ig n~.a $Rl?!o.'A. IV -I : ' &@'*"'' ' ' I 20. Why are 'discantiriuhs ty$e' chips p$e%rFed over the bntinuou's 16.s 4. disc"ss the principal surfaces and planes IS metal cutting.J14f.* . I. snrmt>rr$l .t@l3 F. .. t ' ~~zin -.932 i s i n 60' = 1 17. ai ..qa. ' .F .b*qt 84 s?t&tm: !IS Name the two systems of tool d e s i ~ t i ~ % ~ ( . ' 1.4" =' 135. . . IJ . A kn6 &.&!.F* >1?... '' -d' >!.RXs-0. b- W .. Discuss the two metha& of metal cu@kg. .f.. :I 13.J p 3 ~ r t t w c ~ l l7dr o ~ . : 371.'$l~. &!A? . Which two pressure areas of the cutting tools @e~ k j v. ... . chip velocity. I : . Uncut chip thickness = aiasmun .d0. t . ~. .i. Calcalate. d..26 and K = 5fU Nlm2 L. With of a sketch.?cuttiigtforce N H ~ . .2 mm/rev. .. chip reduction co-efficient..>' .. I f - 1 - I. 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. ~ i ~ i . The following equation for tool life h ' yyOi?3sSg6. . friction and% %R pla&..lt)cl s*. o angle is 75O. 36.. Width of cut '..%~6~1 &la+ .q d 3+m7 . .. & Jf! I nh 9rb.I - .@ctprs tRat cotpiby@ to f@& mar.A % .. Derive an expnssion for optimum val& df itmini . .. (0b 6 *: d ..at200 d d n .e( I 1 I I ~ r t t3rit r..*=' -%-{'. ~ 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 ? .-. In a turning operation.s~ ra%io = 0. . tod w-.--@@I A Textbook of F&&cti~@~ 24. . .: 30..4 mmlmin. Nmpe the. show c ~ a t awear d. + A':+ . 1 The tube is J5 cp'h diametsr and 2.a I.T . . r .. feed and dspth of cut are topher increyd by 25% and also if they are increased individually'by 25%.il@pt3~3 -943 ~ A I ' . .I~I.: 6 . F ~ \ ~ . ' Discuss Taylor's re&ion$hip for cutting qki-tool life. _ 37....tool wpar q d tgRI tifa +pmd.~.~>if1 28. $- t *) 3. '@ w'' " ' 35. Y/V is and a = se<The friction 3 Determine the power in optration.. .r - s 3!3i GI yigd<stim.t ' ... ' "' ' ''. .r. s p i .!. . ' . ..3 . 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.1 a@ - I I & 5 Determine : Shear angle.gtit ~ n r e p .223 &fs7w9. f . 32.fa L+ . .r 25.:!I t=: ~ ! ' J I J . and 180 mlmia r&pe&eIy.md at a feed fate of .'e~.&. . 26.~.> " + : : -wc_... In an orthogonal cutting opw@b.$f&ib b dSi I A 60 min.25 mm. ' .31 'cea2 of the 4o#l onargy consumed. a> \ i 9 " . shear strain rate and the power'fir the cutting operatio34. tool life was obtained using V = 40 dmin.. R'T@fl &j -&XU~ jtiikt11.'-. . 1. f = 0. ~ ' . . ' 'bilrihg m&tnhg b f C-20 steel with an orthogohsll tmi havifrg a d e of 1O0 dt d feed of 0.51 -&. : .< 2.+. -bfri. Discuss various types of tool wears. 1 1 Calculate the effect tool life if speed. . .0 mm >.~~. kame the facton that contribute to crater wear. help \ I . * *I.5 mp thick..Cuttingspeed -.: . Ftnd khr under the same cutting conditions.b* . i I. "' =r j-&:&"4. 29. .-i I &me n = 0.x3 .. The tube meterial obeys the equation. -'> -. r-jc-*7< &i 1. 31. . 3% ..) I I. & $ i ' # ~ ~ W ) ' s h n ' it r . .rI 25.? 4flKfi71?~-.-.. .~~ . . : .. The end of a t@gbh!ing turn& on a +lathe.In an orthogond cutting o& the folkwing d& &e been observed : .x- .j qu j@%wnBli. % f . ' 38. + L . . :a < . d = 2. 4 C > . 27. I L a $6 - bl. shear strain in chip. +w.i %r. L3tip-tkicknc.j<. Enumerate the factors pr~nbigh. If the principal cutting Mge yp%ci$WO. .sgeed: : 33.d ha& wear on a ytting rol. 26 mm with a H. (d) Co-efficient of ofctio" on the &of the tool. $i.muoh ae .75 .4 mm at a cutting g m of 120. 43.. When the use of positive rake angles and negative rak. (g) Specific energy. . .6 6 . What is orthogonal rake angle? S8.'hc fecd isXkl'Omin and thexhip thickness is 0.tools. Why indexable inserts are better than k h d Pdbt fibs. List the various tool angles and discuss their significance.ASA$ystem. 45. r~. i(J) the &p -dw&i011 rm43&nt . What do you understand by the tenn. 1 4 6 9 6 ..*... chip th&@wattio ia Run4 do be 0. Discuss the two construction of tipped tools. The M e angle . ! C . cuttitlg force 2s~1360 ad .127 mm and the d@ sf aut m a l tebe plane of the paper is 2.. The f a thrust f6pee. The cutting s p e d is 4 mls.5 mm wide is made at a speed of 0.of Q. (c) Resultant force. 47. how. .30 mm have been obtaiped. og &e. ' ' 51.58.&ear . Feed is 0. Give the function of each tool element. 39.S. l 49.m/win. V) Shearing . a double carbide cutting tool of 0 10 .15 mmlrev.gles is recommended? Give the sigificape of providing no* +us od todl !ip.S.plane. (4The Sizes k i ~ )or# wd awldo& m 'me ~ mi face.8 .10. 54.. During machining of C-20 steel. p cKip thiqbqcss. The cutting foroe is found tQ ba !l$@. the cutting force is 1400 N and the feed thrust force is 360 N. '~oolbesignation' or 'Tool Signature'. Dcfmeeutting nstfr).i b k t (ii) the &ak angle. 48. Find : ( a ) Chip thickness.25 mni. . 41.cqt of 1.S. (6) The i s S ~ 6 ~4 ‘ d~ 53. {c) The -c@t of @@on on the faoc of $e tool.wd W a r n fbw. (bj The size of the force exerted by the too? on the'chip. I .& mfl') If - - ' r Ki " 8 . tool huviag. Describe thptadrepmmted by. What is the a h x i m a l e fhickness @shear m e in meW cutffng? 52.@:pd the feed tbrust force 900 N. ~the f the is $ 8'. \ \ \ : I % > k 1 (c) The unit power in W/mm'ls.S. . the feed is 0. (b) Shear plane angle. ( i ) back rake (ii) Side &e.is 730*W:hi rake mgfe afthe tool is + lo0.Find : * p r.1 mm (ORS) shape has been used. (e) Friction force and normal force on the chip.a XI0 *. . In orthogonal cutting dperation. % 8 . An orthogonal cut 2.. depth of. Why do carbide tools employ negative rake angles more often than H. 1 * Find : (a) The shear angle.:: d f the principal cutting edge angk is40°. In orthogonal cutting. 42.force and normal force on the shear plane.5 m/s and feed of 0. 44. Calcdate : .win.:z ? r * k~ Ib: 'SkeZCh a &g16-6&t kdng to6t'and sho~-&'itthe various tool and tool angles.54 mm. . ... :O . . I .i<::r).2 r...&Is.>t.J. H.& ...ipraii?irYtnC d&elediagram of U*..q~tis 5 mm. 4.!911f. 1 - ... . '" .?~' ek.:$~.a#&& 'WOW =HI0 rwmw t .:!:' !i 9 . : c w.Eriction 1 2. and the de@$. > ' = l ~ ( .$e: qignificw~.ia. A . 7 ' '' 62.<. . .. . . 4 % ' . . ..#~$~..:. ..i'-'<. 1 $ 1 .. ' 4 .j c..v t t j. I. (ii) aLpe of cut.* . ~ i ~ . (0 Wwk material micro-swture. IT:.2mm .mt&vg @ce ma sun^ .\ooXds reachgd tts Smiting life? . thwlf@m& Pi7 . .J . .' l@~ ~ $ ~ f ! $ ~ m. I r ' i . and a dyn I I ..>.59.rJ..+. Fotlowing data were collected from an orthogonal machine test oit%eef L' -.r.' >Ll Rake angle . .' I . J' ' b ( i'.>+ !'a & % & ~ O ~ ~ & ~ i ~ . .t. k djrip gic&e$:'t.. . ( 3t.!f.is'. i . lo at 33.I ! o 2 . : .lT .. cut 6.~.f:-~zsw ctdtSjng.I. . . > 58.Y. . . !.-~s&a~.* .. E $ t m p Ir$wW@l% ~ ~ ~ : 1 . ..v: . i (c) S b r and normal stresses qn . iflty. . -. .fia]Cutting farce 4. i .<. . .f. *I -61 8 ~ .&e . . What is understood by tog! life? #~$..! -<t. Establish BW (todlrii% cq (Taylor's tool-life equation) : . & ~ # w .s.25 mmlrev.fed!@ hRd .w@ i>i$etp$ed in productivity? What diRnrnf a h d a are u r e s b identify ifid th..~~G:FH&W~~MA~._ . 3 .i r . (4*Shw!p& @tr+n. E . b.: T-!\ .35 mm.: ~ 1 . 57.#ri$iai9.. . . 8 4 * d: . ?.7% . :::.I . { .2 b. A workpiecc is being cut %l?P6rmtk-W 0. t .3 .. .I..1&e ~ (e) Make a rough plot of the d a n reasons l d spa& &.' k If)" ' Clearance angle Width of cut = 3.. . (4 1s the most p r o d ~ t i v e~. to be ! 4 .e:$. (iii) Tool take angle. S .p$y~?mif spe* for vow answer.t~ an e. : . l j f ~ & t f ~ .:+I i$i.' '* - ' - ' I +I ... ' 1 J .clf..' > .. : .~: . .. e .. . shesr wain. :4J# ? . t = ' . ! ' 7 *.. i . ~ . .. Cuttlng speed '* . :i.f.i A tool life of 100 min is obtained from a cutting tool at a cut$&& ~ I D ~ .l~Yms %&%2%fl&%~Jb. d .agsiw chip flm.tH. I .%%M.~3~ *it . force an the rake b. speed for ' +: (c) W i v e rtil e x p r d m for the most economic cutting speed . j i. .2.. k k t ] .'''8 '" i . < ' (6) Co-efficient of friction. W * and . rq.'k feed is .:: ry18 .qJ$$ e ~ .{.!a. ff vel*~tt&tI&6~ forex and evaluate : shear aaelc. .1*:s * ' . . kW. . . . a r ~ ~ g: .! $.. 861.~ i>r~cy?T lrn} c'.36 &'ns'of steel chips with a t&t#'&gth of 50 cm are obtained. Iri is . dmia:kav'e &en.25 In the Taylor quation..8 N/&.36 mm has been obtained.. t i ! t" mm aod ! 16 -. .. I I : b LL i s 4 1 15rlc~ 10 S ~ I I 1ooj ~ e i .speed bu9-t bm ~ f1 . Select the speed in r e v h b fbr tUreing a m d steel bar of diameter 320 man with a H. (A=.2 mmlrev. .(..~:. .* s1 and was the nvtg of the m i for&* s f i a angk and the specific 2:: rrne. The .S. T@ecutting speed b 1.. uwut c & 1 .x>irk ~ r b 1r. I). I rv. 1n:an o&oF9na1 cuFinp operation. the cutti'ng'speekb~>. 14 "$ '5 L ' (c) shear an&~y~m the proeatage ef total energy.L-.u=. . &. :r. lo L n g l ~ r i l t i o4 .. . . is.? i~b(41*1>d ITS: ' (-I 7- 1 ? e !. . hying a M E life o$# minutes. $happ.%. giyen by 63. employed.I 0.h i with a tripple carbide cutting tool having 0 . $$ t.2 rndrev and depth of cut of 2 mrn at the cutting . The cutting speed equation is given in the following Taylorian form. 3 1 .. +i.nQ& ABRS6 pb anlllc 3.. la 311: . degm of rake w l e .r&oN! -1 r i'a~...2 rev. 4.work material t%qptfi4fip p~ltbing. Calculate the chip reduction co-efficient and shear angle. C i c c s arc such that it becomes desirable 69.35 . 68. S h w that &ring osdwgonal cx@@ stra&. A H. ~&f5SIi i .width . . ..1 I! f~ fu:3 3. T.: . € 4 i c ~ is .3%) 66. .8 rm.~&hM ~duittrnL to ? r . ' H. e.2 nun.I tosl.S. thc rake angle of the tool is 10". . . Eqitgate the suitable speed.r .60 15 q!~~unn.. (4080 N.ts d+q. I. 13.rveiageshear stiain Use the relation of Lee and Shaffer to find the shear angle. the specific cutting energy is 4080 N/mmz. :: take "$"':fi" r 1" I .rl.?!t .&.74 mlmin) . rake angle is 8 p d the width "" '' ' hf'thd ad is lo'thtk "Mie uriderfomied chip thidur'dsf 150. 165. . t. .-'1 mp OqS. " .6 . (15..5mls. $ # .+. and . H M cjla:~-Ih ) \)T (So* 30. l%e tool post dynamometer gives cutting and dvust farees i 0 CI %r: ~ ~ M&IdN : a ~ . . 861.q feqi of 0.>r. 07.8'N/1*~.( r 1.I Arr~d * I t . -& : &. ~1.6 . .m.-. tool to run the tdQl for lW0 M&.S. A feed of 0.~ i & A t ~ f f i a O 2 I. i b r qae bwr./min) e t 24 @min.9% 69.& v'7-@5m/mkr>. In ' \ t ! t I = '1 Chip thickness ratio 0.tu3c @ 1331~a t ! . . and depth of cut 1 mm have been chosen. nrnrl~ R 16 arrrwvqo nrrdw rb-bnvrt .lnw (1 q-15 mdr~v.l . the ratio of t k shear %.I f i b I : ' 0 9-u qntJ4113 r. with a zen.S. '"" '''~he%'&h force 6h t I i e ~ k i i 1 d t f i n ~ w ame&dwHI&~'W#sWh. .10 .. Id &$onlrl~rcu%&g of a low carbon steel. l c bnt.44 x 105 5') . Take n = 0. A chip thicJrness of 0. 8 m u ?4 ! lit--) dC wu3A&. *c@~P : (a) the cutting force (b) the aver& shear s t e i n the &ear plane (c) the normal $tress on tfiO shear plhe (d) ihe av&i shear stdin i i ~tiutting ' (e) thd1'. 3.Q.S.i~:5 mm.S. .I!? ...F am and pha chip . ? fanddareinmm. Assuming eo-efiokt of fEieffsnat the taol-ohjp interface as .J d 3 . avait* I mwhine spindle is 3. When turning 19 mm d. i3W) SI 72.53 mm) e e i i cutting an 3i$ mrn and feed &Gi% was 6.25 ~ t i asnQ was 1.>. ~ e r m i k e:'eyPi"" -r JfiR "6 '1.+wddid^@. ".imeer bar on an .73 kW. L i e d 31rfq!7a L Ftrbc:(lbcil#ifm 316 kbl.. The mean I -the cutting ratio.S d is a p p r n ~ i m a t ~ .) &16-~irll1irwl.t b . wm -med N and the *st t k &hipt h k b a was found out . 6. The to k 810 N. If a length of iq 0.' I 1.16 w d r e ~ what is the ?Hween tool ~Rmgces..I oh^ Fcrtd is O. fv:2 I + ~ ~ : : J h . &%/ntia.F C = 54) Wt.5.t r : . (3 MaximumMm' $$--ti .>f!d @ .$$jc&{$&$ life if1 increased t?~JOSmlmin. 16.74 %~4&%. W.25 mm.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. of mtation.32) 7 t 1 71.i. Y .and a life d a ~ e e d .G' . dmin : 4B.?i). & ~ +Tpe ~~~ power fi.7 min.?+. wh ~ 1 1 : ~ ~ : . the tool life at a ouning & $39 mpm is 60 rnin..2P q8. Thg of 100 M w e e i ( d q m ~ t i ~ e ng a4 W d~falrttin. A' HSS tool requires regrinding af€#3 hours and 2(r m b f t s w b a @WMng steel at a cutting . (0. 2 ( r = 21 1 15/kln4) 73.. (122@y/rnin. W M cutt11ag steel with a H.#lvId)?id< g i v ~ tool life of 6 hours.l . Z & 4 9 e i ~ $~etennine rd~~~ ~ t .c ?-gkm$ @f the Tqloa tost a$&quatiom. rev.S.^^^^^& i.. .5 s. ?E. l ' &I .% a $Ws A Textbook of Aimed apprawh aagie iW. The pnnr r w i n d m turn a medium C . (2. ' *b. ?3wiwdMetdm- ~ t y e h 3 83. . & M ~ W J . = q939 m/min. .. : ' 2S+--- 1''- - iius 1 r a-fa? - I--? n.+&~.Y=2QO mhnin. .4O. 6 hm e-td *UP t b * ~ ~ ~ W W ~ &fMgro->$ly!cCbeen H&F nod 3. . Cutting speed :-. n k rw.".' C w of fiic#iW2h i 3 ' 4 I -1 .p2v w a n d the tooi failed in 60 min. . * 1- F: yuiik: PJ t g=. (Aas: 18.765. " ~ 2 3 5 02. Chip thi~knes+q. While turning a &d stel bar on a lathe. .3.~ & ... ~ ~Il T l of cui..l -. h.4 MPa) In h A m i E h i n i n g ~ ~ ._ . -t. ~ . the fiction angle. .25 Ihm . . Determine shear angle.%.. 61.% I ~ . .Cutting force .w&. tool life? 4&$ag a reke angle of lo0. 0. .-.z. !.: .24 N/mm2.:. Rake angle ' lo0 .ww ~1~. .-+. d z .L4. 26. .:#&'5".) min. .i. ultimate shear stress of the mattrial. .iRH di% U w dl$ tukknrdss er$l)b 3tL 0.75 mm ---: - Rake angle.I I .=2. . Determine: shear angle. . . N ' v - 'e Cutting speed . 6rI ? $55 fid . ' kiY.- . I - -'..lmin r ~n!txlo~*k nr OUT r wya e.. fiction angle. ~-fwce.4.<%.+a:. ~ ~ # : *.15 mar and chip t h m ' an& wd (c) the sheat s b.aud &&ace various i force angle relationships.: .'.cP.250 mm . %+ . * o. . of cutting time. 420 N.‘+75.fi? .&=475 .. .5. 94&5 1 kl) awehiggies. . J@WY 3 & 29.@~ Ultimate Shear b t k s ef W d . !..operation.. at the cutting edge is 25 mm.'...~.b2gm $r1tJl:1~> $&emtW .-$3. Ahg@f41(FO @mi&.'." r Width of cut #j$ asnun &&&. (6) the shear plane is 0. ..?^ ? = i02 : tRm . .. .734) . (B .. [...a mm.. nS. :-. the uncut chip thicknrn 87. : SPOE* tbe c i r m ibd c o ~ m ~ r m & ww I 1 C' .. Thrust force osa 23 N *..u . the following oar8 have been observed: !-0. o f i W-atting. the following data were obtained. 125 N? 1 I .. A50mmdimettrbarofstct min. . m .. ' .to % . For ortho$oml w t t i q w i 4 . ! ' t*lsO-MPa. w s . Assuming a straight line reletionship exim what cutting speed should be used to obtain a 30 (Aas: V.-Ddemine (a) the cutting ratio.t'* c ai - ..Jlfi . Rake angle. .I Speed * 170 rev. The speed was changed @. . shear and normal stress on shear f%i*g plaieI khear strain by using the formula dPea. Machined len& c. 86..-: . - I &a ..~~. . - % w h i n i n g time and the total amount of heat ( N a p Universi~) (mg+WkW9 1. .: 0.: .>-. . . * : . * 5 . . -. '1 a08 ' Cutting forceY: a.: ..3 mm EC* 2 If .J *.G0. :- 9 ? . % -..i ~(. &dl@ .~ ~data i n have g been o b m e d : Uneut chip thickness.. = 150mm a r k f A .7 ~ $ M g a > < I A:..ny.574 379. d c i e n t of &on. . .T. .. . -5. I - 36. = .i 12 = 1103.7 N F.LA$ t . .. - . *F. and r 4 ' ! 2 : -8 odr b nrm\ 13: kRl rs bmuf oaw txatz %I rsjsr. L en wrnl . .~ $6 :*#A).? f.: . -. .J i4 ~u~bll.. - i i . . = 154.pp.lt. I ~ . " .r.(a lAns: N.93 N *!.:p:rw rtcnl~t:.-tld. ah& hY'&&g during orthogonal machining of mild steel with an uncut chip ~'r#iid and width of cut being 2.!!. Now fead force component.F. (Eqn. ( a ) Detefmine the various cornpotfenti df the machining force d e f i machining a C.U&g p: Lee and Shaffer relation for shear angle. ~ . 4803 N:8. Principal cutting edge angle = 30°. (a. @=tan-' p = tan-1 0. (Ans: ~ e r i A Th&ry d :Ec= 468.*j. .?SY& h m a bns Hint: Use eqn. ~1 fjqblX (it 1 NOW. * . .*I ->*rar T x F m + m 1A relation fbr s b bmbpd. = 0. p = 0.wit% -.. =f _ .lr qn~li. ~ " " n~ Thrust force.W. . . sin l (normal t&&p wlociQ. : u .r d . 14..?.f.y. Yc: bnfi x i a r t ~f:!. = 139..'* * ~ k * i W . * 8- - along velocity vector). ..IoX c0s4. r .5 mm.Li*Lee and ~ h a f f kt h e w 1 F.eas t=311-85 N mim mud tq the mekined surface) I GCI$I: :rnA) . 1 LI 60 strokedmin.' .i i. = 4 0 ' MPa..7 N .%i~ 20 3!q1!?i ~ S biI pT I:\ -":. b= Now.6...~ ~ 82 O ~~ tJ wW a d " D ~ c & ~ the & average power consumition if $hqht~ration€&s.t. .. Uncut chip Wkness 0. block on a shkqj&'\nifth depth of cut = 4 mm.S.3@'. ".\U Ad*?.'t? bl 4: .?ra~svl. >. 2 2' 90. f& out the results by Uihg relation given by Lee and Shaffer Itlwla WI. =F. i I 340 x 1 ' ~ r ~i s?q-\\ 1 and t .~r!* a'& ~ 1 dt3 .4 . 14. t n h . : jGfq .8 x tan 21' = 423.' '. by psuming the machining cutstant to be 70° in Merchant's second equation for angles. .7 x sin QOO = 366. kr I # . :.. 340 MPa. Take a ... m. . gn (J3 .2 ! .rn? rerr F.a).18~) ?ti.! :.-. i* . ? .6 = 30.25 mnn * c wjdtfiofcut=2m * $ . Solve the above probleh. (.ii! c<Grlz$:~rf&malg ..& r * ..it acts norm4 to tool-job interface. AIM.3.6 N) n!. 7..333) 93.'*tlt i&e angle of the tool . p = 0.09.i tr'1J.OAJ.LT. a s .c o d Qctmnine the cutting f m and the thrust h q e W thE fpaing ratio when machining M.r>I!> fiim tB bhd jool ?hl D f E IIIPJC'. ..-k745O (31 101 = 2y .. .46 N) relation. of 0. feed = 0:rS muW!hr61yd.19 q%h i 1 . . - -1:.I T . sM.2 d msc.25 9igr:t r~rmrit wad2 snrmts...Nr t .960 = 3 t i h . *-I--fil ~ I ~ I ~ I T W Jstuflist ?~ %!ti $. 7. E..it& t .. .. : <. i . 111 '1'111 s7q .iaf ' ..5.-.A Textbook of P~3Wctbn' l ! ? m n g --- +=450 + (I @ t6 i(etermine.I3.pXPin240 = ?LW-. if a 5 .1. 3 m.&? n :i c.. sSE4=423.1s1i141t I (b)the length of the job is 200 mm..5 I ? . . Raks angle = 0' ((!tt$l 5 h y n r ~ r ..~o f 1I~c's34 bit&e tzr~q. 4 . . Take: . .. aad 1. .Tf&". cos C. i :!. @am(aJ.e *!tj rr )g)nrral f m component+ t. 6 5 N . Ff= 310 N). Now MRR = d.(Ans. iYI ff"*.nption for cutting force. 1 r . - and - ' rc>t.P"M-~ ' I .'" C VTn = C - ' > 2 --A- . and a de of cut o 4 mm. A W C cutting tool myhning MS gave a life between ~ g r i n d sof 100 min... and 33 m h when operating at 100 mlmin. x.ili %hitian. ! r v ! ~ .. . tM$u$e$he i f r ~ i ~ m ~a ~ ~ rW pe 0.F. Specific energy eonsumption Power =I 4- rrt. Determine the three +w&ipal cornponeat forees..-: . Calculatethe power consumption. (Ans.P.-. . . . Tht wmlt8#t &We 3 P .76 1000 J 60 .Theory of Metal Cutling (b) Now Work done= Fc x length of stroke = 1103. I - :.&'JV.= ad ! '? r- F . 17n2 97... .-(21. .76 watts 60 D e t Bat.. with depth of cut of 2 mm and a feed of 0.n N 9@. .. I . 0.Log 80+nlog100.---mmhw.f. :.in tool life equatian. horktn$al is 96. LOOO .(1) loglOO+nlog33=logC . + Ff+ (Fn= 635. - 4W'x 0 3 x @ '= o . Determine tfie value of the index and the constant_ -./min.'.5..= logC ./min.owehind at 200 rev.i Which is very n li ble as c m p & to Ppwer consux.is f 200 N and its angle s f inclination to the. Dave. when operating at 80 fiJtnjn. with a feed of 94. I s 1 8 Ff=F.8tub1 psint.. = ' 1080x60 w ds 1000xf@ Power = 1800x n x 2 4 8 x 4 0 =935 1000x60 w>4IQQ Now velociQ ofpfeed afQrce=.= 220.. - Sol. x ' Average Power= 220.t I >a a 54b = 970.3 .5 N..8 N LJd..8 2 ..08 W) Sol: Power consumption = Force ~Telocity r* P" . I ~ I and Velocity = . 935 W. from these two eqw- a = 0.2 and L3 = 28 1.. 5.. -. b 4.5 mmlrev. i .78 x 10' mm3/min) %pb dh$I maohlSik& 8 2 I& &meter bar at 40 rev.. e .76 x 200 = 220. . The approach a g l e to whi& itacts perpendicularly is 26".tan 26" -. the cutting fonx at the tool point was 1800 N and the feed force was 400 N.
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