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Rolling Operation Ace

theory of rolling operation

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Ashish Jha
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468 views25 pages

Rolling Operation Ace

theory of rolling operation

Uploaded by

Ashish Jha
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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materials ch thal sto346% Induced in the material Oe Greater than, Yl6ld Stross than ultimate stress, renin oj 4H = AH, Coloumb friction condition AH yo = HR Asteel billet of 180 mm thickness is rolled to 95 mm thickness in a three high reversible rolling mill. The oll diameter is 650 mm. The coefficient of friction between the rolls and the hot material is assumed as 0,20. How many rolling passes are required? (a) 2 (b)4 (e)5 (a7 Sol: H,=180mm, —H,=95mm D=650mm, — R=325mm (BH) = HER Hy -H, = 0.22x 325 HH, =13 =—_totalreduction No.ofRose5 maxreduetion /pam ol—95 as a7 Scanned with CamScanner NM paint dened inthe deformation if 90 the deformation zone into two 8 Zi sone betweRN tNe ently ond nevi i wl jscalled “lagging zone”, ” zone betweEN Neal Point ond exis feted "eoding zone”, © ent the Velocity ofthe strip is much ess af velocty of the roler referee velocty between roles and the emer. wren we are moving along the deformation + Te because Of INcrease of velocity of ship ppeoratve velocity is reducing, i the neutral point the relative: velocity pecomes equal fo zer0, geyond the neutral point the relative velocity again increasing in the opposite direction ‘ond becomes maximum at the exit, But the maximum relative velocities at the entry and at eit are not equal. , from the above in the deformation zone the relative velocity is reducing first and. then increasing, whereas in lagging zone relative velocity is reducing and in leading zone relative velocity is increasing. As SUP «: Relative velocity + The maximum % of sip taking place in the lagging zone is called as “backward slip”. v-Wo Backward sip = Ye +The maximum % slip taking plac zone is called as “forward sip". M-V Forwardsip= MY = Becouse pressure is inversely proportional to sip. th the deformation zone the pressure is increasing frst and then decreasing. the leading 145 Where os in lagging zone the pressure is only increas avails neteasing and in leading zone pressure is decreasing = Coefficient of friction 9, Ha), 5] (Hoy 7 | (n- v(fe) +1}-(E2) a = Yield stress (or) low stress of the material H, = Thickness in lagging zone where pressure is tobe determined Back tension (It is not compulsory) and itis. an accessory. (aes a) Thickness of strip in leading zone at a distance of x’. Pca Pugos OF€ Pressures in lagging and leading zones at a distance x respectively. 6, = front tension (not compulsory) (Prdoeing =| + Atthe neutral point the pressures are equal (Podoaara = Presa (lo ulte) 1] (los otk + Bysubstituting thickness of stip at neutral point either in the lagging zone pressure equation (or) in the leading zone pressure equation the maximum pressure induced in the rolling operation can be determined. Roll pressure 5 7 mHORC® Length of contact ®t Scanned with CamScanner Typical pressure varlation along the contac! lena! Inrolling. The peak prossure bs located al Ihe neulral Polnt, Th area beneath the curve, represanls toll force, Steel plate 20 mm thick Is to be tolled to mm |g 'N.@ four high mill having roll diamolor 480 mm. If Yleld stress is 120 MPa, determine the followlng 9) Tho Anglo of bite b) Coefficient of tretion, tthe given reduction Is the maximum reduction possiblo €) The Position of neutral axis Q)_ The Forward slip and Backward slip e) The Maximum Pressure kg/cm? Sol: 9) Given: Rolling process, Initial thickness of plate H Final thickness of plate H, Diameter of roller cosa = 98 ROR 4H =H,-H,=20-14=6 Nee OED (#72ertad Deli» thopal« Pune Bhubanernarsuckow = Plas Bengaara Chen D= 480 => R= 240 b) Al maximum feduc, clone Con. ap "ay, Tana = fanf\= vi AH eR 6% #240 0.150 Gonorally Ihe neutral Plane @H, #2010 30% of Ay 114 +95% AH TA (25% 6) * 15.5 0 1 5.3 qd) backward sip = 1= Me) Ih «1a ° 0285.29, Forward slip =(Y-— ') oth 0183 Hote Oey e) Maximum pressure Is OCCUF at th Where (Pring = Proang) ron Flin- nf) +1} ul _ 2% 0.158V240%6 ne ap ng 18 Nev, Pay = 423 fa. 99 -1)(2 = 162-9. = 162 Pounine) = 16.2 £9. Friction In rolling: It dépends on lubrication, wat material and also on the temperature. In cold roling the value of coefiiclent of friction is around 0.1 and In warm working It is around 0.2. In hot rolling iti around 0.4. In hot rolling sticking friction condition’ also seen and then friction coefficient is observed up to 0.7. In sticking the hot wok surface adheres 0 pa « Keka ala Scanned with CamScanner | gue @) a thus the central jane Part of the str gsevere deformation, eee cu er calculations: power (P) =2To (AS two roller considered) T= Torque required per single roller o= lar velocity = 22N. = Angular velocity = 288 Ne=«pm of rollers Torque(") = Avg.force x moment arm T= Fag X 0 Where a = moment arm = aL 2 Fagg X Ab q=arm factor = 0.3 to 0.4 L=Length of deformation zone = RAH £,. = Pog Projected area =P. (bxU) werage force acting by the rollers on to the strip 7 BF (Rol separating force) So. the average force fs also called as roll separating force (RSF). faa =RSFEeOr™ (144g) xb WhereH=—"7 a The average roll pressure can be decreased by reducing the maximum pressure, which is a function of the contact length. So, smaller contact lengths means lesser frictional forces acting. thus, by reducing the contact length, it is possible to decrease the roll separating force. This in turn, can be achieved by reducing the since, smaller rollers would have roller diameter, n larger rollers for the less contact length thar same reduction. The smaller rolls are used for larger reductions ‘and cold rolling where roll separating forces care large. a = pl Pane + Bubace The smaller rollers would not have enough tigidity to support a large rollseparating force. Hence backup rollers are altached to the small rollers to provide the necessary rigidity. Though higher friction between rollers and the metolis requited for increasing the reduction. it aso increases the roll separating force. In cose of back and front tensions, Without opplying thelocdbyrollersitself he compressive stresses are induced in the thickness direction, the force to be applied by the rollers will be reduced. Effect of RSF on rollers: The thickness of the strip produced in the width direction will not be uniform. To get uniform thickness of the strip in width direction the roller will be made like a convex roller (or) cambering of the roller. During roling when roll separating force is ‘acting for the material side the rollers ore deforming elastically and become straight. So that the thickness of the strip in width direction will become uniform ‘A sheet of thickness 40 mm is to be reduced to a thickness of 30 mm by using the rollers of radius 300 mm with a coefficient of friction 0.25. IF the width of the strip is 200 mm and yield stress is 250 MPa, find the average pressure between the rollers and strip. b) Find the total power required for the rollers if RPM of the rollers is 5 r.p.m and take arm factor of 0.4. a) Sol: Inifial thickness H,= 40mm Final thickness H,=30mm Radius of rater = 300 mm Coefficient of friction (1) = ).25 ipa Kapali Kalata Toco Patan Benatar Chena Winona Scanned with CamScanner inner—sechon chy ua ds a Width of slip B= 200 mm, a, = 250 MPa Average pressure Za(i4 tt Po or(1+43) WhereL= RAH = v300%10= 54.77 n= Hes th 35 4 0.25% 54, x280%(1+ 02307) v3 Pog = 318.9 = 317 Power P To where, T= Torque = Fay XAL 2= arm factor= 0,3 10 0.4 L= deformation length = /RAH ‘avg = 317 x 54.77 x 200 = 3472561.015 = 347.2561 kN = 2+76.07686 x 2aN. =2X76.07686 =79.667508 kW Types of Rolling Mills: 1. 2-high rolling Mills: + Uses two rollers only. * Direction of rollers is opposite, * Single reduction possible. 2 Wy 3—high rolling mills: In this + 3-rollers will be used. + 2reductions ore possible. +1, 3rofate In some direction, cenjey opposite direction. 4-high rolling mills (or) back up rolling mils (q) supporting mills + Supporting roller used to avoid the faiive of smaller roller. In 4-high rolling mill the centre of all the four roller must lies on a single vertical line other wise the roling system wil fail, Scanned with CamScanner (MEUM CE Le Mills: ae ih verCO™® of the limitation of the ; + pina : nat Supporting rollers u pnetaryroting Mis: f spreduce the RSF (01) power requirement in vag operation by reducing role racius to the ririmum value the best method is plenetary ‘oling method. risis the best way of reducing RS.F among all process. Sallent points about rolling: 10. u 12. Rolling Is the most extensively used motal forming process and Its share Is roughly 70%. The material to be rolled Is drawn by macns of friction into the two revolving roll gap. The compressive forces applied by the rolls reduce the thickness of the material or changes Its cross sectional area. The geometry of the product depend on the contour of the roll gap. Roll materials are cast iron, cast steel and forged steel because of high slengih and wear resistance requirements . Hot rolls ore generally rough so that they can bite the work, and cold rolls are ground and polished for good finish. Inroliing the crystals get elongated in the rolling direction. In cold rolling crystal more or less retain the elongated shape but in hot rolling they start reforming after coming out from the deformation zone. The peripheral. velocity of rolls at entry exceeds that of the strip, which is dragged in if the interface friction is high enough. In the deformation zone the thickness of the strip gets reduced and it elongates. This increases the linear speed of the at the exit. Thus there exist a neutral point where roll speed ‘and strip speeds are equal. At this point the direction of the friction reverses. When the ar of contact ‘a’ exceeds the friction angle B the rolls cannot draw fresh strip. Roll torque, power etc. increase with increase in roll work contact length or roll radius. WIRE DRAWING Wire drawing process is a cold working process used to produce wires from solid rods by pulling through a stationary die. Scanned with CamScanner ne-1 = 10d cross section areq ‘ection area of zone ~ 1 = Wire cross section areq 2a = die angle Die angle varies trom 12° to 4ge (12° for hard material, 48° for soft material {= length of deformation zone (9) die lana Because of Presence of friction at interface of dle and surface of the wire, fictional force wil be acting in the backward direction, Exit cross 5 Tone ~ 2 Entry (or) lubricating zone: To eliminate (or) minimize the frictional effects taking place due to the friction, the lubricants will be supplied in the wire drawing operation iyderbad- Ba hap No lubricant Condition; gi is observed. Solld powdery. lubricants: shining surface (01) silvery Let, from zone ~2 only co its ng zone", Meg eda lack (or) Bury t Uquid lubricant Condition; The suy fac, Can be observed as dul surface °° 8 hg Commonly sedliguid tricgn rei! Ol kerosene, mineral oi ete, surface, Commonly used sotig lubricants op, Powder. graphite powder, glass Powig® ee Tone ~ 3 Sizing zone: 'tis a constant cross sectioy sectional. area equals to ‘area, 2018 having Wire cross Sctig This is Used for converting elastic eforay Present in plastio'deformation zone into tly Plastic deformation, Zone. toss sec tion areaotioga,) = Fat Cross section of the wire a, Scanned with CamScanner E 151 oS 7 coefficient of elongation at (o TUBE DRAWING he i : wt + tls the method of reducing wea thickness of efioninorea(J) = At! already existing tubes by Pulling through a re ; stationary die with mandrel so that the inside jyawing sess appiied on the front side diometer of tube is maintained remaing 67° (+B) - BY constant or varying. ‘ol 8 ‘Ae oe ot - + tt is. similar 3°" pefficient of friction 20 * Lotdie angle e wing oad = ©, XA, pe above 2, iS CAlCUIGted without Considering % horizontal Component of frictional force acting in the Backward direction, ay considering horizontal compet also then stress to be applied in the. drawing operation at front ides called as total drawing stress (0). oyu? %y* (27 9,0 , = Radius of wire Total drawing load = 6, A) For maximum reduction case «. From above for the given values of die angle and coefficient of friction p, find B and percentage of reduction in area and draft. Under ideal conditions of wire drawing operation the coefficient of friction is assumed to be zero. Therefore, 1=0=B=0 Ao Sr1dean = GV AY ‘ih Bhopal = Pane» Bhubsacew 10 Wire drawing operation, additionally a mancket i used for maintcining the required inside diameter of tube, DIE USED: Same os that of wire drawing Fig. Tube drawing process Types of Mandrels : 1. Cylindrical Mandrel 2. Conical Mandel Cylindrical Mandrel: if there is no need of changing inside diameter Of tube then cylindrical type of mandrels will be. used. ‘Conical Mandrel: inside diameteris to be changed then conical type of mandrels will be used, Based on the movement of mandrel, mandrel are again two types. (i). Stationary mandrel (i). Movable mandrel (i). Floating mandrel Scanned with CamScanner (i). Stationary mandrel: Stationary mandrel will be fixe’ drawing operation. The frictional force at inside and outside surfaces is same. Therefore in this case, pote Tana -Tany din the tube acting . ie 16 Where ais half of the die angle. ys half of th conical angle. For cylindrical mandrel y is zero. (i). Movable mandrel: This type of mandrel will move in forward direction along with the tube at same velocity, hence the friction at the inside surface is equal to zero, Therefore in this case, -l 8 Tong (iil) Floating mandrel: In this the mandrel is moving forward and backward direction inside the tube so that in both cases the direction of frictional forces will be in opposite direction, = Mi=he 8 ="Tana DRAWING STRESS: Drawing stress 14,=Coefficient of friction outside surface of tube and die, h,=coefficient of friction between inside surface of tube and mandrel. In a tube drawing operation, If u, =, and floating type of mandrel if used = B=0 So, drawing stress (c2) =o Ifwe apply L-Hospital rule we Wil St he on20y.l09¢(f) For maximum reduction 0 = Oy -fsth-(8 Limitations of drawing operations are; Because the load application js tensig is necessary to maintain the loag Opp, some more time to convert Elastic Aefom, into total Plastic deformation, + The reduction possible in the tube qq ‘operation are limited. + To overcome the above problems insteg applying front side pulling force, the bac pushing force will be applied in the mates, that the compressive stresses induced in material willbe used for producing deforma, in the component, hence no need to maint the load for some fime. * This process is called as extrusion. ot or A 12.5 mm diameter rod is to be reduced to | mm diameter by drawing in a single pass at « speed of 100 m/min. Assuming a die angle of § Gnd coefficient of friction between the die anc steel rod as 0.15, calculate: (i) The power required in drawing (i) Maximum possible reduction in diameter of the rod (ii) If the rod is subjected to a back pressure of 50 N/mmé, what would be the draw stress? Take yield siress of the work material os 400 N/men?. | Scanned with CamScanner spiameter (Dy) = 12.5 mip girgaret = 10 mm fi in =-100_m yoom/mnin tee = Die land -8C ton = Tans -—8C C= Tans L= 28.62 Sys" (03 -0,) © B=ycota=0.15 cot25 B=3.435 eS Guus = 400+ 4.950 Coa 400.888 = max. possible reduction = o, = 400 MPa Drawing load = og A; = 400x FA10F * 1485.04 N =31.48kN Drawing power = Drawing load x V 2 31.48(kh) x 102( 2) 2.475 KW = 4054 59( 10 us (85) =415 MPa EXTRUSION; Extrusion ig on extensi ted tension Of drawing operat Ucing cross section, 19 Operation for Hence, the advantages of extrusion process over ‘he drawing operation are: 8; Because of compressive force itisnot necessary 10 maintain the load for some time to convert the ED into P.D (Plastic deformation). ED = Elastic deformation PD = Plastic deformation Because of compressive stresses inducedin the material, it behaves as a fui Therefore any amount of deformation is possible to produce. Direction of < | Extruded component Fig. Forward extrusion + In extrusion process the forces are applied ‘on the raw material by using a ram in a rigid closed container such that stresses induced in the material is greater than or equals to flow stress of the material, Scanned with CamScanner Honee due to the behaviour of matetial Is Ike Q fle, the material can ow through the small Opening available in the contalner taklng the Shapo and size remains same as tho openings Types of extrusion: Senn Direct (or) fonvard extrusion Indirect (or) backward extrusion, Hydrostatic extrusion Impact extrusion Dratt(k) = Av ratt(k) 7 K= 40 to 50 (For general case) = 400 (For very soft) Volume before extrusion = Volume after extrusion FAVE A, XV, yi ~~ Backward Extrusion: Direction of ram a STW DY Direction of Extruded component \N “| Fig. Backward extrusion In forward extrusion the direction of movement of ram and extruded component will be in the same direction, Whereas in backward extrusion the direction of the ram and extruded component directions will be in opposite direction. Higher friction will be takes place in forward extrusion. But in backward extrusion the Raw Material can directly deform and flow through die opening to change shape of component. Advantages of backward extusion Hence Ihe force Fequireg lor higher than backware, ols, os Forco Ram travel From the above graph the force forward extrusion Is higher only gy ine of process only, by The design and manufacturing OF tg die combination Is easier in forwar, Ard, whereas in backward extrusion bec, o opening is produced within the ram, tet ram and die must be co-axial, So it is difficult to design and anv, the ram and die combination in ack extrusion, Due to he presence of ction atthe cong wall forward extrusion, hence the semi, and brittle materials cannot be ertudy| by using forward extrusion, whereas yi, backward extrusion with complete absences container wall friction the semi Brittle and Bie materials can be extruded very easly witha any cracks, 8, ‘A 25 to 30% reduction of friction, which alow for extruding larger billets, increasing speed. and an increased ability to extrude sole cross-sections, There is less of a tendency for extrusions 19 crack because there is no heat formed from friction. The container liner will last longer due to les wear. Scanned with CamScanner 4 pooess isn't as versaiie os Grect extrusions » Tl yse the crosssectioncl orec is fmitad by me maemum Sze of the stem, syDROSTATIC EXTRUSION: s {tis similar to forward extrusion. Ihe size of the row material taken is less than the size of container and gop is filed by using hydrostatic liquid. Now the force applied by using ram is acting on fo the hydrostatic liquid, the pressure of liquid is increasing, which is applying uniform force all around the raw material. Therefore raw material is experiencing compressive stress which is greater than flow stress very easily. Because of usage of hydrostatic liquid. the Container wall fiction is completely absent. Hence it can be used for extruding semi brittle, ‘and brittle and high strength super alloys also. line 4 Disetrontoge: * Because of dticuty in cresting te lestoges the hycrostete exon k not sesteectie fF The fuid con be pressurized two Was ; 1. Constentrate extrusion: A rom or plunger 5 used fo pressurize the fuid inside the container 2. Consiant-pressure extrusion: A pump used The advantages of this process include + Noffiction between the coniciner end the blet reduces force requirements. This ultimately ‘cllows for faster speeds. higher reduction ratios. ond lower billet temperatures. + Usually the ductility of the material increases when high pressures are applied. + Uniform flow of material. + Large billets and large cross-sections can be extruded. + Nobillet residue is left on the container walls. DRE) ence er — Scanned with CamScanner The disadvantages are +The bilets must be prepored by tapering one end to match the die entry angle. THs is needed to form a seat at the beginning of the cycle. Usually the entire billet needs to be machined to remove any surface defects. IMPACT EXTRUSION: * Iisa process in which very thin w' be produced from solid rods which are made by very soft materials by using impact load os extrusion load. all tubes will ram. impact] load Z + Impact extrusion is mainly used for producing collapsible tubes like tooth paste tubes. cosmetic tubes etc. Stress Calculations: oo=o,(*52)-(22)]. where, o,= Extrusion stress to be applied on the raw material B=pcote (eFmin = oA, (pan Textron £0410 05 [Forward Bn, 20.5 10 0.6 (Backwarg i) For maximum reduction case ona SQ Nexinusion «= Extrusion constant ideal conditions for extrusions Using direct extrusion Process, 0 round jy let 100 mm length and 50 mm diameter is exing, q Considering an idea! deformation proces fiction and no redundant work) , extrusion oe ‘and average flow siress of material 300 MPa, ) on the ram will be =" — pressure (MPO) 400 mm, c, = 300 MPa, d, = 50mm Sol: L, ‘ Ae. Extrusion Ratio = “A, "A, = 1962.5 mm? 59? = 490mm? Extrusion Force = Aroy in «= Half of Die angle Extrusion force for backward and hydrostatic = 1962.5 300mn(4) extrusion See eae - (a ee )xA, Extrusion pressure = Extrusion Fores. Few = (So * Pio Hoe i =816.2/1962.5% 10° = 416 MPO i i iner 2/1962. P, = pressure required overcoming contait _—_ oe —aginras Coen Vio: Vs pn as Boal Pee ANNE Tanne Scanned with CamScanner RING jadind 5 Perhaps ogg ord MOS KNOWN eye, nan MEtllic tool, pommering. fo9ng #8 basicaly j st meta| lu ere mag, involy, of material between tye, = Paste ce to configuration. Depengn, oe desireg por forging is carieg guy Bene Ot ne closed die forging, Pen dle forging an = 2 re ne forsing, the Metal is. com, ate: Cr - Mechanical hope BY shape is manipulated m, lammer ang anual in closed die forging, ye: obtained by squeezin two shaped between hy ho Stationary die Volume before forging = Volume after forging ' Method of for 0 Hana 8 action ae imeN8 (Prop hammer type) ‘aching forging (Mechanical op hydrostatic forging). itis ne ‘ot sufficient to produce the deformation ir IN Work piece, (i) Machine forging: « Inmachine forging because the force required is obtained from Machine it is Possible to Se either continuous force application or intermittent impact load ‘Opplication, 2. Based on method of shape obtained: (i) Open die forging (i) Closed die forging (ii) Semi closed die forging () Open die forging: * In open die forging operation only dro} hammer type of force application will be used * Press forging is not used since there is no tim for changing the position of component. Features of open die forging: + Repeated impact blows are given on the work * Less dimensional accuracy. + Suitable only for simple shapes of work, + Requires more skill of the operator. (Wace Dan ahpas Paes Babes can Ps Bap iyevad Vitg Tropa» Kulatpaliy« Kllat Scanned with CamScanner ysualy used for a WOrs closed ale forging [toe appr ple and less much easily Dies are sim forging operat itcon be analyzed Itis the simplest of al jons. (i) Closed type or impression ale fora aa itis also called os ‘impression cle ae pecause the shape of the aes is impress jonent. i Smee of forging force appiication may be either press forging type or drop hammer type will be used. Features of Closed Die Forging: Closed die forging involves two or more steps: 1) Blocking Die: Work is rough forged, close to final shape. ii) Finishing Die: Work is forged to final shape and dimensions. Both Blocking Die and Finishing Die are machined into the same die block. ‘More number of dies are required depending ‘on the complexity of the job. Two die halves close-in & work is deformed under high pressure. High dimensional accuracy / close control on tolerances. Suitable for complex shapes. Dies are complex and more expensive. Large production rates are necessary to justify high costs. Semi closed die forging operation: Only drop hammer type of force application is possible, In case of open die and semi closed die forging operations the volume of raw material required is remaining same as that of volume of final finished component. But in case of closed iden Dah Bhopal Foes Bruck aia Beg Chenal-Vinoveda Ving i Pe Kaktpaly «Kota +» To zging operation the Volume » about 10% 10 20% mors thy fo be obtained. Ih die fo Gutter: Flash] accommodate the eres Volum material it is required to provide some o space in dies also called gutter. ‘Because of provision of gutter the foxy produces in the forged component. This fash is unwanted material Which has fy removed by trimming operation. «lash and gutter is used only in COse Of closey die operation but not in open die or seni closed die forging operation. a Significance of Flash in Closed Die Forging: Excess metal is taken initially to ensure that de is completely filled with metal to avoid any voids. Excess metal is squeezed out of the die cavity asa thin strip of metal, called flash. A guiteris provided to reduce the area of flash. Thin flash increases the flow resistance of the system & builds up the pressure to high values which ensures that all intricate shapes of cavity are filled. Flash design is very critical and important step in closed die forging, Extremely thin flash results in very high pressure build. up which may lead to breaking of the dies, Scanned with CamScanner Stationary die xine inl stages of forging operation when ne force 5 applied by using top die the qutword expansion forces and inward fctonal forces ore acting at the interface of die ond row material. gut the magnitude of inward fictional forces gee higher than the outward expansion forces. therefore the materials sticking to the surface of die. When we are moving towards the centre height of component the: effect of frictional forces is reducing. Hence, the material is stored expanding producing barrel shape of the component. So that initial part of forging operation is also called as Barreling operation. Because the material is sticking the surface of die at the ends, the initial process of forging operation can be analysed by using sticking friction model, (On further application of the force using top die and moving the top die, the ‘outward expansion forces are remoining constant ‘At some point the magnitude of outward the inward forces ‘over the forces become greater than hence the material is started siding die producing neatly cylindrical shape of !he component. This pat of operation canbe anolysed by using siding friction model Fro fe ‘ the above the forging operation can be lyzed by using stickin, ing tric cond sic i oe 9 ding friction So the material con be assymed os cylinder which is experiencing intemal pressure so the stresses and strains are same as cylinder component. Hoop stress (circumferential tress) n= 201 Hoop stroin= «= 2c. Force analysis in forging: i) oy X Av (Generally nenese =0.5100.6) 2a hr All = coefficient of friction inal c/s area of the component, Initial c/s orea of the component, Initial height of the component, inal height of the component. final radius of the component, Work done = € = Fae (My Pe» (1) if energy required is obtained from drop hammer type of force application with weight ‘of hammer = ‘W' and height from which it is foling is ‘H’. Then, Work done =2x(W.H) vw (2) From (1) and (2) ora ye Balto MA Scanned with CamScanner A ship of lead with inital dimensions mmx150 mm is forged between two to 0.0 final size of 6 mm x 96 mm x150 Mmm re coefficient of fiction i 0.25, determine maximum forging force. The average yield Hess of lead in tension is 7 N/mm, Sol: Data given: Initial dimensions = 24 x 24 x160 mm Final dimensions = 6 x 96 x150 mm From the above it is evident that 150 mm will be taken as one of the dimension in the cross section and during forging the’ height is reduced, hence height is assumed to be reduced from 24 to 6 mm, Hence, h,= 24mm, hi = 24 mmx24 i ing meth qhis is a forging der = stibuting the Material from ulward non-uniformly, in ths process the length of . - increasing ond dlometer is was uniformly. hh 2, Drawing: This isa forging method used for distibutng, material from centre to outwards uniform, *| + inthisprocesslengthisincreasing and con, | isreducing uniformly. | Up setting: ; + Forging method Used for reducing the lengy andincreasing cross section is called upseting | Edging: =7x 2X 0.25 x 67.7 7 14ago(i + 2X08 677) = 250KN forming operations or types of forging operations | Preliminary operations : a GYOVG 5. Forging method used for collecting the material locally is called edging, Hlattening: Forging operation used for producing fat surfaces is called flattening, Kw irra Deo Scanned with CamScanner # et oe eto used for converting. sha 1%, jorounded comers a a ing method used. for bending the nent into required cngle, 1 swaging. gockit9 O petorging method for producing approximate spope and size it components caled blocking corswaging- finishing operation: forging operation used. for producing exact snape and size of the component rimming: uting operation used for removing the flash present in the forged components: Inthe above given operations 1-6 = open die forging 7-8 = semi closed forging 9 = closed die forging 10 = Trimming operation Find the sequence of operations during conversionof givensolid cylindertoconnecting rod as shown below: 1. Fullering 2. Edging 3, Blocking 4. Finishing 5. Wimming opal one habe 5 Process Die Design Parameters in forging a 5. Parting tine Itis at the lorgest C.S of the part. Itis.a straight line at centre for simple shapes. it may not be in a single plane for complex shape. Flash and Gutter Flosh moterialis alowed to flow into a gutter. It Prevents unnecessary increase of forging load (because of excess/ extra fash) Draft Angles For easy removal of forgings from Ihe die. Similar to draft in costing desion- internal draft angles are lorger—7°- 10 External drafftangles ore ser - > om Fillet: itis a smell radius provided ot comers: 10 ensure smooth flow of metal into die covity. Jo improve die life. ‘Asageneralrule, shouldbe os arse possible small filet radii lead to + Improper metol flow + Rapid wear of die « Fatigue cracking of dies pie material requirements ore Strength ond toughness ot temperature. Hordenabilty ond ability to harden uniformly. Resistance to mechanical and thermal shocks, Wear resistance - to resishiabrasion West due elevated to scales present on work piece- Die materlals used: vintners oa begs Canal, Wer Toot and die steels with Cr, Ni Mo. V von Tepe Rata Wola Scanned with CamScanner Detects in metal forming : Bamboo defect i If the outer cross section of the extruded components ike a bamboo stickits called as Bamboo defect, a It is produced due to Non-uniform: velocity o} Ram in extrusion. Pipe Defect or Center Burst: : In an extruded component some times pin holes are present at the centre of the extruded component called as pipe defect or centre — Center burst (exam This is because of improper cleaning of raw material before placing into container ie. the presence of scales or oxides on the raw material will be forming as a central part of the exiruded component and produces the porous structure in the extruded component called as “pin holes or pipe defect", burst. Surface Cracking: Cause: Excessive working on the surface and too low temperature. High sulfur in fumace leading to hot shoriness Remedy: To increase the work temperature. Cold shut (Fold): Two surfaces of metal fold against each other without welding completely, + Cause: Sharp corer [less fille) chilling, high friction. a) + Remedy: increase filet radius on thy ‘a ' 5, Scale pockets and Underfits; + They are loose scale/ lubricant (Rsidyg accumulate in deep recesses ofthe gM * Cause: Incomplete descaiing ofthe ,.. + Remedy: Proper descaling of wor, forging. Classroom Practice Questions O1, The te stress-true strain curve is given = 1400 <8, where the stress isin MPa, true stress at maximum load in MPa) is (a) 971 a (b) 750 (c) 698 (d) 350 Pro 02. By application of tensile force, the section area of bar 'P'is first reduced by 30% and then by an additional 20%. Another bar '@" of the same material is reduced in cross sectiong crea by 50% in a single step by applying tensie force. After deformation, the true strain in bar ‘P' and bar ‘Q' will, respectively, be (a) 0.5 and 0.5 {b) 0.58 and 0.69 {c) 0.69 and 0.69 (d) 0.78 and 1.00 03. An annealed copper wire of 25 mm dia drawn into a wire of 5 mm dia, The average yield stress in this operation if the flow curve of the materialis given o = 315 <0 Mpq (0) 592 MPa (b) 458 MPa (c) 342 MPa (d) none 04, A metal stip is to be rolled from an initial wrought thickness of 4.5 mm to a final rolled thickness of 2.5 mm in a sinale pass roling mill Scanned with CamScanner ied alls of 450 mm diametey a wide. The average Coeticieny Ship is 450 ¢ rol 99° 1 0.1. Taking pig Hiction i of 180 MPO, Tor the mata « strain low sje spreadin Nd assum: (eghgme pI 19, the rol separa SSUming 'N9 forces groling PrOCeSS, thickness of Ostrip jg & m4mm 103MM using 309 ae a, 'Steduceg eating ot 1OOTPM. The velocity of these misee) ot the Neutral point is reste fo ia (b) 3.14 (jan10 (a) 94.20 inasingle pass rolling ©peration, a 20 ‘i plate with plate width of 100 mm, is ie fo 18 mm thick. The roller radius is 250 nie rotational speed is 10 rom. the average i stress for the plate material is 300 MPa. The power required for the rolling operation in kW iscloser to (a) 15.2 (b) 18.2 (c) 30.4 (d) 45.6 1, The thickness of metallic sheet is reduced from an initial value of 16 mm to a final value of 10 mm in one single pass rolling with a pair of cylindrical rollers each of diameter of 400 mm. The bite angle in degrees will be (a) 5.936 (b) 7.936 {c) 8.936 (0) 9.936 Asteel plate 30 mm thick is to be rolled to 14 diameter 680 mam in a four high mill having roll 00 MPa, if the thickness mm. If yield stress is 2 of the sheet at neutral plone is17.2 mm, the Forward sip and Backward slip ore respectively (0) 23% &43% (b) 48% & 23% (c} 389% (d) none 09. nO two high a ™™.O metal stip of ti om Of thickness 30 mmis red ae ae Of 25 mm with ere ee * The Roll separation distance : (b) 625 (4) 325 ling - 9 Mihov ter dia of 600 (c) 330 . Awire of 0.1 1 mm da i mae po is drawn from a rod of 15 80% ore 19 reductions of 20%, 40% ond Qvailable, For minien F fran uM error in the e » the number of stages and reduction ; ach stage respectively would be (9) 3 stages and 80% reduction for oll three stages (b) 4 stages and 80% reduction for fist three stages followed by a finishing stage of 20%, reduction (c) 5 stages and reduction of 80%, 80%. 40%, 40%, 20% in sequence (d) none of the above . Inwire drawing operation, The folowing data iol wire diameter = 6 mm, final wire is given, diameter = 5.2 mm. die angle = 18°, die land = 4mm, Coefficient of friction =0.15. yield stress= 260 MPa. the drawing stress and total drawing load are (a) 130 MPa & 3.74 kN (b) 3.78 kN & 130 MPa (c) 150 MPa & SKN {d) None Common Data for Questions 12, 13 & 14 In tube drawing process a steel tube having inside dia of 52 mm and wall thickness 2.6 mm & 0 be reduced to 50 mm inside dia and a wall thick ness of 1.8 mm. Taking die angle 24° and # os 0.12. onary mandrel the ratio ox/9, is {b) 0.638 {d) none pega ela aul Scanned with CamScanner 12. With stati (0) 0.367 (c) 0.518 13. With movable mandrel the ratio o,/0,i5 (o} 0.367 (b) 0.638 (c)osi (a) none 14, With floating mandrel the ratio ¢,/, (a) 0.367 (b} 0.638, (c)0518 (a) none ‘Common Data for Questions 15 & 18 In.awire drawing operation dia of é mm is reduced in stages to a dia of 1.34 mm. Assuming ideally rigid plastic material and ideally lubricant condition, 15. The minimum number of passes required by assuming maximum reduction per’ pass is obtained with n= 0.2 and half die’angle as 6°, if the flow strength of the material is 60 MPa. 16. The dia of wire after 2 stage is. ‘Common Data for Questions 17 & 18 A steel wire of length 100 m and dia 12.214 mm is drawn to a final dia of 10 mm. Tensile tests of specimen made befoe and:after,the drawing operation gave yield stress as 200 MPa and 400 MPa respectively, 17. The length of drawn wire is (2) 100m (b) 125m (c) 150m (6) 175m 18. The yield stress of the steel at a true strain 0.2, by assuming linear strain hardening law. (a) 200 MPa (b) 300 MPa (c) 400 MPa (4) 600 MPa 19. In a wire drawing process a 15 mm dia rod is reduced to 6 mm wire, two specimens token one before drawing and other atter drawing and the fensle 12318 shone yo modulus for the respective spe, cin "ia ywPo and 350 MPa, the wie craving yf a {o} hot working process (b) cold working process (c) any of the above (d) none the capacity of the press requied fog, A channels having © CIOSS sectiongy of 3 cm from round billet of 100 mm cfg Yield sess for Ais 10KG/mm? rth pay efficiency for the case of extrusion 0.4 (b) 150 tons (d) none (a} 100 tons (c} 200 tons 21. Around bilef made of brass sto be extnideg (extrusion constant = 250 MPa) at 700°C, The billet diameter is 100 mm and the diomete, of the extrusion is 50 mm. The extrusion force required (in MN) is (a) 1.932 (b) 2.722 (c) 3.423 (d) 4.650 22. ‘Two solid cylinders of equal diameter have different heights. They are compressed plastically by @ pair of rigid dies to create the same percentage reduction in their respective heights. Consider that the die-workpiece interface fiction is negligible. The ratio of the final diameter of the shorter cylinder to that of the'fonger cylinder is A cylincrical billet of 100 mm dia is forged from ‘50mm height fo 40 mm at 100°C. The material has constant flow stress of 80 MPa, The work of deformation is Scanned with CamScanner 2m. i 4 gon mm height to 40 mm at 1999» Poy nas constant fow stress of gg on drop hammers used to compl 8 jon in one BIOW. What will be tn, C. The MPa. if lete the height ft consider the folowing statements jy when lood is fluctuating and impact conditions prevail, forged parts prefered drop forging methods the most extensively used method for forged components Forged components are ideally ‘suitable for applications like crankshaft ond connecting rods, the chief reason being that the cost of production is low, Which of the statements are comect? o)1 82 (b) 283 (1&3 (d)1,283 fa) pe In a two pass wire drawing process, there is a 40% reduction in wire cross sectional area in 1* pass and further 30% reduction in 2% pass. The overall reduction ( in percentage) is 10 mm thick plate is rolled to 7 mm thickness in a rolling mill using 1000 mm diameter rigid tolls. The neutral point is located at an angle of 0.3 times the bite angle from the exit. The thickness (in mm ) of the plate at the neutral Point is. ore dh. 29. 31. 4 found rod of annealed brass 70-30 & being Srawn from a diameter of & mm to 3 mm of © SPeed of 0.6 m/s. Assume that the frictional ond redundant work together constitutes 38% of the ideal work of deformation, (Yield stress 6, = 200 MPa) fi) An aluminium rod, 6.25 mm diameter, is drawn info a wire 5.40 mm diameter. Neglecting fiction between the rod and the dies, the drawing stress and the reduction in area when the Yield stress for aluminium is 35 N/mm? respectively are A piece of lead, size, 25x25x150 mm, having @ yield: strength of 0.7 kg/mm? is pressed between two at dies to a final size of 6.25x100 %150 mm. Assuming that p = 0.25, the forging load required is To obtain 30% reduction in area on a 10mm diameter copper wire, the following data is Given. Yield stress 240 MPa, 2a = 12, 1 = 0.10, the power. of electric motor if the drawing speed is 2.5 m/s. take efficiency of the motor as 95 %is____ (ireenteae eth» Bhopal «Pune + hanivar Lucknow + Fans Bangla Chnnals Whayarada» Via ea Real ab | Scanned with CamScanner

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