Bulk Deformation Processes

Transcription

1 Capter 6 Bulk Deformation Processes Questions Forging 6.1 How can you tell weter a certain part is forged or cast? Describe te features tat you would investigate to arrive at a conclusion. Numerous tests can be used to identify cast vs. forged parts. Depending on te forging temperature, forged parts are generally touger tan cast parts, as can be verified wen samples from various regions of te part are subjected to a tensile test. Hardness comparisons may also be made. Microstructures will also indicate forged vs. cast parts. Grain size will usually be smaller in forgings tan in castings, and te grains will undergo deformation in specific directions preferred orientation). Cast parts, on te oter and, will generally be more isotropic tan forged parts. Surface caracteristics and rougness are also likely to be different, depending on te specific casting processes used and te condition of te mold or die surfaces. 6. Wy is te control of volume of te blank important in closed-die forging? If too large of a blank is placed into te dies in a closed-die forging operation, presses will a) jam, b) not complete teir stroke, and c) subject press structures to ig loads. Numerous catastropic failures in C-frame presses ave been attributed to suc excessive loads. If, on te oter and, te blank is too small, te desired sape will not be completely imparted onto te workpiece. 6.3 Wat are te advantages and limitations of a cogging operation? Of die inserts in forging? Because te contact area in cogging is muc smaller incremental deformation) tan in a regular forging operation, large sections of bars can be reduced at muc low loads, tus requiring lower-capacity macinery, wic is an important economic advantage. Furtermore, various cross sections can be produced along te lengt of te bar by varying te strokes during cogging. A corresponding disadvantage is te time and large number of strokes required to cog long workpieces, as well as te difficulty in controlling deformation wit sufficient dimensional accuracy and surface finis. 6.4 Explain wy tere are so many different kinds of forging macines available. Eac type of forging macine as its own advantages and limitations, eac being ideally suited for different applications. Major factors involved in equipment selection may be summarized as follows: a) Force and energy requirements, b) Force-stroke caracteristics, c) Lengt of ram travel, d) Production rate requirements, e) Strain-rate sensitivity of te workpiece materials, 83

2 f) Cooling of te workpiece in te die in ot forging, and its consequences regarding die filling and forging forces, g) Economic considerations. 6.5 Devise an experimental metod wereby you can measure te force required for forging only te flas in impression-die forging. See Fig. 6.15a.) An experimental metod to determine te forces required to forge only te flas for an axisymmetric part) would involve making te die in two concentric pieces, eac wit its own load cell to measure te force. Te die in te center would only cover te projected area of te part itself, and te outer die ring saped) would cover te projected area of te annular flas. During forging, te load cells are monitored individually and, tus, te loads for te part and te flas, respectively, can be measured independently. Students are encouraged to devise oter possible and practical metods. 6.6 A manufacturer is successfully ot forging a certain part, using material supplied by Company A. A new supply of material is obtained from Company B, wit te same nominal composition of te major alloying elements as tat of te material from Company A. However, it is found tat te new forgings are cracking even toug te same procedure is followed as before. Wat is te probable reason? Te probable reason is te presence of impurities, inclusions, and minor elements suc as sulfur) in te material supplied by Company B. Note tat te question states tat bot materials ave te same nominal composition of te major alloying elements. No mention is made regarding minor elements or impurity levels. 6.7 Explain wy tere migt be a cange in te density of a forged product as compared to tat of te cast blank. If te original material as porosity, suc as from a poor casting wit porosity due to gases or srinkage cavities, its density will increase after forging because te pores will close under te applied compressive stresses. On te oter and, te original blank may be free of any porosity but due to adverse material flow and state of stress during plastic deformation, cavities may develop similar to voids tat develop in te necked region of a tensile-test specimen, see Fig. 3.4 on p. 100). Tus, te density will decrease after forging due to void formation. 6.8 Since glass is a good lubricant for ot extrusion, would you use glass for impression-die forging as well? Explain. Glass, in various forms, is used for ot forging operations. However, in impression-die forging, even tin films because glass is incompressible) will prevent te part from producing te die geometry, and tus develop poor quality, and may prevent successful forging of intricate sapes. If te glass lubricant solidifies in deep recesses of te dies, tey will be difficult and costly to remove. 6.9 Describe and explain te factors tat influence spread in cogging operations on square billets. A review of te events taking place at te dieworkpiece interface in cogging indicates tat te factors tat influence spreading are: a) Friction: te lower te friction, te more te spreading because of reduced lateral resistance to material flow. b) Widt-to-tickness ratio of te workpiece: te iger tis ratio, te lower te spreading. c) Contact lengt in te longitudinal direction)-to-workpiece ratio); te iger tis ratio, te iger te spreading. Recall tat te material flows in te direction of least resistance Wy are end grains generally undesirable in forged products? Give examples of suc products. As discussed in Section 6..5 starting on p. 83, end grains are generally undesirable because corrosion occurs preferentially along grain boundaries. Tus end grains present many grain boundaries at te surface for corrosion to take place. In addition, tey may result in objectionable surface appearance, as well as reducing te fatigue life of te component because of surface rougness tat results from corrosion. 84

3 6.11 Explain wy one cannot produce a finised forging in one press stroke, starting wit a blank. Forgings are typically produced troug a series of operations, suc as edging, blocking, etc., as depicted in Fig. 6.5 on p. 85. Tis is done for a number of reasons: a) Te force and energy requirements on te press are greatly reduced by performing te operations sequentially; b) Te part may ave to be subjected to intermediate annealing, tus allowing less ductile materials to be forged to complicated sapes. c) Reviewing te Arcard wear law given by Eq. 4.6) on p. 145, it can be seen tat low die wear rates can be acieved if te sliding distance and/or te force is low in a stroke. Reviewing Fig. 6.5 on p. 85, it can be seen tat eac operation will involve a large sliding distance between te workpiece and dies, tus causing more wear. 6.1 List te advantages and disadvantages of using a lubricant in forging operations. Te advantages include: a) a reduction in te force and energy required; b) less localization of strain, resulting in improved forgeability; c) te lubricant acts as a termal barrier, so tat te part can remain otter longer and tus ave more ductility; d) te lubricant can protect te workpiece from te environment, especially in ot forging, and also act as a parting agent. Te disadvantages include: a) Te lubricant adds cost to te operation; b) a tick film can result in orange-peel effect on te workpiece; c) lubricants may be entrapped in die cavities, tus part dimensions my not be acceptable; d) te lubricant must subsequently be removed from te part surface, an additional and difficult operation; e) disposal of te lubricant can present environmental sortcomings Explain te reasons wy te flas assists in die filling, especially in ot forging. Te flas is excess metal wic is squeezed out from te die cavity into te outer space between te two dies. Te flas cools faster tan te material in te cavity due to te ig a/ ratio and te more intimate contact wit te relatively cool dies. Consequently, te flas as iger strengt tan te otter workpiece in te die cavity and, wit iger frictional resistance in te flas gap, provides greater resistance to material flow outward troug te flas gap. Tus, te flas encourages filling of complex die cavities By inspecting some forged products suc as a pipe wrenc or coins), you can see tat te lettering on tem is raised rater tan sunk. Offer an explanation as to wy tey are made tat way. Rolling By te student. It is muc easier and economical to macine cavities in a die tus producing lettering on a forging tat are raised from its surface) tan producing protrusions tus producing lettering tat are like impressions on te forged surface). Note tat to produce a protrusion on te die, material surrounding te letters be removed, a difficult operation for most lettering. Recall also similar consideration in cast products It was stated tat tree factors tat influence spreading in rolling are a) te widt-totickness ratio of te strip, b) friction, and c) te ratio of te radius of te roll to te tickness of te strip. Explain ow eac of tese factors affects spreading. Tese parameters basically all contribute to te frictional resistance in te widt direction of te strip by canging te aspect ratio of te contact area between te roll and te strip see also Answer 6.9 above). 85

4 6.16 Explain ow you would go about applying front and back tensions to seet metals during rolling. Front tensions are applied and controlled by te take-up reel of a rolling mill; te iger te torque to tis reel or te iger te rotational speed, te greater te front tension. Back tension is applied by te pay-off reel by increasing te braking torque on te pay-off reel or reducing its rotational speed It was noted tat rolls tend to flatten under roll forces. Wic propertyies) of te roll material can be increased to reduce flattening? Wy? Flattening is elastic deformation of te originally circular roll cross section, and results in a larger contact lengt in te roll gap. Terefore, te elastic modulus of te roll sould be increased Describe te metods by wic roll flattening can be reduced. Roll flattening can be reduced by: a) decreasing te reduction per pass, b) reducing friction, and/or c) increasing te roll stiffness for example, by making it from materials wit ig modulus of elasticity, suc as carbides) Explain te tecnical and economic reasons for taking larger rater tan smaller reductions per pass in flat rolling. Economically, it is always beneficial to reduce te number of operations involved in manufacturing of products. Reducing te number of passes in rolling acieves tis result by lowering te number of required operations. Tis allows less production time to acieve te final tickness of te rolled product. Of course, any adverse effects of ig reductions per pass must also be considered. 6.0 List and explain te metods tat can be used to reduce te roll force. In reviewing te mecanics of a flat rolling operation, described in Section starting on p. 90, it will be apparent tat te roll force, F, can be reduced by: a) using smaller-diameter rolls, b) taking lower reduction per pass, c) reducing friction, d) increasing strip temperature, and e) applying front and/or back tensions, σ f and σ b. 6.1 Explain te advantages and limitations of using small-diameter rolls in flat rolling. Te advantages of using smaller diameter rolls in flat rolling are te following: a) compressive residual stresses are developed on te workpiece surface, b) lower roll forces are required, c) lower power requirements, d) less spreading, and e) te smaller diameter rolls are less costly and easier to replace and maintain. Te disadvantages include: a) larger roll deflections, possibly requiring backup rolls, and b) lower possible drafts; see Eq. 6.46) on p A ring-rolling operation is being used successfully for te production of bearing races. However, wen te bearing race diameter is canged, te operation results in very poor surface finis. List te possible causes, and describe te type of investigation you would conduct to identify te parameters involved and correct te problem. Surface finis is closely related to lubricant film tickness, tus initial investigations sould be performed to make sure tat te film tickness is maintained te same for bot bearing races. Some of te initial investigations would involve making sure, for example, tat te lubricant supply is not reduced wit a larger race size. Also, te iger te rolling speed, te greater te film tickness, so it sould be cecked tat te rolling speed is te same for bot cases. Forward slip sould be measured and te rolling speeds adjusted accordingly. 86

5 6.3 Describe te importance of controlling roll speed, roll gap, temperature, and oter relevant process variables in a tandem-rolling operation. Control of tandem rolling is especially important because te conditions at a particular stand can affect te tose at anoter stand. For example, wit roll gap and rolling speeds, te effect of poor control is te application of too muc or too little front and/or back tension. As is clear from te description on p. 301, tis may result in larger roll forces and torques, or can lead to catter. 6.4 Is it possible to ave a negative forward slip? Explain. It is possible to ave negative forward slip, but only in te presence of a large front tension. Consider tat it is possible to apply a large enoug front tension so tat te rolls slip. A sligtly lower front tension will ave significant slippage, so te workpiece velocity will be muc lower tan te roll velocity. From Eq. 6.4) onp. 91, te forward slip will be negative. Note tat it is not possible to ave negative forward slip if tere are no front or back tensions. 6.5 In addition to rolling, te tickness of plates and seets can also be reduced by simply stretcing. Would tis process be feasible for ig-volume production? Explain. Altoug stretcing may first appear to be a feasible process, tere are several significant limitations associated wit it, as compared to rolling: a) Te stretcing process is a batc operation and it cannot be continuous as in rolling. b) Te reduction in tickness is limited by necking of te seet, depending on its strain-ardening exponent, n. c) As te seet is stretced, te surface finis becomes dull due to te orange-peel effect, and tickness and widt control becomes difficult. d) Stretcing te seet requires some means of clamping at its ends wic, in turn, will leave marks on te seet, or even cause tearing. e) Tere would be major difficulties involved in applying ig temperature during stretcing of less ductile materials. 6.6 In Fig. 6.33, explain wy te neutral point moves towards te roll-gap entry as friction increases. Te best way to visualize tis situation is to consider two extreme conditions. Let s first assume tat friction at te roll-strip interface is zero. Tis means tat te roll is slipping wit respect to te strip and as a result, te neutral no-slip) point as to move towards te exit. On te oter and, if we assume tat friction is very ig, te roll tends to pull te strip wit it; in tis case, te neutral point will tend to move towards te entry of te roll gap. 6.7 Wat typically is done to make sure te product in flat rolling is not crowned? Tere are a number of strategies tat can be followed to make sure tat te material in flat rolling is not crowned, tat is, to make sure tat its tickness is constant across te widt. Tese include: a) Use work rolls tat are crowned. b) Use larger backing rolls tat reduce elastic deformation of te work rolls. c) Apply a corrective moment to te safts of te work rolls. d) Use a roll material wit ig stiffness. 6.8 List te possible consequences of rolling at a) too ig of a speed and b) too low of a speed. Tere are advantages and disadvantages to eac. Rolling at ig speed is advantageous in tat production rate is increased, but it as disadvantages as well, including: Te lubricant film tickness entrained will be larger, wic can reduce friction and lead to a condition were te rolls slip against te workpiece. Tis can lead to a damaged surface finis on te workpiece. Te ticker lubricant film associated wit iger speeds can result in significant orange-peel effect, or surface rougening. Because of te iger speed, catter may occur, compromising te surface quality or process viability. 87

6 Tere is a limit to speed associated wit te power source tat drive te rolls. Rolling at low speed is advantageous because te surface rougness of te strip can matc tat of te rolls wic can be polised). However, rolling at too low a speed as consequences suc as: Production rate will be low, and tus te cost will be iger. Because a sufficiently tick lubricant film cannot be developed and maintained, tere may be a danger of transferring material from te workpiece to te roll pickup), tus compromising surface finis. Te strip may cool excessively before contacting te rolls. Tis is because a long billet tat is rolled slowly will lose some of its eat to te environment and also by conduction troug te roller conveyor. 6.9 Rolling may be described as a continuous forging operation. Is tis description appropriate? Explain. Tis is a good analogy. Consider te situation of forging a block to a tinner cross section troug increments as in incremental forming). As te number of stages increases, te operation eventually approaces tat of te strip profile in rolling Referring to appropriate equations, explain wy titanium carbide is used as te work roll in Sendzimir mills, but not generally in oter rolling mill configurations. Te main reason tat titanium carbide is used in a Sendzimer mill is tat it as a ig elastic modulus, and tus will not flatten as muc; see Eq. 6.48) on p. 99 and te text immediately after tis equation. Titanium carbide is not used for oter roll configuration because of te size of te rolls required and te ig cost of TiC rolls. Extrusion 6.31 It was stated tat te extrusion ratio, die geometry, extrusion speed, and billet temperature all affect te extrusion pressure. Explain wy. Extrusion ratio is defined as te ratio of billet initial) area to final area. If redundant work is neglected, te absolute value of true strain is ɛ = lna o /A f ). Tus, te extrusion ratio affects te extrusion force directly in an ideal situation. Die geometry as an effect because it influences material flow and, tus, contributes to te redundant work of deformation. Extrusion speed as an effect because, particularly at elevated temperatures, te flow stress will increase wit increasing strain rate, depending on te strain-rate sensitivity of te workpiece material. On te oter and, iger temperatures lower te yield stress and tus, reduce forces. 6.3 How would you go about preventing centerburst defects in extrusion? Explain wy your metods would be effective. Centerburst defects are attributed to a state of ydrostatic tensile stress at te centerline of te deformation zone in te die. Te two major variables affecting ydrostatic tension are te die angle and extrusion ratio. Tese defects can be reduced or eliminated by lowering te die angle, because tis increases te contact lengt for te same reduction and tereby increases te deformation zone. Similarly, a iger extrusion ratio also increases te size and dept of te deformation zone, and tus will reduce or eliminate te formation of tese cracks. Tese considerations are also relevant to strip, rod, and wire drawing How would you go about making a stepped extrusion tat as increasingly larger crosssections along its lengt? Is it possible? Would your process be economical and suitable for ig production runs? Explain. If te product as a stepped profile, suc as a round stepped saft wit increasing diameter, te smaller diameter is extruded first. Te die is ten canged to one wit a larger opening and te part is extruded furter. A still larger tird, and furter larger cross sections, can be produced by canging te die to a larger diameter opening. Te process would obviously not be economical at all for ig production runs. For sorter pieces, it is possible to make a die wit a stepped profile, as sown in Fig on p. 317, were te lengt of te stroke is small. 88

7 6.34 Note from Eq. 6.54) tat, for low values of te extrusion ratio, suc as R =, te ideal extrusion pressure p can be lower tan te yield stress, Y, of te material. Explain weter or not tis penomenon is logical. Equation 6.54) on p. 310 is based on te energy principle and is correct. Note tat te extrusion pressure, p, acts on te undeformed) billet area. Consequently, it is not necessary tat its magnitude be at least equal to te yield stress of te billet material In ydrostatic extrusion, complex seals are used between te ram and te container, but not between te extrusion and te die. Explain wy. Te seals are not needed because te leading end of te workpiece, in effect, acts as a seal against te die. Te clearance between te workpiece and te die is very small, so tat te ydraulic fluid in te container cannot leak significantly. Tis may present some startup problems, owever, before te workpiece becomes well-conformed to te die profile List and describe te types of defects tat may occur in a) extrusion and b) drawing. Recognizing tat a defect is a situation tat can cause a workpiece to be considered unsuitable for its intended operation, several defects can occur. Extrusion defects are discussed in Section starting on p Examples include poor surface finis or surface cracking suc as bamboo defect), tailpipe or fistailing, and cevron cracking. In drawing, defects include poor surface finis and cevron cracking. Bot extrusion and drawing also can ave a loss in dimensional accuracy, particularly as attributed to die wear Wat is a land in a die? Wat is its function? Wat are te advantages and disadvantages to aving no land? Te land is sown in Fig on p. 30 for drawing, but is too small to be seen for te figures illustrating extrusion. Te land is te portion of a die tat is parallel to te workpiece travel tat bears against te workpiece. Te land is needed to ensure tat workpiece dimensions are controlled and tat die wear does not affect dimensions, since die wear mainly occurs on te inlet side of te die. Te disadvantage to te land is tat te workpiece surface can be damaged by scratcing against te land; generally, te smaller te land, te better te workpiece surface Under wat circumstances is backwards extrusion preferable to direct extrusion? Wen is ydrostatic extrusion preferable to direct extrusion? Comparing Figs a and 16.47b on p. 309 it is obvious tat te main difference is tat in backward extrusion te billet is stationary, and in direct extrusion it is moving relative to te container walls. Te main advantage becomes clear if a glass pillow is used to provide ot-working lubricant between te workpiece and te die. On te oter and, if tere is significant friction between te workpiece and te camber, energy losses associated wit friction are avoided in backwards extrusion because tere is no movement between te bodies involved) Wat is te purpose of a container liner in direct extrusion see Fig. 6.47a)? Wy is tere no container liner used in ydrostatic extrusion? Te container liner is used as a sacrificial wear part, similar to te pads used in an automotive disk brake. Wen worn, it is far less expensive to replace a liner tan to replace te entire container. In ydrostatic extrusion, te billet doesn t contact te container, and tus wear is not a concern. Drawing 6.40 We ave seen tat in rod and wire drawing, te maximum die pressure is at te die entry. Wy? Te reason is tat at te die entry, te state of stress is plane stress wit equal biaxial compression in te radial direction). Tus, according to yield criteria te state of stress is in te tird quadrant of Fig..36 on p. 67 and ence te pressure as a value of Y. At te die exit, owever, we ave longitudinal tension and biaxial radial) compression due to contact wit te die. According to te yield criteria, because 89

8 of te tensile stress present, te die pressure is lower tan tat at te entry to te die see also Answer 6.43 below) Describe te conditions under wic wet drawing and dry drawing, respectively, are desirable. Wet drawing would be suitable for large coils of wire tat can be dipped fully in te lubricant, wereas dry drawing would be suitable for straigt sort rods. 6.4 Name te important process variables in drawing, and explain ow tey affect te drawing process. Tese are described in Section 6.5 starting on p. 30. Te important variables include: Yield stress, Y ; it directly affects te draw stress and die life. Die angle, α. Te die angle in te deformation zone affects te redundant work; in te entry area, te die angle is important for encouraging lubricant entrainment. Friction coefficient, µ. Te friction coefficient affects te frictional component of work and, ence, te draw stress. See also Eq. 6.68) on p. 3. Reduction in area. As described, tere is a limit to te reduction in area tat can be acieved in drawing. Lubrication condition. Effective lubrication reduces friction, but also may lead to a roug surface due to te orange peel effect Assume tat a rod drawing operation can be carried out eiter in one pass or in two passes in tandem. If te die angles are te same and te total reduction is te same, will te drawing forces be different? Explain. Te drawing forces will be te same, unless te surface of te rod is undergoing some canges wile it is between te two dies, due to external effects suc as te environment or additional lubrication. Te reason wy te forces are not different is tat te drawing process can be regarded as consisting of a series of incremental reductions taking place in one die. Ideally, we can slice te die into a number of segments and, tus, make it a tandem process. Note tat as te distance between te individual die segments decreases, we approac te one-die configuration. Also note tat in a tandem operation, te front tension of one segment becomes te back tension of an adjacent segment Refer to Fig and assume tat reduction in te cross section is taking place by pusing a rod troug te die instead of pulling it. Assuming tat te material is perfectly plastic, sketc te die-pressure distribution, for te following situations: a) frictionless, b) wit friction, and c) frictionless but wit front tension. Explain your answers. Note tat te matematical models developed for drawing and extrusion predict te draw stress or extrusion pressure, but do not sow te die pressure. A quantitative relationsip could be derived for te die pressure, recognizing tat p σ x = Y based on yield criteria, and ten examining Eqs. 6.63) troug 6.67) on p. 31. However, a qualitative sketc of te die pressure can be generated based on te pysical understanding of te friction ill and associated pressure plots in forging and rolling in Sections 6. and 6.3. A qualitative sketc of te die pressures is given below. Note tat te actual position of te curve for te frictionless case wit front tension depends on te level of front tension provided. Dimensionless pressure, p/y Frictionless Wit friction Frictionless wit front tension Position, x 6.45 In deriving Eq. 6.74), no mention was made regarding te ductility of te original material being drawn. Explain wy. Te derivation of Eq. 6.77) on p. 36 is based on te fact tat, at failure, te tensile stress in te wire or rod as reaced te uniaxial yield stress of te material. Tus, it is implicitly assumed tat te material is able to undergo te reduction in cross-sectional area and tat it is ultimately failing under ig tensile stresses. 90

9 Note tat a less ductile material will fail prematurely because of lack of ductility but not lack of strengt Wy does te die pressure in drawing decreases toward te die exit? We refer to Eq. 6.71) on p. 3 wic represents yield criteria in te deformation zone. Note tat as we approac te die exit, te drawing stress, σ, increases; consequently, te die pressure, p, must decrease, as also sown in Fig. 6.6 on p Wat is te magnitude of te die pressure at te die exit for a drawing operation tat is being carried out at te maximum reduction per pass? Te die pressure at te exit in tis case will be zero. Tis is because of te condition set by Eq. 6.73) on p. 34 wic deals only wit uniaxial stress. Note tat tere is a finite die pressure in a normal drawing operation, as depicted in Fig. 6.6 on p. 3, and tat te drawing stress at te exit is lower tan te uniaxial yield stress of te material, as it sould for a successful drawing operation to take place Explain wy te maximum reduction per pass in drawing sould increase as te strainardening exponent, n, increases. Te reason is tat te material is continuously strain ardening as it reaces te die exit. Consequently, at te exit it is stronger and, tus, can resist iger stresses before it yields. Consequently, a strain-ardening material can undergo iger reductions per pass, as can also be seen in Example If, in deriving Eq. 6.74), we include friction, will te maximum reduction per pass be te same tat is, 63%), iger, or lower? Explain. If we include friction, te drawing stress will be iger. As a result, te maximum reduction per pass will be lower tan 63%. In oter words, te cross-sectional area of te exiting material as to be larger tan te ideal case in order to support te increased drawing stress due to friction, witout yielding Explain wat effects back tension as on te die pressure in wire or rod drawing, and discuss wy tese effects occur. Te effect of back pressure is similar to tat of back tension in rolling see Figs on p. 95 and 6.6 on p. 3), namely, te pressure drops. Tis satisfies yield criteria, in tat, as tension increases, te apparent compressive yield stress of te material decreases Explain wy te inomogeneity factor, φ, in rod and wire drawing depends on te ratio, /L, as plotted in Fig By observing Figs. 6.1 on p. 76 and 6.13b on p. 77, we note tat te iger te /L ratio, te more nonuniform te deformation of te material. For example, keeping constant ence te same initial and final diameters), we note tat as L decreases, te die angle as to become larger. Tis, in turn, indicates iger redundant work see Fig on p. 311). 6.5 Describe te reasons for te development of te swaging process. Te major reasons include: a) variety of parts tat can be produced wit relatively simple tooling, b) capacity to produce internal profiles on long workpieces, c) compact equipment, d) good surface finis and dimensional accuracy, and e) improved workpiece properties due to cold working of te material Occasionally, wire drawing of steel will take place witin a seat of a soft metal, suc as copper or lead. Wy would tis procedure be effective? Te main reason tat steel wire drawing takes place in a seat of a softer metal is to reduce te frictional stress. Recall from Eq. 4.5) on p. 140 tat, for te same friction factor, m, te frictional stress is lower if te workpiece ardness is lower. By placing te seat in contact wit te die, te soft metal acts as a solid lubricant and reduces te frictional stresses. Tis, in turn, reduces forces and ence increases drawability. 91

10 6.54 Recognizing tat it is very difficult to manufacture a die wit a submillimeter diameter, ow would you produce a 10 µm-diameter wire? Te most common metod of producing very small wires is troug bundle drawing, werein a large number of wires up to undreds) are simultaneously drawn troug one die. Special care must be taken to provide good lubrication; oterwise, te wires will weld togeter during drawing. Te student sould be encouraged to suggest additional tecniques Wat canges would you expect in te strengt, ardness, ductility, and anisotropy of annealed metals after tey ave been drawn troug dies? Wy? General We would expect tat te yield stress of te material is iger, assuming tat te operation is performed at room temperature. Since it is directly related to strengt, ardness is also iger. Te ductility is expected to decrease, as te material as been strain ardened. Because of preferred orientation during deformation, some anisotropy is also to be expected in cold-drawn rods Wit respect to te topics covered in tis capter, list and explain specifically two examples eac were friction a) is desirable and b) is not desirable. Te student is encouraged to provide several specific examples. For example, friction is desirable in rolling and controlling material flow in forging. It is undesirable in rod and wire drawing except to obtain a burnised surface) and extrusion Coose any tree topics from Capter and wit a specific example for eac, sow teir relevance to te topics covered in tis capter. By te student. For example, a student could discuss yield criteria, and ten sow ow tey are used to develop pressure and force equations for specific operations Same as Question 6.57 but for Capter 3. By te student. For example, a student could select termal effects on mecanical properties, as discussed in Section 3.7 starting on p. 98, and apply it to a discussion of cold versus ot forging List and explain te reasons tat tere are so many different types of die materials used for te processes described in tis capter. Among several reasons are te level of stresses and type of loading involved suc as static or dynamic), relative sliding, temperature, termal cycling, dimensional requirements, size of workpiece, frictional considerations, wear, and economic considerations Wy sould we be interested in residual stresses developed in parts made by te forming processes described in tis capter. Residual stresses and teir significance are discussed in detail in Section.10 starting on p. 59. Te student sould elaborate furter wit specific references to te processes discussed in tis capter Make a summary of te types of defects found in te processes described in tis capter. For eac type, specify metods of reducing or eliminating te defects. By te student; see also Sections 3.8, 4., and 4.3. Problems Forging 6.6 In te free-body diagram in Fig. 6.4b, te in- 9

11 cremental stress dσ x on te element was sown pointing to te left. Yet it would appear tat, because of te direction of frictional stresses, µp, te incremental stress sould point to te rigt in order to balance te orizontal forces. Sow tat te same answer for te forging pressure is obtained regardless of te direction of tis incremental stress. Wewillderivetepressureusingtesameapproac as described in Section 6.. starting on p. 69. Te equivalent version of Fig. 6.4 on p. 69 is sown below. Using te stresses as sown in part b), we ave, from equilibrium and assuming unit widt, σ x + dσ x ) µσ y dx σ x =0 or dσ x µσ y dx =0 For te distortion-energy criterion, it sould be recognized tat σ x is now tensile, wereas in te text it is compressive. Terefore, Eq. 6.11) becomes σ y + σ y = Y = Y 3 Tus dσ x = dσ y Wen substituted into te equilibrium equation, one obtains σ y = Ce µx/ Using te boundary conditions tat σ x = 0 and terefore σ y = Y )atx = 0, gives te value of C as C = Y e µa/ Terefore, substituting into te expression for σ y, σ y = Ce µx/ = Y e µa/ e µx/ = Y e µa x)/ wic is te same as Eq. 6.13) on p Plot te force vs. reduction in eigt curve in open-die forging of a solid cylindrical, annealed copper specimen 5.08 in. cm ig ig and and 1 in..54 in diam- cm in diameter, up toup ato a reduction of of 70%, for te cases of a) no friction between te flat dies and te specimen, b) µ μ =0.5, = and c) µ μ =0.5. = Ignore barreling and use average-pressure formulas. Forannealedcopperweave,fromTable.3on p. on 37, p. 37, K = K 315 = 315 MPa MPa = 46,000 and n psi = and n Te =0.54. flow Te stress flow is stress is Y f Y=46, 000 psi)ɛ 0.54 f = 315 MPa) 0.54 were te absolute value of te strain is ) o ɛ =ln From volume constancy, we ave π 4 r o o = π 4 r or ) o r = ro Note tat r o =0.5= 1.7 inand cm o =in. o = 5.08 Teforg- cm. forging force force is given is given by Eqs. by Eqs. 6.18) 6.18) and and 6.19) 6.19) on p. on 7 p. 7 as: as: F = Y f 1+ µr ) πr ) 3 Some of te points on te curves are te following: Forging Force, kip MN %Red. µ =0 µ =0.5 µ =

12 Te curve is plotted as follows: Forging force, MN Use Fig. 6.9b to provide te answers to Problem were te absolute value of te strain is ) o ɛ =ln From tis, te following forces are calculated recall tat F = p ave A): % F, F, MN kip Red. r, cm in. µ =0.5 µ = Te force required for forging is te product of te average pressure and te instantaneous cross-sectional area. Te average pressure is obtained from Fig. 6.9b on p. 7. Note tat for µ = 0, p ave /Y f = 1, and tus te answer is te same as tat to Problem 6.63 given above. Te following table can be developed were p ave /Y f is obtained from Fig 6.9b, and and r are calculated as in Problem Note tat Fig. 6.9b does not give detailed information for r/ < 10, wic is were te data for tis problem lies. However, te µ =0.5 values interpolated between te µ =0. andµ =0.3 curves) are noticeably above 1 by r/ =5or so, so we give te value 1.5, and all intermediate values are linearly interpolated from tis reading. Similarly, for µ = 0.5, a value between µ =0.3 and sticking suggests p ave /Y is around 1.6orsobyr/ = 3. Tis is te basis for te numbers below. % p ave /Y f Red. r/ µ =0.5 µ = Recall from Problem 6.63 tat Y 46, 000 psi)ɛ f = 317,159 kpa) Te results are plotted below. For comparison purposes, te results from Problem 6.63 are also included as dased lines. As can be seen, te results are fairly close, even wit te roug interpolation done in tis solution. Forging force, MN Calculate te work done for eac case in Problem Te work done can best be calculated by obtaining te area under te curve F vs.. From te solution to Problem 6.63, te force is given by F = Y f 1+ µr ) πr ) 3 were and r = r o ) o [ Y f = 46, 000 psi) o ln Y f = 317,159 kpa) ln o )]

13 Numerous matematical software packages can perform tis calculation. Te results are as follows for te data in Problem 6.63, at 70% re- duction in eigt or Δ = =1.4): cm): µ Work cm-kg) in.-lb) 0 6,445 71, ,065 81, ,685 91, Determine te temperature rise in te specimen for eac case in Problem 6.63, assuming tat te process is adiabatic and te temperature is uniform trougout te specimen. To determine te temperature rise at 70% reduction in eigt, we obtain te work done from Problem 6.65 above. Assuming tere is negligible stored energy, tis work is converted into eat. Tus, we can calculate te temperature rise using Eq..65) on p. 73: T = u total ρc were u is te specific energy, or te energy per volume. Te volume of te specimen is V V = = πr πr = π1.7) = π0.5) 5.08) ) = incm 3 3 Te specific eat eat of of copper copper is given is given in Table in 3.3 Table on 3.3 p. 106 on as p J/kgK as 385 or J/kgK. 0.09Te BTU/lb specific F. Since eat of 1 BTU= copper 780 is ft-lb, tecm-kg/kgk. specific eat Te of copper density isof 861copper in-lb/lbis, F. from Te density te same of copper table, is, 8970 fromkg/m te 3. same Tus, table, using 8970 te kg/m work 3 orvalues 0.34 lb/in obtained 3. Tus, in Problem using 6.65, te work te temperature values obtained rise in is as Problem follows: 6.65, te temperature rise is as follows: µ T ΔT,, KF To determine its forgeability, a ot-twist test is performed on a round bar 5 mm in diameter and 00 mm long. It is found tat te bar underwent 00 turns before it fractured. Calculate te sear strain at te outer surface of te bar at fracture. Te sear strain can be calculated from Eq..) on p. 49 were r =5/ =1.5 mm, l = 00 mm, and φ =π00) = 157 radians. Terefore, te sear strain is γ = 1.5)157) 00 = Derive an expression for te average pressure in plane-strain compression under te condition of sticking friction. Sticking friction refers to te condition were a Tresca friction model is used wit m = 1 [see Eq. 4.5) on p. 140]. Terefore, te following figure represents te applied stresses to an element of forging, wic can be compared to Fig. 6.4b on p. 69. Te approac in Section 6.. starting on p. 69 is followed closely in tis derivation. From equilibrium in te x-direction, σ x + dσ x ) +mkdx σ x =0 Solving for dσ x, Integrating, dσ x = mk dx σ x = mk x + C were C is a constant. Te boundary condition is tat at x = a, σ x =0,sotat Terefore, and 0= mk a)+c C = mk a σ x = mk a x) 95

14 Te die pressure is obtained by applying Eq..36) on p. 64: σ y σ x = Y Note tat in plane strain, Y = 3 Y, and k = Y/ 3 see Section.11.3 on p. 66), so tat k = Y /. Terefore or σ y my a x) = Y σ y = Y [ 1 + m a x) ] Note tat tis relationsip is consistent wit Fig on p. 73 for 0 x a. Since te relationsip is linear, ten we can note tat or σ y = Y [ 1 + ma ) ] + 1) σ y = Y 1 + ma ) For sticking, m = 1 and σ y = Y 1 + a ) 6.69 Wat is te magnitude of µ wen, for planestrain compression, te forging load wit sliding friction is equal to te load wit sticking friction? Use average-pressure formulas. Te average pressure wit sliding friction is obtained from Eq. 6.15) on p. 71, and for sticking friction it is obtained from te answer to Problem 6.68 using m = 1. Equating tese two average pressures, we obtain Y 1 + µa Terefore, µ = 0.5. ) = Y 1 + a ) 6.70 Note tat in cylindrical upsetting, te frictional stress cannot be greater tan te sear yield stress, k, of te material. Tus, tere may be a distance x in Fig. 6.8 were a transition occurs from sliding to sticking friction. Derive an expression for x in terms of r,, and µ only. Te pressure curve for te solid cylindrical case is similar to Fig. 6.5 on p. 70 and is given by Eq. 6.17) on p. 7. Following te same procedure as in Example 6., te sear stress at te interface due to friction can be expressed as τ = µp However, we know tat te sear stress cannot exceed te yield sear stress, k, of te material wic, for te cylindrical state of stress, is Y. Tus, in te limit, we ave te condition or Hence, µy e µr x)/ = Y µr x) ) 1 = ln µ ) ) 1 x = r ln µ µ Note tat tis answer is te same as in te example problem for plane strain Assume tat te workpiece sown in te accompanying figure is being pused to te rigt by a lateral force F wile being compressed between flat dies. a) Make a sketc of te die-pressure distribution for te condition for wic F is not large enoug to slide te workpiece to te rigt. b) Make a similar sketc, except tat F is now large enoug so tat te workpiece slides to te rigt wile being compressed. F Applying a compressive force to te left boundary of te workpiece in Fig. 6.5 on p. 70 raises te pressure at tat boundary. Te iger te force F, te iger te pressure. Eventually te workpiece will slide completely to te rigt, indicating tat te neutral point as now moved all te way to te left boundary. Tese are depicted in te figure below. 96

15 and eigt as Hence, p ave = Y [ 1+ r = V π ) ) )] µ V 1 3 π 3/ 6.7 For te sticking example given in Fig. 6.10, derive an expression for te lateral force F required to slide te workpiece to te rigt wile te workpiece is being compressed between flat dies. Becauseweavestickingonbotdie-workpiece interfaces and a plane-strain case, te frictional stress will simply be Y /. Hence, te lateral force required must overcome tis resistance on bot top and bottom) surfaces. Tus, F =Y a)w) =4Y aw were w is te widt of te workpiece, i.e., te dimension perpendicular to te page in te figure Two solid cylindrical specimens, A and B, bot made of a perfectly-plastic plastic material, are being forged wit friction and isotermally at room temperature to a reduction in eigt of 5%. Originally, specimen A as a eigt of of 5.08 in. and cm and a cross-sectional area of area 1 inof, and specimen cm, and B as specimen a eigt B as of isa 1eigt in. andof a.54 cross-sectional cm and a crosssectional in. area Willof te workcmdone. Will bete work samedone for area of te be te twosame specimens? for te two Explain. specimens? Explain. We can readily see tat specimen B will require iger work because it as a larger dieworkpiece surface area, ence a iger frictional resistance as compared to specimen A. We can prove tis analytically by te following approac. Te work done is te integral of te force and distance: W = Fd were F =p ave )A), and p ave for a cylindrical body is given by Eq. 6.18) on p. 7. Because te volume V of te workpiece is constant, we ave a relationsip between its radius Because it consists of constants, let c = µ/3)v/π), wic results in te following expression: F = Y 1+ c ) ) V 3/ and ence work is o/ [ ) 1 c W = YV + o [ = YV ln c 1 = YV [ 0.88 c 3/ o d ] 0.540) ] 0.809) 3/ o 5/ )] Tus, for tis problem, we ave [ ) ] 1 1 W A A = YV YV c c / 3/ 0.809) = YV 0.88 YV c) 0.86c) [ ) ] 1 1 W B B = YV YV c c / 0.809) = YV 0.88 YV c) 0.809c) Comparing te two sows tat W B >W A In Fig. 6.6, does te pressure distribution along te four edges of te workpiece depend on te particular yield criterion used? Explain. Te answer is yes. Tis is a plane-stress problem, but an element at te center of te edges is subjected not only to a pressure p due to te dies) but also frictional constraint since te body is expanding in all directions. Tus, an element at te center of te edges is subjected to biaxial compressive stresses. Because te lateral stress, σ x, due to frictional forces is smaller tan te normal stress pressure), we note te following: 97

16 a) According to te maximum-sear-stress criterion, te pressure distribution along te edges sould be constant because te minimum stress is zero. Hence, p = Y b) According to te distortion-energy criterion for plane stress, te pressure distribution along te edges sould be as given in Fig. 6.6 on p. 71 and can be sown to obey te following relationsip: p + σ pσ = Y Note tat at te corners p = Y, and tat p is igest at te center along te edges because tat is were te frictional stress is a maximum Under wat conditions would you ave a normal pressure distribution in forging a solid cylindrical workpiece as sown in te accompanying figure? Explain. p Te pressure distribution is similar to te friction ill sown in Fig. 6.5 on p. 70, wit te exception tat tere are two symmetric regions were te pressure is constant. Tese regions sustain pressure but do not contribute to te frictional stress. A trapped layer of incompressible lubricant in grooves macined on te surface of te workpiece, for example, would represent suc a condition. Te grooves would be filled wit te lubricant, wic sustains pressure but would not contribute to sear at te interface because of its low viscosity Derive te average die-pressure formula given by Eq. 6.15). Hint: Obtain te volume under te friction ill over te surface by integration, and divide it by te cross-sectional area of te workpiece.) Y' x Te area, A, under te pressure curve from te centerline to te rigt boundary a) in Fig. 6.5 on p. 70 is given by A = p dx and te average pressure is p ave = 1 p dx a were p = Y e µa x)/ Integrating tis equation between te limits x = 0 and x = a, we obtain p ave = Y ) e µa/ 1 µa/ Letting µa/ = m, and using a Taylor series expansion of te exponent term, e m = 1 + m + m! + m3 3! +... Ignoring tird-order terms and iger as being too small compared to oter terms, we obtain p ave = Y ) 1 + m + m m 1 = Y 1 + m ) = Y 1 + µa ) 6.77 Take two solid cylindrical specimens of equal diameter but different eigts, and compress tem frictionless) to te same percent reduction in eigt. Sow tat te final diameters will be te same. Let s identify te sorter cylindrical specimen wit te subscript s and te taller as t, and teir original diameter as D. Subscripts f and o indicate final and original, respectively. Because bot specimens undergo te same percent reduction in eigt, we can write tf to = sf so and from volume constancy, tf to = ) Dto D tf 98

17 and sf so = ) Dso D sf Because D to = D so, we note from tese relationsips tat D tf = D sf A rectangular workpiece as te following original dimensions: a = 100 mm, = 30 mm and widt = 0 mm see Fig. 6.5). Te metal as a strengt coefficient of 300 MPa and a strainardening exponent of 0.3. It is being forged in plane strain wit µ = 0.. Calculate te force required at a reduction of 0%. Do not use average-pressure formulas. In tis plane-strain problem note tat te widt dimension remains at 0 mm. Tus, wen te reduction in eigt is 0%, te final eigt of te workpiece is = 1 0.)30) = 4 mm = 0.04 m Since volume constancy as to be maintained and we ave a plane-strain situation, we can find te new final) dimension a from 100)30)0) = a)4)0) Tus, a = 6.5 mm = m. Te absolute value of te true strain is ) 30 ɛ = ln = and ence te uniaxial flow stress at te final eigt is Y f = Kɛ n = 400)0.3) 0.3 = 55 MPa and te flow stress in plane strain is Y f = 1.15)55) = 93 MPa. Tus, from Eq. 6.13) on p. 70 te pressure as a function of distance x is p = Y e µa x)/ = 93 MPa)e 0.)0.065 x)/0.04 = 93 MPa)e x To obtain te force required for one-alf of te workpiece per unit widt, we integrate te above expression between te limits x = 0 and x = 0.065, wic gives te force per unit widt and one-alf of te lengt as F = 3. MN/m. Te total force is te product of tis force and te specimen widt times two, or F total = 3.)0.0) = 1.88 MN Note tat if we use te average pressure formula given by Eq. 6.16) on p. 71, te answer will be F total = Y 1 + µa = 93) ) a)w) [ )0.065) 0.04 )0.065)0.0) = 1.11 MN Te discrepancy is due to te fact tat in deriving te average pressure, a low value of µa/ ave been assumed for matematical simplicity Assume tat in upsetting a solid cylindrical specimen between two flat dies wit friction, te dies are rotated at opposite directions to eac oter. How, if at all, will te forging force cange from tat for nonrotating dies? Hint: Note tat eac die will now require a torque but in opposite directions.) From te top view of te round specimen in Fig. 6.8b on p. 7, we first note tat te frictional stresses at te die-specimen interfaces will essentially be tangential. We say essentially because te rotational speed is assumed to be muc iger tan te vertical speed of te dies.) Consequently, te direction of µσ z will be tangential and, because tere will now be no frictional stress in te radial direction, balancing forces in te radial direction will not include friction. Tus, te situation will be basically similar to upsetting witout friction, and te forging force will be a minimum. However, additional work as to be done in supplying torque to te two dies tat are rotating in opposite directions. Note also tat we are assuming µ to be small, so tat it will not cause plastic twisting of te specimen due to die rotation A solid cylindrical specimen, made of a perfectly plastic material, is being upset between flat dies wit no friction. Te process is being carried out by a falling weigt, as in a drop ammer. Te downward velocity of te ammer is at a maximum wen it first contacts te workpiece and becomes zero wen te ] 99