mechanical-properties-and-working-of-metals-and-alloys_compress

488 11 Forging estimated to be 0.28  W2H, in case of press forging, and prediction of working stresses and/or loads in the 0.30  W2H, in case of hammer forging. The volume of the steady-state open-die press forgings of simple-shaped slug, VSLUG, is calculated from the following relation: work-pieces with simple platens. VSLUG = cross-sectional area of hole Figure 11.28 shows a typical open-die forging of a thin  full height of forging in which the hole is punched wide rectangular plate of uniform thickness compressed 3 between a pair of parallel overhanging platens and the ð11:13Þ stresses acting at any instant on this plate under plane strain condition. Suppose, the plate has a thickness of h; an infi- The weight or volume of the selected ingot or rolled stock nitely large width of w and a length of deformation zone, should approximate as nearly as possible to the calculated or L ¼ 2a: The width of the plate is normal to the plane of required weight or volume of the initial work-piece, which is paper, and the thickness of the plate is less as compared with to be forged; the selected one may be more but never less its other dimensions. On this thin plate, the flat platens exert a than the calculated one. normal forging pressure p. The plate is symmetrical about a vertical centre-plane which is called the neutral plane or The degree of working is commonly evaluated by forging neutral surface of the plate, because the material is stationary reduction ratio or factor, called the forging ratio, which is at this plane. The material flows longitudinally outward in defined as the ratio of cross-sectional area of the initial both horizontal directions from the neutral plane when the work-piece before forging to that of the finished forged plate is being forged vertically with the platens. Therefore, product. This ratio should be at least 3–5 for ingots and from the neutral plane is the origin of the initiation of plastic flow 1.1 to 1.5 for rolled stocks. and considered as the origin of the coordinate system, shown by x ¼ 0 in the diagram, where x is the horizontal axis in the 11.7 Plane Strain Forging of Flat longitudinal direction of the plate that increases towards the Rectangular Plate right side of the neutral plane. Outward longitudinal flow of the material perpendicular to the applied normal forging An empirical approach is mainly required for the prediction pressure p; during the process of its compression leads to of forging load and pressure in closed-die forgings and in tangential shearing frictional stresses s at the platen-job problems of metal flow associated with the filling of a contact surfaces. Hence, the friction stresses opposing the complex die shape, which deserve much more detailed study outward longitudinal flow will always act longitudinally than they have received up to this time. The conditions of towards the centre of the plate. As the material is free to flow impact deformation are also difficult to analyse. These will in both directions from its neutral plane, the friction stress s not be considered in this text. Attention will be limited to the opposing the flow will change its direction in both sides of the neutral plane. However, the presence of frictional shear Fig. 11.28 Stresses acting on P a rectangular plate of uniform Moving platen thickness forged in plane strain L = 2a p p [ (σx + dσx)] τ τ dx Work x σ x + dσ x piece h (− σx) dx σx x x= a τ τ Fixed platen x=a p p Neutral plane x=0 x→ P

11.7 Plane Strain Forging of Flat Rectangular Plate 489 stress s leads to horizontal stress rx in the longitudinal approach, the horizontal longitudinal stress on one side the direction of the plate. The stress rx is zero at both edges and vertical slab element at an arbitrary positive distance x from builds up to a peak value at the centre of the plate. To make the origin x ¼ 0; is rx; which changes to rx þ drx on another the analysis simpler, the following assumptions are made: side of this element situated at a positive distance x þ dx from the origin x ¼ 0: Figure 11.28 shows that these lon- (i) Material being forged is assumed to be isotropic and gitudinal stresses along with the normal pressure p and homogeneous. frictional shear stress s are acting on this vertical slab ele- ment. Now taking the horizontal equilibrium of forces acting (ii) Homogeneous deformation is assumed, i.e. plane on this vertical slab element on the right side of the centre vertical section before forging remains plane during line, we get and after forging and hence, internal redundant work is neglected. rxhw À ðrx þ drxÞhw À 2sdxw ¼ 0 ð11:14Þ (iii) Elastic strain is neglected because plastic deforma- Similarly, horizontal equilibrium of forces acting on this tion involved in forging is quite large compared to vertical slab element on the left side of the centre line gives insignificant elastic deformation. rxhw À ðrx þ drxÞhw þ 2sdxw ¼ 0 ð11:15Þ (iv) Since the plate is infinitely wide, i.e. its width is much more than its length of deformation zone, the Equations (11.14) and (11.15) can be simplified and plate is reduced in thickness and extended in length combined as follows: but the width of the plate remains constant during deformation and thus, a condition of plane strain hdrx Æ 2sdx ¼ 0 ð11:16Þ exists. Plane strain compression is also possible if the width of the platens is more than that of the plate. In (11.16), the upper positive sign ‘+’ refers to the right-hand portion and the lower negative sign ‘−’ to the (v) As forging is a hot working operation, so it is quite left-hand portion of the centre line of plate. The applied logical to assume that there is no strain hardening of the work-piece, i.e. the plane-strain flow stress or normal forging pressure p and the horizontal longitudinal deformation resistance r00 of the material being forged stress rx ; which are considered as principal stresses, are is assumed to remain constant. Otherwise, an average shown as compressive stresses in Fig. 11.28. If we consider value of plane-strain flow stress will be assumed, it is the compressive stresses to be positive, then r1 ¼ p and important to remember that in hot forging the flow r2 ¼ rx in Von Mises’ yielding criterion for plane strain stress will depend very much upon the strain rate used. condition, given by (1.88b). Hence, the yielding criterion (vi) The plate is thin enough to assume that the normal can be written as pressure, p, remains constant through the thickness. p À rx ¼ r00 ð11:17Þ (vii) The horizontal longitudinal stress rx is constant across the cross-section of the plate and is a principal where stress since no shear stress act on this cross-section along the vertical direction. r00 ¼ compressive flow stress in plane strain condition  pffiffi (viii) The applied normal forging pressure p is also ¼ 2= 3 Â compressive uniaxial flow stress in assumed to be a principal stress, although this assumption is not really correct, because the shear homogeneous strain condition stresses s act on the planes on which the normal  pffiffi pressure p is acting. ¼ 2= 3 r0: (ix) There is no barrelling of the edges of the work-piece. Since r00 ¼ constant, from (11.17), we get drx ¼ dp; and The barrelling can be eliminated if the width of the on substituting this into (11.16), we get a relationship platens is more than that of the plate between p and rx as follows: (x) The maximum value of the forging load is attained at the end of the operation. Let us consider the state of stresses on a vertical slab hdp Æ 2sdx ¼ 0 element of width dx inside the plate at an arbitrary distance x from the centre line of the plate, which is taken as the origin, Or; dp ¼ Ç 2s ð11:18Þ x ¼ 0 ; of the coordinate system. The value of x may be dx h positive or negative depending on whether the slab element is to the right side or left side of the centre line (neutral In (11.18), the upper negative sign ‘−’ refers to the plane) of the plate. According to free-body equilibrium right-hand portion and the lower positive sign ‘+’ to the left-hand portion of the centre line of plate.

490 11 Forging 11.7.1 Coulomb Sliding Friction thickness L=h of the deformation zone increase, the resis- tance to compressive deformation increases rapidly. Now we Let us consider that low frictional condition exists at the can obtain the horizontal longitudinal stress for Coulomb’s contact surface between the forging dies and the plate, where Coulomb’s law of sliding friction holds good. So, the tan- sliding friction, ðrxÞC:F; from (11.17) and (11.22) as gential shearing frictional stress s is related to the applied normal forging pressure p by Coulomb’s law of sliding & '! friction as s ¼ lp; where l is the coefficient of friction and is 2l ða À jxjÞ assumed to be the same at all points on the surface of con- ðrxÞC:F:¼ pC:F: À r00 ¼ r00 exp h À1 tact. Now (11.18) becomes ð11:24Þ dp ¼ Ç2 l dx ð11:19Þ Equations (11.22) and (11.24) show that the distribution ph of normal deformation pressure as well as horizontal longi- tudinal stress is symmetrical about the central line and rises Integrating both sides of (11.19), we get to a maximum at the centre of the plate, which is the location of the neutral plane. This characteristic rise in normal ln p ¼ Ç 2l x þ C1 ð11:20Þ deformation pressure with distance is often called a friction h hill. From (11.22) and (11.24), the normal deformation pressure pC:F: and the horizontal longitudinal stress ðrxÞC:F: where C1 is the integration constant, which can be evalu- are plotted over the total deformation-zone length 2a of the plate, as illustrated in Fig. 11.29. ated from the following boundary condition. Since at outer The minimum values of pC:F: and ðrxÞC:F: at the edges of free both surfaces of the plate, the horizontal longitudinal the plate, where jxj ¼ a; are, respectively, obtained from (11.22) and (11.24) as stress rx is zero, i.e. at x ¼ Æa; rx ¼ 0; so from (11.17) p ¼ r00: Therefore, from (11.20) &' 2l ða À aÞ ln r00 ¼ À 2l a þ C1; or C1 ¼ ln r00 þ 2l a: ðpC:F:ÞMinimum¼ r00 exp h ¼ r00 ð11:25Þ h h Substituting the value of C1 into (11.20), we get ÂÃ ðpC:F: ÞMinimum Àr00 ¼ 0 ð11:26Þ ðrxÞC:F: Minimum¼ ln p ¼ Ç 2l x þ ln r00 þ 2l a; or, ln p ¼ 2l ða Ç xÞ; On the other hand, the maximum values of pC:F: and h h r00 h ðrxÞC:F at the centre-plane, i.e. at the neutral plane of the Therefore, normal forging pressure for Coulomb’s sliding plate, where x ¼ 0; are, respectively, given by friction is:   2la lL &' ðpC:F:ÞMaximum¼ r00 exp h ¼ r00 exp h ð11:27Þ 2l ða Ç xÞ pC:F: ¼ r00 exp h ð11:21Þ In (11.21), the upper negative sign ‘−’ indicates the right side of the centre line of plate, where x is measured in the positive x-direction and the value of x is positive, whereas p the lower positive sign, ‘+’ refers to the left side of the centre C.F σ0 line, where x is measured in the negative x-direction and the ( σx ) C.F value of x is negative. Thus, for an absolute value of the σ 0 exp 2 μa h distance x; measured from the centre line of plate, (11.21) can be written as &' pC:F: ¼ r00 exp 2l ða À jxjÞ ð11:22Þ 0 h Equation (11.22) can be conveniently written in the fol- lowing way: & ' aa pC:F: ¼ r00 exp lL 1 À 2jxj ð11:23Þ Neutral plane h L Fig. 11.29 Friction hill showing distribution of normal deformation where L ¼ 2 a ¼ the length of the deformation zone. The pressure pC:F: and longitudinal stress ðrxÞC:F: with Coulomb’s sliding above (11.23) shows that as the ratio of the length to friction over the total length 2a of a rectangular plate forged in plane strain

11.7 Plane Strain Forging of Flat Rectangular Plate 491 Âà  ! factor m; i.e. s ¼ m k; as shown in (10.24). Sometimes ðrxÞC:F: Maximum¼ 2la lubricant films of soft materials such as a polymer or lead are r00 exp À1 applied. In such case, the shear stress at the interface will be  h ! constant and given by s ¼ m k; where m is the ratio of the lL shear strength of the lubricant film to that of the work-piece. ¼ r00 exp h À1 ð11:28Þ Hence, substitution of s ¼ m k into (11.18) gives If the value of l is low, then it is possible to simplify (11.22) by using the expansion expðyÞ ¼ 1 þ y þ y2=2! þ 2mk 2prffi0ffi m m h 3 h h y2=3! þ Á Á Á : Hence, exp½f2lða À jxjÞ=hgŠ % 1 þ f2lða À jxjÞ=hg; dp ¼ Ç dx ¼ Ç dx ¼ Çr00 dx ð11:32Þ and (11.22) becomes 1 þ 2l ða À jxjÞ ! where k ¼ pffiffi ¼ r00 =2; according to Von Mises’ yield- h r0= 3 pC:F: ¼ r00 ð11:29Þ ing criterion. In (11.32), the upper negative sign ‘−’ refers to the Since (11.29) describes a straight line, so the curved sides right-hand portion and the lower positive sign ‘+’ to the of the friction hill plotted according to (11.22) will be replaced by straight line when plotted according to (11.29). left-hand portion of the centre line of plate. Integrating both sides of (11.32), we get 11.7.1.1 Average Pressure and Total Load p ¼ Çr00 m x þ C2 ð11:33Þ h The mean height of the friction hill is nothing but the average deformation pressure required for forging, which is where C2 is the integration constant. It is evaluated from the obtained from (11.22) as follows: boundary condition that at x ¼ Æa; rx ¼ 0; and so from (11.17), p ¼ r00 : Therefore, from (11.33) Mean forging pressure due to Coulomb’s sliding friction, pC:F: ¼ Rr20w00aRpR0eaC0xa2:Fprw&200dwe2xhxdl2pxðw&aR2Àh0la djðxxajÞÀ'=jxjÞÀ'd2hxl!a0 r00 ¼ Àr00 m a þ C2; or C2 ¼ r00 þ r00m a : ¼ h h ¼ Substituting the value of C2 into (11.33), the normal deformation pressure considering friction factor is given by m m 1 þ mða Ç xÞ ! h a h pF:F: ¼ Çr00 x þ r00 þ r00 ¼ r00 h a ð11:34Þ ¼ r00 exp½ð2laÞ=hŠ À 1 ¼ r00 exp½ðlLÞ=hŠ À 1 ð11:30Þ For an absolute value of the distance x; measured from ð2laÞ=h ðlLÞ=h the centre line of plate, (11.34) can be written as Now the total forging load for Coulomb’s sliding friction, 1 þ mða À jxjÞ ! PC:F:; can be found from the product of the mean forging h pressure and the area at the die–job interface. pF:F: ¼ r00 ð11:35Þ ) PC:F: ¼mean forgingpressureÂarea atthe dieÀjob interface Now the horizontal longitudinal stress considering fric- tion factor, ðrxÞF:F:; is obtained from (11.17) and (11.35) as ¼ pC:F: ½ð2aÞwŠ ¼ rr0000lwhwexpe½xðp2ll=alhhÞL=hŠÀÀ11! ðrxÞF:F:¼ pF:F: À r00 ¼ r00 mða À jxjÞ ð11:36Þ ¼ h ð11:31Þ The minimum values of pF:F: and ðrxÞF:F: at the edges of the plate, where jxj ¼ a; are, respectively, given by ðpF:F:ÞMinimum ¼ r00 and ½ðrxÞF:F:ŠMinimum ¼ 0; which are same as those shown under Coulomb’s sliding friction. The 11.7.2 Sliding with Shear Friction Factor maximum values of pF:F: and ðrxÞF:F at the centre, i.e. at the and Sticking Friction neutral plane of the plate, where x ¼ 0; are, respectively, Let us consider the case of sliding where the interface given by shearing frictional stress s is described by the shear friction ðpF:F:ÞMaximum¼ h þ mai ¼ r00 ! ð11:37Þ r00 1 h 1 þ mL 2h

492 11 Forging Âà r00 ma ¼ r00 mL ð11:38Þ ÀÁ h ai ¼ r00 ! ð11:41Þ ðrxÞF:F: Maximum¼ h 2h pSticking Maximum¼ r00 1 þ h 1þ L 2h Sliding friction can exist at the die–work-piece interface hi r00 a r00 L h 2h till the shear stress s at the interface does not exceed the ðrxÞSticking ¼ ¼ ð11:42Þ shearing yield stress k of the work-piece. But when the Maximum friction is high, which is usually found in hot forging, and this above limit is reached, then sticking friction occurs at 11.7.2.1 Average Pressure and Total Load the interface and the interface shearing frictional stress is Considering friction factor, the average forging pressure, given by s ¼ k; i.e. the shear friction factor m ¼ 1; in cases pF:F:; is obtained from (11.35) as follows: of sticking friction. Assume that the sticking friction regime Rr20w00aRpR0xaF0a:þ2Frw: 200mdwhxa12dwxþxÀRm0amhðdaxx22Àh!a0jxjÞ!dx extends over the whole interface. Hence, substitution of m ¼ pF:F: ¼ ¼ 1; into (11.35) and (11.36) gives, respectively, the normal ¼ deformation pressure, pSticking; and horizontal longitudinal stress, ðrxÞSticking; for conditions of sticking friction, which are: 1 þ a À jxj ! h pSticking ¼ r00 ð11:39Þ a ! h mai 1 þ mL ðrx ÞSticking ¼ r00 a À jxj ð11:40Þ ¼ r00 1 þ 2h ¼ r00 ð11:43Þ h 4h Equations (11.39) and (11.40) show that the distributions Now the total forging load considering friction factor, of normal deformation pressure and horizontal longitudinal PF:F:; can similarly be found from the product of the mean forging pressure and the area at the die–job interface. stress are not only symmetrical about the central line but also 1 þ mL ! linear with distance from the centre line of the plate (the 4h ) PF:F: ¼ pF:F: ½ð2aÞwŠ ¼ r00 Lw ð11:44Þ location of the neutral plane), as illustrated in the friction hill Equation (11.44) can be used to calculate the forging load diagram of Fig. 11.30. if the value of shear friction factor m at the interface is The minimum values of pSticking and ðrxÞSticking at the available. edges of the plate, where jxj ¼ a; are, respectively, equal to Hence for conditions of sticking friction, the average r00 and 0; which are same as those shown above. The max- forging pressure, pSticking; and the total forging load, PSticking; imum values of pSticking and ðrxÞSticking at the centre, i.e. at can be obtained by substituting m ¼ 1; into (11.43) and the neutral plane of the plate, where x ¼ 0; are, respectively, (11.44), which are: given by pSticking pSticking ¼ h ai ¼ r00 ! ð11:45Þ σ '0 r00 1 þ 2h 1þ L 4h (σ x ) 1þ L ! Sticking 4h ) PSticking ¼ pSticking ½ð2aÞwŠ ¼ r00Lw ð11:46Þ σ '0 a 1+ a h σ'0 h σ '0 11.7.3 Mixed Sticking–Sliding Friction 0 Often a condition exists where sticking friction occurs at the a a central region of the work-piece and sliding friction occurs at the outer regions of the work-piece with coefficient of sliding Neutral plane friction l ¼ constant. This occurs when the friction is somewhat lower than that required to cause sticking friction Fig. 11.30 Friction hill showing linear distribution of normal defor- over the entire interface. Since the normal forging pressure p mation pressure pSticking and longitudinal stress ðrxÞSticking under is the minimum at the edges and increases towards the centre sticking frictional condition over the total length 2a of a rectangular of the work-piece, the magnitude of the frictional stress lp plate forged in plane strain

11.7 Plane Strain Forging of Flat Rectangular Plate 493 will also increase inwards from the outer periphery of the Fig. 11.31. This friction hill diagram shows that on each side work-piece. But at a certain distance away from the edge towards the centre of the work-piece, if the frictional stress of the neutral plane from x ¼ a to x [ xS; the sliding pressure lp reaches a value equal to the shearing yield stress k of the work-piece, then the frictional stress cannot increase further. is lower than the pressure that would be required for sticking Under such condition, the central zone of the work-piece will be controlled by the sticking friction condition. So, the and so, sliding friction prevails there, whereas on each side of equilibrium equation (11.18) is expressed with the following conditions: the neutral plane from x\\xS to x ¼ 0; the sticking pressure is lower than the pressure that would be required for sliding and hence, full sticking friction occurs there. Only at the distance dp ¼ Ç 2s of x ¼ ÆxS from the neutral plane, the slidinÀg pressÁure is dx h equal to the sticking pressure, i.e. ðpC:F:Þx ¼ xS ¼ pSticking x ¼ xS . Normal forging pressure for Coulomb sliding friction given where s ¼ lp if lp\\k; and s ¼ k if lp ! k: Suppose, xS is the distance measured on each side from by (11.22) is repeated below, which is valid from x ¼ a to the neutral plane, i.e. centreline ðx ¼ 0Þ of the work-piece, at x ¼ xS for mixed sticking–sliding frictional conditions: which the transition from sticking to sliding occurs. Hence, frictional condition at the die–job interface will be such that &' Coulomb’s sliding friction prevails from x ¼ a to x ¼ xS; 2l ða À jxjÞ and full sticking friction occurs from x ¼ xS to x ¼ 0 on ðpC:F:Þa to xS ¼ r00 exp h ð11:22Þ each side of the centreline. This can be expressed mathe- matically as follows: Now let us find the solution for the sticking zone. For the sticking zone, the substitution of (11.47b) into (11.18) gives ÀÁ to xS ¼ Ç 2k dx ¼ Ç p2rffiffi0 dx ¼ Ç r00 dx ð11:48Þ dpSticking 0 h 3h h For xS x a; s ¼ lp ð11:47aÞ pffiffi where k ¼ r0= 3 ¼ r00=2; according to Von Mises’ yield- ing criterion. For 0 x xS; s ¼ k ð11:47bÞ In (11.48), the upper negative sign ‘−’ refers to the right-hand portion and the lower positive sign ‘+’ to the For the sliding zone, substitution of (11.47a) into (11.18) left-hand portion of the centre line of plate. and then integration will ultimately result in the solution same as that of (11.22), because the boundary condition for Integrating both sides of (11.48), we get evaluation of the integration constant is the same. Only the validity range of (11.22) is different, which varies from x ¼ a ÀÁ xS ¼ Çr00 x þ C3 ð11:49Þ to x ¼ xS; instead of x ¼ a to x ¼ 0: The variations of the pSticking 0 to h normal forging pressures for conditions of Coulomb’s sliding and full sticking are drawn in the same plot over the total where C3 is the integration constant, which can be evaluated deformation-zone length 2 a of the plate as shown in from the boundary condition of the sticking zone that extends from x ¼ 0 to x ¼ xS on each side of the neutral plane. At the point jxj ¼ xS; Fig. 11.31 Friction hill showing Psticking x=0 xs = a h1 distributions of normal forging PC.F. Neutral plane In pressures, pC:F: and pSticking, respectively, for conditions of 2μ 2μ Coulomb’s sliding and full sticking, resulting in mixed 2μa σ'0 σ '0 a sticking–sliding frictional h 1+ conditions over the total length 2μ 2 a of a rectangular plate in plane σ'0 exp h strain σ'0 0 aa 2 xs

494 11 Forging r00 ; r00 hÀ Á i  ! 2 2l pSticking 0 1 1 sx¼xS ¼ lðpC:F:Þx¼xS ¼ k ¼ or ðpC:F:Þx¼xS ¼ to xS Minimum ¼ r00 2  0:5 1 À ln 2  0:5 ð11:50Þ ¼ r00 Again since at the distance x ¼ ÆxS; ðpC:F:Þx¼xS ¼ ð11:56Þ ðpStickingÞx¼xS ; then substitution of (11.50) into (11.49) gives the value of C3; this is: hÀ Á i ¼ r00 1  À ln  þ ! pSticking 0 ¼ r00 1 2 1 a to xS Maximum h2  0a:i5  0:5 h 1 þ C3 ¼ r00 þ r00 xS ð11:51Þ h 2l h ð11:57Þ Substituting the value of C3 from (11.51) into (11.49), we These minimum and maximum values given by (11.56) and (11.57) are shown in Fig. 11.31. get the normal forging pressure for sticking friction condi- tion from x ¼ 0 to x ¼ xS as ÀÁ xS ¼ r00 ! ð11:52Þ 11.7.3.1 Average Pressure and Total Load pSticking 0 1 þ ðxS Ç xÞ to 2l h For Coulomb’s sliding friction from x ¼ a to x ¼ xS on each side of the neutral plane, the average normal forging pres- For an absolute value of the distance x; measured on each sure, ðpC:F:Þa to xS ; is obtained from (11.22) as follows: side from the centre line of plate, (11.52) can be written as 2RaxwaÀSrðR00pxxaSSCRr:xFaS00:Þ2eaÀwx2topdw2xh&xlSR22xhalwS eddðxxaxpÀ&2jhlxjðÞa'Àdxj ÀÁ 1 ðxS À jxjÞ ! ðpC:F:Þa ¼ pSticking 0 2l h xS ¼ r00 þ ð11:53Þ to xS ¼ ¼ to Again, substitution of (11.50) into (11.22) for ðpC:F:Þx¼xS '!a at jxj ¼ xS; gives a solution for the location of the boundary xjÞ between slipping and sticking friction, which is derived xS below: r00 ¼ r00 exp&2l ða À ' or 2l ða À xSÞ ¼ ln 1 ; ¼ r00 exp½f2lða À xSÞg=hŠ À 1 2l h xSÞ ; h 2l f2lða À xSÞg=h ) xS ¼ a À h ln 1 ð11:58Þ 2l 2l ð11:54Þ Substituting the value of xS; from (11.54) into (11.58), we get Substituting the value of xS, from (11.54) into (11.53), we ðpC:F:Þa ¼ r00 ½ð1=2lÞ À 1Š ð11:59Þ get lnð1=2lÞ ! to xS À Á to xS ¼r00 1 þ a À 1 1 À jxj For sticking friction from x ¼ 0 to x ¼ xS on each side of pSticking 2l  h 2l ln 2l h 0 1 1 þ a À jxj ! ð11:55Þ the neutral plane, the average normal forging pressure, 2l 2l h ¼r00 1 À ln ðpStickingÞ0 to xS ; is obtained from (11.53) as follows: R xS ÀÁ ÀÁ 2w 2pwStRic0kxiSngdx0 dx We can obtain the minimum and maximum values of pSticking 0 to xS ¼ 0 to xS  ðpStickingÞ0 to xS from (11.55) when Coulomb coefficient of ¼ sliding friction l reaches a maximum value under ¼ r00 Z xS 1 þ xS À jxj dx xS 0 2l h h plane-strain sticking frictional condition, where the mini- r00 x þ x xS À x2 !xS mum value of interface normal pressure is considered as xS 2l h 2h 0 pmin ¼ r00. This maximum value of l is ðlmaxÞP:Strain¼ 0:5 r00   [see (10.27)]. Hence, substitution of jxj ¼ a and x ¼ 0 into ¼ xS ð11:60Þ 1þ (11.55) gives, respectively, the minimum and maximum 2l h values of ðpStickingÞ0 to xS as

11.7 Plane Strain Forging of Flat Rectangular Plate 495 Substituting the value of xS; from (11.54) into (11.60), • If xS 0; the Coulomb’s sliding friction exists over the we get entire interface of the rectangular plate with deformation- zone length 2a and the dies, i.e. from x ¼ 0 (centre line) ÀÁ r00  l1þ a 1 1  to x ¼ a (outer edge) on each side of the neutral plane. In pSticking 0 2 h 2l ln2l such cases, forging load is to be calculated from (11.31). to xS ¼ r00 À ¼ 2 • If xS ¼ a; the full sticking friction exists over the entire 1 1 1 þa ! ð11:61Þ interface of the rectangular plate with deformation-zone l 2 2l h length 2a and the dies, i.e. from x ¼ 0 (centre line) to 1 À ln x ¼ a (outer edge) on each side of the neutral plane. In such cases, forging load is to be evaluated from (11.46). Now the total forging load, PTotal; can found from the summation of the forging load for Coulomb’s sliding fric- • If a [ xS [ 0; the mixed sticking–sliding friction condi- tions prevail at the interface of the rectangular plate with tion, ðPC:F:Þa to xS ; over an area from the outer edges of the deformation-zone length 2a and the dies, i.e. full sticking work-piece to the transition boundary of sliding to sticking friction occurs on each side of the neutral plane from x ¼ 0 (centre line) to x ¼ xS (sticking–sliding boundary) on each side of the neutral plane and the forging load for and sliding friction exists on each side of the neutral plane from x ¼ a (outer edge) to x ¼ xS (sliding–sticking sticking friction ðPStickingÞ0 to xS ; over an area from the neutral boundary). In such cases, forging load is to be deter- plane (centre line) of the work-piece to the transition mined from (11.63). boundary of sticking to sliding on each side of the neutral However, if the value of shear friction factor m at the interface is available, it is not required to calculate the value plane. Therefore, considering mean normal forging pressure of xS to determine average pressure, total load, distributions of pressure and longitudinal stress. Hence, (11.44) can be for sliding ðpC:F:Þa to xS ; given by (11.58) and that for sticking used to calculate the forging load. ðpStickingÞ0 to xS ; given by (11.60), we get ÀÁ PTotal ¼ ðPC:F:Þa to xS þ PSticking 0 to xÀS ¼ ðpC:F: Þa to xS&Â2ða À xSÞw þ pSticking Á xS Âð2xS Þw ¼ 2lða À ' !  r00 hw 0 to l exp xSÞ À h 1 þ r00xSw 1 þ xS l h ð11:62Þ If we substitute the value of xS; from (11.54) into (11.62), 11.8 Plane Strain Forging of Strip we get with Inclined Dies PTotal ¼ r00hw   ! Press forging under plane strain condition of a thin strip with l exp ln 1 À1 variation in height from its one end to the other will be  2l  ! considered here. Forging dies overlap the job but are mutu- ally inclined at a small angle. Suppose that forging dies are þ r00 w a À h ln 1 1 1 À 1 ln 1 þa inclined at angles a1 and a2 to the horizontal plane, as shown 2l 2l l 2 2l h in Fig. 11.32. On the strip, the inclined dies exert a pressure p & '  &   '! normal to the die–job interface, which will be called the die h 1 h 1 1 1 1 þa pressure. The vertical plane in the strip, where the material is ¼ r00 w l 2l À 1 þ a À 2l ln 2l l 1 À 2 ln 2l h stationary is called the neutral plane or neutral surface of the work-piece. During forging of the strip with dies, the material ð11:63Þ flows outward from the neutral plane in the longitudinal direction, which leads to tangential shearing frictional stres- 11.7.4 Selection of Proper Equation for Forging ses at the die–job contact surfaces. As the material is free to Load flow in both directions from its neutral plane, friction stress opposing the flow will change its direction in both sides of If the coefficient of friction l at the interface is given for the neutral plane. However, the presence of frictional shear plane strain forging of a rectangular plate, then to select stress s leads to longitudinal stress rx that acts on the vertical proper equations for determinations of average pressure, face of the work-piece. To make the analysis simpler, the total load, distributions of pressure and longitudinal stress, following assumptions are made: the value of xS has to be first evaluated using (11.54) to know the frictional conditions at the interface. Depending on the value of xS ; there may be three different conditions of friction at the die–plate interface, which are as follows:

496 11 Forging Fig. 11.32 Stresses acting on a Die thin strip forged in plane strain with inclined dies and geometry p α1 details of die and job μp p α1 σx μp h2 Job σ x+ dσx x σ x+ dσx σx h+dh dx h dx μp dx x p h1 μp x=0 p α2 x = L α2 Die x Neutral plane (i) Plane strain condition is assumed, where spread of the strip, which is assumed to be the origin of the coordinate work-piece is assumed to be negligible, i.e. width w ¼ system, shown by x ¼ 0 in the diagram, where x is the constant. horizontal axis that increases from the thinnest side to the thickest side of the strip. Figure 11.32 shows that these (ii) Homogeneous deformation is assumed, i.e. plane longitudinal stresses along with the die pressure p and fric- vertical section of the work-piece before forging tion stress lp are acting on the slab elements. remains plane during and after forging and hence, internal redundant deformation is neglected. Let the vertical pressure exerted by the inclined dies on the strip is pv : The vertical pressure pv is assumed to be (iii) Strain hardening of the work-piece is neglected, i.e. uniform throughout a transverse section and considered as a the plane strain flow stress or deformation resistance principal stress. Hence, the vertical equilibrium of forces for r00 of the material being forged is assumed to remain the slab elements of the work-piece will be: constant. For platens inclined at an angle (iv) It is assumed that Coulomb’s law of sliding friction holds good at the contact surface between the inclined a1 : pvdxw ¼ p cos a1 dx w Ç lp sin a1 dx w; dies and the thin strip. cos a1 cos a1 In other words, s ¼ lp; where s is the tangential frictional stress at the contact surface, l is the coef- Or; pv ¼ p Ç lp tan a1 ð11:64aÞ ficient of friction and p is the die pressure. For platens inclined at an angle (v) The coefficient of friction, l ; is constant over the whole surface of contact and assumed to be low. a2 : pvdxw ¼ p cos a2 dx w Ç lp sin a2 dx w; cos a2 cos a2 (vi) The angles of inclination a1 and a2 of the platens with the horizontal plane are assumed to be less than the angle of Or; pv ¼ p Ç lp tan a2 interface friction, i.e. a1\\ tanÀ1 l and a2\\ tanÀ1 l: ð11:64bÞ (vii) The longitudinal stress rx is constant across the cross-section of the strip and is a principal stress since where the upper negative sign ‘−’ indicates the thinner side of no shear stress act on the vertical face of the strip. the neutral plane of the strip and the lower positive sign, ‘+’ is for the thicker side of the neutral plane of the strip. Since l On both sides of the neutral plane within the thin strip and a1 or a2 are small quantities and l tan a1 ( 1; and being forged, let us consider slab elements of thickness dx l tan a2 ( 1, lp tan a1 or lp tan a2 can be neglected, and and width w ; whose height varies from h to h þ dh; as from (11.64a) or (11.64b), it can be written that pv % p: shown in Fig. 11.32. According to free-body equilibrium Hence, p can be considered as the principal stress. approach, the longitudinal stresses are, respectively, rx and rx þ drx at an arbitrary distances x and x þ dx; from Taking equilibrium of forces in the horizontal direction one extreme side of the strip, say the thinnest side of the for the slab element located on the thinner side of the neutral plane of the strip, we get:

11.8 Plane Strain Forging of Strip with Inclined Dies 497 rx hw À ðrx þ drxÞðh þ dhÞw þ p sin a1 dx w where cos a1 r00 ¼ compressive flow stress in plane strain condition dx dx  pffiffi þ p sin a2 cos a2 w þ lp cos a1 cos a1 w ¼ 2= 3  compressive uniaxial flow stress in þ lp cos a2 dx w ¼ 0 homogeneous strain condition cos a2  pffiffi ¼ 2= 3 r0: Since the product drxdh is a very small quantity, so it can be neglected. Further cancelling w from each term, we obtain Since r00 ¼ constant, therefore drx ¼ dp: Now denoting Àrxdh À hdrx þ pðtan a1 þ tan a2Þdx þ 2lpdx ¼ 0 ð11:65Þ 2l ¼ BÃ; we get from (11.69) and (11.70b) ðtan a1 þ tan a2Þ Similarly, horizontal equilibrium of forces for the slab ÀOpr;ÀhZrd00pÁ¼dhrþ00 dhhdpÆÀBZÃppddhhǼpBÀrÃ00dhƼBÃ0p;Ádh; element located on the thicker side of the neutral plane of the strip is given by Àrxdh À hdrx þ pðtan a1 þ tan a2Þdx À 2lpdx ¼ 0 ð11:66Þ Or; r00 dp ¼ dh; Or; r00 Æ BÃp h Now combining (11.65) and (11.66) horizontal equilib- and Z r00 Z rium of forces for both sides of the neutral plane of the strip Z dp þ dh; is given by þ BÃp þ ¼ h dh; dpÀ Z À BÃpÀ ¼ h rxdh þ hdrx À pðtan a1 þ tan a2 Æ 2lÞdx ¼ 0 ð11:67Þ where p þ indicates the forging pressure towards the thinner side of the neutral plane of the strip and pÀ indicates the If we assume that for the platen with inclination angle of a1, the fraction of the elemental difference in height dh over forging pressure towards the thicker side of the neutral plane thickness dx of the elemental slab of the strip is ‘y’, then of the strip. If C þ and CÀ are, respectively, the constants of yðdh=dxÞ ¼ tan a1: Similarly, for the platen with inclination angle of a2, the other fraction of the elemental difference in integration for the thinner and the thicker sides, then height dh over thickness dx is ‘ð1 À yÞ’, and hence ð1 À yÞðdh=dxÞ ¼ tan a2: Addition of these two fractions 1 lnÀr00 þ BÃp þ Á ¼ ln h þ C þ ; gives the total dh as Bà lnÀr00 BÃpÀÁ and À 1 À ¼ ln h þ CÀ; Bà dh ¼ ðtan a1 þ tan a2Þdx ð11:68Þ ) r00 þ BÃp þ ¼ A þ hBà ð11:71aÞ Substituting the value of dh from (11.68) into (11.67), we And r00 À BÃpÀ ¼ AÀhÀBà ð11:71bÞ obtain rxdh þ hdrx À pdh Ç p 2l dh ¼ 0 ð11:69Þ where A þ ¼ expðBÃC þ Þ; and AÀ ¼ expðÀBÃCÀÞ: þ ðtan a1 tan a2Þ Let, h1 the thickness of the plate at the extreme thinnest h2 end and where the upper negative sign ‘−’ indicates the thinner side the thickness of the plate at the extreme thickest of the neutral plane of the strip and the lower positive sign, end ‘+’ is for the thicker side of the neutral plane of the strip. The die pressure p and the longitudinal stress rx, which At the free ends of the strip where h ¼ h1 and h ¼ h2; the are considered as principal stresses, are shown as compres- longitudinal stress rx ¼ 0: Therefore, from (11.70a), the die pressure at both ends of the strip will be given by sive stresses in Fig. 11.32. If we consider the compressive stresses to be positive, then r1 ¼ p and r2 ¼ rx in Von p þ ¼ r00 at h ¼ h1 ð11:72aÞ Mises’ yielding criterion for plane strain condition, given by (1.88b). Hence, the yielding criterion can be written as And pÀ ¼ r00 at h ¼ h2 ð11:72bÞ p À rx ¼ r00 ð11:70aÞ Applying, respectively, the above boundary conditions Or, rx ¼ p À r00 ð11:70bÞ given by (11.72a) and (11.72b) into (11.71a) and (11.71b), we get the values of the integration constants A þ and AÀ; which are:

498 11 Forging Aþ ¼ r00 ð1 þ BÃÞ ð11:73aÞ where, Bà ¼ 2l ; h1Bà ðtan a1 þ tan a2Þ  Bà  Bà AÀ r00ð1 À BÃÞ Þ hNS BÃÞ h2 ¼ And ¼ hÀ2 Bà ð11:73bÞ ) ð1 þ Bà þ ð1 À 2; h1 hNS Or; ð1 þ BÃÞÀhBNÃSÁ2 þ ð1 À BÃÞðh1h2ÞBà ¼ À2h1Bà ÁhNBÃS; Substituting, respectively, the values of A þ and AÀ from Or; ð1 þ BÃÞÀhBNÃSÁ2ÀÀ2h1Bà ÁhBNÃS þ ð1 À BÃÞðh1h2ÞBà ¼ 0; (11.73a) and (11.73b) into (11.71a) and (11.71b), we get  pþ  1¼r¼00ðBrhÃ00B1ðÃþhBÀ2Ã1BÞÀÃh1BÃÞ;hÀBà ) hNS ¼ 8 þ 2h1BÃ Æ q4ffiffiffihffiffi21ffiffiffiBffiffiÃffiffiffiÀffiffiffiffiffi4ffiffiðffiffi1ffiffiffiÀffiffiffiffiffiBffiffiffiÃffiffi2ffiÞffiffiðffiffihffiffi1ffiffihffiffiffi2ffiffiÞffiBffiffiffiÃffi=9B1à r00 Bà r00 þ1 < : and r00 Bà pÀ 2ð1 þ BÃÞ ; r00 À ; Bà pþ ðBà hBà Considering only positive value, we get r00 hB1 à Or; þ 1 ¼ þ 1Þ ; 8 q4ffiffiffihffiffi12ffiffiBffiffiÃffiffiffiÀffiffiffiffiffi4ffiffiðffiffi1ffiffiffiÀffiffiffiffiffiBffiffiffiÃffiffi2ffiÞffiffiðffiffihffiffi1ffiffihffiffiffi2ffiffiÞffiffiBffiffiÃffi=9B1à <2h1Bà Bà pÀ À ¼ ðBà À 1Þ hÀBà ; ) hNS ¼ : þ r00 1  h Bà À h2ÀBà and 1 2ð1 þ BÃÞ ; h1 1 ) pþ ¼ Bà þ Bà ð11:74Þ ð11:77Þ r00 Bà pÀ Bà À 1  Bà 1 1 1 À Bà  Bà It is to be noted that the peak forging pressure occurs at r00 Bà h2 Bà Bà Bà h2 And ¼ þ ¼ À h the neutral plane, where h ¼ hNS: h To find the location of the neutral plane, let us assume ð11:75Þ that it is located at a distance x ¼ xNS; from the thinnest side of the strip, where x ¼ 0: Since at x ¼ 0; h ¼ h1, and at If L is the deformation-zone length of the strip, then x ¼ xNS; h ¼ hNS, hence from (11.68), we can write integration of (11.68), between the limits of x ¼ 0; h ¼ h1 and x ¼ L; h ¼ h2; gives xZ¼xNS 1 hZ¼hNS þ ZL Zh2 dx ¼ ðtan a1 tan a2Þ dh dx ¼ 1 dh x¼0 h¼h1 þ ðtan a1 tan a2Þ ) xNS ¼ hNS À h1 ð11:78Þ ðtan a1 þ tan 0 h1 a2Þ ) L ¼ ðtan h2 À h1 a2Þ ð11:76Þ a1 þ tan 11.9 Forging of Flat Circular Disk 11.8.1 Strip Thickness at Neutral Plane and Its Consider a circular flat disk of uniform height that is being Location press forged in axial direction with flat dies. Let the radius of the circular disk is a and the uniform height is h. The flat The neutral plane is such a plane of the plate where the dies exert a normal pressure p on the flat surfaces of circular disk. The uniaxial compressive flow stress or deformation material is stationary when the strip is being forged with the resistance of the material of the disk is assumed to be r0: The vertical centreline of the circular disk defines its neutral dies. As the material always flows outward from this neutral axis, where the material is stationary but the material flows outward from this neutral line in the radial direction when plane, the friction acts inward and changes its direction in the disk is being forged with the dies. Outward radial flow of the material during the process of its compression leads to both sides of this plane. To find the location of this neutral tangential shearing frictional stresses at the die-disk contact plane, (11.74) is equated to (11.75), i.e. p þ =r00 ¼ pÀ=r00: If surfaces. Therefore, friction stresses opposing the outward the strip thickness h ¼ hNS at the neutral plane, then 1 þ Bà  Bà 1 1 1 À Bà  h2 Bà Bà hNS Bà Bà Bà ; À ¼ À h1 hNS

11.9 Forging of Flat Circular Disk 499 radial flow will always act towards the centre of the circular p disk. The frictional shear stress, s, leads to radial stress, rr; and circumferential stress, rh; in the material. The stresses Neutal σθ rr and rh are zero at the edges and builds up to a peak value axis at the centre of the disk. To simplify our analysis, the fol- lowing assumptions are made: σr + dσr σθ dθ (i) Material being forged is assumed to be isotropic and dθ σr τ homogeneous. h σθ (ii) Homogeneous deformation is assumed, i.e. plane 90°– dθ vertical section before forging remains plane during r dr 2 and after forging and hence, internal redundant work is neglected. σθ a (iii) Elastic strain is neglected because plastic deforma- tion involved in forging is quite large compared to Fig. 11.33 Stresses acting on an elemental sector of a forged flat insignificant elastic deformation. circular disk (iv) As forging is a hot working operation, so it is quite ðrr þ drr Þ hðr þ drÞdh Àrr hr dh þ 2s rdh dr logical to assume that there is no strain hardening of  the work-piece, i.e. the uniaxial compressive flow 90 À dh stress or deformation resistance r0 of the material À 2rh cos 2 h dr ¼ 0 being forged is assumed to remain constant. Other- wise, an average value of flow stress will be Or; rrhr dh þ rrh dr dh þ drrhr dh þ drrh dr dh assumed, it is important to remember that in hot forging the flow stress will depend very much upon À rr hr dh þ 2sr dh dr À 2rh sin dh h dr ¼ 0 ð11:79Þ the strain rate used. 2 (v) There is no barrelling of the edges of the work-piece. For very small angle, assuming sinðdh=2Þ % dh=2 and (vi) Since the height of the work-piece is small, so it can neglecting the term drrh dr dh in (11.79) as it is a very small quantity, it follows from (11.79): be assumed that the normal pressure, p, remains constant through the height of the work-piece and is rrhr dh þ rrh dr dh þ drrhr dh a principal stress; although this assumption is not really correct, because the shear stress s acts on the À rr hr dh þ 2sr dh dr À 2rh dh h dr ¼ 0 ð11:80Þ planes on which the normal pressure p is acting. 2 (vii) The radial stress rr and circumferential stress rh are also considered to be principal stresses. If we consider the compressive stresses to be positive, the (viii) The maximum value of the forging load is attained at the end of the operation. normal pressure p exerted by flat dies on the work-piece is a Let us consider the state of stresses on a vertical slab positive principal stress and so, p ¼ r1: Since radial stress rr element with thickness dr and height h, inside the circular and circumferential stress rh are both shown as compressive disk at an arbitrary distance r from the vertical centre line of in Fig. 11.33, so both of them are positive principal stresses. the disk, which is taken as the origin, r ¼ 0; of the coordi- nate system. According to free-body equilibrium approach, Hence, we can write rr ¼ r2; and rh ¼ r3: From the axial the radial stresses are, respectively, rr and rr þ drr at an symmetry of the disk, since change in the circumferential arbitrary radial distances r and r þ dr; from the centre-line, i.e. r ¼ 0: The radial and circumferential stresses along with strain ¼ deh ¼ de3 ¼ de2 ¼ der ¼ change in the radial the normal pressure p and friction stress s are acting on the strain, so according to the Levy–Mises (1.84b) and (1.84c), slab element, as shown in Fig. 11.33. The slab element makes an angle dh at the centre of the disk. it yields the following (11.81a). With reference to Fig. 11.33, equilibrium of forces in the de 1 ! de 1 ! radial direction gives r 2 rr Þ r 2 deh ¼ rh À ðp þ ¼ der ¼ rr À ðp þ rhÞ Or; rh À 1 rr ¼ rr À 1 rh; or; 3 rh ¼ 3 rr ; 2 2 2 2 ) Circumferential stress rh ¼ Radial stress rr ð11:81aÞ

500 11 Forging Since rh ¼ rr, from (11.80) we get Therefore, normal forging pressure for Coulomb’s sliding friction is: &' 2l ða À rÞ σ rhrdθ + σ rhdrdθ + dσ rhrdθ − σ rhrdθ ð11:81bÞ pC:F: ¼ r0 exp h ð11:86Þ + 2τ rdθdr − σ rhdrdθ = 0 Therefore, radial or circumferential stress for coulomb’s As rdh ¼6 0, so dividing (11.81b) by r dh, we get sliding friction is: s ð11:82Þ ðrrÞC:F: or ðrhÞC:F:¼ pC:F: À &r0 '! hdrr þ 2sdr ¼ 0; or, drr ¼ À2 h dr 2l ða À rÞ The Von Mises’ yield criteria, as shown in (1.64), is: ¼ r0 exp h À1 ð11:87Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffi r0; The variations of pC:F: and ðrrÞC:F: or ðrhÞC:F: obtained ðr1 À r2Þ2 þ ðr2 À r3Þ2 þ ðr3 À r1Þ2 2 from (11.86) and (11.87) are plotted over the diameter 2a of which in the present context becomes the disk as shown in Fig. 11.34. Both stresses build up to a qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðp À rrÞ2 þ fÀðp À rrÞg2 ¼ pffiffi r0; maximum at the centre of the disk and the distributions of 2 stresses are symmetrical about the centreline of the disk. This where r0 is the uniaxial deformation resistance or flow stress characteristic rise of stresses over the length of contact in compression. between the material and the die is called friction hill. In this ) p À rr ¼ r0 ð11:83Þ figure, the vertical centreline of the disk defines the neutral axis. The diagram of friction hill shows that at the outer Since r0 ¼ constant, so from (11.83), we get drr ¼ dp: periphery of the disk, where r ¼ a; the minimum values of Making this substitution into (11.82), we arrive at pC:F: and ðrrÞC:F: or ðrhÞC:F: are, respectively, given by dp ¼ À2 s dr ð11:84Þ ðpC:F:ÞMinimum ¼ r0 and ½ðrrÞC:F:ŠMinimum ¼ ½ðrhÞC:F:ŠMinimum ¼ 0; h and at the centreline, i.e. at the neutral axis of the disk, where r ¼ 0 ; the maximum values of pC:F: and ðrrÞC:F: or ðrhÞC:F: are, respectively, given by 11.9.1 Coulomb Sliding Friction   2la lD ðpC:FÞMaximum ¼ r0 exp h ¼ r0 exp h ð11:88Þ Let us consider that low frictional condition exists at the Âà Âà 2la À ! ðrrÞC:F: Maximum ¼ ðrhÞC:F: Maximum¼ r0 exp 1 contact surface between the forging dies and the circular h disk, where Coulomb’s law of sliding friction holds good.  ! So, the tangential shearing frictional stress s is related to the lD applied normal forging pressure p by Coulomb’s law of ¼ r0 exp h À1 sliding friction as s ¼ lp, where l is the coefficient of friction and is assumed to be the same at all points on the ð11:89Þ surface of contact. Now putting s ¼ lp in (11.84), we get where D ¼ 2a ¼ the diameter of the circular disk or defor- mation zone. dp ¼ À2 l dr; Pmax ph Friction hill (σr ) C.F. or (σθ )C.F. and on integration; ln p ¼ À 2l r þ C1 ð11:85Þ σ 0 PC.F. h 2μa where C1 is the integration constant, which can be evaluated σ 0exp h from the following boundary condition At the outer free surface of the disk, where r ¼ a; rr ¼ 0; σ0 0 Therefore from (11.83), p À 0 ¼ r0 ; or p ¼ r0: Substituting the above into (11.85), ln r0 ¼ À 2l a þ C1; r h aa 2l a; Or, C1 ¼ ln r0 þ h Neutral plane Substituting the value of C1 into (11.85), we get Fig. 11.34 Friction hill showing distributions of normal forging pressure pC:F: and radial stress ðrrÞC:F: or circumferential stress ðrhÞC:F: ln p ¼ À 2l r þ ln r0 þ 2l a; or p ¼ 2l ða À rÞ: with Coulomb friction over the total diameter 2 a of a circular flat disk h h ln r0 h

11.9 Forging of Flat Circular Disk 501 11.9.1.1 Average Pressure and Total Load Integrating both sides of (11.92), we get ð11:93Þ The mean height of the friction hill is the average defor- mation pressure, pC:F:; which is obtained from (11.86) as p ¼ Àr0 p2mffiffi r þ C2 ð11:90Þ follows: 3h Mean forging pressure for Coulomb’s sliding friction is: R0aRp0aC:2Fp: 2rpdrr dr 2p Ra r0 & À ' exp 2l ða rÞ rdr 0 h pC:F: ¼ ¼ 2p a2 2r0 (Z & ' 2Z Z & ' ! )a a2 r 2l rÞ dr d 2l rÞ dr dr ¼ ða À À ða À ¼ 2r0 exp r exp ¼ a2 \"(rÀÀÀ22arhllh2hleexxepphx&ðp0&2Þhl2Àhlða(ðaÀÀrÀÞr'2Þhl'ÀÀ2eZÀxdp2rðhl01ÞÀÀÀ2hÀl2hl2hlhexepx2&epx&2phl2hlð2aðlhaÀaÀrÞ)r'Þ#'!0d!r0a)0a ¼ 2r0 a2 2r0 a2 ¼ 2r0  h 2 À 2la À 1 þ 2la! a2 2l exp hh ) pC:F: ¼ 2r0\" 2lha2  ! 2la À 2la À 1 exp hh# ¼ 2r0 expfð2laÞ=hg À fð2laÞ=hg À 1 fð2laÞ=hg2 Hence, the total forging load for Coulomb’s sliding where C2 is the integration constant. It is evaluated from the friction is: boundary condition that at x ¼ a; rr ¼ 0; and so, from PC:F: ¼ mean pressure  contact area ¼ pC:F: pa2 (11.83) p ¼ r0: Therefore, from (11.93) ¼ r0  h 2  ! p2mffiffi 1 þ p2mffiffia ! 3h 3h exp 2la À 2la À 1 pa2 r0 ¼ Àr0 a þ C2; or C2 ¼ r0 : 2 la hh ! h2  pr0 l 2la À 2la À 1 Substituting the value of C2 into (11.93), the normal ¼ 2 exp deformation pressure considering friction factor is given by hh h2  ! pr0 l lD À lD À 1 ¼ 2 exp ð11:91Þ pF:F: ¼ Àr0 p2mffiffi r þ r0 þ r0 p2mffiffi a hh 3h ! 3h ¼ r0 1 þ 2mpðaffiffiÀ rÞ ð11:94Þ 3h 11.9.2 Sliding with Shear Friction Factor Now the radial and circumferential stresses considering and Sticking Friction friction factor, i.e. ðrrÞF:F: and ðrhÞF:F:; are obtained from Let us consider the case of sliding where the interface (11.83) and (11.94) as shearing frictional stress s is described by the shear friction factor m; i.e. s ¼ m k; as shown in (10.24). Hence, substi- ðrrÞF:F:¼ðrhÞF:F:¼ pF:F: À r0 tution of s ¼ m k into (11.84) gives ¼r0 2mpðaffiffiÀ rÞ ð11:95Þ 3h dp ¼ À 2mk dr ¼ À 2prffi0ffi m dr ð11:92Þ The minimum values of pF:F: and ðrrÞF:F: or ðrhÞF:F: at the h 3h outer periphery of the disk, where r ¼ a; are, respectively, pffiffi given by ðpF:F:ÞMinimum ¼ r0 and ½ðrrÞF:F:ŠMinimum ¼ where k ¼ r0= 3 according to Von Mises’ yielding criterion. ½ðrhÞF:F:ŠMinimum ¼ 0: The maximum values of pF:F: and

502 11 Forging ðrrÞF:F: or ðrhÞF:F: at the centreline, i.e. at the neutral axis of (σ r )sticking or (σθ )sticking σ0 Psticking the disk, where r ¼ 0, are, respectively, given by !! 1 þ p2mffiffia 1 þ pmffiDffi ðpF:FÞMaximum¼ r0 3h ¼ r0 3h ð11:96Þ 2a σ0 1 + 2a σ0 √–3 h √–3 h ÂÃ ÂÃ ðrrÞF:F: Maximum ¼ ðrhÞF:F: Maximum 0 σ0 ¼ r0 p2mffiffia ¼ r0 pmffiDffi ð11:97Þ 3h 3h a a There is a limiting value of the shear stress s at the interface Neutral plane below which sliding friction can exist at the die–work-piece Fig. 11.35 Friction hill showing distributions of normal forging pressure pSticking and radial stress ðrrÞSticking or circumferential stress interface. When the friction is high making the interfacial ðrhÞSticking under sticking frictional condition over the total diameter 2 a of a circular flat disk shear stress s to reach its limiting value, then sticking friction occurs at the die–work-piece interface. This limiting valupe ffiiffis the shearing yield stress of the work-piece, k ¼ r0= 3: Therefore, in cases of sticking friction s ¼ k; i.e. the shear friction factor m ¼ 1: Assume that the sticking friction regime where D ¼ 2a ¼ the diameter of the deformation zone of circular disk. extends over the whole interface. Hence, substitution of m ¼ 1 into (11.94) and (11.95) gives, respectively, the normal forging pressure, pSticking ; and the radial stress, ðrrÞSticking; or 11.9.2.1 Average Pressure and Total Load the circumferential stress,ðrhÞSticking; for conditions of stick- Considering friction factor, the average forging pressure, ing friction, which are: pF:F:; is obtained from (11.94) as follows: 1 þ 2ðpa Àffiffi rÞ ! R0aRp0aF:2Fp: 2rpdrr dr 3h pSticking ¼ r0 ð11:98Þ pF:F: ¼ ðrr ÞSticking ¼ ðrhÞSticking¼ r0 2ðpa Àffiffi rÞ ð11:99Þ ¼ 2pr0 Ra 1 þ p2mffiffia À ! 3h ¼ 3h p2mffiffir r dr 0 3h pa2 p2mffiffi a3! Equations (11.98) and (11.99) show that the distributions 2r0 a2 þ p2mffiffia a2 À 3h 3 a2 2 3h 2 ! of normal forging pressure pSticking and radial stress ðrrÞSticking or circumferential stress ðrhÞSticking are not only ¼ r0 1 þ 6mapÀffiffi4ma symmetrical about the central line but also linear with dis- 3 !3h ! tance from the centre line of the disk (the location of the 1 þ 2pmffiaffi 1 þ mpDffiffi ¼ r0 3 3h ¼ r0 3 3h ð11:102Þ neutral axis), as illustrated in the friction hill diagram of Fig. 11.35. The diagram of friction hill shows that at the Now the total forging load considering friction factor, outer periphery of the disk, where r ¼ a, the minimum PF:F:, can be similarly be found from the product of the mean forging pressure and the area at the die–job interface. values of pSticking and ðrrÞSticking or ðrhÞSticking are, respec- tively, equal to r0 and 0; which are same as those shown ! above and at the centreline, i.e. at the neutral axis of the disk, ) PF:F: ¼ pF:F: pa2 ¼ r0pa2 1 þ 2pmffiaffi ¼ r0pD2 1 þ mpDffiffi ! 3 3h where r ¼ 0, the maximum values of pSticking and ðrrÞSticking or ðrhÞSticking are, respectively, given by ð11:103Þ ÀÁ 1 þ p2ffiaffi ! ! 4 3 3h pSticking Maximum¼ r0 3h 1 þ pDffiffi ¼ r0 Equation (11.103) can be used to calculate the forging 3h load if the value of shear friction factor m at the interface is available. ð11:100Þ Hence for conditions of sticking friction, the average hi hi forging pressure, pSticking, and the total forging load, PSticking, can be obtained by substituting m ¼ 1 into (11.102) and ðrr ÞSticking ¼ ðrhÞSticking Maximum (11.103), which are: Maximum ð11:101Þ ¼ r0 p2ffiaffi ¼ r0 pDffiffi 3h 3h

11.9 Forging of Flat Circular Disk 503 1 þ p2affiffi !! pSticking r=0 3 3h 1 þ pDffiffi p Neutral plane pSticking ¼ r0 ¼ r0 3 3h ð11:104Þ rs =a – h In 1 ð11:105Þ C.F. 2μ √─3 1 þ p2affiffi ! 3 3h ) PSticking ¼ pSticking pa2 ¼ r0pa2 ¼ r0pD2 ! 2 μa σ0 σ0 1+ 2a pDffiffi h √─3 μ √─3h 1þ σ0 exp 4 3 3h σ0 0 11.9.3 Mixed Sticking–Sliding Friction aa 2rs Frequently, a condition exists where sticking occurs at the Fig. 11.36 Friction hill showing distributions of normal forging central region of the work-piece and sliding occurs at the pressures, pC:F: and pSticking, respectively, for conditions of Coulomb’s outer regions of the work-piece with coefficient of sliding sliding and full sticking, resulting in mixed sticking–sliding frictional friction l ¼ constant. This occurs when the friction is somewhat lower than that required to cause sticking friction conditions over the total diameter 2a of a circular flat disk over the whole interface. Since the normal forging pressure p is the minimum at the outer free surfaces and increase Fig. 11.36. This friction hill diagram shows that around the towards the centre of the work-piece, the magnitude of the frictional stress lp will also increase inwards from the outer centreline from r ¼ a to r [ rS; the sliding pressure is lower periphery of the work-piece. But at a certain distance away from the edge towards the centre of the work-piece, if the than the pressure that would be required for sticking and so, frictional stress lp reaches a value equal to the shearing yield stress k of the work-piece, then further increase of the sliding friction prevails there, whereas around the centreline frictional stress is not possible. Under such condition, the central zone of the work-piece will be controlled by the from r\\rS to r ¼ 0; the sticking pressure is lower than the sticking friction condition. So, the equilibrium equation (11.84) is expressed with the following conditions: pressure that would be required for sliding and hence, full dp ¼ À2 s dr sticking friction occurs there. Only at the distance of r ¼ rS h around the neutral axis of the disk, the sliding pressure is where s ¼ lp if lp\\k; and s ¼ k if lp ! k: Suppose, rS is the radius measured from the neutral axis, equal to the sticking pressure, i.e. ðpC:F:Þr ¼ rS ¼ ðpStickingÞr ¼ rS . Normal forging pressure for Coulomb sliding friction given i.e. centreline ðr ¼ 0Þ of the work-piece, at which the tran- sition from sticking to sliding occurs. Hence, frictional con- by (11.86) is repeated below, which is valid from r ¼ a to dition at the die–job interface will be such that Coulomb’s sliding friction prevails from r ¼ a to r ¼ rS, and full r ¼ rS for mixed sticking–sliding frictional conditions: sticking friction occurs from r ¼ rS to r ¼ 0 around the centreline. This can be expressed mathematically as follows: &' 2l ða À rÞ ðpC:F:Þa to rS ¼ r0 exp h ð11:86Þ Now let us find the solution for the sticking zone. For the sticking zone, by substituting (11.106b) into (11.84) and then integrating both sides, we get Z ÀÁ Z À 2k dr ¼ Z p2rffiffi0 dr dpSticking 0 to rS À 3h ¼ h ¼ p2rffiffi0 Z À 3h dr For rS r a; s ¼ lp ð11:106aÞ pffiffi where k ¼ r0= 3; according to Von Mises’ yielding criterion. For 0 r rS; s ¼ k ð11:106bÞ ) ÀÁ ¼ À p2rffiffi0 r þ C3 ð11:107Þ pSticking 0 3h to rS For the sliding zone, substitution of (11.106a) into where C3 is the integration constant, which can be evaluated (11.84) and then integration will ultimately result in the from the boundary condition of the sticking zone that solution same as that of (11.86), because the boundary extends from r ¼ 0 to r ¼ rS around the neutral axis. At the condition for evaluation of the integration constant is the boundary of r ¼ rS, same. Only the validity range of (11.86) is different, which varies from r ¼ a to r ¼ rS; instead of r ¼ a to r ¼ 0: The sr¼rS ¼ lðpC:F:Þr¼rS ¼ k¼ pr0ffiffi ; or, ðpC:F:Þr¼rS ¼ prffiffi0 variations of the normal forging pressures for conditions of 3 3l Coulomb’s sliding and full sticking are drawn in the same plot over the total diameter 2a (or D) of the disk as shown in ð11:108Þ

504 11 Forging Again since at the distance r ¼ rS; ðpC:F:Þr¼rS ¼ We can obtain the minimum and maximum values of ÀÁ pSticking r¼rS ; then substitution of (11.108) into (11.107) ðpStickingÞ0 to rS from (11.112) when Coulomb coefficient of sliding friction l reaches a maximum value under sticking gives the value of C3 ; this is: frictional condition, where the minimum value of interface  pr0ffiffi 1 2 normal pressure is considered as pmin ¼ r0. According to C3 ¼ 3 l þ h rS ð11:109Þ Von Misesp’ ffiyffi ielding criterion, this maximum value of l is Substituting the value of C3 from (11.109) into (11.107), lmax ¼ 1= 3 [see (10.25)]. Hence, substitution of r ¼ a, and r ¼ 0, into (11.112) gives, respectively, the minimum we get the normal forging pressure for sticking friction and maximum values of ðpStickingÞ0 to rS as condition from r ¼ 0 to r ¼ rS as \" ( )# ÀÁ pr0ffiffi 1 2 ! hÀ Á i ¼ pr0ffiffi 1pffiffi 1 À ln pffiffi À1 pffiffiÁ pSticking 0 3 l h pSticking 0 3 3 1= 3 to rS ¼ þ ðrS À rÞ ð11:110Þ to rS Minimum 1= 3  Again, substitution of (11.108) into (11.86) for ðpC:F:Þr¼rS ¼ r0 at r ¼ rS ; gives a solution for the location of the boundary ð11:113Þ between slipping friction and sticking friction, which is hÀ Á i pr0ffiffi \" ( ) # pSticking 0 3 1pffiffi 1 À1 pffiffiÁ 2a derived below. to rS Maximum ¼ À ln pffiffi  1= 3 þ 1= 3 3 h &' ¼ r0 1 þ p2ffiaffi ! prffi0ffi ¼ r0 exp 2l ða À rS Þ ; 3h 3l h ð11:114Þ 2l ln p1ffiffi ; Or, h ða À rS Þ ¼ 3l ) rS ¼ a À h ln p1ffiffi ð11:111Þ These minimum and maximum values given by (11.113) 2l 3l and (11.114) are shown in Fig. 11.36. Substituting the value of rS ; from (11.111) into (11.110), 11.9.3.1 Average Pressure and Total Load we get For Coulomb’s sliding friction from r ¼ a to r ¼ rS around the centreline of the disk, the average normal forging pres- ÀÁ pr0ffiffi 1 2a 1 p1ffiffi 2r ! sure, ðpC:F:Þa to rS ; is obtained from (11.86) as follows: pSticking 0 to rS 3 l h l 3l h ¼ þ À ln À  ¼ pr0ffiffi 1 1 À ln p1ffiffi þ 2ða À rÞ ! 3 l 3l h ð11:112Þ Ra ðpC:F:Þa to rS 2pr dr 2pr0 Ra & ' Ra exp 2l ða À rÞ rdx rS 2pr dr rS Àh Á ðpC:F:Þa to rS ¼ ¼ p a2 À rS2 rS( Z & ' Z Z & ' ! )a 2r0 2l rÞ dr dr exp 2l ¼ a2 À rS2 r ða À À ða À rÞ dr dr ¼ \" exp ða À h dr h '#a rS  & '  h 2 & 2r0 À rh 2l ða À rÞ À À 2l 2l rÞ a2 À rS2 2l exp exp rS h h 2r0    & ' a2 À rS2 ah rSh 2l ¼ À 2l À À 2l exp h ða À rS Þ À rS '# À2hl2 þ  h 2 &2l ða Þ 2l exp h  h 2 2la þ 2lrS exp &'& '! 2r0 2l h h 2l À 1 þ exp 2l rSÞ ¼ a2 À rS2 À  h ða À rSÞ h ða À ¼ 2lrS À r0 Á h2 & ' ! 2l À 2la À 1 2 a2 À rS2 l h þ1 exp h ða À rS Þ h ð11:115Þ

11.9 Forging of Flat Circular Disk 505 For sticking friction from r ¼ 0 to r ¼ rS around the • If rS 0; the Coulomb’s sliding friction exists over the entire interface of the circular disk of radius a and the centreline of the disk, the average normal forging pressure, dies, i.e. from the centre line, where r ¼ 0; to the outer periphery, where r ¼ a: In such cases, forging load is to ðpStickingÞ0 to rS ; is obtained from (11.110) as follows: be evaluated from (11.91). 2p R rS ÀÁ to rS r • If rS ¼ a; the full sticking friction exists over the entire ÀÁ 2ppStiRc0kriSngrd0r r2 À dr interface of the circular disk of radius a and the dies, i.e. pSticking 0 to rS ¼ 0 r2 þ 2rS 2 r3 from r ¼ 0 (centre line) to r ¼ a (outer periphery). In ¼ such cases, forging load is to be calculated from p2ffirffi 0 !rS (11.105). 3 rS2 2l þ h2 h 3 0 ! • If a [ rS [ 0; the mixed sticking–sliding friction condi- ¼ p2ffirffi 0 rS2 rS3 À 2rS3! ¼ 1 þ 2rS tions prevail at the interface of the circular disk of radius pr0ffiffi a and the dies, i.e. full sticking friction occurs around the 3 rS2 2l h 3h 3 l 3h neutral axis from r ¼ 0 (centre line) to r ¼ rS (sticking– sliding boundary) and sliding friction exists around the ð11:116Þ neutral axis from r ¼ a (outer periphery) to r ¼ rS (sliding–sticking boundary). In such cases, forging load Now the total forging load, PTotal ; can be found from the is to be determined from (11.117). summation of the forging load for Coulomb’s sliding fric- tion, ðPC:F:Þa to rS ; over an area around the centreline from the However, if the value of shear friction factor m at the outer periphery of the circular disk to the transition boundary interface is available, it is not required to calculate the value of rS to determine average pressure, total load, distributions of sliding to sticking and the forging load for sticking fric- of pressure and longitudinal stress. Hence, (11.103) can be used to calculate the forging load. tion, ðPStickingÞ0 to rS ; over an area around the centreline from the neutral axis (centre line) of the disk to the transition 11.10 Forging of Circular Disk by Conical Pointed Dies boundary of sticking to sliding. Therefore, considering mean Let us consider the press forging of a circular disk whose top normal forging pressure for sliding ðpC:F:Þa to rS given by and bottom faces are of conical shapes and which has a (11.115) and that for sticking ðpStickingÞ0 to rS given by vertical as well as a horizontal plane of symmetry. The (11.116), we get circular disk is being press forged in axial (vertical) direction with conical pointed dies whose top and bottom faces are of ÀÁ conical shapes. The apices of the conical forging dies form PTotal ¼ ðrðpP20CpC::FF::ÞÞlhaattoo2rrSSÂþp2ÀlhPar2SStiÀcþkinr1gS2Á0þetxoÀprSp&S2tihclkiðngaÁÀ0 torSrSÞÂ'pÀrS2 an angle a with the horizontal direction and lie along the ¼ vertical centre-line of the dies and of the circular disk, as 2la À ! shown in Fig. 11.37a. The radius of the circular disk is R0 ¼ h 1 and the height at its mid-section, i.e. along the vertical centre-line of the dies is h0: The height of the disk increases þ rp0pffirffi S2 1 2rS ! uniformly from its vertical centreline (mid-section) to its 3 l 3h periphery. On the disk, the conical pointed dies exert a þ pressure p normal to the die–job interface, which will be called the die pressure. The uniaxial flow stress or defor- ð11:117Þ mation resistance of the material of the disk is assumed to be r0 : The vertical centreline of the circular disk defines its Now we can substitute the value of rS, from (11.111) into neutral axis, where the material is stationary but the material (11.117) to obtain the total forging load, PTotal, for mixed flows outward from this neutral line in the radial direction sliding–sticking conditions prevailing at the interface of the when the disk is being forged with the dies. Outward radial flow of the material during the process of its compression dies and the circular disk. leads to tangential shearing frictional stresses at the die–job 11.9.4 Selection of Proper Equation for Forging Load If the coefficient of friction l at the interface is given for forging of a circular disk, then to select proper equations for determinations of average pressure, total load, distributions of pressure and longitudinal stress, the value of rS has to be first evaluated using (11.111) to know the frictional condi- tions at the interface. Depending on the value of rS; there may be three different conditions of friction at the interface of the dies and the circular disk, which are as follows:

506 11 Forging (a) (b) σθ p Neutral F σr + dσr axis E dθ Neutral μp axis D Conical die P σr μP C α h0 R0 h dσr h+dh h0 Job segment h h + dh Job σr σr α r dr AB r dr μP R0 Conical die P σθ r=0 r=0 (c) σθ (d) σθ P F E μp dθ Radial Slab element H direction 90º – (dθ/2 ) G σr + dσr σθ σr C D μp h A h + dh B σθ P Fig. 11.37 a Conical pointed dies and circular disk having conical disk with conical faces showing stresses acting on the slab element. end faces, along with slab element showing stresses except the c Only slab element with its geometry and stresses acting on it. d Angle circumferential stress, acting on it. b Segment of the forged circular formed by the circumferential stress rh with the radial direction contact surfaces. Hence, friction stress opposing the outward (iii) It is assumed that Coulomb’s law of sliding friction radial flow will always act towards the centre of the circular holds good at the contact surface between the conical disk. The frictional shear stress, s; leads to radial stress, rr; pointed dies and the circular disk. In other words, and circumferential stress, rh; in the material. To make the s ¼ lp; where s is the tangential frictional shear stress analysis simpler, the following assumptions are made: at the contact surface, l is the coefficient of friction and p is the die pressure on the work piece normal to (i) Homogeneous deformation is assumed, i.e., plane the contact surface. vertical section before forging remains plane during and after forging and hence, internal redundant (iv) The coefficient of friction, l; is constant over the deformation is neglected. whole surface of contact and assumed to be low. (ii) Strain hardening of the work-piece is neglected, i.e., (v) The angle a formed by the apices of the dies with the the flow stress or deformation resistance of the horizontal direction is assumed to be lower than or material being forged is assumed to remain constant. equal to the angle of interface friction, i.e. a tanÀ1 l:

11.10 Forging of Circular Disk by Conical Pointed Dies 507 (vi) Since the height of the work-piece is small, so it can in (11.119), we get be assumed that the die pressure p remains constant through the height of the work-piece σ rrdθ h + σ rdrdθ h + σ rrdθ dh +σ rdrdθ dh + dσ rrdθ h (vii) The radial stress rr and circumferential stress rh are + dσ rdrdθ h + dσ rrdθdh + dσ rdrdθ dh − σ rrdθ h + 2μ prdθ dr considered to be the principal stresses. dθ dθ dh − 2 p tanα rdθ dr − 2σ r 2 drh − 2σ r 2 dr 2 =0 Let us consider a slab element with thickness dr and As dh ¼6 0; so dividing the above equation by dh, we get height varying from h to h þ dh; within the circular disk being forged, as shown in Fig. 11.37a, b. According to rrrdh þ rrdrdh þ drrrh þ drrdrh þ drrrdh þ drrdrdh free-body equilibrium approach, the radial stresses acting on þ 2lprdr À 2p tan ardr À rrdrðdh=2Þ ¼ 0 the slab element are, respectively, rr and rr þ drr at an ð11:120Þ arbitrary radial distances r and r þ dr; from the vertical centre-line of the circular disk, which is taken as r ¼ 0: The The terms rrdrdh; drrdrh; drrrdh; drrdrdh and rrdrðdh=2Þ radial and circumferential stresses along with the die pres- can be neglected from the above (11.120), because they are sure p and friction stress lp acting on the slab element are the products of very small quantities. So, (11.120) can be shown in Fig. 11.37b–c. This slab element makes an angle rewritten as dh at the centre of the disk, as shown in Fig. 11.37b. rrrdh þ drrrh þ 2lprdr À 2p tan ardr ¼ 0; Let the vertical pressure exerted by the conical pointed dies on the circular disk is pv: The vertical pressure pv is assumed As r ¼6 0; so dividing the above equation by r, we get to be uniform throughout a transverse section and considered as a principal stress. Hence, the vertical equilibrium of forces rrdh þ hdrr þ 2lpdr À 2p tan adr ¼ 0 ð11:121Þ for the slab element of the work-piece is given by With reference to Fig. 11.37a, pvr dh dr ¼ p cos ar dh dr þ lp sin ar dh dr ; ðh À h0Þ=2 ¼ tan a; or; h ¼ h0 þ 2r tan a ð11:122aÞ cos a cos a r Or, pv ¼ p þ lp tan a ð11:118Þ Since l and a are small, lp tan a can be neglected and Since h0 ¼ constant and for a given (constant) apex cone from (11.118), it can be written as pv % p: Hence, p can be angle a of the die; differentiating (11.122a), we get considered as the principal stress. dh ¼ 2 tan adr ð11:122bÞ Referring to Fig. 11.37b, c, AC ¼ GE ¼ h ¼ an arbitrary height of disk at any radial distance, r; and BD ¼ FH ¼ Substituting for h and dh from (11.122a) and (11.122b) h þ dh ¼ an arbitrary height of disk at any radial distance, into (11.121), we get r þ dr, from the neutral axis of the disk, i.e. r ¼ 0: rr2 tan adr þ ðh0 þ 2r tan aÞdrr þ 2lpdr À 2p tan adr ¼ 0 With reference to Fig. 11.37b–d, taking equilibrium of ð11:123Þ forces in the radial direction for the slab element: ðrr þ drrÞðarea BDFHÞ À rrðarea ACEGÞ Consider the die pressure p as the principal stress r1, the radial stress rr as the principal stress r2; and the circum- þ 2lp cos aðarea CDFE or ABHGÞ ferential stress rh as the principal stress r3; where rr ¼ rh; (already shown earlier). All stresses are compressive in À 2p sin aðarea CDFE or ABHGÞ nature as shown in Fig. 11.37b–c and consider the com- 90 À dh pressive stresses to be positive. Then the Von Mises’ À 2rh cos 2 ðarea ABDC or GHFEÞ ¼ 0; yielding criterion, as given by (1.64), becomes: Or; ðrr þ drrÞfðr þ drÞdh ðh þ dhÞg  dr À rrðrdh hÞ þ 2lp cos a rdh cos a  dr dh  h þ h þ dh  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi cos a 2 dr ðp À rrÞ2 þ fÀðp À rrÞg2 2 À 2p sin a rdh À 2rh sin ¼0 ¼ r0; 2 since r2 ¼ rr ¼ rh ¼ r3: ð11:119Þ Or, p À rr ¼ r0 ð11:124Þ From the axial symmetry of the disk, since change in the where r0 ¼ the uniaxial compressive flow stress of the circumferential strain ¼ deh ¼ de3 ¼ de2 ¼ der ¼ change in material in homogeneous strain condition. Since the radial strain, so from (11.81a), it can be written that r0 ¼ constant, so from (11.124), we get drr ¼ dp: Making circumferential stress rh ¼ radial stress rr: For very small this substitution into (11.123), we get angle, assuming sinðdh=2Þ % dh=2 and substituting rh ¼ rr

508 11 Forging ðp À r0Þ2 tan adr þ ðh0 þ 2r tan aÞdp normal to the conical surfaces makes an angle equal to the þ 2lpdr À 2p tan adr ¼ 0; angle of friction f with the vertical axis of the circular disk, i.e. if a ¼ f ¼ tanÀ1 l; or l ¼ tan a; then (11.126) reduces to Or; ðh0 þ 2r tan aÞdp ¼Z2ðr0 tan a À lpÞdr; Z dp dr ) r0 tana À lp ¼ 2 h0 þ 2r tan a; p=r0 ¼ 1; or; p ¼ r0 ð11:127Þ Or; À 1 lnðr0 tan a À lpÞ ¼ 2 2 1 a lnðh0 þ 2r tan aÞ þ ln C; Equation (11.127) shows that the effect of friction is l tan vanished, and the forging die pressure is equal to the uni- axial compressive flow stress of the material. This is true for where ln C ¼ constant of integration. a frictionless ideal homogeneous compressive deformation, & 1 'Àl where the specific die pressure p must be equal to the flow stress of the material since the friction is neglected. In case Or, lnðr0 tan a À lpÞ ¼ ln ðh0 þ 2r tan aÞ tan a C ; of compression of cylindrical work-piece having plane end faces with flat dies, the friction force acting inwards resists Or, r0 tan a À lp ¼ ðh0 þ 2r tan aÞÀtaln a A the lateral outward flow of material at both ends of work-piece which are in contact with the surfaces of dies and ð11:125Þ thus, barrelling is developed in the work-piece because there will be unhindered lateral expansion at its central portion. where A ¼ CÀl; and A can be evaluated from the following But if a cylindrical work-piece having conical ends, with a ¼ f (angle of friction), is compressed with conical boundary condition: pointed dies, the material at both ends of the work-piece will Since at the outer free surface of the disk, i.e. at r ¼ be subjected to a simple or homogeneous compression, i.e. a uniaxial state of compression with no frictional effect. As a R0; rr ¼ 0; so from (11.124), we get p ¼ r0: Substitution of result, there will be unhindered lateral uniform expansion p ¼ r0 into (11.125) gives the value of A as throughout the height of the work-piece during its com- pression and barrelling of the work-piece will be completely A¼ r0 tan a À lr0 ; eliminated. l À ðh0 þ 2R0 tan tan a aÞ Substituting the value of A into (11.125), we get À l r0ðtan a À lÞ l r0 tan a À lp ¼ ðh0 þ 2r tan aÞ tan a ; 2 ðh0 þ 2R0 tan À tan a 3 r046tan a75; aÞ &  0  R0 'À l lÞ h0 1 þ 2 r tan a h0 a tan Or; lp ¼ a À ðtan a À h0 h0 1 þ 2 tan a  þ 2ðr=h0Þ a !À l p tan tan a 1 tan tan a ) r0 ¼ l À l À 1 ð11:126aÞ 1 þ 2ðR0=h0Þ tan a If the diameter of the circular disk is D0; i.e. D0 ¼ 2R0; then 11.11 Forging Defects  þ =h0 a!À l Improper forging produces forging defects such as laps, p tan a tan a 1 2ðR0 Þ ðr=R0Þ tan tan a cracks, seams, and improper heating and cooling of the r0 ¼ l À l À 1 forging stocks produce forging defects such as warping 1 þ 2ðR0=h0Þ tan a and cracking of components, decarburization of steel caused by contact of the heated steel with the atmosphere. a  a!À l Apart from the development of such defects, forging tan tan a 1 þ ðD0 =h0 Þðr=R0 Þ tan tan a defects may also be caused due to poor quality of forging ¼ l À l À 1 stock, the source of which is the defects caused by faulty 1 þ ðD0=h0Þ tan a melting practice and ingot defects. Defects resulting from ð11:126bÞ If the conical contact surfaces between the circular disk and the conical pointed dies are shaped in such a way that the

11.11 Forging Defects 509 imperfect melting practice such as dirt, slag or impurities (3) Flash cracking separated from the solid solution in the melting process, pin holes and blow holes caused by liberation of gasses Cracking at the flash of closed-die forgings sometimes pro- during solidification of ingots from molten state, etc., may duces a surface defect since the crack generally penetrates lead to forging defects. Further, ingot defects like pipes, from the root of the flash into the body of the forging during i.e. opening at the centre of the ingot, cracks, scabs or bad the trimming operation of the flash, as shown in Fig. 11.38a. surfaces, segregation, i.e. uneven concentration of elements This type of cracking becomes more common if the ratio of contained in the material, etc., may be one of the reasons the flash thickness to the original thickness of the work-piece for the occurrence of forging defects. Most commonly decreases, because the thinner the flash the greater will be the observed forging defects resulting from improper forging resistance to deformation and the higher will be the die and improper heating and cooling of the forging stocks are pressure in the flash area. Any one of the following measures described below: may be adopted to avoid the flash cracking. (1) Incomplete forging penetration • Increase the ratio of the flash thickness to the original thickness of the work-piece. When light, rapid hammer blows are used to deform work-pieces of large cross section, deformation primarily • Relocate the flash to a less crucial zone of the forging. occurs in the surface layers and does not penetrate up to the • Perform hot trimming of the flash, because higher tem- centre. As a result, a large cast ingot after rapid forging by light hammer at high temperature will show a hot-worked perature relieves the stress. structure at the outside surfaces but the interior will be still • Apply a stress relief annealing treatment to the forging, as cast showing dendritic ingot structure, which has not been broken down by light rapid hammering. Thus, incomplete before the flash is cold trimmed off. forging penetration, which can readily be detected by macroetching a cross section of the forging, results in an (4) Internal cracks inhomogeneous structure and non-uniform distribution of mechanical properties from the outside layers to the centre. Internal cracks are generally observed in open-die forging, To minimize incomplete forging penetration, work-pieces of where high friction at the unlubricated or insufficiently large cross section are usually forged using a forging press. lubricated interface may produce barrelling in the forgings. This barrelling can introduce secondary circumferential (2) Surface cracking tensile stresses, which can develop internal cracks (Semiatin 1984) during open upsetting of a cylindrical or a round Surface cracking can occur due to high thermal gradients work-piece, as shown in Fig. 11.38b. However, this type of between surface and centre, or when the surface of the cracking can be minimized by proper die design. Concave work-piece is excessively deformed at too low a tempera- dies are usually employed in practice in order to minimize ture. The other reason for surface cracking is hot shortness barrelling during open upsetting and the development of that can be produced in nickel and steel by a high sulphur circumferential tensile stresses. To avoid barrelling in concentration in the furnace atmosphere. open-die forging, one may also use conical pointed dies inclined to the horizontal direction with a specific cone angle equal to the angle of friction, as discussed in Sect. 11.10. Internal cracking is less likely to occur in closed-die forging Fig. 11.38 a Cracking at the (a) (b) flash. b Internal cracks due to secondary tensile stresses Flash

510 11 Forging because the application of lateral compressive stress by the Sometimes, buckling of thin web during forging forms lap die wall on the work-piece closes the crack. in the finished part, which may be avoided by increasing the thickness of web. For a given geometry of tooling in open-die (5) Lap, cold shut or fold forging operation, a critical deformation may produce surface laps at the step that separates the deformed from the unde- A common surface defect observed in closed-die forgings is formed segment of the work-piece, because only that portion cold shut or fold. The term cold shut is often applied in of the surface under the bite is being forged at any instant. association with lap. Lap, cold shut or fold appears as a Lap is caused by folding a fin to the surface and squeezing it sharp, elongated surface discontinuity. It is related to flow of in—without welding completely. To prevent formation of material, which is affected by the workability of the material, laps in open-die forging, the ratio of initial to final thickness the details of part and die design. A cold shut is produced by of forgings must not exceed 1.3 (Wistreich and Shutt 1959). folding over of some excess metal from prior forging steps at the surface of forging or folding of material back onto its (6) Seams own surface during its flow in a partly filled-up die cavity. These folds are forged into the surface of forging but are not Seams are elongated surface discontinuity. The crevices in the welded completely to form metallurgical bonding with the surface of the material which have been closed by working surface because of the oxide present between the surfaces. but not welded are seams. Seams can originate from a large As the folds do not combine with the surface of forging number of sources. They can result from cracks formed by smoothly, a boundary layer forms at their intersection, which previous working operation or from a defect in the ingot is called cold shut or lap. This flaw may result when a surface, such as an oxidized hole. During forging, the hole work-piece fails to fill the die cavity during forging, or when cannot weld and is simply elongated producing a crack-like materials with different temperatures flow and meet, or when seam in the surface of work-piece. They can also develop the sequence of filling the die progresses improperly. The from trapped gas pockets, a heavy cluster of non-metallic above can happen due to any of the following reasons: inclusions, or a deep lap that intersects the surface at a large angle. They hardly penetrate to the interior of the stock. They • Flow of material around a sharp die corner, as in the case may look like scratches in the forging and are normally so of a lower rib or an upper rib; tightly closed that it is difficult to detect them visually without some non-destructive inspection techniques such as • Reversal in flow direction of material; magna-flux test. Seams must completely be removed from • Excessive chilling of the material; forging stocks; otherwise, they may cause further cracking in • High friction at the interface between the work-piece and hot forging and quench cracking during heat treatment. the dies. A common thumb rule is to maintain the flow of material (7) Miscellaneous defects in the same direction. The die corner radius must be as high as possible to avoid cold shuts. Figure 11.39 shows These include underfill, forged-in scale, warping and schematically the formation of a lap caused by flow of cracking. Underfill is caused by the pockets of scale or material around a sharp die corner. lubricant that are formed by accumulation of loose scale or Fig. 11.39 Lap formation in a Curve of natural metal flow rib–web forging caused by Reverse flow improper radius in the perform die Forward flow Reverse flow Lap formed Reverse flow causes fold to from

11.11 Forging Defects 511 Fig. 11.40 Flow lines in a 11.12 Solved Problems forged part 11.12.1. A solid circular flat disk of lead with initial diameter of 300 mm and thickness of 50 mm is homogeneously press forged between flat dies to a final thickness of 25 mm. The average shear yield stress of lead is 4 MPa. Determine the total load required to complete the forging operation when the coefficient of friction is: (a) 0.25 and (b) 0.05. Solution lubricant residue in deep recesses of the die. There are many Given that the shearipngffiffi yield stress of lead is k ¼ 4 MPa ¼ reasons for underfill. These include less amount of initial 4N=mm2; *pkffiffi ¼ r0=pffiffi3; where r0 is the uniaxial flow stress, material in the work piece, insufficient preheat temperature, then r0 ¼ 3k ¼ 4 3N=mm2: insufficient forging pressure, improper sequence of filling the material in the die and excessive chilling of die. Forged-in It is further given that the initial radius of a circular flat scale on the finished part develops due to incomplete removal of scale from the work-piece and the die surface. disk is a0 ¼ 150 mm ; the initial thickness is h0 ¼ 50 mm Non-uniform heating and cooling of forgings, such as faster and after forging the final thickness is h ¼ 25 mm : Since cooling of thinner sections than the rest of the forging can cause the development of thermal stress leading to warping the volume remains constant, the final radius of the disk is: or cracking of a forged component. Further, crack can develop due to poor die design, excessive material in the sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi work piece or improper stress distribution in the part. a¼ pð150Þ2 50mm ¼ 212 mm: 25p (a) Coefficient of friction is l ¼ 0:25: According to (11.111), the radius of sticking zone measured from the centre-line of the disk is: (8) Fibre structure rS ¼ a À h ln p1ffiffi ¼ 212 À 2 25 ln pffiffi 1 mm 2l 3l  0:25 3  0:25 Forging deformation produces a certain degree of direc- tionality to the microstructure in which structural disconti- ¼ 170 mm: nuities such as inclusions, cavities, segregation and second phase constituents are oriented parallel to the longitudinal Since a [ rS [ 0; so the mixed sliding–sticking friction (main) direction of mechanical working, which is the condition prevails at the interface of the circular disk of radius direction of maximum strain. This preferred alignment gives a ¼ 212 mm and the dies, i.e. full sticking friction occurs rise to the fibre structure or flow lines, which can be viewed around the neutral axis from r ¼ 0 (centre line) to rS ¼ at low magnification after etching the wrought product. Flow 170 mm (sticking–sliding boundary), from where Coulomb’s lines for a forged part are shown in Fig. 11.40. It is some- sliding friction exists up to a ¼ 212 mm (outer periphery). times desirable to determine whether or not a finished part Hence, the total forging load according to (11.117) is: has been forged, machined or cast into shape, and to note the direction of metal flow, if it was forged. The appearance of r0p h2  & ' ! flow lines is the characteristic of wrought products, espe- 2 l 2lrS 2l À 2la À 1 cially of all forgings, whereas machined or cast parts do not PTotal ¼ h þ1 exp h ða À rSÞ h show flow lines. Hence, the fibre structure is not considered to be a defect of wrought product. However, as mentioned in rp0pffirffi S2 1 2rS ! Sect. 10.7 of Chap. 10, the fibre structure results in lower p3 ffiffi l 3h tensile ductility, impact and fatigue properties in the trans- þ þ verse direction, which is normal to the longitudinal (fibre) ¼ p4 3 direction. To obtain an optimum balance between the lon- &2  25 2   0:25  170  gitudinal and transverse ductility, it is often necessary to 2 '25 1 restrict the extent of deformation ranging from 50 to roughly þ 70% reduction in cross-sectional area. 0:25 ! exp 2  0:25 ð212 À 170Þ À 2  0:25  212 À 1 N 25 25 þ p 4ppffi3ffiðffi1ffi 70Þ2  1  þ 2  170 N 3 0:25 3  25 ¼ ð538917:66 þ 3099034:54ÞN ¼ 3637952:2 N ¼ 3:64 MN:

512 11 Forging (b) Coefficient of friction is l ¼ 0:05: stant after forging at the value of 200 mm, so the deforma- tion is plane strain. According to (11.111), the radius of sticking zone measured The true strain corresponding to the given increase in from the centre-line of the disk is: length of the plate from 50 mm to 100 mm is: rS ¼ aÀ h ln p1ffiffi e1 ¼ lnð100=50Þ ¼ 0:69:  2l 3l Hence, the mean flow stress between initial strain of 0  and strain after deformation, e1 ¼ 0:69; is: ¼ 212 À 2 25 ln pffiffi 1 0:05 mm RRR000e01e:61r9e0ðdd7eRe00:þ69  0:05 3  r0 ¼ 0 ¼ ¼ ð212 À 611:6Þmm ¼ À399:6 mm: Since rS\\0; so the Coulomb’s sliding friction exists over 30e0:33Þde MPa the entire interface, i.e. from r ¼ 0 (centre line) to the outer ede periphery of the circular disk, i.e. radius a ¼ 212 mm: Hence, the total forging load due to Coulomb’s friction ¼ 70  0:69 þ 30  ð0:69Þ0:33 þ 1 MPa 0:69 0:69  ð0:33 þ 1Þ according to (11.91) is: ¼ 70 þ 30 ð0:69Þ0:33 MPa h2  ! 1:33 pr0 l 2la À 2la À 1 PC:F: ¼ 2 exp ¼ 89:96 MPa or N/mm2: hh pffiffi  25 2 pffiffi ¼ p 43 0:05 * r00 ¼ ð2r0Þ= 3; where r00 is the mean flpowffiffi stress in 2 r00 ¼ ð2  89:96Þ= 3 N=mm2 ¼  ! plane strain condition, then 2  0:05  212 À 2  0:05  212 À 1 N 103:87 N=mm2: exp 25 25 (a) Coefficient of friction is l ¼ 0:3: ¼ 1324904:89 N ¼ 1:325 MN: According to (11.54), the distance measured on each side from the neutral surface (centre line) of the plate, at x ¼ 0 ; 11.12.2. A plate of copper with an initial size of 20 mm  up to which sticking friction occurs is: 50 mm  200 mm is homogeneously press forged between flat dies to final dimensions of 10 mm  100 mm  200 mm: xS ¼ a À h 1 The flow curve of copper is described by the Ludwik equation 2l ln 2l of the form r0ðMPaÞ ¼ 70 þ 30e0:33: Determine the total load required to complete the forging operation when the coeffi- ¼ 50 À 2 10 ln 2 1 mm ¼ 41:5 mm: cient of friction is: ðaÞ 0:3; and ðbÞ 0:5 :  0:3  0:3 Solution Since a [ xS [ 0; so the mixed sliding–sticking friction condition prevails at the interface of the plate of length 2a ¼ Given that after the forging operation, the final thickness of the plate is h ¼ 10 mm; and the total length of forged area is 100 mm and the dies, i.e. full sticking friction occurs on each L ¼ 2a ¼ 100 mm; i.e. the forged semilength is a ¼ side of the neutral surface from x ¼ 0 (centre line) to xS ¼ 50 mm: Since the initial width w of the plate remains con- 41:5 mm (sticking–sliding boundary), from where Cou- lomb’s sliding friction exists on each side of the neutral surface up to a ¼ 50 mm (outer edge). Hence, the total forging load according to (11.63) is: & 1 '  h 1 &  1 1  '! h 2l 2l 2l 1 2 2l þa PTotal ¼ r00w l À 1 þ a À ln l 1 À ln h ¼ 103:87  200 & '  &   '! 10 1 10 1 1 1 1 þ 50  0:3 2  0:3 À 1 þ 50 À 2  0:3 ln 2  0:3 0:3 1 À 2 ln 2  0:3 10 N ¼ 6909858:17 N ¼ 6:91 MN:

11.12 Solved Problems 513 (b) Coefficient of friction is l ¼ 0:5: Since the width of the strip does not change after forging, so the deformation is plane strain. It is further given that the According to (11.54), the distance measured on each side homogeneous uniaxial average deformation resistance of from the neutral surface (centre line) of the plate, at x ¼ 0; steel is r0 ¼ 250 MPa ¼ 250 N mm2, and the coefficient of up to which sticking friction occurs is: friction at the strip–platen interface is l ¼ 0:1 : Hence, the xS ¼ a À h ln 1 ¼ 50 À 2 10 ln 2 1 mm ¼ 50 mm: adviteiorangeisdefro00rm¼a2tiorn0=rpes3ffiiffis¼tanÂcðe2 of steel ipn ffipffiÃlane strain con- 2l 2l  0:5  0:5  250Þ= 3 N mm2: The analysis of the problem is based on the assumption that Since xS ¼ a; the full sticking friction exists over the entire a\\ tanÀ1 l; which is true in the present problem because interface of the plate of forged length L ¼ 2a ¼ 100 mm and the dies, i.e. from x ¼ 0 (centre line) to x ¼ a (outer edge) on tanÀ1 l ¼ tanÀ1ð0:1Þ ¼ 5:71: each side of the neutral surface. Hence, the total forging load It is known that the peak forging pressure occurs at the neutral plane. According to (11.77), the strip thickness at the due to sticking friction according to (11.46) is: neutral plane is: 1þ L ! 8 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi91 4h PSticking ¼ r00Lw <2hB1 þ 4h21B À 4ð1 À B2 Þðh1h2ÞB=B : 100 ! hNS ¼ 2ð1 þ BÞ ; 4  10 ¼ 103:87  100  200  1þ N ¼ 7270900 N ¼ 7:3 MN: where B ¼ l cot a ¼ 0:1  cotð2:86Þ ¼ 2: 8 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi91 <2  202 þ 4  202Â2 À 4  ð1 À 22 Þð20  100Þ2=2 11.12.3. An infinitely wide strip of steel varying in thickness )hNS ¼ : ; mm linearly from 20 mm to 100 mm over its length of 800 mm is 2ð1 þ 2Þ homogeneously press forged between inclined platens. Assume that both contact surfaces of the strip are inclined at ¼ 36 mm: the same angle with the horizontal axial plane of symmetry of the strip and Coulomb’s law of sliding friction holds good at (a) If p þ is the forging pressure towards the thinner side of the strip–platen interface with a constant coefficient of friction the neutral surface of the strip and pÀ is the forging pressure value of 0.1. If the width of the strip remains constant after forging and the homogeneous uniaxial average deformation towards the thicker side of the neutral surface of the strip, resistance of steel is 250 MPa, compute the following: then since at the neutral plane, ppþeak ¼ ppÀeak; so the peak forging pressure ppþeak or ppÀeak occurring at the neutral plane (a) Peak forging pressure at the beginning of the press forging. can be obtained from either (11.74) or (11.75). (b) Location at which the peak forging pressure occurs. According to (11.74) at the beginning of the press forging, ppþeak ¼ \" þ 1  B # r00 B B hNS À 1 Solution h1 B Given that the thickness of the strip at its extreme con- ¼ 2 Âp2ffiffi50 \" þ 1 362À # ¼ 1258 MPa 2 1 N=mm2 verging end is h1 ¼ 20 mm; the thickness of the plate at its 3 2 20 2 extreme diverging end is h2 ¼ 100 mm and the deformation-zone length of the strip is L ¼ 800 mm: Since Or according to (11.75) at the beginning of the press both surfaces of the strip are inclined at the same angle with forging, the horizontal axial plane of symmetry of the strip, so the \" À 1  h2 B # r00 B B hNS 1 semicone angle of both platens, i.e. the angle at which both pÀ ¼ þ B platens are inclined with the horizontal central plane of the \" 1002 # strip will be the same and let us ass!ume this angle to be! a: ¼ 2 Âp2ffiffi50 2 À 1 þ 1 N/mm2 a ¼ tanÀ1 ðh2 À h1Þ=2 100 À 20 Hence, L ¼ tanÀ1 2  800 ¼ 3 2 36 2 tanÀ1ð0:05Þ ¼ 2:86; and tan a ¼ 0:05: ¼ 1258 MPa:

514 11 Forging (b) Location of the peak forging pressure is the location of (a) The expected maximum pressure, ðpC:F:ÞMaximum; occurs the neutral plane, which can be obtained from (11.78) as at the centre-plane, i.e. at the neutral plane of the rectangular follows: section. Hence, for the maximum flow stress in plane strain If the required location is at a distance xNS from the thinnest side of the strip, i.e. where h ¼ h1 ¼ 20 mm ; then condition, r0 , we get from (11.27): from (11.78), 0max  2la ðpC:F: ÞMaximum ¼ r0 exp h 0max  xNS ¼ hNS À h1 a2Þ ¼ hNS À h1 ¼ 36 À 20 mm ¼ 2 Âp1ffiffi00  exp 2  0:25  0:2 MN=m2 ðtan a1 þ tan 2 tan a 2  0:05 3 0:4  tan 60 ¼ 160 mm; ½since tan a ¼ 0:05Š: ¼ 133:4 MPa: Hence, the peak forging pressure will occur at a distance (b) According to (11.54), the distance measured on each side 160 mm away from the thinnest side, or ð800 À 160Þmm ¼ from the neutral surface (centre line) of the rectangular 640 mm away from the thickest side of the strip. section, at x ¼ 0; up to which sticking friction occurs is: 11.12.4. A 1.2-m-long cylindrical bloom is required to be xS ¼ a À h ln 1 transformed into hexagonal section with approximately 2l 2l 0.4 m side without changing the length of the bloom in a forging press. The uniaxial flow stress of the bloom is ini- ¼ 0:2 À 0:4  tan 60 ln 2 1 m tially 70 MPa but increases to 100 MPa at the end of forging 2  0:25  0:25 operation. Assume the following two conditions: ¼ 0:2 À 0:96 m ¼ À0:76 m: (i) The coefficient of friction is 0.25 due to partial lubrication of the bloom. Since xS\\0; the sliding friction exists over the entire (ii) There is sticking friction due to the absence of interface of the rectangular section with side length 2a and lubrication. the dies, i.e. from x ¼ 0 (centre line) to x ¼ a (outer edge) on each side of the neutral surface. Hence, the maximum For the above both conditions do the following: total forging load due to Coulomb’s sliding friction according to (11.31) is: (a) Calculate the expected maximum pressure. (b) Evaluate the capacity of the press that would be needed. PC:F:max ¼ mean forging pressure due to maximum flow stress  area at the die-job interface ¼ r0 exp½ð2laÞ=hŠ À 1 ½ð2aÞlŠ ð2laÞ=h 0max Solution exp½ð2laÞ=hŠ À 1 l=h It is reasonable to assume that the load required to transform ¼ r0 l 0max the cylindrical bloom into hexagonal section is approxi- ¼ 2 Âp1ffiffi00  1:2 3 mately equal to that necessary to transform the cylindrical bloom into rectangular section, whose side after forging is exp½ð2  0:25  0:2Þ=ð0:4  tan 60ފ À 1 0:25=ð0:4  tan 60Þ 2a ¼ 0:4m, and thickness is h ¼ 2a tan 60 m ¼ 0:4  tan 60 m:  MN Since the initial length l of the cylindrical bloom remains ¼ 59:6 MN: unchanged after forging at the value of 1.2 m, so the forging is performed under plane strain condition. When the coefficient of friction is 0.25, the required press capacity is: 59.6 MN. It is given that the uniaxial flow stress at the end of forging operation, i.e. the maximum value of uniaxial flow stress of the bloom is r0max ¼ 100 MN=m2; so the maximum (ii) When there is sticking friction, then the interface friction flow stress of the bloom in plane strain condition is: factor is m ¼ 1 : r0 ¼ pffiffi ¼ h  pffiffi i MNm2: (a) The expected maximum pressure, ðpStickingÞMaximum; ð2r0max Þ= 3 ð2 100Þ= 3 occurring at the centre-plane, i.e. at the neutral plane of the 0max (i) The coefficient of friction is l ¼ 0:25: rectangular section, due to the maximum flow stress in plane strain condition, r0 , can be obtained from (11.41): 0max

11.12 Solved Problems 515 ÀÁ r0 h ai Again, from given values: UA ¼ g  UT ¼ 0:4  50 kJ pSticking Maximum h ¼ 20 kJ ¼ 20  103 J ¼ 0max 1þ (a) For a working stroke distance of h ¼ 10 mm ¼ 0:01 m, from (11.128) and given values of UA; we get ¼ 2 Âp1ffiffi00  0:2 ! 3 1 þ 0:4  tan 60 MN=m2 ¼ 148:8 MPa: 103J ¼ 2Ph UA ¼ 20  3 ¼  ¼ (b) The maximum total forging load due to sticking friction 2   À 10À3Á 0:01 6:67 according to (11.46) is: P  P 3 h a i ½ð2aÞlŠ r0 2h 20  103 PStickingmax ¼ 1þ ) P ¼ 6:67  10À N ¼ 3  106N ¼ 3 MPa: 0max ! 3 ¼ 2 Âp1ffiffi00  0:2 3 1 þ 2  ð0:4  tan 60Þ ½0:4  1:2Š MN ¼ 63:4 MN: (b) For a working stroke distance of h ¼ 5 mm ¼ 0:005 m, from (11.128) and given values of UA; we get When there is sticking friction, the required press UA ¼ 20  103J ¼ 2Ph capacity is: 63.4 MN. ¼  3 2 À  10À3Á P If it is required to fulfil both conditions given by (i) and  0:005 ¼ 3:33 (ii), then the press must have a load capacity of 63.4 MN. P 3 11.12.5. For a power forging hammer having a total nominal ) P ¼ 20  103 N ¼ 6  106N ¼ 6 MPa energy of 50 kJ, the forging load P increases from P=3 at the 3:33  10À3 start of the stroke to P at the end of the stroke h: If the blow efficiency is 0.4, compute the total forging load P for a 11.12.6. It is required to make a smooth steel shaft by press working stroke of forging of the following dimensions: diameter D ¼ 800 mm and length L ¼ 5 m: The forging will require two heating. (a) 10 mm and (b) 5 mm. Calculate the weight of an ingot that will be selected as an initial work-piece for the above forging operation. Assume Solution that the density of the steel ¼ 7850 kg=m3: Given that the total nominal energy is UT ¼ 50 kJ, and the Solution efficiency of the blow is g ¼ 0:4: Using (11.11a), we have to calculate the weight of the ingot The useful energy available for deformation UA is the area under the load–stroke ðP À hÞ curve shown in WIN; which will be taken as the initial work-piece. From Fig. 11.41. Therefore, given data, the followings are obtained: UA ¼ P=3 þ P h ¼ 2Ph ð11:128Þ The weight of the forging, 2 3 \"# p ð0:8Þ2  5 WFORG: ¼ 4  7850 Load ¼ 19729:202 kg: P The weight of ingot head discard WH: DISC: ¼ 20% of initial weight of ingot ¼ 0:2  WIN: UA P The weight of ingot bottom discard WB: DISC: ¼ 5% of 3 initial weight of ingot ¼ 0:05  WIN: h Stroke The weight of the scale loss WSC: ¼ 2% of initial weight Fig. 11.41 A load–stroke ðP À hÞ curve in a hammer blow of ingot for first full heating +1.75% of initial weight of ingot for subsequent extra one heating ¼ 0:0375  WIN: The weight of the croppings WCROP: includes only the waste due to trimming scrap, because no hole is required in forging and so, there is no slug waste. The weight of the croppings (excluding slug) only due to trimming scrap from both ends of the said cylindrical press forged steel sections of diameter 0:8 m is:

516 11 Forging hi VIN is the volume of rolled stock, which is used as the initial WCROP: ¼ 2  0:21  ð0:8Þ3  7850 ¼ 1688:064 kg: work-piece. Hence, from (11.11a), Hence, from (11.12b), À VIN ¼ ð11974642:21 þ 0:02  VIN þ 322013:247Þmm3 WIN ¼ 19729:202 þ 0:2 ÂÁWIN þ 0:05  WIN þ 0:0375 ¼ ð12296655:46 þ 0:02  VINÞ mm3; À ÂWIN þ 1688:064 kg Á ¼ 21417:266 þ 0:2875  WIN kg; ) VIN ¼ 12296655:46 mm3 ð1 À 0:02Þ À 21417:266 Á kg ) WIN ¼ 1 À 0:2875 ¼ 30059:32 kg: ¼ 12547607:61 mm3: Therefore, the required length of the rolled stock is 11.12.7. Determine the length of a rolled steel bar of L ¼ VIN cross-sectional area with 200  200 mm2; from which a gear cross-sectal area of rolled stock blank of the following dimensions is to be formed by upset forging in one heat. Outside diameter of the gear blank D ¼ ¼ 12547607:61 mm3 mm 366 mm; height of the gear blank H ¼ 123 mm and diameter ð200  200Þmm2 of hole in the gear blank d ¼ 100 mm. Assume that under the head of croppings, only waste is due to the formation of hole. ¼ 313:69 mm: Solution 11.12.8. A stepped cylindrical steel shaft with its dimensions is shown in Fig. 11.42. This is to be made by hot hammer Under the head of croppings, the only loss of metal incurred forging after heating twice from a rolled stock long enough will be the slug when punching the hole in the gear blank for one shaft only. Calculate the weight as well as the length and so, with the help of (11.13), we get of the rolled stock, if the forging ratio is 1.5. Assume that the density of the steel ¼ 7850 kg=m3: VCROP: ¼ volume of slug; VSLUG # Solution \" ¼ pd2  H ¼ pð100Þ2  123 mm3 \" pð300Þ2  800 þ pð240Þ2 43 43 The volume of the forging VFORG: ¼ 44 ¼ 322013:247 mm3: Â170 þ pð130Þ2  150 þ pð200Þ2  160Š mm3 ¼ 71256819:17 mm3: 44 The volume of the forging is: The volume of the croppings VCROP: includes only the pD2 pd2 \"(4 4 waste due to trimming scrap, because no hole is required in VFORG: ¼ À H forging and so, there is no slug waste. If the shaft is forged ) # from a long rolled stock and then cut off, the cropping losses ¼ pð366Þ2 À pð100Þ2  123 mm3 would be from one end of the rolled stock. As the shaft in the 44 given problem is forged from a rolled stock long enough to make one shaft only, the waste will be cropped off from both ¼ 11974642:21 mm3: ends of the rolled stock. Since the volume of the croppings The volume of the scale loss VSC: ¼2% of initial volume excluding slug from one end of cylindrical hammer-forged of rolled stock for single full heating ¼ 0:02  VIN; where sections of diameter ‘D’ is 0:23  D3; for the above forging, the volume of the croppings from the left-hand end is 130 mm 240 mm 300 mm 200 mm 150 800 mm 160 mm 170 mm mm Fig. 11.42 A stepped cylindrical steel shaft with its dimensions

11.12 Solved Problems 517 hi Exercise VC0 ROP: ¼ 0:23  ð130Þ3 mm3 ¼ 505310 mm3: 11.Ex.1. A cylindrical billet with length of 2 m and diameter Similarly, the volume of the croppings from the 0.5 m is forge-reduced longitudinally by 10% using a press. right-hand end is Allowing 20% for friction over the load for ideal deforma- tion, evaluate the capacity of the press required for such hi forging operation under the following two conditions: VC00ROP: ¼ 0:23  ð200Þ3 mm3 ¼ 1840000 mm3: Therefore, the total volume of metal lost as croppings (a) When forging is carried out at a temperature of 1000 °C, from both ends of forging section is where the average yield stress of the billet is 70 MPa. (b) When forging is carried out at room temperature, where VCROP: ¼ VC0 ROP: þ VC00ROP: the average yield stress of the billet is 350 MPa. ¼ ð505310 þ 1840000Þ mm3 11.Ex.2. A solid circular flat disk of 200-mm diameter and ¼ 2345310 mm3: 10-mm thickness is homogeneously press forged between flat dies to a thickness of 6.4 mm. If a solid foil lubricant, The volume of the scale loss VSC: ¼ 2% of initial volume causing a constant interface shear factor of m ¼ 0:3 is used of rolled stock for first full heating +1.75% of initial volume and the average uniaxial flow stress of the disk material is of rolled stock for subsequent extra one heating ¼ 0:0375  100 MPa, determine the loads required (a) at the com- VIN; where VIN is the volume of rolled stock, which is used mencement and (b) at the completion of forging. as the initial work-piece. 11.Ex.3. It is required to press forge solid circular flat disks Hence, from (11.12b), from diameter to height ratio of 30–40 by using either a viscous lubricant or a solid foil lubricant. The former VIN ¼ ð71256819:17 þ 0:0375  VIN þ 2345310Þ mm3 lubricant would produce a coefficient of friction of l ¼ 0:05 while the latter one would give a constant interface shear ¼ ð73602129:17 þ 0:0375  VINÞmm3; factor of m ¼ 0:25: Which lubricant will you select to reduce the load required to complete the forging operation? ) VIN ¼ 73602129:17 mm3 ð1 À 0:0375Þ 11.Ex.4. A 400-mm-long and 800-mm-diameter cylindrical billet is homogeneously press forged between flat dies to ¼ 76469744:59 mm3: 75% of its original length. Calculate the load required to complete the forging operation carried out homogeneously Therefore, the initial weight of the rolled stock is for the following two conditions: WIN ¼ 76469744:59 mm3 m3  7850 kg=m3 (a) At the room temperature, where resistance of the billet to 109 plane-strain deformation is 600 MPa with a friction coeffi- cient value of 0.15. ¼ 600:287 kg: (b) At 800 °C, where uniaxial homogeneous flow stress is 100 MPa, but there is sticking friction. Given that the forging ratio 11.Ex.5. A 1.2-m-long and 1.2-m-diameter cylindrical ¼ original cross À sectional area of the rolled stock before forging bloom is transformed into a square section of equal area by cross À sectional area of the finished forged shaft homogeneous forging in a hydraulic press, while the length of the bloom remains unchanged. Assuming full sticking ¼ ASTOCK ¼ 1:5: frictional condition, calculate the approximate load required ASHAFT \" to complete the forging operation in units of metric ton if the # uniaxial flow stress of the bloom is 50 MPa. 1:5  pð300Þ2 mm2 ) ASTOCK ¼ 4 11.Ex.6. A rectangular plate with an initial size of 20 mm  30 mm  500 mm is homogeneously press forged at 900 °C ¼ 106028:752 mm2: between flat dies to a final size of 15 mm  40 mm  500 mm: Assuming a coefficient of friction value of 0.4 at Therefore, the required length of the rolled stock is the interface, determine the following: LSTOCK ¼ VIN ASTOCK ¼ 76469744:59 mm3 mm 106028:752 mm2 ¼ 721:217 mm:

518 11 Forging (a) Location of the boundary between slipping and full the total load required to complete the forging operation sticking friction from the centreline of the plate at the end of when the friction factor at the die–plate interface is 0.5. the forging operation. (b) Total load required to complete the forging operation, if 11.Ex.14. Indicate the correct or most appropriate answer the uniaxial average flow stress of the plate material is from the following multiple choices: 80 MPa at 900 °C. (a) In closed-die forging, the penetration of flash crack into 11.Ex.7. In an open-die forging, if the bite ratio is 0.4 and the body of forging is more prevalent if spread ratio is 1.2, what will be the squeeze ratio? (A) flash is hot trimmed; 11.Ex.8. Hot press forging operations are carried out (B) forging is stress relieved prior to cold trimming of the homogeneously under full sticking frictional condition with flash; flat dies on the same material of the following shapes. (C) flash to original metal thickness ratio is increased; (D) flash to original metal thickness ratio is decreased. (a) The height of a flat circular disk is reduced with conse- quent increase in its diameter without barrelling so that after (b) To minimize incomplete forging penetration of a large forging, its height becomes equal to its average diameter. cross-section having dendritic ingot structure, the most (b) An uniformly thick plate of rectangular cross-section is suitable kind of forging equipment is reduced in plane strain condition to generate a square cross-section without barrelling. (A) Mechanical Press; (B) Hydraulic Press; (C) Steam Hammer; (D) Board Hammer. Which one of the above will need a greater peak forging pressure and how many times will it be greater than the other? (c) During upsetting of a cylindrical rod with gripping at its one end, if the unsupported length of the rod exceeds 3 times 11.Ex.9. Calculate the minimum possible values of the its diameter ðdÞ; the operation performed to prevent buckling coefficient of friction at the die–job interface under full of the rod should be sticking frictional condition, according to Von Mises’ yielding criterion, for both (a) and (b) of problem 11.Ex.8, (A) open upsetting; considering the equations developed with the assumptions of (B) closed upsetting with maximum inner die diameter the coefficient of friction to be the same at all points on the ðDmaxÞ ¼ 1:5 d; surface of contact. (C) closed upsetting with Dmax ¼ 3 d; (D) closed upsetting with Dmax ¼ 4:5 d: 11.Ex.10. What will be the ratio of maximum coefficient of friction developed during press forging of a rectangular plate (d) To increase form-filling capacity in closed-die forging under plane strain condition to that of a circular disk of the using hammers, flash land ratio may be increased up to same material, according to Von Mises’ yielding criterion? certain value above which the improvement is small, but the die wear becomes severe. The limiting value is about 11.Ex.11. Two cylinders of the same material with diameter to the height ratios of 50 and 10 are homogeneously ðAÞ 3; ðBÞ 5; ðCÞ 7; ðDÞ 9: open-forged in a press with flat dies. Considering that the friction factor, m ¼ 0:25; at the die–cylinder interface, (e) In closed-die forging, forging pressure will decrease with compute the ratio of the average forging pressure required for the first cylinder to that for the second cylinder. (A) die advance; (B) increase of flash land width; (C) increase of flash thickness; 11.Ex.12. In a power drop hammer, the ram weighs 40 kg (D) decrease of flash thickness. and the diameter of the ram piston is 50 mm. The ram is accelerated by steam pressure of 500 MPa. If the fall height (f) The type of defects observed in closed-die forging is: of the ram is 250 mm, (A) Alligatoring; (B) Flash cracking; (C) Fir-tree cracking; (D) Earing. (a) what is the total energy supplied to the blow? Answer to Exercise Problems (b) what will be the effect of the section thickness of the work-piece on the forging load, if friction is neglected? 11.Ex.1. (a) 18.32 MN; (b) 91.6 MN. 11.Ex.2. (a) 6.77 MN and (b) 15.98 MPa. 11.Ex.13. A plate of lead with an initial size of 20 mm  11.Ex.3. Viscous lubricant, because the forging load 40 mm  300 mm is homogeneously press forged between required with the solid foil lubricant is 1.33 times greater flat dies to final dimensions of 10 mm  80 mm  300 mm: than that with the viscous lubricant. The average shear yield stress of lead is 4 MPa. Determine

11.12 Solved Problems 519 11.Ex.4. (a) 408.68 MN; (b) 106.74 MN. Altan, T., Oh, S.I., Gegel, H.L.: Metal Forming. American Society for 11.Ex.5. 9392 ton. Metals, Metals Park, Ohio, Chap. 11 (1983) 11.Ex.6. (a) Full sticking friction on each side of the cen- treline of the plate exists up to 15.816 mm beyond which Halter, H.W.: Trans. ASME J. Eng. Ind. 105, 270–275 (1983) sliding friction occurs; (b) 3.03 MN. Kamenshchikov, G., Koltun, S., Naumov, V., Chernobrovkin, B.: 11.Ex.7. 0.52. 11.Ex.8. Peak forging pressure for plate is 1.098 times Forging Practice. Peace Publishers, Moscow, Translated from the greater than that for disk. Rassian by Zelokov, L., Chap. VII (1964) 11.Ex.9. (a) 0.366 and (b) 0.333. Lange, K.: Closed-Die Forging of Steel (in German). Springer-Verlag 11.Ex.10. 0.866. OHG, Berlin (1958) 11.Ex.11. 2.3. Lyman, T. (ed.) Forging and Casting, Metals Handbook, 8th edn., vol. 11.Ex.12. (a) 245.5 kJ. (b) No effect, because the forging 5, pp. 12–13, 14–18. American Society for Metals, Metals Park, load is independent of the section thickness if friction is Ohio (1970) reduced to zero. Sabroff, A.M., Boulger, F.W., Henning, H.J., Spretnak, J.W.: Principles 11.Ex.13. 384 kN. of die-forging design (Chapter 3). In: A Manual on Fundamentals of 11.Ex.14. (a) (D) flash to original metal thickness ratio is Forging Practice, Supplement to Technical Documentary Report decreased. (b) (B) Hydraulic Press. (c) (B) closed upsetting No. ML-TDR-64-95, December, 1964, Manufacturing Technology with maximum inner die diameter ðDmaxÞ ¼ 1:5 d : Division, Air Force Systems Command, Ohio, Prepared under (d) (B) 5. (e) (C) increase of flash thickness. (f) (B) Flash Contract No. AF 33(600)-42963 by Battelle Memorial Institute, cracking. Columbus, Ohio (AD No. 460465 in Defence Documentation Center for Scientific and Technical Information, Virginia), pp. 59– References 72 (1964) Semiatin, S.L. (ed.) Forming and Forging, Metals Handbook, 9th edn., Akgerman, N., Becker, J.R., Altan, T.: Metallurgia Metal Form. 40, (from 1993, renamed as ASM Handbook and converted to 135–138 (1973) electronic files in 1998), ASM International Handbook Committee, 14: 66, 73–75, 78, (electronic book) (1988) Altan, T., Boulger, F.W., Becker, J.R., Akgerman, N., Henning, H. J.: Semiatin, S.L. (ed.): Metalworking: Bulk Forming, ASM Handbook, Forging Equipment, Materials, and Practices. Battelle Columbus ASM International, Materials Park, Ohio, 14A: 24, 30, 38, 60 Labs., Metals and Ceramics Information Center, Ohio, (2005) Chap. 1 (1973) Semiatin, S.L.: Workability in forging. In: Dieter, G.E. (ed.) Worka- bility Testing Techniques. American Society for Metals, Metals Altan, T., Henning, H.J.: Metallurgia Metal Form. 39, 83–88 (1972) Park, Cleveland, Ohio, Chap. 8 (1984) Tomlinson, A., Stringer, J.D.: Spread and elongation in flat tool forging. J. Iron Steel Inst Lond. 193, 157–162 (1959) Tool and Manufacturing Engineers Handbook, 4th edn., vol. II, Society of Manufacturing Engineers, Dearborn, Mich, pp. 15–18 to 15–40 (1984) Wistreich, J.G., Shutt, A.: Theoretical analysis of bloom and billet forging. J. Iron Steel Inst. Lond. 193, 163–176 (1959)

Rolling 12 Chapter Objectives • Terminology of rolled product, different methods of rolling and quantities charac- terizing deformation. • Classification of rolling mills. Different types of rolling mills, such as cluster mill, Sendzimir cold-rolling mill and planetary hot-rolling mill, pendulum mill, contact-bend-stretch mill, universal mill. • Hot and cold-rolling practice. • Deformation zone in rolling, angle of bite and artificial methods to increase it, neutral point (no-slip point) and no-slip angle. • Ekelund expression for no-slip angle. • Forward slip, its relation with no-slip angle, its measurement and importance. • Elastic deformation of rolls: roll flattening and roll deflection. • Simplified assessment of rolling load and Ekelund equations for loads in cold and hot rolling. • Theory of rolling: derivation of differential equation of friction hill. • Bland and Ford theory of cold rolling: roll pressure with no external tensions and with back and front tensions, no-slip angle, rolling load and torque, maximum allowable back tension, estimation of friction coefficient. • Sims’ theory of hot rolling: roll pressure, no-slip angle, rolling load and torque, limitations of theory. • Mean strain rate in flat rolling with sticking friction. • Lever arm ratio and its estimation from Sims’ theory, torque and mill power. • Minimum thickness of rolled product. • Factors controlling rolling process. • Gauge control by considering mill modulus line and plastic deformation curve for work-metal and factors affecting outgoing gauge of rolled product. • Defects in rolled products. • Fundamentals of roll-pass design: types and shapes of passes, gap and taper of sides in pass, pass arrangement, pass sequences used in rolling of billets to rods and square bars. • Production of seamless pipe and tube by hot rolling using rotary piercing mill, plug mill, continuous mill, pilger mill and finishing mills, such as reeling, sizing, stretch-reducing and expanding mills. • Problems and solutions. © Springer Nature Singapore Pte Ltd. 2018 521 A. Bhaduri, Mechanical Properties and Working of Metals and Alloys, Springer Series in Materials Science 264, https://doi.org/10.1007/978-981-10-7209-3_12

522 12 Rolling 12.1 Fundamentals of Rolling and columns and beams in buildings. Rolled titanium and aluminium alloys are used to make airplane bodies. Rolled Rolls are solid or grooved cylinders usually made of steel, sheet materials are applied in automobiles and, to make cast iron or tungsten carbide. There are three parts of a roll as beverage cans, many home appliances, and utensils. Rolled shown in Fig. 12.1, namely: bars are used as stating material for hot, warm and cold forging to make many automobile parts. • the roll body or roll barrel or the part on which rolling is done, Depending on the material to be rolled and the application of the rolled product, rolling can be carried out either hot or • the roll necks, which turn in the bearings and support the cold. Hot rolling is very widely used because the change in the roll body and shape of the work-piece can be achieved rapidly and eco- nomically. If it is required to produce good surface finish with • the drive ends, which refer to the ends of the roll where close dimensional tolerance and/or to have improved strength the driving force is applied. and hardness caused by strain hardening, cold rolling is car- ried out. The main objective of conventional rolling, whether The process of a squeezing type of plastic deformation to carried out hot or cold, is to reduce the thickness of the stock. reduce the thickness of the metal by passing it between a pair Ordinarily, the width of the stock increases to a little extent, so of work rolls revolving in opposite directions is known as that the reduction in the thickness results in the elongation of ‘longitudinal rolling’ or simply ‘rolling’. Rolling, first the stock in the rolling direction. developed in the late 1500s, is perhaps the most widely used metal working process because of economy, its rapid rate of On the basis of metal flow during rolling and the geom- continuous production and maintaining close control of the etry of rolled product, rolling of solid sections can be divided final product. Rolling is relatively fast than other methods of into the following categories: metal working, such as forging, and is able to produce a variety of desired shapes and sizes from long work-pieces at • Uniform reduction in thickness without any change in a very high speed in a somewhat continuous fashion. width, where the thickness of the stock is reduced uni- formly by compression causing the length (in the rolling Flat rolling is the most common rolling operation, in direction) to increase, but the width of the stock (trans- which flat work-pieces of uniform cross-section are usually verse to the rolling direction) remains constant. This type rolled. However, grooved rolls are used to produce round of metal flow where the deformation is zero in one rods, and the wires drawn from rolled rods are used in direction is called plane strain condition, and it occurs in fences, cables and elevator ropes. Further, rolling mills have rolling of sheet, strip or foil. Plane strain in rolling exists also been successfully used for rolling of balls and roll when the width of the deformation zone is 20 times the forging operation has been used to reduce the diameters of length of that zone. rods and tubes as well as for tapering their ends, as discussed in Chap. 11. More than 90% of all the copper, aluminium • Uniform reduction in thickness with an increase in width, and steel produced are subjected at least one time to rolling where the thickness of the stock is reduced uniformly process. Rolled steels are used to make cars, railroad tracks, causing both the length and the width of the stock to increase, although the increase in the length is usually (a) much more than that in the width. The spread of the stock in the width direction occurs in hot rolling of ingots into 1 23 blooms or slabs and in subsequent rolling of stock into thick plates. (b) • Non-uniform reduction in thickness with an increase in width, where the thickness of the stock is reduced non-uniformly along the width with an increase in both the length and the width of the stock. For example, in the width direction, metal may flow only towards the edges of the section, as in the case of rolling of an oval section in rod rolling or of an airfoil section. Metal flow along the width may also occur towards the centre. Fig. 12.1 Rolls: a smooth and b grooved. 1 Roll body or roll barrel; Thus, the metal flow in rolling occurs in three directions 2 roll neck; 3 drive end (in the thickness, width, and rolling directions) except in strip rolling. However, to avoid the mathematical com- plexity and simplify the analysis, rolling will be considered

12.1 Fundamentals of Rolling 523 to occur under plane strain condition assuming that the width 12.1.2 Different Methods of Rolling of the stock (transverse to the rolling direction) remains unchanged because the little change in the width can be Depending on the direction of rotation of the rolls with neglected compared to the large change in the length. respect to the direction of motion of the work-piece and the direction of deformation, the following three methods of 12.1.1 Terminology of Rolled Product rolling are distinguished, which are illustrated in Fig. 12.2: Many engineering alloys, such as steels, copper alloys and 1. Longitudinal rolling, aluminium alloys, are often cast into ingots, which are ini- 2. Cross rolling, tially broken down by hot rolling into semifinished products 3. Helical rolling. such as blooms, billets and slabs. These products are further processed by hot rolling into bar, rod, plate, sheet, pipe, rails, The characteristics of the above three methods of rolling or structural shapes. Blooms, billets, slabs, plates and struc- have been compared in the following Table 12.1. tural shapes are usually hot rolled in order to obtain large reductions in thickness with moderate working pressures. Generally, cylindrical rolls with plane surface are used to Sheet, strip and foil are often produced by cold rolling in order produce a flat product such as sheets or strips. The surface of to maintain close thickness tolerance. Since the 1980s, con- the rolls can also be grooved or textured in order to change tinuous casting has been widely used to produce blooms, profile of the work-piece as well as emboss patterns on it. billets and slabs by direct casting of molten steel from steel Most common among rolling processes is the longitudinal melting operation into shapes of these semifinished products. rolling having the characteristics as mentioned in the Table 12.1. Hence, attention in the following text will be The definitions of the terms used to describe rolled limited primarily to the longitudinal rolling and to some products are fairly loose, and based on the terminology, they extent to the helical rolling involved in the manufacture of are traditionally used in the steel metal industry. A bloom is pipes and tubes. a product of nearly square cross-section (generally, having the width equal to its thickness) with an area equal to or 12.1.3 Quantities Characterizing Deformation greater than 23225 mm2 (6 by 6 in.) resulting from the first breakdown of the ingot. A billet is mostly square in cross Let us define certain terms related to the deformation of section with an area varying from a minimum of about stock by rolling. For the purpose of illustration, it is simpler 50 Â 50 mm ð2 Â 2 in:Þ to 127 Â 127 mm ð5 Â 5 in:Þ resulting from the reduction of the bloom by hot rolling. to consider a pair of plain cylindrical rolls, reducing a rect- However, in non-ferrous terminology, the billet may be any ingot that has been hot-rolled, hot-forged, or hot-extruded, or angular stock. Suppose, before rolling, the rectangular stock it may be a casting which can suitably be hot worked, for has a thickness of h1; a width of w1; a length of l1; and a example an extrusion billet. A slab is an oblong hot-rolled cross-sectional area of A1; and after rolling, a thickness of h2; ingot with a cross-sectional area greater than 10322 mm2 a width of w2; a length of l2; and a cross-sectional area of A2: (16 in.2) and a section width of at least twice the section The following terms are defined using the above symbols: thickness. Plate, sheet, strip and foil are subsequently formed by rolling of the semifinished products (blooms, billets and Draft experienced by a rectangular stock due to rolling slabs). The differentiation between plate, sheet or strip and foil is made by the thickness of the product and that between ¼ Dh ¼ h1 À h2 ð12:1aÞ sheet and strip is determined by the width of the product. And relative draft A plate is a rolled product generally thicker than 6.4 mm (0.25 in.), whereas sheet and strip are thinner gauge rolled ¼ Dh ¼ h1 À h2 ð12:1bÞ products with thickness equal to or less than 6.4 mm. h1 h1 A sheet has a very large width-to-thickness ratio and is usually wider than 609.6 mm (24 in.), whereas a strip has a If the cross-section of the stock is not rectangular, the width equal to or less than 609.6 mm. A foil refers to a sheet draft is determined by or strip metal with a thickness of a few thousandths of an inch; for example, aluminium foil produced by rolling may Dh ¼ ðA1=w1Þ À ðA2=w2Þ ð12:1cÞ have a thickness of only about 0.025 mm (0.001 in.). Lateral spread or spread experienced by a stock due to rolling ¼ Dw ¼ w2 À w1 ð12:2aÞ

524 (b) 12 Rolling Top roll (a) Top roll Rolled piece Q (c) Rolled piece Bottom roll Bottom roll Top roll Mandrel Shell Bottom roll Fig. 12.2 a Longitudinal rolling. b Cross rolling. c Helical rolling (Mannesmann mill) And relative spread Percent reduction in area of a stock due to rolling ¼ Dw ¼ w2 À w1 ð12:2bÞ % r ¼ A1 À A2  100 % ð12:4Þ w1 w1 ð12:3Þ A1 Percent elongation of a stock due to rolling The law of constancy of volume states that the volumes of  a metal prior to deformation, V1; and after deformation, V2; are equal, i.e. ¼ Dl  100 % ¼ l2 À l1  100 % l1 l1 V1 ¼ V2; or; h1w1l1 ¼ h2w2l2; or; h2 w2 l2 ¼ 1; h1 w1 l1

12.1 Fundamentals of Rolling 525 Table 12.1 Comparison among the characteristics of three methods of rolling (1) Longitudinal rolling (2) Cross rolling (3) Helical rolling The axes of the rolls are parallel to each The axes of the rolls are parallel to each The rolls are arranged at an angle with other other respect to each other. In Mannesmann mill, which uses the principle of helical rolling, The rolls rotate in the opposite direction The rolls turn (or rotate) in the same the angle between the rolls in the vertical direction plane ranges from 5° to 12° The work-piece or the stock moves in a direction perpendicular to the plane joining The axis of stock is parallel to the axes of the The rolls turn or rotate in the same direction the roll axes rolls. The stock is held in place by means of a special device (force Q) and turns around The rolls impart the work-piece, taken in the The height of the stock is reduced, while the its own longitudinal axis in a direction form of a round billet, both rotary and length and breadth of the stock increase; the opposite to the direction of roll rotation translatory motion. The round billet rotates former usually much more than the latter, so in a direction opposite to the direction of roll a condition of plane strain exists, as shown in This method is used for making shapes rotation along with axial advancement Fig. 12.2a which represent bodies of revolution, i.e. Rolls are solid cylinders shapes having circular cross-section, as This method is used for piercing shells shown in Fig. 12.2b (roughly shaped pipes) in pipe manufacture, as shown in Fig. 12.2c Rolls are solid cylinders The Mannesmann mill uses barrel-shaped rolls. The roll barrel consists of two truncated cones with the larger bases joined together. The roll barrel can also be of more complicated shape where Again, two-high non-reversing mills are of two types: h2=h1 ¼ squeeze ratio ð12:5aÞ (a) Two-high pull-over or drag-over mill and w2=w1 ¼ spread ratio (b) Two-high one-pass mill. ð12:5bÞ l2=l1 ¼ elongation ratio ð12:5cÞ In a two-high non-reversing drag-over mill, as shown in Fig. 12.3, after each pass the stock may be returned to the Since rolling is considered to be a plane strain deforma- entrance of the rolls for re-rolling in the next pass by hand tion condition having no spread (Dw ¼ 0), i.e. w1 ¼ w2 ¼ w carrying or by means of a platform which can be raised to ðsay); the spread ratio ¼ 1; and squeeze ratio ¼ 1=elongation pass the work-piece over the top of the rolls. This type of ratio: mill formerly was used mainly for production of light sheets; it is still used as merchant mills (bar and rod mills) for 12.2 Classification of Rolling Mills rolling of tool and high alloy steels. Rolling mills can be conveniently classified with respect to A two-high non-reversing one-pass mill usually consists the number and arrangements of rolls in a stand. Each set of of several stands, the arrangement of which may be varied. In rolls is called a stand. The three principal types of rolling mills are two-high, three-high and four-high mills. The simplest and most common type of rolling mill is the two-high mill which has two rolls of equal size rotating in the opposite directions, and the rolls are arranged one above the other in a stand. A three-high mill has three rolls of equal size, and a four-high mill has four rolls arranged one above the other in a stand. Two-high mills are of two types: (i) Non-reversing two-high mills and Fig. 12.3 Two-high non-reversing pull-over or drag-over mill (ii) Reversing two-high mills.

526 12 Rolling Fig. 12.4 Two-high tandem continuous mill (a) Stand 1 Stand 2 (b) Stand 1 Stand 2 Stand 3 Fig. 12.6 Two-high reversing mill Fig. 12.5 a Looping mill. b Cross-country mill reversing mill, which is shown in Fig. 12.6. In this mill, the direction of rotation of the rolls can be reversed, and the rolling one common arrangement, as shown in Fig. 12.4, the stands takes place alternately in opposite directions, with work done are in tandem, i.e. arranged in a straight line, the axes of the on the piece while travelling in each direction. There is usually rolls in one stand being parallel to those in the other. In a a long mill table on either side of a two-high reversing mill. tandem two-high continuous mill, the rolling direction These long mill tables make it possible to handle heavy pieces remains constant. The rolled piece passes through the entire in long lengths. The reversing two-high type of mill occupies continuous group of stands successively and is deformed an important position in industry. This type of mill is com- simultaneously in several stands. Since each stand produces a monly used for rolling of ingots to blooms and slabs. different reduction in cross-sectional area of the work-piece, the velocity at which the work-piece moves at each stand will The improvement in productivity is also achieved from be different. The speed of rolls at each stand is synchronized the use of a three-high mill, as shown in Fig. 12.7, which has in such a way that the entry speed of the work-piece into each the advantages of both the two-high reversing and successive stand is equal to the exit speed of the work-piece non-reversing drag-over mills. In a three-high mill, the from the preceding stand. This type of mill is in very common motor-driven top and bottom rolls rotate in the same direc- use for rolling strip, billets, bars, rods, etc. Other arrange- tion and the middle roll rotates by friction in the opposite ments of stand or passes in continuous rolling mills include direction. This mill has lift tables on both sides of the rolls. those as shown in cross-country mill and looping mill, where The work-piece is lifted from the bottom pass to return it to the direction of rolling changes (see Fig. 12.5). the top pass by lift tables. The adjustment of the roll gap between passes in this mill is not possible. Therefore, to In two-high non-reversing drag-over mills, the time achieve different reductions during rolling passes, there must required to transport the work-piece back to the entry side of be grooves or passes cut into the roll body. Three-high mills the rolls limits the speed of production and an improvement in are used for rolling of blooms, slabs, sections, etc. productivity can be achieved from the use of a two-high Four-high mills, as shown in Fig. 12.8, are used for rolling flat materials, such as sheets, strips and plates. They are used for both hot and cold rolling. In a four-high mill, large back-up rolls are employed to reinforce the smaller working rolls. Similar to a tandem two-high continuous mill,

12.2 Classification of Rolling Mills 527 tension. The application of back tension is controlled by adjusting the speed of the uncoiler by a braking system relative to the speed of rolls, while the front tension is controlled by the torque on the windup reel. The advantages of applying a back tension and a front tension will be dis- cussed later. A large decrease in the force and power required for rolling can be achieved by the use of small-diameter work rolls. However, because small-diameter work rolls have less strength and rigidity than large rolls, they deflect due to roll-separating forces and must be supported by large-diameter back-up rolls to maintain control over the dimensions of products. The simplest mill of this type is the four-high mill. 12.2.1 Cluster Mill Fig. 12.7 Three-high mill For rolling very thin strip to very close dimensional tolerance, it is necessary to use work rolls which are so small in diameter TOP BACKING UP ROLL that they can flex in the horizontal plane under the influence TOP WORKING ROLL of tangential forces, as well as they can bend under the action BOTTOM WORKING ROLL of the major vertical roll load. To support such work rolls adequately, two backing rolls are provided for each work roll, with their axes offset before and behind, forming a six-high mill, often called cluster mill, as shown in Fig. 12.10. 12.2.2 Sendzimir Cold-Rolling Mill (Sendzimir 1956) BOTTOM BACKING UP ROLL Sendzimir cold-rolling mill or, simply, Sendzimir mill (fre- quently referred to as Z-mill) is a modification of the cluster Fig. 12.8 Four-high mill mill which is very well adapted to cold-rolling thin sheet or foil from high-strength alloys. When the diameter of work the four-high mill usually consisting of several stands in rolls is very small, say 10 mm, then the diameter of each of tandem, i.e. arranged in a straight line, known as four-high the two back-up rolls required to support each work roll continuous mill, is commonly used to increase the produc- cannot be greater than about 20–30 mm, because the tivity. As discussed before, it is essential to synchronize the diameter of back-up rolls is geometrically restricted to about speed of rolls at each stand so that the entry speed of stock two to three times that of work roll and these small-sized into each stand equals the exit speed of stock from the back-up rolls therefore require further supporting back-up preceding stand. While cold rolling with a four-high con- rolls for themselves. This gives twelve-high Z-mill with tinuous mill, an uncoiler (payoff reel) is used to feed the 1-2-3 type of roll arrangements. The most commonly used stock into the roll gap at the first stand and a power-driven type of Z-mill includes one more stage, with four relatively windup reel (coiler) is used for coiling up the final product large-diameter rolls to provide the final back-up, and thus leaving the last stand, as illustrated in Fig. 12.9. Other makes roll arrangements of 1-2-3-4 type, which is called functions of the uncoiler and a windup reel are to apply, twenty-high Z-mill, as shown in Fig. 12.11. In the 1-2-3-4 respectively, a longitudinal tensile stress to the stock in the arrangement, each work roll is supported throughout its backward direction of rolling, known as a back tension, and entire length by two first intermediate rolls (larger in size that in the forward direction of rolling, known as a front than the work roll) that are, in turn, supported by three second intermediate rolls bigger in size than the first inter- mediate rolls. The three second intermediate rolls, of which outer ones are motor driven, transfer the roll-separating

528 12 Rolling Fig. 12.9 Four-high four-stand tan- dem continuous mill Stand 1 Stand 2 Stand 3 Stand 4 Uncoiler Windup reel Top Backing-up Rolls 12.2.3 Sendzimir Planetary Hot-Rolling Mill Top Working Roll The Sendzimir planetary hot-rolling mill or, simply, planetary mill, first patented by Sendzimir in 1948, was developed to Bottom Working Roll hot reduce a slab to directly coiled strip in a single pass. The principle of this mill is quite simple, but its detailed theory is Bottom Backing-up Rolls complex (Sparling 1962). The planetary assemblies consist of two large-diameter back-up rolls surrounded by a large Fig. 12.10 Six-high cluster mill number of small-diameter planetary work rolls (Giles and Gutteridge 1973) that are mounted in ‘cages’ at their forces to a rigid, one piece cast steel housing through four extremities. The cages are synchronized by external means so backing rolls. In Fig. 12.11, A, B, C and D are top that when the whole system rotates, each pair of opposed backing-up rolls and E, F, G and H are bottom backing-up work rolls passes through the vertical centre line of the mill at rolls and they are of the largest size. The actual work rolls precisely the same time and the axes of the work rolls are are rotated by frictional contact with the first intermediate always parallel to the axes of the back-up rolls. The angular rolls, which are, in turn, driven by frictional contact with the velocity of the cage is somewhat less than half the angular power-driven rolls in the second layer. velocity of the back-up rolls. Commonly, the back-up rolls rotate at 52 rad s−1 (500 r.p.m.) and the cage rotates at Typical diameters of work roll are 6.35 mm (0.25 in.), 24 rad s−1 (232 r.p.m.). The cages rotate in the same direction 28.575 mm (1.125 in.) and 88.9 mm (3.5 in.). The mini- as the corresponding motor-driven back-up rolls, while the mum thickness of strip on rolling a strip of width 152.4 mm work rolls rotate by friction in the opposite direction (i.e. the (6 in.) by the smallest mill with work roll diameter of work rolls turn clockwise when their back-up roll and cage 6.35 mm may be 0.00254 mm (0.0001 in.) with a tolerance turn counterclockwise). Mills are built with different ratios of Æ0:1lm: The minimum thickness of strip on rolling a between back-up roll and work roll diameters and with dif- strip of width 1.22 m (48 in.) by the work roll diameter of ferent numbers of work rolls per back-up roll (18, 20, 24, 26 28.575 mm may be 0.0254 mm (0.001 in.) and that on and 30 work rolls per back-up roll have been employed on rolling a strip of width 5.08 m (200 in.) by the work roll different existing mills). Figure 12.12 shows that the number diameter of 88.9 mm may be 0.0889 mm (0.0035 in.). In of work rolls spaced around one back-up roll is twenty. The order to reduce the elastic flattening of rolls as far as pos- common diameter of work rolls is 76.2 mm (3 in.) and that of sible, the work rolls are usually made from tungsten carbide back-up rolls is 457.2 mm (18 in.). Since only the back-up which has a much greater elastic modulus than steel. These rolls in a planetary mill are directly power-driven and the mills may be used to cold-roll hard materials, such as tita- work rolls actually rotate against the work-piece, there is no nium alloys, stainless steel and nickel alloys. bite. It then becomes necessary to provide a pair of feed rolls for forcing the slab into the roll gap of the mill. These produce compressive force sufficient for suppression of any tensile force created in the rolling operation and, thus, prevent cracking. Hence, a planetary mill can successfully roll rela- tively brittle metals, which are normally not possible to hot-roll. The speed of operation is controlled by the feed rolls. The ingoing speed of the work-piece may be about 0.05 m s−1 (10 ft./min), while the exit speed of the rolled

12.2 Classification of Rolling Mills 529 Fig. 12.11 20-high Sendzimir Backing bearing Housing cold-rolling mill (1–2–3–4 arrangement) Bearing shaft B C A Driven D Driven roll roll 1st Intermediate roll 2nd Intermediate roll Work roll Driven Driven H roll roll G E F Fig. 12.12 Sendzimir planetary BACK-UP PLANISHING hot-rolling mill ROLLS MILL FEED WORK ROLLS ROLLS 500 R.P.M 232 R.P.M product may be about 1 m s−1 (200 ft./min) or more. Since The action in planetary mill is more like forging than very less time is required to complete the whole rolling rolling. As the whole system rotates, small work rolls make operation on a slab, the chance of scaling of hot work-piece is contact with the unworked portion of the hot slab surface and very less than that under ordinary rolling conditions. roll down it. As each pair of planetary work rolls breaks

530 12 Rolling contact with the material after reduction, the next pair of work rolls, one on each side of the work-piece. The work roll rolls contacts the slab and repeats that reduction. The reduc- diameter resembles to that of the planetary mill. The work tion imparted to the slab by each pair of work rolls is almost rolls reciprocate over an arc in contact with the work-piece constant. Each pair of work rolls contributes to a small and remain in contact with the surface of the work-piece reduction in cross-sectional area of the slab along its arc of during complete cycle, as shown in Fig. 12.13. To get this, travel, but the combined effect of all work rolls, i.e. the the work rolls are mounted freely on radial arms driven by summation of a series of small reductions by each pair of rolls, accurately synchronized cranks, which may speed up to as results in a very large overall reduction in area. For example, much as 188 rad s−1 (1800 r.p.m.). Since the work rolls the planetary mill is capable of hot reducing a slab of 50 mm move around a fixed radius, a wave is found to develop thick to a strip of 2 mm in a single pass. This corresponds to across the width of the exit sheet or strip. However, 10–15% 96% reduction in area, which is much higher than a maximum reduction by conventional rolling will cause this wave to of about 50% reduction in area, per pass as achievable in an disappear completely. ordinary mill. A pair of planishing rolls may be needed on the exit side of the mill for improving the surface finish of the The work-piece undergoes a large reduction in area, rolled product. Mill of this type can be designed up to two perhaps 95%, which causes considerable increase in the metres wide. Nowadays, a planetary mill has been directly speed of the work-piece from entry to exit plane of the coupled to a continuous-casting plant (Sendzimir 1963), deformation, and this variable speed governs the revolving where the speed of feed to the planetary mill has been regu- speed of the free work rolls. The speed of the ingoing lated to match the output of continuous casting. work-piece is controlled by an auxiliary driven pair of feed rolls, which must be used to force the work-piece into the Generally speaking, the temperature of the strip is higher at roll gap of the mill, similar to the planetary mill. The the exit side of the planetary mill than the temperature of the work-piece of any hardness can be reduced to the same slab at its entrance, due to the energy of plastic deformation extent by rolling under the same roll load by utilizing the involved in the heavy reductions (as much as 98%) that the variation of feed rate, but the rate of production will of mill is capable of effecting. This permits rolling of slabs at course be different. As in the planetary mill, the rolling lower entering temperatures than in conventional hot-strip mill process is fast and there may be large rise in temperature due practice. Further, as rolling proceeds the appreciable increase to plastic deformation resulting in a lower hardness of the in temperature due to large deformation enables final reduc- finished product after a heavy reduction than after a mod- tions to be made on softer metal than the earlier ones, where erate reduction. In comparison with a modern tandem mill, cooling and consequent hardening would be difficult to avoid. By changing the roll speed in relation to the feed, temperature can reasonably be controlled, since heat from the stock is abstracted by the cold work rolls in the short time during which they are in contact with each other at each revolution. One prominent disadvantage is the formation of back-fin, which is found to be small when the diameter of work rolls is at least 75 mm. As the small-diameter work roll first contacts the hot slab, its forging action displaces some metal back- wards that may build up into a fin. Feed rolls having slight corrugations parallel to their axis, for better gripping, may be used to reduce this defect by forming the depressions in the slab surface. The depressions provide spaces into which the metal can deform instead of piling up in a fin. 12.2.4 Pendulum Mill (Saxi 1963) The pendulum mill is a cold-rolling mill that is capable of Fig. 12.13 Schematic view of the pendulum mill cold reducing a slab into a thin sheet. This mill can cold reduce a 6.35 mm (0.25 in.) thick slab to a thickness of 0.38 mm (0.015 in.) in a single pass, even in hard metals such as titanium alloys. The mill may be coupled with continuous-casting plant for continuous production of thin strip. The pendulum mill uses only two small-diameter work

12.2 Classification of Rolling Mills 531 the low rate of output of the pendulum mill is its main between the floating roll and the bottom work roll, then disadvantage. On the other hand, the capital cost of the successively bends around the bottom work roll and around pendulum mill is low and the time required for interpass the back-up roll, and ultimately emerges from the mill for annealing is avoided. collection by an exit tension reel. The small bend roll with strip wrapped around a portion of its circumference is cra- 12.2.5 Contact-Bend-Stretch Mill (Coffin 1967) dled in the gap between two work rolls. The strip is subject to back tension by the entry tension reel and front tension by In conventional rolling, when a stock is reduced by the the exit tension reel in order to avoid slipping between the contact pressure of rolls while at the same time it is stretched strip and the two work rolls. by longitudinal tensile pulls at the entry and exit plane of the rolls, the operation may be termed as ‘contact-stretch’ roll- The degree of reduction of the strip is a function of the ing. If a rolling process includes plastic bending of the stock pressure or force between each work roll and the bending together with the rolling pressure and the longitudinal ten- roll, the amount of plastic strain produced in bending the sion, it is called ‘contact-bend-stretch’ (C-B-S) rolling pro- strip over the bending roll, the strip tension and the instan- cess. Thus, in C-B-S rolling, invented by L.F. Coffin, the taneous yield stress of the strip material. Reduction taking stock is subjected to simultaneous actions of squeezing, place at two points of contact between the two work rolls bending and pulling. This rolling process is applied to and the bend roll is determined by a ratio of peripheral reduce strip and foil. Coffin’s observation is that less rolling speeds between the work rolls. The reduction is determined pressure and longitudinal tension are required to make the and controlled by utilizing the peripheral speed ratio in place strip thinner and longer if plastic bending is combined with of setting a rigid roll gap as performed conventionally. the forces usually applied in a rolling mill. Longitudinal gauge uniformity is preferably maintained by adjusting the speed ratio. The gap between the two work Figure 12.14 shows one form of C-B-S rolling mill, in rolls decides the relative location of the bend roll, and this which a floating bend roll of smaller diameter is included in gap is smaller than the diameter of bend roll so that the axes a four-high mill; i.e., the mill consists of five rollers around of the bend roll and work rolls are offset. Smaller the offset which the strip travels in four loops. The two work rolls angle, the greater will be the roll-separating force and the rotate in the same direction, rather than in opposite direc- smaller strip tension necessary to maintain the set peripheral tions, as in conventional rolling mills. The two work rolls speed ratio. serve to back up or support the bend roll, while each of them is supported by one bigger back-up roll to prevent roll Some advantages of this rolling process are: bending in the rolling direction. The strip coming from an entry tension reel enters the mill, bends around the top work • Larger reductions in thickness per pass are possible. roll, is reduced between the top work roll and floating bend • Intermediate annealing required is less. Hence, the pro- roll, then bends around the floating roll, is again reduced cess appears to be advantageous for rolling of especially Back-up those materials that are highly strain hardened, such as roll stainless steel. • Crowning of rolls is not needed because the floating roll Floating Work bends and adjusts itself to position. bend roll roll • Reduction across the width of the strip is constant due to absence of the roll bending forces which are usually Tension Work present in conventional rolling mills. reel roll • Gage control and strip flatness are improved due to substantial elimination of roll flattening; i.e., more uni- Tension form strip thickness is achieved. reel • It is possible to roll stronger or wider or thinner materials in this mill than in conventional mills, since deformation Back-up of the strip is easier and the deformation forces involved roll are smaller. • The cost, overall size and weight of this mill are con- Fig. 12.14 Contact-bend-stretch four-high rolling mill siderably less than those of the conventional rolling mill of similar capacity. However, the disadvantage of this mill is that the rate of production is slower than that with four-high mills.

532 12 Rolling Fig. 12.15 Universal mill VERTICAL ROLL TOP HORIZONTAL ROLL FOR EDGE ROLLING BOTTOM HORIZONTAL ROLL 12.2.6 Universal Mill Although rolling of solid sections starts with a cast ingot, metal powders may be directly rolled to produce dense strip A universal mill, as shown in Fig. 12.15, is a combination of or sheet by the technique known as powder rolling (Stur- two rolling mills, one with two large-diameter horizontal geon and Taylor 1972). In this process, a metal powder is fed rolls and the other with a pair of vertical rolls used for edge into the gap of a pair of rolls which compact the powder into rolling. Vertical rolls control the width of stock and maintain a ‘green strip’. Subsequently, the strip is sintered to increase the desired width and preserve the edges of stock, while at its density and strength and may be further hot worked the same time horizontal rolls reduce the thickness of stock. and/or cold worked and heat treated. This process eliminates the initial breakdown step of the ingot at elevated tempera- In case of a two-high reversing mill, the work-piece may ture practised in conventional rolling of solid sections, with a be passed to and fro and turned 90° from 10 to 20 times corresponding large reduction in capital cost required to be during intermediate edging passes through edging grooves in invested for hot-rolling equipment, and this economical the rolls to preserve the edges of stock and maintain the factor is a major advantage of this process. Other advantages desired width. This results in a relatively low rate of pro- include the formation of very fine grain size, the mini- duction. A universal mill may be used to avoid edging mization of preferred orientation and a minimum level of passes and thus increases the rate of production. contamination in hot rolling. There are several other kinds of mills, such as Mannes- 12.3.1 Hot Rolling mann mill, plug mill, pilger mill and reeling mill, etc., which have been discussed in connection with the production of Prior to hot rolling of blooms, billets and slabs, the surface of seamless pipe and tube in Sect. 12.18.1. However, the rolling the material is usually conditioned (prepared) in order to mills can also be classified on the basis of the types of prod- maintain a better surface quality of the product. Conditioning ucts, viz. blooming mill, slabbing mill, plate mill, hot strip is done by grinding, chipping or burning with an oxygen lance mill, cold strip mill, billet mill, merchant mill, section mill etc. to eliminate surface defects such as scabs, seams and silvers. 12.3 Rolling Practice Hot-rolling mill used to breakdown cast ingot into bloom or slab is called cogging mill, also known as blooming mill A cast ingot having coarse-grained, porous and brittle when bloom is produced, or slabbing mill when slab is structure is initially broken down by hot rolling into a produced and it basically acts as primary roughing mill. wrought structure, with refinement of grains and improve- These mills are usually two-high reversing mills with ment of homogeneity, strength and toughness. Since a small grooved rolls, which are used to control the shape changes volume of metal at any given time is subjected to the roll during blooming operation. The roll diameters vary from 0.6 pressure and deformation during rolling, similar to forging to 1.4 m, and the mills are designated by the diameter of the processes but unlike extrusion, hot-rolling equipment of rolls, e.g. a 1.2 m blooming mill. Ingot is reduced often to a moderate loading capacity can be used to deform very heavy small extent during the initial rolling passes. To remove ingots. Various sections such as round, hexagon, angle, initially heavy scale from the surface, the ingot is laid on its channel, I-beam and railroad rails are produced by hot edge and rolled. Thereafter, the ingot is turned 90° so that it rolling using grooved rolls. For such conditions, the rolls are lies on flat surface and then rolled to reduce its thickness. usually designed empirically. The evaluation of working During hot rolling, the width of the ingot spreads appre- loads for these complicated conditions will not be attempted ciably. During intermediate rolling passes, the ingot is turned in the following text, but these working loads may be 90° so that it lies on its edge and then passed through edging roughly approximated from the knowledge of theory for flat grooves in the rolls to preserve the edges and maintain the rolling of comparable cross-sections.

12.3 Rolling Practice 533 desired width. This operation caused by a two-high reversing ground to remove scale from the surface developed during hot primary mill may occur for 10–20 times and reduces the rate rolling or any other surface defects. Hot-rolled strip, usually of production. To improve productivity two-high reversing obtained from continuous hot-strip mill, may be used as the mills may be replaced by universal mill. Three-high mill starting material to produce cold-rolled sheet, although cer- may also be used for rolling of bloom, slab, sections, etc. tain copper alloys may be cold-rolled directly from the cast state. Percentage of rolled products finished by cold rolling is To reduce bloom into a billet by hot rolling, the most usually greater for non-ferrous metals than for steel. commonly used billet mill is a two-high tandem continuous mill. Bars and rods are usually hot rolled from billet using For cold rolling of sheets of steel, copper and aluminium merchant mill. The merchant mill may be two-high tandem alloys, three to five stands of four-high continuous tandem continuous mill, or three-high mill, where both mills must mills or single-stand four-high reversing mils are used. have grooved rolls. For rolling of tool and high alloy steels, Generally, both types of mill have arrangements to apply two-high drag-over mill with grooved rolls may be used as front and back tension to the work-piece. Compared to the merchant mill. The cross-section of the billet is reduced in continuous mill, the reversing mill requires a less capital two directions to produce bars and rods. The billet is reduced investment and is more flexible and used often to produce in only one direction in any single pass and so it is rotated parts that vary widely in dimensions. The continuous tandem 90° on the next pass. Most merchant mills are equipped with mill has higher speed, results in lower labour costs and is guides for feeding the billet into the roll grooves and preferred to the reversing mill, where large-scale productions repeaters for reversing the direction of the bar and feeding it are involved. For example, a five-stand continuous mill can back through the subsequent roll pass. A roughing stand, a deliver product at a speed of 30 m s−1. strand stand and a finishing stand are commonly installed for production of bars and rods. The total reduction produced by cold rolling will nor- mally vary from about 50 to 90%. It is desirable to distribute The plate mill used to produce plates from reheated slab the total reduction as uniformly as possible over various may be a four-high mill used for hot-rolling. The equipment passes, except lass pass, with maintaining closeness to the for producing strip and sheet from reheated slab is called a maximum possible reduction for each pass. The reduction in continuous hot-strip mill. In this production process, the the lass pass is generally kept to a minimum to allow better reheated slab is first passed through a scalebreaker mill, then control of surface finish, gage and flatness of the product. usually through a four-stand four-high continuous mill for For developing cold-rolling schedules, one reasonable roughing operation and finally through a six-stand four-high method (Oliver and Bowers 1965) is to adjust the reduction continuous mill for finishing operation. The width of the in each stand in such a way that a constant rolling load is strip is controlled by using vertical edging rolls during produced. roughing operation, and scale is removed by spraying high-pressure water jets on the strip. When the strip leaves To improve the productivity, two or more layers of flat the last finishing stand, it is either cut into desired lengths or metal are cold rolled together; this type of flat rolling oper- collected in a coiler in the form of continuous long lengths. ation is known as pack rolling. For example, two layers of The planetary hot-rolling mill can also be used to produce aluminium foil are pack rolled, in which one side of each foil coiled strip from a hot slab. is shiny and another side is matte. The side which is in contact with the foil surface will have a dull matte finish, and Seamless tube or pipe is produced by hot rolling using the side which is in contact with the polished roll surface barrel-shaped rolls, such as Mannesmann mill. For most hot looks bright and shiny. rolling of non-ferrous alloys, two- and three-high mills are employed, although four-high continuous mills are used to When annealed mild steel is stretched during deep hot-roll aluminium alloys. Most non-ferrous alloys because drawing or sheet forming operation, it undergoes yield-point of their lower flow stress and smaller ingot sizes allow the elongation, which results in inhomogeneous deformation use of smaller rolling mills. and causes surface markings, known as Lüders bands or stretcher strains, which cannot easily be concealed by 12.3.2 Cold Rolling painting. To avoid the formation of such surface irregulari- ties, the annealed steel sheet is subjected to a small cold Sheet, strip and foil are produced by cold rolling, which reduction of 0.5–2% in area, called temper rolling or skin provides superior surface finish, better dimensional accuracy pass. Temper rolling not only results in an improved surface and improved strength and hardness (imparted due to strain by eliminating the yield-point elongation but also in an hardening in cold reduction), in comparison with hot-rolled improved flatness. When a sheet after rolling exits from the products. Prior to cold rolling, the starting material is pickled roll gap, it may not be always sufficiently flat. To improve with acid, treated mechanically such as blasting with water or the flatness of the rolled sheet, other frequently applied rolling operations are roller levelling and stretcher levelling. In a roller-levelling machine (roller leveller), a series of

534 12 Rolling (a) Sheet (b) Levelling rolls Rod Rollers Fig. 12.16 Roller levelling to a flatten sheets and b straighten round rods (a) (b) Work rest Workpiece Stationary cylindrical Workpiece Force die Reciprocating die plate Stationary die plate Moving cylindrical die Fig. 12.17 Thread rolling with a two flat threaded die plates and b a pair of threaded rolls small-diameter levelling rolls are arranged in two rows in stationary and another is rotating may also be used to form such a way that the axes of the rolls in the top and bottom threads on round objects, whose axis is parallel to the axes of rows are offset, as shown in Fig. 12.16a. Individual electric the roller dies and which is held in place by application of motor is normally used to drive each roll separately. When force by means of a rest work-piece or other special device. the sheet is passed through the set of rollers, it is flexed Three-roller thread-rolling machine is also available. Roll alternatively up and down and subjected to alternating ten- forming (also called contour roll forming) is particularly sile and compressive stresses due to its alternate bending in suitable for production of long irregular-shaped channel upward and downward directions. Due to these alternating sections. In roll forming with a forming speed generally stresses, the strain-hardened sheet undergoes cyclic strain below 1.5 m s−1, strip is cold rolled by passing it through a softening (see Sect. 8.16 in Chap. 8) due to the Bauschinger series of driven rolls, so that it is progressively bent into effect (see Sect. 2.5 in Chap. 2) and the ductility is enhanced complex shapes without causing appreciable change in the resulting in the flattening of sheet as it comes out of the rolls. thickness of the work-piece. The thickness of cold-roll- Roller levelling consisting of a different kind of roller formed sheet typically varies from about 0.125–20 mm arrangements, as shown in Fig. 12.16b, in which the set of rolls in the top row are arranged at an angle to those in the 12.4 Deformation Zone in Rolling bottom row, is used to straighten the round rod. In the stretcher leveller, there are two jaws, by which the edges of In rolling, only a small part of the stock passes through the roll the sheet are griped and pulled with application of a pure gap, which is subject to deformation at any single moment of tensile force in order to straighten the sheet. time. This part of the stock is termed as the ‘deformation zone’, as shown in Fig. 12.18. With reference to Fig. 12.18, the plane There are special types of cold-rolling processes, such as XX0 is called the ‘plane of entry’ and the plane YY0 is called the thread rolling and roll forming. When some threaded objects, ‘plane of exit’. The arcs XY and X0Y0 are called the ‘arcs of like screws and bolts, are to be produced in mass scale, the contact’. The angle between the entrance plane, XX0; and the process of thread rolling, as explained schematically in centre-line of the rolls, OO0; is called the ‘angle of contact’. Fig. 12.17, is applied. In thread rolling process, round rods or Hence, the angle of contact is \\XOY ¼ \\X0O0Y0 ¼ a: work-pieces having sufficient ductility are passed between two flat threaded die plates out of which one is stationary and From Fig. 12.18, OZ ¼ OY À YZ: If R is roll radius, h1 is another is reciprocating. With each stroke of the reciprocating the thickness of stock before rolling and h2 the thickness of die, threads are formed on the round object. One pair of rolled product after reduction, then threaded rolls (cylindrical threaded dies) out of which one is

12.4 Deformation Zone in Rolling 535 But from (12.6a), cos a ¼ 1 À ðDh=2RÞ ð12:10Þ From (12.9) and (12.10), it follows that O \"  Dh2# \" # 1 1 2R 1 ðDhÞ2 L2 ¼ R2 À À ¼ R2 À 1þ Dh À 4R2 R α R–(Δh/2) \"# LZ Δh/2 =(h1 - h2)/2 ¼ R2 Dh À ðDhÞ2 ¼ R Dh À ðDhÞ2 R 4 R2 4 X h2 Y rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffihffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiiffiffi ) L ¼ R Dh À ðDhÞ2=4 h1 Y´ X´ x = 0 ð12:11aÞ x D=2R As Dh is a very small quantity, so ðDhÞ2=4 may be O´ neglected and the length of the deformation zone, L; may be approximated as pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð12:11bÞ L % R Dh ¼ R ðh1 À h2Þ Fig. 12.18 Geometry of deformation zone 12.4.1 Angle of Bite R À ½ðh1 À h2Þ=2Š ¼ R À ðDh=2Þ ¼ R cos a When the stock first contacts the rolls at the points X and X0; Since OZ ¼ R cos a; OY ¼ R; and YZ ¼ ðh1 À h2Þ=2: the forces exerted by the rolls on the stock at the entry plane are the radial force, Pr; and the tangential frictional force, F: ) Draft; Dh ¼ h1 À h2 ¼ 2 R ð1 À cos aÞ ð12:6aÞ This is shown in Fig. 12.19. The horizontal component of Pr (radial force) acting away from the roll gap tends to reject Or; Draft; h  ai 1 1 2 Dh ¼ h1 À h2 ¼ 2 R À À 2 sin2 ¼ 4 R sin2 a ð12:6bÞ 2 rffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffi ) a ¼ 2 sinÀ1 Dh ¼ 2 sinÀ1 h1 À h2 4R 4R ð12:7Þ O Pr sin α α Pr For small angles of contact a; as in cold rolling, where the X F cos α angle of contact seldom exceeds 8°, it may be assumed that sin2 a=2 % ða=2Þ2; where a is expressed in radian. Hence, FY the draft given by (12.6b) may be approximated as Y´ Draft; Dh ¼ h1 À h2 % R a2 ð12:8aÞ X´ pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð12:8bÞ ) a % Dh=R ¼ ðh1 À h2Þ=R The projected length of the arc of contact is called the O´ length of the deformation zone. So the length of the defor- mation zone, L; can be obtained as follows: L ¼ XZ ¼ R sin a ð12:9Þ ) L2 ¼ R2 ð1 À cos2 aÞ Fig. 12.19 Forces exerted by the rolls on the stock at the entry plane

536 12 Rolling the metal and prevent it from entering the rolls, while the ) Angle of bite ¼ Angle of friction: horizontal component of F (friction force) acting towards the roll gap at the entry plane tends to drag the metal inwards, If the coefficient of friction, l ¼ 0; rolling cannot occur, i.e. into the rolls. If the front face of the stock at the plane of but as l increases, progressively larger and larger stocks will entry is located at an angular position a from the line joining be drawn into the roll throat. the roll centres, the net horizontal force, H; acting on the stock in the forward direction of rolling is given by If Dhmax = maximum possible draft in an unaided rolling pass, then from (12.6b), (12.16) and (12.14b) we get H ¼ F cos a À Pr sin a ð12:12Þ Maximum draft; Dhmax ¼ 4R sin2 amax ¼ 4R sin2 f 2 2 When the force, H; is positive, the work-piece moves in ¼ 4R sin2 tanÀ1 l ð12:17aÞ the forward direction of rolling and gets rolled. On the other 2 hand, when this force is negative, the stock moves backward and unaided rolling is not possible, since the rolls refuse to A simplified approximate relation for the mapxiffimffiffiffiffiuffiffimffiffi draft, bite the stock. Therefore, the condition under which the rolls Dhmax; can be found, if we consider L % R Dh from can bite the work-piece and unaided rolling is possible is: (12.11b) and neglect Dh=2; as it is very small, in the fol- lowing derivation. Now, from Fig. 12.18 geometrically we H ! 0; or; F cos a ! Pr sin a; ½from ð12:12ފ can write ) tan a F=Pr ð12:13aÞ pffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi L R Dh ¼ Dh; or; Dh % R tan2 a; The limiting condition for the unaided entry of the tan a ¼ R À Dh=2 % R R work-piece into the throat of the rolls is: ) Maximum draft; Dhmax % R tan2 amax ¼ R tan2 f ¼ R l2 H ¼ 0; or; F cos a ¼ Pr sin a; ½from ð12:12ފ ð12:13bÞ ð12:17bÞ ) tan amax ¼ F=Pr The approximate (12.17b) may be used under low fric- Usually, it is assumed that Coulomb’s law of sliding tional conditions, but it will be grossly in error for high friction holds good at the contact surface between the rolls frictional conditions. and the stock and therefore Hence from (12.17), the conditions to achieve a large draft in a rolling pass are: F ¼ lPr; or; F=Pr ¼ l ð12:14aÞ where l is the coefficient of friction between the rolls and the • Roll radius, R; must be large, and/or stock at the contact arc. • Coefficient of friction, l; between the rolls and the stock From (12.13a) and (12.14a), must be high. tan a l ð12:15aÞ For the same frictional condition, i.e. when l or f is constant, a larger diameter roll will permit a thicker stock to It means the work-piece cannot be drawn into the rolls if enter the rolls than will a smaller diameter roll, provided the the tangent of the contact angle a exceeds the coefficient of thickness of the rolled product is the same in both cases, friction l: From (12.13b) and (12.14a), the limiting condi- because the draft Dh will be higher for higher diameter tion for unaided entry of the work-piece into the rolls is: rolls. tan amax ¼ l ð12:15bÞ Let rmax ¼ the maximum possible reduction of area in an If f is the angle of friction, then unaided rolling pass, and w ¼ the width of the stock, which remains constant throughout the rolling process assuming f ¼ tanÀ1 l or; l ¼ tan f ð12:14bÞ rolling to be a plane strain deformation. Now considering the definition of reduction of area, r; given by (12.4), and the From (12.15a) and (12.14b), relation given by (12.17a), we get the relation for rmax in terms of angle of bite as follows: tan a tan f or; a f ; ð12:16Þ rmax ¼ ðA1 À A2Þmax ¼ ðh1 À h2Þmaxw ) amax ¼ f A1 h1w where amax is the maximum possible angle of contact in ¼ D hmax ¼ 4R sin2 f ¼ 4R sin2 tanÀ1 l ð12:18Þ unaided rolling. amax is called the ‘angle of bite’ or gripping h1 h1 2 h1 2 angle.

12.4 Deformation Zone in Rolling 537 12.4.1.1 Artificial Methods of Increasing Angle pyramids because the indented pyramids on rolls of Bite produce projecting pyramids on surface of rolled material, which are subsequently rolled out without In practice, many roll designers consider the angle of bite in trace in the smooth passes, while the projecting hot rolling as 22.5° to 24°, without artificial roughening. If pyramids on rolls produce indented pyramids on an angle of bite higher than the above is desired, means have surface of rolled material, which may cause the traces to be found to increase the coefficient of friction l: The of ragging to stay on the finished product. Even after following methods may be used for artificial increase in the gradual wearing of the ragged surface, the knurled angle of bite: surface remains rough and, hence, the roll-gripping of metal does not become worse. The maximum angle of • By using external pushing forces to augment the friction bite by ragging of the roll surface with deep grooving force in the direction of rolling so as to effectively is considered to be 34°, which may be exceeded by counter the increased ejection force exerted by the rolls knurling method of roll-ragging. The jamming of due to the larger angle of bite for a thicker stock. billet in the mills with constant high roll speed and the Sometimes, a stock not being rolled at that instant is used jerks in the rolling train may occur with groove-cut to push the stock being rolled. rolls, but they are eliminated if the rolls are knurled. Other advantages of knurling over groove cutting or • By applying inertia forces arising from high speed of welding on the roll surface include increased roll-life stock caused by high-speed roller tables. These forces as knurling strain-hardens the roll surface, and tend to augment the friction force in rolling direction. improved surface quality of the finished product. • By increasing roll roughness, which may be achieved in various ways by: 1. Using ordinary cast iron rolls that acquire a rough 12.4.2 Neutral Point surface after some amount of wear during rolling. When a stock is undergoing reduction in the rolls, as the 2. Adding a roughening agent in the zone of deforma- stock moves further into the roll gap, its cross-sectional area tion. Sand, iron powder, chalk, etc., are sprinkled on is reduced, so that the velocity of the stock must increase as the roll surface. it passes between the rolls from velocity v1 at the plane of entry to velocity v2 at the plane of exit. 3. Ragging of rolls. During ingot breakdown by hot rolling, where it is desired to achieve a large reduction Since equal volumes of metal must pass a given point per in a short time (particularly in cogging mill), the roll unit time, i.e. from the principle of constancy of volume, we surfaces have deep grooves cut in them parallel to the can write roll axis to increase the effective value of l: This is termed as ‘ragging’ of roll surface. Roll ragging is A1 v1 ¼ A2 v2 frequently applied in order to facilitate the gripping of stock by the rolls. Apart from deep groove cutting, where A1 is the cross-sectional area of the stock before other forms of roll surface ragging may be created by rolling and A2 is the cross-sectional area of the product after chiselling on roll surface and welding metal at dif- reduction by rolling. ferent points on roll surface. Since the poor surface finish of the rolled product is the limitation of roll * A1 [ A2; ) v1\\v2; ragging, it is generally applied only in rolling mills producing semifinished products. In case of roll rag- i.e. the exit velocity v2 must be greater than the entrance ging by welding, since places of welding become velocity v1: hardened, they are difficult to machine during recon- ditioning of the rolls. Another method to improve the Suppose the peripheral velocity (or surface velocity) of gripping of stock by the rolls is ragging of roll surface the rolls is vr: It is to be noted that vr ¼ Rx; where R is the by knurling, which has been applied with satisfactory roll radius and x is the angular velocity of rolls expressed in results on rolls in the blooming mill and on the first rad=s: Now, let us consider the following cases: stand of the continuous billet mill. Knurling on the roll surface is made by forming shallow rectangular 1. v1\\v2\\vr; which is shown in Fig. 12.20a. This is the pyramids, which may be either indented into or pro- case of skidding. jecting out of the roll surface. On roll surface, the indented pyramids are preferred to the projecting 2. vr\\v1\\v2; which is shown in Fig. 12.20b. This cannot happen in unaided rolling. This can however happen in drawing.

538 Rolls (b) 12 Rolling v2 (a) v2 vr Velocity Metal Velocity Metal v1 Rolls Roll gap v1 Roll gap Entry plane x vr x=L x (c) Exit plane Entry plane Exit plane x=0 x=L v2 x=0 vr Rolls Velocity Metal v1 Entry plane N or N´ Exit plane x=L Roll gap x=0 x Fig. 12.20 Peripheral velocity of rolls, vr; and entry and exit velocities of metal, v1 and v2; in roll gap. a v1\\v2\\vr: b vr\\v1\\v2: c v1\\vr\\v2 3. v1\\vr\\v2; which is shown in Fig. 12.20c. Since the The plane NN0; obtained by joining two neutral points N rolls have to grip the stock in unaided rolling, vr cannot and N0 on the arc of contact, is called the ‘neutral plane’, as be less than v1; nor greater than v2; and, in general, vr lies shown in Fig. 12.21. The neutral plane, NN0; divides the between v1 and v2 deformation zone into two subzones: the ‘entry zone’, XNN0X0; and ‘exit zone’, NYY0N0: As the velocity of the stock increases progressively from v1 at the entry plane to v2 at the exit plane, there will be a In the entry zone, the stock velocity is less than the point N or N0 on the arc of contact where the velocity of the peripheral velocity of rolls, so this part of the deformation stock equals the velocity of the roll. This point is known as zone is also called the lagging zone, while in the exit zone the ‘neutral point’ or the ‘no-slip point’. So, it is the point on the stock travels faster than the peripheral velocity of the the arc of contact between the roll and the stock at which roll, i.e. it leads the roll, so this part of the zone of defor- mation is also called the forward slip zone. Hence, in the • the surface velocity of the stock equals the peripheral entry or lagging zone, the frictional force on stock acts in the velocity (or, surface velocity) of the rolls, forward direction of rolling so as to draw the stock into the rolls and in the exit or forward slip zone the direction of • there is no slip and frictional force is reversed so that it acts to oppose the • the direction of friction force reverses. delivery of rolled product from the rolls.