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61 views14 pagesRolling and Wxtrusion
Uploaded by
Bharat SharmaCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
/ 14
FrgURE 19.1 the
‘olin process
(epeciealy, ft roling.
ees
‘Rollingisadeformationprocesin which the thicknes ofthe work isreduced by compeesive
forces exerted by two oppexing rolls The ris tate as histated ia Figure 19.1 to pull and
simultancousl sucezc the work betwcca them. The basic proces shown incur igre Mat
rolling used to reduce the thicknes‘ ofa rectangular crosseetion. Aces relate process
shape rolling in which «square cross section fs formes into shape such san beam,
Mest roiling pracees are very capital intrsive, ering massive pieces of p=
‘ment called relling mill to perform them. The highinvesment cost requires the mils tobe
‘se for prouetio in Hage quantities of standard items such ss sheets aad plates Most
rolling i carried out by bot working called hot rolling, owing co the large amount of
{deformation required, Hotrolled metal is generally fee of residual stresses and its
Drcperties are tropic. Disadvantages of ho rolling ae thatthe pret cannot be
held to cowe tolerances, and the surface has a characterni ode sale
Stcelmaking provides the most common appliatica of ling al operations (His:
torial Note 19.)- Lat us follow the sequence of steps tel rolling mil ostate the
‘aro ofpeaducts mace, Similar steps ace inher bse metal indies The work starts
foul ascast soe ingot thts just solute. While ii stl bt, dhe ingot spaced in a
fumace where it remains for many’ hours uni it has reached a uniform temperate
throughout, so that the mil willow consistently during rolling Far stel, the dsied
temperature for sling is around 207°C (2200"F). The heating operations alle soaking,
tnd the fumaces in which its cried ou are called soaking pits
rom soaking the igotismovedto the rolling mil, whet itstollediato one of three
intermediate shapescalled blooms billets or slab A bloom has square eros seston 150
tim» 150mm (6m » Gin) oF laraer. A slab soled frm an ingot ors bloom and as a
‘ectangular cross section of width 250 mm (Dn) or more and thickness 40 mm (15 in) oF
‘more. billers oiled oma oom andissquare with dimensions mam (Sin}on aside
Dr larger. These intermediate shapes are subsequently rolled int fins proc shapes
‘Bloons are riled int tetra shapes and rails for railroad tracks Billesare tiled
mors andreas These shapes are the raw material for machining, wire drawing, opine
and eter metalworking provesses Slabs are rolled into pte sheetsanltepn Hot-olled
Plates are used in shipbuilding. bridges boilers welded structures for various heavy
chines, tubes and pipes andl many other predicts. Figure 192 shows some of these
led steel products. Furher flattening of hot-rlled plates and sete often accom
plished by cold roling inorder to prepare them fr stibquent sheet metal operations
{Chapter 20) Cold elliag strengthens the metal and permis a tighter tolerance on
‘thickness. In dition, the surface of he colle sheet is absent of sale aid peneally
superior tothe corresponding hotell product. These characterises make cok-olled
sheets strips and cods ideal for stampings, exterior panels, and other pasts of products
‘ranging from automobiles to appliances and office furniture
ot
Drover ot werk ow
Wore 19.1.1 FLAT ROLLING AND ITS ANALYSIS
Plat rolling i illustrated in Figures 19.1 and 19.3. It involves the rolling of slabs, strips,
sheets and plates - workpartsof rectangular eros section in which the width is greater
than the thickness, In flat rolling, the work is squeezed between to rolls so that its
thickness is reduced by an amount called the draft
eh (19.1)
Where d= draft, mm (in); Zp = starting thickness, mm (in); and
(in). Drafts sometimes expressed asa fraction of the starting stock
final thickness, mm
sickness called the
Intermediate ood tore Fina oe orm
Strucural shapes
Beem
FIGURE 19.2 Some of
the steel products made ~
2 roling mil (Chapter 19/Bulk Deformation Processes in M
reduction:
(192)
‘where r= reduction, When 2 series of rolling operations are used, reduction is taken as
the sum of the drafts divided by the original thickness
In addition to thickness reduction, rolling usually inereases work width, This is
called spreading. anit tends to he mest pronounced with low widthto-thickness ratios
land low coelficents of fiction, Conservation of materi preserved, so the volume of
metal exiting the rolls equals the volume entering
fotole = rly (93)
Where and wyare the beforeandafter work wis, mm (in):and Land Lyarethe before
‘and afte work lengths. mm (in) Similarly, before and after volume rates of materia low
must be the same, so the before and after veosities canbe related:
move (93)
where and vy are the entering and exiting
The rolls contact the work along an are defined by the angle @ Each ell has rays
and is rotational speed gives ita surface velocity v, This velocity is greater than the
‘entering peed of the work x and Tes than its exiting sped ui Since the metal few is
‘continuous, there isa gradual change in velocity ofthe work between the rll, However,
there one paint along the are where work velocity equals rol velocity. Thisiscalled the
‘no-slip point, also known asthe neutral point. On either ie ofthis point. slipping and
friction ovur between rll and work, The amount of slip between the rolls andthe work
{can be measured by means of the forward slp, a tem used in soling tha s defined
vend
am (os
wheres
rms (Use0),
The tre strain experienced by the workin rolling isbased on heforeandatter Sook
thicknesses. In equation form,
ora sin final (exiting) work velocity ms (sce: anv, = rll peed
aint (09.9)
The te strain ean be used to determine the average flow stess Ty applied to the work
‘material in Hat rolling. Recall from the previous chapter Eq. (182). that
y= Ke.
(9)
Te
“The average flow stres is used to compute estimates of force and power in rolling.
Friction inroling ours witha certain coefficient of friction, athe compression force
‘ofthe oll multiplied by thiscoetficient of ction results ina frtion force betweenthe rolls
andthe work Onthe entrance side ofthe no slip point tion force sinene direction. andon.
the ther sid itisinth opposite direction, However: the twoforcesare nek eal The rion
force cathe entrance sid spreatersathal the net forcepulls the work through therolls ths.
were not the esse, rolling would not be possible. Thore Knit wo the maximum pose
‘daft that can be accomplished in ft rolling with a given coefficient of fiction, dined by
(198) Section 19.1/Rolling 399
where dyjgc= maximum draft, mm (in): 4 — coefficient of frietion; and R = roll radius mm
(in). ‘The equation indicates that if friction were zer0, draft would be zero, and it would be
impossible to accomplish the rolling operation,
Coefficient of friction in rolling depends on lubrication, work material, and
‘working temperature. In cold rolling, the value is around 0.1; in warm working, atypical
value is around 0.2; and in hot rolling, x. is around 0.4 [16]. Hot rolling is often
characterized by a condition called sticking, in which the hot work surface adheres
to the rolls aver the contact arc. This condition often occurs in the rolling of steels and.
high-temperature alloys. When sticking occurs, the coefficient of friction can be as high
a8 0.7. The consequence of sticking is that the surface layers of the work are restricted to.
‘move at the same speed as the roll speed v, and below the surface, deformation is more
severe in order to allow passage of the piece through the roll gap.
Given a coefficient of friction sufficient to perform rolling, roll force F required to
maintain separation between the two rolls can be computed by integrating the unit roll
pressure (shown asp in Figure 19.3) over the roll-work contact area. This can be expressed:
Paw [pdt (199)
where F = rolling force, N (Ib); = the width of the work being rolled, mm (i
roll pressure, MPa (Ib/in’); and = length of contact between rolls and work, mm (in).
The integration requires two separate terms, one for either side of the neutral point
Variation in roll pressure along the contact length is significant. A sense of this
variation ean be obtained from the plot in Figure 19.4. Pressure reaches a maximum
at the neutral point, and ails off on either side to the entrance and exit points. As
friction increases, maximum pressure increases relative to entrance and exit values. AS
friction decreases, the neutral point shifts away from the entrance and toward the exit,
in order to maintain a net pull force in the direction of rolling. Otherwise, with low
friction, the work would slip rather than pass between the rolls.
‘An approximation of the results obtained by Eq, (19.9) can be calculated based
on the average flow stress experienced by the work material in the roll gap. That is,
F=Yywk (19.10)
FIGURE 19.3 _ Side view of flat
rolling, indicating before and after
thicknesses, work velocities, angle
of contact with rolls, and other
features. 400 Chapter 19/Bulk Deformation Process in Metal Working
Example 19.1 Flat
Rolling
No slip oie
Dect fing
FIGURE 19.4 Typical variation in pressure
along the contact length in lat roling. The
peak pressures located atthe neutral point Lo +|
The area beneath the curve, representing the
integration in Eq (199), isthe roll force F nance eat
where Vj = average flow stress from Eq, (19.7), MPa (Ibi?) and the product wis the
foll-work contact area, mm* (in), Contact length can he approximated by
[Rt
a (9)
“The torqucinroling can be esti
the work as it pases hetween the rolls and that itacts with a moment arm of one
contaet length. Thus, torgue for each ral s
tated by assuming thatthe oll force is centeredon
ifthe
r
FL. (19.2)
sured todrive each rolls the product of torgue and angular velocity.
rotational speed ot the roll Thus, the poster fr each
roll is 227. Substituting Eq. (19.12) for torque in this expression for power, and
‘doubling the value to account for the fact that a rolling mill ensits of to powered
rolls, we get the following expression:
(1913)
where P= poner, Js or W (in-bimin): N
force, N (by: and L = contact length, m (i).
A. 300-nm-wide strip 2S-mm thick is fed through a rolling mil with two powered rlls
‘ach ofradivs = 250 mim, The work thickness i to be reduced 0.22 mm in ane passat a
‘oll speed of 30 revimin, The work material has low curve defined by K=275 MPa and
‘n= 0115. and the coetfcient of friction between the rolls and the work is assumed to be
0.12. Determine if the friction is sufficient to permit the rolling operation to. be
accomplished, Ifs0, calculate the rol fore, torque, and horsepower.
Solution: The drat attempted inthis rolling eration
‘
5—22=3mm Section 19.1/Rolling 401
From Eq, (19.8), the maximum possible draft for the given coefficient of fi
(0.12)°(250)
omm
ys
Since the maximum allowable draft exeeeds the attempted reduction, the rolling
‘operation is feasible. To compute rolling force, we need the contact length Zand the
average flow stress V,. The contact length is given by Eq, (1911)
/B025 — 22) = 27.4 mm
Ts
Rolling force is determined from Eq, (19.10):
F=175.1(300)27.4) =
444, 786N,
‘Torque required to drive each roll is given by Eq. (19.12):
T
0.5(1, 444, 786)(27,4)(L0) = 19, 786 N-m_
and the power is obtained from Eq. (19.13):
P = 2x(50)(1, 444, 786)(27.4)(10-) = 12,432,086 N-m/min = 207,201 N-mis(W)
For comparison, let us convert this to horsepower (we note that one horsepower =
745.7 W):
Itcan be seen from this example that large forces and power are required in rolling.
Inspection of Eqs. (19.10) and (19.13) indicates that foree and/or power to roll a strip of a
‘given width and work material can be reduced by any ofthe following: (1) using hot rolling
rather than cold rolling to reduce strength and strain hardening (K and m) of the work
‘material: (2) reducing the draftin each pass; (3) using a smallerroll radius K to reduce force;
and (4) using a lower rolling speed N to reduce power. este
Extrusion isa compression pracessin which the work metals freed to flaw through die
‘openingto produce a desired cross-sectional shape The processcane kenedto squeezing
toothpaste out of a toothpaste tube, Extrusion dates from around 1990 (Historical Note
193), There are several advantages of the modern proces: (1) a vaiely of shapes sre
posible, especially with hot extrusion: (2) grain structure and strength properties ae
enhanced in old and warm extrusion (3) fairly close tolerances are possible. especialy in
coldestrsion:and (4)insome extrusionoperations little orno wasted material isereated
However limitation i that the cross section of the extruded part mist be uniform
‘throwehoot its leneth
Historical Note 19.3
Baesion as an insta
1800 in England, duringthe Indl Revolution when horizontal extrusion pest was bil for exrudig tale
‘hat country was leading the worl intechnoogical with ighermeking pois than kad. Te feature that
innovations The ivertioncondted ofthe ft hyanlic made his possible wasthe use of «dur block that
pres for exruding ead pipes. An important step forward separate the ram fom the work ill
Barusion
process wasinvertedaround yas made Germany around 1880, when the fst
19.5.1 TYPES OF EXTRUSION
[Exirssoniscarredoutin various ways One important stint isbetween direct extrusion
and ingitectextrusion, Another lasiicaion i by working temperature col warm, hot
‘extrusion. Final extrusions performedas itera continuous processor adserete procs
Direct versus indirect Extrusion Direct extrusion (also called forward extrusion) i
ilsteated ia Figure 1930. A metal bilet is loaded into a container. and 4 ram
‘compresses the material forcing i 1o low through ane or more openings ina die at
the opposite endof the container. Asthe ram approaches the die, a small portion ofthe
billet remains that cannot he forced through the die opening. Thisextra portion, called
‘the butt, Separated from the product by cutting it just beyond the exit ofthe die FIGURE 19.20 Direct,
extrusion
FIGURE 19.31
{a} Drect extrusion to
produce aholow or
Semitotow cross
section; hollow and
(© semchollow cross,
sections
Container
(Fal wore tape
‘one
‘Oncoftheproblemsindirestextrasonisthesigifeantrston that exists hetween the
‘work surface and the walls ofthe omtainer asthe billet i force to lide toward the ie
‘opening. Ths friction causes substantial increase in the ram force required in dive,
{extrusion In hot extrusion, the tition problem is aggravated by the presence of an oxide
‘ron the surfice ofthe ill. Tis ide layer can cane defectsnthe extrude product. To
address thse problems, a dummy block isoften used between the ram and the Work bile
‘The diameter ofthe dummy block issiphy smaller than the billet diameter, thatanarcow
ring of work metal (mostly the oxide layer lft inthe container leaving the final product
free of oxides
Hollow sections (eg. tubes) are possible in diet extrusion ty the process setup in
Figure 1931. Testing bletispreparedwithahok parallel toiteanis Mivallowspassageoa
‘mandrel thats tached toe dumany block. Asthe billet iscompressed the materaisforced
{orlow through the clearance between the mandrel and the die opening. The resing eros
sections tubular. Semi-ollow crew scctional shapes ae usually extrudedin the same way.
The tating Dllet in direct extrusion is usally round in eros section, tu the final
shape is determined by the shape ofthe die opening Obvious, the largest dimension of
the die opening must be smaller than the diameter of the billet
In Indirect extrusion, also called backyard exiasion ad reverse extrusion, Fis
‘ure 192{a),the die s mounted tothe ram rather than atthe opposite end ofthe container.
‘Asthe ram penetrates into the work, the meal sored ow tech the clearance in
p-— Corner
— Final no spe
oa
Manel
@
© © roy
hap
Holow ram
Fal wore
‘shape
FIGURE 19.32
19/Bulk Deformation Processes in Metal Working
= a
.
; _
=
Indirect extrusion to produce (a solid cross section and (b) ahollow cross section,
sirection oppesite to the motion of the ram. Since the billet is not forced 10 move relative
{othe container, there sno friction at the container walls andthe am force istherefore lower
than indirect extrusion. Limitations of indirect extrusion ar imposed bythe lower rigicity of
the hollow ram and the dificult in supporting the extruded product as it exits the die.
Indirectextrusion can produce hollow tubular) cross sections. asin Figure 1932(b)-In
this method, the ramis pressed into the billet, forcing the material tolow around the ram and
take acupshape. There are practical limitationson the length ofthe extruded partiatean be
‘male by this method. Support of the ram hecomes problem as work length increases.
Hot versus Cold Extrusion Extrusioncan be performed either hotoreold.dependingon
‘work metal and amount ofstrain owhich itissubjected during deformation, Metalsthat are
typically extruded bot include aluminum, copper, magnesium, zinc tin, and thei alloys
‘These same metals are sometimes extruded cold, Stet alloys are usually extruded hot,
although the softer, more ductile gradesare sometimes cold extruded (eg low carbon steels
and stainlessstel). Aluminumis probably the most ideal metal for extrusion (hot and cold).
and many commercial aluminum products are made by ths proces (structural shapes, door
and window frames, et.
Hot extrusion involves prior heating of the billet to a temperature above its
reerystallization temperature. This reduces strength and inereases ductility of the
‘metal, permitting more extreme size reductions and more complex shapes to be
achieved in the process Additional advantages include reduction of ram force,
increased ram speed, and reduction of grain flow characteristics in the final product.
Cooling of the billet as it contacts the container walls isa problem, and isorhenmal
‘extrusion is sometimes used to overcome this problem, Lubrication is critical in hot
extrusion for certain metals (e. steels). and special lubricants have been developed
that are effective under the harsh conditions in hot extrusion. Glass is sometimes used
asalubricant in hot extrusion; in addition to reducing fiction, italso provides elfective
‘thermal insulation between the billet and the extrusion container.
Cold extrusion ana watm extrusion are generally used to produce scree pats often
infinished (or near finished) form. The term impact extrusion isused toindicate high-speed
coldextrusion,and thismethodisdescribedin more detailinSection 194. Some important
advantages of cold extrusion include increased strength due to strain hardening, close
tolerances improved surface finish, absence of oxide layers andhigh production rates Cold
extrusion at room temperature also eliminates the need for heating the starting bile
Continuous versus Discrete Processing _A true continuous process operatesin steady
sate mode for an indefinite period of time, Some extrusion operations approach this ideal 19.5.2 ANALYSIS OF EXTRUSION
Lets use Figure 19.38 asa reference in discussing some ofthe parameters in extrusion. The
siagram assumes that both billet and extrudate are round in ctess section, One important
parameter is the extrusion ratio, also called the reduetion ratio, The rato is defines!
(19.19)
where —extrusin ratio, —cress sectional ate ofthe stating billet mm? (i2andAy—
{inal cross-sectional area of the extruded section, mm” (in). The ratio applies for both ditect
aningrect extrusion. The vale of, canbe wed todetermine te Sramiextruson en
that ideal deformation ours with frtion an no redundant work
(19.20)
Under the assumption of ideal deformation (no friction and no redundant work), the
‘pressure applied by the ram to compress the billet though the die opening depicted in our
igure can be computed as follows
p=Trlnre (1921)
‘where V = average flow stess during deformation, MPa (Ibn). For convenience, we
restate Eq, (182) ftom th previous chapter:
ke
We Tn
Infact, extrusion isnot a fretioness process andthe previous equations grossly
"underestimate te strain and pressure in an exttsion operation, Friction exists between
the die and the work asthe billet squeezes down and passes though the die opening. In
direct extrusion. friction alo exists between the container wall and the billet surface. The
elicct of fiction i to increase the strain experienced by the metal. Thus, the actual
presure is greater than that given by Eq. (19.21), which assumes no tition.
Remaining bet ength
FIGURE 19.33 Pressure
‘and ather variables in
direct extrusion, Various methods have been suggested to calculate the actual true strain and
associated ram pressure in extrusion [1], [3}. (6). (11), [12]. and [19]. The following
empirical equation proposed by Johnson [11] for estimating extrusion strain has gained
considerable recognition:
eg =atbinr (922)
where €, = extrusion strain; and a and b are empirical constants for a given die angle,
‘Typical values of these constants are: a = 0.8 and b = 1.2 to LS. Values of a and b tend to
increase with increasing die angle.
The ram pressure to perform indirect extrusion can be estimated based on
Johnson's extrusion strain formula as follows:
v=, (19238)
where Y; iscalculated based on ideal strain from Eg, (19.20), rather than extrusion stain
in Eq. (19.22).
In direct extrusion, the effect of friction between the container walls and the billet
causes the ram pressure to be greater than for indirect extrusion, We can write the following
expression which isolates the friction force in the direct extrusion container:
pad?
7
= Hp.xDol.
where py = additional pressure required to overcome friction, MPa (Ib/in?): 2D,2/4
billet cross-sectional area, mm (in’); 1 = coefficient of friction at the container wall; p.
pressure of the billet against the container wall, MPa (Ibfin’); and 7DoL, = area of the
interface between billet and container wall, mm’ (in?)-The right-hand side of thisequation
indicates the billet-container friction force, and the lefi-hand side gives the additional ram
force to overcome that friction. In the worst case, sticking occurs at the container wal so
‘that frietion stress equals shear yield strength of the work metal:
upynDol = ¥inDol.
where ¥; = shear yield strength, MPa (Ibn!) If we assume that ¥, = ¥j/2, then py
redhces to the following:
Based on this reasoning, the following formula can be used to compute ram pressure in
direct extrusion:
(19.23)
where the term 21/D,, accounts for the additional pressure due to fiction atthe container
billet interface. L is the portion of the billet length remaining to be extruded, and Dis the
original diameter of the billet, Note that p is reduced as the remaining billet length decreases
uring the process Typical plots of ram pressure as a funetion of ram stroke for direet and
indirect extrusion are presented in Figure 19.34. Eg, (19.23b) probably overestimates ram
pressure. With good lubrication, ram pressures would be lower than values calculated by this
equation.
“MN am force in indivect or diret extrusion is simply pressue p from Egg. (1928) or
(19.230), respectively, multiplied by billet area Ay:
F=pAy (1928 nple 19.3
usion
sures
Section 19/Extrusion 425
FIGURE 19.34 Typical plots of
‘am pressure versus ram stroke
{and remaiing bilet length) for
ddect and indirect extrusion. The
higher values in direct extrusion
result from fiction at the
container wall. The shape ofthe
inital pressure buildup at the
bginning ofthe plot depends on
de angle higher die angles
cause steeper pressure
buldups). The pressure increase
at the end ofthe stroke is related
to formation ofthe butt. omaining bt ong, |
Dec extrusion
Ram pressure
Butt oration
Indirect extrusion
TT
ual extrusion begins
where F = ram force in extrusion, N (Ib). Power required (o carry out the extrusion
‘operation is simply
Pay (19.25)
where P = power, Js (in-lbvmin): F= ram foree, N (Ib): and
ram velocity, ms (inimin).
A billet 75 mm long and 25 mm in diameter isto be extruded ina direct extrusonoperation
‘with extrusion ratio ry = 40. The extrudate has a round cross section, The die angle (half-
angle) = 90°, The work metal has a strength coefficient = 415 MPa, and strain-hardening
exponent = 0.18. Use the Johason formula with « = 03 and b= LS toestimate extrusion
strain, Determine the pressure applied to the end of the billet as the ram moves forward.
Solution: Letusexamine the ram pressure at billet lengths of £ =75mmn (starting value),
L=50mm, =25mm, and L.=0, We compute the deal tee strain, extrusion strain using,
Johnson's formula, and average flow stress:
L=75 mm: Withadie angle of 4F, the billet metal isassumed to be forced through the die
opening almost immediately; thus our calculation assumes that maximum pressure is
reached atthe billet length of 75 mm. For die angles less than 90°, the pressure Would build
toa maximum as in Figure 19.34 as the starting billet is squeezed into the cone-shaped
portion of the extrusion die. Using Eq. (19.23h),
p= s09(em 2223)