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7/29/2019 Mechanical Working of Metals (Rolling and Forging)
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Rolling and Forging
Mechanical Working of Metals
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Working Processes
Mechanical
Working
Cold Working Hot Working
Done at temp < Trec Done at temp > Trec
Recrystallization temperature :
temperature at which recrystallization occurs
0.3 to 0.6 Tm
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The Effect of Temperature on
Working Processes
Cold Working
metal strain hardens
after deformation
metal hardens
needs high stress to
deform metal
very good surfacefinish
very good
dimensional tolerance
Hot Working
metal recrystallizes
after deformation
metal stays soft
needs little stress to
deform metal
forms SCALES atthe skin
poor dimensional
tolerance
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Rolling
Process of reducing the thickness of
metals by allowing it to pass through a
pair of rotating rolls
More metal is rolled
than the total of all theother working processes
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Products of Rolling
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Roll Stand
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Rolls for I-beams
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Two-high
mill
Two-high
Reversing mill
Three-high
mill
Types of Rolling Mills
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Types of Rolling Mills
Four-high
mill
Six-high
mill
Cluster mill
(Sendzimir Mill)
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Cluster or Sendzimir Mill
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Planetary Mill
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Rolling Sequence
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Rolling Sequence
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Rolling Defects
ALLIGATORING
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Misaligned Rolls
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Wavy Edges
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Edge Cracking
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Rolling Load Calculation
Ideal Rolling Load No friction involved
Actual Rolling Load Friction involved
Friction normally increases the required
load
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Terminologies used in Rolling
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Rolling Load Analysis
h = draft = ho - hfVo = entry speed
Vf = exit speed
R = roll radius
= angle of bite
= angle between
entrance plane and
centerline
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Ideal Rolling Load
Pressure = Load
Area
Rolling Load = Roll Pressure x Area
hRWRL y =
RL = Rolling Load
y = flow stress (yield stress)W = Width
R = Roll Radius
h = draft
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Considerations on
Actual Rolling Load
At any point in
the arc of contact,
2 forces act on
the metal:
Pr = Rolling Load
F = Friction Load
For unaided entry: Fcos = Prsin = tan
and F = Pr
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Maximum Reduction Possible
Rh
hR
hR =
=
2
tan
= tan
(h)max = 2R
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Sample Problem
Determine the maximum reduction possible
on a piece of metal when using 500 mm
diameter rolls and during cold rolling when = 0.1 and hot rolling when = 0.6
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Determine the final height of an aluminumsheet with initial thickness 60 mm if
subjected to cold rolling ( = 0.2) using
300 mm diameter rolls and assumingmaximum reduction (also consider 3%elastic rebound).
Sample Problem
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Observations in Actual Rolling
Exit speed, Vf > Entry speed, Vo
Assuming constant volume passing through
one point at any given time
Vroll
Vo
Vf
A B
Neutral Point
orNo-Slip
point
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Friction Hill
At the Neutral
Point, there is no
slipping betweenthe metal and the
Rolls
At the Neutral
Point, Friction is
at a maximum!
N
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Actual Rolling Load
Condition Condition
Plain Strain 1.155 < 0.5Slipping Friction
HomogenousDeformation
1.0 > 0.5Sticking Friction
1/2
Plane Strain
when
Slipping Frictionis found in cold
working
Sticking Frictionis found in hot
working
+=
22
1
1
hh
hRhRWRL yActual
5>
hR
W
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Sample Problem
Calculate the rolling load required to reducesteel 600 mm wide and 30 mm thick by
20%. The roll diameter used is 800 mm,
= 0.15 and flow stress of steel is 150 MPa.
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Sample ProblemExplain the Design of Cluster Mill.
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Forging
The act of
shaping metal by
hammering or
pressing
Forging is done inspecial molds
called "dies"
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Forging Products
http://www.sbf.net.cn/html/..%5Cimages%5Ccp-a22_r2_c2.jpg7/29/2019 Mechanical Working of Metals (Rolling and Forging)
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Dies used in Forging
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Types of Forging
Press Forging Impact (Smith)
Forging
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Types of Forging
OPEN-DIE
FORGING
CLOSED-DIE
FORGING
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Closed-Die Forging
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Examples of Forging
UPSETTING COGGING
SWAGING
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FULLERING
EDGING
HEADING
Examples of Forging
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Precision Forging
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Precision Forging Operation
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Forging of Bevel Gear
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Forging Defects
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Defects in Forging
Laps Formed in
Forged Metal
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Calculation of Forging Load
UPSET FORGING
P
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Ideal vs Actual
HOMOGENOUS
DEFORMATION
NONHOMOGENOUS
DEFORMATION
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Nonhomogenous Deformation
BARRELING
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Grain Texture in Forged Metal
Eff t f F i ti F i
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Effect of Friction on Forging
Pressure
Pressure builds-up at
the center creating a
pressure hill!
Eff t f F i ti F i
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Effect of Friction on Forging
Load
The build-up ofpressure also creates
a Friction Hill
-Forging Load Calculation-
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Forging Load Calculation
The SLAB METHOD
ASSUME PLANE STRAIN (WIDTH IS CONSTANT)
b = tool bite
Analysis of Stresses using
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Equilibrium of forces in the horizontal direction:
Analysis of Stresses using
Slab Method
( ) 02 =++ hdxhd xyxx
02
=+ dxh
dy
x
Since both x and y are unknown, another equation must beobtained using yield criteria; specifically, the distortion-energy
criterion for plane strain is used.
YYyx
==3
2
Henceyx dd =
and
Analysis of Stresses using
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The two equations for the stressescan now be solved:
( ) hxay eYp
/2
==
hxy
y
yCedx
h
d/2
2
==
The value of the constant, C, is found
using the boundary conditions. At x=a,
x=0 and y= Y`ha
eYC
/2
=The value of the constant, C, is found using the boundaryconditions. At x=a, x=0 and y= Y`. The pressure is then
Analysis of Stresses using
Slab Method
Si lifi ti f F i
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Simplification of Forging
Pressure Equation
( ) hxay eYp
/2 ==
If x
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Sample Problems
A copper slab was upset to give a final
dimension of 60 x 20 x 10 mm (L x W x H). If
the yield stress of copper is 40 MPa and the
coefficient of friction is 0.15, determine theForging Pressure at the following points on
the billet:
i. At the middle
ii. At the ends
iii.(1/3)a distance from the middle
P
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The average pressure canbe approximated from the
mean of the maximum and
the minimum pressure
+h
aYpavg
1
The forging load, F, is
( )( ) ( )widthapF
avg2=
Average Forging Pressure, Pavg
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Reminders on P and Pavg
+
h
aYpavg
1]
)(21['
h
xaYp y
+==
Condition FlowStress Condition
PlaneStrain
Y = 1.155 < 0.5Slipping Friction
HomogenousDeformation
Y > 0.5Sticking Friction
1/2
Plane Strain W/b > 4
when
Slipping Frictionis found in cold
working
Sticking Frictionis found in hot
working
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Sample Problems
A copper slab was upset to give a final
dimension of 60 x 20 x 10 mm (L x W x H). If
the yield stress of copper is 40 MPa and the
coefficient of friction is 0.15, determine theaverage forging pressure of this process.
P
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Sample Problem
Determine the Forging Load required to
reduce the thickness of half of a steel billet
100x40x10 mm (LxWxH) by 40%(assume plane strain). Assume that the
yield stress of steel is 240 MPa and =0.65.
F