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Page 1
TO FIND THE MINIMUM EXTRUSION WORK OF
CIRCULAR BILLET TO CIRCULAR SHAPE
THROUGH DIFFERENT TYPES OF CURVED DIES
BY
GYANA RANJAN PANIGARAHI
BHARAT BHUSHAN
SAMBUDHA KUMAR
SHALINI SHARMA
DYUTI DEY
SHAILESH RANJAN
Guided by
Prof. B.B.MISHRA
MECHANICAL ENGINEERING DEPARTMENT
INSTITUTE OF TECHNICAL EDUCATION AND RESEARCH,
BHUBANESWAR

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INTRODUCTION
• Extrusion is the plastic deformation
process by which a block/billet of
metal is reduced in cross section by
forcing it to flow through a die orifice
under high pressure.
• In general, extrusion is used to produce cylindrical bars
or hollow tubes or for the starting stock for drawn rod,
cold extrusion or forged products.
• Most metals are hot extruded due to large amount of
forces required in extrusion. Complex shape can be
extruded from the more readily extrudable metals such
as aluminium.

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EXTRUSION PRODUCTS
Typical parts produced by extrusion are trim part used in automotive
and construction applications, window frame members, railings,
aircraft structural parts.
• Example: Aluminium extrusions are used in commercial and
domestic buildings for window and door frame systems,
prefabricated houses/building structures, roofing and curtain walling,
shop fronts, etc. Furthermore, extrusions are also used in transport
for road and rail vehicles and in marine applications.
BRASS PRODUCTS ALUMINIUM EXTRUSION

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COLD EXTRUSION
• It is done below recrystallization temperatures.
• Examples of metals: lead, tin, aluminum alloys, copper,
titanium, molybdenum, vanadium, steel.
• Examples of products: collapsible tubes, aluminum cans,
cylinders, gear blanks.
The advantages of cold extrusion are
 No oxidation takes place.
 Good mechanical properties
 Good surface finish with the use of proper lubricants.

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• Direct extrusion is also called forward extrusion.
• Its work operation includes the placement of the billet in
a container, which is heavily walled.
• Ram or screw is used to push the billet through the die.
• In between the billet & ram, there is a dummy block,
which is reusable & is used for keeping them separated.
•
DIRECT EXTRUSION

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Mechanical
properties
condition
Density 11300 kg/m3 298.15 K
Young’s Modulus of
Elasticity
16000MPa
Poisson Ratio 0.44
Table 1. Mechanical properties of lead

Page 7
Thermal Properties condition
Melting point 600.61 K
Boiling point 2022.15 K
Critical temperature 5500 K
Heat capacity 130 J/kg-K 298.15 K
Thermal conductivity 35.3 W/m-K 300 K
Thermal Properties of lead

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VARIABLE AFFECTING EXTRUSION
PROCESS
PERCENTAGE OF AREA REDUCTION
DIE GEOMETRY
PRODUCT GEOMETRY
SPEED OF EXTRUSION
BILLET TEMPERATURE
LUBRICATION

Page 9
DIFFERENT TYPES OF CURVED DIES –
BEING CONSIDERED:

Page 10
THEORETICAL SOLUTION TO EXTRUSION OF CIRCULAR BILLET
TO CIRCULAR SHAPE THROUGH COSINE DIE:
Coordinate for O1:(α1, β1)
O2: (α2, β2)
Distance O1C =r1
O2C=r2
stream function define for portion AC:
(x- α1 )2 +(y- β1)2=r1
2
Y=β1+√( r1
2-(x- α1 )2) (for Y> β1)
Ψ=CY/ {β1+√( r1
2-(x- α1 )2)} (c=constant)
Boundary condition: at the die surface:- Ψ=C
Along the axis:-Y=0, Ψ=0
Stream function defined for portion CB:
(x- α2 )2 +(y- β2)2=r2
2
Y=β2-√( r2
2-(x- α2 )2) (for Y< β2)
Ψ=CY/ {β2-√( r2
2-(x- α2 )2)}
Point of intersection of the two circular profile
Since tangent drawn at the intersection point C is tangent to both the circle
O1CO2 is the straight line so
C divides O1O2 internally in the ratio r1:r2

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Hence,
C(X,Y)=[{(r1α2+ r2α1)/(r1+ r2)},{( r1β2+ r2β1)/(r1+ r2)}]
Point of intersection of two streamline:
At the intersection of two streamline Ψ values are same.
Ψ= CY/ {β1+√( r1
2-(x- α1 )2)} = CY/ {β2-√( r2
2-(x- α2 )2)}
=β1+√( r1
2-(x- α1 )2) =β2-√( r2
2-(x- α2 )2)………(1)
Testing for satisfaction of continuity eqn by Ψ
(dvx/dx)+(dvy/dy)=0
Where stream function vx=dΨ/dy
Vy=-dΨ/dx
Calculation of Mass and Volume flow rate:
Volume flow rate at any section=area×vx
CALCULATION OF WORK DONE(W):
total power consumption can be obtained as
W=WI+Ws

Page 12
Where m is the friction factor ( assume m=0 for cosine die)
Since it is a plane strain state (dvy/dz)=(dvx/dz)=0 and vz=0 so,γyz=γzx=0
εijεij=εxx
2+εyy
2+ εzz
2+0.5γxy+0.5γyz+0.5γzx
where εxx
2=εyy
2
εzz
2
=0
the above equation reduce to
εijεij=2(εxx
2+0.25γxy )
εxx=dvx/dx
γxy=(dvx/dy)+(dvy/dx)
Constrain eqn(selection of α1,α2,β1,β2,r1,r2)
Since the product dimension is fixed,
β2-r2=c1 ……….(1)
(β2-r2)/( β1+r1)= c1 /e ………..(2)
From eqn(1)&(2)
(Β2-β1)= (r1+r2)-c1((1/e)-1) ………..(3)
β1 ≤(r1β2+r2β1)/(r1+r2) =>β1≤β2…….(4)
β2≥(r1β2+r2β1)/(r1+r2) => β1≥β2……(5)

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(r1+r2)2=α2
2+(β2-β1)2
= α2
2+(r1+r2)2+c1
2(1/e-1)-2(r1+r2)×c1×(1/e-1)…..(6)
Above eqn is reduced to α2≤( r1+r2 )
For e=0.5 ,c1=1
β2-r2=1
β1+r1=2
r1+r2=( α2
2+1)/2 & β1≤β2
Solution:-

Page 14
r1 r2 β1 β2 α2 W
1 0 1 1 1 0.752
0.9 0.1 1.1 1.1 1 0.741
0.8 0.2 1.2 1.2 1 0.739
0.7 0.3 1.3 1.3 1 0.730
0.6 0.4 1.4 1.4 1 0.725
0.5 0.5 1.5 1.5 1 0.729
0.4 0.6 1.6 1.6 1 0.735
Choosing independently value of r1&die length(α2=1)

Page 15
TABULAR INFORMATION
SL.NO. TYPE OF
CURVE
WORK OF
DEFORMA
TION
WORK OF
SHEAR
WORK OF
FRICTION
TOTAL
1. Cosine 0.725 0 0 0.725
2. Elliptic 0.9245 0.125 0 1.0495
3. Parabola 1.2534 0.146 0 1.3394
4. Tapered 1.1034 0.129 0 1.2324

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CALCULATION OF DIE PROFILE
FOR 40%REDUCTION:
Raw material dia. =20.4mm
Final product dia. =12.24mm
Hence percentage reduction=40%
From equation of circle with center at (α1 , β1)
(x-α1)2 + (y-β1)2=r1
2………………………. (1)
Here, α1=0, β1=6.12 (considering center perpendicular to inlet point)
=>x2 + (y-6.12)2= (4.08)2
(x-α2)2 + (y-β2)2=r2
2……………………. (2)
=>x= (α2r1+α1β2)/ (r1+r2)……………….. (3)
Since, β2-r2=β1……………………………… (4)
=>r1+r2=√ ((14.86)2 +r2
2)………………… (5)
(r1+r2)2= (14.86)2 + r2
2 ……….from equation 5

Page 17
Therefore r1=4.08 and r2=25.167
From equation 4, (β2-r2) = 6.12
Hence, β2=31.287, α2=14.90
Point of intersection of the two circular profiles, since tangent drawn at
the intersection point C is tangent to the circle, O1CO2 is straight line so
C divides O1O2 internally in the ratio r1:r2.
X = 2.0785 (by putting in equation 4)
Similarly by equation 3
Y = (r1β2+r2β1)/ (r1+r2)
Hence, y=9.6308
From circle 2
(x-α2)2 + (y-β2)2=r2
2
(2.0785-14.90)2+(y-31.287)2=25.1672
(y – 31.287)= -21.657(considering –ve value)
Therefore y=9.630

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DESIGN OF DIE

Page 19
FOR 64% REDUCTION
Raw material dia. =20.4mm
Final product dia. =7.344mm
Hence percentage reduction=64%
From equation of circle with center at (α1 , β1)
(x-α1)2 + (y-β1)2=r1
2………………………. (1)
Here, α1=0, β1=3.672 (considering center perpendicular to inlet point)
=>x2 + (y-3.672)2= (6.53)2
(x-α2)2 + (y-β2)2=r2
2……………………. (2)
=>x= (α2r1+α1β2)/ (r1+r2)……………….. (3)
Since, β2-r2=β1……………………………… (4)
=>r1+r2=√ ((14.9)2 +r2
2)………………… (5)
(r1+r2)2= (14.9)2 + r2
2 ……….from equation 5

Page 20
Therefore r1=6.53 and r2=13.73
From equation 4, (β2-r2) = 3.672
Hence, β2=17.40, α2=14.90
Point of intersection of the two circular profiles, since tangent drawn at the
intersection point C is tangent to the circle, O1CO2 is straight line so C
divides O1O2 internally in the ratio r1:r2.
X = 4.80 (by putting in equation 4)
Similarly by equation 3
Y = (r1β2+r2β1)/ (r1+r2)
Hence, y=8.09
From circle 2
(x-α2)2 + (y-β2)2=r2
2
(4.80-14.90)2+(y-17.40)2=13.732
(y – 17.40)= -9.35(considering –ve value)
Therefore y=8.09

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DESIGN

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PRODUCT BY TAPERED DIE
MACHINE FRAMEDETAILED LOADING PROCESS

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CONTAINER SPECIFICATION
PUNCH SPECIFICATION
CURVE DIES

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CONCLUSION
For extrusion of a circular lead rod of diameter 30mm to
12.24 mm and from 30 mm to 7.34 mm the forces acting on
different curved dies (cosine and tapered) were calculated by
mathematical and experimental method.
It is found that the deformation work as well as shear work
for cosine die found lower than tapered die when friction is
neglected i.e smooth die .
.

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FOR 40%REDUCTION(COSINE DIE):
I.D.-20.4mm
O.D-12.24mm
FORCE-79.7KN
COMPRESSIVE STRENGTH-704.61N/ mm2
FOR 40%REDUCTION(TAPERED DIE):
I.D.-20.4mm
O.D.-12.24mm
FORCE-81.8KN

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FOR 64%REDUCTION(COSINE DIE):
I.D.-20.4mm
O.D-7.34mm
FORCE-128.1KN
FOR 40%REDUCTION(TAPERED DIE):
I.D.-20.4mm
O.D.-7.34mm
FORCE-129.2KN

Page 27

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