Transcript Rolling And Extrusion ME 482 - Manufacturing Systems

Rolling
And
Extrusion
ME 482 - Manufacturing Systems
Rolling Process
Points:
• Significant shape change
• Capital intensive
• Large volume
• Usually hot worked (isotropic)*
• Oxide scale
• Tolerances difficult to hold
* Can be followed by cold rolling to improve
tolerances and directional properties
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Rolling Products
Billet
Blooms (> 6” x 6”)
Slab
Ingots
Billets (> 1.5” x 1.5”)
Slab (> 10” x 1.5”)
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vr
Rolling Model
to
q R
p
vo
vf
tf
L
Assumptions:
• Infinite sheet
vr
• Uniform, perfectly rigid rollers
• Constant material volume:
to wo Lo = tf wf Lf
“rate” (to wo vo = tf wf vf )
where
Lo = initial plate length
Lf = final plate length
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R = roller radius
p = roll pressure
L = contact length
q = contact angle
vr = roll speed
to = initial plate thickness
tf = final plate thickness
vo = plate entry speed
Vf = plate exit speed
Rolling Model
Define draft = d = to - tf
Draft limit = dmax = m2R
m = 0.1 cold
m = 0.2 warm
m = 0.4 – 1.0 hot
Define forward slip = s = (vf – vr)/vr
Does it make sense that vr < vf?
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Point of greatest
contact pressure =
no slip point
Rolling Model – stress, strain, force, power
Define true strain = e = ln(to/tf)
(Note: use to/tf to keep > 0)
Apply average flow stress = Yf = K en /(1 + n)
Approximate roll force = F = Yf w L
where L = R(to – tf)
R
Torque estimated by T = 0.5 F L
Power = P = Tw = 2p w F L (for two rollers)
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q
L
R– (to–tf)/2
Other rolling configurations
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Example 21.1 in text
Roll a 12 inch wide strip, that is 1 inch thick, to 0.875 inch thickness in
one pass with roll speed of 50 rpm and radius = 10 inches. Material has
K = 40,000 psi, n = 0.15 and m = 0.12. Determine if feasible and calculate
F, T, and power if so.
Solution:
Feasible since dmax = (0.12)2 (10) = 0.144 in. > d = 1.0 – 0.875 = 0.125 in.
Contact length = L = 1.118 in.
e = ln(1.0/0.875) = 0.134
Yf = (40,000)(0.134)0.15/1.15 = 25,729 psi
Rolling force = (25,729)(12)(1.118) = 345,184 lb
Torque = (0.5)(345,184)(1.118) = 192,958 in.-lb
Power = P = (2p)(50)(345,184)(1.118) = 121,238,997 in.-lb/min (306 hp)
ME 482 - Manufacturing Systems
Extrusion
Limitation – requires uniform cross-section vs length
Advantages:
• Variety of shapes
• Control mechanical properties in cold and warm extrusion
• Little wasted material
• Good tolerances
Types : Direct extrusion and indirect extrusion
Less
friction!
ME 482 - Manufacturing Systems
Extrusion Model
Assumptions:
L
• Circular cross-section
• Uniform stress distribution p
Do
p = ram pressure
L = remaining billet length
Do = chamber diameter
Df = extrudate diameter
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Df
Extrusion Model – stress and strain
Define extrusion ratio = rx = Ao/Af
Ao = billet (chamber) area
Af = extrudate area
Frictionless model:
ideal true strain = e = ln rx
ideal ram pressure = p = Yf ln rx
With friction:
Johnson eqn ex = a + b ln rx
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a = 0.8
1.2  b  1.5
Extrusion Model – stress and strain
Indirect extrusion ram pressure = p = Yf ex (ex is from Johnson eqn)
and where Yf is found using the the ideal true strain e= ln rx
In direct extrusion, difficult to predict the chamber/billet interactive
friction, so use the shear yield strength ( about Yf /2 ) to estimate the
chamber wall shear force as
pf pDo2/4 = Yf p Do L/2
giving
pf = 2 Yf L /Do
and where pf = additional pressure to overcome wall friction force
Total ram pressure becomes
p = Yf (ex + 2L /Do )
ME 482 - Manufacturing Systems
Extrusion Model – non-circular sections
Apply a shape factor Kx (experimental results):
Kx = 0.98 + 0.02 (Cx / Cc)2.25
where
Cx = perimeter of extruded shape
Cc = perimeter of circle having same area of extruded shape
Applies for
1.2 (Cx / Cc)  1.5
For complex extrudate:
Indirect
Direct
p = Kx Yf ex
p = Kx Yf (ex + 2L /Do )
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Extrusion Model – forces and power
Ram force = F = pAo
Power = P = Fv
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Extrusion example
A billet 3” long and 1” diameter is to be extruded as a round extrudate
in a direct extrusion operation with extrusion ratio of rx = 4. Given die
angle of 90°, strength coefficient of 60,000 psi, and strain hardening
exponent of 0.18, use the Johnson formula with a = 0.8 and b = 1.5 to
estimate extrusion strain, and then determine the pressure applied to
the end of the billet as the ram moves forward.
Solution:
e = ln rx = ln 4 = 1.3863
ex = 0.8 + 1.5(1.3863) = 2.87945
Yf = 60,000(1.386)0.18/1.18 = 53,927 psi
ME 482 - Manufacturing Systems
Extrusion example
A billet 3” long and 1” diameter is to be extruded as a round extrudate
in a direct extrusion operation with extrusion ratio of rx = 4. Given die
angle of 90°, strength coefficient of 60,000 psi, and strain hardening
exponent of 0.18, use the Johnson formula with a = 0.8 and b = 1.5 to
estimate extrusion strain, and then determine the pressure applied to
the end of the billet as the ram moves forward.
Solution continued:
p = Yf (ex + 2L/D) = 53,927 [2.87945 + (2)(3)/1]
p = 478,842 psi
ME 482 - Manufacturing Systems
Rolling and Extrusion
What have we learned?
ME 482 - Manufacturing Systems