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8/10/2019 Handout 8b Metal Rolling
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Module 8 1
Rolling
Principle of the process
Structure
Process modeling
Defects
Design For Manufacturing (DFM)
Process variation
Metalforming
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Rolling Process: Mechanics Analysis
Two opposite rolls and a piece of material flows between
them. The shape of rolls can be designed in a different formto construct a product with different cross sections.
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Objectives of mechanics analysis
a. Physical
phenomenon
b. Torque
c. Powerd. Productivity
System
parameters
Operating
parameters
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Physical phenomenon
Volume flow rate conservation
fff vwtvwt 000
Spreading: Volume before
rolling = the volume after rolling
fff LwtLwt 000L
w
t
fvv 0
fttdDraft 0,
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Vo < Vr < Vf
No-slip point
Work velocity = Roll velocitySlipping Slipping
( ) /f r rForward Slip v v v
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For the rolling process, the true strain is:
ft
t0ln
The average flow stress is the same expression, i.e.
n
kYn
f+
1
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It is the friction between the work and the roll that
drives the workflow between two rolls.
The friction force is developed based on
(1)coefficient of the frictionand(2)compression force of rolls
No-slip pointGreater
Friction Force
Lesser
Friction Force
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Condition to roll- Coefficient of the friction
draft, d = |tf-t0|: dmax
Rd 2
m ax
Max.
Possible
Draft
Friction causes Rolling
If Friction=0, then draft=0, means NO ROLLING
Radius of the rollFriction coefficient
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The pressure varies along thecontact length
Condition to roll- Power to drive the roll and work piece
Roll Force (F)Integrating
unit roll pressureoverroll work contact area
L
pdLwF 0
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Contact length )( 0 fttRL
Contact force
L
pdLwF 0 wLYF f
Power NFLP 2
FLT 5.0Torque N: rotationspeed of the roll,rev / min
d
Power is a function of D. Increase of D leads to increase of P
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Condition to roll- Power to drive the roll and work piece
When the required power (d) is greater than the supplied
power, the rolling of a work piece with d is not possible.
Therefore, the required power = supplied power will lead to
a critical draft d or maximum d.
Criterion 1: Rd 2
m ax
Criterion 2: required power = supplied powerm axd
The actual maximum draft for a rolling system is the smaller
one computed from the two criterions above.
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Example:
A 10-in.wide, 1.0-inthick plate is to be reduced in a
single pass in a two-high rolling mill to a thickness of 0.8
in. The roll has a radius = 20 in., and its speed = 50
ft/min. The work material has a strength coefficient =
35.000 lb/in.2and a strain hardening exponent = 0.2.
Determine (a) roll force, (b) roll torque, and (c) powerrequired to accomplish this operation.
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Given: rolling, t0=1.0 in., tf=0.80 in., w=10.0 in., R=20
in., vr=50 ft/min, flow curve n=0.20 and K=35,000 lb/in2.
Find: (a) F, (b) T, (c) HP.
Draft d=1.0-0.8=0.2 in.
Contact length L = (200.20)0.5 = 2.0 in.True strain = ln (1.0/0.8) = ln 1.25= 0.223
Average flow stress
Yf = 35,000(0.223)0.20/1.20 = 21,607 lb/in2
Rolling force F = 21,607(10)(2) = 423,149 lb
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Torque T = 0.5(432,149)(2.0) = 432,149 in-lb.
N = (50 ft/min)/(220/12) = 4.77 rev/min.
Power P=2 (4.77)(432,149)(2) = 25,929,940 in-lb/min
HP = (25,929,940 in-lb/min)/(396,000) = 65.5 hp
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