Module 8 1

Rolling

Principle of the process

Structure

Process modeling

Defects

Design For Manufacturing (DFM)

Process variation

Metal forming

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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 form to construct a product with different cross sections.

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Objectives of mechanics analysis

a. Physical phenomenon

b. Torque

c. Power

d. 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 000

L

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

kY

n

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 friction and(2)compression force of rolls

No-slip pointGreaterFriction Force

LesserFriction Force

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Condition to roll- Coefficient of the frictionCoefficient of the friction

draft, d = |tf-t0|: dmax

Rd 2max

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 the contact length

Condition to roll- Power to drive the roll and work piecePower to drive the roll and work piece

Roll Force (F) Integrating “unit roll pressure” over roll work “contact area”

LpdLwF

0

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Contact length )( 0 fttRL

Contact force

LpdLwF

0wLYF f

Power NFLP 2

FLT 5.0Torque N: rotation speed of the roll, rev / min

d

Power is a function of D. Increase of D leads to increase of P

Condition to roll- Power to drive the roll and work piecePower 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 2max

Criterion 2: required power = supplied power maxd

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-in – thick 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.2 and a strain hardening exponent = 0.2.

Determine (a) roll force, (b) roll torque, and (c) power

required 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 = (20×0.20)0.5 = 2.0 in.True strain ε = ln (1.0/0.8) = ln 1.25= 0.223Average flow stress

Y f = 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)/(2π×20/12) = 4.77 rev/min.

Power P=2 π (4.77)(432,149)(2) = 25,929,940 in-lb/min

HP = (25,929,940 in-lb/min)/(396,000) = 65.5 hp