MECH934 ADVANCED MANUFACTURING PROCESSES

AUTUMN SESSION 2017

LABORATORY PROJECT – COLD ROLLING EXPERIMENT

Group Members
JANGVIR SINGH (5504752)
ABIN GEORGE (5126903)
VINAY (5044479)
WALEED WASIM(5176694)

1
Contents
AIM..............................................................................................................................................................3
True stress Vs True strain curve...................................................................................................................3
Frictional coefficient....................................................................................................................................9
Forward slip...............................................................................................................................................13
To determine friction coefficient by forward slip......................................................................................15
Pressure distribution Curve.......................................................................................................................15
Gauge meter diagram................................................................................................................................17
Temperature rise of the strip....................................................................................................................22
Automatic Gauge Control in Strip Rolling:.................................................................................................25

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AIM
The aim of this experiment is to conduct the cold rolling of aluminium specimen. This will allow you to
understand the gap control of the rolling mill, and evaluate the rolling force, friction, reduction and
temperature rise of aluminium strip in cold rolling process.

True stress Vs True strain curve

Stress and strain for the given expirements can be found from the following equations.

The initial thickness (H0),length (LO) and width (WO) of the tensile sample are 2mm,
82mm, and 12.5mm respectively. The load F and extension L are measured by instron
materials testing machine and are recorded:

3
 Stress Strain Curve
160

140

120

100

80

60

40

20

0
0 0.05 0.1 0.15 0.2 0.25 0.3

Stress Strain Curve

To determine the parameters YO and XM1 in the strain-stress curve:

𝜎 = 𝑌0(1 + 1000𝜀) M𝑋1:

Assuming y=σ and x=1+1000ε and also re-arranging

𝑙𝑛(𝜎 ) = 𝑙𝑛(𝑌𝑂) + 𝑋𝑀1 ∗ 𝑙𝑛(1 + 1000 ∗ 𝜀 )

Linear fitting by assuming

𝑦 = 𝑙n(𝜎) and 𝑥 = 𝑙n(1 + 1000 𝜀)

From the graphs generated during the tests, we find that

ln(stress)
6

5
f(x) = 0.367557802875329 x + 2.95598768307614
4 ln(stress)
Linear (ln(stress))
3

0
1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

4
 stress
160
140 f(x) = 19.2206973052922 x^0.36755780287533
120
100 stress
Power (stress)
80
60
40
20
0
0 50 100 150 200 250 300

Here xm1=0.3676

Y0= 19.22093

Therefore, the above values of y0 and xm1 can be put in the alexander program.

The chart below shows the average value of force that comes around to be 39.1409 kn. Similarly we can
measure the average values from other charts.

a) Case 1: Reduction 1.87 mm

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 Rolling Force
60
50
40
30
20
10
0
0 200 400 600 800 1000 1200 1400
-10
-20
-30
-40

Average force value ---- 39.1409 kn

Average torque value------ 0.7887 Knm
Speed (rpm)= 20
Therefore,

After calculating, P= 1.65 kW

b) Case 2: Reduction 1.71 mm

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 Rolling Force 2
100

80

60

40

20

0
0 200 400 600 800 1000 1200 1400
-20

-40

Average value of force (as per chart) ---------- 62.05 kn

Average Torque ------- 0.4885 knm

Speed(rpm) = 20

Therefore, Power = 1.02 kW

c) Case 3: Reduction 1.7 mm

Rolling Force 3
60

50

40

30

20

10

0
0 200 400 600 800 1000 1200 1400
-10

-20

-30

Average rolling force (as per chart) ------- 41.335 KN

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 Average torque ----- 0.6699 KN M
Speed (rpm)----30

Therefore, Power = 2.1 kw

d) Case 4: Reduction 1.85mm

Rolling Force 4
100

80

60

40

20

0
0 200 400 600 800 1000 1200 1400 1600
-20

-40

-60

-80

Average rolling force (as per chart) --- 68.311 KN

Average torque ------ 0.1415 KNM

Speed (rpm) – 30
Therefore, power = 0.44 kw

Therefore, from the above results it can be concluded that as the reduction increases,
the power required to do the amount of work also increases.

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 Frictional coefficient
In order to calculate frictional coefficient, the initial width (w1), thickness (h1) and exit
width (w2), thickness (h2), and the temperature of rolling specimens measured during
the lab expiriment by calipers and thermal meter are entered in the given matlab script
alex.m . The edited values in the alex script are as follows,

Calculation 1

r – work roll radius (m) = 0.0225m

xn – work roll speed (rpm) = 20 rpm

h0 – slab thickness (m) = 0.002 m

h1 – entry thickness (m) = 0.002 m

h2 – exit thickness (m) = 0.0187 m

t1 – bad tension stress (Pa) = 0.00 Pa

t2 – front tension stress (pa) = 0.00 Pa

Width – specimen width (m) is the average of the initial and final width before and after
the expirements.

w 1+w 2 0.1013+0.1015
Therfore specimen width = = = 0.1014 m
2 2

Y0 = 19.221 Pa

xm1 = 0.3676

b = 1000

d = 0.00

xm2=0.00

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Calculation 2

r – work roll radius (m) = 0.0225m

xn – work roll speed (rpm) = 20 rpm

h0 – slab thickness (m) = 0.002 m

h1 – entry thickness (m) = 0.002m

h2 – exit thickness (m) = 0.00171m

t1 – bad tension stress (Pa) = 0.00 Pa

t2 – front tension stress (pa) = 0.00 Pa

Width – specimen width (m) is the average of the initial and final width before and after
the expirements.

w 1+w 2 0.1014+ 0.1017

Therfore specimen width = = = 0.1015 m
2 2

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 Y0 = 19.221 Pa

xm1 = 0.3676

b = 1000

d = 0.00

xm2 = 0.00

Calculation 3

r – work roll radius (m) = 0.0325 m

xn – work roll speed (rpm) = 30rpm

h0 – slab thickness (m) = 0.002m

h1 – entry thickness (m) = 0.002m

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 h2 – exit thickness (m) = 0.0017m

t1 – bad tension stress (Pa) = 0.00 Pa

t2 – front tension stress (pa) = 0.00 Pa

Width – specimen width (m) is the average of the initial and final width before and after
the expirements.

w 1+w 2 0.1007+0.101
Therfore specimen width = = = 0.1008 m
2 2

Y0 = 19.221 Pa

xm1 = 0.3676

b = 1000

d = 0.00

xm2 = 0.00

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 The initial value for the coefiicient of friction, xmu is adjusted to 0.1 as per the given
instructions.

In order to find the frictional coefficient, the script file alex.m is run and the result
obtained from the alex_res.dat is evaluated for further calculations. To match the
measured rolling force with the obtained rolling force, the value of xmu is adjusted to 4,
1.2 , 0.05 respectively.

The group didn’t take the case 4 because the value of friction coefficient is coming out to be 4
for which the force value is near to the measured value through alex program. It looks a bit
contradictory.

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Forward slip
From the results data in alex_res.dat, the neautral angle is found to be 0.027516
Radians. The forward slip is then calculated by the following equation.

where, s – forward slip

R’ – Deformed roll radius (m)

H2 – exit thickness (m)

PHN – neutral angle

The deformed roll radius can be calculated from the equation,

where,

P – rolling force (N)

W – strip width (m)

R – roll radius (m)

where,

v r – poisson ’ s ratio=0.3

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 Er – youngs modulus = 207E9

16∗(1−0.32)
CO = = 2.763x10*(-6)
3.14∗207E9

The value of Co is substituted in the equation of deformed roll radius.

2.763 x 10∗(−6)∗62.429
Deformed roll radius, R’ = 0.4563(1+ ¿
0.1015∗(1.3∗10−4 )

= 0.463 m

The obtained value of deformed roll radius is used to calculate the forward slip.

2∗0.463 (1−cos ( 0.0275 ))

Forward slip, s =
0.00187

= 0.495

To determine friction coefficient by forward slip

The obtained values of deformed roll radius and forward slip is substituted in the
equation of frictional coefficient and its respective value is determined.

The frictional coefficient can be calculated by

where,

s – forward slip

H1 – initial thickness before rolling (m)

H2 – final thickness after rolling (m)

0.00299−0.00194
 = 2 0.11951 ( 0.00299−0.00194 )−4 0.001204∗0.11951∗0.00194
√ √

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 = 0.056

Therefore, this value comes out to be similar to the value obtained in the alexander
program.

Pressure distribution Curve

Pressure distribution curve can be plotted using the datas from the data file alex_res.dat
. For plotting the graph the column S (normal pressure) is taken as the y-axis and the
column PHI (angle in the roll bite) is drawn as the x-axis.

Analyzing the two different cases in the alexander program

 In the case no. 2 and case 3 ,the value of roll bite angle is coming out to be
0.094247 and 0.094846.Hence the pressure distribution curve is as follows:

Pressure v Angle
250

200
Nominal Pressure

150

100

50

0
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07

Angle in Roll bite

16
 Pressure V Angle
300

250

200
Nominal Pressure

150

100

50

0
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

Angle in Roll Bite

Pressure V Angle
180
160
140
Nominal Pressure

120
100
80
60
40
20
0
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Roll Bite Angle

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 Findings

For every iteration of the test, it has been observed that at lower values of roll bite
angle, we get higher nominal pressure and as the roll bite angle increases, the pressure
decreases.

Gauge meter diagram

The unloaded roll gap is adjusted by housing screw position at the driver side and
operator side. The strip initial thickness is set to 2mm, by the same time the housing
screw positions LVDTOSO and LVDTDSO at the driver and operator side are recorded. The Co
measured can be determined by the following equation.

Calculation 1

Co measured = 2.00 – ¿ ¿ ¿

[2700+ 2700−( 3110+3110 ) ]

Co measured = 2.00 – 2
1000

= 2.4

to determine the strip plastic deformation curve :

(a) Data point 1: assuming strip exit thickness is equal to entry thickness, there is no

reduction; hence the rolling force is zero. Data point 1 is (h1, 0);

(h1,0) = (0.002,0)

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 H 1−H 2
(b) Data point 2: assuming strip exit thickness is h = H 2 + , rolling force P(h) can
2
be determined by Alexander program by replacing the value of h2 with h. Data point 2 is
(h, P(h) )

H 1−H 2
h=H2+
2

= 0.001935

where, H1 = 0.002 , H2 = 0.00187

(h, P(h) ) = (0.001935,78.924)

(c) Data point 3: Assuming strip exit thickness is H2, rolling force is the measured rolling
force, Pmeasured. Data point 3 is (h2, Pmeasured)

(h2, Pmeasured) = (0.00199,143.907)

Ms – Mill structural stiffness = 674 kN/mm

Pmeasured - measured rollin force

Cocalculated - Initial unload roll gap

143.907
= 1.99 -
674

= 1.776

Calculation 2

Co measured = 2.00 – ¿ ¿ ¿

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 [2700+ 2700−( 3110+3110 ) ]
Co measured = 2.00 – 2
1000

= 2.4

to determine the strip plastic deformation curve :

(a) Data point 1: assuming strip exit thickness is equal to entry thickness, there is no

reduction; hence the rolling force is zero. Data point 1 is (h1, 0);

(h1,0) = (0.002,0)

H 1−H 2
(b) Data point 2: assuming strip exit thickness is h = H 2 + , rolling force P(h) can
2
be determined by Alexander program by replacing the value of h2 with h. Data point 2 is
(h, P(h) )

H 1−H 2
h=H2+
2

= 0.001855

where, H2 = 0.00171, H1 = 0.002

(h, P(h) ) = (0.00255,81.813)

(c) Data point 3: Assuming strip exit thickness is H2, rolling force is the measured rolling
force, Pmeasured. Data point 3 is (h2, Pmeasured)

(h2, Pmeasured) = (0.00171,132.801)

Ms – Mill structural stiffness = 674 kN/mm

Pmeasured - measured rollin force

Cocalculated - Initial unload roll gap

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 132.801
= 2.05 –
674

= 1.853

Calculation 3

[2700+ 2700−( 3110+3110 ) ]

Co measured = 2.00 – 2
1000

= 2.4

to determine the strip plastic deformation curve :

(a) Data point 1: assuming strip exit thickness is equal to entry thickness, there is no

reduction; hence the rolling force is zero. Data point 1 is (h1, 0);

(h1,0) = (0.002,0)

H 1−H 2
(b) Data point 2: assuming strip exit thickness is h = H 2 + , rolling force P(h) can
2
be determined by Alexander program by replacing the value of h2 with h. Data point 2 is
(h, P(h) )

H 1−H 2
h=H2+
2

= 0.00185

where, H2 = 0.0017 , H1 = 0.002

(h, P(h) ) = (0.00185,50.96)

(c) Data point 3: Assuming strip exit thickness is H2, rolling force is the measured rolling
force, Pmeasured. Data point 3 is (h2, Pmeasured)

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 (h2, Pmeasured) = (0.00236,95.42)

Ms – Mill structural stiffness = 674 kN/mm

Pmeasured - measured rollin force

Cocalculated - Initial unload roll gap

95.42
= 2.71 –
674

= 2.5684

Temperature rise of the strip

For calculating the temperature difference we will be considering specimen 3

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 The average yield stress can be taken from the output file of the Alexander programe.

 = 130.03 MPa

The average strain can be calculated by,

= 0.12188

The volume of the sample can be calculated by the equation

V = width * thickness * length

= 100.7*2*400

= 80560 mm3

Uplastic = σ ⋅ε ⋅V = 130.03*80560*0.12188

=1276719.424 N.mm

Frictional coefficient is obtained from the results of matching measured rolling force by
Alexander programme.

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 Frictional coefficient  = 0.056

Average pressure is taken as the average from the output file of the Alexander program

P = 130.03 Mpa

The average relative speed between the roll and strip can be calculated by

= 0.75 * 0.495* Vr

Vr = 30*Pi*d/60

= 102.10 mm/s

Vrelative = 0.75* 102.10*0.495

= 37.90 mm/sec

= √ 32.5 ( 2−1.7 )∗101

= 315.37 mm2

400
=
102.10

= 3.91 s

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 Ufriction= 2*0.056*130.03*315.37*102.10*3.91

= 1833514.98 Nmm

= 1.833 kJ

Therefore, Utotal_in_strip = 0.7*1276719.424 +0.5*1833514.98

= 1810461.08 Nmm

= 1.81 kJ

1.81∗E 3
T = =0.93 oC
2700∗897∗8.0∗E−4

Tmeasured = (21.1-20) oC

= 1.1 oC

From the calculation it can be seen that, the measured value and calculated value are
almost same.

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Automatic Gauge Control in Strip Rolling:

Automatic Gauge Control, nowadays, is being used to make sure that exit strip thickness has a very high
accuracy level, since the quality and accuracy of the exit strip is of prime importance in a Cold-Rolling
operation.

Automatic Gauge Control (AGC) system is installed in the four-high plate mill, and it is the most
important mechanism for dynamic thickness control in conventional rolling mills. Since the AGC system
is responsible for maintaining the dynamic performance of the predicted quality of thickness.

AGC ramp up the mill speed and tension/gap to bring production to target thickness in the shortest
possible time and distance. Once within specification, AGC optimises the mill speed by adjustment of
tension, load or gap. At the end of the coil, Auto-Slowdown ensures the product remains in specification
for as long as possible and brings the mill to a halt with minimum amount of material left on the coil
drum. Target adaptive control monitors the short term product variation and adjusts the setpoint
downwards, so that the lowest point of the product variation is just above the lower specified limit,
providing additional raw material savings (target is increased if selling by weight).

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