FORMING PROCESSES
Example 3.1 (ROLLING)
A strip with a cross-section of 150 mm × 6 mm is
being rolled with 20% reduction in area, using 400-mm-diameter steel rolls.
Before and after rolling, the shear yield stress of the material is 0.35 kN/mm2
and 0.4 kN/mm2, respectively. Calculate (i) the final strip
thickness, (ii) the average shear yield stress during the process, (iii) the
angle subtended by the deformation zone at the roll centre, and (iv) the
location of the neutral point θn. Assume the coefficient of friction
to be 0.1
SOLUTION
(i) As no widening is considered during rolling, 20%
reduction in the area implies a longitudinal strain of 0.2 with consequent 20%
reduction in the thickness. Therefore, the final strip thickness is given as
(ii) The average shear yield stress during the
process is taken to be the arithmetic mean of the initial and the final values
of the yield stress. So,
(iii) From figure, it is clear that
Substituting the values, we get
(iv) To determine θn, first λn
has to be calculated. Since nothing has been mentioned regarding the forward
and the back tension, σxi and σxt will be
assumed zero. Hence,
where
Substituting the values of R, tf,
and θi in the expression for λi, we get λi
= 5.99.
Now using this value of λi, we
find that the value of λn is 1.88
θn can be expressed as
Hence,
EXAMPLE 3.2 (ROLLING)
Assuming the speed of rolling to be 30 m/min,
determine (i) the roll separating force and (ii) the power required in the
rolling process describes in Example 3.1
SOLUTION
From above equation, we see that to find out the
roll separating force F, we need to know the pressure p (as a
function of θ), θn and θi. In
Example 3.1, we have already calculated θn and θi.
For the numerical integration of the above equation, we employ Simpson’s rule,
using four divisions after θn and eight divisions before θn.
Thus,
With these intervals, the pressure at different
station points can be computed, using following equations.
Next, we compile the following data. After the
neutral point:
Station Point
|
θ
(rad)
|
(mm)
|
 |
pafter
(kN/mm2)
|
1
|
0.00000
|
2.40000
|
0.000
|
0.75
|
2
|
0.00575
|
2.40331
|
0.479
|
0.788
|
3
|
0.0115
|
2.4132
|
0.958
|
0.830
|
4
|
0.01725
|
2.4298
|
1.432
|
0.876
|
5
|
0.023
|
2.453
|
1.88
|
0.925
|
Before the neutral point:
Station Point
|
θ
(rad)
|
(mm)
|
 |
Pbefore
(kN/mm2)
|
1
|
0.023
|
2.453
|
1.88
|
0.925
|
2
|
0.02981
|
2.489
|
2.454
|
0.887
|
3
|
0.03662
|
2.534
|
2.997
|
0.855
|
4
|
0.04343
|
2.589
|
3.529
|
0.828
|
5
|
0.05024
|
2.65
|
4.049
|
0.804
|
6
|
0.05705
|
2.725
|
4.555
|
0.786
|
7
|
0.06386
|
2.81
|
5.048
|
0.772
|
8
|
0.07067
|
2.899
|
5.525
|
0.759
|
9
|
0.0775
|
3.000
|
5.99
|
0.75
|
Now, from equation (1), the roll separating force
per unit width is
(i) Since the width of the strip is 150 mm, the roll
separating force is
(ii) The driving torque per unit width for each
roll can be computed, as
Taking the values before and after the neutral point
and using Simpson’s rule, we find
So, the total driving torque for each roll is 100.8
× 150 N-m = 15210 N-m
Now, the total power required to drive the two rolls
is 2 × 15, 120 × ω W, ω being the roll speed in rad/sec. The
rolling speed (same as peripheral speed of rolls) is given to be 30 m/min. So,
Thus, the total power (2PR) = 75.6
kW.
The power required in the rolling
process is 75.6 kW
|
EXAMPLE 3.3
If the diameter of the bearings in Examples 3.1 and
3.2 is 150 mm and the coefficient of friction (µb) is 0.005,
estimate the required mill power.
SOLUTION
So, the mill power is given as
