Rolling
1. (SI units) A series of cold-rolling operations are to be used to reduce
the thickness of a metal plate from 50 mm to 20 mm in a reversing
two-high mill. Roll diameter = 600 mm, and coefficient of friction
between rolls and work = 0.15. The specification is that the draft is
to be equal on each pass. Determine (a) minimum number of passes
required, and (b) draft for each pass?
Solution: (a) Maximum possible draft dmax = 2R = (0.15)2 (300) =
6.75 mm
Minimum number of passes = (to tf)/dmax = (50 20)/6.75 = 4.44
rounded up to 5 passes
(b) Draft per pass d = (50 20)/5 = 6.0 mm
(SI units) In the previous problem, suppose that the percent
reduction rather than the draft were specified to be equal for each
pass. (a) What is the minimum number of passes required? (b) What
is the draft and exiting stock thickness for each pass?
Solution: (a) Maximum draft occurs on first pass: dmax = 2R =
(0.15)2 (300) = 6.75 mm
This converts into a maximum possible reduction x = 6.75/50 =
0.135
Let x = fraction reduction per pass, and n = number of passes. The
number of passes must be an integer. To reduce from to = 50 mm to
to = 20 mm in n passes, the following relationship must be satisfied:
50(1 x)n = 20
(1 x)n = 20/50 = 0.4
(1 x) = 0.41/n
Try n = 5: (1 x) = (0.4)1/5 = 0.8325
Rolling
1. (SI units) A series of cold-rolling operations are to be used to reduce
the thickness of a metal plate from 50 mm to 20 mm in a reversing
two-high mill. Roll diameter = 600 mm, and coefficient of friction
between rolls and work = 0.15. The specification is that the draft is
to be equal on each pass. Determine (a) minimum number of passes
required, and (b) draft for each pass?
Solution: (a) Maximum possible draft dmax = 2R = (0.15)2 (300) =
6.75 mm
Minimum number of passes = (to tf)/dmax = (50 20)/6.75 = 4.44
rounded up to 5 passes
(b) Draft per pass d = (50 20)/5 = 6.0 mm
(SI units) In the previous problem, suppose that the percent
reduction rather than the draft were specified to be equal for each
pass. (a) What is the minimum number of passes required? (b) What
is the draft and exiting stock thickness for each pass?
Solution: (a) Maximum draft occurs on first pass: dmax = 2R =
(0.15)2 (300) = 6.75 mm
This converts into a maximum possible reduction x = 6.75/50 =
0.135
Let x = fraction reduction per pass, and n = number of passes. The
number of passes must be an integer. To reduce from to = 50 mm to
to = 20 mm in n passes, the following relationship must be satisfied:
50(1 x)n = 20
(1 x)n = 20/50 = 0.4
(1 x) = 0.41/n
Try n = 5: (1 x) = (0.4)1/5 = 0.8325
x = 1 0.8325 = 0.1674, which exceeds the maximum possible
reduction of 0.135.
Try n = 6: (1 x) = (0.4)1/6 = 0.8584
x = 1 0.8584 = 0.1416, which exceeds the maximum possible
reduction of 0.135.
Try n = 7: (1 x) = (0.4)1/7 = 0.8773
x = 1 0.8773 = 0.1227, which is within the maximum possible
reduction of 0.135.
(b) Pass 1: d = 50(0.1227) = 6.135 mm, tf = 50 6.135 = 43.86 mm
Pass 2: d = 43.86(0.1227) = 5.38 mm, tf = 43.86 5.38 = 38.48 mm
Pass 3: d = 38.48(0.1227) = 4.72 mm, tf = 38.48 4.72 = 33.76 mm
Pass 4: d = 33.76(0.1227) = 4.14 mm, tf = 33.76 4.14 = 29.62 mm
Pass 5: d = 29.62(0.1227) = 3.63 mm, tf = 29.62 3.63 = 25.99 mm
Pass 6: d = 25.99(0.1227) = 3.19 mm, tf = 25.99 3.19 = 22.80 mm
Pass 7: d = 22.80(0.1227) = 2.80 mm, tf = 22.80 2.80 = 20.00 mm
2. A 300 mm wide strip, 25 mm thick, is fed through a rolling mill
with two powered rolls each of radius 250 mm. The work thickness
is to be reduced to 22 mm in one pass at a roll speed of 50 rev/min.
The work material has a flow curve defined by K = 275 MPa and n
= 0.15, and the coefficient of friction between the rolls and the
work is 0.12. Determine if the friction is sufficient to permit the
rolling operation to be accomplished. If so, calculate the roll force,
and horsepower (or rolling power).
x = 1 0.8325 = 0.1674, which exceeds the maximum possible
reduction of 0.135.
Try n = 6: (1 x) = (0.4)1/6 = 0.8584
x = 1 0.8584 = 0.1416, which exceeds the maximum possible
reduction of 0.135.
Try n = 7: (1 x) = (0.4)1/7 = 0.8773
x = 1 0.8773 = 0.1227, which is within the maximum possible
reduction of 0.135.
(b) Pass 1: d = 50(0.1227) = 6.135 mm, tf = 50 6.135 = 43.86 mm
Pass 2: d = 43.86(0.1227) = 5.38 mm, tf = 43.86 5.38 = 38.48 mm
Pass 3: d = 38.48(0.1227) = 4.72 mm, tf = 38.48 4.72 = 33.76 mm
Pass 4: d = 33.76(0.1227) = 4.14 mm, tf = 33.76 4.14 = 29.62 mm
Pass 5: d = 29.62(0.1227) = 3.63 mm, tf = 29.62 3.63 = 25.99 mm
Pass 6: d = 25.99(0.1227) = 3.19 mm, tf = 25.99 3.19 = 22.80 mm
Pass 7: d = 22.80(0.1227) = 2.80 mm, tf = 22.80 2.80 = 20.00 mm
2. A 300 mm wide strip, 25 mm thick, is fed through a rolling mill
with two powered rolls each of radius 250 mm. The work thickness
is to be reduced to 22 mm in one pass at a roll speed of 50 rev/min.
The work material has a flow curve defined by K = 275 MPa and n
= 0.15, and the coefficient of friction between the rolls and the
work is 0.12. Determine if the friction is sufficient to permit the
rolling operation to be accomplished. If so, calculate the roll force,
and horsepower (or rolling power).
