Example: 1
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 assumed to be
0.12. Determine if the friction is sufficient to permit the rolling operation
to be accomplished. If so, calculate the roll force, torque, and horsepower.
Solution:
The draft attempted in this rolling operation is
d = 25 – 22 = 3 mm
From Eq. (1), the maximum possible draft for the given coefficient of
friction is
0.12
250
3.6mm
Since the maximum allowable draft exceeds the attempted reduction, the
rolling operation is feasible. To compute rolling force, we need the contact
length L and the average flow stress Yf The contact length is given by Eq,
(2):
250 25
22
27.4 mm
Yf is determined from the true strain:
25
22
275 0.128
1.15
0.128
.
175.7 MPa
Rolling force is determined from Eq. (3):
F = 175.7(300)(27.4) = 1,444,786 N
Torque required to drive each roll is given by Eq. (4):
T = 0.5(1,444,786) (27.4) (10-3) = 19,786 N-m
and the power is obtained from Eq. (5):
P = 2π (50) (1,444,786) (27.4) (10-3) = 12,432,086 N –m/min = 207,201
N- m/s (W)
For comparison, let us convert this to horsepower (we note that one
horsepower = 745.7 W):
207,201
745.7
278hp
It can be seen from this example that large forces and power are required
in rolling. Inspection of Eqs. (3) and (5) indicates that force and/or power
to roll a strip of a given width and work material can be reduced by any of
the following: (1) using hot rolling rather than cold rolling to reduce
strength and strain hardening (K and n) of the work material; (2) reducing
the draft in each pass; (3) using a smaller roll radius R to reduce force; and
(4) using a lower rolling speed N to reduce power.
Example: 2
A cylindrical workpiece is subjected to a cold upset forging operation. The
starting piece is 75 mm in height and 50 mm in diameter. It is reduced in
the operation to a height of 36mm. The work material has a flow curve
defined by K = 350 MPa and n = 0.17. Assume a coefficient of friction of
0.1. Determine the force as the process begins, at intermediate heights of
62 mm. 49 mm and at the final height of 36 mm.
Solution:
Work piece volume V = 75π (502/4) = 147,262 mm2. At the moment
contact is made by the upper die, h = 75 mm and the force F = 0. At the
start of yielding, h is slightly less than 75 mm, and we assume that strain =
0.002, at which the flow stress is
y f = K ∈n = 350(0.002)0.17 = 121.7 MPa
The diameter is still approximately D = 50 mm and area A = π (502/4) =
1963.5 mm2. For these conditions, the adjustment factor Kf is computed as.
K f = 1+
0.4(0.1)(50)
= 1 .027
75
The forging force is
F = 1.027 (121.7) (1963.5) = 245,410 MPa
75
= I n(1.21) = 0.1904
62
Y f = 350(0.1904).17 = 264.0MPa
∈= In
Assuming constant volume, and neglecting barre line,
A = 147, 264 / 62 = 2375.2mm2 andD = 55.0mm
0.4(0.1)(55)
K f = 1+
= 1.035
62
F = 1.035(264)9237.2) = 649,303N
Similarly at h = 49 mm, F = 955,642 N. and at h = 36 mm, F = 1,467, 422
N. The load- stroke curve in Figure 19.5 was developed from the values in
this example.
Example 3
Wire is drawn through a draw die with entrance angle = 15o. Starting
diameter is 2.5 mm and final diameter = 2.0 mm. The coefficient of
friction at the work-die interface = 0.07. The metal has a strength
coefficient K = 205 MPa and a strain hardening exponent n = 0.20.
Determine the draw stress and draw force in this operation.
Solution:
The values of D and L, for Eq. (6) can be determined using Eqs. 19.34. D
= 2.25 mm and L, = 1.0 mm. Thus,
0.88
0.12
2.25
1.0
1.15
The areas before and after drawing are computed as Ao = 4.91 mm2 and Af
= 3.14 mm2. The resulting true strain = In (4.91/3.14)= 0.446. And the
average flow stress in the operation is computed:
205 0.446
1.20
.
145.4MPa
Draw stress is given by Eq. (7):
145.4
1
.
1.15 0.446
94.1 MPa
Finally, the draw force is this stress multiplied by the cross-sectional area
of the exiting wire:
F = 94.1(3.14) = 295.5 N.
Maximum Reduction per Pass A question that may occur to the reader
is: Why is more than one step required to achieve the desired reduction in
wire drawing? Why not take the entire reduction in a single pass through
one die, as in extrusion? The answer can be explained as follows. From
the preceding equations, it is clear that as the reduction increases, draw
stress increases. If the reduction is large enough, draw stress will exceed
the yield strength of the exiting metal. When that happens, the drawn wire
will simply elongate instead of new material being squeezed through the
die opening. For wire drawing to be successful, maximum draw stress
must be less than the yield strength of the exiting metal.
It is a straightforward matter to determine this maximum draw stress
and the resulting maximum possible reduction that can be made in one
pass, under certain assumptions.
Let us assume a perfectly plastic metal (n = 0), no friction, and no
redundant work. In this ideal case, the maximum possible draw stress is
equal to the yield strength of the work material. Expressing this using the
equation for draw stress under conditions of ideal deformation, Eq. (8),
and setting Yf = Y (because n = 0),
1
1
This means that In (Ao/Af) = In(l/(l-r)) = 1. Hence, Ao/Af = 1/(1-r) must
equal the natural logarithm base e. That is, the maximum possible strain is
1.0:
1.0
(9)
The maximum possible area ratio is
2.7183
(10)
0.632
(11)
and the maximum possible reduction is
The value given by Eq. (11) is often used as the theoretical maximum
reduction possible in a single draw, even though it ignores the effects of
friction and redundant work, which would reduce the' maximum possible
value, and strain hardening, which would increase the maximum possible
reduction because the exiting wire would be stronger than the starting
metal. In practice, draw reductions per pass are quite below the theoretical
limit. Reductions of 0.50 for single-draft bar drawing and 0.30 for
multiple-draft wire drawing seem to be the upper limits in industrial
operations.
Example 4
A round disk of 150mm diameter is to be blanked from a strip of 3.2mm,
half-hard cold-rolled steel whose shear strength = 310 MPa. Determine (a)
the appropriate punch and die diameters, and (b) blanking force.
Solution:
(a) The clearance allowance for half-hard cold-rolled steel is a = 0.075.
Accordingly,
C = 0.075 (3.2 mm) = 0.24 mm
The blank is to have a diameter = 150 mm, and die size determines blank
size. Therefore,
Die opening diameter = 150.00 mm
Punch diameter = 150 - 2(0.24) = 149.52 mm
(b) To determine the blanking force, we assume that the entire perimeter
of the part is blanked at one time. The length of the cut edge is
150
471.2 mm
And the force is
F = 310(471.2)(3.2) = 467,469 N (This is about 53 tons.)