FORMING PROCESSES (Drawing & Deep Drawing)
EXAMPLE 3.7 (DRAWING)
A steel wire is drawn from an initial diameter of
12.7 mm to a final diameter of 10.2 mm at a speed of 90 m/min. The half-cone
angle of the die is 6° and the coefficient of friction at the job-die interface
is 0.1. A tensile test on the original steel specimen gives a tensile yield
stress 207 N/mm2. A similar specimen shows a tensile yield stress of
414 N/mm2 at a strain of 0.5. Assuming a linear stress-strain
relationship for the material, determine the drawing power and the maximum
possible reduction with the same die. No back tension is applied.
FORMULAE
SOLUTION
To start with, we
determine the average tensile yield stress of the material during the process.
For the given operation, the strain is
Since the stress-strain relationship is linear, the
value of the tensile yield stress at the end of the operation is given by
So, the average yield stress is
From the given data,
For the strain hardening a material, the value of σxf,
strictly speaking, can go up to the value of the tensile yield stress at the
end of the operation (σYf) which may be considerably greater
than the average value σY. So, when (σxf)max
is put equal to σYf,
Now, from equation (1) with Fb =
0,
Using equations (2) and (3), the drawing power we obtain
is
It is interesting to note that the drawing stress is
more than the tensile yield stress of the original material. To find out the
maximum allowable reduction, we use equation (4) with Fb = 0
and ϕ = 2. Hence,
If we use equation (5), then Dmax
= 0.652
EXAMPLE 3.8 (DEEP DRAWING)
A cold rolled steel cup with an inside radius 30 mm
and a thickness 3 mm is to be drawn from a blank of radius 40 mm. The shear
yield stress and the maximum allowable stress of the material can be taken as
210 N/mm2 and 600 N/mm2, respectively.
(i)
Determine the drawing force, assuming
that the coefficient of friction µ
= 0.1 and β = 0.05
(ii)
Determine the minimum possible radius of
the cup which can be drawn from the given blank without causing a fracture.
FORMULAE
SOLUTION
(i) We first calculate the blank holding force Fh
from the given data as
Next, we find the value of σr at r
→ rd by using equation (1). Thus,
Now, using equation (2), we get
It should be noted that this is much less than the
fracture strength though rj – rp = 10 mm =
3.33t, i.e., very close to the limit set by the condition of plastic buckling.
From equation (3), the drawing force is given as
(ii) In this case, σz = 600 N/mm2.
From equation (2), we have
Using equation (1) and taking Fh =
52778 N, we get
Again, it is interesting to note that rj
– rp = 30.8 mm >> 4t, and this goes much beyond
the limit set by the plastic buckling condition.
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