Forming Processes - DRAWING - Numerical Problems with Solutions (solved)

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 σ_xf, strictly speaking, can go up to the value of the tensile yield stress at the end of the operation (σ_Yf) which may be considerably greater than the average value σ_Y. So, when (σ_xf)_max is put equal to σ_Yf,

Now, from equation (1) with F_b = 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 F_b = 0 and ϕ = 2. Hence,

If we use equation (5), then D_max = 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 F_h from the given data as

Next, we find the value of σ_r at r → r_d 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 r_j – r_p = 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, σ_z = 600 N/mm². From equation (2), we have

Using equation (1) and taking F_h = 52778 N, we get

Again, it is interesting to note that r_j – r_p = 30.8 mm >> 4t, and this goes much beyond the limit set by the plastic buckling condition.