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

FORMING PROCESSES (Forging of Strip & Forging of Disc)

EXAMPLE 3.4 (FORGING OF A STRIP)

A strip of lead with initial dimensions 24 mm × 24 mm × 150 mm is forged between two flat dies to a final size of 6 mm × 96 mm × 150 mm. If the coefficient of friction between the job and the die is 0.25, determine the maximum forging force. The average yield stress of lead in tension is 7 N/mm².

SOLUTION

First, let us determine the shear yield stress K for lead by using following equation. Thus,

To find p, the value of x_s is required. From following equation, we have

Now, from the following equations, the expressions for the pressures p₁ and p₂ (for the non-sticking and the sticking zones, respectively) can be found out. Thus,

Using following equation, the force per unit length we get is

Since the length of the strip is 150 mm, the total forging force is

EXAMPLE 3.5

Solve Example 3.4 when the coefficient of friction µ = 0.08

SOLUTION

Using the following equation, we obtain

Since, x_s is more than l, the entire zone is non-sticking, and, as a result, the expression for the pressure throughout the contact surface is given by following equation. Thus,

So,

The corresponding value of the total forging force is

EXAMPLE 3.6 (DISC FORGING)

A circular disc of lead of radius 150 mm and thickness 50 mm is reduced to a thickness of 25 mm by open die forging. If the coefficient of friction between the jib and the die is 0.25, determine the maximum forging force. The average shear yield stress of lead can be taken as 4 N/mm².

FORMULAE:

SOLUTION

Since the volume remains constant, the final radius of the disc is

R = √2 × 150 mm = 212.1 mm.

Þ R = 212.1 mm

Now, using equation (1)

The expressions for p₁ and p₂ using equations (2) and (3) are

Substituting the values of K, h, R, µ and r_s in equation (4), the total forging force we obtain is