Saturday 31 May 2014

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 xs 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 ยต = 0.08

SOLUTION

Using the following equation, we obtain

Since, xs 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.
รž 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, ยต and rs in equation (4), the total forging force we obtain is










8 comments:

  1. HOW YOU GET P2 = 22.93 IN LAST EXAMPLE?
    I CALCULATE AND GETTING 2.6

    PLEASE REPLY ME ASAP.

    ReplyDelete
  2. Solve it by P2=2.608 ....it's correct ..it gives the same final value

    ReplyDelete
  3. I really liked your Information. Keep up the good work. Deep Drawn Metal

    ReplyDelete
  4. What does rs stands for??

    ReplyDelete
  5. A circular disc of 120mm dia. And 64mm height is forged at room temperature between two flat dies to 36mm height. Determine the die load at the end of compression using slab method of analysis. The yield strength of the material is 〖๐œŽ = 15(.01+๐œ€)" " 〗^0.4 〖๐‘/๐‘š๐‘š〗^2 and the coefficient of friction is 0.05. Also Determine ๐‘ƒ_๐‘š ๐‘Ž๐‘›๐‘‘ ๐‘ƒ_๐‘š๐‘Ž๐‘ฅ.

    ReplyDelete