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
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.
ร 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
HOW YOU GET P2 = 22.93 IN LAST EXAMPLE?
ReplyDeleteI CALCULATE AND GETTING 2.6
PLEASE REPLY ME ASAP.
fghjk
DeleteYup bro P2 = 2.608
ReplyDeleteSolve it by P2=2.608 ....it's correct ..it gives the same final value
ReplyDeleteI really liked your Information. Keep up the good work. Deep Drawn Metal
ReplyDeleteWhat does rs stands for??
ReplyDeleteA 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 = 15(.01+๐)" " 〗^0.4 〖๐/๐๐〗^2 and the coefficient of friction is 0.05. Also Determine ๐_๐ ๐๐๐ ๐_๐๐๐ฅ.
ReplyDelete*****
ReplyDelete