# Forging

```S.T.B.S College of Diploma Engineering,
SURAT
Class: Mech-1051(B)
Prepared.BY:Thumar Sanket C.
EN.NO:106470319020
TOPIC : cost estimation of forging
1
Contain for seminar
Definition of forging.
Forging operations.
Different type of forging losses.
Procedure for calculating example.
One example solve.
Definition of forging.
Forging is a manufacturing process
involving the shaping of metal using
localized compressive forces
Forging operations
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Drawing down operation.
Upsetting operation.
Bending operation.
Punching and drafting operation.
Swaging operation.
Hot cutting.
(1) Drawing down operation.
Due to this operation by decreasing the cross-section of
the work-piece, its length can be increased.
(2)Upsetting operation.
This operation increases the cross-section of the bar-stock
thereby reducing its length.
(3) Bending operation.
The bending of a job is done by
holding it between fixtures and
striking the work-piece with the
help of hammer.
(4) Punching and drafting operation
Punching is the operation of
making holes and drifting is the
punched holes.
(5) Swaging operation.
This the operation which reduces &amp; finishes the job to
desired size and shape usually either round or hexagonal.
This operation brings the job into required accurate shape
with the help of a swage.
(6) Hot cutting.
This operation is used to cut the heated metal bars in a
smithy shop.
Different type of forging losses
(1) Scale loss :
It is the loss of material due to surface oxidation in heating
and forging the job. The amount of loss depends upon
surface area, type of material, heating time and forging
technique. Generally it is taken as 5% of the net volume.
(2) Sprue loss :
The connection
between the tong hold
and the forging is
called sprue.
Generally it is taken
as 5% of net volume.
(3) Tong loss :
For forging operations, some length of stock is needed for
holding the job in the tong. This extra length of stock as shown in
fig. Is removed after completion of forging. This loss is generally
taken as 2 to 3 cm. of bar-stock length. In case of hand forging tong
loss is not considered.
(3) Shear loss :
It is also called cut waste loss. The small stock are required for
forging operations are cut from long bar-stock by sawing or
shearing. Some material is lost during cutting, is known as shear
loss or cut waste loss. This loss is generally taken as 5% of the
weight of forging.
(3) Flash loss :
Flash is the excess metal extruded as a thin plate surrounding the
forging at the parting line as shown in fig. for obtaining finished
product, this excess material is trimmed off. The loss of metal due to
trimming the flash is given by
Volume of flash = C*W*T*
Where, C=circumference of product at parting line
W=Average width of the flash
T= Average thickness of the flash
Procedure for calculating example
Step:1
Find out net volume
Step:2
Find out all losses.
2.1 Scale loss=5% of net volume
2.2 Sprue loss=5% of net volume
2.3 Tong loss=∏/4 &times;D2&times;l
2.4 Shear loss=∏/4 &times;D2&times;l
2.5 Flash loss=C&times;W&times;T
Step:3
Gross volume= net volume + All losses
Step:4
Required length of material/piece=Gross volume &divide; C/s area of raw
material
Procedure for calculating example
Step:5
Total required length of material = Required length of material/piece&times;
total quantity
Step:6
Total material cost = Total required length of material &times; Rate of bar stock
Note:
- Volume of hexagonal shape = 3√3&divide;2 &times;(side)2&times;length.
- Raw material = finished part + losses.
OR
- Gross volume = Net volume + losses.
One example solve
Example:
1000 M.S. pin of 4cm diameter and 10cm long are to be prepared by forging
process for this bar stock of 5cm diameter is used. If cost of bar stock is
Rs.20/m than estimate total material cost. Consider all losses.
Solve this example:
Step:1
Net volume= ∏/4 &times;D2&times;l
= ∏/4 &times;(4)2&times;10
Net volume =126.66 Cm3
Step:2
2.1 Scale loss=5% of net volume = 6.28 Cm3
2.2 Sprue loss=5% of net volume = 6.28 Cm3
2.3 Tong loss=∏/4 &times;D2&times;l
= ∏/4 &times;(5)2&times;2
Tong loss = 39.26 Cm3
2.4 Shear loss=∏/4 &times;D2&times;l
= ∏/4 &times;(5)2&times;0.2
Shear loss=3.926 Cm3
2.5 Flash loss=C&times;W&times;T
= 28&times;2&times;0.3
Flash loss= 16.8 Cm3
Step:3
Gross volume= net volume + All losses
= 125.66+72.546
Gross volume= 198.206Cm3
Step:4
Required length of material/piece=Gross volume &divide; C/s area of
raw material
=198.206&divide; ∏/4 &times;(5)2
Required length of material/piece= 10.09 cm
Step:5
Total required length of material = Required length of material/piece&times;
total quantity
= 10.09&times;1000
Total required length of material= 100.94 m
Step:6
Total material cost = Total required length of material &times; Rate of bar
stock
= 100.94&times;20
Total material cost=Rs. 2018.8
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