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S.T.B.S College of Diploma Engineering, SURAT

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. Contain for seminar. Definition of forging. Forging operations. Different type of forging losses.

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S.T.B.S College of Diploma Engineering, SURAT

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  1. 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

  2. Contain for seminar • Definition of forging. • Forging operations. • Different type of forging losses. • Procedure for calculating example. • One example solve.

  3. Definition of forging. Forging is a manufacturing process involving the shaping of metal using localized compressive forces

  4. Forging operations • Drawing down operation. • Upsetting operation. • Bending operation. • Punching and drafting operation. • Swaging operation. • Hot cutting. 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.

  5. (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 operation of enlarging the already punched holes. (5) Swaging operation. This the operation which reduces & 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.

  6. 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.

  7. (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.

  8. (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.

  9. (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

  10. 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 ×D2×l 2.4 Shear loss=∏/4 ×D2×l 2.5 Flash loss=C×W×T Step:3 Gross volume= net volume + All losses Step:4 Required length of material/piece=Gross volume ÷ C/s area of raw material

  11. Procedure for calculating example Step:5 Total required length of material = Required length of material/piece× total quantity Step:6 Total material cost = Total required length of material × Rate of bar stock • Note: • - Volume of hexagonal shape = 3√3÷2 ×(side)2×length. • - Raw material = finished part + losses. • OR • - Gross volume = Net volume + losses.

  12. 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 ×D2×l = ∏/4 ×(4)2×10 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 ×D2×l = ∏/4 ×(5)2×2 Tong loss = 39.26 Cm3

  13. 2.4 Shear loss=∏/4 ×D2×l = ∏/4 ×(5)2×0.2 Shear loss=3.926 Cm3 2.5 Flash loss=C×W×T = 28×2×0.3 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 ÷ C/s area of raw material =198.206÷ ∏/4 ×(5)2 Required length of material/piece= 10.09 cm

  14. Step:5 Total required length of material = Required length of material/piece× total quantity = 10.09×1000 Total required length of material= 100.94 m Step:6 Total material cost = Total required length of material × Rate of bar stock = 100.94×20 Total material cost=Rs. 2018.8

  15. Thank you

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