Analysis and design of tail stock assembly

ANALYSIS OF MACHINE TOOLS

1. ANALYSIS AND DESIGN OF TAILSTOCK ASSEMBLY PRESEN TED BY L SURESH KUMAR R NO: 163506 MANUFACTURING ENGG.

2. TAILSTOCK A tailstock, also known as a foot stock, is a device often used as part of an engineering lathe, wood-turning lathe, or used in conjunction with a rotary table on a milling machine. It is usually used to apply support to the longitudinal rotary axis of a workpiece being machined.

3. PARTS OF TAILSTOCK T 1 =FEED SCREW T 2 =REDUCTION GEAR BOX T 3 =BODY T 4 =ADJUSTABLE BASE T 5 =SPINDLE T 6 =LOCKING PIN

4. ASSEMBLY OF TILSTOCK

5. FORCES AT THE LATHE CENTRES

6. DESIGN OF TAILSTOCK Design of tailsock is associated with the design of the lathe beds. As the tail stock slides on to the lathe beds the forces induced on the lathe bed also transmited to the tailstock. The design of tailstock is preceded by the analysis of forces that are acting on the system due to tool workpiece interaction.

7. BY FINDING THE REACTION FORCES considering the work piece as the SSB AT THE TAILSTOCK CENTRE P x =Thrust force mass of the work piece is = M/2 P z = cutting force THE REACTION FORCES, P y = shear force A' 2 = P y × (L 1 /L) -P x × (d/2L) A'' 2 =P z × (L 1 /L) - (M/2) K 2 = K

8. SPECIFICATIONS OF THE TAILSTOCK Dia of the mandrel (d)= Ch 0.7 Length of the barrel(L)=40 √h Dia of the barrel (D) =1.75d Base length (B) = 1.25h Base width (C) = h A = h+60

9. FORCES IN TAILSTOCK The resultant load at the tailstock centre is given by, F = √(F 2 h +F 2 v ) But F h = P z × (L 1 /L) - (m/2) F v = P y × (L 1 /L) - P x × (d/2L) but if the turning is done at the tailstock i.e, at L 1 = L Further assuming some data i.e, if d = h , and L = 6 to 8h as d is very small (d/ 2* L)<<<L

10. Contd.... thus.., F v ≈ P z F h ≈ P y = 1/8 to 1/4 of P z for conventional cut we know that, F c = cos( β- ٧ ) /{sin( ǿ) × cos( ǿ+β- ٧ )} * ڂ s *A s cos( β- ٧ ) /{sin( ǿ) × cos(ǿ+β- ٧ )} = 4 ڂ s = 40 kg/mm 2 for carbon steels Acc to niccolson and smith standard area at the tailstock centre is assumed as (h 2 /6400)

11. contd... then.., substituting all the values, F = 1.1 P z = 1.1 × (h 2 /6400) × 4 × 40 = h 2 /36 if the mandrel is taken as overhang L 1 = 5 √h L 2 =35√h

12. contd..., δ = QL 2 3 /8EI deflection of cantilever beam with the UDL Q = F × (L 1 /L 2 ) = (h 2 /36) × 5 × √h 35 × √h = h 2 /252 if the clearance b/n the mandrel & the barrel is chosen as easy sliding fit, then, clearance C ≈ 0.002 × d = δ 0.002d = h 2 × (35 √h) 3 × 64 252 × 8 × 2 × 10 4 ×π×d 4

13. contd..., d 5 =10.8 × h (7/2) d = 1.6 × h 0.7 where h is in mm This completes the design of tailstock

14. THANK YOU