The document analyzes and designs a tailstock assembly for a lathe. It describes the parts and assembly of a tailstock. It then analyzes the forces acting on the tailstock during machining operations. Based on the force analysis, it specifies dimensions for the tailstock components like the mandrel diameter, barrel length and diameter, and base dimensions. It also calculates the maximum force on the tailstock and determines the required clearance between the mandrel and barrel.

1. ANALYSIS AND DESIGN
OF
TAILSTOCK ASSEMBLY
PRESENTED 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 T1 =FEED SCREW
T2 =REDUCTION GEAR BOX
T3 =BODY
T4 =ADJUSTABLE BASE
T5 =SPINDLE
T6 =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 Px =Thrust force
mass of the work piece is = M/2 Pz = cutting force
 THE REACTION FORCES, Py = shear force
A'2= Py×(L1/L) -Px×(d/2L)
A''2=Pz×(L1/L) - (M/2)
K2= K

8. SPECIFICATIONS OF THE TAILSTOCK
Dia of the mandrel (d)= Ch0.7
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 = √(F2
h+F2
v)
But Fh = Pz×(L1/L) - (m/2)
Fv = Py×(L1/L) - Px×(d/2L)
but if the turning is done at the tailstock
i.e, at L1 = 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.., Fv ≈ Pz
Fh ≈ Py
= 1/8 to 1/4 of Pz for conventional cut
we know that,
Fc = cos(β-٧) /{sin(ǿ)×cos(ǿ+β-٧)} *‫ڂ‬s*As
cos(β-٧) /{sin(ǿ)×cos(ǿ+β-٧)} = 4
‫ڂ‬ s = 40 kg/mm2 for carbon steels
Acc to niccolson and smith standard area at the tailstock
centre is assumed as (h2/6400)

11. contd...
then..,
substituting all the values,
F = 1.1 Pz
= 1.1×(h2/6400)×4×40
= h2/36
if the mandrel is taken as overhang
L1 = 5√h
L2 =35√h

12. contd...,
δ = QL2
3 /8EI
deflection of cantilever beam with the UDL
Q = F×(L1/L2) = (h2/36)×5×√h
35×√h
= h2/252
 if the clearance b/n the mandrel & the barrel is chosen
as easy sliding fit,
then, clearance C ≈ 0.002×d = δ
0.002d = h2×(35√h)3×64
252×8×2×104×π×d4

13. contd...,
d5 =10.8×h(7/2)
d = 1.6×h0.7 where h is in mm
This completes the design of tailstock