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Turning Operations
L a t h e
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Turning Operations
Machine Tool – LATHE
Job (workpiece) – rotary motion
Tool – linear motions
“Mother of Machine Tools “
Cylindrical and flat surfaces
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Some Typical Lathe Jobs
Turning/Drilling/Grooving/Threading/Knurling/Facing...
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The LatheHead stock
Spindle
Feed rod
Bed
Compound rest
and slide(swivels)
Carriage
Apron
selector
Feed change
gearbox
Spindlespeed
Lead screw
Cross slide
Dead center
Tool post
Guide ways
Tailstock quill
Tailstock
Handle
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The Lathe
Bed
Head StockTail Stock
Carriage
Feed/Lead Screw
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Types of Lathes
Engine Lathe
Speed Lathe
Bench LatheTool Room Lathe
Special Purpose LatheGap Bed Lathe
…
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Size of Lathe
Workpiece Length Swing
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Size of Lathe ..
Example: 300 - 1500 Lathe
Maximum Diameter of Workpiece
that can be machined= SWING (= 300 mm)
Maximum Length of Workpiecethat can be held betweenCenters (=1500 mm)
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Workholding Devices
Equipment used to hold
Workpiece – fixtures
Tool - jigs
Securely HOLD or Support whilemachining
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Chucks
Three jaw Four Jaw
W o r k h
o l d i n g
D e v i c e s . .
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Centers
Workpiece
Headstock center
(Live Centre)
Tailstock center
(Dead Centre)
W o r k h
o l d i n g
D e v i c e s . .
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Faceplates
W
o r k p i e c e
W o r k h
o l d i n g
D e v i c e s . .
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Dogs
W o r k h
o l d i n g
D e v i c e s . .
Tail
Tail
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Mandrels
Workpiece (job) with a hole
W o r k h
o l d i n g
D e v i c e s . .
Workpiece Mandrel
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Rests
W o r k h
o l d i n g
D e v i c e s . .
Jaws
HingeWork Work Jaws
Lathe bed guideways
Carriage
Steady Rest Follower Rest
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Operating/Cutting
Conditions1. Cutting Speed v
2. Feed f3. Depth of Cut d
Workpiece
Tool
Chip
Tool post
S
peripheral
speed (m/min)
N (rev/min)
D
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Operating Conditions
Workpiece
Tool
Chip
Tool post
S
peripheral
speed
N (rev/min)
D
N DS speed peripheral
D
rotation1intraveltoolrelative
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Cutting Speed
The Peripheral Speed ofWorkpiece past the Cutting Tool
=Cutting Speed O p e r a
t i n g C
o n d i t i o n s . .
m/min1000
N Dv
D – Diameter (mm)N – Revolutions per Minute (rpm)
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Feed
f – the distance the tooladvances for every rotation ofworkpiece (mm/rev)
O p e r a
t i n g C
o n d i t i o n s . .
f
Feed
D D 21
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Depth of Cut
perpendicular distance betweenmachined surface and uncutsurface of the Workpiece
d = (D 1 – D 2)/2 (mm)
O p e r a
t i n g C
o n d i t i o n s . .
d Depth
of Cut
D D 21
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3 Operating Conditions
Chip
Machinedsurface
Workpiece
Depth of cutTool
Chuck
N
Feed (f )
Cutting speed
Depth of cut (d
)
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Selection of ..
Workpiece Material
Tool Material
Tool signatureSurface Finish
AccuracyCapability of Machine Tool
O p e r a
t i n g C
o n d i t i o n s . .
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Material Removal Rate
MRR Volume of material removed in one
revolution MRR = D d f mm3 • Job makes N revolutions/min
MRR =
D d f N (mm3
/min) In terms of v MRR is given by
MRR = 1000 v d f (mm3 /min) O p e r a
t i o n s o n L a t h e . .
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MRR
dimensional consistency bysubstituting the units
O p e r a
t i o n s o n L a t h e . .
MRR : D d f N (mm)(mm)(mm/rev)(rev/min)
= mm3 /min
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Operations on Lathe
Turning
Facingknurling
Grooving
Parting
Chamfering
Taper turningDrilling
Threading
O p e r a
t i o n s o n L a t h e . .
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Turning
O p e r a
t i o n s o n L a t
h e . .
Cylindrical job
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Turning ..
Cylindrical job
Cutting
speed
Chip
Workpiece
Depth of cut (d)
Depth of cutTool
FeedChuck
N
Machinedsurface
O p e r a
t i o n s o n L a t
h e . .
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Turning ..
Excess Material is removedto reduce Diameter
Cutting Tool: Turning Tool
a depth of cut of 1 mm willreduce diameter by 2 mm
O p e r a
t i o n s o n L a t
h e . .
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Facing
Flat Surface/Reduce length
Depth ofcut
Feed
WorkpieceChuck
Cutting
speed
Tool
d
MachinedFace
O p e r a
t i o n s o n L a t
h e . .
i
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Facing ..
machine end of job Flat surfaceor to Reduce Length of Job
Turning Tool
Feed: in direction perpendicular toworkpiece axis
Length of Tool Travel = radius of
workpieceDepth of Cut: in direction parallel to
workpiece axis O p e r a
t i o n s o n L a t
h e . .
i
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Facing ..
O p e r a
t i o n s o n L a t
h e . .
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Eccentric Turning
Axis of job
A
Eccentric peg(to be turned)
4-jawchuck
Cuttingspeed O
p e r a
t i o n s o n L a t
h e . .
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Knurling
Produce rough textured surfaceFor Decorative and/or
Functional PurposeKnurling Tool
A Forming ProcessMRR~0
O p e r a
t i o n s o n L a t
h e . .
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Knurling
O p e r a
t i o n s o n L a t
h e . .
Knurling toolTool post
Feed
Cutting
speed
Movement
for depth
Knurled surface
K li
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Knurling ..
O p e r a
t i o n s o n L a t
h e . .
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Grooving
Produces a Groove onworkpiece
Shape of tool shape ofgroove
Carried out using GroovingTool A form tool
Also called Form Turning O p e r a
t i o n s o n L a t
h e . .
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Grooving ..
O p e r a
t i o n s o n L a t
h e . .
Shape produced
by form tool Groove
Grooving
toolFeed or
depth of cutForm tool
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Parting
Cutting workpiece into Two
Similar to grooving
Parting Tool
Hogging – tool rides over – at
slow feedCoolant use
O p e r a
t i o n s
o n L a t
h e . .
ti
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Parting ..
O p e r a
t i o n s
o n L a t
h e . .
FeedParting tool
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Chamfering
O p e r a
t i o n s
o n L a t
h e . .
Chamfering tool
Feed
Chamfer
Ch f i
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Chamfering
Beveling sharp machined edges
Similar to form turning
Chamfering tool – 45° To
Avoid Sharp Edges
Make Assembly Easier
Improve Aesthetics O p e r a
t i o n s
o n L a t
h e . .
T T i
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Taper Turning
Taper:
C B
A
L
D90°
2D1
O p e r a
t i o n s
o n L a t
h e . .
L D D
2tan 21
T T i
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Taper Turning..
Methods Form Tool Swiveling Compound Rest
Taper Turning Attachment Simultaneous Longitudinal and
Cross Feeds O p e r a
t i o n s
o n L a t h e . .
Conicity L D D K 21
Taper Turning
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Taper Turning ..By Form Tool
O p e r a
t i o n s
o n L a t h e . .
Taper Workpiece
Straight
cutting edge
Directionof feed
Form
tool
Taper Turning
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Taper Turning ,,
By Compound Rest
O p e r a
t i o n s
o n L a t h e . .
Face plate
Dog
Tail stock quill
Tail stock
Mandrel
Direction of feed
Compound restSlide
Compound rest
Hand crank
Tool post &
Tool holder
Cross slide
D illi
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Drilling
Drill – cutting tool – held in TS – feed from TS
O p e r a
t i o n s
o n L a t h e . .
Feed
Drill
Quill
clamp movingquill
Tail stock clamp
Tail stock
S
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Process Sequence
How to make job from rawmaterial 45 long x 30 dia.?
20 dia
40
15
O p e r a
t i o n s
o n L a t h e . .
Steps:•Operations•Sequence
•Tools•Process
Process Sequence
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Process Sequence .. Possible Sequences
TURNING - FACING - KNURLING
TURNING - KNURLING - FACING
FACING - TURNING - KNURLING FACING - KNURLING - TURNING
KNURLING - FACING - TURNING
KNURLING - TURNING – FACING
What is an Optimal Sequence? O p e r a
t i o n s
o n L a t h e . .
X
X
X X
M hi i Ti
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Machining Time
Turning Time
Job length L j mm
Feed f mm/revJob speed N rpm
f N mm/min
min N f
Lt
j
O p e r a
t i o n s
o n L a t h e . .
M f t i Ti
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Manufacturing Time
Manufacturing Time
= Machining Time
+ Setup Time
+ Moving Time+ Waiting Time
O p e r a t i o n s
o n L a t h e . .
E l
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Example
A mild steel rod having 50 mmdiameter and 500 mm length isto be turned on a lathe.
Determine the machining timeto reduce the rod to 45 mm inone pass when cutting speed is
30 m/min and a feed of 0.7mm/rev is used.
Example
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Example
calculate the required spindlespeed as: N = 191 rpm
m/min1000
N Dv
Given data: D = 50 mm, L j = 500 mmv = 30 m/min, f = 0.7 mm/rev
Substituting the values of v and D in
E l
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Example
Can a machine has speed of191 rpm?
Machining time:min
N f
Lt
j
t = 500 / (0.7191) = 3.74 minutes
E l
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Example
Determine the angle at which thecompound rest would be swiveled
for cutting a taper on a workpiecehaving a length of 150 mm andoutside diameter 80 mm. The
smallest diameter on the taperedend of the rod should be 50 mmand the required length of thetapered portion is 80 mm.
E l
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Example
Given data: D 1 = 80 mm, D 2 = 50mm, L j = 80 mm (with usual
notations)tan = (80-50) / 280
or = 10.620
The compound rest should beswiveled at 10.62o
E l
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Example
A 150 mm long 12 mm diameterstainless steel rod is to be reduced
in diameter to 10 mm by turning ona lathe in one pass. The spindlerotates at 500 rpm, and the tool is
traveling at an axial speed of 200mm/min. Calculate the cuttingspeed, material removal rate andthe time required for machining thesteel rod.
E l
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Example
Given data: L j = 150 mm, D 1 = 12mm, D 2 = 10 mm, N = 500 rpm
Using Equation (1) v = 12500 / 1000
= 18.85 m/min.
depth of cut = d = (12 – 10)/2 = 1mm
E l
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Example
feed rate = 200 mm/min, we getthe feed f in mm/rev by dividingfeed rate by spindle rpm. That is
f = 200/500 = 0.4 mm/rev From Equation (4),
MRR = 3.142120.41500 =
7538.4mm3/min
from Equation (8),
t = 150/(0.4500) = 0.75 min.
E l
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Example
Calculate the time required tomachine a workpiece 170 mm long,
60 mm diameter to 165 mm long50 mm diameter. The workpiecerotates at 440 rpm, feed is
0.3 mm/rev and maximum depth ofcut is 2 mm. Assume totalapproach and overtravel distanceas 5 mm for turning operation.
E l
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Example
Given data: L j = 170 mm, D 1 =60 mm, D 2 = 50 mm, N = 440rpm, f = 0.3 mm/rev, d = 2 mm,
How to calculate the machiningtime when there is more than oneoperation?
Example
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Example Time for Turning:
Total length of tool travel = job length + lengthof approach and overtravel L = 170 + 5 = 175 mm Required depth to be cut = (60 – 50)/2 = 5
mm Since maximum depth of cut is 2 mm, 5 mm
cannot be cut in one pass. Therefore, we
calculate number of cuts or passes required. Number of cuts required = 5/2 = 2.5 or 3(since cuts cannot be a fraction)
Machining time for one cut = L / (f N )
Total turning time = [L / (f N )] Number of
Example
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Example Time for facing:
Now, the diameter of the job isreduced to 50 mm. Recall that incase of facing operations, length oftool travel is equal to half thediameter of the job. That is, l = 25mm. Substituting in equation 8, we
get t = 25/(0.3440)
= 0.18 min.
Example
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Example
Total time: Total time for machining = Time
for Turning + Time for Facing
= 3.97 + 0.18 = 4.15 min.
The reader should find out the total
machining time if first facing isdone.
Example
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Example
From a raw material of 100 mmlength and 10 mm diameter, acomponent having length 100 mmand diameter 8 mm is to beproduced using a cutting speed of31.41 m/min and a feed rate of 0.7mm/revolution. How many times
we have to resharpen or regrind, if1000 work-pieces are to beproduced. In the taylor’s expression
use constants as n = 1.2 and C =
Example
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Example
Given D =10 mm , N = 1000 rpm,v = 31.41 m/minute
From Taylor’s tool life expression,we have vT n = C
Substituting the values we get,
(31.40)(T )1.2 = 180 or T = 4.28 min
Example
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Example
Machining time/piece = L / (f N ) = 100 / (0.71000)
= 0.142 minute.
Machining time for 1000 work-pieces= 1000 0.142 = 142.86 min
Number of resharpenings = 142.86/4.28
= 33.37 or 33 resharpenings
Example
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Example 6: While turning a carbon steel
cylinder bar of length 3 m and diameter0.2 m at a feed rate of 0.5mm/revolution with an HSS tool, one ofthe two available cutting speeds is to beselected. These two cutting speeds are100 m/min and 57 m/min. The tool lifecorresponding to the speed of 100
m/min is known to be 16 minutes withn =0.5. The cost of machining time,setup time and unproductive timetogether is Rs.1/sec. The cost of one
tool re-sharpening is Rs.20.
Example
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Example Given T 1 = 16 minute, v 1 = 100
m/minute, v 2 = 57 m/minute, D =200mm, l = 300 mm, f = 0.5 mm/rev Consider Speed of 100 m/minute N 1 = (1000 v ) / ( D ) =
(1000100) / (200) = 159.2 rpm t 1 = l / (f N ) = 3000 / (0.5 159.2) =
37.7 minute Tool life corresponding to speed of 100
m/minute is 16 minute. Number of resharpening required = 37.7 /
16 = 2.35
Example
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Example
Total cost =
Machining cost + Cost of
resharpening Number ofresharpening
= 37.7
60
1+ 20
2 = Rs.2302
Example
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Example
Consider Speed of 57 m/minute Using Taylor’s expression T 2 = T 1 (v 1 / v 2)2 with usual notations
= 16 (100/57)2 =49 minute
Repeating the same procedure we
get t 2 = 66 minute, number ofreshparpening=1 and total cost =Rs. 3980.
Example
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Example
Write the process sequence tobe used for manufacturing the
component
from raw material of 175 mm length
and 60 mm diameter
Example
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Example
40 Dia50
50 40
20
50
20 Dia
Threading
Example
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Example
To write the process sequence, first listthe operations to be performed. Theraw material is having size of 175 mmlength and 60 mm diameter. Thecomponent shown in Figure 5.23 ishaving major diameter of 50 mm, step
diameter of 40 mm, groove of 20 mmand threading for a length of 50 mm.The total length of job is 160 mm.Hence, the list of operations to be
carried out on the job are turning,
Example
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Example
A possible sequence for producingthe component would be:
Turning (reducing completely to 50mm)
Facing (to reduce the length to 160mm)
Step turning (reducing from 50 mmto 40 mm)
Thread cutting.