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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


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


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


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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


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


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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.

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5 Lathe