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Turning Operations

L a t h e

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Turning Operations

Machine ToolLATHE

Job (workpiece) rotary motion

Tool linear motionsMother 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 Lathe

Tool Room Lathe

Special Purpose Lathe

Gap 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 Workpiece that can be

held between Centers (=1500 mm)

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WorkholdingDevices

Equipment used to hold

Workpiecefixtures

Tool -jigs

SecurelyHOLD orSupportwhile machining

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Chucks

Three jaw Four Jaw

Workh

olding

Devices..

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Centers

Workpiece

Headstock center

(Live Centre)

Tailstock center

(Dead Centre)

Workh

olding

Devices..

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Faceplates

W

orkpiece

Workh

olding

Devices..

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Dogs

Workh

olding

Devices..

Tail

Tail

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Mandrels

Workpiece (job) with a hole

Workh

olding

Devices..

Workpiece Mandrel

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Rests

Workh

olding

Devices..

Jaws

HingeWork Work Jaws

Lathe bed guideways

Carriage

Steady Rest Follower Rest

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Operating/Cutting Conditions

1. Cutting Speed v

2. Feedf

3. 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

NDSspeedperipheral

D

rotation1intraveltoolrelative

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Cutting Speed

The Peripheral Speed of Workpiece past theCutting Tool

=Cutting SpeedOpera

tingC

onditions..

m/min1000

NDv

D Diameter (mm)N Revolutions per Minute (rpm)

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Feed

f the distance the tool advances for everyrotation of workpiece (mm/rev)

Opera

tingC

onditions..

f

Feed

DD 21

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Depth of Cut

perpendicular distance between machinedsurface and uncut surface of the Workpiece

d= (D1D2)/2 (mm)

Opera

tingC

onditions..

dDepth

of Cut

DD 21

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3 Operating Conditions

Chip

Machined

surface

Workpiece

Depth of cutToolChuck

N

Feed (f )

Cutting speed

Depth of cut (d)

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Selection of ..

Workpiece Material

Tool Material

Tool signature Surface Finish

Accuracy

Capability of Machine Tool

Opera

tingC

onditions..

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Material Removal Rate

MRRVolume of material removed in one

revolution MRR = Ddf mm3 Job makes N revolutions/min

MRR = DdfN (mm3/min)

In terms ofv MRR is given by

MRR = 1000 v d f (mm3/min)Opera

tionsonLathe..

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MRR

dimensional consistency bysubstituting the units

Opera

tionsonLathe..

MRR: DdfN(mm)(mm)(mm/rev)(rev/min)

= mm3/min

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Operations on Lathe

Turning

Facing knurling

Grooving

Parting

Chamfering

Taper turning Drilling

Threading

Opera

tionsonLathe..

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Turning

Cylindrical job

Opera

tionsonLat

he..

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Turning ..

Cylindrical job

Cutting

speed

Chip

Workpiece

Depth of cut (d)

Depth of cutTool

FeedChuck

N

Machinedsurface

Opera

tionsonLat

he..

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Turning ..

Excess Material is removed toreduce Diameter

Cutting Tool: Turning Tool

a depth of cutof 1 mm willreduce diameter by 2 mm

Opera

tionsonLat

he..

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Facing

Flat Surface/Reduce length

Depth ofcut

Feed

WorkpieceChuck

Cutting

speed

Tool

d

MachinedFace

Opera

tionsonLat

he..

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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

workpiece Depth of Cut: in direction parallel to

workpiece axisOpera

tionsonLat

he..

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Facing ..

Opera

tionsonLat

he..

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Eccentric Turning

Axis of job

A

Eccentric peg(to be turned)

4-jawchuck

CuttingspeedO

pera

tionsonLat

he..

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Knurling

Produce rough textured surface For Decorative and/or Functional Purpose

Knurling Tool

A Forming Process

MRR~0

Opera

tionsonLat

he..

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Knurling

Opera

tionsonLat

he..

Knurling toolTool post

Feed

Cutting

speed

Movement

for depth

Knurled surface

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Knurling ..

Opera

tionsonLat

he..

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Grooving

Produces a Groove onworkpiece

Shape of tool shape ofgroove

Carried out using Grooving Tool

A form tool

Also called Form TurningOpera

tionsonLat

he..

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Grooving ..

Opera

tionsonLat

he..

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 feed

Coolant use

Opera

tions

onLat

he..

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Parting ..

Opera

tions

onLat

he..

FeedParting tool

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Chamfering

Opera

tions

onLat

he..

Chamfering tool

Feed

Chamfer

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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

Opera

tions

onLat

he..

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Taper Turning

Taper:

CB

A

L

D90

2D1

Opera

tions

onLat

he..

LDD

2tan 21

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Taper Turning..

Methods Form Tool

Swiveling Compound Rest

Taper Turning Attachment Simultaneous Longitudinal and Cross

FeedsOpera

tions

onLathe..

ConicityLDDK 21

T T i

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Taper Turning ..

By Form Tool

Opera

tions

onLathe..

TaperWorkpiece

Straight

cutting edge

Directionof feed

Form

tool

T T i

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Taper Turning ,,

By Compound Rest

Opera

tions

onLathe..

Face plate

Dog

Tail stock quill

Tail stock

Mandrel

Direction of feed

Compound restSlide

Compound rest

Hand crank

Tool post &

Tool holder

Cross slide

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Drilling

Drill cutting tool held in TS feed from TS

Opera

tions

onLathe..

Feed

Drill

Quill

clamp movingquill

Tail stock clamp

Tail stock

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Process Sequence

How to make job from raw material 45 longx 30 dia.?

20 dia

40

15

Opera

tions

onLathe..

Steps:OperationsSequenceTools

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?Opera

tions

onLathe..

X

X

XX

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Machining Time

Turning Time

Job length Lj mm

Feedf mm/rev

Job speed N rpm

fN mm/min

minNf

Lt

j

Opera

tions

onLathe..

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Manufacturing Time

Manufacturing Time= Machining Time

+ Setup Time

+ Moving Time

+ Waiting Time

Operations

onLathe..

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Example

A mild steel rod having 50 mm diameter and500 mm length is to be turned on a lathe.Determine the machining time to reducethe rod to 45 mm in one pass when cutting

speed is 30 m/min and a feed of 0.7mm/rev is used.

E l

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Example

calculate the required spindle speed as: N =

191 rpm

m/min1000

NDv

Given data: D= 50 mm, Lj= 500 mmv= 30 m/min, f= 0.7 mm/rev

Substituting the values ofvand D in

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Example

Can a machine has speed of 191rpm?

Machining time:

minNf

Lt

j

t= 500 / (0.7191) = 3.74 minutes

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Example

Determine the angle at which thecompound rest would be swiveled forcutting a taper on a workpiece having a

length of 150 mm and outside diameter80 mm. The smallest diameter on thetapered end of the rod should be 50 mm

and the required length of the taperedportion is 80 mm.

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Example

Given data: D1 = 80 mm, D2 = 50 mm, Lj= 80 mm (with usual notations)

tan = (80-50) / 280

or = 10.620

The compound rest should be swiveled at10.62o

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Example

A 150 mm long 12 mm diameter stainlesssteel rod is to be reduced in diameter to10 mm by turning on a lathe in one pass.

The spindle rotates at 500 rpm, and thetool is traveling at an axial speed of 200mm/min. Calculate the cutting speed,

material removal rate and the timerequired for machining the steel rod.

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Example

Given data: Lj = 150 mm, D1 = 12 mm, D2= 10 mm, N = 500 rpm

Using Equation (1)

v= 12500 / 1000

= 18.85 m/min.

depth of cut = d= (12 10)/2 = 1 mm

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Example feed rate = 200 mm/min, we get the feedf

in mm/rev by dividing feed rate by spindlerpm. That is

f= 200/500 = 0.4 mm/rev From Equation (4), MRR = 3.142120.41500 = 7538.4

mm3/min from Equation (8), t= 150/(0.4500) = 0.75 min.

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Example

Calculate the time required to machine aworkpiece 170 mm long, 60 mmdiameter to 165 mm long 50 mm

diameter. The workpiece rotates at 440rpm, feed is 0.3 mm/rev and maximumdepth of cut is 2 mm. Assume total

approach and overtravel distance as 5mm for turning operation.

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Example

Given data: Lj = 170 mm, D1 = 60 mm, D2 = 50mm, N = 440 rpm,f= 0.3 mm/rev, d= 2 mm,

How to calculate the machining time whenthere is more than one operation?

Example

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Example

Time for Turning:

Total length of tool travel = job length + length ofapproach 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 cutsor passes required.

Number of cuts required = 5/2 = 2.5 or 3 (since cutscannot be a fraction)

Machining time for one cut = L / (fN) Total turning time = [L / (fN)] Number of cuts = [175/(0.3440)] 3= 3.97

min.

Example

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Example

Time for facing:

Now, the diameter of the job isreduced to 50 mm. Recall that in case offacing operations, length of tool travel isequal to half the diameter of the job. Thatis, l= 25 mm. 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 forTurning + Time for Facing

= 3.97 + 0.18 = 4.15 min.

The reader should find out the total

machining time if first facing is done.

Example

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a p e

From a raw material of 100 mm lengthand 10 mm diameter, a component havinglength 100 mm and diameter 8 mm is tobe produced using a cutting speed of31.41 m/min and a feed rate of 0.7mm/revolution. How many times we haveto resharpen or regrind, if 1000 work-pieces are to be produced. In the taylorsexpression use constants as n = 1.2 and C= 180

Example

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p

Given D =10 mm , N = 1000 rpm, v=31.41 m/minute

From Taylors 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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p

Machining time/piece = L / (fN) = 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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p

6: While turning a carbon steel cylinder bar oflength 3 m and diameter 0.2 m at a feed rate of 0.5mm/revolution with an HSS tool, one of the twoavailable cutting speeds is to be selected. These twocutting speeds are 100 m/min and 57 m/min. Thetool life corresponding to the speed of 100 m/min isknown to be 16 minutes with n=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.

Which of the above two cutting speeds shouldbe selected from the point of view of the total costof producing this part? Prove your argument.

Example

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p

Given T1 = 16 minute, v1 = 100 m/minute, v2= 57 m/minute, D = 200mm, l= 300 mm,f=0.5 mm/rev

Consider Speed of 100 m/minute N1 = (1000 v) / (D) = (1000100) /

(200) = 159.2 rpm t1 = l/ (fN) = 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 or number of resharpenings = 2

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Example

Total cost =

Machining cost + Cost of resharpening Number of resharpening

= 37.7601+ 202 = Rs.2302

Example

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p

Consider Speed of 57 m/minute

Using Taylors expression T2 = T1 (v1 /v2)2 with usual notations

= 16 (100/57)2 = 49 minute

Repeating the same procedure we get t2 =

66 minute, number of reshparpening=1 andtotal cost = Rs. 3980.

The cost is less when speed = 100 m/minute.Hence, select 100 m/minute.

Example

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p

Write the process sequence to be used for

manufacturing the component

from raw material of 175 mm length

and 60 mm diameter

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Example

40 Dia50

50 40

20

50

20 Dia

Threading

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Example

To write the process sequence, first list theoperations to be performed. The raw materialis having size of 175 mm length and 60 mmdiameter. The component shown in Figure

5.23 is having major diameter of 50 mm, stepdiameter of 40 mm, groove of 20 mm andthreading for a length of 50 mm. The totallength of job is 160 mm. Hence, the list of

operations to be carried out on the job areturning, facing, thread cutting, grooving andstep turning

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Example A possible sequence for producing the

component would be:

Turning (reducing completely to 50 mm)

Facing (to reduce the length to 160 mm) Step turning (reducing from 50 mm to 40

mm)

Thread cutting.

Grooving