turning operations l a t h e. turning operations machine tool – lathe job (workpiece) – rotary...

Turning Operations

L a t h e

Turning Operations

• Machine Tool – LATHE • Job (workpiece) – rotary motion• Tool – linear motions

“Mother of Machine Tools “Cylindrical and flat surfaces

Some Typical Lathe JobsTurning/Drilling/Grooving/Threading/Knurling/Facing...

The Lathe

Head stock

Spindle

Feed rod

Bed

Compound restand slide(swivels)

Carriage

Apron

selector

Feed changegearbox

Spindlespeed

Lead screw

Cross slideDead center

Tool postGuide ways

Tailstock quillTailstock

Handle

The Lathe

Bed

Head StockTail Stock

CarriageFeed/Lead Screw

Main Parts

• Bed• Headstock• Feed and lead screws• Carriage• Tailstock

6

Lathe Bed

• Heavy, rugged casting

• Made to support working parts of lathe

• On top section are machined ways

– Guide and align major parts of lathe

7

Lathe Bed

8

Headstock

• Clamped on left-hand end of bed• Headstock spindle

– Hollow cylindrical shaft supported by bearings• Provides drive through gears to work-holding

devices– Live center, faceplate, or chuck fitted to spindle nose

to hold and drive work• Driven by stepped pulley or transmission gears• Feed reverse lever

– Reverses rotation of feed rod and lead screw

9

Headstock

10

Headstock

11

Back Gear arrangement

Headstock belt drive

Quick-Change Gearbox

• Contains number of different-size gears• Provides feed rod and lead-screw with various

speeds for turning and thread-cutting operations– Feed rod advances carriage when automatic

feed lever engaged– Lead screw advances the carriage for thread-

cutting operations when split-nut lever engaged

12

Quick-Change Gearbox

13

Carriage

• Used to move cutting tool along lathe bed• Consists of three main parts

– Saddle• H-shaped casting mounted on top of lathe ways,

provides means of mounting cross-slide and apron

– Cross-slide– Apron

14

15

Carriage

< Saddle

< Apron

Carriage

16

Carriage

17

18

Apron

• The apron attached to the front of the carriage, holds most of the control levers. These include the levers, which engage and reverse the feed lengthwise (Z-axis) or crosswise (X-axis) and the lever which engages the threading gears.

• The apron is fastened to the saddle, houses the gears and mechanisms required to move the carriage and cross-slide automatically.

• The apron hand wheel can be turned manually to move the carriage along the Lathe bed. This hand wheel is connected to a gear that meshes in a rack fastened to the Lathe bed.

• The automatic feed lever engages a clutch that provides the automatic feed to the carriage

Cross-slide

• Mounted on top of saddle• Provides manual or automatic cross movement

for cutting tool• Compound rest (fitted on top of cross-slide)

– Used to support cutting tool– Swiveled to any angle for taper-turning– Has graduated collar that ensure accurate

cutting-tool settings (.001 in.) (also cross-slide)

19

Cross-slide

20

21

Top Slide (Compound slide)

• Fitted to top of Cross slide• Carries tool post and cutting tool• Can rotate to any angle• Is used to turn tapers

22

Tailstock

• Upper and lower tailstock castings• Adjusted for taper or parallel turning by two

screws set in base• Tailstock clamp locks tailstock in any position

along bed of lathe• Tailstock spindle has internal taper to receive

dead center– Provides support for right-hand end of work

23

24

Tailstock

Supports long workpieces when machining.

60 degree rotating center point.

Drill Chuck

Turn the tailstock handwheel to advance the ram.

Tailstock

25

Lead Screw and Feed Rod

26

< Lead Screw

< Feed Rod

Types of Lathes

• Engine Lathe• Speed Lathe• Bench Lathe

• Tool Room Lathe• Special Purpose Lathe

• Gap Bed Lathe…

Size of LatheWorkpiece Length Swing

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)

Workholding Devices

• Equipment used to hold– Workpiece – fixtures– Tool - jigs

Securely HOLD or Support while machining

Chucks

Three jaw Four Jaw

Work

hold

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D

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Chucks

• Used extensively for holding work for lathe machining operations– Work large or unusual shape

• Most commonly used lathe chucks– Three-jaw universal– Four-jaw independent– Collet chuck

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Three-jaw Universal Chuck

• Holds round and hexagonal work• Grasps work quickly and accurate

within few thousandths/inch• Three jaws move simultaneously when

adjusted by chuck wrench– Caused by scroll plate into which all three

jaws fit• Two sets of jaw: outside chucking and

inside chucking

34

Three-jaw Universal Chuck

35

Three jaw self centering chuck

36

Four-Jaw Independent Chuck

• Used to hold round, square, hexagonal, and irregularly shaped workpieces

• Has four jaws– Each can be adjusted independently by

chuck wrench

• Jaws can be reversed to hold work by inside diameter

37

Four-Jaw Independent Chucks

Four-Jaw Independent Chucks

38

• With the four jaw chuck, each jaw can be adjusted independently by rotation of the radially mounted threaded screws.

• Although accurate mounting of a workpiece can be time consuming, a four-jaw chuck is often necessary for non-cylindrical workpieces.

MandrelsWorkpiece (job) with a hole

Work

hold

ing

D

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Workpiece Mandrel

Mandrels

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• Holds internally machined workpiece between centers so further machining operations are concentric with bore

• Several types, but most common• Plain mandrel• Expanding mandrel• Gang mandrel• Stub mandrel

41

Mandrels to Hold Workpieces for Turning

Figure 23.8 Various types of mandrels to hold workpieces for turning. These mandrels usually are mounted between centers on a lathe. Note that in (a), both the cylindrical and the end faces of the workpiece can be machined, whereas in (b) and (c), only the cylindrical surfaces can be machined.

RestsW

ork

hold

ing

D

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Jaws

HingeWork Work Jaws

Lathe bed guideways

Carriage

Steady Rest Follower Rest

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Steadyrest• Used to support long work held in chuck

or between lathe centers– Prevent springing

• Located on and aligned by ways of the lathe

• Positioned at any point along lathe bed• Three jaws tipped with plastic, bronze or

rollers may be adjusted to support any work diameter with steadyrest capacity

44

Steadyrest

45

Follower Rest

• Mounted on saddle• Travels with carriage to prevent work

from springing up and away from cutting tool– Cutting tool generally positioned just

ahead of follower rest– Provide smooth bearing surface for two

jaws of follower rest

46

Follower Rest

Operating/Cutting Conditions

1. Cutting Speed v2. Feed f3. Depth of Cut d

Workpiece

Tool

Chip

Tool post

S peripheralspeed (m/min)

N (rev/min)

D

Operating Conditions

Workpiece

Tool

Chip

Tool post

S peripheralspeed (m/min)

N (rev/min)

D

NDSspeedperipheral

D

rotation1intraveltoolrelative

Cutting Speed

The Peripheral Speed of Workpiece past the Cutting Tool

=Cutting SpeedOp

era

tin

g

Con

dit

ion

s..

m/min1000

NDv

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

Feedf – the distance the tool advances for every

rotation of workpiece (mm/rev)

Op

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g

Con

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s..

f Feed

DD 21

Depth of Cutperpendicular distance between machined

surface and uncut surface of the Workpiece d = (D1 – D2)/2 (mm)

Op

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g

Con

dit

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s..

d Depth of Cut

DD 21

3 Operating Conditions

Chip

Machined surface

Workpiece

Depth of cutTool

Chuck

N

Feed (f )

Cutting speed

Depth of cut (d)

Selection of ..

• Workpiece Material • Tool Material• Tool signature • Surface Finish• Accuracy • Capability of Machine Tool

Op

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g

Con

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s..

Material Removal Rate

MRRVolume of material removed in one

revolution MRR = D d f mm3

• Job makes N revolutions/minMRR = D d f N (mm3/min)

• In terms of v MRR is given byMRR = 1000 v d f (mm3/min)O

pera

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s o

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Lath

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MRR

dimensional consistency by substituting the units

Op

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s o

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Lath

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. MRR: D d f N (mm)(mm)(mm/rev)(rev/min) = mm3/min

Operations on Lathe

• Turning• Facing• knurling• Grooving• Parting

• Chamfering• Taper turning• Drilling• Threading

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Turning

Cylindrical job

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Turning ..

Cylindrical job

Cutting speed

Chip

Workpiece

Depth of cut (d)

Depth of cutTool

FeedChuck

NMachined surface

Op

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Turning ..

• Excess Material is removed to reduce Diameter

• Cutting Tool: Turning Tool

a depth of cut of 1 mm will reduce diameter by 2 mm

Op

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FacingFlat Surface/Reduce length

Depth of cut

Feed

WorkpieceChuck

Cutting speed

Tool

dMachined Face

Op

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s o

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Lath

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Facing ..

• machine end of job Flat surfaceor to Reduce Length of Job

• Turning Tool• Feed: in direction perpendicular to

workpiece axis–Length of Tool Travel = radius of

workpiece• Depth of Cut: in direction parallel to

workpiece axisOp

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Facing ..O

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Eccentric Turning

Axis of job

Axis of lathe

Eccentric peg(to be turned)

4-jaw chuck

Cutting speed

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Knurling

• Produce rough textured surface– For Decorative and/or Functional Purpose

• Knurling Tool

A Forming ProcessMRR~0

Op

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KnurlingO

pera

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Knurling toolTool post

Feed

Cutting speed

Movement for depth

Knurled surface

Knurling ..O

pera

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Grooving

• Produces a Groove on workpiece

• Shape of tool shape of groove• Carried out using Grooving Tool A form tool

• Also called Form Turning

Op

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Grooving ..O

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. Shape produced by form tool Groove

Grooving tool

Feed or depth of cutForm tool

Parting

• Cutting workpiece into Two• Similar to grooving• Parting Tool• Hogging – tool rides over – at slow feed• Coolant use

Op

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Parting ..O

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FeedParting tool

ChamferingO

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Chamfering toolFeed

Chamfer

Chamfering

Beveling sharp machined edges Similar to form turning Chamfering tool – 45° To

• Avoid Sharp Edges• Make Assembly Easier• Improve Aesthetics

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Taper Turning

• Taper:

CB

A L

D90°

2D1

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L

DD

2tan 21

Taper Turning..

Methods • Form Tool• Swiveling Compound Rest• Taper Turning Attachment• Simultaneous Longitudinal and

Cross FeedsOp

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L

DDK 21

Taper Turning ..By Form Tool

Op

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. TaperWorkpiece

Straight cutting edge

Direction of feed

Form tool

Taper Turning ,,By Compound Rest

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Face plate

Dog

Tail stock quill

Tail stock

Mandrel

Direction of feed

Compound rest Slide

Compound rest Hand crank

Tool post & Tool holder

Cross slide

Drilling

Drill – cutting tool – held in TS – feed from TS

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Feed

Drill

Quill clamp moving

quill

Tail stock clamp

Tail stock

Process Sequence• How to make job from raw material 45 long

x 30 dia.?

20 dia

40

15

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Steps:•Operations

•Sequence•Tools•Process

Process Sequence .. Possible Sequences

• TURNING - FACING - KNURLING• TURNING - KNURLING - FACING• FACING - TURNING - KNURLING• FACING - KNURLING - TURNING• KNURLING - FACING - TURNING• KNURLING - TURNING – FACINGWhat is an Optimal Sequence?

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X

XXX

Machining Time

Turning Time• Job length Lj mm• Feed f mm/rev• Job speed N rpm• f N mm/min

min Nf

Lt jO

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Manufacturing Time

Manufacturing Time = Machining Time + Setup Time + Moving Time + Waiting Time

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ExampleA mild steel rod having 50 mm diameter and

500 mm length is to be turned on a lathe. Determine the machining time to reduce the rod to 45 mm in one pass when cutting speed is 30 m/min and a feed of 0.7 mm/rev is used.

Example

calculate the required spindle speed as: N = 191 rpm

m/min1000

NDv

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

Substituting the values of v and D in

ExampleCan a machine has speed of 191

rpm?Machining time:

min Nf

Lt j

t = 500 / (0.7191) = 3.74 minutes

Example

• Determine the angle at which the compound rest would be swiveled for cutting a taper on a workpiece having a length of 150 mm and outside diameter 80 mm. The smallest diameter on the tapered end of the rod should be 50 mm and the required length of the tapered portion is 80 mm.

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

at 10.62o

Example

• A 150 mm long 12 mm diameter stainless steel rod is to be reduced in diameter to 10 mm by turning on a lathe in one pass. The spindle rotates at 500 rpm, and the tool is traveling at an axial speed of 200 mm/min. Calculate the cutting speed, material removal rate and the time required for machining the steel rod.

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

Example• feed rate = 200 mm/min, we get the feed f in

mm/rev by dividing feed rate by spindle rpm. 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.

Example

• Calculate the time required to machine a workpiece 170 mm long, 60 mm diameter to 165 mm long 50 mm diameter. The workpiece rotates at 440 rpm, feed is 0.3 mm/rev and maximum depth of cut is 2 mm. Assume total approach and overtravel distance as 5 mm for turning operation.

Example

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

• How to calculate the machining time when there is more than one operation?

Example

• Time for Turning:• Total length of tool travel = job length + length of

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 / (fN)• Total turning time = [L / (fN)] Number of cuts• = [175/(0.3440)] 3= 3.97

min.

Example

• Time for facing:• Now, the diameter of the job is

reduced to 50 mm. Recall that in case of facing operations, length of tool travel is equal to half the diameter of the job. That is, l = 25 mm. Substituting in equation 8, we get

• t = 25/(0.3440)• = 0.18 min.

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 is done.

Example

• From a raw material of 100 mm length and 10 mm diameter, a component having length 100 mm and diameter 8 mm is to be produced using a cutting speed of 31.41 m/min and a feed rate of 0.7 mm/revolution. How many times we have to resharpen or regrind, if 1000 work-pieces are to be produced. In the taylor’s expression use constants as n = 1.2 and C = 180

Example

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

• From Taylor’s tool life expression, we havevT n = C

• Substituting the values we get,• (31.40)(T)1.2 = 180• or T = 4.28 min

Example

• 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

• 6: While turning a carbon steel cylinder bar of length 3 m and diameter 0.2 m at a feed rate of 0.5 mm/revolution with an HSS tool, one of the two available cutting speeds is to be selected. These two cutting speeds are 100 m/min and 57 m/min. The tool life corresponding to the speed of 100 m/min is known to be 16 minutes with n=0.5. The cost of machining time, setup time and unproductive time together is Rs.1/sec. The cost of one tool re-sharpening is Rs.20.

• Which of the above two cutting speeds should be selected from the point of view of the total cost of producing this part? Prove your argument.

Example

• 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

Example

• Total cost =• Machining cost + Cost of resharpening

Number of resharpening• = 37.7601+ 202 • = Rs.2302

Example

• Consider Speed of 57 m/minute• Using Taylor’s 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 and total cost = Rs. 3980.

•• The cost is less when speed = 100

m/minute. Hence, select 100 m/minute.

Example

• Write the process sequence to be used for manufacturing the component

from raw material of 175 mm length and 60 mm diameter

Example

40 Dia50

50 40

20

50

20 Dia

Threading

Example

• To write the process sequence, first list the operations to be performed. The raw material is having size of 175 mm length and 60 mm diameter. The component shown in Figure 5.23 is having major diameter of 50 mm, step diameter of 40 mm, groove of 20 mm and 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, facing, thread cutting, grooving and step turning

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