What is the working of SHAPER MACHINE and Lathe

Lathe-2.ppt

Full slides of working of lathe and shaper machine with parts

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 Jobs
Turning/Drilling/Grooving/
Threading/Knurling/Facing…

267.dwg

The Lathe

268.dwg

The Lathe
Bed
Head Stock
Tail Stock
Carriage
Feed/Lead Screw

269.dwg

Types of Lathes
Engine Lathe
Speed Lathe
Bench Lathe
Tool Room Lathe
Special Purpose Lathe
Gap Bed Lathe
…

Size of Lathe
Workpiece Length
Swing

270.dwg

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
Workholding Devices ..

271.dwg

Centers
Workholding Devices ..

272.dwg

Faceplates
Workholding Devices ..

273.dwg

Dogs
Workholding Devices ..

274.dwg

Mandrels
Workpiece (job) with a hole
Workholding Devices ..

275.dwg

Rests
Workholding Devices ..
Steady Rest Follower Rest

276.dwg

Operating/Cutting Conditions
Cutting Speed v
RPM
Depth of Cut d

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Operating Conditions

278.dwg

Cutting Speed
The Peripheral Speed of Workpiece past the Cutting Tool
=Cutting Speed
Operating Conditions..
D – Diameter (mm)
N – Revolutions per Minute (rpm)

Feed
f – the distance the tool advances for every rotation of workpiece (mm/rev)
Operating Conditions..

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Depth of Cut
perpendicular distance between machined surface and uncut surface of the Workpiece
d = (D1 – D2)/2 (mm)
Operating Conditions..

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3 Operating Conditions

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Selection of ..
Workpiece Material
Tool Material
Tool signature
Surface Finish
Accuracy
Capability of Machine Tool
Operating Conditions..

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)
Operations on Lathe ..

MRR
dimensional consistency by substituting the units
Operations on Lathe ..
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
Operations on Lathe ..

Turning
Operations on Lathe ..
Cylindrical job

Turning ..
Cylindrical job
Operations on Lathe ..

285.dwg

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
Operations on Lathe ..

Facing
Flat Surface/Reduce length
Operations on Lathe ..

286.unknown

Facing ..
machine end of job  Flat surface
or 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 axis
Operations on Lathe ..

Facing ..
Operations on Lathe ..

Eccentric Turning
Operations on Lathe ..

288.dwg

Knurling
Produce rough textured surface
For Decorative and/or Functional Purpose
Knurling Tool

A Forming Process
MRR~0
Operations on Lathe ..

Knurling
Operations on Lathe ..

289.dwg

Knurling ..
Operations on Lathe ..

Grooving
Produces a Groove on workpiece
Shape of tool  shape of groove
Carried out using Grooving Tool  A form tool
Also called Form Turning
Operations on Lathe ..

Grooving ..
Operations on Lathe ..

291.dwg

Parting
Cutting workpiece into Two
Similar to grooving
Parting Tool
Hogging – tool rides over – at slow feed
Coolant use
Operations on Lathe ..

Parting ..
Operations on Lathe ..

292.dwg

Chamfering
Operations on Lathe ..

293.dwg

Chamfering
Beveling sharp machined edges
Similar to form turning
Chamfering tool – 45°
To
Avoid Sharp Edges
Make Assembly Easier
Improve Aesthetics
Operations on Lathe ..

Taper Turning
Taper:
Operations on Lathe ..

294.dwg

Taper Turning..
Methods
Form Tool
Swiveling Compound Rest
Taper Turning Attachment
Simultaneous Longitudinal and Cross Feeds
Operations on Lathe ..
Conicity

Taper Turning ..
By Form Tool
Operations on Lathe ..

297.dwg

Taper Turning ,,
By Compound Rest
Operations on Lathe ..

298.dwg

Drilling
Drill – cutting tool – held in TS – feed from TS
Operations on Lathe ..

299.dwg

Process Sequence
How to make job from raw material 45 long x 30 dia.?
Operations on Lathe ..
Steps:
Operations
Sequence
Tools
Process

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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?
Operations on Lathe ..
X
X
X
X

Machining Time
Turning Time
Job length Lj mm
Feed f mm/rev
Job speed N rpm
f N mm/min
Operations on Lathe ..

Manufacturing Time

Manufacturing Time
= Machining Time
+ Setup Time
+ Moving Time
+ Waiting Time
Operations on Lathe ..

Example
A 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
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

Example
Can a machine has speed of 191 rpm?
Machining time:
t = 500 / (0.7191)
= 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) / 280
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 = 12500 / 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 = 3.142120.41500 = 7538.4 mm3/min
from Equation (8),
t = 150/(0.4500) = 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 / (fN)
Total turning time = [L / (fN)]  Number of cuts
= [175/(0.3440)]  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/(0.3440)
= 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 have vT n = C
Substituting the values we get,
(31.40)(T)1.2 = 180
or T = 4.28 min

Example
Machining time/piece = L / (fN)
= 100 / (0.71000)
= 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) = (1000100) / (200) = 159.2 rpm
t1 = 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

or number of resharpenings = 2

Example
Total cost =
Machining cost + Cost of resharpening  Number of resharpening
= 37.7601+ 202
= 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 Dia

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

Head stock
Spindle
Feed rod
Bed
Compound rest
and slide(swivels)
Carriage
Apron
selector
Feed change
gearbox
Spindle
speed
Lead screw
Cross slide
Dead center
Tool post
Guide ways
Tailstock quill
Tailstock
Handle

Workpiece
Headstock center
(Live Centre)
Tailstock center
(Dead Centre)
Workpiece

Tail

Tail
Workpiece
Mandrel
Jaws
Hinge
Work
Work
Jaws
Lathe bed guideways
Carriage
Workpiece
Tool
Chip
Tool post
S
peripheral
speed (m/min)
N (rev/min)
D
Workpiece
Tool
Chip
Tool post
S
peripheral
speed
(m/min)
N (rev/min)
D
N
D
S
speed
peripheral
D
p
p
=
=
=
rotation
1
in
travel
tool
relative
m/min
1000
N
D
v
p
=

Feed

D
D
2
1

d
Depth
of Cut

D
D
2
1

Chip

Machined
surface
Workpiece
Depth of cut
Tool
Chuck
N
Feed (
f
)
Cutting speed
Depth of cut (
d
)
Cutting
speed

Chip

Workpiece
Depth of cut (d)
Depth of cut
Tool
Feed
Chuck
N
Machined
surface
Depth of
cut
Feed
Workpiece
Chuck
Cutting
speed
Tool
d
Machined
Face
Axis of job
Axis of lathe
Eccentric peg
(to be turned)
4-jaw
chuck
Cutting
speed
Knurling tool
Tool post
Feed
Cutting
speed
Movement
for depth
Knurled surface
Shape produced
by form tool
Groove
Grooving
tool
Feed or
depth of cut
Form tool
Feed
Parting tool
Chamfering tool
Feed
Chamfer
a
C
B
A
L
D
90°
a
2
D
1
L
D
D
2
tan
2
1
–
=

a
L
D
D
K
2
1
–
=
a
Taper
Workpiece
Straight
cutting edge
Direction
of feed
Form
tool
Face plate
Dog
a
Tail stock quill
Tail stock
Mandrel
Direction of feed
Compound rest
Slide
Compound rest
Hand crank
Tool post &
Tool holder
Cross slide
Feed
Drill
Quill
clamp
moving
quill
Tail stock clamp
Tail stock

20 dia
40
15
min

N
f
L
t
j
=
40
Dia
50
50
40
20
50
20
Dia
Threading

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