Chip Formation Mechanics - Power, Energy, Forces, Temperature

Mechanics

Jul 18, 2012 (5 years and 11 months ago)

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Chip Formation Mechanics -
Power, Energy, Forces,
Temperature
Temperature
ver. 1
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
1
Overview
Overview
•Chip formation mechanics
–Power
–Energy
–Forces

Tor
q
ues
q
–Temperature

Obliquecutting
Oblique

cutting

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
2
IdealizedChip
formationProcess
Idealized

Chip
-
formation

Process
chip
cuttingtool
shear zone
cutting

tool
workpiece
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
3
Chi
p
-formation Geometr
y
p
y
chip
primary shear zone
tc
B
tool
to
φ
ζ
α
tool
V (cutting velocity)
ζ

䵅M㘲㈲㨠䵡湵晡捴畲楮朠偲潣敳獥猠慮搠卹獴敭猠=

4
Cutting Energy, Power, Forces &
Torques
•To get forces and torques,
calculate
p
ower
(
ener
gy),
and
p(gy),
back calculate
P = F x s
PT
P
=
T
x
ω
䵅M㘲㈲㨠䵡湵晡捴畲楮朠偲潣敳獥猠慮搠卹獴敭猠=

5
SpecificCuttingEnergy
(u)
Specific

Cutting

Energy

(u)
chip
of
volume
chipformto(work)energy
u=
Vl
Work
Vl
Energy
u
chip
of
volume
=
=
V
o
l
ume
V
o
l
ume
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
6
CuttingWork
(W)
Cutting

Work

(W)
Work
(W)
donebycuttingforce
(F)
=
Work

(W)
done

by

cutting

force

(F)
=
cutting power (P)x cutting time (t)
W = P t

=
Volume
Power
Volume
Work
MRR
time
Volume
=

time
Volume
Volume
time
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
7
CuttingPower
(P)
Cutting

Power

(P)
MRR
P
V
W
u
V
E
===
PMRR
or
P
= u x
MRR
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
8
Cuttin
g
Power
(
P
)
g
()
•So, we can calculate
P = u x MRR
•Hence, the force or torque from:
PF
P
=
F
x s
P = T x ω
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
9
uis com
p
osed of:
p

u
s
,theshearenergyperunit
u
s
,

the

shear

energy

per

unit

volume

u
frictionalenergyperunit

u
f
,
frictional

energy

per

unit

volume
chipcurlenergy

chip

curl

energy
•chip acceleration kinetic energy
•surface energy of new surfaces
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
10
Wheredoescuttingenergygo?
Where

does

cutting

energy

go?
•90% to chip

5%totool
5%

to

tool
•5% to workpiece
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
11
u
canbeobtainedintwoways
u
can

be

obtained

in

two

ways
•Tabulated

Estimated

Estimated
–u ~ HB
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
12
CuttingEnergy
(u)
Cutting

Energy

(u)
Asafirstapproximation:
As

a

first

approximation:
u ≈us
+ uf
075
08
u
s

0
.
75
-
0
.
8
u
hence
u ≈1.5 us
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
13
Estimation of u
•Assume a simple, rigid-perfectly
plasticmaterial:
plastic

material:

σ
σ
0
HB
τεσ
d
?0?1u
fmos

=

=
⋅=

γ
ε
0
4
2
γ
2
1
ε
3
HB
σ
o

for heavily cold-worked metals
4
2
γ
2
1
ε

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
14
Estimationof
u
Estimation

of

u
HB
6
1
2
σ
τ
o
fm
≈≈
6
2
HBHB
3
2
3
1
u
s
→≈
3
3
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
15
Whatnumbertouse?
What

number

to

use?
From above
u ≈1.5 us
us
≈(1/3 to 2/3) HB
u ≈(1/2 to 1) HB
so, if no data, u ≈HB
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
16
Comparison of Tensile and Cutting
(Ex. 1-1)
•304 stainless steel rod

d
=
05ind
f
=
048inl
=
6in
d
o

0
.
5

in
,
d
f

0
.
48

in
,
l
o

6

in
Whtithidi

Wh
a
t

i
s
th
e energy requ
i
re
d
us
i
ng
tension?
•What is the energy required using
cutting?
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
17
Tension:Ex1
2
Tension:

Ex
.
1
-
2
σ
σ
1
1
1

=
=⋅
+

K
u
d
n
ε
ε
σ
ε
ε1
1
1
+

n
u
d
ε
σ
䕮敲杹癯汵浥×
=
×
K = 185,000 psi; n = 0.45
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
18
Tension
Ex1
3
Tension
-
Ex
.
1
-
3
l
A
K
Energy
n1
1

=
+
ε
0816.0
48
0
5.0
ln2
1
=

=
ε
()()
(
)
(
)
l
A
n
Energy
oo
6
25
0
0816.0000,185
1
2
45.1

=
+
=
π
48
.
0
1

(
)
(
)
Jlbfin449970,3
6
25
.
0
45.1
=−=
=
π
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
19
Cuttin
g
-Ex. 1-4
g
Energy = (specific cutting energy) *
(volume removed)
(
)
l
D
D
u
E
f
i

×=
22
4
π
uss
≈1.5 hp min/in3
x 550 x 60 x 12
=594000in
lbf/in
3
(
)
f
i
4
=

594
,
000

in
-
lbf/in
3
(
)
648.05.0
4
000,594
22
⋅−×=
π
E
= 54,650 in-lbf (6,176 J)
(
)
4
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
20
Comparison
Ex1
5
Comparison
-
Ex
.
1
-
5
•Tension: E = 3,940 in-lbf (449 J)
•Cutting: E = 54,650 in-lbf (6,176 J)
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
21
TurningvsOrthogonalCutting
Turning

vs
.
Orthogonal

Cutting
Terminologyusedinaturning

Terminology

used

in

a

turning

operation on a lathe, where f is
the feed rate (in./rev or mm/rev)
and d is the depth of cut.

Inturningthe

orthogonality

is

In

turning
,
the

orthogonality

is

to the left in the drawing, hence
a change of coordinate system
is needed. If you were doing a
diametral cut-off
(p
lun
g
e cut
)

(pg)
operation, no change would be
needed.
•Note that feed in turning is
equivalent to the depth of cut in
thlttidth
or
th
ogona
l
cu
tti
ng, an
d

th
e
depth of cut in turning is
equivalent to the width of cut in
orthogonal cutting.
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
22
Relationbetweenuandt
Relation

between

u

and

t
0

A
s the de
p
th of cut decreases
,
the
p,
surface area to volume ratio increases,
hence friction
(
ener
gy)
increases
(gy)
•Since
u
u
u
+

ε敮捥
1
fs
u
u
u
+

0
1
t
u

ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
23
TurningPowerandForce
Ex2
1
Turning

Power

and

Force
-
Ex
.
2
-
1
TurningTitanium:
Turning

Titanium:
•speed = 1 m/s = 200 sfpm
fdt01/00004”/

f
ee
d
ra
t
e =
0
.
1
mm
/
rev =
0
.
0004”

/
rev
•depth of cut = 3 mm = 0.1”
What is cutting power and cutting force?
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
24
Power
-
Ex.2
-
2
Power

Ex.

2
2
u

0.06kW/cm
3
/min
=
3.6W/mm
3
/sec
u

0.06

kW/cm
/min

3.6

W/mm
/sec
MRR=Q=sfd=1000x01x3
MRR

=Q

=

sfd

=

1
,
000

x

0
.
1

x

3

= 300 mm3/sec
P =
u
x MRR =
3
.
6
x
300
= 1
,080
W
u36300
,080
=1.45 hp
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
25
Force
Ex2
3
Force
-
Ex
.
2
-
3
F= P/s = 1,080 W ÷1 m/s
=
1080N(243lbf)
1
,
080

N

(243

lbf)
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
26
uvst
inTurning
Ex3
1
u

vs
.
t
0
in

Turning

Ex
.
3
-
1

•If
(
)
3
4.0
0
/
1
2.2mmsW
t
u−≈
•And
1/
0
t

s =
1
m
/
s
–f (depth of cut) = 0.01, 0.02, 0.05, 0.1
mm/rev
mm/rev

–d (width of cut) = 3 mm
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
27
uvst
inTurning
Ex3
2
u

vs
.
t
0
in

Turning

Ex
.
3
-
2

Fixedspeed(1m/s),widthofcut(3mm)
Fixed

speed

(1

m/s),

width

of

cut

(3

mm)
feedrate
u
MRR
Power
1600
1800
P(W)
feed

rate
(mm/s)
u

(W-s/mm
3)
MRR
(mm3/s)
Power
(W)
0.0113.930416
002
106
60
631
600
800
1000
1200
1400
1600
P
ower
(W)
MRR (mm3/s)
0
.
02
10
.
6
60
631
0.057.31501094
0.15.53001658
0
200
400
00.020.040.060.080.1
feedrate(mm/s)
feed

rate

(mm/s)
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
28
uvst
inTurning
Ex3
3
u

vs
.
t
0
in

Turning

Ex
.
3
-
3
•For a constant power = 1700 W, width of cut = 3 mm
feed rateuSpeedMRR
(mm/s)(
W
-s/mm3)(mm/s)(mm3/s)
0.0113.94082122
002
105
2693
162
0
.
02
10
.
5
2693
162
0.057.31554233
01
55
1025
308
0
.
1
5
.
5
1025
308
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
29
Force
Ex4
1
Force

Ex
.
4
-
1
•Drilling a hole into aluminum
–hole diameter = 1 inch
–feed rate = 5 in/min

speed = 200 rpm
•Determine power, torque, force on drill
bit, force on operato
r
•u = 0.18 hp/in
3/min
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
30
Force

Ex. 4-2
•MRR different than turning

here, drill s
p
eed onl
y
matters for determinin
g
feed
pyg
rate and rotational speed
–MRR = Q = VA = feed rate x cross sectional area
(
)
min/93.3
4
1
5
4
3
2
2
in
d
fMRR=⋅=⋅=
ππ
4
4
hpMRRuPower71.093.318.0
=

=
×=
lbf
ft
h
?
sec
60
550
71
0
lbfft
hp
lbf
ft
h
p
power
Torque−=

==6.18
2
i
200
min
sec
60
sec
550
71
.
0
π
ω
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
31
revm
i
n
Force
Ex4
3
Force

Ex
.
4
-
3
Fdillbit
F
orce on
d
r
ill

bit
lbf
in
ft
in
lb
f
f
t
Force4.446
12
1
5.0
6.18
Torque
=
×

==
in
12
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
32
Force
Ex4
4
Force

Ex
.
4
-
4
Force on o
p
erato
r
p
lbf
ft
in
lbfft
Force2.37
1
6
6.18
point napplicatio todistance
Torque
=
×

==
in
in
12
6
×
6 in
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
33
Force
Ex5
1
Force

Ex
.
5
-
1
•Slab milling titanium
–speed of mill (V) = 500 rpm
–mill diameter (D) = 3 in

mill width = 4 in
–work piece width = 4 in
dfki[fdt]()5i/i

spee
d
o
f
wor
k
p
i
ece
[f
ee
d
ra
t
e
]

(
v
)
=
5

i
n
/
m
i
n
–depth of cut (d) = 0.2 in

u=12hp/in
3
/min
u

=

1
.
2

hp/in
/min
•Determine power, torque, and force
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
34
Force
Ex5
2
Force

Ex
.
5
-
2
•MRR is different from turnin
g
.
g
–Here, speed of mill only matters to
determine feed rate and rotational velocity.
–MRR = Q = velocity x cross-sectional area
3
min/4542.0
3
invbd
M
RR
=
×
×
=
×
×
=
hp
MRR
u
Power
8
4
4
2
1
u
u
hp
MRR
u
Power
8
.
4
4
2
.
1
=
×
=
×
=
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
35
Force
Ex5
3
Force

Ex
.
5
-
3
lbfft
rev
hp
lbfft
hp
power
Torque−=

==4.50
min
sec
60
sec
5508.4
rev
rev

2
min
500
π
ω
lbf
ft
lbffttorque
Force2.403
12
5.1
4.50
=

==
12
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
36
CuttingForces
Cutting

Forces
•Cutting (power) force: Fc

doestheworkofcutting(P
=
F
c
xV)
does

the

work

of

cutting

(P

F
c
x

V)
•Thrust force: Ft
perpendiculartocuttingvelocity

perpendicular

to

cutting

velocity
–pushes tool and work piece apart
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
37
CuttingForceDiagram
Cutting

Force

Diagram
R
V
β
α
R
Fc
F
N
F
φ
R
F
t
Fs
Ns
N
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
38
Nomenclature
•F
t
= thrust force (⊥to cutting velocity)

F
c
=
cuttingforce(
||

F
c
=

=

=
(
=
||

=

=

s
Ⱐ,
s
㴠乯牭慬Ⱐ瑡湧敮琠獨敡爠景牣攠潦=

α= ra
k
e ang
l
e
•φ= shear angle
β


β

=

b‽⁷楤瑨⁯映捨楰 楮瑯⁢潡牤=
䵅M㘲㈲㨠䵡湵晡捴畲楮朠偲潣敳獥猠慮搠卹獴敭猠=

Merchant

sForceCircle
Merchants

Force

Circle
F
s
α
N
s
α
φ
β

α
V
Fc
N
s
Ft
R
α
β
α
β
N
F
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
40
M. Eugene Merchant
1913 -2006
ForcesatToolFace
Forces

at

Tool

Face
F
=
F
c
sin
α
+
F
t

α
F

F
c

α
=
F
t

α

c

t
sinα
F/N = μ
(friction coefficient)
: tan-1μ= β
α
μ
tan
ct
FF
F
+
=
=
α
μ
tan
tc
FFN−
=
=
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
41
ForcesonShearPlane
Forces

on

Shear

Plane
Fs
= Fc
cosφ-F
t
sinφ
N
=
F
sin
φ

t

φ
N
s
=
F
c

φ
+
=
F
t

φ
䵅M㘲㈲㨠䵡湵晡捴畲楮朠偲潣敳獥猠慮搠卹獴敭猠=

AverageStressesonShearPlane
Average

Stresses

on

Shear

Plane
A
F
s
s
s
=
τ
A
N
s
s
=
σ
area
plane
shear
bt
A
A
o
s
s
=
=
area
plane
shear
A
s
I
sin
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
43
Average Stresses at Chip-Tool
Interface
•Difficult to determine, as there is no
g
ood method to estimate chi
p
-tool
gp
contact length. It must be measured
after the cut
,
which is difficult because
,
of chip curl.
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
44
ShearAngle
(
φ
)
Shear

Angle

(
φ
)
B慳敤⁯渠䵥牣桡湴鉳a慮慬祳楳
䙩dhliih


n
d

h

l

i
m
i

h

⊥獳畭楮朠楤敡氠灬慳瑩捩瑹⁡湤s
β independent of
φ

nottrue
not

true
s
s
s
A
F
=
τ
()
α
β
φ

+
=cosRF
s
s
(
)
β
F
R
(
)
α
β

=sec
c
F
R
0
bt
A
=
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
45
φ
sin
A
s
Shearangle
(
φ
)
Shear

angle

(
φ
)
(
)
(
)
sincosseca
F
F
c
s
φ
β
φ
α
β

+

(
)
(
)
τ
d
?
?
?
?
0
btA
c
s
s
s
φ
β
φ
β
τ
==
(
)
(
)
0sinsincoscos
=

+

+=
φ
β
φ
φ
β
φ
φ
τ
aa
d
d
s
hence
()
(
)
φ
φ
α
β
φ
−°
=
=
−+90tancottan
thf
th
ere
f
ore
2
2
45
βα
φ
−+°=
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
46
2
2
Shearangle
(
φ
)
Shear

angle

(
φ
)
䱥攠慮搠卨慦晥爠⠱㤵ㄩ
α
–perfectly plastic material
–slip line analysis

Normalstressrequired
π

B
π
/4
Normal

stress

required

to cause flow
–σs
= k

Shearstressrequired
π

α
η

Shear

stress

required

to cause flow
–τs
= k
S
τ
s
k
k
φ
C
D

S
o, σs
= τs
σ
s
䵅M㘲㈲㨠䵡湵晡捴畲楮朠偲潣敳獥猠慮搠卹獴敭猠=

C
Shearangle
(
φ
)
Shear

angle

(
φ
)
䙲潭⁍敲捨慮璒s=捩牣汥
(
)
1tan
=
=

+
=
k
k
s
α
β
φ
σ
(
)
k
s
β
φ
τ
(
)
(
)
1
(
)
(
)
α
β
φ

+
=

1tan
1
β
α
φ

+
°=45
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
48
Shearangle
(
φ
)
Shear

angle

(
φ
)
䙯Fa汩湥慲獴牡楮
䙯F
=
a
=

Ⱐ,

=

p
k
‽=s
h

t

⽳楮φ
w = thickness of
p
s
shear zone
ps
p
k
max
k
w
s
ps
Δ−
w
p
k
0
s
φ
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
49
Shearangle
(
φ
)
Shear

angle

(
φ
)
N
1
Average normal stress on shear plane
k
w
s
p
A
N
s
s
s
s
Δ−==
2
1
σ
0max
kkk

=
Δ
½ to give average
pressure
Ratio of average pressure (
σs) to shear
stress on the shear plane (
τs
= kmax
)
()
max
max
2
1
tan
k
k
w
s
k
p
F
N
s
s
s
s
s
Δ
−=−+==
αβφ
τ
σ
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
50
max
max
s
s
Shearangle
(
φ
)
Shear

angle

(
φ
)

1
max
=
k
ps
and
0
=
Δ
k
()
1tan
=
−+
α
β
φ
LeeandShaffer

sresult
()()
αβφ
−+=

1tan
1
β
α
φ

+
°=45
Lee

and

Shaffers

result
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
51
3DCutting(Oblique)
3D

Cutting

(Oblique)
Z
Vc
A
αn
X, V
αe
i
O
Y
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
52
3-D Cuttin
g
g
•i = inclination of tool to

αn
= normal rake angle between
z
-
axisandOA
z
axis

and

OA
•αe
= effective cutting angle
betweenV
andline

=
σ
c

=

=

=
σ
=

=

c

1
⡳楮
2

2

)
α
e
=
sin
-
1
(sin
2
i

+

cos
2
i
x
sin
α
n
)
䵅M㘲㈲㨠䵡湵晡捴畲楮朠偲潣敳獥猠慮搠卹獴敭猠=

TemperatureRise
Temperature

Rise
Assume: average chip-tool interface
temperature rise (
Δ

s灥捩晩挠捵瑴楮朠敮敲杹p⡵(
c畴瑩湧⁳灥敤×⡖(


p

(
t

)
=
(
㴠晥敤⁦潲⁴畲湩=

⤨g)
t桥牭慬⁣潮摵捴楶楴礠潦⁷潲歰楥捥 欩


÷
⁯映睯牫
p

(
ρ
)

ρ
)
s灥捩晩挠桥慴⁯映睯牫灩散攠⡣p
䵅M㘲㈲㨠䵡湵晡捴畲楮朠偲潣敳獥猠慮搠卹獴敭猠=

Temperature Rise Dimensional
Analysis
•Two dimensionless groups result:
cT⋅⋅Δ
ρ
ρ
oo
tV
k
ctV⋅
=
⋅⋅
&
u
α
k
k
c
k

=
ρ
α
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
55
TemperatureRiseEquation
Temperature

Rise

Equation
Plotting data for many metals:
1
3
1
4.0

⋅=Δ
α
ρ
o
Vt
c
u
T

α
ρ
c
works for b/to
> 5
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
56
TemperatureRise
Ex6
1
Temperature

Rise
-
Ex
.
6
-
1
Estimatethechip
-
toolinterface
Estimate

the

chip
tool

interface

temperaturefor:

302stainlesssteel;HB=200kg/mm
2

302

stainless

steel;

HB

=

200

kg/mm
2
•Cut on a lathe:

t
o
= f (for turning) = 0.38 mm
•feed rate = 0.38 mm/rev

b = d =2.54 mm
–V = 1.5 m/s
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
57
TemperatureRise
Ex6
2
Temperature

Rise
-
Ex
.
6
-
2
•k = 16.2 W/m-K

c
=
500J/kg
-
K
c

500

J/kg
K
•ρ= 8000 kg/m3
40510
6
2
/
•α=
4
.
05
x
10
-
6
m
2
/
s
•u = 200 kgf/mm
2
= 1.96 x 10
9
N/m2
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
58
Tem
p
erature Rise -Ex. 6-3
p
to use ΔT equation, check
567.6
38
0
54.2
>==
t
b
Vt
1

.
0
0
t
Vt
c
u
T
o
4.0
3

⋅=Δ
αρ
1038.05.11096.1
4.0
3
1
6
39

××

×
⋅=

K1020
10 4.05
5008000
6
-
=

×

×
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
59
TemperatureRise
Ex6
4
Temperature

Rise
-
Ex
.
6
-
4
Assuming room temperature = 20oC
Then:
T = ΔT + 20 = 1040
oC = 1313 K
and
Tmelt
= 1421oC = 1694 K
So we are at 80% of Tmelting, very HOT.
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
60
LimitsonTemperature
Limits

on

Temperature
•Tool

mosttoolshaveauselimit,abovewhich
most

tools

have

a

use

limit,

above

which

Workpiece
Work

piece
–absolute limit is melting point

otherlimitsincludecombustionofchips

other

limits

include

combustion

of

chips
,
warpage of work piece
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
61
Modellimitations
Model

limitations

A
ssumptions
–slow, orthogonal cutting
–constant material properties
–constant temperature
–simple sliding friction

no strain hardening
•Use analysis for:
–trends and building intuition
–basis for detailed study
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
62
Summary
Summary
•Chip formation mechanics

Power
Power
–Energy

Forces
Forces
–Torques

Temperature

Temperature
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
63
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2009
64