me 303 – manufacturing engineering course notesme.metu.edu.tr/courses/me303/me 303 - course notes...
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ME 303 – Manufacturing EngineeringCourse Notes
Prof. Dr. Ömer ANLAĞANProf. Dr. S. Engin KILIÇ
Prof. Dr. A. Erman TEKKAYAAssoc. Prof. Dr. Melik DÖLEN
Middle East Technical UniversityDepartment of Mechanical EngineeringAnkara 06531, TURKEY
2/22/2014 ME 303 - Section 01 1.1
What is Manufacturing?
Meaning: Manufacture (of Latin origin)Manu → HandFact → Made/fashionedManufactus → Hand-made
Definition: Manufacturing is the making of the goods from raw materials into a suitable form.
Materials
Design
Processes
2/22/2014 ME 303 - Section 01 1.2
Definition of Manufacturing
Manufacturing is a blend of art, science, and economics with very broad social implications.
Three Pillars of Manufacturing Scientific Sense Engineering (Practical) Sense Economic Sense
2/22/2014 ME 303 - Section 01 1.3
Historical Developments
Manufacturing started some 6000 years ago. Early humanbeings used sharp and strong materials
as cutting tools and their bodies as machine tools. Hands: Workpiece- and tool holding devices Muscles: Actuators for primary and secondary motions Eyes: Position transducers (sensors) Brain: Control computer
2/22/2014 ME 303 - Section 01 1.4
History (Cont’d)
First industrial revolution(1760-1830) Introduction of the steam
engine (replacement of muscle power)
Early mechanization
History (Cont’d)
Second industrial revolution(1950-present) Extensive use of electric power
Introduction of the computer and solid-state electronics (replacement of eyes and partly brain operations)
Development of new materials and manufacturing techniques
2/22/2014 ME 303 - Section 01 1.5
2/22/2014 ME 303 - Section 01 1.6
Economic Role of Manufacturing
Prosperity of a nation is closely related to the capabilities of its manufacturing industry.
Even after the second industrial revolution, one cannot create wealth without manufacturing industry.
Competitiveness can be achieved through high level of productivity. Nations falling behind in this respect find their living standards gradually diminishing.
2/22/2014 ME 303 - Section 01 1.7
Manufacturing and GNP
PakistanIndia
Bangladesh
MexicoTurkey
Brazil
Argentina
SwitzerlandUSA
Germany
JapanKuwait
5 30
Contribution of Manufacturing (%)
Log
of G
NP
Gross National Product: Value of all goods and servicesproduced in a national econony.
2/22/2014 ME 303 - Section 01 1.8
Automation Meaning: Automation (of Greek origin)
Auto → SelfMatos → Acting/moving
Definition: Automation is a form of manufacturing in which production, movement, and inspection are performed by self-operating machine without human intervention. The skill of a human operator is essentially transfered to a specialized machine.
2/22/2014 ME 303 - Section 01 1.9
Automation (2) Mechanization implies an
operation being performed by a mechanical system (not by hand). It is generally based on open-
loop (no feedback) control strategy.
Ex: Use of a cam mechanism to move the cross-slide of a lathe.
2/22/2014 ME 303 - Section 01 1.10
Automation (3) Automation implies closed-loop control and
adaptive control in an advanced context. Automation utilizes programmable devices with different flexibility.
Hard Automation constitutes control methods that require considerable effort to reprogram a certain machine for different operations.
Soft/Flexible Automation consists of control methods that allow easy reprogramming through simply changing the software.
2/22/2014 ME 303 - Section 01 1.11
Spectrum of Specialization
Manufacturing EngineeringApplication of science, engineering, technology, and economics to manufacture products of a quality, quantity, and cost competitiveness in the market place.
ProcessEngineers
ProcessEngineers Tool
Engineers
ToolEngineers Standarts
Engineers
StandartsEngineers
MaterialHandlingEngineers
MaterialHandlingEngineers
QualityControl
Engineers
QualityControl
EngineersFacility
Engineers
FacilityEngineers
Manufacturing EngineersManufacturing Engineers
2/22/2014 ME 303 - Section 01 1.12
Process Engineers
Develop a logical sequence of manufacturing operations for each assembly and for the final assembly of all sub-assemblies into a finished product. Need a detailed knowledge of Machining process Forming process Assembly process Plant capabilities
Existing equipment
2/22/2014 ME 303 - Section 01 1.13
Tool Engineers
Provide proper tools for manufacturing. Following tools are commonly used in manufacturing industry: Conventional Tools
Standard tools: Cutting tools, dies Non-standard tools: Workpiece holding
devices
Non-traditional Tools Electro-discharge machining (EDM):
electrodes Laser
2/22/2014 ME 303 - Section 01 1.14
Standards Engineers
Synchronize the entire manufacturing process. For this purpose, they determine the time requirement for each operation and their sequences utilizing work standards and standard times.
2/22/2014 ME 303 - Section 01 1.15
Material Handling Engineers Plans for efficient transfer of
materials from one point to another. They must avoid the following: Disruption of production due to
lack of batches Unnecessary level of in-process
stock High degree of congestion
2/22/2014 ME 303 - Section 01 1.16
Facility Engineers
Provide efficient equipment layout. They must avoid the following: Higher material handling
cost Longer manufacturing time High level of in-process
stock High degree of congestion
in the process flow. In some cases, facility
engineers and material handling engineers combine in a single group called plant engineers.
2/22/2014 ME 303 - Section 01 1.17
Quality Control Engineers
Maintain the described level of product quality by employing a quality assurance system. Quality can be established at: Product design stage Design stage of production system Production stage by inspection.
2/22/2014 ME 303 - Section 01 1.18
Manufacturing Quality
Quality is defined at the design stage and is built into the product during manufacturing.
Design quality sets the specifications on the material and tolerances.
Manufacturing quality is the degree of comformity to the design specifications.
2/22/2014 ME 303 - Section 01 1.19
Quality vs. Cost
Quality ↑ Cost ↑ Demand ↓
Quality ↑ Quantity ↓ Cost ↑
2/22/2014 ME 303 - Section 01 1.20
Manufacturing Quality (2)Low Reliability Good
Enough
High Reliability
High CostLow Cost
Quality Level
Cost
and
Val
ue
GoodEnough
Value tocustomer
Cost tocustomer
Quality Level
Cost
and
Val
ue
GoodEnough
Totalcost
Originalcost
Runningcost
2/22/2014 ME 303 - Section 01 1.21
Classification of Manufacturing Processes (DIN 8530)
Creation of Cohesion
Maintenance of Cohesion
Destruction of Cohesion
Increase of
Cohesion
Shape (Form) Modification
2. Deforming 3. Seperating 4. Joining
6. Changing Material Properties
1. Primary Forming
Rearrangement
of Particles
Removal of
Particles
Addition of Particles
5. Coating
2/22/2014 ME 303 - Section 01 1.22
Manufacturing Processes
1. Primary Forming: Casting, powder metallurgy2. Deforming: Metal forming processes (bulk and sheet
forming)3. Separating: Machining
• Conventional machining (turning, milling, grinding, etc.)• Non-traditional machining (EDM, ECM, EBM, LBM, etc.)
4. Joining: Welding, brazing, riveting, etc.5. Coating: Painting, electroplating, etc.6. Changing Material Properties: Heat treatments.
2/22/2014 ME 303 - Section 02 2.1
Molecular Structure
Primary bonds: Strong atom-to-atom attractions by exchange of valence electrons. Ionic bond Covalent bond Metalic bond
Secondary bonds: Weak attraction between molecules (van der Waals forces) Permanent Dipole Bond Fluctuating Dipole Bonds
2/22/2014 ME 303 - Section 02 2.2
Ionic Bonds
Large inter-atomic forces are created by the “Coulomb”effect produced by positively and negatively charged ions.
2/22/2014 ME 303 - Section 02 2.3
Covalent Bonds
Large inter-atomic forces are created by the sharing of electrons to form directional bonds.
The atoms have small differences in electro-negativity & close to each other in the periodic table.
2/22/2014 ME 303 - Section 02 2.4
Metalic Bonds
Atoms loose their outer shell electrons and become (+) ions surrounded by free electron cloud. Free electrons act like “cement”to hold atoms together.
2/22/2014 ME 303 - Section 02 2.5
Properties
Property Ionic Bond Covalent Bond Metalic Bond
Hardness High Low → Very High Low → High
Ductility Brittle Brittle Ductile
Melting Temperature
High Low → Very High Low → High
Electrical and Thermal
Conductivity
Low Low High
2/22/2014 ME 303 - Section 02 2.6
Crystaline Structure
Body Centered Cubic (BCC)
Face Centered Cubic (FCC)
Hexagonal Closed Packed Cubic (HCP)
[*] Adapted from Groover (1996).
2/22/2014 ME 303 - Section 02 2.7
Structures of Common Metals*
BCC Chromium (Cr), Iron (Fe), Molybdenum (Mo), Tantalum (Ta), Tungsten (W)
FCC Aluminum (Al), Copper (Cu), Gold (Au), Lead (Pb), Silver (Ag), Nickel (Ni)
HCP Magnesium (Mg), Titanium (Ti), Zinc (Zn)
[*] At room temperature (20oC).
2/22/2014 ME 303 - Section 02 2.8
Imperfections and Defects
Imperfections or defects refer to the deviations in the regular pattern of the crystalline lattice structure.
Imperfections / DefectsImperfections / Defects
Point Defects♦ Vacancy♦ Ion-pair Vacancy♦ Interstitialcy♦ Displaced Ion
Line Defects♦ Edge Dislocation♦ Screw Dislocation
Surface Defects♦ Grain Boundaries
2/22/2014 ME 303 - Section 02 2.9
Point Defects
2/22/2014 ME 303 - Section 02 2.10
Line Defects
[*] Adapted from Groover (1996).
2/22/2014 ME 303 - Section 02 2.11
Edge Dislocation
2/22/2014 ME 303 - Section 02 2.12
Screw Dislocation
2/22/2014 ME 303 - Section 02 2.13
Maximum Shear Stress for Perfect Crystals
Consider a perfect crystal that does not contain any defects whatsoever.
Now, a shear stress (τ) is applied to deform the crystal lattice by x.
2/22/2014 ME 303 - Section 02 2.14
Maximum Shear Stress (Cont’d)
γτ ⋅= G
ax
≈≈ γγtan
Assuming that γ is small, the shearstress can be expressed as
where G is the shear modulus [MPa]of the material. Since x<< a (or b),
Hence,
axG=τ
2/22/2014 ME 303 - Section 02 2.15
Max. Shear Stress (Cont’d)
bb/2 x
ττm
b/4
⋅
=b
xm
πττ 2sinIn theory,
bx
m⋅
≈πττ 2
As x << b, sin θ ≈θ. Therefore
Substituting this into the previousone yields
abGm ⋅
=π
τ2
2/22/2014 ME 303 - Section 02 2.16
Analogy
The closest analogy for the shear stresses observed while deforming a perfect crystal would be the push of a ball up the hill.
2/22/2014 ME 303 - Section 02 2.17
Discrepancy betweenTheory and Practice
As a numerical example, let us consider steel. When a = b; G = 80 [GPa], the maximum shear stress to be applied so as to plastically deform this perfect crystal becomes roughly 8000 MPa!
The shear stress ranging between 8 to 80MPa is sufficient to deform steel in practice.
There is a huge discrepancy (by 100 ... 1000folds) between the theory and practice! Why?
2/22/2014 ME 303 - Section 02 2.18
Reason
The reason is the line defects (i.e. dislocations) present in the crystal lattice. No such thing as a perfect crystal!
Such defects tend to reduce the energy required to deform the material plastically.
2/22/2014 ME 303 - Section 02 2.19
Deformation Mechanism
Practical plastic deformation of metals takes placeprimarily by the movement of dislocations.
Movement of dislocations takes place in the most closely packed plane and along the closest packed crystallographic direction.
CRYSTALZONE
Slip
Rotation
Slip
Slip plane
Edge dislocation
Slipplane
Extra atomplane
[*] Adapted from Schey (1987).
Twinning
A second mechanism of plastic deformation in which atoms on one side of a plane (called the twinning plane) are shifted to form a mirror image of the other side.
2/22/2014 ME 303 - Section 02 2.20
3D Defects - Grains
A block of metal may contain millions of individual crystals, called grains. Such a structure is referred to as
polycrystalline.
Each grain has its own unique lattice orientation; but collectively, the grains are randomly oriented in the block.
2/22/2014 ME 303 - Section 02 2.21
2/22/2014 ME 303 - Section 02 2.22
Effects of Grain Boundaries
Grain boundaries are sources of dislocations.
At the same time, grain boundaries represent obstacles to dislocation propagation.
dKY+σ=σ 0
Hall-Petch Relationship:
where
σ is the yield strength,d is the average grain size,σ0, KY are material constants.
GrainBoundaries
Direction ofplanes
2/22/2014 ME 303 - Section 02 2.23
Deformation Mechanisms
Strength Ductility
Small grain Good Poor
Large grain Poor Good
2/22/2014 ME 303 - Section 02 2.24
Uniaxial Tension Experiment
2/22/2014 ME 303 - Section 02 2.25
Stress and Strain
0AP
eng =σ
0
0
−=engε
where P is load; A0 refers to initial(undeformed) area.
Hooke’s Law (Elastic Region):
engeng E εσ ⋅=
Proportional Limit
Fracture Strength
Ultimate Strength
Yield Strength
Elastic Region Plastic Region
STRAIN (ε )eng0.2% offsetor0.002 (mm/mm)
Engineering Stress - Engineering Strain Curve
UniformElongation Necking Fracture
Engineering stress:
Engineering strain:
2/22/2014 ME 303 - Section 02 2.26
Stress and Strain (2)
Strain (ε, εeng)
Stre
ss (
σ, σen
g)
x
x
σ, ε curve
σeng, εeng curve
yield stress
True Stress:AP
=σ
True Strain Increment:
dd =ε
True Strain (Total):
== ∫
0
ln0
dε
2/22/2014 ME 303 - Section 02 2.27
Additive Property of True Strain
Assume that a specimen of length l0 is elongated to l1then to l2:
=
→0
220
ln
ε
⋅=
→0
1
1
220
ln
εor
+
=
→0
1
1
220
lnln
ε hence211020 →→→
+= εεε
2/22/2014 ME 303 - Section 02 2.28
Example 1
A uniform bar of 100 mm initial length is elongated to a length of 200 mm in three stages: Stage 1: 100 mm to 120 mm Stage 2: 120 mm to 150 mm Stage 3: 150 mm to 200 mm.
Calculate the engineering and true strains for each stage and compare the sums of the three with the overall values of the strains.
2/22/2014 ME 303 - Section 02 2.29
Strain Relationships
Constant volume during plastic deformations implies:
≈
=⇒≈→⋅≈⋅
AA
AAAA 0
00
000 lnln
ε
Relationship between true and engineering strains:
( )engengeng εεε +=
→+=→−=
−= 1lnln11
0000
0
Hence ( ) 11ln −=+= εεεε eengeng or
2/22/2014 ME 303 - Section 02 2.30
Stress Relationships (Cont’d)
Relationship between true and engineering stresses:
0
0
0
engAA
AP
AP σσ ≈⋅==
+
−≈ 1
0
0
engσσ
with 0
0
−=engε
We obtain: ( ) εσσεσσ eengengeng ≈+≈ or1
or
2/22/2014 ME 303 - Section 02 2.31
Strain Hardening
b
a c
de
g
h
f
Load
Strain, εeng
When metals are plasticallydeformed, their strength increase.This is called strain- or work hardening.
Power (Ludwik or Hollomon)Law for the flow curve:
nplf Kεσ =
where K, n are material constants.n is called strain hardeningexponent.
2/22/2014 ME 303 - Section 02 2.32
Strain Hardening (Cont’d)AP f ⋅= σ
0=⋅+⋅= dAdAPd ff σσnecking
const.=⋅= AVεdAdAdAdAdAdV ⋅−=⋅−=→=⋅+⋅=
0
εσε
εεσεσ dndKndK fn
fn
f ⋅⋅=⋅⋅⋅≈→⋅≈ −1
0=⋅−⋅⋅= εσεσε
ddnPdu
necking un ε=∴
(1)
(2)
(3)
2/22/2014 ME 303 - Section 02 2.33
(Cold) Flow Curves
Courtesy of Prof. Erman Tekkaya
2/22/2014 ME 303 - Section 02 2.34
DuctilityDuctility is the ability to deform without damage.
Elongation is not uniform after necking.Engineering “strain” at fracture:
0
0
−= fracfracengε
Another measure of ductility is the reduction of area measured on the fractured test piece:
0
0
AAA
q frac−=
l
Necking
2/22/2014 ME 303 - Section 02 2.35
Yield Strength
b
a c
de
g
h
f
Load
Strain, εeng
Assume material is cold worked to point “d”:
c
dc A
PYS =
For small elastic strains: Ac ≈ Ad
nd
d
ddflow K
AP εσ ⋅≈=
nd
dflowc KYS εσ ⋅≈≈∴
2/22/2014 ME 303 - Section 02 2.36
Ultimate Tensile Strength
b
a c
de
g
h
f
Load
Strain, εeng
Cold Worked Material:
c
ec
APUTS =
Annealed Material:
a
ea
APUTS =
( ) ( )c
aac
AAUTSUTS ⋅= annealedhardenedwork
2/22/2014 ME 303 - Section 02 2.37
Effects of Cold Work
2/22/2014 ME 303 - Section 02 2.38
Example 2
A tensile specimen with 12 mm diameter and 50 mm gage length is subjected to a load of 32.2 kN. At this instant, the gage length is 60 mm. Assuming uniform deformation at this point, Compute all stresses and strains (true and engr.) Determine the final diameter Assuming that the plastic behavior of the specimen material can be
characterized as σf = 800ε0.5 [MPa], estimate the ultimate tensile strength of the deformed specimen.
Plot the yield strength and ultimate tensile strength of the specimen as a function of amount of cold work inflicted.
Example 2 - Plots
2/22/2014 ME 303 - Section 02 2.39
0 5 10 15 20 25 30 35 400
100
200
300
400
500
600
Amount of Cold Work (%)
Stre
ss [M
Pa]
YSUTS
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50
100
200
300
400
500
600
True Plastic Strain (εpl)
Stre
ss [M
Pa]
YSUTS
Example 2 – Matlab Code
2/22/2014 ME 303 - Section 02 2.40
K = 800; n = .5; c = K*(n/exp(1))^n;eps = linspace(0,n,100)';q = 1 - exp(-eps);
YS = K*eps.^n;UTS = c*exp(eps);
YS_q = K*(-log(1-q)).^n;UTS_q = c*(1./(1-q));
figure(1)plot(q*100,[YS_q UTS_q]);xlabel('Amount of Cold Work (%)')ylabel('Stress [MPa]'); legend('YS','UTS')
figure(2)plot(eps,[YS UTS]);xlabel('True Plastic Strain (\epsilon_{pl})')ylabel('Stress [MPa]'); legend('YS','UTS')
2/22/2014 ME 303 - Section 02 2.41
Strain Hardening Hypothesis
The amount of plastic deformation in any forming operation can be characterized through an equivalent true strain εeq which is specific for the respective state of deformation. The new yield stress in the deformed material can be determined from the uni-axial flow curve utilizing the equivalent strain instead of the uni-axial strain.
2/22/2014 ME 303 - Section 02 2.42
Equivalent Strain
2/22/2014 ME 303 - Section 02 2.43
Equivalent Strain – Wire Drawing
For wire drawing, the equivalent plastic strain is defined as
Since
Flow stress (~ yield strength) of thedrawn-wire becomes
2/22/2014 ME 303 - Section 02 2.44
Example 3
The strain hardening behavior of an annealedlow-carbon steel is σ = 700⋅ε0.3 [MPa]. The bar made from this metal is initially cold worked by 10%, followed by an additional cold work of 15%. Determine the probable yield strengths Find the (ultimate) tensile strengths of the bar for the
initial- and final case. If the ductility of the annealed material is 50%,
calculate the ductility for each case.
2/22/2014 ME 303 - Section 02 2.45
Homologous Temperature Scale
Behaviour of metals and polymers can be normalized by means of the homologous temperature scale (TH):
( ) ( )( )Kelvinin
Kelvinindimensionunit
mH T
TT =
0.5 Tm corresponds to the recrystallization temperature of metals and separates cold (high strength, low ductility) and hot (low strength, high ductility) behaviours.
2/22/2014 ME 303 - Section 02 2.46
Effect of Temperature
2/22/2014 ME 303 - Section 02 2.47
Example 4
Find the temperature where hot working range starts for copper. Note that the melting temperature of copper is Tm= 1100oC.
Solution:
TH = (1100+273)/2 = 687 K ≡ 414 oC
2/22/2014 ME 303 - Section 02 2.48
Hot Working Hot working is performed at TH > 0.5. Restoration and slip processes occur simultaneously. Hot working is strain-rate sensitive as
dtdC m
fε
=εε=σ ,)(
The strain-rate sensitivity exponent m takes the following values: Cold working: -0.05 < m < 0.05 Hot working: 0.05 < m < 0.30 Super-plasticity: 0.30 < m < 0.70 (very fine grain metals)
Higher the value of m the larger the elongation.
2/22/2014 ME 303 - Section 02 2.49
(Warm/Hot) Flow Curves
Courtesy of Prof. Erman Tekkaya
2/22/2014 ME 303 - Section 02 2.50
Definition of Annealing
Annealing is the process of heating a material to some elevated temperature, holding it at that temperature, and cooling it back to room temperature.
Annealing is applied to: Reduce hardness and brittleness (recovery) Recrystallize cold worked metals and obtain a specific
microstructure (recrystallization) Relieve residual stresses induced by prior shaping
processes (stress relief annealing)
2/22/2014 ME 303 - Section 02 2.51
Annealing
Although the cold-worked dislocation cell structure is mechanically stable, it is NOT thermodynamically stable!
Recovery Annealing:No change in grain structure, just rearrangements of dislocations
Primary Recrystallization:New grains are built at nuclei produced by cold workingSecondary Recyrstallization:Grain growth by coalescence of grains. Hydrogen embrittlement leads to a decrease in ductility.
2/22/2014 ME 303 - Section 02 2.52
Annealing (2)
Time and temperature have interchangeable effects!
2/22/2014 ME 303 - Section 02 2.53
Annealing (3)
Recyrstallization can occur only with nuclei produced by cold working.
The more the prior cold work, the more the nuclei and the smaller the grain size.
If there is no prior cold work, there will be no recyrstallization at all.
2/22/2014 ME 303 - Section 02 2.54
Stress States
σxx
σxx
τyx
σyyσyy
τxy
xy
z
τxz
σzz
σzz
τyz
τzyτzx
General 3-D State of Stress:
σ1
σ1
σ2
σ2
σ3σ3
σ1 > σ 2 > σ 3
Principal State of Stress:
Independent Cartesian Stress Components:
σxx, σyy, σzz, τxy, τyz, τzx
2/22/2014 ME 303 - Section 02 2.55
Stress States (Cont’d)
σxx
σxx
τyx
σyy
σyy
τxy
xy
z
x
y
σyy
σyy
τyx
σxxσxx
τyx
τxy
τxy
Plane StressState:
Uniaxial StressState:
σxx
σxx
x
y
σxxσxx
2/22/2014 ME 303 - Section 02 2.56
Transformation of Stress States: Mohr’s Circle
2/22/2014 ME 303 - Section 02 2.57
Yield Criteria
Uniaxial Yield Criterion:
Generalized Yield Criteria:
Tresca’s Yield Criterion:
von MisesYield Criterion:
YSx =σ
k=−
=2
31max
σστ YS=− 31 σσ
( ) ( ) ( )[ ] YSeq =−+−+−= 213
232
2212
1 σσσσσσσ
or
Yielding starts if
2/22/2014 ME 303 - Section 02 2.58
Example – Hydrostatic Pressure
Consider a small metal cube submerged into the ocean by 10 km.Would it plasticallydeform (or crack) when subjected to this huge hydrostatic pressure?
2/22/2014 ME 303 - Section 02 2.59
Hydrostatic Pressure (Cont’d)Principal stress componentsare σ1 = σ2 = σ3 = -100 MPa.
Applying Tresca’s yield criteriongives
NO deformation whatsoever!
2/22/2014 ME 303 - Section 02 2.60
Elastic Stress-strain Relationships
[ ]
[ ]
[ ])(1
)(1
)(1
yyzzzzzz
zzxxyyyy
zzyyxxxx
E
E
E
σσνσε
σσνσε
σσνσε
+−=
+−=
+−=
zxzx
yzyz
xyxy
G
G
G
τγ
τγ
τγ
1
1
1
=
=
=
Recall that for (small-strain) elastic behaviour, Hooke’s law dictates ”engineering” stress-strain relationships as
where shear modulus is)1(2 ν+
=EG
2/22/2014 ME 303 - Section 02 2.61
Plastic Stress-strain Relationships
λτγ
τγ
τγ
σε
σε
σε ddddddd
zx
zx
yz
yz
xy
xy
z
zz
y
yy
x
xx ====′
=′
=′
However, expressing a general “flow-rule” for plastic deformations is a bit more challenging. Levy-Mises Rule is often-times employed for the sake of its simplicity:
In these equations, the deviatoric stresses are defined as
hzzz
hyyy
hxxx
σσσ
σσσσσσ
−=′
−=′−=′
where the hydrostatic(or mean) stress is 3
zzyyxxh
σσσσ
++=
2/22/2014 ME 303 - Section 02 2.62
Notes on Flow Rule Levy-Mises Rule states that a small increment of
plastic-strain depends on the current deviatoric stress-state. NOT on the stress (increment) which is required to
bring it about! This rule ignores elastic deformations.
dλ is an instantaneous (non-negative) constant of proportionality. Usually varies throughout the “straining” process!
2/22/2014 ME 303 - Section 02 2.63
Elementary Theory of Plasticity
3 displacements + 6 stresses + 1 hydrostatic stress = 10 unknowns
Columns of the Mathematical Theory of PlasticityColumns of the Mathematical Theory of Plasticity
EquilibriumEquations
EquilibriumEquations Yield Criterion
(Flow Condition)
Yield Criterion(Flow Condition) Stress-Strain (or Str.-rate)
Relationship (Flow Rule)
Stress-Strain (or Str.-rate)Relationship (Flow Rule)
6 Equations1 Equation3 DifferentialEquations
A Total of 10 Equations
2/22/2014 ME 303 - Section 02 2.64
Basic Assumptions
Deformation is homogeneous (plane section remain plane)
Material is homogeneous and isotropic Principal directions of the stress and strain state are
known All body forces are neglected Friction is described by the Coulomb’s law:
( )constant=⋅= µσµτ normalfriction
2/22/2014 ME 303 - Section 02 2.65
Methods for Theory of Plasticity
Methods of Elementary Theory of PlasticityMethods of Elementary Theory of Plasticity
Energy Methods(Seibel, 1925)
Energy Methods(Seibel, 1925) Equilibrium (or Slab) Methods
(Sachs, 1927)
Equilibrium (or Slab) Methods (Sachs, 1927)
2/22/2014 ME 303 - Section 02 2.66
Energy Methods
Introduced by the renowned German scientist Siebel in 1925.
These methods can be applied to stationary processes where the process force is quasi-stationary in time.
2/22/2014 ME 303 - Section 02 2.67
Slab Methods
Roots back to Sachs (1927). Based on equilibrium consideration of
infinitesimal slab-, pipe- or disk elements which remain plane during deformation.
Can be applied to both stationary (SS) or transient (non-SS) type of processes.
2/22/2014 ME 303 - Section 03 3.1
Classification of Forming Processes
Metal Forming ProcessesMetal Forming Processes
Bulk Forming ProcessesBulk Forming Processes Sheet Forming ProcessesSheet Forming Processes
• Spatial workpieces• Large changes in cross-section
& large changes in thicknesses• Material flows in all directions• Generally multi-axial compressive
stress states• Larger relative forces
• Planar workpieces (sheets, plates)• Hollow pieces with almost constant
thickness• Generally two-axial stress states:
tensile-tensile/tensile-compressive
2/22/2014 ME 303 - Section 03 3.2
Bulk Forming Processes
Bulk Forming ProcessesBulk Forming Processes
Steady-state ProcessesSteady-state Processes Non-steady-state ProcessesNon-steady-state Processes
Deformation patterns are independent of time.
Example:Extrusion, rolling, drawing
Deformation patterns are dependent on time.
Example:Upsetting, forging, extrusion
2/22/2014 ME 303 - Section 03 3.3
Steady-State Processes
All parts of the workpiece are subjected to same mode of deformation.
Once the deformation zone is analyzed, the analysis remains valid throughout the process. Rolling (Flat, Ring, Tube, Shape) Drawing (Wire, Bar, Tube)
2/22/2014 ME 303 - Section 03 3.4
Non-Steady-State Processes
Geometry of the part changes continually.
The analysis must be repeated at instances (in time). Upsetting Forging
2/22/2014 ME 303 - Section 03 3.5
Transitional Processes Some processes do have transitional
character. In extrusion, the deformation is non-
steady state at the beginning and the end of the process.
It acquires steady state characteristics when the greater part of the billet is extruded.
2/22/2014 ME 303 - Section 03 3.6
Processes (Cont’d)
(Schey, 1983)
2/22/2014 ME 303 - Section 03 3.7
Metal Forming Processes
Metal Forming ProcessesMetal Forming Processes
Cold FormingCold Forming Hot FormingHot FormingWarm FormingWarm Forming
Forming at room temp.(everyday sense)
Forming below therecrystallization temp.
Steel @ Room Temp.
Forming above roomtemp. (everyday sense)
Forming below or aroundthe recrystallization temp.
Steel @ 750-950oC
Forming above roomtemp. (everyday sense)
Forming above the recry-stallization temperature
Steel @ 1150-1250oC
2/22/2014 ME 303 - Section 03 3.8
Bulk Workability
Ductility is the ability to deform without fracture in uni-axial tension.
Bulk workability is the generalization of the term ductility to multi-axial stress states. Hence, the ability to deform without fracture under general stress states is termed as bulk workability.
Good bulk workability requires good ductility. But the amount ofdeformation a material can sustain depends also on the stress state of the forming process.
As a general rule, the more compressive the hydrostatic stress the higher is the bulk workability.
2/22/2014 ME 303 - Section 03 3.9
Bulk Workability Criteria
Other Criteria:
Cockroft & Latham Criterion: For fracture, the work done by thehighest local tensile stress must reach a critical value.Datko’s Criterion: For fracture, a natural tensile strain must be equal to the natural strain at fracture in a simple tensionspecimen.
The most practical bulk workability criterion is based on the ideal equivalent strain of a forming process. For each process there exists a limiting ideal equivalent strain until fracture.
2/22/2014 ME 303 - Section 03 3.10
Forces in Bulk FormingRelevance of Forming Forces:
1. To design the tools (tool pressures, stresses)2. To select the appropriate forming press3. To evaluate the elastic deformations in the forming system
Methods of Force ComputationMethods of Force Computation
EmpiricalEmpirical AnalyticalAnalyticalExperimentalExperimental NumericalNumerical
2/22/2014 ME 303 - Section 03 3.11
Friction in Metal Forming
Note that k = 0.5·Y (Tresca) or k = 0.577·Y (von Mises)
Coulomb Model:Ffriction = µ ·Fnormal
τfriction = µ ·σnormal
Friction coefficient µ is
0 ≤ µ ≤ 0.5 (0.577)Best suited for cold forming.
Maximum possible frictional shear stress is: kyieldmaxfriction =τ=τ
Shear Model:
τfriction = m·k
Friction factor m is
0 ≤ m ≤ 1Best suited for hot forming.
2/22/2014 ME 303 - Section 03 3.12
Friction in Metal Forming
Normal Stress0*Y 1*Y 2*Y 3*Y 4*Y
Fric
tiona
l She
ar S
tress
τfri
ctio
n
0.0k
0.2k
0.4k
0.6k
0.8k
1.0k
1.2k
plasticelast.
0 Y 2Y 3Y 4Y
(τfriction)max= k = Y/2 (Tresca)µ = 1.0
µ = 0.5µ = 0.2
µ = 0.1
Normal Stress0*Y 1*Y 2*Y 3*Y 4*Y 5*Y 6*Y
Max
imum
Fric
tion
Coe
ffici
ent
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
plasticelast.
0 Y 2Y 3Y 4Y 5Y 6Y
µ max
2/22/2014 ME 303 - Section 03 3.13
Friction in Metal FormingMetal working lubricants are used:
To prevent die/workpiece adhesion To control surface finish To cool system during cold working To help prevent heat loss in hot working
Metal working lubricants must be: Non-toxic Easy to use & remove Inactive (non-reactive)
2/22/2014 ME 303 - Section 03 3.14
Mean Flow Stress
Flow stress (σf) is defined as the stress needed to maintain plastic deformation at the temperature, strain, and strain-rates prevailing in the process.
For non-steady state processes (like forging), instantaneous flow stress at the end of deformation is utilized in the computations.
For steady-state processes (such as rolling and extrusion), mean flow stress is often-times employed.
2/22/2014 ME 303 - Section 03 3.15
Mean Flow Stress (Cont’d)
0+ εeqε
σσf = Κε n
σfmean
∫+
+−==
eq
dfeq
fmmeanf
ε
εεσε
σσ0
)(0
1
Average value of the flow stress in cold-forming can be expressed as
eqeq
nKdK
n
eq
n
eq
εεε
εεε
ε ++
+
==+
∫0
1
0 11
1+=
nK n
eqfm
εσ where 0+ ≈ 0.
2/22/2014 ME 303 - Section 03 3.16
Mean Flow Stress (Cont’d)
UTSfm ≈σ
For convenience, manufacturing engineers frequently employ the tensile-strength (UTS) of a material as the mean flow stress:
nnuUTS nKK )()( == εσ
Recall that the true stress correspondingto the UTS (an engineering quantity) is
nUTSeng eUTSe ⋅=⇒= σσσ
ε
Since
n
fm enKUTS
=≈σHence
For n = 0.1 → 0.5, the difference between the integrated expression and the UTS is less than 10%.
2/22/2014 ME 303 - Section 03 3.17
Forging
Forging ProcessesForging Processes
Open DieForging
Open DieForging Closed Die
Forging
Closed DieForgingImpression
Forging
ImpressionForging
Open-die Forging
2/22/2014 ME 303 - Section 03 3.18
Closed-die Forging
2/22/2014 ME 303 - Section 03 3.19
2/22/2014 ME 303 - Section 03 3.20
Plain Strain UpsettingAssumptions:
Width of the workpiece doesnot change (plane strain).
σx is uniform over the planenormal to the x-direction.
σx and σy= -p are principalstresses, although friction is present.
Plane sections remain planealthough friction is included.
Assume no hardening.
p
h
p
dx
2a
x
yUpper
Die
Lower Die
Workpiece
2/22/2014 ME 303 - Section 03 3.21
Plane Strain Upsetting (2)
τ
h
τ
p
dx
p
σx
σ
σx+dσx
∑ = 0xF ( ) 01211 =⋅⋅τ⋅−⋅⋅σ+σ+⋅⋅σ−∴ dxhdh xxx
02 =⋅⋅−⋅ dxdh x τσ
( ) kpx ⋅=−− 2σ
dpd x =σ−
02 =⋅⋅−⋅− dxdph τ
02=⋅
⋅+ dx
hdp τ
Tresca:
2/22/2014 ME 303 - Section 03 3.22
Plane Strain Upsetting (3)p⋅= µτ
∫∫ ⋅⋅
−= dxhp
dp µ2 *2ln Ch
xp +⋅⋅
−=µ
hx
eCxp⋅⋅−
⋅=µ2
)(
Assuming Coulomb friction:
Integrating:
Hence: where
Boundary condition: axx == at0σ kp 2=
Therefore: ha
ekC⋅⋅
⋅=µ2
2( )xa
heYxp
kxp −⋅
⋅
==µ2)(
2)(
02=⋅
⋅⋅+ dx
hpdp µ
C = ln C*
2/22/2014 ME 303 - Section 03 3.23
Plane Strain Upsetting (4)
2a
x
p(x)
The average interface pressure can be found as:
dxkxp
akp a
av ⋅⋅= ∫0 2
)(12
( )122
2 −⋅⋅
== ⋅⋅ haavav ea
hYp
kp µ
µ
Taylor series expansion including the quadratic term yields
ha
Yp
kp avav ⋅+≈= µ12
2/22/2014 ME 303 - Section 03 3.24
Plain Strain Upsetting (5)k=τ 02
=⋅+ dxhkdp
∫∫ ⋅−= dxhkdp 2 **2 Cx
hkp +⋅−=
**
2)( C
hx
kxp
+−=
Assuming sticking friction:
Integrating:
Hence:
Boundary condition: axx == at0σ kp 2=
Therefore:haC += 1**
hxa
Yxp
kxp −
+== 1)(2
)(
2/22/2014 ME 303 - Section 03 3.25
Plain Strain Upsetting (6)
2a
x
p(x)
(1+
a/h)
Y
The average interface pressure can be found as:
dxkxp
akp a
av ⋅⋅= ∫0 2
)(12
ha
Yp
kp avav ⋅+== 5.012
2/22/2014 ME 303 - Section 03 3.26
Plain Strain Upsetting (7)
Assume that we have in the core sticking friction up to x = x1 andoutside sliding:
)(, 11 xpkxxat ⋅=== µτSince the general solution for the sliding friction case is still valid, we get:
( ) ( )
⋅⋅=⋅=
−⋅⋅
12
1 2xa
hekxpkµ
µµ( )1
2
21xa
he−⋅
⋅
⋅⋅=µ
µ
Taking the logarithm of both sides:
( ) ( )122ln0 xa
h−
⋅+⋅=
µµ
⋅⋅
−=µµ 2
1ln21
hax
frictionCoulombaxxfrictionstickingxx
≤≤≤≤
1
10
2/22/2014 ME 303 - Section 03 3.27
Plane Strain Upsetting (8)
Pressure at x1:( )
+−⋅
⋅−⋅
⋅
==== µµµµµ
21ln2
1ln2
2211 1)(
2)( eee
Yxp
kxp
haahxa
h
µ⋅==
21)(
2)( 11
Yxp
kxp
Pressure at 0 ≤ x ≤ x1:µ⋅
=+−=2
12
)( **11 Chx
kxp
µ⋅+=
211**
hxC
−
⋅+
−==
µµ 21ln1
21)(
2)(
hxa
Yxp
kxp
2/22/2014 ME 303 - Section 03 3.28
Plane Strain Upsetting (9)
Condition for total surface slide:
021ln
201 ≤
−→≤
µµhax
≤
µµ 21ln12
ha
2/22/2014 ME 303 - Section 03 3.29
Example
An aluminium slab of 750 mm x 80 mm x 36 mm is forged to thethickness of 25 mm. Find the upsetting force if the flow curve ofaluminium is σf=140ε0.25 MPa and the coefficient of friction is assumed to be 0.2.
750 mm
80 mm36 mm
2a
750 mm
25 mm
2/22/2014 ME 303 - Section 03 3.30
SolutionFrom volume constancy, we can determine the unknown dimension 2a:
23 2527503680750 mmamm ××=×× mma 2.1152 =
=
=
mmmm
hh
eq 2536lnln
1
0ε 36.0=eqε
( ) 25.036.0140 ⋅=⋅= MPaneq
finalf K εσ MPa4.108=final
fσ
Ideal equivalent strain:
Effective flow stress:
2/22/2014 ME 303 - Section 03 3.31
Solution (Cont’d)
Check sticking or sliding friction:
⋅⋅
−=µµ 2
1ln21
hax
⋅⋅−=
2.021ln
2.0225
22.115
1mmmmx
mmx 33.01 = Negligible! Therefore, we have only sliding friction.
( ) ( ) dxwxp2dAxpFA
a
0
⋅== ∫ ∫The upsetting force is given by:
( ) ( )a
xah
xah
a
ehYwdxeYwF0
22
0 222
⋅
−⋅⋅⋅=⋅⋅=
−−⋅⋅
∫µµ
µ
2/22/2014 ME 303 - Section 03 3.32
Solution (Cont’d)
−⋅⋅⋅=
−⋅
⋅⋅⋅⋅=
⋅⋅⋅⋅
112
222
ha
ha
ehYwehYwFµµ
µµ
−⋅⋅⋅=
⋅⋅
12.0
254.108750 252/2.1152.02
2mm
mm
emmmm
NmmF
MNNF 4.15996,378,15 ==
2/22/2014 ME 303 - Section 03 3.33
Axisymmetric UpsettingForce equilibrium in radial direction:
0)()(2
22
=+⋅θ⋅⋅σ+σ−
θ⋅⋅⋅σ+
⋅θ⋅⋅⋅µ⋅+θ⋅⋅⋅σθ
drrdhd
ddrhdrdrpdrh
rr
r
Neglecting higher order product of differentials:
02
=σ⋅⋅−⋅σ⋅−⋅σ⋅+⋅⋅⋅µ⋅ θ
rr drhdrhdrhdrrp
2/22/2014 ME 303 - Section 03 3.34
Axisymmetric Upsetting (Cont’d)
Nrl ⋅
=π2
To derive strain-relationships, let us assume that a circle with radius r was initially drawn on the surface of this disk and that it was divided into Nequal segments. The length of each segment then becomes
An incremental change in the radius of this circle yields dr
Ndl π2
=
Hence .2
2
rdr
rdr
ldl
N
N ==π
πBy definition,
rdrd
ldld r == ˆˆ εεθ
Consequently, rdd εεθ =
2/22/2014 ME 303 - Section 03 3.35
Axisymmetric Upsetting (Cont’d)
hr
r
h
ddσσ
εσσ
ε
θ
θ
−=
−
Similary, the application of Levy-Mises flow rule (see Chapter 1) in the principal directions gives
where the hydrostatic stress is
)(31
zrh σσσσ θ ++=
Since ⇒= rdd εεθ rσσθ =
On the other hand, the Tresca yield criterion reads
dpdYkp rr −=⇒==+ σσ 2
Hence Ch
rpdprhdrrp +⋅⋅
−=⇒⋅⋅−=⋅⋅⋅⋅µµ 2ln2
2/22/2014 ME 303 - Section 03 3.36
Axisymmetric Upsetting (Cont’d)
From the Tresca yield criterion and the boundary condition:
Rrr == at0σ
It can be shown that at r = R: Ykp == 2
So that)(2)(
2)( rR
heYrp
krp −
==µ
−
⋅⋅−⋅
⋅
=⋅⋅⋅⋅
==⋅⋅
∫ 12212)(1
2
22
02 h
ReR
hdrrYrp
RYp
kp h
RRaveave µ
µπ
π
µ
For purely sticking friction: hR
kpave
⋅+=
31
2
2/22/2014 ME 303 - Section 03 3.37
Plane Strain Indentation
p
heig
ht h L
Assumptions: Indenter is narrow
(local deformation). w >> L, so that plane
strain condition can be assumed.
Pressure at the indenter-workpiece interface, p, can be assumed constant over the contact area.
2/22/2014 ME 303 - Section 03 3.38
Plane Strain Indentation (2)Case I:
Semi-Infinite Workpiece
p
L
h ∝ ∞
8≥Lh
UTSp f ⋅≈⋅≈ 33 σ
Case IIThick Workpiece
L
h
p
p
81 << Lh
fiQp σ⋅=
Case IIIThin Workpiece
L
h
p
p
Lh≥1
fp σ⋅≈ 15.1
2/22/2014 ME 303 - Section 03 3.39
Plain Strain Indentation (3)
Hill (1950)
Pressure multiplying factor due to inhomogeneousdeformation:
Or, approximately:
7.03.0 +⋅≈LhQi
2/22/2014 ME 303 - Section 03 3.40
Rolling
Kinematic ClassificationKinematic Classification
Longitudinal RollingLongitudinal Rolling
Oblique RollingOblique Rolling
Transverse RollingTransverse Rolling
Classification by Die (Roll) GeometryClassification by Die (Roll) Geometry
Flat RollingFlat Rolling
Shape RollingShape Rolling
2/22/2014 ME 303 - Section 03 3.41
Shape Rolling
Transverse Rolling: Oblique Rolling:
2/22/2014 ME 303 - Section 03 3.42
Flat Rolling
+
+
Angle of contact
Friction force
Radial force
Direction of travel
Radial force
UpperRoll
LowerRoll
α
The process of reducing the thickness of a slab to produce a thinner and longer but only slightly wider product is referredto as flat rolling.
Primary forming process. Hot rolling (not tight tolerances) Cold rolling (tight tolerances) Normal anisotropy is induced in
the sheet or band.
2/22/2014 ME 303 - Section 03 3.43
Deformation Zone
Deformation Zone and Nomenclature in Flat Rolling
2/22/2014 ME 303 - Section 03 3.44
Kinematics of Flat Rolling
111000 whvwhvwhv ⋅⋅=⋅⋅=⋅⋅
200 ≥Lw
www =≈ 10
hhvv 0
0 ⋅=
Volume constancy requires:
For
so that
Resulting
βcosrollrollx vv = vvroll
x =Horizontal roll velocity at any β:
Neutral point:
2/22/2014 ME 303 - Section 03 3.45
Initial GraspTo have initial grasp:
Or, for small α:
(in most cases)
(for thin sheet)
where
2/22/2014 ME 303 - Section 03 3.46
Roll ForcesRoll Forces: fmr QwLP σ⋅⋅⋅⋅= 15.1
Case I - Partial Plastification: hmean/L>1
7.03.0 +⋅=L
hQ mean (Hill, 1950)
Case II - Full Plastification: hmean/L ≤ 1
−
⋅µ=
⋅µ
1eL
hQ meanhL
mean (Slab Method)Length of region: ( )10 hhRL −=
Mean thickness:2
10 hhhmean+
=
2/22/2014 ME 303 - Section 03 3.47
Mean Flow StressCold Rolling:
1)(
+≅
nK n
eqfm
εσ ( )10ln hheq =εwhere
Hot Rolling:
meqfm C )(εσ ≅
( )sheetmean
eqeq vL
hht
10ln≈
∆
∆≈
εεwhere
Note that the mean flow stress in hot rolling depends on many different parameters including temperature, reduction ratio, and strain-rate (least significant!) (Sims, 1954).
2/22/2014 ME 303 - Section 03 3.48
Torque and Power Requirements
Remark: The torque is obtained from the normal forces alone, since frictionalforces can be considered to cancel each other.
Torque per roll:
2LPT r ⋅≈
Total Power Requirement:
( )R
vLPTPower rollrroll ⋅
⋅⋅=⋅⋅≡
222 ω
RvLPPower roll
r ⋅⋅≡
2/22/2014 ME 303 - Section 03 3.49
Elastic Flattening of Rolls
Application of front and back tension
Reducing friction by good lubrication
Reducing the flow stress by annealing
Using smaller diameter rolls backed up by a cluster of rolls
Using rolls with high elastic modulus (such as sintered carbide)
Placing the material between layers of soft material, artificially increasing the thickness
Precautions to ReduceElastic Flattening:
2/22/2014 ME 303 - Section 03 3.50
Back and Front Tension
When a front tension (σft)or a back tension (σbt) is applied, the compressive stress in effectwill be lowered:
where
2/22/2014 ME 303 - Section 03 3.51
Support by Cluster of RollsHousing
Using cluster of rolls to reduce elastic flattening of working rolls.
2/22/2014 ME 303 - Section 03 3.52
Effect of Cambered Rolls
Use of cambered rollsto compensate forroll bending.
Thickness variation of the sheet due to rollbending (uncambered)
2/22/2014 ME 303 - Section 03 3.53
Defects in Flat Rolling*
Edge Wrinkling
Warping
Centerline Cracking
Residual Stresses:Edge – CompressiveCenter – Tensile
Here are some some common defectswhen roll camber is insufficient:
[*] After Hosford & Caddell (1983)
2/22/2014 ME 303 - Section 03 3.54
Defects* (Cont’d)
Centreline Wrinkling
Splitting
Edge Cracking
Residual Stresses:Edge – TensileCenter – Compressive
[*] After Hosford & Caddell (1983)
Here are some some common defectswhen roll camber is excessive:
2/22/2014 ME 303 - Section 03 3.55
Classification of Extrusion
1. Reducing: Rods, Tubes. 2. Extrusion
a. Extrusion with rigid tools i. Forward (direct): Rods, Tubes. ii. Backward: Rods, Tubes. iii. Transverse: Rods, Tubes.
b. Extrusion with non-rigid tools i. Hydrostatic forward: Rods.
3. Impact Extrusion a. Impact extrusion with rigid tools
i. Forward (direct): Rods, Tubes, Canes. ii. Backward: Rods, Tubes, Canes. iii. Transverse: Rods, Tubes.
b. Impact extrusion with non-rigid tools i. Hydrostatic forward: Rods, Tubes.
2/22/2014 ME 303 - Section 03 3.56
ExtrusionForward Rod Extrusion Forward Tube Extrusion
2/22/2014 ME 303 - Section 03 3.57
Transverse/Lateral Extrusion
2/22/2014 ME 303 - Section 03 3.58
Classification of DrawingDRAWINGDRAWING
Drawing of wires, rods and slabsDrawing of wires, rods and slabs
Drawing of tubes without a mandrelDrawing of tubes without a mandrel
Drawing of tubes with a stationary mandrelDrawing of tubes with a stationary mandrel
Drawing of tubes with a free mandrelDrawing of tubes with a free mandrel
Drawing of tubes with a moving mandrelDrawing of tubes with a moving mandrel
IroningIroning
Drawing of tubesDrawing of tubes
2/22/2014 ME 303 - Section 03 3.59
DrawingWire Drawing Tube Drawing
2/22/2014 ME 303 - Section 03 3.60
Forward Rod Extrusion Force by the Elementary Theory of Plasticity
Bendingshearfrictionidealtotal WWWWW +++=
Bendingshearfrictionidealtotal FFFFF +++=
The total external work for deformation composes as:
For quasi-stationary processes the total forming force is:
ntdisplacemepunchtot
totWF =
yielding:
2/22/2014 ME 303 - Section 03 3.61
Forward Rod Extrusion – Ideal Work
initial position
final position
Ideal work:
( ) ( )( ) zr
zrideal
zdAdAzzdAdrrzdW
σσσσπ
⋅∆⋅+⋅⋅∆=⋅∆⋅+⋅⋅⋅∆= 2
constant=∆⋅=∆ zAV
( )zrideal AdAVdW σσ −⋅⋅∆=
( )zrideal
ideal AdA
VdWdw σσ −⋅=
∆=
Using volume constancy
we obtain
The infinitesimal specific ideal work:
2/22/2014 ME 303 - Section 03 3.62
Specific Ideal Work
frz σσσ =−
AdAdw fideal ⋅−= σ
∫∫ ⋅−=≡1
0
A
Afidealideal A
dAdww σ
⋅=−= ∫
1
0ln1
0AA
AdAw fm
A
Afmideal σσ
2)(1 10
1
0
ffffm d
σ+σ≈ε⋅εσ
ε∆=σ ∫
ε
ε
Using the Tresca yield criterion
leads to
Integrating over the deformation zone:
Utilizing the mean value theorem:
The mean flow stress is defined as:
2/22/2014 ME 303 - Section 03 3.63
Total Ideal Work
The work to deform a material of volume V is simply:
⋅⋅=⋅=
1
0lnAAVVwW fmidealideal σ
The ideal equivalent strain for extrusion (drawing) is defined as:
=
1
0lnAA
eqε
Finally, we obtain: eqfmideal VW εσ ⋅⋅=
2/22/2014 ME 303 - Section 03 3.64
Assumptions on Pressures
It is difficult to determine local stresses in the scope of the elementary theory of plasticity.
Pressure distributions in the container and die-shoulder must be assumed.
Siebel assumes that the internal pressures are in the order of the local yield stresses: Interface pressure in container ~ σf0 (Y0) Interface pressure at die-exit ~ σf1 (Y1) Interface pressure in shoulder ~ σfm (Ym)
2/22/2014 ME 303 - Section 03 3.65
Frictional Work for Die Shoulder
A0
initial position
final position
Ym
µYm dz ∆s∆s
cos α
dzcos αA1
d
α
Area: A
rz
shoulderdiefriction
containerfrictionfriction WWW −+=
Frictional Work:
ααπσµ
coscosdzsddW fm
shoulderdiefriction ⋅
∆⋅⋅⋅⋅=−
dAdzd −=⋅⋅⋅ απ tan
AVs ∆
=∆
The decrease in the cross-sectional area is:
Also:
2/22/2014 ME 303 - Section 03 3.66
Frictional Work (Cont’d)
αασµ
sincos1⋅
⋅⋅∆
⋅⋅−=− dAAVdW fm
shoulderdiefriction
ααµσ
sincosln
1
0
⋅⋅⋅⋅=−
AAVW fm
shoulderdiefriction
αµεσ2sin
2⋅⋅⋅=−
eqfmshoulderdie
friction VW
Hence:
Integrating between A0and A1 results:
Utilizing the concept of the equivalent strain:
2/22/2014 ME 303 - Section 03 3.67
Frictional Work for Container
rz
dz
dsinitial
positionhh 0
final position
s
d0
Y0
µY0
( ) ( ) dshddW fcontainerfriction ⋅⋅⋅⋅⋅= 00 σµπ
hhs −= 0
( )220002
1 hhdW fcontainerfriction −⋅⋅⋅⋅= σµπ
Utilizing the relationship
and integrating between h0 and hyields
2/22/2014 ME 303 - Section 03 3.68
Forward Rod Extrusion – Shearing Work
dV
drr
r0
2αγ
τmax0
τmax1
∆s
s
r
z
dVdW fshear ⋅⋅= γσ tan21
00
srsdrsrsrW f
r
fshear /312
21 3
000
00
0
⋅⋅∆⋅=⋅∆⋅⋅⋅=∆ ∫ σππσ
ασ tan31
00
0 ⋅⋅=∆
∆=∆ f
shearshear V
Ww
ασ tan31
11 ⋅⋅=∆ fshearw
ασ tan32
⋅⋅⋅= fmshear VW
Infinitesimal shearing work at inlet:
Integrating over the whole inlet shear zone:
Specific shearingwork at inlet:
Specific shearingwork at outlet:
Total shearing workfor a volume V:
2/22/2014 ME 303 - Section 03 3.69
Shearing Work (Cont’d)
τmax0
∆s
r
s
∆s tan γ
γ
dAγ
∫ ⋅⋅∆=∆0
00max0 )2()tan(
r
shear drrsW πτγ
srs
drrsrsW
f
r
fshear
/31
221)(
300
000
0
⋅⋅∆⋅=
⋅⋅⋅∆=∆ ∫
σπ
πσ
⇒==∆
=∆ 20
0 ,tan, rAsr
AVs πα
ασαπσ tan31tan)(
31
02
000 VAVrW ffshear ∆=⋅
∆⋅=∆
2/22/2014 ME 303 - Section 03 3.70
Forward Rod Extrusion Force
( )220002
12sin
21tan32 hhdVW feqfmtotal −⋅⋅⋅⋅⋅+
⋅
++⋅⋅⋅= σµπε
αµασ
000 2sin21tan
32
feqfmcontainerfriction hdAFFF σµπε
αµασ ⋅⋅⋅⋅+
⋅
++⋅⋅⋅=+=
⋅
++⋅⋅⋅= eqmYAF ε
αµα2sin
21tan32
0 00 YhdF Containerfriction ⋅⋅⋅⋅= µπ
The total forming work required for a volume of V by Siebel (1925):
with and
The total average forming force can be determined as Siebel (1925):
2/22/2014 ME 303 - Section 03 3.71
Example - Extrusion
Example 1:A 110×Ø75 mm billet made of C15 steel
(annealed) is cold extruded to Ø45 mm in a die with a die angle of 2α = 90o. If the flow curve
of C15 can be represented by
with an initial flow stress of
Determine the extrusion force. Assume that thecoefficient of Coulomb friction is 0.06.
MPa700 24.0eqf εσ ⋅=
MPa2400 =fσ
2/22/2014 ME 303 - Section 03 3.72
Extrusion Force by Empirical Methods
1
0
AARe =
==
1
0lnlnAAReeqε
eqpunch
m dddv
εα
ε 31
30
20 tan6
−⋅⋅⋅
=
00 AQApF fmee ⋅⋅=⋅= σ eqeQ ε⋅+= 2.18.0
⋅⋅⋅+=+= frictioncontainerfriction dFFFF τπ 00
Extrusion ratio: Equivalent strain:
Mean strain rate:
Extrusion force(without container friction)
with
For forward extrusion:
(required in hot extrusion)
For backward can extrusioncheck also the piercing force:
)()53( 10 AAtoF fmpiercing −⋅⋅= σ
2/22/2014 ME 303 - Section 03 3.73
Drawing Force by Empirical Methods
210 ddh +
=
( )1
01 lncot1
AA
fmz ⋅⋅⋅+⋅= ϕαµσσ
Lh
⋅+= 2.08.0ϕLh
⋅+= 12.088.0ϕ
11 AF z ⋅= σ
where
Drawing stress at the exit:
where (axisymmetrical) or (plane strain)
Drawing force:
Remark: Centerburst (Chevroning) occurs if h/L ≥ 2.
2/22/2014 ME 303 - Section 03 3.74
Example - Drawing
Example 2:A shaped wire is drawn from annealed, 3-mm-diameter 302 stainless steel wire. The cross-sectional area of the shape is
5.0 mm2. A commercial oil-based lubricant is used, the dies have 12o
included angle, and drawing speed is 2 m/s. Calculate the force and power
requirements. Take the coefficient of friction 0.04 and the material constants
in the Ludwik expression asK = 1300 MPa and n = 0.3.
d0
d1
Ftotal
α
2/22/2014 ME 303 - Section 03 3.75
Backward Can Impact Extrusion Force
33
2 1 2
2
0
10
e
iA d
dA
AA=
−=ε
6.0to5.0≥Aε
The relative area reduction is defined as
The solution by Dipper (1949) is valid for:
2/22/2014 ME 303 - Section 03 3.76
Backward Can Impact Extrusion - Zone 2
Upsetting in Zone 2: ( )5.021
1 +⋅= µµ
( )[ ] ( ) πσµπσσσ ⋅⋅+
⋅⋅⋅⋅=−⋅⋅+− 22
24 2
22222
sddzddd ifiezzz
dzsd fz ⋅⋅⋅=⋅− 22 σµσ
( )zbs
fz −⋅
⋅⋅−= 2
2
2 σµσ s
bfmz ⋅⋅−= 22 σµσ
⋅+⋅−=−=
sb
ffmzmr µσσσσ 12222
The effective coefficient of friction:
Force equilibriumin z-direction:
Integrating:
Simplifying yields:
Mean stress:
Using Tresca’s yield criterion:
2/22/2014 ME 303 - Section 03 3.77
Backward Can Impact Extrusion - Zone 1
Upsetting in Zone 1:
( )[ ] drrbrd frrr ⋅⋅⋅⋅⋅⋅=⋅⋅⋅⋅+− πσµπσσσ 222 11111 drbd f ⋅⋅⋅=⋅− 111 2 σµσ
mrif
r rdb 2
111 2
2σ
σµσ +
−⋅
⋅⋅−=
mri
frm bd
21131 σσµσ +⋅⋅⋅−=
punchfi
ffrmzm psb
bd
−=
⋅+⋅−
⋅⋅+⋅−=−= µσµσσσσ 1
311 2111
=→ b
heqf
011 lnεσ ( )
+⋅=+=→
sdbh i
eqeqtotaleqf 81ln 021,2 εεεσ
Let us consider radial force equilibrium:
Integration:
Mean stress:
Applying Tresca’s yield criterion:
With:
2/22/2014 ME 303 - Section 03 3.78
Piercing Force by Empirical Methods*
d0
dp
d0
dp
d0
dp
d0
dp
Constrained Piercing: Unconstrained Piercing:
( ) pfmpunch AtoF ⋅⋅= σ53 ( )( ) ( )3ddforAddF
3ddforA3F
p0pfmp0punch
p0pfmpunch
<⋅σ⋅=
≥⋅σ⋅=
[*] Adapted from Schey (1987).
2/22/2014 ME 303 - Section 03 3.79
Example – Backward Impact Extrusion
Example 3:A C15 steel can of 70 mm outer diameter
and 58 mm inner diameter is to be produced by the back extrusion of 70-mm diameter
annealed slugs (initial length = 35mm). The can base is 5 mm thick. If the flow curve of
C15 can be represented by
with an initial flow stress of
Determine the extrusion force. Assume that the coefficient of Coulomb friction is 0.06.
MPa700 24.0eqf εσ ⋅=
MPa2400 =fσ
Summary
2/22/2014 ME 303 - Section 03 3.80
2/22/2014 ME 303 - Section 03 3.81
Remarks on Extrusion and Drawing - Remark 1
For the reducing type extrusion (a.k.a. open-die extrusion) process, the friction force in the container must be omitted since during reducing there is no contact between the billet and the container.
2/22/2014 ME 303 - Section 03 3.82
Remark 2
In the case of backward (impact) extrusion, the frictional force in the container must be omitted.
2/22/2014 ME 303 - Section 03 3.83
Remark 3
α'Dead-metal
Zone
Rod extrusion is usually performed with a die-angle of 2α=180o.
A dead zone is built at the corners of the die leading to an artifical cone of 2α’=90o.
In that case, the friction force in the die shoulder must be taken as Fideal with µ = 0.5 α’= α = 45o
2/22/2014 ME 303 - Section 03 3.84
Remark 4
The computations of the forces for tube drawing process assume that the inner diameter of the tubes do NOT change.
If the inner diameters change, bending force components must be taken into consideration.
Here, bending force occurs!
2/22/2014 ME 303 - Section 03 3.85
Remark 5 In case of forward
impact extrusion of tubes, the interface friction (µpr) at the mandrel-exit can be assumed as 10...12 MPa.
In this region, the material is not plastic anymore!
2/22/2014 ME 303 - Section 03 3.86
Remark 6 The negative sign in the
friction force at the mandrel-die interface must be used in the case of ironing process.
In that case, the force at the base of the tube is obtained.
To compute the pressing force (Fp), this frictional force must be omitted!
2/22/2014 ME 303 - Section 04 4.1
Process Classification
Sheet Forming ProcessesSheet Forming Processes
BendingBending
NeckingNecking
Deep DrawingDeep Drawing
HydroformingHydroforming
ShearingShearing
FlangingFlanging
Stretch FormingStretch Forming
SpinningSpinning
2/22/2014 ME 303 - Section 04 4.2
Formability
Formability in sheet metal working is limited by the following failure modes:
• Orange Peel Effect• Lüder’s Lines or Stretcher-Strain Marks• Localized Necking• Fracture
2/22/2014 ME 303 - Section 04 4.3
Orange Peel Effect
Grainy appearance is observed. Individual grains oriented in different
crystallographic directions deform toslightly differing degrees.
Structural integrity of the material is not effected but aesthetic appearance is not satisfactory.
To solve this problem, one must use finergrained material .
Orange Peel Effect (Cont’d)
2/22/2014 ME 303 - Section 04 4.4
2/22/2014 ME 303 - Section 04 4.5
Localized Necking
Due to excessive tensile deformations, local necks may form. Load carrying capacity is reduced and appearance is poor. High n-value delays onset of necking and high m-value
spreads out necking.
Rupture of the sheet:
If further deformation occurs after localized necking, fractureoccurs.
Post-necking strain is a function of the m-value.
2/22/2014 ME 303 - Section 04 4.6
Formation of Lüder’s Lines or Stretcher-Strain Marks
Structural integrity of the material is not effected but aesthetic appearance is not satisfactory.
Reason for this is yield-point elongation or negative strain rate sensitivity.
To solve this problem, one must use non-aging materials or roller levelling sheet.
Highly localized initial yielding results visible bands (Lüder’s lines). These are successive micro-necking regions.
Stretcher-strain marks are incomplete Lüder’s lines.
yield pointelongation
2/22/2014 ME 303 - Section 04 4.7
Lüder’s Lines
Strecher-Strain Marks
2/22/2014 ME 303 - Section 04 4.8
2/22/2014 ME 303 - Section 04 4.9
Yield Phenomena
Some small solute atoms (such as N and C in iron) can fit into spaces in the crystal lattice of the solvent metal to form interstitial solid solution.
These generally migrate to dislocation sites to form a condensed atmosphere (Cottrell or Suzuki clouds), which in turn pins the dislocations.
A larger stress must be applied before dislocations can break away from the condensed atmosphere of interstitial atoms.
After the dislocations break away, they multiply and move in large groups in the direction of maximum shear stress (45o to applied force) resulting localized yielding (Lüder’s bands).
Strain bands cover entire surface at a low stress (yield-point elongation). After this strain hardening behaviour becomes evident.
2/22/2014 ME 303 - Section 04 4.10
Strain Aging
• Strain aging: is the recondensation of soluteatoms to dislocation sites.
• Strain aged material has reduced ductility and stillthe yield-point elongation phenomena.
2/22/2014 ME 303 - Section 04 4.11
Stepwise or Serrated Yielding
Related to negativestrain-rate sensitivity. Example: substitutional
aluminum alloys. Leads to stretcher-strain
marks.
2/22/2014 ME 303 - Section 04 4.12
Material Properties: Texture (Anisotropy)
During tensile straining slip planes rotate toward thedirection of straining and duringcompressive straining acrossthe direction of straining.
The alignment of the slip planes in preferred directions is named as texture. Textured materials exhibit an anisotropic behavior.
For different crystalinestructures, the slip directions/systems greatly vary:• BCC → 48 slip systems• FCC → 12 slip systems• HCP → 3 slip systems
2/22/2014 ME 303 - Section 04 4.13
Texture (Anisotropy) (2)
wt
r0 r45
r90
rolling direction
111000 ⋅⋅=⋅⋅ wtwt
Volume constancy requires:
10
1
0
1
0
1 =⋅⋅
ww
tt
0=++
εεε wt
or
=
0
1ln
ε (lengthstrain)
0)1ln(ln0
1
0
1
0
1 ==
⋅⋅
ww
ttTaking the
logarithm:
yields where:
=
0
1lntt
tε
=
0
1lnww
wε(thicknessstrain)
(widthstrain)
2/22/2014 ME 303 - Section 04 4.14
Texture (Anisotropy) (3)
Definition of the r-value:strainthickness
strainwidth≡=
t
wrεε
Isotropic material: directions1∀=r Anisotropic material: 1≠r
Normal anisotropy:
Planar anisotropy:
190450 ≠== rrr
90450 rrr ≠≠
( )45900 241 rrrr ⋅++=Mean r-value: Planar
anisotropy:( )45900 2
21 rrrr ⋅−+=∆
2/22/2014 ME 303 - Section 04 4.15
Interpretation of Planar Anisotropy Parameter*
( )45900 221 rrrr ⋅−+=∆
[*] Adapted from Lange (1980).
2/22/2014 ME 303 - Section 04 4.16
Texture (Anisotropy) (5)
REMARK: Annealing does not necessarily restore isotropy.
Material HCP withhigh c/a
HCP withlow c/a
FCC BCC
Plastic strain to produce anisotropy
20 to 30% > 50%
Mean r-value
Very small (typically
0.2)
Very large(but < 6.0)
0.4 to 0.8 0.8 to over 2.0
2/22/2014 ME 303 - Section 04 4.17
Texture (Anisotropy) (6)
High r-valueLow r-value
2/22/2014 ME 303 - Section 04 4.18
Texture (Anisotropy) (6)
2/22/2014 ME 303 - Section 04 4.19
Classification of ShearingShearing is the physical process of cutting sheet and strip intoappropriate shapes and involves the following types:
1. Shearing: Cutting a sheet along a straight line2. Slitting: Cutting a long strip into narrower widths between rotary blades3. Blanking: Cutting a contoured part between a punch and a die in a press4. Punching: Cutting out a contoured shape from a part between a punch
and a die in a punch 5. Notching: Cutting out a part of the sheet edge6. Lancing: Partially cut a hole with no material removed7. Nibbling: Cutting a contoured part by repeated small cuts8. Trimming: Finish cutting of the excess material in drawn products
2/22/2014 ME 303 - Section 04 4.20
Sub-processes in Shearing
Rounding of the edge (roll-over) Extrusion-like plastic deformation
(burnished zone) Crack formation at die Crack formation at punch Meeting of the cracks Pushing out the part over the die land
Shearing Operation
Schematic illustration of shearing with a punch and die, indicating some of the process variables.
Characteristic features of (b) a punched hole and (c) the slug.
2/22/2014 ME 303 - Section 04 4.21
2/22/2014 ME 303 - Section 04 4.22
Shearing Operation (Cont’d)
a) Blanking with optimum clearance.
b) Small clearance (skirt of torn edge).
c) Excessive clearance (burr is produced).
2/22/2014 ME 303 - Section 04 4.23
Shearing Force
a) Crack initiation for soft materials(roughly at half the sheet thickness).
b) Crack initiation for hard materials(earlier than halfthe sheet thickness).
c) Burr produced inblanking reduceselongation attainable in subsequent tensile deformation.
2/22/2014 ME 303 - Section 04 4.24
Shearing Force (2)
( ) ( )areashearingstressshearing ⋅=sP
⋅≈⋅≈
valueaverage an as 0.70materials ductile lessfor 0.65
materials ductilefor 0.85UTSUTSstressshearing 1C
n
annealed enK
≈UTS
Shearing force:
2/22/2014 ME 303 - Section 04 4.25
Types of Shearing
Circular Punching: Straight Cutting (no blade angle):
2/22/2014 ME 303 - Section 04 4.26
Straight Cutting with Angle
hareashearing ⋅=
αtanh
=with:
α
Ps
h
2/22/2014 ME 303 - Section 04 4.27
Shearing Areas
Shearing force can be reduced by having an angle between the edges (rake or shear).
Guillotine Blanking Punching (piercing)
2/22/2014 ME 303 - Section 04 4.28
Shearing Energy
curvent displaceme-forceunder Area =sEShearing energy:
hPCE ss ⋅⋅≈ 2
=materials hardfor 35.0materialssoft for 50.0
2C
Or, approximately:
with
2/22/2014 ME 303 - Section 04 4.29
Finish Blanking & PunchingThe standard shearing process produces a fracture surface which is not perfectly perpendicular to the sheet surface and exhibits some roughness. However, there is a great demand to produce very clean-cut edges, perpendicular to the sheet surface and of a surface finish sufficiently smooth.
Negative-clearanceblanking
Precision (fine)blanking
Counter-blanking
Shaving
2/22/2014 ME 303 - Section 04 4.30
Example
A carbon steel sheet metal of 2 mm thickness is to be cut along its length of 1.5 m with a shearing machine. The annealed UTS of the material is 700 MPa.a) Find the shearing force of a 0o blade angle.b) Find the shearing force of a 10o blade angle.
Solution:a)
tons150N101.47m1.5m002.0MN/m7007.0
UTS
6
21
≈⋅=
⋅⋅⋅=
⋅⋅⋅= hCPs
ton1N 11,11610tanm002.0MN/m7007.0
tanhhUTSChUTSCPo222
11s
≈=⋅⋅=
α⋅⋅⋅=⋅⋅⋅= b)
2/22/2014 ME 303 - Section 04 4.31
Bendingaxis neutral of length =NA
fiberouter of length =OF
hRR b
OFh
b
NAb +
=+
=
2
α
Bending Radius:
Bending Strain:
121
+⋅=
−=
hRbNA
NAOFtensileOF
ε
2/22/2014 ME 303 - Section 04 4.32
Bending (2)
Rb: radius of bending dieαb: bending angleβb: included die angle
Rf: radius of bent partαf: final bend angleβf: final included
part angle
o180=+ bb βαo180=+ ff βα
Note:
2/22/2014 ME 303 - Section 04 4.33
Remarks for Bending
If Rb is generous, the neutral line is in the center.
If Rb is tight, the neutral line shifts towards the compressive side.In this case the centerline is elongated and volume constancy is preserved by thinning! The increased length of the centerline is taken into account for Rb < 2h by assuming that the neutral lineis located at one-third of the sheet thickness.
When the sheet is relatively narrow ( w/h < 8 ), there is also a contraction in width w.
2/22/2014 ME 303 - Section 04 4.34
Limitations in Bending
1) Orange Peel
3) Minimum Bend Radius:
112
1, −≈<
+⋅
= nueng
b
tensileeng e
hR εεFor no necking: ( ) h
eeR n
n
b ⋅−
−>
122
For no fracture: 2.0for 121
<⋅
−> qh
qRb
( ) 2.0for 21
2
2
≥⋅
−−
> qhqq
qRb
(less ductile materials)
(more ductile materials)
2) Crushing on inside surface
2/22/2014 ME 303 - Section 04 4.35
Springback in Bending
b
f
R
Rb
f
h/2
h/2
f
bαα
..
b
f
b
f hR
hRαα ⋅
+
+=
2
2
3
431
⋅⋅+
⋅⋅−≈
EY
hR
EY
hR
RR bb
f
b
For a gentle bend and a high Y/E:
From constancy ofneutral axis length:
< 1 (always)
2/22/2014 ME 303 - Section 04 4.36
Factors Influencing Springback
↑∴↓⇒↑ springbackf
bb
RR
hR
↑∴↓⇒↑ springbackf
b
RRY
↓∴↑⇒↑ springbackf
b
RRE
2/22/2014 ME 303 - Section 04 4.37
Residual Stresses in Bending
a) Bending stresses b) Residual stresses
2/22/2014 ME 303 - Section 04 4.38
Reduction of Springback in Bending
a) Overbending c) Counterpunchb) Indentation
d) Stretching e) Warm/hot bending
2/22/2014 ME 303 - Section 04 4.39
Bending Force
2 UTSb
hPw
⋅ ⋅=
: length of bend: sheet thickness
UTS : ultimate tensile strength: width of the die (see Fig.)
h
w
Bending force for bending:
where
2/22/2014 ME 303 - Section 04 4.40
Deep DrawingConverting a circular sheet metal of diameter d0 into a flat-bottomed cup of depth several times the thickness of the sheet metal through a draw die with the aid of a punch of diameter Dp is named as deep-drawing (cupping).
2/22/2014 ME 303 - Section 04 4.41
Stress States in Deep Drawing*
[*] Adapted from Schey (1987).
2/22/2014 ME 303 - Section 04 4.42
Parameters of Deep Drawing
0
0Reductiond
Dd p−≡
pDd0 RatioDrawing ≡
( )pDmax0d
LDR Ratio DrawingLimiting =
2/22/2014 ME 303 - Section 04 4.43
Common Defects in Deep Drawing
Wrinklingin flange
Wrinkling in wall
Tearing Earing Surfacescratches
2/22/2014 ME 303 - Section 04 4.44
Common Defects (Cont’d)
2/22/2014 ME 303 - Section 04 4.45
Deep Drawing: Modes of Operation
Mode I: Drawing without a blankholder
Wrinkling can be avoided for shallow draws:
2.10 <pD
d Thick blanks give higherdrawing ratios:
2/22/2014 ME 303 - Section 04 4.46
Deep Drawing: Modes of Operation (2)
Mode II: Drawing with a blankholder
Y⋅≈ 015.0 pressurer blankholde
−⋅⋅⋅⋅= 7.0UTS 0
ppd D
dhDP π
Drawing force:
2/22/2014 ME 303 - Section 04 4.47
Forming Limit Diagram*
[*] Adapted from Goodwin (1968).
2/22/2014 ME 303 - Section 04 4.48
FLD (Cont’d)
2/22/2014 ME 303 - Section 04 4.49
Limiting Drawing Ratio (LDR)The draw force is composed of the forces required to• compress the sheet in the flange circumferentially• overcome friction between blank and blank holder and die surfaces• bend and unbend the sheet around draw radius• overcome friction around draw radius
Factors effecting the limiting drawing ratio:• High n strengthens the cup but also increases draw force.• High m strengthens an incipient neck while not affecting the draw force• High r-value resists thinning. Also the yield locus is improved.• Tight punch and die radii impose severe bending. But too large radii cause puckering
(wrinkling between punch and die). Therefore, optimal radii are: R > 4 h for thick (h > 5 mm) and R > 8 h for thin (h < 1 mm) sheet.
• Friction between blank holder, die and flange surfaces adds to the draw force.• Friction on the punch decreases force which is supported by the cup base.• For thin sheet (d0/h > 50) friction forces dominate, so that with increasing d0/h LDR
drops.
2/22/2014 ME 303 - Section 04 4.50
Effect of r-value on LDR*
Planar anisotropy leads to earing.
[*] Adapted from Schey (1987).
2/22/2014 ME 303 - Section 04 4.51
Further Drawing
Redrawing Ironing Reverse redrawing
2/22/2014 ME 303 - Section 05 5.1
Terms and Definitions
Machining: Removal of material in the form of chips from the workpieceby shearing with a sharp tool.
Resultant Cutting Motionin Cylindrical Turning
2/22/2014 ME 303 - Section 05 5.2
Oblique Cutting
Terms and Definitions
Orthogonal Cutting
2/22/2014 ME 303 - Section 05 5.3
Orthogonal Cutting Analogy in Turning (for κ=00)
Terms and Definitions
2/22/2014 ME 303 - Section 05 5.4
Relative Motion between tool and workpiece
Primary motion
Cutting motion
Cutting speed Feed rate
Feed motion
Depth of cut
Secondary motion Depth of cut
adjustment
Terms and Definitions
2/22/2014 ME 303 - Section 05 5.5
Chip Formation
2/22/2014 ME 303 - Section 05 5.6
Continuous Chip
Common in machining ductile materials
2/22/2014 ME 303 - Section 05 5.7
Discontinuous Chip
Machining brittle materials
Small rake angle Large depth of cut Machining ductile
materials at low cutting speed high feed
2/22/2014 ME 303 - Section 05 5.8
Continuous Chip with BUE Occurs in machining
ductile materials with high friction at tool-chip interface
Chip welds to tool face Destroys accuracy and
surface finish Increases tool wear Can be reduced by
decreasing depth of cut increasing cutting speed
2/22/2014 ME 303 - Section 05 5.9
Cutting Forcesac
a0
γn
FF
F
FF
F
F
c
t
n
f
s
ns r
β
φ
γn
β − γn
γn
cuttingtool
cuttingtool
Fc: tangential (main) cutting forceFt: thrust (feed) cutting forceFf: frictional force on rakeFn: normal force on rakeFs: shear force on shear planeFns
: normal force on shear planeFr: resultant force
φ: shear angleγn: normal rake angle
2/22/2014 ME 303 - Section 05 5.10
βµ
γγγγ
φφφφ
γβ
tan= FF =
sinF-cos
cosF + sin
cos sinsinF- cos
FF = )-tan(
F
n
f
tn
ntn
t
c
tn
222222r
2
ncn
cf
tcn
cs
fnnstc
FFFF
FFFFF
FFFFFF
s
s
=
=
+==
+=+=+=
Cutting Forces
2/22/2014 ME 303 - Section 05 5.11
Orthogonal Cutting Geometry
( )
( )
nc
nc
cc
nc
n
cs
rr
raa
aa
aal
γγφ
γφφ
γφφ
sin1costan
cossin
cossin
0
0
0
−=
=
−=
−==
2/22/2014 ME 303 - Section 05 5.12
Theoretical Models
Only two of the simple thin shear-zone models will be covered: Ernst and Merchant’s model Lee and Shaffer’s model
2/22/2014 ME 303 - Section 05 5.13
Common assumptions:• Sharp tool tip ⇒ no rubbing or ploughing between tool andw.p.
• Two dimensional deformation⇒ no side spread
• Uniform stress distribution on shear plane
• Resultant force on shear planeequal and opposite to res.force at chip-tool interface.
Ernst and Merchant Model
2/22/2014 ME 303 - Section 05 5.14
Shear angle would take up such a value as to reduce the work done in cutting to a minimum.
For given cutting conditions, work done in cutting is proportional to Fc, it is necessary to develop an expression for Fc in terms of φand then to obtain the value of φ for which Fc is a minimum:
( )
φττ
γβφ
sin
cos
cssss
nrs
AAF
FF
==
−+=
Main Assumption (EM Model)
2/22/2014 ME 303 - Section 05 5.15
angle rake normal friction of anglemean
chipuncut theof area sectional-cross Aplaneshear theof area
planeshear on the material work theofstrength shear
c
=====
n
s
s
A
γβ
τ
( )( )
( )( )n
ncsc
nrc
n
csr
AF
FF
AF
γβφγβ
φτ
γβγβφφ
τ
−+−
=∴
−=−+
=
coscos
sin
coscos
1sin
22
F minimize to0 c
πγβφ
φ
=−+⇒
=∂∂
n
cF
Shear Angle (EM Model)
2/22/2014 ME 303 - Section 05 5.16
Strain
Stre
ss
Lee and Shaffer’s Model
Work material is rigid plastic. Elastic strain is negligible.
Behaviour of work material is independent of the rate of deformation.
Temperature effects are neglected.
Inertia effects are neglected. Uniform stress distribution at the
chip-tool interface.
2/22/2014 ME 303 - Section 05 5.17
Main Assumptions (LS Model) A slip-line field is formed in the triangular plastic
zone extending from the shear plane to the interface between the tool and the chip where no deformation takes place except for the transmission of forces from tool-chip interface to shear plane and for the material being stressed to its yield point.
All the deformation takes place in the plane (Shear Plane) extending from the tool cutting edge to the point of intersection of the free surfaces of the work and the chip.
2/22/2014 ME 303 - Section 05 5.18
Assumptions (Cont’d)
Maximum shear stress throughout the zone is τs, shear stress on the shear plane and two directions of this max. shear stress are indicated by two orthogonal sets of lines (slip lines).
Top surface of the triangular plastic zone then becomes a free surface across which no stresses are transmitted. Therefore between this surface and the max. shear stress plane (shear plane) there is an angle of π/4.
Principal stresses act on the chip-tool interface (secondary def. Zone) at angles β and β+π/2.
Directions of max. shear stress lie at π /4 to the dir. of principal stress
2/22/2014 ME 303 - Section 05 5.19
Lee-Shaffer Theory (Cont’d)
2ϕ = 90o
σ1σ2
τmax
σ
τ
(Stress) Free Surface
Shear Plane
Chip-T
ool I
nterfa
ceβ
Φ
45o
Slip Lines
σ1σ1
σ2
σ2
γn
90o − γn
45o + β
Fr
ϕ = 45o
Hence, Φ + β − γn = π/4
2/22/2014 ME 303 - Section 05 5.20
Metal Removal Rate
w = v.f.dwhere, w: metal removal ratev: cutting speed f: feed rate cutting conditionsd: depth of cutAc: f.d = undeformed chip cross sectional area
a
d
c
f
ac
f
κ
ac = f cosκ κ: Side Cutting Edge Angle
2/22/2014 ME 303 - Section 05 5.21
Power Requirement
2/22/2014 ME 303 - Section 05 5.22
Power Requirement (Cont’d)
2/22/2014 ME 303 - Section 05 5.23
Power RequirementPower required at the spindle:P = Fc v = E w where Fc = cutting force
v = cutting speedw = chip removal rate = v f dE = spec. cut. energy
Fc v = E v f d E = (Fc/fd)E is a function of material property and the undeformed chip thickness
If Eo is the specific cutting energy for an undeformed chip thickness of 1 mm, then the specific cutting energy for a chip thickness ac :
E = Eo . (ac )− α => so ac EE = Power / Material Removal Rate
E = (Fc.v) / (v.f.d) where ac = f . cos κ
E = (Fc . cos κ) / (ac . d)
2/22/2014 ME 303 - Section 05 5.24
log E
log ac
log E1
logE2
log a c1 log a c2
α
-α = ln (E1/E2) / ln (ac1/ac2)
E1/E2 = (ac1/ac2) -α
α ranges between 0.2 - 0.4
Power Requirement
2/22/2014 ME 303 - Section 05 5.25
Power Requirement
2/22/2014 ME 303 - Section 05 5.26
EXAMPLE:
E = 3.8 (W-sec/mm3) v = 36 (m/min) f = 0.25 (mm/rev)Power available at the spindle = 5 hpFind maximum metal removal rate and corresponding depth of cut.(1 hp = 746 W)
[ ]
[ ]
[ ]
w = PowerE = 5 x 746
3.8
w = 982
F = PowerV
F = 5 x 746
3660
F = 6216
= wV f =
982
0.25
d = 6.55
WW
mmsecmm
sec
N - msec
msec
Nmmsec
mmsec mm
mm
3
3
3
d 3660 103
Power Requirement
2/22/2014 ME 303 - Section 05 5.27
Specific Cutting Energy
2/22/2014 ME 303 - Section 05 5.28
Problem 1
In an orthogonal cutting test on mild steel, the following resultswere obtained:
ac= 0.25 mm Fc = 900Na0= 0.75 mm Ft = 450Nd = 2.5 mm γn = 100
a) Calculate the mean angle of friction on the tool rakeb) Calculate the cutting ratioc) Calculate the shear angle
2/22/2014 ME 303 - Section 05 5.29
Problem 2
For an orthogonal turning operation it was found that power consumption of lathe when idle = 325 Wpower consumption of lathe when cutting = 2580 W
For the following conditions: spindle speed, N = 124 rpm; cutting speed, v = 24.5 m/mindepth of cut, d = 3.8 mm; feed rate, f = 0.2 mm/rev
Find:a) specific cutting energy of the work material,b) torque at the spindle,c) cutting force,d) specific cutting energy for 1 mm undeformed chip thickness
assuming α = 0.4.
2/22/2014 ME 303 - Section 05 5.30
Problem 3
φ25 mm holes will be drilled on a steel workpiece, having a hardness of Rc=45 {specific cutting energy, E = 77 W/(cm3/min)} using an HSS twist drill at the following conditions:
cutting speed, v = 24.5 m/minfeed rate, f = 0.2mm/rev
Find:a) the motor power if the efficiency of the transmission is 85%,b) torque required.
2/22/2014 ME 303 - Section 05 5.31
Tools must be so selected that they can cut properly and efficientlyunder the selected cutting conditions which may lead to a harshcutting environment due to high cutting temperatures and high cutting pressures. In selection two principal aspects must be considered:
a) tool geometry b) tool material.The geometry ⇒ the optimum performance for the given
tool material and the operation. The tool material ⇒ highest possible strength, resistance and durability
against forces, temperatures and wearing action during machining
Cutting Tools
2/22/2014 ME 303 - Section 05 5.32
Cutting Tools - Geometry
2/22/2014 ME 303 - Section 05 5.33
Side Rake Angle (gs):
great influence on chip formation
less deformation of removed layer of w.p. as increased ⇒ less resistance to chip formation
lower cutting forceslower power consumption
less heat transfer area on rake substantial increase ⇒ weakening of cutting edge
catastrophic tool failure
Cutting Tools - Geometry
2/22/2014 ME 303 - Section 05 5.34
Side Rake Angle (γs)
+-Side rake angle0
Side rake angle0 +-
Tool
life H3
H2H1
H1>H2>H3g1<g2<g3
Increased w.phardness
Tool
life
Side rake angleg1 g2 g3
Effect of Side Rake Angle on Tool Performance
2/22/2014 ME 303 - Section 05 5.35
Side Clearance Angle (SRA)less rubbing between flank and w.p.
• as increased ⇒ reduced heat generation at tertiary def. zonereduced flank wear
• excessive clearance ⇒ reduction in strength of wedge• larger for soft and ductile w.p., smaller for hard and brittle w.p.
Tool
life
Side clearance anglea1 a2 a3
f1
f2
f3
f1> f2>f3a1<a2<a3
2/22/2014 ME 303 - Section 05 5.36
Side Cutting Edge Angle (κ)
• Determines thickness and width of uncut chip layer:ac= f cos kaw= d/cos k
Hence, for the given feed and depth of cut, an increase in k causesa decrease in chip thickness and an increase in chip width.
• As increased: i) interface temp. θ decreases since ii) distribution of heat generated over a longer cutting
edge, hence longer tool life and higher permissible feed
ca∝θ
2/22/2014 ME 303 - Section 05 5.37
Side Cutting Edge Angle (Cont’d)iii) increase in radial force component ⇒ high lateral
deflections, poor dim. accuracy, severe vibrationand chatter especially in long and slender w.p.
iv) better surface finish since Rmax= f/(tank+cotke)where Rmax= max. surf. rough., f = feed rate
k = side cut. edge angle, ke= end cut. edge angle
Am
plitu
de
of c
hatte
r
k
Cut
ting
spee
d fo
r a
fixed
tool
life
k∼300
HSS
Cemented carbide
2/22/2014 ME 303 - Section 05 5.38
End Cutting Edge Angle (κe)• as increased: i) less cutting at the end cutting edge, hence less
friction between end flank and finished w.p. surface and higher tool life
ii) lower surface quality• substantial increase ⇒ decrease in tool included angle
⇒ poorer heat transfer from the nose ⇒ shorter tool life
50-100 ke
Cut
ting
spee
d fo
r a
fixed
tool
life
2/22/2014 ME 303 - Section 05 5.39
Back Rake Angle (γb)
γb=0γb=(+) γb=(-)
Back Rake Angle basically affects chip flow.
2/22/2014 ME 303 - Section 05 5.40
Back Rake Angle (Cont’d)It also affects specific cutting energy and chip-tool interface temperature.
Note that for machining hard materials with cemented carbides, γb ≅ 200
2/22/2014 ME 303 - Section 05 5.41
Cutting Tool Materials
Requirements:
Hot hardness Wear resistance Toughness Low friction Favorable cost
Classification:
Carbon Tool Steels Medium Alloy Steels High Speed Steels Cast Cobalt Based Alloys Cemented Carbides Ceramics and Ultra-hard Materials
2/22/2014 ME 303 - Section 05 5.42
Carbon Tool Steels
Martensite based / tempered & hardened. Operate at low cutting speeds. Poor hot hardness (max. temp. 200oC) Typical composition:
0.8 - 4.3 % C 0.1 - 0.4 % Si 0.1 - 0.4 % Mn
Hardness wear resistance at narrow temperature increase with increased C% up to ∼0.8 - 1%.
Generally used for cutting wood and plastics.
2/22/2014 ME 303 - Section 05 5.43
Medium Alloy Steels
Improved hardenability due to small additions of Cr & Mo.
Addition of up to 4% of W improves wear resistance.
Poor hot hardness is NOT satisfactory for high speed turning or milling.
2/22/2014 ME 303 - Section 05 5.44
High Speed Steels (HSS) First introduced in 1900 by Taylor and White. Superior hot hardness and wear resistance (max. temp 600oC). Typical compositions of various High Speed Tool Steels:
American Institute for Steel Industry (AISI) classifies them based on composition: Tungsten based: T1 – T9, T15 Molybdenum based: M1 – M10 Molybdenum + Cobalt based: M30 – M46
Designation Type W Cr V Mo C FeT-1 W 18 4 1 - 0.7 Bal.M-1 Mo 1.5 4 1 8.5 0.8 Bal. M-2 W-Mo 6 4 2 5 0.8 Bal.
2/22/2014 ME 303 - Section 05 5.45
HSS (Cont’d)
Tungsten and Molybdenum behave in the same general way, however Mo is twice as effective as Wto improve hot hardness. They tend to increase hot hardness by forming strong
complex carbides. Addition of Co to the structure in 4, 8 or 12% further
improves the hot hardness. E.g. T-4 has the same composition as T-1 but 4% Co is added. It increases hot hardness by going into the solution in ferrite
matrix increasing the recrystallisation temperature.
2/22/2014 ME 303 - Section 05 5.46
High Speed Steels (Cont’d)
Vanadium inhibits grain growth at high temperatures required in heat treatment and increases the wear resistance. Vanadium steels are very difficult to grind and
used machining highly abrasive stock. Co & Mo have a tendency to promote
decarburization. Such steels should be ground to a greater depth in finishing to remove decarburized layer.
2/22/2014 ME 303 - Section 05 5.47
High Speed Steels (Cont’d)
HSS tools are manufactured in wide range of sizes and shapes but mostly in the form of solid tools. Solid tools are then ground to required geometry.
There is tendency of clamping, brazing or welding HSS tool to a cheaper low alloy or carbon steel body.
Recent developments are Powder metal HSS Coated HSS
2/22/2014 ME 303 - Section 05 5.48
Powder Metal HSS
Powder Metal HSS has the following advantages: Superior structure. Free from segregation. Ensure good and nearly uniform properties in
all directions. Lower incidence of premature failure. Ability to produce steels with higher alloy
content.
2/22/2014 ME 303 - Section 05 5.49
Coated HSS
Coated HSS has the following features: Tools can be coated with thin layers of refractory metal
carbide or nitride using physical or chemical vapor deposition techniques (less than 10µm thickness).
Titanium nitrate (TiN) coating has a distinguishing gold colorwhereas titanium carbide (TiC) has a black color.
Life may be as high as 300% or 400% of the life of uncoated tools
Built-up edge formation is nearly eliminated. Regrinding must be followed by a careful polishing and
recoating.
2/22/2014 ME 303 - Section 05 5.50
Cobalt Based Alloys Known as stellites, composed of a number of nonferrous alloys
high in cobalt. Representative composition:
Co: 40-50% Cr: 27-32% W: 14-29% C: 2-4%
Cannot be heat treated and are used as cast. Ground to its final shape (single-edge tools or saw blades)
If compared to HSS, stellites can retain hardness at a much higher temperature. They can be used at higher cutting speeds than HSS tools (25% ...
200%)
2/22/2014 ME 303 - Section 05 5.51
Cobalt Based Alloys (Cont’d)
Not as hard as carbon tool steels at room temperature but retain hardness to much higher temperatures.
Used when tools are required to be used for a wide range of cutting speeds, i.e. facing very large diameter parts at constant rotational speed.
Not widely used. Expensive due to shortages on strategic materials (Co, W,
Cr, etc.) Brittle, hence need proper support.
2/22/2014 ME 303 - Section 05 5.52
Cemented Carbides
Became popular during WWII. Increased cutting speed by four- or fivefolds (if compared to HSS!).
Produced by Powder Metallurgy Mixtures of transition metal carbides and metals in which the
metal, usually the cobalt, binds carbides together. Contains carbides of tungsten, titanium and tantalum at least
80% by volume. Strongly metalic in character
Good electrical and thermal conductivities Metalic appearance
2/22/2014 ME 303 - Section 05 5.53
Cemented Carbides (Cont’d) Characteristics of Cemented Carbide Tools:
High hardness over a wide range of temperatures. Maintain hardness up to 1200oC.
Very stiff (Young's modulus three times that of steel). Very brittle No plastic flow even at very high stresses (up to 3.5
GPa). Low thermal expansion. Relatively high thermal conductivity (especially K-
Grade). Strong tendency to form pressure welds at low cutting
speeds.
2/22/2014 ME 303 - Section 05 5.54
Cemented Carbides (Cont’d)
Carbide grades are classified according to codes developed by various organizations: C1 – C2 grades (AISI) or K grade (ISO): used for machining cast
iron (CI) C4 – C8 grades or P grade (ISO): used for machining steel. M-grade: general purpose
C1-C3 grades contain only tungsten carbide and cobalt. Effective in machining CI and certain abrasive nonferrous alloys. Not suitable for steels due to their high affinity towards that alloy Addition of titanium carbide and/or tantalum carbide lowers that
affinity, hence making steel cutting possible (steel P-grades).
2/22/2014 ME 303 - Section 05 5.55
Carbide Inserts
2/22/2014 ME 303 - Section 05 5.56
TiC Based Tools
Commonly used in Automotive Industry Employed as throw-away tool tips.
Difficult to braze. Bonding metal is Ni instead of Co (in 10-20%). Resistant to diffusion wear in steel cutting. Cutting speed is 2 ... 5 times that of HSS. Can be used at higher cutting speeds than the
conventional WC based tools. Low toughness. Lack reliability and consistency of performance
2/22/2014 ME 303 - Section 05 5.57
Coated Tools About 10 µm thick coatings of TiC,
Al203, TiN can be coated on the surface of a tough WC substrate using physical or chemical vapourdeposition technique. First layer: TiC (Strength and wear
resistance) Second layer: Al203 (Chemical
stability at high temp and resistance to abrasive wear)
Final layer: TiN (Low coefficient of friction)
Reduce tool wear. Increase in cutting speed is possible.
2/22/2014 ME 303 - Section 05 5.58
Ceramics
Basic material is Alumina (Al203) and may contain MgO, TiO and other additions to promote densification and grain size stability.
Major advantages are Retention of hardness and compressive strength
to higher temperatures than with carbides. Practically inert to steel up to melting point.
Lower toughness and tensile strength than that of carbides.
Non-metallic in character, hence electrical insulator with poor thermal conductivity.
2/22/2014 ME 303 - Section 05 5.59
Ceramics (Cont’d)
Can be used to cut steels at much higher speeds than possible with carbides (a cutting speed of 600-750m/min at a feed of 0.25 mm without excessive wear when cutting CI and steels).
Negative rake throw-away tools are used. Main usage in cutting grey CI with very good surface
finish to eliminate subsequent grinding operation (e.g. clutch facings and brakes).
Alumina containing up to 30% TiC is suitable for turning and milling operations on CI continuous machining of steel.
Not suitable for machining Al-alloys.
2/22/2014 ME 303 - Section 05 5.60
Sialons
Si-Al-O-N's are silicon nitrade-based materials with aluminum and oxygen additions.
Tougher than alumina. In interrupted cutting, higher feeds and speeds without
fracture are possible than those attainable with alumina ceramics.
In machining aerospace alloys, Ni-based gas turbine discs are faced with sialon tips at 180-300 m/min at a feed of 0.2 mm/rev, whereas carbide tools can be used at only 60 m/min.
Higher thermal conductivity. Lower coefficient of expansion. Increased resistance to shock and thermal fatigue compared to
alumina.
2/22/2014 ME 303 - Section 05 5.61
Cubic Boron Nitrade
Next hardest substance to diamonds. Two commercially available products: BZN (GEC) and
Amborite (DeBeers) different in character. BZN-laminated tool tip, consolidated CBN (0.5 mm
thick) on cemented WC-Co substrate. Amborite-entirely of consolidated CBN. In early stages of development. 20-23 times as costly as cemented carbide tools. Advantage over diamonds, its stability at high
temperatures (over 1000oC) in air or in contact with iron and other metals.
2/22/2014 ME 303 - Section 05 5.62
Cubic Boron Nitrade (Cont’d)
CBN can be most economically used in machining hardened steel (60-68 Rc) and chilled CI at speeds (45-60 m/min) and feeds of 0.2-0.4 mm/rev.
Long tool life so that rolls may be machined to a dimensional tolerance and surface finish which eliminate grinding operation.
High hot hardness value. Excellent abrasive resistance and resistance to react with
ferrous materials. Good toughness when used with negative rake and
chamfers can be used for interrupted cutting of hardened steel.
2/22/2014 ME 303 - Section 05 5.63
Diamonds
Hardest of all materials. Used in operations where other tools cannot perform effectively. Have much lower wear rate and longer tool life than carbides and
ceramics where abrasion is dominant wear mechanism. Single crystal natural diamonds are used to produce surfaces of
extremely high accuracy and finish. (e.g. optical instruments and gold jewellery).
Deficient in toughness, easy chipping of cutting edges. Polycrystalline diamond tools are made with a layer of consolidated
synthetic diamonds (0.5 - 1 mm thick) bonded on cemented carbide substrates (2 - 2.5 mm thick).
2/22/2014 ME 303 - Section 05 5.64
Diamonds (Cont’d)
Cost 20-30 times the equivalent carbide tool. Maintain exceptional wear resistance. Recommended for machining aluminium alloys (speeds can be
over 500 m/min with long life). Also used in machining copper and copper alloys and cemented
carbides in pre-sintered condition. Not used for high speed machining of steel and nickel because of
excessive wear. Diamond does not revert to graphitic form in the absence of air at
temperatures below 1500oC. In contact with iron, graphitization begins just over 730oC.
2/22/2014 ME 303 - Section 05 5.65
Cutting Speed Comparison Carbon Tool Steel: 1 (unit) HSS: 1.5 ... 2 Cast Cobalt: 2 Carbides
WC + Co: 4 Ti/W 6 ... 7
Ceramics: 8 Coated Carbide
TiC/TiN: 12 Al2O3: 16 ... 17 TiC/Al2O3/TiN: 20
2/22/2014 ME 303 - Section 05 5.66
Hardness of Tool Materials
2/22/2014 ME 303 - Section 05 5.67
Tool Life - Definitions
Criterion to mark the end point for the tool’s life is not necessarily to correspond to a state that the tool is unable to cut the workpiece, but that it is merely unsatisfactory for the purpose.
Such tools which have consumed their lives maybe Resharpened (if applicable) and used again Used on less restrictive operations Disposed off
A tool that no longer performs the desired function is said tohave reached the end of its useful life.
2/22/2014 ME 303 - Section 05 5.68
Tool Wear
Failure of the tool to perform a desired function may be due to Catastrophic failure
Fracture Temperature
Gradual (progressive) failure Forms of wear
Crater wear Flank wear Workpiece
Tool
Flank
RakeChip
Craterwear
Flank wear
2/22/2014 ME 303 - Section 05 5.69
Tool Wear Mechanisms
2/22/2014 ME 303 - Section 05 5.70
Tool Wear (2)
VB
250200150100500
0.8
0.70.60.5
0.4
0.20.1
0.3
C
D
B
Rapidbreakdown
Uniform wear rate
Initial breakdown
Wid
th o
f fla
nk w
ear l
and
VB
, mm
Cutting time, s
2/22/2014 ME 303 - Section 05 5.71
Tool Wear (3)
KT
KBKM
A
A
+rε
CraterRake
Section A-AZone Zone Zone
C B N
VNVC
bb/4rε
VBVBmax
Wear notch
2/22/2014 ME 303 - Section 05 5.72
Tool Life Criteria
A tool-life criterion is defined as a predetermined threshold value of a tool-wear measure or the occurrence of a phenomenon.
For HSS or Ceramic Tools Catastrophic failure, or VB = 0.3 mm (uniformly worn flank in zone B) or VBmax = 0.6 mm (irregularly worn flank)
VB = 0.75 mm (HSS – finish)VBmax = 1.5 mm (HSS – rough)
ISO
Schey’s recommendation
2/22/2014 ME 303 - Section 05 5.73
Tool Life Criteria (2)
For Sintered Carbide Tools: VB = 0.3 mm (uniformly worn flank in zone B), or VBmax = 0.6 mm (irregularly worn flank), or KT = 0.06 + 0.3f mm (f : feed)
VB = 0.4 mm (uniformly worn flank in zone B), orVBmax = 0.6 mm (irregularly worn flank)
ISO
Schey’s recommendation
2/22/2014 ME 303 - Section 05 5.74
Tool Life
Tool life is defined as the cutting time required to reach a tool-life criterion.
Most important factor affecting tool life (for a given work material-tool combination) is the cutting speed.
2/22/2014 ME 303 - Section 05 5.75
Taylor’s Tool Life Model
Frederick W. Taylor(1856-1915)
2/22/2014 ME 303 - Section 05 5.76
Taylor’s Equation
vTn = C v : cutting speed [m/s] T: tool life [min] C: Constant [m/s] (cutting speed for 1 min. tool life)
Typical values of n are 0.1 HSS 0.25 Cemented Carbides 0.3 Coated Carbides 0.4 Ceramics
For an approximate value of C: take recommended value and multiply it by 1.75 (HSS) or 3.5 (Carbides).
2/22/2014 ME 303 - Section 05 5.77
Extended Taylor’s Equation
21 2
11
11nnn
dfvKT
nnn<<=
For HSS: 0.1 < 0.17 < 0.25
2/22/2014 ME 303 - Section 05 5.78
Machinability
Machinability is defined as the relative ease and economy with which a material (usually a metal) can be machined using appropriate tooling and cutting conditions.
There are various criteria to evaluate machinability: Tool life Forces and power Surface finish Ease of chip disposal
2/22/2014 ME 303 - Section 05 5.79
Material Properties Affecting Machinability
Ductility Strain hardening ability Strength Bonding tendency
between tool and workpiece
Thermal conductivity Melting temperature
Abrassive particle content of workpiece
Existence of second-phase particles (lead, sulphur, phosphorus, etc.) which are soft or softened at high temperatures (favors machinability!)
2/22/2014 ME 303 - Section 05 5.80
Measure of Machinability
Tool life: Maximum cutting speed to obtaina given tool life (i.e. vT)
Tool wear: Time required to develop a givenamount of flank wear.
Surface finish: Surface quality obtained at standard speedand feed.
Most popular measures of performance in machinability testing are
Machinability performance of a test material is measured relative to that of a base (standard) material.
Cutting Fluids
Advantages of using cutting fluids: Increase in tool life Improvement of surface finish Reduction in cutting forces and power
consumption Washing chips free from cutting region Reduction in thermal distortion of workpiece Protection of finished surface from corrosion
2/22/2014 ME 303 - Section 05 5.81
Cutting Fluids (2)
Cutting fluids affect cutting process through their two basic actions: Lubrication
Reduces tool friction
Cooling
2/22/2014 ME 303 - Section 05 5.82
1) Lubrication
Forms boundary lubrication and a layer of soft chemical compound with chip and workpiece at very low cutting speeds: Built-up edge (BUE) formation is reduced. Reduction in coefficient of friction, hence increase in
shear angle and reduction in cutting forces. Improved surface finish. Increase in tool life.
2/22/2014 ME 303 - Section 05 5.83
2) Cooling
Reduces temperature in cutting area by removing heat: Tool wear is reduced and tool life is increased; tool
material retains its hardness at reduced temperature: Tool gains more resistance to abrasion. Diffusion wear rate is reduced due to reduced temperature
However, shear flow stress may increase, resulting reduced tool life.
Reduced thermal expansion and distortion of workpiece (especially in grinding)
2/22/2014 ME 303 - Section 05 5.84
Secondary Function
An important subsidiary function of a cutting fluid is to wash away swarf. Third-body abrasion is avoided.
Or at least reduced!
Surface finish is improved.
2/22/2014 ME 303 - Section 05 5.85
Effects of Cutting Fluids
2/22/2014 ME 303 - Section 05 5.86
ConditionCoefficient of Friction
Heat from friction (kJ/kg of metal
removed)
Heat from deformation (kJ/kg of metal removed)
Dry 1.0 102.3 222.7
Cutting fluid
0.9 93.9 210.4
% decrease 10 19.1 12.8
Effects of Cutting Fluids (2)
2/22/2014 ME 303 - Section 05 5.87
Type of Cutting Fluid Tool Life (min) Power (kW)
Dry 12 1.36
Sulphurised oil 27 1.3
Sulphurised and chlorinated oil 33 1.35
Material cut: Alloy steelShear strength: 690 MPa Effective rake: 15o
Types of Cutting Fluids
Cutting fluids are mainly required to carry away heat and swarf are generally based on water: Oils and emulsifying agents are added:
to inhibit rusting, to lubricate, to a certain extent.
Concentrations: Grinding (oil/water) ≅ 1/40 ... 1/50 Turning and milling ≅ 1/20
2/22/2014 ME 303 - Section 05 5.88
Types of Cutting Fluids (2) Straight cutting oils are blended from two types of oils:
Fatty oils: Organic oils of animal or vegetable origin.
Mineral oils: Paraffin and other petroleum oils.
2/22/2014 ME 303 - Section 05 5.89
Fatty Oils:• Have good lubricating properties• Promote good finishes• Less stable than mineral oils
• May decompose in prolonged use.
Mineral Oils:• Cheaper• More stable
Sulphurized Oil:• Lubricates via chemicallycombined sulphur:
• Prevents pressure welding ofchip to tool (BUE)
Types of Cutting Fluids (3)
Straight cutting oils have wide application for two major reasons:
Employed in automatic machines because water-base coolants may find their way into headstock, contaminate lubricating oil and cause serious deterioration of mechanism.
Promote a superior surface finish in gear cutting, honing, threading and broaching. Paraffin is sometimes used on aluminum alloys instead of soluble oils because of its wetting property.
Greases are more convenient than oils in tapping and reaming operations
2/22/2014 ME 303 - Section 05 5.90
2/22/2014 ME 303 - Section 05 5.91
Economics of Machining
Production cost and production rate are vitally important for a manufacturer.
Increasing production rate means producing more from the available resources.
Decreasing production cost means less expenditure for thesame volume of production.
If the conditions are so selected to maximize the production rate and minimize the production cost, his profit can be maximized.
It is however not possible to find a common set ofmanufacturing conditions to satisfy this.
2/22/2014 ME 303 - Section 05 5.92
Economics of Machining (2)
A common practice is to treat each case separately and find the corresponding conditions, then to make a compromise in between.
Production of a component involves several machining operations using a variety of machine tools.
Only one operation to be performed on one machine will be considered and the conditions leading to min. production cost and max. production rate (min. prod. time) will be evaluated.
2/22/2014 ME 303 - Section 05 5.93
Assumptions
One operation-one machine, hence prep times between operations and transportation between machines are ignored.
Components are ready at the side of the machine before operation and to be stacked at the side of the machine after operation.
Appropriate tool and cutting fluid have been chosen. Depth of cut has been selected at maximum value
(limited by the total stock to be removed, chatter vibrations, cutting force, etc.)
2/22/2014 ME 303 - Section 05 5.94
Optimization Criteria
Unit production time: average time taken to produce one component.
Unit production cost: total average cost of producing one component: As cutting speed and feed rate are increased, both tend to decrease
due to an increase in metal removal rate. At the same time, they may tend to increase due to increased
frequency of tool changes (increasing tool wear rate). Optimum conditions do exist where the measure of the selected
criterion is minimum. Profit rate: it is very difficult to attribute revenue to individual
operations and profit rate cannot be expressed as an explicit function of cutting conditions.
2/22/2014 ME 303 - Section 05 5.95
Choice of Cutting Conditions
Effects of depth of cut, feed rate and cutting speed on metal removal rate are the same.
Cutting speed has the highest adverse effect on tool life amongst the cutting conditions.
Common practice is to select depth of cut at its maximum value, then choose feed rate as high as possible considering the limitations such as the available feeds on the machine, surface finish requirement, force level which the cutting edge can withstand, etc.
Cutting speed will be selected as the one to optimize the selected criterion.
2/22/2014 ME 303 - Section 05 5.96
Unit Production Time
Unit production time = {time for machining} + {time for tooling}
tpr = tf + ttp
tf = tl + tc ttp = nt tch
where tl = loading and unloading time/piecetc= actual machining time /piecetch= tool changing time (time/edge)nt = fraction of tool life to cut one piece (edge/piece)
2/22/2014 ME 303 - Section 05 5.97
Unit Production Time
tpr = tl + tc + nt tch
For turning operation:
wVtc =
=
=
=
piece time timemachining
timevolume rate removal chip
piecevolume removed be tomaterial of volume
ct
w
V
2/22/2014 ME 303 - Section 05 5.98
Actual Machining Time
di d0
l
( )
speed cuttingfeed
2cut ofdepth
..
.4
0
220
==
−==
=
−=
vf
ddd
vfdw
lddV
i
iπ
2/22/2014 ME 303 - Section 05 5.99
Actual Machining Time (2)
( )( )( )
( )
vBtl
fddB
vf
ldd
vfdd
lddddt
ci
i
i
iic
=⇒⋅+π
=
⋅
⋅+π
=⋅⋅
−
⋅+−π
=
if 0
0
0
00
2
2
2
4
2/22/2014 ME 303 - Section 05 5.100
Actual Machining Time (3)
=
=
=
=
piecetime timemachiningt
edgetime life toolt
edge cutting one of life during machined be topieces of no.
edgepieces
c
t
ct
N
ttN
2/22/2014 ME 303 - Section 05 5.101
Unit Production TimeTaylor’s Equation:
( )
( )
nr
nn
chlpr
r
nn
nt
vCt
vBc
Nt
r
nn
C
Bt
vtvBtt
tv
C
Bntt
n
tvCtCvt
nrt
1
1
1
11
1
1
:speed cutting theoffunction a as timeproductionUnit
1
⋅⋅+
+=
=⇒===
⋅
=⇒=
−
−
2/22/2014 ME 303 - Section 05 5.102
Unit Production Cost
Unit production cost = {cost of machining} + {cost of tooling}
Cpr = Cf + Ctp [cost/piece]
Cf = ( R0 + Rm ) ( tl + tc )
cost of machiningper unit time(cost of time)
time for machining
2/22/2014 ME 303 - Section 05 5.103
Unit Production Cost (2)
R0 = Operator’s wage rate + overheads [money paid/time]
Rm= Machine’s cost (depreciation) rate + overheads [cost/time]
Ctp = nt [ tch ( R0 + Rm ) + Ct ]
tool changing cost
toolcost
Ct = cost of each sharp cutting edge [cost/edge]tch = tool changing time [time/edge]
2/22/2014 ME 303 - Section 05 5.104
Cost Parameters
rate mach. of % as rate overhead machineon timeamortizati
machine ofcost R
rate wagesop.' of % as rate overheadlabor rate wagesoperator'R
edge cuttinggrinding ofcost usable edges cutting of no.
toolofcost C
m
0
t
+=
+=
+=
Cost of time: R0 + Rm , usually expressed in cost/hour
2/22/2014 ME 303 - Section 05 5.105
Unit Production Cost
( )( ) ( )[ ]tm0chtclm0pr CRRtnttRRC +++++=
But from the earlier analysis:
vBtc =
( )
r
nn
nt t
v
C
Bn−
=1
1and
2/22/2014 ME 303 - Section 05 5.106
Unit Production Cost (2)
Substituting these in the main equation, unit production cost can be expressed as a function of the cutting speed:
( ) ( )[ ]nr
nn
tmchlmpr
C
Bt
vCRRtvBtRRC 1
1
001
⋅⋅+++
++=
−
2/22/2014 ME 303 - Section 05 5.107
Optimum Conditions
In order to get the optimum conditions (i.e. the optimum cutting speed), one needs to optimize the objective function according to the selected criterion.
It is necessary to find the point where the slope of the objective function is zero, (i.e. the partial derivative of the objective function w.r.t. the cutting speed) is zero.
Solving for the cutting speed to satisfy that condition will give the optimum cutting speed for the selected criterion.
2/22/2014 ME 303 - Section 05 5.108
Minimization of Unit Production Cost
Solving for the cutting speed gives the cutting speed for minimum cost, vmc and the corresponding tool life tmc:
0=∂
∂
vC pr
n
tmoch
rmomc CRRt
tRRn
nCv
++
+−
=])([
)(1
+
+−
=mo
tchmc RR
Ctn
nt 1
n
mcrmc v
Ctt/1
=
2/22/2014 ME 303 - Section 05 5.109
Maximization of Production Rate
Solving for the cutting speed gives the cutting speed for maximum production rate, vmp and the corresponding tool life tmp:
0=∂
∂
vt pr
n
ch
rmp t
tn
nCv
⋅
−=
1
chmp tn
nt −=
1
n
mprmp v
Ctt/1
=
2/22/2014 ME 303 - Section 05 5.110
Machining at High Efficiency
unit prod. cost
production rate
Uni
t pro
duct
ion
cost
Prod
uctio
n ra
te
vmc vmp
high efficiency range
Speed
2/22/2014 ME 303 - Section 05 5.111
Problem 4
A coated carbide cutting tool has 60 min tool life when cutting mild steel at 1.2 m/s. If Taylor’s tool-life exponent is 0.25.
a) find Taylor’s tool-life constant
b) find the tool life if the cutting speed is doubled.
2/22/2014 ME 303 - Section 05 5.112
Problem 5A mill roll that is 2 m long and 50 cm in diameter is to be rough turned to a diameter of 49.2 cm. Work material is alloy steel with a hardness of 300 BHN. The labour and the machine costs including the overheads are$20.00 per hour for each. The loading and unloading time is 10 min. There are two alternatives for the tool material to be used:i) an HSS tool which costs $2.00 and can be ground 10 times with a cost
of $1.20 per grind. Removing and resetting the tool requires 4 min.ii) a triangular throwaway carbide tool having 3 cutting edges costs
$6.00. Indexing the tool takes 2 min.
a) Which type of tool should be used?b) What would be the unit production cost if cost minimization criterion
is applied when WC tool is used? Assume that the recommended cutting speeds correspond to a tool life of 90 min and Taylor’s tool life exponent is 0.25 for WC tools.
2/22/2014 ME 303 - Section 05 5.113
Problem 6
A cylindrical component of ∅60mm x 120mm is to be rough turned orthogonally from a carbon steel bar stock of ∅66mmusing a WC throw-away type insert tool. The following data apply for the operation:• Time taken to load and unload work parts = 60 s• Time taken to index a cutting edge = 20 s• Cutting speed = 2.5 m/s• Specific cutting energy = 1.3 GN/m2 (const. for all chip
thickness values)• Motor power = 5 kW• Machine efficiency = 80%
2/22/2014 ME 303 - Section 05 5.114
Problem 6 (Cont’d)
a) Find the feed rate in mm/rev for maximum power utilization.b) If the cutting force is not allowed to exceed 1300 N, find the
allowable feed rate.c) For the feed rate found in (b), it has been experimentally found
that the tool life becomes 2400 s when the cutting speed is 4 m/sand the tool life exponent is 0.25. Using this information and the feed rate found in (b), calculate the optimum cutting speed to maximize the production rate.
d) Check if the available power is enough to perform the operation at the conditions set in (c). If not, adjust the cutting speed to make the operation feasible.
e) Find the total production time/part for the conditions found in (d).
References
1. Groover M.P., Fundamentals of Modern Manufacturing - SI Version, John Wiley, 4th Edition, 2011.
2. Schey, John A., Introduction to Manufacturing Processes, McGraw Hill, 2nd Edition, 1987.
3. Hosford, W. F., Caddell R. M., Metal Forming: Mechanics and Metallurgy, 4th Edition, Cambridge Univ. Press, 2011.
4. Mielnik, E. M., Metal Working Science Engineering, McGraw Hill, 1991.
5. Tlusty, G., Manufacturing Process and Equipment, Prentice Hall Inc., 2000.
