Stress strain behavior of ductile materials
σ
DEFORMING
εe
ε
εp
Elastic-plastic with strain hardening
Elastic-perfectly plastic
Rigid perfectly plastic
Rigid plastic with strain hardening
Topics for today’s lecture
Process characteristics
Deformation processes
Material/continuum changes during processing
~
Deforming process characteristics
~
Machining
= Local, concentrated
~
Deforming physics
~
Deforming
= Over large volume
~
Common deforming processes
~
DFM
Force
~
Machining:
~10s - 100s of lbs
~
Stamping:
~10s - 100s of tons
Materials - Virtually all ductile materials
Shapes
- Limited by strain/flow
Size
- Limited by force/equipment
Advantages and disadvantages
Important physics
Advantages
Stress/strain
~
Parallel process: rapid bulk formation
~
Affect CQFR
~
Overall material properties improved
~
Affect deforming force -> equipment & energy requirements
Disadvantages
~
Cost of equipment and dies
~
Limited flexibility in shapes and sizes (i.e. compared to machining)
~
Accuracy
~
Repeatability
Friction
~
Affect on deforming force -> equipment & energy requirements
~
Why lubrication is needed and can help
~
Friction is not repeatable… quality….
1
Stress strain behavior of ductile materials
Example: 12L14 Steel in Duratec
σ
12L14 Steel True Stress-Strain Behavior
600
80
Intercept
εe
ε
εp
Tangent modulus
0.41 GPa [0.06 Mpsi]
400
70
60
50
300
40
Young’s modulus
190 GPa [27.4 Mpsi]
30
200
Stress, kpsi
Stress, MPa
500
20
100
10
0
0.00
Elastic-plastic with strain hardening
0.01
Elastic-perfectly plastic
Rigid perfectly plastic
0.02
0.03
mm
/mm
Strain,
0.04
0
0.05
Rigid plastic with strain hardening
Forging force and friction
Forging force and friction
Axisymmetric upsetting
Fz
Purpose
h
~
Find Fz(µ)
~
Sensitivity
Eqxn: A – Equilibrium in r direction
dFinner arc
dFfriction top & bottom
dFhoop
Σ dFr = 0 = −σ r ⋅ h ⋅ r ⋅ dθ − 2 ⋅ µ ⋅ p ⋅ r ⋅ dθ ⋅ dr − 2 ⋅ σ θ ⋅ h ⋅ dr ⋅
dFouter arc
dθ
+ (σ r + dσ r ) ⋅ (r + dr ) ⋅ h ⋅ dθ
2
Assumptions:
2R
dσ r 2 ⋅ µ ⋅ p
2 ⋅ µ ⋅σ z
=
=−
dr
h
h
~
Tresca flow
~
Constant friction coefficient
~
Plastic deformation
⎡ 2µ
(R − r )⎤⎥
⎣ h
⎦
σ z = −Y ⋅ exp ⎢
Eqxn: B - Tresca Yield Criterion
σr −σz = Y
Forging force and friction
Forging force and friction cont.
Affect of friction on contact pressure
0
Y=75000 psi
h = ½ inch
0
0.05
0.1
0.15
0.2
0.25
mu = 0
mu = .1
mu = .2
mu = .5
Contact pressure [psi]
-25000
-50000
-75000
0.3
2
R
1 ⎛ h ⎞
⎡ ⎛ 2⋅µ ⋅R ⎞ ⎛ 2⋅µ ⋅ R ⎞ ⎤
Fz = ∫ σ z ⋅ 2 ⋅ π ⋅ r ⋅ dr = (π ⋅ R 2 )⋅ ⋅ ⎜⎜
⎟ ⋅ Y ⋅ ⎢exp⎜
⎟−⎜
⎟ − 1⎥
0
2 ⎝ µ ⋅ R ⎟⎠
⎣ ⎝ h ⎠ ⎝ h ⎠ ⎦
Now use Taylor’s series expansion (3 terms) to approximate the exponential function
-100000
Expand about 0, makes this approximation valid for small values of 2µR/h
-125000
Increasing µ
-150000
⎡ ⎛ 2 µ ⋅ R ⎞⎤
Fz = (π ⋅ R 2 )⋅ Y ⋅ ⎢1 + ⎜ ⋅
⎟⎥
⎣ ⎝ 3 h ⎠⎦
Distance from center line [inches]
2
Forging force and friction
Common processes
Y=75000 psi
h = ½ inch
Affect of part size and friction on forging force
Sheet metal
Forging force [tons]
10
mu = 0
mu = .1
mu = .2
mu = .5
8
Forging
6
Extrusion
4
Increasing µ
2
Rolling
0
0
0.05
0.1
0.15
0.2
0.25
Part size [inches]
Bending and shearing
Forging
Rolling and extrusion
Affect of grain size
Properties affected by grain size
~
Strength
~
Hardness
~
Ductility
~
For example (Ferrite)
σ y = σo +
ky
d grain
We desire smaller grains for better properties
3
Recrystallization
DFM for deforming processes
Parting line and flash
Draft angle
Radii
Lubrication
http://www.mmat.ubc.ca/other/courses/apsc278/lecture11.pdf
Mechanics of spring back
Example: Elastic spring back
Q: Why do we see spring back?
Example: Bending wire
A: Elastic recover of material
σ
Ro
ε
εSB
Ro < R1 ⇒
1
1
>
Ro R1
R1
Quantifying elastic spring back
Elastic recovery of material leads to spring back
~
Y = Yield stress
E = Young’s Modulus
3
~
Ri
⎛R Y ⎞
⎛R Y ⎞
= 4 ⋅ ⎜ i ⋅ ⎟ − 3⋅ ⎜ i ⋅ ⎟ +1
Rf
⎝ t E⎠
⎝ t E⎠
~
With increase in Ri / t or Y
t = thickness
2
1
1 3 ⎛ Y ⎞ 4 Ri ⎛ Y ⎞
−
= ⋅⎜ ⎟ − 3 ⋅⎜ ⎟
Ri R f t ⎝ E ⎠
t
⎝E⎠
OR
3
decrease in E spring back increases
c
In
Titanium
re
as
Y/E
ing
Rf
Steel
Ri
4