forging analysis - 2ramesh/courses/me649/forming_3.pdf · prof. ramesh singh, notes by dr. singh/...
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Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging Analysis - 2
ver. 1
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Overview• Slab analysis
– frictionless– with friction– Rectangular– Cylindrical
• Strain hardening and rate effects• Flash• Redundant work
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging – cylindrical partsliding region
h
p
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Equilibrium in r direction
( ) ( ) θσσθσ
θµθσ
θ dhdrrdddrh
drdrpdrhdF
rr
rr
⋅⋅+⋅++⋅⋅⋅⋅−
⋅⋅⋅⋅⋅−⋅⋅⋅−==∑2
2
20
neglecting HOTs
02 =⋅−⋅−⋅+⋅ rr dhrdrhdrhdrpr σσσµ θ
22sin θθ dd
=
N.B.
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Axisymmetric flow and yield
For axisymmetric flow
By Tresca
flowflowr kp τσσ 22 ===+
dpd r −=σ p σr
σflow/2 = τflow
( )
θθ
θ
σσεεπ
ππεε
==
=−+
==
rr
r rdr
rrdrr
rdr
;2
22;
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Stress in z direction
dphrdrpr ⋅−=⋅µ2
02 =⋅+⋅−⋅+⋅ dphrdrhdrhdrpr rr σσµsubstituting
or
rearranging
drhp
dp µ2−=
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging pressure - sliding
drhp
dp R
rp
flow
r
⋅−= ∫∫µ
τ2
2
( )
−= rRh
pflow
r µτ
2exp2
for rk < r < R
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Average forging pressure –sliding
( ) ( ) ( ) drrrRhrR
drrprR
p R
rk
R
r flow
r
kflow
ave
kk
∫∫
−
−=⋅⋅
−=
µπτπτ
2exp222
12 2222
( )R
rkflow
ave
k
hr
hr
hRh
rRp
−−
⋅
−
−
= 122exp2exp2
22
2
22µµµ
µτ
( )
−−
⋅
−−
−−
⋅
−
−
= 122exp122exp2exp2
22
2
22 hr
hr
hR
hR
hRh
rRp kk
kflow
ave µµµµµµτ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Average forging pressure –sliding
( )( )
−
−
+⋅
−
⋅
−= 12122exp
22
2
2
22 hR
hr
hrRh
rRp kk
kflow
ave µµµµτ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging force – sliding
( )22kaveaveforging rRpApF −⋅⋅=⋅= π
( ) ( )222
12122exp2
4 kkk
flowforging rRhR
hr
hrR
RhF −⋅⋅
−
−
+⋅
−
⋅= πµµµ
µτ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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• Taking the first four terms of a Taylor’s series expansion for the exponential about 0 for
Average forging pressure –all sliding approximation (rk = 0)
1≤x
yields
( ) ∑=
=+++++=n
k
kn
kx
nxxxxx
0
32
!!!3!21exp L
+=hRp
flow
ave
321
2µ
τ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging force –all sliding approximation
2
3212 RhRF flowforging πµτ ⋅
+⋅=
2RpApF aveaveforging ⋅⋅=⋅= π
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Transition sticking / sliding
• Set τflow = µp and solve for rk
−
=hrRp k
flowµ
τ2exp
2
−
=⋅ h
rRp
p kµµ
2exp2
−
=
hrR kµ
µ2
21ln
−=
µµ 21ln
2hRrk
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging pressure - sticking region
• Use the same method as for sliding • Substitute µp = τflow, • Assume Tresca yield criterion
drh
dp flowτ2−=
dphrdrpr ⋅−=⋅µ2
dphrdrrflow ⋅−=⋅τ2
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging pressure - sticking regiondr
hdp
r
r
flowp
p k
r
kr
∫∫ −=τ2
( )kflowrr rr
hpp
k−−=−
τ2
( )hrrpp k
flow
rr k −=
−
τ2
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging pressure - sticking region
( ) ( )hrrrR
hp k
kflow
r −+
−=µ
τ2exp
2
for 0 < r < rk
prk determined from sliding equation
( )
−= k
flow
r rRh
pk µ
τ2exp
2
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Average forging pressure -sticking
( ) rdrhrrrR
hrdrrp
rp kk r
kk
k
r
rkflow
ave ⋅
−+
−=⋅⋅= ∫∫
02
02
2exp2212
µππτ
( ) drrhh
rrrRh
rr
p krk
kkflow
ave ⋅
−
⋅+
−⋅= ∫
0
22
12exp22
µτ
( )kr
kk
kflow
ave
hr
hrrrR
hr
rp
0
322
2 322exp
22
2
−
⋅+
−⋅=µ
τ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Average forging pressure - sticking
( )
−+
−⋅=
hr
hrrR
hr
rp kk
kk
kflow
ave
322exp
22
2
332
2µ
τ
( )
+
−=
hrrR
hp k
kflow
ave
32exp
2µ
τ
( )kr
kk
kflow
ave
hr
hrrrR
hr
rp
0
322
2 322exp
22
2
−
⋅+
−⋅=µ
τ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Forging force – sticking region
2kaveaveforging rpApF ⋅⋅=⋅= π
( ) 2
32exp2 k
kkflowforging r
hrrR
hF ⋅⋅
+
−⋅= πµτ
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Sticking and sliding
• If you have both sticking and sliding, and you can’t approximate by one or the other,
• Then you need to include both in your pressure and average pressure calculations.
( ) ( )stickingaveslidingaveforging ApApF ⋅+⋅=
stickingslidingforging FFF +=
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Strain hardening(cold - below recrystallization
point)
nflow KY ετ ==2
Tresca
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Strain rate effect(hot – above recrystallization point)
Tresca
( )mflow CY ετ &==2
heightousinstantanevelocityplaten
hv
dtdh
h1ε ===&
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Flash for closed die forging(plane strain)
• Say we have a typical flash with thickness h/20 and length w/4
ww/4
h/20h
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Average forging pressure
• in forging (Tresca)
• in flash (Tresca)
+⋅=
hwp flowave 2
12 µτ
+⋅=
hwp flowave
µτ 512
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Flash
• Flash’s high deformation resistance results in filled mold
• Process wouldn’t work without friction
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Deformation WorkIn general, work done in bulk deformation processes
has three components
Total work, W = Wideal + Wfriction + Wredundant
Work of ideal plastic deformation, Wideal
= (area under true stress-true strain curve)(volume)
= (volume)For a true stress-true strain curve :
∫ ttdt
εσε
0
ntt Kεσ =
( ) ( ) tf
nt
ideal YnKW εε volume
1volume
1
=
+
=+
stress flow Avg. =fY
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Deformation WorkFriction between dies and workpiececauses inhomogeneous (non-uniform) deformation called barreling
Barreling Effect
Frictional Forces
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Deformation Work
Internal shearing of material requires redundant work to be expended
Ideal Deformation Redundant Deformation
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Redundant Zone
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Closed/Impression Die Forging• Analysis more complex due to large
variation in strains in different parts of workpiece
• Approximate approaches
– Divide forging into simple part shapes e.g. cylinders, slabs etc. that can be analyzed separately
– Consider entire forging as a simplified shape
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Closed/Impression Die ForgingSteps in latter analysis approach• Step 1: calculate average height from
volume V and total projected area At of part (including flash area)
• Step 2:
LwV
AVht
avg ==
==
avg
iavg h
hlnstrain avg.ε
avgavg h
v== ratestrain avg.ε&
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Closed/Impression Die Forging• Step 3: calculate flow stress of material Yf
for cold/hot working
• Step 4:
Kp = pressure multiplying factor= 3~5 for simple shapes without flash= 5~8 for simple shapes with flash= 8~12 for complex shapes with flash
tfpavg AYKF == load forging Avg.
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Other Analysis Methods• Complex closed die forging simulated
using finite element software
Source: http://nsmwww.eng.ohio-state.edu/html/f-flashlessforg.html
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Upper Bound Theorem
• Any estimate of the collapse load of a structure made by equating the internal rate of energy dissipation to the rate at which external forces do work in some assumed pattern of deformation will be > or = to the correct load.
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Upper Bound Theorem Assumptions
• Isotropic and homogeneous• Neglect strain hardening and strain rate• Frictionless or constant shear stress
condition exists at tool-work piece interface
• 2-D, plane strain with all deformation occurring by shear on a few planes. Elsewhere, material is rigid.
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Upper Bound Theorem
• k = shear flow stress• Si = length of shear plane• Vι
* = velocity of shear
∑=
∗=n
iiiVkSdt
dW
1
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Upper Bound Theorem
• Indentation of a plate (slip-line analysis)
wL
F
h = ∞
pA
BC
D
C’
A’ D’ vovBA
vCA vDC
vBC
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Work, shearing force
• Work is done by shearing along AB, BC, AC, and CD.– Lengths calculated from figure
at right.• Shearing force along any
boundary, per unit length, w, is k (shear yield stress) times the length of the boundary, L.
A
BC
D
C’
A’ D’
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Shearing velocities
vovBA
vCA vDC
vBC
2
2
2
oDC
oCA
oBC
oBA
vv
vv
vvvv
=
=
==
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Motions
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Total power delivered
• each term has been counted twice– due to symmetry
• Simplifyingp = 6k
+++=
22222 0 oo
ooLvLvLvLvkLpv
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Total power deliveredp = 6k
• using von Mises
• hence YYk ⋅== 577.0
3
YYp ⋅== 46.33
6
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Exact solution
p = 5.14 k = 2.97 Y
• Solution abovep = 6 k = 3.46 Y
• so we can see the effect of constraint– redundant work: higher pressure
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Non-homogeneous deformation and Redundant work
• If the slab is thick or friction:– non-homogeneous deformation– redundant work
• If the slab is thin or unconstrained: (e.g., open die forging without friction)– no redundant work
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Indenting at h/L =1
L
h
F
F
v1v1
vo
vo
AB
CE
v1
v1
vo
vovCE
vBC
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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v1v1
vo
vo
AB
CE
2LCEBC ==
oCEBC vvv ×== 2
v1
v1
vo
vovCE
vBC
222
22 kLvLpv
oo ××=
Analysis - power delivered
p = 2k = 1.15 Y (plane strain result)
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Redundant work limit (∆ = h/L)(plane strain)
• h/L < 1: no redundant work– p = 1.15 Y
• 1 < h/L < 8.7: some redundant work– 1.15 Y < p < 2.97 Y
• h/L > 8.7: redundant work– same as infinite plate – p = 2.97 Y
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Redundant work correction factor (Qr)
• Can be characterized by:p = Qr Y
orQr = p/σy = p/2τy (by Tresca)
• where Qr = correction factor for redundant work
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Redundant work factor (Backofen)(frictionless)
Qr =
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Redundant work factor (Kalpakjian) -
(friction)
Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton
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Summary• Slab analysis
– frictionless– with friction– Rectangular– Cylindrical
• Strain hardening and rate effects• Flash• Redundant work