Forging Analysis - 2ramesh/courses/ME649/Forming_3.pdf · Prof. Ramesh Singh, Notes by Dr. Singh/...

Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton

1

Forging Analysis - 2

ver. 1


2

Overview• Slab analysis

– frictionless– with friction– Rectangular– Cylindrical

• Strain hardening and rate effects• Flash• Redundant work


3

Forging – cylindrical partsliding region

h

p


4

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.


5

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;


6

Stress in z direction

dphrdrpr ⋅−=⋅µ2

02 =⋅+⋅−⋅+⋅ dphrdrhdrhdrpr rr σσµsubstituting

or

rearranging

drhp

dp µ2−=


7

Forging pressure - sliding

drhp

dp R

rp

flow

r

⋅−= ∫∫µ

τ2

2

( )

−= rRh

pflow

r µτ

2exp2

for rk < r < R


8

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


9

Average forging pressure –sliding

( )( )

−

−

+⋅

−

⋅

−= 12122exp

22

2

2

22 hR

hr

hrRh

rRp kk

kflow

ave µµµµτ


10

Forging force – sliding

( )22kaveaveforging rRpApF −⋅⋅=⋅= π

( ) ( )222

12122exp2

4 kkk

flowforging rRhR

hr

hrR

RhF −⋅⋅

−

−

+⋅

−

⋅= πµµµ

µτ


11

• 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


12

Forging force –all sliding approximation

2

3212 RhRF flowforging πµτ ⋅

+⋅=

2RpApF aveaveforging ⋅⋅=⋅= π


13

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


14

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


15

Forging pressure - sticking regiondr

hdp

r

r

flowp

p k

r

kr

∫∫ −=τ2

( )kflowrr rr

hpp

k−−=−

τ2

( )hrrpp k

flow

rr k −=

−

τ2


16

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


17

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


18

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


19

Forging force – sticking region

2kaveaveforging rpApF ⋅⋅=⋅= π

( ) 2

32exp2 k

kkflowforging r

hrrR

hF ⋅⋅

+

−⋅= πµτ


20

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 +=


21

Strain hardening(cold - below recrystallization

point)

nflow KY ετ ==2

Tresca


22

Strain rate effect(hot – above recrystallization point)

Tresca

( )mflow CY ετ &==2

heightousinstantanevelocityplaten

hv

dtdh

h1ε ===&


23

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


24

Average forging pressure

• in forging (Tresca)

• in flash (Tresca)

+⋅=

hwp flowave 2

12 µτ

+⋅=

hwp flowave

µτ 512


25

Flash

• Flash’s high deformation resistance results in filled mold

• Process wouldn’t work without friction


26

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


27

Deformation WorkFriction between dies and workpiececauses inhomogeneous (non-uniform) deformation called barreling

Barreling Effect

Frictional Forces


28

Deformation Work

Internal shearing of material requires redundant work to be expended

Ideal Deformation Redundant Deformation


29

Redundant Zone


30

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


31

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&


32

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.


33

Other Analysis Methods• Complex closed die forging simulated

using finite element software

Source: http://nsmwww.eng.ohio-state.edu/html/f-flashlessforg.html


34

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.


35

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.


36

Upper Bound Theorem

• k = shear flow stress• Si = length of shear plane• Vι

* = velocity of shear

∑=

∗=n

iiiVkSdt

dW

1


37

Upper Bound Theorem

• Indentation of a plate (slip-line analysis)

wL

F

h = ∞

pA

BC

D

C’

A’ D’ vovBA

vCA vDC

vBC


38

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’


39

Shearing velocities

vovBA

vCA vDC

vBC

2

2

2

oDC

oCA

oBC

oBA

vv

vv

vvvv

=

=

==


40

Motions


41

Total power delivered

• each term has been counted twice– due to symmetry

• Simplifyingp = 6k

+++=

22222 0 oo

ooLvLvLvLvkLpv


42

Total power deliveredp = 6k

• using von Mises

• hence YYk ⋅== 577.0

3

YYp ⋅== 46.33

6


43

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


44

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


45

Indenting at h/L =1

L

h

F

F

v1v1

vo

vo

AB

CE

v1

v1

vo

vovCE

vBC


46

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)


47

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


48

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


49

Redundant work factor (Backofen)(frictionless)

Qr =


50

Redundant work factor (Kalpakjian) -

(friction)


51

Summary• Slab analysis

– frictionless– with friction– Rectangular– Cylindrical

• Strain hardening and rate effects• Flash• Redundant work

Forging Analysis - 2ramesh/courses/ME649/Forming_3.pdf · Prof. Ramesh Singh, Notes by Dr. Singh/...

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Transcript of Forging Analysis - 2ramesh/courses/ME649/Forming_3.pdf · Prof. Ramesh Singh, Notes by Dr. Singh/...