Forging Analysis - 1ramesh/courses/ME649/Forming...Prof. Ramesh Singh, Notes by Dr. Singh/ Dr....

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

Forging Analysis - 1

ver. 1


Overview

• Slab analysis– frictionless– with friction– Rectangular– Cylindrical

• Strain hardening and rate effects• Flash• Redundant work


Slab analysis assumptions• Entire forging is plastic

– no elasticity• Material is perfectly plastic

– strain hardening and strain rate effects later

• Friction coefficient (µ) is constant– all sliding, to start

• Plane strain– no z-direction deformation

• In any thin slab, stresses are uniform


Open die forging analysis –rectangular part

F

F

h

w

dx

yx

unit depthinto figure


Expanding the dx slice on LHS

• p = die pressure• σx, dσx from material on

side• τfriction = friction force = µp

p

p

σx σx + dσx

τf

τf


Force balance in x-direction

p σx

k

dxh

d

dxhd

frictionx

frictionx

τσ

τσ

202

−=

=+

Mohr’s circle

(distortion energy (von Mises) criterion, plane strain)

flowflowx kp σσσ ⋅===+ 15.1322

N.B. all done on a perunit depth basis


Force balanceDifferentiating, and substituting into Mohr’s

circle equation

noting: τfriction = µp

pdxh

dp µ2= dx

hpdp µ2

=

( ) ( )

dxh

dpdxh

d

ddppdkd

frictionfrictionx

xx

=∴−=

−=∴+=

ττσ

σσ22

2


Sliding region

• Noting: @ x = 0, σx = 2k = 1.15 σflow

dxhp

dp xp

k

x

∫∫ =02

2µ


Forging pressure – sliding region

( )

⋅⋅=

=

=−

hxp

hx

kp

hxkp

flowx

x

x

µσ

µ

µ

2exp15.1

2exp2

22lnln

Sliding region result (0 < x < xk)

N.B done on a per unit depth basis


Forging pressure –approximation

• Taking the first two terms of a Taylor’s series expansion for the exponential about 0, for

+=

hx

kpx µ212

1≤x

yields

( ) ∑=

=+++++=n

k

kn

kx

nxxxxx

0

32

!!!3!21exp L

00

+⋅⋅=

hxp flowx

µσ 2115.1


Average forging pressure –all sliding approximation

• using the Taylor’s series approximation

2

22

2

21

2

22

2

0

22

0

2

0

whxx

w

dxhx

w

dxkp

kp

www

x

ave

+

=

+

==∫∫

µµ

N.B done on a per unit depth basis

+=

hw

kpave

21

2µ

+⋅⋅=

hwp flowave 2

115.1 µσ


Forging force –all sliding approximation

depthwhwF flowforging ⋅⋅

+⋅⋅=

2115.1 µσ

depthwidthpF aveforging ⋅⋅=


Slab - die interface• Sliding if τf < τflow• Sticking if τf ≥ τflow

– can’t have a force on a material greater than its flow (yield) stress

– deformation occurs in a sub-layer just within the material with stress τflow

slidingregion

stickingregion


Sliding / sticking transition• Transition will occur at xk

• using k = µp, in:

• hence:

µµ 21ln

21

=hxk

=hx

kpx µ2exp2

=hx

kk kµµ

2exp2


Sticking region

• Using p = k/ µ

pdxh

dp µ2=

( )kxx xxhkpp

k−=−

2dxhkdp

x

x

p

p k

x

kx

∫∫ =2

dxkh

dpµ

µ2=


Sticking region

We know that • at x = xk, pxk

= k/µ

• and

µµ 21ln

21

=hxk


Forging pressure - sticking region

hx

kpx +

−=

µµ 21ln1

21

2

Combining (for xk < x < w/2)

+

−⋅⋅=

hxp flowx µµ

σ21ln1

2115.1


Forging pressure –all sticking approximation

• If xk << w, we can assume all sticking, and approximate the total forging force per unit depth (into the figure) by:

x=0 x=w/20

pedge

∆p


Forging pressure –all sticking approximation

kpedge 2=

+⋅⋅=

+=∴

hxp

hx

kp

flowx

x

115.1

12

σ

( )xhkkpx22 =−dx

hkdp

xp

k

x

∫∫ =02

2


Average forging pressure –all sticking approximation

22

2

1

22

2

2

0

22

0

2

0

whxx

w

dxhx

w

dxkp

kp

www

x

ave

+

=

+

==∫∫

+=

hw

kpave

41

2

+⋅⋅=

hwp flowave 4

115.1 σ


Forging force –all sticking approximation


+⋅⋅=

4115.1 σ

depthwidthpF aveforging ⋅⋅=


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 +=


Material Models

nflow KY εσ ==

Strain hardening (cold – below recrystallization point)

Strain rate effect (hot – above recrystallization point)

( )mflow CY εσ &==

heightousinstantanevelocityplaten

hv

dtdh

h1ε ===&


Forging - Ex. 1-1• Lead 1” x 1” x 36”• σy = 1,000 psi• hf = 0.25”, µ = 0.25• Show effect of friction on total forging force.• Use the slab method.• Assume it doesn’t get wider in 36” direction.• Assume cold forging.

1”

1”

36”


Forging - Ex. 1-2• At the end of forging:hf = 0.25”, wf = 4” (conservation of mass)

• Sliding / sticking transition

"347.025.02

1ln25.0225.021ln

21

=××

=

=

k

k

x

hx

µµ


Forging - Ex. 1-3

• Sliding region:

( )xhxpf

flowx

2exp1150

2exp15.1

⋅=

⋅⋅=

µσ


Forging - Ex. 1-4

• Sticking region

( )xhxpf

flowx

46.01150

21ln1

2115.1

+⋅=

+

−⋅⋅=

µµσ


Forging - Ex. 1-5

0

200040006000

800010000

12000

0 0.5 1 1.5 2 2.5 3 3.5 4

Distance from forging edge (in)

Forg

ing

pres

sure

(psi

)

P(sticking)P(sliding)

stickingsliding slidingfriction hill

xkxk


Forging - Ex. 1-6

• Friction hill– forging pressure must be large (8.7x) near the

center of the forging to “push” the outer material away against friction


Forging - Ex. 1-7

• Determine the forging force from:

• since we have plane strain

dApForce ∫∫ ⋅=

dxpdepthunitF x

x∫=0


Forging - Ex. 1-8

• We must solve separately for the sliding and sticking regions

sticking

w

xx

sliding

x

xforging depthdxpdepthdxpFk

k

+

= ∫∫ .2.2

2/

0


Forging - Ex. 1-9

Sliding first

k

k

x

k

x

flow

ave

hxh

x

dxhx

depthunit

p

0

0 2exp20

2exp15.1

=

−

=∫ µ

µ

µσ

( )0

12exp2

−

−

=k

k

xhxh µ

µ


Forging - Ex. 1-12Substituting values

( ) 44.10347.0

125.0

347.025.02exp25.0225.0

15.1=

−

−

××

×=

depthunit

pflow

ave

σ

3.5

347.024

347.044

25.021347.0

24

25.021ln1

25.021

15.12

2

=

−

−

×+

−

×−

×=

depthunit

pflow

ave

σ

sliding

sticking


Forging - Ex. 1-10

Sticking next

k

w

xflow

ave

xw

dxhx

depthunit

p

k

−

+

−

=∫

2

21ln1

21

15.1

2

µµσ

22

2

221ln1

21

w

x

k

k

xwhxx

−

+⋅

−

=µµ


Forging - Ex. 1-11

( )k

kkflow

ave

xw

xwh

xw

depthunit

p

−

−+−

−

=2

421

221ln1

21

15.12

2

µµσ


Forging - Ex. 1-13

Now calculate the area/unit depth

31.3347.024

69.02347.0

=×−=

=×=

sticking

sliding

A

A


Forging - Ex. 1-14Now calculate the forces

( ) ( )( )stickingaveslidingaveflow ApApdepthunitF

⋅+⋅= σ15.1

( ) ( )( )

inchlbdepthunitF

/110,21

31.33.569.044.11150

=

×+××=


Forging - Ex. 1-15

• Now, assume all sticking, so:

depthinchlb

hw

wlengthunitF

f

ffflow

/000,2325.044141150

4115.1

=

⋅+⋅⋅=

+⋅⋅= σ


Forging - Ex. 1-16or since the part is 36” deep:F(both) = 759,960 lbs = 380 tons

F(all sticking) = 828,000 lbs = 414 tons

F(all sliding) = 496,800 lbs = 225 tons

All sticking over-estimates actual value.


+⋅⋅=

4115.1 σ


+⋅⋅=

2115.1 µσ


Forging – Effect of friction• Effect of friction coefficient (µ) – all sticking

• Friction is very important

Friction coefficient Fmax (lbf/in depth) xk Stick/slide0 4600 2 slide

0.1 11365 2 slide0.2 19735 0.573 both

0.25 21331 0.347 both0.3 22182 0.213 both0.4 22868 0.070 both0.5 23000 0 stick


Forging - Ex. 1-17Forging force vs. stroke – all sticking

0

5000

10000

15000

20000

25000

0 0.2 0.4 0.6 0.8

Stroke (in)

Forg

ing

forc

e (lb

f/in

dept

h)

mu=0mu=0.1mu=0.2mu=0.25mu=0.3mu=0.4mu=0.5


Forging - Ex. 1-19Maximum forging force vs. friction coefficient (µ)

all sticking

0

5000

10000

15000

20000

25000

0 0.1 0.2 0.3 0.4 0.5 0.6

Friction coefficient

Max

forg

ing

forc

e (lb

f/in

dept

h)


Summary• Slab analysis

– frictionless– with friction– Rectangular– Cylindrical

• Strain hardening and rate effects• Flash• Redundant work

Forging Analysis - 1ramesh/courses/ME649/Forming...Prof. Ramesh Singh, Notes by Dr. Singh/ Dr....

Documents

Transcript of Forging Analysis - 1ramesh/courses/ME649/Forming...Prof. Ramesh Singh, Notes by Dr. Singh/ Dr....