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

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

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

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

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

Open die forging analysis –rectangular part

F

F

h

w

dx

yx

unit depthinto figure

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

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

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

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

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

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

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

Sliding region

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

dxhp

dp xp

k

x

∫∫ =02

2µ

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

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

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

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

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

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

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

Forging force –all sliding approximation

depthwhwF flowforging ⋅⋅

+⋅⋅=

2115.1 µσ

depthwidthpF aveforging ⋅⋅=

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

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

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

Sliding / sticking transition• Transition will occur at xk

• using k = µp, in:

• hence:

µµ 21ln

21

=hxk

=hx

kpx µ2exp2

=hx

kk kµµ

2exp2

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

Sticking region

• Using p = k/ µ

pdxh

dp µ2=

( )kxx xxhkpp

k−=−

2dxhkdp

x

x

p

p k

x

kx

∫∫ =2

dxkh

dpµ

µ2=

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

Sticking region

We know that • at x = xk, pxk

= k/µ

• and

µµ 21ln

21

=hxk

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

Forging pressure - sticking region

hx

kpx +

−=

µµ 21ln1

21

2

Combining (for xk < x < w/2)

+

−⋅⋅=

hxp flowx µµ

σ21ln1

2115.1

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

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

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

Forging pressure –all sticking approximation

kpedge 2=

+⋅⋅=

+=∴

hxp

hx

kp

flowx

x

115.1

12

σ

( )xhkkpx22 =−dx

hkdp

xp

k

x

∫∫ =02

2

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

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

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

Forging force –all sticking approximation

depthwhwF flowforging ⋅⋅

+⋅⋅=

4115.1 σ

depthwidthpF aveforging ⋅⋅=

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

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

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 ===&

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

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”

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

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

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

Forging - Ex. 1-3

• Sliding region:

( )xhxpf

flowx

2exp1150

2exp15.1

⋅=

⋅⋅=

µσ

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

Forging - Ex. 1-4

• Sticking region

( )xhxpf

flowx

46.01150

21ln1

2115.1

+⋅=

+

−⋅⋅=

µµσ

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

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

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

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

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

Forging - Ex. 1-7

• Determine the forging force from:

• since we have plane strain

dApForce ∫∫ ⋅=

dxpdepthunitF x

x∫=0

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

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

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

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

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

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

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

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

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

Forging - Ex. 1-11

( )k

kkflow

ave

xw

xwh

xw

depthunit

p

−

−+−

−

=2

421

221ln1

21

15.12

2

µµσ

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

Forging - Ex. 1-13

Now calculate the area/unit depth

31.3347.024

69.02347.0

=×−=

=×=

sticking

sliding

A

A

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

Forging - Ex. 1-14Now calculate the forces

( ) ( )( )stickingaveslidingaveflow ApApdepthunitF

⋅+⋅= σ15.1

( ) ( )( )

inchlbdepthunitF

/110,21

31.33.569.044.11150

=

×+××=

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

Forging - Ex. 1-15

• Now, assume all sticking, so:

depthinchlb

hw

wlengthunitF

f

ffflow

/000,2325.044141150

4115.1

=

⋅+⋅⋅=

+⋅⋅= σ

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

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.

depthwhwF flowforging ⋅⋅

+⋅⋅=

4115.1 σ

depthwhwF flowforging ⋅⋅

+⋅⋅=

2115.1 µσ

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

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

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

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

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

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)

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

Summary• Slab analysis

– frictionless– with friction– Rectangular– Cylindrical

• Strain hardening and rate effects• Flash• Redundant work