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
unit depth
into figure
y
h
x
dx
w
F
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Expanding the dx slice on LHS
p
τf
σx
σx + dσx
• p = die pressure
• σx, dσx from material on
side
• τfriction = friction force = µp
τf
p
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Force balance in x-direction
hdσ x + 2τ friction dx = 0
dσ x = −
2τ friction
h
Mohr’s circle
dx
k
p
2
σ x + p = 2k =
σ flow = 1.15 ⋅ σ flow
3
(distortion energy (von Mises) criterion,
plane strain)
σx
N.B. all done on a per
unit depth basis
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Force balance
Differentiating, and substituting into Mohr’s
circle equation
d (2k ) = d (σ x + p )
dσ x = −
2τ friction
h
dx
∴ dp = − dσ x
 2τ friction 
∴ dp = 
dx
 h 
noting: τfriction = µp
2µ
dp =
pdx
h
dp 2 µ
dx
=
p
h
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Sliding region
px
2µ
dp
∫2k p = ∫0 h dx
x
• Noting: @ x = 0, σx = 2k = 1.15 σflow
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging pressure – sliding
region
x
ln px − ln (2k ) = 2 µ
h
Sliding region result (0 < x < xk)
px
 2 µx 
= exp

2k
 h 
 2 µx 
p x = 1.15 ⋅ σ flow ⋅ exp

 h 
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
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 x ≤ 1
0
2
0
3
n
x
x
x
+
+L+
=
exp(x ) = 1 + x +
2! 3!
n!
n
∑
k =0
xk
k!
yields
p x  2 µx 
= 1 +

h 
2k 
 2 µx 
p x = 1.15 ⋅ σ flow ⋅ 1 +

h 

Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Average forging pressure –
all sliding approximation
• using the Taylor’s series approximation
w
2
w
2
w
2
2


2
x
µ
px
 2 µx 

dx ∫ 1 +
dx  x +
∫
2h  0
h 
pave 0 2k

0
=
=
=
w
w
w
2k
2
2
2
pave  µw 
= 1 +

2k 
2h 
pave
 µw 
= 1.15 ⋅ σ flow ⋅ 1 +

2h 

Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
N.B done on a
per unit depth basis
Forging force –
all sliding approximation
F forging = pave ⋅ width ⋅ depth
F forging
µw 

= 1.15 ⋅ σ flow ⋅  1 +
 ⋅ w ⋅ depth
2h 

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
sliding
region
sticking
region
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Sliding / sticking transition
• Transition will occur at xk
• using k = µp, in:
px
 2 µx 
= exp

2k
 h 
 2 µxk 
= exp

2 µk
 h 
k
• hence:
xk
1
1
=
ln
h 2µ 2µ
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Sticking region
2µ
dp =
pdx
h
• Using p = k/ µ
px
x
2k
∫p dp = x∫ h dx
x
k
k
2µ k
dp =
dx
h µ
2k
( x − xk )
p x − p xk =
h
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Sticking region
We know that
• at x = xk, px = k/µ
k
• and
xk
1
1
=
ln
h 2µ 2µ
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging pressure - sticking
region
Combining (for xk < x < w/2)
 1  x
px
1 
1 − ln 

=
+


 h
2k 2 µ 
2
µ


 1 
 1  x 
p x = 1.15 ⋅ σ flow ⋅  1 − ln 
  + 
 2 µ 
 2 µ   h 
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:
∆p
pedge
0
x=0
x=w/2
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging pressure –
all sticking approximation
pedge = 2k
px
x
2k
∫2k dp = ∫0 h dx
2k
(x )
p x − 2k =
h
px 
x
∴
= 1 + 
2k  h 
x

p x = 1.15 ⋅ σ flow ⋅ 1 + 
 h
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Average forging pressure –
all sticking approximation
w
2
w
2
px
x

dx ∫ 1 + dx
∫
h
pave 0 2k
0
=
=
=
w2
w2
2k

x 
 x + 
2h  0

w2
pave 
w
= 1 + 
2k  4h 
pave
w
2
w

= 1.15 ⋅ σ flow ⋅ 1 + 
 4h 
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
2
Forging force –
all sticking approximation
F forging = pave ⋅ width ⋅ depth
F forging
w

= 1.15 ⋅ σ flow ⋅ 1 +  ⋅ w ⋅ depth
 4h 
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.
Fforging = Fsliding + Fsticking
F forging = ( pave ⋅ A)sliding + ( pave ⋅ A)sticking
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Material Models
Strain hardening (cold – below recrystallization point)
σ flow = Y = Kε
n
Strain rate effect (hot – above recrystallization point)
σ flow = Y = C (ε& )
m
1 dh v
platen velocity
ε& =
= =
h dt h instantaneous height
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-1
•
•
•
•
•
•
•
36”
Lead 1” x 1” x 36”
1”
σy = 1,000 psi
hf = 0.25”, µ = 0.25
1”
Show effect of friction on total forging force.
Use the slab method.
Assume it doesn’t get wider in 36” direction.
Assume cold forging.
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
xk
1
1
=
ln
h 2µ 2µ
0.25
1
xk =
ln
= 0.347"
2 × 0.25 2 × 0.25
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-3
• Sliding region:
 2 µx 

px = 1.15 ⋅ σ flow ⋅ exp
 h 
 f 
= 1150 ⋅ exp(2 x )
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-4
• Sticking region
 1 
 1  x
p x = 1.15 ⋅ σ flow ⋅  1 − ln
  +
 2 µ 
 2µ   h f
= 1150 ⋅ (0.6 + 4 x )
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton



Forging - Ex. 1-5
Forging pressure (psi)
sliding
sliding
sticking
friction
hill
12000
10000
8000
xk
xk
6000
4000
P(sticking)
2000
P(sliding)
0
0
0.5
1
1.5
2
2.5
3
Distance from forging edge (in)
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
3.5
4
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:
Force = ∫∫ p ⋅dA
• since we have plane strain
x
F
= ∫ p x dx
unit depth 0
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-8
• We must solve separately for the sliding
and sticking regions
F forging
  w/ 2

  xk



+ 2  ∫ p x dx .depth 
= 2  ∫ p x dx .depth 


 0

 x





 sliding
 k
 sticking
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-9
Sliding first
xk
 2 µx 
∫0 exp h dx h  2µx  xk
=
exp
=

2µ
unit depth
xk − 0
 h 0
pave
1.15σ flow
=
h
2µ
  2 µxk
exp
  h
(xk − 0)
 
 − 1
 
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-12
Substituting values
sliding
0.25   2 × 0.25 × 0.347  
exp
 − 1

0.25
2 × 0.25  
 
= 1.44
=
(0.347 − 0)
unit depth
pave
1.15σ flow
sticking
1 
1  4
1  42

2
 − 0.347 
 − 0.347  +
1 − ln
2 × 0.25 
2 × 0.25  2
 2 × 0.25  4

=
4
unit depth
− 0.347
2
= 5.3
pave
1.15σ flow
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-10
Sticking next
w
 1 
1  x
∫x  2µ 1 − ln 2µ  + h dx
= k
w −x
unit depth
2 k
pave
1.15σ flow
2
w
2
 1 
x2 
1 
1 − ln
 ⋅ x +


2µ 
2µ 
2h 

=


w −x
k
2




xk
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-11
(
)
1 
1 w
1  w2
2
− xk 

1 − ln
 2 − xk +
2µ 
2µ 
2h  4

=
w −x
unit depth
2 k
pave
1.15σ flow
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-13
Now calculate the area/unit depth
Asliding = 0.347 × 2 = 0.69
Asticking = 4 − 2 × 0.347 = 3.31
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging - Ex. 1-14
Now calculate the forces
(
F
= 1.15σ flow ( pave ⋅ A)sliding + ( pave ⋅ A)sticking
unit depth
F
= 1150 × ((1.44 × 0.69 ) + (5.3 × 3.31))
unit depth
= 21,110 lb / inch
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
)
Forging - Ex. 1-15
• Now, assume all sticking, so:
F
= 1.15σ flow ⋅ w f
unit length

wf
⋅ 1 +
 4h
f

4 

= 1150 ⋅ 4 ⋅ 1 +

 4 ⋅ 0.25 
= 23,000 lb / inch depth
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton




Forging - Ex. 1-16
or since the part is 36” deep:
F(both) = 759,960 lbs = 380 tons
F forging
w

= 1.15 ⋅ σ flow ⋅ 1 +  ⋅ w ⋅ depth
 4h 
F(all sticking) = 828,000 lbs = 414 tons
 µw 
F forging = 1.15 ⋅ σ flow ⋅ 1 +
 ⋅ w ⋅ depth
2h 

F(all sliding) = 496,800 lbs = 225 tons
All sticking over-estimates actual value.
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
Forging – Effect of friction
• Effect of friction coefficient (µ) – all sticking
Friction coefficient Fmax (lbf/in depth)
0
4600
0.1
11365
0.2
19735
0.25
21331
0.3
22182
0.4
22868
0.5
23000
• Friction is very important
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
xk
2
2
0.573
0.347
0.213
0.070
0
Stick/slide
slide
slide
both
both
both
both
stick
Forging - Ex. 1-17
Forging force (lbf/in
depth)
Forging force vs. stroke – all sticking
25000
mu=0
mu=0.1
mu=0.2
mu=0.25
mu=0.3
mu=0.4
mu=0.5
20000
15000
10000
5000
0
0
0.2
0.4
Stroke (in)
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
0.6
0.8
Forging - Ex. 1-19
Maximum forging force vs. friction coefficient (µ)
all sticking
Max forging force
(lbf/in depth)
25000
20000
15000
10000
5000
0
0
0.1
0.2
0.3
0.4
Friction coefficient
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton
0.5
0.6
Summary
• Slab analysis
– frictionless
– with friction
– Rectangular
– Cylindrical
• Strain hardening and rate effects
• Flash
• Redundant work
Prof. Ramesh Singh, Notes by Dr.
Singh/ Dr. Colton