forging analysis - 1ramesh/courses/me649/forming...prof. ramesh singh, notes by dr. singh/ dr....
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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