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4. Yield Criterion
Form of Initial Yield Surface
Isotropic Metals
Denote the initial yield surface by
0)(0 =ijf
with f0
a function of stress only. Assume material is isotropic in its
reference state (t
= 0), i.e., it has no preferred directions. Then f0
is independent of the particular
(Cartesian) coordinate system used. At a fixed point, take ;
axes along the principaldirections of the stress tensor; the value
of f
0depends on
1 , 2 , 3 the principal
directions, i.e.,
),,(),,()( 3213210 IIIFgf ij == (1)
where 1I , 2I and 3I are the invariants of ij and i satisfy
032
2
1
3 =+ III
The invariants can be written as
kkI =1
)(2
12 ijijjjiiI =
)det(3 ijI = .
Also
),,(),,(),,( 312132321 ggg ==
etc. That is, g is completely symmetric in 1 , 2 , 3 .
Define hydrostatic pressure p by
13
1
3
1Ip kk ==
In general F depends on 1I but this dependency is found
(experimentally) to be very
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slight for metals.
Assume (for metals) that yield surfaces, initial and subsequent,
are independent ofhydrostatic stress. Then
a) A hydrostatic stress alone does not cause plastic flow, i.e.,
ijij p = lies inside
the yield surface for all p .
b) If ij is a plastic state, so is ijijij pq = for all p .
NOTE : This is not true for soils and metal powders.
Define deviatoric stressij by
ijijijkkijij p +==3
1
then
031 == kkmmkkkk
ij
is unchanged by addition of a hydrostatic stress toij and also
ij defines ij
uniquely to within a hydrostatic stress. Therefore for isotropic
metals, the yield stresscan be defined in terms of the invariants
of
ij .
Define
01
==kk
J
)(2
1
2
1 23
2
2
2
12 ++== ijijJ
)(3
1
3
1 33
3
2
3
13 ++== kljkijJ
Then the initial yield surface can be written as
( ) 0, 32 = JJF (2)
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Assume : No Bauschinger effect initially-yield strength is same
in tension as
compression. If ij is a plastic state as is ij . Then F in (2)
must be evenin
3J .
Yield Locus in -plane (isotropic metals)
Consider the form of (1) of initial yield surface
0),,(321
=g
Instead of 9-D space, we need only 3-D space for representing
stress state in terms
of 1 , 2 , 3 If ),,( 321 is on yield surface, as is ),,( 321 +++
.
Hence yield surface is a cylinder with generators parallel to
the line 321 ==
[(1, 1, 1) vector] and perpendicular to the -plane, where
-plane : 0321 =++
The yield surface is determined by its intersection with the
-plane.
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perspective view
If ),,( 321 is on yield surface, so is ),,( 231 , etc., by
isotropy. So yield
surface is symmetric w.r.t. 1 , 2 3 axes (in -plane).
No Bauschinger effect locus is symmetric w.r.t.origin.
Therefore locus is symmetric about 6 equally spaced lines. Then
B, C, D, E, F all lie
on locus. Locus must lie outside or on hexagon ABCDEFA since
yield surface must
be convex. Likewise it must lie inside or on A'B'C'D'E'F'A'.
Therefore it is bounded by these two hexagons and the location
of point A, the yield
stress in tension, severely restricts its locations.
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C DD
Specific Yield Criteria - Surfaces
1) Tresca's Criterion (1864)
Plastic yielding occurs when the maximum shear stress reaches a
critical value, say
Tk . If 231 the maximum shear stress is )(
2
121 . Thus the Tresca's
criterion is
Tjiji
k2max,
=
Note thatTk221 = is normal to (1, 1, 0) and parallel to (1, 1,
1) and (0, 0, 1).
projectingTk2)( 21 = , we have
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Therefore, it is shown that Tresca's criterion is represented by
a hexagon in the -
plane. Also note that this Tresca's yield criterion satisfies
all the conditions for initial
yield criterion.
For uniaxial tension
0,0, 321 === Y
From the Tresca's yield criterion
TkY 2121 ===
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For pure shear
0, 321 === shearkThe Tresca's criterion states
Tshear kk 22)( 1121 ===
Therefore, kT in the Tresca's yield criterion in the same as the
yield stress in shear,
sheark .
Considering the results from uniaxial tension and pure shear,
Tresca's criterion
implies
shearT kkY 22 ==
2) Von Mises yield criterion (1913)
Plastic yielding occurs when J2
reaches a critical value
2
22
1Mijij kJ ==
or
22
3
2
2
2
12 )(2
1MkJ =++=
If we use, ijkkijij 3
1= , then 2J can be represented by
[ ]22
31
2
23
2
12
21133
23322
222112
)(
)()()(6
1
Mk
J
=+++
++=
Consider a point ),,( 321 in the principal stress space. The
component parallel
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to (1, 1, 1) is
)(3
1
),,(3
1,3
1,3
1ON
321
321
++=
=
Since -plane is perpendicular to (1, 1, 1), the projection of OP
onto -plane will have
length OP when
( ) ( )
[ ]213
2
32
2
21
2
113
2
3
2
332
2
2
2
221
2
1
133221
2
3
2
2
2
1
133221
2
3
2
2
2
1
2
3
2
2
2
1
2321
2
3
2
2
2
1
22
3
2
2
2
1
2
)()()(3
1
)222(3
1
)222222(3
1
)222(3
1
)(
3
1
ONPO
++=
+++++=
++=
+++++++=
++++=
++=
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Therefore, distance from an arbitrary point ),,( 321 in the
principal stress space to
(1, 1, 1) line is
[ ]213232221 )()()(3
1distance ++=
Comparing this with J2
expression in the stress space
[ ]3
)()()(6
12
2213
232
2212
YkJ M ==++=
Von Mises criterion is represented by a circle with its origin
at (0, 0, 0) on -plane. On
-plane, the radius of the circle is Mk2 , or Y3
2.
For uniaxial tension.
MM kYkY
J
Y
33
1=
)(61
0,
22
22
212
321
==
+=
===
For pure shear
[ ]22
21
22
212
321
)(
)2(61
0,
Mshear
shear
kk
J
k
==
++=
===
Therefore, Von Mises condition gives
shearkY 3=
In a famous experiment, Taylor & Qunney concluded that the
Mises condition was
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better than the Tresca condition. However, if we set
TM kk =
So that the yield surface agree in the pure shear state,
then
MMTMTT YYkkkY 155.13
2222 =====
If we take
TM kk =08.1
then the yield loci of Tresca and Mises in the -plane differ
everywhere by less than
8%.
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C D
3
2
9
4
9
1
9
1|| =++=n
=
=
3
2,
6
1,
6
1
|| n
nn
Now m is contained in -plane 0321 =++ mmm
m is perpendicular to n 02 321 =+ mmm
From the two equations, we have
21
3 0
mm
m
=
=
Therefore
= 0,
2
1,
2
1m .
For a general point ),,( 321 , the projected coordinate system
(a, b) can be
calculated by
321
321
21321
3
2
66),,(
22),,(
+==
+==
n
m
b
a
Then
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[ ]
2
23
22
21
213
232
221
2
213
221
222
2
)()()(
)()()(3
1
)2(6
1
2
)(
J
bar
ijij==
++=
++=
+
=
+=
or
22
2
3
22 == Jr
( )
( )
=
=
12
21311
3
2tantan
a
b
or
C C)(
)2(tan312
2131
=
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Projection of Tresca's Yield Criterion onto -planeCC
When 1 3 2> > CCCCCC
1 2 2 = =k YT C C C C CCCCCDC
C
CDCCCCDCDCCCCCDCDCDC
C
C
C
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C D
C
CC
EC
C
32
3 1 2
2 1tan
( )
( )
=
C
C
C = 0,C
C
2or02 213213
+== C
C
m
= + + = + +
+
= +1 2 31 2
1 2
1 2
32
3 2C
C
0
2
2
3
1222
2111
=
==
==
m
m
C C
C
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C DD
CCCCCCC
C
C
CC = 30 , or tan =
1
3C
C
23 1 2 2 1
= C
2 2 03 2
= C
= 3 2
C
C
m =
+ +
=
+1 2 3 1 2
3
2
3 C
C
C
33
2
3
)(2
3
2
2121222
2121111
=
+==
=
+==
m
m
C
C
CCC
C
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C D
CC
C
CCYield Criteria under Plane Strain ConditionCC
CCCCCC
C
1 2 3
0 0 0 =, , C
C
CMises criterionC
C
= + + =J k2 1 22
2 3
2
3 1
2 216
( ) ( ) ( ) C
C
C3
0= , we haveC
C
= + + =J k2 1 22
2
2
1
2 21
6( ) C
2221
2
2
2
1 3 Yk ==+ C
CCCCCCC
C
( )1
3
2
)(
2
)(2
22
2
21 =
+
YY
C
C
C 1 and 2 are the coordinate axes of the principal stress space
rotated 45
from
1 and 2, respectively.C
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C D
C
CCC
C
Y
Y
Y
Y
Y
Y
=
=>>
=>> C
C
CCCCCCCCCCCCC
ECCC ( , , ) 1 2 3 CC -plane. Recall thatC
C
a
b
=
=
2 1
3 2 1
2
2
3
1
6
1
6
C
C
CC
C
C
tan ( )
= =
= ba
32
32 1
3 2 1
C
C
C
C
= 3tan C
C
C
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C D
CC
=
23 1 2
1 2
C
C
C = 0 ,C
C
3
1 2
2=
+C
23
22
3
2
3
23
22
3
2
3
12
2112
31232122
21
2121
32132111
=
+
=
=++
=
=
+
=
=++
=
C
03 = Csince 0=kk C
321
2 =+=m C
C
C = 0 corresponds to pure shear. Also tan = 0CC = 0C
C
C = 1,C
1 2 3 1 2
2 = + + C
= 2 3
C
C
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CCCC( , , )1 0 0 CC( , , ) + 1 2 2 2 CAlso
note that = 30 .C
CC = 1,C
1 2 3 1 2
2 = C
1 3
C
C
CCCC( , , )0 02 CEC = 30
.C
C
CCCCCCCCCCC
CCCCCCCC 1 1 .C
C
CCCCCC
C
3
1 2 1 2
2 2=
++
C
C
C3
expression into J2
or von Mises criterion, we haveC
C
[ ]
221
2
21
2121221212
221
213
232
221
2
)(4
3
)22
()22
()(2
1
)()()(2
1
+
=
+
++
++=
++=Y
C
2
21
3
2
+=
YCCCCCCC
C
CCCC
C
121 =
Y
C
C
CCC
C
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CCCCCCCCCCCC
CC
C
C
