q q q
p
p
p
σ1
σ1
σ1
σ2
Cone type
σ3
σ2
Cap type
σ3 σ2
2 yield surface
σ3
（3）hardening law f (σ ) = k
ij
σ
k2 k1
After yield, k changes, How does k change? Which factor causes k change? k increases — hardening k decreases — softening k constant — theoretical
σ1
Variation of yield surface
if f > k , k changes, yield surface moves
σ
k2 k1
q
σ2
f =k
σ3
σ1
ε
p
σ2
σ3
Loading and unloading
Current stress state — on yield surface, A new stress increment is applied. * unloading
σ1
Failure surface —— locus of the points in stress space which represent failure
Failure criterion
σ2
σ3
Trasca criterion
= kf 2 σ σ 2 σ σ 1 σ σ 3 σ1 σ 2 σ σ 1 σ σ 3 k f 2 k f 2 k f 3 k f 3 k f 1 kf = 0 2 2 2 2 2 2
g (σ δε = δλ σ
p ij
ij
)
g (σ
ij
)
δε
δε 3p
p
2
ij
Strain space is overlapped with stress space. Plastic strain increment is perpendicular to plastic potential surface
σ
k1 k2
hardening
ε
softening
k = F (H ) f (σ ij ) = F (H ) f (σ ij , H ) = 0
ε
σ
H — hardening parameter, a physical variant which causes k change For a given value of H, yield surface is defined.
σ
σ ij
σ ij
δσ ij
σ ij
δε p
σ ij
δW p
ε
δσ ij
f
σ1
σ ij = σ ij σ + δσ ij
ij
in loading unloading ,
δW = σ ijδε > 0
p
(σ
ij
σ ij )δε ijp + δσ ijδε ijp > 0
p ij
σ ij
α
δε ijp
f f
g= f
σ2 ε2
σ2 ε2
Non-associated flue rule
p ij
σ 1 ε1
σ ij
σ ij
α
δε ijp
f
(σ
ij
σ ij )δε ijp < 0 σ ij
σ2 ε2
α ≥ 90°
is perpendicular to yield surface f if not,
α ≥ 90°
ε σ1 1 σ ij
σ 1 ε1
σ ij
δε ijp
α
σ ij
4. Elasto-plastic model
ε =εe +ε p ε e —— recoverable strain, elastic ε p —— irrecoverable strain, plastic
σ
{ε } = {ε e }+ {ε p }
Plastic strain
ε
p
εe
ε
。failure criterion, yield criterion 。hardening law 。flue rule
σ
π
plane
σ1
Mohr-Coulunb Trasca Mises
σ1
σ2
σ3
σ2
σ3
σ1 σ3
2
1 2 1 q= 2 q=
=
σ1 +σ3
2
sin + c cos
σ 2 = σ 3 + b(σ 1 σ 3 )
(σ 1 σ 2 )2 + (σ 2 σ 3 )2 + (σ 1 σ 3 )2 (σ 1 σ 3 )2 (1 b )2 + b2 (σ 1 σ 3 )2 + (σ 1 σ 3 )2
σ1 σ 3
Hexagonal column Saturated soil, undrained
τ
c
σ 1 σ 2 = 2c
kf = c
σ1
σ
σ2
σ3
Mises criterion
q = kf
q= 1 2
σ1
(σ 1 σ 2 )2 + (σ 2 σ 3 )2 + (σ 3 σ 1 )2
Circular column surface Extensive Mises criterion
（1）failure criterion
σ < kf σ = kf
elastic failure
f
σ
kf
simple stress state
f f
(σ ) = k complicate stress state (σ ) —— failure function
ij ij
ε
variables are stress components
[
]
σ2
—— second deviator stress invariant 1 J2 = q I1 = 3 p 3
σ3
Circular cone surface
Cambridge university q = Mp
q M
q = M (p + pr )
q= 1 2
1 3
(σ 1 σ 2 )2 + (σ 2 σ 3 )2 + (σ 3 σ 1 )2
df = f dσ ij < 0 σ ij
σ1
dσ ij
n
α
2 vectors multiply elastic
σ2 σ1
n
α > 90 °
* loading
f df = dσ ij > 0 σ ij
α
dσ ij
α < 90 °
plastic
* neutral loading
f df = dσ ij = 0 σ ij
σ2
σ 1 ε1
p ij
If σ on yield surface,
ij
δσ ijδε > 0
(σ
σ ij )δε ijp > δσ ijδε ijp ij
(σ
ij
σ ij )δε ijp ≥ 0
σ ij
σ ij
α
δε ijp
f
α ≤ 90°
σ2 ε2
derivation
*All the points which represent the stress σ ij must be on the other side of the plane perpendicular to δε ijp yield surface f must be convex. if concave * δε
σ
k
( ) ( )
ε
τ =τ f τ <τf
p
theoretical material, yield = failure geotechnical material, yield ≠ failure
τ
τ f = σ tan + c
τ =0
σ
εv
p
εv
Yield surface
f (σ ij ) — yield function, corresponding to yield surface in stress space yield surface — locus of the points in stress space which reach yield
δε
— plastic strain increment Direction of δε p determines each component of the plastic strain increment. Flue rule gives direction of δε p
p
δε1p
δε
p
Conceive a plastic potential function
dε vp deijp = dε ijp δ ij 3
1, (i = j ) δ ij = 0, (i ≠ j )
4.
ε p = ∫ dε p = ∫ dε ijp dε ijp
5.
H = f ε vp , ε sp