河海大学殷宗泽高等土力学3(英文)Elasto-plastic model

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
(
)
（4）flue rule
How the plastic strain develops among the strain components? How to determine the proportion of the strain components?
ε1 σ
1
δε p
g (σ ij )
g δε = δλ p
p v
g δε = δλ q
p s
ε2
σ
σ ε3
3
2
Associated flue rule
g (σ ij ) = f (σ ij )
Drucker’s postulation — an element exits initial stress state, loading slowly, and then unloading, during loading, work done by external agency is positive. And during loading and unloading, work done by external agency is not negative.
σ1
n
σ2
α
dσ ij
α = 90 °
limit of elastic
σ2
Yield surface for geo-material
f (σ ij ) = k σ ij → σ 1 , σ 2 , σ 3 I1 , I 2 , I 3 J2, J3 p, q, b
Independent of direction of coordinate system x,y,zλq Nhomakorabeaq0
σ2
σ3
Lade - Duncan criterion
I1 = kf I3 I1 = σ 1 + σ 2 + σ 3 I 3 = σ 1σ 2σ 3
3
σ1
σ2
σ1
σ3
σ2
σ3
（2）yield criterion
Concept of yield
simple stress elastic σ <k plastic, yield σ =k complicate stress elastic f σ ij < k plastic, yield f σ ij = k
q = f (p, k f
)
σ2
f ( p, q ) = k f
—— first
σ3
σ1
Geotecnical material Drucker-Prager
βI1 + J2 = J2 = k f I1 = σ 1 + σ 2 + σ 3
stress invariant
1 (σ 1 σ 2 )2 + (σ 2 σ 3 )2 + (σ 3 σ 1 )2 6
theoretical
k 2 k1
ε
H →W p
1.
ε vp
ε sp
εp
ε vp & ε sp
W p = ∫ σ ij dε ijp W p = ∫ pdε vp + qdε sp
2.
3.
ε p = ∫ dε 1p +dε 2p + dε 3p
v
ε sp = ∫ dε sp = ∫
2 p p deij deij 3
p=
(σ 1 + σ 2 + σ 3 )
pr
2 3
p
σ1
q
σ1
3p
σ2
σ3
σ2
σ3
Mohr-Coulomb criterion
τ = σ tan + c σ1 σ 3 σ1 + σ 3
2 = 2 sin + c cos
τ
c cos
c
σ1 + σ 3
2
sin
Hexagonal cone with equal edges but unequal angles
σ1
b=0 b =1
q1
q = 1 b + b 2 (σ 1 σ 3 ) = λ (σ 1 σ 3 )
(σ 1 σ 3 ) = q
λ
λ＝ 1－b + b 2
1 2b
(σ 1 + σ 3 ) = 2 p +
6λ sin ( p + c cot ) q= 3 (1 2b) sin 6λ sin ( p + c cot ) b = 0, q0 = 3 sin 6λ sin ( p + c cot ) b = 1, q1 = 3 + sin