Introduction to second gradient theory of elasticity - Arjun Narayanan

An Introduction to Higher Gradient Theories of Elasticity
Arjun Narayanan, Ali Javili, Christian Linder
Outline
Introduction
Mathematical Preliminaries
Variational Structure of Gradient Elasticity
FE Discretization of Variational Equation
Numerical Examples
Deformation Measures
Invariants
Computational Micromechanics of Materials Lab Group Meeting

Part I Introduction
Gradients in Classical Field Theories
Newtonian Gravity
Vh
Vh
M
m
m
• Background absolute space.
• Euclidean parallelism is preserved.
narayanan, javili, linder Stanford University Group Meeting – 2 · 26 · 2016 1

Part I Introduction
Gradients in Classical Field Theories
Einsteinian Gravity
Vh
Vh
M
m
m
• Objects follow geodesics.
• Gradient of gravitational potential → Connection → (non-Euclidean) Parallelism.

Part I Introduction
Higher Gradients in Continuum Mechanics
Where are the higher gradients hiding?

Part I Introduction
Where are the higher gradients hiding?
ALL
information is in
φ

Part I Introduction
Look closer at the Taylor expansion of the deformation ﬁeld.

Part II Mathematical Preliminaries
Basic Geometric Notions
Induced bases
• Tangent Space: Basis of contravariant vectors:
xi =
∂x
∂ξi
• Cotangent (Dual) Space: Basis of linear functionals on xi
dξj
(xi) = dξj ∂x
∂ξi
= δ
j
i

Gradients
• The gradient of {·} is deﬁned as:
Grad{·} :=
∂{·}
∂ξi
⊗ dξi

Gradients
• The gradient of {·} is deﬁned as:
Grad{·} :=
∂{·}
∂ξi
⊗ dξi
An arbitrary vector is written as a linear combination of the contravariant basis v = vjxj.
Grad{·}(v) =
∂{·}
∂ξi
⊗ dξi
(vj
xj) = vj ∂{·}
∂ξi
δi
j = vj ∂{·}
∂ξj

Divergences
• The identity is deﬁned as
I = xi ⊗ dξi
I(vj
xj) = xi ⊗ dξi
(vj
xj) = vi
xi

Divergences
• The identity is defined as
I = xi ⊗ dξi
I(vj
xj) = xi ⊗ dξi
(vj
xj) = vi
xi
• The divergence is defined as
Div{·} = Grad{·} .
. I
Specifically,
Div{·} =
∂
∂ξj
{·} · dξj
• The surface divergence is defined as
S(v) = Grad(v · I ) .
. I
= Div (v) + Kv · N

Ingredients For Higher Gradient Elasticity
The Product Rule d(fg) = gdf + fdg
• Consider A = Ajkxj ⊗ xk and v = vixi
Div(v · A) =
∂
∂ξj
(v · A) · dξj
=
∂
∂ξj
(v) · A · dξj
+ v ·
∂
∂ξj
(A) · dξj
= A
.
.
∂
∂ξj
(v) ⊗ dξj
+ v ·
∂
∂ξj
(A) · dξj
Div(v · A) = A
.
. Gradv + v · DivA

The Product Rule d(fg) = gdf + fdg
• Consider A = Ajkxj ⊗ xk and v = vixi
Div(v · A) =
∂
∂ξj
(v · A) · dξj
=
∂
∂ξj
(v) · A · dξj
+ v ·
∂
∂ξj
(A) · dξj
= A
.
.
∂
∂ξj
(v) ⊗ dξj
+ v ·
∂
∂ξj
(A) · dξj
Div(v · A) = A
.
. Gradv + v · DivA
• Generalizable to higher contractions,
B
.
.
. GradA = Div(A
.
. B) − A
.
. DivB

Stokes’ Theorem and The Fundamental Theorem of Calculus
F(x)
f(x)
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2-2
-1.5
-1
-0.5
0
0.5
1
1.5
0
0.2
0.4
0.6
0.8
1
2
b
a
dF
dx
= F(b) − F(a)

Integral Theorems
• Integrate A .. Gradv = Div(v · A) − v · DivA on B0
B0
A
.
. Gradv =
∂B0
v · A · N −
B0
v · DivA (1)
• Integrate A .. Gradv = Div(v · A) − v · DivA on ∂B0
∂B
A
.
. Gradv =
∂2B
v · A · M −
∂B
v · S(A) − GradNv · (A · N) (2)
• We will repeatedly apply equations 1 and 2 in a variational context to obtain the strong form
of the equilibrium equations.

Part III Variational Structure of Gradient Elasticity
Higher Gradient Elasticity
Obtaining the Strong Form
• ψtotal = ψtotal
internal + ψtotal
external
δψtotal !
= 0
• ψinternal = ψinternal(Gradφ, Grad2
φ)
• Integrate ψinternal and vary φ
δ
B0
ψinternal(Gradφ, Grad2
φ) =
B0
∂ψinternal
∂Gradφ
.
. Gradδφ +
B0
∂ψinternal
∂Grad2
φ
.
.
. Grad2
δφ

internal + ψtotal
external
δψtotal !
= 0
φ)
δ
B0
φ) =
B0
∂ψinternal
∂Gradφ
.
. Gradδφ +
B0
∂ψinternal
∂Grad2
φ
.
.
. Grad2
δφ
δ
B0
φ) =
B0
P1
.
. Gradδφ
∂B0
δφ·P1·N− B0
δφ·DivP1
+
B0
P2
.
.
. Grad2
δφ
∂B0
Gradδφ
.
.P2·N− B0
Gradδφ
.
.DivP2

internal + ψtotal
external
δψtotal !
= 0
φ)
δ
B0
φ) =
B0
∂ψinternal
∂Gradφ
.
. Gradδφ +
B0
∂ψinternal
∂Grad2
φ
.
.
. Grad2
δφ
δ
B0
φ) =
B0
P1
.
. Gradδφ
∂B0
δφ·P1·N− B0
δφ·DivP1
+
B0
P2
.
.
. Grad2
δφ
∂B0
Gradδφ
.
.P2·N− B0
Gradδφ
.
.DivP2
• The decomposition of B0
P2
... Grad2
δφ can be further expanded,
∂B0
Gradδφ
.
. P2 · N =
∂2B0
δφ · P2
.
. (M ⊗ N) −
∂B0
δφ · S(P2 · N)
+
∂B0
GradN δφ · P2
.
. (N ⊗ N)

• Combine the coeﬃcients of each independent variation in each domain,
δ
B0
ψinternal Gradφ, Grad2
φ =
B0
δφ · (Div(DivP2) − DivP1)
+
∂B0
δφ · (P1 · N − S(P2 · N) − DivP2 · N) + GradN δφ · P2
.
. (N ⊗ N)
+
∂2B0
δφ · P2
.
. (M ⊗ N)

• Combine the coeﬃcients of each independent variation in each domain,
δ
B0
ψinternal Gradφ, Grad2
φ =
B0
δφ · (Div(DivP2) − DivP1)
+
∂B0
δφ · (P1 · N − S(P2 · N) − DivP2 · N) + GradN δφ · P2
.
. (N ⊗ N)
+
∂2B0
δφ · P2
.
. (M ⊗ N)
• The structure of the above equation gives us the structure of the variation of external work
δψtotal
external
δψtotal
external = −
B0
δφ · b −
∂B0
δφ · t −
∂B0
GradN δφ · c −
∂2B0
δφ · l
• Add the two equations and demand the satisfaction of δψtotal = 0 point-wise to get the
strong form.

The Strong Form of the Governing Equilibrium Equations
Div(DivP2 − P1) − b = 0 in B0
(P1 − DivP2) · N − S(P2 · N) − t = 0 on ∂B0
P2
.
. (N ⊗ N) − c = 0 on ∂B0
P2
.
. (M ⊗ N) − l = 0 on ∂2
B0

Part IV FE Discretization of Variational Equation
The Path To a Numerical Implementation
Finite Element Discretization
• For a given deformation field φ, we have the following residual,
R(φ) =
B0
P1
.
. Gradδφ +
B0
P2
.
.
. Grad2
δφ −
B0
δφ · b −
∂B0
δφ · t
−
∂B0
GradNδφ · c −
∂2B0
δφ · l
• φ is an admissible deformation field if R(φ) = 0.
• Use a finite dimensional approximation to the variation in deformation to approximate the
residual,
δφ(ξ, η, ζ) = ∑
I
NI
(ξ, η, ζ)δφI
R(φ) =
B0
δφI
· P1 · Grad(NI
) +
B0
δφI
· P2
.
. Grad2
(NI
) −
B0
(δφI
NI
) · b
−
∂B0
(δφI
NI
) · t −
∂B0
(GradNI
· N)(δφI
· c) −
∂2B0
(δφI
NI
) · l

Finite Element Discretization
• Get rid of the displacement variations, which are arbitrary, to obtain the work-conjugate of
δφI.
RI
(φ) =
B0
P1 · Grad(NI
) +
B0
P2
.
. Grad2
(NI
) −
B0
NI
b
−
∂B0
NI
t −
∂B0
(GradNI
· N)c −
∂2B0
NI
l
• Newton - Raphson solution scheme RI(φn+1) = RI(φn) + ∂RI
∂φJ · ∆φJ !
= 0
KIJ
=
∂RI
∂φJ
=
B0
∂P1
∂F
.
. (I ⊗ GradNJ
) · GradNI
+
B0
∂P2
∂GradF
.
.
. (I ⊗ Grad2
NJ
) .
. Grad2
NI
• Deﬁne material tangents
A =
∂P1
∂F
=
∂
∂F
∂ψinternal
∂F
=
∂2ψinternal
∂F∂F
B =
∂P2
∂GradF
=
∂
∂GradF
∂ψinternal
∂GradF
=
∂2ψinternal
∂GradF∂GradF

Constitutive Equations
• We use the following strain energy density function,
ψinternal =
1
2
µ(F
.
. F − 3 − 2ln(J)) +
1
2
λ
1
2
(J2
− 1) − ln(J) +
1
2
µl2
EAB,CEAB,C
• We have the following component-wise expression for P1 and P2,
[P1]bL = µ([F]bL − [F−T
]bL) +
λ
2
(J2
− 1)[F−T
]bL
+
1
2
µl2
[GradF]bJK([GradF]aJK[F]aL + [F]aJ[GradF]aLK)
[P2]bLM =
µl2
2
[F]bJ([GradF]aLM[F]aJ + [F]aL[GradF]aJM)

Constitutive Equations
• We have the following component wise expression for A and B
[A]bLgM = µ δb
gδL
M + [F−T
]gL[F−T
]bM + λJ2
[F−T
]gM[F−T
]bL
−
λ
2
(J2
− 1)[F−T
]gL[F−T
]bM +
µl2
2
[GradF]bJK [GradF]gJKδL
M + [GradF]gLKδJ
M
[B]bLMgPQ =
µl2
4
[F]bJ[F]gJ(δL
PδM
Q + δL
QδM
P ) + [F]bP[F]gLδM
Q + [F]bQ[F]gLδM
P

Element Formulation
(-1,-1) (+1,-1)
(+1,+1)(-1,+1)
Quadrature Space
Reference Element Deformed Element
Parameter Space

Element Formulation
Computing GradF
X = X(ξ1
, ξ2
) x = x(ξ1
, ξ2
)
∂xi
∂XP
=
∂xi
∂ξα
∂ξα
∂XP

Element Formulation
Computing GradF
X = X(ξ1
, ξ2
) x = x(ξ1
, ξ2
)
∂xi
∂XP
=
∂xi
∂ξα
∂ξα
∂XP
∂2xi
∂XP∂XQ
=
∂
∂XQ
∂xi
∂ξα
∂ξα
∂XP
∂2xi
∂XP∂XQ
=
∂2xi
∂ξβ∂ξα
def. hessian
∂ξβ
∂XQ
∂ξα
∂XP
+
∂xi
∂ξα
∂2ξα
∂XQ∂XP
inv. ref. hessian

Element Formulation
“Inverting” the Hessian
∂XP
∂XQ
=
∂XP
∂ξα
∂ξα
∂XQ
= δP
Q
∂2XP
∂XR∂XQ
=
∂
∂XR
∂XP
∂ξα
∂ξα
∂XQ
= 0
=
∂2XP
∂ξβ∂ξα
∂ξβ
∂XR
∂ξα
∂XQ
+
∂XP
∂ξα
∂2ξα
∂XR∂XQ
= 0
∂2ξγ
∂XR∂XQ
= −
∂2XP
∂ξβ∂ξα
∂ξβ
∂XR
∂ξα
∂XQ
∂ξγ
∂XP

Part V Numerical Examples
Finally, some pictures.
Example 1
reﬁnement no. of elements
1 4
2 16
3 64
4 256
5 1024
length scale
0.00
0.10
0.25
1.00
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
L = 1.0
D=1.0
0.2
0.2
applied
displacem
ent
Figure: Boundary conditions
λ = 8.0, µ = 392.0
How are displacements
and stresses localized?

Example 1: Deformed mesh under reﬁnement
Figure: Length scale =
0.0
0.1
0.25
1.0
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2

Example 1: Von Mises Stress
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
200
400
600
800
1000
1200
1400
1600
0.0
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
5
10
15
20
25
30
35
40
0.1
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
5
10
15
20
25
0.25
0 0.2 0.4 0.6 0.8 1 1.2
0
0.2
0.4
0.6
0.8
1
1.2
4
6
8
10
12
14
16
18
20
22
1.0

Refinement level
1 1.5 2 2.5 3 3.5 4 4.5 5
MaxvonMisesStress
0
200
400
600
800
1000
1200
1400
1600
Figure: Length scale = 0.0
Refinement level
1 1.5 2 2.5 3 3.5 4 4.5 5
MaxvonMisesStress
0
5
10
15
20
25
30
35
40
45

Refinement level
1 1.5 2 2.5 3 3.5 4 4.5 5
MaxvonMisesStress
0
5
10
15
20
25
Refinement level
1 1.5 2 2.5 3 3.5 4 4.5 5
MaxvonMisesStress
0
5
10
15
20
25

Part VI Deformation Measures
In search of a better constitutive relation
Can we do better than 1
2 µl2EAB,CEAB,C ?
• The challenge lies in identifying better (what do we mean by that?) measures of deformation
that can be constructed from the second gradient of deformation.
• First gradient theories have a well deﬁned tensorial measure of deformation – E
(Green-Lagrange strain). The invariants of E describe local stretch. A general constitutive
law is formulated in terms of the invariants of E.
Figure: The ﬁrst gradient of deformation describes changes in lengths and angles

Can we do better than 1
2 µl2EAB,CEAB,C ?
• Better =⇒ geometric invariance.
• Example: n2 numbers Aα
β in ξα coordinate system. n2 numbers A
p
q in φp coordinate system.
A is once contravariant once covariant (i.e. type (1,1)) tensor if,
A
p
q = Aα
β
∂φp
∂ξα
∂ξβ
∂φq
• We want a tensorial measure of deformation S that is constructed from the second gradient
of deformation – GradF. The most general second gradient constitutive law must be
expressible in terms of the invariants of S .
Figure: The second gradient of deformation describes changes in the tangent space under “parallel transport”

Connections
Reference
Aα =
∂X
∂ξα
Aα,β =
∂2X
∂ξβ∂ξα
=
A
Γ
µ
βα
∂X
∂ξµ
A
Γ
γ
βα = dξγ
(Aα,β)
Deformed
aα =
∂x
∂ξα
aα,β =
∂2x
∂ξβ∂ξα
=
a
Γ
µ
βα
∂x
∂ξµ
a
Γ
γ
βα = dξγ
(aα,β)
• The coefficients
A
Γ
γ
βα and
a
Γ
γ
βα describe the “connection” in the reference and deformed
configurations.
• They describe local (i.e. infinitesimal) parallel transport of vectors.

Are Connections Tensorial?
Transform coordinates ξα → φp
Reference
Aα =
∂X
∂ξα
=
∂X
∂φp
∂φp
∂ξα
Aα,β =
∂
∂ξβ
∂X
∂φp
∂φp
∂ξα
=
∂2X
∂φq∂φp
∂φq
∂ξβ
∂φp
∂ξα
+
∂X
∂φp
∂2φp
∂ξβ∂ξα
dξγ
(Aα,β) =
∂ξγ
∂φr
dφr ∂2X
∂φq∂φp
∂φq
∂ξβ
∂φp
∂ξα
+
∂X
∂φp
∂2φp
∂ξβ∂ξα
A
Γ
γ
βα =
A
Γr
qp
∂ξγ
∂φr
∂φq
∂ξβ
∂φp
∂ξα
+
∂2φr
∂ξβ∂ξα
∂ξγ
∂φr
Deformed
aα =
∂x
∂ξα
=
∂x
∂φp
∂φp
∂ξα
aα,β =
∂
∂ξβ
∂x
∂φp
∂φp
∂ξα
=
∂2x
∂φq∂φp
∂φq
∂ξβ
∂φp
∂ξα
+
∂x
∂φp
∂2φp
∂ξβ∂ξα
dξγ
(aα,β) =
∂ξγ
∂φr
dφr ∂2x
∂φq∂φp
∂φq
∂ξβ
∂φp
∂ξα
+
∂x
∂φp
∂2φp
∂ξβ∂ξα
a
Γ
γ
βα =
a
Γr
qp
∂ξγ
∂φr
∂φq
∂ξβ
∂φp
∂ξα
+
∂2φr
∂ξβ∂ξα
∂ξγ
∂φr

Are Connections Tensorial?
No, but the component-wise diﬀerence of connection coeﬃcients does
transform like a type (1,2) tensor!
S
γ
βα =
A
Γ
γ
βα −
a
Γ
γ
βα =
∂ξγ
∂φr
∂φq
∂ξβ
∂φp
∂ξα
A
Γr
qp −
a
Γr
qp

Metrics
• We have access to the Euclidean inner product from the embedding X(ξ1, ξ2) and x(ξ1, ξ2).
The Euclidean inner product is by definition symmetric and positive definite.
• How do the components of this inner product transform under ξα → φp?
Reference
A
gαβ =
∂X
∂ξα
·
∂X
∂ξβ
A
gαβ =
∂X
∂φp ·
∂X
∂φq
∂φp
∂ξα
∂φq
∂ξβ
A
gαβ =
A
gpq
∂φp
∂ξα
∂φq
∂ξβ
Deformed
a
gαβ =
∂x
∂ξα
·
∂x
∂ξβ
a
gαβ =
∂x
∂φp ·
∂x
∂φq
∂φp
∂ξα
∂φq
∂ξβ
a
gαβ =
a
gpq
∂φp
∂ξα
∂φq
∂ξβ
• The components
A
gαβ and
a
gαβ transform as a (0,2) tensor.
• Moreover,
A
gαβ and
a
gαβ inherit the properties of symmetry and positive definiteness from the
Euclidean structure.
•
A
gαβ and
a
gαβ are valid Riemannian metrics induced by the parametrization.

Metric: Bilinear form or
linear transformation?
• Either of
A
gαβ or
a
gαβ deﬁne the components of a valid metric tensor for the problem.
• We need only one of them – let us choose
A
gαβ
• The metric tensor is deﬁned as
A
g =
A
gαβdξα ⊗ dξβ

Metric: Bilinear form or
linear transformation?
• Either of
A
gαβ or
a
gαβ deﬁne the components of a valid metric tensor for the problem.
• We need only one of them – let us choose
A
gαβ
• The metric tensor is deﬁned as
A
g =
A
gαβdξα ⊗ dξβ
• Consider u = uµAµ and v = vνAν
TX B × TX B → R
A
g(u, v) =
A
gαβdξα
⊗ dξβ
(uµ
Aµ, vν
Aν)
=
A
gαβuα
vβ
=
∂X
∂ξα
·
∂X
∂ξβ
uα
vβ
= uα ∂X
∂ξα
· vβ ∂X
∂ξβ
A
g(u, v) = (u · v)
TXB → T∗
X B
A
g(u) =
A
gαβdξα
⊗ dξβ
(uµ
Aµ)
=
A
gαβuβ
uα
dξα
= uαdξα
The components of the metric (and its
inverse) can be used to lower (and raise)
the indices of tensorial components.

Part VII Invariants
Invariant contractions of the S tensor
• Geometric invariance – geometric objects have well deﬁned transformation laws. All tensors
are geometric invariants under diﬀeomorphisms.
• A scalar (real valued function) is a type (0,0) tensor. Scalars are invariant – one simply
changes the functional dependence of the scalar. f(φp) = f(φp(ξα))
• Respecting variance is good – Covariant-Contravariant contractions are always tensorial. For
example,
S
γ
βα = Sr
qp
∂ξγ
∂φr
∂φq
∂ξβ
∂φp
∂ξα
Multiply by δα
γ,
δα
γS
γ
βα = Sr
qp
∂ξγ
∂φr
∂φq
∂ξβ
∂φp
∂ξα
δα
γ
S
γ
βγ = Sr
qp
∂ξγ
∂φr
∂φp
∂ξγ
δ
p
r
∂φq
∂ξβ
S
γ
βγ = Sr
qr
∂φq
∂ξβ
• We are looking for the scalar invariants of S. We want all functionally independent scalar
contractions of S.

Part VII Invariants
• To contract indices occupying positions of same variance, one needs a mechanism to
counteract the matching variances. This is achieved by using the metric tensor. For example,
S
γ
βγ = Sr
pr
∂φp
∂ξβ
A
gαβ
=
A
gpq ∂ξα
∂φp
∂ξβ
∂φq
A
gαβ =
A
gpq
∂φp
∂ξα
∂φq
∂ξβ
S
γ
βγS
µ
αµ
A
gαβ
= Sr
pr
∂φp
∂ξβ
St
qt
∂φq
∂ξα
A
gαβ
= Sr
prSt
qt
A
guv ∂φp
∂ξβ
∂ξβ
∂φv
δ
p
v
∂φq
∂ξα
∂ξα
∂φu
δ
q
u
S
γ
βγS
µ
αµ
A
gαβ
= Sr
prSt
qt
A
gqp

Part VII Invariants
• Example invariants
SI = S
γ
βγS
µ
αµ
A
gαβ
SII = S
γ
µα Sν
ρβ
A
gµα A
gρβ A
gγν
SIII = S
γ
µα Sν
ρβ
A
gρµ A
gβα A
gγν
SIV = S
γ
µα Sν
γβ S
β
ξη S
ρ
νρ
A
gµα A
gξη
But we need a more systematic way of doing this. We need to answer the following questions,
• How many independent invariants do we have?
• What geometric signiﬁcance can we attach to them?

Part VII Invariants
Visualizing Invariants
Length Scale = 0.0
Figure: SI Figure: SII
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
5000
10000
15000
20000
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
#104
-1
-0.5
0
0.5
1
1.5

Part VII Invariants
Length Scale = 0.0
Figure: SIII Figure: SIV
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
#104
0
1
2
3
4
5
6
7
8
9
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
#107
-1.5
-1
-0.5
0
0.5
1

Part VII Invariants
Length Scale = 0.25
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-0.5
0
0.5
1
1.5
2
2.5
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5

Part VII Invariants
Length Scale = 0.25
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1.5
-1
-0.5
0
0.5
1
1.5

Part VII Invariants
Length Scale = 1.0
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6

Part VII Invariants
Length Scale = 1.0
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.5
1
1.5
2
2.5
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-0.1
-0.05
0
0.05
0.1

Part VII Invariants
Questions?
Thank You!

Introduction to second gradient theory of elasticity - Arjun Narayanan

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