Prof. Dr. Eleni Chatzi Dr. Giuseppe Abbiati, Dr ... · Understanding limitations of geometrically linear analysis Understanding diﬀerent nonlinear strain measures Understanding

Introduction to geometrical non-linearities

Prof. Dr. Eleni ChatziDr. Giuseppe Abbiati, Dr. Konstantinos Agathos

Lecture 4 - 19 October, 2017

Institute of Structural Engineering, ETH Zürich

October 19, 2017

Institute of Structural Engineering Method of Finite Elements II 1

Outline

1 Introduction

2 An example of geometrical non-linearity

3 The deformation gradient

4 Non-linear strain measures

5 Stress measures

6 Constitutive equations

7 Weak form of equilibrium equations

8 Linearization of the weak form


Learnig goals

Understanding limitations of geometrically linear analysis

Understanding different nonlinear strain measures

Understanding different definitions of stress

Understanding how the weak form is linearized


Significance of the lecture

Applications:

Structures undergoing large displacements and/or rotationssuch as cables, arches and shells

Materials such as elastomers and biological/soft tissue

Modeling of plastically deforming materials


Large displacements of a rigid beam

Small displacements

Example taken from: “Nonlinear Finite Element Methods” by P. Wriggers,Springer, 2008



Small displacements

Equilibrium equation:

Fl = cϕ




Small displacements


Fl = cϕ

Large displacements




Small displacements


Fl = cϕ

Large displacements


Fl cos (ϕ) = cϕ




0 20 40 60 80

ϕ

0

2

4

6

8

10

Fl c

Small displacements

Large displacements

Force versus rotation




In the previous example the beam was considered rigid

Non linear strain measures would have to be used to take intoaccount the beam flexibility

In the following non linear strain measures are introduced


Reference/current configuration

We consider a deformable body described by a set of material points.We define two states of the body:

Reference configuration → Configuration of the body in thebeginning of the deformation (usually undeformed).

Current configuration → Configuration of the body (usuallydeformed) at a given time t.


Description of motion

x (X, t) = X + u (X, t)

where

x (X, t) location incurrent configuration

X location in referenceconfiguration

u (X, t) displacementvector


The deformation gradient

We define the deformation gradient as:

F = ∇x (X, t) = ∇ (X + u (X, t))⇒ F = I +∇u (X, t)

⇒ Fij = δij +duidXj



Infinitesimal element in the deformed configuration:

dx = d (X + u (X, t))= dX +∇u (X, t) dX= (I +∇u (X, t)) dX

⇒ dx = FdX⇒ dxi = FijdXj

→ F maps infinitesimal line elements from the reference to thedeformed configuration



Expression for the deformation gradient:

F = ∇x =

x1,1 x1,2 x1,3x2,1 x2,2 x2,3x3,1 x3,2 x3,3

It follows that:

J = detF 6= 0

dX = F−1dx


Transformation of area and volume elements

Area is transformed as:

da = nda = JF−T NdA = JF−T dA

where:

da = nda is the area element in the current configurationdA = NdA is the area element in the reference configuration

And volume as:

dv = JdV

where:

dv is the volume element in the current configurationdV is the volume element in the reference configuration


Strain measures

We consider the square of an infinitesimal element in thecurrent configuration:

dx · dx = (FdX) · (FdX) = dX ·(

FT F)· dX

C = FT F → right Cauchy-Green deformation tensor


Green-Lagrange strain

Difference of squares of lengths of infinitesimal elements:

dx · dx− dX · dX =dX ·(

FT F)· dX− dX · dX =

=dX ·(

FT F− I)· dX

E = 12(

FT F− I)

= 12 (C− I) → right Green-Lagrange strain tensor

E = 12[(I +∇u)T (I +∇u)− I

]= 12

(∇u +∇uT +∇uT∇u

)

In 1D: E = l2 − L22L2 , where L is the reference length, l is the current


Strain measures

We consider the square of an infinitesimal element in thereference configuration:

dX · dX =(

F−1dx)·(

F−1dx)

= dx ·(

F−T F−1)· dx

c = F−T F−1 → Cauchy deformation tensor


Euler-Almansi strain

Difference of squares of lengths of infinitesimal elements:

dx · dx− dX · dX =dx · dx− dx ·(

F−T F−1)· dx =

=dx ·(

I− F−T F−1)· dx

e = 12(

I− F−T F−1)

= 12 (I− c) → Euler-Almansi strain tensor

In 1D: e = l2 − L2

2l2 , where L is the reference length, l is the current


Comparison of strain measures

Green-Lagrange strain tensor: E = 12 (C− I)

Almansi strain tensor: e = 12(−1)(C−1 − I

)Both of the above are members of the Seth-Hill family of strainmeasures: E(m) = 12m (C

m − I)

Another special case is the logarithmic strain: E(0) = 12 ln C


ExampleBar under rotation and stretch

Deformation gradient

F = ∇x =

lL cos (θ) − sin (θ)lL sin (θ) cos (θ)

Deformed configuration

x =[

x1x2

]=

lL cos (θ) X1 − sin (θ) X2lL sin (θ) X1 + cos (θ) X2


Example

Displacements

u =[

u1u2

]=

( l

L cos (θ)− 1)

X1 − sin (θ) X2lL sin (θ) X1 + (cos (θ)− 1) X2

Displacement gradient

∇u =

lL cos (θ)− 1 − sin (θ)lL sin (θ) cos (θ)− 1


Example

Linear strain measure

� = 12(∇u +∇uT

)=

l cos (θ)− LL (1− L) sin (θ)2L(1− L) sin (θ)2L cos (θ)− 1

Green-Lagrange strain

E = 12(∇u +∇uT +∇uT∇u

)=

l2 − L22L2 00 0


Cauchy stress tensor

Force per unit of deformed area: t(n) = lima→0

f(n)a =

df(n)

da

Cauchy stress tensor: t(n) = σσσ · n


1st Piola-Kirchoff stress tensor

Force per unit of undeformed area: t(N) = limA→0

f(N)A =

df(N)

dA

⇒ f(N) in deformed configuration

1st Piola-Kirchoff stress tensor: t(N) = P ·N

t(N)dA = t(n)da⇒ P ·NdA = σσσ · nda

nda = JF−T ·NdA

⇒ P = JσσσF−T


2nd Piola-Kirchoff stress tensor

Force per unit of undeformed area: T(N) = limA→0

F(N)A =

dF(N)

dA

⇒ F(N) in undeformed configuration

⇒ dF(N) = F−1df(N) transformation of the force vector

2nd Piola-Kirchoff stress tensor: T(N) = S ·N

⇒ dF(N) = F−1df(N) ⇒ S ·NdA = F−1P ·NdA⇒ S = F−1P


Comparison of stress measures

Stressmeasure

Cauchy stress(σ)

1st Piola-Kirchoffstress (P)

2nd Piola-Kirchoffstress (S)

Area Current Reference ReferenceForce Current Current ReferenceTransform - P = JσσσF−T S = F−1P =

JF−1σσσF−T


St. Venant materials

St. Venant materials:

A linear relationship between Green-Lagrange strain and secondPiola-Kirchhoff stress is assumed

The relationship corresponds to Hooke’s law for infinitesimaldisplacements

Good approximation when displacements and rotations are largebut deformations are small



Stress strain relationship:

S = DE

Using Voigt notation:

S11S22S33S23S13S12

=

D1111 D1122 D1133 D1123 D1113 D1112D2211 D2222 D2233 D2223 D2213 D2212D3311 D3322 D3333 D3323 D3313 D3312D2311 D2322 D2333 D2323 D2313 D2312D1311 D1322 D1333 D1323 D1313 D1312D1211 D1222 D1233 D1223 D1213 D1212

·

E11E22E33

2E232E132E12

where Dijkl are material parameters



In terms of E and ν the above can be written as:

S11S22S33S23S13S12

= E(1 + ν) (1− 2ν)

1− ν ν ν 0 0 0ν 1− ν ν 0 0 0ν ν 1− ν 0 0 00 0 0 1− 2ν2 0 0

0 0 0 0 1− 2ν2 0

0 0 0 0 0 1− 2ν2

·

E11E22E33

2E232E132E12


Hyperelastic materials

The above relation is not accurate for elastic materialsundergoing large deformations

Those materials are identified as hyperelastic

Examples of such materials are rubber and soft tissue

For more details we refer to: Lecture Notes by Carlos A. FelippaNonlinear Finite Element Methods (ASEN 6107)


https://www.colorado.edu/engineering/CAS/courses.d/NFEM.d/NFEM.Ch08.d/NFEM.Ch08.pdfhttps://www.colorado.edu/engineering/CAS/courses.d/NFEM.d/NFEM.Ch08.d/NFEM.Ch08.pdf

Weak form

Current configuration:∫v

σσσ : δedv =∫v

f · δudv +∫a

t · δuda

Reference configuration:∫V

S : δEdV =∫V

F · δUdV +∫A

T0 · δUdA


Weak form

Reference to current configuration:

∫V

S : δEdV =∫V

(JF−1σσσF−T

): δ(

FT eF)

dV =∫v

σσσ : δedv


Incremental solution

Incremental solution procedures are used due to the nonlinearnature of the problem

Quantities are decomposed to their values at each step plus anincrement

Two alternatives exist for the definition of the referenceconfiguration


Total/updated Lagrangian formulation

Total Lagrangian (TL)

The initial configuration isthe reference configuration.

Stress & Strain measures atthe target configuration(increment i + 1) arecomputed with respect tothe initial configuration.

Derivatives and Integrals aretaken with respect to V(initial conf.).

Updated Lagrangian (UL)

The previous increment isthe reference configuration.

Stress & Strain measures atthe target configuration(increment i + 1) areevaluated with respect tothe previous configuration(increment i).

Derivatives and Integrals aretaken with respect to v i .


Total Lagrangian formulation

∫V

S : δEdV =∫V

F · δUdV +∫A

T0 · δUdA

︸︷︷︸R

Displacements at increment i + 1:

ui+1 = ui + ∆u


Incremental decomposition of strains

By substituting in the definition of the Green-Lagrange strain:

Ei+1 =12

[∇ui+1 +∇

(ui+1

)T+∇

(ui+1

)T∇ui+1

]= 12

[∇ui +∇

(ui)T

+∇(

ui)T∇ui

]︸︷︷︸

Ei

+

+ 12

[∇∆u +∇∆uT +∇

(ui)T∇∆u +∇∆uT∇ui

]︸︷︷︸

ei

+

+ 12[∇∆uT∇∆u

]︸︷︷︸

ηηη

Here ei is not to be confused with the Euler-Almansi strainInstitute of Structural Engineering Method of Finite Elements II 35

Incremental decomposition of strains

In the above the strain is decomposed as:

Ei+1 = Ei + ∆E = Ei + ei + ηηη

where:

Ei The strain at increment i∆E The strain increment

ei Linear with respect to ∆uηηη Nonlinear with respect to ∆u


Decomposition of virtual strains and stresses

We observe that the variation of Ei+1 is equal to the variation of theincrement:

δEi = 0⇒ δEi+1 = δ∆E = δei + δηηη

Applying the incremental decomposition to stresses we obtain:

Si+1 = Si + ∆S


Weak form

Applying the above decompositions to the weak form we obtain:∫V

∆S : δ∆EdV +∫V

Si : δηηηdV = Ri+1 −∫V

Si : δeidV

∫V

Si : δeidV is known and has been moved to the RHS


Weak form




Si : δeidV

∫V

Si : δeidV is known and has been moved to the RHS (why?)


Weak form




Si : δeidV

∫V

Si : δeidV is known and has been moved to the RHS (why?)∫V

Si : δηηηdV is linear with respect to ∆u


Weak form




Si : δeidV

∫V


Si : δηηηdV is linear with respect to ∆u (why?)


Weak form




Si : δeidV

∫V


Si : δηηηdV is linear with respect to ∆u (why?)∫V

∆S : δ∆EdV is non linear with respect to ∆u


Weak form




Si : δeidV

∫V


Si : δηηηdV is linear with respect to ∆u (why?)∫V

∆S : δ∆EdV is non linear with respect to ∆u (why?)


Linearization of stresses

Applying the above decompositions to the weak form we obtain:

For the variation of ∆E we assume:

δηηη = 0⇒ δ∆E = δei

since δηηη is higher order

For the stresses we employ the Taylor expansion:

∆S = ∂∆S∂∆E ·∆E =

∂∆S∂∆E ·

(ei + ηηη

)


Linearized Weak Form

Assuming linear elastic material response: ∂∆S∂∆E = D

Then the first term of the weak form becomes:

∫V

∆S : δEdV =∫V

(ei + ηηη

)T·D · δeidV





∫V

∆S : δEdV =∫V

(ei + �ηηη

)T·D · δeidV





∫V

∆S : δEdV =∫V

(ei)T·D · δedV





∫V

∆S : δEdV =∫V

(ei)T·D · δedV

The linearized weak form is formulated as:∫V

ei ·D · δeidV

︸︷︷︸linear w.r.t. ∆u

+∫V

Si : δηηηdV

︸︷︷︸linear w.r.t. ∆u

= Ri+1 −∫V

Si : δedV

︸︷︷︸known


IntroductionAn example of geometrical non-linearityThe deformation gradientNon-linear strain measuresStress measuresConstitutive equationsWeak form of equilibrium equationsLinearization of the weak form

Documents

Prof. Dr. Eleni Chatzi Dr. Giuseppe Abbiati, Dr ... · Understanding limitations of geometrically linear analysis Understanding diﬀerent nonlinear strain measures Understanding