Method of Finite Elements II - Introduction

Prof. Dr. Eleni ChatziKonstantinos Tatsis, Dr. Konstantinos Agathos

Lecture 1 - 17 September, 2020

Chair of Structural Mechanics & Monitoring, ETH Zurich

September 17, 2020

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Outline

1 Course information

2 Learning goals

3 Linear vs. Nonlinear operators - Strong vs. Weak form

4 A simple finite element example with Matlab implementation -FEM I review

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Course Information

InstructorsProf. Dr. Eleni Chatzi, email: chatzi@ibk.baug.ethz.chDr. Konstantinos Agathos, email: agathos@ibk.baug.ethz.chKonstantinos Tatsis, email: tatsis@ibk.baug.ethz.ch

AssistantsKostas Vlachas, HIF B21, Sergio Nicoli, HIL E19.4

Course WebsiteLecture Notes and Demos will be posted in: http://www.chatzi.ibk.ethz.ch/education/method-of-finite-elements-ii.html

Help us Structure the Course!Participate in our online survey: http://goo.gl/forms/ws0ASBLXiY

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Course Information

Computer LabsComputer Labs take place online, as announced in the class schedule.They require your interactive participation!

Performance Evaluation - Final Project (100%)You can work with a partner (max groups of 2 allowed).Topics to be announced on 15.10.2020

Option for PhD students: You are free to shape your own project!

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Course Information

Suggested reading:Finite Element Procedures” by K.J. Bathe, Prentice Hall, 1996, ETHZurich Library Portal

“Nonlinear Finite Elements for Continua and Structures”, by T.Belytschko, W. K. Liu, and B. Moran, John Wiley and Sons, 2000

“The Finite Element Method: Linear Static and Dynamic FiniteElement Analysis”, by T. J. R. Hughes, Dover Publications, 2000

“Nonlinear finite element analysis of solids and structures”,Crisfield,M.A., Remmers, J.J. and Verhoosel, C.V., John Wiley & Sons, 2012

“Computational methods for plasticity: theory and applications”, DeSouza Neto, E.A., Peric, D. and Owen, D.R., 2011

Lecture Notes by Carlos A. Felippa Nonlinear Finite Element Methods(ASEN 6107):http://www.colorado.edu/engineering/CAS/courses.d/NFEM.d/Home.html

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Course Outline

FEM II - Lecture ScheduleIntroduction - From Linear to Nonlinear ConsiderationsBlock #1 - Geometrical NonlinearitiesGeometrical Nonlinearities I - Nonlinear stress and strain measuresGeometrical Nonlinearities II - FE formulationGeometrical Ninlinearities III - Nonlinear solution methods

Computer Lab I

Block #2 - Material NonlinearitiesMaterial Nonlinearities I - Constitutive ModelingMaterial Nonlinearities II - FE Implementation

Computer Lab II

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Course Outline

FEM II - Lecture Schedule

Block #3 - Nonlinear DynamicsNonlinear Dynamics I - The Newmark methodNonlinear Dynamics II - Modelling of Hysteresis

Computer Lab III

Block #4 - FractureFracture I - General principlesFracture II - The eXtended Finite Element MethodFracture III - Elastic/Plastic fracture

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Today’s Lecture

Learning goalsBy the end of today’s lecture, the goal is to:

Be able to distinguish between linear and nonlinear behavior

Understand what a weak formulation of a problem is and why itis useful

Review basic concepts from Linear Finite Elements that will beof use

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Linear maps

How can we in general define linear?

To help us answer the question we will use the followingmathematical definition:

DefinitionGiven two vector spaces V and W, a linear map is a functionf : V→W, such that for every u, v ∈ V and a scalar c, thefollowing conditions apply:

f (u + v) = f (u) + f (v)f (cu) = cf (u)

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Linear maps

But what is a vector space?

DefinitionA vector space is a collection of objects for which addition andmultiplication with a scalar are defined.

The above is quite general and applies in the same way for e.g.:

Vectors in Rn

Functions defined in Rn, for instance all polynomials of order pdefined in R

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Linear maps

Examples of linear maps:

An m × n matrix:

- Transforms elements from the vector space of vectors in Rn toelements of the vector space of vectors in Rm

- Preserves addition and multiplication by a scalar

Differentiation:

- Transforms elements from the function space of differentiablefunctions in R

- Preserves addition and multiplication by a scalar

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Strong form

Consider the simple problem of a bar with:

Constant cross section A

Length L

A fixed left end

A distributed axial load P

A load R at the right end

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Strong form

The equilibrium equation canbe obtained by:

Considering the stressesapplied at a segment ofinfinitesimal length

Considering the appliedforce constant in thatinterval

σ (x) A =[σ (x) + dσ (x)

dx dx]

A + P (x) dx A

⇒ dσdx + P = 0

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Strong form

Additionally we consider:

Linear1elastic materialbehavior:

σ = Eε

where ε are the strains

A linear1 strain measure:

ε =(du

dx

)where u are thedisplacements

By substituting in the equilibrium equation we obtain:

E d2udx2 + P = 0

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Strong form

Finally, the boundary conditions are also considered:

E d2udx2 + P = 0

u = 0 for x = 0

Aσ = AE(du

dx

)= R for x = L

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Linear equations

We can easily verify that the equilibrium equation is linear!

In general, differential equations of the form:

a0 (x) u (x) + a1 (x) u′ (x) + a2 (x) u′′ (x) + · · ·+ an (x) u(n) (x) = 0

are linear.

However, equations with products of derivatives of the same ordifferent degree are not, for instance:

u (x) u′′ (x) = 0

is not linear.

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Sources of nonlinearity

The linear equilibrium equation can be turned into a nonlinear one iffor instance:

An alternative materialbehavior is considered:

σ = E1ε+ E2ε2

corresponding equation:

E1d2udx2 +2E2

dudx

d2udx2 +P = 0

An alternative strainmeasure is considered:

ε = dudx + 1

2

(dudx

)2

corresponding equation:

E[

d2udx2 + du

dxd2udx2

]+ P = 0

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Sources of nonlinearity

Such occurrences are common in practice, and in general thefollowing types of nonlinearity can be identified:

Material nonlinearity, i.e. nonlinear material laws

Geometrical nonlinearity, i.e. nonlinearity associated with thegeometry of the deformation

Boundary associated nonlinearity, e.g. contact

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Strong form solution

The solution of the differential equation would satisfyequilibrium at every point

Thus, the equation is called the “strong form” of the problem

Approximate solutions could also be obtained by:

- Assuming a specific form for the solution, e.g. polynomial of acertain degree

- Enforcing the equation and boundary conditions at selectedpoints to obtain unknown coefficients

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Weak form

An alternative to the above is posing the problem in “weak form”.Then:

The equations are not satisfied at every point

Instead they are satisfied in an “average” (weak) sense

In the following, alternative ways of deriving such a “weak form” willbe presented.

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Weak form - Galerkin method

Starting with the equilibrium equation:

Eu′′ + P = 0

we employ an arbitrary function w (x), called the weight function,and we demand that:∫ L

0

∫A

w(Eu′′ + P

)dAdx =

∫ L

0w(EAu′′ + q

)dx = 0

regardless of the values of the parameters involved in the definitionof w . In the above AP = q

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Weak form - Galerkin method

Then by integrating the first term by parts we obtain:

−∫ L

0EAw ′u′dx +

(wu′)∣∣L

0 +∫ L

0wqdx = 0

By further assuming that w (0) = 0,w (L) = 0 and rearranging weobtain the weak form as:∫ L

0EAw ′u′dx −

∫ L

0wqdx = 0

We observe that the highest derivative appearing in the equation is 1as opposed to 2 for the initial equation.

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Weak form - Potential energy minimization

The same result can be obtained by following a different route. Westart with the total potential energy functional for the bar:

Π =∫ L

0

12EA

(u′)2 dx −

∫ L

0qudx

but what is a functional?

DefinitionFunctional is a mapping from a function space to the real numbers.

In other words a functional maps functions to numbers, i.e. it is afunction of functions!

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Weak form - Potential energy minimization

At the point of equilibrium, the total potential energy should beminimized.

To minimize a functional we need to define variations:

DefinitionThe variation of a function u (x) is defined as an arbitrary andsufficiently smooth function η (x) that vanishes at the points whereboundary conditions are applied:

δu = η

For the derivatives of η the following should apply:

dnη

dxn = dnδudxn = δ

(dnudxn

)

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Weak form - Potential energy minimization

Then for a functional we have:

DefinitionThe variation of a functional F of a function u and its derivatives(u′, u′′, . . . , un) is defined as:

δF = limε→0

F[u + εη, (u + εη)′ , (u + εη)′′ , . . . , (u + εη)n

]− F [u, u′, u′′, . . . , un]

ε

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Weak form - Potential energy minimization

The variational operator has several common properties todifferentiation, for instance:

δ (F + Q) = δF + δQ, δ (FQ) = (δF ) Q + (δQ) F

Also for functionals including integrals:

δ

∫F (x) dx =

∫δF (x) dx

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Weak form - Potential energy minimization

Similar to functions, the stationary conditions for functionals is thattheir variations should vanish.

For the total potential energy functional, this yields:

δΠ =∫ L

0EAu′δu′dx −

∫ L

0qδudx = 0

For comparison, the Galerkin method yields:

∫ L

0EAw ′u′dx −

∫ L

0wqdx = 0

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Weak form

We can observe:

Minimizing potential energy is equivalent to the principle ofvirtual work if δu are considered as the virtual displacements

Both of the above formulations are equivalent to the Galerkinmethod if δu are considered as the weights

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Weak form

For all the formulations presented we can also remark the following:

The highest derivative of the displacement in the weak form isof degree 1 as opposed to 2 for the strong form

The above also relaxes the smoothness requirements forapproximate solutions

Similar to the strong form, the weak form is linear with respectto the displacements

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Strong vs. Weak form

To better understand the concepts lets consider an example:

We set:R = 0

P = P0 ⇒ q = q0 = AP0

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Strong vs. Weak form

Using the strong form:

Eu′′ + P0 = 0u = 0 for x = 0

Aσ = AE(du

dx

)= 0 for x = L

we can easily solve the problem by assuming the displacements are aquadratic polynomial:

u (x) = a0 + a1x + a2x2

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Strong vs. Weak form

Then plugging displacements into the equilibrium equation yields:

E (a0 + a1x + a2x2)′′ + P0 = 0 ⇒ E (2a2) + P0 = 0 ⇒ a2 = P02E

and the boundary conditions:

u (0) = 0 ⇒ a0 = 0AEu′ (L) = 0 ⇒ a1 + 2a2L = 0 ⇒ a1 = −2a2L

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Strong vs. Weak form

Putting everything together yields the solution:

u (x) = −P0LE x + P0

2E x2 = P02E x (2L− x)

or in terms of q:

u (x) = q02EAx (2L− x)

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Strong vs. Weak form

To solve the same problem using potential energy minimization westart with the same assumption for the displacements:

u (x) = a0 + a1x + a2x2

Then the variations of the displacements and their derivatives are:

δu = δa0 + δa1x + δa2x2, δu′ = δa1 + 2δa2x

Notice that it is of the same form as the assumed displacements.

Also δu should vanish at x = 0, thus:

δu = δa1x + δa2x2

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Strong vs. Weak form

Plugging these expressions into the weak form we obtain:

δΠ =∫ L

0EAu′δu′dx −

∫ L

0qδudx = 0

⇒δa1

[EAL (a1 + a2L)− L2q

2

]+

+δa2

[EAL2

(a1 + 4

3a2L)− L3q

3

]= 0

Since δa1 and δa2 are arbitrary parameters, both expressions inbrackets should vanish.

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Strong vs. Weak form

By solving the system resulting form the above equations we obtain:

a1 = LqEA , a2 = − q

2EAApplying the boundary condition at x = 0 yields:

a0 = 0

By putting everything together, the same solution as with the strongform is obtained:

u (x) = q02EAx (2L− x)

We observe that the boundary condition at the right end is satisfiedautomatically!

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Strong vs. Weak form

The same result as with the previous methods can be obtained usingthe Galerkin method if:

A quadratic polynomial is assumed for the displacements

The same form is assumed for the weights

The weights are forced to vanish at the locations whereboundary conditions are applied

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A finite element example

A bar FE formulation can be obtained in exactly the same way by:

Removing the Dirichlet boundary conditions

Assuming the displacements to be FE shape functions

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A finite element example

Weak form:

δΠ =∫ L

0EAu′δu′dx −

∫ L

0qδudx −

2∑i=1

δuiRi = 0

Shape functions:

u = Nun =[

xL

L− xL

] [ u1u2

]where u1, u2 are nodal displacements

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A finite element example

By substituting and rearranging we obtain:

δΠ =[δu1 δu2

] ∫ L

0

xL

L− xL

′

EA[

xL

L− xL

]′dx[

u1u2

]′

−[δu1 δu2

] ∫ L

0

xL

L− xL

qdx

−[δu1 δu2

] [ R10

]−[δu1 δu2

] [ 0R1

]= 0

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A finite element example

By carrying out the integrations and further rearranging:

EAL

[1 −1−1 1

]︸ ︷︷ ︸

K

[u1u2

]︸ ︷︷ ︸

un

= q0L2

[11

]+[

R1R2

]︸ ︷︷ ︸

f

We obtain the bar stiffness matrix and load vector!

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Implementation example

Problem data:L = 1E = 1A = 1

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Implementation example

Input:

Nodal coordinates

Element connectivities

Material parameters

Constraints

Nodal loads

Output:

Nodal displacements

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