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Method of Finite Elements II - Introduction
Prof. Dr. Eleni ChatziDr. Giuseppe Abbiati, Dr. Konstantinos Agathos
Lecture 1 - 19 September, 2019
Institute of Structural Engineering, ETH Zurich
September 19, 2019
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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: [email protected]. Konstantinos Agathos, email: [email protected]
AssistantsSergio Nicoli, HIL E19.4, email: [email protected]
Course WebsiteLecture Notes and Demos will be posted in:http://www.chatzi.ibk.ethz.ch/education/method-of-finite-elements-ii.html
Performance Evaluation - Final Project (100%)Topics to be announced on 17.10.2019Computer Labs are already be pre-announced and it is advised to bring alaptop along for those sessions
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Course Information
Suggested reading:
“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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Learning goals
By the end of the 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 the most common types of nonlinearities
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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
Institute of Structural Engineering Method of Finite Elements II 39
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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!
Institute of Structural Engineering Method of Finite Elements II 40
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Implementation example
Problem data:
L = 1E = 1A = 1
Institute of Structural Engineering Method of Finite Elements II 41
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Implementation example
Input:
Nodal coordinates
Element connectivities
Material parameters
Constraints
Nodal loads
Output:
Nodal displacements
Institute of Structural Engineering Method of Finite Elements II 42