Lecture No. 5 Lecture Notes_β¦Β Β· CE 60130 FINITE ELEMENT METHODS - LECTURE 5 - updated 2018- 01-...
Transcript of Lecture No. 5 Lecture Notes_β¦Β Β· CE 60130 FINITE ELEMENT METHODS - LECTURE 5 - updated 2018- 01-...
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Lecture No. 5
πΏπΏ(π’π’) β ππ(π₯π₯) = 0 in πΊπΊ
πππΈπΈ(π’π’) = πππΈπΈ on π€π€πΈπΈ
ππππ(π’π’) = ππππ on π€π€ππ
β’ For all weighted residual methods
π’π’ππππππ = π’π’π΅π΅ + οΏ½πΌπΌππππππ
ππ
ππ=1
β’ For all (Bubnov) Galerkin methods
π€π€ππ = ππππ (test = trial)
Summary of Conventional Galerkin Method
πππΈπΈ(π’π’π΅π΅) = πππΈπΈ and πππΈπΈ(ππππ) = 0, ππ = 1,ππ on π€π€πΈπΈ
ππππ(π’π’π΅π΅) = ππππ and ππππ(ππππ) = 0, ππ = 1,ππ on π€π€ππ
ΤπΌπΌ = πΏπΏοΏ½π’π’πππππποΏ½ β ππ(π₯π₯)
< ΤπΌπΌ ,π€π€ππ > = 0, ππ = 1,ππ
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Fundamental Weak Weighted Residual Galerkin
β’ Only satisfy the essential b.c.βs. This makes things a lot easier!
πππΈπΈ(π’π’π΅π΅) = πππΈπΈ and πππΈπΈ(ππππ) = 0 ππ = 1,ππ on π€π€πΈπΈ
β’ Only require a limited degree of functional continuity. Therefore we can piece together
functions to make up ππππ. Again this makes the problem much easier.
β’ Since weβre not satisfying the natural b.c.βs, we also consider the error associated with the
locations where the natural b.c.βs is violated:
ΤπΌπΌ = πΏπΏοΏ½π’π’πππππποΏ½ β ππ(π₯π₯)
Τπ΅π΅,ππ = βπππποΏ½π’π’πππππποΏ½ + ππππ
< ΤπΌπΌ ,π€π€ππ > +< Τπ΅π΅,ππ ,π€π€ππ > = 0 ππ = 1,ππ
This establishes the fundamental weak form. Thus we have relaxed the admissibility
conditions such that only βessentialβ b.c.βs must be satisfied.
β’ The Interior and Boundary Error (for the natural boundary) must go to zero by requiring π€π€ππ
to the orthogonal to these combined errors. Hence we are no longer βclampingβ the natural
b.c. down at the time we set up our sequence but we are βdrivingβ the natural b.c. error to
zero as the number of trial functions increases.
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β’ Expanding out the fundamental weak form by substituting ΤπΌπΌ , Τπ΅π΅,ππ and π’π’ππππππ.
< πΏπΏ(π’π’π΅π΅) β ππ,π€π€ππ > +οΏ½πΌπΌππ < πΏπΏ(ππππ),ππ
ππ=1
π€π€ππ >πΊπΊ +< ππππ β ππππ(π’π’π΅π΅),π€π€ππ >π€π€ππ
βοΏ½πΌπΌππ <ππ
ππ=1
ππππ(ππππ),π€π€ππ >π€π€ππ= 0 ππ = 1,ππ
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Symmetrical Weak Weighted Residual Form
β’ Letβs continue to relax admissibility by integrating the fundamental weak form by parts.
The βhalfway pointβ of the integration represents the symmetrical weak formβ.
β’ This integration by parts procedure:
β’ Reduces the required functional continuity on the approximating functions. It reduces
functional continuity requirements on the trial functions while increasing them on the
test functions. At the symmetrical weak form these requirements are balanced. Thus
this is the optimal weak form.
β’ Furthermore the unknown functions evaluated at the boundary disappear.
β’ In general the natural b.c. is only satisfied in an average sense of ππ β β.
β’ When the operator is self-adjoint, the matrix generated will still be symmetrical for both the
symmetrical and fundamental weak forms.
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Definition and Use of Localized Functions
β’ Split the domain up into segments over which trial/test functions are defined.
Motivation, itβs easier to satisfy b.c.βs for complex 2D/3D geometries since these can be
satisfied locally (for weak formulations we need only satisfy the essential b.c.βs). In addition
the matrix properties will in general be much better.
β’ For the finite element method (FEM), we use functions defined over small subdomains, i.e.
localized functions.
β’ However we must ensure that we satisfy the correct degree of functional continuity. For
example, consider:
πΏπΏ(π’π’) =ππ2π’π’πππ₯π₯2
ππ(2) space is required for fundamental weak form (continuity of the function and the first
derivative and a finite second derivative).
ππ(1) space is required for the symmetrical weak form (continuity of the function and a
finite first derivative).
β’ Split the domain up into segments and define trial functions within each element.
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Option 1
No functional continuity but finite function β ππ(0) space
(a) The simplest approximation is a histogram (i.e. constant value of the function over the
element). This defines only one unknown per βfinite elementβ.
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(b) We can define more complex functions over the element leading to more unknowns per
element (to represent an increased order of polynomial or function within the element).
However we still have no functional continuity between the elements and thus these
functions still are from the ππ(0) space.
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Option 2
The function is continuous which defines the ππ(1) = πΆπΆππ space. This space requires no
continuity of the derivative. We define the discrete variables equal to the functional values at
the nodes. Nodes must always join up at the inter element boundaries!
(a) The simplest approximation consists of a linear approximation over each element
which involves 2 unknowns per element and 1 unknown per node: Thus over each
element:
π’π’ = ππππ + ππ1π₯π₯
Thus weβve gone from an element measure to a nodal measure.
An interpolating function defined with nodal functional values and ππ(1) = πΆπΆππ continuity is called a Lagrange interpolating function.
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(b) We can use higher degree functions within each element (i.e. more nodes per element)
however functional continuity will still only be ππ(1). For example:
π’π’ = ππππ + ππ1π₯π₯ + ππ2π₯π₯2 + ππ3π₯π₯3
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Option 3
The function and first derivative continuous defines the ππ(2) = πΆπΆ1 space. In this case we
must define the function and the first derivative as nodal variables.
(a) Simplest approximation involves 2 nodes with 2 unknowns per node. Thus over each
element:
π’π’ = ππππ + ππ1π₯π₯ + ππ2π₯π₯2 + ππ3π₯π₯3
Therefore u and π’π’,π₯π₯ are the unknowns. This involves 4 unknowns per element, 2 unknowns
per node.
Interpolating functions defined with nodal functional and first derivative values with
ππ(2) = πΆπΆ1 continuity are called Hermite Interpolating functions.
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1st order Hermite: π’π’ and π’π’,π₯π₯ continuous
2nd order Hermite: π’π’, π’π’,π₯π₯ and π’π’,π₯π₯π₯π₯ continuous β 3 unknowns per node
Note 1
For the Galerkin method, π€π€ππ = ππππ
β’ The fundamental weak form requires different order functional continuity for ππππ as
trial functions than the π€π€ππ (or ππππ as test) functions.
β’ The symmetrical weak form is obtained by integration by parts until the space
requirements for ππππ and π€π€ππ are the same. This is the most attractive form to use. We
note that the number of unknowns has been minimized.
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Example
β’ Fundamental weak form:
οΏ½οΏ½ππ2π’π’πππππππππ₯π₯2 + π’π’ππππππ β π₯π₯οΏ½π€π€πππππ₯π₯ + οΏ½οΏ½ππ β
πππ’π’πππππππππ₯π₯ οΏ½π€π€πποΏ½
π₯π₯=1= 0 ππ = 1,ππ
1
0
π€π€ππ β πΏπΏ2 and π’π’ππππππ (or ππππ) β ππ(2), ππ = 1,ππ
Therefore we must choose π€π€ππ and ππππ β ππ(2)
The functional and first derivative must be continuous) leading to a cubic approximation
over the element with 2 unknowns per node (function value and its 1st derivative). Thus we
require Hermite interpolation
β’ Symmetrical Weak Form
οΏ½οΏ½βπππ’π’πππππππππ₯π₯
πππ€π€πππππ₯π₯ + π’π’π€π€ππ + π₯π₯π€π€πποΏ½πππ₯π₯ + οΏ½πππ€π€πποΏ½π₯π₯=1 = 0
1
0
ππ = 1,ππ
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β’ The opposite of the fundamental weak form is obtained by interchanging π’π’ππππππ and π€π€ππ and
leads to the B.E.M form
οΏ½πππ€π€πππππ₯π₯ π’π’πππππποΏ½
0
1
+ οΏ½πππ€π€πποΏ½π₯π₯=1 + οΏ½οΏ½π’π’π€π€ππ + π₯π₯π€π€ππ + π’π’ππππππππ2π€π€πππππ₯π₯2 οΏ½ πππ₯π₯ = 0
1
0
ππ = 1,ππ
Now π€π€ππ β ππ(2) and π’π’ππππππ (or ππππ) β πΏπΏ2
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Note 2
Alternative strategy for deriving the symmetrical weak form:
β’ Start with standard Galerkin formulation
β’ Integrate by parts
β’ Substitute in for the specified natural b.c.
Example
β’ Standard Galerkin formulation
οΏ½οΏ½ππ2π’π’πππππππππ₯π₯2 + π’π’ππππππ + π₯π₯οΏ½π€π€πππππ₯π₯ = 0
1
0
β’ Integrate by parts:
οΏ½οΏ½βπππ’π’πππππππππ₯π₯
πππ€π€πππππ₯π₯ + π’π’π€π€ππ + π₯π₯π€π€πποΏ½ πππ₯π₯ + οΏ½
πππ’π’πππππππππ₯π₯ π€π€πποΏ½
π₯π₯=0
π₯π₯=1
= 01
0
β
οΏ½οΏ½βπππ’π’πππππππππ₯π₯
πππ€π€πππππ₯π₯ + π’π’π€π€ππ + π₯π₯π€π€πποΏ½ πππ₯π₯ + οΏ½
πππ’π’πππππππππ₯π₯ π€π€πποΏ½
π₯π₯=1β οΏ½
πππ’π’πππππππππ₯π₯ π€π€πποΏ½
π₯π₯=0= 0
1
0
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β’ However for our example case:
πππππππππππππ₯π₯
οΏ½π₯π₯=1
β ππ (specified natural b.c.)
β’ Substituting the above equation as well as π€π€πποΏ½π₯π₯=0 = 0 (since ππππ satisfies the
homogeneous form of the essential b.c.), leads to:
οΏ½οΏ½πππ’π’πππππππππ₯π₯
πππ€π€πππππ₯π₯ + π’π’π€π€ππ + π₯π₯π€π€πποΏ½ πππ₯π₯ + οΏ½πππ€π€πποΏ½π₯π₯=1 = 0
1
0
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Note 3
For collocation we cannot use an integration by parts procedure to lower the required degree
of functional continuity as we could for Galerkin methods. This is due to the form of the
weighting functions, π€π€ππ = πΏπΏοΏ½π₯π₯ β π₯π₯πποΏ½ (the dirac delta function) which is not differentiable.
πππΏπΏοΏ½π₯π₯ β π₯π₯πποΏ½πππ₯π₯
does not exist. Therefore it is not possible to find any weak forms!
Therefore for a 2nd order differential operator πΏπΏ we must use Hermite type interpolation
compared to the Galerkin Symmetrical weak form for which we only needed Lagrange type
interpolation.
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Note 4
β’ Summary of Advantages of Weak Forms:
β’ Fundamental weak form has relaxed b.c. requirements. Itβs much easier to find trial/test
functions which satisfy the homogeneous form of the essential b.c. than both the essential
and natural b.c.βs
β’ In addition when we define local trial functions on finite elements, it will be much
simpler to satisfy the essential b.c.βs locally on the element level rather than on a global
level.
When using the FE method where we simply go into the matrix and set the essential b.c.
u and we need not worry about βstrictlyβ enforcing the natural boundary condition πππ’π’/πππ₯π₯
β’ Symmetrical weak form has relaxed functional continuity requirements. This is very
important when defining functions over split domains (FE method).
The symmetrical weak form lowers the inter-element functional continuity requirements
and thus lowers the number of unknowns.
β’ In addition, part of the natural b.c. error falls out due to the way in which it was defined.
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Notes on a 4th derivative operator problem
Consider the equilibrium equation for a beam on an elastic foundation
πΈπΈπΈπΈππ4π£π£πππ₯π₯4 + πππ£π£ = ππ(π₯π₯)
where
v = beam displacement
E = modulus of elasticity
I = moment of inertia
k = foundation constant
p = distributed load on the beam
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Standard Galerkin
β¨πΏπΏ(π£π£) β ππ, πΏπΏπ£π£β© = 0
β
οΏ½οΏ½πΈπΈπΈπΈππ4π£π£πππ₯π₯4
+ πππ£π£ β πποΏ½πΏπΏπ£π£πππ₯π₯ = 0ππ
0
β’ We can establish what the essential and natural b.c.βs are by using the integration by parts
procedure to find:
β¨πΏπΏ(π£π£), πΏπΏπ£π£β© = β¨πΏπΏβ(πΏπΏπ£π£), π£π£β© +
οΏ½ [πΉπΉ1(πΏπΏπ£π£)πΊπΊ1(π£π£) + πΉπΉ2(πΏπΏπ£π£)πΊπΊ2(π£π£) β πΉπΉ1(π£π£)πΊπΊβ1(πΏπΏπ£π£) β πΉπΉ2(π£π£)πΊπΊβ2(πΏπΏπ£π£)]πππ€π€
β’ We perform 2 integration by parts to find the bcβs (assuming EI constant):
β¨πΏπΏ(π£π£), πΏπΏπ£π£β© = οΏ½οΏ½πΈπΈπΈπΈππ2π£π£πππ₯π₯2
ππ2πΏπΏπ£π£πππ₯π₯2 + πππ£π£πΏπΏπ£π£οΏ½πππ₯π₯
ππ
0
+ οΏ½βπΈπΈπΈπΈππ2π£π£πππ₯π₯2
πππΏπΏπ£π£πππ₯π₯ + πΈπΈπΈπΈ
ππ3π£π£πππ₯π₯3 πΏπΏπ£π£οΏ½π₯π₯=0
π₯π₯=1
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β’ This is halfway point of the integration by parts procedure and represent the symmetrical
weak form:
β’ Hence the essential b.c.β are: πΉπΉ1(π£π£) = π£π£ β displacement
πΉπΉ2(π£π£) = πππππππ₯π₯β slope
β’ the natural b.c.βs are: πΊπΊ1(π£π£) = βπΈπΈπΈπΈ ππ3ππ
πππ₯π₯3= ππ β applied shear
πΊπΊ2(π£π£) = +πΈπΈπΈπΈ ππ2ππ
πππ₯π₯2= ππ β applied moment
β’ Note that the weighting functions that are to be used in establishing the fundamental
weak form fall out of this integration by parts procedure used in establishing self-
adjointness of the operator (which also establishes b.c.βs). You want terms to drop later
on!
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β’ Natural boundary errors
The error in the moment is:
Τπ΅π΅β² = οΏ½πΈπΈπΈπΈππ2π£π£πππ₯π₯2 β πποΏ½ οΏ½
πππΏπΏπ£π£πππ₯π₯
οΏ½
A error in the shear is:
Τπ΅π΅" = οΏ½βπΈπΈπΈπΈππ3π£π£πππ₯π₯3 β πποΏ½ {πΏπΏπ£π£}
β’ Weighted natural boundary errors
A moment error is weighted by a rotation:
β¨Τπ΅π΅β² ,πππΏπΏπ£π£πππ₯π₯
β© = οΏ½ οΏ½πΈπΈπΈπΈππ2π£π£πππ₯π₯2 β πποΏ½
Ξππ
οΏ½πππΏπΏπ£π£πππ₯π₯ οΏ½ ππΞ
A shear error is weighted by a displacement
β¨Τπ΅π΅" , πΏπΏπ£π£β© = οΏ½ οΏ½βπΈπΈπΈπΈππ3π£π£πππ₯π₯3 β πποΏ½
Ξππ
{πΏπΏπ£π£} ππΞ
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β’ Note that we must specify essential b.c.βs somewhere!
β’ Boundary conditions must be specified in pairs for this problem
β’ Displacement and rotation are paired
β’ Shear and moment are paired
β’ The error/weighting products fall out of the integration by parts process. They are also
dimensionally consistent with all terms in the total error equation.
β’ The total error equation for this problem is the fundamental weak form
β¨ΤπΌπΌ , πΏπΏπ£π£β© + β¨Τπ΅π΅β² ,πππΏπΏπ£π£πππ₯π₯
β© + β¨Τπ΅π΅" , πΏπΏπ£π£β© = 0
where
ΤπΌπΌ = πΈπΈπΈπΈππ4π£π£πππ₯π₯4 + πππ£π£ β ππ
β’ The symmetrical weak form is established by integrating the fundamental weak form by
parts.