Me Em 5150 Chapter 8

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MEEM 5150 22.

Rayleigh-Ritz Method

Parenthesis : what is anessential boundary condition?

Examples

1) axially loaded bar

f(x) P

x

GDE: E A u( ) = f x( ) for 0 x L

BC :u 0

( )= 0

EA u L( ) = P

m = ?

?

?

If the GDE is of order2m, then

- an essential bc is of order 0tom-1

- a natural bc is of order mto2m-1

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MEEM 5150 32.


2) beam bending

p(x)

V

xM

GDE: E I w( ) = p x( ) for 0 x L

BC :

w 0( ) = 0w 0( ) = 0

EI w L( ) =MEI w L( ) = V

m = ?

?

?

?

?

End of parenthesis : Back to RRM

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Substitute the approximate displacement field into the total potential energy to get

Then apply the PMPE (or PVW)

Thus we end up with3nequations for the3nunknown coefficients (qi,ri,si).

This is an approximate method since (unless you are very lucky), the basis functions are

not correct, thus u, vand wwill be approximate. The closer the basis functions are

to the exact spatial variation of the displacements, the better the approximation.

= q

1,..., q

n, r

1,..., r

n,s

1,..., s

n( )

qi= 0 i = 1,...,n( )

ri

= 0 i = 1,...,n( )

si= 0 i = 1,...,n( )

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RRM: applications

1)

2)

P

0 LE, I

x

3 L / 4

z

0 LE, A

x, ukpo=c

st

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MEEM 5150 62.

Application 1 - Exact Solution

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.2 0.4 0.6 0.8 1

x/L

E Au x( )poL2

= kL/EA = 0

= 0.5

= 1

= 10= 10,000

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Application 1 - First attempt

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.2 0.4 0.6 0.8 1

x/L

E Au x( )poL

2

= 0

= 1

Exact

RRM

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Application 1 - Second attempt

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.2 0.4 0.6 0.8 1

x/L

E Au x( )poL

2

= 0

= 1

Exact

RRM

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Application 2 - First attempt

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0 0.2 0.4 0.6 0.8 1

EIw(x)/(P L^3)

x/L

w1(x)wexact(x)

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RRM: final notes

Conclusions

Advantages

Simplicity

keep old terms when adding new ones

Disadvantages

Basis functions are hard to find for complicated geometries (especially in 2-D and

3-D cases)

qihave no physical significance

Convergence is hard to quantify

These deficiencies are addressed by the finite element method

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3. Basic concepts of FEM: solution of 1-D bar problem

Table of contents

3.1 Basic concepts : mesh, nodes, elements, interpolation, ...

3.2 FEA of axially loaded bar

3.3 Notes : direct method, higher-order elements,

3.4 Principle of Virtual Work (PVW) approach

3.5 Galerkin Weighted Residual (GWR) method

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3.1 Basic concepts

The FEM is also based on the RRM, but

the basis functions are easy to find : interpolation

the qihave a physical significance : nodal displacements

Basic idea

Discretize the domain with a finite element meshcomposed of nodesand elements

Compute the best values of the nodal displacements(based, for example, on the

PMPE) Use interpolationto find the solution everywhere else in the discretized domain

There are many elements of different types and geometries : 1-D, 2-D, 3-D, plane

stress, plane strain, plates, shells, structural, thermal, fluid mechanics, electromagnetic,

elastic, plastic, static, dynamic,

NODE

ELEMENT

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3.2 FEA of axially loaded bar

In this section, we introduce the 6 basic steps of a FEA by solving the following simplestructural problem

po(N/m)

x

0 LStiffness ECross-section A(x)=Ao(1-x/aL) with a > 1

(we will take a=10)Exact solution

GDE: d

dxEAo 1

x

aL

du

dx

= po for 0 x L

BC :

u 0( )= 0EA L( )du

dxL

= 0

Solution :

EAouex x( )=poL2 ax

L+ a2 a( )ln 1 x

aL

ex x( )=Eduex

dx=

poL

Ao

1x

L

1 x

aL

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FEA of axially loaded bar

Finite element solution

We will use the PMPE. The total potential energy for this problem is

= U + V =1

2EA

du

dx

2

dx po u x( )dx0

L

0

L

Every FEA consists of 6 steps

Step 1 : Discretization

Lets choose 3 elements of equal length with constant cross-section

(computed at element center:x=L/6,x=L/2andx=5L/6)

L/3 L/3 L/3

1 2 3 41 2 3

4 nodes3 elements

E,59Ao

60E,

57Ao

60E,

55Ao

60

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Step 2 : Property of an individual (generic) element

Lets look at a generic 2-node element of length l, stiffnessEand constant cross-sectionA

local coordinate s

E, A, la b

local numbering

The basic idea of the FEM is to write the approximate displacement in an element as the

interpolationbetween the (so far unknown) nodal values Uaand Ub

u s( )

For a two-node element, the interpolation is linear

s

E, A, la b

Ua

Ub

u(s)~

u s( ) = Ua 1s

l

+ Ubs

l= Ua Na s( )+UbNb s( )

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Na(s) andNb(s)play an important role in FEA and are called interpolationor shape

functions

Since we have the approximate expression of u(s) on the generic element, we can

compute the (approximate) contribution of the element to the total potential energy

=u (s)( )

e e =

1

2E A

du

ds

2

ds po u s( )ds0

l

0

l

o( )

la b

Nb(s)=s/lNa(s)=1-s/l

11

but u s( ) = UaNa s( )+UbNb s( )

thus

du s( )ds = Ua

dNa s( )ds + Ub

dNb s( )ds

Substitute in o( ) e = 12

EA UadNa

ds+Ub

dNb

ds

2

ds po UaNa s( ) +UbNb s( )[ ]ds0

l

0

l


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Expand the square term :

e

=1

2 EA Ua2 dNa

ds

2

+ 2UaUbdNa

ds

dNb

ds +Ub2 dNb

ds

2

ds po UaNa s( )+UbNb s( )[ ]ds0

l

0

l

Put in a matrix form :

e =12

UaUb

EA dNa

ds

2

ds0

l

EA

dNa

ds

dNb

ds

ds

0

l

EA

dNa

ds

dNb

ds

ds0

l

EA dNb

ds

2

ds0

l

UaUb

Ua Ub

po

Na

s( )ds0

l

poNb s( )ds

0

l

local stiffness matrix

[k]

local load vector

{r}

In short notation :

e =

1

2Ua Ub k[ ]

Ua

Ub

Ua Ub r{ }

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Since we know that

Na s( ) = 1s

l

and Nb s( ) =s

l

we can compute [k] and {r}. For example,

k12 = k21 = EAdNa

ds

dNb

dsds = EA

1l

1

l

ds = EA

l0

l

0

l

Find

k[ ] = EAl

1 11 1

r{ }= po l

1/ 2

1/ 2

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Step 3 : Assemble the elements (i.e., put the pieces together)

Now that we know the properties of a generic element, we can go back to the global level

L/3 L/3 L/3

1 2 3 41 2 3

4 nodes

3 elements

The objective is to find the 4 nodal displacements D = U1 U2 U3 U4

Once these are determined, we can use the shape functions to interpolate the displacement

field inside each element

1 2 3 4

U1 U2 U3 U4

u(s)~

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Note : the approximate displacement solution is continuous withinand between

elements

Withinan element :why? Betweenelements : why?

u s( )

To solve for the nodal displacement vector {D}, we need a system of 4 equations.

These equations will be obtained through the PMPE for the whole structure.

The first thing we need to do is to write the (approximate) expression of the total potentialenergy of the structure by adding the contribution of all three elements (using the result of Step 2).

Recall that, for a generic 2-node element of length l:

e =

1

2Ua Ub k[ ]

Ua

Ub

Ua Ub r{ }

k[ ] = EAl

1 11 1

r{ }= po l

1/ 2

1/ 2

with

Objective: write the (approximate) totalpotential energy as

TOT = e

e

=1

2D K[ ] D{ } D R{ }

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Lets add the contribution of all three elements to the total potential energy

Element #1 : l=L/3 a=1 b=2 A=59Ao/60

(1) =1

2U

1 U

2

59EAo20L

59EAo20L

59EAo20L

59EAo20L

U1

U2

U1 U2

poL

6poL

6

TOT =1

2U1 U2 U3 U4

59EAo

20L

59EAo

20L0 0

59EAo20L

59EAo20L

0 0

0 0 0 00 0 0 0

U1

U2

U3

U4

U1 U2 U3 U4

poL

6poL

6

00

Its contribution to the total

potential energy is :

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FEA of axially loaded barElement #2 : l=L/3 a=2 b=3 A=57Ao/60

(2 )

=

1

2U

2

U3

57EAo

20L

57EAo

20L

57EAo20L

57EAo

20L

U2

U3

U

2

U3

poL

6

poL6

TOT =1

2U1 U2 U3 U4

59EAo20L

59EAo20L 0 0

59EAo20L

59EAo20L

+57EAo20L

57EAo20L

0

0 57EAo

20L

57EAo20L

0

0 0 0 0

U1

U2

U3

U4

U1 U2 U3 U4

poL

6poL

6+

poL

6poL

60



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FEA of axially loaded barElement #3 : l=L/3 a=3 b=4 A=55 Ao/60

(3 ) =1

2

U3 U4

55EAo

20L

55EAo

20L

55EAo20L

55EAo20L

U3

U4

U3 U4

poL

6

poL6

TOT =1

2U1 U2 U3 U4

59EAo20L

59EAo20L

0 0

59EAo20L

59EAo20L

+ 57EAo20L

57EAo20L

0

0 57EAo20L

57EAo20L

+55EAo20L

55EAo20L

0 0 55EAo20L

55EAo20L

U1

U2

U3

U4

U1 U2 U3 U4

poL

6poL

6+

poL

6poL

6+

poL

6poL

6



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We have thus achieved our objective :

TOT

=

1

2 D K[ ] D{ } D R{ }where [K] is the GLOBAL STIFFNESS MATRIX

{R} is the GLOBAL LOAD VECTOR

D = U1 U2 U3 U4

K[ ]= EAo20L

59 59 0 059 116 57 0

0 57 112 550 0 55 55

R{ }= poL

1/ 6

1/ 3

1/ 3

1/ 6

Here

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At this point, we apply the PMPE by choosing the nodal displacements which minimize

the approximate total potential energy of the structure :

TOT

U1= TOT

U2= TOT

U3= TOT

U4= 0

or, in matrix form,

TOT

D{ }= 0{ }

Since

we get the linear system

Note: in practice, one does not need to compute the total potential energy in (*): just go

directly to (**) by assembling [k] and {r} into [K] and {R}, respectively.

TOT =

1

2D K[ ] D{ } D R{ } (*)

K[ ] D{ } = R{ } (**)

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Step 4 : Apply load and displacement boundary conditions

First, we impose the nodal(concentrated) loads. In this case, we only have the reactionRoat the wall (node 1)

1 2 3 4

Ro

All the nodal loads are added to the globalload vector {R}

EAo

20L

59 59 0 059 116 57 0

0 57 112 550 0 55 55

U1

U2

U3

U4

= poL

1/ 6

1/ 3

1/ 3

1/ 6

+

Ro0

0

0

Next, we apply the displacementboundary conditions. The linear system above is singular

(i.e., det[K]=0). To remove the singularity, we have to impose the fact that

U1 =0

We can thus remove the first row and column from our 4*4 system and reduce it to a 3*3 system.

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Step 6 : Post-processing

In this step, we

plot the deformed shape of the structureobtain the stresses and strains in each elementcompute the nodal reactions at the supports

1) Deformed shape

Using the nodal values and the

interpolation functions, we get :

Note: the nodal values are NOT exact!

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2) Axial stress and strain in each element

Lets go back to the generic element :

E, A, la b

Ua Ub

Since we have used a linear interpolation between Uaand Ub, the approximate axial strain

is constant in the element

xx =du s( )

ds= Ua

dNa

ds+Ub

dNb

ds= Ua

1l

+Ub1

l=

1

lUb Ua( )

Similarly, the approximate stress is also constant in the element

xx = E xx =E

lUb Ua( )

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Lets review the elements one by one and use the nodal displacements we just obtained in Step 5.

Element #1 : l=L/3 a=1 b=2

Ua = U1 = 0 Ub = U2 = 0.28249poL

2

EAo

xx(1) =

E

L/ 3U2 U1( )= 0.84746

poL

Ao


Ua = U2 = 0.28249poL

2

EAoUb = U3 = 0.45792

poL2

EAo

xx(2 ) =

E

L/ 3U3 U2( )=0.52632

poL

Ao


Ua = U3 = 0.45792poL

2

EAoUb = U4 = 0.51853

poL2

EAo

xx(3 ) = E

L/ 3U

4 U3( )= 0.18182poLAo

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3) Reactions at supports

Lets reconsider the equation we deleted from the original system in Step 4when we applied

the displacement boundary conditions:

Now that we know the nodal displacements, we can use that equation to solve for the unknown

reactionRo. We find

Ro =

1

6poL +

59EAo

20L 0.28249

poL2

EA = poL as expected!

EAo

20L

59 59 0 059 116 57 0

0 57 112 55

0 0 55 55

U1

U2

U3

U4

= poL

1/ 6

1/ 3

1/ 3

1/ 6

+

Ro0

0

0

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3.3 Notes

Conclusions :

The finite element method

is a special form of the RRMis based on the interpolation (using shape functions) of nodal displacement valuescontains 6 basic steps

1) discretization: mesh, nodes, elements

2) property of a generic element: local stiffness matrix [k]and load vector {r}

3) assembling: global load matrix [K]and load vector {R}

4) apply the load and displacement boundary conditions

5) solvethe linear system [K] {D} = {R}

6) post-processing: deformed shape, element stresses and support reactions

Important notes :

1) What if we use more elements?(Nof them)

The property of a generic 2-node element remains unchanged :

k[ ] =E Al

1 11 1

r{ } =pol

1/ 21/ 2{

But now, we have l=L/N

(N+1) unknown nodal displacements : U1, U2, , UN+1

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N = 10

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N = 10

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Notes (Contd)

2) Why do we call [k] the element stiffnessmatrix?

To understand the physical significance of [k], lets examine the behavior of a generic elementsubjected to nodal forces

E, A, la b

Fa

Fb

Ua Ub

(1) Lets fix Ua>0and Ub=0, and compute the nodal forces Faand Fbto achieve these displa-

cements:xx =

Ub Ual

= Ua

l

xx

=Exx

= EUa

l(compressive)

We have thus

Fa = E AUa/ lFb = E AUa/ l

a b

Fa>0Ua

Fb

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Notes (Contd)

(2) Now lets fix Ua=0and Ub>0, and compute the corresponding nodal forces Faand Fb

xx =Ub Ua

l

=Ub

l

xx = Exx = EUb

l

We have thus

Fa = E AUb/ l

Fb = E AUb/ l

a b

Fa0

(3) In the general case, Uaand Ub0. We can combine (1) and (2) to get:

Fa

Fb

=EA /l EA /l

EA/ l EA /l

Ua

Ub

= k[ ]Ua

Ub

Thus [k] represents the stiffness of the element

[K] represents the stiffness of the whole structure

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Notes (Contd)

3) How to incorporate a concentrated load?

Concentrated (nodal) loads must be added in Step 4.

po(N/m)x

0 L

P(N)RoE.g.,

EAo

20L

59 59 0 059 116 57 0

0 57 112 550 0 55 55

U1

U2

U3

U4

= poL

1/ 6

1/ 3

1/ 3

1/ 6

+

Ro0

0

P

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Notes (Contd)

4) Higher-order elements

Instead of increasing the number of elements, we could increase the order of the shape

functions from order 1 (linear) to order 2 (quadratic).

Then each element has 3 nodes and 3 shape functions:

u s( ) = UaNa s( ) + Ub Nb s( ) + Uc Nc s( )

l/2a b c

UaUb

Uc

parabola

l/2

where

l/2a

b c

1

l/2

s

Na s( ) = 2

l2

s l

2 s l( )

l/2

ab c

1

l/2

s

Nb s( ) =

4

l2

s s l( )

Na(s) = ?

Nb(s) = ?

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Two applications

x

x = 0x = L

x = L/2

J

mo

1) Use the finite element method (direct and PMPE approaches) to solve the following problem:

2) Define a suitable mesh and derive the 2- and 3-node finite element formulation ([k]and {r})

for the following linearly elastic beam torsion problem:

1 2 3 4 5

k1 k2 k3 k4

k5

k6

QP

1 2 3 4

5

6

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Notes (Contd)

5) Other methods to derive the FE formulation

So far, we have seen 2 ways to derive the FE formulation (i.e., to derive [k] and {r}) for the

1D axially loaded bar problem :

the direct methodthe PMPE approach

But, as indicated earlier, there are two other ways to obtain the same results:

the method based on the Principle of Virtual Work (PVW)

the Galerkin Weighted Residual Method (GWRM)

These two methods are discussed in the next two sections.

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3.4 PVW approach

For the axially loaded bar problem examined earlier, the PVW is written as

W = Win +Wex = Edudx dudx A dx + po u dx = 0 u0

L

0

L

Let us introduce in Wthe interpolation-based approximation

u x( ) u x( ) = N x( ) D{ }

where, for aN-element &N+1-node mesh

D = U1 U2 U3 ... U N UN+1

N x( ) = N1 x( ) N2 x( ) N3 x( ) ... N N x( ) NN+1 x( )

1 2 3 4 5 6 7

N3(x) N5(x) N7(x)

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Then, we havedu

dx

du

dx=

dN x( )dx

D{ }

The PVW becomes

W= EAdu

dx

dN x( )dx dx

0

L

D{ } + po u dx0

L

= 0 u

W= EAdNi x( )

dx

dN x( )dx

dx0

L

D{ } + po Ni x( ) dx0

L

= 0 for i = 1,..., N+ 1

At this point, let us choose u = Ni(x) (i=1, 2, 3, ., N, N+1)

In a matrix form :

K[ ] D{ } + R{ } = 0{ }

where

Kij = E AdNi x( )

dx0

L

dNj x( )

dxdx

Ri = poNi x( ) dx0

L

Just like we did before, we construct the global stiffness matrix [K] and global load vector {R}

by computing [k] and {r} at the element level and then assembling into [K] and {R}.

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3.5 Galerkin Weighted Residual approach

For physical problems which do not have a functional form, from which the differential

equations and non-essential (i.e., natural) boundary conditions can be derived (e.g., when the

DEQ contains odd-order terms).

We need a different approach which starts from the DEQ itself, or, more precisely, from anintegral formulation (or weak form) of the problem. This approach is called the weighted

residual method (WRM).There are many types of WRM.

3.5.1. Presentation of the WRM

Let us write the DEQ and natural BC as

Du f =0 in VBu g =0 onS V

where uis the unknown function ofx,DandBare differential operators andfand gare known

functions ofx.

E.g., for beam bending problems: D E Id4

dx4 fq x( )

B E Id2

dx2 g M

B E I d3

dx 3 g V

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Look for approximation satisfying essential boundary conditions and choose

the unknown coefficients

u x,a( )ai i =1, n( ) such that u x,a( )is "close to" u x( )

But u x,a( ) is not exact DEQ are not satisfied.

Let us denote the residues by RD = RD a, x( ) =Du fRB = RB a, x( ) =Bu g

The objective is thus to find the coefficients aisuch that the residues are small.

There are many different options to compute the residues:

* Collocation method

RD a, xi( ) = 0 i = 1, 2,..., j 1RB a, xi( ) = 0 i = j, j +1,...,n

* Subdomain (collocation) method

RD a, x( )Vi dV = 0 i = 1, 2,..., j 1

RB a, x( )dSSi

= 0 i = j, j + 1,...,n

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MEEM 5150 472.

* (Continuous) least squares method

I

ai= 0 with I= RD a, x( )[ ]

V

2dV+ RB a, x( )[ ]

S

2dS

where is a scalar multiplier for dimensional homogeneity and to stress the relative

importance ofRDvs. RB.

* Least squares collocation method

I

ak=0 k= 1,...,n( ) with I= RD a, xi( )[ ]2

i=1

j1

+ RB a, xi( )[ ]2i=j

m

Simple collocation if m = n.Overdetermined collocation if m > n.

* Galerkin method

Let vi x( )denote some weight functions (chosen as u a i - see example) .Then find the coefficients ai such that

Ri = vi x( )RD a, x( )dVV

= 0 i = 1, 2,...,n( )

I.e., RD is made orthogonal to the shape functions vi x( ).

Before solving (*), use integration by parts to balance the continuity requirements

and to introduce the natural bc (throughRB).

Note: if a variational principle is available, the Galerkin and RR methods yield identical

solutions with the same approximating function u

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MEEM 5150 482.

WRM: application

Problem description:

E, A, L

p(x) = po * x/L

P

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MEEM 5150 492.

WRM: application

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.2 0.4 0.6 0.8 1

u(x)/cL^3

x/L

Residual methods for axially loaded bar with b/cL^2=1

exactpoint collocation at x=L/3

subdomain and continuous least squaresGalerkin

Solution:

b

c L2=

P

poL=1

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MEEM 5150 502.

WRM: application

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

u(x)/cL^3

x/L

Residual methods for axially loaded bar with b/cL^2=-0.1

exactpoint collocation at x=L/3

subdomain and continuous least squaresGalerkin

b

c L2=

P

poL= 0.1

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MEEM 5150 512.

As indicated earlier, we start from the GDE itself :

E A u x( ) = po 0 x L( )

The basic idea of the GWRM is as follows : instead of satisfying the GDE at every point in the

domain (0 x L), we will only satisfy it in a weighted average sense (weak form) :

E A u x( ) = po( ) v x( ) dx0

L

where v(x)is anyweight functiondefined on the domain vanishing where essential boundary

conditions are applied on u(here, the essential BC is u(0) = 0 and thus we must have v(0) = 0).

E A u x( ) v x( ) dx0

L

= po v x( ) dx0

L

(*)

In (*), the continuity requirements for u(x)and v(x)are uneven: u(x)must be at least ??while

v(x)has to be ?? only.

3.5.2. Galerkin finite element method

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MEEM 5150 522.

= 0 = 0Thus (*)becomes

E A u x( ) v x( ) dx0

L

= po v x( ) dx0

L

(**)

E A u v dx0

L

= E A u v[ ]0L E A u v dx

0

L

= E A u L( )v L( ) E A u 0( )v 0( ) E A u v dx0

L

To balance these requirements, let us integrate the left-hand side by parts :

E AdN x( )

dx0

L

v x( ) dx D{ } = po v x( ) dx0

L

At this point, we introduce into (**)the interpolation-based approximation for u(x):

and choose our weight functions to be the shape functions : v(x) = Ni(x) (i=1, , N+1)

E AdN x( )

dx

dN x( )dx

dx0

L

K[ ]1 24 4 4 44 34 4 4 4 4

D{ } = po N x( ){ }dx0

L

R[ ]1 24 4 34 4

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Or, in component form :

Kij = E AdNi x( )

dx0

L

dNj x( )

dxdx

Ri = poNi x( ) dx0

L

Me Em 5150 Chapter 8

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