mae 323: chapter 2 - padtinc.com

2011 Alex Grishin

The Basics: FEA 101MAE 323: Chapter 2

MAE 323 Chapter 2 1

Matrix Structural Analysis (MSA):

•Legacy Structural Methods : The immediate advantage of the MSA

developed by Duncan and Collar was that one could now estimate structural

behavior for much more complex systems than was possible before. The key

to making this work lay in the assumption(s) used to connect the structural

primitives together. Historically, there were two options:

1. The Flexibility or Force Method (FM), in which forces (fe) and

moments are the primary variables of interest. In this method, you

assemble structural members, or “elements” according to:

2. The Direct Stiffness Method (DSM), in which displacements (de) are

the primary variable of interest. In this method, you assemble

structural members according to:

e e e eE= +d Q f d

e e e eE= +f K d f

e e e

e e

= +∑ ∑Q f d d

e e e

e e

= +∑ ∑K d f f

where

where

2011 Alex Grishin


MAE 323 Chapter 2 2

•Legacy Structural Methods (Cont.) :

Symbol Meaning

ed

eK

ef

eQ

eEd

eEf Element external force while de=0

Element external displacement while fe=0

Element internal forces

Element internal displacements

Element flexibility matrix

Element Stiffness matrix

The Stiffness Method (DSM) eventually “won”*, and still provides the

rationale for the algebraic matrix assembly today. The major advance

in today’s structural FE codes lies in how Ke is calculated (in fact, this

may be thought of as a major distinguishing feature of FEM)

*For a fascinating account of this history, see http://www.colorado.edu/engineering/CAS/Felippa.d/FelippaHome.d/Publications.d/Report.CU-CAS-00-13.pdf

2011 Alex Grishin


MAE 323 Chapter 2 3

Example 1 – Linear Springs (Pre-FEM):

•Let’s look at a simple example of Matrix Structural Analysis (MSA – here used

generically to refer to numerical structural analysis before FEM) to develop

some of the key themes which we will revisit later in the course

u1,F1

u2,F2

1

2

kx

yWe’ll start by considering a one-

dimensional spring which can

elongate in only one direction (it’s

length, which is parallel to the x-

axis).

Apply force in positive X-

direction at location 2

Fix deflection

u=u1 at location 1

Hooke’s Law states that the

springs extension, ∆u =(u2-u1)

should be equal to the spring

constant k, times force F

2011 Alex Grishin


MAE 323 Chapter 2 4

Example 1 (Cont.):

•Now let’s calculate the reaction force, F1 at location 1

•By Newton’s Third Law, we should have:

x

y

OR:

F1+F2=0 or F1=-F2

•Substituting Hooke’s Law gives:

F1=-∆u x k

F1=k (u1-u2)

•Note in particular the relation of the

sign of F1 to sign of the extension, ∆u.

This is a convention we will use to add

springs to this system and not get lost

in bookkeeping (keeping track of signs)

•A note on the color coding: Green

refers to fixed (known) displacements

while red refers to fixed (known)

forces

(1)

u1,F1

u2,F2

1

2

2011 Alex Grishin


MAE 323 Chapter 2 5

Example 1 (Cont.):

•Let ‘s use what we’ve learned and apply it to a series of three springs

x

y•The fixed end implies that we have set u0=0.

Since this displacement is zero, we just

eliminate it from subsequent calculations

•Start by calculating the sum of forces

at location 1:

•It will be the sum of forces associated

with extension u1 (represented by )

and u2-u1 (represented by )

u1

u2

u3,F3

k2

k1

k3

1

2

3u1

k1

k2

u2

1f e

2fe x∑F @ 1

1 2 1 1 2 1 2: f f ( ) 0e e

k u k u u+ = + − =

2011 Alex Grishin


MAE 323 Chapter 2 6

Example 1 (Cont.):

•Proceed in the same way to calculate forces at locations 2 and 3:

x

y

k2

k3

u2

2 3 2 2 1 3 2 3: f f ( ) ( ) 0e ek u u k u u+ = − + − =

3 3 3 23: f ( )e

k uF u= = −

x∑F @ 2

x∑F @ 3

u3,F3

2011 Alex Grishin


MAE 323 Chapter 2 7

Example 1 (Cont.):

•We now have three equations and three unknowns:

2 2 1 3 2 3( ) ( ) 0k u u k u u− + − =

3 3 2( )k u u− =F3

1 1 2 1 2( ) 0k u k u u+ − =

•This can be written in matrix form:

1 2 2 1

2 2 3 3 2

33 3 3

0 0

0

0

k k k u

k k k k u

k k u

+ −

− + − = − F

(2)

2011 Alex Grishin


MAE 323 Chapter 2 8

A more automated way (?):

•We can see a pattern. If we were to add a fourth spring, we would have:

4

1 2 2 1

2 2 3 3 2

3 3 4 4 3

4 4 4

0 0 0

0 0

0 0

0 0

k k k u

k k k k u

k k k k u

k k u

+ −

− + − = − + − − F

•The pattern seems to imply that this matrix can be built up by adding sub-

matrices along their diagonal:

4

1 2 2 1

2 2 3 3 2

3 3 4 4 3

4 4 4

0 0 0

0 0

0 0

0 0

k k k u

k k k k u

k k k k u

k k u

+ −

− + − = − + − − F

2011 Alex Grishin


MAE 323 Chapter 2 9

A more automated way(Cont.):

•And let’s look at one of the sub-matrices:

•This represents the spring k2 detached from the system

2 2 1

2 2 2

0

0

k k u

k k u

− =

−

u1

u2

1

2

k2

•This is equivalent to a linear longitudinal spring element

(though we didn’t use FEM to derive it)!* It has no

boundary conditions (that’s why the matrix is singular)

and no external loads, but it is ready to receive both

•For example, if we want to fix u1=0, we would strike out

(eliminate) the first row and column. If we want to add a

force at location 2, we simply place the force value in the

bottom entry of the column vector on the RHS

*In fact, let the student beware! We haven’t DERIVED anything. We’ve only made a very

clever observation (the kind engineers make) to help understand a very useful concept

2011 Alex Grishin


MAE 323 Chapter 2 10

A more automated way (Cont.):

•Let’s re-construct the matrix equation (2) knowing what we’ve learned

•We’ve got three element equations, one for each spring:

•We’ll deal with the boundary condition in the next step…

u0

u1

0

1

k1

01 1

1 1 1

0

0

k k

k k u

u− =

−

u1

u2

1

2

k2

u2

u3,F3

2

3

k3

2 2 1

2 2 2

0

0

k k u

k k u

− =

−

3

3 3

3 2

3 3

0k k u

k k u

− =

− F

2011 Alex Grishin



A more automated way (Cont.):

•Now, add the matrices according to the following formula:

1. Initialize an n x n matrix with zeroes (where n is the number of locations in the series.

Also initialize an n x 1 displacement and force vector

2. Place all displacement variables into their associated location in the initialized

displacement vector (the mapping of indices is arbitrary but you must be consistent)

3. Add the element matrices to the initialized matrix such that each component

corresponds to it’s proper location w/r to the displacement vector (for example, if u2

corresponds to the third row of the displacement vector, then all entries associated with

u2 go in the third row and third column)

4. Add all the force vectors (if any) to the initialized force vector according to the same

principle

5. Eliminate all rows and columns associated with zero fixed displacements

3

0

1

2

u

u

u

u 3

0

0

0

F

1 1

1 1

k k

k k

−

− 2 2

2 2

k k

k k

−

− 3 3

3 3

k k

k k

−

−

=+

+0

0

2011 Alex Grishin



Review 1:

•The method of assembling element matrices (sub-matrices) in a larger system of

equations is called the Stiffness Method , or Direct Stiffness Method (DSM)*

•Recall from earlier that for a structural member (element), it’s internal equilibrium may

be expressed as

where the subscript e is now understood to stand for “element”

•The element equilibria are used to assembly the global system of equations according

to:

•In spite of it’s simplicity, this is an extremely powerful method of solving equations

numerically. The reason is that ANY element matrix may be added to ANY other

element matrix as long as their Degrees of Freedom (DoF) are compatible

•Some more explanations and definitions follow…

e e e eE= +f K d f

*The other method – the Force Method (FM) has fallen into obscurity. It is so

seldom used, I am aware of only one book containing the subject still in print

e e eE= +∑ ∑ ∑f K d f

2011 Alex Grishin



Review 1 (Cont.):

•The term “Degrees of Freedom” refers to the number of discrete variables to be solved

(in structural problems, usually representing displacement)

•Each variable to be solved refers to a solution at a particular spatial location. These

locations are called nodes

ui

uj

I

J

kj

A single

spring

element

The element has

two nodes, with

one DoF each

1 1

1 1

i i

i

j j

u fk

u f

− =

−

It’s corresponding matrix

•Here are some examples of other element types (nodes not numbered)

•The element and it’s nodes are depicted in most commercial

systems graphically by it’s nodes connected by lines

2011 Alex Grishin



Review 1 (Cont.):

•So far, we have constructed a spring element and built a simple model using spring

elements. The spring element we derived is a mathematical idealization of Hooke’s Law

in one spatial dimension and two DoF’s. In numerical terms, it carries an infinite stress

within the context of a continuum because it transmits a load across a zero area

•This is not a problem because stress and strain are not quantities we store, use, or

require in a pure spring problem.

•This is equivalent to saying that the spring solution (the displacements) does NOT

approximate any differential equation within the domain of it’s element support -rather,

it is a pointwise solution which transmits force between two nodes. The solution at

nodes may in fact approximate a differential equation.

•This may become a problem as soon as we connect a spring to an element that IS the

solution to a differential equation over a higher dimensional volume (i.e.: The spring is

one dimensional. We can expect infinite stresses if we attach this to a continuum

element of dimension 2 or higher)

•This can easily happen because of the flexibility of the SM. The nodes of a spring with

DoF’s in a given spatial direction will connect to the nodes of any other element having

nodes with the same DoF’s

2011 Alex Grishin



A bar or truss element

•Although we’re getting warm, we still haven’t encountered FEM. To get there, we have

to make the transition from a point-wise matrix solution (springs) to continuum

elements via a “reduced-continuum” element type

•The usual meaning of “continuum elements” is: elements that approximate the solution

to a continuum problem. In the context of structural mechanics, that means a solution

to either a 2D or 3D problem in elasticity. In this definition, the spatial dimension of the

solution is equal to that of the space in which the element is used. This sets them apart

from beams and trusses, for example (and springs, for that matter, which may be

thought of as “no-continuum” elements), in that those are 1D elements which are often

used in 2 and 3 dimensional models

•We will invent the term “reduced-continuum” * to describe this latter type of element.

So, a reduced-continuum element will be one which has a smoothly varying solution

over it’s support and the spatial dimension of the solution is less than that of space in

which it is used. By “smoothly varying”, I mean that the solution has finite derivatives

over the support, up to the order of the associated differential equation

*or “reduced-order” elements, because this is sometimes the way certain commercial FE codes refer to

these element types. Creating models using these elements is sometimes referred to as Reduced-Order

Modeling (ROM)

2011 Alex Grishin



A bar or truss element :•We will develop the element formulation for a bar or truss member. This will provide our first

example of a reduced-continuum element type

•We start with a verbal description of this element’s behavior: A truss element is one which

represents material loaded along a single axis, reacting only along that axis, having a constant cross-

section, and obeying the laws of linear elasticity.

•From the verbal description, we can immediately deduce that such an element will be a constant

strain element (6a). We can integrate this equation to obtain a linear deflection relationship (3a)

•We’ll use the same coordinate reference we used previously, so that the bar or truss is aligned with

the x-axis, and the displacement variable is designated as u. Turning the previous statements into

mathematics:

x

y u1,F1

1

2u2,F2

X X

( ) bu x ax= +

du Fa

dx EAε = = =

Linear deflection

Linear elasticity

Ax-x=A Constant Area

(3a)

(4a)

(5)L

FE c

Aσ ε= = =

Constant strain(6a)

2011 Alex Grishin



A bar or truss element :

•Like the spring element, the truss element has only two nodes and two DoF’s. We

know that it has a linear displacement field (equation (3a)), but we need a local

coordinate reference for this equation. The coordinate system we have been using was

a global reference. For the local reference, we’ll just translate the global reference and

set x=0 to coincide with node 1. We will call this the element coordinate system.

2 11

( )( )

u uu x x u

L

−= +

2 1( )=

u udu F

dx EA Lε

−= =

Linear deflection

Linear elasticity

Constant strain

(3b)

(4b)

(6b)

u1,F1

1

2

u2,F2

y

x

•We can solve for the coefficients a and b in equation

(3) because we know that u(x)=u1 at node 1 (x=0) and

u(x)=u2 at node 2 (x=L)

•Thus, a=(u2-u1)/L and b=u1

•So, substituting these into equations (3a) , (4a), and

(6a):

2 1( )

F Eu u

A Lσ = = −Note:

We have also assumed that

no body loads exist for this

element

2011 Alex Grishin




•Now, multiplying equation (4b) by A, yields:

2 1( )

EAF u u

L= −

•And , equation (6b) tells us that (u2-u1)/L (the strain) is constant.

So, over a truss element of length L , we can recover Hooke’s

Law:EA

F u k uL

= ∆ = ∆

•So, the truss element has constant stiffness:

EAk

L=

•From our work with springs, this implies the

element equilibrium equation must be:

1 1

2 2

1 1

1 1

u FEA

u FL

− =

−

(7)

(8)

2011 Alex Grishin




•Finally, if we collect terms involving u1 and u2 in equation (3b):

•This produces an equation of the form:

1 2( ) 1

x xu x u u

L L

= − +

1

( ) ( )n

e

i i

i

u x N x c=

=∑Where n is the order of the polynomial used to

approximate the solution and are the nodal

solution values for element e

e

ic

(8)

•Equation (8) is defined only over the element support (in this case, x1≤x≤x2)

defined by nodes 1 and 2 (it is zero elsewhere). This makes it a piecewise

polynomial. The functions, Ni are called shape functions. They are very useful in

determining the solution anywhere in an assembled structure (not just nodes).

Because of displacement compatibility at nodes (referred to in the literature as C0

continuity), we can use (9) to construct an interpolating displacement field over the

entire structure as:

(9)

1

( ) ( )nx

i i

i

u x N x c=

=∑Where ci are now all DoF’s associated with

all elements in the same spatial direction as

u (x in this case).

(10)

2011 Alex Grishin



A bar or truss element (Cont.) :

•Shown below are the shape functions of two truss elements which share a node

(node 2):

1 2 3

•Here’s a curve generated by equation (10) with c={1,2,5} and the shape

functions, N above:

L

L

1

2N 1

1N 2

2N 2

1N

2011 Alex Grishin




•But this truss element is confined to coordinate directions parallel to it’s length

•In order to account for more arbitrary orientations, we need to perform a coordinate

transformation

x

y

x

y

θ •The element coordinate system (x’-y’) makes a an

angle, θ with the global (x-y) . Let u be the deflection

in the x-direction and v be the deflection in the y-

direction.

•The element deflections and node 1 and 2 in the

primed coordinate system are:

u1

v1

u2

v2

'

1 1 1cos sinu u vθ θ= +'

2 2 2cos sinu u vθ θ= +

•Now, to simplify notation, we make

the substitution:

cos

sin

c

s

θ

θ

=

=

(11)

1

2

2011 Alex Grishin




•Now, re-write (11) in vector-matrix notation as:

' =u Rd

Where d is the displacement vector in global coordinates and R is the

“global-to-local” rotation matrix. Expanding this yields:

1

'

11

'22

2

0 0

0 0

u

vc su

uc su

v

=

•This transformation can be used to represent the element equilibrium

in any arbitrarily orientated global coordinate system as:*

( ' )

e e eor

=

=

TR k R d f

k d f

*This represents a tensor rotation, which can be derived by at least two methods. For

details, see:

T. Yang, Finite Element Structural Analysis. Prentice-Hall, Inc., 1986.

(12a)

2011 Alex Grishin




•Expanding (12a) results in the full matrix equation for an arbitrarily oriented truss

element :

2 211

2 211

2 222

2 222

x

y

x

y

Fuc cs c cs

Fvcs s cs sEA

FuL c cs c cs

Fvcs s cs s

− −

− − = − − − −

•Note that we have gone from an element with 2 DoF’s to 4 (even

though the element doesn’t behave any differently)

•The same transformation works for spring elements (simply replace

EA/L by k. A similar process may be used to transform spring and truss

elements into a 3D space

•These element equations are now in a form ready to be assembled into

a global system via the technique (DSM) we have already seen

(12b)

2011 Alex Grishin



What else is there? :

•Actually, there’s a lot more. For now, the student should be aware that any numerical

procedure in mechanics involves “discretizing” an equation in order to approximate

solutions to it at discrete points. This means the domain must be split into finite pieces,

or elements. This is called meshing in the literature. Methods of doing this in an

automated way are not trivial and will be discussed in future lectures

•Equation (8) and (12) were derived according to the “Direct Method”*. In other words,

the element matrices were derived based on our pre-existing knowledge of the solution

to the associated differential equations. Many, if not all element types encountered in

structural analysis may be obtained this way. For example, an Euler-Bernoulli beam

element may be derived this way (for details see the reference below):

•Without actually giving the derivation, we will present the element stiffness matrix for

the Euler-Bernoulli beam element with DoF’s shown below

* R. Cook, D. Malkus, and M. Plesha, Concepts and Applications of Finite Element Analysis,

3rd ed. New York, NY, USA: John Wiley & Sons, 1989.

1 2

v1,y

x

v2

θ2θ1

2011 Alex Grishin



The Euler-Bernoulli Beam Element :

•The equilibrium equation for the Euler-Bernoulli beam element:

1 1

2 3

1 1

3

2 2

2 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

e e e

v FL L

ML L L LEI

v FL LL

ML L L L

θ

θ

−

− = = = − − − −

k d f

•We can further augment this beam element to include extensions (truss

behavior) by adding the two DoF’s from the truss element as follows:

1

1

2 2

1

3

2

2

2 2

2

1 0 0 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 12 6 0 12 6

0 0 0 0 0 0 0 6 4 0 6 2( )

1 0 0 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 12 6 0 12 6

0 0 0 0 0 0 0 6 2 0 6 4

e e e e

T B

u

vL L

L L L LAE EI

uL L

vL L

L L L L

θ

θ

−

− −

+ = = + −

− − − −

k k d f

1

1

1

2

2

2

x

y

x

y

F

F

M

F

F

M

=

(13)

2011 Alex Grishin



What else is there? :

•At this point, we have outlined a matrix-based method of solving structural problems.

However, it still cannot be said that we have introduced the finite element method.

What’s missing is a general procedure for determining element matrices given a

differential equation (maybe one we’ve never seen before)

•Such a procedure (actually there is a family of related procedures) does exist and is

based on a minimization of total energy, or virtual work. This is a crucial part of the

modern finite element method

2011 Alex Grishin



Energy Methods and the Weak Form :

•The heart of FEM is it’s connection to energy methods for solving differential equations.

Although it was long understood that structural problems could often be more easily

treated by energy methods, their use as a generic, semi-automated means of generating

matrix equations wasn’t firmly established until the early sixties

•The first energy method to be used with FEM was the Rayleigh-Ritz method. This

method is based on the idea of minimizing the total energy in a system by minimizing it’s

variation, δ. This can be seen in the example below

x0 x2

u(x0,0)

u(x2,0)

x1

u(x1,δ)=u(x1,0)+δ u(x1,δ)

u(x1,0)

A smoothly varying

function, u over the

domain x0≤x≤x2 may

be found by

minimizing all

smooth functions

u(x,δ) which equal

u(x,0) at x0 and x2

u(x1,0)

δ

2011 Alex Grishin



Energy Methods and the Weak Form (Cont.) :

The Rayleigh-Ritz Method

•The parameter, δ may be thought of as a “virtual function” (or variation), and the

function, u is a function of this parameter (thus u(x,δ)=δu(x) may be thought of as a

virtual displacement), as well as distance x

•Mathematically, the problem may be stated as:

Find the function, u for which the variation of the line integral, I for fixed x0,x2 is zero:2

0

( ) ( , ',... ) 0x

xI x L u u x dxδ δ= =∫

OR:

0

0dI

d δδ =

=

(13)

(14)

•Applying (14) to (13) and integrating by parts leads to:

2

0 '

x

x

dI L d L udx

d u dx uδ δ

∂ ∂ ∂ = −

∂ ∂ ∂ ∫ (15)

where ' /u u x=∂ ∂

2011 Alex Grishin





•Now, without going into details (this is not a course on variational calculus), the

Fundamental Lemma of the Calculus of Variations states that in order for (15) to hold,

given the other requirements, the following (the first part of the integrand in (15)) must

also hold:

•This is the equation that guarantees (13)

•Now, if L is not explicitly a function of u’, then (16) reduces to

0'

L d L

u dx u

∂ ∂− =

∂ ∂

(17)

(16)

0L

u

∂=

∂

•In structural problems, L (called the “Lagrangian”) is the

net energy, given by

L T V= −

•Where T is kinetic energy, and V is potential energy

2011 Alex Grishin





•If there is no kinetic energy and L is not an explicit

function of u’, then

(18)0L V

u u

∂ ∂= =

∂ ∂

•This last form is often known as Castigliano’s Second Theorem

•Equations (16) thru (19) may be used directly to solve for u (usually

the primary variable of interest).

•If the primary unknown, u is displacement, then ∂L/∂u is equal to the

generalized force, F (in structural problems, this comes in the form of

external loads). So, for structural static problems (elastostatics), we

usually use*:

0V

Fu

∂− =

∂(19)

*The external force, F is absent from the derivation of (16) thru (18) because these equations are usually

derived in the absence of constraints. This is not necessary, but we want to keep things simple for now

2011 Alex Grishin





•To see how equation (19) is used, consider the simple case of a

spring, fixed at one end with an applied load F on the other

•First, write the expression for the potential energy:

•Now, substitute into equation (19):

21

2V ku=

0v

F ku Fu

∂− = − =

∂

•Hooke’s Law is recovered, and we can solve for u. In continuum and

reduced continuum problems, equation (19) will be an integral equation

(the energy distribution in a continuum is continuous!)

2011 Alex Grishin





•So, the Rayleigh-Ritz method is used by first finding an expression for the

potential energy of the discrete or continuous problem under investigation.

Then you plug into equation (19) and solve for u

•Below are some common expressions of potential energy found in structural

mechanics

2

0 2

L M dxV

EI= ∫

2

0 2

L S dxV

GA= ∫

2

0 2

L F dxV

EA= ∫

2

0 2

L T dxV

GJ= ∫

Symbols:M->Bending moment

S->Shear force

F->Tensile force

T->Torsion

A->Cross section area

E->Younng’s Modulus

G->Shear Modulus

{ε}->strain tensor in vector form*

{σ}->stress tensor in vector form*

A beam in bending A beam in shear A bar in tension A bar in torsion

{ } { }0 0 0

TW L T

dzdydxε σ∫ ∫ ∫

An elastic continuum*

* Here, second rank stress and strain tensors are expressed as vectors by

exploiting symmetry (see “Voigt Notation”)

2011 Alex Grishin




The Rayleigh-Ritz Method – An example

•Let’s try to reproduce Equation (8) – The equilibrium equations for a truss

element by using the Rayleigh-Ritz Method

•Start with the expression for potential energy in a bar (truss)2

0 2

L F dxV

EA= ∫

•Now, substitute the expressions for stress and strain (4a and 6a):

2

0 2

L

EAdudx

dxV

EA

= ∫

•Substitute your trial or shape functions*:

1 2

1

( ) 1n

i i

i

x xu x N u u u

L L=

= = − +

∑

*This is a major distinguishing feature of approximating energy methods. Instead of solving a

differential equation for u, we assume a polynomial solution a priori. This particular set of shape

functions happens to be the exact solution for the truss, but this is not a general requirement.

2011 Alex Grishin




The Rayleigh-Ritz Method – An example

•Apply Castigliano’s Theorem (19) and the chain rule to obtain:

•Now evaluate using:

( ) ( )( )' ' ' '

1 1 1 1 2 2 10

1

LVEA N N u N N u dx F

u

∂= + =

∂ ∫

( ) ( )( )' ' ' '

2 1 1 2 2 2 20

2

LVEA N N u N N u dx F

u

∂= + =

∂ ∫

'

1

'

2

1

1

NL

NL

−=

=

•Making the substitution:

1 1

2 2

1 1

1 1

u FEA

u FL

− =

−

2

1 21 2

0 2

L

N NEA u u dx

x xV

EA

∂ ∂ +

∂ ∂ = ∫

2011 Alex Grishin





•At this point, the student may not think anything has been gained (is

the Rayleigh-Ritz procedure really easier than the direct method?).

However, we can omit some steps if we observe that Castigiano’s

Theorem always produces an equation of the form*:

for one-dimensional elements, where C is the constitutive law, Ni, and Nj are

shape functions, and uj are the nodal coefficients

•This now provides a rule which can be easily (naively) used to generate algebraic

equations, which can then be solved numerically. Most importantly for us, it is

easily programmed

' '

1

n

i j j i

ji L

VN CN u F

u =

∂= =

∂∑∫ (20)

*Note that there are no body loads or surface distributed loads in this formulation.

We’ll deal with these when we talk about the Galerkin Method next

2011 Alex Grishin



T

V

V∂= ⋅ ⋅ =

∂ ∫∆ C ∆ d Fu



•In the most general case – that of an elastic continuum, we start with a

full strain matrix, and so, for a given element, Castigliano’s Theorem has

the form:

where, for 3 dimensional isotropic materials, B , C, u and F are given by*:

0 0

0 0

0 0

0

0

0

x

y

z

y x

z y

z x

∂ ∂

∂ ∂

∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

u

v

w

x

y

z

F

F

F

C ∆∆∆∆ d F

1 0 0 0

1 0 0 0

1 0 0 0

0 0 0 (1 2 ) / 2 0 0(1 )(1 2 )

0 0 0 0 (1 2 ) / 2 0

0 0 0 0 0 (1 2 ) / 2

E

ν ν ν

ν ν ν

ν ν ν

νν ν

ν

ν

−

− −

++ − −

−

(21)

2011 Alex Grishin




The Galerkin Method

•As we’ve seen, the Rayleigh-Ritz Method provides a formula, or template, for

generating stiffness matrices. For structural problems, it will always work.

However, for other types of boundary value problems, we may not always be able

to generate a Lagrangian (L) with all the required mathematical properties to

satisfy equations (16) thru (18)

•For these types of problems, another type of approximate energy method is

available. This method is more general than the Rayleigh-Ritz Method and is

equivalent to it in such situations when a proper Lagrangian function CAN be

obtained. This method involves constructing the weak form of the governing

differential equation.

•Because of it’s broad applicability and popularity, we will discuss this method next.

2011 Alex Grishin




The Galerkin Method

•Consider an arbitrary one-dimensional differential equation given by:

on x≤x≤L

Where u(p) stands for the pth derivative of u with respect to x, and C is a scalar

constitutive relation. Let’s assume the boundary conditions are u=0 at x=0, and

Cu(p+1)=P at x=L. The function b(x) may be considered a body load over L. Next,

multiply both sides by a trial function, w(x) which equals zero at essential boundary

conditions:

•Integrate over L:

( ) ( ) ( )pCu x b x= −

( ) ( ) ( ) ( )pCu x w x bw x= −

( )

0 0( ) ( ) ( ) ( )

L Lp

Cu x w x dx b x w x dx= −∫ ∫

(22)

(23)

•Equation (23) is the Weak Form of equation (22). Solving this equation instead of

(22) offers several advantages because it weakens the restrictions on admissible

solutions. In particular, the definite integral implies that solutions only have to

solve (22) in an average sense – smoothing over discontinuities and singularities.

2011 Alex Grishin




The Galerkin Method

•Next, use integration by parts:

(24)

•But, by the Fundamental Theorem of Calculus, the first term above is just equal to

the surface loads or tractions (remembering that w=0 at x=0):

( )( ) ( 1) ( 1) ( 1)( ) ( ) ( ) ( ) ( )p p p pdu x dx w x u x dx w x u x dx

dx

+ + += −∫ ∫ ∫

•Substitute (24) into (23):

( )( 1) ( 1) ( 1)

0 0 0( ) ( ) ( ) ( ) ( ) ( )

L L Lp p pd

C w x u x dx Cw x u x dx b x w x dxdx

+ + +− = −∫ ∫ ∫

( )( 1) ( 1)

00( ) ( ) ( ) ( ) ( )

L Lp pd

C w x u x dx Cw x u x w L Fdx

+ += =∫•Finally, this leaves us with:

( 1) ( 1)

0 0( ) ( ) ( ) ( ) ( )

L Lp p

Cw x u x dx F x w x dx w L F+ + = +∫ ∫ (25)

2011 Alex Grishin




The Galerkin Method

•For the discretized solution, we therefore substitute our chosen shape functions

for BOTH u and w:

Here, we go back to the notational shortcut, f(x) = f:

1

( ) ( )n

i i

i

u x N x u=

=∑

( 1) ( 1)

0 01 1

( )n nL L

p p

i j i i

i i

N u CN u dx bN u dx N L P+ +

= =

= +∑ ∑∫ ∫

•In the Galerkin Method, the trial function, w is assumed to take the same form as

the solution, u

( 1) ( 1)

0 01 1

( )n nL L

p p

i j j i i i

i i

N CN u dx bN u dx N L F+ +

= =

= +∑ ∑∫ ∫If there is no body load and (22) was a first order equation :

' '

01

( )n L

i j j i

i

N CN u dx N L F=

=∑∫ (26)

2011 Alex Grishin




The Galerkin Method

•Thus, we see that the Galerkin procedure does indeed produce algebraic systems

equivalent to the Rayleigh-Ritz Method (at least for first-order differential

equations in one dimension).

•However, note a subtle difference produced by our assumptions. Let’s put the two

equations side-by-side:

•The difference appears in the external force vector on the RHS. In the Rayleigh-Ritz

Method, we started out with the assumption that all external tractions occur as nodal

point loads. Although we did not have to make this assumption, a comparison of the

two solutions reveals the Galerkin form to be more general in that it can accommodate

any force vector, F distributed anywhere within the domain (not just at nodes)!

•Multiplying such a vector by the shape function automatically weights it (lumps it) at

nodes. This is convenient for higher order, or higher dimensional problems

' '

01

( )n

L

i j j i

i

N CN u dx N L F=

=∑∫

' '

1

n

i j j i

ji L

VN CN u F

u =

∂= =

∂∑∫ Rayleigh-Ritz

Galerkin

2011 Alex Grishin



mae 323: chapter 2 - padtinc.com

Documents