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Fundamentals and ProceduresMAE 323: Lecture 2

2011 Alex Grishin 1MAE 323 Lecture 2 Fundamentals and Procedures

Part I: FEM – The Basic Idea



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 where

e e e

e e

= +∑ ∑K d f f where



•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



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



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

F1=-∆u x k

F1=k (u1-u2)

•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

•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)



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:u1

u2

u3,F3

k2

k1

k3

1

2

3u1

k1

k2

u2

1f

e

2f

e x∑F @ 11 2 1 1 2 1 2

: f f ( ) 0e e

k u k u u+ = + − =

•It will be the sum of forces associated

with extension u1 (represented by )

and u2-u1 (represented by )



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 ( ) ( ) 0

e ek u u k u u+ = − + − =

3 3 3 23: f ( )

ek uF u= = −

x∑F @ 2

x∑F @ 3

u3,F3



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)



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



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



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



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

00

0

0 0



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)

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) in the absence of body loads. 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:



2 1

1

( )( )

u uu x x u

L

−= +

2 1( )

=u udu F

dx EA Lε

−= =

Linear deflection

Linear elasticity

Constant strain

(3b)

(4b)

(6b)

•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

u1,F1

1

2

u2,F2

y

x

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.




•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)



•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 u=

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

approximate the solution and are the nodal

solution values for element e

e

iu

(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 u=

=∑ (10)


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



1 2 3

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

functions, N above:

L

L

A bar or truss element (Cont.) :

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

(node 2):

1

2N

1

1N

2

2N

2

1N




•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




•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)




•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)



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 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



The Euler-Bernoulli Beam Element :

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

1 1

2 2

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)



Part II: The Procedure



•Every finite element analysis consists of the following steps

1. Formulation of Physical Problem

2. Analysis Type Selection

3. Physical Reduction (Idealization)

4. Element Type Selection

5. Model Creation

a) Loads and Boundary Conditions

b) Material Properties

c) Mesh

6. Solution

7. Post-processing

8. Validation



1. Formulation of the Physical Problem

•In the case of MAE 323, this is a slam-dunk, because we will always be

performing a linear static thermal, structural, or thermal/structural

problem

•However, the student should note that in general, this isn’t always as

obvious as it seems. The questions that must answered here include:

•What forces are involved?

•By what mechanism is energy dissipated? Remember, in our case

it’s always some variant of Hooke’s Law. This will always reveal

the constitutive law(s)

•Is the problem linear or nonlinear?

•Is the problem time-dependant?



1. Formulation of the Physical Problem

•As an example, suppose your boss (or customer) asks you to perform

a vehicle crash study. At first, you may be (rightly) bewildered. Start

by answering the questions on the previous slide:

•What forces are involved? Mostly momentum-driven impacts of

the form F=-m∆V/∆t

•By what mechanism is energy dissipated? Mostly mechanical

deformation and damping

•Is the problem linear or nonlinear? Highly nonlinear

•Is the problem time-dependent? Yes



2. Analysis Type Selection

•In the case of the vehicle crash study, answering the

questions in the last step reveals that we have to do a

nonlinear-transient time-history study

•Once we have determined the physical formulation (which should

emerge naturally from answering the previous questions), we

determine what analysis type is to be performed. This is like making a

selection from a menu. The analyst has to be aware which finite

element solver/element-type combinations exist in order to match

the physical formulation with a finite element procedure capable of

approximating it . This may not be easy, and in many cases may

amount to asking “Does our commercial software product have this

capability?”



•Even in the case of problems in MAE 323 (all physics may be characterized as

thermal/mechanical elastostatics), questions will arise as to whether the

problem may be best characterized as a full continuum, reduced continuum

(dominated by bending or membrane behavior), spring connections, or a

combination of all three

•This problem is multiplied when we consider the analysis of assemblies (as

you often will in your professional careers). In many cases, you often analyze

for predefined performance points. For example, a pressure regulator will be

analyzed for predefined (design) pressure conditions which represent either

typical or extreme performance conditions (more about that in later

lectures). The analyst has to make assumptions about the load paths in each

of these conditions, and these will determine how the model is built (and

how many models must be built!)

3. Physical Reduction (idealization)



4. Element Type Selection

•Element types may be categorized in terms of:

•The physical field they approximate (in our case, it will always be

material displacement or temperature)

•The type of field behavior: continuum, multiple-continuum

(coupled field), mixed-variable(for example, in hyper-elastic

applications), reduced continuum, or special connection (such as

contact elements, but more on these in later lectures)

•So back to the vehicle-crash example: the engine-block and other internal

heavy components may be modeled as full-continuum or mass-elements,

depending on whether we care about stresses and strains in those

components. If we don’t, then we just need to model the mass and possibly

use special elements to connect the mass to the components we DO care

about. What we care about mostly is the chassis and body. The mechanical

behavior of these components will most certainly be dominated by bending

(and eventually buckling and crumpling), and thus we should use shell

elements for the body and possibly beams for the chassis



•Finally, the vehicle will impact another object – maybe another vehicle or

something stationary – so contact must be defined between the vehicle surfaces

and what the vehicle will impact. At the basic mathematical level, contact is

treated as an additional (nonlinear) constitutive law which connects element

surfaces with either a penetrability condition, or some other law relating how the

surfaces move relative to one another. We refer here to contact between

impacting surfaces, but that is just one type of contact or special connection we

might define. For example, we might use rotational springs, or DoF couples to

attach the doors, hood, and trunk to the chassis. Numerically, there are several

ways to impose such a condition. Some commercial software does so with special

elements (and others simply add the appropriate matrix terms to the assembled

system matrices internally). This (contact) is a very large topic and the details are

handled many different ways by different commercial software



5. Model Creation

•With all the previous selections out of the way, we can begin building the

model…a) Loads and Boundary Conditions: The choices made in steps 2 and 3 dictate

where and how the loads and boundary conditions will be distributed. In

the case of our example, one vehicle/object might be fixed (stationary) or

free-free but motionless (In some types of solvers, you would need to attach

light springs to such an object to prevent matrices from becoming singular.

Other types don’t require this). The other vehicle would have an initial

velocity imposed upon it

b) In MAE 323, the materials will usually be well-defined. However, in highly

nonlinear applications, the choices get more difficult. For example, in the

vehicle crash study, we will probably need not only the full stress/strain

curves (obtained from tensile tests), but also strain-rate behavior (hard to

come by)

c) We can now mesh the model. We need to know what type and size mesh we

need (more on this topic in the next lecture). In our case, let’s say we’ve

settled on a quadrilateral mesh for the shells (body), and tetrahedra for solid

internals. Most commercial software can automatically mesh these bodies,

but some of our connections may require manual construction



5. Solution

•Solvers are generally categorized by exactly how the system

equations are solved. In the vehicle crash example, options include

implicit vs. explicit time-integration, sparse vs. iterative, etc. Vehicle

crash studies usually involve very short timeframes (relative to the

period of the system natural frequencies), and thus explicit solvers

are usually used (these have other enormous advantages, but the

discussion takes us too far from the class goals)

6. Post-processing

•In this step, the user has to extract the metrics of the material

response. These are usually in the form of displacement, stress, and

strain contour plots at a given time in the solution (or temperature in

the case of thermal analyses). But in our example, we might also want

accelerations and velocities at specific points (nodes). These would be

in the form of tables (vs. time)



7. Validation

•Much can go wrong in an analysis. Not only can we impose incorrect

boundary conditions and loads, but we might have constructed the

mesh so poorly that it yields incorrect results (especially for derived

quantities such as stress and strain). In the case of the vehicle crash

study, we might have also chosen a numerical integration time-step

that is too large (in the case of implicit solvers), which would also yield

incorrect results, or perhaps used the wrong material properties

(maybe the density we used for the heavy internal components was

way off)

•This is why we always have to validate our models. In MAE 323, this is

much easier than the vehicle crash study offered here as an example.

We can use the techniques we learned in Lecture 1. But we can also

consult standard tables (for things like stress concentration factors) for

more complicated problems



A Simple Example: The Bolted Flange

•Suppose you have a pressure vessel and you want to know the

maximum (von Mises) stress at a bolted flange. To keep things simple,

we’ll already start with a simplifying assumption: Namely that the

flange around each bolt may be represented as below (pressure vessels

have many bolts):

Two mating plates

(0.5” thick) Mating

surfaces see 2000

psi pressure

Bolt

Washer 1

Washer 2Nut

Thru-

hole




•Let’s go through the check-list to see if we know how to model this

1. Formulation of physical problem: This is a linear static

structural problem

2. Analysis type: Linear static problems only have one analysis

type (linear static)

3. Physical Reduction: This will be a full continuum problem

because we are looking for full 3D stress fields produced by

surface interactions. There may be some geometric

simplification. For example, we don’t even have to model the

bolt if all we care about is flange stresses (more on this later)

4. Element type selection. Since this is a 3D structural continuum

problem, we can accept the workbench default element types

for this type of problem (tetrahedral solid elements with mid-

side nodes)




5. Model Creation

a) Loads and Boundary Conditions. The loads are simply

pressure loads on the flange surfaces. The boundary

conditions are a little more involved (we’ll discuss that

next)

b) Material Properties. In this problem, structural steel is

used for all parts

c) Mesh. The mesh will need to have enough resolution to

adequately capture the peak stresses. One way to do this

is to require a certain maximum percentage change in the

results of two successive mesh densities (eg: max. stress

must not change by more than 4 percent when reducing

element size by 50 percent)

6. Solution

We will accept default settings for now (later lectures will

address solution settings)




7. Post-Processing

Since this is a pressure vessel, we will want to look at both

deflections and stress (with pressure vessels, flange

separation is often of interest)

8. Validation

Since there are equal and opposite pressures on each flange

surface, we will want to ensure that the total reaction force in

the pressure direction is zero (this should be determined at

the region that is fixed in that direction). We might also want

to ensure that the reaction at each flange is P*A where P is

the pressure on that flange and A is the surface area over

which it is applied




Loads and Boundary Conditions

•The only thing we have not yet determined is what kind of

boundary condition this model should receive. That will require

some discussion

•First, it helps to understand something about bolted flanges.

Many books on Mechanical Engineering Design offer a

simplification* involving an estimate of how the load is transmitted

through the flanges. The simplification assumes the load is entirely

concentrated within the frustum produced by the plane of the

flange interface intersecting a hypothetical cone of a prescribed

angle, also intersected by the plane of the underside of the bolt

head

*see, for example: J. Shigley and C. Mischke, Mechanical

Engineering Design, 5th ed. McGraw-Hill, 1989.





•The simplification is depicted graphically below

θ

Values of

∼30° are

typically used

for bolt load

frusta

t

db

dp

2 tanp bd d t θ= +

The flanges are

bonded only within

the frustum

diameter dp

No load is

transmitted across

these surfaces and

so no contact is

assumed





•We expect the following type of material response due to the

pressure on each flange surface:





•Suppose we are interested only in the stress at the bolt flange

interface (the diameter of the frustrum dp). If this is the case, we

can remove the bolt (we don’t have to model it)

•The annular region whose outer diameter is represented by dp

must still be modeled (we need a circular “patch” on the interface

with which we can define the bonded region)

dp

What kind of

boundary condition

should go on these

surfaces?





•Remember, this bolted flange model already has an assumption

built into it: Namely that it is a repeated feature which reacts in the

same way (and to the same extent) as all the others

•The bolted flange model is effectively “cut” out of a larger model

(along the dotted green line above). The cut boundaries behave as

Cartesian symmetry planes

Symmetry PlaneSymmetry Plane





•So, what’s a symmetry plane?

•A symmetry boundary condition is a surface on which primary solution variables

(displacement in structural models) are set to zero normal to the surface. Such a

boundary condition usually reflects a theoretical dividing line through a domain – both

sides of which behave identically. The figure below shows a planar model (one type of

symmetry), which is identically loaded at opposite ends. The geometry and loading admit

a plane of symmetry which can be used to “cut” the model in half

x

y

ux=0





•But our simplified bolt flange model has even more symmetry

than the planar symmetry we’ve identified at the cut faces. It could

be represented as a quarter symmetry model.

Symmetry on all

these faces

part i: fem –the basic idea€¦ · part i: fem –the basic idea. mae 323: lecture 2...

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