Notes 62 - Basic Finite Element Theory

8/8/2019 Notes 62 - Basic Finite Element Theory

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Basic Finite Element Theory - Page 1 of 1

Basic Finite Element Theory

Objectives:1) Define a shape function2) Know the three basic assumptions for finite elements

3) Describe the main element types available for structures

The next step in structural analysis is to be able to model more complex structures. Theability to include the effects of shear walls, floor diaphragms, complex connections and

model shell structures is required. This is accomplished by using additional types of

elements in the modeling process. The Finite Element Method (FEM) allows morecomplex element behavior to be modeled. The FEM was originally just an extension of

matrix structural analysis, developed by structural engineers. It has since been used in

just about every field where differential equations define the problem behavior. Theresult of the FEM is to create a stiffness matrix and a set of loads. After that, the solution

process is identical to that covered in this text. There are many excellent books covering

the FEM, this section is intended only as an introduction.

The basic idea of the finite element method is to break up a continuum into a discrete

number of smaller "elements". These elements can be modeled mathematically by a

stiffness matrix and are connected by nodes that have degrees of freedom. This isidentical to what we have done with bending and truss elements. However, beams and

trusses have natural locations at which to define nodes. In addition, the derivation of

their stiffness matrices can be done on a physical basis.

Simple FEM Theory

More general finite elements require slightly more complicated procedures than used forbeams in order to derive the stiffness matrix. The basic procedure is to assume a shape

function that describes how the nodal displacements are distributed throughout the

element based. From the differential equation, we form an operator matrix that willconvert the displacements within the element into strains. Next the internal and external

virtual work can be formed and equated to develop the stiffness matrix. The last step isidentical to that used for truss and bending elements.

As an example, we will develop the stiffness matrix for a truss element, an axial member.

The truss element has two nodal displacements, v1 and v2, one at each end. For any given

set of displacements at the ends, a function is required to convert these into displacementsalong the length of the element. The obvious selection for the functions is the linear set

given below.


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Notice how the given functions distribute the end displacements throughout the element.

These distribution functions are called the shape functions. The shape functions can beput into matrix form along with the end displacements to form an equation that describes

the displacement within the element. This displacement anywhere within the element isdescribed by the following matrix equation:

=2

11)(

v

v

L

x

L

xxu

Notice that the displacement is a function of x, the position in the element. Also note that

the displacement anywhere in the element, u(x), is the sum of the displacements caused

by both end displacements distributed throughout the element. Equation 8.1 can be re-written as:

v*H(x)=u(x)

The matrix H(X) is called the shape function matrix and v are the element nodaldisplacements. We now need a differential equation that converts the displacement into

strain. For our one-dimensional example this is:

x

u(x)=x

It is useful to re-write equation 8.3 into the form of two matrices. This alternate formputs the differential operator into the form of an operator matrix, D. Equation 8.3 is re-

written to:

u(x)*D=x


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Where the operator matrix has the form;

x

D =

If we apply this operator to the displacement, u(x) equation, we can find the strain as afunction of the element displacements, v, and the shape function matrix, H(X). This

gives the form for the strain in terms of the shape function matrix H(x).

v*H(x)*D=x

Note that the nodal displacements, v, are constants with respect to X and need not be

operated on, differentiated. Therefore, only the derivatives of the shape functions need tobe taken. For our case of the axial element, the strain can then be written by substituting

the shape functions, H(x), and applying the D operator giving:

=2

1

v

v

L

1

L

1(x)

Typically, the differential operator times the shape function matrix is called B, the strain-displacement matrix. The strain is then commonly written in the shorter form:

v*B=x

We also need the relationship of Hooks law that converts strain into stress.

*E=

Therefore, if we calculate the internal strain energy, as was done in previous chapters inorder to develop a stiffness matrix, virtual strain times stress, with substitutions we have:

Vv*B*E*B*vTT

volume

T

volume

i ==W

Equating internal to external virtual work and removing the arbitrary virtual

displacements, vT , we get:

( ) v*VB*E*BT =Svolume

Looking at the equation we see that this is the familiar stiffness form of the element

relationship between forces and displacements. As a result, we can see that the integral is

just the element stiffness. Taking that portion out of the equation we have:


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( ) VB*E*BT =Kvolume

e

This is the classic form of the finite element stiffness formulation. For our axial element

example we substitute forB from before and E is just the familiar Young's Modulus. Inthe general case, E is the constitutive matrix. For the linear case E is just the 3-D

representation of Hooks Law. Integrating over the Y and Z coordinates for part of the

volume integral we get the area of the cross section. Multiplying the matrices after thepartial integration for the area and removing the constants from the integral we get:

dx*

L

1

L

1-

L

1-

L

1

*AE=K

22

22

length

e

Integrating the matrix term by term over the length we get the familiar form for truss

stiffness as:

11-

1-1*

L

AE=Ke

This is the final result we are looking for. Of course this is identical to the standard truss

stiffness matrix developed by traditional stiffness methods. However the described shape

function process can be extended to other types of elements where traditional stiffness by

definition methods are not possible.

Available Elements

Energy derivations (here virtual work) are commonly used to form the stiffness for avariety of element types. The most common elements are the membrane (planar), plate,

shell and solid elements. Each of these elements has a given set of nodes and

displacements associated with those nodes. The common forms of these elements aregiven below.


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These elements have additional restrictions on their behavior that depend on theirderivation. However, the result is always a stiffness matrix that can then be treated like

any other stiffness matrix and may be rotated and transformed as desired. When

combining these elements, the same concerns about boundary conditions and matching

DOF at the nodes must be accounted for. Additional concerns are also generated sincethe shape function assumption can affect the accuracy if the results. The standard beam

element can be derived in a similar fashion using the cubic beam functions given in thesection on Consistent Geometric Stiffness.

Element derivation has become increasingly complex. Techniques that include

nonlinearities while still reducing the number of unknowns in the element have becomevery theoretically demanding. However the use of these sophisticated elements is

identical to their simpler counterparts.

The three elements most commonly used by structural engineers are the membrane,

plate/shell and solid elements. Each of these three will be discussed briefly here and then

in detail in subsequent chapters. The membrane element is a two dimensional flat

extensional element. The common versions are triangular and rectangular elements. Thetriangular elements vary from three to six nodes. The rectangular elements vary from

four to nine nodes. There are two in plane displacement DOF's at each node of the

element. The elements can be used to model two dimensional elasticity problems, planestrain and plane stress. It can reproduce the two normal and one shear stress in the plane

of the element.


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The membrane element has no rotational stiffness or stiffness normal to the plane of theelement. It can be situated arbitrarily in space but the resultant forces must lie in the

plane of the element. This is similar to the three-dimensional truss element. The three

node triangular element can model constant stress. The nine node rectangular elementcan model linear variations of stress. Triangular elements are popular with automatic

mesh generation and adaptive mesh generation schemes. Loads on membrane elements

can only consist of in-plane loads.

The flat plate element is a two dimensional element that acts like a flat plate. It is found

in triangular and rectangular versions. There are two out of plane rotations and thenormal displacement as DOF. These elements model plate-bending behavior in two

dimensions. The element can model the two normal moments and the cross moment in

the plane of the element. Some versions will also give the transverse shear as a result.

The three node triangular version models constant moment. The higher node elementscan model linear variation of moment across the element. This element has no rotational

stiffness normal to the plane and no in plane stiffness. Superimposing the membrane and

plate elements on top of one another creates flat shell elements. Loading on plateelements can consist of any combination of forces normal to the plate and out of plane

moments. Loading on shell elements can consist of the combination of plate and

membrane loadings.


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The solid element is a three dimensional extensional element. Versions found vary fromthe four-node tetrahedron to the 27-node brick element. The most common version is theeight-node brick element.

The solid element has three translations at each node for DOF. This element can model a

full three-dimensional stress state. The eight-node element has some linear variation ofstress throughout the element. The solid element has no rotational stiffness.

Tetrahedrons are popular for use in mesh generation and adaptive mesh refinementschemes.

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Notes 62 - Basic Finite Element Theory