ME 323 – Mechanics of Materials

Lectures 27-29: Introduction to finite element methods

Lecture Book: Chapter 17

Joshua Pribe

Fall 2019

Outline

• What is FEM? Why is it useful? How does it fit in with this class?

• FEM for axially-loaded rods• Derivation using energy methods (see pre-week video for details) →

• Choose nodes and elements

• Stiffness matrix, force vector, and displacement boundary conditions

• Reminder: FEM is not just “another way to solve axial bar problems”

• Solving problems with a basic Matlab FEM code• Computers can efficiently solve large systems of equations

2

{ } [ ]{ }F K u=

FEM Overview

So far in ME 323: idealize physical systems as connected rods, shafts, beams

What happens when we want a closer approximation of the real-world structure (e.g. complex shapes, connections/joints, inhomogeneous material)

3

C. Felippa, Introduction to Finite Element Methods

FEM Overview

Steps for coming up with the finite element equations for linear elastic rods:

• Discretize a body with a finite element mesh of nodes and elements

• Calculate the strain energy in each element in terms of the displacements of each node

• Add in the potential energy from forces acting on the body

• Find the displacement field based on the Principle of Minimum Potential Energy and the displacement boundary conditions

4

[ }1

2{ } ]{TU u K u=

F i iU U Fu= −+=

0iu = for any nodal displacement ui { } [ ]{ }F K u=

FEM Derivations

5

Discretize the body into nodes and elements

Lecture Book: Ch. 17, pg. 13

FEM Derivations

6

Strain energy in terms of nodal displacements—key outcome is the stiffness matrix [K]

Lecture Book: Ch. 17, pg. 10

FEM Derivations

7

Strain energy in terms of nodal displacementsFor N elements and N+1 nodes: Lecture Book: Ch. 17, pg. 12

FEM Derivations

8

Potential energy for applied forces (the forces are conservative since no energy is dissipated):(Look up the “Principle of Virtual Work” for a more general approach)

1

1

{ } { }N

i

T

F i iFu F u+

=

= = − − where Fi is the force at node i and ui is the displacement at node i

The total potential energy in the system is then [1

{ }2

]{ } { } { }T T

F K u F uU u= = + −

Principle of Minimum (or “stationary”) Potential EnergyFor all admissible displacement fieldsfor a conservative system,the actual displacement field, which satisfies equilibrium, minimizes the potential energy

Key outcome for the finite element method:Last step: enforce displacement BCs (best illustrated using an example)

Lecture Book: Ch. 17, pg. 3

0iu = for any nodal displacement ui

{ } [ ]{ }F K u=

Lecture Book: Ch. 17, pg. 2

See proof on Ch. 19, pg. 15

Example 17.2

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Use a three-element finite element model to determine the displacements at C and D.

Example 17.3

10

Use a two-element finite element model to determine the displacements at B and C.

Example 17.5

11

Set up and solve for the displacements at C, D, and H using a four-element finite element model.

Example 17.6

12

Use finite element models with different numbers of elements to find the displacement at end B of the rod

Matlab code

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This part solves the problem with the given BCs, force vector, and element properties

Lecture Book: Ch. 17, pg. 18

This part defines the properties of each element, the force vector, and the displacement BCsEdit this part based on the problem you are solving

Summary for FEM

1. Define the nodes and elements. There must be a node at every location where there is an external force.

a) For N elements, there are N+1 nodes

2. Construct the stiffness matrix [K]

a) Recall the form of the stiffness matrix; e.g. for 3 elements:

3. Construct the force vector {F}

a) This is the external force on each node

4. Enforce the displacement BCsa) For fixed displacement at node i, eliminate:

The ith row and column of the stiffness matrix, and the ith row of the force vector

5. Solve the system of equations for the unknown displacements:

6. Find the average stress in each element from the displacements

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1 1

1 1 2 2

2 2 3 3

3 3

0 00

00 0

k kk k k k

k k k kk k

− − + − − + − −

( ),avg j

j

j

Ak

E

L=where

( )1i

i i i i i

i

Eu u

LE += = −

{ } [ ]{ }F K u=

for the jth element