Lecture Notes: Introduction to Finite Element Method

Chapter 1. Introduction

x

i

j

fi

u

k

uj fj

i

Two nodes:

i, j

Nodal displacements:

ui,

uj (in, m, mm)

Nodal forces:

fi,

fj (lb, Newton)

Spring constant (stiffness):

k

(lb/in, N/m, N/mm)

Spring force-displacement relationship:

F = k∆

with ∆ = uj

−ui

© 1997-2003 Yijun Liu, University of Cincinnati

14

Lecture Notes: Introduction to Finite Element Method

Chapter 1. Introduction

In matrix form,

k

− k ui

=

fi

− k

k uj

f j

or,

© 1997-2003 Yijun Liu, University of Cincinnati

15

Lecture Notes: Introduction to Finite Element Method

Chapter 1. Introduction

1

2

3

u1, F1

u2, F2

u3, F3

k1

− k1

u1

f

1

=

11

− k1

k1 u2

f2

k

− k

u

2

2

2

2

= f

12

− k2

k2 u3

f2

© 1997-2003 Yijun Liu, University of Cincinnati

16

Lecture Notes: Introduction to Finite Element Method

Chapter 1. Introduction

That is,

F1 = k1u1 − k1u2

F2 = −k1u1 +(k1 + k2 )u2 − k2u3

F3 = −k2u2 + k2u3

In matrix form,

k1

− k1

0

u1

F1

− k k

+ k

2

− k

2

u

=

F

1

1

2

2

0

− k

2

k

2

u

F

3

3

k

1

− k

1

0 u

f 1

1

1

− k

1

k

1

0 u

=

f 1

2

2

0

0

0 u

3

0

0

0

0

u

0

0

k

2

− k

2

u1

=

f 2

2

1

0

− k

2

k

2

u

f 2

3

2

© 1997-2003 Yijun Liu, University of Cincinnati

17

Lecture Notes: Introduction to Finite Element Method

Chapter 1. Introduction

k

1

− k

1

0

u

f 1

1

f 1

1

− k

1

k

1

+ k

2

− k

2

u

=

+ f 2

2

2

1

0

− k

2

k

2

u

f 2

3

2

Assuming,

u1

= 0

and

F2

= F3

= P

we have

k1

− k1

0

0

F1

− k

1

k

1

+ k

2

− k

2

u

=

P

2

0

− k

2

k

2

u

3

P

which reduces to

k

1

+ k

2

− k

2

u

2

P

=

− k2

k2 u3

P

Unknowns are

U =

u

2

and the reaction force F1 (if desired).

u3

© 1997-2003 Yijun Liu, University of Cincinnati

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