Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

Stiffness Matrix in the 2-D Space

In the local coordinate system, we have

EA		1 −1 u'				f	'
					i	=	i
L			1		'		'
	−1			uj		f j

Augmenting this equation, we write

1		0	−1
	0	0	0
EA
L −1		0	1
	0	0	0

or,

k'u' = f '

0 ui'			fi '
0		'
	vi		0
			=
0 u		'j	f j'
		'		0
0 v j

Using transformations given in (29) and (30), we obtain

k' Tu = Tf

Multiplying both sides by TT and noticing that TTT = I, we obtain

TT k'Tu = f

(31)

Thus, the element stiffness matrix k in the global coordinate system is

k = TT k'T

(32)

which is a 4×4 symmetric matrix.

© 1997-2002 Yijun Liu, University of Cincinnati

42

Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

Explicit form,

			ui		vi			uj	vj
		l2			lm		−l2		−lm
			lm m			2	−lm −m			2
k =	EA
	L		−l	2	−lm		l	2	lm
		−lm			− m2		lm		m2

Calculation of the directional cosines l and m:

l = cosθ =	X j − Xi	,	m = sinθ =	Yj −Yi
				L
	L

(33)

(34)

The structure stiffness matrix is assembled by using the element stiffness matrices in the usual way as in the 1-D case.

Element Stress

ui

'

1

1

l m

0 0

ui

vi

σ = Eε = EB

'

= E −

0 0

L

u

j

L

l m

uj

v j

That is,

ui

E

vi

σ =

[− l

− m

l m]

(35)

L

u j

v j

© 1997-2002 Yijun Liu, University of Cincinnati

43

Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements

Example 2.3

3

A simple plane truss is made

45o

of two identical bars (with E, A, and

L), and loaded as shown in the

2

P2

figure. Find

1) displacement of node 2;

Y

P1

2

2) stress in each bar.

1

Solution:

45o

X

This simple structure is used

1

here to demonstrate the assembly

and solution process using the bar element in 2-D space.

In local coordinate systems, we have

'

EA 1

−1

'

k1 =

= k

2

L −1

1

These two matrices cannot be assembled together, because they are in different coordinate systems. We need to convert them to global coordinate system OXY.

Element 1:

θ = 45o , l = m = 22

Using formula (32) or (33), we obtain the stiffness matrix in the global system

© 1997-2002 Yijun Liu, University of Cincinnati

44

Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements

u1

v1

u2

v2

1

1

−1 −1

k1 =

T '

=

EA

1

1

−1 −1

T1

k1T1

2L

−1

1

−1

1

−1

1

1

−1

Element 2:

θ = 135o , l = −

2

, m =

2

2

2

We have,

u2

v2

u3

v3

1

−1

−1

1

k2 =

T '

=

EA −1

1

1

−1

T2

k2T2

1

1

−

2L −1

1

−1

−1

1

1

Assemble the structure FE equation,

u1

v1

u2

v2

u3

v3

1

1 −1 −1

0

0 u1

F1X

1

1 −1 −1

0

0 v

F

1

1Y

EA −1

−1 2

0

−1 1 u2

F2 X

−1 0

2

=

2L −1

1 −1 v2

F2Y

0

0 −1 1

F3 X

1 −1 u3

0

0

1 −1

−1 1 v3

F3Y

© 1997-2002 Yijun Liu, University of Cincinnati

45

Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

Load and boundary conditions (BC):

u1 = v1 = u3 = v3 = 0,					F2 X = P1 , F2Y = P2
Condensed FE equation,
EA 2	0 u2			P1
2L 0	2 v	2		= P
				2

Solving this, we obtain the displacement of node 2,

u2	=	L	P1
v2		EA P2

Using formula (35), we calculate the stresses in the two bars,

0

0

(P

+ P )

σ

=

E 2

[

−1

−1 1 1

L

=

2

1

L 2

]

EA

P

2 A 1

2

1

P

2

P

= E 2

1

2 (P − P )

σ

2

[

1

−1

−1 1

L P2

=

L 2

]

0

2 A

1

2

EA

0

Check the results:

Look for the equilibrium conditions, symmetry, antisymmetry, etc.

© 1997-2002 Yijun Liu, University of Cincinnati

46