Finite Element Method:Beam Element Beam Elementocw.snu.ac.kr/sites/default/files/NOTE/10530.pdf ·...

Mechanics and Design

Finite Element Method: Beam Element

SNU School of Mechanical and Aerospace Engineering

Beam Element

¨ Beam: A long thin structure that is subject to the vertical loads. A beam shows more evident bending deformation than the torsion and/or axial deformation.

¨ Bending strain is measured by the lateral deflection and the rotation -> Lateral deflection and rotation determine the number of DoF (Degree of Freedom).




¨ Sign convention

1. Positive bending moment: Anti-clock wise rotation.

2. Positive load: $y-direction.

3. Positive displacement: $y-direction.




¨ The governing equation

- + = = -wdx dV or w dVdx

$$

0

V dx dM or V dMdx

$$

+ = =0

ˆ 0 orˆ

ˆ 0 orˆ

dVwdx dV wdx

dMVdx dM Vdx

- - = = -

- = =




¨ Relation between beam curvature (k ) and bending moment

k MEI

= =1r or

d vdx

MEI

2

2

$$

=

¨ Curvature for small slope (q = dv dx$ / $ ): kd vdx

=2

2

$$

r: Radius of the deflection curve, $v: Lateral displacement function along the $y-axis direction

E : Stiffness coefficient, I : Moment of inertia along the $z-axis




Solving the equation with M ,

ddx

EI d vdx

w x2

2

2

2$$$

$FHG

IKJ = - a f

When EI is constant, and force and moment are only applied at nodes,

EI d vdx

4

4 0$$

=




Step 1: To select beam element type

Step 2: To select displacement function

Assumption of lateral displacement

3 21 2 3 4ˆ ˆ ˆ ˆ ˆ( )v x a x a x a x a= + + +

- Complete 3-order displacement function is suitable, because it has four degree of

freedom (one lateral displacement and one small rotation at each node)

- The function is proper, because it satisfies the fundamental differential equation of a beam.

- The function satisfies continuity of both displacement and slope at each node.

Representing $v with functions of $d y1 , $d y2 , $f1, 2̂f

$ ( ) $v d ay0 1 4= =

dvdv

a$$

$01 3

a f= =f

$ $v L d a L a L a L aya f = = + + +2 13

22

3 4

dv Ldx

a L a L a$$

$a f= = + +f 2 1

22 33 2




Replacing a1~a4 with $d y1 ,

$d y2 , $f1, 2̂f

$ $ $ $ $ $

$ $ $ $ $ $ $ $

vLd d

Lx

Ld d

Lx x d

y y

y y y

= - + +LNM

OQP

+ - - - +LNM

OQP + +

2 1

3 1 2

3 1 2 2 1 23

2 1 2 1 22

1 1

e j d i

e j d i

f f

f f f

Representing it in matrix form, $ $v N d= o t

$

$$$$

d

d

d

y

yo t =

RS||

T||

UV||

W||

1

1

2

2

f

f N N N N N= 1 2 3 4

NL

x x L L NLx L x L x L

NL

x x L NLx L x L

1 33 2 3

2 33 2 2 3

3 33 2

4 33 2 2

1 2 3 1 2

1 2 3 1

= - + = - +

= - + = -

$ $ $ $ $

$ $ $ $

c h c h

c h c h

where N1, N2, N3, N4 : shape functions of the beam element




Step 3: To define strain-displacement relation and stress-strain relation

Assume that the equation of strain-displacement relation is valid

e x x ydudx

$, $$$b g = , $ $ $

$u y dv

dx= - => e x x y y d v

dx$, $ $ $

$b g = -2

2

Basic assumption: Cross-section of the beam sustains its shape after deformation by bending,

and generally rotates by degree of dv dx$ / $b g .

Bending moment-lateral displacement relation and shear force-lateral displacement relation

$ $ $$

$ $$

m x EI d vdx

V EI d vdx

b g = =2

2

3

3




Step 4: To derive an element stiffness matrix and governing equations by direct stiffness

method

¨ Element stiffness matrix and governing equations

$ $ $$

$ $ $ $

$ $$$

$ $ $ $

$ $ $$

$ $ $ $

$ $$$

$

f V EId vdx

EIL

d L d L

m m EId vdx

EIL

Ld L d L

f V EId v Ldx

EIL

d L d L

m m EId v Ldx

EIL

L

y y y

y y

y y y

1

3

3 3 1 1 2 2

1

2

2 3 12

1 22

2

2

3

3 3 1 1 2 2

2

2

2 3

012 6 12 6

06 4 6 2

12 6 12 6

6

= = = + - +

= - = - = + - +

= - = - = - - + -

= = =

b g e jb g e jb g e j

b g

f f

f f

f f

d L Ld Ly y12

1 22

22 6 4+ - +$ $ $f fe j

- Matrix Form

$

$$

$

$$$$

fmfm

EIL

L LL L L L

L LL L L L

d

d

y

y

y

y

1

1

2

2

3

2 2

2 2

1

1

2

2

12 6 12 66 4 6 212 6 12 6

6 2 6 4

RS||

T||

UV||

W||=

--

- - --

L

N

MMMM

O

Q

PPPP

RS||

T||

UV||

W||

f

f

$k EIL

L LL L L L

L LL L L L

=

--

- - --

L

N

MMMM

O

Q

PPPP3

2 2

2 2

12 6 12 66 4 6 212 6 12 6

6 2 6 4




Step 5: To constitute a global stiffness matrix using boundary conditions

Assemble example

Assume EI of the beam element is constant.

1000 lb load and 1000 lb ft- moment are applied at the center of the beam.

Assume load and moment were only applied at nodes.

Left end of the beam is fixed and right end is pin-connected.

The beam is divided into two elements (nodes 1, 2, and 3 as shown above figure).




k EIL

L LL L L L

L LL L L L

d dy y

( )13

2 2

2 2

12 6 12 66 4 6 212 6 12 6

6 2 6 4

1 1 2 2

=

--

- - --

L

N

MMMM

O

Q

PPPP

f f

k EI

L

L LL L L L

L LL L L L

d dy y

( )23

2 2

2 2

12 6 12 66 4 6 212 6 12 6

6 2 6 4

2 2 3 3

=

--

- - --

L

N

MMMM

O

Q

PPPP

f f

FMFMFM

EIL

L LL L L L

L L L LL L L L L L L L

L LL L L L

d

d

d

y

y

y

y

y

y

1

1

2

2

3

3

3

2 2

2 2 2 2

2 2

1

1

2

2

3

3

12 6 12 6 0 06 4 6 2 0 012 6 12 12 6 6 12 6

6 2 6 6 4 4 6 20 0 12 6 12 60 0 6 2 6 4

R

S

|||

T

|||

U

V

|||

W

|||

=

--

- - + - + -- + + -- - -

-

L

N

MMMMMMM

O

Q

PPPPPPP

´

R

S

|||

T

|||

U

V

|||

W

|||

f

f

f




Boundary conditions and constrains at the node 1(fixed) and node 3(pin-connected) are

f1 1 30 0 0= = =d dy y

-RS|T|

UV|W|=

L

NMMM

O

QPPP

RS|T|

UV|W|

10001000

0

24 0 60 8 2

6 2 43

2 2

2 2

2

2

3

EIL

LL L

L L L

d y

ff

(5.2.5)




Example: Beam analysis using direct stiffness method

Cantilever beam supported by roller at the center




Load P is applied at node 1.

Length: 2L ,Stiffness: EI

Constrains: (1) Roller at node 2, (2) Fixed at node 3.

- Global stiffness matrix

K EIL

L LL L L

L L LL L L L

LSymmetry L

d d dy y y

=

--+ - + -

+ --

L

N

MMMMMMM

O

Q

PPPPPPP

3

2 2

2 2 2

2

12 6 12 6 0 04 6 2 0 0

12 12 6 6 12 64 4 6 2

12 64

1 1 2 2 3 3f f f




FMFMFM

EIL

L LL L L L

L LL L L L L

L LL L L L

d

d

d

y

y

y

y

y

y

1

1

2

2

3

3

3

2 2

2 2 2

2 2

1

1

2

2

3

3

12 6 12 6 0 06 4 6 2 0 012 6 24 0 12 6

6 2 0 8 6 20 0 12 6 12 60 0 6 2 6 4

R

S

|||

T

|||

U

V

|||

W

|||

=

--

- - --

- - --

L

N

MMMMMMM

O

Q

PPPPPPP

R

S

|||

T

|||

U

V

|||

W

|||

f

f

f

Boundary conditions d dy y2 3 30 0 0= = =f

-RS|T|

UV|W|=

L

NMMM

O

QPPP

RS|T|

UV|W|

PEIL

L LL L LL L L

d y

00

12 6 66 4 26 2 4

32 2

2 2

1

1

2

ff

d PLEIy1

3712

= - f f1

2

2

234 4

= =PLEI

PLEI




¨ Substituting the obtained values to the final equation,

FMFMFM

EIL

L LL L L L

L LL L L L L

L LL L L L

PLEI

PLEI

PLEI

y

y

y

1

1

2

2

3

3

3

2 2

2 2 2

2 2

3

2

2

12 6 12 6 0 06 4 6 2 0 012 6 24 0 12 6

6 2 0 8 6 20 0 12 6 12 60 0 6 2 6 4

712

34

0

400

R

S

|||

T

|||

U

V

|||

W

|||

=

--

- - --

- - --

L

N

MMMMMMM

O

Q

PPPPPPP

-R

S

||||

T

||||

U

V

||||

W

||||

F P M F P

M F P M PL

y y

y

1 1 2

2 3 3

0 52

0 32

12

= - = =

= = - =

F Py1 = - : Force at node 1

F y2 , F y3 , M3 : Reacting forces and moment at nodes

M1, M2 : Zero




¨ Calculating local nodal loads

Force at the element 1.

When $ $ $f kd=

$

$$

$

fmfm

EIL

L LL L L L

L LL L L L

PLEI

PLEI

PLEI

y

y

1

1

2

2

3

2 2

2 2

3

2

2

12 6 12 66 4 6 212 6 12 6

6 2 6 4

712

34

0

4

RS||

T||

UV||

W||=

--

- - --

L

N

MMMM

O

Q

PPPP

-R

S

|||

T

|||

U

V

|||

W

|||

$ $ $ $f P m f P m PLy y1 1 2 20= - = = = -

Shear moment curve




Homework: Distributed load

¨ Equivalent force

Finite Element Method:Beam Element Beam Elementocw.snu.ac.kr/sites/default/files/NOTE/10530.pdf ·...

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