15 marzec 2012id 16158 Nieznany (2)

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Bernoulli-Euler Beam Theory

Hermite Cubics

Małgorzata Stojek

Cracow University of Technology

March 2012

MS

(L-53 CUT)

FEM

03/2012

1 / 24

Assumptions & Nomenclature I

c

Felippa

In tro d u ctio n to FEM

Pl an e Be am Te rm i n ol ogy

z

Beam cross
section

Symmetry
plane

Neutral
surface

Neutral axis

x, u

y, v

y, v

q(x)

L

MS

(L-53 CUT)

FEM

03/2012

2 / 24

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Assumptions & Nomenclature II

c

Felippa

In tro d u ctio n to FEM

B e rn ou l l i -Eu l e r Ki n e m ati cs

of Pl an e Be am El e m e n t

1

x, u

2

v

1

v

2

1

2

y, v

P'(x+u,y+v )

P(x,y)

x

E, I

MS

(L-53 CUT)

FEM

03/2012

3 / 24

Strong Formulation

RECALL: (S)

d

2

dx

2



EI

d

2

w

dx

2



=

q

on Ω

= (

0, L

)

2 BCs

at 0

2 BCs

at L

Types of Boundary Conditions:

essential

BCs:

transverse displacement

w ;

slope (of the deflection curve)

w

.

natural

BCs (κ — curvature):

prescribed moment

EI κ

=

M;

prescribed shear

(

EI κ

)

=

Q.

NOTE: "conjugate" or "dual" pairs:

generalized displacement

generalized force

w

Q

w

M

MS

(L-53 CUT)

FEM

03/2012

4 / 24

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Weak Formulation I

Z

L

0

d

2

dx

2



EI

d

2

w

dx

2



v dx

=

Z

L

0

qv dx,

EI

d

2

w

dx

2

=

M



for

−→



first integration by parts

Z

L

0

d

2

(−

M

)

dx

2

v dx

= −

Z

L

0



d

(−

M

)

dx

dv
dx



dx

+



d

(−

M

)

dx

v



L

0

second integration by parts

Z

L

0



d

(−

M

)

dx

dv
dx



dx

=

Z

L

0



(

M

)

d

2

v

dx

2



dx



(−

M

)

dv
dx



L

0

finally we get LHS equal to

Z

L

0

d

2

dx

2



EI

d

2

w

dx

2



v dx

=

Z

L

0



EI

d

2

w

dx

2

d

2

v

dx

2



dx

− [

Qv

]

L
0



(−

M

)

dv
dx



L

0

MS

(L-53 CUT)

FEM

03/2012

5 / 24

Weak Formulation II

Terms due to BCs:

[

Qv

]

L
0

+



(−

M

)

dv
dx



L
0

=

v

(

0

)

v,

x

(

0

)

v

(

L

)

v,

x

(

L

)



Q

0

M

0

Q

L

M

L

Q

0

M

0

Q

L

M

L

=⇒

at x

=

0

end loads

at x

=

L

y

M

0

Q

0

[

beam

]

Q

L

y

M

L

RECALL:

global

coordinate system

−→

=⇒

(

sign convention

:

y

M,

Q

q

(

x

)

and

κ

=

d

2

w

dx

2

MS

(L-53 CUT)

FEM

03/2012

6 / 24

background image

Weak Formulation III

Definition

(W) for Bernoulli-Euler beam

a

(

w , v

) = (

q, v

) +

v

(

0

)

Q

0

+

v,

x

(

0

)

M

0

+

v

(

L

)

Q

L

+

v,

x

(

L

)

M

L

where:

a

(

w , v

) =

Z

L

0

EI

d

2

w

dx

2

d

2

v

dx

2

dx,

(

q, v

) =

Z

L

0

qv dx

An additional

discretization

requirement —

C

1

interpolant

,

i.e. continuity of both the deflection and the slope.

We need

two DOFs

per each

node

: deflection, w , and slope, θ

=

w,

x

.

d

i

=



w

i

θ

i



MS

(L-53 CUT)

FEM

03/2012

7 / 24

Introduction to FEM

Degrees of Freedom of Plane Beam Element

v

1

1

2

c

Felippa

d

e

=

d

1

d

2

d

3

d

4

=

w

1

θ

1

w

2

θ

2

,

for

−→

y

y

MS

(L-53 CUT)

FEM

03/2012

8 / 24

background image

c

Felippa

Introduction to FEM

interpenetration

gap

(a)

(b)

Piecewise cubic interpolation

provides required C continuity

1

Piecewise linear interpolation

gives unacceptable C continuity

0

(deflections grossly exaggerated for visibility)

Physical Interpretation of C Continuity

1

w(x)

w(x)

MS

(L-53 CUT)

FEM

03/2012

9 / 24

Hermite Cubics

Natural Coordinates

deflection

— shape functions

-1

0

1

0

1

N

e

1

=

1
4

(

1

ξ

)

2

(

2

+

ξ

)

-1

0

1

0

1

N

e

3

=

1
4

(

1

+

ξ

)

2

(

2

ξ

)

slope

— shape functions

(

h

=

2

)

-1

0

1

0.0

0.4

N

e

2

=

1
8

h

(

1

ξ

)

2

(

1

+

ξ

)

-1

0

1

-0.4

0.0

N

e

4

= −

1
8

h

(

1

+

ξ

)

2

(

1

ξ

)

MS

(L-53 CUT)

FEM

03/2012

10 / 24

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Element Stiffness Matrix

Localization & Curvature-Displacement Matrix

element shape functions and DOFs

a

(

w

e

,

v

e

) =

Z

h

0

EI

d

2

w

e

dx

2

d

2

v

e

dx

2

dx,

v

e

∈ {

N

e

1

,

N

e

2

,

N

e

3

,

N

e

4

}

local interpolant

w

e

(

x

) =

N

e

1

N

e

2

N

e

3

N

e

4



w

1

θ

1

w

2

θ

2

=

N

e

d

e

curvature-displacement

matrix, B

e

=

d

2

N

e

dx

2

d

2

w

e

(

x

)

dx

2

=



d

2

N

e

1

dx

2

d

2

N

e

2

dx

2

d

2

N

e

3

dx

2

d

2

N

e

4

dx

2



w

1

θ

1

w

2

θ

2

=

B

e

d

e

MS

(L-53 CUT)

FEM

03/2012

11 / 24

Element Stiffness Matrix

Natural Coordinates & Change of Variables

natural coordinates,

x

∈ (

0, h

)

,

ξ

∈ (−

1, 1

)

ξ

=

2

x

h

1,

d ξ

=

2
h

dx

RECALL: change of variables - rules of calculus for any f

(

x

)

Z

h

0

f

(

x

)

dx

=



h
2



Z

1

1

f

(

ξ

)

d ξ

;

df

(

x

)

dx

=

df

(

ξ

)

d ξ

d ξ

dx

=



2
h



df

(

ξ

)

d ξ

;

d

2

f

(

x

)

dx

2

=

2
h

d

dx



df

(

ξ

)

d ξ



=



2
h



2

d

2

f

(

ξ

)

d ξ

2

MS

(L-53 CUT)

FEM

03/2012

12 / 24

background image

Element Stiffness Matrix

Curvature-Displacement Matrix

chain rule

B

e

=

d

2

N

e

dx

2

=



2
h



2

d

2

d ξ

2

N

e

1

N

e

2

N

e

3

N

e

4



symbolic derivation

B

e

=



2
h



2

d

2

d ξ

2

1
4

(

ξ

1

)

2

(

ξ

+

2

)

1
8

h

(

ξ

1

)

2

(

ξ

+

1

)

1
4

(

ξ

+

1

)

2

(

ξ

2

)

1
8

h

(

ξ

1

) (

ξ

+

1

)

2

T

results in

B

e

=

1
h

6
h

ξ

3ξ

1

6
h

ξ

3ξ

+

1



MS

(L-53 CUT)

FEM

03/2012

13 / 24

Beam Element Stiffness Matrix I

Fact

For B

=

B

1

B

2

B

3

B

4



B

1

B

2

B

3

B

4

B

1

B

2

B

3

B

4



=

B

2

1

B

1

B

2

B

1

B

3

B

1

B

4

B

1

B

2

B

2

2

B

2

B

3

B

2

B

4

B

1

B

3

B

2

B

3

B

2

3

B

3

B

4

B

1

B

4

B

2

B

4

B

3

B

4

B

2

4

Definition

Beam Element Stiffness Matrix

K

e

4

×

4

=

Z

h

0

(

EI

)

B

T

B dx

MS

(L-53 CUT)

FEM

03/2012

14 / 24

background image

Beam Element Stiffness Matrix II

EI=const

RECALL:

B

=

1
h

6
h

ξ

3ξ

1

6
h

ξ

3ξ

+

1



change of variables

K

e

=

Z

h

0

(

EI

)

B

T

B dx

=



h
2



Z

1

1

(

EI

)

B

T

B d

ξ

prismatic beam, i.e. EI

=

constant

K

e

=

EI

h
2



1
h



2

Z

1

1

6
h

ξ

3ξ

1

6
h

ξ

3ξ

+

1

6
h

ξ

3ξ

1

6
h

ξ

3ξ

+

1



d ξ

MS

(L-53 CUT)

FEM

03/2012

15 / 24

Beam Element Stiffness Matrix III

EI=const

symbolic derivation results in

K

e

=

EI
2h

h

2

2

24
h

2

12

h

24
h

2

12

h

12

h

8

12

h

4

24
h

2

12

h

24
h

2

12

h

12

h

4

12

h

8

K

e

=

EI
h

3

12

6h

12

6h

4h

2

6h

2h

2

12

6h

symm

4h

2

MS

(L-53 CUT)

FEM

03/2012

16 / 24

background image

Element Load Vector I

RECALL:

(

q, v

e

) =

Z

h

0

qv

e

dx ,

v

e

∈ {

N

e

1

,

N

e

2

,

N

e

3

,

N

e

4

}

element load vector due to internal load, q

(

x

)

:

F

e

q

=

Z

h

0

qN

T

dx

change of variables

F

e

q

=



h
2



Z

1

1

q

1
4

(

ξ

1

)

2

(

ξ

+

2

)

1
8

h

(

ξ

1

)

2

(

ξ

+

1

)

1
4

(

ξ

+

1

)

2

(

ξ

2

)

1
8

h

(

ξ

1

) (

ξ

+

1

)

2

d ξ

MS

(L-53 CUT)

FEM

03/2012

17 / 24

Element Load Vector II

RECALL:

F

e

q

=

h
2

Z

1

1

q

1
4

(

ξ

1

)

2

(

ξ

+

2

)

1
8

h

(

ξ

1

)

2

(

ξ

+

1

)

1
4

(

ξ

+

1

)

2

(

ξ

2

)

1
8

h

(

ξ

1

) (

ξ

+

1

)

2

d ξ

q

(

x

) =

constant

F

e

q

=

qh

2

1

1
6

h

1

1
6

h

=

qh

2

qh

2

12
qh

2

qh

2

12

MS

(L-53 CUT)

FEM

03/2012

18 / 24

background image

Element Load Vector III

RECALL: Dirac delta (generalized) function δ

(

x

)

δ

(

x

) =



+

for x

=

0

0

for x

6=

0

;

Z

+

δ

(

x

)

dx

=

1

"continuation"

of the concentrated force

P

q

(

x

) =

P δ

x

(

x

h
2

)

or

q

(

ξ

) =

P δ

ξ

(

ξ

)

Since

R

+

f

(

x

)

δ

(

x

x

0

)

dx

=

f

(

x

0

)

,

F

e

q

=

Z

h

0

qN

T

dx

leads to

F

e

q

=

P N

T

x

=

h
2

or

F

e

q

=

P N

T

ξ

=

0

F

e

q

=

P

1
4

(

0

1

)

2

(

0

+

2

)

1
8

h

(

0

1

)

2

(

0

+

1

)

1
4

(

0

+

1

)

2

(

0

2

)

1
8

h

(

0

1

) (

0

+

1

)

2

=

1
2

P

1
8

Ph

1
2

P

1
8

Ph

MS

(L-53 CUT)

FEM

03/2012

19 / 24

Nonhomogeneous Natural BCs

RECALL:

y

M

0

Q

0

[

beam

]

Q

L

y

M

L

a

(

w , v

)

=

(

q, v

)

+

v

(

0

)

Q

0

+

v,

x

(

0

)

M

0

+

v

(

L

)

Q

L

+

v,

x

(

L

)

M

L

Kd

=

F

=

F

q

+

W

W

i

=

N

i

(

0

)

dN

i

dx

(

0

)

N

i

(

L

)

dN

i

dx

(

L

)



Q

0

M

0

Q

L

M

L

MS

(L-53 CUT)

FEM

03/2012

20 / 24

background image

Finite Element Program

Physical
system

Discrete
model

Mathematical
model

IDEALIZATION

DISCRETIZATION

FEM

Preprocessing:

defining the FEM model;

Assembler

Element

Stiffness

Internal

Force Recovery

Element Int

Forces

Built in

Equation

Solver

BC

Application

Element Library

local DOFs

Processing:

setting up the stiffness equations and solving for unknown
nodal DOFs;

Postprocessing:

recovery of derived quantities and graphical presentation

of results.

MS

(L-53 CUT)

FEM

03/2012

21 / 24

Postprocessing

Siły przyw ˛ezłowe W

c

Felippa

Elem ent Library

Interna l

Element Int

Forces

d

is known

y

M

1

Q

1

[

beam element

]

Q

2

y

M

2

K

e

d

e

=

F

e

=

F

e

q

+

W

e

W

e

=

Q

1

M

1

Q

2

M

2

=

K

e

d

e

F

e

q

MS

(L-53 CUT)

FEM

03/2012

22 / 24

background image

c

Felippa

Introduction to FEM

Total Potential Energy of Beam Member

U

W

U

1

2

V

xx

e

xx

dV

1

2

L

0

M dx

1

2

L

0

EI

2

x

2

2

dx

1

2

L

0

EI

2

dx

W

L

0

q dx

Internal

External

External energy due to transverse load q

Internal energy due to bending

c

Felippa

Introduction to FEM

Element Stiffness and Consistent Node Forces

e

1

2

u

e T

K

e

u

e

u

e T

f

e

K

e

0

EI B

T

B dx

1

1

EI B

T

B

1

2

d

f

e

0

N

T

q dx

1

1

N

T

q

1

2

d

Varying the element TPE

we get


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