Advanced Finite Element Method, Übungen von Maschinenbau / Maschinenkonstruktionslehre

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Advanced Finite Element Methods for Engineers

Exercise 2

Prof. Dr.-Ing. Mikhail Itskov

Department of Continuum Mechanics, RWTH Aachen University, Germany

WiSe 2020/

Overview

Repetition

Coordinate transformation

Matrix reduction due to boundary conditions

Procedure for the tasks in this exercise

Task 1

Task 2

2 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Coordinate transformation

The work (scalar product of force and way) has to be the same in both coordinate

systems, therefore

r ¯

eT

¯a

e

= r

eT

a

e

4 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Coordinate transformation

The work (scalar product of force and way) has to be the same in both coordinate

systems, therefore

r ¯

eT

¯a

e

= r

eT

a

e

and through inserting of a¯

e

= L

e

a

e

follows

r ¯

eT

L

e

a

e

= r

eT

a

e

4 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Coordinate transformation

The work (scalar product of force and way) has to be the same in both coordinate

systems, therefore

r ¯

eT

¯a

e

= r

eT

a

e

and through inserting of a¯

e

= L

e

a

e

follows

r ¯

eT

L

e

a

e

= r

eT

a

e

respectively

r ¯

eT

L

e

= r

eT

With (AB)

T

= B

T

A

T

and

(

A

T )T

= A follows

L

eT

r ¯

e

= r

e

4 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Coordinate transformation

Right now we know the equilibrium equations of element e in both coordinate

systems

r

e

= K

e

a

e

, (1) r¯

e

= K¯

e

a¯

e

and the relations between a

e

and ¯a

e

respectively r

e

and r¯

e

¯a

e

= L

e

a

e

, (3) r

e

= L

eT

r ¯

e

5 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Coordinate transformation

Example

We consider a rod with a rotation of α.

PSfrag replacements

x

y

x ¯

y ¯

6 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Coordinate transformation

Example

We consider a rod with a rotation of α.

frag replacements

s = sin α

c = cos α

F^ ¯

x

Fy

Fx

Fy

F^ ¯

xc

Fxs

F^ ¯

ys

F^ ¯

yc

6 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Coordinate transformation

Example

We consider a rod with a rotation of α. For the forces in both nodes holds

F 1 ,x = ¯F 1 ,x cos α − F¯ 1 ,y sin α, F 2 ,x = ¯F 2 ,x cos α − F¯ 2 ,y sin α,

F 1 ,y = F¯ 1 ,x sin α + F¯ 1 ,y cos α, F 2 ,y = F¯ 2 ,x sin α + F¯ 2 ,y cos α,

which leads to the matrix form (see eq. (4))

   

F 1 ,x

F 1 ,y

F 2 ,x

F 2 ,y

    ︸ ︷︷ ︸ re

   

cos α − sin α 0 0

sin α cos α 0 0

0 0 cos α − sin α

0 0 sin α cos α

    ︸ ︷︷ ︸ Le T    

F^ ¯

1 ,x

F^ ¯

1 ,y

F^ ¯

2 ,x

F^ ¯

2 ,y     ︸ ︷︷ ︸ r ¯e

6 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Coordinate transformation

Example

With

L

eT

( cos α − sin α 0 0 sin α cos α 0 0 0 0 cos α − sin α 0 0 sin α cos α )

, L

e

( cos α sin α 0 0 − sin α cos α 0 0 0 0 cos α sin α 0 0 − sin α cos α )

and the element stiffness matrix of the rod (see exercise 1)

K^ ¯e^ =

EA

L

( 1 0 − 1 0 0 0 0 0 − 1 0 1 0 0 0 0 0 )

follows directly

K

e

= L

e T

K¯eLe^ =

EA

L

( cos(α)^2 cos(α) sin(α) − cos(α)^2 − cos(α) sin(α) cos(α) sin(α) sin(α)^2 − cos(α) sin(α) − sin(α)^2 − cos(α)^2 − cos(α) sin(α) cos(α)^2 cos(α) sin(α) − cos(α) sin(α) − sin(α)^2 cos(α) sin(α) sin(α)^2 )

7 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Matrix reduction due to boundary conditions

An assembled system r = Ka can be reduced, if because of given boundary

conditions some displacement degrees of freedeom ai are zero.

The columns and rows corresponding to said 0-displacement are therefore deleted

from the stiffness matrix and the two vectors.

The reduced system is named r˜ = K˜˜a.

8 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Matrix reduction due to boundary conditions

Example

        F 1 x F 1 y F 2 x F 2 y F 3 x F 3 y         =         K 1 x 1 x K 1 x 1 y K 1 x 2 x K 1 x 2 y K 1 x 3 x K 1 x 3 y K 1 y 1 x K 1 y 1 y K 1 y 2 x K 1 y 2 y K 1 y 3 x K 1 y 3 y K 2 x 1 x K 2 x 1 y K 2 x 2 x K 2 x 2 y K 2 x 3 x K 2 x 3 y K 2 y 1 x K 2 y 1 y K 2 y 2 x K 2 y 2 y K 2 y 3 x K 2 y 3 y K 3 x 1 x K 3 x 1 y K 3 x 2 x K 3 x 2 y K 3 x 3 x K 3 x 3 y K 3 y 1 x K 3 y 1 y K 3 y 2 x K 3 y 2 y K 3 y 3 x K 3 y 3 y                 a 1 x a 1 y a 2 x a 2 y a 3 x a 3 y         =         K 1 x 1 x · a 1 x + K 1 x 1 y · a 1 y + K 1 x 2 x · a 2 x + K 1 x 2 y · a 2 y + K 1 x 3 x · a 3 x + K 1 x 3 y · a 3 y K 1 y 1 x · a 1 x + K 1 y 1 y · a 1 y + K 1 y 2 x · a 2 x + K 1 y 2 y · a 2 y + K 1 y 3 x · a 3 x + K 1 y 3 y · a 3 y K 2 x 1 x · a 1 x + K 2 x 1 y · a 1 y + K 2 x 2 x · a 2 x + K 2 x 2 y · a 2 y + K 2 x 3 x · a 3 x + K 2 x 3 y · a 3 y K 2 y 1 x · a 1 x + K 2 y 1 y · a 1 y + K 2 y 2 x · a 2 x + K 2 y 2 y · a 2 y + K 2 y 3 x · a 3 x + K 2 y 3 y · a 3 y K 3 x 1 x · a 1 x + K 3 x 1 y · a 1 y + K 3 x 2 x · a 2 x + K 3 x 2 y · a 2 y + K 3 x 3 x · a 3 x + K 3 x 3 y · a 3 y K 3 y 1 x · a 1 x + K 3 y 1 y · a 1 y + K 3 y 2 x · a 2 x + K 3 y 2 y · a 2 y + K 3 y 3 x · a 3 x + K 3 y 3 y · a 3 y        

• Boundary conditions: a 2 y = a 3 y = 0 and F 1 x = F 1 y = 0 and F 2 x = F 3 x = 100.

9 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Matrix reduction due to boundary conditions

Example

        0 0 100 F 2 y 100 F 3 y         =         K 1 x 1 x K 1 x 1 y K 1 x 2 x K 1 x 2 y K 1 x 3 x K 1 x 3 y K 1 y 1 x K 1 y 1 y K 1 y 2 x K 1 y 2 y K 1 y 3 x K 1 y 3 y K 2 x 1 x K 2 x 1 y K 2 x 2 x K 2 x 2 y K 2 x 3 x K 2 x 3 y K 2 y 1 x K 2 y 1 y K 2 y 2 x K 2 y 2 y K 2 y 3 x K 2 y 3 y K 3 x 1 x K 3 x 1 y K 3 x 2 x K 3 x 2 y K 3 x 3 x K 3 x 3 y K 3 y 1 x K 3 y 1 y K 3 y 2 x K 3 y 2 y K 3 y 3 x K 3 y 3 y                 a 1 x a 1 y a 2 x 0 a 3 x 0         =         K 1 x 1 x · a 1 x + K 1 x 1 y · a 1 y + K 1 x 2 x · a 2 x + 0 + K 1 x 3 x · a 3 x + 0 K 1 y 1 x · a 1 x + K 1 y 1 y · a 1 y + K 1 y 2 x · a 2 x + 0 + K 1 y 3 x · a 3 x + 0 K 2 x 1 x · a 1 x + K 2 x 1 y · a 1 y + K 2 x 2 x · a 2 x + 0 + K 2 x 3 x · a 3 x + 0 K 2 y 1 x · a 1 x + K 2 y 1 y · a 1 y + K 2 y 2 x · a 2 x + 0 + K 2 y 3 x · a 3 x + 0 K 3 x 1 x · a 1 x + K 3 x 1 y · a 1 y + K 3 x 2 x · a 2 x + 0 + K 3 x 3 x · a 3 x + 0 K 3 y 1 x · a 1 x + K 3 y 1 y · a 1 y + K 3 y 2 x · a 2 x + 0 + K 3 y 3 x · a 3 x + 0        

• Boundary conditions: a 2 y = a 3 y = 0 and F 1 x = F 1 y = 0 and F 2 x = F 3 x = 100.
• The full system contains 6 equations and 6 unknowns.

9 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe

Matrix reduction due to boundary conditions

Example

        0 0 100 F 2 y 100 F 3 y         =         K 1 x 1 x K 1 x 1 y K 1 x 2 x K 1 x 2 y K 1 x 3 x K 1 x 3 y K 1 y 1 x K 1 y 1 y K 1 y 2 x K 1 y 2 y K 1 y 3 x K 1 y 3 y K 2 x 1 x K 2 x 1 y K 2 x 2 x K 2 x 2 y K 2 x 3 x K 2 x 3 y K 2 y 1 x K 2 y 1 y K 2 y 2 x K 2 y 2 y K 2 y 3 x K 2 y 3 y K 3 x 1 x K 3 x 1 y K 3 x 2 x K 3 x 2 y K 3 x 3 x K 3 x 3 y K 3 y 1 x K 3 y 1 y K 3 y 2 x K 3 y 2 y K 3 y 3 x K 3 y 3 y                 a 1 x a 1 y a 2 x 0 a 3 x 0         =         K 1 x 1 x · a 1 x + K 1 x 1 y · a 1 y + K 1 x 2 x · a 2 x + 0 + K 1 x 3 x · a 3 x + 0 K 1 y 1 x · a 1 x + K 1 y 1 y · a 1 y + K 1 y 2 x · a 2 x + 0 + K 1 y 3 x · a 3 x + 0 K 2 x 1 x · a 1 x + K 2 x 1 y · a 1 y + K 2 x 2 x · a 2 x + 0 + K 2 x 3 x · a 3 x + 0 K 2 y 1 x · a 1 x + K 2 y 1 y · a 1 y + K 2 y 2 x · a 2 x + 0 + K 2 y 3 x · a 3 x + 0 K 3 x 1 x · a 1 x + K 3 x 1 y · a 1 y + K 3 x 2 x · a 2 x + 0 + K 3 x 3 x · a 3 x + 0 K 3 y 1 x · a 1 x + K 3 y 1 y · a 1 y + K 3 y 2 x · a 2 x + 0 + K 3 y 3 x · a 3 x + 0        

• Boundary conditions: a 2 y = a 3 y = 0 and F 1 x = F 1 y = 0 and F 2 x = F 3 x = 100.
• The full system contains 6 equations and 6 unknowns.
• The system can now be reduced by deleting the rows/columns 4 and 6. ( 0 0 100 100 ) = ( K 1 x 1 x K 1 x 1 y K 1 x 2 x K 1 x 3 x K 1 y 1 x K 1 y 1 y K 1 y 2 x K 1 y 3 x K 2 x 1 x K 2 x 1 y K 2 x 2 x K 2 x 3 x K 3 x 1 x K 3 x 1 y K 3 x 2 x K 3 x 3 x ) ( a 1 x a 1 y a 2 x a 3 x )

9 of 21 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |

Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe