5 lect6 beam students

background image

Computational Methods

1D Examples

Małgorzata Stojek

Cracow University of Technology

March 2013

MS

(L-53 CUT)

Beam & Truss

03/2013

1 / 44

Beam Example

M = qL

2

2L

L

P = qL

q

e1

e2

θ

1

w

1

θ

θ

w

w

2

2

3

3

d

1

d

2

d

3

d

4

d

5

d

6

=

w

1

θ

1

w

2

θ

2

w

3

θ

3

MS

(L-53 CUT)

Beam & Truss

03/2013

2 / 44

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Beam Element Library

Prismatic Beam

Stiffness Matrix

:

K

e

=

EI
h

3

12

6h

12

6h

4h

2

6h

2h

2

12

6h

symm

4h

2

Load Vector due to q

(

x

)

:

constant

distributed load

concentrated

force

q

(

x

) =

q

q

(

x

) =

P

δ

x

(

x

h
2

)

F

e

q

=

q

h

1
2

1

12

h

1
2

1

12

h

F

e

q

=

P

1
2

1
8

h

1
2

1
8

h

MS

(L-53 CUT)

Beam & Truss

03/2013

3 / 44

First Element

Stiffness Matrix

DOFs: 1, 2, 3, 4; h

=

2L

K

e

1

4

×

4

=

EI

(

2L

)

3

12

6

(

2L

)

12

6

(

2L

)

6

(

2L

)

4

(

2L

)

2

6

(

2L

)

2

(

2L

)

2

12

6

(

2L

)

12

6

(

2L

)

6

(

2L

)

2

(

2L

)

2

6

(

2L

)

4

(

2L

)

2

=

EI

2L

3

3

3L

3

3L

3L

4L

2

3L

2L

2

3

3L

3

3L

3L

2L

2

3L

4L

2

MS

(L-53 CUT)

Beam & Truss

03/2013

4 / 44

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First Element

Load Vector due to

q

(

x

)

F

e

1

q

=

q

(

2L

)

1
2

1

12

(

2L

)

1
2

1

12

(

2L

)

=

qL

1

1
3

L

1

1
3

L

y

1
3

qL

2

x

1
3

qL

2

[

element 1

]

qL

qL

MS

(L-53 CUT)

Beam & Truss

03/2013

5 / 44

Second Element

DOFs: 3, 4, 5, 6; h

=

L

Stiffness Matrix

K

e

2

4

×

4

=

EI

(

L

)

3

12

6

(

L

)

12

6L

6

(

L

)

4

(

L

)

2

6

(

L

)

2

(

L

)

2

12

6

(

L

)

12

6

(

L

)

6

(

L

)

2

(

L

)

2

6

(

L

)

4

(

L

)

2

Load Vector due to P

=

qL

F

e

2

q

=

P

1
2

1
8

L

1
2

1
8

L

y

1
8

qL

2

x

1
8

qL

2

[

element 2

]

1
2

qL

1
2

qL

MS

(L-53 CUT)

Beam & Truss

03/2013

6 / 44

background image

Assembly

Global Stiffness Matrix

K

=

0

6

×

6

(+)

K

e

1

4

×

4

(+)

K

e

2

4

×

4

K

=

EI

(

L

)

3

3
2

3
2

L

3
2

3
2

L

0

0

3
2

L

2L

2

3
2

L

L

2

0

0

3
2

3
2

L

3
2

+

12

3
2

L

+

6L

12

6L

3
2

L

L

2

3
2

L

+

6L

2L

2

+

4L

2

6L

2L

2

0

0

12

6L

12

6L

0

0

6L

2L

2

6L

4L

2

K

=

EI

(

L

)

3

3
2

3
2

L

3
2

3
2

L

0

0

3
2

L

2L

2

3
2

L

L

2

0

0

3
2

3
2

L

27

2

9
2

L

12

6L

3
2

L

L

2

9
2

L

6L

2

6L

2L

2

0

0

12

6L

12

6L

0

0

6L

2L

2

6L

4L

2

MS

(L-53 CUT)

Beam & Truss

03/2013

7 / 44

Assembly

Global Load Vector

F

q

=

0

6

×

1

(+)

F

e

1

q

4

×

1

(+)

F

e

2

P

4

×

1

0

0

0

0

0

0

qL

1
3

qL

2

qL

1
3

qL

2

0

0

qL

1
3

qL

2

qL

+

1
2

qL

1
3

qL

2

+

1
8

qL

2

1
2

qL

1
8

qL

2

=

qL

1
3

qL

2

3
2

qL

5

24

qL

2

1
2

qL

1
8

qL

2

MS

(L-53 CUT)

Beam & Truss

03/2013

8 / 44

background image

Assembly

Generic System of Linear Equations

Kd

=

F

q

EI

(

L

)

3

3
2

3
2

L

3
2

3
2

L

0

0

3
2

L

2L

2

3
2

L

L

2

0

0

3
2

3
2

L

27

2

9
2

L

12

6L

3
2

L

L

2

9
2

L

6L

2

6L

2L

2

0

0

12

6L

12

6L

0

0

6L

2L

2

6L

4L

2

w

1

θ

1

w

2

θ

2

w

3

θ

3

=

qL

1
3

qL

2

3
2

qL

5

24

qL

2

1
2

qL

1
8

qL

2

NOTE:

det K

=

0,

rank

(

K

) =

4

.

MS

(L-53 CUT)

Beam & Truss

03/2013

9 / 44

Boundary Conditions

kinematic constraints

natural

BCs

(essential "BCs")

(nonhomogeneous)

w

1

=

0,

θ

1

=

0,

w

2

=

0

at x

=

3L,

M

=

qL

2

w

1

θ

1

w

2

θ

2

w

3

θ

3

0
0
0

θ

2

w

3

θ

3

qL

1
3

qL

2

3
2

qL

5

24

qL

2

1
2

qL

1
8

qL

2

qL

1
3

qL

2

3
2

qL

5

24

qL

2

1
2

qL

1
8

qL

2

+

qL

2

NOTE:
rank

(

K

) =

4

&

3

support constraints

←→

beam is

statically indeterminate

.

MS

(L-53 CUT)

Beam & Truss

03/2013

10 / 44

background image

Solution I

EI

(

L

)

3

3
2

3
2

L

3
2

3
2

L

0

0

3
2

L

2L

2

3
2

L

L

2

0

0

3
2

3
2

L

27

2

9
2

L

12

6L

3
2

L

L

2

9
2

L

6L

2

6L

2L

2

0

0

12

6L

12

6L

0

0

6L

2L

2

6L

4L

2

0

0

0

θ

2

w

3

θ

3

=

qL

1
3

qL

2

3
2

qL

5

24

qL

2

1
2

qL

7
8

qL

2

EI

(

L

)

3

6L

2

6L

2L

2

6L

12

6L

2L

2

6L

4L

2

θ

2

w

3

θ

3

=

5

24

qL

2

1
2

qL

7
8

qL

2

+

EI

(

L

)

3

3
2

L L

2

9
2

L

0

0

12

0

0

6L

0

0

0

MS

(L-53 CUT)

Beam & Truss

03/2013

11 / 44

Solution II

RECALL:

EI

(

L

)

3

6L

2

6L

2L

2

6L

12

6L

2L

2

6L

4L

2

θ

2

w

3

θ

3

=

5

24

qL

2

1
2

qL

7
8

qL

2

Solution is:

θ

2

w

3

θ

3

=

7

12

qL

3

EI

19
16

qL

4

EI

41
24

qL

3

EI

d

=

0

0

0

θ

2

w

3

θ

3

=

0

0

0

7

12

qL

3

EI

19
16

qL

4

EI

41
24

qL

3

EI

MS

(L-53 CUT)

Beam & Truss

03/2013

12 / 44

background image

Postprocessing

Siły Przyw ˛ezłowe

RECALL:

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

y

M

1

Q

1

[

beam element

]

Q

2

y

M

2

MS

(L-53 CUT)

Beam & Truss

03/2013

13 / 44

Siły Przyw ˛ezłowe

Element 1

y

M

1

Q

1

[

beam element

]

Q

2

y

M

2

W

e

1

=

K

e

1

d

e

1

F

e

1

q

W

e

1

=

EI

2L

3

3

3L

3

3L

3L

4L

2

3L

2L

2

3

3L

3

3L

3L

2L

2

3L

4L

2

0

0

0

7

12

qL

3

EI

qL

1

1
3

L

1

1
3

L

Q

e

1

1

M

e

1

1

Q

e

1

2

M

e

1

2

=

1
8

qL

1
4

qL

2

15

8

qL

3
2

qL

2

y

1
4

qL

2

y

3
2

qL

2

[

element 1

]

1
8

qL

15

8

qL

MS

(L-53 CUT)

Beam & Truss

03/2013

14 / 44

background image

Siły Przyw ˛ezłowe

Element 2

y

M

1

Q

1

[

beam element

]

Q

2

y

M

2

W

e

2

=

K

e

2

d

e

2

F

e

2

q

W

e

2

=

EI
L

3

12

6L

12

6L

6L

4L

2

6L

2L

2

12

6L

12

6L

6L

2L

2

6L

4L

2

0

7

24

qL

3

EI

43
48

qL

4

EI

17
12

qL

3

EI

qL

1
2

1
8

L

1
2

1
8

L

Q

e

2

1

M

e

2

1

Q

e

2

2

M

e

2

2

=

qL

3
2

qL

2

0

qL

2

x

3
2

qL

2

y

qL

2

[

element 2

]

qL

0

MS

(L-53 CUT)

Beam & Truss

03/2013

15 / 44

Reactions at Supports I

1
4

qL

2

y

y

3
2

qL

2

[

element 1

]

1
8

qL

15

8

qL

3
2

qL

2

x

y

qL

2

[

element 2

]

qL

0

15

8

qL

+

R

2

=

qL

R

2

=

qL

+

15

8

qL

=

23

8

qL

8

1

1
4

2

qL

23
8

P = qL

q

2L

L

qL

qL

M = qL

2

MS

(L-53 CUT)

Beam & Truss

03/2013

16 / 44

background image

Reactions at Supports II

8

1

1
4

2

qL

23
8

P = qL

q

2L

L

qL

qL

M = qL

2

W

=

Kd

F

q

=

K

0

0

0

7

12

qL

3

EI

19
16

qL

4

EI

41
24

qL

3

EI

qL

1
3

qL

2

3
2

qL

5

24

qL

2

1
2

qL

1
8

qL

2

=

1
8

qL

1
4

qL

2

23

8

qL

0
0

qL

2

MS

(L-53 CUT)

Beam & Truss

03/2013

17 / 44

Equations of Equilibrium

8

1

1
4

2

qL

23
8

P = qL

q

2L

L

qL

qL

M = qL

2

i

P

i

y

?

=

0

1
8

qL

q

·

2L

+

23

8

qL

qL

=

0

i

M

i

x

=

0

?

=

0

1
4

qL

2

+

q

·

2L

·

L

23

8

qL

·

2L

+

qL

·

2L

+

L
2

+

qL

2

=

0

MS

(L-53 CUT)

Beam & Truss

03/2013

18 / 44

background image

Wykresy Sił Przekrojowych

8

1

1
4

2

qL

23
8

8

1

8

15

qL

2

qL

1
4

2

3

qL

2

P = qL

q

2L

L

qL

qL

M = qL

2

Q(x)

qL

qL

M(x)

2

qL

MS

(L-53 CUT)

Beam & Truss

03/2013

19 / 44


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