v12

Uvod v osnove ra~una dinamike in stabilnosti gradbenih konstrukcij Vaja 34

1

Za konstrukcijo na sliki dolo~i odziv na podano obte`bo.

M1=20000 kg M2=22000 kg M3=25000 kg EI=425 MPa F1(t)=12500 N⋅sin(4.3 t)

RE[ITEV 1. Dolo~itev {tevila in lege prostostnih stopenj

2. Dolo~itev ~lenov masne matrike

EI

M1

15.0 m

M2M3

u2

EI

M1

15.0 m

M2

M3

u3

Kineti~na energija je:

20.0 m EI M1

20.0 m

15.0 m

M2 M3

F1(t)

20.0 mEIM1

20.0 m

15.0 m

M2M3

u1

u2

u3

20.0 mEI

M1

20.0 m

15.0 m

M2M3

u1


2

{ } { }

( )

TM u M u M u M u M u

Tu

M uM u

ddt

Tu

M u M u u u

Tu

M u M uM M u

ddt

T

=⋅

+⋅

+⋅

+⋅

+⋅

=⋅ ⋅

= ⋅

= ⋅ = ⋅

=⋅ ⋅

+⋅ ⋅

= + ⋅

1 12

2 22

3 22

2 32

1 32

1

1 11 1

11 1 1 1 2 3

2

2 2 3 22

2 3 2

2 2 2 2 22

2

0 0

22

22

& & & & &

&

&&

&&& , , && ,&& ,&&

&

& &&

∂∂

∂∂

∂∂

( ) ( ){ } { }

( )

( ) ( ){ } { }

∂∂

∂∂

∂∂

Tu

M M u M M u u u

Tu

M u M uM M u

ddt

Tu

M M u M M u u u

T

T

&&& , , && ,&& ,&&

&

& &&

&&& , , && ,&& ,&&

22 3 2 2 3 1 2 3

3

2 3 1 31 2 3

31 2 3 1 2 1 2 3

0 0

22

22

0 0

= + ⋅ = + ⋅

=⋅ ⋅

+⋅ ⋅

= + ⋅

= + ⋅ = + ⋅

[ ]M M= = ++

=

MM M

M M

1

2 3

1 2

0 00 00 0

20000 0 00 47000 00 0 42000

3. Podajnostna matrika:

[M1]

M1

15.0 m

M2M3

x1=1

--

-

-20.0

-20.0

-40.0

M1

15.0 m

M2M3

x2=1

-

[M2]

-20.0

[M3]M1

15.0 m

M2M3 x3=1

-

-

-20.0

-15.0


3

d

d

d

d

d

d

11 6 6 65

12 65

13 6 65

22 66

23 66

33 6 65

2666 666666425 10

6000425 10

18666 666666425 10

6 431372548 10

6666 66666425 10

15686274515 10

2250425 10

9000425 10

2 647058824 10

2666 666666425 10

6 274509805 10

3000425 10

7 058823529 10

4500425 10

1125425 10

1323529412 10

=⋅

+⋅

+⋅

= ⋅

=⋅

= ⋅

=⋅

+⋅

= ⋅

=⋅

= ⋅

=⋅

= ⋅

=⋅

+⋅

= ⋅

−

−

−

−

−

−

. . .

. .

.

. .

.

.

[ ]d = =⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅

− − −

− − −

− − −

d6 431372548 10 15686274515 10 2 647058824 1015686274515 10 6 274509805 10 7 058823529 102 647058824 10 7 058823529 10 1323529412 10

5 5 5

5 6 6

5 6 5

. . .. . .. . .

4.0 Dinami~na matrika izra~una se kot:

[ ] [ ]DM = ⋅

=⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅

⋅

=

− − −

− − −

− − −

d m

6 431372548 10 15686274515 10 2 647058824 1015686274515 10 6 274509805 10 7 058823529 102 647058824 10 7 058823529 10 1323529412 10

20000 0 00 47000 00 0 42000

5 5 5

5 6 6

5 6 5

. . .. . .. . .

1.2862745096 0.737254902205 1.111764706080.3137254903 0.2949019648535 0.296470588218 0.5294117648 0.331764705863 0.55588235304

5. Izra~un lastnih frekvenc Uporabimo lahko iterativni postopek Stodolla-Vianello (metodo vektorske iteracije) ali pa direktni postopek, saj je rang problema 3. ω ν1 1= → =0.7128730949705181 0.1134572768617761rad s Hz/

$Φ1 =− ⋅− ⋅− ⋅

5.649557436991221 101.497278542782818 102.470219735227371 10

-3

-3

-3


4

ω ν2 2= → =3.161066612962625 0.503099376895757rad s Hz/

$Φ 2 =⋅⋅⋅

- 2.878731490332587 104.20457854476734 10

2.832425089503154 10

-3

-3

-4

ω ν3 3= → =3.801246336665464 0.6049871443902676rad s Hz/

$Φ 3 =⋅⋅

− ⋅

3.129761296441738 101.16459090280028 104.198489246186126 10

-3

-3

-3

Izra~un lastnih frekvenc s pomo~jo metode kon~nih elementov Pri sestavi togostnih matrik posameznih elementov moramo upo{tevati, da smo v prej uporabljenem ra~unskem modelu upo{tevali palice kot osno nedeformabilne. Zato moramo za pre~ni prerez A upo{tevati zelo veliko vrednost, npr. A=1 106 m2, medtem ko za vztrajnostni moment in modul elasti~nosti uporabimo dejanski vrednosti. Tako izra~unane lastne frekvence so: νννννν

1

2

3

4

5

6

======

0.1134572729416623 Hz0.5030993637443831 Hz0.6049862396418373 Hz464.0129065592229 Hz716.8000245766392 Hz783.1350381686085 Hz

Vidimo, da je ujemanje rezultatov dovolj dobro. 6. Matrika lastnih vektorjev

[ ]$Φ =− ⋅− ⋅− ⋅

− ⋅⋅⋅

⋅⋅

− ⋅

5.649557436991221 101.497278542782818 102.470219735227371 10

2.878731490332587 104.20457854476734 10

2.832425089503154 10

3.129761296441738 101.16459090280028 104.198489246186126 10

-3

-3

-3

-3

-3

-4

-3

-3

-3

Izvedemo lahko kontrolo:

[ ] [ ] [ ] [ ]I MT

= ⋅ ⋅$ $Φ Φ

7. Izpeljava nevezanih ena~be gibanja


5

Vezane ena~be gibanja se glasijo: [ ] { } [ ] { } ( ){ }M u K u P t⋅ + ⋅ =&&

( ){ } ( )P tMMM

u tg= −

⋅

1

2

3

&&

Vpeljemo transformacijo { } [ ] { }u y= ⋅$Φ in dobimo

[ ] [ ] { } [ ] [ ] { } ( ){ }M y K y P t⋅ ⋅ + ⋅ ⋅ =$ && $Φ Φ

Ena~bo z leve pomnoìmo s [ ]$Φ T in dobimo:

[ ] [ ] [ ] { } [ ] [ ] [ ] { } [ ] ( ){ }[ ] { } [ ] { } [ ] ( ){ }

$ $ && $ $ $

&& $

Φ Φ Φ Φ Φ

Φ

T T T

T

M y K y P t

I y y P t

⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ = ⋅

⋅ + ⋅ = ⋅ω 2

kjer je

[ ]ωω

ωω

21

2

22

32

0 00 00 0

=

Ker sta obe matriki na levi strani ena~be diagonalni, je jasno da imamo sedaj sistem nevezanih ena~b, ki jih re{imo vsako posebej. V obravanavanem primeru dobimo:

1 0 00 1 00 0 1

0 00 00 0

1

2

3

1

2

3

⋅

+

⋅

&&

&&

&&

yyy

yyy

0.50818804953284539.992342131587005

14.44947371201261

=− ⋅− ⋅

⋅

− ⋅⋅⋅

− ⋅⋅

− ⋅

5.649557437 102.8787314 10

3.129761296 10

1.4972785423 104.204578545 101.164590903 10

2.4702197352 102.8324250895 104.1984892462 10

-3

-3

-3

-3

-3

-3

-3

-4

-3

( )⋅

⋅ ⋅

00

125004 3sin . t

( )=−

−

⋅ ⋅

30.877746690342143.54053136187894352.48111557732659

sin .4 3 t

Sistem sedaj zapi{emo kot tri nevezane diferencialne ena~be, ki jih re{imo vsako posebej:


6

( )

( )

( )

&& sin .

&& sin .

&& sin .

y y t

y y t

y y t

1 1

2 2

3 3

4 3

4 3

4 3

+ ⋅ = − ⋅ ⋅

+ ⋅ = ⋅ ⋅

+ ⋅ = − ⋅ ⋅

0.5081880495328453 30.87774669034214

14.44947371201261 52.48111557732659

9.992342131587005 3.540531361878943

Splo{na re{itev diferencialne ena~be

( ) ( )&& sin cosy y A t B t+ ⋅ = ⋅ ⋅ + ⋅ ⋅ ≠ω ω ω ω ω2 je

( ) ( ) ( ) ( ) ( )y tA

t tB

t t=−

⋅ ⋅ − ⋅ ⋅

+ −

⋅ ⋅ − ⋅ ⋅

ω ω

ω ωωω ω ω

ω ωωω2 2 2 2sin sin cos cos

Tako dobimo odziv v posplo{enih koordinatah:

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

y t t ty t t ty t t t

1

2

3

4 33

4 3

= ⋅ ⋅ − ⋅ ⋅

= ⋅ ⋅ − ⋅ ⋅

= ⋅ ⋅ − ⋅ ⋅

1.717162875465498 10.357805304259 0.71287306022880.56676599081384 3.161066275799987 0.41664764182139 412.98876376686 14.692979208751 3.8012497944754959

sin . sinsin sin .

sin . sin

Sedaj izra~unamo odziv v glavnih koordinatah

{ } [ ] { } [ ]u y= ⋅ = ⋅$ $Φ Φ

( ) ( )( ) ( )( ) ( )

1.71716287547 10.357805304 0.71287306022880.56676599081 3.1610662758 0.4166476418214 4

12.9887637669 14.69297921 3.801249794475496

⋅ ⋅ − ⋅ ⋅⋅ ⋅ − ⋅ ⋅⋅ ⋅ − ⋅ ⋅

sin . sinsin sin .sin . sin

4 33

4 3

t tt t

t t

Del izrazov v oklepajih pripada lastnemu nihanju, ki pa se kmalu izdu{i. ê ta del odziva zanemarimo, lahko stacionarni odziv zapi{emo kot:

{ } [ ] { } [ ]u y= ⋅ = ⋅$ $Φ Φ ( )1.717162875470.416647641821412.9887637669

−

⋅ ⋅sin .4 3 t

{ } ( )u tst =−

⋅ ⋅

0.032149936519417640.0108036972580798

0.05889296694340836sin .4 3

Celoten odziv pa je:


7

{ } ( ) ( )

( )

( )

u t t

t

t

st =−

⋅ ⋅ +

⋅ ⋅

+−

⋅ ⋅

+−−

⋅ ⋅

0.032149936519417640.0108036972580798

0.0588929669434084

0.05851701598758550.01550851963238957 0.02558605507622417

0.71287306023

0.0016315671054053570.002383012124879683 0.000160532221225824

3.1610662758

0.045985517656974390.01711130992154595

0.061688315202380573.8012497944755

sin . sin

sin

sin

4 3

8.0 Stabilnost - dolo~itev interpolacijskih funkcij 8.1 Prva interpolacijska funkcija

Robni pogoji so: v11(0)=0 pomik v11'(0)=0 zasuk v11(20.0)=0 pomik v12(0.0)=0 pomik v12(15.0)=0 pomik v13(0.0)=0 pomik v13(20)=u1 pomik v13''(20)=0 moment Pogoji zveznosti v11'(20.0)=v12'(0) zasuk EI1⋅v11''(20.0)=EI2⋅v12''(0) moment v12'(15.0)=v13'(0) zasuk EI2⋅v12''(15.0)=EI3⋅v13''(0) moment

Splo{ne re{itve diferencialne ena~be ( )v x' ' ' ' = 0 so:

u1

v12

v11

v13


8

( )v xC x C x

C x C111

32

2

3 46 2=

⋅+

⋅+ ⋅ + , ( )v x

C x C xC x C12

53

62

7 86 2=

⋅+

⋅+ ⋅ + in

( )v xC x C x

C x C139

310

2

11 126 2=

⋅+

⋅+ ⋅ +

Ob upo{tevanju robnih pogojev in pogojev zveznosti dobimo:

( )

( )

( )

v x = -3

212000u x +

310600

u x

v x =1

15900u x -

35300

u x - u x

v x = -1

26500u x +

31325

u x +21

1060u x

11 13

12

12 13

12

1

13 13

12

1

⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

3530

8.2 Druga interpolacijska funkcija

Robni pogoji so: v21(0)=0 pomik v21'(0)=0 zasuk v21(20.0)=u2 pomik v22(0.0)=0 pomik v22(15.0)=0 pomik v23(0.0)=u2 pomik v23(20.0)=0 pomik v23''(20.0)=0 moment Pogoji zveznosti v21'(20.0)=v22'(0) zasuk EI1⋅v21''(20.0)=EI2⋅v22''(0) moment v22'(15.0)=v23'(0) zasuk EI2⋅v22''(15.0)=EI3⋅v23''(0) moment


( )v xC x C x

C x C211

32

2

3 46 2=

⋅+

⋅+ ⋅ + , ( )v x

C x C xC x C22

53

62

7 86 2=

⋅+

⋅+ ⋅ + in

u2v22

v21

v23


9

( )v xC x C x

C x C239

310

2

11 126 2=

⋅+

⋅+ ⋅ +


( )

( )

( )

v x = -1

6784u x -

23142400

u x

v x =7

159000u x -

92650

u x + 87

2120u x

v x =1

42400u x -

33120

u x - 33

1060u x + u

21 23

22

22 23

22

2

23 23

22

2 2

⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

8.3 Tretja interpolacijska funkcija

Robni pogoji so: v31(0)=0 pomik v31'(0)=0 zasuk v31(20)=0 pomik v32(0)= 0 pomik v32(15)=u3 pomik v33(0.0)=0 pomik v33(20)=0 pomik v33''(20)=0 moment Pogoji zveznosti v31'(20)=v32'(0) zasuk EI1⋅v31''(20)=EI2⋅v32''(0) moment v32'(15)=v33'(0) zasuk EI2⋅v32''(15)=EI3⋅v33''(0) moment


( )v xC x C x

C x C311

32

2

3 46 2=

⋅+

⋅+ ⋅ + , ( )v x

C x C xC x C32

53

62

7 86 2=

⋅+

⋅+ ⋅ + in

( )v xC x C x

C x C339

310

2

11 126 2=

⋅+

⋅+ ⋅ +


u3v32

v31

v33


10

( )

( )

( )

v x =17

159000u x

177950

u x

v x = -32

178875u x +

173975

u x +34795

u x

v x =1

15900u x -

32265

u x +8

159u x

31 33

32

32 33

32

3

33 33

32

3

⋅ ⋅ − ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

8.4 Izra~un togostne in geometrijske togostne matrike, ter masne matrike Diagram osnih sil

[N]

Pukl

-

- Pukl-- Pukl

Iz diagrama vidimo, da nastopa tla~na osna sila samo v elementih [1] in [3], zato integriramo samo po teh elementih.

k v v dx v v dxg i jx

L

i jx

L

i j,' ' ' '= ⋅ ⋅ + ⋅ ⋅

=

=

=

=

∫ ∫1 10

20

3 30

201 3

Tako dobljena geometrijska togostna matrika je:

[ ]k g =

− −

−

−

15093280900

1472809

47270225

1472809

605156180

5914045

472

7022559

1404528448

1896075

^lene togostne matrike sedaj izra~unamo s pomo~jo inverzije è znane podajnostne matrike ali pa s pomo~jo izra~unanih interpolacijskih funkcij kot:

k EI v v dx EI v v dx EI v v dxi j i jx

i jx

i jx

, " " " " " "= ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅= = =∫ ∫ ∫1 1 1

0

20

2 2 20

15

3 3 30

20


11

Tako dobljena togostna matrika je:

[ ]k =

96226.41509433968 - 60141.50943396229 -160377.358490566- 60141.50943396229 436025.9433962264 -112264.1509433962-160377.358490566 -112264.1509433962 456184.4863731657

ki se seveda ne razlikuje od togostne matrike, dobljene z inverzijo podajnostne matrike. ^lene masne matrike dobimo kot:

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )

M M v v M v v M M v v

M v vi j i j i j i j

i j

, , , , , , ,

, ,

= ⋅ ⋅ + ⋅ ⋅ + + ⋅ ⋅

+ ⋅ ⋅1 3 3 2 3 3 1 2 2 2

3 1 1

20 20 0 0 20 20

20 20

Zavedati se je potrebno, da je mogo~e s pomo~jo interpolacijskih funkcij v mansi matriki zajeti tudi zvezno porazdeljeno maso na elasti~nih elementih. 8.5 Izra~un kriti~ne uklonske sile Pukl Problem dolo~itve kriti~ne uklonske sile se prevede v obliko:

[ ] [ ]( )K P kuuu

ukl g− ⋅ ⋅

=

1

2

3

000

Kriti~no uklonsko silo dobimo iz ena~be:

[ ] [ ]K P K

P P P

ukl g

ukl ukl ukl

− ⋅ =

− ⋅ ⋅ − ⋅ ⋅

+ ⋅ =

0

10533750010

10 0

3 2 9

15

4517+1595.300819192462 6.928352327257719

2.896816042021733

Kriti~na uklonska sila je: Pukl=467883.0158490324 GN Izra~un z metodo kon~nih elementov da rezultat: P=467792.995731081 GN. Vidimo, da je ujemanje rezultatov dobro.

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v12