Download pdf - Curs - 04 - Cfdp

11/13/2013

1

Metoda Elementului Finitcurs

As. Dr. Ing. Crișan Andreidep. de Construcții Civile și Mecanica Construcțiilor

Universitatea POLITEHNICA Timiș[email protected]

2013 - 2014 Curs 4Analiza unei grinzi cu zabrele cu MEF

• Datele problemei

• Elementul de tip zăbrea (TRUSS)

• Transformarea: coordonate locale – coordonate globale

• Rezolvarea unei grinzi cu zabrele simple

Curs 4Analiza unei grinzi cu zabrele cu MEF

• Recapitulare Cursul 3– Tipuri de sisteme in MEF


• Recapitulare Cursul 3– Rezolvarea unei aplicații simple – sistem de resoarte

– Pasul 1: Discretizarea sistemului

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– Pasul 2: Alegerea / determinarea funcţiilor de interpolare

F = Δ × k





– Pasul 3: Alegerea / determinarea proprietăților elementelor

Equația elementului:Ke

. δe = fe






– Pasul 4: Asamblarea elementelorCompatibilitatea

Echilibrul

Într-un nod, valorea necunoscutăa deplasărilor trebuie să fieaceeași pentru toate elementeleconectate în acel nod






– Pasul 4: Asamblarea elementelorCompatibilitatea

Echilibrul

Suma forțelor interne aplicate înnodul i trebuie să fie egală curezultanta forțelor exterioare înnod

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3






– Pasul 4: Asamblarea elementelor

ElementNod

Local Global

112

12

212

23

312

23

412

34







ElementNod

Local Global

112

12

212

23

312

23

412

34

�� −��

−�� + �� + �� −�� − ��

−�� − �� + ��

δ�

δ�

δ�

=

F�

F�

F�

�� −��

−�� −�� − ��

−�� − �� + �� + �� −��

�� + �� + ��

δ�

δ�

δ�

δ�

=

F�

F�

F�

F�

Compatibilitatea



– Matricea de rigiditate

k1

k1

0

0

k1

k1

0

0

0

0

0

0

0

0

0

0

1

2

3

4

F1

F2

0

0

1

0

0

0

0

0

k2

k2

0

0

k2

k2

0

0

0

0

0

1

2

3

4

0

F2

F3

0

2

0

0

0

0

0

k3

k3

0

0

k3

k3

0

0

0

0

0

1

2

3

4

0

F2

F3

0

3

0

0

0

0

0

0

0

0

0

0

k4

k4

0

0

k4

k4

1

2

3

4

0

0

F3

F4

4



– Matricea de rigiditate

K

k1

k1

0

0

k1

k1

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

k2

k2

0

0

k2

k2

0

0

0

0

0

0

0

0

0

0

k3

k3

0

0

k3

k3

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

k4

k4

0

0

k4

k4

kN

m

f

R1

R2

R3

R4

F1

F2

0

0

1 0

F2

F3

0

2

0

F2

F3

0

3

0

0

F3

F4

4

11/13/2013

4







ElementNod

Local Global

112

12

212

23

312

23

412

34

�� −�� −�� + �� + �� −�� − ��

� −�� − �� + �� + �� −��

� � �� − �� + ��

δ�

δ�

δ�

δ�

=

F�

F�

F�

F�

Compatibilitatea







ElementNod

Local Global

112

12

212

23

312

23

412

34

�� −�� −�� + �� + �� −�� − ��

� −�� − �� + �� + �� −��

� � �� − �� + ��

δ�

δ�

δ�

δ�

=

F�

F�

F�

F�

Echilibru



– Matricea de rigiditate – proprietăți

• Simetrică• Singulară, nu are inversă• Suma elementelor pe linii și coloane este 0Lătimea de bandă = 2m - 1







– Pasul 5: Aplicarea condiţiilor de margine

�� −�� −�� + �� + �� −�� − ��

� −�� − �� + �� + �� −��

� � �� − �� + ��

δ�

δ�

δ�

δ�

=

R�

��

1 0

11/13/2013

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� � � �� + �� + �� −�� − �� −�� − �� + �� + �� −��

� � �� − �� + ��

�δ�

δ�

δ�

=

R�

��

1 0








– Pasul 6: Rezolvarea sistemului de ecuații

Ke. δe = fe

Ke-1 . Ke

. δe = Ke-1 . fe

I . δe = Ke-1 . Fe

δe = Ke-1 . fe

| Ke-1 (înmulțire la stânga) 2

3

4

Kmod1

0

0

P









Kmod

k1 k2 k3 k2 k3

0

k2 k3 k2 k3 k4

k4

0

k4

k4









Kmod1

1

k1

1

k1

1

k1

1

k1

k1 k2 k3

k1 k2 k3

k1 k2 k3

k1 k2 k3

1

k1

k1 k2 k3

k1 k2 k3

k1 k2 k1 k3 k1 k4 k4 k2 k4 k3

k1 k4 k2 k3

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2

3

4

1

k1

1

k1

1

k1

1

k1

k1 k2 k3

k1 k2 k3

k1 k2 k3

k1 k2 k3

1

k1

k1 k2 k3

k1 k2 k3

k1 k2 k1 k3 k1 k4 k4 k2 k4 k3

k1 k4 k2 k3

0

0

P

2

3

4

Kmod1

0

0

P









P 100 kN

k

1500

2000

2200

1200

kN

m

2

3

4

0.06667

0.09048

0.17381

m

2

3

4

Kmod1

0

0

P









f

k1

k1

0

0

k1

k1 k2 k3 k2 k3

0

0

k2 k3 k2 k3 k4

k4

0

0

k4

k4

kN

m

1

2

3

4

fT 100 5.821 10

14 5.821 10

14 100 kN



– Grade de liberate în noduri?

AE

L

1

1

1

1

u1

u2

Fe1

Fe2

y

x?

k

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– Grade de liberate în noduri

y

x

LocalGlobal



– Necunoscute?

y

x

În nod: 2 deplasări (x, z) => u şi v

Element: 4 deplasări necunoscute4 forţe nodale

) Lo x2 x1 2 y2 y1 2

angle x.2 x.1 y.2 y.1



– Necunoscute?

y

x

În nod: 2 deplasări (x, z) => u şi v

Element: 4 deplasări necunoscute4 forţe nodale

.u u.2 u.1 cos ( )

.v v.2 v.1 sin ( )

.u .v

. u.2 u.1 cos ( ) v.2 v.1 sin ( )



– Forma matriceală

y

x

F.1a

F.2a

0

F.2a AE

L.o

.eT u.1 v.1 u.2 v.2

f.eT F.1 F.2 F.3 F.4

KeA E

Lo

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2

11/13/2013

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• Obținerea matricei de rigiditate

KeA E

Lo

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2

Sistem de coordonale

global


local

A E

L.

1

1

1

1

1

2

F.a1

F.a2

K.e .e f.e

.e

u.1

v.1

u.2

v.2

f.e

F.1

F.2

F.3

F.4



KeA E

Lo

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2


global

�� = cos � �� − si� � ��

�� = sin � �� + cos � ��

x1

y1

cos ( )

sin ( )

sin ( )

cos ( )

x1

y1

x1

y1

cos ( )

sin ( )

sin ( )

cos ( )

x1

y1

Transformarea local - global

Transformarea global - local


local



KeA E

Lo

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2


global

�� = cos � �� − si� � ��

�� = sin � �� + cos � ��Sistem de

coordonale local

x1

y1

cos ( )

sin ( )

sin ( )

cos ( )

x1

y1

x1

y1

cos ( )

sin ( )

sin ( )

cos ( )

x1

y1

Transformarea global – local pentru 2 puncte

Transformarea global - local

x1

y1

x2

y2

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

x1

y1

x2

y2


• Transformarea coordonatelor – global – local


global


local

K.e .e f.e

.e

u.1

v.1

u.2

v.2

f.e

F.1

F.2

F.3

F.4

.e

.1

0

.2

0

f.e

Fa.1

0

Fa.2

0

e e

fe fe

K.e .e f.e

K.e .e f.e

K.eA E

L

1

0

1

0

0

0

0

0

1

0

1

0

0

0

0

0

11/13/2013

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global


local

e e

fe fe

K.e .e f.e

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

K.e .e f.e




global


local

e e

fe fe

K.e .e f.e

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

K.e .e f.e

1K.e .e f.e

K.e .e f.e

K.e 1K.e



K.e 1K.e

KeA E

L

1

0

1

0

0

0

0

0

1

0

1

0

0

0

0

0

1

T

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

T

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )



K.e 1K.e

K

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

T1

0

1

0

0

0

0

0

1

0

1

0

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

K.e

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2

cos ( )2

cos ( ) sin ( )

cos ( )2

cos ( ) sin ( )

cos ( ) sin ( )

sin ( )2

cos ( ) sin ( )

sin ( )2

11/13/2013

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K.e 1K.e

K

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

T1

0

1

0

0

0

0

0

1

0

1

0

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

0

0

0

0

cos ( )

sin ( )

0

0

sin ( )

cos ( )

� =��

�

�� −�� −�� −�� −��

−�� −�� −�� −��

� = cos (�)� = sin (�)


• Exemplu de analiză cu element finit – Grindă cu zăbrele

Modul de elasticitateE = 210e9 Pa

Lungime elementeL = 4m

Diametre bareBare orizontale d1 = 0,05 mBare înclinate d2 = 0,10 m

F = 5e4 N (50 kN)

y

x



Pasul 1: Discretizarea structuriiNoduri: 1, 2, 3

X Y1 0 0

2 2 � �3 4 0

Elemente: 1, 2, 3Start Final

1 1 22 1 33 2 3

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x



Pasul 1: Discretizarea structuriiPasul 2: Alegerea funcţiilor de interpolarePasul 3: Proprietăţile elementelor

�� = �� − �� + �� − ��

�

�� =(4 4 4)

�� = ��(�� − ��, �� − ��)

�� = (60 0 300)

�� = (�� )

�� =��

�

�� =

��

�

� = cos (�)� = sin (�)

� =

�� −�� −�� −�� −��

−�� −�� −�� −��

��

��= (4.123�8 1.03�8 4.123�8)

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

11/13/2013

11




� =

�� −�� −�� −�� −��

−�� −�� −�� −��

�1 =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = 600

� = cos 600 =1

2

� = sin 600 =3

2

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x




�1 =

0.25 0.433 −0.25 −0.4330.433 0.75 −0.433 −0.75−0.25 −0.433 0.25 0.433

−0.433 −0.75 0.433 0.75

� =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = cos 600 =1

2

� = sin 600 =3

2

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x




�1 =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = 600

�2 =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = 00

� =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = cos 00 = 1� = sin 00 = 0

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x




�1 =

0.25 0.433 −0.25 −0.4330.433 0.75 −0.433 −0.75−0.25 −0.433 0.25 0.433

−0.433 −0.75 0.433 0.75

�2 =

1 0 −1 00 0 0 0

−1 0 1 00 0 0 0

� =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = cos 00 = 1� = sin 00 = 0

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

11/13/2013

12




�1 =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = 600

�2 =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = 00

�3 =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = 3000

� =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = cos 3000 =1

2

� = sin 3000 = −3

2

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x




�3 =

0.25 −0.433 −0.25 0.433−0.433 0.75 0.433 −0.75−0.25 0.433 0.25 −0.4330.433 −0.75 −0.433 0.75

� =

�� −�� −�� −�� −��

−�� −�� −�� −��

� = cos 3000 =1

2

� = sin 3000 =3

2

�1 =

0.25 0.433 −0.25 −0.4330.433 0.75 −0.433 −0.75−0.25 −0.433 0.25 0.433

−0.433 −0.75 0.433 0.75

�2 =

1 0 −1 00 0 0 0

−1 0 1 00 0 0 0

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x




F = 5e4 N (50 kN)

1

2

3

2

1 3

�3 =

0.25 −0.433 −0.25 0.433−0.433 0.75 0.433 −0.75−0.25 0.433 0.25 −0.4330.433 −0.75 −0.433 0.75

� =

�� −�� −�� −�� −��

−�� −�� −�� −��

�1 =

0.25 0.433 −0.25 −0.4330.433 0.75 −0.433 −0.75−0.25 −0.433 0.25 0.433

−0.433 −0.75 0.433 0.75

�2 =

1 0 −1 00 0 0 0

−1 0 1 00 0 0 0

��2

�

��1

�

��2

�

��

��= (4.123�8 1.03�8 4.123�8)

y

x




�3 =

1.031 −1.785 −1.031 1.785−1.785 3.092 1.785 −3.092−1.031 1.785 1.031 −1.7851.785 −3.092 −1.785 3.092

�1 =

1.031 1.785 −1.031 −1.7851.785 3.092 −1.785 −0.75

−1.031 −1.785 1.031 1.785−1.785 −3.092 1.785 3.092

�2 =

1.03 0 −1.03 00 0 0 0

−1.03 0 1.03 00 0 0 0

� 108

� 108

� 108

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

11/13/2013

13



Pasul 1: Discretizarea structuriiPasul 2: Alegerea funcţiilor de interpolarePasul 3: Proprietăţile elementelorPasul 4: Asamblarea matricei globale

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x




k1

1.031

1.785

1.031

1.785

0

0

1.785

3.092

1.785

3.092

0

0

1.031

1.785

1.031

1.785

0

0

1.785

3.092

1.785

3.092

0

0

0

0

0

0

0

0

0

0

0

0

0

0

k2

1.03

0

0

0

1.03

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1.03

0

0

0

1.03

0

0

0

0

0

0

0

k3

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1.031

1.785

1.031

1.785

0

0

1.785

3.092

1.785

3.092

0

0

1.031

1.785

1.031

1.785

0

0

1.785

3.092

1.785

3.092

F = 5e4 N (50 kN)

1

2

3

2

1 3

� 108

� 108� 108

y

x




K

2.061

1.785

1.031

1.785

1.03

0

1.785

3.092

1.785

3.092

0

0

1.031

1.785

2.062

0

1.031

1.785

1.785

3.092

0

6.184

1.785

3.092

1.03

0

1.031

1.785

2.061

1.785

0

0

1.785

3.092

1.785

3.092

� = �1 + �2 +�3

F = 5e4 N (50 kN)

1

2

3

2

1 3

� 108

y

x



Pasul 1: Discretizarea structuriiPasul 2: Alegerea funcţiilor de interpolarePasul 3: Proprietăţile elementelorPasul 4: Asamblarea matricei globalePasul 5: Aplicarea condiţiilor de margine

ÎncărcăriRezemări

� =

00___0

� =

__

5000000_

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

11/13/2013

14



Pasul 1: Discretizarea structuriiPasul 2: Alegerea funcţiilor de interpolarePasul 3: Proprietăţile elementelorPasul 4: Asamblarea matricei globalePasul 5: Aplicarea condiţiilor de marginePasul 6: Rezolvarea sistemului de ecuaţii

F = 5e4 N (50 kN)

1

2

3

2

1 3

� = ��

�� = ��

Matricea de rigiditate- Simetrică şi singulară

y

x




F = 5e4 N (50 kN)

1

2

3

2

1 3

�� = ��

�1�

�1�

�2�

�2�

�3�

�3�

= ��

�1�

�1�

�2�

�2�

�3�

�3�

00

�2�

�2�

�3�

0

= ��

�1�

�1�

5000000

�3�

y

x




F = 5e4 N (50 kN)

1

2

3

2

1 3

�� = ��

00

�2�

�2�

�3�

0

= ��

�1�

�1�

5000000

�3�

K

2.061

1.785

1.031

1.785

1.03

0

1.785

3.092

1.785

3.092

0

0

1.031

1.785

2.062

0

1.031

1.785

1.785

3.092

0

6.184

1.785

3.092

1.03

0

1.031

1.785

2.061

1.785

0

0

1.785

3.092

1.785

3.092

� 108

y

x




F = 5e4 N (50 kN)

1

2

3

2

1 3

�� = ��

00

�2�

�2�

�3�

0

= ��

�1�

�1�

5000000

�3�

K

2.061

1.785

1.031

1.785

1.03

0

1.785

3.092

1.785

3.092

0

0

1.031

1.785

2.062

0

1.031

1.785

1.785

3.092

0

6.184

1.785

3.092

1.03

0

1.031

1.785

2.061

1.785

0

0

1.785

3.092

1.785

3.092

� 108

y

x

11/13/2013

15




F = 5e4 N (50 kN)

1

2

3

2

1 3

�� = ��

00

�2�

�2�

�3�

0

= ��

�1�

�1�

5000000

�3�

y

x

K.mod

2.062

0

1.031

0

6.184

1.785

1.031

1.785

2.061

108

K.mod1

0.728

0.14

0.485

0.14

0.243

0.28

0.485

0.28

0.971

108




F = 5e4 N (50 kN)

1

2

3

2

1 3

�� = ��

00

�2�

�2�

�3�

0

= ��

�1�

�1�

5000000

�3�

y

x

d.mod

0.364

0.07

0.243

103

d

0

0

3.638 104

7.004 105

2.427 104

0




F = 5e4 N (50 kN)

1

2

3

2

1 3

K

2.061

1.785

1.031

1.785

1.03

0

1.785

3.092

1.785

3.092

0

0

1.031

1.785

2.062

0

1.031

1.785

1.785

3.092

0

6.184

1.785

3.092

1.03

0

1.031

1.785

2.061

1.785

0

0

1.785

3.092

1.785

3.092

108

d

0

0

3.638 104

7.004 105

2.427 104

0

� = � � �

y

x

f

50.004

43.282

49.993

9.214 103

0.011

43.273

103




F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

f

50.004

43.282

49.993

9.214 103

0.011

43.273

103

y

-50.004

y-43.282 43.273

49.993

� = � � �

11/13/2013

16



Pasul 1: Discretizarea structuriiPasul 2: Alegerea funcţiilor de interpolarePasul 3: Proprietăţile elementelorPasul 4: Asamblarea matricei globalePasul 5: Aplicarea condiţiilor de marginePasul 6: Rezolvarea sistemului de ecuaţiiPasul 7: Calcule suplimentare

Determinarea eforturilor în bare

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

f

50.004

43.282

49.993

9.214 103

0.011

43.273

103

y

-50.004

y-43.282 43.273

49.993





Echilibrul în noduri

F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

f

50.004

43.282

49.993

9.214 103

0.011

43.273

103

y

-50.004

y-43.282 43.273

49.993

-50004

-43282Ne2

�1� + ��1 �� 600 + ��2 � �� 00 = 0

�1� + ��1 �� 600 + ��2 � �� 00 = 0






F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

f

50.004

43.282

49.993

9.214 103

0.011

43.273

103

y

-50.004

y-43.282 43.273

49.993

-50004

-43282Ne2

−50004 + ��1 �

�

�+ ��2 � 1 = 0

−43282 + ��1 �

�

�+ ��2 � 0 = 0






F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

f

50.004

43.282

49.993

9.214 103

0.011

43.273

103

y

-50.004

y-43.282 43.273

49.993

-50004

-43282Ne2

−50004 + ��1 �

�

�+ ��2 � 1 = 0

��1 =

43282

32

= 49978

11/13/2013

17






F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

f

50.004

43.282

49.993

9.214 103

0.011

43.273

103

y

-50.004

y-43.282 43.273

49.993

-50004

-43282Ne2

49978 ��

�+ ��2 � 1 = 50004

��1 =

43282

32

= 49978






F = 5e4 N (50 kN)

1

2

3

2

1 3y

x

f

50.004

43.282

49.993

9.214 103

0.011

43.273

103

y

-50.004

y-43.282 43.273

49.993

-50004

-43282Ne2

��2 = 50004 −49978

2= 25015

��1 =

43282

32

= 49978