probleme partial grinzi cu zabrele

Tip 1 probleme partial MECANICA Problema 6

Să se determine torsorul sistemului de forţe din figură în raport cu punctul O. Să se determine sistemul echivalent cel mai simplu. Rezolvare

F6

F1

F2

F3

F4

F5

O C

DE

x

y

∑ ⋅−=⋅−°⋅⋅−°⋅⋅−⋅−°⋅⋅=⋅=

=−+=+=

−=°⋅−−°⋅−°⋅==°⋅+°⋅++°⋅−°⋅−=

mkNa,aFsinaFcosaFaFcosaFdFM

kN,),(,YXR

kN,sinFFsinFsinFY

kN,cosFcosFFcosFcosFX

iiO 78516023024304

451956103316

5610603060

331660303060

65432

2222

5621

54321

Sistemul echivalent cel mai simplu este dat de rezultanta R, aflată pe dreapta suport

ce întâlneşte axele în punctele de coordonate ),(),( BA yOBsiOxA .

ma,,

a,

R

Md

ma,,

a,

X

My

ma,,

a,

Y

Mx

O

O

B

O

A

6624519

7851

1733316

7851

945610

7851

===

=−−=−=

=−

−==

kNF

kNF

kNF

kN,F

kNFF

7

12

10

58

5

6

5

4

3

21

====

==

Ox

yM o

A

BR

d

Să se determine torsorul în O şi să se determine sistemul echivalent cel mai simplu pentru figura următoare.

y

xO

20 kN

15 kN 15 kN

10 kN

Ox

yMo

A

BR

d

Problema 10


y

xO

25 kN

15 kN

10 kN

15 kN

A

B

Rd

Rezolvare

kN44.2420sin1040sin2015Y

kN92.2020cos1040cos2015X

−=°⋅+°⋅−−==°⋅−°⋅+=

kN17.32R

96.1034)44.24(92.20YXR 2222

=⇒=−+=+=

mkN3.114

40cos5.12020sin410415315Mo

⋅−==°⋅⋅−°⋅⋅+⋅−⋅−=

m55.317.32

3.114

R

Md

m46.592.20

3.114

X

My

m68.444.24

3.114

Y

Mx

O

O

B

O

A

=−==

=−−=−=

=−−==

Rezolvare

kN34.191535sin2510Y

kN48.535cos2515X

=+°⋅+−==°⋅+−=

kN10.20R

07.40434.1948.5YXR2222

=⇒=+=+=

mkN52.54

35cos125310215515Mo

⋅==°⋅⋅−⋅−⋅+⋅=

m71.210.20

52.54

R

Md

m95.948.5

52.54

X

My

m82.234.19

52.54

Y

Mx

O

O

B

O

A

===

−=−=−=

===

Problema 11


y

x

10 kN

50 kN

30 kN

10 kN

O

Problema 12


x

y

50 kN

10 kN

15 kN

20 kN

O

Rezolvare

kN78.030sin5025sin1030Y

kN24.4430cos5025cos1010X

=°⋅−°⋅−=−=°⋅−°⋅+−=

kN25.44R

79.1957)78.0()24.44(YXR 2222

=⇒=+−=+=

mkN29.17

25cos5.2105.210

30sin55030cos5.1505.230Mo

⋅==°⋅⋅−⋅+

+°⋅⋅−°⋅⋅+⋅=

m39.025.44

29.17

R

Md

m39.024.44

29.17

X

My

m17.2278.0

29.17

Y

Mx

O

O

B

O

A

===

=−

−=−=

===

Rezolvare

kN15.30

25sin1030cos2035sin15Y

kN78.36

5025cos1030sin2035cos15X

−==°⋅−°⋅−°⋅−=

−==−°⋅−°⋅+°⋅=

kN56.47R79.2261

)15.30()78.36(YXR 2222

=⇒=

=−+−=+=

mkN32.123

25cos10225sin10230sin202

35cos15435sin152450Mo

⋅==°⋅⋅+°⋅⋅−°⋅⋅−

−°⋅⋅−°⋅⋅−⋅=

m59.256.47

32.123

R

Md

m35.378.36

32.123

X

My

m09.415.30

32.123

Y

Mx

O

O

B

O

A

===

=−

−=−=

−=−

==

Problema 13


y

xO

10 kN

15 kN

20 kN

25 kN

mkN96.125

52525cos105.125sin10535sin155.25.120Mo

⋅==⋅+°⋅⋅+°⋅⋅−°⋅⋅−⋅=

m8.422.26

96.125

R

Md

m62.522.23

96.125

X

My

m35.1017.12

96.125

Y

Mx

O

O

B

O

A

===

−=−=−=

===

Tip 2 Probleme partial MECANICA

Să se determine reacţiunile pentru următoarele structuri plane: Problema 1

Rezolvare

kN17.12

2525sin1035sin15Y

kN22.23

25cos102035cos15X

==+°⋅−°⋅−=

==°⋅−+°⋅=

kN22.26R28.687

)17.12()22.23(YXR 2222

=⇒=

=+=+=

( )

( )

∑

∑

∑

∑

=+⋅−⇔=

=⇒

⋅+⋅⋅=

⇔=⋅−⋅⋅−⋅

⇔=

=

⇒⋅−⋅+⋅⋅=

⇔=⋅−⋅⋅−⋅+⋅

⇔=

=⇒=−⇔=

0V8pV0V

:Verificare

kN87.41V8

354810V

03548p8V

0M

kN13.38V

8

65354810V

03548p6H8V

0M

kN5H05H0H

BA

BB

B

A

A

A

AA

B

AA

OK087.418013.38 =+−⇒

Problema 5

Problema 6

( )

( )

kN07.26V

7

3060305.122V

0305.3753205.1207V

0M

kN93.8V

7

305.12230V

0305.3755.1207V

0M

kN20H020H0H

B

B

B

A

A

A

A

B

BB

=⇒

−++=⇒

=−⋅⋅−⋅−⋅+⋅

⇔=

=⇒

−+−=⇒

=+⋅⋅−⋅+⋅

⇔=

=⇒=−⇔=

∑

∑

∑

Verificare:

( )OK03507.2693.8

075VV0VBA

=−+⇔

=⋅−+⇒=∑

( )

( )

kNm125.98M

0M30460cos2075.35.15

0M

kN10V

060cos20V

0V

kN82.24H

05.1560sin20H

0H

A

A

0

A

A

0

A

A

0

A

=⇒

=+−⋅⋅−⋅⋅−

⇔==

⇒=⋅−

⇔==⇒

=⋅−⋅−

⇔=

∑

∑

∑

Problema 7

Problema 8

( )

( )

kN42.8V

5.4

5.7375.30V

05.2325.25.435.4V

0M

kN08.5V

5.4

5.4375.3012V

025.25.435.13435.4V

0M

kN3H03H0H

B

B

B

A

A

A

A

B

AA

=

⇒+=⇒

=⋅−⋅⋅−⋅

⇔==

⇒++−=⇒

=⋅⋅−⋅−⋅+⋅

⇔=

=⇒=−⇔=

∑

∑

∑

Verificare:

( )OK05.1342.808.5

05.43VV0VBA

=−+⇔

=⋅−+⇒=∑

( )

( )kN25.5V

4

183V

036134V0M

kN75.0V4

283V

012124134V0M

0H0H

BB

BA

AA

AB

A

=⇒+=⇒

=⋅⋅−−⋅⇔=

=⇒−+−=⇒

=⋅⋅+⋅⋅−+⋅⇔=

=⇔=

∑

∑

∑

Verificare:

( )OK0625.575.0

061VV0VBA

=−+⇔

=⋅−+⇒=∑

Problema 9

Problema 10

( )

( )

kN25V

3

60135V

012305.13303V

0M

kN125V

3

375V05.25303V

0M

0H0H

B

B

B

A

A

AA

B

B

=

⇒−=⇒

=⋅⋅+⋅⋅−⋅

⇔==

⇒=⇔=⋅⋅−⋅

⇔=

=⇔=

∑

∑

∑

Verificare:

( )OK015025125

0530VV0VBA

=−+⇔

=⋅−+⇒=∑

( )

( )

kN57.40V

5.3

41.263.791.89V

025sin5.22525cos5.32505.41.1205.3V

0M

kN09.4V

5.3

41.261.12V

025sin5.22555.01.1205.3V

0M

kN57.10H025sin25H0H

B

B

B

A

A

A

A

B

AA

=

⇒−+=⇒

=°⋅⋅+°⋅⋅−⋅⋅−⋅

⇔==

⇒+−=⇒

=°⋅⋅−⋅⋅+⋅

⇔=

=⇒=°⋅−⇔=

∑

∑

∑

Verificare:

( )OK066.222257.4009.4

025cos251.120VV0VBA

=−−+⇔

=°⋅−⋅−+⇒=∑

Problema 11

Problema 12

( )

( )

kN89,38V

5

47,194V

036156015sin2035V

0M

kN21,58V

5

06,291V

036156015sin20315cos2055V

0M

kN82,84H

061515sin20H0H

B

B

B

A

A

A

A

B

A

A

−=

⇒−=⇒

=⋅⋅+−°⋅⋅−⋅

⇔==

⇒=⇔

=⋅⋅−+°⋅⋅+°⋅⋅−⋅

⇔==

=⋅−°⋅+⇔=

∑

∑

∑

Verificare:

( )OK032,1989,3821,58

015cos20VV0VBA

=−−⇔

=°⋅−−⇒=∑

( )

( )

kN66,20V

6

02,124V

036156060sin5,12060cos6206V

0M

kN38V

6

94,227V

036156060sin5,42060cos6206V

0M

kN80H061560cos20H0H

B

B

B

A

A

A

A

B

AA

−=

⇒

−=⇒

=⋅⋅+−°⋅⋅−°⋅⋅−⋅

⇔==

⇒=⇒

=⋅⋅−+°⋅⋅−°⋅⋅+⋅

⇔=

=⇒=⋅−°⋅+⇔=

∑

∑

∑

Verificare:

( )OK034,1766,2038

060sin20VV0VBA

=−−⇔

=°⋅−−⇒=∑

Problema 13

10 kN/m 10 kN/m

20 kN

Problema 14

10 kN/m

20 kN

Probleme – Grinzi cu zabrele

Să se determine eforturile din barele următoarelor grinzi cu zăbrele: 1.

100 kN50 kN 50 kN100 kN 100 kN

2.00 2.00 2.00 2.00

2.00

45°

100 kN50 kN 50 kN100 kN 100 kN

1

2

3

4

5

6

7

8

9

100 kN50 kN 50 kN100 kN 100 kN

V0

H0

V8

0

Rezolvare: Determinarea reacţiunilor cu ajutorul grinzii echivalente:

( )

( )

OKVVV

kNV

VM

kNV

VM

HH

04002002000503100500

2008

200400600400

021004100610085080

2008

200400600400

021004100610085080

00

80

8

80

0

08

0

=−+⇔=−×−−+⇒=

=+++=⇒

=×−×−×−×−×⇒=

=+++=⇒

=×−×−×−×−×⇒=

=⇒=

∑

∑

∑

∑

Metoda izolării nodurilor

1

50 kN

N10

N13

0

50

200

N02

N03

45°

N241502

N23

3N350

100

N340212.13

45° 45°

N572005

100

N54

4N46150

100 N47

45°

70.72

45°

Metoda secţiunilor

00

500500

13

1010

=⇒=

−=⇒=+⇒=

∑

∑NH

kNNNV

kNNcos.NNcosNH

kN.Nsin

NsinNV

15045172120450

1321245

150045200500

02020203

030303

=⇒°⋅=⇒=+°⋅⇒=

−=⇒

°−=⇒=°⋅−−⇒=

∑

∑

kNNNH

kNNV

15001500

00

2424

23

=⇒=+−⇒=

=⇒=

∑

∑

kNNNcossin

cos.H

kN.sin

NsinNsin.V

20004545

5045172120

727045

5004545132121000

3535

3434

−=⇒=+°⋅°

+°⋅⇒=

=°

=⇒=°⋅+°⋅−⇒=

∑

∑

kNNNH

kNNNV

20002000

10001000

5757

5454

−=⇒=+⇒=

−=⇒=+⇒=

∑

∑

kNNNcossin

cos.H

kN.sin

NsinNsin.V

15004545

504572701500

727045

500454572701000

4646

4747

=⇒=+°⋅°

+°⋅−−⇒=

=°

=⇒=°⋅+°⋅−⇒=

∑

∑

100 kN50 kN 50 kN100 kN 100 kN

2.00 2.00 2.00 2.00

2.00

45°

1

2

3

4

5

6

7

8

9

0

b

b

a

a

c c

d

d

e e

Secţ a-a

00

500500

13

1010

=⇒=

−=⇒=+⇒=

∑

∑NH

kNNNV

Secţ b-b

kNNN)M(

kN..

N.N)M(

1502

3000225022000

77212411

300041125022000

02023

03032

==⇒=⋅−⋅−⋅⇒=∑

−=−=⇒=⋅+⋅−⋅⇒=∑

Secţ c-c

kNNNH

NV

15001500

00

2424

23

=⇒=−⇒=

=⇒=

∑

∑

Secţ d-d

kN..

N.NN)M(

kNNN)M(

9270411

1000411225022000

2002

40002210045042000

3434352

35354

==⇒=⋅+⋅+⋅−⋅⇒=∑

−=−=⇒=⋅+⋅−⋅−⋅⇒=∑

Secţ. e-e

kNNNH

kNNV

20002000

1001000

5757

54

−=⇒=+⇒=

−=+⇒=

∑

∑

20 kN

40 kN

40 kN

20 kN

0

1

2 4

3

5

6 8

7

10

9

111.50 1.50 1.50 1.50 1.50 1.50

0.7

5 1.5

0 2.25

20 kN

011

40 kN 40 kN 20 kN

20 kN

0

40 kN 40 kN 20 kN

V11V0

H0


( )

( )

OK0120309002024020VV0V

kN309

6012090V

05.1403405.4209V0M

kN909

90240300180V

05.4206405.7409209V0M

0H0H

110

11

110

0

011

0

=−+⇔=−×−−+⇒=

=++=⇒

=×−×−×−×⇒=

=+++=⇒

=×−×−×−×−×⇒=

=⇒=

∑

∑

∑

∑

894.0cos

447.0sin

=α=α

N01

N020

27°

20

90

N24140

2

N21

kN140N

894.06.156894.0NN0cosNN0H

kN6.156447.0

70N0sinN90200V

02

01020102

0110

=⇒

⋅=⋅−=⇒=α⋅+⇒=

−=−=⇒=α⋅++−⇒=

∑

∑

kN140N0N1400H

0N0V

2424

21

=⇒=+−⇒=

=⇒=

∑

∑

40 N13

156.6N14

1

140

N43

4

44.75

N46

40

3

111.8620

N35

N36

20

5

67.11 N57

N56

6100

56.5540

N67

N68

45°45°

7

N79

67.11

0 N78

45°

860

N89

N8-10

0

kN86.111N11.67N75.44)1(

kN75.44N94.89N2)2()1(

)2(6.156NN

0cosNcosNcos6.1560H

)1(11.67sin

30NN

0sinNsinNsin6.156400V

1313

1414

1314

1314

1314

1314

−=⇒=−−⇒−=⇒−=⇒+

−=+⇔

=α⋅+α⋅+α⋅⇒=

=α

=−⇔

=α⋅+α⋅−α⋅+−⇒=

∑

∑

kN100N0Ncos75.441400H

kN20447.075.44N0Nsin75.440V

4646

4343

=⇒−+α⋅+−⇒=

=⋅=⇒=+α⋅−⇒=

∑

∑

kN55.56N)11.67(632.014.14N)1(

kN11.67N100N894.0N447.010

100N894.0)N632.014.14(707.0

100N894.0N707.0

0cosN45sinNcos86.1110H

)1(N632.014.14N10N447.0N707.0

0sinN45cosNsin86.11120400V

3636

353535

3535

3536

3536

35363536

3536

−=⇒−×+−=⇒

−=⇒−=++−−=++−×⇒

⇒−=+⇔

=α⋅+°⋅+α⋅⇒=+−=⇒−=−⇔

=α⋅+°⋅−α⋅+−−⇒=

∑

∑

kN40N0sin11.67sin11.67N200V

kN11.67N0cosNcos11.670H

5656

5757

=⇒=α⋅+α⋅+−−⇒=

−=⇒=α⋅+α⋅⇒=

∑

∑

kN60N

0N045cos55.561000H

kN0N

045sinN45sin55.56400V

68

68

67

67

=⇒

=++°⋅+−⇒==⇒

=°⋅+°⋅−⇒=

∑

∑

kNN

sin.Nsin.V

kN.NcosNcos.H

0

0116711670

1167011670

78

78

7979

=⇒

=⋅+−⋅−⇒=

−=⇒=⋅+⋅⇒=

∑

∑αα

αα

kNNNH

kNNsinNV

600600

000

108108

8989

=⇒=+−⇒=

=⇒=⋅⇒=

−−∑

∑ α

10

N10-9

N10-1160

11

30

N11-9

N11-10

N01 N02 N21 N24 N13 N14 N43 N46 N36 N35 N56

-156.6 140 0 140 -111.86 -44.75 20 100 -56.55 -67.11 40

N57 N67 N68 N78 N79 N89 N8-10 N9-10 N10-11 N9-11

-67.11 0 60 0 -67.11 0 60 0 60 -67.11

20 kN

40 kN

40 kN

20 kN

0

1

2 4

3

5

6 8

7

10

9

111.5 1.5 1.5 1.5 1.5 1.5

0.8

1.5

2.3

a

a

b

b

c

c

d

d

e

e

f

f

Metoda sectiunilor Sect a-a

20 kNN01

N02

90

1

227°

kNNNH

NV

600600

00

11101110

109

=⇒=+−⇒=

=⇒=

−−

−

∑

∑

kNNNcos.H

kN.NsinNV

60011670

11670300

10111011

911911

=⇒=−⋅⇒=

−=⇒=+⋅⇒=

−−

−−

∑

∑α

α

kN6.156N05.1sinN5.1205.1900)M(

kN140N075.0N5.1205.1900)M(

01012

02021

−=⇒=⋅α⋅+⋅−⋅⇒=

=⇒=⋅−⋅−⋅⇒=

∑

∑

Sect b-b

20

40

0

1

290 N24

N14

N13

4

d1

3d2

3.00

27° d1

d2

3

40

11.68

1.68

341.1d2

d68.1

2

d68.1A

m341.1447.03sin3d

2

21

134

1

=⇒⋅=⋅=

=⋅=α⋅=

Sect c-c

20 kN

40 kN

40 kN

0

1

2 4

3

5

6N46

N36

N35

d3d4

45°

90

m2d2

d68.1

2

d12.2A

m59.1d25.2d2

4

43

356

3

22

3

=⇒×=×=

=⇒=⋅

( )( )( ) kN6.56N0dN25.21005.1403405.4205.4900M

kN5.67N0dN5.1403405.4205.4900M

kN100N05.1N5.1403203900M

363365

354356

46463

−=⇒=⋅−⋅−⋅−⋅−⋅−⋅⇔=

−=⇒=⋅+⋅−⋅−⋅−⋅⇔=

=⇒=⋅−⋅−⋅−⋅⇔=

∑

∑

∑

kN86.111N0341.1N5.1403203900)M(

kN140N075.0N5.1205.1900)M(

13134

24241

−=⇒=⋅+⋅−⋅−⋅⇒=

=⇒=⋅−⋅−⋅⇒=

∑

∑

( ) kN74.44N0dN5.1N5.1403203900M14214243

−=⇒=⋅+⋅−⋅−⋅−⋅⇒=∑

Sect d-d

8

7

10

9

30

6

5

N75

N76

N86

d5

d6

45°

m2dd

m59.1dd

64

53

====

( )( )( ) 0N0dN25.2605.4300M

kN5.67N0dN5.4300M

kN60N05.1N3300M

765765

756756

86867

=⇒=⋅+⋅+⋅−⇔=

−=⇒=⋅−⋅−⇔=

=⇒=⋅+⋅−⇔=

∑

∑

∑

Sect e-e

8

7

10

9

30

N97

N98

N10-8

d7

d8

m341.1dd

m341.1dd

28

17

====

( )( )( ) 0N0dN5.1603300M

kN11.67N0dN3300M

kN60N075.0N5.1300M

987987

978978

8108109

=⇒=⋅+⋅+⋅−⇔=

−=⇒=⋅−⋅−⇔=

=⇒=⋅+⋅−⇔=

∑

∑

∑ −−

Sect f-f

10

9

30

N11-9

N11-10

11

( )( ) kN11.67N0sin5.1N5.1300M

kN60N075.0N5.1300M

91191110

101101119

−=⇒=α⋅⋅−⋅−⇔=

=⇒=⋅+⋅−⇔=

−−

−−

∑

∑

Pentru simplificare, determinarea eforturilor din tiranti se face utilizand izolare de noduri.

20 kN 20 kN

80 kN 80 kN

100 kN

3 5 791

2

4

6

8

10

1.5

0

3.0

0

2.50 2.50 2.50 2.50

20 kN 20 kN80 kN80 kN 100 kN

20 kN 20 kN80 kN80 kN 100 kN

V1

H1

V9

50 kN

42°

17°


( )

( )

OK03005.1575.1420100280220VV0V

kN5.15710

75500800200V

05.1505.28051005.780102010V0M

kN5.14210

75500800200V

05.1505.28051005.780102010V0M

kN50H0H

91

9

91

1

19

1

=−+⇔=−×−×−+⇒=

=+++=⇒

=×−×−×−×−×−×⇒=

=−++=⇒

=×+×−×−×−×−×⇒=

=⇒=

∑

∑

∑

∑

957017

287017

743042

669042

.coscos

.sinsin

.coscos

.sinsin

=°≈=°≈

=°≈=°≈

ββα

α

Metoda sectiunilor + metoda izolarii nodurilor

20 kN 20 kN

80 kN 80 kN

100 kN

3 5 791

2

4

6

8

1050 kN

42°

17°

a

a

b

b

c

c

d

d

Sect a-a

N13

N14

N24

1

2

20 kN50 kN

50

142.5

3

4

d1

d2

d3

20 kN50 kN

N24

N212

( )

kN24.161N

m67.1669.05.2sin5.2d

m15.2957.025.2cos25.2d

0dNdN5.1505.2205.25.1420M

kN44.169N

25.2N25.2505.25.1425.22075.0500)M(

kN08.52N

m44.1957.05.1cos5.1d

0dN5.1500)M(

14

3

2

3142243

13

134

24

1

1241

−=⇒

=⋅=α⋅==⋅=β⋅=

=⋅+⋅+⋅+⋅−⋅⇒=

=⇒

=⋅−⋅+⋅+⋅−⋅−⇒=−=⇒

=⋅=β⋅=

=⋅+⋅⇒=

∑

∑

∑

kN44.33N0sinNN200V212421

−=⇒=β⋅+−−⇒=∑

Sect b-b

N35

N45

N46

1

2

20 kN50 kN

50

142.5

3

4

d2

d3

80 kN

5

6

d5

170N34

N35

3 Sect c-c

N75

N85

N86

9

10

20 kN

157.5

7

8

d2

d3

80 kN

5

6

d5

( ) kN88.12N0dNdN5.2205.25.1570M

kN86.169N0dN5.28052055.1570)M(

kN78.152N025.2N5.2205.25.1570)M(

853852867

865865

75758

=⇒=⋅+⋅+⋅−⋅⇒=

−=⇒=⋅+⋅−⋅−⋅⇒=

=⇒=⋅−⋅−⋅⇒=

∑

∑

∑

( )

( )kN61.9N

0dNdN5.1505.2205.25.1420M

kN86.169N

0dN5.1505.28052055.1420M

kN44.169N

025.2N25.2505.25.1425.22075.0500)M(

45

4452463

46

5465

35

354

−=⇒

=⋅+⋅+⋅+⋅−⋅⇒=

−=⇒

=⋅+⋅+⋅−⋅−⋅⇒=

=⇒

=⋅−⋅+⋅+⋅−⋅−⇒=

∑

∑

∑

m87.2957.03cos3d

m67.1dd

m67.1669.05.2sin5.2d

m15.2957.025.2cos25.2d

5

34

3

2

=⋅=β⋅===

=⋅=α⋅==⋅=β⋅=

0N0V34

=⇒=∑

m87.2957.03cos3d

m67.1dd

m67.1669.05.2sin5.2d

m15.2957.025.2cos25.2d

5

34

3

2

=⋅=β⋅===

=⋅=α⋅==⋅=β⋅=

N68

100 kN

6N64

N65

Sect d-d

7

8

N97

N98

N10-8

9

10

20 kN

157.5

d1

d2

d3

20 kNN10-8

N10-9 10

N24 = -52.08 kN N13 = 169.44 kN N14 = -161.24 kN N21 = -33.44 kN N35 = 169.44 kN N46 = -169.86 kN N45 = -9.61 kN N34 = 0 N75 = 152.78 kN N86 = -168.86 kN N85 = 12.88 kN N65 = -100 kN N97 = 152.78 kN N8-10 = 0 N98 = -205.84 kN N9-10 = -20 kN N78 = 0

adev0cosNcosN0H

kN100N0100N0V

6864

6565

=β⋅+β⋅−⇒=∑

−=⇒=−−⇒=∑

( )kN84.205N

0dNdN5.2205.25.1570M

0N0dN0dN0)M(

kN78.152N

025.2N5.2205.25.1570)M(

98

39821087

10811085869

97

978

−=⇒

=⋅+⋅+⋅−⋅⇒=

=⇒=⋅⇒=⋅⇒=

=⇒

=⋅−⋅−⋅⇒=

−

−−

∑

∑

∑

kN20N0sinN20N0V109810109

−=⇒=β⋅+−−⇒=∑ −−−

probleme partial grinzi cu zabrele

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