1)Metoda matricore e forcave

1. Analiza statike e skemes

L=3K-SH=3*2-3=3. Pra skema eshte 3 here statikisht e pacaktuar.

Ndertoj sistemin baze:

1

2. Diskretizimi

Ndajme strukturen ne nyje dhe elemente:

3. Ekuivalentimi i ngarkeses ne forca nyjore:Ne elementin 6 forcen uniformisht te shperndare q e ekuivalentoj ne force nyjore duke u bazuar ne sistemin baze te M.D:

Pacaktueshmeria statike: n❑=nr+ns=2+2=4

2

R3=R4=ql2

=20 ∙82

=80kN

M 3=M 4=q l2

12=20 ∙8

2

12=106.7kNm

Ngarkesa uniformisht e shperndare zvendesohet me ngarkesa nyjore, te cilat jane te barabarta ne vlere, por me kah te kundert me ato qe percaktuam me siper.

Skema ekuivalente:

4. Sistemet koordinativ te forcave te jashtme dhe te brendshme.a) Sistemi koordinativ i forcave te jashtme:

3

{P }={P1P2P3P4

}={ 10080106.7100

}(kN )ose (kNm )

b) Sistemi koordinativ i forcave te brendshme:

4

{F }={F1F2F3F4F5F6F7F8

}5. Formulimi I matricave te fleksibiliteti te elementeve dhe te struktures me

elemente te palidhur.a. Matricat e fleksibiliteti te elementeve:

Matrica e fleksibilitetit per elementet 1, 2 ,4 (gjatesi L, ngurtesi seksioni EI, me njerin skaj te inkastruar dhe tjetrin cerniere):

[δ ]g=lg

3 E I g[1 ](1× 1)

[δ ]1=[δ ]2¿ [δ ]4=l3 EI

[1 ](1× 1)=1EI

[1.333 ]= 1E

[246.852 ]

I=b∗h3

12=0.3∗0.6

3

12=0.0054m3

Matrica e fleksibilitetit per elementin 5 (gjatesi L, ngurtesi seksioni EI, me te dy skajet te lidhur ne nyjet e ngurta):

[δ ]g=lg

6 E I g [ 2 −1−1 2 ]

(2× 2)

[δ ]5=l6 EI [ 2 −1

−1 2 ](2× 2)

= 1EI [ 1.333 −0.668

−0.668 1.333 ]= 1E [ 246.852 −123.704

−123.704 246.852 ]

Matrica e fleksibilitetit per elementin 6 (gjatesi 2L, ngurtesi seksioni EI, me te dy skajet te lidhur ne nyjet e ngurta):

[δ ]6=2 l6 EI [ 2 −1

−1 2 ](2× 2)

= 1EI [ 2.666 −1.333

−1.333 2.666 ]= 1E [ 493.704 −246.852

−246.852 493.704 ]Matrica e fleksibilitetit per elementin 3 (gjatesi 2L, ngurtesi seksioni EI, me te dy skajet te lidhur me cerniera):

[δ ]g=l gEA

[1 ](1×1 )

[δ ]3=2lEA

[1 ](1×1)=8EA

[1 ]= 1E

[44.444 ]

5

A=b∗h=0.6∗0.3=0.18m2

b) Matrica e fleksibilitetit te struktures me elemente te palidhur:

[ f ]=[[δ ]1 0 0 0 0 00 [δ ]2 0 0 0 00 0 [δ ]3 0 0 00 0 0 [δ ]4 0 00 0 0 0 [δ ]5 00 0 0 0 0 [δ ]6

]¿ 1E [246.8520000000

0246.852000000

00

44.44400000

000

246.8520000

0000

246.852−123.704

00

0000

−123.704246.85200

000000

493.704−246.852

000000

−246.852493.704

](8x 3)

6. Llogaritja e te panjohurave te teperta.

Ndertimi i matrices [b]0 dhe [b]X:Per ndertimin e ketyre matricave, ne sistemin baze ndertojme epjyrat e momenteve perkulese per shkak te te panohurave X1, X2, X3 dhe te sistemit te forcave te jashtme P, ne vlere njesi.

6

7

8

P1 P2 P3 P4

[b ]0=

F1F2F3F4F5F6F7F8

[0 0 0 00 0 0 00 0 0 00 0 0 40 −8 0 −4−4 8 0 80 0 1 00 8 −1 0

](KNmose KN )

X1 X2 X3

[b ]x=

F1F2F3F4F5F6F7F8

[0 4 00 0 −40 0 10 0 48 −4 0−8 0 00 −4 4

−8 4 −4] (KNmoseKN )

Te panjohurat [X ] llogariten me formulen:

[X ]=− [D ]−1 [b ] xT [ f ] [b ]0=−( [b ]x

T [ f ] [b ]x)−1 ( [b ]x

T [ f ] [b ]0 )

[b ]xT [ f ] [b ]x=

1E [ 7.903∗10

4 −3.557∗104 2.372∗104

−3.557∗104 3.161∗104 −2.372∗104

2.372∗104 −2.372∗104 3.166∗104 ]

( [b ]xT [ f ] [b ]x )

−1=E[2.606∗10

−5 3.352∗10−5 5.588∗10−6

3.352∗10−5 1.154∗10−4 6.133∗10−5

5.588∗10−6 6.133∗10−5 7.334∗10−5][b ]x

T [ f ] [b ]0=1E

¿

[X ]=[−0.243 1 0.011 0.64−0.169 −1.017∗10−3 0.122 0.0370.055 −1.117∗10−3 −0.104 −0.576]

9

[X ]= [X ] {P }=[−0.243 1 0.011 0.64−0.169 −1.017∗10−3 0.122 0.0370.055 −1.117∗10−3 −0.104 −0.576]{

10080106.7100

}[X ]={120.874−0.264

−63.286}(KN )

7. Llogaritja e forcave te brendshme dhe ndertimi i epiurave te tyre.

Percaktimi i forcave te brendshme per shkak te veprimit te forcave te jashtme njesore:

[b ]= [b ]0+ [b ]x [X ]

[b ]=[0 0 0 00 0 0 00 0 0 00 0 0 40 −8 0 −4−4 8 0 80 0 1 00 8 −1 0

]+[0 4 00 0 −40 0 10 0 48 −4 0−8 0 00 −4 4

−8 4 −4]∗[−0.243 1 0.011 0.64

−0.169 −1.017∗10−3 0.122 0.0370.055 −1.117∗10−3 −0.104 −0.576 ]

[b ]=[−0.676 −4.068∗10−3 0.488 0.148−0.22 4.468∗10−3 0.416 2.304

0.0550.22

−1.268−2.0560.8961.048

−1.117∗10−3

−4.468∗10−3

4.068∗10−3

0−4∗10−4

4∗10−4

−0.104−0.416−0.4

−0.0880.096

−0.184

−0.5761.6960.9722.88

−2.452−2.668

] Percaktimi I forcave te brendshme nga veprimi I te gjithe ngarkeses se dhene

nyjore: {F }N=[b ] {P }

10

{F }N=[−0.676 −4.068∗10−3 0.488 0.148−0.22 4.468∗10−3 0.416 2.304

0.0550.22

−1.268−2.0560.8961.048

−1.117∗10−3

−4.468∗10−3

4.068∗10−3

0−4∗10−4

4∗10−4

−0.104−0.416−0.4

−0.0880.096

−0.184

−0.5761.6960.9722.88

−2.452−2.668

]{ 10080106.7100}=

F1F2F3F4F5F6F7F8

{−1.056253.145−63.286146.855−71.95573.01

−145.389−181.601

} Percaktimi I forcave te brendshme perfundimtare:

{F }={F }N+ {F }R={−1.056253.145−63.286146.855−71.95573.01

−145.389−181.601

}+{000000

−106.7106.7

}={−1.056253.145−63.286146.855−71.95573.01

−252.089−74.901

}(KNmose KN )

Ndertimi i epyres perfundimtare:

8. Llogaritja e zhvendosjeve te nyjeve te struktures.

11

Llogaritja e matrices se elasticitetit [F] e te gjithe sistemit referuar drejtimit te forcave te dhena (kur kerkohen zhvendosje sipas drejtimit te forcave te jashtme):

[F ]=[b ]0T [ f ] [b ]=[ 4.667∗10

−6 6.71∗10−9 3.701∗10−7 −7.877∗10−6

1.126∗10−7 −3.23∗10−8 2.864∗10−8 −1.307∗10−7

−3.752∗10−7

−7.843∗10−61.937∗10−9

−4.151∗10−86.912∗10−7

5.424∗10−75.332∗10−7

2.289∗10−5]

{u }=[F ] {P }={ 3.59∗10−4

1.33∗10−6

8.94∗10−5

1.559∗10−3}m9. Kontrolli I zhvendosjeve ne drejtim te te panjohurave te teperta.

Llogaritja e matrices se elasticitetit [F] e te gjithe sistemit referuar drejtimit te te panjohurave:

[F ]=[b ]xT [ f ] [b ]=[−1.26∗10

−7 3.23∗10−8 −2.864∗10−8 1.307∗10−7

5.821∗10−8 −1.888∗10−8 1.284∗10−8 −9.465∗¿10−8

−4.451∗10−8 ¿1.092∗10−8 4.632∗10−8]

Llogaritja e zhvendosjeve sipas drejtimit te te panjohurave:

{u }=[F ] {P }={ 1.33∗10−6

−3.783∗10−6

−1.652∗10−6}m

Zhvendosjet ne drejtim te te panjohurave kane vlera shume te vogla, te cilat pranohen ne menyre te perafert te barabarta me vleren zero. Pra, kontrolli plotesohet.

12

2) Metoda matricore e deformimeve

1. Diskretizimi

2. Ekuivalentimi i ngarkeses ne forca nyjore

13

Ne elementin 2 forcen uniformisht te shperndare q e ekuivalentoj ne force nyjore duke u bazuar ne sistemin baze te M.D:

M 2=q l2

8=20∗4

2

8=40KNm;

R2=5ql8

=5∗20∗48

=50KN ;

R3=3ql8

=3∗20∗48

=30KN

Ngarkesa uniformisht e shperndare zvendesohet me ngarkesa nyjore, te cilat jane te barabarta ne vlere, por me kah te kundert me ato qe percaktuam me siper.

14

{P }={P1P2P3P4P5

}={100−50−40−30−100

}(KN ose KNm)

3. Sistemet koordinativ dhe indeksimi i shkalleve te lirise

15

4. Formulimi i matricave te ngurtesise te elementeve dhe te struktures

Formulimi matricave te ngurtesise te elementeve kundrejt akseve lokale:

elementi 1:

1 2 3 4 5 6

16

[k ]1¿=

1¿2¿34

¿ 5¿6 [

EA2 L

0 0−EA2 L

0 0

012 EI8L3

6EI4 L2

0−12 EI8L3

6 EI4 L2

0−EA2 L00

6 EI4 L2

0−12 EI8L3

6 EI4 L2

4 EI2 L0

−6 EI4 L2

2 EI2 L

0EA2 L00

−6 EI4 L2

012EI8 L3

−6 EI4 L2

2EI2L0

−6 EI4 L2

4 EI2L

]=105[ 67.5 0 0 −67.5 0 00 0.38 1.52 0 −0.38 1.520

−67.500

1.520

−0.381.52

8.10

−1.524.05

067.500

−1.5200.38

−1.52

4.050

−1.528.1

]elementi 2:

4 5 6 7 8 9

[k ]2¿=¿

elementi 3:

17

10 11 12 7 8 9

[k ]3¿=¿

elementi 4:

13 14 15 10 11 12

18

[k ]4¿=

13¿14¿1510

¿ 11¿12[

EAL

0 0−EAL

0 0

012 EIL3

6 EIL2

0−12EI

L36 EIL2

0−EAL00

6 EIL2

0−12 EI

L3

6 EIL2

4 EIL0

−6 EIL2

2 EIL

0EAL00

−6 EIL2

012EIL3

−6 EIL2

2 EIL0

−6 EIL2

4 EIL

]=105[ 135 0 0 −135 0 00 3.04 6.08 0 −3.04 6.080

−13500

6.080

−3.046.08

16.20

−6.088.1

013500

−6.0803.04

−6.08

8.10

−6.0816.2

]elementi 5:

10 11 12 16 17 18

[k ]5¿=

10¿11¿1216

¿ 17¿18 [

EAL

0 0−EAL

0 0

012 EIL3

6 EIL2

0−12 EI

L36 EIL2

0−EAL00

6 EIL2

0−12 EI

L3

6 EIL2

4 EIL0

−6 EIL2

2 EIL

0EAL00

−6 EIL2

012 EIL3

−6 EIL2

2EIL0

−6 EIL2

4 EIL

]=105[ 135 0 0 −135 0 00 3.04 6.08 0 −3.04 6.080

−13500

6.080

−3.046.08

16.20

−6.088.1

013500

−6.0803.04

−6.08

8.10

−6.0816.2

]elementi 6:

19

19 20 21 16 17 18

[k ]6¿=

19¿20¿2116

¿ 17¿18 [

EAL

0 0−EAL

0 0

03 EI

L30 0

−3 EIL3

3 EI

L2

0−EAL00

00

−3 EIL3

3 EIL2

0000

0EAL00

003 EIL3

3 EIL2

003 EI

L2

3 EIL

]=105[ 135 0 0 −135 0 00 0.759 0 0 −0.759 3.040

−13500

00

−0.7593.04

0000

013500

0 00 0

0.7593.04

3.0412.15

] Formulimi matricave te ngurtesise te elementeve kundrejt akseve globale:

Per elementet 2 dhe 5θ=0, [R ]=[1 ] ; [k ]¿=[k ]gl; Pra:

[k ]gl2 =105[

135 0 0 −135 0 00 0.759 3.04 0 −0.759 00

−13500

3.040

−0.7590

12.1503.040

013500

3.040

0.7590

0000]

20

[k ]gl5 =105[

135 0 0 −135 0 00 3.04 6.08 0 −3.04 6.080

−13500

6.080

−3.046.08

16.20

−6.088.1

013500

−6.0803.04

−6.08

8.10

−6.0816.2

]Per elementet 1; 3; 4; 6[k ]gl=[R ]T [k ]¿ [R ] ; θ=90° ❑⇒ cosθ=0 ;sinθ=1;

[R ]=[cosθ sinθ 0 0 0 0

−sinθ cosθ 0 0 0 00000

0000

1000

0cosθ

−sinθ0

0sinθcosθ0

0001]=[

0 1 0 0 0 0−1 0 0 0 0 00000

0000

1000

00

−10

0100

0001]

[k ]gl1 =105[

0.38 0 −1.52 −0.38 0 −1.520 67.5 0 0 −67.5 0

−1.52−0.380

−1.52

00

−67.50

8.11.5204.05

1.520.3801.52

0067.50

4.051.5208.1

][k ]gl

3 =105[0.759 0 −3.04 −0.759 0 00 135 0 0 −135 0

−3.04−0.75900

00

−1350

12.15−3.0400

−3.040.75900

001350

0000]

[k ]gl4 =105[

3.04 0 −6.08 −3.04 0 −6.080 135 0 0 −135 0

−6.08−3.040

−6.08

00

−1350

16.26.0808.1

6.083.0406.08

001350

8.16.08016.2

]21

[k ]gl6 =105[

0.759 0 0 −0.759 0 −3.040 135 0 0 −135 00

−0.7590

−3.04

00

−1350

0000

00.7590

−3.04

001350

0−3.040

12.15]

Formulimi matrices se ngurtesise te struktures

22

23

5. Modifikimi i matrices se ngurtesise te struktures

Ne matricen e ngurtesise se struktures heq rreshtat dhe shtyllat e zhvendosjeve 0 dhe marr matricen e ngurtesise te modifikuar te struktures:

24

6. Llogaritja e zhvendosjeve dhe forcave te panjohura Llogaritja e zhvendosjeve:

[K ]m {u }={P }

{u }=[K ]m−1 {P }=105[135.38 0 1.520 68.259 3.041.52 3.04 20.25

−135 0 00 −0.759 00 3.04 0

0 0 0 0 00 0 0 0 00 0 0 0 0

−135 0 00 −0.759 3.040 0 0

135.759 0 −0.7590 135.759 0

−0.759 0 138.799

0 −3.04 0 0 0−135 0 0 0 00 3.04 −135 0 0

0 0 00 0 0000

000

000

0 −135 0−3.04 0 3.04000

000

−13500

273.04 6.08 0 −3.04 6.086.08 44.55 0 −6.08 8.10

−3.046.08

0−6.088.1

135.7590

−3.04

0138.04−6.08

−3.04−6.0828.35

]−1

{100−50−400

−30000

−10000

}=10−5 {121.873−0.242−11.057121.351−0.2

−10.76−0.2279.795

−11.5240.253

−3.931

}(mose rad) Llogaritja e forcave te panjohura

25

=(−29.5116.34140.47100−50−400

−30000092.330.6513.9−1000020.7

−34.20

)(KN )

7. Llogaritja e forcave te brendshme dhe ndertimi i epiurave te tyre

Elementi 1:

{P }= [k ]1¿ [R ] {u }

26

{N1

Q1

M 1

N2

Q2

M 2

}=[67.5 0 0 −67.5 0 00 0.38 1.52 0 −0.38 1.520

−67.500

1.520

−0.381.52

8.10

−1.524.05

067.500

−1.5200.38

−1.52

4.050

−1.528.1

][0 1 0 0 0 0

−1 0 0 0 0 00000

0000

1000

00

−10

0100

0001]{

000

121.873−0.242−11.057

}={16.33529.505140.466−16.335−29.50595.685

}Elementi 2:

{N2

Q2

M 2

N3

Q3

M 3

}=[135 0 0 −135 0 00 0.759 3.04 0 −0.759 00

−13500

3.040

−0.7590

12.1503.040

013500

3.040

0.7590

0000][1 0 0 0 0 00 1 0 0 0 00000

0000

1000

0100

0010

0001]{121.873−0.242−11.057121.351−0.20

}={70.47

−33.645−135.686−70.47−33.581

0}

Elementi 3:

{N4

Q4

M 4

N3

Q3

M 3

}=[135 0 0 −135 0 00 0.759 3.04 0 −0.759 00

−13500

3.040

−0.7590

12.1503.040

013500

3.040

0.7590

0000] [0 1 0 0 0 0

−1 0 0 0 0 00000

0000

1000

00

−10

0100

0001]{

−10.76−0.2279.795121.351−0.20

}={−3.64570.049

−217.1873.645

−70.4950

}Elementi 4:

{N5

Q5

M 5

N4

Q4

M 4

}=[135 0 0 −135 0 00 3.04 6.08 0 −3.04 6.080

−13500

6.080

−3.046.08

16.20

−6.088.1

013500

−6.0803.04

−6.08

8.10

−6.0816.2

][0 1 0 0 0 0

−1 0 0 0 0 00000

0000

1000

00

−10

0100

0001]{

000

−10.76−0.2279.795

}={30.64592.313.919

−30.645−92.393.258

}Elementi 5:

27

{N4

Q4

M 4

N6

Q6

M 6

}=[135 0 0 −135 0 00 3.04 6.08 0 −3.04 6.080

−13500

6.080

−3.046.08

16.20

−6.088.1

013500

−6.0803.04

−6.08

8.10

−6.0816.2

][1 0 0 0 0 00 1 0 0 0 00000

0000

1000

0100

0010

0001]{

−10.76−0.2279.795

−11.5240.253

−3.931}={

103.1434.194123.92

−103.14−34.19412.739

}Elementi 6:

{N7

Q7

M 7

N6

Q6

M 6

}=[135 0 0 −135 0 00 0.759 0 0 −0.759 3.040

−13500

00

−0.7593.04

0000

013500

0 00 0

0.7593.04

3.0412.15

][0 1 0 0 0 0

−1 0 0 0 0 00000

0000

1000

00

−10

0100

0001]{

000

−11.5240.253

−3.931}={

−34.15520.6970

34.155−20.697−12.729

} Ndertimi i epiures perfundimtare:

Forcat e brendshme perfundimtare: {F }={F }N+ {F }R

Epiura perfundimtare:

28

29