Matrična analiza konstrukcija

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    PRIMER:

    Sraunati nepoznata pomeranja i sile u presecima sistema datog na slici, a zatim nacrtati

    dijagrame presenih sila M, T i N. Sistem je izloen uticaju optereenja.

    Napomena: promena krivine greda prema kvadratnoj paraboli.

    7 23 10E kN m= , 51 10 1t C =

    Proraunska ema konstrukcije:

    Koordinate sistema:

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    Lokalne koordinate elemenata:

    Koordinate vorova:

    vor x y

    1 12,0 8,5

    2 32,0 8,5

    3 44,0 8,5

    4 0,0 8,5

    5 3,5 0,06 40,5 0,0

    Geometrija tapova:

    ( ) ( )2 2

    k i k il x x y y= + , ( ) ( )sin , cosk i k is y y l c x x l = = = =

    Tabela 1

    tapkraj tapa

    xk-xi yk-yil

    [m]

    c

    (cos)

    s

    (sin)i k

    1 1 4 -12,0 0,0 12,0 -1 0

    2 1 2 20,0 0,0 20,0 1 0

    3 2 3 12,0 0,0 12,0 1 0

    4 1 5 -8,5 -8,5 12,0208 -0,7071 -0,7071

    5 2 6 8,5 -8,5 12,0208 0,7071 -0,7071

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    - MATRICE KRUTOSTI TAPOVA U LOKALNIM KOORDINATAMA:

    Matrica krutosti tapa i-k sa promenljivim poprenim presekom:

    ( ) ( )

    ( ) ( )

    2 2

    2 2

    0 0 0 0

    0 0

    0 0

    0 0 0 0

    0 0

    0 0

    1 1

    1 1

    ik ik

    ik ki ik ik ki ki

    ik ik ik ik

    ik ik

    ik ki ik ik ki ki

    ki ik ki ki

    c c l c l c c l c l

    c l a c l b

    c c l c l c c l c l

    c l b c l a

    + + =

    + +

    k

    Matrica krutosti tapa i-g sa promenljivim poprenim presekom :

    2 2

    2 2

    1 0 0 1 0

    0 0

    0 01 0 0 1 0

    0 0

    ig ig

    ig ig ig

    ig ig ig g

    ig ig

    ig ig ig

    d l d l d l

    d l d d l

    d l d l d l

    =

    k

    Matrica krutosti tapa i-g sa konstantnim poprenim presekom :

    3 2 3

    2 2

    3 2 3

    0 0 0

    0 3 3 0 3

    0 3 3 0 3

    0 0 0

    0 3 3 0 3

    ig ig ig ig

    ig ig ig ig ig ig

    ig ig ig ig ig ig ig

    ig ig ig ig

    ig ig ig ig ig ig

    EF l EF l

    EI l EI l EI l

    EI l EI l EI l

    EF l EF l

    EI l EI l EI l

    =

    k

    - Proraun matrica krutosti:

    Matrica krutosti tapova 1 i 3:

    Aksijalno naprezanje - reenje numerikom integracijom:

    10

    00 0 0

    1 1 1 1 1,3

    l l l

    ig x m m

    mx x x

    dx dx dxEF bE h bE bE h=

    = = = = =

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    x hx =1/hx

    0,0 0,600 1,666667 1 1,66667

    1,2 0,609 1,642036 4 6,56814

    2,4 0,636 1,572327 2 3,14465

    3,6 0,681 1,468429 4 5,87372

    4,8 0,744 1,344086 2 2,68817

    6,0 0,825 1,212121 4 4,84848

    7,2 0,924 1,082251 2 2,16450

    8,4 1,041 0,960615 4 3,84246

    9,6 1,176 0,85034 2 1,70068

    10,8 1,329 0,752445 4 3,00978

    12,0 1,500 0,666667 1 0,66667

    = 36,17393

    10

    0

    1 1 1,2 36,17393 36,173933 0,4 3

    ig m

    m

    EbE E=

    = = =

    3

    11 14 41 44

    127,644220 10

    36,17393ig

    Ek k k k E

    = = = = = =

    Aksijalno naprezanje teoretsko reenje:12

    2

    0 0

    1 1 1 1 1 20 6 6 36,173947arctan ,

    0,4 0,00625 0,6 0,4 3 2 E

    l

    ig

    x

    dx dxbE h E x E

    = = = = +

    Popreno savijanje numerikom integracijom:

    ( )12

    i x

    x xM

    l= =

    ( )

    2 2

    ( ) ( )

    ( )3

    ( )0 0 0

    2

    ( )

    3

    ( )

    12 12

    12,

    3

    x

    l l li x i x

    ig x

    x

    ki x

    ig m m

    m i x

    M Mdx dx dx

    EI bE bEh

    M

    bE h=

    = = = =

    = =

    x hx Mix = (Mix)2/( hx)3

    0,0 0,600 0,00000 0,00000 1 0,000001,2 0,609 0,10000 0,04427 4 0,17710

    2,4 0,636 0,20000 0,15549 2 0,31097

    3,6 0,681 0,30000 0,28497 4 1,13989

    4,8 0,744 0,40000 0,38851 2 0,77702

    6,0 0,825 0,50000 0,44522 4 1,78089

    7,2 0,924 0,60000 0,45634 2 0,91268

    8,4 1,041 0,70000 0,43435 4 1,73742

    9,6 1,176 0,80000 0,39351 2 0,78702

    10,8 1,329 0,90000 0,34507 4 1,38029

    12,0 1,500 1,00000 0,29630 1 0,29630

    = 9,299565

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    12 12 1,2 111,5947849,299565

    3 0,4 3

    k

    ik m

    m ibE E E =

    = = =

    318,960992 10ig

    ig

    d E= =

    Popreno savijanje teoretsko reenje:

    ( )

    2 2 12 2( ) ( )

    3 2 3

    ( )0 0 0

    12 12 ( 12)

    0,4 (0,00625 0,6)x

    l lg x g x

    ig

    x

    M M xdx dx dx

    EI bE h E x = = =

    +

    12 125 6 6 10 1arctan 111,594298

    0,4 81 2 27ig

    E E

    = + =

    318,961031 10ig

    ig

    d E= =

    Sada su, prema Error! Reference source not found., matrice krutosti za tapove 1 i 3,

    3

    1 3 10 E

    = =

    k k

    27,644220 0 0 -27,644220 0

    0 0,062229 0,746749 0 -0,062229

    0 0,746749 8,960992 0 -0,746749

    -27,644220 0 0 27,644220 0

    0 -0,062229 -0,746749 0 0,062229

    Matrica krutosti tapa 2 :

    Aksijalno naprezanje - reenje numerikom integracijom:

    ( )

    ( ) ( )0 0 0

    1 1 1l l l

    ik x

    x x

    dxdx dx

    EF bE h bE = = =

    ( )

    1 1,

    3

    k

    ik m m

    m i xbE h=

    = =

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    x hx =1/hx

    0,0 1,500 0,66667 1 0,66667

    2,0 1,320 0,75758 4 3,03030

    4,0 1,180 0,84746 2 1,69492

    6,0 1,080 0,92593 4 3,70370

    8,0 1,020 0,98039 2 1,96078

    10,0 1,000 1,00000 4 4,00000

    12,0 1,020 0,98039 2 1,96078

    14,0 1,080 0,92593 4 3,70370

    16,0 1,180 0,84746 2 1,69492

    18,0 1,320 0,75758 4 3,03030

    20,0 1,500 0,66667 1 0,66667

    = 26,112746

    10

    0

    1 1 226,112746 43,5212433 0,4 3

    ik m

    mEbE E=

    = = =

    3

    11 14 41 44

    122,977285 10

    43,521243ik

    Ek k k k E

    = = = = = =

    Aksijalno naprezanje teoretsko reenje:20

    2

    0 0

    1 1 1 1 1 2 43,52098820 2 arctan

    0,4 0,005 0,1 1,5 0,4 2 E

    l

    ik

    x

    dx dxbE h E x x E

    = = = = +

    Popreno savijanje numerika integracija Simpsonovim pravilom:

    ( ) 120

    i x

    xM =

    ( )

    20

    k x

    xM =

    ( )

    2 2

    ( ) ( )

    ( )3

    ( )

    12 12

    x

    i x i x

    ik x

    x

    M Mdx dx dx

    EI bE h bE = = =

    ( )

    2

    ( )

    3

    12,

    3x

    ki x

    ik m m

    m i

    M

    bE h=

    = =

    Zbog simetrije tapa je ik = ki

    ( )

    ( ) ( ) ( ) ( )

    ( )3

    ( )

    12 12

    x

    i x k x i x k x

    ik x

    x

    M M M Mdx dx dx

    EI bE h bE = = =

    ( )

    ( ) ( )

    3

    12,3

    x

    ki x k x

    ik ki m m

    m i

    M M

    bE h=

    = = =

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    m x hm Mi Mk m m m

    0 0,0 1,50 1,0 0,0 0,29630 0,00000 1

    1 2,0 1,32 0,9 0,1 0,35218 0,03913 4

    2 4,0 1,18 0,8 0,2 0,38952 0,09738 2

    3 6,0 1,08 0,7 0,3 0,38898 0,16670 4

    4 8,0 1,02 0,6 0,4 0,33924 0,22616 2

    5 10,0 1,00 0,5 0,5 0,25000 0,25000 4

    6 12,0 1,02 0,4 0,6 0,15077 0,22616 2

    7 14,0 1,08 0,3 0,7 0,07144 0,16670 4

    8 16,0 1,18 0,2 0,8 0,02435 0,09738 2

    9 18,0 1,32 0,1 0,9 0,00435 0,03913 4

    10 20,0 1,50 0,0 1,0 0,00000 0,00000 1

    m =6,371847 3,940839

    12 12 2 127, 4369426,371847

    3 0, 4 3

    k

    ik ki m

    m ibE E E =

    = = = =

    12 12 2 78,816784

    3,9408393 0, 4 3

    k

    ik ki m

    m ibE E E =

    = = = =

    ( )2 2 2 22 21 1

    127,436942 78,816784 10.028,088794ik ik

    E E = = =

    23127,436942 12,707999 10

    10.028,088794

    kiik ki

    Ea a E

    E

    = = = =

    2

    378,816784 7,859602 1010.028,088794

    ikik ki

    Eb b E

    E

    = = = =

    3 3 312, 707999 10 7, 859602 10 20, 567601 10ik ki ik ik c c a b E E E = = + = + =

    Popreno savijanje teoretsko reenje:

    ( )

    2 2 20 2( ) ( )

    3 2 3

    ( )0 0 0

    12 12 (1 0,05 )

    0, 4 (0, 005 0,1 1,5)x

    l li x i x

    ik

    x

    M M xdx dx dx

    EI bE h E x x

    = = =

    +

    12 25 2 2 55 127, 435185arctan0, 4 8 2 36

    ikE E

    = + =

    ( )

    8( ) ( ) ( ) ( )

    3 2 3

    ( )0 0 0

    12 12 (1 0,005 )(0, 005 )

    0, 3 (0, 005 0,1 1, 5)x

    l li x k x i x k x

    ik

    x

    M M M M x xdx dx dx

    EI bE h E x x

    = = =

    +

    12 5 2 2 25 78,820370arctan

    0,3 8 2 12ik

    E E

    = + =

    312,709108 10ik kia a E= =

    37,860754 10ik kib b E

    = = 320,569861 10ikc E

    =

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    Sada je, prema Error! Reference source not found., matrica krutosti za tap 2,

    2

    =

    k

    022,977285 0 -22,977285 0 0

    0 0,102838 1,028380 0 -0,102838 1,028380

    0 1,028380 12,707999 0 -1,028380 7,859602

    -22,977285 0 0 22,977285 0 0

    0 -0,102838 -1,028380 0 0,102838 -1,0283800 1,028380 7,859602 0 -1,028380 12,707999

    E

    -310

    Matrica krutosti tapova 4 i 5: 12,0208l m= , 20,24F m= , 40,0072I m=

    4 5 E

    = =

    k k-3

    19,965368 0 0 -19,965368 0

    0 0,012435 0,149481 0 -0,012435

    100 0,149481 1,796883 0 -0,149481

    -19,965368 0 0 19,965368 0

    0 -0,012435 -0,149481 0 0,012435

    - MATRICE KRUTOSTI TAPOVA U GLOBALNIM KOORDINATAMA:

    - Globalne koordinate tapova

    - Matrice transformacije

    Prema Error! Reference source not found.a, Error! Reference source not found.itabeli 1, imamo:

    1

    =

    T

    -1 0 0 0 0

    0 -1 0 0 0

    0 0 1 0 0

    0 0 0 -1 0

    0 0 0 0 -1

    , 2

    =

    T

    1 0 0 0 0 0

    0 1 0 0 0 0

    0 0 1 0 0 0

    0 0 0 1 0 0

    0 0 0 0 1 0

    0 0 0 0 0 1

    , 3

    =

    T

    1 0 0 0 0

    0 1 0 0 0

    0 0 1 0 0

    0 0 0 1 0

    0 0 0 0 1

    ,

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    4

    =

    T

    -0,70711 -0,70711 0 0 0

    0,70711 -0,70711 0 0 0

    0 0 1 0 0

    0 0 0 -0,70711 -0,70711

    0 0 0 0,70711 -0,70711

    5

    =

    T

    0,70711 -0,70711 0 0 0

    0,70711 0,70711 0 0 0

    0 0 1 0 0

    0 0 0 0,70711 -0,70711

    0 0 0 0,70711 0,70711

    - Matrice krutosti u globalnom koordinatnom sistemu

    * 3

    1 1 1 1 10T E

    = =

    k T k T

    2 5 7 1 10

    2

    5

    7

    1

    10

    27,644220 0 0 -27,644220 0

    0 0,062229 -0,746749 0 -0,062229

    0 -0,746749 8,960992 0 0,746749

    -27,644220 0 0 27,644220 0

    0 -0,062229 0,746749 0 0,062229

    * 3

    2 2 2 2 10T E

    = =k T k T

    2 5 7 3 6 8

    22,977285 0 0 -22,977285 0 0

    0 0,102838 1, 028380 0 -0,102838 1, 028380

    0 1,028380 12,707999 0 -1,028380 7,859602

    -22,977285 0 0 22,977285 0 0

    0 -0,102838 -1,028380 0 0,102838 -1,0283800 1,028380 7,859602 0 -1,02 8

    2

    5

    7

    3

    6

    8380 12,707999

    * 3

    3 3 3 3 10T E

    = =

    k T k T

    3 6 8 4 9

    3

    6

    8

    4

    9

    27,644220 0 0 -27,644220 0

    0 0,062229 0,746749 0 -0,062229

    0 0,746749 8,960992 0 -0,746749

    -27,644220 0 0 27,644220 0

    0 -0,062229 -0,746749 0 0,062229

    * 3

    4 4 4 4 10T

    E= =k T k T

    2 5 7 11 12

    9,988902 9,976466 0,105699 -9,988902 -9,9764669,976466 9,988902 -0,105699 -9,976466 -9,988902

    0,105699 -0,105699 1,796883 -0,105699 0,105699

    -9,988902 -9,976466 -0,105699 9,988902 9,976466

    -9,976

    2

    5

    7

    11

    12466 -9,988902 0,105699 9,976466 9,988902

    * 3

    5 5 5 5 10T E

    = =k T k T

    3 6 8 13 14

    9,988902 -9,976466 0,105699 -9,988902 9,976466

    -9,976466 9,988902 0,105699 9,976466 -9,988902

    0,105699 0,105699 1,796883 -0,105699 -0,105699

    -9,988902 9,976466 -0,105699 9,988902 -9,976466

    9,97646

    3

    6

    8

    13

    146 -9,988902 -0,105699 -9,976466 9,988902

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    - MATRICA KRUTOSTI SISTEMA:* *

    *

    * *

    ss so

    os oo

    =

    K KK

    K K

    *

    ss E=K

    1 2 3 4 5 6 7 8

    -3

    27,64422 -27,644220 0 0 0 0 0 0-27,64422 60,610407 -22,977285 0 9,976466 0 0,105699 0

    0 -22,977285 60, 610407 -27,64422 0 -9,976466 0 0,105699

    0 0 -27,644220 27,64422 0 0 0 010

    0 9,976466 0 0 10,153969 -0,102838 0,175932 1,

    6

    7

    8

    1

    2

    3

    4

    5028380

    0 0 -9, 976466 0 -0,102838 10,153969 -1,028380 -0,175932

    0 0,105699 0 0 0,175932 -1,028380 23,465874 7,859602

    0 0 0,105699 0 1,028380 -0,175932 7,859602 23,465874

    *

    0 0 0 0 00 0 0

    0 0 0

    0 0 0 0 0

    0 0

    0 0 0

    0 0

    0 0 0

    so E

    =

    K

    9 10 11 12 13 14

    -3

    00 -9,988902 -9,976466

    0 -9,988902 9,976466

    010

    0 -0,062229 -9,976466 -9,988902

    -0,062229 9,976466 -9,988902

    0 0,746749 -0,105699 0,105699

    -0,746749 -0,105699 -0,105699

    6

    7

    8

    1

    2

    3

    4

    5

    *

    os E

    =

    K

    1 2 3 4 5 6 7 8

    -3

    0 0 0 0 0 -0,062229 0 -0,746749

    0 0 0 0 -0,062229 0 0,746749 0

    0 -9,988902 0 0 -9,976466 0 -0,105699 010

    0 -9,976466 0 0 -9,988902 0 0,105699 0

    0 0 -9,988902 0 0 9,976466 0 -0,105699

    0 0 9,976466 0 0 -9,988902 0 -0,105699

    9

    10

    11

    12

    13

    14

    *

    oo E

    =

    K

    9 10 11 12 13 14

    9

    10

    11

    12

    13

    14

    -3

    0,062229 0 0 0 0 0

    0 0,062229 0 0 0 0

    0 0 9,988902 9,976466 0 0

    10 0 0 9,976466 9,988902 0 0

    0 0 0 0 9,988902 -9,976466

    0 0 0 0 -9,976466 9,988902

    * 1

    ssE

    =K3

    2,057554 2,021380 1,999633 1,999633 -1,974472 1,952773 0,068042 0,069374

    2,021380 2,021380 1,999633 1,999633 -1,974472 1,952773 0,068042 0,069374

    1,999633 1,999633 2,021380 2,021380 -1,952773 1,974472 0,069374 0,

    10

    068042

    1,999633 1,999633 2,021380 2,057554 -1,952773 1,974472 0,069374 0,068042

    -1,974472 -1,974472 -1,952773 -1,952773 2,027603 -1,905999 -0,065594 -0,072383

    1,952773 1,952773 1,974472 1,974472 -1,905999 2,027603 0,072383 0

    ,065594

    0,068042 0,068042 0,069374 0,069374 -0,065594 0,072383 0,050610 -0,013846

    0,069374 0,069374 0,068042 0,068042 -0,072383 0,065594 -0,013846 0,050610

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    A) UTICAJ ZADATOG OPTEREENJA:

    - Vektori ekvivalentnog optereenja tapova:

    Proraun momenata punog ukljetenja

    tap 1:

    ( )

    2

    ( )

    1( )

    ( )

    3,

    160 5

    ,12

    112

    2

    x

    x

    o x

    xh

    xM

    M p x x

    = +

    =

    =

    2 212 12 101( ) 1( )

    11 30( ) ( )0 0

    2

    1

    3

    12 12,

    3

    1,2

    x x

    m

    mx x

    mm

    m

    M ME dx dx

    I b bh

    Mm

    h

    =

    = = =

    = =

    12 12 101( ) ( ) 1( ) ( )

    10 30( ) ( )0 0

    1

    3

    12 12

    3

    x o x x o x

    m

    mx x

    m om

    m

    m

    M M M ME dx dx p

    I b bh

    M M

    h

    =

    = = =

    =

    m x hm M1 Mo m m m

    0 0,00 0,600 0,0 0,00 0,00000 0,00000 1

    1 1,20 0,609 0,1 6,48 0,04427 2,86895 42 2,40 0,636 0,2 11,52 0,15549 8,95594 2

    3 3,60 0,681 0,3 15,12 0,28497 14,36256 4

    4 4,80 0,744 0,4 17,28 0,38851 16,78359 2

    5 6,00 0,825 0,5 18,00 0,44522 16,02805 4

    6 7,20 0,924 0,6 17,28 0,45634 13,14253 2

    7 8,40 1,041 0,7 15,12 0,43435 9,38205 4

    8 9,60 1,176 0,8 11,52 0,39351 5,66657 2

    9 10,80 1,329 0,9 6,48 0,34507 2,48452 4

    10 12,00 1,500 1,0 0,00 0,29630 0,00000 1

    m = 9,299565 269,601769

    10

    11

    0

    12 12 1,29,299565 111,594784

    3 0, 4 3m

    m

    Eb

    =

    = = =

    10

    10

    0

    12 12 1,0269,601769 3.235,221228

    3 0,4 3m

    m

    E p pb

    =

    = = =

    2

    1 14

    3.235, 22122828,990793 1,6106 463,852681

    111,594784 8

    p plX M p kNm= = = = =

    2 2

    4 14

    1 1 16 12463,852681 57,34561

    2 12 2

    plV M kN

    l

    = + = =

    1 4 16 12 57,34561 134,65439V pl V kN = = =

  • 8/10/2019 Matrina analiza konstrukcija

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    1

    kN

    kN

    kNm

    kN

    kN

    =

    Q

    0

    134,65439

    463,85268

    0

    57,34561

    *

    1 1 1

    T

    = =

    Q T Q

    2

    5

    7

    1

    10

    0

    -134,65439

    463,85268

    0

    -57,34561

    tap 2:

    2

    ( )

    1( )

    2( )

    3, 2,0

    200 10 2

    1 ,20

    20

    x

    x

    x

    x xh m

    x

    M

    xM

    = + =

    =

    =

    220 101( )

    11 30( )0

    12 12

    3

    x

    m

    mx

    ME dx

    b bh

    =

    = =

    20 101( ) 2( )

    12 30( )0

    12 12

    3

    x x

    m

    mx

    M ME dx

    b bh

    =

    = =

    220 102( )

    22 3 0( )0

    12 12

    3

    x

    mmx

    ME dx

    b bh

    =

    = =

    20 101( ) ( )

    10 30( )0

    12 12

    3

    x o x IV

    m

    mx

    M ME dx

    b bh

    =

    = =

    20 102( ) ( )

    20 30( )0

    12 12

    3

    x o x V

    m

    mx

    M ME dx

    b bh

    =

    = =

    m x hm M1 M2 Mo m m = m IV V

    m m = m

    0 0,0 1,50 0,0 1,0 0 0,00000 0,00000 0,00000 1

    1 2,0 1,32 0,1 0,9 40 0,00435 0,03913 1,73915 4

    2 4,0 1,18 0,2 0,8 80 0,02435 0,09738 9,73809 2

    3 6,0 1,08 0,3 0,7 120 0,07144 0,16670 28,57796 4

    4 8,0 1,02 0,4 0,6 160 0,15077 0,22616 60,30863 2

    5 10,0 1,00 0,5 0,5 200 0,25000 0,25000 100,00000 4

    6 12,0 1,02 0,6 0,4 160 0,33924 0,22616 90,46294 2

    7 14,0 1,08 0,7 0,3 120 0,38898 0,16670 66,68191 4

    8 16,0 1,18 0,8 0,2 80 0,38952 0,09738 38,95238 2

    9 18,0 1,32 0,9 0,1 40 0,35218 0,03913 15,65239 4

    10 20,0 1,50 1,0 0,0 0 0,29630 0,00000 0,00000 1

    m = 6,371847 3,940839 1249,529748

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    10

    11 22

    0

    12 12 26,371847 127,436942

    3 0, 4 3m

    m

    E Eb

    =

    = = = =

    10

    12

    0

    12 12 23,940839 78,816784

    3 0, 4 3m

    m

    Eb

    =

    = = =

    10

    10 20

    0

    12 12 21249,529748 24.990,594953

    3 0, 4 3

    IV

    m

    m

    E Eb

    =

    = = = =

    1

    2

    127,436942 78,816784 24.990,5949530

    78,816784 127,436942 24.990,594953

    X

    X

    + =

    1 12 121,164332X M= =

    2 21 121,164332X M= =

    2

    kN

    kN

    kNm

    kN

    kN

    kNm

    =

    Q

    0

    -20,00000

    -121,16433

    0

    -20,00000

    121,16433

    *

    2 2 2

    8

    T

    = =

    Q T Q

    2

    5

    7

    3

    6

    0

    -20,00000

    -121,16433

    0

    -20,00000

    121,16433

    - Vektor ekvivalentnog optereenja sistema tapova:

    * * *

    *

    * * *

    0

    s s s

    o o

    = = +

    S Q RS

    S Q R

    Ovde je: - * 0s =R vektor sila zadatih u vorovima, u pravcima slobodnih koordinata sistema,

    - *oR vektor sila u pravcima vezanih koordinata, odnosno reakcije oslonaca i ukljetenja

    { }* * *1 2 3 4 5 6 7 8

    0 0 0 0 154,65439 20 342,68835 121,16433 T

    s s s= + = S Q R

    { }*10 11 12

    0 57,34561 0 0 0 0 T

    o = Q

    9 13 14

    - Odreivanje pomeranja u pravcima slobodnih koordinata sistema:

    Polazei od sistema uslovnih jednaina

    * * *=K q S ,

    odnosno* * * *

    * * * *

    ss so s s

    os oo o o

    =

    K K q S

    K K q S,

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    dobijamo* * * * *

    ss s so o s+ =K q K q S ,

    odakle, uz uslov da su pomeranja u pravcima vezanih stepeni slobode jednaka nuli, tj. * 0o =q , imamo

    * * *

    ss s s=K q S .

    Sada je vektor pomeranja u pravcima slobodnih koordinata sistema

    * * 1 *

    s ss s

    =q K S ,

    odnosno,

    *

    sE

    =q3

    2,057554 2,021380 1,999633 1,999633 -1,974472 1,952773 0,068042 0,069374

    2,021380 2,021380 1,999633 1,999633 -1,974472 1,952773 0,068042 0,069374

    1,999633 1,999633 2,021380 2,021380 -1,952773 1,974472 0,069374 0,068

    10

    042

    1,999633 1,999633 2,021380 2,057554 -1,952773 1,974472 0,069374 0,068042

    -1,974472 -1,974472 -1,952773 -1,952773 2,027603 -1,905999 -0,065594 -0,072383

    1,952773 1,952773 1,974472 1,974472 -1,905999 2,027603 0,072383 0,06

    0

    0

    0

    0

    154,65439

    20,00000

    342,68835

    121,16

    5594

    0,068042 0,068042 0,069374 0,069374 -0,065594 0,072383 0,050610 -0,013846

    0,069374 0,069374 0,068042 0,068042 -0,072383 0,065594 -0,013846 0,050610 433

    3* 10s

    E

    = =

    q

    1

    2

    3

    4

    298, 02802 9, 934 mm

    298, 02802 9, 934 mm

    294,53343 9,818 mm

    294,53343 9,818 mm

    -306,70618 -10,224 mm

    286,97138 9,566 mm

    24,36251 0,00081 rad

    11,26959 0,00038 rad

    6

    7

    8

    5

    - Odreivanje reakcija oslonaca i oslonakih ukljetenja *oR :

    Iz sistema uslovnih jednaina imamo, * * * * * * *0os s oo o o o+ = = +K q K q S Q R ,

    Odakle je, za * 0o =q ,* * * *

    o os s o= R K q Q

    *

    o

    E

    =

    R 3

    0 0 0 0 0 -0,062229 0 -0,746749

    0 0 0 0 -0,062229 0 0,746749 0

    0 -9,988902 0 0 -9,976466 0 -0,105699 0

    0 -9,976466 0 0 -9,988902 0 0,105699 010

    0 0 -9,988902 0 0 9,976466 0 -0,105699

    0 0 9,976466 0 0 -9,988902 0 -0,105699

    310

    E

    298,02802

    298,02802 0

    294,53343 -57,346

    294,53343 0

    -306, 70618 0

    286,97138 0

    24,36251 0

    11,26959

    *

    o

    = = =

    = =

    =

    =

    R

    9

    10

    11

    12

    13

    14

    3

    4

    5

    5

    6

    6

    V -26,2735 kN

    V 94,6243 kN

    H 80,2963 kN

    V 92,9664 kN

    H -80,2963 kN

    V 70,6828 kN

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    - Vektori generalisanih pomeranja tapova u lokalnom koordinatnom sistemu:

    3 3*

    1 1 1

    10 10

    E E

    = = =

    q T q

    -1 0 0 0 0 298,02802 -298,02802

    0 -1 0 0 0 -306,70618 306,70618

    0 0 1 0 0 24,36251 24,36251

    0 0 0 -1 0 298,02802 -298,028020 0 0 0 -1 0 0

    2

    5

    7

    1

    10

    3 3*

    2 2 2

    8

    10 10

    E E

    = = =

    q T q

    1 0 0 0 0 0 298,02802 298,02802

    0 1 0 0 0 0 -306,70618 -306,70618

    0 0 1 0 0 0 24,36251 24,36251

    0 0 0 1 0 0 294,53343 294,53343

    0 0 0 0 1 0 286,97138 286,9713

    0 0 0 0 0 1 11,26959

    2

    5

    7

    3

    6

    8

    11,26959

    3 3*

    3 3 3

    10 10

    E E

    = = =

    q T q

    1 0 0 0 0 294,53343 294,53343

    0 1 0 0 0 286,97138 286,971380 0 1 0 0 11,26959 11,26959

    0 0 0 1 0 294,53343 294,53343

    0 0 0 0 1 0 0

    3

    6

    8

    4

    9

    3 3*

    4 4 4

    10 10

    E E

    = = =

    q T q

    -0,7071 -0,7071 0 0 0 298,02802 6,13639

    0,7071 -0,7071 0 0 0 -306,70618 427,61165

    0 0 1 0 0 24,36251 24,36251

    0 0 0 -0,7071 -0,7071 0 0

    0 0 0 0,7071 -0,7071 0 0

    2

    5

    7

    11

    12

    3 3*

    5 5 5

    10 10

    E E

    = = =

    q T q

    0,7071 -0,7071 0 0 0 294,53343 5,34717

    0,7071 0,7071 0 0 0 286,97138 411,18599

    0 0 1 0 0 11,26959 11,26959

    0 0 0 0,7071 -0,7071 0 0

    0 0 0 0,7071 0,7071 0 0

    3

    6

    8

    13

    14

    - Vektori generalisanih sila na krajevima tapova u lokalnom koordinatnom sistemu:

    Za tap j , j j j j= R k q Q

    3

    1 3

    10

    10

    EI

    EI

    =

    R

    27,6442 0 0 -27,6442 0 -298,02802 00 0, 0622 0, 7467 0 -0, 0622 306, 70618 134,6544

    0 0,7467 8,9610 0 -0,7467 24,36251 463,8

    -27,6442 0 0 27,6442 0 -298,02802

    0 -0,0622 -0,7467 0 0,0622 0

    =

    0-97,376

    527 -16,508

    0 0

    57,3456 -94,624

    3

    2 3

    10

    10

    EI

    EI

    =

    R

    022,9773 0 -22,9773 0 0 298,02802

    0 0,1028 1,0284 0 -0,1028 1,0284 -306,706

    0 1, 0284 12, 7080 0 -1, 0284 7, 8596

    -22,9773 0 0 22,9773 0 0

    0 -0,1028 -1, 0284 0 0,1028 -1, 0284

    0 1, 0284 7, 8596 0 -1, 0284 12, 7080

    =

    0 80,296

    18 -20,0000 -4, 409

    24,36251 -121,1643 -91,189

    294,53343 0 -80,296

    286,97138 -20,0000 44,409

    11,26959 121,1643 -396,997

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    3

    3 3

    10

    10

    EI

    EI

    = =

    R

    27,6442 0 0 -27,6442 0 294,53343 0

    0 0,0622 0,746749 0 -0,0622 286,97138 26,274

    0 0,7467 8,9610 0 -0,7467 11,26959 315,282

    -27,6442 0 0 27,6442 0 294,53343

    0 -0,0622 -0,7467 0 0,0622 0

    0

    -26,274

    3

    4 3

    10

    10

    EI

    EI

    = =

    R

    19,9654 0 0 -19,9654 0 6,13639 122,515

    0 0,0124 0,1495 0 -0,0124 427,61165 8,959

    0 0,1495 1,7969 0 -0,1495 24,36251 107,696

    -19,9654 0 0 19,9654 0 0 -122,51

    0 -0,0124 -0,1495 0 0,0124 0

    5

    -8,959

    3

    5 3

    10

    10

    EI

    EI

    = =

    R

    19,9654 0 0 -19,9654 0 5,34717 106,758

    0 0,0124 0,1495 0 -0,0124 411,18599 6,798

    0 0,1495 1,7969 0 -0,1495 11,26959 81,715

    -19,9654 0 0 19,9654 0 0 -106,758

    0 -0,0124 -0,1495 0 0,0124 0

    -6,798

    Sile na krajevima tapova:

    - Pomeranja vorova, reakcije oslonaca:

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    - Dijagrami presenih sila: