Otpornostvezbe-1b

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  • | 10

    Zadatak 5.

    Za datu sliku, R=4cm, odrediti moment inercije i glavne momente inercije kao

    elipse inercije.

    1 =2 2

    2= 2 2 = 2 42 = 32 = 100.53 2 1 0,

    4 2

    3= 0; 3.395

    2 =2

    2= 25.133 2, 2 ;

    4

    3= 4;1.697

    3 =2

    2= 25.133 2, 2 ;

    4

    3= 4; 1.697

    = = 1 2 + 3 = 22

    2

    2+

    2

    2= 2 2 = 100.53 2

    = = 1 1 2 2 + 3 3 = 22 4 2

    3

    2

    2

    4 2

    3+

    2

    2

    4 2

    3

    =16 3

    3

    4 3

    3= 4 3 = 256 3

    = = 1 1 2 2 + 3 3 = 22 0

    2

    2+

    2

    2= 3

    = 3 = 256 = 804.247 3

    = =3

    2 2=

    2=

    4

    2= 2 2

    = =4 3

    2 2=

    256

    100.53= 2.546 2

    1 =4 2

    3=

    8

    3=

    8 4

    3= 3.395

    1 =2 4

    8=

    84

    8=

    4096

    8= 512 = 1608.49 4

    1 = 24

    8

    8

    9= 84

    8

    8

    9= 449.74 4

    =2 4

    8=

    84

    8=

    4096

    8= 512 = 1608.49 4

    1 1 = 0

  • | 11

    1 =4

    3=

    4 4

    3= 1.697

    2 =4

    8=

    44

    8=

    256

    8= 100.53 4

    2 =4

    8

    8

    9= 44

    8

    8

    9= 28.188 4

    =4

    8=

    44

    8=

    256

    8= 100.53 4

    2 2 = 0

    1 =4

    3=

    4 4

    3= 1.697

    3 =4

    8=

    44

    8=

    256

    8= 100.53 4

    3 =4

    8

    8

    9= 44

    8

    8

    9= 28.188 4

    =4

    8=

    44

    8=

    256

    8= 100.53 4

    2 2 = 0

    = 1 2 + 3 = 512256

    8+

    256

    8= 1 = 512 = 1608.49

    4

    = 1 2 +2

    1 + 3 +2

    3 = 1 = 512 = 1608.494

    = 1 1 2 2 +4

    3 1+ 3 3 +

    4

    3 3

    = 0 0 +4

    3 1+ 0 +

    4

    3 3= 0

    c odnosno od y ose za xc

    = 2 = 1608.49 2.546 2 100.53 = 956.840 4

    = 2 = 1608.49 2 2 100.53 = 1206.37 4

    = = 0 2 2.546 100.53 = 511.904 4

    tan 2 =2

    =2 511.904

    956.84 1206.37= 4.1029455 2 = 76.3025 = 38.1512

    12 =1

    2+

    1

    2

    2+ 4 2 =

    12 =1

    2956.84 + 1206.37

    1

    2956.84 1206.37 2 + 4 511.9042 =

    12 = 1081.605 526.888

    1 = 1608.494 4

    2 = 554.716 4

    1 =1 =

    1608.494

    100.531= 15.999 = 4

  • | 12

    2 =2 =

    554.716

    100.531= 5.5225 = 2.35

  • | 13

    Zadatak 6.

    Za datu sliku, R=2cm, odrediti moment inercije i glavne momente inercije kao i

    elipse inercije.

    1 =5 4

    2= 10 2 = 40 2,

    15

    3;

    4

    3= 3.333;2.667

    2 =2 2

    2= 2 2 = 8 2,

    22

    3;

    5

    3= 1.333;3.333

    3 =2 2

    2= 2 2 = 25.133 2, 3

    4 2

    3; 2 = 1.697;4

    4 =2

    2= 6.283 2, 4

    4

    3; 2 = 0.848; 4

    = = 1 2 + 3 4 = 40 8 + 25.137 6.283 = 50.854 2

    = = 1 1 2 2+ 3 3 4 4

    =40 3.333 8 1.333+25.133 1.697 6.283 0.848

    50.854=

    90.662

    50.854= 1.678

    = = 1 1 2 2+ 3 3 4 4 =

    40 2.666 8 3.333+25.137 4 6.283 4

    50.854=

    155.392

    50.854= 3.056

    Za trouglove 1 i 2 sopstveni momenti inercije

    =3

    36

    =3

    36

    = 2 2

    72

    Za trougao 1 sopstveni momenti inercije

    = 5 = 5 2 = 10

    = 4 = 4 2 = 8

    1 =3

    36=

    5 4 3

    36=

    320 4

    36=

    10 83

    36= 142.222 4

  • | 14

    1 =3

    36=

    5 3 4

    36=

    103 8

    36= 222.222 4

    1 1 =2 2

    72=

    5 2 4 2

    72=

    102 82

    72= 88.888 4

    Za trougao 2 sopstveni momenti inercije

    = 2 = 2 2 = 4

    = 2 = 2 2 = 4

    2 =3

    36=

    2 2 3

    36=

    4 43

    36= 7.111 4

    2 =3

    36=

    2 3 2

    36=

    43 4

    36= 7.111 4

    2 2 =2 2

    72=

    2 2 2 2

    72=

    42 42

    72= 3.555 4

    Za polovine krugova 3 i 4 sopstveni momenti inercije

    1 =4

    3

    4 =4

    8= 6.283 4

    4 =4

    8

    8

    9= 0.1098 4

    4 4 = 0

    Za polovinu kruga 3 3 = 2 = 2 2 = 4

    1 =4 2

    3=

    4 4

    3= 1.697

    3 =34

    8=

    2 4

    8=

    44

    8= 100.531 4

    3 = 34

    8

    8

    9= 2 4

    8

    8

    9= 0.1098 2 4 = 28.108 4

    3 3 = 0

    Za pololovinu kruga 4 4 = = 2

    1 =4

    3=

    4 2

    3= 0.848

    1 =4

    3=

    4 2

    3= 0.848

    4 =44

    8=

    4

    8=

    24

    8= 6.2831 4

    4 = 44

    8

    8

    9= 4

    8

    8

    9= 0.1098 4 = 1.7568 4

    3 3 = 0

    = 1 + 12

    1 2 + 22

    2 + 3 + 32

    3 4 + 42

    4

    =320

    12+

    16

    910

    16

    36

    25

    92 + 16

    8

    8

    9+ 2 4

    8

    8

    9

    4

    24 =

    = 142.22 + 2.6662 40 7.11 + 3.3332 8 + 100.53 + 42 25.137 6.283 + 42 6.283 =

    = 726.507 4

    = 1 + 12

    1 2 + 22

    2 + 3 + 32

    3 4 + 42

    4

    = 222.22 + 3.3332 40 7.111 + 1.3332 8 + 28.108 + 1.6972 25.137

    1.7568 + 0.8482 6.283 =

    = 739.581 4

    = 1 1 + 1 1 1 2 2 + 2 2 2 + 3 3 + 3 3 3 4 4 + 4 4 4

  • | 15

    = 88.888 + 3.333 2.67 40 3.556 + 1.333 3.333 8 + 0

    + 1.697 4 25.133 0 + 0.848 4 6.283 = 255.13 4

    = 12 = 726.507 3.056 2 50.854 = 251.574 4

    = 12 = 739.581 1.678 2 50.854 = 596.392 4

    = = 78.222 1.678 3.056 50.854 = 182.555 4

    2 =2

    =2 182.555

    251.574 596.392= 1.05885 2 = 46.637 = 23.318

    12 =1

    2+

    1

    2

    2+ 4 2

    12 =1

    2251.574 + 596.392

    1

    2251.574 596.392 2 + 4 182.5552

    12 = 423.983 251.099

    1 = 423.983 + 251.099 = 675.082 cm4

    2 = 423.983 251.099 = 172.884 cm4

    1 =1 =

    675.082

    50.854= 3.643

    2 =2 =

    172.884

    50.854= 1.843

  • | 16

    Zadatak 7.

    Za datu sliku, a=1cm, odrediti moment inercije i glavne

    momente inercije kao

    1 = 22 = 4 2 = 4 2 1 , = 1; 1

    2 =2

    4= 0.785 2, 2

    4

    3;

    4

    3= 0.424; 0.424

    3 = 2= 0.5 2, 2

    5

    3;

    5

    3= 1.666; 1.666

    = = 1 2 3 =

    = = 4 22

    4

    2

    2= 2.715 2 = 2.715 2

    = = 1 1 2 2 3 3

    = 422

    4

    4

    3

    2

    2

    5

    3=

    17

    63 = 2.8333 3

    = = 1 1 2 2 3 3

    = 422

    4

    4

    3

    2

    2

    5

    3=

    17

    63 = 2.8333 3

    = =2.8333 3

    2.715 2= 1.0436 1.0436

    = =2.8333 3

    2.715 2= 1.0436 1.0436

    Iz tablica

    Za kvadrat

    1 = 1 =2 4

    12=

    16 4

    12= 1.3333 4 = 1.3333 4

    1 1 = 0

    2 = 2 =4

    16

    4

    9= 0.054878 4 = 0.05878 4

    2 2 =4 4

    9

    1

    8= 0.016471 4 = 0.016471 4

    3 =3

    36=

    3

    36=

    4

    36= 0.027777 4 = 0.027777 2

    3 =3

    36=

    3

    36=

    4

    36= 0.027777 4 = 0.027777 2

    3 3 =2 2

    72=

    2 2

    72=

    4

    72= 0.013888 4 = 0.013888 2

    i

    inercije

    1 = 42 = 4 2 1 1.0436 ; 1.0436 = 0.0436 ; 0.0436 = 0.0436; 0.0436

    2 =2

    4= 0.785 2, 2 1.0436 +

    4

    3; 1.0436 +

    4

    3= 0.6192; 0.6192

  • | 17

    3 = 2= 0.5 2, 2

    5

    31.0436 ;

    5

    31.0436 = 0.623 ; 0.623 = 0.623; 0.623

    = 1 + 12

    1 2 + 22

    2 3 + 32

    3 =

    = 1.333 + 0.04362 4 0.0549 + 0.61922 0.785 0.02777 + 0.19406 4 =

    = 0.76308 4 = 0.76308 4

    = 1 + 12

    1 2 + 22

    2 3 + 32

    3 =

    = 1.333 + 0.04362 4 0.0549 + 0.61922 0.785 0.02777 + 0.6232 0.5 4 =

    = 0.76308 4 = 0.76308 4

    = 1 1 + 1 1 1 2 2 + 2 2 3 3 31 + 3 3 3 =

    = 0 + 0.04362 4 0.0165 + 0.61922 0.785 0.01389 + 0.6232 0.5 4 =

    = 0.3241 4 = 0.3241 4 o, a aksijalni momenti za ose i su jednaki

    jedna glavna os je osa simetrije a druga glavna osa je upravna

    na nju.

    2 =2

    =2 0.3241

    0.76308 0.76308

    12 =1

    2+

    1

    2

    2+ 4 2

    1 = 0.763084 + 0.3241 4 = 1.0872 4 = 1.0872 4

    12 = 0.763084 0.3241 4 = 0.43898 4 = 0.43898 4

    1 =1 =

    1.0872 4

    2.715 2= 0.40044 2 = 0.6328 = 0.6328

    2 =2 =

    0.43898 4

    2.715 2= 0.1616869 2 = 0.4021 = 0.4021