Otpornostvezbe-1b
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Transcript of 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
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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
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| 12
2 =2 =
554.716
100.531= 5.5225 = 2.35
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| 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
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| 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
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| 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
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| 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
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| 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
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