Расчет стержневых систем на устойчивость и динамическую нагрузку. Ерастов В.В. - 32 стр.

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26
()
()
()
;
I
288,60
943,29447,12500,7136,3847,4415,2
I
1
2736,2
3
2
736,22
2
1
I
1
2764,1
3
2
764,12
2
1
I
1
764,129,1229,129,12764,1764,12
6
6
I
1
446,1
3
2
3446,1
2
1
I
1
182,1764,12182,1
182,12764,1764,12
6
6
I
1
554,1
3
2
3554,1
2
1
I
1
dx
EJ
MM
1
13
32
21
1
6
1
l
0
11
11
=
=+++++=+
++++
++++
+=
=δ
()
()
()
;
I
886,12
052,6292,0351,0710,2739,0742,2
I
1
223,1
3
2
23,12
2
1
I
1
227,0
3
2
27,02
2
1
I
1
114,027,02114,0114,0227,027,02
6
6
I
1
344,1
3
2
344,13
2
1
I
1
426,027,0227,0
27,02426,0426,02
6
6
I
1
656,1
3
2
656,13
2
1
I
1
dx
EJ
MM
1
13
32
21
1
6
1
l
0
22
22
=
=+++++=+
+++++
+++++
+=
=δ
l
()
()
()
.
I
22
462,13905,1768,0915,2378,0573,2
I
1
2736,2
3
2
23,12
2
1
I
1
2764,1
3
2
27,02
2
1
I
1
27,029,1114,0764,129,1114,02764,127,02
6
6
I
1
446,1
3
2
344,13
2
1
I
1
182,127,0426
,0764,1426,0182,1227,0764,12
6
6
I
1
554,1
3
2
656,13
2
1
I
1
dx
EJ
MM
113
32
21
1
6
1
0
21
12
=+++++=+
++++
++++
+=
=δ
l
4.3. Записываем вековое уравнение в развернутом виде, т. е.
()
(
)
0mmδδδλmδmδλ
21
2
1211222111
2
22
=++
или
0
I
68,109826
λ
I
49,1700
λ
2
11
2
=+
.
Отсюда:
          6       l
                    M1 ⋅ M1     1 1             2
δ 11 = ∑ ∫                  dx = ⋅ ⋅ 1,554 ⋅ 3 ⋅ ⋅ 1,554 +
          1       0   EJ        I1 2            3
     1 6                                                                 1 1              2
+      ⋅ ⋅ (2 ⋅ 1,764 ⋅ 1,764 + 2 ⋅ 1,182 ⋅ 1,182 − 2 ⋅ 1,764 ⋅ 1,182 ) + ⋅ ⋅ 1, 446 ⋅ 3 ⋅ ⋅ 1, 446 +
     I1 6                                                                I2 2             3
     1 6                                                              1 1             2
+      ⋅ (2 ⋅ 1,764 ⋅ 1,764 + 2 ⋅ 1, 29 ⋅ 1, 29 − 2 ⋅ 1,29 ⋅ 1,764 ) + ⋅ ⋅ 2 ⋅ 1,764 ⋅ ⋅ 1,764 ⋅ 2 +
     I2 6                                                             I3 2            3
     1 1              2             1
+      ⋅ ⋅ 2 ⋅ 2,736 ⋅ ⋅ 2,736 ⋅ 2 = (2, 415 + 4,847 + 3,136 + 7 ,500 + 12 , 447 + 29 ,943 ) =
     I3 2             3             I1
     60 , 288
=             ;
        I1


              6       ll
                           M2 ⋅M2      1 1               2
δ 22 =    ∑∫  1              EJ
                                  dx =   ⋅ ⋅ 3 ⋅ 1, 656 ⋅ ⋅ 1, 656 +
                                       I1 2              3
                      0

     1 6                                                                           1 1              2
+      ⋅ ⋅ (2 ⋅ 0 , 426 ⋅ 0 , 426 + 2 ⋅ 0 , 27 ⋅ 0 , 27 + 2 ⋅ 0 , 27 ⋅ 0 , 426 ) +   ⋅ ⋅ 3 ⋅ 1,344 ⋅ ⋅ 1,344 +
     I1 6                                                                          I2 2             3
     1 6                                                                      1 1               2
+      ⋅ (2 ⋅ 0 , 27 ⋅ 0 , 27 + 2 ⋅ 0 ,114 ⋅ 0 ,114 + 2 ⋅ 0 , 27 ⋅ 0 ,114 ) +   ⋅ ⋅ 2 ⋅ 0 , 27 ⋅ ⋅ 0 , 27 ⋅ 2 +
     I2 6                                                                     I3 2              3
     1 1              2              1
+      ⋅ ⋅ 2 ⋅ 1, 23 ⋅ ⋅ 1, 23 ⋅ 2 =    (2 ,742 + 0 ,739 + 2 ,710 + 0 ,351 + 0 , 292 + 6 ,052 ) =
     I3 2             3              I1
     12 ,886
=            ;
        I1



                      6    l
                             M1 ⋅ M 2     1 1             2
     δ 12 = ∑ ∫                       dx = ⋅ ⋅ 3 ⋅ 1,656 ⋅ ⋅ 1,554 +
                      1    0   EJ         I1 2            3
         1 6                                                                           1 1             2
     +     ⋅ ⋅ (2 ⋅ 1,764 ⋅ 0,27 − 2 ⋅ 1,182 ⋅ 0,426 + 1,764 ⋅ 0,426 − 0,27 ⋅ 1,182 ) + ⋅ ⋅ 3 ⋅ 1,344 ⋅ ⋅ 1,446 +
         I1 6                                                                          I2 2            3
         1 6                                                                       1 1            2
     +     ⋅ (2 ⋅ 0,27 ⋅ 1,764 − 2 ⋅ 0,114 ⋅ 1,29 + 1,764 ⋅ 0,114 − 1,29 ⋅ 0,27 ) + ⋅ ⋅ 2 ⋅ 0,27 ⋅ ⋅ 1,764 ⋅ 2 +
         I2 6                                                                      I3 2           3
         1 1             2             1                                                    22
     +     ⋅ ⋅ 2 ⋅ 1,23 ⋅ ⋅ 2,736 ⋅ 2 = (2,573 + 0,378 + 2,915 + 0,768 + 1,905 + 13,462 ) =    .
         I3 2            3             I1                                                   I1

         4.3.                  Записываем вековое уравнение в развернутом виде, т. е.
         λ2 − (δ11m1 + δ22m2 ) ⋅ λ + (δ11δ − δ122 ) ⋅ m1m2 = 0
                                               22




         или
                           1700,49    109826,68
         λ2 −                      λ+           = 0.
                              I1         I12

         Отсюда:

26