Расчет стержневых систем на устойчивость методом перемещений. Себешев В.Г. - 39 стр.

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38
Рис. 3.3
F
F
F
1
,
5
F
1
,
5
F
Z
1
=
1
A B
C
D
L
F
F
F
1
,
5
F
1
,
5
F
A B
C
D
L
F
F
F
1
,
5
F
1
,
5
F
A
B
C
D
L
Z
2
=
1
r
11
r
21
r
31
r
C,
1
=
0
k
=
1
r
12
r
C,
2
=
0
r
32
r
C,3
=
c
0
r
13
r
2
r
Z
3
= 1
3,
0
AA
rR
+
3,
0
BB
rR
+
2
3
2
3
33,
l
ir
B
ν
=
3,A
r
Δ
1,3
=
1,25
Δ
2,3
=
0,75
1
k
=
2
k
=
3
A,B
=
0 D,L
C
1
0,75
1,25
O
1,5i
0
ϕ
4
(
0,75
ν
)
2i
2
ϕ
3
(
ν
2
)= 4i
0
ϕ
3
(0,75
ν
)
4i
2
ϕ
2
(
ν
2
)
= 8i
0
ϕ
2
(
0,75
ν
)
3i
1
ϕ
1
(
ν
1
)
= 2,4i
0
ϕ
1
(
0,8839
ν
)
2i
2
ϕ
3
(
ν
2
)= 4i
0
ϕ
3
(0,75
ν
)
4
i
2
ϕ
2
(
ν
2
)
= 8i
0
ϕ
2
(
0,75
ν
)
i
4
ν
4
tg
ν
4
=
=
0,707i
0
ν
tg
(0,3535
ν
)
=Δ )(
6
243,2
2
2
νϕ
l
i
1,5i
0
ϕ
4
(0,75
ν
)
=Δ )(
3
113,1
1
1
νϕ
l
i
0,6i
0
ϕ
1
(0,8839
ν
)
М
1
М
2
М
3
r
22
                                                   F
        Z1= 1                                                                                      2i2ϕ3(ν2)= 4i0ϕ3(0,75ν)
                  r11                                              r21
                          F                    F L                        3i1ϕ1(ν1)= 2,4i0ϕ1(0,8839ν)
  rC,1= 0                 1,5F         1,5F
                                               D
                                                                    r31
                  C                                                         М1
                              k=1                                                         4i2ϕ2(ν2)= 8i0ϕ2(0,75ν)
   A                                           B


                                                               F
                                                                                                   4i2ϕ2(ν2)= 8i0ϕ2(0,75ν)
           r12                                             L
                          F                                         r22
                              Z2= 1            F                          2i2ϕ3(ν2)= 4i0ϕ3(0,75ν)
 rC,2= 0
                                                                    r32
                    C 1,5F 1,5F D
                                                                                                       i4ν4 tgν4 =
                                                                            М2              = 0,707i0ν tg (0,3535ν)
                              k=2
   A                                           B

                                                   O
                                                                                     6 i2
  A,B = 0 1             D,L                                                                 Δ 2 , 3 ϕ 4 (ν 2 ) = 1,5i0ϕ4(0,75ν)
                                                                                     l2
           1,25           0,75
                      C                                    F               3i1
                      1                                            r23           Δ1, 3 ϕ1 (ν 1 ) = 0,6i0ϕ1(0,8839ν)
                                 Δ1,3 = 1,25                   L           l1
              r13
                                                       F
Δ2,3 = 0,75                F              1,5F
                                                                          r33
                                                               D
 rC,3 = c0                C         1,5F
rA′ ,3                                                 Z3= 1                М3                     1,5i0ϕ4(0,75ν)
                                           2
    A                                     ν3           B
                        rB′ ,3   = i3 ⋅    2
                                          l3
    0
   RA   + rA′′,3
                                                   RB + rB′′,3
                                                           0
                              k=3


                                                               Рис. 3.3


                                                                     38