Metoda Sił - rysowanie wykresów sił wewnętrznych w belkach statycznie niewyznaczalnych z przegubem.

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Zadanie: Narysuj wykresy sił N, T, M.

Zadanie rozwiąż metodą sił.

q =4kN/m

P=18kN P=24kN₁

4 2 3 1 3

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Określenie stopnia statycznej niewyznaczalności :

2

3

1

6

3 =

−

=

−

=

_r _P s

l

n

q =4kN/m P=18kN P=24kN₁ 4 2 3 1 3 EJ EJ EJ EJ Układ dwukrotnie statycznie niewyznaczalny

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Dobór schematu podstawowego:

X 1

X 2

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Dobór schematu podstawowego:

X 1

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Schemat alternatywny:

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Schemat podstawowy:

X 1

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Wykres X1=1 X =1 1 X =11 4 2 4 2 1 1/4 3/4 1/2 2 1 2 M₁

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Wykres X2=1 X =1 2 X =12 4 2 4 2 1 1/4 3/4 M₂ 1/2 1

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Wykres od obciążenia zewnętrznego q =4kN/m P=18kN P=24kN₁ 4 2 3 1 2 1 R =27kN E0 R =17kN D0 R =18kN C0 R =20kN A0 M =48kNm A0 A _B C D E 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów:       ⋅ ⋅ ⋅ ⋅ = 2 3 2 4 2 2 1 1 11 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów:       ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = 1 3 2 2 1 2 1 2 3 2 4 2 2 1 1 11 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów:       ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = 1 3 2 4 1 2 1 1 3 2 2 1 2 1 2 3 2 4 2 2 1 1 11 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów: EI EI 3 22 1 3 2 4 1 2 1 1 3 2 2 1 2 1 2 3 2 4 2 2 1 1 11  =      ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów: EI EI 3 2 1 3 1 4 1 2 1 1 12  =      ⋅ ⋅ ⋅ ⋅ = δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów:       ⋅ ⋅ ⋅ ⋅ = 1 3 2 4 1 2 1 1 22 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów:       ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = 1 3 2 2 1 2 1 1 3 2 4 1 2 1 1 22 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów: EI EI 2 1 3 2 2 1 2 1 1 3 2 4 1 2 1 1 22  =      ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów:       ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ = 2 2 1 4 8 4 4 3 2 2 3 2 4 48 2 1 1 2 10 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów:       ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ = 1 2 1 2 8 2 4 3 2 2 2 1 4 8 4 4 3 2 2 3 2 4 48 2 1 1 2 2 10 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2

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Całkowanie wykresów: ( )   ⋅ ⋅ ⋅ ⋅ + + ⋅ ⋅ ⋅ ⋅ ⋅ +             ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ = 2 1 1 3 4 2 1 1 3 1 25 , 0 3 2 3 24 2 1 1 2 1 2 8 2 4 3 2 2 2 1 4 8 4 4 3 2 2 3 2 4 48 2 1 1 2 2 2 10 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2 1 0,25

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Całkowanie wykresów: EI 3 1 1 25 , 0 3 2 3 24 2 1 1 2 1 2 8 2 4 3 2 2 2 1 4 8 4 4 3 2 2 3 2 4 48 2 1 1 2 2 10             ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ = δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2 1 0,25

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Całkowanie wykresów:       ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = 0,75 2 1 3 8 3 4 3 2 75 , 0 3 2 3 24 2 1 1 2 20 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2 1 0,25 0,75

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Całkowanie wykresów: ( )      + ⋅ ⋅ ⋅ ⋅ ⋅ +       ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = 0,75 1 2 1 1 8 1 4 3 2 1 3 1 75 , 0 3 2 1 24 2 1 75 , 0 2 1 3 8 3 4 3 2 75 , 0 3 2 3 24 2 1 1 2 2 20 EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2 1 0,25 0,75

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Całkowanie wykresów: ( ) 1 1 1 1 75 , 0 2 1 1 8 1 4 3 2 1 3 1 75 , 0 3 2 1 24 2 1 75 , 0 2 1 3 8 3 4 3 2 75 , 0 3 2 3 24 2 1 1 2 2 20 ⋅ ⋅ ⋅ ⋅ −       + ⋅ ⋅ ⋅ ⋅ ⋅ +       ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = EI δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2 1 0,25 0,75

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Całkowanie wykresów: ( ) EI 2 0,75 1 1 1 8 1 4 3 2 1 3 1 75 , 0 3 2 1 24 2 1 75 , 0 2 1 3 8 3 4 3 2 75 , 0 3 2 3 24 2 1 1 2 2 20       + ⋅ ⋅ ⋅ ⋅ ⋅ +       ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ = δ 4 2 4 2 1 1 2 M₁ M₂ 1 48 =2 24 18 M [kNm] 0 4 2 8 2 1 0,25 0,75

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Układ równań kanonicznych metody sił dla schematu dwukrotnie statycznie

niewyznaczalnego:    = + ⋅ + ⋅ = + ⋅ + ⋅ 0 0 20 2 22 1 21 10 2 12 1 11 δ δ δ δ δ δ X X X X kNm X kNm X EI X EI X EI EI X EI X EI 97 , 6 59 , 17 0 3 77 2 3 2 0 3 401 3 2 3 22 2 1 2 1 2 1 − = − = ↓       = + ⋅ + ⋅ = + ⋅ + ⋅

cych:

q =4kN/m P=18kN P=24kN₁ 4 2 3 1 2 1 12,82 17,59 _17,59 6,97 6,97 A _B R =11,2kN_A T =4,8kNBL C T =4,8kN_BP T =12,8kNCL T =4,8kNBP T =4,8kNBL _D T =16,65kNCP T =23,35kNDL T =12,8kNCL T =16,65kNCP R =29,45kNC T =5,52kN_DP R =23,52kN_E E T =5,52kN_DP T =23,35kN_DL D R =17,83kN M [kNm] 18 6,97 17,59 14,38 12,82

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Rysowanie wykresu sił tn