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70<br />

2<br />

K34<br />

d34 = = 0,<br />

6<br />

K<br />

Probleme rezolvate. Sisteme cu 1 GLD – vibraţii forţate<br />

Pab 1 ⋅ 2l<br />

⋅ l 2<br />

M23 = = = l = 1,<br />

11111Nm<br />

2<br />

2<br />

l ( 3l)<br />

9<br />

2<br />

2<br />

Pa b 1 ⋅ ( 2l)<br />

⋅ l 4<br />

M32 = =<br />

= l = 2,<br />

22222Nm<br />

2<br />

2<br />

l ( 3l)<br />

9<br />

Verificarea diagramei M:<br />

∑ ∫<br />

3<br />

2<br />

M1M l<br />

2l<br />

dx = − ⋅ 1,<br />

111067 ⋅ 1 + ( −2<br />

⋅ 1,<br />

111067 ⋅ 1 +<br />

EI 3EI<br />

6EI<br />

l<br />

+ 2 ⋅ 1,<br />

975324 ⋅ + 1 ⋅ 1,<br />

975324 −<br />

3<br />

1<br />

l<br />

1<br />

− ⋅ 1,<br />

111067)<br />

+ ( 2 ⋅ 1,<br />

975324 ⋅ −<br />

3<br />

6EI<br />

3<br />

l<br />

= ( 7,<br />

901296 − 7,<br />

90094)<br />

6EI<br />

ε<br />

T<br />

1.11067<br />

1.975324<br />

M,M<br />

1<br />

⋅ 1,<br />

481481)<br />

=<br />

3<br />

1.111067 1.481481<br />

1<br />

2l l<br />

Fig.2.13<br />

1.481481<br />

0.7074<br />

7,<br />

901296 − 7,<br />

90094<br />

−3<br />

% = ⋅ 100 = 4,<br />

5 ⋅ 10 < 0,1<br />

7,<br />

901296<br />

Calculăm coeficientul de flexibilitate:

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