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Apostila de Treliças 2

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HB<br />

1º Passo Condição <strong>de</strong> Isostática<br />

2.n = b+ν<br />

2.7 = 11+3<br />

14 = 14 OK<br />

2º Passo Reações <strong>de</strong> Apoio<br />

ΣFx = 0 ΣFy = 0 ΣM = 0 (Momento fletor)<br />

HA+HB =0 VB = 0 -HA.3+2.6+2.4+2.2=0<br />

HA = -HB HA = 24÷3<br />

HB = -8 t HA = 8 t<br />

3º Passo Método dos Nós<br />

Decomposição das forças<br />

Nó “5” Forças Verticais (V) Forças Verticais (H)<br />

VB<br />

= 0<br />

0<br />

N56<br />

8÷cos26,57°<br />

VB<br />

N51 N51 N56<br />

ΣFV = 0 ΣFH = 0<br />

VB-N51-N56.sen26,57° = 0 -HB+N56.cos26,57°<br />

HB<br />

N56<br />

6-N51-8,94.sen26,57° = 0 -8+N56.cos26,57° =<br />

-N51+6-4 = 0 N56 =<br />

N51 = 2 t N56 = 8,94 t<br />

14

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