pdfcoffee com_exercicios-resolvidos-lista-4-pdf-free (1)_0028-0028

Prévia do material em texto

( ) ( ) ( )
( ) ( ) ( )
0VV32:0V
0HH:0H
64HV302H6V432:0M
16HV02H2V216:0M
91
91
99991
9999dir6
=++−=Σ
=−=Σ
=+⇒=++−=Σ
=−⇒=−+−=Σ
 
 
resolvendo resulta em: 
 
kN20V
kN4H
kN12V
kN4H
9
9
1
1
=
=
=
=
 
Na barra 9 – 10: 
 
( ) 0M24:0M
0H4:0H
0V20:0V
109
10
10
=+−=Σ
=−=Σ
=+−=Σ
 
 
logo: 
m.kN8M
kN4H
kN20V
10
10
10
=
=
=
 
 
Cálculo das barras submetidas a força axial: 
 
9
8765
4
3
2
1
10
12 kN
4 kN
4 kN
20 kN
8 kN.m
-4 kN -8.9 kN
0
-4 kN
0
-8.9 kN
0 -4.0 kN 0
-20.0 kN
-20.0 kn
-4.0 kN