AULA2011RMI_3a_TRELICA

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Nae = -40 kN | compressão
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FDG=
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3 m
d/4 d/4=0,75m
d/4 d/2=1,5m
d/4 3d/4=2,25m
d/4 d=3m
Barra FE = 3 m
Barra DG = 3,75m
Barra CH = 4,5m
Barra BI = 5,25m
Barra AJ = 6m
CÁLCULO DOS ÂNGULOS
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3 m
θ
θ= tan–1 3 = 10,62°
 16 
16 m
CÁLCULO DOS ÂNGULOS
α = 
β = 
γ = 
δ = 
θ = 10,62°
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3 m
α
αi = tan–1 4 = 53,13°
 3 
 α = 90° – αi = 36,87°
4 m
αi
CÁLCULO DOS ÂNGULOS
α = 36,87°
β = 
γ = 
δ = 
θ = 10,62°
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3,75 m
β
βi = tan–1 4 = 46,85°
 3,75 
 β = 90° – βi = 43,15°
4 m
βi
CÁLCULO DOS ÂNGULOS
α = 36,87°
β = 43,15°
γ = 
δ = 
θ = 10,62°
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4,5 m
γ
γi = tan–1 4 = 41,63°
 4,5 
 γ = 90° – γi = 48,37°
4 m
γi
CÁLCULO DOS ÂNGULOS
α = 36,87°
β = 43,15°
γ = 48,37°
δ = 
θ = 10,62°
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5,25 m
δ
δi = tan–1 4 = 37,30°
 5,25 
 δ = 90° – δi = 52,70°
4 m
δi
CÁLCULO DOS ÂNGULOS
α = 36,87°
β = 43,15°
γ = 48,37°
δ = 52,70°
θ = 10,62°
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Como não há forças atuando nas barras, FE = FG = 0 
α = 36,87°
β = 43,15°
γ = 48,37°
δ = 52,70°
θ = 10,62°
Nó F:
FG
F
FE
AB =
AJ = 
BC =
BI =
BJ =
CD =
CH =
CI =
DE =
DG =
DH =
EF = 0
EG =
FG = 0
GH =
HI =
IJ =
RJx =
RJy =
RAx =
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ΣFx = 0 → EG cos α + ED cos θ = 0
	EG cos(36,87°) + ED cos(10,62°) = 0
	0,8 EG + 0,98 ED = 0
ΣFy = 0 →	 EG sen α – ED sen θ + EF – 4kN = 0
	EG sen(36,87°) – ED sen(10,62°) + 0 = 4kN
	0,6 EG – 0,185 ED = 4kN
 
0,8 EG + 0,98 ED = 0
0,6 EG – 0,185 ED = 4kN 	 
0,8 EG = –0,98 ED
EG = –0,98 ED = –1,228 ED
 0,8
0,6 (–1,228 ED) – 0,185 ED = 4kN
–0,735 ED – 0,185 ED = 4kN
–0,92 ED = 4kN
ED = 4 kN
 –0,92
ED = –4,341 kN
EG = –1,228 ED
EG = –1,228 x 4,348 kN
EG = 5,333 kN
α
θ
EF
ED
EG
E
4kN
Nó E:
α = 36,87°
β = 43,15°
γ = 48,37°
δ = 52,70°
θ = 10,62°
AB =
AJ = 
BC =
BI =
BJ =
CD =
CH =
CI =
DE = –4,341 kN
DG =
DH =
EF = 0
EG = 5,333 kN
FG = 0
GH =
HI =
IJ =
RJx =
RJy =
RAx =
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α
GH
EG
G
GD
FG
ΣFy = 0 →	 EG sen α + GD = 0
	 5,333 kN sen(36,87°) + GD = 0
	GD = –3,2000kN 
ΣFx = 0 → FG – GH + EG cos α = 0
	0 – GH + 5,333 kN cos(36,87°) = 0 
	GH = 4,2666kN 
 
Nó E:
AB =
AJ = 
BC =
BI =
BJ =
CD =
CH =
GH = 4,2666 kN
HI =
IJ =
RJx =
RJy =
RAx =
α = 36,87°
β = 43,15°
γ = 48,37°
δ = 52,70°
θ = 10,62°
CI =
DE = –4,341 kN
DG = –3,200 kN
DH =
EF = 0
EG = 5,333 kN
FG = 0
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ΣFx = 0 → DH cos β + CD cos θ + DE cos θ = 0
	DH cos(43,15°) + CD cos(10,62°) +
	4,341 kN x cos(10,62°) = 0
	 0,73 DH + 0,98 CD = – 4,2667kN
ΣFy = 0 →	 DH sen β – DE sen θ – CD sen θ – DG = 0
	 DH sen(43,15°) – 4,341 kN x sen(10,62°) – CD sen(10,62°) – 3,2kN = 0
	0,68 DH – 0,18 CD – 0,8 – 3,2 = 0
	0,68 DH – 0,18 CD = 4kN
 
0,73 DH + 0,98 CD = – 4,2667kN 
0,68 DH – 0,18 CD = 4kN 	 
– 0,18xCD = 4 – 0,68 DH
CD = 4 – 0,68 DH
 – 0,18
CD = 3,7112 DH – 21,70
0,73 DH + 0,98(3,7112 DH – 21,70) = –4,2667
0,73 DH + 3,6477 DH + 12,8 = –4,2667
(3,6477 + 0,73)DH = –4,2667 + 21,33
4,3772 DH = 17,0667
DH = 3,8990
0,68 (3,8990) – 0,18 CD = 4
2,6667 – 0,18 CD = 4
 0,18 CD = 4 – 2,6667
 - 0,18 CD = 1,3333
CD = 1,3333 = - 7,2350
 - 0,18
DG
D
Nó D:
AB =
AJ = 
BC =
BI =
BJ =
CD = – 7,2350
CH =
GH = 4,2666 kN
HI =
IJ =
RJx =
RJy =
RAx =
α = 36,87°
β = 43,15°
γ = 48,37°
δ = 52,70°
θ = 10,62°
CI =
DE = – 4,341 kN
DG = – 3,200 kN
DH = – 3,8990
EF = 0
EG = 5,333 kN
FG = 0
DE
CD
DH
θ 
β
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6
7
8
5
2
1
3
4
P2
P1
2
2
30°
Cos 30° = √3/2
Cos 60° = √1/2
P1 = P2 = P
H
H x cos30°= 2
H = 2 = 2,3094
 cos30° 
22 + C12 = H2
C12 = H2 – 22
C1 = √ H2 – 22 = 1,1547
C2 = 2 x C1 = 2,3094
C1
ΣM4 = 0
 2xP + 2xP – F7 = 0
F7 = 4P
C2
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