aula2011rmi_3a_trelica
DESCRIPTION
TRELICA METALICATRANSCRIPT
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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=1,5md/4 3d/4=2,25md/4 d=3mBarra FE = 3 m
Barra DG = 3,75mBarra CH = 4,5mBarra BI = 5,25mBarra AJ = 6m
CÁLCULO DOS ÂNGULOS
3 m
θ
θ= tan–1 3 = 10,62° 16
16 m
CÁLCULO DOS ÂNGULOS
α = β = γ = δ = θ = 10,62°
3 mα
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°
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°
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°
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°
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:
FGF
FE
AB =AJ = BC =BI =BJ =CD =CH =
CI =DE =DG =DH =EF = 0EG =FG = 0
GH =HI =IJ =
RJx =RJy =RAx =
ΣFx = 0 → EG cos α + ED cos θ = 0EG cos(36,87°) + ED cos(10,62°) = 00,8 EG + 0,98 ED = 0
ΣFy = 0 → EG sen α – ED sen θ + EF – 4kN = 0EG sen(36,87°) – ED sen(10,62°) + 0 = 4kN0,6 EG – 0,185 ED = 4kN
0,8 EG + 0,98 ED = 0
0,6 EG – 0,185 ED = 4kN
0,8 EG = –0,98 EDEG = –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 = 4kNED = 4 kN –0,92ED = –4,341 kN
EG = –1,228 EDEG = –1,228 x 4,348 kNEG = 5,333 kN
αθ
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 kNDG =DH =EF = 0EG = 5,333 kNFG = 0
GH =HI =IJ =
RJx =RJy =RAx =
αGH
EG
G
GD
FG
ΣFy = 0 → EG sen α + GD = 0 5,333 kN sen(36,87°) + GD = 0GD = –3,2000kN
ΣFx = 0 → FG – GH + EG cos α = 00 – GH + 5,333 kN cos(36,87°) = 0 GH = 4,2666kN
Nó E:
AB =AJ = BC =BI =BJ =CD =CH =
GH = 4,2666 kNHI =IJ =
RJx =RJy =RAx =
α = 36,87°β = 43,15°γ = 48,37°δ = 52,70°θ = 10,62°
CI =DE = –4,341 kNDG = –3,200 kNDH =EF = 0EG = 5,333 kNFG = 0
ΣFx = 0 → DH cos β + CD cos θ + DE cos θ = 0DH 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 = 00,68 DH – 0,18 CD – 0,8 – 3,2 = 00,68 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 DHCD = 4 – 0,68 DH – 0,18CD = 3,7112 DH – 21,70
0,73 DH + 0,98(3,7112 DH – 21,70) = –4,26670,73 DH + 3,6477 DH + 12,8 = –4,2667(3,6477 + 0,73)DH = –4,2667 + 21,334,3772 DH = 17,0667DH = 3,8990
0,68 (3,8990) – 0,18 CD = 42,6667 – 0,18 CD = 4- 0,18 CD = 4 – 2,6667 - 0,18 CD = 1,3333CD = 1,3333 = - 7,2350 - 0,18
DG
D
Nó D:
AB =AJ = BC =BI =BJ =CD = – 7,2350CH =
GH = 4,2666 kNHI =IJ =
RJx =RJy =RAx =
α = 36,87°β = 43,15°γ = 48,37°δ = 52,70°θ = 10,62°
CI =DE = – 4,341 kNDG = – 3,200 kNDH = – 3,8990EF = 0EG = 5,333 kNFG = 0
DE
CD
DH
θ
β
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52 1
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P2 P1
2 2
30°
Cos 30° = √3/2Cos 60° = √1/2
P1 = P2 = PH
H x cos30°= 2H = 2 = 2,3094 cos30°
22 + C12 = H2
C12 = H2 – 22
C1 = √ H2 – 22 = 1,1547C2 = 2 x C1 = 2,3094
C1
ΣM4 = 0 2xP + 2xP – F7 = 0F7 = 4P
C2
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