Armaduras

DETERMINAR LOS ESFUERZO EN TODAS LA BARRAS

2000 kg.
2000 kg.
2000 kg.
2000 kg.
2000 kg.
2.6
2.5
2.5
2.5
2.5
2.5
A
C
E
G
I
K
F
B
L
D
H
J

B

2000 kg.
2000 kg.
2000 kg.2000 kg.
2000 kg.
2.6
2.5
2.5
2.5
2.5
2.5
A
C
E
G
I
K
F
B
L
D
H
J
DIAGRAMA DE CUERPO LIBRE
VAy

VAx

RLy

∑ MA = 0
-2000(2.5)-2000(5)-2000(7.5)-2000(10)-2000(12.5)+ RLY(15)=0
RLY (15)=75000
RLY =5000 kg.
∑ FY = 0
(-2000*5)+ RLY +VAY =0
VAY =5000 kg.
∑ FX = 0
HAx =0

Analizamos en nudo “A”:
∑ FY = 0
RAY - FAB*sen (45) = 0
FAB = 7071.07 kg. TRACCION
∑FX = 0
-FAC-FAB*cos (45)= 0
FAC = -5000 kg. COMPRESION
A
FAC
FAB
5000 kg


∑ FY = 0
FBC+FBE*sen45+FAB*sen(45)= 0
-2000+5000+FBE*sen(45) = 0
FBE = -4242.64 kg. COMPRESION
∑ FX = 0-FBD-FBE*cos45+FAB*cos(45) = 0
-FBD-FBE*cos45+5000= 0
FBD = 8000 kg. TRACCION
Analizamos en nudo “B”:
B
FCB
FAB
FBE
FBD

∑ FY = 0
- FCB -2000 = 0
FCB = -2000 kg.COMPRESION
∑ FX = 0
-FCE+ FAC=0
-FCE= -FAC
FCE = -5000 kg. COMPRESION
FCE
FAC
2000 kg
FCB
CC

∑ FY = 0
FDE = 0
∑ FX = 0
FBD - FDF = 0
FBD = FDF
FDF = 8000 Kg. TRACCION

D
FDEFDF
FBD

∑ FY = 0
-2000- FED - FEF sen (45) - FEB sen(45) =0
FEF sen(45) = 1000
FEF = 1414.21 kg. TRACCION
∑ FX = 0
FCE - FEG - FEF cos (45) + FBE cos (45) =0
-FEG -5000 -3000- FEF cos(45) = 0
FEG =-9000 kg. COMPRESION
E
FED
2000 kg
FEG
FEF
FBE
FCE

∑ FY = 0
FGF + FFI sen (45) + FFE sen (45) =0
-2000+FFIcos⁡(45)+ 1000 = 0
FFI = 1414.21 kg. TRACCION
∑ FX = 0
-FFI cos (45) - FFH + FDF -FFE cos (45) =0
-FFI cos (45) - FFH = -7000
FFH = 6000 kg. TRACCION

F
FDF
FFE
FGF
FFI
FFH

FAC

FGE
G
∑ FY = 0
FGF = -2000 kg.COMPRESION
∑ FX = 0
FEG = FGI
FGI = -9000 kg. TRACCION

FGJ

FGF

∑ FY = 0
FHI = 0
∑ FX = 0
FFH - FHJ = 0
FHJ = 6000 kg. TRACCION

H
FHI
FHJ
FFH

∑ FY = 0
- 2000 - FFI sen (45)...