tp2 faltan flechas
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Transcript of tp2 faltan flechas
![Page 1: tp2 faltan flechas](https://reader036.fdocuments.ec/reader036/viewer/2022082416/550087f44a7959da6c8b5365/html5/thumbnails/1.jpg)
ESTRUCTURAS 1 - CATEDRA ARQ. GLORIA DIEZ
TRABAJO PRACTICO Nº 2
COMPOSICION Y DESCOMPOSICION DE FUERZAS
1) Componer gráfica y analíticamente los siguientes sistemas de fuerzas
a) Fuerzas concurrentes
Fuerza Módulo Angulo Coordenadas(t) (Grados) (x,y)
1 4 30 0;02 5 135 0;03 2 240 0;04 6 180 0;0
Fuerza Módulo Angulo Cos a Sen a P = P.Cos a
P =P.Cos b
1 4 30 0.866 0.5 3.44 22 5 135 -0.75 0.7 -3.75 3.53 2 240 -0.5 -0.86 -1 -1.724 6 180 -1 0 -6 0
-7.31 3.78
R2 = (-7.31)2 + (3.78)2
R = √(49.84 + 14.28)R = 8
Tan = 3,78/ -7.31 = -0.53arTan = -27º55’
b) Fuerzas no concurrentes
Fuerza Módulo Angulo Coordenadas(t) (Grados) (x,y)
1 2 120 1;32 5 270 -2;-13 1 30 3;1
c) Fuerzas paralelas
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Fuerza Módulo Angulo Coordenadas(t) (Grados) (x,y)
1 2 90 1;32 5 270 -2;-13 1 90 3;1
R= P1 + P2 + P3R= 2t+5t+1tR= -2t
D1= 0mD2= 3mD3= 2m
R.dr = P1.D1 + P2.D2 + P3.D3R.dr = 2t.0m + 5t.3m + 1t.2mR.dr = -3tmDr= 6.5m
d) Trasladar la fuerza
Fuerza Módulo Angulo Coordenadas(t) (Grados) (x,y)
1 5 90 2;0
Al punto B de coordenadas (-1,1)
M = f x d
M = 5t x 3m
M = 15tm
e) Componer la fuerza
Fuerza Módulo Angulo Coordenadas(t) (Grados) (x,y)
1 4 135 2;0
Con el ar M= -15 tm
M=F.d
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M=15tm-15tm=4t.d-15tm/4t=d-3,75=d
2) Descomponer gráfica y analíticamente la siguiente fuerza
Fuerza Módulo Angulo Coordenadas(t) (Grados) (x,y)
1 6 90 2;0
En las direcciones dadas
Ejer. Direc.
Ang. Coord
Direc Ang Coord
Direc Ang Coord
(º) (x;y) (º) (x;y) (º) (x;y)a) 1 90 2;3 2 120 2;3 --- --- ---b) 1 30 -2;0 2 90 7;0 --- --- ---c) 1 90 0;0 2 90 5;0 3 0 2
a)
Px= Fax + Fbx P.Cos αp = Fa . Cos αa + Fb. Cos αbPy= Fay + Fby P.Sen αp = Fa . Cos αa + Fb. Cos αb
6t . cos90º = Fa . Cos 30º + Fb. Cos120º6t . sen90º = Fa . Cos 30º + Fb. Cos120º
6t . 0 = Fa . 0.86 + Fb . (-0.5)6t . 1 = Fa . 0.5 +Fb . 0.86
0t = Fa . 0.86 + Fb . (-0.5)6t= Fa . 0.5 +Fb . 0.86
Fa = -Fb . (-0.5) / 0.866t= (Fb . 0.5 / 0.86) . 0.5 + Fb . 0.866t= Fb . 0.5 . 0.58 + Fb . 0.866t= Fb . 0.29 + Fb . 0.866t= Fb . 1.156t / 1.15 = FbFb= 5,21t
Fa= (6t – 5.2 . 0.86) / 0.5Fa= 3,06t
b)
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M = f x d
R x dA = fa x da
6t x 2m = fa x (-2m)
fa = 2.4t
R = fa + fb
6t = 2.4t + fb
fa = 3.6 t
c)
- R x d = - fa x dfa
- (6t x 3m) = - fa x 0.7679m
-18tm / -0.7679m = fa
fa = 23.44t
- R x d = - fb x dfb
-(6t x 3m) = - fb x 0.8868m
-18tm / -0.8868m = fb
fb = 20.29t
R x d = - fc x dfc
20.29t x 0.8453m = - fc x 3m
17.15tm / -3m = fc
fc = - 5.71t
M1= 6t . 3m = Fa . 0 + Fb . 0 + Fc . 5m18tm = Fc . 5m18tm/5m = FcFc= 3,6t
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M2 = 6t . 3m = Fa . 2.1m + Fb . 0 + Fc . 018tm = Fa . 2.1m18tm/2.1m = FaFa= 8.57t
M3 = 6t . 3m = Fa . 0 + Fb . 8.3m + Fc . 018tm = Fb . 8.3mFb = 18tm / 8.3m
3) Equilibrar gráfica y analíticamente la fuerza del punto 2
________________________________________R= P1 + P2 + P3R= 2t+5t+1tR= -2t
D1= 0mD2= 3m
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D3= 2m
R.dr = P1.D1 + P2.D2 + P3.D3R.dr = 2t.0m + 5t.3m + 1t.2mR.dr = -3tmDr= 6.5m
________________________________________________Algo Con el ar M= -15 tm
M=F.dM=15tm-15tm=4t.d-15tm/4t=d-3,75=d
Px= Fax + Fbx P.Cos αp = Fa . Cos αa + Fb. Cos αbPy= Fay + Fby P.Sen αp = Fa . Cos αa + Fb. Cos αb
6t . cos90º = Fa . Cos 30º + Fb. Cos120º6t . sen90º = Fa . Cos 30º + Fb. Cos120º
6t . 0 = Fa . 0.86 + Fb . (-0.5)6t . 1 = Fa . 0.5 +Fb . 0.86
0t = Fa . 0.86 + Fb . (-0.5)6t= Fa . 0.5 +Fb . 0.86
Fa = -Fb . (-0.5) / 0.866t= (Fb . 0.5 / 0.86) . 0.5 + Fb . 0.866t= Fb . 0.5 . 0.58 + Fb . 0.866t= Fb . 0.29 + Fb . 0.866t= Fb . 1.156t / 1.15 = FbFb= 5,21t
Fa= (6t – 5.2 . 0.86) / 0.5Fa= 3,06t
_______________________________
M1= 6t . 3m = Fa . 0 + Fb . 0 + Fc . 5m18tm = Fc . 5m18tm/5m = Fc
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Fc= 3,6t
M2 = 6t . 3m = Fa . 2.1m + Fb . 0 + Fc . 018tm = Fa . 2.1m18tm/2.1m = FaFa= 8.57t
M3 = 6t . 3m = Fa . 0 + Fb . 8.3m + Fc . 018tm = Fb . 8.3mFb = 18tm / 8.3m