Postes de Concreto h 2.40m
12
FLEXION CASO 1 ancho (b) 12 cm altura (h) 12 cm recubrimiento "r" 2.5 cm peralte efect. (d) 9.5 cm peso prop 34.56 kg/m carga W 103.68 kg/m longitud (L 2.90 m M = 108.99 kg-m Mu = 152.59 kg-m Acero Asmin = 0.28 cm2 Asmax = 1.82 cm2 As req = 0.52 cm2 coloco: 2Ø1/4" 0.64 cm2 ambas caras CORTANTE V 150.34 kg Vu = 210.47 kg ØVc = 744.23 kg ØVc > Vu no estribo se colocara estribos para armar CASO 2 L w H b d r F
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diseño estructural de diseño de postes para un cerco de un condominio
Transcript of Postes de Concreto h 2.40m
FLEXION CASO 1
ancho (b) 12 cmaltura (h) 12 cmrecubrimiento "r" 2.5 cmperalte efect. (d) 9.5 cm
peso prop 34.56 kg/mcarga W 103.68 kg/m
longitud (L) 2.90 m
M = 108.99 kg-mMu = 152.59 kg-m
Acero
Asmin = 0.28 cm2Asmax = 1.82 cm2
As req = 0.52 cm2
coloco: 2Ø1/4" 0.64 cm2 ambas caras
CORTANTE
V 150.34 kgVu = 210.47 kg
ØVc = 744.23 kg
ØVc > Vu no estribo se colocara estribos para armar el fierro longitudinal
CASO 2
b
hd
r
L
w
H
1.50
b
hd
r
F
ACERO As min 0.28 cm2As max 1.82 cm2
As req 0.61 cm2
As colocado 2 Ø 1/4" 0.64 cm2 ambas caras
Mu = 210 kg-m
se colocara estribos para armar el fierro longitudinal
H
1.50
b
hd
r
F
FORMULAS
0.0619971
coloco estribos: Ø6mm [email protected], [email protected], [email protected] c/e
b = 12 cm
b
hd
r
L
w
H
1.50
b
hd
r
F
h = 12 cmr = 2.5 cmd = 9.5 cm
FLEXION
F = 100 kgM= 150 kg-m
Vu = 140 kgMu = 210 kg-m
CORTANTE
Vu = 140 kgØVc = 744.23ØVc > Vu no hay estribos
Vu = 140 kg
coloco estribos: Ø6mm [email protected], [email protected], [email protected] c/e
H
1.50
b
hd
r
F
FLEXION CASO 1
ancho (b) 10 cmaltura (h) 10 cmrecubrimiento "r" 2 cmperalte efect. (d) 8 cm
peso prop 24 kg/mcarga W 72 kg/m
longitud (L) 2.90 m
M = 75.69 kg-mMu = 105.97 kg-m
Acero
Asmin = 0.19 cm2Asmax = 1.27 cm2
As req = 0.40 cm2
coloco: 4 Ø 1/4" 1.28 cm2
CORTANTE
V 104.40 kgVu = 146.16 kg
ØVc = 522.27 kg
ØVc > Vu no estribo se colocara estribos para armar el fierro longitudinal
CASO 2
b
hd
r
L
w
H
1.50
b
hd
r
F
H = 2.40 m
ACERO As min 0.19 cm2As max 1.27 cm2
As req 0.54 cm2
As colocado 2 Ø 1/4" 0.64 cm2 ambos lados
Mu = 210 kg-m
se colocara estribos para armar el fierro longitudinal
H
1.50
b
hd
r
F
FORMULAS
coloco estribos: Ø6mm [email protected], [email protected], [email protected] c/e
b = 10 cm
b
hd
r
L
w
H
1.50
b
hd
r
F
h = 10 cmr = 2 cmd = 8 cm
FLEXION
F = 100 kgM= 150 kg-m
Vu = 100 kgMu = 150 kg-m
CORTANTE
Vu = 140 kgØVc = 522.27ØVc > Vu no hay estribos
Vu = 140 kg
coloco estribos: Ø6mm [email protected], [email protected], [email protected] c/e
H
1.50
b
hd
r
F
