Esfuerzos Admisibles
Enviado por oscarrubio19 • 25 de Agosto de 2013 • 642 Palabras (3 Páginas) • 458 Visitas
ESFUERZOS ADMISIBLES
ESFUERZOS Y SOLICITACIONES ADMISIBLES
σAdm: Esfuerzo Admisible = σRotura ó Falla/ F.S
SAdm: Solicitación Admisible = Solicitación Rotura/ F.S
Ejemplo
f´m = 10 Mpa, t = 0.12 m, h = 2.4 m
Se arranca considerando el rango elástico.
σ
"σmax " "P" /" A " " ± " "Mc" /"I"
Mc/I=(Mx h/2)/((t x h^3)/12)= 6M/( th^2 )
σ_□((T+)/(C-)) " "= P/th ± 6M/(th^2 )
SON TRES METODOS
COMPRESION
TENSION
CORTANTE
POR COMPRESION
Fa = Resistencia Mamposteria
Fa = 0.2 f`m Re Re = Reduction de Esbeltez
Fa = 0.2 x 10 Mpa X 0.9 = 1.8 Mpa
Re= 〖(h`/40t)〗^3 h’ = Altura libre muro
Re= 〖(2.0/40(0.12))〗^3= 0.90
Fa ≥ fa
Resistencia Mamposteria Lo que está actuando
Fa = 1.8 Mpa ≥ fa = P/th
Fa=0.2 x 10 Mpa X 0.9=1.8 Mpa
fa=(540000 N)/(120 mm x 2400mm)=1.875 Mpa
Como no cumple, aumentamos el f’m
Aumentamos a 11 Mpa
Fa=0.2 x 11 Mpa X 0.9=1.99 Mpa ok
Fb ≥ fb
Fb = 0.33 f’m ≤ 14 Mpa
Fb=0.33 x 11 Mpa = 3.63 Mpa
fb=P/th+ 6M/〖th〗^2 = (540000 N)/(120mm 2400mm) X (6 x 280 X 〖10〗^6)/(120mm x 〖2400mm〗^2 )= 4.3 Mpa
Fb = 3.63 Mpa ≥ fb = 4.3 Mpa
Como no cumple aumentamos f’m
F’m = 14 mpa
Fb=0.33 x 14 Mpa = 4.62 Mpa
Fb= 4.62 Mpa ≥fb=4.3 Mpa Ok
σT=P/th- 6M/〖th〗^2
σT=540000N/(120mm x 2400mm)- (6 x 280 x 〖10〗^6)/〖120mm x 2400mm〗^2
σT= -0.55 Mpa
T = As x fs
fs = 0.5Fy ≤ 170 Mpa
Fy = 420 Mpa
fs = 210 Mpa
fs = 170 Mpa
(4.33+0.55)/(2.4)=(0.55)/x
x = (0.55 x 2.4)/(4.88) =0.27 m
T = (270mm x 0.55 Mpa)/2 x 120mm=8910N=8.9 KN
T = As x fs despejando
As = T/fs= (8910 N)/(170 Mpa )= 〖52.41mm〗^2=〖0.52cm〗^2
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