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diseño estructural
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P d = 3.1P l = 0.53fyd= 4200fcd= 210
6.9
a= 0.35b= 0.35
h= 0.35b= 0.35
b*h^2 = bo*ho^2bo= 0.3ho= 0.37
Por lo tanto b=nuestra seccion: h=
Nd= 1.4Pd+1.7Pl= 5.241As= Area de acero
0.125As 9Por lo tanto nuestra area de acero sera mayor o igual a 9 cm23/8`` 0.711/2`` 1.275/8`` 1.98
61/2``+23/8``= 9.04
Por lo tanto se usaran 61/2``+23/8`
l (distancia columnas)=
As
TonTonKg/cm2Kg/cm2m
mm
0.300.40
Ton
cm2cm2
cm2cm2cm2
9.0 cm2OK
61/2``+23/8`
Hoja1