Diseño De Losa

Diseñar la siguiente loza de entrepiso de esquina en dos lados adyacentes discontinuos, colada monolíticamente.

Datos
W=653kg/cm2
F´c= 250 kg/cm2
Fy= 4200 kg/cm2
r= 2 cm

1. Coeficientes por momentos
m=a1a2=7.558.46=0.90
2. Determinacion delmomentos flexionante

M=(a12)*Wu*Coef10000=1.4653(7.55)210000*Coef=5.21*Coef

Mc (-)= 5.21*371= 1933
= 5.21*219 =1141

Ml (-) = 5.21*360= 1876
=5.21*206= 1073

M (+) c = 5.21*176= 917
L = 5.21*138= 719

3.Determinación del peralte

Pb=4800*f´´cfyfy+6000=4800170420010200=0.019
P=0.75*Pb=0.75*0.014

q=pfyf´´c=0.014*4200170=0.35ku=Frf´´c*q1-0.5q=0.9(170*0.351-0.50.35=44.18

Ru=FR*fy1-0.5q=0.9*4200*1-0.5*0.35=3119

d=M max*100Ku*100=1933*10044.18*100=6.61cm
4. Verificando d min según RCDF

h min=7.55+8.461.25+(7.55+8.46)270=13.34 cm

h min = 14 cm
Usamos d= 12h= 14
5. Áreas de acero

As -= 1933*1003119*12=5.16 cm2

As += 917*1003119*12=2.45 cm2

As -= 1141*1003119*12=3.05cm2

As -= 719*1003119*12=1.92 cm2
6. Separación de armado
Proponemos var # 3 as = 0.71 cm2

S -= 0.71*1005.16=13 cm

S-= 0.71*1003.05=23 cm

S += 0.71*1002.45=28 cm

S += 0.71*1001.92=36 cm
7. S max = 3.5 d = 3.5(12)= 42 cm

8. Diseño