File

Diseño de una viga simplemente
armada
• Datos
F´c = 200 kg/cm2
Fy = 4200 kg/cm2
F.C = 1.40
F.r= 0.90
• Vmax=
• Mmax=
Mu= Mmax * F.C
Mu= 13.5 ton * 1.40
Mu= 18.9 toneladas
Mu>Mr
Bminimo= L/32
Bminimo= 600/ 32
h= 2b
B = 20
cm
h= 2*20
h= 40cm
d= h-5
d= 40-5
d= 35 cm
F*c= 0.8 x f´c
F*c= 0.8 x 200kg/cm2
F*c= 160 kg/cm2
F”c= 0.85 x f*c
F”c= 136 kg/cm2
• MR= FrF”c bd2 q (1 – 0.5q)
• MR= Mu
• MR= 18.9 (100)(1000)
• MR= 18.9x105 kg.cm
FrF”c bd2=
FrF”c bd2= (0.9)(136kg/cm2)(20cm)(352)
Mr
FrF”c bd2=
Mr
FrF”c
bd2=
18.9x105
2.9988x106
0.6320
0.63= q – 0.5q2
0.5q2 –q + 0.63= 0
ax2 + bx+ c= 0
a= 0.5
b= -1
C= 0.63
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
X=…….
se propone:
• b=25cm
• d= 45cm
FrF”c bd2=
FrF”c bd2= (0.9)(136kg/cm2)(25cm)(452)
Mr
FrF”c bd2=
Mr
FrF”c
bd2=
18.9x105
6.1965x106
0.3050
0.305= q – 0.5q2
0.5q2 –q + 0.305= 0
ax2 + bx+ c= 0
q=
a= 0.5
b= -1
C= 0.305
−𝑏± 𝑏2 −4𝑎𝑐
2𝑎
q=
−1± −12 −4(.5)(.305)
2(.5)
q1= 1.62
q2= 0.38
P= q
P= 0.38
𝑓"𝑐
𝑓𝑦
P= q
130𝑘𝑔/𝑐𝑚2
4200𝑘𝑔/𝑐𝑚2
P= 0.0123
P= 0.01176
𝑓"𝑐
𝑓𝑦
•As= Pbd
As= (.0123)(25)(45)
As= 13.8 cm2
5 #6
2 #8
Elaborado por Abel Alejandro Espíndola
Martínez
Instituto Tecnológico de Pachuca
Estructuras de concreto
Catedrático
Ing. Martin Silva Badillo