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