1. Disear por flexin el momento positivo de diseo. Considere ancho = 1.60m, s/c=600 kg/m2, fc=280kg/cm2, fy=4200 kg/cm2.0.17

CP=0.170.17

Solucin:Dimensionamiento:

Metrado de Cargas:P.P = 0.2804 1.60 2.4 1.2=1.29t/mAcabado = 0.10 1.60 1.2=0.19 t/ms/c = 0.6 1.60 1.6=1.54 t/mwu2 = 3.02P.P = 0.15 1.60 2.4 1.2=0.69 t/mAcabado = 0.10 1.60 1.2=0.19 t/ms/c = 0.6 1.60 1.6=1.54 t/mwu1 =2.42 t/m

Rb = 5.92 t L1 = 1.70 + 0.15/2 = 1.775 mL2 = 2.10 + 0.40/2 = 2.30mHallamos el momento mximo:Vx = RB wu2Xo = 0 Xo = 1.96 m

Diseo: 1/2 = 1.27 cmd = 17 - (2 + 1.27/2) = 14.37 cm

a = 2 cm As = 9.14 cm2 a = 1.01 cm As = 8.82 cm2 a = 0.97 cm As = 8.80 cm2 a = 0.97 cm conforme#varillas =8.8/1.27=6.97Usar :

Respuesta: 7 1/2" @ 0.25

2. Disear la escalera E1 que se muestra en la figura adjunta. Considere fc=175 kg/cm2; fy=4200 kg/cm2, s/c=400 kg/m2.

Solucin:Dimensionamiento:Primer Tramo

Segundo Tramo

Diseo Primer Tramo:P.P. Tramo Inclinado:

Metrado de Cargas:Peso Propio = 0.22 1.00 2.4 1.2= 0.63t/mAcabado = 0.1 1.00 1.2= 0.120 t/ms/c = 0.4 1.0 1.6= 0.640 t/m wu1 = 1.39 t/m

wu2 P.P. = 0.12 2.4 1.2 1.00 = 0.35 t/m Acabado= 0.12 t/ms/c= 0.64 t/mwu2 = 1.11 t/m

Diseo:Considerando 3/8" d =12 2 = 9.52cmMdiseo = 0.9 (+)Mmax = 1.17 t-m

As = 3.28cm2 a = 0.93 cm As = 3.27 cm2 a = 0.92 cmconforme min = 0.0018 As min = min b d = 0.0018 100 9.52 = 1.71cm2

Para b = 1.20 m; As = 1.20 3.27 = 3.92 cm2 #varillas =3.92/0.71=5.56 Usar 6 3/8" @ 0.18

Espaciamiento Mximo: smax = 3t = 36 cmsmax = 45 cmCONFORME

Refuerzo Negativo

Asmin = 1.71 cm2 1.2 =2.052 cm2

Se considera la mitad del refuerzo de momento positivo debido a que el apoyo de muro de albailera portante continuo es un apoyo rgidoUsar3 3/8" @ 0.35 s = 0.56 msmax = 3t = 36cmsmax = 45 cmAs transversal = Ast = 0.0018 (100) 12 = 2.16 cm2 3/8" @ 0.33Usar 3/8" @ 0.30 mDISEO DEL 2do. TRAMOTramo Inclinado:Cos (a) = 0.827

Metrado de Cargas:Peso propio : 0.327 1.0 2.4 1.2 = 0.94 t/mAcabado : 0.10 1.0 1.2= 0.12 t/ms/c : 0.4 1.0 1.8 = 0.64t/m wu1 = 1.70 t/m

wu2 = p.p = 0.20 1.0 2.4 1.2 = 0.58 t/mAcabado= 0.12 t/ms/c= 0.64 t/m wu2 = 1.34 t/m

Vx = R1 W2x0 (w1 w2) (x0 1.375) = 0

3.51 1.34 x0 (0.36) x0 + 0.495 = 0 x0 = 2.36 m

(+)Mdiseo= 0.9(+)Mmax = 3.88 t-m

Diseo:Usando 1/2" d = 20 - 2 - = 17.36cm

a = 1.97cmconforme

As min = 0.0018 bd = 0.0018 100 17.36 = 3.12 cm2

@ 0.17 m

Para b = 1.20 As = 1.2 6.96 = 8.35 cm2

7 1/2" @ 0.18 mAs transversal = As temp = 0.0018 bt = 0.0018 (100)(20)= 3.6 cm2 3/8 @ 0.20 mRefuerzo para momento negativo en los apoyos:

Para b = 1.20m -As = 4.18cm2 6 3/8" @ 0.22 m

Los espaciamientos son menores que los mximos permisibles.

VI. BIBLIOGRAFIA http://biblioteca.sena.edu.co/exlibris/aleph/u21_1/alephe/www_f_spa/icon/construccion/4/9.html