Ejercicio 1.Calcular el momento resistente de una viga de seccin rectangular doblemente armada. Determinar si el acero en tensin y compresin se encuentran en estado de sedencia.DatosFc= 200kg/cmFy= 4200kg/cmd= 50cmb= 25cmd= 4cmAs= 30.42cm 6 barras N. 8As= 11.48cm 4 barras N. 6Solucin

Pmx .75 pbPmx= .75 .85B1 Fc 6 000 fy 6 000 + fy

Pmx= .75 .85 x .85 x 200 6 000 4,200 6 000 + 4,200Pmx= .75 x .0344x .5882Pmx=.0152

P= As = 30.42 = 0.0243 Bd 25 x 50

P Pmx0.0243 0.0151P P= As As = As- As = 30.42-11.48 = 0.0151 Bd Bd Bd 25 x 50

.75 pb= 0.01510.0152 0.0151

P-P .85 B1 fc d 6 000 Fy d 6 000 fy

.00151 .85 (.85) 200x4 6 000 4200x50 6 000 4200

0.7225x 3.80 x 10 x 3.33= 0.009160.0151 0.00916

a= (As-As) fy = (30.42 11.48) 4200 = 18.71 .85 fc b .85x 200 x 25

Mu= .90[(30.42-11.48)4200(50-.5(18.71))+11.48x4200(50-4)] .90[18.94(170709)+(2217936)]Mu= 4,906,048.014 kg-cm

Ejercicio 2.

Calcular el momento resistente de una viga de seccin rectangular doblemente armada. Determinar si el acero en tensin y compresin se encuentran en estado de sedencia.DatosFc= 200kg/cmFy= 4200kg/cmd= 50cmb= 25cmd= 4cmAs= 11.48cm 4 barras N. 6As= 5.74cm 2 barras N. 6SolucinPmx 0.75 pbPmx = 0.75 85B1 Fc 6 000 fy 6 000 + fy

Pmx= .75 .85 x .85 x 200 6 000 4,200 6 000 + 4,200

Pmx= .75 x .0344x .5882Pmx=.0152

P= As = 11.48 = 0.009184 Bd 25 x 50

P < Pmax 0.0152 0.009184 La viga trabaja como simplemente armada.

W=Pfy = 0.009184x4200 = 0.1928 fc 200

Mu= b d fc w (1-0.59w) Mu= 0.90 x 25 x 50 x 200 x .1928 (1- 0.59(.1928)) = 1,922,271.912 kg cm

Ejercicio 3.Disear por flexin una viga rectangular doblemente armada.DatosWL=20 ton/m = 20,000 kg/mClaro= 5mFc= 200kg/cmFy= 4200kg/cmd= 50cmb= 30cmd= 5cmWL= 20,000 kg/m

5 m

SolucinWu= 1.4 D + 1.7 L

Wu= 1.4 x 0.3 x 0.55 x 2300 + 1.7 x 20 000 kg/ m = 34, 531. 3 kg / mMu= Wl= 34 531.3 x 5 = 107,910.31 kg-m = 10,791,031.5 kg-cm 8 8

Pmx 0.75 pbPmx = 0.75 85B1 Fc 6 000 fy 6 000 + fy

Pmx= .75 .85 x .85 x 200 6 000 4,200 6 000 + 4,200

Pmx= .75 x .0344x .5882Pmx=.0152

As = pbd = 0.0152 x 30 x 50 = 22. 8 cma= As fy = 22.8 x 4200 = 18.78 0.85 fc b .85 x 200 x 30

M = As fy ( d 0.5a)M = 0.90 x 22.8 x 4200 (50-0.5 (18.78))= 3,499,932.24 kg-cmM= 3,499,932.24 kg-cm < 10,791,031 kg-cmMu= M + M M = Mu - MM = 10791,031 kg-cm 3,499,932.24 kg-cm = 7,291,098.76 kg-cmM = As fy ( d d) As = Mu fy ( d d)

As= 7,291,090.76 = 42.86 cm 0.90 x 4200( 50-5) As = As +As =22.8 cm + 42.86 cm = 65.66 cm = 7 barras # 11 = 66.96 cmAs = 42.86 cm = 7 barras # 9 = 44.97 cm

30 cm5 cm50 cm

Ejercicio 4.Disear la trabe 3-B-C

6m7m7m6m

A54321DCB4m8.5mm4m

Datos Columnas de 40 x 40 cmTrabe de 25 x 50 cmLosa de vigueta y bovedillaTecho de entrepiso para bibliotecafc= 300kg/cm2fy= 4200kg/cm2d= 5cm r=5cm

d=45cmb=25cmL= 8.5mrea de la Losa= (7)(8.5)= 59.5m2WL= 365kg/m2WL= 350kg/m2 = (350)(59.5)= 20825kg/m2= 20825 = 2,450kg/m 8.5

Vigueta de 20-6 y bovedilla de WD= 365kg/m2= 365x59.5= 21717.5 = 2,555 8.5Peso propio de la losa 365kg/m2 Acabado inferior 36 kg/m2 Firme de nivel 54 kg/m2 Piso + Adhesivo 35 kg/m2 Sobrecarga 40 kg/m2 530 kg/m2WD= (530) (59.5) = 31535 kg/m2 = 31535 = 3,710 kg/m 8.5

SolucinWu = 1.4 D + 1.4 D + 1.7 LWu= 1.4 (0.25 x 0.50 x 2300) + 1.4(3710) + 1.7(2450) = 9, 761.5kg/mMu= Wl 10Mu= 9,761.5x8.5 = 70, 526.83kg-m = 7, 052, 683.75kg-cm 10

Pmx 0.75 pbB1= 1.05 fc = 1.05 300 = .8357 1400 1400

Pmx = 0.75 .85B1 Fc 6 000 fy 6 000 + fy

Pmx= .75 .85 x .8357 x 300 6 000 4,200 6 000 + 4,200

Pmx= .75 x .0507 x .5882Pmx=.022384

As1 = pbd= (0.022384) (25) (45) = 25.182cm2a= As fy = 25.182 x 4200 = 16.5904 0.85 fc b .85 x 300 x 25

M = 0.90 x 25.182 x 4200 (45 -0.5 (16.5904))= 3,888,656.617 kg-cmM= 3,888,656.617 kg-cm < 7, 052, 683.75kg-cm

M = Mu - MM = 7, 052, 683.75kg-cm 3,888,656.617 kg-cm 3, 164, 027.133 kg-cmM = As fy ( d d) As = Mu fy ( d d)

As= 3, 164, 027.133 = 20.92 cm 0.90 x 4200( 45-5)

As = As +As = 25.182 + 20.92= 46.102 cm= 6 barras # 10=47.65 cmAs = 20.92cm = 8 barras #6= 22.92 cm

50 cm5 cm25 cm