concreto presforzado ejemplo 1

5
FACULTAD DE INGENIERIA CIVIL PREGUNTA 1 : Datos: L = 12 m Ancho Trib = 5m g con = 2.4 Ton/m3 f'c = 350 Kg/cm2 f'ci = 280 Kg/cm2 s/c 0.25 Ton/m2 Alig = 0.30 Ton/m2 Carga Muerta = 0.15 Ton/m2 Cota fondo = 0.075 m R = 1.15 K = 100% fpu = 18900 Kg/cm2 A acero = cm2 Propiedades seccion Viga: b = 0.40 m h = 0.60 m A = 0.240 m2 I = 0.00720 m4 Yb = 0.300 m Yt = 0.300 m Zb = 0.0240 m3 Zt = 0.0240 m3 exc = 0.225 m PARABOLA 2 PARABOLA 1 y1 = 0.300 m y = k*x^2 y0 = 0.075 m dy = 0.450 m 0.225 m y2 = 0.525 m dx= 0.6 7.2 0.4 4.8 e1= 0.000 m k = dy/dx^2 = 0.009 0.010 e2= 0.225 m x = 0.1 1.2 e3=(e1+e2)/2= 0.1125 m y(L/2) = y0+k*x^2= 0.088 m e=yb-y(L/2)= 0.213 m a 0.12435 rad 0.09348 rad 7.1250 g 5.3558 g f = e + e3 = 0.3250 m Carga Actuantes Wpp = 0.576 Ton/m Walig = 1.500 Ton/m Wcm = 0.750 Ton/m Ws/c = 1.250 Ton/m Calculo de Fuerza Final Efectiva (Pe): Wbalanc = Wpresf= 2.826 Ton/m Reacc.=Wbalanc*L/2 = 16.96 Ton 19.41 Ton 14.61 Ton Despejando Pe la ecuación de Carga Equivalente Pe = 156.5 Ton 9.20687783 Excentricidad en el extremo = 0.000 m Mto en los Extremos = 0.000 Ton.m UNIVESIDAD NACIONAL DE INGENIERIA Departamento de Estructuras CONCRETO PRESFORZADO 2010-1 SOLUCIONARIO DE PRACTICA DIRIGIDA balanc cm alig pp K = ) ( * 2 8 L f P e presf =

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  • FACULTAD DE INGENIERIA CIVIL

    PREGUNTA 1 :

    Datos:

    L = 12 m

    Ancho Trib = 5 m

    g con = 2.4 Ton/m3f'c = 350 Kg/cm2

    f'ci = 280 Kg/cm2

    s/c 0.25 Ton/m2

    Alig = 0.30 Ton/m2

    Carga Muerta = 0.15 Ton/m2

    Cota fondo = 0.075 m

    R = 1.15

    K = 100%

    fpu = 18900 Kg/cm2

    A acero = cm2

    Propiedades seccion Viga:

    b = 0.40 m

    h = 0.60 m

    A = 0.240 m2

    I = 0.00720 m4

    Yb = 0.300 m

    Yt = 0.300 m

    Zb = 0.0240 m3

    Zt = 0.0240 m3

    exc = 0.225 m

    PARABOLA 2 PARABOLA 1

    y1 = 0.300 m y = k*x^2

    y0 = 0.075 m dy = 0.450 m 0.225 m

    y2 = 0.525 m dx= 0.6 7.2 0.4 4.8

    e1= 0.000 m k = dy/dx^2 = 0.009 0.010

    e2= 0.225 m x = 0.1 1.2

    e3=(e1+e2)/2= 0.1125 m y(L/2) = y0+k*x^2= 0.088 m

    e=yb-y(L/2)= 0.213 m a 0.12435 rad 0.09348 rad

    7.1250 g 5.3558 g

    f = e + e3 = 0.3250 m

    Carga Actuantes

    Wpp = 0.576 Ton/m

    Walig = 1.500 Ton/m

    Wcm = 0.750 Ton/m

    Ws/c = 1.250 Ton/m

    Calculo de Fuerza Final Efectiva (Pe):

    Wbalanc = Wpresf= 2.826 Ton/m

    Reacc.=Wbalanc*L/2 = 16.96 Ton 19.41 Ton 14.61 Ton

    Despejando Pe la ecuacin de Carga Equivalente

    Pe = 156.5 Ton 9.20687783

    Excentricidad en el extremo = 0.000 m

    Mto en los Extremos = 0.000 Ton.m

    UNIVESIDAD NACIONAL DE INGENIERIA

    Departamento de Estructuras

    CONCRETO PRESFORZADO 2010-1

    SOLUCIONARIO DE PRACTICA DIRIGIDA

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    balanccmaligppK = )(*

    2

    8

    L

    fPepresf =

  • DIAGRAMA DE MOMENTOS

    VERIFICANDO ESFUERZOS ADMISIBLES:

    ESTADO FINAL

    Ubicacin: 0.4 L1 Ubicacin: 1.0 L1

    Momentos SAP s t s b Momentos SAP s t s bTon.m kg/cm2 kg/cm2 Ton.m kg/cm2 kg/cm2

    pp+alig 20.930 -87.2 87.2 pp+alig -37.350 155.6 -155.6

    cm 7.560 -31.5 31.5 cm -13.490 56.2 -56.2

    s/c 17.220 -71.8 71.8 s/c -22.490 93.7 -93.7

    Axial -65.2 -65.2 Axial -65.2 -65.2

    -29.270 122.0 -122.0 50.690 -211.2 211.2

    Hiperestticos 6.208 Hiperestticos 15.520

    Los esfuerzos Finales sern: Los esfuerzos Finales sern:

    Combinacin s t s b Combinacin s t s bpresf 56.7 -187.2 presf -276.4 146.0

    pp+alig -87.2 87.2 pp+alig 155.6 -155.6

    cm -31.5 31.5 cm 56.2 -56.2

    s/c -71.8 71.8 s/c 93.7 -93.7

    S = -133.7 3.3 S = 29.1 -159.5

    OK OK OK OK

    Esfuerzos Admisibles:

    s adm (comp.) = -210.0 kg/cm2

    s adm (traccin) = 29.9 kg/cm2

    Estados de carga

    pretensado

    Estados de carga

    pretensado

    0

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    0.6

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    m)

    ESF (kg/cm2)

    DIAGRAMA DE ESFUERZOS FINALES

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    DIAGRAMA DE ESFUERZOS FINALES

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    cfcadm '6.0_ =s cftadm '6.1_ =s

  • ESTADO INICIAL:

    Pi = 180.0 Ton

    Ubicacin: 0.4 L1 Ubicacin: 1.0 L1

    Combinacin s t s b Combinacin s t s bpresf 65.3 -215.2 presf -317.9 167.9

    pp+alig -87.2 87.2 pp+alig 155.6 -155.6

    S = -22.0 -128.0 S = -162.3 12.3

    OK OK OK OK

    Esfuerzos Admisibles:

    s adm (comp.) = -168.0 kg/cm2

    s adm (traccin) = 13.4 kg/cm2

    ES

    0

    -22.0

    -128.0

    0

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    DIAGRAMA DE ESFUERZOS INICIALES

    -0.1

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    DIAGRAMA DE ESFUERZOS INICIALES

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    cifcadm '6.0_ =s icftadm '8.0_ =s

  • PREGUNTA 2 :

    -3.26

    16.302

    Colocando estos valores en el SAP2000 se obtiene

    DIAGRAMA DE MOMENTOS

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  • VERIFICANDO ESFUERZOS ADMISIBLES:

    ESTADO FINAL

    Ubicacin: 0.4 L1 Ubicacin: 1.0 L1

    Momentos SAP s t s b Momentos SAP s t s bTon.m kg/cm2 kg/cm2 Ton.m kg/cm2 kg/cm2

    pp+alig 20.930 -87.2 87.2 pp+alig -37.350 155.6 -155.6

    cm 7.560 -31.5 31.5 cm -13.490 56.2 -56.2

    s/c 17.220 -71.8 71.8 s/c -22.490 93.7 -93.7

    Axial -65.2 -65.2 Axial -65.2 -65.2

    -31.340 130.6 -130.6 44.920 -187.2 187.2

    Hiperestticos 3.880 Hiperestticos 9.700

    Los esfuerzos Finales sern: Los esfuerzos Finales sern:

    Combinacin s t s b Combinacin s t s bpresf 65.4 -195.8 presf -252.4 122.0

    pp+alig -87.2 87.2 pp+alig 155.6 -155.6

    cm -31.5 31.5 cm 56.2 -56.2

    s/c -71.8 71.8 s/c 93.7 -93.7

    S = -125.1 -5.3 S = 53.2 -183.6

    OK OK NO CUMPLE OK

    Esfuerzos Admisibles:

    s adm (comp.) = -210.0 kg/cm2

    s adm (traccin) = 29.9 kg/cm2

    ESTADO INICIAL:

    Pi = 180.0 Ton

    Ubicacin: 0.4 L1 Ubicacin: 1.0 L1

    Combinacin s t s b Combinacin s t s bpresf 75.2 -225.2 presf -290.2 140.2

    pp+alig -87.2 87.2 pp+alig 155.6 -155.6

    S = -12.0 -138.0 S = -134.6 -15.4

    OK OK OK OK

    Esfuerzos Admisibles:

    s adm (comp.) = -168.0 kg/cm2

    s adm (traccin) = 13.4 kg/cm2

    ES

    0

    -12.0

    -138.0

    0

    Estados de carga Estados de carga

    pretensado pretensado

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    DIAGRAMA DE ESFUERZOS INICIALES

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    cfcadm '6.0_ =s cftadm '6.1_ =s

    cifcadm '6.0_ =s icftadm '8.0_ =s