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• Eduardo Suárez

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Diseño Zapata Aislada - Nudo 13 M ateriales kgf f'c ≔ 210 ⋅ ―― cm 2

Resistencia de Suelo kgf fy ≔ 4200 ⋅ ―― cm 2

Cargas de Servicio P ≔ 24.87 ⋅ tonnef Mx ≔ 0.013 ⋅ tonnef ⋅ m My ≔ 0.58 tonnef ⋅ m

tonnef Qadm ≔ 10 ⋅ ――― m2

Cargas Últim as Pu ≔ 32.20 ⋅ tonnef Mux ≔ 0.018 ⋅ tonnef ⋅ m Muy ≔ 0.76 tonnef ⋅ m

Geom etria de la sección: Bx ≔ 1.8 ⋅ m By ≔ 1.8 ⋅ m bx ≔ 0.40 ⋅ m by ≔ 0.40 ⋅ m

h1 ≔ 0.2 ⋅ m h2 ≔ 1.0 ⋅ m h3 ≔ 0.35 ⋅ m

Azap ≔ Bx ⋅ By = 3.24 m 2 Df ≔ h1 + h2 + h3 = 1.55 m My ex ≔ ―― = 0.023 m P Mx ⎛ ey ≔ ―― = ⎝5.227 ⋅ 10 -4⎞⎠ m P

Bx = 0.3 m ―― 6 By = 0.3 m ―― 6

Com probación de que la carga esta en eltercio m edio de la zapata: ⎛ ⎞ Bx if ⎜ex < ―― , “Ok” , “Chequear”⎟ = “Ok” 6 ⎝ ⎠ ⎞ ⎛ By if ⎜ey < ―― , “Ok” , “Chequear”⎟ = “Ok” 6 ⎝ ⎠ ⎛ P 6 ⋅ ex 6 ⋅ ey ⎞ kgf qmax ≔ ――― ⋅ ⎜1 + ――+ ――⎟ = 0.829 ―― Bx ⋅ By ⎝ Bx By ⎠ cm 2 Se verifica que la presión transm itida alsuelo es m enor a la adm isible if ((qmax < Qadm , “Cumple” , “No Cumple”)) = “Cumple” Diagram a de reacciones delsuelo bajo cargas últim as en la cim entación Muy exu ≔ ――= 0.024 m Pu Mux ⎛ eyu ≔ ――= ⎝5.59 ⋅ 10 -4⎞⎠ m Pu

Bx = 0.3 m ―― 6 By = 0.3 m ―― 6

Com probación de que la carga esta en eltercio m edio de la zapata: ⎞ ⎛ Bx if ⎜exu < ―― , “Ok” , “Chequear”⎟ = “Ok” 6 ⎝ ⎠ ⎞ ⎛ By if ⎜eyu < ―― , “Ok” , “Chequear”⎟ = “Ok” 6 ⎝ ⎠ Calculo de volum en de reacciones en la cim entación: ⎛ 6 ⋅ exu 6 ⋅ eyu ⎞ Pu tonnef q1 ≔ ――― ⋅ ⎜1 + ―― + ――⎟ = 10.739 ――― Bx ⋅ By ⎝ Bx By ⎠ m2 ⎛ ⎞ 6 ⋅ exu 6 ⋅ eyu Pu tonnef q2 ≔ ――― ⋅ ⎜1 - ―― + ――⎟ = 9.175 ――― Bx ⋅ By ⎝ Bx By ⎠ m2 ⎛ ⎞ 6 ⋅ exu 6 ⋅ eyu Pu tonnef q3 ≔ ――― ⋅ ⎜1 + ―― - ――⎟ = 10.702 ――― Bx ⋅ By ⎝ Bx By ⎠ m2 ⎛ ⎞ 6 ⋅ exu 6 ⋅ eyu Pu tonnef q4 ≔ ――― ⋅ ⎜1 - ―― - ――⎟ = 9.138 ――― Bx ⋅ By ⎝ Bx By ⎠ m2 Diseño a corte La seccion critica alcortante se encuentra a una distancia "d" de las caras de las colum nas, se analiza en las dos direcciones "x" e "y". d ≔ h3 - 7.5 ⋅ cm = 0.275 m

Dirección X

⎛ 6 ⋅ exu ⎞ Pu tonnef qmax1 ≔ ――― ⋅ ⎜1 + ――⎟ = 10.72 ――― Bx ⋅ By ⎝ Bx ⎠ m2 ⎛ 6 ⋅ exu ⎞ Pu tonnef qmin1 ≔ ――― ⋅ ⎜1 - ――⎟ = 9.156 ――― Bx ⋅ By ⎝ Bx ⎠ m2 Para una distancia "d" delborde de la colum na se tiene un esfuerzo de: x1 ≔ 0 ⋅ m

x2 ≔ Bx = 1.8 m

qmax1 - qmin1 tonnef m ≔ ――――― = 0.869 ――― x2 - x1 m3 Bx bx x3 ≔ ―― + ― + d = 1.375 m 2 2 tonnef qcrit ≔ qmax1 - m ⋅ ((x2 - x3)) = 10.351 ――― m2 ((qmax1 + qcrit)) Vux ≔ ―――――― ⋅ By ⋅ ((Bx - x3)) = 8.06 tonnef 2 gon es : Elesfuerzo cortante que actua sobre la sección y la capacidad resistente delhorm i 1 Vux tonnef σu ≔ ――――= 19.156 ――― 0.85 ⋅ By ⋅ d m2

vc ≔ 0.5 ⋅

2

― 2

tonnef ‾‾‾ f'c = 22.913 ――― m

if ⎛⎝σu < vc , “OK” , “Falla a corte”⎞⎠ = ?

OK

Dirección Y ⎛ 6 ⋅ eyu ⎞ Pu tonnef qmax2 ≔ ――― ⋅ ⎜1 + ――⎟ = 9.957 ――― Bx ⋅ By ⎝ By ⎠ m2 ⎛ 6 ⋅ eyu ⎞ Pu tonnef qmin2 ≔ ――― ⋅ ⎜1 - ――⎟ = 9.92 ――― Bx ⋅ By ⎝ By ⎠ m2 Para una distancia "d" delborde de la colum na se tiene un esfuerzo de: x11 ≔ 0 ⋅ m

x22 ≔ By = 1.8 m

qmax2 - qmin2 tonnef m1 ≔ ――――― = 0.021 ――― x22 - 0 m3 By by x33 ≔ ―― + ― + d = 1.375 m 2 2 tonnef qcrity ≔ qmax2 - m1 ⋅ ((x22 - x33)) = 9.948 ――― m2 ((qmax2 + qcrity)) Vuy ≔ ―――――― ⋅ Bx ⋅ ((By - x33)) = 7.614 tonnef 2 Elesfuerzo cortante que actua sobre la sección y la capacidad resistente delhorm i gon es : 1 Vuy tonnef σu ≔ ――――= 18.095 ――― 0.85 ⋅ Bx ⋅ d m2

vc ≔ 0.5 ⋅

2

― 2

tonnef ‾‾‾ f'c = 22.913 ――― m

if ⎛⎝σu < vc , “OK” , “Falla a corte”⎞⎠ = ?

OK

Diseño a Cortante por Punzonamiento d Xcrit ≔ ―= 0.138 m 2 Pu tonnef q ≔ ――― = 9.938 ――― Bx ⋅ By m2 Vup ≔ q ⋅ ((((Bx ⋅ By)) - ((bx + 2 ⋅ Xcrit)) ⋅ ((by + 2 ⋅ Xcrit)))) = 27.672 tonnef Cortante actuante Esfuerzo cortante por punzonam iento que actua en la sección: Vup kgf σup ≔ ――――――――――――――――― = 4.385 ―― 0.85 ⋅ ((bx + Xcrit ⋅ 2)) ⋅ 2 ⋅ d + 0.85 ⋅ ((by + Xcrit ⋅ 2)) ⋅ 2 ⋅ d cm 2 Esfuerzo resistente a corte por punzonam iento:

Vc ≔

2

1 ― 2

kgf ‾‾‾ f'c = 14.491 ―― cm

if ⎛⎝σup < Vc , “OK” , “Falla”⎞⎠ = ?

OK

Diseño a flexion: Dirección X: Elrefuerzo requerido por flexion sera m ayor en la franja en que se encuentra elm áxim o esfuerzo espacialde reaccion delsuelo (q1-q2). kgf q1 = 1.074 ―― cm 2 kgf x2 = 1.8 m q2 = 0.917 ―― cm 2 q1 - q2 ⎛ kgf mx ≔ ――― = ⎝8.688 ⋅ 10 -4⎞⎠ ―― x2 - x1 cm 3

x1 = 0 m

Bx bx x3 ≔ ―― + ― = 1.1 m 2 2

kgf qcrit3 ≔ q1 - mx ⋅ ⎛⎝x2 - x3⎞⎠ = 1.013 ―― cm 2

kgf qx1 ≔ q1 - qcrit3 = 0.061 ―― cm 2

bx1 ≔ Bx - x3 = 0.7 m

M om ento flector en zona critica en una franja de 1 m : ⎛ qcrit3 ⋅ bx1 2 ⎛ qx1 ⋅ bx1 ⎞ ⎛ 2 ⋅ bx1 ⎞⎞ Mux ≔ ⎜―――― + ⎜―――⎟ ⋅ ⎜――― ⎟⎟ ⋅ 100 cm = 2.581 tonnef ⋅ m 2 2 ⎝ ⎝ ⎠ ⎝ 3 ⎠⎠ Acero de refuerzo: ⎞ 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ 2 ⋅ Mux 0.85 ⋅ f'c ⋅ 100 cm ⋅ d ⎛ ⎟ = 2.51 cm 2 Asxf ≔ ――――――― ⋅ ⎜1 - 1 - ――――――――― 2 fy ⎜⎝ 0.85 ⋅ 0.90 ⋅ f'c ⋅ 100 cm ⋅ d ⎟⎠ Asmin ≔ 0.0033 ⋅ 100 cm ⋅ d = 9.075 cm 2 Asx ≔ if ⎛⎝Asxf < Asmin , Asmin , Asxf⎞⎠ = 9.075 cm 2 /m

Acero de diseño

Dirección Y: Elrefuerzo requerido por flexion sera m ayor en la franja en que se encuentra elm áxim o esfuerzo espacialde reaccion delsuelo (q1-q3). kgf q3 = 1.07 ―― cm 2 kgf By = 1.8 m q1 = 1.074 ―― cm 2 q1 - q2 ⎛ kgf my ≔ ――― = ⎝8.688 ⋅ 10 -4⎞⎠ ―― By - x1 cm 3

x1 = 0 m

By by yy ≔ ―― + ― = 1.1 m 2 2 kgf qy1 ≔ q1 - qcrity = 0.061 ―― cm 2

kgf qcrity ≔ q1 - my ⋅ ⎛⎝By - yy⎞⎠ = 1.013 ―― cm 2 by1 ≔ By - yy = 0.7 m

M om ento flector en zona critica en una franja de 1 m :

⎛ qcrity ⋅ by1 2 ⎛ qy1 ⋅ by1 ⎞ ⎛ 2 ⋅ by1 ⎞⎞ Muy ≔ ⎜―――― + ⎜―――⎟ ⋅ ⎜――― ⎟⎟ ⋅ 100 cm = 2.581 tonnef ⋅ m 2 2 ⎝ ⎝ ⎠ ⎝ 3 ⎠⎠ Acero de refuerzo: ⎞ 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ 2 ⋅ Muy 0.85 ⋅ f'c ⋅ 100 cm ⋅ d ⎛ ⎟ = 2.51 cm 2 ⋅ ⎜1 - 1 - ――――――――― Asxf ≔ ――――――― 2 fy ⎜⎝ 0.85 ⋅ 0.90 ⋅ f'c ⋅ 100 cm ⋅ d ⎟⎠ Asmin ≔ 0.0033 ⋅ 100 cm ⋅ d = 9.075 cm 2 Asx ≔ if ⎛⎝Asxf < Asmin , Asmin , Asxf⎞⎠ = 9.075 cm 2 /m

Acero de diseño