Ejemplo De C_lculo (asce 7-10) 2v4k27

EJEMPLO CÁLCULO ACCIÓN DEL VIENTO ASCE7-10 GEOMETRÍA

STEP 1 : Determinar la categoría de riego Categoría II - Uso vivienda

Table 1.4-1 ASCE7

STEP 2 : Determinar la velocidad básica del viento m V ≔ 49 ― s

MRI=700 años ; Exposición C

Figure 26.5-1A ASCE7

STEP 3 : Parámetros de viento kd ≔ 0.85

Factor de direccionabilidad

Categoría de exposición : B

Table 26.6-1 ASCE7-10 Section 26.7 ASCE7-10

kzt ≔ 1

Factor topográfico

Section 26.8 ASCE7-10

G ≔ 0.85

Factor de ráfaga

Section 26.9 ASCE7-10

Gis ≔ −0.18

Coeficiente de presión interno succión

Gip ≔ 0.18

Coeficiente de presión interno presión

STEP 4 : Coeficiente Kz o Kh kz30 ≔ 0.99

kh ≔ 0.99

Table 27.3-1 ASCE7-10

kz20 ≔ 0.87 kz10 ≔ 0.72 STEP 5 : Velocidad de presión

Section 27.3.2 ASCE7 10

Section 27.3.2 ASCE7-10

STEP 5 : Velocidad de presión

2

s C ≔ 0.613 N ⋅ ―― 4 m

2

qh ≔ C ⋅ kh ⋅ kzt ⋅ kd ⋅ V = 1.24 kPa qz30 ≔ qh = 1.24 kPa 2

qz20 ≔ C ⋅ kz20 ⋅ kzt ⋅ kd ⋅ V = 1.09 kPa 2

qz10 ≔ C ⋅ kz10 ⋅ kzt ⋅ kd ⋅ V = 0.9 kPa

STEP 6.1 : Viento X+. Succión interior

Gis ≔ −0.18

H ≔ 30

L ―= 0.333 B

L ≔ 25

Fachada barlovento

B ≔ 75

pwx_30 ≔ qz30 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 1.07 kPa pwx_20 ≔ qz20 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 0.96 kPa pwx_10 ≔ qz10 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 0.84 kPa

Fachada sotavento

plx ≔ qh ⋅ G ⋅ (−0.7) − qh ⋅ Gis = −0.51 kPa

Cubierta - 0 to H/2

pc ≔ qh ⋅ G ⋅ (−1) − qh ⋅ Gis = −0.83 kPa

Cubierta - > H/2

pc ≔ qh ⋅ G ⋅ (−0.7) − qh ⋅ Gis = −0.51 kPa

Cubierta - 0 to H/2

pc ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gis = 0.03 kPa

Cubierta - > H/2

pc ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gis = 0.03 kPa

H ―= 1.2 L

STEP 6.2 : Viento X+. Presión interior Fachada barlovento

Gip ≔ 0.18

pwx_30 ≔ qz30 ⋅ G ⋅ 0.8 − qh ⋅ Gip = 0.62 kPa pwx_20 ≔ qz20 ⋅ G ⋅ 0.8 − qh ⋅ Gip = 0.52 kPa pwx_10 ≔ qz10 ⋅ G ⋅ 0.8 − qh ⋅ Gip = 0.39 kPa

Fachada sotavento

plx ≔ qh ⋅ G ⋅ (−0.7) − qh ⋅ Gip = −0.96 kPa

Cubierta - 0 to H/2

pc ≔ qh ⋅ G ⋅ (−1) − qh ⋅ Gip = −1.28 kPa

Cubierta - > H/2

pc ≔ qh ⋅ G ⋅ (−0.7) − qh ⋅ Gip = −0.96 kPa

Cubierta - 0 to H/2

pc ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gip = −0.41 kPa

Cubierta - > /2

pc ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gip = −0.41 kPa

STEP 7.1 : Viento Y+. Presión interior Fachada barlovento

Gip ≔ 0.18

pwy_30 ≔ qz30 ⋅ G ⋅ 0.8 − qh ⋅ Gip = 0.62 kPa pwy_20 ≔ qz20 ⋅ G ⋅ 0.8 − qh ⋅ Gip = 0.52 kPa pwy_10 ≔ qz10 ⋅ G ⋅ 0.8 − qh ⋅ Gip = 0.39 kPa

Fachada sotavento

ply ≔ qh ⋅ G ⋅ (−0.25) − qh ⋅ Gip = −0.49 kPa

Cubierta - succión máxima 0 to 15m

pcy_15 ≔ qh ⋅ G ⋅ (−0.9) − qh ⋅ Gip = −1.17 kPa

15 to 30m

pcy_30 ≔ qh ⋅ G ⋅ (−0.9) − qh ⋅ Gip = −1.17 kPa

30 to 60m

pcy_60 ≔ qh ⋅ G ⋅ (−0.5) − qh ⋅ Gip = −0.75 kPa

60 to 75m

pcy_75 ≔ qh ⋅ G ⋅ (−0.3) − qh ⋅ Gip = −0.54 kPa

Cubierta - succión mínima 0 to 15m

pcy_15 ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gip = −0.41 kPa

15 to 30m

pcy_30 ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gip = −0.41 kPa

30 to 60m

pcy_60 ≔ qh ⋅ G ⋅ (−0.18)) − qh ⋅ Gip = −0.41 kPa

60 to 75m

pcy_75 ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gip = −0.41 kPa

STEP 7.2 : Viento Y+. Succión interior Fachada barlovento

Gis ≔ −0.18

pwy_30 ≔ qz30 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 1.07 kPa pwy_20 ≔ qz20 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 0.96 kPa pwy_10 ≔ qz10 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 0.84 kPa

Fachada sotavento

ply ≔ qh ⋅ G ⋅ (−0.25) − qh ⋅ Gis = −0.04 kPa

Cubierta - succión máxima 0 to 15m

pcy_15 ≔ qh ⋅ G ⋅ (−0.9) − qh ⋅ Gis = −0.72 kPa

15 to 30m

pcy_30 ≔ qh ⋅ G ⋅ (−0.9) − qh ⋅ Gis = −0.72 kPa

30 to 60m

pcy_60 ≔ qh ⋅ G ⋅ (−0.5) − qh ⋅ Gis = −0.3 kPa

60 to 75m

pcy_75 ≔ qh ⋅ G ⋅ (−0.3) − qh ⋅ Gis = −0.09 kPa

Cubierta - succión mínima 0 to 15m

pcy_15 ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gis = 0.03 kPa

15 to 30m

pcy_30 ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gis = 0.03 kPa

30 to 60m

pcy_60 ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gis = 0.03 kPa

60 to 75m

pcy_75 ≔ qh ⋅ G ⋅ (−0.18) − qh ⋅ Gis = 0.03 kPa

STEP 8 : Resumen y estrategia

Estrategia y simplificaciones 1. No consideramos carga en cubierta. Edificio muy masivo 2. No consideramos carga en las paredes laterales 3. Realizamos el ejemplo con Viento en dirección X. El análisis en dirección Y se haría de manera análoga STEP 9. Casos de carga Caso 1. Viento en dirección X

Fachada barlovento pwx_30 ≔ qz30 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 1.07 kPa pwx_20 ≔ qz20 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 0.96 kPa pwx_10 ≔ qz10 ⋅ G ⋅ 0.8 − qh ⋅ Gis = 0.84 kPa

Fachada sotavento plx ≔ qh ⋅ G ⋅ (−0.7) − qh ⋅ Gis = −0.51 kPa

Caso 2. Viento en dirección X con reducción y momento torsor Fachada barlovento

Fachada sotavento

pwx_30 ⋅ 0.75 = 0.8 kPa

plx ⋅ 0.75 = −0.39 kPa

pwx_20 ⋅ 0.75 = 0.72 kPa pwx_10 ⋅ 0.75 = 0.63 kPa Momento torsor aplicado a cada piso Bx ≔ 75 m

ex ≔ 0.15 ⋅ Bx = 11.25 m

h≔3 m MT_30 ≔ 0.75 ⋅ ⎛⎝pwx_30 + plx⎞⎠ ⋅ h ⋅ Bx ⋅ ex = 1046.3 kN ⋅ m MT_20 ≔ 0.75 ⋅ ⎛⎝pwx_20 + plx⎞⎠ ⋅ h ⋅ Bx ⋅ ex = 852.5 kN ⋅ m MT_10 ≔ 0.75 ⋅ ⎛⎝pwx_10 + plx⎞⎠ ⋅ h ⋅ Bx ⋅ ex = 610.3 kN ⋅ m

Caso 3. Viento en dirección X e Y con reducción Dirección +X Fachada barlovento

Dirección +Y Fachada barlovento

pwx_30 ⋅ 0.75 = 0.8 kPa

pwy_30 ⋅ 0.75 = 0.8 kPa

pwx_20 ⋅ 0.75 = 0.72 kPa

pwy_20 ⋅ 0.75 = 0.72 kPa

pwx_10 ⋅ 0.75 = 0.63 kPa

pwy_10 ⋅ 0.75 = 0.63 kPa

Fachada sotavento plx ⋅ 0.75 = −0.39 kPa

Fachada sotavento ply ⋅ 0.75 = −0.03 kPa

Caso 4. Viento en dirección X e Y con reducción y momento torsor

Dirección +X

Dirección +Y

Fachada barlovento

Fachada barlovento

pwx_30 ⋅ 0.563 = 0.6 kPa

pwy_30 ⋅ 0.563 = 0.6 kPa

pwx_20 ⋅ 0.563 = 0.54 kPa

pwy_20 ⋅ 0.563 = 0.54 kPa

pwx_10 ⋅ 0.563 = 0.47 kPa

pwy_10 ⋅ 0.563 = 0.47 kPa

Fachada sotavento plx ⋅ 0.563 = −0.29 kPa

Fachada sotavento ply ⋅ 0.563 = −0.02 kPa

Momento torsor Bx ≔ 75 m

By ≔ 25 m

ex ≔ 0.15 ⋅ Bx = 11.25 m ey ≔ 0.15 ⋅ By = 3.75 m En cada uno de los diafragmas se aplicarán los siguientes momento torsores h≔3 m MT_30 ≔ 0.563 ⋅ ⎛⎝pwx_30 + plx⎞⎠ ⋅ Bx ⋅ ex ⋅ h + 0.563 ⋅ ⎛⎝pwy_30 + ply⎞⎠ ⋅ By ⋅ ey ⋅ h = 947.7 kN ⋅ m MT_20 ≔ 0.563 ⋅ ⎛⎝pwx_20 + plx⎞⎠ ⋅ Bx ⋅ ex ⋅ h + 0.563 ⋅ ⎛⎝pwy_20 + ply⎞⎠ ⋅ By ⋅ ey ⋅ h = 786.1 kN ⋅ m MT_10 ≔ 0.563 ⋅ ⎛⎝pwx_10 + plx⎞⎠ ⋅ Bx ⋅ ex ⋅ h + 0.563 ⋅ ⎛⎝pwy_10 + ply⎞⎠ ⋅ By ⋅ ey ⋅ h = 584 kN ⋅ m