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Practica calificada de estabilidad de taludes

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mecanica de taludes

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Practica calificada de estabilidad de taludes

  1. 1. PRACTICA CALIFICADA DE ESTABILIDAD DE TALUDES PROBLEMA 01: De conformidad con los datos P = 50º, γr = 0.0027MPa/m = 2.7TM/m3, F= 60º, H = 30 m g = 9.81KN/m3 Φ = 35º, C = 10TM/m2, Z= 8 m. ¿ Encontrar la variación del F.S. cuando esta totalmente seco a saturado V=0 Y V≠ 0 U = 0 U≠ 0  Cálculo del valor de Z/H 𝑍 𝐻 = 8 30 = 0.27  CALCULO DEL AREA DE LA SUPERFICIE 𝐴 = 𝐻 − 𝑍 𝑠𝑒𝑛 50° = 28.72 𝑚.  Calculo del U 𝑈 = 1 2 .  𝑟. 𝑍 𝑊 (𝐻 − 𝑍) 𝑠𝑒𝑛 50° 30 m Z F P
  2. 2. 𝑈 = 1 2 . 1 𝑇𝑛 𝑚3 .8𝑚 28.72𝑚 𝑼 = 𝟏𝟏𝟒. 𝟖𝟖𝑻𝒏/𝒎  Calculo del volumen 𝑽 = 𝟏 𝟐 ( 𝒁 𝒘) 𝟐 𝑽 = 𝟏 𝟐 .1. 𝟖 𝟐 𝑽 = 𝟑𝟐 𝑻𝒏 𝒎 𝑾 = 𝟏 𝟐  𝐻 2 ⌊ 1 − ( 𝑍 𝐻) 2 𝑡𝑔  𝑝 − 1 𝑡𝑔 𝑓 ⌋ 𝑾 = 𝟏 𝟐 (2.7) 30 2 ⌊ 1− (0.27) 2 𝑡𝑔 50° − 1 𝑡𝑔60° ⌋ 𝑾 = 𝟐𝟒𝟓. 𝟓𝟑 Factor de seguridad en estado seco: Zw = 0 U = 0 V = 0 𝑭𝑺 = 𝑪𝑨 + (𝑾 𝑪𝒐𝒔 𝑝 − 𝑈 − 𝑉 𝑆𝑒𝑛  𝑝) 𝑻𝒈 ∅ 𝑾 𝑺𝒆𝒏  𝑝 + 𝑉 𝐶𝑜𝑠  𝑝 𝑭𝑺 = 𝟒 𝑻𝒏 𝒎 𝟐 (𝟐𝟖. 𝟕𝟐𝒎 𝟐 ) + 245.53(𝐶𝑜𝑠50°)𝑻𝒈 𝟑𝟎° 𝟐𝟒𝟓. 𝟓𝟑 (𝑺𝒆𝒏 50°) FS = 1.095 El factor de seguridad completamente saturado
  3. 3. 𝑭𝑺 = 𝑪𝑨 + (𝑾 𝑪𝒐𝒔 𝑝 − 𝑈 − 𝑉 𝑆𝑒𝑛  𝑝) 𝑻𝒈 ∅ 𝑾 𝑺𝒆𝒏  𝑝 + 𝑉 𝐶𝑜𝑠  𝑝 𝑭𝑺 = 𝟒(𝟐𝟖. 𝟕𝟐) + (𝟐𝟒𝟓. 𝟓𝟑 𝑪𝒐𝒔50° − 114.88 − 32 𝑆𝑒𝑛 50°)𝑻𝒈 𝟑𝟎° 𝟐𝟒𝟓. 𝟓𝟑 𝑺𝒆𝒏 50°+ 32 𝐶𝑜𝑠 50° FS = 0.60

mecanica de taludes

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