Sofia Martinez Tarea 4

Geotecnia II\n1. Datos\nB = 5m\nl = 7m\nq0 = 200 kN/m2\nz = 7m\n\nSolución\nPaso 1 B1 B2 l1 y l2\nB1 = B2 = 5/2 = 2.5 m\nl1 = l2 = 7/2 = 3.5 m\n\nPaso 2 Define m, n\nm = B1 = B2/z\nm = 2.5/7 = 0.3\n\nPaso 3\nΔq = q0 (4.1)\nAe = 200 x 4 x 0.055973\nAe = 44.744 kN/m2 Bulbo de tensiones\nB = 3\nz = 2.5\n(-4m a -6m) Arena -> C = 0\nOr = q0 Na Sq + 1/2 y BNySy\nNspt = 6 -> y = 17 kN/m3\nq = 3.17 = 5/ kPa\nNspt medio = (18 + 22 + 20)/2\nθ = 28 + 0.4 (2.0) = 36°\nNc = 50.5\nNa = 37.75\nNy/Nc = 0.75\ntg θ = 0.73\n\nB = 1\nSa = 4 + tg 36 = 1.73\nSy = 0.60\n\nσz' = 51.37 75 . 1.73 + 1/2 . 3.56 31 0.60\nσz' = 3.38 MPa\n\nσa' = 3.38/3 = 1.127 MPa Bulbo de tensiones\nB = 3\n(De = 5m a -10m) (arcilla saturada)\nNspt = (22 + 8 + 4 + 11 + 11 = 12.2)/5\nC* = 10. Nspt = 10. 12.2 = 122 kPa\nC* = 2/3 C = 2/3. 12.2 = 21.33 kPa\nCmed = (12.2 + 12.33)/2 = 102.665 kPa\nφ = 0\nNc = 5.14\nNa = 1\nNy = 0\nNay/Nc = 0.10\ntg θ = 0\nq = Sa y + 2 (y sat - 10)\nq = 1.17 + 2(21 - 10)\nσ' = 39 kPa\n\nσx' = Cmed Nc Sc + q Na Sa\nσx' = 101.66 5 5.14 1.20 + 32.1.1\nσz' = 0.666 MPa σ1 = 3.38 MPa\nσ2 = 0.666 MPa\nσ12 = (3.38 + 0.666) / 2 = 2.023 MPa\nΔσ = σ12 . B . L / (B+2)(L+ε)\nΔσ = 2.023 . 3 . 3 / (3+4)(3+4) = 10.371 MPa\nz = 4\nΔσ = 0.37 MPa ≤ σ12 = 0.666 MPa (σ2)\nσa = 0.37 / 3 = 0.123 MPa\n\nMétodo: semiemperico\n\nBalzo (-4m a -6m) B = 3\nNspTmed = 20\nσa = 20 + 0.051 MPa = 0.451 MPa\n\nBulbo (-6m a -10m) B = 3\nNspT = 12.2\nσr = 12.2 / 50 - 0.051 MPa = 0.295 MPa b) semi-emperico\n\nDb = 1.5m (-8a - 11m)\nNspTmed = (-9 + 9 + 7) / 3 = 8.308 → y = 17 kN/m³\nq = 8 x 17 = 136 kN/m² = 0.136 MPa\nσa = NspT + q (MPa)\nσu = 8 / 50 + 0.136 MPa = 0.29 MPa\nσa = NspT x q (MPa)\n= 8 / 30 = 0.17 MPa\nDb = 3m (-8a - 14m)\nNspTprom = (-9 + 7 + 9 + 11) / 6 = 8.7 → y = 17 kN/m³\nq = 9 x 17 = 163 kN/m² = 0.15 MPa\nσa = NspT + q = 9 + 0.75 = 0.33 MPa\nσg = NspT - q = 0.3 MPa\n\nb) Pilotes Excavados\n\nAux. vel. (Arena Arcilosa, NspT = cota de apoyo: 75 s)\nσi = k . NspT F1 = 3; k = 0.60 MPa; NspT: 9 en la...\nσ1 = 0.60 MPa x a = 1.8 MPa\nσa = σr + 1.3, 0.6 MPa Decourt Quarenta (c = 0.40 MPa, NP = media de los tres valores de NspT = cota de apoyo anterior y posterior/3\nF = 5=4)\nArena → q = 0.50\nσ1 = q . c . NP\nNP = (q + 9 + 7) / 3 = (8 + q + 7) / 3\nσr = 0.950 x 0.40 x 8 = 1.6 MPa\nσg = σr - 1.6 = 0.4 MPa