Geotechnics 2 Repetition PDF

GEOTECHNICS 2 REPITITION Dr. Ákos Wolf What do you need to know from Geo1? 2  soil densities and phase relationships  stresses in soil  in-situ stress state (vertical and horizontal stress)  total, effective and neutral stress  ov...

GEOTECHNICS 2 REPITITION Dr. Ákos Wolf What do you need to know from Geo1? 2  soil densities and phase relationships  stresses in soil  in-situ stress state (vertical and horizontal stress)  total, effective and neutral stress  overconsolidation ratio  strength parameter of soil:  shear strength  Coulomb criteria  friction angle (j) and cohesion (c)  laboratory testing of strength  soil compression  oedometer modulus – compression curve  consolidation Soil bulk densities 3 air air water water particle particle particle dry wet saturated rd = md / V rn = mn / V rt = mt / V md = mass of particle mn = mass of particle and water mt = mass of particle and water 1,6-1,9 g/cm3 1,8-2,0 g/cm3 1,9-2,1 g/cm3 as a density indicator pressure calculation pressure calculation under GWL  Phase relationships 4 The following definitions have been established to describe the proportion Vl of each constituent in the soil Vv Vh e e∙S r V 1+e Vs 1 mv Vv. ρv e. Sr. ρv w= = = ms Vs. ρs 1. ρs  Stresses in soil – in-situ stress state 5 sz0 sz0 sz0 z sx0 sx0 𝜎z0 = 𝑝 + 𝑧 ⋅ 𝜌 ⋅ 𝑔 𝜎z0 = 𝜎′𝑧0 + 𝑢0 𝑢 = ℎ𝑣 ⋅ 𝜌𝑣 ⋅ 𝑔 Stresses in soil – in-situ stress state 6 sz0 sx0 sz0 sx0 sz0 z sx0 sx0 𝜎z0 = 𝑝 + 𝑧 ⋅ 𝜌 ⋅ 𝑔 𝜎x0 ′ = 𝜎𝑧0 ′ ⋅ 𝐾0 𝜎z0 = 𝜎′𝑧0 + 𝑢0 𝐾0 = (1 − sin 𝜙′) ⋅ 𝑂𝐶𝑅  Example 7 Calculate the in situ stresses at points A, B and C! 𝜎z0 = 𝑝 + 𝑧 ⋅ 𝜌 ⋅ 𝑔 = 𝑝 + ෍ ℎ𝑖 ∙ 𝜌𝑖 ∙ 𝑔 rn = 1,90 g/cm3 2,0 m clay j = 24° c = 30 kPa 𝑢0 = ℎ𝑣 ⋅ 𝜌𝑣 ⋅ 𝑔 A rn = 1,95 g/cm3 𝜎z0 = 𝜎′𝑧0 + 𝑢0 1,4 m B GWL j = 30° 𝜎x0 ′ = 𝜎𝑧0 ′ ⋅ 𝐾0 c = 0 kPa 2,6 m sand rt = 2,00 g/cm3 𝐾0 = (1 − sin 𝜙′) ⋅ 𝑂𝐶𝑅 C  Example 8 Calculate the in situ stresses at points A, B and C! sz0 𝜎z0𝐴 = 2,0 ∙ 1,9 ∙ 10 = 38 𝑘𝑃𝑎 rn = 1,90 g/cm3 𝜎z0B = 𝜎z0𝐴 + 1,4 ∙ 1,95 ∙ 10 = 65,3 𝑘𝑃𝑎 2,0 m j = 24° clay 𝜎z0C = 𝜎z0B + 2,6 ∙ 2,0 ∙ 10 = 117,3 𝑘𝑃𝑎 c = 30 kPa A 38 𝑢𝐴 = 𝑢𝐵 = 0 𝑘𝑃𝑎 rn = 1,95 g/cm3 𝑢𝐶 = 2,6 ∙ 1,0 ∙ 10 = 26 𝑘𝑃𝑎 1,4 m sz0’ B GWL 65,3 j = 30° ′ c = 0 kPa 𝜎z0 𝐴 = 𝜎z0𝐴 = 38 𝑘𝑃𝑎 𝜎z0B ′ = 𝜎z0B = 65,3 𝑘𝑃𝑎 2,6 m sand rt = 2,00 g/cm3 ′ sz0’ 𝜎z0C = 𝜎z0C − 𝑢𝑐 = 91,3 𝑘𝑃𝑎 C u0 117,3  Example 9 Calculate the in situ stresses at points A, B and C! sz0 sx0 𝐾0,𝑐 = 1 − 𝑠𝑖𝑛24 = 0,593 rn = 1,90 g/cm3 𝐾0,𝑠 = 1 − 𝑠𝑖𝑛30 = 0,50 2,0 m clay j = 24° c = 30 kPa A 𝜎𝑥𝐴,𝑐 ′ = 𝐾0,𝑐 ∙ 𝜎𝑧𝐴 ′ = 22,5 𝑘𝑃𝑎 38 22,5 𝜎𝑥𝐴,𝑠 ′ = 𝐾0,𝑠 ∙ 𝜎𝑧𝐴 ′ = 19,0 𝑘𝑃𝑎 rn = 1,95 g/cm3 19 1,4 m sz0’ B sx0’ GWL 65,3 32,65 j = 30° ′ 𝜎𝑥𝐵 ′ = 𝐾0,𝑠 ∙ 𝜎𝑧𝐵 = 32,65 𝑘𝑃𝑎 c = 0 kPa 𝜎𝑥𝐶 ′ = 𝐾0,𝑠 ∙ 𝜎𝑧𝐶 ′ = 45,65 𝑘𝑃𝑎 2,6 m sand rt = 2,00 g/cm3 ′ sz0’ sx0’ 𝜎𝑥𝐶 = 𝜎𝑥𝐶 + 𝑢0 = 71,65 𝑘𝑃𝑎 C u0 u0 117,3 71,65 Soil strength 10  shear strength (tf) of a soil is the maximum internal resistance to applied shearing force  effective friction angle (j) is a measure of the shear strength of soils due to friction  cohesion (c) is a measure of the intermolecular forces, soil tension and cementation  undrained shear strength (su) is the shear strength of a soil when sheared at constant volume  Soil strength 11 Coulomb failure criteria For cohesionless soil t 𝜏 = 𝜎 ∙ 𝑡𝑔𝜑 + 𝑐 j tt For cohesive soil 2 𝜑 𝜑 𝜎1 = 𝜎3 ∙ 𝑡𝑔 45 + + 2 ∙ 𝑐 ∙ 𝑡𝑔 45 + 2 2 2 𝜑 𝜑 𝜎3 = 𝜎1 ∙ 𝑡𝑔 45 − − 2 ∙ 𝑐 ∙ 𝑡𝑔 45 − c 2 2 s3 st s1 s  Soil strength 12 s1 Coulomb failure criteria t 𝜏 = 𝜎 ∙ 𝑡𝑔𝜑 + 𝑐 s3 s3 s3 s1 j 45+j/2 tt s3 s1 s3 s3 90+j c s3 st s1 s s1  Soil strength – triaxial test 13 Dsz pc – cell pressure Dsz – failure pressure t pc pc j pc pc pc pc c s3=pc s1=pc+Dsz s Dsz  Soil strength – triaxial test 14 Dsz pc – cell pressure Dsz – failure pressure t pc pc j pc pc pc pc c s3=pc s1=pc+Dsz s Dsz  Soil strength – direct shear 15 N → s = N/A – normal stress on failure plane T → t = N/A – shear stress on failure plane N t 𝜏 = 𝜎 ∙ 𝑡𝑔𝜑 + 𝑐 T Failure plane T N j c s Soil Strength - example 16 Calculate the soil strength’s parameters based on the following direct shear test results! - a = 6 cm t − s1 = 100 kPa and s2 = 200 kPa - T1 = 212,5 N and T2 = 415 N j c s Soil Strength - example 17 Calculate the soil strength’s parameters based on the following direct shear test results! - a = 6 cm A = 0,0036 m2 150 − s1 = 100 kPa and s2 = 200 kPa shear stress, t [kPa] - T1 = 212,5 N and T2 = 415 N 100 50 t1 = T1/A = 59 kPa t2 = T2/A = 115 kPa 0 0 50 100 150 200 250 normal stress, s [kPa] Soil Strength - example 18 Calculate the soil strength’s parameters based on the following direct shear test results! - a = 6 cm A = 0,0036 m2 150 150 − s1 = 100 kPa and s2 = 200 kPa [kPa] stress, tt [kPa] - T1 = 212,5 N and T2 = 415 N 100 t2-t1 shear stress, j 50 s2-s1 shear t1 = T1/A = 59 kPa t2 = T2/A = 115 kPa 0 0 50 100 150 200 250 normal stress, s [kPa] 𝜏2 − 𝜏1 𝑡𝑔𝜑 = 𝜎2 − 𝜎1 j = 29,25° Soil Strength - example 19 Calculate the soil strength’s parameters based on the following direct shear test results! - a = 6 cm A = 0,0036 m2 150 − s1 = 100 kPa and s2 = 200 kPa shear stress, t [kPa] - T1 = 212,5 N and T2 = 415 N 100 t2-t1 j j = 29,25° 50 s2-s1 t1 = T1/A = 59 kPa 0 0 50 100 150 200 250 normal stress, s [kPa] 𝜏 = 𝜎 ∙ 𝑡𝑔𝜑 + 𝑐 c = 3 kPa Compressibility of soil 20  Consolidation is the time dependent settlement of soils resulting from the explusion of water from the soil pores  Primary consolidation is the change in volume of a fine-grained soil caused by the explusion of water from the voids and the transfer of thress from the excess porewater pressure to the soil particles  types of strains  shear strain (g):  volumetric strain (e) Compressibility of soils 21 The compression of soils is time and stress dependent process (De = f(s,t)) Compression curve - test 22 oedometer test:  linear strain condition  load: eg. 10-50-100-200-400-800 kPa vertical stress, sz [kN/m2]  loading time: 24 hours / consolidation vertical volumetric strain, ez [%] Compression curve - linearization 23 Compression modulus (oedometric modulus, modulus of volume compressibility) ∆𝜎𝑧 𝐸𝑠 = vertical stress, sz [kN/m2] ∆𝜀𝑧 sz0’ sz0’+Dsz’ vertical volumetric strain, ez [%] Dsz Changing with - initial stress level - increased stress level ez0 Dez ez0+Dez  Compression – example 24 Calculate compression of the clay layer (the settlement of the enbankment)! compression curve of clay vertical stress, sz [kPa] enbankment 5m 0 50 100 150 200 r = 2 g/cm3 0,0 0,5 vertical strain, ez [%] clay 5m 1,0 r = 2 g/cm3 1,5 2,0 2,5  Compression – example 25 Calculate compression of the clay layer (the settlement of the enbankment)! compression curve of clay vertical stress, sz [kPa] enbankment 5m 0 50 100 150 200 r = 2 g/cm3 0,0 0,5 2,5 m vertical strain, ez [%] clay 5m 1,0 r = 2 g/cm3 1,5 2,0 𝜎z0 = 𝜌 ∙ 𝑧 ∙ 𝑔 = 2,0 ∙ 2,5 ∙ 10 = 50 𝑘𝑃𝑎 2,5 ∆𝜎z = 𝜌 ∙ 𝐻 ∙ 𝑔 = 2,0 ∙ 5,0 ∙ 10 = 100 𝑘𝑃𝑎  Compression – example 26 Calculate compression of the clay layer (the settlement of the enbankment)! compression curve of clay sz0’ vertical stress, sz [kPa] enbankment s ’+Dsz’ 200 5m 0 50 100 150 z0 r = 2 g/cm3 0,0 0,5 Dsz 2,5 m vertical strain, ez [%] ez0 clay 5m 1,0 r = 2 g/cm3 Dez 1,5 2,0 ez0+Dez 𝜎z0 = 𝜌 ∙ 𝑧 ∙ 𝑔 = 2,0 ∙ 2,5 ∙ 10 = 50 𝑘𝑃𝑎 2,5 ∆𝜎z = 𝜌 ∙ 𝐻 ∙ 𝑔 = 2,0 ∙ 5,0 ∙ 10 = 100 𝑘𝑃𝑎  Compression – example 27 Calculate compression of the clay layer (the settlement of the enbankment)! ∆𝜎𝑧 100 𝐸𝑠 = = = 9,1 𝑀𝑃𝑎 ∆𝜀𝑧 (0,021 − 0,01) compression curve of clay sz0’ vertical stress, sz [kPa] enbankment s ’+Dsz’ 200 5m 0 50 100 150 z0 r = 2 g/cm3 0,0 0,5 Dsz 2,5 m vertical strain, ez [%] ez0 clay 5m 1,0 r = 2 g/cm3 Dez 1,5 2,0 ez0+Dez 𝜎z0 = 𝜌 ∙ 𝑧 ∙ 𝑔 = 2,0 ∙ 2,5 ∙ 10 = 50 𝑘𝑃𝑎 2,5 ∆𝜎z = 𝜌 ∙ 𝐻 ∙ 𝑔 = 2,0 ∙ 5,0 ∙ 10 = 100 𝑘𝑃𝑎  Compression – example 28 Calculate compression of the clay layer (the settlement of the enbankment)! ∆𝜎𝑧 100 𝐸𝑠 = = = 9,1 𝑀𝑃𝑎 ∆𝜀𝑧 (0,021 − 0,01) enbankment 5m r = 2 g/cm3 2,5 m clay ∆𝜎𝑧 100 ∆ℎ = ∙ℎ = ∙ 5,0 = 0,055 𝑚 = 5,5 𝑐𝑚 5m r = 2 g/cm3 𝐸𝑠 9100 𝜎z0 = 𝜌 ∙ 𝑧 ∙ 𝑔 = 2,0 ∙ 2,5 ∙ 10 = 50 𝑘𝑃𝑎 ∆𝜎z = 𝜌 ∙ 𝐻 ∙ 𝑔 = 2,0 ∙ 5,0 ∙ 10 = 100 𝑘𝑃𝑎 Compressibility of soils – unloading-reloading 29 𝐸𝑈𝑅 = 3 ÷ 5 ∙ 𝐸𝑠