Licture_4 - Subsurface Exploration Subsurface 2.1. Purpose...

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Unformatted text preview: Subsurface Exploration Subsurface 2.1. Purpose of Subsurface Exploration. 2.2. Subsurface Exploration Program.. 2.3. Exploratory Boring in the Field. 2.3. 2.4. Procedures of soil sampling and SPT. 2.6. Vane Shear test. 2.5. Observation of Water Table 2.7. Cone penetration Test (CPT). 2.8. Coring of Rocks. 2.9. Preparation of Boring Logs. Subsurface Exploration (Con’t) Split Spoon Sampler and SPT Start of the next Lecture • It is a steel tube that split longitudinally into half. • At a selected depth the sampler is lowered to the bottom of the hole with the drilling rod. • Then it driven into the soil by a hammer which has standard weight and falls from standard height. a) Split Spoon sampler • The number of blows required to penetrate 3-(6 3in) intervals are recorded. b) Spring core catcher 1 Standard Penetration Test Consistency of clay cu = undrained strength OCR = Over consolidation Ratio Dr = relative density φ’ = effective friction angle N = N2 + N3 CLAY N SAND = N1 = N2 = N3 • Standard Penetration Number is the summation of the number of blows required for the last two intervals. • It referred to as N. • This tests is called Standard Penetration Test Standard • N60 is the field correction of N Correction factors are: hammer efficiency, borehole diameter, rod length, and sampler correction. SPT useful correlations • N60 is The standard Penetration Number. For Clay soil 1. Undrained shear strengths, cu: cu kN / m 2 = K N 60 ⎛N OCR = 0.193⎜ 60 ⎜ σ' ⎝o ⎞ ⎟ ⎟ ⎠ 0.689 ( ) K = 3.5 to 6.5 2. Overconsolidation ration, OCR: N 60 = Nη Hη Bη Sη R 60 Values of η’s are from Table 2.2 ' σ o = effective vertical stress MN/m2. 2 For Sand soil 1. Relative density, Dr: 1.7 ⎡ ⎛ 0.06 ⎞ ⎤ ⎜ 0.23 + ⎟⎥ ⎢ N60 ⎜ D50 ⎟ ⎥ ⎢ ⎝ ⎠ Dr = ⎢ ' ⎥ ⎛σ ⎞ ⎢ ⎥ 9⎜ o ⎟ ⎜p ⎟ ⎢ ⎥ ⎝ a⎠ ⎣ ⎦ 0.5 2. The Peak Friction angle, φ’: ⎤ ⎡ ⎥ ⎢ N 60 ' −1 ⎢ ⎥ φ = tan ' ⎢ ⎛σ ⎞⎥ ⎢ 12.2 + 20.3⎜ o ⎟ ⎥ ⎜p ⎟ ⎢ ⎝ a ⎠⎥ ⎦ ⎣ (kN/m2) 2 0. 5 0.34 ' Dr (%) = 11.7 + 0.76 222 N60 + 1600 − 53σ o − 50 (Cu ) ( ) φ ' (deg) = 27.1 + 0.39( N 1 )60 − 0.00054 (( N 1 )60 )2 D50 = size through which 50% of the soil pass (mm) size Cu = uniformity coefficient of the sand D60/D10 uniformity Pa = atmospheric pressure (100 kN/m2) atmospheric Where: (N1)60 is corrected number for the effective overburden pressure . ( N 1 )60 = C N N 60 Correction factor Corrected Value value from the field • CN is calculated as follow: CN = 2 ⎛σ' ⎞ 1+⎜ o ⎟ ⎜p ⎟ ⎝ a⎠ SPT advantages and disadvantages ADVANTAGES 1. Obtain both a sample & a number 2. Simple 3. Suitable in many soil types 4. Can perform in weak rocks DISADVANTAGES 1. Obtain both a sample & a number 2. Disturbed sample (index tests only) 3. Crude number for analysis 4. Not applicable in soft clays & silts 5. High variability and uncertainty σ = effective vertical stress. ' o Pa = atmospheric pressure (100 kN/m2) atmospheric Important notes about SPT • The equations are approximate. • The value of N60 obtained from a given borehole can vary widely due to soil non homogeneity. • For soils contains boulders and gravel Standard Penetration Number may be unreliable. 3 Example_1 A soil profile is shown in the flowing figure with the standard penetration numbers in clay layer. Determine the variation of cu and OCR with depth G.S. (use K = 6) 2m W.T 2m 1.5 m 1.5 m 6 1.5 m 4 1.5 m 6 Sand N60 5 8 Dry sand Solution For Clay soil 1. Undrained shear strengths, cu: γ = 17 kN/m Sand 3 cu kN / m 2 = K N 60 ⎛N OCR = 0.193⎜ 60 ⎜ σ' ⎝o ⎞ ⎟ ⎟ ⎠ 0.689 ( ) K=6 γsat = 19 kN/m 3 2. Overconsolidation ration, OCR: Clay γsat = 17 kN/m3 ' σ o = effective vertical stress MN/m2. Example_2 The Following table gives the field standard penetration numbers (N60) in a sand deposit. The ground water table located at 6 m depth. The dry unite weight of sand is 18 kN/m3. Estimate an average peak soil friction angle φ’ using the two proposed Equations. Depth (m) 1.5 3.0 4.5 6.0 N60 6 8 9 8 Solution For sandy soil The Peak Friction angle, φ’: ⎤ ⎡ ⎥ ⎢ N 60 ' −1 ⎢ ⎥ φ = tan ' ⎢ ⎛σ ⎞⎥ ⎢ 12.2 + 20.3⎜ o ⎟ ⎥ ⎜p ⎟ ⎢ ⎝ a ⎠⎥ ⎦ ⎣ 0.34 φ ' (deg) = 27.1 + 0.39( N 1 )60 − 0.00054 [( N 1 )60 ] 2 4 Where: (N1)60 is corrected number for the effective overburden pressure . ( N 1 )60 = C N N 60 Correction factor Corrected Value value from the field • CN is calculated as follow: CN = 2 ⎛σ' ⎞ 1+⎜ o ⎟ ⎜p ⎟ ⎝ a⎠ σ 'o = effective vertical stress. Pa = atmospheric pressure (100 kN/m2) atmospheric 5 ...
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