Thủy Lực Và Cơ Học Đất: Soil Mechanics Experiment Report

Ho Chi Minh city University of technology Programme de Formation d’Ingénieurs d’Excellence au Vietnam SOIL MECHANICS EXPERIMENT REPORT Instructor: Dr Đỗ Thanh Hải Ms Tô Lê Hương Students: All members of class VP13XDC Contents Experiment 1…………………………………… Page Experiment 2…………………………………… Page Experiment 3…………………………………… Page 12 Experiment 4…………………………………… Page 17 Experiment 5…………………………………… Page 22 Experiment1: LABORATORY ANALYSIS GRAIN SIZE COMPOSITION SIEVE METHODS (d ≥ 0,074mm - ≠ 200) DEPOSITION METHOD (d < 0,074mm) I Purpose: -Laboratory analysis of particle size (grain size):determine the relative proportion as a percentage of the different groups in the soil particles -Based On grain composition and graded roads to assess the level of uniformity and gradation;waterproof permanence; select building materials; predict the mechanical properties change determines the magnitude group of particle size; the distribution and soil classification II Laboratory instrument: Using the sieve method (d >0.074mm) -Sieve ministry: sieve cap, sieve, sieve bottoms Dry Sieve Wash Sieve Sieve size / Number 4” (Sieve size) Diameter d (mm) 2” 50,8 1” 25,4 3/4” 19,1 1/2” 12,7 3/8” 9,51 #4 (Number) 4,76 #6 3,36 #10 2,00 #20 0,84 #40 0,42 #60 0,25 #100 0,149 101,6 #200 0,074 -Scales (accuracy 1g for large scales; 0,1g for small scales) -Divided land instruments, spoon,pestle, oven (105oC), sifting machines … Using the method of deposition (d 0.074mm) -Get the M (g) land after drying and separating rubber particles in baseball, for into the sieve sieve getting ranked in order descending from the top down, proceed sieve until volume set on each sieve unchanged (about 10 minutes) Land mass balance on each sieve (or cumulative weight) Volume of soil were taken as follows: Fine-grained soils: 100 –200g Sandy soil: 300 –500g largest soil particles 3/8”: 1000g largest soil particles 1/2”: 3kg largest soil particles 3/4”: 5kg largest soil particles 1”: 10kg -After the end of the dry sieve, taking m (g) of land at the bottom of the sieve through the sieve No.200 lavage until water no longer beads through a sieve No.200 land Part of the land is located on the No.200 sieve (No.20, No.40, No.80, No.100, No.140, No.200) The land No.200 pass through the sieve washed in water to be condensed 1000ml also to perform deposition experiments -Land cumulative balance (balance volume of soil from large sieve, sieve followed by small incremental weight), concentration loss type of soil: clay, sandy clay, sandy clay soil IP 17 The name of soil Sand Cát pha Sét pha Sét - determine liquid indicators, IL :(viscosity B) IL=(W-Wp)/IP => The State of the land: malleable plastic sticky liquid, liquid, Type of soil Sandy clay Clay Name and status Hard Plastic Liquid Hard Semisolid soft plastic plastic liquid liquid IL IL≤ 0≤IL≤1 IL>1 IL < 0 < IL ≤ 0.25 0.25 < IL ≤ 0.5 0.5 < IL ≤ 0.75 0.75 < IL ≤ IL > II Laboratory instruments: (For experimental pastes limit) + Instruments Casagrande (lifting height cap demand is cm) + Trench cutter + The mixing knife, mixing glasses, spoon, sieve N40 (register 0, 42 mm grains), water, cans of sample containers, weight (precision 0, g), drying III Sequencing experiments: A) Liquid Limit test : - Use about 100 g ground through a sieve to N40, mix with enough water - Use a putty knife, mixing a land grab about 2/3 of the globe calotte - Use the cutting knife, divide the land in the globe calotte into equal portions (distance 2mm gap, thickness 8mm) - Get the globe calotte lifted and fell to hr = cm, velocity v = times/s, count the times of the fall( N)until the ground in parts of the globe calotte closed IV RESULTS OF EXPERIMENT A) Table 1: B) RESULTS: Liquid limit of soil: WL = 66.4% 72 71.4549 70 Humidity W% 68 66 64.6583 64 62 60 58.8954 58 15 20 25 30 35 40 number of falling - Plastic limit WP = (36.944+35.119)/2 = 36.03% - Plasticity index : IL = (W - WP)/(WL - WP )=(61.5-36.03)/(66.4-36.03)=0.83 With W = 61.5% - Liquid index : IP = WL - WP= 66.4 – 36.03 = 30.37% 10 C) REASON: - IP = 30.37% > 17%  it’s a clay - 0.75 < IL = 0.83 ≤  it’s a plasticity of clay D) COMMENT -Based on the value of IP and IL Named and notation the soil ; determine the status of the soil base on TCVN 9362 2012 -We can classify the land from the experimental results based on a line graph A straight-line 11 Exercise 3: COMPACTION TEST OF SOIL I          II Objective : For construction of highways, airports, and other structures, it is often necessary to compact soil to improve its strength Objective of Standard Compaction test : To determine relation between water content and dry density of soil To determine optimum water content and corresponding maximum dry density for soil To determine relation between penetration resistance and water content for compacted soil Importance of Standard Compaction test Compaction increases the shear strength of the soil Compaction reduces the voids ratio making it more difficult for water to flow through soil This is important if the soil is being used to retain water such as would be required for an earth dam Compaction can prevent the build up of large water pressures that cause soil to liquefy during earthquakes Equipments  Drying oven  Weighing balance, accuracy 0,01g  Standard Proctor Compaction Mold, V = 944cm3  Standard Proctor Hammer (the height of fall is h = 30,48cm ; Q = 2,5kg)  IS sieve, N4 4.75 mm  Steel straight edge  Moisture cans  Mixing tools, spoons, trowels, etc 12 III Procedure Obtain about kg of air-dried soil and break the soil lumps Sieve the soil through a N4 sieve Collect all the soil specimen passing N4 sieve in a large pan Add water to the soil specimen and mix thoroughly  Determine the quantity of water added: 0.01𝑚 𝑞= (𝑤 − 𝑤𝑡 ) + 0.01𝑤𝑡 𝑠 with: q: the quantity of water added (g) ws: the moisture contend required (%) wt: the moisture of soil before adding water (%) m: the weight of soil before adding water ( dry soil) (ws – wt): the increase of the moisture (about 2-3%) Determine the weight of the Proctor Mold + base plate (not extension) Weigh it to the nearest gram Attach the extension to the top of the mold Pour the moist soil in three equal layers Compact each layer uniformly with the hammer 25 times before each additional layer of loose soil is poured At the end of the three-layer compaction, the soil should extend slightly above the top of the rim of the compaction mold 13 Remove the extension carefully Then, trim excess soil with a straight edge Determine the weight of the Proctor Mold + base plate + compacted moist soil Remove the base plate from the mold Extrude the compacted moist soil cylinder using a jack 10 Take a moisture can and determine its mass 11 From the moist soil extruded in step 9, collect a moist sample in a moisture can (step 10) and determine the mass of moist soil + can 12 Place the moisture can with soil in the oven to dry to a constant weight 13 Break the rest of the soil cylinder by hand and mix with leftover moist soil Add more water and mix thoroughly 14 Repeat steps 6-12 In this process, the weight of the mold + base plate + moist soil will first increase with the increase in moisture content and then decrease Continue the test until at least one successive decreased readings are obtained 15 The next day, determine the mass of the moisture cans + soil samples (from step 12) 14 IV Calculation DATA SHEET FOR COMPACTION TEST Observation: Unit Data Ordinal experiment A- Mass of mold, compacted soil and base plate of mould and base B-Mass V-Volume of mold Moist unit weight, γ Serial number of can A- mass of moisture can + moist soil B- mass of moisture can + dry soil C-Mass of can g 3692.5 g cm3 g/cm3 1995.5 944 1.80 3765.5 3862.0 1995.5 944 1.88 1995.5 944 1.98 3895.0 3871.5 1995.5 944 2.01 1995.5 944 1.99 g 156.055 140.783 207.177 128.761 149.637 g 146.357 129.107 185.334 112.638 127.510 g 3.112 3.102 3.142 3.043 3.090 Compaction moisture content, w % 6.77 9.27 11.99 14.71 17.78 Dry unit weight, γd g/cm3 1.69 1.77 1.75 1.69  Moist unit weight: γ = 1.72 𝐴−𝐵 𝑉 with : A- Mass of mold, compacted soil and base plate B- Mass of mould and base V- Volume of mold  Compaction moisture content: w = 𝐴−𝐵 𝐵−𝐶 × 100% with : C- Mass of can  Dry unit weight: γd = γ 1+0.01𝑤 15 Graph between w and gd 1.78 Dry unit weight, γd (g/cm3) 1.77 1.76 1.75 1.74 1.73 1.72 1.71 1.7 1.69 1.68 10 12 14 16 18 20 Compaction moisture content w (%)  Results: Maximum dry density (from plot): γdmax ≈ 1.77(g/cm3) Optimum water content (from plot): Wopt ≈ 12 % 16 EXPERIMENT 4: DIRECT SHEAR TESTS I Experimental purposes: Direct shear tests to determine the basic characteristics of the soil (mechanical properties; c, φ), which reviews: Soil shear strength: S = σtanφ + c Load bearing capacity of the ground: R tc  m ( A b  tc  B h  tc *  D ctc ) A, B, C are coefficients dependent on c, φ Aside from c, φ can identify with the other experiments: + Unconfined compression test : applied to the soil sticky, simple, direct result, the destruction would be the weakest + Direct shear test : applies to soil and land left stick, simple, direct result, the destruction is between horizontal surface of the cutting board cutting boxes are fixed in advance + Compression Triaxial test shall apply to all types of soil, but complex experiments to complete the targets, with experimental methods; Undrained - Unconsolidated (UU), Undrained - Consolidated (CU), Drained - Consolidated (CD) II Laboratory instruments: Direct shear box apparatus Round knife to create experimental soil samples: diameter 6,3cm (A = 31.17 cm2), height 2cm Gauges horizontal displacement, gauges horizontal strain; / 1000mm: bar = 0,01 mm - gauges horizontal displacement Knife, water bottle, weights to apply pressure or undisturbed soil sample was prepared 17 III Process of experiment Firstly, soak the two samples of stone into a water container until they’re saturated Use a steel rope to cut the cylinder-shape soil sample that is approximately cm high Create a experiment sample by pressing down the round knife and whittling around the sample Then, whet the surface moothly Cover sides of sample with pieces of wet-filter paper 18 We set the soil sample into the box shear apparatus that is between pieces of filter stones, lock the bolts carefully Set that box into shear machine, set the measurer back to zero, get the bolt out of the shear machine Put the load suitably with loading level 19 Let the machine cut within velocity of 3mm/min until the sample is destroyed completely, record the maximum values of shearing stress 20 IV Result of experiment and discussion Sample height Section Read-around ratio Compressive Pressure (kN/m^2) 50 100 150 (cm) 30 (cm2) 1.653 (kN/m2 per div) Maximum deflection 28.5 46.4 65.9 Shear Stress τ (kN/m^2) 47.1105 76.6992 108.9327 Perform the data on chart, we have the graph below:  Prolong the line, it meets vertical exis at C=15.759 kPa is cohension force and the slope of line is angle of internal friction ϕ=34.290 Discussion:  We can measure shear resistance of a specific soil thanks to the test  Graph drawn is linear and similar to theory leant 21 Eperiment 5: Consolidation Compression Test I Objective Consolidation compression tests are used for determining several parameters, such as subsidence compression coefficient a, coefficient of volume changes mv, compression index cc, expansion index cs, coefficient of permeability k, modulus of deformation E, coefficient of consolidation cv, void ratios for each load level, etc with the purpose of calculating the deformation (subsidence) of the ground level Land subsidence is the process of shrinking pore volume, also known as compression Under the impact of external loads, solid particles are folded, leading to the reduction of pore volume Hence, land is compressed When the land is put under loads, water in the pores in the soil is absorbed and has tendancy to drain out By consequence, pore water pressure tends to plunge, leading to the gradually increase of effective pressure Once the process of water drainage completely happens, soil particles will suffer from all of the pressure of external loads The phenomenon of soil compression due to the steady water drainage from soil pores is called consolidation II Instrument  Consolidation compressor  Modeling tools (sharp metal ring with 2cm height and cross-sectional area of 20cm2, trimming tool, steel wire can be used for cutting soft clay samples)  Stopwatch, loads, scale, drying oven, etc III Experiment steps     Use metal ring and trimming tools for putting sample into shape Put sample in the compressor, right between two pieces of pumice Balance the lever using water Put loads on the lever: 0.25, 0.5, 1, 2, 4,…(kg / cm2) It might take at least 24 hours for the sample to reach its stable compression state under the pressure of loads  Observe and record the figures on the deformation meter for each level of load after first 15 seconds until the deformation reached its stable state The times between each two records double respectively: 30s, 1m, 2m, 4m, 8m, 15m, … 1h, 24h  After the stablization of the sample at the last load level, begin to remove loads, vice versa Record the figures displayed on the deformation meter 22 IV Result and comment P(kg/cm2) Δh(mm) e 0 0.88 0.25 0.96 0.5 429.45946 0.001128 973 0.0004841 2104.0777 4.23E-05 22477.778 0.000235 4071.2 0.0002256 4313.75 0.0003384 2899.1667 0.39026 5.09 0.25 0.0027824 0.36676 5.21 0.5 378.63636 0.3583 5.46 0.0033088 0.45512 5.55 364.58333 0.56792 4.52 0.0036096 0.70704 3.32 E(KPa) 0.78976 1.84 a(1/KPa) 0.40154 0.41 CONSOLIDATION CAUSED BY WATER DRAINAGE (e-P) 0.9 0.8 0.7 e 0.6 0.5 0.4 0.3 0.2 0.1 0 50 100 150 200 250 300 350 400 450 P (KPa) 23 CONSOLIDATION CAUSED BY WATER DRAINAGE (e-logP) 0.9 0.8 0.7 e 0.6 0.5 0.4 0.3 0.2 0.1 10 36.1 100 1000 logP (KPa)  Determine compression index: e200 − e400 0.45512 − 0.3583 = = 0.3216 400 log 400 − log 200 log 200  Determine expansion index: Cc = er 200 − er 400 0.36676 − 0.3583 Cc = = = 0.0281 400 log 400 − log 200 log 200  Determine pre-consolidation pressure:  From e-log p chart, choose the starting point of the curve and draw lines: parallel to the horizontal axis, tangent to the curve and the bisector of two above lines  Extend the end of the curve e-log p to the intersection with the bisector at a point whose horizontal coordinate is the pre-consolidation pressure  Comments:  Better understand the bearing capacity of the soil through each level of load and the consolidation process  Identify the highest pressure that soil can bear and pressure that soil has suffered in the past  e-log p illustrated that there are linear regions whose intersection is the maximum vertical stress caused by the weight of the upper layer of soil in the past That is the pre-consolidation pressure When the pressure exceeds the consolidation value, the particles are rearranged and voids reduce 24 ... of soil were taken as follows: Fine-grained soils: 100 –200g Sandy soil: 300 –500g largest soil particles 3/8”: 1000g largest soil particles 1/2”: 3kg largest soil particles 3/4”: 5kg largest soil. ..Contents Experiment 1…………………………………… Page Experiment 2…………………………………… Page Experiment 3…………………………………… Page 12 Experiment 4…………………………………… Page 17 Experiment 5…………………………………… Page 22 Experiment1 :... evaporate, doing the experiment again - Doing the same experiment about 3-4 times, determine times of the fall N