Soil Mechanics Exercises - Math Problem Example

?Question No a) (i) Total mass=231 g Sieve size (mm) Mass retained (g) % retained in each sieve Percent finer 10 0 0 100 6.3 11 4.76 95.24 2.0 51.4 22.25 73 1.0 46.2 20 53 0.6 44 19.04 33.95 0.3 34.6 14.98 19 0.15 25.4 11 8 0.063 18.4 8 0 (ii) Here is the particle size distribution curve (iii) From the chart we have D10 = 0.2 D30 = 0.45 D60 = 1.20 So Coefficient of uniformity is = D60/D30 Cu = 1.20/0.2 = 6.0 And Coefficient of curvature of the soil is = D302/(D10 x D60) Cc = 0.452/(0.2 x 1.2) = 0.84375 (iv) From the particle size distribution chart we can see that the particles are distributed over a wide range. So, this is a well graded soil. (b) As Plasticity index Ip = liquid limit – plastic limit So Ip of soft sandy clay= 36 – 15 = 21 Ip of Firm silty clay = 47 - 18 = 29 Ip of Stiff Clay = 67 – 19 = 48 And Liquidity Index IL = So IL of soft sand clay = 27-15/21 = 0.57 IL of Firm silty clay = 29-18/29 = 0.38 IL of Stiff clay = 25-19/48 = 0.125 Question No. 2 (a) Coefficient of permeability k is given by the relation K= Where V= volume of water L= length of soil column A= area of soil column H= head t= time required to get V volume Now here L= 10.8 cm Dia of permeameter= 8.9 cm Area of permeameter= A = 3.14 x 8.92/4 = 62.04 cm2 Test number Volume of water (V) (cm3) Time (t) (s) Head (h) (mm) Coefficient of permeability K= (cm/s) 1 420 84 509 0.0171 2 400 80 449 0.0194 3 380 76 385 0.0226 (b) Diameter of stand pipe = 2.82 cm Area of standpipe = a = 6.24 cm2 Diameter of sand sample= 3.696 cm Area of sand sample = A= 10.72 cm2 Length of sample = L= 16.28 cm Initial head= h1= 160.2 cm Final head= h2 = 80.1 cm Time interval for the head to fall from h1 to h2 =90s Now permeability k of the sample is given by the relation K= K = 0.073 cm/s (c) Steady state flow: Steady-state flow refers to the condition where the fluid properties at a point in the system do not change over time. Transient flow: Transient flow is such a flow where the velocity and pressure changes over time. (d) (i) Coefficient of consolidation is given by the relation Cv = Where Cv= coefficient of consolidation K= coefficient of permeability Mv= coefficient of volume compressibility ?w = unit weight of water So Cv = = 1.02 x 10-6 m2/s (ii) Settlement of clay layer Final settlement = 0.4 x 3 x 80/1000 = 96 mm (iii) Time for 70 % settlement= Tv x h2/Cv =0.403 x 32/1.02 x 10-6 = 6.76 years Question No. 3 (a) At 0 m Total stress= ?= ?d hw +q = 16 (0) + 10 =10 Kpa Pore pressure = u = ?w x (h – hw) = 9.81 (0)=0 Effective stress = ?’= ? – u = 10 – 0 =10 Kpa At 1 m Total stress= ?= ?d hw +q = 16 (1) + 10 = 26 Kpa Pore pressure = u = ?w x (h – hw) = 9.81 (0) = 0 Effective stress = ?’= ? – u = 26 – 0 = 26 Kpa At 3 m Total stress= ?= ?d hw +q + ?sat (h – hw) = 16 + 10 + 20 (2) = 66 KPa Pore pressure = u = ?w x (h – hw) = 9.81 x (2) = 19.62 Kpa Effective stress = ?’= ? – u = 66 – 19.62 = 46.38 Kpa At 7 m Total stress= ?= ?d hw +q + ?sat (h – hw) = 16 + 10 + 20 (6) = 146 KPa Pore pressure = u = ?w x (h – hw) = 9.81 x (6) = 58.86 Kpa Effective stress = ?’= ? – u = 146 – 58.86 = 87.14 Kpa (b) Unit weight of silty sand = ?s= porosity * specific gravity * 9.81 = 0.54 x 2.61 x 9.81 = 14 kN/m3 Saturated weight of clay =?c= = = 17.82 kN/m3 At 0 m Total stress= ?= ?s h = 14 (0) = 0 Pore pressure = u = ?w x (h – hw) = 9.81 (0) =0 Effective stress = ?’= ? – u = 0 At 2.5 m Total stress= ?= ?s h = 14 (2.5) = 35 KPa Pore pressure = u = ?w x (h – hw) = 9.81 x (0) = 0 Effective stress = ?’= ? – u = 35 - 0 = 35 Kpa At 5 m Total stress= ?= ?s h = 14 x 5 = 70 Kpa Pore pressure = u = ?w x (h – hw) = 9.81 x (2.5) = 24.5 kPa Effective stress = ?’= ? – u = 70 – 24.5 = 45.5 Kpa At 9 m Total stress= ?= ?s h +?c (h – 5) = 14 x 9 + 17.82 (4) = 197.3 KPa Pore pressure = u = ?w x (h – hw) = 9.81 x (6.5) = 63.76 Kpa Effective stress = ?’= ? – u = 197.3 – 63.76 = 133.54 Kpa Question No. 4 (a) Shear box apparatus The soil is contained in a box which has a separate top half and bottom half. A normal stress is applied onto the soil by placing weights on the lid of the box. The horizontal shear force needed to cause failure of the soil is measured. Measurement of shear strength We carry out shear box tests on a soil with different normal stresses. We than draw a graph of shear stress at failure against normal stress The shear strength of the soil (??) in the shear box is simply the angle under the graph as shown in Figure. It is usual to calculate the angle from the slope of the graph rather than to measure it, since tan?? = slope of graph Sometimes the graph does not pass through the origin and the soil appears to have some shear strength at zero normal effective stress. This value of shear stress is called the cohesion c? of the soil, measured in kPa. Soil cohesion should be use with caution in geotechnical design, it adds a lot of apparent strength to the soil but in many situations, particularly long-term, the cohesion may be less than apparent in the laboratory test and may even be zero. Properties measured from the shear box test: The shear box measures the shear strength of a soil in a direct way. Further it also easy to visualize the behavior in this test. (b) In the shear box test we can also measure the vertical movement of the lid of shear box. If the lid moves up during the test, the volume of the soil is increasing (dilation). If the lid moves down during the test, the volume of the soil is decreasing (compression). Usually dense soils dilate during shear and loose soils compress during shear. The reason for this behavior is described below. The changing strength and volume of soils during a shear box test is best shown with typical graphs of shear box test results. Figure below shows two graphs from a typical shear box test on a dense and loose sample of a soil. Both graphs have the horizontal travel of the top half of the shear box (x) on the horizontal axis. The first graph shows the shear force T needed to shear the soil during the test. In the dense soil, peak shear strength is reached after which the shear strength of the soil reduces. The loose soil requires only a small shear force at first but the shear force increases as the test continues. Eventually, both graphs reach the same shear force and will remain at this shear force if the test is continued. Both the dense and the loose soil have reached the same critical state. The second graph shows the vertical movement of the lid of the shear box. We can see for the dense soil the lid may move down a little at the start if the soil is not perfectly dense but later the lid moves upwards as the soil dilates. The dilation does not continue forever but stops when the soil reaches the critical state. For the loose soil, the lid moves down as the soil compresses. Again, this does not continue forever but stops when the soil reaches the critical state. Both the dense and the loose soils will continue to shear at the critical state without changing volume (i.e. without the lid moving up or down). (c) This is preferred for the following reasons: i) Direct measurement of volume change of saturated soils ii) Shear failure occurs on preferred planes in the soil rather than forced on a particular plane as in a shear box iii) More control of parameters, such as pore pressure and normal stress, is possible (d) ?3 = 100 kPa at failure (?1 – ?3) = 150 kPa So, ?1= 150 + 100= 250 Now angle of friction ? is given by the relation ?= sin-1 = sin-1 = 25.37 Question No. 5 (a) Coefficient of permeability k is given by the relation K= Where V= volume of water L= length of soil column A= area of soil column H= head difference t= time required to get V volume Now here L= 15 cm Dia of permeameter= 10 cm Area of permeameter= A = 3.14 x 102/4 = 78.5 cm2 t = 60 sec Volume of water collected in 1 min (V) (cm3) Difference in standpipe levels (h) (mm) Coefficient of permeability K= (cm/s) 270 75 0.1146 220 60 0.1167 160 45 0.1132 110 30 0.1167 Sample calculation for V=270 cm3 and h= 75 mm K= = 0.1146 cm/s (b) (i) Bulk density = dry mass/ total volume = 140/82 = 1.7 g/cm3 (ii) Dry density = Bulk density = dry mass/ total volume = 140/82 = 1.7 g/cm3 (iii) water content = wet mass/dry mass – 1 = 180/140 – 1 = 28.57 % (iv) Mass of water in the sample= wet mass- dry mass = 180 – 140 = 40 g The volume of water in the soil= mass of water/ density of water = 40/1 = 40 cm3 (v) Volume of solids in the soil= dry mass/ specific gravity = 140/2.7 =51.85 cm3 (vi) Volume of voids in the sample = total volume - volume of solids = 82 – 51.85 = 30.15 cm3 (vii) Voids ratio of the sample = volume of voids/volume of solids = 30.15/51.85 = 0.58 (viii) degree of saturation= volume of water/ volume of voids = 40/30.15 =1.32 Question No. 6 (a) The following is the compaction curve From graph we see that maximum dry unit weight is 20.8 and optimum water content is 16 % (b) ?3 = 100 kPa at failure (?1 – ?3) = 140 kPa So, ?1= 140 + 100= 240 Now angle of friction ? is given by the relation ?= sin-1 = sin-1 = 24.31 (c) In the shear box test we can measure the vertical movement of the lid of shear box. If the lid moves up during the test, the volume of the soil is increasing (dilation). If the lid moves down during the test, the volume of the soil is decreasing (compression). Usually dense soils dilate during shear and loose soils compress during shear. The reason for this behavior is described below. The changing strength and volume of soils during a shear box test is best shown with typical graphs of shear box test results. Figure below shows two graphs from a typical shear box test on a dense and loose sample of a soil. Both graphs have the horizontal travel of the top half of the shear box (x) on the horizontal axis. The first graph shows the shear force T needed to shear the soil during the test. In the dense soil, peak shear strength is reached after which the shear strength of the soil reduces. The loose soil requires only a small shear force at first but the shear force increases as the test continues. Eventually, both graphs reach the same shear force and will remain at this shear force if the test is continued. Both the dense and the loose soil have reached the same critical state. The second graph shows the vertical movement of the lid of the shear box. We can see for the dense soil the lid may move down a little at the start if the soil is not perfectly dense but later the lid moves upwards as the soil dilates. The dilation does not continue forever but stops when the soil reaches the critical state. For the loose soil, the lid moves down as the soil compresses. Again, this does not continue forever but stops when the soil reaches the critical state. Both the dense and the loose soils will continue to shear at the critical state without changing volume (i.e. without the lid moving up or down). Read More