SOIL MECHANICS/GEOTECH
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Problem 2.2 solved, Coefficient of soil permeability
Coefficient of soil permeability: The results of a constant head permeability test of a fine sand are as follows: area of the soil sample 180sqcm, length of the specimen 32cm, constant head maintained 46cm, flow of water through the specimen 200ml in 5 minutes. Determine the coefficient of permeability in cm/s Solution Coefficient of soil permeability NOTE The coefficient of permeability k is a measure of the rate of flow of water through saturated soil under a given hydraulic gradient i, cm/cm, and is defined in accordance with Darcy’s law as: Assumption: Flow is laminar When the flow through soil is laminar, Darcy’s law (Darcy, 1856) applies: Darcy’s law states that for laminar flow the velocity of flow, V is proportional to the hydraulic gradient i. V=Ki (k is the constant of proportionality) q = v . A = k i A Where q = rate of flow, cm3/s. A = cross-sectional area of soil conveying flow, cm2 The rate of flow under laminar flow conditions through a unit cross-sectional are of the porous medium under a unit hydraulic gradient is defined as the ]coefficient of permeability. Permeability in soil mechanics is a measure of how easily a fluid (water) can flow through a porous medium (soil). The flow of water through soils, called seepage, occurs when there is a difference in the water level (energy) on the two sides of a structure such as a dam or a sheet pile.
Problem 6.8, solved Compaction of soil
Compaction of soil: The maximum and the minimum unit weight of a sand collected from the field were determined in the laboratory as 18.38kn/m^3, respectively. it is required that the sand in the field be compacted to a relative density of 85%. Determine what would be the relative compaction in the field. Solution Compaction of soil NOTE What is compaction? Soil compaction occurs when soil particles are pressed together, reducing pore space between them. Heavily compacted soils contain few large pores, less total pore volume and, consequently, a greater density. A compacted soil has a reduced rate of both water infiltration and drainage. What are the three main reasons for compaction? According to Proctor (1933) who developed the procedures for compaction, the degree of compaction a given soil can achieve depends on three factors: (1) water content, (2) compactive effort , and (3) soil type (coarse-grained versus fine-grained; grain size distribution; amount and type of clay minerals)
Problem 3 solved, Water content and void ratio in soil
Water content and void ratio in soil: Question Three A sample of saturated clay was placed in a container and weighed. The weight was 6N. The clay in its container was placed in an oven for 24 hours at 105o The weight reduced to a constant value of 5N. The weight of the container is 1N. If Gs = 2.7, determine the; – water content void ratio bulk unit weight dry unit weight Effective unit weight. Solutions Water content NOTE The water content (w), also known as natural water content or natural moisture content, is the ratio of the weight of water to the weight of the solids in a given mass of soil. This ratio is usually expressed as percentage. When voids are completely filled with air, water content is equal to zero (dry soil) Moisture content is one of the most important index properties used for the correlation of soil behavior and its index properties. The moisture content of the soil is used to express the phase relationships of water, air, and solids in a given volume or weight of the material. Water content determination purpose: This test is performed to determine the water (moisture) content of soils. The water content is the ratio, expressed as a percentage, of the mass of “pore” or “free” water in a given mass of soil to the mass of the dry soil solids. The most crucial factor in civil engineering designs is the proper assessment of the water content of on-site soil because its mechanical reaction to weight loading ultimately depends on its moisture content at the time of building.
Problem 4 solved, Effective stresses
Effective stresses: In a deposit of fine sand the water table was at 3m below the ground surface but the sand up to a height of 1m above the water table was saturated by capillary rise. The sand above this height can be considered as dry. For the sand G_s = 2.65 and n = 40%. Calculate the effective stress at a depth 8m below the ground surface. Solution Effective stresses NOTE What is the pore water pressure? Pore water pressure is the pressure of groundwater held between soil or rocks in the gaps (or ‘pores’) between particles. It is affected by the soil type, water flow conditions and level of the water table. Effective stress can be defined as the stress that keeps particles together. In soil, it is the combined effect of pore water pressure and total stress that keeps it together. It can also be defined in equation form as the total stress minus the pore pressure. Total stress at a point is defined as the sum of the stress exerted by the total weight (solids plus water) of the soil/rock and the load on the foundation.
Problem 3 solved, Water pore pressure
Water pore pressure: For the soil profile given below determine the total and effective stresses at point A. What will be the total, neutral (water pore pressure ) and effective stresses at point if the ground water table rises to the ground surface?, Will there be any change effective stress at point A if the water table rise 2m above the ground surface ? Solution Water pore pressure NOTE What is the pore water pressure? Pore water pressure is the pressure of groundwater held between soil or rocks in the gaps (or ‘pores’) between particles. It is affected by the soil type, water flow conditions and level of the water table. Effective stress can be defined as the stress that keeps particles together. In soil, it is the combined effect of pore water pressure and total stress that keeps it together. It can also be defined in equation form as the total stress minus the pore pressure. Total stress at a point is defined as the sum of the stress exerted by the total weight (solids plus water) of the soil/rock and the load on the foundation.
Problem 2 solved, Pore pressure in soil
Pore pressure in soil For the given soil profile, How high should the water table rise so that the effective stress at C is 190kn/m^3, assume saturation to be the same for both layers as 19.25kn/m^3. Solution Pore pressure in soil NOTE What is the pore water pressure? Pore water pressure is the pressure of groundwater held between soil or rocks in the gaps (or ‘pores’) between particles. It is affected by the soil type, water flow conditions and level of the water table. Effective stress can be defined as the stress that keeps particles together. In soil, it is the combined effect of pore water pressure and total stress that keeps it together. It can also be defined in equation form as the total stress minus the pore pressure. Total stress at a point is defined as the sum of the stress exerted by the total weight (solids plus water) of the soil/rock and the load on the foundation.
Problem 1 solved, effective stress in soil, Calc
effective stress in soil For the given soil profile, Calculate the distribution of total stress, pore pressure, and effective stress along depth and report their values at point A, B, and C. effective stress in soil NOTE Effective stress can be defined as the stress that keeps particles together. In soil, it is the combined effect of pore water pressure and total stress that keeps it together. It can also be defined in equation form as the total stress minus the pore pressure. What is the pore water pressure? Pore water pressure is the pressure of groundwater held between soil or rocks in the gaps (or ‘pores’) between particles. It is affected by the soil type, water flow conditions and level of the water table. Total stress at a point is defined as the sum of the stress exerted by the total weight (solids plus water) of the soil/rock and the load on the foundation.
Solved, Q2 Calculate the dry and saturated unit weight of the material.
Saturated Unit Weight A soil has a voids ratio of 0.7. Calculate the dry and saturated unit weight of the material. Assume that the solid material occupies 1 m3. Assuming Gs = 2.65, the distribution by volume and weight is as follows. Phase Volume (m3) Dry Weight (kN) Saturated Weight (kN) Voids 0.7 0 0.7 X 9.81m = 6.87 Solids 1.0 2.65 X 9.81 = 26.0 = 26.0 Answer [pms-restrict subscription_plans=”9,10″]
Solved, calculate the effective stress at a depth of 5 m below the surface
Effective stress A uniform layer of sand 10 m deep overlays bedrock. The water table is located 2 m below the surface of the sand which is found to have a voids ratio e = 0.7. Assuming that the soil particles have a specific gravity Gs = 2.7 calculate the effective stress at a depth of 5 m below the surface. Answer More Effective stress problems