Geotechnical study pdf

Log in with Facebook Log in with Google. Remember me on this computer. Enter the email address you signed up with and we'll email you a reset link. Need an account? Click here to sign up. Download Free PDF. A short summary of this paper. Download Download PDF. Translate PDF. Zeng1 and 2 and J. There will be no further reduction of volume by reducing water content.

The space left behind due to reduction of water content will be filled by air. Shrinkage ratio is defined as ratio of given change in volume expressed as percentage of dry volume, to the corresponding change in water content.

Degree of Shrinkage. This is change in volume per unit original volume. Determine a The liquidity index b Activity number c Consistency and nature of soil. Determine the plasticity index of the soil and comment about the nature of the soil. What is its natural water content? What is the name of this soil? Example Liquid limit tests on a given sample of clay were carried out. The data obtained are as given below. Test No. Solution Following figure gives the flow curve for the given sample of clay soil.

Following data were obtained. Soil A Soil B No. On oven drying, the weight reduces to 18 g. The volume of dry specimen as determined by displacement of mercury is 9. Determine shrinkage limit and specific gravity.

The origin of a soil may refer either to its constituents or to the agencies responsible for its present state. Based on constituents, soils may be classified as : 1. Inorganic soils 2. Organic soils i Plant life ii Animal life Based on the agencies responsible for their present state, soils may be classified as : 1.

Residual soils 2. This system is based on both grain size and plasticity propertied of soil. This system is also used by Bureau of Indian Standards. In this system soils are divided in two divisions 1. Soils are classified into two categories based on percent passing IS sieve No. Coarse grained soils are further sub-divided into gravels and sands. The soils of these groups are represented by symbols starting with G and S respectively.

These are further classified on the basis of uniformity coefficient Cu and coeff. If fines are present, the soil is classified according to procedure for fine grained soils. If the soil is silty then GM or SM is used as the symbol.

Fine grained soil is classified into inorganic silt M , inorganic clay C and organic silts and clap O by plotting the liquid limit x — axis and PI on y — axis.

Soil with LL more than 50 is high plastic, and represented by symbol H. A soil lying below A - line is silt M and lying above is clay. Hence ML would mean inorganic silt with low plasticity.

CL would mean inorganic clay with low plasticity. The coordinates plot above A-line on the plasticity chart, in the CH region. It is, thus, an inorganic clay of high compressibility. Values of liquid limit and plastic limit are unusually high. It is a fine grained soil. Hence, the soil can be assigned the dual symbol CL- CI.

It is an inorganic clay of low to intermediate compressibility. This will be the same as the per cent passing 4. Also, if the values of wL and Ip are plotted on the plasticity chart, the point falls above A-line. Hence the soil is to be classified as SC, as per IS classification. Even according to Unified Classification System, this will be classified as SC, which may be checked easily.

In a two phase solid liquid system, these voids are completely filled by the liquid, which is water in most of the cases of soil mechanics. Water in these tubes is free to flow when a potential difference is created in a soil mass. Water flows from zones of higher potential to lower potential.

Gravels are more pervious than sands, sands are more pervious than silts and silts are more pervious than clays. A loose sand is much more pervious then when it is dense. Constant head permeaters are specially suited to the testing of pervious, coarse grained soil. In highly impervious soils the quantity of water that can be collected will be small and, accurate measurements are difficult to make.

Therefore, the constant head permeameter is mainly application cable to relatively pervious soils, although, theoretically speaking, it can be used for any type of soil. The soil sample is contained in a Perspex cylinder. Water is allowed to flow through the sample from a reservoir designed to keep the water level constant by over flow. The quantity of water flowing out of the soil or discharge Q during a given time t is collected in a vessel and weighed.

Several such tests at varying rates of flow can be performed and the average value of k is determined. This method is used to determine the permeability of fine grained soils such as fine sands, silts, and clays.

In such soils, the permeability is too small to enable accurate measurement of discharge using constant head permeaters. A cylinder containing the soil sample is placed on a base perforated disc fitted with a fine gauge. The cylinder is fitted with a rubber stopper on top. A graduated standpipe of known diameter is inserted into the rubber stopper.

The test is conducted by fitting the standpipe with desired water and allowing flow to take place through the sample. During the test, the water level will continuously drop and the height of water in the standpipe is recorded at several time intervals during the test. Any one pair of measurement, namely the time taken for the head to fall h1 to h2, will yield one value of k. The average value of k can be computed from several such readings.

The head drops by 5 cm in 10 minutes. Calculate the time required to run the test for the final head to be 20 cm. The area of the standpipe was 0. During the test, the head fell from 50 to 30 cm in 4.

The test conditions are shown in figure. Here the acquifer is underlain by an impermeable stratum and the test well extends to the bottom of the permeable stratum.

The pumping generates a radial flow of water towards the filter well and as a result, the water table assumes a curved surface called drawdown water table. Here, the drawdown surface is, for all values of r, above the upper surface of the acquifer.

Two observation wells were sunk at horizontal distances of 16 m and 34m respectively from the pumping well, The initial position of the water table was 2.

Calculate the coefficient of permeability of the sand. A constant discharge of 2 cubic metres per minute is pumped out of the aquifer through a tubewell till the water level in the tubewell becomes steady. Two observation wells at distances of 15 m and 70 m from the tubewell show falls of 3 m and 0.

Find the permeability of the aquifer. Answer: 1. Viscosity effects are more important. Greater the viscosity, lower the permeability. Other things being constant, calculate the percentage change in coefficient of permeability. Example A cohesionless soil has a permeability of 0. Make predictions of the permeability of this soil when at a void ratio of 0.

The internal diameter of the permeameter is 5 cm and the measured difference in head between two gauging points 15 cm vertically apart is 40 cm. Calculate the coefficient of permeability. If the dry weight of the 15 cm long sample is 4. Estimate its permeability at a void ratio of 0.

Hn, and let k1,k The hydraulic gradients, on the other hand, change from layer to layer. What are the values of the horizontal and vertical coefficients of permeability of the three layers, and what is their ratio.

The attraction between the water and capillary tube, or the tendency of water to wet the walls of the tube affects the shape of the air-water interface at the top of the column of water. For water and glass, the shape is concave as seen from top, that is, the water surface is lower at the centre of the column than at the walls of the tube. This angle, incidentally, is zero for water and glass see figure. The column of water in the capillary tube rises, against the pull of gravity, above the surface of the water source.

For equilibrium, the effect of the downward pull of gravity on the capillary column of water has to be resisted by surface tension of the water film adhering to the wall of the tube to hold the water column.

This equation helps one in computing the capillary rise of water in a glass capillary tube. Capillary Rise in Soil The rise of water in soils above the ground water table is analogous to the rise of water into capillary tubes placed in a source of water. However, the void spaces in a soil are irregular in shape and size, as they interconnect in all directions. Thus, the equations derived for regular shaped capillary tubes cannot be, strictly speaking, directly applicable to the capillary phenomenon associated with soil water.

However, the features of capillary rise in tubes facilitate an understanding of factors affecting capillarity and help determine the order of a magnitude for a capillary rise in the various types of soils. Equation 1 indicates that even relatively large voids will be filled with capillary water if soil is close to the ground water table.

As the height above the water table increases, only the smaller voids would be expected to be filled with capillary water. The larger voids represent interference to an upward capillary flow and would not be filled.

The soil just above the water table may become fully saturated with capillary water, but even this is questionable since it is dependent upon a number of factors. The larger pores may entrap air to some extent while getting filled with capillary water. In both these zones constituting the capillary fringe, even absorbed water contributes to the pore water see figure.

In the zone of partial saturation due to the capillary phenomenon, capillary movement of water may occur even in the wedges of the capillary V formed wherever soil grains come into contact see figure.

Since void spaces in soil are of the same order of magnitude as the particle sizes, it follows hat the capillary rise would be greater in fine-grained soils than in coarse-grained soils. Temperature plays an important role in the capillary rise in soil. At lower temperature capillary rise is more and vice versa. Capillary flow may also be induced from a warm zone towards a cold zone. It is to be noted that the negative pressures in the pore water in the capillary zone transfers a compressive stress of equal magnitude on to the mineral skeleton of the soil.

What is the water pressure just under the meniscus in the capillary tube? If the capillary rise in the first soil is 72 cm, what would be the capillary rise in the second soil? What is the height to which water will rise in this tube by capillarity action? What is the pressure just under the meniscus? The capillary rise of water in the first sand is mm. What is the capillary rise in the second sand, if the void ratio is the same for both sands? Many researchers have attempted to use Reynolds' concept to determine the upper limit of the validity of Darcy's law.

The values of R for which the flow in porous media become turbulent have been measured as low as 0. The probable reason that porous media do not exhibit a definite critical Reynold's number is because soil can be no means be accurately represented as a bundle of straight tubes.

There is overwhelming evidence which shows that Darcy's law holds in silts as well as medium sands and also for a steady state flow through clays. For soils more pervious than medium sand, the actual relationship between the hydraulic gradient and velocity should be obtained only through experiments for the particular soil and void ratio under study.

The smaller the magnitude of the selected stress, the greater the depth upto which an isobar extends. Since the vertical stress on a given horizontal plane is the same in all directions at points located at equal radial distances from the axis of loading, an isobar is really a spatial, curved surface of the shape of an onion bulb. Hence, in literature, an isobar is often termed the pressure bulb. Within a pressure bulb, a soil mass will be stressed to stresses higher than the designated stress of the pressure bulb, while the soil mass beyond the pressure bulb will be stressed to lower stresses.

Solution Let the circular area of radius a be loaded uniformly with q per unit area as shown in figure. Let us consider an elementary ring of radius r and thickness dr of the loaded area. The legs rest on piers located at the corners of a square 6 m on a side. What is the vertical stress increment due to this loading at a point 7 m beneath the centre of the structure?

The load can be approximated to a point load acting at the corners of a square of 6 m side. Using the point load approximation with four equivalent point loads, calculate the stress increment at a point in the soil which is 4 m below the centre of the loaded area.

Dividing the loaded area into four equal squares of 2 m size, as shown in figure, the load from each small square may be taken to act through its centre. The radius of the circle is 3 m.

Calculate the resultant vertical stress produced by these loads on a horizontal plane 1. Determine i dry unit weight ii void ratio iii porosity iv degree of saturation. Calculate its void ratio, dry unit weight, saturated unit weight and submerged unit weight. Answer: 0. The combined mass of the clay and the wax was The volume of the clay and the wax was found to be ml.

The sample was broken open and the moisture content and specific gravity were found. The specific gravity of the wax was 0. Determine the bulk density, void ratio and degree of saturation of the soil.

Under an oven dried state, the dry mass of the sample is 2. The specific gravity of the solid is 2. Determine its bulk density, water content, void ratio, porosity, degree of saturation and air content. After drying in an oven, its weight is reduced to 2. The specific gravity of solids and the mass specific gravity are 2. Determine the water content, void ratio, porosity, and degree of saturation. Assume the gram specific gravity as 2. Laboratory tests on a dried sample indicated values of 0.

Calculate the degree of saturation and relative density. AMIE S08, 8 marks : What would be the maximum porosity for a uniformly graded sand having perfectly spherical particles. Determine the volume of soil in compacted embankment on the assumption that there is no change in dry mass and moisture content of soil. The relative density of the soil is 2. Determine the degree of saturation and air content of the soil in the borrow pit and in the embankment.

Answer: 59 litres Q. Determine dry unit weight, void ratio and specific gravity of solid particles. Answer: 2. Take specific gravity of solids as 2. If the specific gravity of solids be 2.

If the soil is saturated, what will be its saturated unit weight? After drying in an oven, its weight is reduced to 3 N.

The specific gravity of solids and the mass specific gravity arc 2. Determine the water content, void ratio, porosity, and the degree of saturation. Find bulk unit weight, saturated unit weight and submerged unit weight. Find the dry density, void ratio and specific gravity of soil solid particles. What will be the volume of compacted embankment, assuming no change in dry mass and moisture content of soil?

Relative density, G, is 2. Also, find the degree of saturation and air content of the borrow pit and the embankment. AMIE S14, 5 marks : A sample of sand has a volume of ml in natural state Its maximum volume when compacted is ml. When gently poured in a measuring cylinder, its maximum volume is ml. Determine the relative density. After oven drying, its weight is reduced to