Geotechnical Engineering

Geotechnical Engineering (A50120) Geotechnical Engineering (A50120) Unit 1 Dr. PVSN Pavan Dr. PVSN Pavan Kumar Associate Professor

Dr. PVSN Pavan Kumar

Syllabus (Unit 1) • INTRODUCTION: Soil formation – clay mineralogy and soil structure – moisture content – weight ‐ volume relationship – Relative density. • INDEX PROPERTIES OF SOILS: Grain size analysis – Sieve analysis, Principle of Hydrometer methods – Consistencyy limits and indices – I. S. classification of soils Dr. PVSN Pavan Kumar

Soil • Soil is originated from Latin word solium. • It has h different diff t meaning i in i different diff t scientific i tifi fields fi ld such as Civil Engineering, Geology and agronomy. • For Civil Engineers: It is a loose unconsolidated inorganic material on the earth crust produced by the disintegration of rocks overlying hard rock with or without itho t organic matter. matter • Foundation of all structures should be on or below such soils. Civil Engineers are thus interested in Engineering behavior of soil. Dr. PVSN Pavan Kumar

Soil • For Geologist: Loose material on earth crust consisting of disintegrated rock with an admixture of organic matter that supports plant life. • For Agricultural scientist: Loose material on earth g rock with an crust consistingg of disintegrated admixture of organic matter supporting plant life. • Top soil may ma contain large quantities q antities organic matter and is not suitable as a construction material or as a foundation for structures. Top soil is removed before construction of structure. Dr. PVSN Pavan Kumar

Soil

• Unconsolidated Unconsolidated material material composed of solid particles produced by i l d db disintegration of rocks. • Voids between the particles is occupied by particles is occupied by air, water or both.


Soil

Dr. PVSN Pavan Kumar Cross section of Earth

Soil Mechanics • Soil Mechanics was coined by Dr. K. Terzaghi in 1925 in his book Erdbaumechanic in German. Terzaghi g is regarded as Father of soil mechanics. • According to Terzaghi, soil mechanics is the A di t T hi il h i i th application of laws of mechanics and hydraulics to engineering problems dealing with sediments and other unconsolidated accumulations of solid particles produced by the mechanical and chemical disintegration of rock regardless of whether or not disintegration of rock regardless of whether or not they contain an admixture of organic constituents. Dr. PVSN Pavan Kumar

Applications ‐ Geotechnical Engineering

Pile Foundations

Shallow Foundations


Applications ‐ Geotechnical Engineering

Stability of earth dams and y embankments

Stability of Excavation Other applications Oth li ti Retaining and sheet pile walls Underground structures Pavement Design Pavement Design Dr. PVSN Pavan Kumar

Soil Formation • Soil Soil is formed by the process of weathering of rocks is formed by the process of weathering of rocks due to mechanical disintegration or chemical decomposition. decomposition • Rock surface exposed to atmosphere for a longer period disintegrates and decomposes into smaller period disintegrates and decomposes into smaller particles. • Physical disintegration y g is caused byy – Unequal expansion and contraction of rock minerals due to temperature change. – Wedging action of ice: Disintegration of rock due to formation of ice. – Spreading of roots Spreading of roots of trees in cracks formed in rock. of trees in cracks formed in rock – Abrasion and scouring due to flow of water, wind and glacier over the surface of rock. Dr. PVSN Pavan Kumar

Soil Formation • Soils formed due to physical disintegration has the properties of parent rock. Gravel and sand form due to physical disintegration. • Chemical decomposition – Original rock minerals are changed to new minerals due to chemical reactions. Soil formed do not have properties of parent rock • Hydration: water combines with rock mineral forms a new chemical compound. Changes volume and rock decomposes to small particles. • Carbonation: Carbon dioxide in atmosphere combines with water to form carbonic acid. This reacts with rock and decomposes the rock. Dr. PVSN Pavan Kumar

Soil formation • Oxygen ions combine with rock and decomposes similar i il to t rusting ti off steel t l – Oxidation. O id ti • Some rock minerals dissolve in water to form solutions Eg: Lime stone • Hydrogen cat ions, ions H+ in water replace the metallic ions such as calcium, sodium and potassium in rock minerals i l to form f soilil with i h a new chemical h i l combination. Hydrolysis • Chemical decomposition forms clay minerals and impart plastic properties to soil Dr. PVSN Pavan Kumar

Soil deposits • Soils transported and deposited by water are known as alluvial deposits. deposits • Deposits made in lake are lacustrine deposits. • Soil formed in ocean or sea by flowing water is called marine deposits. • Soil deposited by wind is called aeolian deposits. Ex: Sand dunes • Loess is a silt deposit made by wind. • Drift is a deposit made by glaciers. Till is a deposit formed due to melting of ice. • Colluvial soils such as Talus are formed due to gravity. Dr. PVSN Pavan Kumar

Soil deposits in India India is covered with five major d deposits it alluvial deposits bla deposits, bla ck cotton soil, lateritic soil, desert soils and marine i deposits Dr. PVSN Pavan Kumar

Soil deposits in India

Black cotton soil

Desert Soil

Lateritic Soil

Marine Soil

Alluvial Deposits Dr. PVSN Pavan Kumar

Soil vs other materials Soil

Other material (Steel or Concrete) ( )

Soil is formed due to natural process without any control and is highly ith t t l d i hi hl heterogeneous and unpredictable.

Steel is manufactured in accurately controlled environment. Properties of t ll d i t P ti f concrete controlled to some extent.

p g p Properties of soil change from place to place and with depth. Change in environment, loading and drainage conditions conditions. Properties of soil depend on number of factors and an elaborate testing is necessary. necessary

Major properties of soil j p p are modulus of elasticity, tensile and compressive strength.

Soil is transported for limited distance and Can be transported for large distance local available material is used for construction. Foundation soils are not open for inspection Bore holes are to be drilled to inspection. Bore holes are to be drilled to view the samples.

Steel and concrete can be inspected before use before use.


Basic Definitions • Soil Soil mass mass consists of solids and voids. and voids. • Voids may be filled with water filled with water or air partly or fully. y Block diagram

Three constituents are blended together to form a complex material. For convenience solid particles are placed at the bottom and water, air separately. This three phase diagram is known as block diagram. Dr. PVSN Pavan Kumar

Dry and saturated soil

Saturated soil

Dry soil y

V = Total volume of soil; Vs = Volume of solids; Vw = Volume of water; Va = Volume of air; V = Vs + Vw + Va M = Total mass of soil; Ms = Mass of solids; M M = Total mass of soil; M = Mass of solids; Mw = Mass of water; = Mass of water; Ma = Mass of air = 0; M = Ms + Mw Relative proportion of three constituents will govern the behavior of soil. Mass may be replaced with weight of constituent, W Dr. PVSN Pavan Kumar

Volume relationship •

Volume of voids, vv V id ti Void ratio, e = volume of solids, vs

Expressed in decimal, e for coarse grained soil is Expressed in decimal e for coarse grained soil is smaller than that for fine grained soil. For some soils e may be greater than unity soils e may be greater than unity. • Porosity, n =

Volume of voids, vv Total volume v

Porosity is expressed in percentage. Porosity can not exceed 100% as vv will not exceed v. exceed 100% as v will not exceed v Porosity and void ratio are a measure of denseness or y looseness of soil. As soil becomes more and more dense these values decrease. Dr. PVSN Pavan Kumar

Relation between n and e

1 V Vv + Vs Vs 1 e +1 = = =1+ =1+ = n Vv Vv Vv e e

e ⇒n= 1+ e

Relation between e and n 1 V Vv + Vs Vs 1 = = =1+ =1+ n Vv Vv Vv e

1 1 1− n ⇒ = −1 = e n n

Volume of water ,Vw • Degree of saturation, S = Volume of Voids,Vv Expressed in percentage. For dry soil, S = 0 and for fully saturated soil, S = 100% Dr. PVSN Pavan Kumar

n ⇒e= 1− n

Volume relationship Volume of air , Va P Percentage air voids, (n t i id ( a) = ) Total Volume, V

• Expressed in percentage

• Air content, (ac) = Volume of

air , Va Volume of voids, Vv

Percentage air voids na = Va = Va × Vv = a × n Percentage air voids, n c V

Vv

Moisture or water content = w = Moisture or water content = w =

V

Mass of water , M w Mass of solids , M s

Expressed in percentage, moisture content of some Expressed in percentage moisture content of some fine grained soils such as silt and clay is more than that of coarse grained soil such gravel and sand that of coarse grained soil such gravel and sand. Some fine grained soils have m.c > 100% Dr. PVSN Pavan Kumar

Volume Mass Relationship Total mass of soil , M • Bulk Mass density, ρ = Total volume, V

This is also called wet mass density. Units are kg/m This is also called wet mass density Units are kg/m3, g/ml g/ml • Dry mass density, ρd = Mass of solids, M s T l volume Total l ,V

Dry density indicates denseness of soil. • Saturated Mass density, ρsat = •

M Mass off saturated t t d soil il , M sat Total volume, V

M Mass off submered b d soil il , M sub Submerged mass density, ρsub = Total volume, V

When soil mass exists below water it is under submerged condition. A g submerged/buoyancy force will act on the soil solids. According to Archimedes principle, buoyant force equal to weight of According to Archimedes principle, buoyant force equal to weight of water displaced by the solids. Dr. PVSN Pavan Kumar

Submerged density Mass of saturated soil, Msat = M Ms + V + V v γw Buoyance force, U = V γw Submerged mass of soil, Msub = Msat ‐ U

M sub M sat − U = V V ρ wV = ρ sat − = ρ sat − ρ w V

ρsub =

ρ sub = ρ sat − ρ w

Submerged unit weight is half of saturated unit eight Submerged unit weight is half of saturated unit eight Dr. PVSN Pavan Kumar

Volume mass relationship • Density of solids, ρs =

Ms Vs

I t d f Instead of mass if weight is used if i ht i d • • • •

W Bulk unit weight, γ g , γ = V Ws Dry unit weight, γd = V

Wsat Saturated Unit weight, γsat = V Wsub = γ sat − γ w S b Submerged unit weight, γ d it i ht sub = V

Ws g , γs = V • Unit weight of soil solids, γ s Density of water, ρw is 1000 kg/m3 = 1 g/cc Unit weight of water, γw is 9.81 kN/m Unit weight of water, γ is 9.81 kN/m3 Dr. PVSN Pavan Kumar

1 kg = 9.81 N

Specific gravity Specific gravity of solid particles is the ratio of mass of a certain volume of solids to the mass of a equal volume of water at 4°C. ρs G= ρw

• It is used to determine void ratio and particle size. • Specific gravity of water ‐1.0 S ifi Specific gravity of soil – it f il 2.65 to 2.80, 2 65 t 2 80 Specific gravity of mercury – p g y y 13.6


Three phase diagram in terms of void ratio Assume Volume of solids Vs = 1.0 Volume of solids, V =10 Cross section of soil = 1.0 Vv ⇒ Volume of voids = Vv = e e= Vs Vw , volume of water = Vw = ew = eS S= Vv

Volume of air, Va = ea = e – es = e (1‐s) Mass of air, Ma = 0; Mass of water, Mw = esρw Mass of solids Ms = ρ Mass of solids, M = ρs 1 = Gρ 1 = Gρw Dr. PVSN Pavan Kumar

Three phase diagram in terms of void ratio Vv e Porosity = n = V = 1 + e

e (1 − S ) Percentage air voids = na = 1 + e

Va e(1 − S ) Air content, ac = V = e = 1 − S v

Bulk Density, ρ =

Dry Density ρd = Dry Density, ρ

M Ms + Mw = V V Gρ w + eSρ w (G + eS )ρ w = = 1+ e 1+ e

M s Gρ w = V 1+ e

Gρ w + eρ w M sat or S = 1 1 ⇒ ρ sat = 1 + e Saturated Density, ρsatt = V or S Saturated Density, ρ (G − 1)ρ w Dr. PVSN Pavan Kumar ( G + e )ρ w = ρsub = ρsat ‐ ρw = 1 + e 1+ e

Three phase diagram in terms of porosity Assume total volume of l l f soil, V = 1 Vv Porosity, n = V ⇒ Vv = n Volume of solids, V , s = 1‐n Vv n Void ratio, e = = Vs 1 − n

Vw Degree of saturation, S = V v

Volume of water, n , w = Sn Mass of water, Mw = Snρw ; Mass of solids = Ms = (1‐n) Gρw Vv n = Void ratio = e = Vs 1 − n

M (G (1 − n) + S n) ρ w = = G ρ w (1 − n) + S nρ w B lk D i Bulk Density, ρ bulk = V 1 Dr. PVSN Pavan Kumar

Three phase diagram in terms of porosity Dry density, ρd = Gρw (1‐n) Saturated Density, ρsatt = G ρ w (1 − n) + nρ w Saturated Density, ρ

Submerged density, ρsub = ρsat ‐ ρw = G ρ w (1 − n) + ρ w (n − 1)

Relation between e, S, W, G M w ρ wVw ⎛ 1 ⎞⎛ Vw Vv ⎞ e × S = = ⎜ ⎟⎜⎜ × ⎟⎟ = Water content, W = Ms ρ sVs ⎝ G ⎠⎝ Vv Vs ⎠ G ⇒ e S =W G


Mass density in terms of water content

( G + es )ρ w ρ= 1+ e

(G + WG ) ρ w (1 + W )Gρ w = = ⎛ WG ⎞ ⎛ WG ⎞ 1+ ⎜ 1+ ⎜ ⎟ ⎟ ⎝ S ⎠ ⎝ S ⎠

If soil is fully saturated, S = 1 ρ sat

ρ sub = ρ sat − ρ w Gρ w ρd = = 1+ e

(G + WG ) ρ w = 1 + WG

( G + WG )ρ w = −ρ 1 + WG

Gρ w ρ ⇒ ρd = 1+W ⎛ WG ⎞ 1+ ⎜ ⎟ ⎝ S ⎠ Dr. PVSN Pavan Kumar

w

( G − 1)ρ w = 1 + WG

Relation between dry density and % air voids V = Va + Vw + Vs Va Vw Vs 1= + + V V V ⎛ Mw ⎞ ⎛ Ms ⎟+⎜ 1 = na + ⎜⎜ ⎟ ⎜ρV ρ V ⎝ w ⎠ ⎝ s

⎞ ⎟ ⎟ ⎠

⎛ mM s 1 − na = ⎜⎜ ⎝ ρ wV

⎞ ⎟ ⎟ ⎠

⎛ mρ d 1 − na = ⎜⎜ ⎝ ρw

⎞ ⎛ ρd ⎟+⎜ ⎟ ⎜ρ ⎠ ⎝ s

⎞ ⎛ ρd ⎞ ⎛ 1 ⎞ ρ d (Gm + 1) ρ d ⎟+⎜ ⎟ = ⎜m + ⎟ = ⎟ ⎜ Gρ ⎟ ⎝ G ⎠ ρw G ρw w⎠ ⎠ ⎝

Gρ w (1 − na ) ρd = Gm + 1


Moisture content determination • • • • • • •

Oven drying method Infrared Torsion balance moisture meter Infrared Torsion balance moisture meter Pycnometer Sand bath method Alcohol method Alcohol method Calcium carbide method Radiation method


Moisture content – Oven drying method IS 2720 Part II (1973)

Balance Container

Oven

Mass of container + lid = M1 Mass of container + lid + wet soil = M2 Mass of container + lid + Dry soil = M3

( Mass of water , M w M2 − M3) = Water Content, W = MassDr. PVSN Pavan Kumar of dry soil or solids, M s M 3 − M1

Moisture content – Oven drying method IS 2720 Part II (1973)

• Quantity of sample taken for the test depend on the gradation, particles and degree g of wetness of soil maximum size of p • If sample is dry more quantity is required. • Soil sample in the container is oven dried at a temperature of 110 110° ± 5° C for 24 hours. Sl. Size of particles more Minimum No

than 90% passing

quantity (gm)

• Temperature lower than above Temperature lower than above 1 425 Micron 25 may not completely evaporate the water. 2 2 mm 50 • Temperature more than above 3 4.75 mm 200 may cause the break down of 4 10 mm 300 crystalline structure of soil and crystalline structure of soil and chemically bound water may lost. 5 20 mm 500 • When soil contains gypsum and 6 40 mm 1000 organic matter a temperature of 60 i tt t t f 60 Minimum quantity of soil sample for test to 80° is recommended. Dr. PVSN Pavan Kumar

Moisture Content – Infrared lamp, torsion balance moisture meter R di Reading

• Equipment consists of a infra red lamp and a torsion balance.

Thermometer

Top Cover

Adjustable scales j Energy regulator for lamp


Moisture Content ‐ Pycnometer • A glass jar of 1 liter capacity, capacity fitted with brass conical cap having 6 mm diameter. diameter • Rubber or a fiber washer is placed between cap and a jar. • This method is used only if specific gravity, G of solids is known. • Wet soil sample of 200 to 400 g is taken in pycnometer and weighed. • Water is added to the soil to make it half full. Dr. PVSN Pavan Kumar

Moisture Content ‐ Pycnometer • More water is added to make the pycnometer More water is added to make the pycnometer full and weighed. full and weighed

M2

M1

Ms

M3

MS

ρs

M1 = Mass of pycnometer; M2 = Mass of pycnometer + wet soil; M3 = Mass of pycnometer = Mass of pycnometer + wet soil + remaining water; + wet soil + remaining water; M4 = Mass of pycnometer + water M 4 = M 3 – MS +

MS

ρs

ρw

MS 1 ⇒ M 4 = M 3 – MS + ⇒ M 4 = M 3 − MS (1 − ) G G

G M 2 − M1 − M S M s = (M 3 − M 4 ) Water Content , m = × 100 (G − 1) MS Dr. PVSN Pavan Kumar

M4

Moisture content ‐ Sand bath method • SSand d bath b th is i a large l open vessell containing t i i sand d filled filled to a depth of 3cm or more. • Sample is taken in a tray, crumbled, placed loosely, few pieces of white paper are placed. placed Tray is weighed. • Tray is placed l d on sand d bath. b h Sand d bath b h placed l d on a stove and heated. Specimen is turned with a spatula and heated till white paper turns to brown colour. • Tray is removed from sand bath cooled and weighed. i h d Dr. PVSN Pavan Kumar

Moisture Content ‐ Alcohol method • Soil sample taken in an evaporating dish. • Large lumps of soil shall be broken and mass of wet sample noted. • Sample p is mixed with methylated y spirit p ((Alcohol). ) • One ml of spirit is needed for 1 gm of soil. • Spirit and soil are mixed to a uniform color. color • Spirit is ignited and allowed to burn completely. • Dish is cooled, mass of dry soil is noted. • Spirit is volatile and can not be used if it contains large proportion of clay, organic matter, gypsum or any other calcareous material. Dr. PVSN Pavan Kumar

Moisture content – Calcium carbide method • M Moisture content is determined by the fact that when water reacts i i d i db h f h h with calcium carbide (cac2), acetylene gas is released. Cac2 + 2 H + 2 H2O ÆC O ÆC2 H2 + Ca (OH) + Ca (OH)2


Moisture Content – Calcium carbide method • W Wet sample is placed in a sealed container t l i l di l d t i containing calcium carbide. • Sample of sand is ground and pulverized. • Cohesive and plastic soils are tested after addition Cohesive and plastic soils are tested after addition of steel balls in pressure vessel. Test requires 6g of soil.l • Pressure of gas released acts on diaphragm of Pressure of gas released acts on diaphragm of moisture tester and moisture content is recorded based on wet mass of soil based on wet mass of soil


Moisture content – Radiation method Itt co consists s sts o of ttwo o bo boree holes o es with t a stee steel cas casing. g In o onee o of tthem e a capsule of radioactive isotope of Cobalt 60 is placed. Casing A has a small opening through which rays will come out. Detector is placed in other casing B of bore hole facing the capsule. Two casings will have openings facing each other. Hydrogen atoms in water of soil cause scattering of neutrons and they h lose l energy. Energy lost is proportional to quantity of water present in soil. Method should be used with proper precaution to shield against radiation. Dr. PVSN Pavan Kumar

Specific gravity determination Following apparatus/methods are used to determine the specific gravity of solids depending on the size p g y p g of particles 1 Density bottle method 1. D it b ttl th d 2. Pycnometer y 3. Measuring Flask 4 Gas Jar method 4. G J h d Principle of determination is same for all apparatus. p pp


Specific gravity – Density bottle method IS 2720 Part 11 ‐ 1980

• Density bottle of 50 ml capacity with a stopper and a h l at top is used hole d to determine specific gravity. • Density bottle is cleaned, dried at a temperature of 105° and d cooled. l d • Mass of bottle and stopper = M ass o bott e a d stoppe 1 • 5 – 10 g of oven dry soil passing 2 mm sieve is taken in a bottle and weighed. Mass of bottle, stopper and dry soil = M2 Distilled water is added to cover the soil, allowed to soak in water for 2 hours, more water is added until Dr. PVSN Pavan Kumar the bottle is half full.


• Air entrapped in density bottle is expelled by applying vacuum pressure for one hour or entrapped air is removed by heating. More water added dd d to make k the h bottle b l full. f ll Mass of bottle,, stopper, pp , soil and water = M3 • Bottle is emptied, washed and refilled with water. Mass of bottle, stopper and water = M4



M2

M1

Ms

MS

ρs

M 4 = M 3 – MS +

M3

Ms

ρs

ρw

1⎞ ⎛ M s ⎜1 − ⎟ = M 3 − M 4 ⎝ G⎠

⎛ M3 − M4 ⎞ 1 ⎛ M3 − M4 ⎞ 1 ⎟ ⎟ 1 − = ⎜⎜ = 1 − ⎜⎜ ⎟ ⎟ G ⎝ M 2 − M1 ⎠ G M − M 1⎠ ⎝ 2 ( 1 (M 2 − M1 ) − (M 3 − M 4 ) M 2 − M1 ) ⇒ G= = (M 2 − M1 ) G Dr. PVSN Pavan Kumar (M 2 − M 1 ) − (M 3 − M 4 )

M4


• Specific gravity is reported at 27°C or at 4°C. G27

specific ifi gravity it off water t att t °C = Gt × specific gravity of water at 27°C

• Kerosene is a better wetting agent and is some times used in place of water times used in place of water. • Method is suitable for fine grained soils with more than 90% passing 2 mm sieve.


Specific gravity – Pycnometer method • Capacity of pycnometer is large. • 200 – 300g of oven dry medium grained soil is used to determine specific gravity. l d d f • Procedure is similar to density bottle method. method Measuring flask of 250 to 500 ml is suitable for determination of specific gravity of for determination of specific gravity of medium and fine grained soil.

1 liter glass jar with a rubber bung is used to determine specific gravity of soil Dr. PVSN Pavan Kumar

Dry density determination Bulk density, ρbulk = M V Mass of a soil sample can be determined with precision. It is difficult to determine the volume of sample accurately. Methods discussed below differ in procedure to determine p volume. ρbulk ρ = • Dry density obtained from bulk density. Dry density obtained from bulk density. d 1+ w Methods to determine density of soil Water displacement method Water displacement method Submerged density method Core cutter method C h d Sand replacement method Water ballon method

• • • •


Dry Density determination ‐ Water displacement method

• Volume of soil sample is determined by water displacement. • Soil mass disintegrates in contact with water, sample is coated with paraffin wax to make it impervious. • Sample S l trimmed i d to a regular l shape, h weighed i h d and d dipped di d in i molten l wax, allowed to cool and again weighed. Difference of two weights give weight of wax coated. • Waxed specimen is immersed in water, volume of waxed specimen = volume Dr. PVSN Pavan Kumar of water displaced.

Dry Density determination ‐ Water displacement method Volume of soil mass = Volume of waxed specimen – Volume of wax peeled off l d ff V = V V Vt –

(M t − M ) ρP

V = Volume of specimen; Vt = Volume of waxed specimen; Mt = Mass of waxed specimen; M = Mass of specimen ρP = Density of wax Density of wax A representative sample of soil is taken from middle for water content determination content determination. Bulk density, moisture content and dry density of soil is d determined. i d Dr. PVSN Pavan Kumar

Dry Density determination – Submerged Mass density method • Mass of wax coated specimen in water, M1 is determined by a special type of is determined by a special type of balance. • Mass of wax coated specimen = M Mass of wax coated specimen = Mt Volume of soil mass = Volume of waxed specimen Volume of wax peeled off specimen – Volume of wax peeled off V =

M t − M1

( Mt − M ) −

ρw ρP V = Volume of specimen; Vt = Volume of waxed specimen; p ; M1 = Mass of waxed specimen in water; M = Mass of specimen

ρP = Density of wax = Density of wax Dr. PVSN Pavan Kumar

Dry Density determination – Submerged Mass density method Mass density method

Arrangement to weigh the waxed specimen in water specimen in water Dr. PVSN Pavan Kumar

Dry Density determination – Core cutter method • Core cutter consists of an open steel cylindrical barrel with h d hardened sharp cutting edge. d h d • Core cutter is driven into the soil by a hammer. To avoid damage at the top a dolly is placed. • Cutter containing soil is taken out of ground and soil at edges i is cut. t Mass of cutter with soil = M2; Mass of empty cutter = M1 V l Volume off soil, il V in i core cutter tt is i determined d t i d by b measuring i th the dimensions or filling the cutter with water and finding mass of water. ρbulk

soft fine grained soils where surface is M 2 − M1 Method is suitable for soft, = exposed. It cannot be used for stoney or gravelly soil. V Dr. PVSN Pavan Kumar

Dry density determination – Sand replacement method • Test consists of two stages T i f • Calibration of apparatus • Measurement of volume soil hole Calibration of apparatus

2. Sand pouring cylinder is placed 2 Sand pouring cylinder is placed on calibrating cylinder and Calibrating Container shutter is opened till it fills the Sand Pouring Cylinder (SPC) Metal tray with hole at center calibrating cylinder and cone in Uniform dry, clean Sand passing 600 μ and the SPC. Mass of SPC is M2. 1

1. 2. 3. 4.

2

3

1. Sand pouring cylinder is filled 1 Sand pouring cylinder is filled with sand and its mass is M1

retained on 300 μ is used.

3. SPC is refilled and placed on a 3 SPC i fill d d l d 4. Volume of calibration cylinder is level surface and shutter is determined, V. opened. Sand occupies the p p M1 − M 2 − M c conical portion. Mass of sand in ρ = Density of sand, sand V Dr. PVSN Pavan Kumar cone, M c

Dry density determination – Sand replacement method • Tray with a central hole is placed on the level ground surface. p • Hole of 100 mm diameter and 150 mm deep is excavated in soil. Let weight of soil determined be M. • Sand pouring cylinder filled with sand and weighed. Let the mass be M1. • Filled sand pouring cylinder is placed on the hole, shutter opened, sand fills the hole and conical portion sand fills the hole and conical portion. Volume of hole =

M1 − M 3 − M c

ρ sand

M3 = Mass of SPC after filling the hole and cone Mc = Mass of sand in cone f di M Wet density of soil, ρwet = Volume of hole

Method is applicable to both fine grained and coarse grained soils. Dr. PVSN Pavan Kumar

Dry density determination – Rubber balloon method • Volume Volume of hole is determined by using of hole is determined by using rubber balloon method or by filling water in the hole covered with a plastic sheet. • Apparatus consist of a density plate, graduated cylinder enclosed in an aluminum case. aluminum case. • Cylinder is partly filled with water. There is a hole at bottom of casing covered by balloon. • Balloon is pulled up into the cylinder or pushed down through the hole A pump pushed down through the hole. A pump is attached to the cylinder for this purpose. • Balloon comes into the hole through the plate by applying pressure and is pulled into the cylinder by applying vacuum into the cylinder by applying vacuum. Dr. PVSN Pavan Kumar

Dry density determination – Rubber balloon method • Balloon deflates against the surface of soil by applying pressure. • Difference between volume of water in the cylinder before and after making a hole will give the volume before and after making a hole will give the volume of hole. • Bulk and dry density, void ratio and degree of g saturation can be determined from following equations if specific gravity, G is known. ρbulk

Mass off soil excavated = Volume of hole

Gρ w ρd = 1+ e

⇒ eS = WG Dr. PVSN Pavan Kumar

Properties of soil

Index Properties

Engineering Properties

• Engineering properties of soil are permeability, compressibility and shear strength. strength • Permeability is required to determine seepage of water through dam. • Compressibility is required to determine the settlement of foundations. • Shear strength is the ability to resist shear stress and is needed in stability of slopes, bearing capacity of soil and earth pressure on retaining structures. • Above tests are elaborate, elaborate time consuming and require undisturbed sample. sample • Rough assessment of soil properties is possible without conducting elaborate testing by determining index properties. • Index d properties are an indicative d off Engineering properties. • Index properties of coarse grained soil are particle size and relative density. • Index properties of fine grained soil are Atterberg limits and consistency. • Correlation between index and Engineering properties may not be perfect. Dr. PVSN Pavan Kumar

Mechanical analysis • Mechanical Mechanical analysis also called particle size analysis is a analysis also called particle size analysis is a method of separation soil into different fractions. This is represented graphically by a particle size distribution p g p y y p curve. • Mechanical analysis is done in two stages : Sieve y g analysis and sedimentation analysis • Sieve analsysis – Coarse grained soil with particles of size greater than 75 μ. • Sedimentation or wet analysis – Fine grained soil with particles of size smaller than 75 μ. i l f i ll h • Any soil is a combination of coarse and fine grained particles and both types of analysis is required. ti l d b th t f l i i i d • Particles of size smaller than 0.2 mm are determined by electron microscope or X ray diffraction electron microscope or X‐ray diffraction. Dr. PVSN Pavan Kumar

Sieve Analysis • Sieve is made of brass and phosphor bronze cloth. cloth • Sieve is designated by size of square opening in mm or microns. • Sieves of size ranging g g from 80 mm to 75 μ are available. Diameter of sieve is between 15 to 20 cm • Sieve analysis is carried out for coarse grained soil. Coarse grained soil is classified as gravel fraction (> 4.75 4 75 mm) and sand fraction ( 4.75 mm to 75 μ) • Coarse sieves are 80 mm, 40 mm, 20 mm, 10 mm and 4.75 mm are required. • Fine sieves consist of 2 mm, 1 mm, 600 μ, 425 μ, 212 μ, 150 μ and 75 μ. μ • Sieves are stacked one above the other with size decreasing from top to bottom. Sieve of largest opening is kept at the top, lid or top cover is i placed l d above b l largest sieve i size. i • Pan is placed at the bottom of sieve. Dr. PVSN Pavan Kumar

Sieve Analysis

Sieves

Sieve Pan

Dr. PVSN Pavan Kumar Sieve lid

Sieve Shaker

Dry sieve analysis • Dry Dry sieve analysis is suitable for dry cohesion less soil with little or no sieve analysis is suitable for dry cohesion less soil with little or no fines. • Quantity of soil taken for sieve analysis is Q y y Maximum size , mm 80 80 20 4.75

Quantity (kg) 60 6.5 0.5

• Soil should be oven dried and not to contain any lumps. Soil should be pulverized. • Sample is sieved through 4.75 mm and gravel fraction is sieved through a set of coarse sieves manually or by mechanical shaker. W i ht f il Weight of soil on each sieve is obtained. Similarly sand fraction is h i i bt i d Si il l d f ti i sieved through a set of fine sieves. • Normally 10 minutes of shaking is sufficient. Normally 10 minutes of shaking is sufficient • Mass of soil retained on each sieve and pan is obtained to 0.1g. Dr. PVSN Pavan Kumar

Wet Sieve Analysis • If If soil contains more than 5% fine particles wet sieve analysis is required. soil contains more than 5% fine particles wet sieve analysis is required • Representative soil sample of required quantity is taken using riffler and oven dried. • Dried sample is taken in a tray and soaked with water, for deflocclation sodium hexameta‐phosphate at a rate of 2g per liter of water is added. • Slurry is sieved through 4.75 mm with a jet of water, material retained is gravel fraction. This is oven dried and sieved through coarse sieves. • Material passing 4.75 mm sieve is sieved through 75μ sieve. Material is washed till wash water is clear washed till wash water is clear. • Material retained on 75μ is collected, dried, sieved through set of fine sieves. Mass of soil retained on each sieve is determined each sieve is determined. Dr. PVSN Pavan Kumar

Soil Riffler Soil Riffler


Sieve analysis Total mass of sample = 900 g p g IS Sieve, mm

Mass of soil retained, g

Percentage retained

Cumulative percentage retained

Percentage Finer

20

35

3.89

3.89

96.11

10

40

4 44 4.44

8 33 8.33

91 67 91.67

4.75

80

8.89

17.22

82.78

2.0

150

16.67

33.89

66.11

1.0

150

16.67

50.56

49.44

0.6

140

15.56

66.11

33.89

425 μ

115

12.78

78.89

21.11

212 μ

55

6.11

85.00

15.00

150 μ

35

3.89

88.89

11.11

75 μ

25

2.78

91.67

8.33

Pan

75

8.33

100.00

0.00


Particle size distribution 100 90 80

% finer

70 60 50 40 30 20 10 0 0.01

0.1

1 Sieve Size mm Sieve Size, mm Dr. PVSN Pavan Kumar

10

Stokes law • Soil particles finer than 75μ can not be sieved. sieved • Particle size distribution of such soil is determined by sedimentation analysis. analysis • Stoke’s law gives the terminal velocity of a small sphere settling in a fluid of infinite extent. extent • When a small sphere settles in a fluid of infinite extent, its velocity first increases under the action of gravity Drag force increases under the action of gravity. Drag force comes into action and retards the velocity. After an initial adjustment period steady conditions an initial adjustment period steady conditions are attained and velocity becomes constant. This velocity is terminal velocity. • Following forces act on sphere 4 3 r γs ∏ Weight (↓) = g ( ) 3

Buoyant force, U (↑) =

Drag force, FD (↑) = 6πηrv 4 ∏ r 3γ w Dr. PVSN Pavan Kumar 3

Stokes law From equilibrium of vertical forces q

W = U + FD 4 3 4 3 πr γ s = πr γ w + 6πηrV 3 3 4 3 4 3 πr gρ s = πr gρ w + 6πηrV 3 3 1 gD D 2 (G − 1) ρ w V= 18 η

He

Soil solution

D = Diameter of sphere; G = Specific gravity of sphere; η= viscosity of fluid in poise, (dyne‐sec/cm2) Assume particle falls through a height He centimeters in t minutes Assume particle falls through a height, H centimeters in t minutes He 1 gD 2 (G − 1) ρ w V= = ⇒D= 60t 18 η

0.3η × H e g (G − 1) ρ w × t


Stokes law gg = 981 cm/sec = 981 cm/sec2 ; G = 2.67; ρ ; G = 2 67; ρw = 1 g/cc; = 1 g/cc; viscosity of water, η = 10.09 × 10‐3 Poise 1 gD D 2 (G − 1) ρ w V= = 9020 D 2 18 η He = 9020 D 2 H e in cm; t in sec; D in cm t Time required for a spherical particle to settle by 100 mm Time required for a spherical particle to settle by 100 mm Sl No

Diameter in mm

Time

1

0 075 0.075

19 72 sec 19.72 sec

2

0.02

4.62 min

3

0.006

51.36 min

4

0.002

7.70 hr

5

0.001

30.81 hr


Preparation of suspension for sedimentation analysis • 50 g off oven dried d d soill is transferred f d to evaporating dish. • 100 mll dispersion di i solution l ti is i added dd d in i evaporating ti dish di h to cover the soil. • Dispersion solution sol tion is prepared by b addition of 33 g sodium hexa meta phosphate and 7 g sodium carbonate to distilled water. water • Mixture is warmed for 10 minutes and transferred to the cup of mechanical stirrer; distilled water is added to make the cup three‐fourth full. • Suspension p stirred for 15 minutes,, washed on 75μ μ sieve with jet of distilled water. Solution passing the sieve is transferred to 1000 ml measuring jar to conduct sedimentation di i analysis. l i Dr. PVSN Pavan Kumar

Sedimentation analysis ‐ Hydrometer

Container

High speed stirrer Hydrometer

Stop watch Dr. PVSN Pavan Kumar

Sodium hexa metaphospate & Sodium Carbonate

Pretreatment of soil • Soil containing organic matter and calcium compounds should be pretreated before adding dispersion agent. agent • Organic matter is removed by addition of hydrogen peroxide to the soil at a rate of 100 ml for 100g of soil, mixture warmed to oxidise organic matter. • Calcium compounds are removed by adding hydrochloric acid at the rate of 100 ml for every 100 g of soil.


Theory of sedimentation

• At commencement of sedimentation the soil particles are dispersed throughout the suspension. • After a time period at a given depth only those particles remain suspension which are not settled.

Before sedimentation

After time t

At level DD, 2 each of small, intermediate and large size Dr. PVSN Pavan Kumar particles exist.

• Assume soil is composed of particles of three sizes of particles of three sizes only. • Small size particles settle by h by h • Intermediate particle settle by 2h • Large size particles by 3h. L i ti l b 3h • At level BB only 2 small size particles exist. Their concentration is same as before. • At level CC 2 each of small and intermediate particles exist.

Theory of sedimentation • Particles of same size will have same velocity. • Particles of a size less than D will be present at depth He in the same degree of concentration as at th beginning. the b i i P ti l off size Particles i larger l th D will than ill settle below the depth, He. mD % finer = N = m × 100 s

mD = Mass M off particles i l off per mll off suspension i at depth, He after time t. mS = Mass of particles of per ml of suspension at beginning of sedimentation Dr. PVSN Pavan Kumar

Sedimentation analysis – Pipette method • 500 ml soil suspension p is prepared in sedimentation tube as explained. • Apparatus consists of 10 ml pipette for extraction of sample. 10 ml

• Pipette is fitted with a suction inlet. Calibration of pipette • Nozzle of pipette is immersed in distilled water • Stop cock S is closed and three way stop cock, T is opened. Distilled water is sucked k d into i pipette. i • Stop cock is connected to water outlet to drain excess water. •Stop cock is turned to release the water into glass weighing bottle. Dr. PVSN Pavan Kumar Mass of water in bottle in grams gives volume in ml.

Sedimentation analysis – Pipette method Sedimentation test • Sedimentation tube containing suspension is placed in a constant temperature bath at 27°c for one hour.

10 ml

• Tube is bath, rubber the top and by inverting times.

removed from bung is placed on solution is mixed the tube several

• Bung is removed and tube is placed back in the temperature bath. • Sedimentation commences f from the h instance i the h tube b is i placed in the bath. p watch is started to • Stop record the time. • Pipette is lowered into the specimen, sample collected from a Dr. PVSN Pavan Kumar depth of 100 mm. First sample collected after 2 mins.

Sedimentation analysis – Pipette method Sedimentation test • Pipette is lowered slowly 20 sec before collection of sample. • More samples are taken after 4, 8, 15 and 30 minutes and 1, 2, 4, 8 and 24 hours. 10 ml

• To collect sample, stop cock T is opened, suspension is drawn into pipette. pipette Stop cock T is closed. Excess suspension drawn into safety bulb is drained through wash outlet. • Safety bulb is flushed out with distilled water in the bulb f funnel. l • Stop cock turned to transfer g g the solution into weighing bottle. • Distilled water used to transfer any solid particles adhering to Dr. PVSN Pavan Kumar glass bottle.

Sedimentation analysis – Pipette method • Collected Collected sample is oven dried at 105 – sample is oven dried at 105 110 110° C for 24 hours C for 24 hours • Mass of soil solids per ml of suspension, mD = mD‘ ‐ ‘ m mD‘ = mass of solids/ml obtained from sample m = mass of dispersing agent/ ml of suspension Particle size distribution curve of fine grained soils is plotted from the particle size and % finer. D=

0.3η × H e g (G − 1) ρ w × t

mD % finer = N = × 100 ms

This method is accurate but apparatus is delicate and expensive. p Dr. PVSN Pavan Kumar

Tutorial • The The following observations were taken during a pipette following observations were taken during a pipette analysis for the determination particle size distribution of a soil sample. (a) Depth below the water surface at which the sample was taken = 100 mm (b) C (b) Capacity of pipette = 10 ml it f i tt 10 l (c) Mass of sample when dried = 0.3 g (d) Time of taking sample = 7 minutes after the start (d) Time of taking sample = 7 minutes after the start (e) Volume of soil suspension in the sedimentation tube = 500 ml (f) Dry mass of soil used in making the suspension = 25 g Determine the coordinates of point on the particle size p p distribution curve corresponding to above observations G = 2.70; η = 10.09 millipoise; ρw = 1 gm/ml Dr. PVSN Pavan Kumar

Sedimentation analysis – Hydrometer method • Hydrometer is a instrument used to measure the specific gravity of liquids. q Stem • Special type of hydrometer has a stem and bulb. • Stem is marked from 0.995 at top to Neck 1.030 at bottom. • At the commencement of Bulb sedimentation, specific gravity is uniform f at allll depths. d h • As coarse particles settle deeper th than fi fine particles, ti l specific ifi gravity it becomes non‐uniform with depth. Detailed view of scale • Casagrande showed that hydrometer gives specific gravity at the center of immersed volume. Dr. PVSN Pavan Kumar

Sedimentation analysis – Hydrometer method • If volume of stem is neglected and center of immersion is center of bulb. Thus hydrometer ggives the specific p ggravityy at center of its bulb. Calibration of hydrometer • Calibration of hydrometer is done determine the depth at which specific gravity is measured. • Volume of hydrometer, VH is determined by immersing it in a partly filled jar and noting the h increase in volume l off water. • Effective depth, He is the depth at which specific ifi gravity it off solution l ti is i determined. d t i d This Thi is distance between layer AA from free surface BB. Dr. PVSN Pavan Kumar

Sedimentation analysis – Hydrometer method • As soon hydrometer inserted level rises to level A’A’ by VH . 2A

as is AA i.e.

A = Cross section area of jar • Level BB rises to B’B’. Effective depth Effective depth, h VH VH − He = H + + 2 2A A h VH =H+ − 2 2A H = Depth from free surface B’B’ to neck (lowest mark on the stem) Dr. PVSN Pavan Kumar

Sedimentation analysis – Hydrometer method Preparation of calibration chart Preparation of calibration chart • Obtain d1, d2, d3, d4, d5, d6, d7 and h by measuring with a scale and h by measuring with a scale. • Effective depth, He is determined corresponding to determined corresponding to each hydrometer reading. Rh = (G‐1) × 1000

• As sedimentation p progresses, g , specific p gravity of suspension decreases, hydrometer goes deeper and deeper.


Sedimentation analysis – Hydrometer method Sedimentation test • 1000 ml soil solution is prepared as per the procedure mentioned procedure mentioned earlier. • Suspension is mixed thoroughly by placing bung on top of jar and turning upside down turning upside down. • Jar is placed on a level surface and stop watch is started. • Hydrometer inserted in suspension and first reading is taken after ½ , 1, 2 and 4 minutes from commencement of sedimentation. • Hydrometer is removed from jar rinsed with distilled water and placed in a comparison cylinder filled with water and dispersion agent in same concentration t ti as in i soilil solution. l ti R di Readings are taken t k after ft 8, 8 15, 15 30 mins, 1, 2, 4, 8 and 24 hours. Dr. PVSN Pavan Kumar

Sedimentation analysis – Hydrometer method Temperature correction, correction Ct • Hydrometer is calibrated at 27°C • If temperature is more than 27°C, solution becomes lighter and a low value of hydrometer reading is recorded. C Correction ti is i positive. iti • Temperature is less than 27°C 27 C, correction is negative. negative Meniscus correction, Cm = Reading at top – Reading at level of suspension top Reading at level of suspension

Corrected hydrometer reading, Rh = R = Rh ‘ + C + Cm

Dispersion agent correction, Cd • Addition of dispersion agent increases the specific gravity of suspension. C Correction ti is i negative. ti

R = Rh ‘ + Cm ± Ct ‐ Cd Dr. PVSN Pavan Kumar

Relation between Percentage Finer and Hydrometer Reading • LLet Ms be b the h mass off dry d soilil in i volume, l V • At commencement of sedimentation, soil water suspension is uniform mass of solids per unit volume, uniform, volume Ms/V is same at all depth. depth • Initial density of suspension, ρi = M s + mass of water in suspension V Ms M Mass off water t ρi = + V V

Ms Volume l off solids l d / Volume l off suspension = Gρ wV Ms Volume of water / Volume of suspension = 1 − Gρ wV

⎛ Ms ⎞ ⎟ρw M Mass off water t / Volume V l off suspension i = ⎜⎜1 − ⎟ G ρ V w ⎠ ⎝ Ms ⎡ Ms ⎤ Ms ⎡ 1 ⎤ M s ⎡ G − 1⎤ Initial density = ρi = + ⎢1 − ρ = ρ + 1 − = ρ + ⎥ w w w ⎢ G⎥ ⎢ G ⎥ V G ρ V V V ⎣ ⎦ ⎣ ⎦ w ⎦ ⎣ Dr. PVSN Pavan Kumar

Relation between Percentage Finer and Hydrometer Reading

• If If M MD is the mass of solids in volume, V at a depth, is the mass of solids in volume V at a depth He after time, t M D ⎛ G −1⎞ Density of suspension at that depth, ρ = ρ w + ⎜ ⎟ V ⎝ G ⎠

mD N × ms MD Ms % Finer, N = ×100⇒ mD = mD = ms = ms 100 V V ms ⎛ G − 1 ⎞ 100 ⎛ G ⎞ ( ) ρ = ρw + N × ⎜ ⎟ × ρ − ρw × ⎟⇒ N =⎜ 100 ⎝ G ⎠ ms ⎝ G −1⎠

G R V R = (ρ − ρ w ) × 1000 N= × × × 100 G − 1 1000 M s If Ms is mass of solids for 1000 ml of solution G R % Finer , N = × × 100 G − 1 M s Dr. PVSN Pavan Kumar

Tutorial • A dry sample of mass 50 g is mixed with distilled water to prepare a suspension of 1000 ml for p p p hydrometer analysis. The reading of hydrometer taken after 5 minutes was 25 and the depth of the taken after 5 minutes was 25 and the depth of the center of bulb below the water surface when the h d hydrometer was in the jar was 150 mm. The volume t i th j 150 Th l of the hydrometer was 62 ml and the area of cross section of jar was 55 cm2. Assuming G = 2.68 and η p = 9.81 millipoise. Determine the coordinates of point corresponding to the above observation. Dr. PVSN Pavan Kumar

Limitations of sedimentation analysis • SSoilil particles i l are not spherical h i l and d their h i diameter di i close is l to thickness hi k rather than length or width. • Particles of different size in solution have different specific gravity and use of an average value is incorrect. Error is negligible. • Stokes is law is applicable to fluids of infinite extent, extent walls of jar make the fluid finite. • Stokes law applicable to settlement of one sphere. Simultaneous settlement of number of spheres and their interference is a source of error. • Sedimentation analysis can not be used for particles larger than 0.2 mm as turbulent conditions develop and stokes law is not applicable. • Sedimentation analysis is not applicable for particles smaller than 0.2 μ. Particles do not settle as per stokes law. • Sedimentation method can not be used for chalky soil. Dr. PVSN Pavan Kumar

Grading of soil

•

Soil consist of coarse and fine grained soil soil. Particle size analysis is carried by conducting sieve and sedimentation analysis. l i

•

Grading of soil means distribution of particles of different sizes. sizes

• Flat S curve containing particles of different sizes in good proportion is called well

graded soil. •Soil with some particle sizes missing will have a curve with hump is called gap graded or skip graded soil. soil • Steep curve indicates a soil containing particles of same size. Such soils are known as Dr. PVSN Pavan Kumar uniform soils.

Grading of soil D

• Uniformity U if i off soilil is i expressed d by b uniformity if i coefficient, ffi i Cu = 60 D10 D60 = Particle size such that 60% of the soil is finer than this size D10 = Particle size such that 10% of the soil is finer than this size Cu < 2 – Uniform soil D30 2 C ffi i t f Coefficient of curvature, t Cc = Cu > 4 – Well graded gravel D60 × D10 Cu > 6 – Well graded sand D30 = Particle size corresponding to 30% finer Denseness

Relative Density

Very loose

50%

Liquid limit – Cone penetrometer method • Apparatus pp consists of a steel cone with an apex angle of 30° ± 1° and length 35 mm. • Cone attached to sliding rod to mass is 80 g. • Soil sample prepared as explained in Casagrande method. • Soil pat is placed in a cup of 50 mm diameter and 50 mm height. • Cup is filled with soil by not entrapping air and excess soil removed. • Cone is lowered to just touch the surface of soil. • Graduated scale is adjusted to zero. • Cone released and allowed to penetrate for 30 secs. • Water content at which penetration is 25 mm is liquid limit. Dr. PVSN Pavan Kumar

Stand

Graduated Scale

Control of movement of cone Cone Cup filled with soil

Base

Liquid limit – Cone penetrometer method

• Liquid limit is determined by chart or from following equation wl = w = wy + 0.01 (25‐y) (w + 0 01 (25 y) (wy +15) • Shear strength of soil at liquid limit is 1.76 kN/m2 • Method is easy to perform, applicable for wide range of soils and do y p , pp g not depend on judgment of operator. Dr. PVSN Pavan Kumar

Plastic limit

• Soil is air dried and sieved through g 425 μ. • 30g of soil taken in an evaporating dish. dish • Mixed with water till it becomes plastic and molded with fingers. p g • 10g of plastic soil taken in one hand and ball formed. • Ball B ll rolled ll d with ith fingers fi on glass l plate l t to form a soil thread of uniform diameter. • Rate of rolling 80 to 90 strokes per minute. • If diameter of thread is smaller than 3 mm without cracks moisture content is more than plastc limit. Dr. PVSN Pavan Kumar

Plastic limit • Soill is kneaded k d d further f h to reduce d water content. • Soil S il rerolled ll d into i t a thread th d off 3 mm diameter till it crumbles. • Water content at which hich soil can be rolled into a thread of 3mm diameter without crumbling is known as plastic limit. • Test repeated by taking fresh sample. • Shear strength g of soil at p plastic limit is 100 times that at liquid limit. Dr. PVSN Pavan Kumar

Shrinkage limit

• Shrinkage limit is the lowest water content t t att which hi h the th soilil is i saturated. t t d • Maximum water content at which reduction of water content will not decrease the volume of soil. age limitt iss tthee water ate co content te t at • SShrinkage stage II.

Shrinkage dish filled with wet soil

Air drying of wet soil decreases volume Dr. PVSN Pavan Kumar from V1 to V2

Glass cup Porcelain evaporating dish Measuring Jar Steel Shrinkage dish Steel Shrinkage dish 30 to 40 mm dia 15 mm height Plain glass plate, glass plate with prongs glass plate with prongs

Oven dried soil

Shrinkage limit

Mass of water at stage I = M1 – MS Loss of mass of water from stage I to stage II = (V1 – V2)ρw (M − M S ) − (V1 − V2 )ρw Shrinkage g limit is the water content at stage g II WS = 1 MS V1 = Volume at stage 1 = Volume of shrinkage dish is determined by filling the h dish di h with i h mercury and d removing i the h excess with i h a plain l i glass l plate. l = Mass of mercury in shrinkage dish 3

(

Density of mercury 13600 kg / m

Shrinkage dish filled with wet soil

)

Air drying of wet soil decreases Dr. PVSN Pavan Kumar volume from V1 to V2

Oven dried soil

Shrinkage limit • Mass of empty shrinkage dish is determined. 50g of soil passing 425 μ sieve is mixed with water to a creamy paste. Water content of soil should be above liquid limit; sample placed in the dish in 3 layers by tapping. Mass of wet sample l in i shrinkage hi k di h is dish i M1. • Soil in shrinkage dish is air dried till colour changes from dark to light. light Mass of dry pat of soil is MS. V2 = Volume of oven dried soil pat = Volume of oven dried soil pat =

Mass of mercury overflown by complete immersion of dry pat in glass cup density of mercury (13600 kg / m3 )


Glass cup Porcelain evaporating dish Measuring Jar Steel Shrinkage dish Steel Shrinkage dish 30 to 40 mm dia 15 mm height Plain glass plate, glass plate with prongs glass plate with prongs

Shrinkage Parameters 1.

Shrinkage Index = Plastic limit – Shrinkage limit

2.

Shrinkage ratio, SR =

(V1 − V2 )

Vd (W1 − W2 )

W1 and W2 are water contents at stage 1 and stage 2 respectively. 3.

⎛ V1 − Vd Volumetric Shrinkage, VS = ⎜ ⎜ V d ⎝

4.

Linear Shrinkage, LS =

⎞ ⎟ × 100 ⎟ ⎠

(Initial length − Final length ) × 100 Initial length

Mould of 140 mm length and semi circular section of 25 mm diameter is used. Mould of 140 mm length and semi circular section of 25 mm diameter is used. w − wp

• Liquidity Index, Il =

Ip

× 100

It indicates nearness of water content of wet sample to the liquid limit. • Consistency Index, I y , c =

wl − w × 100 Ip

It indicates nearness of water content of wet sample to the plastic limit. Dr. PVSN Pavan Kumar

w1 − w2 • Flow Index, If = ⎛ N1 ⎞ ⎟⎟ logg10 ⎜⎜ ⎝ N2 ⎠

Lower shear strength Larger shear strength

N1 = Number of blows required at water content, W content, W1 N2 = Number of blows required at water content, W2 • Flow Fl i d is index i the h rate at which hi h soilil loses its shear strength with increase of water content • Soil with steeper slope has a small increase of number of blows ( or shear strength) compared to a flatter slope for same water content. • Soil 1 has larger shear strength and 2 has smaller shear strength. Toughness Index, I t =

Plasticity Index, I P Flow Index, I f

Toughness index is a measure of shear strength at plastic limit


Sl No

Consistency

Consistency Index (%)

Unconfined compressive strength, qu in kN/m2

Characteristics of soil

1

Very soft

0 – 25

100 > 100

200 – 400 200

soil can be soil can be indented with indented with thumb nail

6

Hard

> 100

> 400

Soil can be indented with difficulty by thumb nail difficulty by thumb nail

25 – 50

Thumb can be pressed into soil soil

• Cohesive soil has certain structure in its natural state and is disturbed when soil is remoulded. • Sensitivity indicates weakening of soil due to remolding. remolding

Unconfined compressive strength of undisturbed clay, (q u ) u Sensitivity = Unconfined compressive strength of remoulded clay, clay (q u ) r • Both the above strengths are determined at same water content Dr. PVSN Pavan Kumar

Unconfined compressive strength of soil

C li d i l il Cylindrical soil sample l


Sensitivity Soil type 16

Quick

• Thixotropy : If remoulded soil is allowed to stand without loss of water, soil regains some lost strength with time. time • Strength increases due to gradual reorientation of water molecules in adsorbed water layer and re‐establishment of chemical equilibrium equilibrium.

Plasticity Index, I P Activity, A = P Percentage t off clay l fraction f ti ( < 2 μ))

Activity is a measure of water holding capacity of clay capacity of clay


Indian standard classification system IS 1498 ‐ 1970 Soil divided into 18 groups Coarse grained soil, > 50% material μ sieve retained on 75 μ 8 soil groups

Soil Components Soil

Coarse grained Components

Fine grained Fine grained Components

Fine grained soil, > 50% material passing through 75 μ sieve

Organic soil, Peat 1 group

9 soil groups

Soil Components

Particle size

Boulder

Average diameter more than 300 mm

Cobble

Average diameter less than 300 mm but retained on 80 diameter less than 300 mm but retained on 80 mm sieve

Gravel (G)

Passing 80 mm retained on 4.75 mm

Sand (S)

Passing 4.75 mm and retained on 75 μ

Silt (M)

Size smaller than 75 μ, slightly plastic or non plastic , exhibits little or no strength when air dried exhibits little or no strength when air dried

Clay (C)

Size smaller than 75 μ, exhibits plastic properties and considerable strength when air dried

Organic matter (O)

Exists in various sizes and stages of decomposition Dr. PVSN Pavan Kumar

Classification of fine grained soil ML – Inorganic silts with none to low plasticity; CL ML Inorganic silts with none to low plasticity; CL – Inorganic clays of low plasticity; OL Inorganic clays of low plasticity; OL – Organic silts of low plasticity; MI ‐ Inorganic silts with medium plasticity; CI – Inorganic clays of medium plasticity; OI – Organic silts of medium plasticity; MH ‐ Inorganic silts with high compressibility; CH – Inorganic clays of high plasticity; OH compressibility; CH Inorganic clays of high plasticity; OH – Organic clay of medium to high Organic clay of medium to high plasticity;


Plasticity Chart

Classification of coarse grained soil Dual symbols GM – GC and SM – SC are used if Atterberg limits fall above A line, line IP between 4 and 7.

Coarse grained soil ((More than half of material is g larger than 75 μ)

Fines between 5 and 12 require dual symbols GP – GM; SW – SC etc.

Gravel (G) ( ) (more than half of coarse fraction is l larger than h 4.75 mm

Sand (S) (more than h lf f half of coarse fraction is smaller than 4 75 mm 4.75 mm

Clean gravel Clean gravel (Fines less than 5%)

Gravel with appreciable i bl fines (Fines more than 12%)

Clean sand (Fines less than 5%)

Sands with appreciable amount of fines f fi (Fines more than 12%) Dr. PVSN Pavan Kumar

Well graded gravel, GW; Cu > 4 and Cc between 1 and 3 and 3 Poorly graded gravel, GP; Not meeting gradation requirements of GW requirements of GW Silty gravels, GM; Atterberg limits below A line or IP less than 4 Clayey gravels, GC; Atterberg limits above A line and IP greater than 7 line and I greater than 7 Well graded sand, SW; Cu > 6 and Cc between 1 and 3 Poorly graded sand, SP; Not meeting gradation requirements of SW Silty sand, SM; Atterberg limits below A line or IP less than 4 Clayey sand, SC; Atterberg limits above A line with IP greater than 7

Assignment 1 1 Describe briefly the origin of soils and bring out the 1. factors which control their formation 2. Define i) porosity ii) degree of saturation iii) air content iv) submerged density 3. Derive from fundamentals an expression for saturated unit weight in terms of void ratio and specific gravity 4. Explain in detail all the three consistency limits & indices. indices 5. Explain Indian Standard classification of soils 6. Explain about hydrometer method and grain size distribution curve. Dr. PVSN Pavan Kumar

Geotechnical Engineering (A50120) Geotechnical Engineering (A50120) Unit 2 Dr. PVSN Pavan Dr. PVSN Pavan Kumar Associate Professor


Syllabus • PERMEABILITY: Soil water – Capillary rise – Flow of water through soils – Darcy Darcy’ss law – Permeability – Factors affecting permeability – Laboratory determination of coefficient of permeability – Permeability of layered soils – In‐situ permeability tests (pumping in & pumping out test) • EFFECTIVE STRESS & SEEPAGE THROUGH SOILS: T t l neutral Total, t l and d effective ff ti stress t – Principle P i i l off effective stress – Quick sand condition – Seepage through soils – Flow nets: characteristics and uses. Dr. PVSN Pavan Kumar

Soil water

Free water

Held Water

Structural water

Adsorbed water

Capillary water

• Free water moves in pores of soil under gravity. Water flows due to difference of total head. • Held water is of three types and retained in pores of soil. It can not move due to gravitational force. • Structural water forms an integral part of soil solid can not be removed from crystal structure of mineral. • Adsorbed Ad b d water is i the h water held h ld due d to electrochemical l h i l forces f existing i i on the soil surface depending on clay fraction of soil, chemical composition and environment surroundingg the p particle. It exists in clayy soil and is negligible g g for coarse grained soil. Dr. PVSN Pavan Kumar • Capillary water – Water held in the voids of soil due capillary force.

Surface Tension Fu At Surface

Below Surface Fd

Water molecule is in equilibrium as molecular attraction tt ti actt all around

• At free surface pull from air above is smaller than pull from water molecules below. Equilibrium quilibrium is disturbed. • Pull from air, Fu is smaller than pull due to water , Fd below. • Net force decreases the air liquid interface to a minimum. • Surface S f assumes a curved d shape h t maintain to i t i equillibrium. illib i • Surface behaves like a stretched membrane and surface tension exists at the interface.

• Surface tension is the force acting per unit l th off a line length li drawn d on the th surface. f • Surface Tension of water at normal temperature p of 20°C,, TS = 0.073 N/m. / • Surface tension decreases with increase of Dr. PVSN Pavan Kumar temperature

Capillary rise in small diameter tubes • Water rises in small diameter tubes as the effect of adhesion between glass and water is more significant than cohesion between water molecules. • In mercury, cohesion is significant than adhesion. Liquid level is d depressed. d Fu = Upward pull due to surface tension = Ts cosθ πd Fd = Downward force due to mass of water = γ w F For equilibrium ilib i Fu = Fd ⇒ Ts cosθ π d = γ w Capillary rise = hc =

π

d 2 hc

4 4TS cosθ

gρ wd

π

d 2 hc

4 For a clean glass tube and pure water, meniscus is hemispherical i.e. θ = 0

4TS ⇒Dr. PVSN Pavan Kumar gρ wd

Capillary rise in tubes of non uniform diameter • A tube with large bulb of diameter d2 is dipped in water. • Capillary rise is limited to hc1 and it may not rise past the bulb of diameter, d2 when flow is upward. • If same tube is filled by pouring water from above and equilibrium g , hc2 can be maintained at a height, . Upward flow

Capillary rise do not depend on inclination of tube

T1 and T2 are upward and downward surface tension forces Dr. PVSN Pavan Kumar

Downward flow

Capillary tension • Water in a capillary tube is under negative pressure or under tension. • Pressure at point, point C and D is zero. zero • Pressure at point B, which is at a height hc above free surface, height, surface pB = ‐ γw hc =

• Pressure P at point i B is i negative i or is i under d tension. i • At point E, pressure pE = ‐ γw h • Capillary C ill tension i increases i li linearly l with i h height h i h above b water surface f and d is i positive below. 4T • Difference in pressure on either side of meniscus γw hc = s d Capillary water is like hanging a weight inside the walls of chimney = weight of suspended column of water, F =

π

d 2 hcγ w

4Dr. PVSN Pavan Kumar

Capillary rise in soil • Water which falls on the ground as rain flows under gravity, passes through the soil and reaches a surface known as ground water table or phreatic surface. • Ground water is a moving stream flowing under gravitational force. force • Water drawn above the water table is due to capillary action. • Soil mass consist of interconnected voids forming capillary tubes of circular cross section of varying i di diameter and d inclination.

Zone of aeration

Zone of capillary saturation


Capillary rise in soil • Capillary rise in soil depends on size and grading of soil. Capillary rise in soil depends on size and grading of soil • Diameter of channel, d = 0.2 D10 D10 = Effective diameter = Effective diameter • Capillary rise is large (may be 30m) in fine grained soil and small in coarse grained soil. coarse grained soil. C Maximum height of capillary fringe, hc max = eD10

C C C = Constant considering shape of grain and impurities, e = Void ratio id i h f i di ii V id i • Capillary water between solid particles l is under d tension and d solid ld particles are under compression. • Negative pressure of capillary water is known as soil suction. Soil suction is given by PF . PF = logg10 hc • Tenacity with which the soil holds the capillary water is capillary Dr. PVSN Pavan Kumar potential.

Frost heave • Water migrating upward from water table to the capillary fringe may freeze if atmospheric temperature falls to the freezing point and ice is formed. • This increases the volume of soil of soil when h water t is i converted t d to t ice. i • Due to frost heave soil at ground surface is lifted and this may lift light structures on the ground. • Water film from adjacent soil particles is removed, this disturbs soil suction equilibrium, more water is drawn. • Soils prone to frost action are silts and fine sand as they have large capillary rise due to relatively fine particles and good permeability. permeability • In coarse grained soils frost heave is limited due to low capillary rise. • In clay soil capillary rise is large but their permeability is low and frost heave is limited. If clay has cracks and fissures water moves easily and frost Dr. PVSN Pavan Kumar heave is large.

Frost heave • Foundation ou dat o o of st structures uctu es sshould ou d be ca carried ed be below o frost ost dept depth to avoid frost heave after completion of structure • Frost line: If the temperature persists below freezing point for a long period, frost penetrates deeper and the depth of affected soil increases. Depth p to which water mayy freeze is called frost line.

Frost boil • After frost heave if temperature increases frozen soil thaws, free water is liberated. • Liberated water is trapped in upper layer while lower layers are still frozen. Soften upper pp layer y has a reduced strength. g • Frost boil is important for highway pavements as a hole is formed due to extrusion of soft soil and water under the action of wheel loads. • Soils most prone to boiling Dr. PVSN Pavan Kumar are silt.

Prevention of frost action • Frost heave and boil cause great difficulties in maintenance of hi h highways and d runways. Following ll i are measures to mitigate ii – Replace frost susceptible soil by coarse grained soil such as gravel or coarse sands. sands Due D e to large quantities q antities of soil required, req ired method is not economically feasible. – Provision of insulating blanket between water table and ground surface. A blanket of 15 to 30 cm thickness decreases capillarity. – Good drainage system – Addition of dispersion agent such as sodium polyphospahte permeabilityy of soil. decreases the p


Capillary siphoning • Loss of water from storage reservoir to the downstream of dam due to capillary action

Bulking of sand • A damp sand deposited loosely has more volume than same sand

deposited in dry state. • Phenomena of increase in volume is known as bulking of sand. • In damped state cohesion develops between particles due to capillary water t and d increases i th volume. the l • Bulking is predominant at a water content of 4 to 5% and volume increases by 20 to 30%. • If damp sand is saturated by adding more water, effect of capillary Dr. PVSN Pavan Kumar action is eliminated and volume of sand decreases.

Slaking of clay

• Disintegration of clay suddenly immersed in water is knows as slaking of clay. g limit have some voids filled with • Clayy dried below shrinkage water. water • Explosion of air filled voids due to immersion in water.


Permeability of soil • Any material having voids is called porous, if pores are interconnected and allow water/liquid to flow through voids are called permeable material. material • Size, cross section and orientation of inter connected voids differ. differ • Property of soil to permit flow of water is called permeability. permeability • Water flows easily through soil – Highly p pervious • Water can not easily flow ‐ impervious • Permeability of soil is important for determining settlement of buildings, yield of wells, seepage through and below earth structures and design of filters. Dr. PVSN Pavan Kumar

Permeability of soil

Total head

V2 + +Z • Total Total head at any point in a flowing fluid = head at any point in a flowing fluid = γ 2g p V2 ‐ Pressure head ‐ Velocity head Z ‐ Datum head γ 2g p • Pressure head, at a point is the head indicated γ p

by a piezometer with its tip at that point • Velocity V l it off flow, fl v through th h soilil is i smallll and d velocity l it V2 head, can be neglected. 2g

• Elevation head, Z is the vertical distance between point and datum. Downstream water is generally taken as datum p

Piezometric head = Pressure head, + Datum head, Z γ Dr. PVSN Pavan Kumar

Various heads and Hydraulic gradient

L

• Consider two vessels A and B filled with water at different levels connected by a small tube filled with soil sample of length, L. • Flows takes from vessel A to B. Datum at water surface in vessel B. Point

Elevation Pressure head head

Total head

1

‐z1

h1

h

2

‐z2

h2

h'

3

‐z3

h3

0

h = hydraulic head = Difference between water t level l l in i vessell A and dB h l Line joining hydraulic head at different points along the sample is hydraulic gradient line or piezometric Dr. PVSN Pavan Kumar surface

Hydraulic gradient, i =

Darcy’s law • In 1856, Darcy demonstrated experimentally that for a laminar flow in a homogenous soil, velocity of flow is proportional to hydraulic gradient v = k i k =coefficient of permeability, permeability i = hydraulic gradient • Discharge, q = v A = k i A A = includes both solids and void • Coefficient of p permeabilityy ((L T‐1 ) is equal q to velocityy of flow if hydraulic y gradient is unity.

•

S No

Soil type

Coefficient of permeability, mm/sec

Drainage Properties

1

Clean Gravel

10 to 100

Very good

2

Coarse and medium sand Coarse and medium sand

0 01 to 10 0.01 to 10

Good

3

Fine sand, loose silt

10‐4 to 10‐2

Fair

4

Dense silt, Clayey silt

10‐5 to 10‐4

Poor

5

Silty clay, clay

10‐8 to 10‐5

Very poor

ρ vD Darcy law is valid if Reynolds number number, η < 1 . It is valid for flow in clay, clay silt

and fine sand as flow is laminar. In coarse sand, gravel and boulders flow is Dr. PVSN Pavan Kumar turbulent.

Determination of Coefficient of permeability • Laboratory Laboratory method method – Constant head permeability test : suitable for pervious soil – Variable head permeability test : For less pervious soil Variable head permeability test For less pervious soil • Field method – Pumping out test P i tt t – Pumping in test • Indirect method I di h d – Computation from particle size – Computation from consolidation test f ld • Capillarity – permeability test


Constant head permeability test, IS 2720 Part 17 • Coefficient of permeability of permeable soil is determined from constant head method. • Soil sample is compacted to the required d i in density i the h mould. ld • Soil sample in mould should be saturated by pouring soil in mould filled with water or attaching constant head reservoir to the drainage base and allowing upward flow of water or by applying vacuum to drainage cap.

Inlet Air release valve

Drainage Cap

metallic mould 100 mm diameter, 127.3 mm height, 1000 ml capacity Outlet Dr. PVSN Pavan Kumar

Drainage Base

Porous Stone

Constant head permeability test, IS 2720 Part 17 • Constant head reservoir is filled with water and connected to drainage cap. cap • Water is allowed to flow out from drainage base for some time till steady state is established. t bli h d • chamber is filled with water at the beginning of test. Water flowing through sample falls in chamber and is collected in a graduated jar for a convenient period. h Discharge through soil, q = k i A = k A l

⇒k=

ql Ah

q = Discharge =

Drinage Cap

V l Volume off water t collected ll t d in i jar j ,v time, t

l = length of mould A = Area of mould h = Constant head = Distance between water level in reservoir and chamber reservoir and chamber Dr. PVSN Pavan Kumar

Drinage base

Variable head permeability test • Quantity of water collected ll t d in i the th jar j is very small an cannot be measured accurately. For such soils variable head permeability test is conducted. • Mould to conduct the test test, preparation and saturation of sample are same as constant head test. • Vertical graduated stand pipe of known diameter is fitted to drainage g cap. p

Porous Stone

St d i Stand pipe

Drainage Cap


Drainage Base

Mould

Variable head permeability test • Stand pipe if filled with water. • Test started by allowing water in stand pipe to flow through sample. • Time, t required for water level to fall from initial head, h1 to final head, h2 is d t determined. i d Head H d is i measured d from f water level in chamber. • Consider the instance when water level in stand pipe is at a head h above water level in chamber. • Assume that in a small time ‘dt’ head falls by ‘dh’. • From continuity of flow a = area of stand p pipe p a dh = ‐ q dt h a dh = ‐ (k i A) dt ⇒ a dh = ‐ (k A) dt ⇒ L Dr. PVSN Pavan Kumar

A K dt − dh = aL h

AK aL

t2

∫ t1

Variable head permeability test h 1 d =− dt h

2

∫

⎛ h1 ⎞ AK (t2 − t1 ) = loge ⎜⎜ ⎟⎟ aL ⎝ h2 ⎠

dh

h1

⎛ h1 ⎞ aL K= loge ⎜⎜ ⎟⎟ At ⎝ h2 ⎠

⎛ h1 ⎞ 2.3aL log10 ⎜⎜ ⎟⎟ K= At ⎝ h2 ⎠

Discharge and seepage velocity Discharge Velocity, V =

q A

q = Discharge; A = Area of sample q = Discharge; A = Area of sample • Water flows only through voids and velocity of water through voids is called seepage velocity q vA ⇒ v = q = vA = v s Av Seepage Velocity, Vs = s Av Av

vA × l v k i ⇒ vs = = = = k p Dr. PVSN Pavan Kumar × i kp = coefficient of percolation n Av × l n

Factors affecting permeability of soil FFollowing ll i equation i is i obtained b i d for f coefficient ffi i off permeability bili off soilil considering laminar flow through soil ⎛γ k = C ⎜⎜ w ⎝ μ

⎞ ⎛⎜ e 3 ⎞⎟ 2 ⎟⎟ D ⎜ ⎟ ⎠⎝ 1 + e ⎠

C is i a constant t t which hi h considers id shape h off the th channel h l through th h voids id γw unit weight of water; μ coefficient of viscosity of water; e = void ratio D = Diameter Di t off particle. ti l • Particle size: k ∝ D2, k of coarse grained soil is one million times more than fine grained soil. soil • Permeability is more for • Structure of soil mass: flocculated structure than dispersed

Flocculated Structure

structure. • Permeability of stratified soil deposits is more parallel to plane of stratification than in perpendicular Dr. PVSN Pavan Kumar Dispersed Structure direction

Factors affecting permeability of soil Structure of soil mass: • Permeability of soil should be determined on a undisturbed sample Disturbance destroys the original structure and affects the sample. permeability. Effect of disturbance is more in case of fine grained soil than in coarse grained soil. Shape of particles: • Angular particles have more specific surface (i.e. (i e surface/unit mass) than rounded particles. Soil with angular particles have less permeabilityy than rounded p p particles as p permeabilityy is inverselyy proportional to the specific surface. • Void ratio of soil with angular g particles is ggreater than that for p rounded particles, soil with angular particles is more permeable. Void ratio: For a given soil k increases with void ratio; Clay has more void ratio but voids are not connected and permeability is less. Dr. PVSN Pavan Kumar

Factors affecting permeability of soil

Variation of K with void ratio Dr. PVSN Pavan Kumar

Factors affecting permeability of soil Effect of water k ∝ γw ; k ∝

1

μ

• As temperature increases viscosity decreases and permeability increases. μt • Coefficient of permeability is reported at 27°c k27 = kt μ27 Degree of saturation • Permeability of a partially saturated soil is considerably smaller than permeability of fully saturated soil. • Presence of air blocks the passage. Adsorbed water • Presence of adsorbed water in fine grained soil obstructs flow of water and reduces permeability of soil. Void ratio occupied by adsorbed d b d water t is i about b t 0.1. 01 Dr. PVSN Pavan Kumar

Factors affecting permeability of soil IImpurities in water ii i • Any foreign matter in water has a tendency to plug the flow passage and reduce effective voids and permeability. permeability

Coefficient of absolute permeability • Coefficient of permeability, k depends on properties of soil and water. • Coefficient of absolute permeability, K considers properties of soil only, dimensions are L2 , units are darcy, 1 darcy dimensions are L units are darcy 1 darcy = 0.987 x 10 0 987 x 10‐8 cm2. ⎛ γ w ⎞ ⎛⎜ e 3 ⎞⎟ 2 ⎟⎟ k = C ⎜⎜ D ⎟ ⎜ ⎝ μ ⎠⎝ 1 + e ⎠

⎛ e3 ⎞ 2 ⎟D K = C⎜ ⎜1+ e ⎟ ⎝ ⎠ Dr. PVSN Pavan Kumar

Pumping out test • • • • • • • • •

Laboratory methods may not give reliable results. results Samples are disturbed and may not represent insitu structure. Field tests may be in the form of pumping out test or pumping in test. test Pumping out test is useful for a homogenous, coarse grained deposits in which it is difficult to obtain undisturbed sample. sample Pumping out test is influenced by large area, costly, justified for large projects. Aquifers: Reservoirs of ground water that can be easily drained or pumped p p out. Aquiclude: Soil deposits that are impervious Unconfined aquifer: An aquifer do not have an aquiclude at top and water table is in the aquifer itself. Confined aquifer: q Aquifer q confined between two aquicludes, q one at top and other at bottom Dr. PVSN Pavan Kumar

Aquiclude


Pumping out test – Unconfined aquifer • Tube well penetrates unconfined aquifer and reaches impervious strata. • Tube is surrounded by a screen called strainer to check the flo flow of soil particles. Impervious strata • Water is pumped out of tube well till steady state is reached.

d, Drawdown

D, Thickness of aquifer D Thickness of aquifer rw , radius of well

• Water level which is horizontal is depressed near the well. well Water table near the well forms an inverted cone known as cone of depression. • Maximum depression of water table is known as drawdown. • Expression for coefficient of permeability is derived from Dupuit’s Dr. PVSN Pavan Kumar assumption

Dupuit’s assumption 1. Flow is laminar and Darcy’s law is valid. 2. Soil is homogenous and isotropic 3. Well penetrates the entire thickness of aquifer. aquifer 4. Flow is steady. 5 Coefficient 5. of permeability remains constant. 6. Flow towards well is radial and horizontal. 6 horizontal 7. Natural ground water regime remains constant. p of hydraulic y ggradient line is small. It is taken as tangent g of angle g 8. Slope in place of sine of angle. i = dz dr Dr. PVSN Pavan Kumar

Pumping Out test – Unconfined aquifer • Flow through g a cylindrical surface of height z at a radial distance r from center. • From F D Darcy’s ’ law l q=kiA d dz =k (2πrz) dr dr 2 π k z dz = r q r2

∫ r1

dr d = r

z2

∫

z1

⎛ r2 2 π k z dz d ⇒ log e ⎜⎜ q r2 ⎝ r1 k=

2.3q log g10

(

(

⎞ 2 π k z 2 2 − z12 ⎟= ⇒k= ⎟ 2 2 q 2 π z − z ⎠ 2 1

r1

π z2 2 − z12

)

)

r2 q log e r1


(

)

Pumping out test – Unconfined acquifer • Near ea test well e tthere e e iss a rapid ap d d drop op o of head ead aand d assu assumption pt o 8 iss not valid. • Two observation wells are drilled in mutually perpendicular direction. • Circle of influence over which the effect of pumping is observed is called radius of influence, R varies between 150 to 300 m. R = 3000 d √k R 3000 d √k R = radius of influence (m); d = drawdown (m); k = coefficient of permeability (m/sec) q R k= loge 2 2 rw π D −h

(

)

q = Discharge; D = depth of aquifer below water table; h = depth of water t in i well; ll R = radius di off influence; i fl rw = radius di off wellll Dr. PVSN Pavan Kumar

Pumping out test – Confined aquifer • Confined aquifer of thickness ‘b’ is lying between two aq icl des aquicludes. • Peizometric surface is above the top of aquifer. • Water pressure is indicated by peizometric surface. • Peizometric surface is water table equivalent in confined aquifer. • Initially peizometric surface is horizontal. When pumping is started from well cone of depression forms. • K is derived based on same assumptions as unconfined aquifer. • Consider discharge through a cylindrical surface at a radial distance r C id di h th h li d i l f t di l di t from center and of height z Dr. PVSN Pavan Kumar

Pumping out test – Confined aquifer From Darcy From Darcy’ss law law q = k i A ⎛ dz ⎞ q = k ⎜ ⎟ 2 π r b ⎝ dr ⎠ dr 2 π k b dz = r q r2

∫ r1

dr = r

z2

∫

z1

2 π k b dz q

⎛ r2 ⎞ 2 π kb (z2 − z1 ) loge ⎜⎜ ⎟⎟ = q k= ⎝ r1 ⎠

r2 q log l e r1 2πb (z 2 − z1 )

q R k= logge 2πb(D − h ) rw Dr. PVSN Pavan Kumar

r2 2.3q log10 r1 k= 2πb (z 2 − z1 )

Pumping‐In test • Pumping in test is conducted to determine the permeability of an individual stratum through which hole is drilled. • Economical than pumping out test; less reliable. • Pumping in test give permeability close to the hole whereas pumping out test over a large area around the hole. p g in test is of two types yp 1)) open p end test 2)) Packer • Pumping test • In open end test water flows out of hole through its end whereas in packer test water flows through the sides of section enclosed between packers. packers • K is determined from quantity of water accepted by hole, water should be clean, clean free of impurities; slightly higher temperature than ground water is used. Dr. PVSN Pavan Kumar

Open end test ‐ Pumping‐In test • A cas casingg iss inserted se ted into to tthee bo boree hole o e to tthee desired depth. Hole is cleaned out. • Hole should be filled with water below WT to avoid squeezing of soil . • Water is passed into the hole and constant rate of flow, q for steady state conditions are determined. q k= 5.5 r H r = inside radius of casing; H = difference of levels between inlet of casing and water table. • Bottom of hole should be at a distance more than 10r from top and bottom of layer. Dr. PVSN Pavan Kumar

Casing

Bore hole

test section

Open end test ‐ Pumping‐In test

• Water can be pumped with pressure, p and head causing flow is H+

P

γw • Test T t can be b conducted d t d above b water t table, t bl H = Difference Diff b t between inlet of casing to bottom. Dr. PVSN Pavan Kumar

Pumping in test – Single Packer test • W Water is i pumped d into i the h soilil and comes out of sides of hole below packer. packer • Pumping continued till steady state conditions are attained. Discharge of flow = q q ⎛L⎞ k= loge ⎜ ⎟ L > 10r 2πLH ⎝r⎠ q −1⎛ L ⎞ k= sinh ⎜ ⎟ 10r > L ≥ r 2πLH ⎝ 2r ⎠

Test below Test below water table

Test above water table

r = Inside radius of hole; L = Length of hole tested; ; g ; H = Distance between water level at entry to water table (below wt) For holes tested above water table, H = distance between water level at entry and moddle of test section. Dr. PVSN Pavan Kumar

Pumping in test – Double packer test

Indirect methods to determine k 1. Allen Hazen formula k (cm/s) = C D10 2 k (cm/s) = C D C = Constant 100 to 150 D10 = Effective size in cm 2. Kozeny – Carman equation gρ w e3 × k (cm / s ) = 2 2 1+ e Cs μ S T

Test below water table Test below water table

Test above water table

g = 981 cm/sec2; ρw = Density of water (g/cc) Cs = Shape factor, = 2.5 for granular soil μ= Coefficient of Viscocity (Poise) S = Specific surface S Specific surface = Surface area/volume Surface area/volume (cm2/cm3) Dr. PVSN Pavan Kumar T = Tortuosity e = Void ratio

Permeability of stratified soil deposits • Stratified soil deposits have layers with different permeabilities. p y p • Average permeability parallel and perpendicular to plane of stratification is to be determined. Flow parallel to plane of stratification • Consider two layers of thickness H y 1 and H2 . Total thickness is H • Loss of head, h is same for two layers over a length of sample, L h • Hydraulic gradient for layer 1 and 2, i = i1 = i2 = L Total discharge, q = q1 + q2

khi (H ×1) = kh1i1 (H1 ×1) + kh2i2 (H2 ×1) kh1H1 + kh 2 H 2 kh = H


Permeability of stratified soil deposits Flow perpendicular to plane of stratification p p p • Consider two layers of thickness H1 and H2. • Coefficient of permeability of layer 1 and 2 is kv1 and kv2 respectively. kv is average coefficient of permeability of soil deposits g p y p Discharge per unit width is same through each deposit q = q1 = q2 kv iv (L x 1) = kv1 iv1 (L x 1) = kv2 iv2 (L x 1)

kviv ⇒ iv1 = kv1

iv 2

kviv = kv 2


Permeability of stratified soil deposits Flow perpendicular to plane of stratification p p p • Total loss of head, h = sum of loss of heads of individual layers i.e. h = h1 + h h = h + h2 iv H = iv1 H1 + iv2 H2

kviv kviv H1 + iv H = H2 kv1 kv 2 kv =

H ⎛ H1 ⎞ ⎛ H 2 ⎞ ⎟ ⎜ ⎜ ⎟ ⎜k ⎟+⎜k ⎟ ⎝ v1 ⎠ ⎝ v 2 ⎠

kv kv H1 + H= H2 kv1 kv 2

Kh > Kv i.e coeffient of permeability parallel to plane of stratification is close to most permeable layer and coeffient of permeability normal to plane of stratification is close to most impermeable layer Dr. PVSN Pavan Kumar

Effective stress • Effective stress principle is enunciated by K.Terazaghi in 1936 • Consider a saturated soil mass, a prism of soil of cross sectional area, A W i ht off soilil in Weight i prism, i P = γsat h A P Total stress, σ = = γ sat h A

Pore water pressure is the pressure due to pore water filling the voids, u = γw h Effective stress,σ = Total stress,σ − Pore water pressure, u

σ = γ sat h − γ wh = γ sub h

Effective stress is the normal stress transmitted through solid particles Effective stress is the normal stress transmitted through solid particles Compression and shear strength of depend on effective stress, Dr. PVSN Pavan Kumar σ

Effect of fluctuation of water table on effective stress • Water table is at a depth p H1 below surface • Soil above WT is wet and bulk unit H weight is considered. • Soil below WT is saturated and saturated unit weight is considered. • Consider X1 X1 X1 Total stress, σ = γbulk H1 + γsat H2

GS H1

γbulk

WT

γsat H2

X2

Pore water pressure, u = γ p , γw H2 Effective stress, = γ σ bulk H1 + γsat H2 ‐ γw H2 = γbulk H1 + γsub H2 If water table rises to ground surface If water table rises to ground surface Effective stress, = γ σ sat H ‐ γw H = γsub H If water table below section X1X2 Effective stress, = γ σ bulk H


Change off location Ch l i off water table changes pore water pressure,, as WT moves down p effective stress increases.

Effective stress under hydrostatic condition Water level remains constant

• Water level in piezometer rises to same elevation when connected to different soils.

Water table is above ground surface Section AA

σ = γ wH − γ wH = 0

S ti BB Section BB

σ = γ w H + γ sat1H1 − γ w H − γ w H1 = γ sub1H1

Section CC

σ = γ w H + γ sat1H1 + γ sat 2 H 2 − γ w H − γ w H1 − γ w H 2 = γ sub1H1 + γ sub 2 H 2 Dr. PVSN Pavan Kumar


• Water level above AA, H is reduced to zero.

Water table at soil surface AA Section AA Section BB Section CC

σ =0

σ = γ sat1H1 − γ w H1 = γ sub1H1 σ = γ sat1H1 + γ sat 2 H 2 − γ w H1 − γ w H 2 = γ sub1H1 + γ sub 2 H 2

A water column of H above level AA does not change the effective stress, A water column of H above level AA does not change the effective stress, marine soil deposits which are at a large depth have a low effective stress and Dr. PVSN Pavan Kumar correspondingly low shear strength


Water table at level DD at a depth H1 ‘

σ =0 Section DD σ = γ1H1' Section AA

Section BB Section CC

σ = γ1 H1'+γ sat1 H1''−γ w H1'' =γ1 H1'+γ sub1 H1'' σ = γ1 H1'+γ sat1 H1''+γ sat2 H2 −γ w H1''−γ w H2 =γ1 H1'+γ sub1 H1''+γ sub2 H2



Water table at level EE at a depth H2 ‘ Section AA

σ =0

Section BB

σ = γ1 H1

Section EE

σ = γ1 H1 + γ2 H2'

Section CC Section CC

σ = γ1 H1 + γ 2 H2'+γ sat2H2''−γ wH2''= γ1 H1 + γ 2 H2'+γ sub2H2''



Section AA

σ =0

Section BB

σ = γ1 H1

Section CC

σ = γ1 H1 + γ 2 H2

Water table is below CC Water table is below CC

• Effective Eff i stress at any section i increases i as water table bl goes down. d • Effective stress is calculated using γbulk above water table and γsat below water table Dr. PVSN Pavan Kumar

Increase of effective stress due to surcharge • Soil surface is subjected to a surcharge, q • Water table is at level BB at a depth p H1 below surface. Section AA Section AA

σ =q

Section BB

σ = q + γ1H1

Section CC

σ = q + γ1H1 + γ sat1H2 −γ wH2 = q + γ1H1 + γ sub1H2

Effective stress throughout the depth increases by surcharge, q


Effective stress in soils saturated by capillary action • Pore water pressure is negative above water table. Soil saturated above upto Soil saturated above upto surface level surface level AA Section AA

σ = 0−(−γ wH1) = γ wH1

S ti DD Section DD

σ = γ sat1H1'−(−γ wH1'') = (γ subb1H1'+γ wH1') + γ wH1''= γ subb1H1'+γ wH1

Section BB

σ = γ sat1H1 −(0) = γ sub1H1 + γ wH1

Section CC

σ = γ sat1H1 + γ sat2H2 −(γ wH2) = γ sub1H1 + γ wH1 + γ sub2H2 + γ wH2 −(γ wH2) = γ sub1H1 + γ sub2H2 + γ wH1

Effective stress at all levels below plane of saturation, AA increases by γ Effective stress at all levels below plane of saturation AA increases by γw H1. γw H1 acts as a surcharge at all levels Dr. PVSN Pavan Kumar

Effective stress in soils saturated by capillary action Soil saturated upto level DD or capillary rise is H1’’ Section AA Section DD

σ =0 σ = γbulk1 H1'−(−γ wH1'') σ = γbulk1 H1'+γ wH1''

Section BB

σ = γbulk1 H1'+γ sat1H1''−(0) =γbulk1 H1'+γ sub1H1''+γ wH1'' Section CC

σ = γbulk1 H1'+γ sat1H1''+γ sat2H2 −γ wH2 =γbulk1 H1'+γ sub1H1''+ + γ sub2H2 + γ wH1''

Effective stress increases by γw H1’’ below level of saturation, DD.


Seepage Pressure

• Force exerted by a flowing water t on soilil is i known k as seepage force and the pressure in known as seepage p g p pressure. • Consider a upward flow of water through a soil sample of area A, area, A length, length L under a hydraulic head, h. • Boundary pressures u1 and u2 at top and bottom of soil sample are shown below. They can be divided into hydrostatic and hydrodynamic pressures • Hydrodynamic y y pressure is due to head, h • Seepage S f force, J = γw h A

Dr. PVSN Pavan Kumar Hydro Static Pressure Boundary pressure

Hydro dynamic Pressure

Seepage pressure • Seepage Pressure =

J γ whA = = γ wh = γ wiL A A

• Seepage force/Unit Volume, j =


γ whA AL

= iγ w

Effective stress under steady seepage condition Flow is downwards • Head causing the flow is h • Pore water pressures at section p AA and Section BB are indicated by piezometer. Section AA Section BB

σ = γ wHw −γ wHw = 0

σ = γ wHw + γ sat1H1 −γ wHw1 = γ wHw + γ sub1H1 + γ wH1 −γ wHw1 = γ subb1H1 + γ w(Hw + H1 − Hw1)

Hw + H1 > Hw1 thus effective stress increases with depth. Section CC Section CC

σ = γ wHw + γ sat1H1 + γ sat2H2 −0 = γ wHw + γ sub1H1 + γ wH1 + γ sub2H2 + γ wH2 = γ sub1H1 + + γ sub2H2 + γ w(Hw + H1 + H2) = γ sub1H1 + γ sub2H2 + γ wh

At section CC effective stress increased by γ h compared with hydrostatic Dr. PVSN Pavan Kumar w condition. Effective stress increases due to downward flow.

Effective stress under steady seepage condition Flow is upwards p • Piezometer at various elevations indicate pore water. Section AA

σ = γ wHw −γ wHw = 0

Section BB σ Section BB

= γ wHw + γ sat1H1 −γ wHw1 = γ wHw + γ sub1H1 + γ wH1 −γ wHw1 = γ sub1H1 + γ w(Hw + H1 − Hw1)

Hw + H1 < Hw1 Section CC

σ = γ wHw + γ sat1H1 + γ sat2H2 −γ wHw2 = γ wHw + γ sub1H1 + γ sub2H2 + γ wH1 + γ wH2 −γ wHw2 = γ sub1H1 + γ sub2H2 + γ w(Hw + H1 + H2 − Hw2) = γ sub1H1 + γ sub2H2 −γ wh


Quick Sand Condition σ = γ sat L = (γ w + γ sub )L u = γ w H w1 = γ w ( L + h) σ = σ − u = γ sub L − γ wh = γ sub L − γ wiL • Effective stress is reduced due to upward d flow fl off water. • A stage is reached when effective stress becomes zero. ero This condition is called quick sand condition. • A specimen of length, L is subjected j to an upward p pressure. p Stress developed at section C – C

γ subb σ = 0 ⇒ γ sub L − γ wiL = 0 ⇒ i = γw Dr. PVSN Pavan Kumar

Quick Sand Condition

• Hydraulic gradient at which effective stress, σ becomes zero is called critical i i l hydraulic h d li gradient, di i c. i=

γ sub G −1 = γw 1+ e

• Shear strength of cohesion less soil is given by, S = σ tan φ • When effective stress, σ is zero; shear strength of soil becomes zero. Soil is said to be quick or alive and this is a hydraulic condition. • If critical i i l gradient di i exceeded is d d soilil particles i l move upward d and d soilil appears boiling. Discharge suddenly increases due to increase of coefficient of permeability. permeability • Weight placed on the soil surface such as humans or animals sink down due to zero shear strength. • Quick sand will not suck the victims below the surface. • Quick sand behaves like a highly viscous liquid of unit weight about twice of water. Person may die due to drowning if he is tired and falls Dr. PVSN Pavan Kumar his head due to panic.

Quick Sand Condition • Person caught in quick sand condition should keep his head above soil surface and slowlyy move towards the bank. Tryy to catch a tree and pull him out of quick sand. • Q Quick sand condition mayy develop p in ggravel if hydraulic y ggradient,, i exceeds critical gradient, ic . Discharge required to maintain quick sand condition is very large and it may not be available. • Shear strength of cohesive soil, τ = c + σ tan ϕ will have shear strength due to cohesion intercept, c though effective stress is zero. • Quick sand condition is most likelyy to occur in silt and fine sand.


Seepage pressure approach for Quick sand condition At Section XX At Section XX σ= γsat L u = γw (h+L)

σ = γ sat L − γ wh − γ w L = γ sub L − γ wh x

x

Effective force =

γ sub AL − γ w Ah

For quick sand, effective stress/force =0 Downward Force, Wsub = Upward Force, J γsub A L = γw h A

h γ sub =i= L γw Dr. PVSN Pavan Kumar

Effect of submergence and surcharge on quick sand condition

A At section CC i CC

σ = γ w H w + q + γ sat L, u = γ w H w1 = γ w (h + H w + L) γ sub L + q σ = q + γ sub L − γ wh = 0 h = γw Compared with the case with no surcharge, head required for quick Dr. PVSN Pavan Kumar sand condition increases by q/γw

Failure of hydraulic structures by piping • Hydraulic structures such as weirs Hydraulic structures such as weirs and dams built on pervious foundations fail by formation of a pipe shaped channel. • Failure occurs if water flowing through foundation has hydraulic through foundation has hydraulic gradient, i > ic . (Exit gradient > critical gradient) Piping failure below earth dam • Two types of failures occur •Backward erosion piping failure • Heave piping failure Downstream Upstream • Backward erosion failure – Soil particles are removed at exit points i l d i i due to flowing water. • Scour hole forms at exit, extends , to upstream and approaches u/s Percolation below weir Dr. PVSN Pavan Kumar reservoir. Downstream

Failure of hydraulic structures by piping

Seepage through earth dam

• Backward erosion piping may also occur due to flow of water through the earth dam. Phreatic line cuts the downstream face of dam and seepage pressure is high. • Soil S il at downstream d toe off earth h dam d erodes. d • Backward erosion piping may occur along weak bedding plane in foundation of earth dam or alongg the p periphery p y of a conduit in earth dam Dr. PVSN Pavan Kumar

Failure of hydraulic structures by piping • Heave piping H i i

Sheet pile walls

• Rise Ri or heave h off a large l mass off soilil due d to t seepage pressure. • When seepage force due to upward flow of water at any level is greater than submerged weight of soil above the level, level the entire mass of soil in that zone heaves up and is blown out by percolating water. Dr. PVSN Pavan Kumar

• Heave piping occurs over a distance of D/2, D is depth of sheet h pile l below b l g.s. • This occurs in the zone marked abcd. abcd • Heave piping occurs if Upward p seepage p g force is greater than submerged weight of soil Flow lines Impervious strata

Equipotential lines

D Upward seepage force, U = γw (0.35 h) × 1 2 Flow below sheet pile wall

Submerged weight of soil, W = γsub D x D × 1 2

W Dr. PVSN Pavan Kumar U

Factor of safety against piping failure, F =

Measures for prevention of piping failure Piping failure is catastrophe and following measures adopted to control. Piping failure is catastrophe and following measures adopted to control • Increase length of path of percolation – Increase base width, Increase base width – Provide vertical cutoff wall below hydraulic structure – Provide an impervious blanket Provide an impervious blanket

Cutoff trench Grouting

• Provide Provide an impervious core in the dam an impervious core in the dam • Drainage filter changes the direction of flow of water from downstream d ect o a d a o s to o t oug direction and allows to flow through filter. te • Provision of a loaded filter at the exit points. Dr. PVSN Pavan Kumar

Flow net

Flow lines or stream lines Phreatic line

H Equipotential lines

Discharge through the dam, q = k h

Nf Nd

k coefficient of permeability, h hydraulic head Nf Number of flow lines; Nf Number of potential drops; Characteristics of flow net • Every intersection between a flow line and an equipotential Every intersection between a flow line and an equipotential line should be at line should be at right angles. • Discharge, Δq between two adjacent flow lines is constant. Drop of head, Δh between two equipotential lines is constant. • Flow net consists of approximate square. Dr. PVSN Pavan Kumar

Seepage below sheet pile wall

Loss of head from one potential line to other is h/n Loss of head from one potential line to other is h/nd • Total head at any point, hp = h – n x (h/nd ) p p p p point P n is potential drops upto • Hydraulic gradient = Δh Δs


Assignment ‐ 2 1. What is Darcy’s law ? 1 2. What are the factors effecting permeability 3 A constant head 3. h d permeability bili test was run on a sand d sample 30cm in length and 20cm2 in area. When a loss of head was 60cm the quantity of water collected in 2 min was 250ml, determine coefficient of permeability of the soil. 4. What are the different types of soil waters and discuss them. 5. What is quick sand condition? 6. What is flow net. Explain its characteristics and uses. 7. Explain pumping in and pumping out tests for permeabilityy of soil ((5 Marks)) determination of p Dr. PVSN Pavan Kumar

Unit III Unit III STRESS DISTRIBUTION IN SOILS Dr. PVSN Pavan Kumar Dr. PVSN Pavan


Syllabus • B Boussinesq’s i ’ and d Westergaard’s W t d’ theories th i for f point i t load, uniformly loaded circular and rectangular areas, pressure bulb, variation of vertical stress point load alongg the vertical and horizontal under p plane, and Newmark’s influence chart for irregular areas. areas


Introduction • At At a depth z below ground surface a depth z below ground surface there will be stress due to weight of soil above and due to weight of g applied loads such as – – – –

Buildings Apartments Embankments R n a Runway

B ildi Building

z

• Stresses and displacements are needed for stability analysis of soil needed for stability analysis of soil mass, settlements of buildings and determination of earth pressure. • Stresses developed in soil depend on stress – strain characteristics of foundation soil. This is complex and depends on drainage conditions, water content, void ratio, rate of loading, load level and stress path. Dr. PVSN Pavan Kumar

Introduction

• Simplifying p y g assumptions p are made such as soil is homogenous, g , isotropic and stress – strain relation is linear. • Theory of elasticity is used to determine the stresses in soil. This requires modulus of elasticity, E and Poisson ratio, ν. •Triaxial compression test or unconfined fi d compression test are conducted on undisturbed samples to obtain stress vs strain response.

Deviator stress

strain

• Secant modulus or initial tangent modulus is determined. Secant modulus or initial tangent modulus is determined Dr. PVSN Pavan Kumar

Modulus of elasticity and Poisson ratio Type of Soil

Soft clay H d Cl Hard Clay Silty Sand Loose sand Dense sand Dense Gravel

Modulus of elasticity, E (MN/m2 ) 1.5 – 4.0 6 0 15 6.0 ‐ 6.0 – 25.0 10.0 – 25.0 40.0 – 80.0 100 – 200

Type of soil

Poisson ratio, ν ti Saturated clay 0.4 – 0.5 , Unsaturated clay 0.1 – 0.3 Poisson ratio = Laterall strain ε x Longitudinal strain, ε y Silt 0.3 – 0.35 Loose sand 0.3 – 0.5 Dr. PVSN Pavan Kumar Dense sand 0.2 – 0.3

Geostatic stress • When ground surface is horizontal properties of soil do not change along a horizontal plane, stresses due to self weight are called geostatic stresses. • Vertical V ti l stress, t σz actt perpendicular to horizontal plane and horizontal stress, stress σx and σy act perpendicular to vertical plane. plane • Stress due to self weight of soill is more than h that h due d to Dr. PVSN Pavan Kumar imposed load.

Vertical stress due to concentrated load • Boussinesq gave a theoretical solution for an elastic medium subjected to a concentrated load at surface. • Stress in a soil due to externallyy applied pp load is determined. Assumptions , •Soil is an elastic continuum, modulus of elasticity is constant • Soil is homogenous and isotropic • Soil mass is semi infinite • Soil is weightless Consider ground surface subjected to a point load, Q at point O. Boussinesq p , p q gave that vertical stress at point at a depth z

3Q cos 3 β σz = 2π R 2


GS

x

Vertical stress due to concentrated load β = Angle between line joining the origin, origin O and point, point P with vertical axis through O R = x2 + y2 + z 2 = r 2 + z 2

3Q cos 3 β σz = 2π R 2 3 z 3Q R 3Q ⎛ z 3 ⎞ ⎜⎜ 5 ⎟⎟ = = 2 2π R 2π ⎝ R ⎠

x GS

( )

⎛ ⎞ ⎛ ⎞ 5 3 z z 3Q ⎜ ⎟ 3Q ⎜ ⎟ = = 5 ⎟ 5 ⎟ 2 ⎜ ⎜ 2π ⎜ 2 2 π z 2 2 ⎟ ⎜ r2 + z2 2 ⎟ r + z ⎝ ⎠ ⎝ ⎠ 3Q 1 Q IB = Boussinesq = 3 = = IB 2 5 / 2 2 2 2πz ⎡ z influence factor 2π ⎛r⎞ ⎤ ⎢1 + ⎜ ⎟ ⎥ Dr. PVSN Pavan Kumar z ⎝ ⎠ ⎢⎣ ⎥⎦

(

)

(

)

1 ⎡ ⎛ r ⎞2 ⎤ ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ z ⎠ ⎥⎦

5/ 2

Observations on the above equation

• Vertical stress,, σz does not depend p on modulus of elasticity, y, E and Poison ratio, ν. Stress distribution is same for all linear, elastic material. Q • Vertical stress, σz below point load is 0.477 2 (r=0) z

• At surface i.e. z = 0, vertical stress, σz is infinite. But in actual case, stress is finite and acts on a finite area. • Vertical stress, σz decreases rapidly with increase of r/z. At r/z = 5 vertical stress, σz becomes extremely small and neglected. • In actuall practice foundation f d l d are not applied loads l d directly d l on ground d surface. • Vertical V ti l stress t d decreases d to due t excavation ti or removall off load. l d • Field measurements indicate that actual stress is smaller than theoretical solutions. solutions • In deep deposits of sand E increases with depth. Dr. PVSN Pavan Kumar

Other stress components

Similarly other stress components σx , σ Similarly other stress components, σ σy , ττxy , ττyz , ττzx may be required. may be required Dr. PVSN Pavan Kumar

Isobar Diagram • Isobar Isobar is a curve joining points of equal stress intensity. is a curve joining points of equal stress intensity. • It is in a shape of electrical bulb or an onion. • Following are calculations for an isobar of intensity, 0.1 Q per unit g y p area. Q σ z = IB 2 ; z

Q 0.1Q = I B 2 ⇒ z

I B = 0.1z

2

IB =

3 1 2 5/ 2 2π ⎡ ⎛r⎞ ⎤ ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ z ⎠ ⎦⎥

• For different depths, z influence factor, IB is determined, r/z is determined,, then radial distance r is determined. z IB r/z r

0.25 0.5 0.75 0 0062 0 0.0062 0.025 025 0 0.0562 0562 1.5 1.16 2.16 0.54 0.75 0.87

1 1.25 1.5 1.75 01 0 0.1 0.1562 1562 0 0.225 225 0 0.3062 3062 0.93 0.75 0.59 0.44 0.93 0.94 0.89 0.77


2.0 04 0.4 0.27 0.54

z 0.25

r 0.54

0.5

0.75

0.75

0.87

1

0.93

1.25

0.94

1.5

0.89

1.75

0.77

Isobar diagram

2.0 0.54 • Join points marked at different radial distance. This curve is isobar. • Shape of isobar curve is lemiscate curve. • Isobars of higher intensities 0.2Q, 0.3Q 1 0.75 0.5 0 0.5 0.75 1 will lie inside the isobar of intensity 0.1Q • Zone Z within i hi which hi h the h stresses have h significant i ifi effect ff on the h settlement l off structures are known as pressureDr. PVSN Pavan bulb. Isobar of 0.1Q forms a pressure bulb Kumar and area outside has negligible stresses.

Vertical stress on a horizontal plane • V Vertical i l stress on a horizontal plane at a depth, z = 2m is determined as follows Q Q σ z = I B 2 = I B 2 = 0.25I B Q z 2 • Maximum stress occurs below the load and decreases to either ends.

r r/z IB 0 0 0.477465 0.5 0.25 0.410317 1 0.5 0.273317 1.5 0.75 0.156456 2 1 0.084405 2.5 1.25 0.045424 3 1.5 1 5 0.025075 0 025075 4 2 0.008541


σz 0.119366Q 0.102579Q Q 0.068329Q 0.039114Q 0.021101Q 0.011356Q 0 006269Q 0.006269Q 0.002135Q

Influence line diagram • Vertical Vertical stress distribution diagram on a horizontal plane at a depth z stress distribution diagram on a horizontal plane at a depth z due to unit load at ground surface. • When a number of concentrated loads are acting on ground surface g g Q2

Q3

A’

σBA B

σAA A

σCA C

Vertical stress at Point A = Q Vertical stress at Point A = Q1σAA + Q + Q2σBA + Q + Q3σCA σAA = vertical stress at A due to unit load A’ σBA = vertical stress at B due to unit load A’ σCA = vertical stress at C due to unit load A’ Dr. PVSN Pavan Kumar

Vertical stress on a vertical plane • Consider a vertical plane at a radial distance, r = 1m. z r/z 0.25 4 05 0.5 2 1 1 1 5 0.667 1.5 0 667 2 0.5 2.5 0.4 5 0.2

IB 0.000401 0 008541 0.008541 0.084405 0 19041 0.19041 0.273317 0.329455 0.432871

σz 0.006411Q 0 034165Q 0.034165Q 0.084405Q 0 084627Q 0.084627Q 0.068329Q 0.052713Q 0.017315Q

• V Vertical ti l stress t i plotted is l tt d horizontally h i t ll along radial axis. Vertical stress first increases then decreases. Dr. PVSN Pavan Kumar

Vertical stress due to line load q‘ δy

• Vertical Vertical line load of intensity, q line load of intensity q’ per per meter length is acting along the y – axis. • Vertical stress at point P due to line load is to be determined. • Consider load acting on a small length δy. Point load is q’ δy

3q ' δy 1 • Vertical stress, Δσz = 5 2 2 2πz 2 ⎡ ⎤ ⎛r⎞ 1 + ⎢ ⎜ ⎟ ⎥ ⎢⎣ ⎝ z ⎠ ⎥⎦ 3q' δy 1 3q' δy 1 3 = = 5 5 3 q ' z δy 1 2 2 2 2 2 2 2 2 = π 2πz ⎡ 2 z ⎡x + y + z ⎤ ⎛ x2 + y2 ⎞ ⎤ 2π (x 2 + y 2 + z 2 )5 2 ⎟ ⎥ ⎢1 + ⎜ ⎢ ⎥ 2 z ⎣ ⎦ ⎟ ⎥ ⎢ ⎜ z Dr. PVSN Pavan Kumar ⎠ ⎦ ⎣ ⎝

Vertical stress due to line load q‘ δy

+∞

3q' z 3 1 σz = ∫ 2π x 2 + y 2 + z 2 −∞

(

)

5

dy 2

Assume x 2 + z 2 = u 2 Assume +∞

3q ' z 3 1 σz = ∫ 2π u 2 + y 2 −∞

(

3

3q ' z 1 σz = ∫ 2π u 2 + y 2 −∞

(

3

3q ' z = 2πu 4

π

2

dyy 2

dy = u sec2 θ dθ

Let y = u tanθ +∞

)

5

3q ' z dy = 5 2 2π

)

3

3q ' z cos θ d θ = 4 ∫ 2 π u −π 3

2

π

3

π

2

∫π

−

2

u sec 2 θ d θ u 5 sec5 θ

∫ (1 − sin θ )cosθ dθ 2

−π

2

2

Sinθ = t cosθ dθ = dt 3 1

3q ' z = 2πu 4

(

)

3

3q ' z 2 1 − = t dt ∫−1 2πu 4

1

⎡ t ⎤ 3q' z 3 ⎛ 1 1 ⎞ 3q ' z 3 2 2q ' z 3 1− +1− ⎟ = = ⎢t − ⎥ = 4 ⎜ 4 4 u u 3 π 3 π 3 π u 2 3 Dr. PVSN Pavan Kumar ⎠ ⎝ ⎣ ⎦ −1 3

Vertical stress due to line load 2q ' z 3 2q ' z 3 σz = = 4 πu π x2 + z 2

(

)

2

=

2q ' z 4

πz (x + z 2

)

2 2

=

2q ' ⎛ ⎛ x⎞ πz ⎜⎜1 + ⎜ ⎟ ⎝ ⎝z⎠


2

⎞ ⎟ ⎟ ⎠

2

Vertical stress under a strip load Point P lies below the center of the strip Point, P lies below the center of the strip

qdx (line load)

• Strip load of width B (= 2b) and intensity is ‘q’ intensity is q kN/sq.m. kN/sq m • Consider a line load acting on a small width ‘dx’ small width dx at a distance x at a distance x from the center of load. • Infinitesimal vertical stress, Δσ z =

2(qdx ) ⎛ ⎛ x⎞ πz ⎜⎜1 + ⎜ ⎟ ⎝ ⎝z⎠

2q z 4 dx σz = πz −∫b x 2 + z 2

2

⎞ ⎟ ⎟ ⎠

2

b

(

x = z tan u

)

2

dx

=

2qz 3

π

b

∫ (x

−b

dx 2

= zsec2u du

+z

)

2 2

θ

θ

2qz 3 z sec 2 u du 2q 2 σDr. PVSN Pavan Kumar = = cos udu z 4 ∫ 4 ∫ πz −θ sec u π −θ

Vertical stress under a strip load θ

θ

2q 1 + cos 2u q ⎡ sin 2u ⎤ q⎛ sin 2θ ⎛ sin 2θ ⎞ ⎞ σz = du = ⎢u + = ⎜⎜θ + − ⎜ −θ − ⎟ ⎟⎟ ∫ ⎥ π −θ 2 π⎣ 2 ⎦ −θ π ⎝ 2 2 ⎠⎠ ⎝ q = (2θ + sin 2θ )

π

Stress at a point P not below the strip

• Point P is not blow the center of strip • Extremities of strip make an angle β1 and β2 with Point P. and β with Point P • Load qdx will act as a line load at a distance x from origin. 2(qdx ) Δσ z = 2 2 ⎛ ⎛ x⎞ ⎞ πz ⎜1 + ⎜ ⎟ ⎟ ⎜ ⎝z⎠ ⎟ ⎝ ⎠

x From Fig . tan β = z dx = z sec 2 β dβ


Δσ z =

σz =

β2

∫β

1

2 qz sec 2 β d β

(

π z 1 + (tan β )

)

2 2

2 q cos β d β 2

π

=

2 q cos 2 β d β

π β2

⎛ 1 + cos 2 β ⎞ = ⎜ ⎟ dβ ∫ π β1 ⎝ 2 ⎠ 2q

β2

sin i 2β ⎤ sin i 2 β 2 − sin i 2 β1 ⎤ q ⎡ q ⎡ = ⎢β + = ⎢ (β 2 − β 1 ) + ⎥ ⎥ 2 ⎦ β1 π ⎣ 2 π ⎣ ⎦ =

q

[(β 2 − β 1 ) + (sin

π If β 1 + β 2 = 2ϕ

σz =

q

π

β 2 cos β 2 − sin β 1 cos β 1 )]

[(2θ ) + (sin 2θ cos 2ϕ )] Dr. PVSN Pavan Kumar

Vertical stress under a strip load z

θ (radians)

θ (angle)

σz

0.5b

1.107149

63.43495

0.959q

b

0.785398

45

0.818q q

1.5b

0.588003

33.69007

0.668q

2b

0.463648

26.56505

0.550q

3b

0.321751

18.43495

0.396q

4b

0.244979

14.03624

0.306q

5b

0 197396 0.197396

11 30993 11.30993

0 248 0.248q

6b

0.165149

9.462322

0.208q

7b

0.141897

8.130102

0.179q

8b

0.124355

7.125016

0.158q

9b

0.110657

6.340192

0.140q

10b

0.099669

5.710593

0.126q

11b

0.09066

5.194429

0.115q

12b

0 083141 0.083141

4 763642 4.763642

0 106 0.106q

tan θ =

b z

σz =

q

π

(2θ + sin 2θ )


Vertical stress under a under a circular area • Load applied to soil surface are not concentrated loads but are usually spread over a finite area of the footing. • Footing is assumed to be flexible and contact pressure is distributed uniformly. • Following F ll i is i the h procedure d to determine the vertical stress at point P at a depth z below the center of circular area. q = stress per unit area R = Radius of loaded area Consider a ringg of radius ‘r’ of width ‘dr’. Concentrated load of q 2πr dr acts at a radial distance r from point ‘O’. Dr. PVSN Pavan Kumar

Vertical stress under a under a circular area

3(q 2πr dr ) 1 3(q 2πr dr ) 1 Δσ z = = 5 5 ∫0 2π z 2 2 2 2 2 2π z 2 ⎡ ⎤ ⎡ ⎤ ⎛r⎞ ⎛r⎞ + + 1 1 ⎜ ⎟ ⎢ ⎥ ⎢ ⎜ ⎟ ⎥ ⎢⎣ ⎝ z ⎠ ⎥⎦ ⎢⎣ ⎝ z ⎠ ⎥⎦ R

r dr 3q 3q z 5 r dr = 2∫ = 2∫ 5 z 0 ⎡ r 2 + z 2 ⎤ 2 z 0 r 2 + z 2 52 ⎢ z2 ⎥ ⎣ ⎦ R

R

(

)

Assume r 2 + z 2 = u

2r dr = du

R2 + z2

2

R +z

= 3qz 3

∫

z2

2

⎡ ⎤ −3 3qz 3 ⎢ u 2 ⎥ du ⎢ ⎥ = 5 2 ⎢⎛ − 5 ⎞ ⎥ 2u 2 + 1⎟ ⎢⎣ ⎜⎝ 2 ⎠ ⎥⎦ z 2

⎡ 1 q ⎢ = − qz ⎢⎣ R 2 + z 2 3

(

)

3

2

2

3qz 3 ⎡ −3 2 ⎤ R + z = u ⎢ ⎥⎦ z 2 2× − 3 ⎣ 2

2

⎡ ⎤ ⎢ ⎥ ⎡ 1 ⎤ ⎢ ⎥ 1 ⎤ 1 1 3 ⎥ = qqz ⎢ ⎥ = q ⎢1 − ⎥ − − 3 3 3 3 ⎢⎣ z 2 2 R 2 + z 2 2 ⎥⎦ ⎢ ⎛ ⎛ R ⎞2 ⎞ 2 ⎥ z 2 2 ⎥⎦ ⎢ ⎜1 + ⎜ ⎟ ⎟ ⎥ Dr. PVSN Pavan Kumar ⎢⎣ ⎜⎝ ⎝ z ⎠ ⎟⎠ ⎥⎦

( )

( )

(

)

Vertical stress under a under a circular area σ

z

⎡ ⎢ ⎢ = q ⎢1 − ⎢ ⎢ ⎢⎣

1 2 ⎛ ⎞ R ⎛ ⎞ ⎜1 + ⎜ ⎟ ⎟ ⎜ ⎝ z ⎠ ⎟⎠ ⎝

3

2

⎤ ⎥ ⎡ ⎥ 1 ⎥ = q ⎢1 − ⎢ ⎥ 1 + (tan θ ⎣ ⎥ ⎥⎦

(

)

2

)

3

2

If θ = 90 If θ 90° or R or R = ∞ ∞ or z or z = 0; 0; Verticalstress at point P, σz = q

• When loaded area is large g compared with depth z vertical stress at point, P is approximately q. • Isobar of 0.1q (Pressure bulb) cuts the axis of load at a depth of about 4R below the loaded area. • This is shallow compared Dr. PVSN Pavan Kumar with that of strip load.

⎤ ⎛ 1 ⎤⎞ 3 ⎥ = q ⎜⎜ ⎡1 − ⎟ = q 1 − cos θ 3 ⎟ ⎢ ⎥ sec θ ⎦ ⎠ ⎥ ⎝⎣ ⎦

(

)

Vertical stress under corner of rectangular area • Consider a uniformly loaded rectangular area carrying a stress ‘q’ q on the surface of dimensions L & B. • Stress at a depth z is given by taking dQ = q dA

3 qz 5 dA 1 dσ z = 2π z 2 x 2 + y 2 + z 2

(

3 qz 3 σz = 2π

L B

∫∫ 0 0

(x

dxdy 2

+y +z 2

2

)

)

5

5

2

2

N Newmark k performed the integration and following is the solution f d h i i d f ll i i h l i ⎞⎤ q ⎡ mn m2 + n2 + 2 mn −1 ⎛ ⎟⎥ σz = + sin ⎜⎜ ⎢ 2 2 2 2 ⎟ 2 2 2 2 2π ⎢⎣ m 2 + n 2 + 1 m + n + m n + 1 m n m n + + + 1 ⎝ ⎠ ⎥⎦ m=B n=L σ = I N q Dr. PVSN Pavan Kumar N z z

Point anywhere below the rectangular area ABCD

• Rectangular area is divided into four rectangles AEPH, BEPF, FPGC and HPGD such that each of them will have a corner at point P. P • Vertical stress at point, P σz = q [IN1 + IN2 + IN3 + IN4] where IN1 , IN2 , IN3 , IN4 are influence factor corresponding to each rectangle area. • If point is at center of loaded area, σz = 4 IN


Point is outside the loaded rectangular area ABCD

• • • • • •

In this case point ‘P’ is located outside the loaded area. Vertical e t ca st stress ess at po point, t, P σz = q [[IN1 ‐ IN2 ‐ IN3 + IN4] IN1 is due to loaded area 1, i.e. AEPF, IN2 is due to loaded area 2 i.e. BEPH IN3 is due to loaded area 3 i.e. DGPF IN4 is due to loaded area 4 i.e. i e CGPH Dr. PVSN Pavan Kumar

Point below the edge of rectangular area ABCD

In this case point ‘P’ is below the edge of the loaded area ABCD. Vertical stress at p point,, P σz = q [[IN1 + IN2] • IN1 is due to loaded area 1, i.e. APED, • IN2 is due to loaded area 2 i.e. PBCE


Newmark’s Influence Chart • Vertical stress due to other shapes can be obtained from Newmark’s Influence chart. • This is based on the vertical stress below center of circular area. • Consider a circular area of radius, R1 is divided into 20 equal sectors. • Vertical stress at point ‘P’ P at depth z below the center of loaded area 1 due to a hatched area i e due to a hatched area i.e. of full circle. of full circle 20 Assume vertical stress due to hatched area is 0.005q 3 ⎡ ⎧ ⎫ 2⎤ ⎢ ⎪ ⎥ ⎪ q ⎢ ⎪ 1 R1 ⎪ ⎥ 0.005q = ⇒ = 0.27 7 1− ⎨ 2⎬ ⎥ ⎢ z 20 R ⎛ ⎞ ⎪ ⎪ 1 1+ ⎜ ⎢ ⎟ ⎪ ⎥ z ⎪ ⎠ ⎭ ⎥Dr. PVSN Pavan Kumar ⎢⎣ ⎩ ⎝ ⎦

Each sector of a circle of radius, R1 = 0.27z will exert a stress of 0.005q at center.

Newmark’s Influence Chart • Consider another concentric circle off radius di R2 is i also l divided di id d into i 20 sectors. • Two hatched areas exert a stress of 2 x 0.005q. 3 ⎡ ⎧ ⎫ 2⎤ ⎢ ⎪ ⎥ ⎪ q ⎢ ⎪ R2 1 ⎪ ⎥ 2 × 0.005q = 1− ⎨ ⇒ = 0.40 2⎬ ⎥ ⎢ 20 z R ⎛ ⎞ ⎪ ⎪ 2 1+ ⎜ ⎢ ⎟ ⎪ ⎥ z ⎪ ⎠ ⎭ ⎥ ⎢⎣ ⎩ ⎝ ⎦

• Radii R dii off third hi d to ninth i h circle i l is i determined. d i d Values V l are 0.52 0 52 z, 0.64 0 64 z, 0.77 z, 0.92 z, 1.11 z, 1.39 z and 1.91 z. Radius of 9.5 circle is 2.54 z. Radius of tenth circle is infinite. infinite


Newmark’s Influence Chart Circle No

Radius

R (if z = 2cm)

1

0.27 z

0.54

2

0.40 z

0.8

3

0.52 z

1.04

4

0.64 z

1.28

5

0 77 z 0.77 z

1 54 1.54

6

0.92 z

1.84

7

1.11 z

2 22 2.22

8

1.39 z

2.78

9

1.91 z

3.82

9.5

2.54 z

5.08

10

∞

∞

Consider some depth, depth z AB i.e. ie 2cm and draw 10 concentric circles. Plan of loaded area is drawn on a tracing paper to a scale such that Dr. PVSN Pavan Kumar length AB = depth, z of point P below the surface.

Newmark’s Influence Chart • If pressure is required at a depth, depth z = 1m plan is drawn to a scale of 2 cm = 1 m. • Traced plan of loaded area is placed over the Newmark chart such that the point P where the pressure is required coincides with the center of chart. Vertical stress at point, P = σ z = I × n × q I = Influence coefficient = 0.005 n = number of area units covered by the plan. (One unit is the area between the two radial lines and two circles.) q = Intensity of load


Comparison of stress due to loads of different shapes • At surface vertical stress is same as stress. • Vertical stress decreases rapidly with ith depth, d th z. • For circular and square loads vertical stress is about 10% q at a depth 2B. • For strip loads stress is much greater. • For a strip load at a depth z= 3B vertical stress is about 20% q Dr. PVSN Pavan Kumar

Westergaard Solution – Concentrated load Vertical stress, stress σz at a depth z below the ground surface and a radial distance, r is

σ z = Iw

Q 1 Q = 3 2 2 2 z z2 π ⎡1 + 2 r z ⎤ ⎢⎣ ⎥⎦

( )

Q = Concentrated load 2 For r/z OMC, water fills the volume of solid particles and density decreases.

Below OMC


At OMC

Above OMC

Standard Proctor Compaction test (IS 2720 Part 7) Theoretical maximum dry density: Dry density corresponding to zero air Theoretical maximum dry density: Dry density corresponding to zero air voids at a given water content. Gρ w Gρ w = ρ dry = 1 + e 1 + wG S Air voids are zero this implies all voids filled with water and degree of saturation, S = 100% Gρ w ρ dtheo max = 1 + WG


Standard Proctor Compaction test (IS 2720 Part 7) (ρd)theomax

Moisture (ρd)theomax Content

Zero air void line or 100% saturation line

0.12 2.04 0.14 1.96 0.16 1.89 0.18 1.82 0.20 1.75 0.22 1.69 Gρ w ρ = Similarly for 10% air voids or 90% saturation , dry density is dry wG 1 1+ + and corresponding air void lines can be plotted. d di i id li b l tt d 0.9

( 1 − na )Gρ w ρd = 1 + wG

na = Percentage of air voids = Va

V


Standard Proctor Compaction test (IS 2720 Part 7) Soil is compacted p in the field byy addingg optimum p moisture content water to the soil and the amount of compaction in the field should be same as that in the lab.

Modified Proctor Compaction test (IS 2720 Part 8) • • • • • •

Developed to represent heavy compaction. compaction Mould is same as Standard Proctor test (100 mm or 150 mm) Mass of rammer = 4.89 4 89 kg Free drop = 450 mm No of layers in which soil is compacted in the mould = 5 No. Compactive effort in modified Proctor test = 2700 kJ/m3 standard Proctor test = 592 kJ/m3


Curve 2 for modified Proctor test is higher and to the left of curve 1 standard Proctor compaction test i.e. OMC decreases and MDD increases. Li joining Line j i i MDD is i called ll d line li off optimum ti it is i parallel ll l to t zero air i void id line. Dr. PVSN Pavan Kumar

Factors affecting compaction Moisture Content: At low water content soil is stiff, stiff offers resistance to compaction. As water is increased soil particles are lubricated and particles are arranged p g in a close p packing. g • Dry density of soil increases with increase of water content, volume of air voids decrease and attains a constant volume at OMC. • Increase of water content beyond OMC replaces the solids by water and decreases the dry density. Amount of Compaction: As compaction effort increases MDD increases and OMC decreases. Increase of MDD is limited to certain increase of compaction effort. • Dry Density may not increase with compaction effort beyond a certain stage. Dr. PVSN Pavan Kumar

Factors affecting compaction Type of soil Type of soil

• Well graded sand has more dry density and less optimum moisture content. • As soil becomes plastic OMC increases and MDD decreases. Cohesive soil will have more voids.


Factors affecting compaction Type of compaction Dry density achieved by a soil depends on the type of compaction i.e. kneading action, dynamic action or static action. Admixtures Dry density achieved depends on the type and amount of admixtures such as lime, cement and bitumen.

Effect of compaction on soil properties

SSoil structure il • Soils compacted dry of optimum will have a flocculated structure and those on the wet side of optimum have a dispersed structure those on the wet side of optimum have a dispersed structure. • On the dry side of optimum water content is less the attractive forces between particles is predominant than repulsive forces. On wet side of optimum repulsive f forces increase and the h particles l get oriented into a dispersed structure. Dr. PVSN Pavan Kumar Flocculated Structure

Dispersed Structure

Effect of compaction on soil properties • Permeability: Permeability decreases with increase of water content on dry side of optimum as the volume of voids decrease. As compaction effort increases permeability of soil decreases due to increased dry density. • Swelling: Soil compacted dry of optimum has a higher water deficiency and swells more than soil compacted wet of optimum. p • Pore water pressure: Sample on dry side of optimum will have low p pore water p pressure compared p with soil on wet side of optimum. • Shrinkage: g Soil compacted p dryy side of optimum p shrink less compared with soil on wet side of optimum. • Co Compressibility: p ess b ty So Soil co compacted pacted o on d dryy sside de o of opt optimum u compresses less compared to soil on wet side of optimum. Dr. PVSN Pavan Kumar

Effect of compaction on soil properties Stress – Strain relation: Soil compacted on dry side of optimum has steeper slope than soil compacted on wet side.

Shear strength: Soil compacted dry of optimum will have more shear strength than a sample compacted wet of optimum. optimum Dr. PVSN Pavan Kumar

Methods of Compaction used in field

H dT Hand Tampers

Block iron stone of 3 to 5 kg attached to a wooden rod and dropped by 0.3m.

Mechanical Rammers Mechanical Rammers

Operated by gasoline.

Tampers or rammers are used to compact a narrow confined area such as trenches and behind the bridge abutments. Output is low due to large Dr. PVSN Pavan Kumar quantities of soil

Rollers

Smooth wheel roller • This consists of three wheels, two large in the rear and one small wheel in the front in the front. • Mass of smooth wheel roller is between 2 to 15 T between 2 to 15 T • These rollers are used to seal the surface of fill at the end of day’ss work surface of fill at the end of day work to provide smooth surface. y rollers Pneumatic tyred • Consist of 9 to 11 wheels fixed to two axles. Mass of roller varies from 5 to 200 T

Smooth wheel roller Smooth wheel roller

Pneumatic tyred roller Dr. PVSN Pavan Kumar

Sheep foot roller • In ancient days it is a usual practice to pass a flock of sheep on the newly formed soil fill to cause compaction. ti • Sheep foot roller consists of hollow drum with a large number of small projections on the surface. • Sheep foot rollers are suited for Sheep foot roller compaction of cohesive soil. • These projections penetrate the soil layer and lower portion is compacted in the first pass. In the successive passes compaction of middle ddl and d top layers l off soill occur. • Small rollers compact layers of thickness 15 cm and heavy rollers of 30 cm thickness. thickness Dr. PVSN Pavan Kumar

Vibratory Compactors • Vibrations are induced in the soil during compaction. • Vibrator is mounted on the drum. • Motor drives two eccentric weights i ht and d creates t up and d down oscillations.

Pneumatic type

Smooth wheel type Vibratory plate compactor


Relative Compaction or Degree of compaction Field dry density × 100 = Maximum Dry Density

• FFor embankments this should be > 95% to ensure proper compaction b k t thi h ld b > 95% t ti of soil layers. Compaction quality control is ensured in the field by addition of appropriate water content to the soil and checking the field dry density.


Assignment 3 1 A concentrated load of 1000 kN acts vertically at the ground surface. 1. surface Determine the vertical stress at a point which is at (i) a depth of 2.5 m and a horizontal distance of 4.0m (ii) at a depth of 5.0m and a radial distance of 2.5 m. 2. Derive the expression for vertical stress at a depth due to a line load off q’.’ 3. Draw Newmarks Influence Chart and explain how it is used to determine the vertical stress due to a stress, stress q. q 4. Explain the effect of compaction on soil properties. 5. A sample of soil was prepared by mixing a quantity of dry soil with 10% by mass of water. Find the mass of this wet mixture required to produce a cylindrical, compacted specimen of 15 cm diameter and 12 5 cm deep 12.5 d and d having h i 6% air i content. t t Find Fi d also l the th void id ratio ti and d the dry density of the specimen if G = 2.68 6. Write a note on Standard Proctor and Modified Proctor compaction test Dr. PVSN Pavan Kumar

Unit IV Unit IV Consolidation Dr. PVSN Pavan Kumar Dr. PVSN Pavan


Syllabus Types of compressibility – Immediate settlement, settlement primary and secondary consolidation – stress hi history off clay, l e‐p and d e‐log l p curves – normally ll consolidated soil, over consolidated soil and under consolidated soil – preconsolidation pressure and its determination – Terzaghi Terzaghi’ss 1 1‐D D consolidation theory – coefficient of consolidation: square root time and logarithm of time fitting methods – computation of total settlement and time rate of settlement. l Dr. PVSN Pavan Kumar

Compressibility, Consolidation • Decrease in volume of soil due to application of compressive load is known as compressibility. • Compression p of soil occurs due to – Compression of solid particles and water in the voids – Compression and expulsion of air in the voids – Expulsion of water in the voids.

• Compression of solid particles and water is small. • Compression of air and expulsion occurs in partially saturated soil due to application of load. This not relevant for saturated soil. • In a fully f ll saturated d soill compression occurs due d to expulsion l off water and adjustment of particles. • Compression C i off a saturated t t d soilil under d a steady t d static t ti pressure due d to t expulsion of water from voids is known as consolidation. • Small volume changes occur due to bending, bending distortion and fracture. fracture Dr. PVSN Pavan Kumar

Compressibility, Consolidation Steady static load Steady static load

Expulsion of water from voids is consolidation


• Due to consolidation, settlement of structure occurs . Procedure to determine the magnitude and time required for settlement is presented. d Consolidation of a soil deposit occurs in three stages: ¾ Initial Consolidation: Decrease in volume immediately after application of load due to expulsion and compression of air in the voids Due to compression of solid particles. voids. particles ¾ Primary Consolidation: • After initial consolidation further reduction of volume occurs due to expulsion of water from voids. • When a saturated soil is subjected to a pressure, initially all the applied pressure is taken up by water as an excess pore water pressure. • A hydraulic gradient develops and water starts flowing out and a decrease in volume occurs. • This decrease depends on permeability of the soil and is time dependent. • In fine grained soil primary consolidation occurs over a long time whereas in coarse grained d soill consolidation ld occurs immediately. d l • As water escapes from the soil applied pressure gradually transfers from water Dr. PVSN Pavan Kumar in voids to the solid particles.

Tower settled by about 2.5 meters into the ground due to consolidation off foundation f d soil.l Tilted by 5.5° over a period of 800 years. years

Leaning Tower of Pisa Dr. PVSN Pavan Kumar

Primary Consolidation: • As consolidation process progress pore water pressure decreases and effective stress increases. Secondary Consolidation: This occurs after primary consolidation and occurs at a very slow rate due to plastic readjustment of solid particles and the adsorbed water t th to the new stress system. Likely to occur in organic soils. t t Lik l t i i il Analogy of consolidation • Consolidation C lid i i the is h action i off squeezing i off water from f a saturated d sponge under pressure. Soil behaves as a saturated sponge. Spring Analogy: Analogy • A cylinder fitted with tight fitting piston has a valve. Cylinder is filled with water. water • Initial length of spring be 100 mm and stiffness of spring be 10 mm/N. mm/N • Load, P of 1 N is applied to the piston with valve closed. Dr. PVSN Pavan Kumar

Spring Analogy

• Initially entire load is taken by water and no load is taken by spring.

Pw + Ps = P

Pw = Load taken by water; Ps = Load taken by spring

At time, t = 0, when valve is closed , P Dr. PVSN Pavan Kumar s = 0 and Pw = 1

Spring Analogy

• When valve is ggraduallyy opened, p , water starts escaping p g from cylinder, y , spring shares a part of load length of spring decreases. • If ΔP is the load transferred from water to spring, PS = ΔP and PW = 1 ‐ ΔP. As more and more water escapes, load carried by the spring increases. Dr. PVSN Pavan Kumar

Spring Analogy • At time, t = tf water stops escaping ; Ps = 1 and Pw = 0. • Load causes a decrease in length of the spring. Stiffness of spring is 10 mm/N, mm/N final length is 90 mm. mm • If load is increased to 2N, load transfer repeats and length of spring will be 80 mm.

Load transfer from water to spring Load transfer from water to spring


Spring Analogy Spring Water in the cylinder Valve in the piston Valve in the piston Load carried by water, Pw Load carried by spring, P carried by spring PS Total load, P

Consolidation of soil Solid Particles Solid Particles Water in the voids Permeability of soil Permeability of soil Pore water pressure, u Effective stress σ Effective stress, Applied pressure, σ

Applied pressure, σ = Effective stress, σ + Pore water pressure, u • When a pressure, σ1 is applied to a saturated soil sample, initially pressure is taken by pore water. t = 0 u = σ1 • A hydraulic gradient is set up and pore water escapes from voids. Finally applied pressure is transferred to solid particles. t = tf σ = σ 1 • Volume or Void ratio of soil decreases with time Dr. PVSN Pavan Kumar

Consolidation • For a given pressure increment, increment σ1 the void ratio decreases finally to ef1. • As pressure further increases to σ2 , σ3 so on void ratio decreases to ef2 , ef3 so on

e vs t

ef vs σ


Consolidation Compaction Gradual process of reduction of Compaction is a rapid process of volume under sustained static volume reduction by mechanical loading. means such as tamping, g p g, rollingg and vibration. Reduction in volume of a Volume of a p partiallyy saturated saturated soil due to expulsion of soil is decreased due to expulsion water. of air from voids.

Occurs naturally when saturated Artificial process carried out in soil is subjected to static load due field to increase the density. Dr. PVSN Pavan Kumar to weight of building.

Consolidation test • Consolidation test is carried out by consolidometer or an oedometer. • It consists of a loading device and cylindrical container. (Consolidation cell) • Steel ring is pushed into soil, porous stones are placed on the top and bottom of sample.

Dial gauge g g hanger Consolidation Cell Lever arm

Water Jacket

Top plate

Ring

Water jacket with soil with soil sample

Top porous Bottom stone Pressure porous stone pad

Consolidation Cell Consolidation Cell

Weight

• Diameter of cell is 60 mm, thickness of sample is 20 mm • Loading pad is placed on sample along with porous stones and entire set up is placed in water jacket. • Water jacket is filled with water to allow flow of water into and out of sample. Dr. PVSN Pavan Kumar

Consolidation test Water Jacket

Dial gauge g g hanger Consolidation Cell Lever arm

Top plate

Ring

Water jacket with soil with soil sample

Top porous Bottom stone Pressure porous stone pad

Consolidation Cell Consolidation Cell

Weight

• Soil S il sample l is i subjected bj d to load l d axially i ll by b placing l i a hanger h on pressure pad and placing weights on bottom hanger. • Dial gauge is placed on the sample to measure the change in thickness of sample as consolidation takes place. • Initial seating pressure of 5 kN/m2 is applied on sample, load is maintained till there is no change in dial gauge reading or 24 hours Dr. PVSN Pavan Kumar whichever is less.

Consolidation test • First increment of load to give a pressure of 10 kN/m2 is applied on the specimen. Dial gauge reading are taken after 0.25, 1, 2.25, 4, 6.25, 9, 12.25, 16, 20.25, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 289, 324, 400, 500 600 and 1440 Min (24 hours). 500, hours) • Primary consolidation is assumed to be completed within 24 hours. • Second increment of load is applied and successive pressures applied are 20, 40, 80, 160, 320 and 640 kN/m2 till the desired maximum pressure is applied. • Maximum pressure applied is actual loading on the soil in the field after the structure is constructed. • After Aft consolidation lid ti under d the th final fi l load l d increment i t is i complete, l t load l d is i reduced by one‐fourth i.e. 160 kN/m2 and allowed to swell. • Dial gauge reading is noted after 24 hours and pressure reduced to 40 kN/m2. • Dial gauge reading is noted after 24 hours and pressure reduced to 10 kN/m2. Final dial gauge reading is noted after 24 hours. Through out the test consolidation cell is filled with water. Dr. PVSN Pavan Kumar

Consolidation test Dial ggauge g readings g are converted into void ratio from followingg methods Height of solids methods H − Hs e= Hs

H = Height of soil sample under each load increment Ms Gρ w

Hs =

Ms = Mass of dry soil; G = Specific gravity; ρw = Density of water Change in void ratio method 1 + el Δe = ΔH Hl

el = Void ratio at end of test Hl = Thickness of sample at end of test ΔH Change in thickness of sample ΔH = Change in thickness of sample Dr. PVSN Pavan Kumar

Consolidation test result

• Final void ratios corresponding to • Thickness of sample is more at final thickness of sample are obtained b i i beginning off test t t and d decreases d corresponding to each applied with increase of time for a pressure. pressure increment. p • Sand is less compressible compared • For sand change in thickness with clay. occurs very quickly. • This is used to determine the • Above graph is required to magnitude of consolidation determine the coefficient of settlement. Dr. PVSN Pavan Kumar consolidation


− de 1) C ffi i t f 1) Coefficient of compressibility, kN/m ibilit av = kN/ 2 dσ

• Coefficient of compressibility is the slope of e vs Coefficient of compressibility is the slope of e vs σ curve. curve • Coefficient of compressibility decreases with increase of effective stress. • Soil becomes stiffer at a higher effective stress and curve becomes flatter. Dr. PVSN Pavan Kumar

Consolidation test result − ΔV

2) Coefficient of volume change, mv =

V0

Δσ ΔV = Change in volume; V0 = Initial volume of sample ΔV = Change in volume; V = Initial volume of sample Δσ = Increase of effective stress

Δe ΔV ΔH = = V0 H 0 1 + e0 av mv = 1 + e0

Where ΔH is the change in thickness; H0 is the initial thickness of sample; is the initial thickness of sample; Δe is the change in void ratio; e0 is the initial void ratio

mv depends d d on the h effective ff stress at which h h it is determined. d d It decreases with increase of effective stress. Dr. PVSN Pavan Kumar

Consolidation test result 3) Compression index, Cc is the slope of e vs log σ Cc =

− Δe ⎛ σ + Δσ log10 ⎜⎜ 0 ⎝ σ0

⎞ ⎟⎟ ⎠

Δe = Change in void ratio I iti l ff ti t σ 0 = Initial effective stress Δσ = Increase of effective stress • Compression index, Cc is used to determine settlement in field • For undisturbed clays, C For undisturbed clays Cc = 0.009 (LL – = 0 009 (LL – 10) LL = Liquid limit in % 10) LL = Liquid limit in % • For remoulded clays, Cc = 0.007 (LL – 10) Dr. PVSN Pavan Kumar


Loading Reloading Curve Cc • Increase of effective stress Normal or virgin compression from σ 1 to σ 2 decreases the void ratio. Curve AB is known Cr as virgin compression curve or Ce normal consolidation. consolidation Slope is Cc Over Consolidation compression index, cc • Decrease of effective stress from σ 2 to σ 1 slightly increases the void ratio. Curve BC indicates swelling or Loading Reloading Curve unloading. Slope of this curve i expansion is i index, i d Ce. • Increase of effective stress from σ 1 to σ 2 is known as reloading or recompression CB. recompression, CB Slope of this curve is called recompression index, index Cr • Curve BG is known as virgin compression curve or normal Dr. PVSN Pavan Kumar consolidation. Slope is Cc

Consolidation test result Pre Consolidation pressure • Maximum pressure to which a over consolidated clay layer is subjected bj d in i the h past is i known k as pre consolidation pressure. Over Consolidation ratio • Ratio

of

pre

consolidation

pressure to the present existing σ2 pressure, σ 1 • Over Consolidation ratio can be obtained for different existing pressures,σ 1 Settlement of structures under normal consolidation condition is more than the soil under recompression condition. Dr. PVSN Pavan Kumar

Δσ

Consolidation settlement H

Settlement of a soil deposit subjected to a pressure, Δσ Δ H = mv Δ σ H mv = Coefficient of volume compressibility at middle of clay layer ∆σ = Pressure at middle of clay layer y y H = thickness of clay layer or Soil is divided into number of layers Soil is divided into number of layers n

S f = ∑ mvi Δσ i Δz i i =1

mvi = Coefficient of volume compressibility of layer Δσ i = Pressure at middle of each layer iddl f hl Δzi = thickness of each layer Dr. PVSN Pavan Kumar

Consolidation settlements In case of normally consolidated clays

cc H σ 0 + Δσ log ΔH = 1 + e0 σ0 where cc = compression index =Slope of e versus log p plot. = 0.009 (LL – 10) LL = Liquid limit (%) eo = initial void ratio at the middle initial void ratio at the middle of clay stratum,

W.T B Df

q

2V

H

∆σ

1H

H = Stratum thickness, Thick stratum is divided into several layers σo = Effective overburden pressure at mid height of clay layer ∆σ = Average increase in pressure at middle of clay layer from the Dr. PVSN Pavan Kumar foundation loads.

Consolidation settlements Preconsolidated soil Final settlements are small in case of preconsolidated soil as recompression index, cr is smaller than compression index, cc

σ 0 + Δσ cr H ΔH = log 1 + e0 σ0 Above equation is applicable when σ 0 + Δσ < preconsolidation pressure, σ c If σ 0 + Δσ > preconsolidation pressure, σ c

ΔH =

σ cH σ + Δσ cr H log c + c log 0 1 + e0 σ 0 1 + e0 σc


Terzaghi’s theory of consolidation Terzaghi (1925) gave a theory to determine the rate of consolidation of a saturated soil mass subjected to steady static load based on the following assumptions. 1. Soil is homogenous and isotropic, fully saturated. Solid particles and water in voids are incompressible. Consolidation is due to expulsion of water from the voids – Assumptions are generally satisfied if volume of air is zero. 2. Coefficient of permeability of soil is same at all points and period of consolidation – remains constant duringg the entire p Coefficient of permeability, k varies at different points in a deposit. K decreases as consolidation progresses. 3. Darcy’s law is valid throughout the consolidation process – At y ggradients Darcy’s y law is not strictlyy applicable. pp low hydraulic Dr. PVSN Pavan Kumar

Terzaghi’s theory of consolidation Assumptions p 4. Soil is laterally confined and consolidation takes place in axial direction only. Drainage of water also occurs only in the vertical direction – This is largest error and consolidation occurs in 3 dimensions. 5. Time lag in consolidation is entirely due to low permeability of soil – Secondary consolidation also occurs along with primary. 6. There is a unique relation between void ratio, e and effective stress, i.e. av and mv are constant – use of actual relation complicates the expression.


Terzaghi’s theory of consolidation

• Consider a clay layer of height, H = 2d sandwiched between two sand layers. W.T is at top of clay layer. • When a uniform pressure of Δσ is applied on the surface of the top sand layer, total stress developed at all points in the clay layer is increased by Δσ. • Initially the whole pressure is taken up by water and a pore water pressure of Δσ/γw develops. Dr. PVSN Pavan Kumar


At various points along the thickness of clay layer are connected by flexible tubes to piezometers. At time, A i t = 0 pore water pressure is i Δσ/γ Δ / w throughout h h the h clay l layer. l Represented by line AB. Water starts escaping p g towards the upper pp and lower sand layers y due to excess pore water pressure. At points C and E pore water pressure drops to zero. In Dr. PVSN Pavan Kumar the middle of clay layer pore water pressure is more indicated by D.


• At time, time t = tf pore water pressure dissipates indicated by CFE. CFE • Curves joining points of equal pore water pressure are known as Isochrones. Dr. PVSN Pavan Kumar

Terzaghi’s theory of consolidation B i diff Basic differential equation for one dimensional consolidation is ti l ti f di i l lid ti i

d 2u du cv 2 = dz dt

k

= cv is coefficient of consolidation obtained from consolidation mvγ w test k = Coefficient of permeability of soil mv = Coefficient of volume change C ffi i t f l h γw = Unit weight of water Above equation is solved by Fourier series ∞

2 2 ⎞ −⎛⎜ n π 2 ⎟ cv t H ⎠ ⎝

2ui z ⎞ ⎛ (1 − cos nπ )⎜ sin nπ ⎟e u =∑ H⎠ ⎝ n =1 nπ n = Intezers Dr. PVSN Pavan Kumar

Terzaghi’s theory of consolidation 2d

1 Ui = Avg. initial excess pore water pressure over entire depth = ui dz ∫ 2d 0 2d

1 U t = Avg. pore water pressure over entire depth after t = u dz ∫ 2d 0 Average degree of consolidation of the whole layer after time t, Dissipated pore water pressure at time t U i − U t = f (Tv ) U = = Ui Initial pore water pressure

cv t Time factor , Tv = 2 d t is the time measured from the instant load is applied on the soil layer d drainage d, d i path th the th maximum i di t distance water t should h ld travel t l to t reach ha free draining surface.


Terzaghi’s theory of consolidation Tv =

π

U2

4 Tv = −0.933 log10 (1 − U ) − 0.085

for U < 0.6 for U > 0.6

Li it ti Limitations of Terzaghi fT hi theory of Consolidation th fC lid ti • Coefficient of consolidation, cv is assumed to be constant but in reality i changes it h with i h consolidation lid i pressure. • Difficulty in determining the drainage path, d • In field consolidation is three dimensional and lateral drainage is significant • Initial consolidation and secondary compression are neglected and in some cases they are important. • Load is not applied instantaneously, effect of loading period is to be considered. id d Dr. PVSN Pavan Kumar

Determination of coefficient of consolidation • A curve between b t di l gauge reading dial di and d time, ti t obtained bt i d from f consolidation test on a soil sample at a pressure increment. • There are two methods to determine coefficient of consolidation • Square root of time method • Logarithm of time method


Square root of time method • A curve is plotted between the dial gauge reading, R as ordinate and t as abscissa. (Curve ABCDE shown in Fig.) • Curve C b i at R0 begins at time t0 indicated byy p point A. • As load increment is applied there is an i iti l compression initial i obtained by producingg back initial linear p p part of the curve to intersect the dial ggauge g reading at point A’. This corresponds to corrected zero reading, Rc . • Consolidation between the dial gauge reading R0 and Rc is the initial compression compression. • A line A’C is drawn such that its Dr. PVSN Pavan abscissa Kumar is 1.15 times that of the initial linear portion A’B of the curve.

Square root of time method • Point C indicates 90% of degree of consolidation. • Dial gauge reading at C is R90 and corresponding absicca bi i t . is • For U = 90%, Tv = 0.848. use the following equation to determine coefficient off consolidation lid ti from f 90

Tv d 2 equation cv = equation, t

Point D for 100% primary consolidation is obtained from following equation

Rc − R100 10 = . Find R . DE is secondary compression. Dr. PVSN Pavan Kumar 100 Rc − R90 9

Square root of time method • Thickness of sample = 2cm • Double drainage t 90 = 6 . 55 ⇒ t 90 = 42 . 90 min 2 Tv d 2 0.848 × 1 cm cv = ⇒= = 0.0198 min t 42.9


Logarithm of time method Initial Compression

Primary Consolidation

Secondary Consolidation

• Curve plotted between dial gauge reading R and logt. reading, logt • Consists of three parts ‐Initial parabolic portion ‐ Middle portion linear ‐ Last portion to which horizontal axis is asymptote.

R0 be initial dial gauge reading g g g Rc be the corrected reading obtained from the initial parabolic portion. This is point A’ located such that two points B and C are at some arbitrary time t1 and 4t1 will give A’B = a and BC = a. Rc corresponds to zero consolidation. Dr. PVSN Pavan Kumar

Logarithm of time method Initial Compression

Primary Consolidation

Secondary Consolidation

• Point F corresponding to 100% consolidation is obtained from intersection of tangent, DF and EF. • Rc R100 is primary consolidation. • R100 Rf secondary compression

is

• Point M corresponding p g to 50% consolidation is obtained from followingg Rc − R50 1 = t , time corresponding to 50% consolidation equation. Rc − R100 2 50 0.196 d 2 U = 50%; Tv = 0.196; cv = t50 Dr. PVSN Pavan Kumar

Preconsolidation pressure

• Preconsolidation pressure or over consolidation pressure, σ c is the maximum pressure to which soil is subjected in the past. • Weight of overburden on a soil specimen is removed before the sample collection. • Soil S il specimen i expands d due d t a to reduction of pressure. Specimen is preconsolidated or over consolidated. • When the specimen is loaded in the consolidation test, initial portion AB of the compression curve ABCD is a Dr. PVSN Pavan Kumar recompression curve.

Preconsolidation pressure

• Initial portion AB is flat and d straight i h line, li CD with a steep slope σc indicates a normal consolidation or virgin compression. • In the transition range BC slope changes gradually. • Determine point E on the curve where radius of curvature is minimum. minimum • Draw tangent EF to the curve at E. • Draw horizontal line, EG at E. Bisect the angle between tangent EF and horizontal EG. Draw bisector EH. • Produce back the straight line portion, CD of the curve to intersect the bisector at P. Point J indicates preconsolidation pressure. ABK is a recompression curve and KCD isDr. PVSN Pavan Kumar virgin compression curve.

Causes of preconsolidation in soils Due to demolition of buildings or other structures. structures Due to overburden which is removed by erosion. Melting of glaciers which covered the soil deposit in the past. past Capillary pressure acted in the past it is destroyed due rise of water table. • Preconsolidation is due to desiccation (drying) of clay. • Sustained seepage force cause an increase of effective stress and soil is preconsolidated. • Preconsolidation is due to p past tectonic forces. • • • •


Assignment 1. 2. a) b) c) d) e) f) g) 3.

Explain the difference between consolidation and compaction p p Explain following Coefficient of compressibility, av Coefficient of Volume Change, mv Compression Index, cc D Degree of consolidation, U f lid ti U Coefficient of Consolidation, cv Time Factor Tv Time Factor, T Preconsolidation pressure Write the equations to determine the magnitude of consolidation q g settlement. 4. What are the assumptions in Terzaghi theory of consolidation 5. A clay layer 4m thickness has a final settlement of 6 cm. Layer has double drainage. Coefficient of consolidation is 0.02 cm2/min, determine a) the time required for 50% consolidation b) Settlement Dr. PVSN Pavan Kumar after 2 years

Unit V Unit V Shear Strength g Dr. PVSN Pavan Kumar Dr. PVSN Pavan

Dr.PVSN Pavan Kumar

Syllabus IImportance t off shear h strength t th – Mohr’s M h ’ ‐ coulomb l b Failure theories – Types of laboratory tests for strength parameters – Strength tests based on g conditions – strength g envelops p ‐ Shear drainage strength of sands – dilatancy – critical void ratio‐ Liquefaction – shear strength of clays. clays

Dr.PVSN Pavan Kumar

Shear strength ‐ Importance • Shear strength of a soil is its maximum resistance to shear stress just Shear strength of a soil is its maximum resistance to shear stress just before failure. Normal stress, σ

Normal stress σ Normal stress, σ Shear stress τ Shear stress, τ

Soil sample subjected to direct shear

• Soils are rarely subjected to direct shear. Shear stress developed in a soil is due to direct compression. p • Shear failure occurs if induced shear stress > shear strength g of soil. Failure occurs due to relative movement of particles. • Shear strength governs the following ‐ Bearing capacity of soil, Stability of soil slopes, Earth pressure against retaining structures Dr.PVSN Pavan Kumar

Soil sample subjected to compressive stress

Stress system with principal planes parallel to the coordinate axis • Soil mass is subjected to a three dimensional stress system. • In Engineering problems stress in the y – di direction i i not relevant. is l S Stress system is i simplified to two dimensional. • Plane strain conditions are assumed in which strain in y – direction is zero. point in stressed body, y there are • At everyy p three planes on which shear stress is zero. These planes are called Principal planes. • Plane l with h a maximum compressive stress, σ1 is called major principal plane. σ1 is major principal p p stress. • Plane with a minimum compressive stress, σ3 is called minor principal plane. σ3 is minor principal stress. Dr.PVSN Pavan Kumar

y x

z

Minor Prin. Stress

Minor Prin. Plane

Major Prin. Stress Major Prin. Plane

Stress system with principal planes parallel to the coordinate axis Major Prin. Stress Major Principal plane – Horizontal plane Major Prin. Minor Principal plane – Vertical plane Plane • Compressive stress, σ1 acts on a horizontal plane, σ3 acts on a vertical plane. • Consider a plane AB making an angle θ withMinor Prin. h i t l (major horizontal ( j principal i i l plane, l AC) in i Stress counter clock wise direction. Sum of forces acting in x‐direction Sum of forces acting in x direction on wedge on wedge Minor Prin. ABC Plane σ3 (BC x 1) (BC x 1) ‐ σ (AB x 1) sin θ (AB x 1) sin θ + τ + τ (AB x 1) cosθ (AB x 1) cosθ = 0 =0 σ3

BC − σ sin θ + τ cos θ = 0 AB

σ 3 sin θ − σ sin θ + τ cos θ = 0

‐‐‐‐‐‐‐‐ Eq. 1

Sum of forces acting in y‐direction on wedge ABC σ1 (AC x 1) ‐ σ (AB x 1) cos θ ‐ τ Dr.PVSN Pavan Kumar (AB x 1) sinθ = 0

Stress system with principal planes parallel to the coordinate axis σ1 cosθ ‐ σ cos θ ‐ τ sinθ = 0 =0

Major Prin. Stress Major Prin. Plane

‐‐‐‐‐‐‐‐ Eq. 2 Eq. 2

Eq. 1 cosθ ‐ Eq. 2 sinθ = 0

σ 3 sin θ cos θ − σ sin θ cos θ + τ cos 2 θ 1 cos θ sin θ + σ cos θ sin θ + τ sin θ = 0 −σ 2

Minor Prin. Stress

(σ 1 − σ 3 ) τ = (σ 1 − σ 3 ) sin θ cos θ = sin 2θ 2 Minor Prin. σ sin θ − σ sin θ + τ cos θ = 0 From Eq.1 Plane 3 ⎛ σ1 − σ 3 ⎞ σ 3 sin i θ − σ sin i θ +⎜ i θ cos θ cos θ = 0 ⎟2 sin ⎝ 2 ⎠ ⎛ 1 + cos 2θ ⎞ σ 3 = σ − (σ 1 − σ 3 )⎜ ⎟

2 ⎝ ⎠ ⎛ 1 + cos 2θ ⎞ σ = σ 3 + (σ 1 − σ 3 )⎜ ⎟ 2 ⎝ ⎠ σ σ cos 2θ σ 3 σ 3 cos 2θ σ 1 + σ 3 (σ 1 − σ 3 ) cos 2θ σ = σ3 + 1 + 1 − − Dr.PVSN Pavan Kumar = + 2 2 2 2 2 2

Stress system with principal planes parallel to the coordinate axis Major Prin. Stress Major Prin. Plane

• σ and τ are normal and shear stress on the plane AB inclined at angle θ in the anti clockwise direction. MOHR’s Circle • Otto Mohr developed a graphical method for determination of stresses on a plane inclined to the principal planes.

Minor Prin. Stress

Minor Prin. Plane

• Compressive stress is positive. positive Shear stress is positive if an anti clockwise moment acts at a point inside the wedge ABC.

Dr.PVSN Pavan Kumar

Stress system with principal planes parallel to the coordinate axis • Origin O is selected • Normal N l stress, t σ is i plotted l tt d along horizontal axis. • Shear stress,, τ alongg vertical axis. • Point E represents the minor principal stress, stress σ3. • Point F represents the major principal stress, σ1. • Circle is drawn with center C and EF as diameter. • Each point on the circle gives the stress σ and τ on a particular plane. • Point E is known as pole (P) or origin of planes. Stress at point D σ1 + σ 3 σ1 − σ 3 σ1 − σ 3 + σ= cos 2θ τ = sin 2θ 2 2 2 • ED is parallel to AB. Similarly stress on any other plane is Dr.PVSN Pavan Kumar obtained by drawing a line parallel to plane through E.

Stress system with principal planes parallel to the coordinate axis • CD makes an angle 2θ with horizontal. • Line OD represents resultant, R making an angle β with horizontal. • Angle of obliquity of resultant, R is β with the normal to AB.

Dr.PVSN Pavan Kumar

Characteristics of Mohr’s Circle • Maximum shear h stress, τ max

( σ1 − σ 3 ) = 2

this h occurs on plane l inclined l d at angle l

45° to the principal planes. • Point D on Mohr circle represents the stresses (σ,τ) on a plane inclined at angle θ with major Principal plane. Resultant stress on the plane is −1 ⎛ τ ⎞ R = σ 2 + τ 2 at an angle of obliquity, β = tan ⎜ ⎟ ⎝σ ⎠ • Maximum angle g of obliquity, q y, ⎛ (σ 1 − σ 3 ) ⎞ ⎟ −1 ⎜ −1 ⎛ (σ 1 − σ 3 ) ⎞ 2 ⎟⎟ β max = sin ⎜ ⎟ = sin ⎜⎜ ( ) σ + σ 3 ⎜ 1 ⎟ ⎝ (σ 1 + σ 3 ) ⎠ 2⎠ ⎝ • Shear stress, τf on a plane of maximum obliquity is τf which is less than τmax. • Mohr circle is symmetrical about σ Mohr circle is symmetrical about σ axis axis and only top half of circle can be drawn. Dr.PVSN Pavan Kumar

Stress system with principal planes inclined to the coordinate axis σ1 Major Principal stress σ3 Minor Principal stress Major Principal σ

τ

(σ,τ)

σ3

plane

Minor Principal Principal plane

σ1

Pole, P is located from intersection of lines EP and FP. PD is drawn making an Dr.PVSN Pavan Kumar angle θ with PF, major principal plane.

Stress system with vertical and horizontal planes not principal planes

• Soil element ABC has vertical and horizontal planes BC and AC are subjected to shear stress in addition to normal stress . • These planes are not principal planes • Stress S on a plane l AB inclined i li d at an angle, l θ to plane l AC is i determined d i d as follows Sum of forces in x direction Sum of forces in x direction

σ x BC + τ xy AC + τ cos θ AB − σ sin θ AB = 0 E 1 σ x sin i θ + τ xy cos θ + τ cos θ − σ sin i θ = 0 Eq. 1 Dr.PVSN Pavan Kumar


Sum of forces in y direction

σ y AC + τ xy BC − τ sin θ AB − σ cos θ AB = 0 σ y cos θ + τ xy sin θ − τ sin θ − σ cos θ = 0

Eq. 2

Eq 1 x sin θ + Eq 2 x cos θ =0

σ x sin 2 θ + τ xy sin θ cos θ + τ sin θ cos θ − σ sin 2 θ + σ y cos 2 θ + τ xy sin θ cos θ − τ sin θ cos θ − σ cos 2 θ = 0 σ = σ x sin i 2 θ + σ y cos 2 θ + 2τ xy sin i θ cos θ Dr.PVSN Pavan Kumar

Stress system with vertical and horizontal planes not principal planes ⎛ 1 − cos 2θ ⎞ ⎛ 1 + cos 2θ ⎞ σ = σ x⎜ ⎟ +σ y⎜ ⎟ + τ xy sin 2θ 2 2 ⎝ ⎠ ⎝ ⎠ ( σ x + σ y ) (σ y − σ x )cos 2θ σ= + + τ xy sin 2θ Eq. 3 2 2 Eq 1 x cos θ ‐ Eq 2 x sin θ =0

σ x sin θ cos θ + τ xy cos 2 θ + τ cos 2 θ − σ sin θ cos θ − σ y cos θ sin θ − τ xy sin 2 θ + τ sin 2 θ + σ cos θ sin θ = 0 ⎛σ y −σ x ⎞ ⎟⎟ sin 2θ − τ xy cos 2θ Eq. 4 τ = ⎜⎜ ⎝ 2 ⎠ 2 2 2 ( ) ( ) ⎤ ⎡ σ σ σ σ θ σ σ cos 2 + − − ⎡ ⎡ y ⎤ ⎛ y x y ⎤ x x ⎞ 2 ⎟⎟ sin 2θ − τ xy cos 2θ ⎥ ⎜ σ τ τ sin 2 θ − + = + + xy ⎢ ⎥ ⎢ ⎥ ⎢⎜ 2 2 ⎣ ⎦ ⎣ ⎦ ⎣⎝ 2 ⎠ ⎦

( σ =

y −σ x )

2

4

+ τ 2 xy

Dr.PVSN Pavan Kumar


( ( σ x + σ y )⎤ σ y −σ x ) ⎡ 2 2 σ − + τ = + τ xy ⎥ ⎢ 2 4 ⎦ ⎣ 2 ⎛ σ x +σ y ⎞ σ − σ ⎛ ⎞ ⎜⎜ ,0⎟⎟ d di ⎜ y x Ab Above equation is for a circle with center and radius i i f i l ih 2 ⎟ + τ xy ⎜ 2 ⎟ ⎝ 2 ⎠ ⎠ ⎝ • Normal stress, σx and σy are represented t d by b points i t Q and d R respectively on σ axis. Represents • QS = τxy represents p shear stress on plane BC p Represents planes BC. plane AB • RT = ‐ τxy represents shear stress on p Represents planes AC prin plane • Point C is at middle of QR and middle of QR and coordinates are Represents ⎛ σ x +σ y ⎞ ⎜⎜ ,0⎟⎟ plane AC plane AC ⎝ 2 ⎠ 2

2

Dr.PVSN Pavan Kumar

Stress system with vertical and horizontal planes not principal planes • Radius of circle is ⎛⎜ σ y − σ x ⎞⎟ + τ 2 xy ⎟ ⎜ 2

⎝

2

⎠

Represents plane BC

• Point S represents the stress on plane Point S represents the stress on plane BC and T on plane AC.

Represents plane AB

(σ, τ) Represents prin plane

• Draw line CD making an angle 2θ with line CT. • Point D gives stresses on plane AB. L Location of Principal planes i fP i i l l • Principal plane will not carry any shear stress, τ = 0 ⎛ σ y −σ x ⎞ ⎟⎟ sin 2θ p − τ xy cos 2θ p = 0 ⎜⎜ ⎝ 2 ⎠ 2τ xy θp is the angle which principal i h l hi h i i l tan 2θ p = plane makes with AC. Dr.PVSN Pavan Kumar σ −σ

(

y

x

)

Represents plane AC

Stress system with vertical and horizontal planes not principal planes Major principal stress is given by point U σ 1

Minor principal stress given by point V σ 3

( σ =

( σ =

x +σ y )

2

x +σ y )

2

+

−

(σ

y −σ x )

2

4

(σ

y −σ x )

+ τ 2 xy

2

4

+ τ 2 xy

CU at an angle 2θp represents major principal plane and CV at an angle 180 + 2θp represents minor principal plane. plane

Major j P.P Minor P.P

Dr.PVSN Pavan Kumar

Revised Mohr Coulomb equation • Shear strength g of soil at a p point on a particular plane is expressed as τ f = c' + σ tan ϕ ' c' = Cohesion intercept ϕ‘ = angle ϕ g of shearingg resistance. σ = Effective normal stress • Failure occurs when Mohr circle touches the failure envelope at point B. • If stresses are plotted by point A below failure envelope it represents a stable non failure condition. • State of stress represented by point C is not possible. • c’ and ϕ ϕ’ depend on number of factors such as water content, drainage conditions and conditions of testing. Dr.PVSN Pavan Kumar

Different types of tests and drainage conditions • Followingg are different tests to measure the shear strength g of soil – Direct shear test, Triaxial compression test, Unconfined compression test, Vane shear test

• Shear tests are conducted to following drainage conditions that simulate actual field problem – Consolidation stage in which confining pressure is applied to the specimen and allowed to consolidate. – Shear stage in which shear stress (or deviator stress) is applied on the specimen to shear.

• • • •

Unconsolidated undrained test (UU test) or quick test Consolidated undrained test (CU test) or R test Consolidated drained test (CD test) or slow test Strain controlled test: Test is conducted at a constant strain rate. Force is

measured with proving ring. • Stress St controlled t ll d test: t t Shear Sh stress t i increased d att a uniform if rate. t Shear Sh displacements are obtained byDr.PVSN Pavan Kumar dial gauge.

Direct shear test Upper half of shear box Upper half of shear box Locking Pins Porous Spacing Screws

Base Plate

Porous Stone

Non Porous Gi Gripper Plate Pl

Lower half of shear box Pressure Pad P P d Shear box split into two halves along a horizontal plane at middle. Box of size Dr.PVSN Pavan Kumar 60 mm x 60 mm x 50 mm

Direct shear test Fixed Fi d Support

Loading Yoke

g Proving Dial Gauge Proving Ring

Shear force application Container Roller

Gear

Loading Frame Loading Frame On/off

Strain rate

L Lever arm

Weights

Dr.PVSN Pavan Kumar

Motor

Direct shear test Di l G Dial Gauge Proving Ring Shear Box

Container On/off

Frame

Weight

Dr.PVSN Pavan Kumar

Direct shear test

Procedure: • Base plate is placed in the lower half of the shear box. • Bottom porous stone is placed in the box. box • For undrained test a plain grid is kept on the porous stone keeping segregations at right angles to the direction of shear. • For drained test perforated grids are used instead of plain grids. box • Soil sample is placed in the shear box. • Upper grid plate, porous stone and pressure pad are placed on the Dr.PVSN Pavan Kumar sample.

Direct shear test

Procedure: • Shear box is placed in a large container and mounted on the loading frame. • Upper half of box brought in contact with proving ring. • Loading yoke is mounted on the steel ball placed on pressure pad. • Locking pins are removed. Gap is created between to halves with help p of spacing p g screws. Dr.PVSN Pavan Kumar

Direct shear test Procedure: • Normal stress of 25 kN/m2 is applied on the sample. Shear load is applied at a constant rate of strain. • For undrained test – Rate of loading 1.0 mm/min to 2.0 mm/min. • For drained test ‐ sand 0.2 mm/min, Clay 0.005 to 0.02 mm/min. • Sample shears along a horizontal plane between to halves. • Readings of proving ring and dial gauge are taken at every 30 second interval. Test is continued till specimen fails. • Failure indicated by receding of proving ring reading after reaching a maximum value or for a shearing strain of 20%. • Test repeated under a normal stress of 50, 100, 200 and 400 kN/m / 2 till it covers the range of normal stress applied in the field. Dr.PVSN Pavan Kumar

Direct shear test Area of sample, A =6cm x 6cm = 36 cm2 Normal stress, σ l = 25 kN/m k / 2 Time, t in sec

Dial gauge reading, Shear Horizontal Horizontal strain = = displacement = ΔH/L ΔH, (mm)

Length of specimen, L = 6 cm h f

Proving Ring readings

Shear force, F (N) F, (N)

Shear stress, τ = F/A = F/A, (kN/m2)

0 30 60 90 120

Normal stress, σ Shear stress, τ

• Generally CD test is conducted on cohesion less soil. • Occasionally used to determine the strength parameters of silt and clay. • Triaxial T i i l test t t is i better b tt suited it d for f silt ilt and d clay. Dr.PVSN Pavan Kumar

Results of Direct shear test • In dense sand shear stress attains a maximum value at a low strain this is called peak stress. • Shear stress decreases slightly and becomes more or less constant (ultimate stress) with further increase of strain. • In loose sand shear stress increases gradually with strain and d attains an ultimate l stress.

Peak Peak stress Ultimate stress

• Stress strain curve for medium dense sand is also shown in Figure. • Failure il strain i for f dense d sand d is i 2 to 4%. % Loose sand d 12 to 16%. %

Dr.PVSN Pavan Kumar

Results of Direct shear test • Volume of dense sand decreases initially and further increases with shear strain. strain • In loose sand volume decreases with increase of shear strain. • Void ratio decreases with increase of shear strain for loose sand. • Void ratio of an initial dense sand increases with increase of shear strain. • Void ratio at which there is no change in it with increase of shear strain is Dr.PVSN Pavan Kumar known as critical void ratio i.e. no change in volume of soil.

Results of Direct shear test • Failure envelope is obtained by plotting points of different shear strength corresponding to different normal stress. • Inclination of failure envelope with horizontal is angle of shearing resistance, ϕ its intercept with vertical axis is cohesion intercept, intercept c. c • For dense sand failure envelop is drawn for peak and ultimate stress. stress

Dr.PVSN Pavan Kumar

Merits of Direct Shear test • Sample preparation is easy, simple and convenient. • Sample thickness is small, drainage is quick, pore water pressure dissipate easily. CD and CU tests take relatively small time. • Suited to conduct drained tests on cohesion less soils. • Apparatus is cheap.

Demerits of Direct Shear test f h • Stress conditions are known only at failure, conditions prior to failure and intermediate are not known. • Orientation of failure plane is fixed and cannot be changed. • Measurement of pore water is not possible. • Side walls of shear box cause lateral restraint on the specimen and do not allow it to deform laterally not allow it to deform laterally.

Dr.PVSN Pavan Kumar

Different types of soil Soils are of three types • Cohesion less soils – Soils do not have cohesion, c’=0, derive shear strength from angle of shearing resistance, ϕ. Ex: Sand, Gravel. • Purely cohesive soil – Soils exhibit cohesion and angle of shearing resistance, ϕ = 0. Ex: Saturated clays and silt under drained condition. • Cohesive Frictional soil – Soil have both c’ and ϕ’. Ex: Clayey sand, Silty sand, Sandy clay etc.

Dr.PVSN Pavan Kumar

Triaxial compression test σ σ c

d

= σ1

+ σc = σ3

σc = σ3

+ C Consolidation Stage lid i S

Shearing Stage

All around the soil sample a confining • Pressure on the sides of l remains i same as σc = σ3 sample pressure, σc = σ3 is applied • Deviator stress, σd is applied vertically on top and bottom plane of sample till it fails in shear shear. σ1 = Major Principal Stress = σ M j P i i l St c + σd Dr.PVSN Pavan Kumar σ3 = Minor Principal Stress = σ c

Result of triaxial compression test c = Cohesion C h i

τ

ϕ = Angle of internal friction

Sample 3 Sample 2 p

Sample 1

ϕ

c σ31

σ32

σ11

σ33 σ12

Dr.PVSN Pavan Kumar

σ13

σ

Triaxial Compression test Triaxial Cell

Sampling Sampling tube

Membrane stretcher

Loading Cap Porous and non porous stones

Triaxial cell Dr.PVSN Pavan Kumar

Triaxial Compression test

Triaxial cell Dr.PVSN Pavan Kumar

Triaxial Compression test

St l t b Steel tube Extrusion of soil sample soil sample

Undisturbed U di t b d sampling tube

Collection of undisturbed samples Diameter of sample = 37.5 mm Diameter of sample = 37 5 mm Height of sample = 75 mm Dr.PVSN Pavan Kumar

Triaxial Compression test Pore P pressure gauge

Volume V l Change measurement

Proving Ring Dial gauge Dial gauge Triaxial Cell with sample and filled with water

Load frame

Constant pressure system pressure system Dr.PVSN Pavan Kumar

Triaxial Compression test Procedure: • A porous stone is placed on the pedestal. Filter paper is placed on the porous stone. p • Soil specimen of 37.5 mm diameter and 75 mm height is placed on the filter paper. • A porous stone is placed on the specimen. Sample is enclosed in a rubber membrane slid over the specimen with the help of membrane stretcher. • Membrane is sealed to the specimen with O – rings. • Triaxial cell is placed over the base and fixed to it by tightening the nuts. • Cell is filled with water by connecting it to pressure supply. Some portion on top of cell is filled by injecting oil through the oil valve. • When Wh excess oilil begins b i to t spill ill through th h the th air i ventt valve l both b th valves l are closed. Dr.PVSN Pavan Kumar

Triaxial Compression test • Pressure is applied to the water by connecting triaxial cell to constant pressure system. As soon as pressure acts on the specimen it starts consolidating. g • Specimen connected to the burette through pressure connections for measurement of volume changes. Consolidation is complete when there is no more volume change. • In case of undrained test non porous stones are to be used for no consolidation of sample. • When consolidation is complete specimen is ready for to be sheared. • In case of undrained test, drainage valve is closed and soil sample connected to pore water pressure measurement device. • In case of drained test the drainage valves are kept open. • Force applied on the sample is measured by proving ring and axial d f deformation ti by b a dial di l gauge. Dr.PVSN Pavan Kumar

Triaxial Compression test • Sample is sheared by applying deviator stress by loading machine. machine • Proving ring readings are taken corresponding to axial strains of 1/3%, 2/3% 1%, 2/3%, 1% 2%, 2% 3%, 3% 4%, 4% 5% …. until failure or 20% axial strain. strain • Mass, length and water content of sample post shear is to be determined. Merits • Complete control on the drainage conditions. • Pore pressure changes and volumetric changes can be measured directly. y • Sample is free to fail on the weakest plane, stress distribution is uniform along failure plane. • State of stress at all intermediate stages is known up to failure. Mohr circle can be drawn at any stage of shear. • Apparatus is suitable to conduct extension test and tests for different Dr.PVSN Pavan Kumar stress paths.

Triaxial Compression test Demerits: • Apparatus is elaborate, costly and bulky. • Drained tests take a longer period compared to a direct shear test. test • Strain conditions are not uniform due to frictional resistant produced by loading cap and the pedestal. pedestal This leads to formation of dead zones at the end. Non uniform distribution of stress is eliminated by lubrication of end surfaces. If length/diameter g / ratio ≥ 2. • At large strains specimen may not remain cylindrical. • Test simulates onlyy axis – symmetrical y problem. In field the p p problem is three dimensional. • Consolidation of specimen is isotropic. In field consolidation is anisotropic.

Dr.PVSN Pavan Kumar

Triaxial Compression test Results

O.C.C ‐ Over Consolidated clay; N.CC – Normally consolidated clay

Dr.PVSN Pavan Kumar

Unconfined compression test • It is special form of triaxial compression test in which confining pressure, σc = 0. • Test conducted on intact saturated clayy samples. • There are two plates with conical seatings for the specimen. • Specimen placed on the bottom plate so that it makes a contact with upper plate. Dial • Dial gauge and proving ring are set to zero. ggauge g • As the machine is switched on bottom plate moves upward to compress the soil. • Strain rate is 1/2% to 2% per min. • Shearing is continued till the specimen fails or 20% strain t i occurs. Dr.PVSN Pavan Kumar

Proving ring

Soil Sample

Unconfined compression test • Compression force is determined by proving ring and axial strain from dial gauge reading. • Axial stress at which a specimen p fails is known as unconfined compressive strength, qu. • Stress strain curve can be plotted. • Corrected area of sample is determined as follows,

A0 A= 1− ε A0 = Initial area, ε = Axial strain • Mohr circle is drawn passing through origin. • Minor principal stress, σ3 = 0. p is horizontal, ϕu = 0 • Failure envelope q • Cohesion intercept, cu = u 2 Dr.PVSN Pavan Kumar

Failure envelope

Vane shear test (IS 2720 Part 30 1980)Torque application • Undrained shear strength of soft clay is determined by a laboratory vane shear test. • Field vane shear test is conducted in the bottom of a bore hole. • Apparatus consists of a vertical steel rod having four thin stainless steel blades (vanes) fixed to its bottom end • Height of vane is twice the diameter. • For conducting d laboratory test specimen of size 38 mm diameter and 75 mm height is taken in a container.

Torque Torque Applicator

Rod Vanes Soil Cylinder

Springs Dr.PVSN Pavan Kumar

Steel rod

Steel blades (Vanes)

Vane shear test (IS 2720 Part 30 1980) • Vane is gradually lowered into the specimen till the top is at a depth of 10 to 20 mm below the soil. • Torque is applied gradually to the upper end of rod at the rate of about 6° per minute. • Torque acting on the specimen is i di indicated d by b a pointer i fi d to the fixed h spring. • Torque is indicated till the specimen fails in shear. • Shear strength of soil is constant on the cylindrical sheared surface and top and y bottom faces of a sheared cylinder. Dr.PVSN Pavan Kumar

Vane shear test (IS 2720 Part 30 1980) Torque, T applied Torque T applied = Resisting torque at sides (T1) + Torque at top and bottom (T2) D/2

T = (sπDH )× D + 2 ∫ s 2πrdr r 2 0

T s= ⎡ D 2 H D3 ⎤ π⎢ + ⎥ 2 6 ⎣ ⎦ S = Shear strength; T = Torque; D = Diameter of Vane; H = Height of vane • Part of top of vane is above the sample and a depth of H1 is inside the sample sample. T

s=

⎡ D 2 H1 D 3 ⎤ π⎢ + ⎥ 2 12 ⎣ ⎦

Dr.PVSN Pavan Kumar

Vane shear test (IS 2720 Part 30 1980) • Vane shear test is used to determine the sensitivity of the soil. y • After an initial test vane is rotated rapidly through several revolutions and soil becomes remoulded. Undisturbed shear strength sensitivity = Re moulded shear strength

Merits • Test is simple and quick, used to determine in situ shear strength of p q , g non fissured fully saturated clay. • Convenient to determine the sensitivity of soil. Demerits • Test cannot be conducted on fissured clay or clay containing sand or silt.

Dr.PVSN Pavan Kumar

Geotechnical Engineering

Geotechnical Engineering

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