ÖÙNG SUAÁT TRONG NEÀN ÑAÁT Baøi giaûng A. Prof. Dr Chaâu Ngoïc AÅn ÖÙNG SUAÁT TRONG NEÀN ÑAÁT

ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG Haït theùp nöôùc Ñaát baõo hoøa nöôùc

ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn ÖÙNG SUAÁT HÖÕU HIEÄU VAØ AÙP LÖÏC NÖÔÙC LOÃ ROÃNG Haït theùp nöôùc Ñaát baõo hoøa nöôùc * bình beân traùi theâm vaøo treân maët lôùp ñaát caùc haït theùp taïo moät aùp löïc p, maãu ñaát bò luùn xuoáng. AÙp löïc p coù aûnh höôûng leân öùng suaát khung neân laø öùng suaát höõu hieäu, kyù hieäu laø ’ * bình beân phaûi theâm nöôùc treân maët ñeå taïo aùp löïc p, maãu ñaát khoâng luùn xuoáng vì nöôùc theâm vaøo thoâng vôùi nöôùc trong loã roãng taùc ñoäng leân ñaùy bình chöùa, p do coät nöôùc khoâng aûnh höôûng leân khung haït (trung hoøa).

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn ÖÙng suaát  taïi moät ñieåm trong neàn ñaát goàm “öùng suaát giöõa caùc haït” hay öùng suaát höõu hieäu ’ vaø aùp löïc nöôùc trong loã roãng u, theo ñònh ñeà Terzaghi  = ’ + u

TÍNH ÖÙNG SUAÁT TRONG ÑAÁT NEÀN DO TROÏNG LÖÔÏNG BAÛN THAÂN Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn TÍNH ÖÙNG SUAÁT TRONG ÑAÁT NEÀN DO TROÏNG LÖÔÏNG BAÛN THAÂN ÖÙng suaát toång do troïng löôïng baûn thaân ñaát theo phöông thaúng ñöùng kyù hieäu laø bt hay v taïi moät ñieåm baát kyø trong ñaát caùch maët ñaát moät chieàu saâu baèng H, coù theå tính nhö laø troïng löôïng khoái ñaát beân treân truyeàn xuoáng.

trong ñoù  laø heä soá Poisson. Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn vôùi K0 laø heä soá aùp löïc ngang ôû traïng thaùi tónh cuûa ñaát coá keát thöôøng (ñaëc ñieåm coá keát thöôøng vaø coá keát tröôùc seõ ñöôïc phaân tích trong caùc chöông sau). Khoaûng nöûa theá kyû tröôùc, heä soá aùp löïc ngang ñöôïc vay möôïn töø lyù thuyeát ñaøn hoài vôùi kyù hieäu laø  vaø coù daïng : trong ñoù  laø heä soá Poisson. Vôùi toång keát töø raát nhieàu keát quaû thí nghieäm vaø ño ñaïc giaùn tieáp, Jaky ñaõ ñöa ra moät coâng thöùc ñeå tính heä soá aùp löïc ngang ôû traïng thaùi tónh (cuûa ñaát coá keát thöôøng) nhö sau : K0 = 1 - sin’ Vôùi ’ laø goùc ma saùt trong ñieàu kieän caét thoaùt nöôùc (seõ phaân tích roõ trong chöông choáng caét). Coâng thöùc cuûa Jaky phuø hôïp cho ñaát rôøi hoaëc ñaát loaïi caùt. Neáu goùc ma saùt ’= 350 thì K0 = 1 – sin350 = 0,426 Ñoái vôùi ñaát dính hoaëc ñaát loaïi seùt coá keát thöôøng, Alpan ñeà nghò moät coâng thöùc thöïc nghieäm. K0 = 0,19 + 0,233logIP Neáu moät maãu seùt coù chæ soá deûo IP = 20, thì heä soá aùp löïc ngang ôû traïng thaùi tónh K0 theo coâng thöùc Alpan laø : K0 = 0,19 + 0,233log20 = 0,493

Seùt, c2 Seùt, c1 Caùùt, s2 Caùùt, s1 V hW hS1 hC1 hS2 hC2 ’V u Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Seùt, c2 Seùt, c1 Caùùt, s2 Caùùt, s1 V hW hS1 hC1 hS2 hC2 ’V u

’ z u hS1 hW1 BS Caùùt, S1 BC hC1 Seùt, C1 hS2 hW2 Caùùt, S2 hC2 Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn ’ z u hS1 hW1 BS Caùùt, S1 BC hC1 Seùt, C1 hS2 hW2 Caùùt, S2 hC2 Seùt, C2

Nguyeân lyù ño aùp löïc nöôùc loã roãngtrong ñaát Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Nguyeân lyù ño aùp löïc nöôùc loã roãngtrong ñaát

Soil loaded by an applied weight W Soil loaded by water weighing W W W

Soil loaded by an applied weight W Soil loaded by water weighing W W W Compression No deformation

Definition of Total and Effective Stress (1)

Definition of Total and Effective Stress (1) Effective vertical stress (2)

Definition of Total and Effective Stress (1) Effective vertical stress (2)

Definition of Total and Effective Stress (1) Effective vertical stress (2)

Effective Stress N T Fig 2 Two Pieces of Rock in Contact Water pressure uw Fig 2 Two Pieces of Rock in Contact

Effective Stress Effective Force N (3a) T U = uw ( A - Ac ) Water pressure uw (3a) U = uw ( A - Ac )

Effective Stress Effective Force N (3a) T U = uw ( A - Ac ) Water pressure uw (3a) U = uw ( A - Ac ) Frictional Failure

Effective Stress Effective Force N (3a) T U = uw ( A - Ac ) Water pressure uw (3a) U = uw ( A - Ac ) Frictional Failure Failure in terms of stress (3b)

Calculation of Effective Stress Surcharge q d1 Layer 1 Layer 2 z d2 Layer 3 d3 Fig 3 Soil Profile

Calculation of Total Vertical Stress Elevation q Force on base = Force on top + Weight of soil z d1 z d2 Plan A

Calculation of Total Vertical Stress Elevation q Force on base = Force on top + Weight of soil A sv = A q + A g1 d1 + A g2 d2 + A g3 ( z - d1 - d2 ) z d1 z d2 Plan A

Calculation of Total Vertical Stress Elevation q Force on base = Force on top + Weight of soil A sv = A q + A g1 d1 + A g2 d2 + A g3 ( z - d1 - d2 ) sv = q + g1 d1 + g2 d2 +g3 ( z - d1 - d2 ) z d1 z d2 (4) Plan A

Calculation of pore water pressure Water table H P (5) Fig 4 Soil with a static water table

Calculation of pore water pressure Water table H P (5) Fig 4 Soil with a static water table The water table is the level of the water surface in a borehole.

Calculation of pore water pressure Water table H P (5) Fig 4 Soil with a static water table The water table is the level of the water surface in a borehole. It is the level at which the pore water pressure uw = 0

Example: determining the effective stress Step 1: Draw ground profile showing soil stratigraphy and water table Dry 2 m Saturated 3m Fig 5 Soil Stratigraphy

Example Step 2: Calculation of relevant bulk unit weights Voids Solid Vv=e Vs = 0.7m3 Voids Solid Vs= 1m3 Distribution by Volume

Example Step 2: Calculation of relevant bulk unit weights Voids Solid Vv=e Vs = 0.7m3 W = 0 Voids Solid Vs= 1m3 Distribution by Volume Distribution by weight for the dry soil

Example Step 2: Calculation of relevant bulk unit weights Voids Solid Vv=e Vs = 0.7m3 W = 0 Voids Solid Vs= 1m3 Distribution by Volume Distribution by weight for the dry soil Distribution by weight for the saturated soil

Example Step 2: Calculation of relevant bulk unit weights Voids Solid Vv=e Vs = 0.7m3 Ww=0 Voids Solid Vs= 1m3 Distribution by Volume Distribution by weight for the dry soil Distribution by weight for the saturated soil 26 . 46 kN G g g = = 15 . 56 kN / m 3 = s w dry 1 . 70 m 3 1 + e

Example Step 2: Calculation of relevant bulk unit weights Voids Solid Vv=e Vs = 0.7m3 Ww=0 Voids Solid Vs= 1m3 Distribution by Volume Distribution by weight for the dry soil Distribution by weight for the saturated soil

Example Step 3 Calculate total stress 2 m 3m

Example Step 3 Calculate total stress 2 m Step 4 Calculate pore water pressure 3m

Example Step 3 Calculate total stress 2 m Step 4 Calculate pore water pressure 3m Step 5 Calculate effective stress

Vertical stress and pore pressure variation 50 100 150 kPa 0m 2m 4m Total Stress (5m) pore water pressure 6m Effective stress Depth 8m

Stresses acting on a soil element z z y x x Fig 7 Definition of Stress Components

Principle of Effective Stress Effective stress relations for general stress states (10)

Example 1 m Initial GWL 3 m Lowered GWL z Clay Rock aquifer

Example 1 m Initial GWL 3 m Lowered GWL z Clay Rock aquifer

Example Effective stress increases - soil compresses - ground surface settles Effective stress decreases- soil swells - ground surface heaves. The following problems may then occur surface flooding flooding of basements built when GWL lowered uplift of buildings failure of retaining structures failures due to reductions in bearing capacity

ÖÙNG SUAÁT TRONG NEÀN ÑAÁT DO TAÛI NGOAØI Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn ÖÙNG SUAÁT TRONG NEÀN ÑAÁT DO TAÛI NGOAØI

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Tröông töï, taïi M1 caùch M moät ñoaïn dR, coù chuyeån vò S1 Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn c1 P a1 a c R  d R Döôùi taùc duïng cuûa P ñieåm M chuyeån vò moät ñoaïn S theo phöông baùn kính R. M caøng xa O thì S caøng nhoû. Maët khaùc, vôùi R = const, goùc  caøng lôùn thì S cuõng caøng nhoû. Xuaát phaùt töø nhaän xeùt ñoù, ta coù theå vieát bieåu thöùc S coù daïng : Tröông töï, taïi M1 caùch M moät ñoaïn dR, coù chuyeån vò S1

Bieán daïng töông ñoái R cuûa ñoaïn dR laø: Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Bieán daïng töông ñoái R cuûa ñoaïn dR laø: Boû qua R.dR vì raát nhoû so vôùi R2 Theo giaû thuyeát quan heä giöõa öùng suaát vaø bieán daïng laø tuyeán tính do ñoù öùng suaát xuyeân taâm R gaây neân bieán daïng R ñöôïc xaùc ñònh nhö sau Trò soá A.B coù theå xaùc ñònh döïa theo ñieàu caân baèng tónh hoïc. Xeùt ñieàu kieän caân baèng tónh hoïc cuûa baùn caàu (O; R)

Trong ñoù: dF – dieän tích maët ñai troøn caa1c1 dF = 2(Rsin)(Rd) Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Trong ñoù: dF – dieän tích maët ñai troøn caa1c1 dF = 2(Rsin)(Rd)

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn

Chuyeån vò theo chieàu caùc truïc : Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Chuyeån vò theo chieàu caùc truïc :

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn

Baøi toaùn Mindlin y A x R1 O z M(x,y,z) r h R2 P Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Baøi toaùn Mindlin y A x R1 O z M(x,y,z) r h R2 P

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ÑÖÔØNG LÖÏC HAY LÖÏC ÑÖÔØNG THAÚNG Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn ÑÖÔØNG LÖÏC HAY LÖÏC ÑÖÔØNG THAÚNG ÖÙng suaát do aûnh höôûng cuûa löïc daïng ñöôøng thaúng phaân boá ñeàu (kM/m) nhö: ñöôøng raây; töôøng chòu löïc trong neàn ñaát, … ñöôïc Flamant phaùt trieån töø baøi toaùn Boussinesq (1892) baèng caùch chia ñöôøng löïc thaønh voâ soá löïc taùc ñoäng pdy leân moät ñoaïn thaät ngaén dy, aùp duïng coâng thöùc Boussinesq cho löïc nhoû naøy roài tích phaân leân caû chieàu daøi taùc ñoäng löïc ñeå coù ñöôïc caùc coâng thöùc sau:

Taûi phaân boá ñeàu treân dieän tích baêng (Flamant) Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Taûi phaân boá ñeàu treân dieän tích baêng (Flamant) Taûi phaân boá treân dieän tích baêng laø daïng raát thöôøng gaëp trong neàn moùng coâng trình nhö: moùng baêng, ñöôøng, ñeâ,… Khaûo saùt moät ñoaïn dx trong phaïm vi töø -b/2 ñeán +b/2, giaù trò taûi töông öùng laø pdx töông töï moät ñöôøng löïc, tính öùng suaát dz do ñöôøng löïc pdx gaây ra vaø ñoåi bieán soá sang goùc nhìn töø M veà ñaùy moùng. p M

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Trò soá 2 laáy vôùi daáu döông khi ñieåm M naèm ngoaøi phaïm vi hai ñöôøng thaúng ñi qua hai meùp cuûa taûi troïng

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Tröôøng hôïp ñôn giaûn nhaát laø ñoái vôùi caùc ñieåm naèm treân maët chöùa Oz (ñi qua truïc taâm taûi troïng). Vì tính ñoái xöùng cho neân : 1 = 2 =  2

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Taûi phaân boá ñeàu treân dieän tích chöõ nhaät Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn Taûi phaân boá ñeàu treân dieän tích chöõ nhaät ÖÙng suaát thaúng ñöùng Z cuûa ñieåm naèm treân truïc thaúng ñöùng ñi qua taâm dieän chòu taûi, ôû ñoä saâu z:

Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn ÖÙng suaát thaúng ñöùng z do taûi phaân boá ñeàu treân dieän chòu taûi chöõ nhaät, doïc truïc thaúng ñöùng beân döôùi ñieåm goùc dieän chòu taûi. Hoaëc

PHÖÔNG PHAÙP ÑIEÅM GOÙC. Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn PHÖÔNG PHAÙP ÑIEÅM GOÙC. A B C D M A1 B1 C1 D1 öùng suaát (z) cuûa dieän chöùa taûi ABCD döôùi ñieåm M laø toång caùc öùng suaát goùc M cuûa caùc dieän MA1AD1; MB1BA1; MC1CB1; MC1DD1. Z,M = p[kg(MA1AD1) + kg(MB1BA1) + kg(MC1CB1) + kg(MC1DD1)] p laø aùp löïc phaân boá ñeàu treân dieän chòu taûi ABCD.

Z,M = p[kg(MD2AB2) - kg(MD2DC2) - kg(MA2BB2) + kg(MA2CC2)] Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn A B C D M B2 C2 D2 A2 öùng suaát (z) cuûa taïi M do dieän chöùa taûi ABCD taùc ñoäng laø toång ñaïi soá caùc öùng suaát goùc M cuûa caùc dieän MA2CC2; MD2DC2; MD2AB2; MA2BB2. Trong ñoù chæ coù dieän tích ABCD laø chöùa taûi phaân boá ñeàu p, phaàn dieän tích coøn laïi khoâng coù taûi. Z,M = p[kg(MD2AB2) - kg(MD2DC2) - kg(MA2BB2) + kg(MA2CC2)] p laø aùp löïc phaân boá ñeàu treân dieän chòu taûi ABCD.

TÍNH ÖÙNG SUAÁT THEO PHÖÔNG PHAÙP THAÙP LAN TOÛA Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn TÍNH ÖÙNG SUAÁT THEO PHÖÔNG PHAÙP THAÙP LAN TOÛA Ñôn giaûn hôn thaùp lan toûa coù ñoä doác 2:1 öùng suaát z ôû ñaùy thaùp lan toûa coù daïng:

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Baøi giaûng Prof. Dr. Chaâu Ngoïc AÅn z M 2 1 A B C x 2b p z  M A B C x

Ñoái vôùi ñaát caùt thöôøng coù caáu truùc haït vaø ôû ba traïng thaùi rôøi, chaët trung bình vaø chaët. Duø ôû traïng thaùi naøo, dieän tích tieáp xuùc giöõa caùc haït heát söùc nhoû so vôùi dieän tích xung quanh cuûa caùc haït. Caùc dieåm tieáp xuùc haït trong caáu truùc haït (ñaát caùt) Nhö tyû leä dieän tích tieáp xuùc thöôøng gaëp cuûa caùt laø 0,03%, khi taùc duïng moät öùng suaát phaùp 100kPa thöøng öùng suaát thöïc taùc ñoäng leân dieän tích tieáp xuùc laø 330 Mpa. Vaø vôùi aùp löïc naøy thì nöôùc lieân keát taïi caùc ñieåm tieáp xuùc seõ bò ñaåy khoûi voû nöôùc, do vaäy coù ñieåm tieáp xuùc raát toát giöõa caùc haït.

Stresses in a Soil Mass

How are stresses produced? Geostatic stresses total stress effective stress pore water pressure Additional stresses surface loads foundation embankment vehicle

Effects of stresses Geostatic stresses Additional stresses soil compresses or consolidates based on geostatic stress levels Additional stresses produces additional strain in soil which causes settlement under point of load.

Normal and Shear Stresses on a Plane Use methods learned in Mechanics of Materials The normal stress on any plane is The shear stress on any plane is

Normal and Shear Stresses on a Plane Use Mohr’s circle method to graphically depict stresses: orientation, and magnitude.

Normal and Shear Stresses on a Plane The major principal stress is The minor principal stress is

Pole Method of Finding Stresses on a Plane Also called “Origin of Planes” Draw a line from a known point on the Mohr’s circle parallel to the plane on which the state of stress acts. The point of intersection of this line with the Mohr’s circle is called the pole. To find the state of stress on any other plane, draw a line parallel to the plane of interest through the pole. State of stress is the intersection of this line with Mohr’s circle.

Basics of Surface Loads Categorize into groups based on areal extent infinite extent (fills, surface surcharge) finite extent point load line load strip load linearly increasing load uniformly loaded circular area rectangularly loaded area

Basics of Surface Loads Load produces stress and strain Stress and strain occur in all directions Commonly focus only on vertical stress increase Analysis based on elastic theory isotropic, homogeneous material linear elastic behavior (spring-like) material comprises a half-space

Stress Caused by a Point Load First solved by Boussinesq (1883) Stress is maximum nearest applied load and diminishes at distances away.

Stress Caused by Line Load Line load of magnitude q/unit length acts on half-space surface Vertical stress increase is:

Superposition This principle works because this system is linear elastic, isotropic, and homogeneous. This procedure greatly simplifies subsurface stress analysis. Merely add up each separate component.

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