Soil Mechanics II 土の力学II

Hiroyuki Tanaka

田中洋行

Soil Mechanics

• Geotechnical Engineering

• Meaning of “Geo” The earth, or Ground

• Geology

• Geo-sciences, chemistry, graphy, so on.

• This lecture is proceeding based on “土質力学入門”, written by Prof. 三田地利之.

Shear strength

Evaluation of strength

Compression Small tensile strength

Concrete

Bending Moment

Steel

P 119

Page number of the text book

Criteria for Soil Vertical Force: N

Shear Force: S

Normal Stress: σn = N/A A: Cross Area Shear Stress: τ=S/A

τ=c+σntanφ Coulomb’s Criteria

c

φ

Stable

Unstable

Impossible

Boundary Criteria

τ

σ

C: cohesion, φ: friction angle

P129

Shear and Normal Stresses

σ and τ are changed according to an angle

We can find α for τ = 0 When τ = 0, we call this plane “Principal” Plane. Principal stresses: The maximum and minimum principal stresses: σ1 and σ3

P121-122

Mohr’s Stress Circle

Principal Stress

σ1

σ1

σ1

σ3

σ3 σ3

σ

σ

σ τ

τ

τ

P122-123

τ

σ

P125

300

100 100

σ

τ

50

50

300

50

50

Positive σ: compression τ: anti-clockwise

How to draw the Mohr’s circle

Measuring parameters c, φ

• Laboratory Test – Sampling from a borehole

– Direct Shear Test

– Triaxial (Unconfined compression) Test

• In situ Test – No sample

– Vane Test

– Standard Penetration Test (N value)

P129, 8-12

Direct Shear Test

Failure envelopment τ=c+tanφ

σ τ

Merit: Easily understand Demerit: stress and strain are not uniform Control of drainage is difficult

P132

Triaxial Test

Tri: Three Axial: Axis Piston

Cell

Specimen

Rubber Membrane

Pressure Gauge

bure

tte

No shear stress because of water Principal plane Lateral Pressure, Cell pressure

Deviator stress (σ1-σ3)

P134

σ2=σ3

σ1

σ3

α

σ1

σ3

σ1

σ3

Ⅰ Ⅰ

Plane acting the maximum principal stress

Pole

α α

Failure point

P127

Failure criterion of Mohr and Coulomb In stead of using σ and τ, the failure criterion is presented by σ1 and σ3

τ

σ σ1 σ3

φ c

(σ1-σ3)/2

(σ1+σ3)/2 c cotφ

σ1-σ3=(σ1+σ3)sinφ+2c・cosφ P99 Mohr and Coulomb’s criteria

Three conditions by Drainage

• Unconsolidated Undrained (UU) Consolidation Shear

• Consolidated Undrained (CU)

• Consolidated Drained (CD)

Principle of Effective Stress

The behavior including the strength is governed by the effective stress. σ’=σ-u σ’: Effective stress, σ: Total stress

P130-132

Performance of UU Test

σ’3

τ

σ’1

σ’: Effective stress

σ: Total stress

τ Failure envelop : φ=0

Cu or Su Apparent cohesion, or undrained shear strength

P135

Fully Saturated

Unconfined Compression Test Sometimes called Uniaxial Test

σ: Total stress

τ Failure envelop : φ=0

σ3=0

σ1: At failure, we call this strength qu unconfined compression strength

σ3 qu

cu

Cu=qu/2 P140-142 In practice, cu is called “cohesion”, or apparent cohesion.

Young Modulus, E50 and Sensitivity

Sensitivity = qu/qr

P140-141

Performance of CU Test

σ’

τ φ’

C’

After consolidation, σ3 = σ’3

Pore pressure generated by shearing

Undrained shear strength

su

P136-137

Incremental Undrained Shear Strength

σ’, p

su

After consolidation, σ3 = σ’3

If Normally consolidated Su/p: constant

For Japanese clays, su/p=0.3～0.35

P136-137

Performance of CD Test

σ’

τ φ’

C’

σ’3 does not change during shearing

Total and effective stresses are always the same because of no excess pore water pressure

P138-139

Effective stress and Total stress analysis

• The effective stress analysis (ESA) seems more reasonable.

• Permeability is high (sandy soil), the ESA is applicable. Sand

• For clayey soil, effective stress or pore water pressure is unknown. Total stress analysis, in an other word, φ=0 method Clay

P137

Shear strength for Total stress analysis

• Undrained shear strength

• For low OCR, i.e., negative dilatancy, undrained shear strength (UC or UU test) always is smaller than the drained shear strength.

• For long consolidation, the increase in the undrained shear strength can be expected (CU test)

Su measured by in situ test Vane test

P142

Dilatancy

• Volume change during shearing • Performance is changed under undrained or

drained conditions – Undrained:

• No volume change ∆V=0 • Pore water pressure change ∆u

– Positive: Negative Negative: Positive

– Drained: • No Pore water pressure u=0 • Volume change ∆V

– Positive: Expand Negative: Compressive

P144

Critical Void Ratio Void ratio, e

loose，γ :small

large small

dense，γ: unit weight large

Dilatancy Positive Negative

Drained：Volume change

Undrained：Pore water pressure

Positive

Compression

Negative

Expand

ecrit

P 144-147

Pore water Pressure

σ3

σ1

∆u=B{∆σ3+A(∆σ1-∆σ3)}

Excess pore water pressure: Deviator Stress: shear

Skempton’s A and B Coefficient: B=1 for saturated soil, A: dependent on Dilatancy

P151

Dilatancy for sand and clay

e: large, loose e: small, dense

Dilatancy Positive Negative

Drained：Volume change

Undrained：Pore water pressure

Positive

Compression

Negative

Expand

ecrit

Sand

Clay OCR: Over-Consolidation Ratio OCR=pc/pvo: pc=preconsolidation pressure pvo= the current pressure

OCR: High Heavily OverConsolidated

OCR:1～2 Normally Slightly

Skempton’s A: Low or negative A: Hight

P 144-147, 150-151

Drained and Undrained strength

σ’

τ φ’

C’

CD CU: Positive Dilatancy

CU: Negative Dilatancy

Liquefaction Liquid

σ3

σ1

σ’3=σ3-u

increase constant

If σ’3=0, fully liquefied

P147-149

Counter measurements for liquefaction

• Densification – Positive dilatancy. No positive water pressure

– Vibration

• Lowering the ground water table – No water

• Stabilized with cement – Cohesion τ=c’+(σ-u)tanφ

Earth Pressure

Earth Pressure Relating the movement of a wall

Passive Earth Pressure

Active Earth Pressure

Earth pressure at rest

Eart

h Pr

essu

re

Retaining wall P160, 169

How to calculate the earth pressure?

• Rankine’s Method – Plastic equilibrium

– Theoretically, limitation of its application

• Coulomb’s Method – Stability of soil mass

– Trial calculation, more extensive application

P161, 170

Rankine method

σ

τ φ’

C’

σv=γz: γ=unit weight of soil, z=depth

Passive State

Active State

σh

σv

σa: Active σp: Passive

P161

Rankine method

σ

τ φ’

C’=0

σa: Active σp: Passive

γz

(σp-γz)/2

(σp+γz)/2

21 sin tan (45 )1 sin 2

ppK

zσ φ φγ φ

−= = = +

+sin p

p

zz

σ γφσ γ

−=

+

2tan (45 )2

aaK

zσ φγ

= = −

Using half angle formulae

Important Value: φ=30°Ka=1/3, Kp=3.0

Earth Pressure at Rest

σh

σv For conventional ground

σv>σh

ho

vK σ

σ= Ko: Coefficient of earth pressure at rest

Depending on OCR. For NC (OCR=1） Ko=1-sinφ

P169

Earth Pressure acting on the wall

Total Pressure

Center of Gravity

2 2 21 1tan (45 ) tan (45 )2 2 2 2

a t tP H H Hφ φγ γ= ° − = ° −

P166

Rankine method φ=0

σ

τ

C=(qu/2)

σa: Active

σp: Passive

γz

2c

2c

σa=γz-2c σp=γz+2c

For Clay

With surcharge

Change in properties or existence of water table

Hydraulic pressure Earth pressure (Effective stress)

Coulomb’s Method

Force Polygon

δ: Friction between the ground and the wall in Rankine’s theory, cannot take account.

β

Pa

Max. Pa

P170-174

0 m

φ=30ﾟ

Water Level

9m

3m

12m

φ=30ﾟ

γ=20kN/m3

γ’(γsub)=10kNm3

Sheet pile

Tie Rod Force

Embedded Depth

Active Earth Force

Passive Earth Force

①

②

③

la

lp

② x la<③ x lp

Check for Embedded Depth

0 m

φ=30ﾟ

Water Level

9m

3m

12m

Active Earth Pressure

Passive Earth Pressure

The embedded depth is enough?

①

② ③

Pa1=(20 x 3)/2=30kN/m

Pa2=(20 x 9)=180kN/m Pa3=(30 x 9)/2=135kN/m

Pp=(90 x 3)/2=135kN/m

Center of Gravity

la3=3 + (2/3)x9=9m

lp=9 + (2/3)x3=11m

la3=3 + 9/2=7.5m

la1=(2/3)x3=2m

Pp x lp=135 x 11 = 1485 kN < Pa x la= 30x2+180x7.5+135x9=2625

Unstable

Stability of slope

Evaluation of the slope

Slip Slip surface

SF or Fs : Safety Factor = Resistance Drive Force

Resistance: Shear Strength

Drive Force: Gravity, Seismic

Safe or Danger?

Resistance

Drive

Stability of infinitive slope 1

h

W=1 x cosi x h x γt N=W x cosi S=W x sini Driving Force R=N x tanφ Coulomb’s criteria: Resistance Force

N

S

FS= R S

= cos2ihγttanφ cosihγtsini

= tanφ tani FS=1, i=φ φ: angle of repose

P188

Circle Failure method φ=0: Short term

SF= Moment for Resistance Moment for Drive

Resistance moment = (FE x su or cu)xR

Drive Moment=W x x P192

Circle Failure method Slice method

SF= ΣMoment for Resistance ΣMoment for Drive

Resistance moment = (FE x su or cu)xR

Drive Moment=W x x P196

Search for the smallest SF

Critical Circle

P193

Tailor’s Chart

P194

Stability Factor Ns

t cs

HNc

γ= No dimension

N/m3 x m

N/m2

P194

Toe Failure Base Failure Slope Failure

Stability Number Slope angle

Failure pattern

Base Failure Toe Failure Slope Failure

P192

Bearing Capacity

Shallow Foundation

Deep Foundation

Pattern of Failure Load

Sett

lem

ent

General Failure

Local Failure

Bearing Capacity: Ultimate: Qu Allowable : Qal=Qu/SF SF: Settlement, uncertainty for soil parameters

P209

Theoretical Value of Prandtl

III Passive Zone

I Active Zone

II Transition Zone

α:dependent on the roughness of the footing. Smooth:

452φα = ° +

Important value: φ=0, Qu=2bqu, qu=(2+π)c=5.14c

P210

Calculation of Q

B

Q=qB=?

Terzaghi’s equation

Shape Factor

Factors General Failure Local Failure

Continuous Square Rectangular Circle

1 2u c f qq acN BN D Nγβγ γ= + +Cohesion (Cu, Su)

Friction Surcharge

B

Df

Qu=Bqu

γ2

γ1

P211, 212

Deep Foundation (Piles)

Types of Pile classified by support system

Skin

Fri

ctio

n

Point Resistance

Pointed Pile Friction Pile

P217

Design of Pile

• Using N value (Standard Penetration Test).

• Empirical equation

1 1(40 )x9.85 2

u p s s c cR NA N A N A= + +

Ap: cross sectional area As: surface area of the sand layer Ac: surface area of the clay layer

P218

Standard Penetration Test (SPT)

Raymond sampler

Hammer (Mass = 63.5 kg)

Height = 76 cm

Knocking Head

Definition of N value How many blows for penetration of 30 cm

P10

Pile Group

Pressure Bulb

RT = E・n・Ru

RT: Bearing capacity of the pile group Ru: Bearing capacity of the single pile n: Number of the Pile

E: Efficiency of the pile group <1.0 P221

Negative Skin Friction

Conventional case Reclaimed Ground

Positive Friction

Negative Friction

P223