ECGD 4122 – Foundation Engineering Lecture 4 Faculty of Applied Engineering and Urban Planning...

ECGD 4122 – Foundation EngineeringECGD 4122 – Foundation Engineering

Lecture 4Lecture 4

Faculty of Applied Engineering and Urban PlanningFaculty of Applied Engineering and Urban Planning

Civil Engineering DepartmentCivil Engineering Department

22ndnd Semester 2008/2009 Semester 2008/2009

ContentContent

• Shear StrengthShear Strength

• Subsoil ExplorationSubsoil Exploration

Mohr-Coulomb Failure CriterionMohr-Coulomb Failure Criterion

Total Stresses:Total Stresses:

Coulomb approximation for shear stress on the Coulomb approximation for shear stress on the failure plane:failure plane:

Shear Strength MeasurementShear Strength Measurement

Based on Mohr’s circle criteria, Based on Mohr’s circle criteria, twotwo ways are ways are possible to cause failure:possible to cause failure:

FirstFirst::

Increase the normal stress in one direction Increase the normal stress in one direction


SecondSecond::

Apply shear directlyApply shear directly


3=0

1

Direct ShearDirect Shear

Uniaxial CompressionUniaxial Compression


• Unconfined compression Unconfined compression test is used when test is used when = 0 = 0 assumption is validassumption is valid

• Triaxial compression is a Triaxial compression is a more generalized versionmore generalized version

• The soil sample is first The soil sample is first compressed isotropically compressed isotropically and then sheared by axial and then sheared by axial loadingloading

1

3

Triaxial CompressionTriaxial Compression


Direct Shear TestDirect Shear Test

Unconfined Compression TestUnconfined Compression Test

• For clay soilsFor clay soils

• Cylindrical specimenCylindrical specimen

• No confining stressesNo confining stresses

(i.e. (i.e. 3 3 = 0)= 0)

• Axial stressAxial stress = = 11

3 = 0

1


3=0

1



Unconfined Compression Test DataUnconfined Compression Test Data

c

c

A

P

AA

l

l

10

0


2u

u

u

qStrngthShearUndrainedS

StrengthnCompressioUnconfinedq

3=0

1


1

Horiz. plane

Max. shear plane


Triaxial Compression TestTriaxial Compression Test

Triaxial Compression TestsTriaxial Compression Tests

• Unconsolidated Undrained (UU-Test); Unconsolidated Undrained (UU-Test);

Also called “Undrained” TestAlso called “Undrained” Test

• Consolidated Undrained Test (CU- Consolidated Undrained Test (CU-

Test)Test)

• Consolidated Drained (CD-Test); Also Consolidated Drained (CD-Test); Also

called “Drained Test” called “Drained Test”

Consolidated Undrained Triaxial Test Consolidated Undrained Triaxial Test for Undisturbed Soilsfor Undisturbed Soils

Shear StrengthShear Strengthin terms of Total Stressin terms of Total Stress

• Shear Strength in terms of Shear Strength in terms of effective effective stressstress

• Shear strength in terms of Shear strength in terms of total total

stressstress tancs

u tancs

hydrostatic pore pressure

Shear StrengthShear StrengthTotal Stress - Total Stress - = 0 condition = 0 condition

• Shear strength in terms of Shear strength in terms of total total stressstress

• For cohesive soils under saturated For cohesive soils under saturated conditions, conditions, = 0. = 0.

tancs

csu

Shear Shear Strength,SStrength,S

Normal Stress, Normal Stress,

CC

= 0= 0


• BoringBoring

• SamplingSampling

• Standard Penetration Test (SPT)Standard Penetration Test (SPT)

• Vane Shear TestVane Shear Test

• Cone Penetration Test (CPT)Cone Penetration Test (CPT)

• Observation of Water TableObservation of Water Table

Subsoil ExplorationSubsoil Exploration

BoringBoring

Boring SpacingBoring Spacing

ASTM MethodASTM Method

BoringBoring

ASTM Method for Calculating Depth of BoringASTM Method for Calculating Depth of Boring

BoringBoring

AugerAuger

• 3-5 m deep3-5 m deep

• Hand-drivenHand-driven

Boring MethodsBoring Methods


Flight AugerFlight Auger

• 1-3 steps1-3 steps

• TractorTractor

mountedmounted


• Rotary drillingRotary drilling

• Wash boringWash boring

• Standard Split SpoonStandard Split Spoon

The sampler is driven into the soil at the The sampler is driven into the soil at the bottom of the borehole by means of hammer bottom of the borehole by means of hammer blows.blows.

SamplingSampling

Common Sampling MethodsCommon Sampling Methods

SamplingSampling

• Standard Split SpoonStandard Split SpoonThe number of blows required for driving the The number of blows required for driving the sampler through three 152.4 mm (6sampler through three 152.4 mm (6””) ) intervals is recorded. The sum of the intervals is recorded. The sum of the number of blows required for driving the last number of blows required for driving the last two 152.4 mm (6 in.) intervals is defined as two 152.4 mm (6 in.) intervals is defined as the standard penetration number (N).the standard penetration number (N).

SamplingSampling

Common Sampling MethodsCommon Sampling Methods

Standard Penetration NumberStandard Penetration Number

Vane Shear TestVane Shear Test


K

TCu

Measures: Measures:

• The Cone ResistanceThe Cone Resistance

• The Frictional ResistanceThe Frictional Resistance

Cone Penetration TestCone Penetration Test

Cone Penetration TestCone Penetration Test

Assignment # 1Assignment # 1

Determine the effective stress at point Determine the effective stress at point XX for the for the two cases shown below. The drawing is not to two cases shown below. The drawing is not to scale. Take scale. Take satsat = 19 kN/m = 19 kN/m33..


(a)(a) Static condition:Static condition:

= (9.81)(0.6) + (19)(1) = 24.89 kPa= (9.81)(0.6) + (19)(1) = 24.89 kPa

uu = (0.6 + 1)(9.81) = 15.70 kPa = (0.6 + 1)(9.81) = 15.70 kPa

’’ = = – – uu = 9.19 kPa = 9.19 kPa

• Flow-down condition:Flow-down condition:

i i = = h/L = 0.6/3.0 = 0.2h/L = 0.6/3.0 = 0.2

= (9.81)(0.6) + (19)(1) = 24.89 kPa= (9.81)(0.6) + (19)(1) = 24.89 kPa

uu = (0.6 + 1)(9.81) - (0.2)(1.0)(9.81) = 13.73 kPa = (0.6 + 1)(9.81) - (0.2)(1.0)(9.81) = 13.73 kPa

’’ = = – – uu = 11.16 kPa = 11.16 kPa


A soil deposit is composed of clay with A soil deposit is composed of clay with SatSat = 20 = 20

kN/mkN/m33. The GWT is at the ground surface. . The GWT is at the ground surface.

Calculate the shear strength on a horizontal Calculate the shear strength on a horizontal

plane at a depth of 10m, if plane at a depth of 10m, if cc’’ = 5 kPa and = 5 kPa and ’’ = =

30º.30º.


= = cc’’ + + ’’tantan’’

’’ = (20)(10) – (9.81)(10) = 101.9 kPa= (20)(10) – (9.81)(10) = 101.9 kPa

= 5 + (101.9)(tan30º) = 63.83 kPa= 5 + (101.9)(tan30º) = 63.83 kPa


A drained direct shear test is performed on a A drained direct shear test is performed on a

normally consolidated silty soil. The soil sample normally consolidated silty soil. The soil sample

is 75 mm in diameter and 25 mm in height. The is 75 mm in diameter and 25 mm in height. The

vertical load is 883 N and the shear force at vertical load is 883 N and the shear force at

failure is 618 N. Calculate failure is 618 N. Calculate cc and and ..


Drained test Drained test = = cc + + tantan

Normally consolidated soil Normally consolidated soil cc = 0 = 0

= = cc + + tantan = = tantan = tan = tan-1-1((//))

= S/A = 618/[= S/A = 618/[(37.5(37.51010-3-3))22]]

= 139.89 kPa= 139.89 kPa

= N/A = 883/[= N/A = 883/[(37.5(37.51010-3-3))22]]

= 199.87 kPa= 199.87 kPa

= tan= tan-1-1(139.89/199.87) (139.89/199.87) 35º 35º

Quiz # 1Quiz # 1

Determine the height of Determine the height of

water level (H) above the water level (H) above the

surface of gravel, given surface of gravel, given

that the flow rate (q) is to that the flow rate (q) is to

be maintained at 40 liters be maintained at 40 liters

per second. per second.

Quiz # 1Quiz # 1qq11 = = qq22 = = qq = 40 = 401010-3-3 m m33/s/s

aa11 = = aa22 = = aa = = rr22 = = (0.75)(0.75)22 = 1.767 m = 1.767 m22

vv11 = = vv22 = = vv = = qq//aa = 2.264 = 2.2641010-2-2 m/s m/s

ii11 = = vv11//kk11 = 2.264 = 2.2641010-2-2/10/101010-2-2 = 0.2264 = 0.2264

ii22 = = vv22//kk22 = 2.264 = 2.2641010-2-2/1/11010-2-2 = 2.264 = 2.264

ii11 = 0.2264 = = 0.2264 = hh11//LL11 = ( = (HH + 1 – + 1 – zz)/(1.0))/(1.0)

ii22 = 2.264 = = 2.264 = hh22//LL22 = ( = (zz + 1)/(1.0) + 1)/(1.0)

Solving the equations for Solving the equations for zz and and HH

zz = 1.264 m = 1.264 m

HH = 0.4904 m = 0.4904 m

Quiz # 1Quiz # 1

Quiz # 2Quiz # 2

For a clay deposit: For a clay deposit: CCcc = 0.4 and = 0.4 and CCrr = 0.05, = 0.05,

determine the final void ratio for each of the determine the final void ratio for each of the

following loading conditions:following loading conditions:

a) Initially a) Initially = 50 kPa, e = 1.2, Finally = 50 kPa, e = 1.2, Finally = 90 kPa, = 90 kPa,

and given that and given that pcpc = 100 kPa = 100 kPa

b) Initially b) Initially = 50 kPa, e = 1.2, Finally = 50 kPa, e = 1.2, Finally = 190 = 190

kPa, and given that kPa, and given that pcpc = 100 kPa = 100 kPa

Quiz # 2Quiz # 2

CC = slope = = slope = ee//loglog’ ’ = (= (eeii – – eeff)/log()/log(’’ff//’’ii))

(a)(a) 0.05 = (1.2 – 0.05 = (1.2 – eeff)/log(90/50))/log(90/50)

eef f = 1.187= 1.187

(b)(b) Two stages: Two stages:

0.05 = (1.2 – 0.05 = (1.2 – eefifi)/log(100/50))/log(100/50)

eefifi = 1.185= 1.185

0.4 = (1.185 – 0.4 = (1.185 – eeff)/log(190/100))/log(190/100)

eeff = 1.073= 1.073

ECGD 4122 – Foundation Engineering Lecture 4 Faculty of Applied Engineering and Urban Planning...

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