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SKAA 2722

GEOTECHNICS 1

MUHAMMAD AZRIL HEZMI

M47 – SURVEY LAB

0137511595

CIVIL.UTM.MY/AZRIL

AZRIL@UTM.MY

Week Lecture Topic / Content

1

1

2

INTRODUCTION – schedule and overview of SKAA 2722 (Geotechnics I) and review of SAB 1713 (Soil Mechanics) e.g. vertical stress, Terzaghi’sprinciple of effective stress, soil properties and shear strength.

STRESSES IN SOILS – Vertical stresses due to applied load: Simple method (2:1 method); Boussinesq Method - Point load, Line load, Uniform surcharge load.

234

Fadum ChartPressure Bulb Chart.

356

Lateral earth pressure at –rest, active and passive conditions.Rankine Method and Coulomb Method.

4

7

8

Lateral pressure diagram and location of resultant force against sheet pile & gravity retaining wall.

Soil pressures against & force in tie-back rod for anchored tie-back sheet piles.

5

9

10

COMPRESSIBILITY AND CONSOLIDATION – Introduction, Definition of ‘consolidation’ and ‘settlement’.

Curve of dial gauge reading versus time, and void ratio versus log effective stress curve (consolidation or oedometer test).

6

11

12

Compressibility parameters of soils; Terzaghi’s Consolidation Theory –One-dimensional consolidation.Test 1 – Stresses in Soils & Earth Retaining Walls.

7

13

14

Pre-consolidation pressure and definition of ‘normally consolidated soil’, ‘overconsolidated soil’, and underconsolidated soil’.

Components of total settlement= immediate (elastic) settlement + consolidation settlement + secondary settlement (creep).

8 MID SEMESTER BREAK

COURSE OUTLINE: SKAA 2722

Week Lecture Topic / Content

9

15

16

Calculation of ultimate or consolidation settlement

Degree of consolidation (U) – relationship between U and settlement (∆H), U and void ratio (e), and U and excess pore-water pressure (u

e).

1017

18

Average consolidation vs. Time Factor (Tv) Chart; and Isochrone (y/Hdp, Uy, Tv) Chart. Problem solving exercises.

11

19

20

Test 2 – Compressibility & Consolidation

SLOPE STABILITY – Infinite & Finite Slopes, Stability of infinite slope: dry slope, saturated slope and partially-submerged slope (water table or seepage parallel to surface).

122122

φU=0 analysis and Taylor’s ChartMethods of Slices: Fellenius and Simplified Bishop

132324

Stability of natural slope, embankment, and excavation

14

25

26

Methods of slope stabilization. Problem-solving exercises for slope stability

Student’s self-study and work on term paper / project on Topics 1&2 (Retaining Structures & Consolidation of Soils)Group discussions amongst the students

152728

Term Paper & Poster / Project Report write-upTerm Paper & Poster / Project Report write-up and submission

16-18 REVISION WEEK AND FINAL EXAMINATION

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GRADING

No.

Assessment Number % each % total Dates

1. Assignments 1 5 5 Week 7

2. Project 1 10 10 Week 14-15

3. Quizzes 2 2.5 5 Week 2 & 9

4. Presentation - - - -

5. Test 2 15 30Week 6 and

11

6. Final Exam 1 50 50 Week 16

Overall Total 100

ATTENDANCE

The student should adhere to the rules of attendance as stated in the University Academic Regulation :-

1.

Student must attend not less than 80% of lecture hours asrequired for the subject.

2.

The student will be prohibited from attending any lecture and assessment activities upon failure to comply the above requirement. Zero mark will be given to the subject.

Stresses in SoilMUHAMMAD AZRIL HEZMI

Stresses in Soil

� = Total Stress

� = Effective Stress

� u = Pore water pressure

u+′= σσ

σσ ′

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In-Situ (Overburden) Stress

•Determine pore water pressure, effective and total stress along

XY

In-Situ (Overburden) Stress

In-Situ (Overburden) Stress

� A soil profile consists of a surface layer of loose sand 3.5 m thick overlying a layer of stiff clay. Draw the distribution of total, effective and pore water pressure in the soil and calculate the stresses at depth of 3.5 m and 5.5 m below the surface.

Loose sand, = 16.5

kN/m3

Stiff clay, = 18.5

kN/m3

3.5 m

5.5 m

GWT

bγ

satγ

Ground Level

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In-Situ (Overburden) Stress

3.5 m

5.5 m

In-Situ (Overburden) Stress

FIND THE SOLUTIONS

WITHIN 15 MINUTES

In-Situ (Overburden) Stress� A soil profile existed 2

m below from a river

stream as shown in

the figure. Draw the

distribution of total,

effective and pore

water pressure in the

soil at 8 m from the

ground level.

Clay, = 18.88 kN/m3

0 m

8 m

Ground Level

satγ

-2 m

River Stream

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3sat 18.88kN/mγ =

(9.81 x 2)

(18.15 x 2) + 19.62

(18.15 x 2) + 55.92

(18.15 x 4) + 92.21 (9.81 x 10)

In-Situ (Overburden) Stress

250kN/m

Stresses Due to Apply Load

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( )( ) z)z)(L(B

σBL

zLzB

PΔσ z ++

=++

=

UNIFORM LOAD (SIMPLE METHOD)

Where

P = concentrated load = stress

B = width of the foundation

L = length of the foundation

z = depth

σ

• One of the simplest methods to calculate

stress in a soil element at depth due to

load distribution under rectangular or

strip foundation.

� For a rectangular footing 3 x 4 m in size carrying a

column load 1500 kN, determine the vertical stress at

depth of 2 m using a 2:1 load distribution.

UNIFORM LOAD (SIMPLE METHOD)

UNIFORM LOAD (SIMPLE METHOD)

� Q = 1500 kN, B = 3, L = 4, z = 2

( )( )2

z /50)24)(23(

1500

zLzB

PΔσ mkN=

++=

++=

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25

22

z

)zr

(1z 2π

3PΔσ

+

=

2z z

PKΔσ =

( )22 yxr +=

where,

POINT LOAD (Boussinesq)

• Boussinesq derived equations due to a vertical point load only

Expression for vertical

stress

Point Load (Westergaard)

( ) ( )( ) ( ) ( )[ ]2

322

z

r/z2ν2/2ν12ππ

2ν2/2ν1PΔσ

+−−

−−=

( )[ ]2

322

z

r/z21πz

PΔσ

+=

2wz z

PIΔσ =

Expression for vertical

stress

v = Poisson’s ratio, if v = 0, hence

POINT LOAD (Boussinesq &

Westergaard)

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Line Load

�222

3

z)zπ(r

2QzΔσ

+=

Line Load

A long concrete wall

fence induced a line

load of Q = 18 kN/m

on a soil (Figure

P2.5). Calculate the

stress induced at

distance 1 m from

the fence at a depth

of 2 m below the

base

Line Load

( )2

222

3

222

3

z /67.321

2182

)zπ(r

2QzΔσ mkN=

+

××=+

=π

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Line Load

� Determine the increase of stress

at point A

14.6kN/mq2 = 7.3kN/mq1 =

mz 2.1=

A

3m

1.5m

7.3kN/mq1 =

mz 2.1=

A

+

14.6kN/mq2 =

mz 2.1=

A

( )2

222

3

222

3

z,1 /59.02.15.1

2.13.72

)zπ(r

2QzΔσ mkN=

+

××=+

=π

( )2

222

3

222

3

z,2 /147.02.13

2.16.142

)zπ(r

2QzΔσ mkN=

+

××=+

=π

2/737.0147.059.0Δσ mkNz =+=

Line Load

Uniform Load (Fadum)

Steinbrenner evolved equations

due to a vertical uniform load

� Expression for vertical stress

� Where

� Is = Influence factor depend on L,

B and z

� The values of this influence factor

were prepared by Fadum (1948), where depend upon the two

coefficients (See Figure)

� m = B/z

� n = L/z

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Uniform Load (Fadum)

Uniform Load (Fadum)

� For point E:

� B = 2, L= 3 , z = 2; m = 1, n = 1.5

� I = 0.195, hence

� For point C:

� B = 2, L =1.5, z = 2; m = 1, n = 0.75

� I = 0.165, hence

2/78200195.022σ mkNIqE =××=×=

2c /132200165.044σ mkNIq =××=×=

Uniform Load (Pressure Bulb)

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[ ]

+−= 3/22 1(R/z)

11qΔσ

Uniform Load (Pressure Bulb)

� An oil tank is carrying a uniform load of 117 kPa.

Determine the vertical stresses at depth 2 m below a)

the center b) the sides of the tank c) point C at distance

4 m from the center. The diameter of the tank is 3.9 m.

Uniform Load (Pressure Bulb)

� Radius of the tank, a = 3.9/2 = 1.95

� a) Below the center of the tank, z/a = 2/1.95 = 1.02

� r/a = 0, then Iz = 0.63 σz = qxIz = 117 x 0.63 = 74 kPa

� b) Below the sides of the tank, z/a = 2/1.95, r/a= 1.0,

then Iz = 0.33, σz = qxIz = 117 x 0.33 = 39 kPa

� c) Below point C, z/a = 2/1.95, r/a= 4/1.95 = 2.04, then

Iz = 0.045, σz = qxIz = 117 x 0.045 = 5.2 kPa

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Uniform Load (Newmark)

� Can be used for irregular shaped foundations and due to a vertical uniform load

� Expression for vertical stress

� Where

� I = Influence factor = 1 / No of box =1/200 = 0.005 (See Figure)

� n = number of unit contained the foundation

� P = concentrated load

Uniform Load (Newmark)

Uniform Load (Newmark)