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SKAA 2722
GEOTECHNICS 1
MUHAMMAD AZRIL HEZMI
M47 – SURVEY LAB
0137511595
CIVIL.UTM.MY/AZRIL
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)