Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show. Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.

1. 1
Geotechnical Engineering–II [CE-321]
BSc Civil Engineering – 5th Semester
by
Dr. Muhammad Irfan
Assistant Professor
Civil Engg. Dept. – UET Lahore
Email: mirfan1@msn.com
Lecture Handouts: https://groups.google.com/d/forum/geotech-ii_2015session
Lecture # 6
22-Sep-2017

2. 2
STRESS DISTRIBUTION IN SOIL
What causes stress in soil?
Two principle factors causing
stresses in soil.
1. Self weight of soil
2. External loads (Structural loads,
external load, etc.)
v
h


3. 3
STRESSES DUE TO SELF WEIGHT OF SOIL
Vertical geostatic stress at any depth
= weight of the soil above that depth.
(A) Unit weight of the soil is
constant with depth
𝜎 = 𝛾. 𝑧
Where,
z = depth
γ = unit weight of soil
v
h

4. 4
STRESSES DUE TO SELF WEIGHT OF SOIL
Soil density generally increases with depth because
of compression caused by geostatic stress.
(B) Unit weight of soil varies
continuously with depth
𝜎 =
0
𝑧
𝛾. 𝑑𝑧
(C) Stratified soil; having unit
weight different for each layer;
𝜎 = 𝛾. ∆𝑧
v
h

5. 5
Practice Problem #1
The water table in a deposit of uniform sand is located at 1
meter below the ground surface. The specific gravity of soil
solids is 2.7 and voids ratio of sand deposit is 0.8.
a. Assuming the soil above the water table to be dry. Find out
the effective stress at a depth of 6m below the ground
surface.
b. If the soil above the water table is saturated by capillary
action, sketch the total stress, pore water pressure and
effective stress diagrams for the sand deposit for a depth of
6m below the ground surface.

6. 6
STRESS DUE TO EXTERNAL LOAD
Contact Pressure
Pressure developed at the
contact point of foundation and
soil.
𝜎 𝑜 =
𝑃
𝐵 ∙ 𝐿
Where,
σo = contact pressure
P = Point load
B x L = Contact area
P
𝝈 𝒐 =
𝑷
𝑩 ∙ 𝑳

7. 7
Stress Distribution in Soil with Depth
• Intensity of stress decreases with depth.
• Intensity of stress decreases radially from the point load.
STRESS DUE TO EXTERNAL LOAD

8. 8
STRESS INCREASE (∆q) DUE TO
EXTERNAL LOAD
Determination of stress due to external load at any
point in soil
1. Approximate (2:1) Method
2. Boussinesq’s Theory
3. Westergaard’s Theory

9. 9
APPROXIMATE METHOD
Use of 2:1 (V:H) stress
distribution.
𝜎 𝑧 =
𝑄
(𝐵 + 𝑧) ∙ (𝐿 + 𝑧)
Where,
σz = Stress at depth ‘z’
Q = Point load
B x L = Footing dimensions
𝝈 𝒛

10. 10
For rectangular foundation
For strip footing
Where,
σz = Stress at depth ‘z’
Q = Point load
B x L = Footing dimensions
𝝈 𝒐 =
𝑸
𝑩𝑳
𝝈 𝒛 =
𝑸
𝑩 + 𝒛 . (𝑳 + 𝒛)
APPROXIMATE METHOD
𝜎 𝑧 =
𝑄
(𝐵 + 𝑧) ∙ (𝐿 + 𝑧)
𝜎 𝑧 =
𝑄
(𝐵 + 𝑧) ∙ 1

11. 11
STRESS INCREASE (∆q) DUE TO
EXTERNAL LOAD
Determination of stress due to external load at any
point in soil
1. Approximate Method
2. Boussinesq’s Theory
3. Westergaard’s Theory

12. 12
Boussinesq’s Theory for Point Load
Q
Boussinesq (1885) solved the problem of stress produced by any point
load on following assumptions;
• The soil mass is elastic, isotropic, homogeneous and semi-infinite.
• The soil mass is weightless.
• The load is a point load acting on the surface.
Q

13. 13
Boussinesq’s Theory for Point Load
Q
 
 
 
  































23
2
2
22
5
2
23
2
2
22
5
2
21
3
2
21
3
2
rL
zx
zLLr
xy
L
zyQ
rL
zy
zLLr
yx
L
zxQ
y
x






5
3
2
3
L
Qz
z

 
22
yxr 
Where,
  2522
3
2
3
zr
zQ



22222
zrzyxL 
 = Poisson’s
ratio

14. 14
Q
  2522
3
5
3
2
3
2
3
zr
zQ
L
Qz
z




The above relationship for
z can be re-written as
   








 2522
1
1
2
3
zrz
Q
z


where
   252
1
1
2
3


zr
IB

QBI
z
Q
2

Independent of all material properties
Boussinesq’s Theory
for Point Load

15. 15
Practice Problem #2
A concentrated load of 1000 kN is applied at the ground
surface. Compute the vertical stress
(i) at a depth of 4m below the load,
(ii) at a distance of 3m at the same depth.
(A) Use Boussinesq’s equation
(B) Use Westergaard’s equation P

16. 16
Vertical Stress caused by Line Load
x
z
z
z
Q/unit length
x
A
y
By integrating the point load equation along a line, stress due
to a line load (force per unit length) may be found.
Lz I
z
q

 
2
2
/1
12








zx
IL

Where,
q is line load in “per unit length”

17. 17
Practical Problem #3
Following figure shows two line loads and a point load acting
at the ground surface. Determine the increase in vertical stress
at point A, which is located at a depth of 1.5 m.
P = 30 kN
z
2 m
A
1.5 m
2 m
3 m
q2 = 10 kN/m q1 = 15 kN/m

18. 18
STRESS INCREASE (∆q) DUE TO
EXTERNAL LOAD
 Point load
 Line Load
• But engineering loads typically act on areas and
not points or lines.
• Bousinesq solution for line load was thus
integrated for a finite area
Bz I
z
Q
2

Lz I
z
q

Uniformly Loaded
Circular Area
Uniformly Loaded
Rectangular Area
Trapezoidal,
Triangular, etc.

19. 19
CONCLUDED
REFERENCE MATERIAL
An Introduction to Geotechnical Engineering (2nd Ed.)
Robert D. Holtz & William D. Kovacs
Chapter #10