The settlement is divided into two categorizes:

Elastic (Immediate settlement).

Consolidation settlement.
In some calculations of settlement it is required to find the increase in stress at any depth of soil mass, so we will discuss the calculation of increase in stress.
1-
 
due to concentrated load:
  
3 P
2
 z
2 



1
 r z
2



5
2 r
 x
2  y
2
X, Y and Z are the coordinates of point under consideration.
2-
 
due to circularly loaded area load: q : Load per unit area on circle. o
B: Diameter of circle. r: Distance from center of circle to point under consideration. z: Depth of point under consideration.
Go to table 5.1 . find
 
/ q o by determining the terms:
 r/(B/2)
 z/(B/2)
  
3q o
I
 
below rectangular loaded area: q o
: Load per unit area on rectangle.
I: influence factor.
1

The point should be under corner of rectangle, if not we have to divide the rectangle to sub rectangles for which the point is in corner for each part.
Go to table 4.2 P.224-225 find I, by determining the terms:

 m

B
Z n

L
Z
   q o
I
   q o

I
1

I
2

I
3

I
4

For simplification, if the point is in the center of the rectangle:
   q o
I c
Go to table 4.3 P.227 to find I by determining the terms: c

 m
1

L
B n

1 B z
2
2

4Approximate (2:1) method:
This method consider that the increase in stress will be distributed in the soil below the foundation by slope 1 Horizontal : 2 Vertical as shown in the figure below:
Load

P
 q o

B

L
  at depth z below the foundation


B
 z
P

L
 z

3

5Average vertical increase in stress due to rectangular loaded area:
Sometimes it is required as we calculate consolidation settlement to calculate the average increase in stress in specific layer that start from depth z=H
1
to depth z=H
2
, use the following formula:

  avg

H 2 / H 1
 q o


H
2
I a ( H 2 )
H
2
I a ( H 2 )
; I a
Take H

H
2
I a ( H 1 )
; I a
Take H

H
1


H
1
I a ( H 1 )
H
1

 m

2 n
2

B
H
L
H
4

ةطقن يه تاداهجلاا يف ةطسوتملا ةدايزلا باسح ديرن يتلا ةطقنلا نوكت امدنع طقف حلصي قباسلا نوناقلا
لثمي ثيحب ءازجأ ةدعل ليطتسملا ميسقتل ىعسن )
Corner
( ةطقنلا نكت مل اذا يلاتلاب و )
Corner
( ةدعاقلا ىلع ةيفرط
.)
Corner
( ءزج لك
Determine the average stress increase below the center of the square foundation (3m x
3m) in the layer bounded between z=3m and z=5m.
We will divide the area for 4 parts because the relation used to calculate the average stress requires that the point is to be corner and out point is center.
Also as it is center we will calculate for one part and then multiply the result by 4 because all parts are equal.

  avg

H 2 / H 1
 q o


H
2
I a ( H 2 )
H
2
For I a ( H 2 )
: m
2

B

1 .
5
5

0 .
3 n
2

From
H
L
H

1 .
5
5

0 .
3 fig 5 .
7 text book

  avg

H 2 / H 1

 q o


H
2
I a ( H 2 )
H
2

H
1
I a ( H 1 )

H
1
I a ( H 2 )



0 .
126

H
1
I a ( H 1 )

H
1


For I a ( H 1 )
: m
2

B
H

1 .
5
3

0 .
5 n
2

From
L
H

1 .
5
3

0 .
5 fig 5 .
7 text book

I a ( H 1 )

0 .
175

  avg

H 2 / H 1


100 
5

0 .
136
5


3
3

0 .
175



5 .
5 kN / m
2 
* 4
For Embankment loads and Newmark chart, please review Chapter 9 SOIL MECHANICS
5

  e avg

A
1
A
2 q o
B
E s
Go to figure 5.14 to find A
1
by determining the terms (H/B and L/B).
Go to figure 5.14 to find A
2
by determining the term (D f
/B).
Es = β . cu
See Table 5.7 to get a typical value of β = f (OCR , PI)
Refer to the following figure, determine settlement of the foundation .
Solution:




H
B
L
B
  e avg

A
1
A
2 q o
E s


1
1 .
5
.
3
3
5


2
2





B
A
1

0 .
66
D f
S
B e


1 .
2

0 .
8

A
2

0 .
93
1 .
5
0 .
66

0 .
96

150

1 .
5
600

0 .
24 m

24 cm
E
S

600 KN / m
2
Saturated clay q o

150 KN / m
2
6

S e

C
1
C
2
 q
'  q z  z
 z 

0
2 I z
E s
 z
C : Correction factor for depth of embedment
1

C
1

1

0 .
5 q
' q
 q
C : Creep correction factor
2
C
2

1

0 .
2 log
Time in years
0 .
1
' q : Stress at the level of foundation including external loads and soil weight. q : Effective vertical overburden pressure. In the absence of water table
 
D f
.
ىلع هتمس قب موقنف )
Net load
( انيطعي دق و .) q
' – q
( ربتعت يهف )
Net stress
( ةرشابم لاؤسلا يف اناطعا اذإ
.تاداهجإ ىلإ هليوحتل ةدعاقلا ةحاسم
E s
: Modulus of elasticity of soil below the foundation which is variable.

Why Es is various?
Due to nonhomogenuity of soil the value of Es is varying, we can estimate the value of
Es from field tests as Standard Penetration Number (N) or Cone Penetration Resistance
  c
7

E s

2 .
5 q c
For circular or square
1 US
3 .
5 q c
For strip footing
tons / ft
2    
Ib / ft
2 footing
1 t / ft
2    
KN / m
2
1 Ib
   
N
I z
: Strain influence factor, it is given as shown below:

For square or circular foundation 
For foundation with L/B 10
Z Iz
Z Iz
0
Z
2
=2B
0.1
Z
1
=0.5B 0.5
0
0
Z
Z
1
2
=B
=4B
0.2
0.5
0
.ةيعيبطلا ضرلاا حطس نم سيل و ةدعاقلا لفسأ نم اهباسح متي z
1 and z
2
ةفاسملا :ادج مهم
For L/B between 1 and 10, we have to do interpolation for each depth, for example for
L/B = 4 at z=0: at L/B = 1→ Iz = 0.1 at L/B = 10→ Iz = 0.2 at L/B = 4 → Iz = ??? by linear interpolation Iz = 0.1333

Z : divisions in soil layer below the foundation.
8

ديدحت للاخ نم ةرشابم ةاطعم نوكت دق ةفاسملا هذهو ،ةدعاقلا لفسأ ةبرتلل
Es
ةميق فلاتخا نم جتنت 
Z
ةفاسملا
.مسرلاب حضوم وه امك ةفلتخملا
Es
ميقل لاؤسلا



Z

2 m
Z

1 m
Z

2 m
9

A continuous foundation on a deposit of sand layer of soil, Assume
 
115 Ib / ft 3 and time in years for C
2
= 10yr.
Use the strain influence factor to find the elastic settlement.
q
S e


C
1
C
2
 q '  q z  z



0 z
2 I z
E s

D f

115

5

 z
575 Ib / ft
2
.
C
1

1

0 .
5 q
' q
 q

1

0 .
5
575
4000

575

0 .
916
C
2

1

0 .
2 log
Time in years

1

0 .
2 log
10

1 .
4
0 .
1 0 .
1
S e

0 .
916

1 .
4

4000

575
 z z

 z

0
2 I z
E s
 z
 
E s

Ib / ft
2

 z

S e

4392 .
22 z z

 z

0
2 I z
E s z
 
to the center of each part
 z
I z
at the center of each part
6
14
20
126,000
250,560
208,800
3
13
30
0.3125
0.3958
0.04166

Ib / in
2   
Ib / ft
2
Iz is given by interpolation.
S e

4392 .
22

4 .
098

10

5 
0 .
18 ft

55 mm
I z
 z
 ft
3
/ Ib

E s
1 .
488

10

5
2 .
211

10

5
3 .
99

10

6
4 .
098

10

5
10

4Consolidation settlement:
The allowable settlement of foundation may control the allowable bearing capacity, so that: q u
/ FS q all
 minimum of q allowable settlement q
 Meyerhof’s Theory: net
 q all
 
D f
The maximum allowable settlement Se = 25.4mm = 1in
This value is appropriate so that the differential settlement is 19mm which is acceptable.
Based on this theory, Meyerhof set some empirical equations to calculate the net allowable settlement:
Meyerhof thinks that he was too conservative when he calculated the net allowable bearing capacity so that he has increased its value by 50%:
 q net ( all )
 used

 q net ( all )
 calculated

1 .
5
Look book for relations in English units (US)

Bowles theory:
F d
: Depth factor
11

F d

1

0 .
33
D f
B
Look book ,l for relations in English units (US)
Example
Find the size of the square footing that carry allowable load of 1000 KN given that:
N

10
S
60
  e all

25 .
4 mm
Use Bowles theory.
Solution
As the load is relatively large, we will solve assuming B>1.22m q net ( all ) q net ( all )

N
60
0 .
08

N
60
0 .
08
B
B
B
0 .
3
0 .
3
2
F d


S e
25


2
F d


S e
25


1000

B
2
10
0 .
08
B
B
0 .
3
2
 
1 0 .
33
1 .
5
B

25
25
By trial and error, B = 2.3m
12

Ultimate bearing capacity باسح نم نكمتن رابتخلاا اذه للاخ نم
قمع ل ايواسم قمع اهل ةرفح يف عضوي ةنيعم داعبأ و ةكامسب يرئاد وأ عبرم صرق نع ةرابع مدختسملا زاهجلا و
.زاهجلا ضرع فاعضأ ةعبرلأ ايواسم ضرع و رفحلا
طوبهل ا و داهجلإا بس حن ةرم لكو ةتوافتم تارتفبو ةيجراخ لاقثأ قيرط نع يجراخ لمحب هيلع رثؤن صرقلا اذه
.طوبهلا و داهجلإا نيب ةقلاع مسرن و لمحلا نم جتانلا
For clay soil:
    u f u p
For sand soil:
    u f u p
B f
  u
  u f p
B
P
: Ultimate bearing capacity for foundation
: Ultimate bearing capacity for plate
B f
: Width of foundation
B p
: Width of plate

Settlement relationships from plate load test:
For clay:
S f

S p
B f
B p
For sand:
2
S f

S p
B f
2 B f

B p
S f
: Settlement of foundation
S p
: Settlement of plate

Determination of load bearing capacity using two plates in plate load test:
1Determine the areas of two plates (A
1
and A
2
) and perimeters of them (P
1
and P
2
).
2From the plate load test curve find the loads related to
  e all
that are ( Q
1
andQ
2
).
3Find the factors m and n by solving the two equations:
Q
1
Q
2


A
1
A
2 m m


P
1
P
2 n n
Q o
4Find the design load of the foundation:

Am

Pn
A: Foundation area
P: Foundation perimeter.
13

Example
From plate load test the following results were given:

Square plate (305mm x 305mm)

Allowable load on foundation = 2500KN.
   e all
= 25mm

Sand soil.
Determine the size of square foundation.
Relation between Settlement and stress
0
10
20
30
40
50
60
70
0 200
Stress (KN/m2)
400 600 800
Solution
S f

S p
2
B f
2 B f

B p q o

Q
A
 q o

2500
B
2
: Stress on plate.
q o q o

KN / m 2
 S p
given
B(m)
14
7 from curve
2.5
3.1
400
260
So that the appropriate width of foundation is 3.1m
S f
given from equation
44
23.2
S f
Check

25 mm
Not OK
OK
14