Soils that effect of external load are subjected to stress. The vertical stress increase in soil due to various type of loading. At any point in soil the stress applied from own weight of soil which called effective stress, and from external load which called net stress, the net stress which applied must be determined.

The stress on element:

          As shown in fig. (1) the stress on element are as follow:

On plan       XY    the stress are         σz, τzx ,τ  zy

                   YZ     the stress are         σx, τxy, τxz

                   ZX     the stress are         σy, τyz, τyx

          From this σz, σx, σy its stresses called normal stresses, but the other is called shear stresses which is τxy, τyx, τzy, τyz, τxz, τzx.

Shape of external load:

1 .

 Point load on concentrated load applied from coloumn, wheel of machine, load called point load, because it is effect  in point.

2 .

Line load which the load is effect on line as the load on Rail way and Dimension ton/m'.

3 .

Uniform load which the load effect on area, whereas the load uniform and dimension ton/m2

4 .

Triangular load which applied from embankment, so from the dam, the load dimension is ton/m'

5 .

         The loads mention before are uniform load, but the area which load is effect is changed two of them are regular as circular and strip but the other is irregular area although the load is uniform.

Polar coordinate system:

As we know the cartizian coordinate system which shown in fig. (2) any point (A) in this system defined as three dimension x, y, z.

 

 

In polar coordinate system point in this system defined as (r, φ, Z) as shown in fig. (3).

 

 

The relation between polar and cartizian coordinate as shown in fig. (4).

         r        =      

          R       =    =

          Cos α =   

 

 

 

1)    Stress distribution under vertical concentrated load:

Boussinesq’s Method:

When a point load Q acting on the surface of a semi infinite solid, a vertical stress σz produces at any point in addition to lateral and shear stress.

Assumptions of Boussinesq theory:

a.                For soil, the soil mass is elastic, isotropic, homogeneous and semi-infinite.

b.               The soil is weightless.

c.                For load, the load is vertical, concentrated acting on the surface.

d.               Hook’s Low Applied, it is mean that the constant ratio between stress and strain.

Boussinesq’s Formula:

Boussinesq’s solved the problem of stresses produced at any point (A) due to point load Q.

At point (A) …

 

σz

=

×

σz  

=

×

 
           
 

=

×

  R5   = (r2 + Z2)5/2  
           
 

=

 

×

 
           
 

=

 

×

 
 

=

 

×  
           
 

=

×

 

 
           

σz  

=

×

I p

=

I p

 

     

Where: 

σz      : Vertical stress at point A  as shown in fig.(5)

Z       : Vertical dimension for point A at load

Ip       : Influence factor depend on ()  = F()   

Q       : Point load

The variation of Ip for various value of  is given in table (1).

Table (1)

Example (1):

For concentrated vertical load equal 100 KN determine vl stress σz at point A (r, θ, z) = (1, 20 o, 2) and point B (x, y, z) = (0, 2, 4)

Solution:

For point A:

      =    =  0.5             From table        Ip = 0.2733

σz      =  Ip

=

 × 0.2733  =   6.833   KN/m2

For point B:

          r        =  =  2

      =    =  0.5                              Ip = 0.2733

σz      =  Ip

              =

 × 0.2733  =   1.71   KN/m2


Example (2):

Determine point load which make vl stress at point A (2, 0, 1). Don’t increase on 6 ton/m2.

Solution:

r        =  =  2                                          

      =    =  2                                  Ip = 0.0084

σz      =  Ip

Q       =  =   714.3    ton

a.     Stress Distribution under vl load on vl plan:   

σz      =  Ip

  At vl plan under load Q assume     r = 0

 = 0                    from table (1)

Ip       = 0.4775

σz      =  Ip

 

Z 0 0.1 0.2 0.3 0.4 0.5 1 2 4 6
σz 47.8 Q 12 Q 5.3 Q 3 Q 2 Q 0.48 Q 0.12 Q 0.03 Q 0.01 Q

b.     Stress Distribution under vl load on hl plan:   

At hl plan

Z = Constant

σz      =  Ip

assume        Z = 1 m

 

r 0 0.5 1.0 1.5 2.0 4.0
r/z 0 0.5 1.0 1.5 2.0 4.0
Ip 0.478 0.275 0.087 0.025 0.01 0.0004
σz 0.478 Q 0.275 Q 0.087 Q 0.025 Q 0.01 Q 0.0004 Q

 

c.     Distributed Load: 

Under distributed load q we can divided foundation to square small area and calculate concentrated load as consideration.

 

   

Assume the uniform distributed load q  and the small area Σ A from that produce   dQ = q.DA (as point load) from small concent- rated load.

z    =  Ip

σz      =  

 

Example:

Determine the vertical stress at depth Z = 6m under the centre of Raft 8m ×8m  with uniform load q = 20 ton/m2.

Solution:

b        =    =    = 2

ΣA     =  2 ´ 2 = 4 m2

dQ    = q.DA  = 20 ´   80 = 4ton

group (I)      1 – 4 – 13 – 16.

(II)            2 – 3 – 5 – 8 – 9 – 12 – 14 – 15.

(III)         6 – 7 – 10 – 11.

For group (I):

      =    =   = 0.71

Ip       = 0.165

 For group (II):

      =    =   = 0.53

Ip       = 0.27

 For group (III):

      =    =   = 0.236

Ip       = 0.4

 σz      =  

          =  [ 0.1654 ´ 4.0+ 8 ´ 0.27 + 4 ´ 0.4 ]

          = 9.82  ton/m2

 

2)    Vertical stress under circular uniform load:

Circular uniform load shown in Fig(8)

At point A we can calculate the vertical stress.

Assume small element with area rdφ.dr

of the uniform load q from Boussinesq’s theory

dQ = qdr.rdφ

 

σz

= q

  σz   = q . A

This equation when the point A lies under C.G of uniform load

To calculate vl stress to point I which has distance equal r (see fig. (8) and table (2).

σz      = q (A + B)

Where:

σz       : Vertical stress at point I

q        : Uniform circular load

A,B    : Partially influence factor depend on (,)       

Table (2)

Example:

Circular foundation diameter 10 m with uniform load q = 150 KN/m2. Determine σz at point I, II at depth 10 m.

Solution:

At point I

  =    = 2

   = 0

\ A   = 0.106      &       B   = 0.179

 σZI    = q (A + B)

          = 150 [0.106 + 0.179] =  42.75 KN/m2

At point II

  =     = 2

   =    = 1.5

from table      A     = 0.063        &       B   = 0.064

σZII    = q (A + B)

          = 150 [0.063 + 0.064] =  19.05 KN/m2

 

3)    Vertical stress under uniform load at corner [Fadum Method]:

The load is uniform load q effected on rectangular area B*L at vertical dimension Z to calculate σz at point (A) under the corner of load for point load. As we known the vertical stress σz is as follow

 

   
   

from this

σz      = q IP

    where:

   
   IP =
   m  =
    n  =

The value IP directly calculated from table as a function in ,

As shown in table (3)

Table (3)

Example:

Determine vl stress at depth 4.0 m under vl uniform load equal  150 KN/m2 under point K.

Solution:

       Divided the area at point K. The point (K) must be at corner

q        = 150 KN/m2

 

        σZ      = σZI + σZII + σZIII + σZIV

At region I

L = 2           B = 2               Z = 4

      = 0.5

      = 0.5

\ IP   = 0.084

  σZI   = qIP  = 150 × 0.084  = 12.6

 At region II

L = 4           B = 2             Z = 4

      = 1

      = 0.5

\ IP   = 0.1202

σZII    = qIP  = 150 × 0.1202  = 18.03

 At region III

        L = 2           B = 6             Z = 4

      =  = 0.5  

      =  = 1.5

\ IP   = 0.131

σZIII    = qIP  = 150 × 0.131  = 19.65

 At region IV

   L = 6           B = 4             Z = 4

      = 1.5

      = 1

\ IP   = 0.1934

          σZIV    = qIP  = 150 × 0.1934  = 29.01

 σZ      = σZI + σZII + σZIII + σZIV

          = 12.6 + 18.03 + 19.65 + 29.01

          = 79.29 KN/m2

 

4)    Stress duo to a uniform line load:

Boussinesq determine this load at point A (r, z) as shown in fig (9)

σz ×

 

 

σz = Q I L  

where:

 Q      : Line load intensity

IL      : Influence factor due to line load      =  F ()

See table (4)

 

Table (4)

  Example:

Determine vl stress σz at point A(r,z) = (2, 2 ) where Q = 100 KN/m.

Solution:

      = 1               from table

IL       = 0.159                          

σz      = 15.9          KN/m2

 

5)      Stress distribution due to triangular strip load:

 As shown in fig (10)

To calculate stresses distribution under triangular strip load must be known the distances C, Z, X.

The distance X measured from zero load to direction of point (A).

  We can determine the angle of β, α from that

σz      =                     à (1)

where:

β, α determined from cross section dimension.

 Equation (1) can get the form

σz      = q IT

Where:

σz      : vl stress at point A

q        : intensity of triangular load

IT       : Influence factor for triangular strip load

          = F( ), F ()   see table (5)

The distance x must be –ve x.

If the distance x measured in the – ve direction.

Table (5)

6)     Stress Distribution Under Uniform Strip Load:

Uniform contact pressure  effected on Strip area with dimension 2b to study

The vl stress at point A which lies at Z dimension.

For determining the stress distribution under uniform strip load. Must be calculate the angle β, α and the distances b, x, z at cross section from fig. (11).

The vertical stress Distribution at point A become

σz      = q IS

Where:

σz      : vl stress at point A

q        : intensity of strip uniform load

IS       : Influence factor for strip load

          = F( ), F( )  see table (6)

x      : Measured from centre line of load.

 

 

 

Table (6)

Example:

if q = 500 KN/m2 determine σz at point A where x = 3m from c.l at vl distance.Z= 5.0m if the width of strip footing B= 4.0m

Solution:

      = 2.5

      = 1.5

\ IS   = 0.285

σz      = 500 ´ 0.285 = 142.5  KN/m2.

New Mart Chart Method

If the shape of loaded area is irregular and different from shapes discussed before , the stress under any point can be computed by using the New mark chart .

The stress under uniform load at center

 
1 -

  Y (1)

             

=

-1   Y (2)
      =

   For example at r = 2cm and Z = 5 cm, =0.20, with another mean, if there circular load radius equal 2 cm and under the centre of circle at depth Z = 5 cm, the Vl stress σz at this point equal 0. 2qS  by making the following table from equ. (2) Where as Z = 5 cm.

          Z = 5 cm

Circular No. σz/q r/z r
1 0 0 0
2 0.1 0.27 1.35
3 0.2 0.4 2.0
4 0.3 0.518 2.5
5 0.4 0.637 3.2
6 0.5 0.766 3.85
7 0.6 0.918 4.6
8 0.7 1.11 5.55
9 0.8 1.387 6.95
10 0.9 1.908 9.55
1.0

Construct the circles and divided by rayes, every element caused when loaded vl stress at centre of circles.

         σz      =       1/20  ´ 1/10 q

σz      =       q. . N

          =       q. IF. N

  Where

σz      =       vl stress at any point.

q        =       uniform load for irregular area

IF       =       influence factor depend on m, n

m       =       number of circles

n        =       number of Rays

N       =       Symma area contained load.

Fig (13): Show standard chart for New mark with depth equal 1 inch = 2.54 cm.

Fig (13 )

Approximate Method:

          In this method load transmit during soil with inclined slop. 2 : 1.

1.     For squar area with dimension ( B × B )

σz      =        

 

 

2.     For rectangular B.L

σz      =        

 

3.     For strip take 1.0 m length

σz      =        

 
   

4.     For circular with Diameter D

σz      =      

Table (1)

Table (2)

 

Table (3)

Table (4)

 

Table (5)

 

Table (6)

 

Fig (13)