Definitions:

1.     Ultimate B.C ( qult )

It's the gross pressure at the base of foundation at which the soil fails in shear. It's not used for design because it has a big value

2.     Net ultimate B.C ( qu net )

It's net increase in pressure at the base of foundation cause the failure…

     qu net   =         qult   -   γDF

Where:

     γDF    =       Over burden pressure at foundation level

     qult     =       Ultimate B.C

qu net   =       Net ultimate B.C

3.     Net safe Bearing Capacity ( qn.s ):

It's the pressure at which foundation designed.

     qn.s     =        

Where:

     F.S    = Factor of safety equal   from(2 to 5)

4.     Safe Bearing Capacity ( qs ):

It means the gross safe Bearing Capacity which used in design.

     qs       =       qgs      =       qns +   γD

                                  =        + γD

5.     Net safe settlement pressure qn.p

It's the net pressure which the soil carry without increase in allowable settlement.


6.     Net allowable B.C qn.all

It's the net pressure which can be used for the design of foundation, which ensure that there is not shearing failure, or the settlement within reach the limit, to choose the allowable B.C ( qn.all ).

If the net safe settlement pressure more than net safe B.c

     qn.p    >       qn.s

     qall     =       qn.s

If the net safe B.C more than the net safe settlement pressure the the allowable B.C equal the net safe settlement pressure.

     qn.s     >       qn.p

     qall     =       qn.p

where:

     qn.s     =       Net safe B.C

     qn.p    =       Net settlement pressure

     qall     =       Net allowable B.C (Design allowable B.C)

Shear Failure:

          Failure soil at foundation level due to shear strength happened when increased the foundation load or decreased the resistance of soil for shear.

 Shear failure happened on many stages:

I)                  Stage I: The soil in the elastic case and behave as the part of foundation it still that, and by increasing the load performed the region I which called active zone.

II)              Stage II: At this stage the foundation load effect on the active zone and neighboring soil so perform the region which called arc of logarithmic spiral zone.

III)           Stage III: By increasing the load performed the third part curve in which part the soil became in the passive case it make to resist the failure.

The soil fails when load on foundation increase and became more than the soil resist or B.C of soil. See Fig. (1).

    In this case there are three component produced to resist the failure of soil.

I)                  (Pp)γ   = Component produced by the weight of shear zone II, III.

II)               (Pp)c   = Component produced by the cohesive stress.

III)            (Pp)q = Component produced by the surcharge q.

Terzaghi's Bearing Capacity Theory:

Terzaghi Assumptions:

          The failure for Terzaghi's theory as shown in Fig (2):

1.     The base of foot is rough, to prevent the shear displacement.

2.      The foot is shallow foundation, i.e. the depth of foundation is less than the width of foot…

Df      ≤        B

3.     Shear strength above the level of the base of foot is negligible. i.e.      C = 0 above ( F.L ).

4.     Consider only the surcharge which produced as uniform pressure        q = γDF at foundation level.

5.      The load on foundation is vertical and uniform.

6.     The foot is long strip foot.

As mentioned before ….

      qult     =       (Pp)γ +   (Pp)c  +   (Pp)q

          (Pp)γ   =       Component produced by cohesive stress.

(Pp)c   =       Component produced by surcharge  q = γDF

(Pp)q =       Component produced by the weight of soil in zone II, III.

qult     = C Nc  + q Nq  + 0.50 γ B Nγ

Nc, Nq, Nγ    =   Bearing Capacity factor which are dimensionless depend on angle of shear resistance φ.

        Nq      =      

          a        =   e

          Nc      =      

          Nγ      =      

          Kp      =              = coefficient of passive earth pressure.

 

A.R.E Bearing Capacity Equation:

          من كتاب المواصفات المصرية يحدث الانهيار تحت الأساس لعدت أسباب منها :

1.     زيادة حمل الأساس.

2.     تناقص مقاومة تربة الأساس

3.     حفر التربة المجاورة للأساس  

4.  تقدر قدرة تحمل التربة القصوى إما بالحساب أو بيانياً وهو بدراسة الجزء الموضح في الرسم وباستخدام المعادلات.

وذلك في حالة تربة متجانسة إلى العمق d ومنسوب رسوب المياه الجوفية أوطي من أي أسفل المنسوب d أي أن التربة  جافة فإنه ...

I.       Under concentrated vertical central load …

qult   =   CNcλc  +  qNqλq +  γ2BNγλγ

Where:

 qult         =       Ultimate Bearing Capacity.

      C       =       Cohesive stress.

      q        =       Over burden pressure above (F-L)  

                =       γ1DF

      γ1       =       Unit weight of soil above (F-L) 

      γ2       =       Unit weight of soil at the base of foundation

      B        =       Width of foundation

 Nc, Nq, Nγ = Bearing Capacity (B.C) factors depend on φ (angle of internal friction)             

              Nq      = e πtanφ . tan2 ( 45 +  )

              NC     = ( Nq - 1 ) cotφ    

              Nγ      = ( Nq - 1 ) tanφ

              Nc, Nq, Nγ = F (φ )                    see table (1)

         λC, λq, λγ =

factors depend on the shape of foundation Shape  dimension [B, L].

 

Foundation λC  -  λq λγ

Strip

1.0

1.0

Rectangular

1 + 0.3 B/L

1 – 0.3 B/L

Square & Circle 1.3 0.7

Table (2) The value of shape Factor

          DF   = Depth of foundation , Show Fig (4)

Fig(4) , Cases of depth of foundation

 II.            Eccentric Vertical Load:

      1.  Eccentricity in direction L = eL

        

        A = area =

           The shape factors become

        λC, λq, λγ = F ( B / L-)

 

     2.     Eccentricity in direction B = eB as the mentioned before ..

    B- = B – 2eB

    A = B-.L

   And the shape factor became

    λC, λq, λγ = F ( B / L)

   qu = CNc λC + qNq λq + γB-Nγ λγ

 

     3.     Eccentricity in direction (B, L).

B-  = B – 2eB

            L-  = L – 2eL

           And the shape factors become

           λC, λq, λγ = F ( B- / L- )

            A = B- . L-

 

III.                   Central inclined…

In the case of inclined load R the resultant can analyses at two components H & V where :

H=

V=

 

1.     By increasing the angle of δ the value of Bearing Capacity decreased where δ = tan-1 H/V.

2.     H       ≤            

Where A = Area of foundation

3.     B.C equation become

qult     = C NC λC iC + q Nq λq iq + γB-Nγλγiγ

Where:

iC, iq, iγ = Inclination Factors

iq        =

iγ        =

iC       = iq -

H       = Horizontal component of load

V       = Vertical component of foot

                  φ       = Angle of Internal Friction

    Special Case:

          When          φ       = 0    

                   à      cotφ   = α

                             iq        =   iγ   =   iC  =  1

          When          C       = 0

                             iq        =  

                                      =  

          iγ        =

                                      =  

                             iC       = iq -

      IV.          Eccentric inclined on foot is inclined and eccentricity so we make it as mentioned before in II & III.  


V .    Effect of Ground surface inclination:

The Bearing capacity of soil decrease when the foot lies near from inclination of ground surface. See Fig.(4).

From fig (5) note that …

1.     The surcharge decrease from q to q- so that value of Nq becomes Nq-.

2.     The surface which produced to resist the failure L decrease and become L- so Nc decrease to Nc-.

N γ = as before

    Where:

,  = B.C factors in case of inclined G.S this function

 F (b/B, D/B, β, φ). See table (3) and B.C equation become

                   qult     = CλC + qλq + γB-Nγ λγ

Table (3) From Code

Table (3)

 Example:

          Calculate the allowable B.C (qall) for surcharge foundation 36 m wheares depth of foundation 1.5 m for C – φ soil where φ = 10o ,C = 4 t/m2 and unit weight of soil 1.8 t/m3, and compare the results if there back fill inclinations with β = 60o, b = 0.

 Solve:

        1. φ = 10o    from table    .................»     Nc      = 8.5

                                                                       Nq      = 2.5

                                                                       Nγ      = 0.5

        2.                      

        3. q = γDF = 1.8 ´ 1.5 = 2.7 ton/m2

        4. qult = CNCλC + qNqλq + γB-Nγλγ

           = 4 ×8.5 × 1.15 + 2.7 × 2.5×  1.15+  1.8× 3 ×0.5  × 0.85

            = 49.16 t/m2

    qn ult = 49.16 – 2.7 = 46.46 t/m2

      qall  =            = 15.48 t/m2

                                    = 1.5 Kg/cm2

  For Inclination:

          Φ       = 10

          β        = 60o

                =  = 0.5     .............................»    From table

                = 0

               = 6.33

               = 0.5

               = 0.5

  qult   = CλC + qλq + γBNγλγ

                   = 4 × 6.33 ×1.15 + 2.7 × 0.5 ×1.15 + 1.8 × 3× 0.5 × 0.85

                   = 32.96 ton/m2

  qnet ult = 32.96 – 2.7  = 30.27

   qall      = 30.3 / 3         = 10.1 ton/m2

                                      = 1 Kg / cm2

   % decrease              =

 

VI . Effect of ground water table ( G. W. T):

  qult     = CNCλC + qNqλq + γBNγλγ

   1.)   G.W.T under G.S and above the base

                   q = γDF = γsat . dw + γsub . h1

                   γB      = γsub . B

 

 

    2.)   G.W.T under foundation level:

   a.     If dw > , it means that the water is far from the shear failure plan and his effect so …

   q = γDF

 γB     à as the case of   soil  dry or bulk.

 

      b.   If dw < , and the ( G.W.T ) lies between the base and shear failure plan.

      q   = γ1DF  

      γD = [γsub+ Fw (γ1 - γsub ) ]

  where:

     Fw   = Coefficient depend on φ and  as shown in fig (6).

 

 

Fig (6)

VII.     Effect of Multi Layer:

     To calculate B.C for Multi layer soil for foot ( B . L ).

   1.     Calculate the B.C for first layer by using the properties of this soil (q1all).

   2.     Calculate the B.C for second layer (q2all) by using the properties of second layer γ2, φ2, C2 where B- become ( B + h ) and  = DF + h1. After that calculate the equavelent B.C

 q2equ       =

   3.     Compare the values of q1all  and  q2equ

   If                 q1all > q2equ

   Design         qall = q2equ

   If                 q1all < q2equ

   Design         qall = q1all

   

 

Example:

          As shown in fig the B.C at F.L for clay layer = 1.0 kg/cm2 under clay soil lies organic clay soil at 3.0 m under (F.L) wheares B.C = 0.2 kg/cm2 determine the allowable B.C if the foot …

a.      Strip ( B = 2 ).

b.     Square foot ( 2 ×2 ) m

 

Solve :

          a. for strip foot:

qequ         =

              =       = 0.5 kg/cm2

 q1all        = 1 kg/cm2             qequ    = 0.5 kg/cm2               

qall design  = 0.5 kg/cm2

        

        b. For square:

qequ         =

              =  = 1.25 kg/cm2

qall = 1 kg/cm2                      qequ    = 1.25 kg/cm2            

qall design  = 1 kg/cm2