Allowable Bearing Capacity and Settlement by wku51683

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									Shallow Foundations
Allowable Bearing
Capacity and Settlement
           Introduction

Apart from bearing pressure, the other major
design consideration for shallow footings is
settlement
Excessive settlement (primarily differential
settlement) can cause a number of
problems
               Settlement
• Calculate vertical stress increase
• Calculate settlement due to the stress
  increase
• Bearing capacity based on settlement
  criterion
  Components of Settlement
• Initial or elastic settlement, Se
• Primary consolidation settlement, Sc
• Secondary compression or creep, Sc(s)
  It occurs at constant effective stress   (i.e. no drainage of pore water
  occurs) and is irreversible.

• The total settlement, S = Se + Sc + Sc(s)
              Elastic Settlement

                                         Q
           


                                              E
H         E
                        Generalized stress
               z        and strain field



                                     
    Se = H /E = H.z     Se =       z .dz
                                 0
Stress Due to a Concentrated Load
• Boussinesq solution                                    P


                 3P
                            5/ 2             R
                                                        z
             r      2
                           
       2 z 1   
            2
                                          z
             z
                          
                           
  where                                             r

  r  x2  y 2                        x
                                                    y
       Stress due to a Circularly Loaded Area
For the incremental area shown settlement at the centre can be calculated as:
Load on incremental area
                                         load, q
                 q o rd  dr
d                                5/2                          dr
              r            2

        2 z 1    
             2

              z                                            d
                     
 After the double integration                      a
                                                           r
 At Centre :
                                                     z
                            
                   1        
       q 1              
                      2 3/ 2
               B                                     z
             1     
               2z   
                      
                Distribution of Stress
                    (from a vertical line load)
The stresses at point X due to a line load of Q
per unit length on the surface are as followings:
                                                             Q/m
        2Q    Z3
   z 
         (x2  Z 2 )2
                     2
        2Q     xZ                                             z
   x 
         ( x 2  Z 2 )2                            z

          2Q     xZ 2                        x
    xy                                                 x
           ( x 2  Z 2 )2
       Square             B/2          Strip
                B/2
1.5B   1B                         1B       2B   3B
                            .9
                      0.9 0

                  0.7
                           0.5                   1B
                  0.5

                           0.3
1B                0.3                            2B
                           0.2
                  0.2
                                                 3B
                            0.1
2B               0.1                             4B


                                                 5B
                           0.05

3B                                               6B
                0.05

                                                 7B
      Stress below a Rectangular Area
  • Solution after Newmark for
    stresses under the corner of a
                                                         L
    uniformly loaded flexible
    rectangular area:                               B
  • Define m = B/z and n = L/z
  • Solution by charts or
    numerically
                                                    z
    = qo.I
      (It must be emphasised that the chart
       gives the stresses under the corner of   z
       uniformly load flexible rectangular
       area.)

                                               2 2   1/2
I = 1 2mn(m2+n2+1)1/2 . m2+n2+2 + tan-1 2mn(m +n +1)
    4 m2+n2-m2n2+1     m2+n2+1          m2+n2-m2n2+1
                             Stress Influence Chart
  0 .2 6
I                                                                              m = oc

  0 .2 4                           A rea covered           m = 2.0
                                                                     m = 3 .0
                                                                m = 2.5                             The influence
                                   w ith u nifo rm       m = 1 .8

  0 .2 2
                                  n orm al load,
                                  q
                                                        m = 1.6
                                                       m = 1 .4
                                                                                m = 1 .2            factor is
                       nz

  0 .2 0       y
                             mz
                                     z
                                           x                                    m = 1.0
                                                                                m = 0.9
                                                                                                    shown in
  0 .1 8
                                      z =
                                         q.I
                                               
                                                                                m = 0.8
                                                                                m = 0 .7
                                                                                                    Table 5.2 (Pgs
  0 .1 6
                         z
                                                                                m = 0 .6
                                                                                                    208, 209)
           N ote: m and n are interch ang eable                                 m = 0 .5
  0 .1 4

  0 .1 2                                                                        m = 0.4


  0 .1 0
                                                                                m = 0.3

  0 .0 8
                                                                                m = 0.2
  0 .0 6

  0 .0 4                                                                        m = 0 .1

  0 .0 2
                                                                                m = 0 .0
       0
       0.0 1       2    3 4 5     0 .1     2   3 4 5        1.0      2    3 45             10
                                                                                                n
                       V E R T IC AL S T R E S S BE LO W A C O R N E R
                        O F A U N IF O R M LY L O A D E D F L E X IB L E
                                  R E C TA N G U LA R AR E A .
2:1 method




          qo BL
 
     ( B  z )( L  z )
Average Vertical Stress Increases
Due to a Rectangularly Loaded Area




                           qo I a
                         I a  f (m2 , n2 )
                              B
                         m2 
                              H
                              L
                         n2 
                              H
Average Vertical Stress Increases
Due to a Rectangularly Loaded Area




                      qo I a
                    I a  f (m2 , n2 )
                         B
                    m2 
                         H
                         L
                    n2 
                         H
Average Vertical Stress Increases
         in a given layer




                     H 2 I a ( H 2 )  H1 I a ( H1 ) 
              qo                                  
                              H 2  H1               
     Elastic settlement Based on the
            Theory of Elasticity
1. Determine vertical strains:
                                                    Q
2. Integrate strains:

 z = 1 [z -  (x + y )]
      E
                                           R        z
             
  Se =       z .dz
         0                            z

                                               r

                                 x
                                               y
  Elastic Settlement of Rectangular footings

           )(1-s2)IsIf/Es
Se = qo (B’
where
qo = net applied pressure on the foundation
s = Poisson’ ratio of soil
            s
B’= B/2 for center of foundation
   =B for the corner of foundation
Is = shape factor = F1 + (1 -2s)(1-s)F2 (see Tables 5.4 and 5.5)
If = depth factor (see Fig. 5.15; pg 227)
  Elastic Settlement on Saturated Clay
  The average vertical immediate displacement under a
  flexible area carrying a uniform pressure q is given by
              q0 B
   se  A1 A2
               Es
  where
  A1 depends on the shape of the loaded area
  A2 depends on the depth of footing


The above solutions for vertical displacement are used mainly to estimate the
immediate settlement of foundation on saturated clays; such settlement occurs
                                                             s
under undrained conditions, the appropriate value of Poisson’ ratio being 0.5.
The value of the undrained modulus Es is therefore required.
A2




A1




Fig. 5.17; Pg. 231
         Multi-layer systems
         q
             Se = Se(H1,E1) + Se(H1+H2,E2) - Se(H1,E2)


     B
H1                               E1

H2                               E2


             “Rigid”
             Flexible vs Rigid

              F                       F


          stress
                              stress

      deflection          deflection

centre               0.93 centre   RF = 0.93
        Settlement in Sandy Soil
                (Schmertmann, 1970)
The method was developed as a means to computing the
settlement of spread footings on sandy soils
It has a physical base and calibrates with empirical data
CPT results are often used with this method
Schmertmann conducted extensive research on the
distribution of vertical strain below the spread footings
He found that the greatest strains do not occur
immediately below the footing but at a depth of ½ B to B
depends on the footing shape
The distribution of vertical strain is idealized as two
straight lines
Idealized Distribution of Z
Peak Vertical Strain Influence
          Factor Izp


              q       q




                     q q
    I zp  0.5  0.1
                        
                       vp
Settlement in Sandy Soil
        (Schmertmann, 1970)
                   layer  n
                           Iz
Se  C1C2 (q  q )           z
                  layer 1 Es

              q 
C1  1  0.5     
             q q
                     t 
C 2  1  0.2 log10      
                     0.1 
t  0.1 (in years)
q  stress at the level of the foundation
q =  Df
 Range of Material Parameters for
  Computing Elastic Settlement
• Elastic modulus can be determined:
   – Triaxial tests
   – Unconfined compression tests
   – In-situ tests
      • SPT: Es = N60
      • CPT, Es = 2.5 qc (Square footing), 3.5 qc (Wall footing)
      • PMT, DMT,…
• Es (sand): 10 ~ 55 MPa
• Es (clay): 4 ~ 100 MPa

Elastic modulus can also be estimated using Table 5.8
      Primary Consolidation
• A phenomenon occurs in both sands and clays
• Expulsion of water from soils accompanied by
  increase in effective stress and strength
• Amount can be reasonably estimated based on
  lab data, but rate is often poorly estimated
   Consolidation Settlement
This method makes use of the results of the
conventional oedometer test where the consolidation
parameters of the soil are measured
To compute the stress changes within the soil mass.
The stress changes are computed using a Boussinesq-
type approach assuming elasticity
The important parameter for consolidation settlement
calculation is the net effective stress change in the soil
Usually the settlements are calculated for the soil
divided into a number of sub-layers and the final total
settlement is the sum of individual sub-layer settlements
          Use Compression Index
The compression index (Cc) is used to calculate
the settlement for net stress change beyond the
preconsolidation pressure, 'c

The recompression index (Cs) is used to calculate
the settlement for net stress change from initial
stress to the preconsolidation pressure

            Cs        c   Cc         
                                       o   av 
Sc( p )          log           log              Hc
           1  eo       
                        o 1  eo         c     
     Secondary Consolidation
• At the end of primary settlement,
  settlement may continue to develop due to
  the plastic deformation (creep) of the soil
• The stage of consolidation is called
  secondary consolidation
Secondary Compression Index




Empirical relation: C  0.04 Cc for inorganic clays and silts
                   C  0.05 Cc for organic clays and silts
                   C  0.075 Cc for peats
             Field Load Test
• Ultimate load,
  allowable load, and
  settlement can be
  determined from a
  field load test
  (ASTM D-1194-72)
• The test plates are
  either square (12” or
  18”) or circular (12”
  to 30”)
  Allowable Bearing Capacity for
  Sandy Soil based on Settlement
• Meyerhof proposed a correlation for the
  net allowable bearing capacity for
  foundations with SPT N value
                          Se
  qnet    20 N 60 Fd (      )       B  1.22 m
                        25mm
                    B  0.3 
                                 2
                                      Se
  qnet  12.5 N 60           Fd (      )        B  1.22 m
                    B             25mm
                    Df
  Fd  1  0.33
                     B
  Allowable Bearing Capacity for
  Sandy Soil based on Settlement
• Meyerhof proposed a correlation for the
  net allowable bearing capacity for
  foundations with CPT qc value
            qc   Se
    qnet    (      )       B  1.22 m
            15 25mm
            qc  B  0.3 
                        2
                             Se
    qnet                (    )    B  1.22 m
            25  B  25mm
       Reliability of Settlement
            Computations
• The predication is quite satisfactorily in
  general
• The predication is better for inorganic,
  insensitive clays than for others
• The time rate of consolidation settlement
  is not well-predicted
                         s
   Example of Schmertmann’ method
A footing 2.5 m square carries a net foundation pressure
   of 150 kN/m2 at a depth of 1 m in a deep deposit of fine
   sand. The water table is at a depth of 4 m. Above the
   water table the unit weight of the sand is 17 kN/m3 and
   below the water table the saturated unit weight is 20
   kN/m3. The variation of cone penetration resistance
   with depth is given in following table.
    Depth (m)    1-1.9   1.9-2.4   2.4-8
   qc (MN/m2)     2.3     3.6       5

Estimate the settlement of the footing using
                  s
  Schmertmann’ method.
             2B
             Iz
S = C1C2  Z
           0 E


• Estimate the maximum Izp
• Plot influence factor curve on top of the qc/z plot
• Divide the soil profile into layers assuming qc is
  constant for each sub-layer
• Estimate the E based on qc for each layer
• Determine Iz at the center of each layer
• Calculate the ratio Iz*z/E
• Estimate C1 and C2
• Calculate settlement
B = 2.5 m
Square footing 2B(5 m)
Depth of influence is from 1m to 6m
Izp at B/2 (1.25m)
                         
       Izp = 0.5 + 0.1  '
                          v p
 = 150 kPa
v’ = 2.25 x 17 = 38.25
    p
Izp = 0.698
Plot qc versus z to obtain sub-layers
 Layer         z           qc          E            Iz       Iz*z/E
    1          0.9         2.3        5.75        0.316       0.0495
    2         0.35         3.6           9        0.601       0.0234
    3         0.15         3.6           9        0.684       0.0114
    4          3.6           5        12.5        0.335       0.0965

                                                               0.181
C1 = 1 – 0.5*(1*17)/150 = 0.943; The correction factor for creep will be
taken as unity since there is no time given.
S = 0.943 * 150 * 0.181 = 25.57 ~ 26 mm

								
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