Soil Mechanics II

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Lecture Note

Transcript of Soil Mechanics II

  • Addis Ababa University Faculty of Technology Department of Civil Engineering

    1 Lecture Note (Instructor: Amsalu Gashaye) Academic Year: 2001E.C

    CHAPTER ONE

    STRESS DISTRIBUTION IN SOIL 1.1 Introduction

    Stress in soil is caused by one or both of the following:

    a) Self weight of soil b) External loading (unloading) = structural loading

    The vertical stress in soil owing to self weight is given by: 'v = z,

    where 'v = the vertical stress in the soil at depth z below the surface = unit weight of soil

    Foundations are designed with an adequate factor of safety against shear failure of soil; thus it is safe to presume that the operating stresses in soil are small enough to assume stress-strain proportionality. Fortunately, the order of magnitudes of stress transmitted in to soil from structural loading are also small and hence the application of elastic theory for the determination of stress distribution in soil gives reasonably valid results. Hence stresses due to applied structural loadings are determined by using elastic theory. The resultant stress can be obtained by superposition of structural load stress and overburden pressure. 1.2 Stress Distribution for a Point Load Bossinesq (1885) has given the solution for the stresses caused by the application of a point load at the surface of a homogenous, elastic, isotropic and semi-infinite medium.

    zy

    A (x,y,z)

    Q

    Yz

    X

    zx

    zyxz

    xyx

    yzyx

    Z

    r

    Q

    R

    z

    y

    xr

    R

    rz

    r

    Fig. 1.1 Point Load on the surface

    With reference to Fig. 1.1, the expressions for the increase in stress at point A due to a point load Q at its surface are

    ratiospoissonvRz

    Rr

    zyxzrR

    yxr

    '

    cos;sin

    22222

    22

    ===

    ++=+=+=

  • Addis Ababa University Faculty of Technology Department of Civil Engineering

    2 Lecture Note (Instructor: Amsalu Gashaye) Academic Year: 2001E.C

    In rectangular coordinates

    )6.1(23

    )5.1(23

    )4.1()(

    )2(321

    23

    )3.1()(

    )2()(

    1321

    23

    )2.1()(

    )2()(

    1321

    23

    )1.1(23

    5

    2

    5

    2

    235

    323

    2

    5

    2

    323

    2

    5

    2

    5

    3

    RyzQRxzQ

    zRRxyzRv

    RxyzQ

    Rz

    zRRyzR

    zRRv

    RzyQ

    Rz

    zRRxzR

    zRRv

    RzxQ

    RzQ

    yz

    xz

    xy

    y

    x

    z

    =

    =

    +

    +=

    +++

    +=

    +++

    +=

    =

    In cylindrical coordinates

    )10.1(23

    )9.1()(

    1)21(2

    )8.1()(

    2132

    )7.1(23

    5

    2

    3

    5

    2

    5

    3

    RrzQ

    Rz

    zRRvQ

    zRRv

    RzrQ

    RzQ

    rz

    r

    z

    =

    +=

    +

    =

    =

    In most foundation problems it is very necessary to be acquainted with the increase in vertical stresses (for settlements) and the increase in shear stresses (for shear strength analysis). The equation for the vertical stress can be rewritten as follows

    25

    2

    2

    25

    222522

    3

    5

    3

    )(11

    23

    )11.1(infdim

    )(11

    23

    )(23

    23

    +=

    ==

    +=+==

    zrI

    factorluenceensionalnonIwherezQI

    zrzQ

    zrzQ

    RzQ

    z

    zz

    z

    Values of Iz for different values of r/z ratio can be tabulated (Table 1.1) or plotted as Iz versus r/z ratio (Fig 1.2) and hence can be used for routine stress calculation. Note that the maximum vertical stress is observed directly below the load (r=0).

  • Addis Ababa University Faculty of Technology Department of Civil Engineering

    3 Lecture Note (Instructor: Amsalu Gashaye) Academic Year: 2001E.C

    Table 1.1r/z I z0 0.4775

    0.2 0.43290.4 0.32950.6 0.22140.8 0.13861.0 0.08441.2 0.05131.4 0.03171.6 0.02001.8 0.01292.0 0.00852.2 0.00582.4 0.0042.6 0.00282.8 0.00213.0 0.0015 Fig 1.2 Non dimensional influence factor

    00.050.1

    0.150.2

    0.250.3

    0.350.4

    0.450.5

    0 0.5 1 1.5 2 2.5 3 3.5r/z

    Iz

    The equation for the shear stress can be rewritten as follows

    )11.1(

    23

    2

    5

    2

    abovegivenfactorInfluenceIwherezQ

    zrI

    zr

    RrzQ

    zzrz

    zrz

    ==

    ==

    Pressure distributions Graphical vertical stress distribution on a horizontal plane at any depth z below the ground surface can be drawn as shown here under. The vertical stress on a horizontal plane at a depth z is given by

    depthspecifiedabeingzz

    QzIz 2

    =

    For several assumed values of r, r/z is calculated and Iz is found for each, the value of z is then computed. As an example consider the following cases Case 1: Let z =c (where c= constant number) and r be varied as 0, 0.25c, 0.5c, 0.75c, c, 1.25c, 1.5c, 1.75c, 2c, etc

    Case 2: Let z=2c and r be varied as case 1 Case 3: Let z=4c and r be varied as case 1

    Now the stresses are calculated (tabulated) and plotted together below.

  • Addis Ababa University Faculty of Technology Department of Civil Engineering

    4 Lecture Note (Instructor: Amsalu Gashaye) Academic Year: 2001E.C

    r r/z I z (zc2)/Q r r/z I z (zc2)/Q r r/z I z (zc2)/Q0 0 0.4775 0.4775 0 0 0.4775 0.1194 0 0 0.4775 0.0298

    0.25c 0.25 0.4103 0.4103 0.25c 0.125 0.4593 0.1148 0.25c 0.0625 0.4728 0.02960.5c 0.50 0.2733 0.2733 0.5c 0.250 0.4103 0.1026 0.5c 0.125 0.4593 0.0287

    0.75c 0.75 0.1565 0.1565 0.75c 0.375 0.3436 0.0859 0.75c 0.1875 0.4380 0.0274c 1.00 0.0844 0.0844 c 0.500 0.2733 0.0683 c 0.25 0.4103 0.0256

    1.25c 1.25 0.0454 0.0454 1.25c 0.625 0.2094 0.0523 1.25c 0.3125 0.3782 0.02361.5c 1.50 0.0251 0.0251 1.5c 0.750 0.1565 0.0391 1.5c 0.375 0.3436 0.0215

    1.75c 1.75 0.0144 0.0144 1.75c 0.875 0.1153 0.0288 1.75c 0.4375 0.3082 0.01932c 2.00 0.0085 0.0085 2c 1.000 0.0844 0.0211 2c 0.5 0.2733 0.0171

    Case 1 (z=c) Case 2 (z=2c) Case 3 (z=4c)

    0.000.050.100.150.200.250.300.350.400.450.50

    -3 -2 -1 0 1 2 3r/c

    zc2 /Q

    z=cz=2cz=4c

    From the above curves one can see that the stress diminishes as we move down from the ground surface and also as we move away from the point of load application. Similarly vertical stress distribution on a vertical plane at any radial distance r from the load can be drawn as shown here under. For several assumed values of z, r/z is calculated and Iz is found for each, the value of z is then computed. As an example consider the following cases Case 1: Let r=a (where a= constant number) and z be varied as 0, 0.5a, a, 2a,

    5a, 10a, etc Case 2: Let r=2a and z be varied as case 1 Case 3: Let r=4a and z be varied as case 1

    Now the stresses are calculated and plotted below.

    z r/z I z (za2)/Q z r/z I z (za2)/Q z r/z I z (za2)/Q0 - - indet. 0 - - indet. 0 - - indet.

    0.5a 2.00 0.0085 0.0342 0.5a 4.0 0.0004 0.0016 0.5a 8 0.0000 0.0001a 1.00 0.0844 0.0844 a 2.0 0.0085 0.0085 a 4 0.0004 0.0004

    2a 0.50 0.2733 0.0683 2a 1.0 0.0844 0.0211 2a 2 0.0085 0.00215a 0.20 0.4329 0.0173 5a 0.4 0.3295 0.0132 5a 0.8 0.1386 0.005510a 0.10 0.4657 0.0047 10a 0.2 0.4329 0.0043 10a 0.4 0.3295 0.0033

    Case 1 (r=a) Case 2 (r=2a) Case 3 (r=4a)

    0

    2

    4

    6

    8

    10

    12

    0.00 0.02 0.04 0.06 0.08 0.10

    z/a

    za2/Q

    r=ar=2ar=4a

  • Addis Ababa University Faculty of Technology Department of Civil Engineering

    5 Lecture Note (Instructor: Amsalu Gashaye) Academic Year: 2001E.C

    Stress isobar or Pressure Bulb An isobar is a stress contour or a line which connects all points below the ground surface at which the vertical pressure is the same. In fact, an isobar is a spatial curved surface and resembles a bulb in shape; this is because the vertical pressure at all points in a horizontal plane at equal radial distances from the load is the same. Thus, the stress isobar is also called the bulb of pressure or simply the pressure bulb. The vertical pressure at each point on the pressure bulb is the same. An isobar diagram, consisting of a system of isobars appears as shown in Fig 1.3.

    Z

    Isobars

    Q

    Fig 1.3 Isobar diagram

    The procedure for plotting an isobar is as follows. Let it be required to plot an isobar for which z=0.2Q per unit area (20% isobar) From 22.0

    22.0

    2

    2z

    Q

    zQzI

    Q

    zz

    zIz

    QzIz ====

    Assuming various values for z, the corresponding Iz- values are computed; for these values of Iz, the corresponding r/z-values are obtained; and, for the assumed values of z, r-values are got. It is obvious that, for the same value of r on any side of the z-axis, or line of action of the point load, the value of z is the same; hence the isobar is symmetrical with respect to this axis. When r=0, Iz=0.4775; the isobar crosses the line of action of the load at a depth of: unitszIzzIzzzI 545.12.0/4775.02.0/2.0/2.0

    22 ===== The calculations are best performed in the form of a table as given below. The plot is shown too.

    z Iz r/z r (unit) z0.5 0.0500 1.211 0.605 0.2Q1.0 0.2000 0.645 0.645 0.2Q1.5 0.4500 0.155 0.232 0.2Q

    1.5452 0.4775 0.0 0.000 0.2Q

    0

    0.4

    0.8

    1.2

    1.6

    2

    -1.0 -0.5 0.0 0.5 1.0r

    z

    Westergaard (1938) has obtained an elastic solution for stress distribution in soil under a point load. He has assumed the soil to be lateral