Geotechnical Engineering

DHARM

360 GEOTECHNICAL ENGINEERING

The calculations are best performed in the form of a table as given below:

Table 10.3 Data for isobar of σz = 0.1 Q per unit area

Depth z (units) Influence r/z r (units) σz

Coefficients KB

0.5 0.0250 1.501 0.750 0.1 Q

1.0 0.1000 0.932 0.932 0.1 Q

1.5 0.2550 0.593 0.890 0.1 Q

2.0 0.4000 0.271 0.542 0.1 Q

2.185 0.4775 0 0 0.1 Q

In general, isobars are not circular curves. Rather, their shape approaches that of the

lemmiscate.

10.2.4 Westergaard’s Solution

Natural clay strata have thin lenses of coarser material within them; this accentuates the non-

isotropic condition commonly encountered in sedimentary soils, which is the primary reason

for resistance to lateral strain in such cases.

Westergaard (1938) has obtained an elastic solution for stress distribution in soil under

a point load based on conditions analogous to the extreme condition of this type. The material

is assumed to be laterally reinforced by numerous, closely spaced horizontal sheets of negligi-

ble thickness but of infinite rigidity, which prevent the medium from undergoing lateral strain;

this may be viewed as representative of an extreme case of non-isotropic condition.

The vertical stress σz caused by a point load, as obtained by Westergaard, is given by:

σz =

Q

z r

z

2 2 32

1

2

12

22

12

22

. /

π

υ

υ

υ

υ

−

−

−

−

F

HG

I

KJ

+F

HG

I

KJ

L

N

M

M

O

Q

P

P

...(Eq. 10.13)

The symbols have the same meaning as in the case of Boussinesq’s solution; v is Poisson’s
ratio for the medium, and may be taken to be zero for large lateral restraint. (The gives, in
fact, the flattest curve for stress distribution, as shown in Fig. 10.6, a flat curve being the
logical shape for a case of large lateral restraint). Then the equation for σz reduces to:

σz =

Q

zrz^2232

1

12

.

/

[(/)]/

π

+

...(Eq. 10.14)

or σz = K

Q

w z

. 2 ...(Eq. 10.15)

where Kw =

1

12 232

/

[(/)]/

π

+ rz

...(Eq. 10.16)