Geotechnical Engineering

DHARM

LATERAL EARTH PRESSURE AND STABILITY OF RETAINING WALLS 509

smin

b/3

b/3

b/3

N

R

smax

b/3

b/3

b/3

N

R

smax

(i) e < b/6 (ii) e = b/6

smin=0

b

b/3 b/3 3b¢

N

b¢

smin

(Tension)

smax

b/3

(iii) e > b/6 (iv) b reduced to 3b for case (iii)¢

Fig. 13.51 Distributions of base pressure for different values of eccentricity

of the resultant force on the base

Equations 13.73 and 13.74 apply for the first case—e <

b

6

. For case (ii) e = b/6,

σmax =

2 N

b

...(Eq. 13.75)

and σmin is zero.

For case (iii) e > b/6, tension is supposed to have developed as shown.
Since soil is considered incapable of resisting any tension, the pressure is taken to be
redistributed along the intact base of width 3b′, where b′ is the distance of the line of action of
R (or N) from the toe. σmax is then given by:

σmax =

2

3

N

b′

...(Eq. 13.76)

or σmax =

2

3

2

N

Fb−e

HG

I

KJ

...(Eq. 13.77)

since b′ =

b e

2

F −

HG

I

KJ ...(Eq. 13.78)

σmax should not be greater than the allowable bearing capacity of the soil. (More of the concept
of bearing capacity will be seen in Chapter 14).

(b) The condition of no tension is also easily verified. If tension occurs, there are two
choices, one is to increase the trial base width and go through the calculations again. Another
is to consider that only that part of the base width equal to 3b′ (called the ‘effective’ base width)
is useful in resisting the pressure and recompute σmax as given by Eqs. 13.76 and 13.78 and
verify if criterion (a) is now satisfied.