Wall in cohesive soil

The active pressure at any depth z may be expressed as

Where

= vertical pressure,

Z = depth form the surface of the backfill

The passive pressure at any depth y below the dredge line may be expressed as

The soil is supposed to be in tension up to a depth of and the pressure on the wall is zero in this zone. The net pressure distribution on the wall is shown by the shaded triangle.

At the dredge line (at point A)

(a) The active pressure acting towards the left is

When

Where = unconfined compressive strength of the clay soil (b) The passive pressure acting towards the right at the dredge line is

since The resultant of the passive an active pressures at the dredge line is

The resultant of the passive and active pressures at any depth y below the dredge line is

Passive pressure

Active pressure, The resultant pressure is

It indicates that resultant pressure remains constant at at all depths.If the passive pressure is developed on the backfill side at the bottom of the pile (point B), then

acting towards the left

acting toward the rightThe resultant is

For static equilibrium, the sum of all the horizontal forces must be equal zero, that is,

Simplifying,

, therefore,

Also, for equilibrium, the sum of the moments at any point should be zero. Taking moments about the base,

Substituting for h and simplifying,

Where,

Maximum bending moment The maximum bending moment may occur within the depth (D-h) below the

dredge line. Let this depth be below the dredge line for zero shear. We may write,

Or The expression for maximum bending moment is,

Where The section modulus of the sheet pile may now be calculated as before.

DESIGNIn carrying out the design, the following points should be considered:

A qs=10kN/m2 surcharge must be applied behind the wall.The excavation pitch will be drainaged during excavation, we will consider the case with water table at great depth.The surcharge pressure can be considered as a layer of depth s

10% of the wall height is 0.5m, so the design height will be 6+0.7=6.7mThe pressure distribution is assumed as shown in figure.

For

of wall

For the determination of h, equate the summation of all horizontal forces to zero, thus

Therefore, For the determination of D, taking moments of all the forces about the base of the wall, we have

Or Substituting for h and simplifying, we have

or 14.6D2 - 7.252D - 4.176 = 0 (by using equation for C)Solving D = 0.838m; increasing D by 40%, we have D = 1.4 x ( 0.838 ) = 1.1732mMaximum bending moment

m

of wall