Earth Pressure
Earth Pressure
Q: Explain active earth pressure, passive earth pressure and earth pressure at rest
Soil is neither a solid nor a liquid, but it exhibits some of the characteristics of both. One of the characteristics similar to that of a liquid is its tendency to exert a lateral pressure against any object in contact.
This important property influences the design of retaining walls, abutments, bulkheads, sheet pile walls, basement walls and underground conduits which retain or support soil, and, as such, is of very great significance.
Retaining walls are constructed in various fields of civil engineering, such as hydraulics and irrigation structures, highways, railways, tunnels, mining and military engineering.
Lateral earth pressure is the force exerted by the soil mass upon an earth-retaining structure, such as a retaining wall
In 1929 Terzaghi (The Father of Soil Mechanics) conducted experiments on the retaining wall and showed the relation of pressure on the wall. Wall changes its position i.e to move inwards to the backfill, outwards of it or remain at its place.
There are three types of earth pressures on the basis of the movement of the wall.
Active earth pressure:
The wall moves away from the backfill
there is a decrease in the pressure on the wall and this decrease continues until a minimum value is reach
After which there is no reduction in the pressure and the value will become constant.
This kind of pressure is known as active earth pressure.
Passive earth pressure:
The wall moves towards the backfill
there is an increase in the pressure on the wall and this increase continues until a maximum value is reach
After which there is no increase in the pressure and the value will become constant.
This kind of pressure is known as passive earth pressure.
Pressure at rest:
When the wall is at rest and the material is in its natural state then the pressure applied by material is known as Earth Pressure at Rest.
It is represented by Po.
Active pressure = Passive pressure
Explain Coefficient of active, passive pressure and earth pressure at rest
Coefficient of active earth pressure at rest:
When the retaining wall is at rest then the ratio between the lateral earth pressure and the vertical pressure is called the co-efficient of the earth pressure at rest,
for normally consolidated soils
K0(NC)=1-sinφ
Co-efficient of active earth pressure:
When the retaining wall is moving away from the backfill the ratio between lateral earth pressure and vertical earth pressure is called coefficient of active earth pressure.
Ka=1-sinφ1+sinφ= 45o- 2
Coefficient of passive earth pressure:
When the retaining wall is moving towards the backfill, then the ratio between the lateral earth pressure and the vertical earth pressure is called the Coefficient of passive earth pressure.
Ka=1+sinφ1-sinφ= 45o+ 2
Q: State concept of Rankine’s and Coulomb’s theory of earth pressure
Q: Differentiate between Rankine’s and Coulomb’s theory of earth pressure
Q: Explain Rankine’s and Coulomb’s theory of earth pressure
Rankine (1857) developed his theory of lateral earth pressure when the backfill consists of dry, cohesion less soil.
The following are the important assumptions in Rankine’s theory:
The soil mass is semi-infinite, homogeneous, dry and cohesion less.
The ground surface is a plane which may be horizontal or inclined.
The face of the wall in contact with the backfill is vertical and smooth. (In other words, the friction between the wall and the backfill is neglected).
The wall yields about the base sufficiently for the active pressure conditions to develop; if it is the passive case that is under consideration, the wall is taken to be pushed sufficiently towards the fill for the passive resistance to be fully mobilised.
The following are the assumptions in Coulomb’s theory:
1. The backfill soil is considered to be dry, homogeneous and isotropic; it is elastically un-deformable but breakable, granular material, possessing internal friction but noncohesion.
2. The rupture surface is assumed to be a plane for the sake of convenience in analysis. It passes through the heel of the wall. It is not actually a plane, but is curved and this is known to Coulomb.
3. The sliding wedge acts as a rigid body and the value of the earth thrust is obtained by considering its equilibrium.
4. The position and direction of the earth thrust are assumed to be known. The thrust acts on the back of the wall at a point one-third of the height of the wall above the base of the wall and makes an angle δ, with the normal to the back face of the wall. This is an angle of friction between the wall and backfill soil and is usually called ‘wall friction’.
5. The problem of determining the earth thrust is solved, on the basis of two-dimensional case of ‘plane strain’. This is to say that, the retaining wall is assumed to be of great length and all conditions of the wall and fill remain constant along the length of the wall. Thus, a unit length of the wall perpendicular to the plane of the paper is considered.
6. The back face of the wall is a plane.
(Sample example) Retaining wall with soil
Out of syllabus (new course)
Q: Explain Rebhann’s Graphical Method
Let AB represent the back face of the wall and AD the backfill surface
Draw BD inclined at φ with the horizontal from the heel B of the wall to meet the backfill surface in D.
Draw BK inclined at ψ (= α – δ) with BD, which is the ψ-line.
Through A, draw AE parallel to the ψ-line to meet BD in E. (Alternatively, draw AE at (φ + δ) with AB to meet BD in E).
Describe a semi-circle on BD as diameter.
Erect a perpendicular to BD at E to meet the semi-circle in F.
With B as centre and BF as radius draw an arc to meet BD in G
Through G, draw a parallel to the ψ-line to meet AD in C.
With G as centre and GC as radius draw an arc to cut BD in L; join CL and also draw a perpendicular CM from C on to LG.
BC is the required rupture surface.
Q: Explain Culmann’s Graphical Method
Draw the ground line, φ-line, and ψ-line, and the wall face AB.
Choose an arbitrary failure plane BC1. Calculate weight of the wedge ABC and plot it as B-1 to a convenient scale on the φ-line.
Draw 1 – 1′ parallel to the ψ-line through 1 to meet BC1 in 1′. 1′ is a point on the Culmann-line.
Similarly, take some more failure planes BC2, BC3, ..., and repeat the steps (ii) and (iii) to establish points 2′, 3′, ...
Join B, 1′, 2′, 3′, etc., smoothly to obtain the Culmann curve.
Draw a tangent t-t, to the Culmann line parallel to the φ-line.
Let the point of the tangency be F′
Draw F′F parallel to the ψ-line to meet the φ-line in F.
Join BF′ and produce it to meet the ground line in C.
BF′C represents the failure surface and FF ′ represents Pa to the same scale as that chosen to represent the weights of wedges.
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