Terzaghi gave the theory for the determination of the rate of consolidation of a saturated soil mass subjected to a static, steady load. The theory is based on the following assumption:
The soil is homogeneous and isotropic.
The soil is fully saturated.
The solid particles and water in the voids are incompressible. The consolidation occurs due to expulsion of water from the voids.
The coefficient of permeability of the soil has the same value at all the points, and it remains constant during the entire period of consolidation.
Darcys law is valid throughout the consolidation process.
Soil is laterally confined, and the consolidation takes place only in axial direction. Drainage of water also occurs only in the vertical direction.
The time lag in consolidation is entirely due to the low permeability of the soil.
There is a unique relationship between the void ratio and effective stress, and this relationship remains constant during the load increment. In other words, the coefficient of compressibility and the coefficient of volume change are constant.
Let us consider a saturated clay layer of thickness 2D sandwiched between two layer of sand is shown in the figure. When a uniform pressure of `triangle sigma` is applied on the surface of the top sand layer ,the total stress developed at all points in the clay layer is increased by `triangle sigma`.
As explained in the spring analogy model, initially the whole of the pressure is taken up by water, and the hydrostatic excess pressure of `(triangle sigma)/gamma_w` develops. At time `t=0`,just after the application of the load, the excess hydrostatic pressure `bar(u_i)` is equal to `(triangle sigma)/gamma_w` throughout the clay layer. This is represented by the original line `AB`. The excess hydrostatic pressure is independent of the position of the water table for convenience, the water table is assumed at the level of the surface of the clay layer.
The hydrostatic excess pressure at the top and the bottom of the clay layer, indicated by point C and D in the pressure diagram drops to zero.However, the excess hydrostatic pressure in the middle portion of the clay layer at `D` remains high. The curve indicating the distribution of excess hydrostatic pressure are known as isochrones. The isochrone `CDE` indicates the distribution of excess hydrostatic pressure at time t.
As the consolidation progress, the excess hydrostatic pressure in the middle of the clay layer decreases. Finally at time `t=t_f`, the whole of the excess hydrostatic pressure has been dissipated, and the pressure distribution is indicated by the horizontal isochrone `CFE`.
Let us consider the equilibrium of an element of the clay at a depth of Z from its top at time t. The consolidation pressure `triangle sigma` is partly carried by water and partly by solid particles as:
`triangle sigma=triangle bar(sigma)+bar(u)`
Where,
`triangle bar(sigma)=`the pressure carried by the solute particles
`bar(u)=`the excess hydrostatic pressure
Now,hydraulic gradient `i` at that depth is,
`i=(delh)/(delz)=(del(bar(u)/gamma_w))/(del z) `
Or,`i=1/gamma_w*(del baru)/(del z)`
Where, `bar u` is the excess hydrostatic pressure at depth `z`.
Let us consider a thin slice of clay layer `triangle z` at depth `z`. Now, the pressure difference across this thickness is given by:
`triangle baru=(bar u+(delbar u)/(del z)*del z)-bar u`
`=(del bar u)/(del z)*dz`
The unbalanced head across the thickness is given by,
`triangle h=(triangle bar u)/gamma_w`
`=1/gamma_w(del bar u)/(del z)*dz`
Thus, the hydraulic gradient becomes,
`i=(triangle h)/dz=1/gamma_w*(del bar u)/(del z)`
Again from darcy law, the velocity of flow at depth `z` is,
`v=k*i=k*1/gamma_w*(del bar u)/(del z)`
The velocity of flow at the bottom of the element of thickness `triangle z` can be written as,
`=v+(del v)/(del z)*dz`
`=v+del/(del z)(k/gamma_w*(del bar u)/(del z))*dz`
Or,`v+(del v)/(delz)dz=v+ k/gamma_w(del^2baru)/(del z^2)*dz`
Or,`(del v)/(del z)=k/gamma_w(del^2baru)/(del z^2)`
The discharge entering the element is,
`Q_(Inp)=v*(trianglex*triangley)`
Where `trianglex` and `triangley` are the dimension of the element.
The discharge leaving the element is,
`Q_(out)=v+(del v)/(del z)*dz(trianglex*triangley)`
Therefore, the net discharge squeezed out of the element is:
`triangle Q=Q_(out)-q_(Inp)`
`=[v+(del v)/(del z)*dz-v](trianglex*triangley)`
`=(del v)/(del z)(trianglex*triangley* triangle z)`
As the water is squeezed out, the effective stress is increased and the volume of the soil mass decreases.
We have,
`triangle V=-m_vV_0 triangle bar(sigma)`
Where,
`V_0=`initial volume of soil mass
`=trianglex*triangley* triangle z`
`triangle bar (sigma)=`Increase in effective stress.
The decrease in volume of soil per unit time is,
`del /(del t)(triangle V)=-m_v*trianglex*triangley* triangle z*del/(del t)(triangle bar sigma)`
As the decrease in volume of soil mass per unit time is equal to the volume of water squeezed out per unit time,We have,
`(del v)/(del z)(trianglex*triangley* triangle z)==-m_v*trianglex*triangley* triangle z*del/(del t)(triangle bar sigma)`
`(del v)/(delz)=-m_v*del/(del t)(triangle bar sigma)`
We know that,
`triangle bar sigma=triangle sigma-bar u`
`del/(delt)(triangle bar sigma)=del/(delt)(triangle sigma)-del/(delt)(bar u)`
For a given pressure increment,`del (triangle sigma)=0`,so,
`del/(delt)(triangle bar sigma)=-del/(delt)(bar u)`
Using this, we get,
`(del v)/(del z)=m_v (del bar u)/(del t)`
Equating the two values of `(delv)/(delz)`, we get,
`k/gamma_w*(del^2bar u)/(del z^2)=m_v*(del bar u)/(del t)`
Or, `c_v*(del^2bar u)/(del z^2)=(del bar u)/(del t)`...........(i)
Where, `c_v=`coefficient of consolidation and is given by,
`c_v=k/(gamma_wm_v)=k/(g rho_w m_v)`
Equation (i) is the basic differential equation of one-dimensional consolidation. It gives the distribution of hydrostatic excess pressure `bar u` with depth `z` and time `t`.
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