Derive Terzaghi 1D consolidation theory

Assumptions

1. The clay layer is homogeneous and fully saturated.
2. Soil grains and water are incompressible.
3. Compression and flow are one-dimensional (vertical).
4. Darcy's law is valid; $k$ is constant during consolidation.
5. The coefficient of volume compressibility $m_v$ is constant (linear $e$–$\sigma'$ relation).
6. The load is applied instantaneously and is uniform over a large area, so that excess pore pressure at $t=0$ equals the load.
7. Strains are small and secondary compression is neglected.

Derivation

Consider a clay layer of thickness $H$ with drainage upward. Take an element of thickness $dz$ and unit area at depth $z$, with excess pore pressure $u$ (a function of $z$ and $t$).

   ▼ ▼ ▼ ▼  load Δσ
  ---------------  drainage (top)
   z ↓   +---+  ← element dz
         +---+
  ================ impervious base

Flow through the element. The excess head is $u/\gamma_w$, so the gradient is $i = \dfrac{1}{\gamma_w}\dfrac{\partial u}{\partial z}$ and by Darcy's law the upward velocity is $v = \dfrac{k}{\gamma_w}\dfrac{\partial u}{\partial z}$ (z measured downward).

Water leaves through the top face (velocity $v$) and enters through the bottom face (velocity $v + \frac{\partial v}{\partial z}dz$). The net outflow per unit area is

$$-\frac{\partial v}{\partial z}dz = -\frac{k}{\gamma_w}\frac{\partial^2 u}{\partial z^2}dz$$

Volume change. The decrease in volume of the element per unit time equals the net outflow. With constant total stress, a fall in $u$ is an equal rise in effective stress, $\partial\sigma'/\partial t = -\partial u/\partial t$, so the rate of decrease in volume is

$$m_v\,dz\,\frac{\partial\sigma'}{\partial t} = -m_v\,dz\,\frac{\partial u}{\partial t}$$

Equating:

$$m_v\frac{\partial u}{\partial t} = \frac{k}{\gamma_w}\frac{\partial^2 u}{\partial z^2}$$ $$\boxed{\frac{\partial u}{\partial t} = c_v\frac{\partial^2 u}{\partial z^2}}, \qquad c_v = \frac{k}{m_v\gamma_w}$$

This is Terzaghi's one-dimensional consolidation equation, $c_v$ being the coefficient of consolidation (m²/s).

Solution. With $u = \Delta\sigma$ at $t=0$ and $u = 0$ at drainage faces, the solution is

$$u = \sum_{m=0}^{\infty}\frac{2u_0}{M}\sin\!\left(\frac{Mz}{H_{dr}}\right)e^{-M^2T_v}, \quad M = \frac{\pi}{2}(2m+1), \quad T_v = \frac{c_vt}{H_{dr}^2}$$