---
title: "Derive Terzaghi 1D consolidation theory"
subject: "Soil Mechanics"
chapter: 9
key: "semester_4/soil_mechanics:terzaghi-1d-consolidation-derivation"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/terzaghi-1d-consolidation-derivation
site: BCT Engineering
---

# Derive Terzaghi 1D consolidation theory

Soil Mechanics · Chapter 9: Compressibility of Soil

- Asked in: 2075 Bhadra, 7 marks; 2074 Bhadra, 7 marks (asked 2 times)
- Video: [Terzaghi's 1D consolidation Derivation](https://www.youtube.com/watch?v=Tv40F8Ye7gY) — H.M VIJAYA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/terzaghi-1d-consolidation-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9

## Question

Derive an expression (governing differential equation) for the one dimensional consolidation theory suggested by Terzaghi.

## Short answer

For a saturated element, net outflow −(k/γ_w)∂²u/∂z² dz equals volume decrease −m_v dz ∂u/∂t. Equating gives ∂u/∂t = c_v ∂²u/∂z² with c_v = k/(m_v γ_w). This is Terzaghi’s one-dimensional consolidation equation.

## Answer

#### 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 $t$).

```text
   ▼ ▼ ▼ ▼  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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the derivation the velocity of flow is obtained from
   A. Hooke's law
   B. the hydraulic gradient i = (1/gamma_w) du/dz and Darcy's law
   C. Bernoulli's equation only
   D. the effective stress
   Answer: B. Excess head is u/gamma_w, so Darcy's law gives v = (k/gamma_w) du/dz. (Key point: Darcy flow)

2. The rate of volume decrease of the element is expressed as
   A. k dz du/dt
   B. m_v dz d(sigma')/dt = -m_v dz du/dt
   C. gamma_w dz du/dz
   D. e dz dt
   Answer: B. At constant total stress, a fall in u is an equal rise in effective stress, so volume change rate = -m_v dz du/dt. (Key point: Volume change)

3. The governing equation of Terzaghi's theory is
   A. du/dt = cv d^2u/dz^2
   B. du/dz = cv d^2u/dt^2
   C. u = cv t
   D. d^2u/dt^2 = cv du/dz
   Answer: A. Equating net outflow with volume decrease gives du/dt = cv d^2u/dz^2. (Key point: Governing equation)

4. The coefficient of consolidation is
   A. k m_v gamma_w
   B. k/(m_v gamma_w)
   C. m_v/(k gamma_w)
   D. gamma_w/(k m_v)
   Answer: B. cv = k/(m_v gamma_w). (Key point: Coefficient of consolidation)

## Related in Soil Mechanics

- [Derive 1D primary consolidation settlement](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Asettlement-equation-1d-primary) — Soil Mechanics, chapter 9
- [Terzaghi limitations; primary vs secondary](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aterzaghi-theory-limitations-primary-secondary) — Soil Mechanics, chapter 9
- [Consolidation test, 90% time, k, Cc](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aconsolidation-test-50-90-percent-time) — Soil Mechanics, chapter 9

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