---
title: "General state of plastic equilibrium"
subject: "Foundation Engineering"
chapter: 3
key: "semester_5/foundation_engineering:general-plastic-equilibrium"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/general-plastic-equilibrium
site: BCT Engineering
---

# General state of plastic equilibrium

Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls

- Asked in: 2076 Asoj, 3 marks
- Video: [Rankine Theory of Earth Pressure | Elementary Engineering](https://www.youtube.com/watch?v=3CnW3mHvJ2Q) — Elementary Engineering
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/general-plastic-equilibrium
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3

## Question

What do you understand by "General State of Plastic Equilibrium"?

## Short answer

Every point in the soil mass is on the verge of shear failure: shear stress equals c+σ tanφ, and the Mohr circle touches the failure envelope. In the active state horizontal stress reduces to Kaσv−2c√Ka; in passive it becomes Kpσv+2c√Kp. The concept forms the basis of Rankine’s earth pressure theory.

## Answer

A soil mass is in a **state of plastic equilibrium** when every point in it is on the verge of shear failure; the shear stress at each point equals the shear strength given by the Mohr-Coulomb criterion $\tau_f = c + \sigma\tan\phi$. The Mohr circle of stress at every point then just touches the failure envelope.

- **Rankine's general state:** this condition exists throughout the whole semi-infinite soil mass, which happens when the soil is stretched or compressed horizontally enough.
- **Active state:** the soil expands horizontally (e.g. the wall moves away). The vertical stress $\sigma_v$ is the major principal stress and the horizontal stress falls to the minimum, $\sigma_h = K_a\sigma_v - 2c\sqrt{K_a}$.
- **Passive state:** the soil is compressed horizontally. The horizontal stress becomes the major principal stress, $\sigma_h = K_p\sigma_v + 2c\sqrt{K_p}$.
- The failure planes make an angle of $45^\circ + \phi/2$ with the major principal plane. This is $45^\circ + \phi/2$ to the horizontal in the active state and $45^\circ - \phi/2$ to the horizontal in the passive state.

It is the basis of Rankine's earth pressure theory.

## Check yourself

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

1. A soil mass is in plastic equilibrium when:
   A. Every point is on the verge of shear failure, with the Mohr circle touching the failure envelope
   B. It is at rest
   C. Only the base fails
   D. Stress is zero
   Answer: A. Shear stress equals strength tau_f = c + sigma tan(phi) at every point. (Key point: Definition)

2. In the active state:
   A. Horizontal stress is the major principal stress
   B. Vertical stress is the major principal stress and horizontal stress falls to the minimum, Ka sigma_v - 2c sqrt(Ka)
   C. Horizontal stress is Kp sigma_v
   D. No failure planes form
   Answer: B. The soil expands horizontally and sigma_h drops to its minimum. (Key point: Active state stress)

3. Failure planes make which angle with the major principal plane?
   A. 45 degrees +/- phi/2
   B. phi
   C. 90 - phi
   D. 0
   Answer: A. Failure planes are at 45 degrees +/- phi/2. (Key point: Failure plane angle)

4. In the passive state the horizontal stress is:
   A. Minor principal stress, Ka sigma_v
   B. Major principal stress, Kp sigma_v + 2c sqrt(Kp)
   C. Equal to vertical stress
   D. Zero
   Answer: B. Horizontal compression makes sigma_h the major principal stress. (Key point: Passive state)

## Related in Foundation Engineering

- [General, local and punching shear failure](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Amodes-shear-failure) — Foundation Engineering, chapter 6, asked 4×

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