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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.

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

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 τf=c+σtan⁡ϕ\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 σv\sigma_v is the major principal stress and the horizontal stress falls to the minimum, σh=Kaσv−2cKa\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, σh=Kpσv+2cKp\sigma_h = K_p\sigma_v + 2c\sqrt{K_p}.
  • The failure planes make an angle of 45∘+ϕ/245^\circ + \phi/2 with the major principal plane. This is 45∘+ϕ/245^\circ + \phi/2 to the horizontal in the active state and 45∘−ϕ/245^\circ - \phi/2 to the horizontal in the passive state.

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

✓ Answer checked by Claude Sonnet 5.5

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