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Terzaghi's bearing capacity theory

Short answer

Terzaghi's theory gives ultimate bearing capacity for a shallow strip footing with a rough base, considering equilibrium of a failure wedge. Assumptions: strip footing (plane strain), Df ≤ B, rough base, homogeneous isotropic soil obeying Mohr-Coulomb, shear strength above base ignored (replaced by surcharge q = γ Df), general shear failure, vertical central load. The equation is qu = cNc + γ Df Nq + 0.5 γ B Nγ. For local shear, c and tanφ are reduced to two-thirds.

Discuss Terzaghi's bearing capacity theory. Also state the assumptions.

Terzaghi's theory (1943) gives the ultimate bearing capacity of a shallow footing by considering the equilibrium of a failure wedge. It is the most widely used classical theory and extends Prandtl's and Reissner's work by including soil weight and a rough base.

Assumptions

  1. The footing is a long strip (plane strain), and Df≤BD_f \le B (shallow foundation).
  2. The base is rough, so the soil wedge directly below it stays elastic and moves down with the footing.
  3. The soil is homogeneous and isotropic, and obeys Mohr–Coulomb's law τ=c+σtan⁡ϕ\tau = c + \sigma\tan\phi.
  4. The shear strength of the soil above the base level is ignored. It is replaced by a uniform surcharge q=γDfq = \gamma D_f.
  5. Failure occurs in general shear (for local shear, strength is reduced to two-thirds).
  6. The load is vertical and central.

Failure mechanism

   q_u  ____________
       |  I  |
  II  /\    /\  II
 III /  \  /  \ III

Under the footing, zone I is an elastic (triangular) wedge with base angles ϕ\phi. Zone II is a radial shear zone bounded by a logarithmic spiral. Zone III is a passive Rankine zone.

Derivation (outline)

Considering vertical equilibrium of the wedge: quB+W=2Pp+2Csin⁡ϕq_u B + W = 2P_p + 2C\sin\phi, where C=cB2cos⁡ϕC = \dfrac{cB}{2\cos\phi} is the cohesive force on each inclined face, WW the weight of the wedge, and PpP_p the passive resistance from zones II and III (made of three parts, due to cc, qq and γ\gamma). This gives

qu=cNc+γDfNq+0.5γBNγ(strip)q_u = cN_c + \gamma D_f N_q + 0.5\gamma B N_\gamma \quad (\text{strip})

For other shapes: square qu=1.3cNc+γDfNq+0.4γBNγq_u = 1.3cN_c + \gamma D_fN_q + 0.4\gamma BN_\gamma; circular qu=1.3cNc+γDfNq+0.3γBNγq_u = 1.3cN_c + \gamma D_fN_q + 0.3\gamma BN_\gamma. Nc,Nq,NγN_c, N_q, N_\gamma depend only on ϕ\phi (e.g. for ϕ=35∘\phi = 35^\circ: 57.8,41.4,42.457.8, 41.4, 42.4). For local shear use c′=23cc' = \tfrac{2}{3}c, tan⁡ϕ′=23tan⁡ϕ\tan\phi' = \tfrac{2}{3}\tan\phi.

✓ Answer checked by Claude Sonnet 5.5

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