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GEO-04

Rankine active earth pressure

Active resultant on a smooth vertical wall in c–φ soil.

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Earth pressureRankine 1857Eurocode 7

Governing equation

Pa=12γH2Ka−2cHKa,Ka=tan⁡2(45∘−φ/2)P_a = \tfrac12 \gamma H^2 K_a - 2 c H \sqrt{K_a},\quad K_a=\tan^2(45^\circ-\varphi/2)

H∈[1,20]H \in \left[1, 20\right]

where

\gamma
Unit weight (kN/m³)
H
Height (m)
\varphi
Friction angle (°)
c
Cohesion (kPa)
K_a
Active coefficient (—)
P_a
Active thrust (kN/m)

The figure, part by part

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①

Governing law — red

Pa=12γH2Ka−2cHKa,Ka=tan⁡2(45∘−φ/2)P_a = \tfrac12 \gamma H^2 K_a - 2 c H \sqrt{K_a},\quad K_a=\tan^2(45^\circ-\varphi/2)

H∈[1,20]H \in \left[1, 20\right]

H=1H = 1Ka=−2.774K_a = -2.774

②

Domain of the variable — blue

Hmin⁡=1,Hmax⁡=20H_{\min} = 1,\quad H_{\max} = 20

H∈[1,20]H \in \left[1, 20\right]

H=1H = 1H=1H = 1

③

Response curve — green

Pa=12γH2Ka−2cHKa,Ka=tan⁡2(45∘−φ/2)P_a = \tfrac12 \gamma H^2 K_a - 2 c H \sqrt{K_a},\quad K_a=\tan^2(45^\circ-\varphi/2)

H∈[1,20]H \in \left[1, 20\right]

H=1H = 1Ka=−2.774K_a = -2.774

④

Rate of change — amber

dKadH\frac{\mathrm{d}K_a}{\mathrm{d}H}

H∈[1,20]H \in \left[1, 20\right]

H=1.119H = 1.119Ka′=0.943K_a^{\prime} = 0.943

Lecture brief

Historical brief

Soil mechanics became a quantitative laboratory after Karl von Terzaghi’s 1925–1943 work on effective stress, consolidation and bearing. These sheets still size shallow foundations, retaining walls, piles and drainage in SI. This sheet (GEO-04 — Rankine active earth pressure) is the form associated with Rankine 1857 · Eurocode 7. Working symbols: γ\gamma, HH, φ\varphi, cc →\rightarrow KaK_a, PaP_a. Rankine assumed conjugate plastic states with failure planes at 45°±φ/2. Ka = tan²(45°−φ/2) and the cohesion intercept is 2c√Ka.

Purpose

Purpose: compute KaK_a, PaP_a from γ\gamma, HH, φ\varphi, cc in Geotechnical engineering via Pa=12γH2Ka−2cHKa,Ka=tan⁡2(45∘−φ/2)P_a = \tfrac12 \gamma H^2 K_a - 2 c H \sqrt{K_a},\quad K_a=\tan^2(45^\circ-\varphi/2) Active resultant on a smooth vertical wall in c–φ soil. Use it when a real geotechnical engineering question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given γ=18.000 kN/m3\gamma = 18.000\,\mathrm{kN/m^{3}}, H=6.000 mH = 6.000\,\mathrm{m}, φ=30.000 ∘\varphi = 30.000\,\mathrm{^{\circ}}, c=5.000 kPac = 5.000\,\mathrm{kPa}, the governing relation Pa=12γH2Ka−2cHKa,Ka=tan⁡2(45∘−φ/2)P_a = \tfrac12 \gamma H^2 K_a - 2 c H \sqrt{K_a},\quad K_a=\tan^2(45^\circ-\varphi/2) yields Ka=0.3333 —K_a = 0.3333\,\mathrm{—}, Pa=73.359 kN/mP_a = 73.359\,\mathrm{kN/m}. Horizontal backfill, no wall friction. Pa includes the 2c term. Move a slider: the numbers are this situation, not a canned story.

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Inputs

Outputs

  • Active coefficient K_a0.3333 —
  • Active thrust P_a73.359 kN/m
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Narration of this film

Horizontal backfill, no wall friction. Pa includes the 2c term.

Rankine assumed conjugate plastic states with failure planes at 45°±φ/2. Ka = tan²(45°−φ/2) and the cohesion intercept is 2c√Ka.

Reading speed

Watch on YouTube