Learn: Triangular active earth thrust
For a dry cohesionless backfill, the ideal active pressure grows linearly from zero at the top to a maximum at the base. Its total horizontal thrust is the area of that triangular pressure diagram.
What the formula is saying
The base pressure is Ka γ H. A triangle has area one half times base times height, giving Pa = 0.5 Ka γ H² per metre length of wall.
Read the symbols in plain language
- Ka
- Active coefficient
Active coefficient. The effective vertical overburden at depth H is scaled by the active coefficient.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- γ
- Soil unit weight
Weight force per unit volume, including gravity. Mass density in kg/m³ cannot be entered directly in a kN/m³ field.
kN/m³Use kN/m³ as the base unit shown here. γ is kN/m³, H is m and Ka is dimensionless. Base pressure is kPa; integrating over height gives kN/m of wall. This is not the force on the entire wall unless multiplied by a justified wall length.
- H
- Retained height
Retained height. The effective vertical overburden at depth H is scaled by the active coefficient.
mMetres measure length; 1 m = 1000 mm.
- Pa
- Result to find
Triangular active earth thrust. Integrating the linear pressure profile gives half the base pressure times wall height.
kN/m
Sort out the units first
γ is kN/m³, H is m and Ka is dimensionless. Base pressure is kPa; integrating over height gives kN/m of wall. This is not the force on the entire wall unless multiplied by a justified wall length.
Assumptions before calculating
Assume level, homogeneous, cohesionless backfill with the selected active state fully mobilized and a triangular pressure distribution. Ka and unit weight refer to the same effective-stress model.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find Pa and explain the result in the stated output unit.
- Ka · Active coefficient
- 0.333
- γ · Soil unit weight
- 18 kN/m³
- H · Retained height
- 5 m
Find active pressure at the base
The effective vertical overburden at depth H is scaled by the active coefficient.
(0.333) × (18) × (5) = 29.97 kPaTake the triangular pressure area
Integrating the linear pressure profile gives half the base pressure times wall height.
0.5 × (29.97) × (5) = 74.925 kN/m
Does this worked answer make sense?
The triangular resultant acts H/3 above the base in this model. Doubling wall height quadruples thrust when Ka and γ are unchanged.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- Ka · Active coefficient
- 0.3
- γ · Soil unit weight
- 19 kN/m³
- H · Retained height
- 4 m
Find active pressure at the base
The effective vertical overburden at depth H is scaled by the active coefficient.
(0.3) × (19) × (4) = 22.8 kPaTake the triangular pressure area
Integrating the linear pressure profile gives half the base pressure times wall height.
0.5 × (22.8) × (4) = 45.6 kN/m
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- Ka · Active coefficient
- 0.4
- γ · Soil unit weight
- 18 kN/m³
- H · Retained height
- 6 m
Find: Learn: Triangular active earth thrust
A hint, not the answer
The base pressure is Ka γ H. A triangle has area one half times base times height, giving Pa = 0.5 Ka γ H² per metre length of wall.
γ is kN/m³, H is m and Ka is dimensionless. Base pressure is kPa; integrating over height gives kN/m of wall. This is not the force on the entire wall unless multiplied by a justified wall length.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find active pressure at the base
The effective vertical overburden at depth H is scaled by the active coefficient.
(0.4) × (18) × (6) = 43.2 kPaTake the triangular pressure area
Integrating the linear pressure profile gives half the base pressure times wall height.
0.5 × (43.2) × (6) = 129.6 kN/m
Avoid the common trap
Do not omit the one-half factor or use H only once. Do not confuse the base pressure with total thrust or place the triangular resultant at midheight.
When this method applies — and when it does not
Surcharge, groundwater, compaction, cohesion, layering, wall friction and seismic effects are omitted. For submerged conditions, effective soil pressure and water pressure require separate consistent treatment.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
One idea understood. Keep going.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Triangular active earth thrust. Read the relevant soil phase, seepage, earth-pressure, settlement or foundation topic. Effective stress, drainage and idealized geometry determine whether the relationship applies.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.