For each step, place the known project values into the correct parts of the equation. When all entries are correct, the game performs the arithmetic and reveals the result.
Conceptual training case only: level backfill, drained soil, no surcharge, no groundwater, no seismic loading, and no passive resistance.
Landscape ArchitectEngineer
Landscape Architect mode: interpretive, encouraging, and unnecessarily celebratory.
Step 1: Soil friction angle and active pressure coefficient
Soil unit weighti120γpcf
Friction anglei30°φφ
Wall unit weighti150γwallpcf
Base frictioni0.50μμ
Values calculated so far
New results stay here for use in later equations.
Complete Step 1 to begin your reference list.
Active pressure coefficientKₐ = 0.33
Total active earth forcePₐ = 720 lb/ft
Wall cross-sectional areaA = 18.00 ft²
Wall weightW = 2,700 lb/ft
Centroid from toex̄ = 2.44 ft
Sliding factor of safetyFSslide = 1.88
Overturning factor of safetyFSOT = 4.58
Bearing pressuresqmax = 765 psf qmin = 585 psf
Calculation path
Step 1 of 8
Score: 0 / 8
1. Active earth pressure coefficient
Technical determination 1.1 — active earth-pressure coefficient.
Under the stated idealized Rankine conditions, including level retained backfill,
negligible wall-soil interface friction, homogeneous granular material, and sufficient
wall movement to mobilize the active state, determine Kₐ from φ using
Kₐ = tan²(45° − φ/2). This dimensionless coefficient transforms vertical geostatic
stress into horizontal active stress and is subsequently used in the lateral-force,
sliding, overturning, and bearing-pressure evaluations.
What we are determining: how much of the soil’s vertical weight becomes horizontal pressure. Why: Kₐ scales the soil weight into the lateral pressure used in every later stability check.
Kₐ = tan²(45° − φ/2)
Kₐ = tan²(
°
−
° ÷
)
Coefficient verified
2. Total active earth force
Technical determination 2.1 — resultant active lateral force.
Assume a linearly increasing horizontal stress distribution with zero pressure at the
retained grade and maximum pressure at the base. Integrate the triangular distribution
over wall height H to obtain Pₐ = ½KₐγH² per unit length of wall. The resultant is
applied horizontally at H/3 above the base and represents the principal destabilizing
action for this simplified analysis.
What we are determining: the total sideways push from the retained soil on one linear foot of wall. Why: this is the force trying to slide and overturn the wall.
Pₐ = ½KₐγH²
Pₐ = ½ ×
×
× ²
Lateral force contained
3. Wall cross-sectional area
Technical determination 3.1 — gross wall section area.
Model the gravity wall as a trapezoidal section having base width B, top width t, and
vertical height H. Determine gross area from A = (B + t)H/2. The result is expressed
per transverse wall section and forms the geometric basis for computing dead load,
centroid location, resisting moment, and bearing stress.
What we are determining: the area of wall material visible in cross section. Why: area lets us convert the wall geometry into wall weight.
A = (B + t)H/2
A = (
+
) ×
÷
Geometry accepted
4. Wall weight
Technical determination 4.1 — dead load per unit wall length.
Multiply gross section area A by the assumed wall material unit weight γwall and a
one-foot longitudinal strip. The resulting W is a vertical dead load in lb/ft. This
load provides both the normal force used in the base-friction calculation and the
primary stabilizing moment resisting rotation about the toe.
What we are determining: the downward weight of a one-foot-long slice of wall. Why: wall weight creates sliding friction and the resisting moment against overturning.
W = Aγwall × 1 ft
W =
×
×
Dead load confirmed
5. Wall centroid from the toe
Technical determination 5.1 — composite centroid and lever arm.
Decompose the trapezoid into elementary rectangular and triangular subareas. Compute
x̄ = ΣAᵢxᵢ/ΣAᵢ measured from the toe. The centroid establishes the line of action of
the wall dead load. Its horizontal distance from the toe is the stabilizing lever arm
used in the overturning analysis and in locating the base resultant.
What we are determining: the point where the wall’s total weight can be treated as acting. Why: its distance from the toe is the lever arm used to calculate resisting moment.
x̄ = (A₁x₁ + A₂x₂)/(A₁ + A₂)
x̄ = (
×
+ ×
) ÷ (
+
)
Centroid located
6. Factor of safety against sliding
Technical determination 6.1 — translational stability.
Evaluate the nominal sliding factor of safety as FSslide = μW/Pₐ, where μW is the
available frictional resistance and Pₐ is the horizontal driving force. This simplified
expression excludes passive resistance, cohesion, shear keys, foundation embedment,
hydrostatic pressure, seismic effects, and reductions in effective normal force.
Compare the computed ratio with the stated conceptual acceptance benchmark.
What we are determining: how much base-friction resistance exists relative to the lateral soil force. Why: the wall must have substantially more resistance than driving force; 1.5 is a common conceptual benchmark.
FSslide = μW/Pₐ
FSslide =
×
÷
Sliding denied
7. Factor of safety against overturning
Technical determination 7.1 — rotational stability about the toe.
Compute the stabilizing moment Mᵣ = Wx̄ and the overturning moment Mₒ = Pₐ(H/3).
Evaluate FSOT = Mᵣ/Mₒ. The toe is treated as the potential instantaneous center of
rotation. This calculation is conceptual and does not replace project-specific
geotechnical evaluation, structural design, load combinations, or applicable code
requirements.
What we are determining: the resisting moment from wall weight compared with the overturning moment from soil pressure. Why: a wall can rotate about its toe even when it does not slide.
FSOT = (Wx̄)/(PₐH/3)
FSOT = (
×
) ÷ (
×
÷
)
Overturning embarrassed
Final bonus check: Is the wall still pushing down safely?
Technical determination 8.1 — maximum toe pressure.
The soil pushes the wall toward the toe, so more of the wall’s weight is carried there.
For this final check, calculate the pressure at the toe and make sure the soil can handle it.
What we are determining: the greatest contact pressure beneath the wall base after accounting for the shifted resultant. Why: the supporting soil must safely carry this pressure, and the base should remain in compression.
