Gravity Wall Stability Challenge

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.
Base width B = 4.0 ft H = 6.0 ft Top width t = 2.0 ft TOE HEEL φ = 30° Pₐ → toe acts at H/3 Wall area A W x̄ from toe Pₐ drives μW resists Toe pivot Pₐ: CCW ↺ driving W: CW ↻ resisting qmax qmin Rᵥ toward toe
Step 1: Soil friction angle and active pressure coefficient
Soil unit weight120γpcf
Friction angle30°φφ
Wall unit weight150γwallpcf
Base friction0.50μμ

Values calculated so far

New results stay here for use in later equations.

Complete Step 1 to begin your reference list.
Active pressure coefficient Kₐ = 0.33
Total active earth force Pₐ = 720 lb/ft
Wall cross-sectional area A = 18.00 ft²
Wall weight W = 2,700 lb/ft
Centroid from toe x̄ = 2.44 ft
Sliding factor of safety FSslide = 1.88
Overturning factor of safety FSOT = 4.58
Bearing pressures qmax = 765 psf
qmin = 585 psf
Calculation path
Step 1 of 8
Score: 0 / 8

1. Active earth pressure coefficient

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( ° − ° ÷ )

2. Total active earth force

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

3. Wall cross-sectional area

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 = ( + ) × ÷

4. Wall weight

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 = × ×

5. Wall centroid from the toe

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 = ( × + × ) ÷ ( + )

6. Factor of safety against sliding

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 = × ÷

7. Factor of safety against overturning

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 = ( × ) ÷ ( × ÷ )

8. Maximum bearing pressure

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.
xR = (Wx̄ − PₐH/3)/W; e = B/2 − xR; qmax = (W/B)(1 + 6e/B)

First, enter the values used to find xR:

xR = ( × − × ÷ ) ÷

Then, enter the base values used in qmax:

qmax = ( ÷ ) × [1 + 6e ÷ ]

Wall evaluation complete

Sliding FS1.88passes 1.5
Overturning FS4.58passes 2.0
Eccentricity0.09 ftwithin B/6
Maximum pressure765 psfcompare to allowable

Your wall works! What are you, an engineer?

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