Cantilever Retaining Wall Design: Overturning, Sliding, and Bearing Checks with a Worked Example

By Engr. Ruel H. Cepeda, Structural Engineer

A cantilever retaining wall holds back soil through its own weight plus the backfill on its heel slab, using far less concrete than a gravity wall of the same height. Before the stem, heel, and toe are sized for reinforcement, the trial section must survive three stability checks: overturning, sliding, and bearing/eccentricity. This article covers proportioning, Rankine active pressure, when passive resistance may be counted, the checks, and one full example that fails sliding on first trial and shows the fix.

Proportioning Rules of Thumb

Pick trial dimensions from ranges that converge quickly to a passing design. These are starting points only, not code minimums, and each must still be verified by calculation.

If sliding later fails, widening the base is not always the cheapest fix — a shear key engaging deeper soil often restores the required factor of safety with less concrete, as shown below.

Rankine Active Earth Pressure

For a vertical virtual back plane through the heel, level backfill, and cohesionless soil (c′=0, no wall friction), the active coefficient is:

K_a = tan²(45° − φ/2)

Total active thrust per metre, a triangular pressure over the full height H, acts at H/3 above the base:

P_a = ½·K_a·γ·H²

A uniform surcharge q adds a rectangular block over the same height, resultant at H/2:

P_a,q = K_a·q·H

A sloped backfill or significant wall friction calls for Coulomb's theory instead.

Passive Resistance in Front of the Toe — Count It or Not?

Soil in front of the toe resists sliding through passive pressure:

K_p = tan²(45° + φ/2)    P_p = ½·K_p·γ·D²

D is the soil depth in front, from finished grade to the footing (or key) bottom. Two concerns limit P_p: it needs far more wall movement to mobilize than active pressure needs, and that soil is the least reliable on site — excavation or erosion can remove it. Practice is either (a) neglect P_p entirely, the conservative default, or (b) count on it only over an embedment that will demonstrably remain, typically a shear key shown on the drawings.

Stability Check 1 — Overturning

Sum moments about the toe, the assumed pivot point:

FS_overturning = ΣM_resisting / ΣM_overturning ≥ 2.0

ΣM_resisting comes from the self-weight of stem, footing, and soil on the heel; ΣM_overturning from P_a and P_a,q at their heights above the base. The 2.0 is a practice value: textbooks most commonly cite FS ≥ 2.0, while several building codes and references accept 1.5 as the floor, particularly with the middle-third check below also enforced. Use the value your governing code or geotechnical report specifies.

Stability Check 2 — Sliding

FS_sliding = (μ·ΣV + P_p) / P_a ≥ 1.5

The 1.5 is the usual practice minimum. μ ≈ tanδ, where the base friction angle δ is commonly taken between ½φ and ⅔φ (up to φ for concrete cast directly against soil) — in practice about 0.35–0.55 for sands/silts, lower for clays; confirm against the geotechnical report. ΣV is the same vertical load used below; P_p belongs here only under the conditions discussed above.

Stability Check 3 — Bearing Pressure and the Middle-Third Rule

Locate the resultant from the toe, x̄ = (net moment about toe)/ΣV, and its eccentricity from the centerline:

e = B/2 − x̄

For linear contact pressure with no separation, the resultant must fall within the middle third, e ≤ B/6. Then:

q_max/min = (ΣV/B)·(1 ± 6e/B) ≤ q_a

q_a is the allowable bearing capacity from the geotechnical report — see the sister site's bearing capacity guide on RHCES for how it's derived. If e exceeds B/6, an effective-width formula applies instead (see the FAQ).

Designing the Stem, Heel, and Toe as Cantilevers

Once stable, each element is designed as its own RC cantilever on a 1 m-wide strip:

• Stem — base moment from active pressure (plus surcharge) over the stem height alone, not full H.
• Heel — loaded down by backfill and self-weight, offset by the upward soil reaction under it (often taken conservatively as just q_min, or ignored).
• Toe — loaded up by net bearing pressure, less its own self-weight.

Compute each factored moment, then use the RC beam design tool (ACI 318 / NSCP 2015 method) to size steel, as in the isolated footing worked example.

Worked Example — 4.0 m Cantilever Retaining Wall

Given: H = 4.0 m; γ = 18 kN/m³, φ = 30°, level backfill; surcharge q = 10 kPa; μ = 0.5; q_a = 150 kPa; γ_c = 24 kN/m³. Per metre length of wall.

1. Trial geometry

t_f = H/10 = 400 mm; h_s = H − t_f = 3.6 m; t_stem (uniform) = 300 mm; B = 0.6H = 2.4 m; L_t = B/3 = 0.8 m; L_h = B − L_t − t_stem = 2.4 − 0.8 − 0.3 = 1.3 m.

2. Earth pressure

K_a = tan²(45−15) = tan²30° = 0.333. K_p = tan²(45+15) = tan²60° = 3.0.
P_a(soil) = 0.5(0.333)(18)(4.0²) = 48.0 kN/m at H/3 = 1.333 m above the base.
P_a(surcharge) = (0.333)(10)(4.0) = 13.33 kN/m at H/2 = 2.0 m above the base.
Total P_a = 48.0 + 13.33 = 61.33 kN/m.

3. Weights, ΣV

Heel surcharge weight is not added to ΣV, only its thrust, since a live surcharge should not count as stabilizing weight (a permanent fill could add 1.3×10 = 13.0 kN/m).

4. Overturning check

ΣM_overturning = 48.0(1.333) + 13.33(2.0) = 64.0 + 26.67 = 90.67 kN·m/m.
FS_overturning = 199.69/90.67 = 2.20 ≥ 2.0 — OK.

5. Sliding check

No passive resistance: FS_sliding = μΣV/P_a = 0.5(133.20)/61.33 = 66.60/61.33 = 1.09 < 1.5 — NOT OK. Widening the base alone (keeping the toe at B/3) needs B ≈ 0.84H (≈3.4 m), outside 0.5–0.7H. Add a shear key instead, with finished grade in front at the top of the footing so the front embedment is D = 1.0 m (0.4 m footing + 0.6 m key):
P_p = 0.5(3.0)(18)(1.0²) = 27.0 kN/m.
FS_sliding = (66.60+27.0)/61.33 = 93.60/61.33 = 1.53 ≥ 1.5 — OK. Size the key for 27.0 kN/m shear and show the embedment on the grading plan.

6. Bearing pressure check

x̄ = (199.69−90.67)/133.20 = 109.02/133.20 = 0.8185 m.
e = 1.20−0.8185 = 0.3815 m (say 0.382 m). B/6 = 0.40 m. e < B/6 — within middle third.
ΣV/B = 55.50 kPa; 6e/B = 0.954.
q_max = 55.50(1.954) = 108.4 kPa; q_min = 55.50(0.046) = 2.6 kPa.
108.4 ≤ q_a = 150 kPa — OK (≈72% utilized); q_min > 0 confirms no uplift. (Adding the 13.0 kN/m surcharge weight on the heel for the bearing case pulls the resultant toward the centre, e = 0.299 m, and gives q_max ≈ 106 kPa, so the case above governs.)

7. Stem design moment

Over h_s = 3.6 m alone: P_a,stem(soil) = 0.5(0.333)(18)(3.6²) = 38.88 kN/m at 1.2 m; P_a,stem(surcharge) = (0.333)(10)(3.6) = 12.0 kN/m at 1.8 m. Service moment = 38.88(1.2)+12.0(1.8) = 68.3 kN·m/m. Applying the 1.6 load factor that the NSCP 2015 / ACI 318 strength combinations place on lateral earth pressure and live surcharge gives M_u ≈ 1.6(68.3) ≈ 109 kN·m/m for the beam design tool above.

Assumptions & Limitations

• Rankine theory throughout (vertical virtual back, horizontal backfill, cohesionless soil, no wall friction); use Coulomb's theory for sloped backfill or significant wall friction.
• μ, q_a, and any passive embedment must come from a project-specific geotechnical investigation, not assumed.
• Hydrostatic pressure, seismic increment (Mononobe–Okabe), and drainage (weep holes) are outside this scope but mandatory in a complete design.
• Only the stem base moment is carried numerically; full flexural design uses the linked beam design tool.
• Uses NSCP 2015 load-combination families in general terms; confirm the exact combination adopted by the building official of record.

Frequently Asked Questions

Should passive pressure always be included in the sliding check?

Not by default. It needs far more wall movement to mobilize than active pressure, and the soil in front of the toe is the most easily disturbed on site. Many designers neglect P_p, or count on it only through a key shown on the drawings, as above.

What happens if the eccentricity e exceeds B/6?

Contact pressure is no longer linear. The fix is an effective-width formula, q_max = 2ΣV / [3(B/2−e)]; better still, re-proportion so e stays within B/6.

Why did the 2.4 m base (already 0.6H) fail sliding?

μ = 0.5 with K_a = 0.333 backfill plus surcharge pushed the driving force above what friction alone could resist — common once surcharge is present. Reaching FS = 1.5 by width alone (toe kept at B/3) needed B ≈ 0.84H; a key was the cheaper fix.

Overturning, sliding, and bearing must all pass before a single bar goes into the stem, heel, or toe — and sliding often governs once a surcharge is in the picture. Once stable, carry the moments into the concrete beam design tool, or browse all free web tools for footing, slab, and column calculators. Offline spreadsheets are on the download page.