--- title: "Stability of cantilever wall, 8.6 m" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:stability-cantilever-wall-8-6m" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/stability-cantilever-wall-8-6m site: BCT Engineering --- # Stability of cantilever wall, 8.6 m Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2082 Bhadra, 8 marks - Video: [Cantilever wall basics: what every engineer needs to know](https://www.youtube.com/watch?v=RnRJBF5JMlU) — NOVACK ENGINEERING LTD - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/stability-cantilever-wall-8-6m - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question Check different stability conditions for the given retaining wall. Assume appropriate value where necessary. [Figure: cantilever retaining wall, stem height 8.6 m, top width 0.4 m, backfill sloping at $10^\circ$ with $c = 0$, $\phi = 32^\circ$, $\gamma = 18\ \text{kN/m}^3$; base slab total width 4.7 m, base slab thickness 0.6 m, toe 0.8 m, embedment 1.5 m, stem inclined at $84^\circ$ at the base. Dimensions as read from the scan.] ## Short answer Overturning factor 2.42 and bearing capacity are safe with eccentricity within the middle third. Sliding safety factor is only 1.12, so a shear key is needed to reach the required 1.5. ## Answer **Assumptions (figure read from the scan):** stem height 8.6 m, top width 0.4 m, front face vertical, back face inclined at $84^\circ$ to the horizontal, so the stem thickness at the base is $0.4 + 8.6/\tan84^\circ = 1.30$ m. Base slab 0.6 m thick and 4.7 m wide, toe 0.8 m, so heel $= 4.7-0.8-1.30 = 2.60$ m. Concrete 24 kN/m$^3$; backfill $\gamma=18$, $\phi=32^\circ$, $c=0$, sloping at $\beta=10^\circ$; foundation soil taken the same; base friction $\delta = \tfrac23\phi$. Rankine active thrust on a vertical plane through the heel, parallel to the slope. #### 1. Active thrust Height of the plane: $H' = 9.2 + (4.7-1.2)\tan10^\circ = 9.82$ m. $$ K_a = \cos\beta\,\frac{\cos\beta-\sqrt{\cos^2\beta-\cos^2\phi}}{\cos\beta+\sqrt{\cos^2\beta-\cos^2\phi}} = 0.3210 $$ $$ P_a = \tfrac12K_a\gamma H'^2 = 0.5\times0.3210\times18\times9.82^2 = 278.4\ \text{kN/m} $$ at $H'/3 = 3.27$ m above the base: $P_h = 274.2$, $P_v = 48.3$ kN/m (acting at the heel end, $x=4.7$ m). #### 2. Vertical loads (about the toe) | Part | Force (kN/m) | Arm (m) | Moment | |---|---|---|---| | Stem rectangle 0.4 x 8.6 x 24 | 82.6 | 1.00 | 82.6 | | Stem triangle | 93.3 | 1.50 | 140.0 | | Base slab 4.7 x 0.6 x 24 | 67.7 | 2.35 | 159.0 | | Soil on heel and above back face (27.29 m$^2$ x 18) | 491.3 | 3.18 | 1562.0 | | $P_v$ | 48.3 | 4.70 | 227.2 | | **Total** | **783.1** | | **2170.8** | $M_o = P_h\times H'/3 = 274.2\times3.27 = 897.2$ kN m/m. #### 3. Stability checks - **Overturning:** $FS_{ot} = 2170.8/897.2 = 2.42$ (>2). Safe. - **Sliding:** $\mu=\tan21.3^\circ = 0.391$, $$ FS_s = \frac{783.1\times0.391}{274.2} = 1.12\ (<1.5) $$ Not safe. The passive pressure on the 1.5 m embedment ($P_p = \tfrac12K_p\gamma D^2 = 65.9$ kN/m with $K_p=3.25$) gives $FS_s = 1.36$, still below 1.5, so a **shear key** under the stem (or a wider/longer heel) is required. - **Base pressure:** $\bar x = (2170.8-897.2)/783.1 = 1.626$ m, $e = 2.35-1.626 = 0.724$ m $< B/6 = 0.783$ m (no tension). $$ q_{max} = 321\ \text{kN/m}^2\ (\text{toe}),\qquad q_{min} = 13\ \text{kN/m}^2\ (\text{heel}) $$ - **Bearing capacity** (Vesic, $\phi=32^\circ$: $N_q=23.18$, $N_\gamma=30.21$, $B' = 4.7-2e = 3.25$ m, $D_f=1.5$ m): $q_u = \gamma D_fN_q + \tfrac12\gamma B'N_\gamma = 1510$ kN/m$^2$; $FS_b = 1510/321 = 4.7$ (>3). Safe. - **Overall stability:** check a deep slip circle with Bishop's method ($FS\ge1.5$) if soft soil lies below. **Answer:** overturning ($FS = 2.42$), eccentricity and bearing are safe, but sliding ($FS = 1.12$) is not; add a shear key. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which Ka is used for the 10 degree sloping backfill? A. tan^2(45 - phi/2) = 0.307 B. Rankine sloping-backfill Ka = cos(beta)(...)/(...) = 0.3210 C. 1 - sin phi D. Passive coefficient Answer: B. Slope beta changes Ka to 0.3210 and the thrust is parallel to the slope. (Key point: Ka for sloping backfill) 2. Which check fails for this wall? A. Overturning B. Sliding (FS = 1.12, still 1.36 with passive) C. Bearing D. Eccentricity Answer: B. Overturning 2.42, bearing 4.7 and e < B/6 pass, but sliding is below 1.5. (Key point: Critical failure mode) 3. What remedy is recommended? A. Reduce the heel B. Add a shear key under the stem or widen the base C. Remove the embedment D. Increase backfill slope Answer: B. A key mobilises passive resistance deeper, increasing sliding resistance. (Key point: Remedy for sliding) 4. Why is Pv included with arm 4.7 m? A. Sloping-backfill thrust has a vertical component acting at the heel end that adds to the resisting moment B. It reduces weight C. It is uplift D. It is water pressure Answer: A. Pv = 48.3 kN/m acts downward at the heel end of the plane, adding stabilising moment. 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