--- title: "Anchor reduces cantilever sheet pile depth" subject: "Foundation Engineering" chapter: 5 key: "semester_5/foundation_engineering:anchor-reduces-embedment" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/anchor-reduces-embedment site: BCT Engineering --- # Anchor reduces cantilever sheet pile depth Foundation Engineering · Chapter 5: Flexible Retaining Structures and Coffer Dams - Asked in: 2080 Bhadra, 4 marks - Video: [Cantilever Sheet Pile | Analysis of Pressure & Force | Actual Method | cohesionless soil |Foundation](https://www.youtube.com/watch?v=BXq5kmiWUX8) — Sudip Dhakal - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/anchor-reduces-embedment - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-5 ## Question Assume suitable condition to prove that use of anchor can reduce the depth of embedment of cantilever sheet pile wall. ## Short answer For a 5 m high retained sand (φ=30°), cantilever analysis yields embedment depth 4.63 m. Adding an anchor at 1 m depth using free earth support reduces required embedment to 1.90 m, a 59% decrease, because the anchor supplies horizontal reaction that the cantilever would otherwise obtain from deeper passive resistance. ## Answer **Assumed condition:** sheet pile wall retaining $H = 5$ m of dry sand ($\phi = 30^\circ$, same on both sides), $\gamma$ the same on both sides, passive resistance with $FS = 1$ for the comparison, anchor at 1 m below the top (free earth support). $$ K_a = \tan^2 (45^\circ-15^\circ) = 0.3333,\qquad K_p = \tan^2(45^\circ+15^\circ) = 3.0 $$ **(a) Cantilever wall:** the moment about the toe must balance: $$ \tfrac12K_a\gamma(H+D)^2\frac{(H+D)}{3} = \tfrac12K_p\gamma D^2\frac{D}{3}\ \Rightarrow\ \frac{D}{H+D} = \left(\frac{K_a}{K_p}\right)^{1/3} = 0.4807 $$ $$ D_{cant} = \frac{5\times0.4807}{1-0.4807} = 4.63\ \text{m} $$ **(b) Anchored wall (free earth support):** take moments about the anchor point (the anchor force $T$ drops out): $$ \tfrac12K_a\gamma(H+D)^2\left[\tfrac23(H+D) - 1\right] = \tfrac12K_p\gamma D^2\left[H + \tfrac23D - 1\right] $$ Solving, $D_{anch} = 1.90$ m, and the anchor force from $\sum H = 0$ is $$ T = \tfrac12K_a\gamma(H+D)^2 - \tfrac12K_p\gamma D^2 \ (\text{per unit }\gamma) = 7.94\gamma - 5.43\gamma = 2.51\gamma $$ (with $\gamma = 18$ kN/m$^3$, $T \approx 45$ kN/m). **Comparison:** $D_{anchored} = 1.90$ m $\ll D_{cantilever} = 4.63$ m, about 59% less. The anchor supplies the horizontal reaction that the cantilever wall must get from the passive resistance of a deep embedment, so the embedment, and also the pile length and bending moment, are reduced. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the cantilever wall, what does D/(H+D) equal? A. (K_a/K_p)^(1/3) B. (K_a/K_p)^(1/2) C. K_a/K_p D. (K_p/K_a)^(1/3) Answer: A. Equating moments about the toe gives D/(H+D) = (K_a/K_p)^(1/3) = 0.4807. (Key point: Cantilever moment equation) 2. What is the cantilever embedment for H = 5 m, phi = 30? A. 1.90 m B. 3.0 m C. 4.63 m D. 2.5 m Answer: C. D = 5 x 0.4807/(1 - 0.4807) = 4.63 m. (Key point: Cantilever embedment) 3. Why are moments taken about the anchor in the anchored (free earth) case? A. Anchor force T passes through that point and drops out B. To include the water pressure C. To ignore passive pressure D. Because the toe is fixed Answer: A. T has zero moment about its own point, so D can be found without T. (Key point: Anchored wall moment point) 4. Why does the anchor reduce the embedment (1.90 m vs 4.63 m)? A. It increases active pressure B. It supplies the horizontal reaction that a cantilever must get from deep passive resistance C. It removes passive resistance D. It increases the pile weight Answer: B. The anchor gives the support, so less passive resistance and shorter embedment is needed, also lowering length and moment. 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