--- title: "Toe width for overturning, submerged backfill" subject: "Foundation Engineering" chapter: 3 key: "semester_5/foundation_engineering:gravity-wall-width-a-overturning" source: https://www.bctengineer.com/q/semester_5/foundation_engineering/gravity-wall-width-a-overturning site: BCT Engineering --- # Toe width for overturning, submerged backfill Foundation Engineering · Chapter 3: Lateral Earth Pressure Theories and Retaining Walls - Asked in: 2081 Bhadra, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/gravity-wall-width-a-overturning - Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-3 ## Question The retaining wall shown in figure is to be designed to retain a granular backfill. Determine the width of the wall in front of the wall, i.e. dimension 'a' in order to provide factor of safety 2 against overturning alone for the case that the drainage system of backfill gets clogged and backfill gets submerged for the depth of 1 m from the surface and soil is fully saturated above it. The property of retaining wall and backfill is, $G = 2.65$, void ratio $e = 0.594$, angle of internal friction $\phi = 30^\circ$ and unit weight of concrete $= 23\ \text{kN/m}^3$. [Figure: wall 6 m high, base slab thickness 0.75 m, stem width 0.75 m, heel 1.5 m, toe width 'a' to be found.] ## Short answer Using saturated unit weight 19.96 kN/m³ and submerged active thrust plus water pressure, the overturning moment is 375.83 kN m/m. Setting resisting moment equal to twice that value gives toe width a = 1.46 m (use about 1.5 m). ## Answer **Reading of the data and assumptions:** the wall is 6 m high with a 0.75 m thick base slab (so the stem is 5.25 m high and 0.75 m wide), heel 1.5 m and toe length $a$ (unknown). The water table is taken 1 m below the top of the backfill; the soil above it is saturated by capillary rise. The backfill is therefore at $\gamma_{sat}$ throughout, with water pressure below the 1 m level. Rankine active pressure on a vertical plane at the heel; passive resistance in front and base uplift are neglected; $\gamma_w = 9.81$. #### 1. Soil unit weights $$ \gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{(2.65+0.594)\times9.81}{1.594} = 19.96\ \text{kN/m}^3,\quad \gamma' = 10.15\ \text{kN/m}^3 $$ $K_a = \tan^2(45^\circ-15^\circ) = 1/3$. #### 2. Active thrust and overturning moment (about the toe, arms measured above the base) - At 1 m depth: $\sigma_v' = 19.96$, $p = 6.65$ kN/m$^2$. - At the base: $\sigma_v' = 19.96 + 10.15\times5 = 70.74$, $p = 23.58$ kN/m$^2$. - Water: $0$ at 1 m to $9.81\times5 = 49.05$ kN/m$^2$ at the base. | Part | Force (kN/m) | Arm (m) | Moment | |---|---|---|---| | Triangle (top 1 m) | 3.33 | 5.333 | 17.75 | | Rectangle (5 m) | 33.27 | 2.500 | 83.19 | | Triangle (5 m) | 42.31 | 1.667 | 70.52 | | Water | 122.62 | 1.667 | 204.38 | | **Total** | **201.54** | | **375.83** | #### 3. Resisting moment as function of $a$ (base width $B = a + 2.25$) - Stem: $0.75\times5.25\times23 = 90.6$ kN/m at $a+0.375$. - Slab: $(a+2.25)\times0.75\times23$ at $(a+2.25)/2$. - Soil on heel: $1.5\times5.25\times19.96 = 157.2$ kN/m at $a+1.5$ from the toe. Requiring $FS = M_r/M_o = 2$, i.e. $M_r = 751.7$ kN m/m, and solving for $a$ gives $$ a = 1.46\ \text{m}\quad (B = 3.71\ \text{m}) $$ Check at $a=1.46$ m: stem 90.6 x 1.84 + slab 64.1 x 1.86 + soil 157.2 x 2.96 gives $M_r = 751.7$, and $M_r/M_o = 2.00$. **Answer: toe width $a \approx 1.46$ m (provide about 1.5 m)** for $FS = 2$ against overturning with the backfill saturated and submerged. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is gamma_sat computed from G and e? A. (G + e) gamma_w/(1 + e) B. G gamma_w/(1 + e) C. (G - 1) gamma_w/(1 + e) D. G gamma_w (1 + e) Answer: A. gamma_sat = (2.65 + 0.594) x 9.81/1.594 = 19.96 kN/m^3. (Key point: Saturated unit weight) 2. Why is the thrust so large (201.5 kN/m)? A. Submerged backfill adds water pressure (122.6 kN/m), the biggest part B. Ka is large C. Surcharge D. Passive Answer: A. Clogged drainage means full hydrostatic pressure acts on the wall, along with effective soil pressure. (Key point: Water pressure adds thrust) 3. How is the toe length a found? A. Set Mr/Mo = 2 with Mr written as a function of a and solve B. Set Ph = Pv C. Set e = B/6 D. Use Rankine passive Answer: A. Mr must be 2 x 375.83 = 751.7, giving a = 1.46 m. (Key point: Condition for a) 4. Which unit weight is used for soil pressure below the water table? A. gamma_sat B. gamma' = gamma_sat - gamma_w = 10.15, with water pressure added separately C. gamma_dry D. gamma_w Answer: B. Effective stress uses the submerged weight, and water pressure is added. 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