Question 8 of 9: Gravity retaining wall — overturning, sliding, and the effect of a rising water table

Question 8: Gravity retaining wall — overturning, sliding, and the effect of a rising water table (24 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The factors of safety against overturning about the toe and against sliding along the base, first with a very deep water table and then with the water standing 4 m above the base in the backfill, together with the pressure distribution diagram for the flooded case.

[Figure not reproduced: Figure 3 (redrawn from Figure 1) — The gravity wall. The base is 5.0 m wide and 1.2 m thick; the stem tapers from 2.5 m at the base to 1.0 m at the top over a height of 8.0 m; the heel projects 2.0 m and the toe 0.5 m. The back face (shown in red) leans towards the toe by 1.0 m over 8.0 m, so . See the official exam paper.]

Approach. The free body is the wall, its base and the wedge of backfill standing on the heel inboard of the back-face plane; Coulomb's coefficient is evaluated for that plane and the resulting thrust resolved into horizontal and vertical components; moments are then taken about the toe for overturning and horizontal forces summed for sliding. The flooded case repeats the same bookkeeping with submerged unit weights below the water line, a separate hydrostatic thrust, and uplift beneath the base.

1. Fix the geometry of the back face. Measuring from the toe at the underside of the base, the stem runs from x = 0.5 m to x = 3.0 m at its foot and from x = 1.0 m to x = 2.0 m at its crest. The back face therefore rises 8.0 m while moving 1.0 m towards the toe, so its foot lies further into the backfill than its top: the soil wedge overhangs the face and part of its weight bears down on it. Coulomb's β is measured from the horizontal, with β = 90° for a vertical face, and for this sense of batter it is $$\beta=90^{\circ}-\tan^{-1}\!\left(\frac{1.0}{8.0}\right)=90^{\circ}-7.125^{\circ}=82.875^{\circ}$$ Because β is less than 90° the coefficient comes out above the vertical-face value: the face must carry part of the overhanging wedge, so the resultant it has to supply is larger, not smaller. The sign of this batter correction is easy to get backwards, so it is worth proving rather than asserting. An independent Coulomb trial-wedge solution — a force polygon on the wedge bounded by the face, the failure plane and the ground surface, maximised over the inclination of that plane. It returns K_a = 0.3527 for this geometry, and when the same routine is run on a vertical face with δ′ = 0 it reproduces the exact Rankine values (1 − sin φ′)/(1 + sin φ′) = 0.3333 and a critical plane at 45° + φ′/2, which is what licenses the reading.
2. Evaluate Coulomb's active earth pressure coefficient. With a horizontal backfill (α = 0), φ′ = 30° and δ′ = 0.6(30°) = 18°, $$K_a=\frac{\sin^{2}(\beta+\phi')}{\sin^{2}\beta\,\sin(\beta-\delta')\left[1+\sqrt{\dfrac{\sin(\phi'+\delta')\sin(\phi'-\alpha)}{\sin(\beta-\delta')\sin(\alpha+\beta)}}\right]^{2}}$$ Substituting β = 82.875° gives sin²(112.875°) = 0.84890, sin²β = 0.98462, sin(64.875°) = 0.90538, and a bracket of [1 + √0.41360]² = 2.69983, so $$K_a=\frac{0.84890}{0.98462(0.90538)(2.69983)}=\boxed{0.3527}$$ For comparison, the same soil against a vertical face would give K_a = 0.2986, so this batter costs an 18 per cent increase in thrust rather than buying a reduction.
3. Assemble the vertical forces on the free body. The free body is the concrete plus the backfill standing on the heel inboard of the back-face plane, and moment arms are measured from the toe:

   Component	Area (m2)	Unit weight (kN/m3)	Force (kN/m)	Arm x (m)	Moment (kN·m/m)
   Base slab, 5.0 × 1.2	6.00	24	144.0	2.500	360.0
   Stem (trapezoid, 2.5 m to 1.0 m over 8.0 m)	14.00	24	336.0	1.643	552.0
   Backfill on the heel	20.00	20	400.0	3.733	1493.3
   Sum of weights	—	—	880.0	—	2405.3

   The stem centroid follows from splitting the trapezoid into a 0.5 m × 8 m triangle, a 1.0 m × 8 m rectangle and a 1.0 m × 8 m triangle; the soil block from a 2.0 m × 8 m rectangle plus the 1.0 m × 8 m triangle between the leaning face and the vertical through the foot of the stem.
4. Compute the active thrust and resolve it. Taking the full wall height, since the base forms part of the structure, $$P_a=\tfrac{1}{2}K_a\gamma H^{2}=\tfrac{1}{2}(0.3527)(20)(9.2)^{2}=298.5\ \text{kN/m}$$ On a face inclined at β, the thrust acts at δ′ to the face normal, so its inclination below the horizontal is (90° − β) + δ′ = 7.125° + 18° = 25.125°. Hence $$P_h=298.5\cos 25.125^{\circ}=270.3\ \text{kN/m},\qquad P_v=298.5\sin 25.125^{\circ}=126.8\ \text{kN/m (downward)}$$ acting at H/3 = 3.067 m above the base underside, where the pressure plane stands at x = 2.767 m from the toe.
5. Factor of safety against overturning about the toe. The resisting moment is the weight moment plus the moment of the vertical thrust component; the overturning moment is that of the horizontal component alone: $$\sum M_R=2405.3+126.8(2.767)=2756.0\ \text{kN}\!\cdot\!\text{m/m},\qquad \sum M_O=270.3(3.067)=828.9\ \text{kN}\!\cdot\!\text{m/m}$$ $$\text{FS}_{\text{overturning}}=\frac{2756.0}{828.9}=\boxed{3.32>2.0\ \checkmark}$$
6. Factor of safety against sliding on the base. The available friction uses a reduced interface friction angle, conventionally k₁φ′₂ with k₁ = 2/3 for concrete cast against soil, and the founding soil is the same cohesionless material (c′ = 0). With the vertical thrust component included in the normal force, $$\sum V=880.0+126.8=1006.8\ \text{kN/m},\qquad F_R=\sum V\tan\!\left(\tfrac{2}{3}\times 30^{\circ}\right)=1006.8(0.36397)=366.4\ \text{kN/m}$$ $$\text{FS}_{\text{sliding}}=\frac{366.4}{270.3}=\boxed{1.36<1.5\ \times}$$ The wall does not meet the usual minimum of 1.5 against sliding even with a very deep water table, and that is the finding of this part rather than an arithmetic slip — the batter on the back face raises K_a by 18 per cent, and a gravity wall this slender against a 9.2 m retained height has little margin in sliding to give away. Passive resistance in front of the wall has been neglected, which is the conservative and usual choice because the soil over the toe may be excavated for services. Including it over the 2.0 m depth, with K_p = tan²(45° + 15°) = 3.0, adds P_p = 0.5(3.0)(20)(2.0²) = 120 kN/m and lifts the factor of safety to 1.80; taking full interface friction instead of two-thirds would give 2.15. So the wall is only adequate in sliding if credit is taken for the passive wedge, or for full friction on the concrete-soil interface, and a shear key or a wider base is the proper remedy.
7. Check the base pressures while the geometry is in hand. The eccentricity of the resultant is $$e=\frac{B}{2}-\frac{\sum M_R-\sum M_O}{\sum V}=2.5-\frac{2756.0-828.9}{1006.8}=0.586\ \text{m}<\frac{B}{6}=0.833\ \text{m}$$ so the whole base is in compression, and $$q_{\max,\min}=\frac{\sum V}{B}\left(1\pm\frac{6e}{B}\right)=201.4(1\pm 0.7031)\Rightarrow q_{\max}=343\ \text{kPa},\ q_{\min}=60\ \text{kPa}$$ The whole base remains in compression, but 343 kPa is a substantial pressure on a sand of φ′ = 30° at only 2.0 m embedment and should be checked against the bearing capacity of the founding soil with the load-inclination factors included. So of the three modes the dry wall passes overturning, needs a bearing check, and fails sliding.
8. Now raise the water to 4 m in the backfill: the pressure diagram. Above the water line the backfill remains at γ = 20 kN/m³; below it the effective unit weight is γ′ = 20 − 9.81 = 10.19 kN/m³, and a hydrostatic pressure acts in addition. Working down from the backfill surface, the effective vertical stress at the water line (5.2 m down) is 20(5.2) = 104.0 kPa and at the base 104.0 + 10.19(4.0) = 144.8 kPa, so the effective active pressure ordinates on the plane are $$\sigma'_a(0)=0,\qquad \sigma'_a(5.2\ \text{m})=0.3527(104.0)=36.68\ \text{kPa},\qquad \sigma'_a(9.2\ \text{m})=0.3527(144.8)=51.06\ \text{kPa}$$ to which is added a hydrostatic triangle reaching u = 9.81(4.0) = 39.24 kPa at the base. The total pressure at the base is therefore 90.3 kPa against 64.9 kPa in the dry case.

9. Resolve the flooded-case forces. The effective active thrust is the sum of three parts — the upper triangle, the rectangle carried down from the water line, and the lower submerged triangle: $$P_1=\tfrac{1}{2}(0.3527)(20)(5.2)^{2}=95.37,\quad P_2=0.3527(104.0)(4.0)=146.73,\quad P_3=\tfrac{1}{2}(0.3527)(10.19)(4.0)^{2}=28.75$$ in kN/m, acting 5.733 m, 2.000 m and 1.333 m above the base respectively, so P′_a = 270.85 kN/m at 3.244 m. Resolving at 25.125°, $$P'_h=245.2\ \text{kN/m},\qquad P'_v=115.0\ \text{kN/m}$$ The water thrust is P_w = 0.5(9.81)(4.0)² = 78.5 kN/m at 1.333 m above the base, contributing 77.9 kN/m horizontally and 9.7 kN/m vertically on the slightly inclined plane. Uplift beneath the base, taken to vary linearly from the full head at the heel to zero at the drained toe, is $$U=\tfrac{1}{2}(39.24)(5.0)=98.1\ \text{kN/m}\ \text{at}\ x=\tfrac{2}{3}(5.0)=3.333\ \text{m from the toe}$$
10. Recompute both factors of safety. The driving force rises while the normal force falls: $$\sum H=245.2+77.9=323.1\ \text{kN/m},\qquad \sum V=880.0+115.0+9.7-98.1=906.6\ \text{kN/m}$$ $$\sum M_O=245.2(3.244)+77.9(1.333)+98.1(3.333)=1226.3\ \text{kN}\!\cdot\!\text{m/m}$$ $$\sum M_R=2405.3+115.0(2.745)+9.7(2.983)=2750.0\ \text{kN}\!\cdot\!\text{m/m}$$ $$\text{FS}_{\text{overturning}}=\frac{2750.0}{1226.3}=\boxed{2.24},\qquad \text{FS}_{\text{sliding}}=\frac{906.6(0.36397)}{323.1}=\boxed{1.02}$$
11. Answer the question and explain the mechanism. Both factors of safety decrease — overturning from 3.32 to 2.24, a fall of 33 per cent, and sliding from 1.36 to 1.02, a fall of 25 per cent, so a wall that was already short of the 1.5 minimum against sliding ends up barely in equilibrium. Three separate effects act in the same direction and none of them cancels. First, water is much heavier laterally than submerged soil: below the water line the soil contributes only K_aγ′ = 0.3527(10.19) = 3.59 kPa per metre of depth while the water contributes the full γ_w = 9.81 kPa per metre, so below the water line the lateral pressure accumulates at 3.59 + 9.81 = 13.40 kPa per metre of depth instead of the 7.05 kPa per metre that dry soil would give — 1.9 times the rate, and the kink in Figure 4 is the visible expression of it. Second, the water resultant acts low on the wall, at one third of 4.0 m, but its magnitude is large enough that the total overturning moment still rises by 48 per cent, from 829 to 1226 kN·m/m. Third, uplift beneath the base removes 98 kN/m of normal force, cutting the frictional resistance by 10 per cent, and adds a further overturning moment of 327 kN·m/m about the toe — which by itself is 82 per cent of the whole 397 kN·m/m increase in overturning moment. The engineering conclusion is that this wall does not have an adequate sliding margin in any condition, and that drainage of the backfill is what keeps it from being marginal to being unsafe. The remedy is therefore twofold: a shear key beneath the base (or a wider base with a longer heel) to bring the drained sliding factor of safety above 1.5, and a granular drainage blanket against the back face, weep holes at 1.5 to 2 m centres through the stem with a filter behind them, and a longitudinal collector drain at the heel so that the flooded case never becomes the operating case. Neither measure is an optional refinement; the wall as dimensioned relies on both.