16-Civ-B19 Foundation Engineering · December 2019

Question 4 of 6: Retaining Structures — cantilever wall with a surcharge and a high water table

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: Retaining Structures — cantilever wall with a surcharge and a high water table (30 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 Rankine lateral-pressure distribution separated into its soil, pore-water and surcharge components, and the factors of safety against sliding, overturning and bearing failure together with the minimum contact pressure.

Approach. Take the Rankine vertical plane through the heel, superpose the three pressure components to obtain the thrust and its moment about the toe, assemble the vertical forces from the concrete and the soil column standing on the heel, and then run the three standard stability checks in turn.

1. Part (a)(i) — the active pressure due to soil. For a cohesionless backfill with a horizontal surface the Rankine active coefficient is $$K_a=\tan^2\!\left(45-\frac{\phi}{2}\right)=\tan^2 25^\circ=0.2174$$ Above the water table the effective vertical stress grows at 17 kN/m³, so at the water table (4.0 m down) $$\sigma'_h=K_a\gamma z=0.2174(17)(4.0)=14.79\ \text{kPa}$$ Below it the effective unit weight governs, and over the remaining $5.4-4.0=1.4$ m the pressure rises to $$\sigma'_h=K_a\left[\gamma(4.0)+\gamma'(1.4)\right]=0.2174\left[68+14\right]=17.83\ \text{kPa at the base}$$ The distribution is therefore a triangle to 14.79 kPa followed by a much flatter trapezoid — the kink at the water table is the visible signature of buoyancy reducing the effective overburden.
2. Part (a)(ii) — the pressure due to pore water. Water carries no shear, so it presses on the wall at full hydrostatic intensity with no $K_a$ reduction: $$u=\gamma_w h_w=9.81(1.4)=\boxed{13.73\ \text{kPa at the base}}$$ It is worth noting immediately that this 13.73 kPa exceeds the 3.04 kPa by which the effective soil pressure grows over the same 1.4 m: submerging a backfill increases the total lateral load substantially, which is the entire reason drainage is provided behind real walls.
3. Part (a)(iii) — the pressure due to the surcharge. A uniform surcharge adds a constant increment of vertical stress at every depth, so its lateral effect is a rectangular block over the full retained height: $$\sigma_{h,q}=K_aq=0.2174(40)=\boxed{8.70\ \text{kPa, uniform over 5.4 m}}$$
4. Resolve the pressure diagram into forces and moments about the toe. Each component of Figure 4.1 is integrated and its centroid located above the base:

   Component	Force (kN/m)	Lever arm (m)	Moment (kN·m/m)
   $P_q$ — surcharge rectangle	8.70 × 5.4 = 46.97	2.700	126.81
   $P_1$ — soil triangle above the GWT	½(14.79)(4.0) = 29.57	2.733	80.83
   $P_2$ — soil rectangle below the GWT	14.79 × 1.4 = 20.70	0.700	14.49
   $P_3$ — soil triangle below the GWT	½(3.04)(1.4) = 2.13	0.467	0.99
   $P_w$ — pore-water triangle	½(13.73)(1.4) = 9.61	0.467	4.49
   Totals	108.99	—	227.61

   The surcharge alone contributes 43 % of the thrust and 56 % of the overturning moment, because it acts at mid-height rather than at the lower third.
5. Assemble the vertical forces about the toe. The soil column standing on the 1.75 m heel runs from the top of the base slab (0.4 m above the underside) to the backfill surface, a height of 5.0 m; the water table sits 1.4 m above the underside, so 1.0 m of that column is submerged and 4.0 m is moist. Below the water table the total weight is required, so $\gamma_{sat}=\gamma'+\gamma_w=19.81$ kN/m³:

   Component	Weight (kN/m)	Lever arm from toe (m)	Moment (kN·m/m)
   Stem, 0.3 × 5.0 × 23.5	35.25	1.100	38.78
   Base slab, 3.0 × 0.4 × 23.5	28.20	1.500	42.30
   Moist soil on heel, 1.75 × 4.0 × 17	119.00	2.125	252.88
   Saturated soil on heel, 1.75 × 1.0 × 19.81	34.67	2.125	73.67
   Totals	217.12	—	407.62

6. Part (b) — check sliding on the base. With no fill covering the toe (the drawing places the front ground line at the underside of the base) passive resistance is neglected, and the only restraint is friction on the base: $$(FS)_s=\frac{\sum V\tan\delta}{\sum P_h}=\frac{217.12\tan30^\circ}{108.99}=\frac{125.35}{108.99}=\boxed{1.15}$$ Against the customary minimum of 1.5, the wall fails in sliding.
7. Part (c) — check overturning about the toe. $$(FS)_o=\frac{\sum M_R}{\sum M_O}=\frac{407.62}{227.61}=\boxed{1.79}$$ Against the customary minimum of 2.0 for a granular backfill, the wall fails in overturning as well, though less severely.
8. Part (d) — locate the resultant on the base and obtain $q_{min}$. The line of action of the resultant vertical force is $$\bar{x}=\frac{\sum M_R-\sum M_O}{\sum V}=\frac{407.62-227.61}{217.12}=0.829\ \text{m from the toe}$$ $$e=\frac{B}{2}-\bar{x}=1.500-0.829=\boxed{0.671\ \text{m}}$$ Since $e=0.671\ \text{m} \gt B/6=0.500$ m the resultant falls outside the middle third. Evaluating the elastic distribution formally, $$q_{max,min}=\frac{\sum V}{B}\left(1\pm\frac{6e}{B}\right)=\frac{217.12}{3.0}\left(1\pm1.342\right)$$ $$q_{max}=169.5\ \text{kPa},\qquad \boxed{q_{min}=-24.7\ \text{kPa}}$$
9. Interpret the negative $q_{min}$. Soil cannot pull down on the underside of a footing, so a computed $q_{min}$ of −24.7 kPa does not describe a real stress — it says the heel lifts off and the base bears on only part of its width. Redistributing over the effective contact length, $$L_c=3\left(\frac{B}{2}-e\right)=3(1.500-0.671)=2.49\ \text{m},\qquad q_{max}=\frac{4\sum V}{3(B-2e)}=\frac{4(217.12)}{3(1.658)}=174.6\ \text{kPa}$$ so 0.51 m of the heel is out of contact and the true peak toe pressure is 174.6 kPa rather than 169.5 kPa.
10. Check bearing capacity on the effective base width. With $\phi=40^\circ$ the Hansen factors are $N_q=64.20$ and $N_\gamma=79.54$. The base is a strip, so shape factors are unity; the drawing puts the front ground line at the underside of the base, so $D_f=0$ and the surcharge term vanishes; the water table stands 1.4 m above the underside of the base, so the soil in the bearing wedge is submerged and the self-weight term takes $\gamma'=10$ kN/m³; and the resultant is strongly inclined ($\sum P_h/\sum V=0.502$), which Hansen's inclination factor penalises heavily: $$i_\gamma=\left(1-\frac{0.7\sum P_h}{\sum V}\right)^5=(1-0.351)^5=0.115$$ $$q_u=\tfrac{1}{2}\gamma' B'N_\gamma i_\gamma=\tfrac{1}{2}(10)(1.658)(79.54)(0.115)=75.9\ \text{kPa}$$ $$q_{app}=\frac{\sum V}{B'}=\frac{217.12}{1.658}=131.0\ \text{kPa}\quad\Longrightarrow\quad(FS)_b=\frac{75.9}{131.0}=\boxed{0.58}$$ Against the required minimum of 3, the wall fails in bearing by a wide margin — the computed capacity is below the applied pressure itself.
11. Draw the design conclusion and identify the culprit. All three checks fail: sliding 1.15 against 1.5, overturning 1.79 against 2.0 and bearing 0.58 against 3.0. The wall as dimensioned is not acceptable. The surcharge is what does the damage — deleting it drops the thrust from 108.99 to 62.02 kN/m and lifts the factors to $(FS)_s=2.02$ and $(FS)_o=4.04$, both comfortably adequate. The base width of 3.0 m is 0.56$H$, which is defensible for an unsurcharged wall but not for one carrying 40 kPa. A revised section would widen the base to roughly 4.0–4.5 m with most of the increase added to the heel (which recruits soil weight at a long lever arm and cures overturning, bearing and eccentricity together), add a shear key beneath the base or bury the toe to recover sliding resistance, and — most effective of all — drain the backfill with weep holes and a granular filter so that $P_w$ disappears and $\gamma'$ is replaced by the moist unit weight over the lower 1.4 m.