Concrete Canal Wall — Middle-Third Stability Check: 04-BS-7, December 2019, Q3

Question 3: Concrete Canal Wall — Middle-Third Stability Check (5 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.

Check: the digit for the footing (heel-slab) thickness is missing from the printed paper; only its unit, m, is printed. A round t = 1.0 m is assumed. Over t = 0.5–2.0 m the resultant's location barely moves (x̄ stays between 1.56 m and 1.99 m from the toe) and the safe/unsafe conclusion is unchanged, so the missing digit does not change the graded answer.

Given.

Quantity	Value
Total base width, B (from the Note: B/3 = 2 m)	6 m
Stem width (flush with the toe)	2 m
Overall wall height = water depth, H	7 m
Footing (heel-slab) thickness, t (illegible digit — assumed)	1.0 m
Density of concrete, $\rho_c$ (Constants, p.12)	2400 kg/m³

L-shaped wall: 2 m stem flush with the toe on a 6 m-wide footing; water stands on the exposed 4 m heel and against the stem's wetted face down to the heel top.

Find. Whether the resultant of all forces crosses the base within its middle third (2–4 m from either edge).

Approach. Take moments about the toe (per the hint) of every force acting on the wall: the horizontal hydrostatic thrust on the stem's exposed wetted face, the weight of the stem, the weight of the footing slab, and the weight of the water standing on top of the exposed heel. Divide the net moment by the total vertical force to locate the resultant, then compare against the middle-third band.

Horizontal hydrostatic thrust on the stem's wetted face. The wetted face runs from the water surface down to the top of the heel, a height of $H-t=6.0$ m: $$F_H=\tfrac12\rho_w g(H-t)^2=\tfrac12(1000)(9.81)(6.0)^2=\boxed{176.6\ \text{kN/m}}$$ acting at $t+(H-t)/3=1.0+2.0=3.0$ m above the base.
Weight of water on the heel. Over the 4 m-wide exposed heel, water stands to the same 6.0 m depth as the stem's wetted height: $$W_{water}=\rho_wg\,b_{heel}(H-t)=(1000)(9.81)(4.0)(6.0)=235.4\ \text{kN/m, at }4.0\text{ m from the toe}$$
Weight of the stem and the footing slab. $$W_{stem}=\rho_cg\,b_{stem}(H-t)=(2400)(9.81)(2.0)(6.0)=282.5\ \text{kN/m, at }1.0\text{ m from the toe}$$ $$W_{footing}=\rho_cg\,Bt=(2400)(9.81)(6.0)(1.0)=141.3\ \text{kN/m, at }3.0\text{ m from the toe}$$
Resultant location about the toe. Summing vertical forces and moments (each weight's moment arm is its own distance from the toe; $F_H$'s arm is its height above the base): $$\Sigma V=235.4+282.5+141.3=659.2\ \text{kN/m}$$ $$\bar{x}=\frac{\Sigma(W_i\,x_i)-F_H z_{FH}}{\Sigma V}=\frac{(235.4)(4.0)+(282.5)(1.0)+(141.3)(3.0)-(176.6)(3.0)}{659.2}=\boxed{1.70\ \text{m from the toe}}$$
Middle-third check. The middle third spans from 2.0 m to 4.0 m from either edge. Since $\bar{x}=1.70\ \text{m}\lt2.0\ \text{m}$, the resultant falls SHORT of the middle third, toward the toe: $\boxed{\text{the base is NOT fully under compression}}$. The eccentricity $e=|B/2-\bar{x}|=1.30$ m exceeds $B/6=1.0$ m, so the linear pressure distribution goes negative (tension) at the heel: $\sigma_{toe}=253.1$ kPa, $\sigma_{heel}=-33.4$ kPa (tension) — confirming the wall is unsafe as sized.

Quantity	Value
Hydrostatic thrust, F_H	176.6 kN/m
Total vertical force, ΣV	659.2 kN/m
Resultant location from toe	1.70 m
Middle-third band	2.00 – 4.00 m from either edge
Base pressure (toe / heel)	253.1 kPa / −33.4 kPa (tension)
Verdict	NOT safe — resultant falls short of the middle third

Concept & Theory

Topic. Cantilever (heel-and-toe) retaining-wall stability: hydrostatic thrust, self-weight, and the middle-third rule.

Key relations. Hydrostatic force on a plane vertical surface $F=\tfrac12\rho gh^2$ acting at $h/3$ above its own base; resultant location $\bar{x}=\Sigma M_{toe}/\Sigma V$; base pressure $\sigma=\tfrac{\Sigma V}{B}\left(1\pm\tfrac{6e}{B}\right)$.

Why this works. Unlike a solid gravity dam, a cantilever wall relies partly on the WEIGHT OF WATER standing on its heel to help resist overturning; here that water weight (235 kN/m at a favourable 4 m arm) is the single largest resisting contribution, yet it is still not enough to pull the resultant into the middle third once the toe-ward overturning moment from $F_H$ is included — illustrating why a heel this size is genuinely under-designed for a 7 m retained depth.

Common pitfall. Forgetting that water sitting on the heel counts as a VERTICAL load contributing to $\Sigma V$ and the resisting moment, in addition to the horizontal thrust on the stem — omitting it would understate stability, not overstate it.

Source. White, Fluid Mechanics, Ch. 2 (Hydrostatic forces on plane surfaces; dam/wall stability).

Question 3 of 13: Concrete Canal Wall — Middle-Third Stability Check

Question 3: Concrete Canal Wall — Middle-Third Stability Check (5 marks)