Middle-Third (No-Tension) Check on a Concrete Canal Wall: 04-BS-7, May 2016, Q3

Question 3: Middle-Third (No-Tension) Check on a Concrete Canal Wall (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.

Given.

Quantity	Value
Toe (footing) block	4 m wide × 2 m tall
Stem block	2 m wide × 7 m tall
Total base width B	4 + 2 = 6 m
Water depth (submerges the toe and part of the stem)	6 m
Concrete density ρ_c	2400 kg/m³

Find. Whether the resultant force on the base lies within the middle third (2 m ≤ x ≤ 4 m, measured from either edge of the 6 m base).

Fig. Q3 — free body of the canal wall: self-weight W, hydrostatic thrust Fℎ, moments taken about the right edge; the resultant at 1.41 m falls outside the shaded middle third.

Approach. Compute the wall's self-weight W (through its own centroid) and the horizontal hydrostatic thrust F_h from the 6 m water depth (acting at h/3 above the base), take moments of both about the extreme right-hand edge as hinted, and locate where the resultant crosses the base.

Self-weight, per metre of wall length. $$W_{toe} = \rho_c g (4)(2) = 2400(9.81)(8) = 188.35\ \text{kN/m}, \qquad W_{stem} = \rho_c g(2)(7) = 329.62\ \text{kN/m}$$ $$W = W_{toe}+W_{stem} = \boxed{517.97\ \text{kN/m}}$$
Horizontal hydrostatic thrust. The water's vertical projection is 6 m regardless of the step in the wall face, so $$F_h = \tfrac{1}{2}\rho_w g H_w^2 = \tfrac12(1000)(9.81)(6^2) = \boxed{176.58\ \text{kN/m}}$$ acting at $H_w/3 = 2.0$ m above the base.
Moments about the extreme right-hand edge (the toe of the stem, per the hint). Restoring moment from self-weight (each block's own centroid distance from the right edge is B−W_i/2 for the toe and W_s/2 for the stem): $$M_r = W_{toe}(4.0) + W_{stem}(1.0) = 753.4 + 329.6 = 1083.0\ \text{kN}\cdot\text{m/m}$$ Overturning moment from the water thrust (acting at 2.0 m above the base): $$M_o = F_h(2.0) = 353.2\ \text{kN}\cdot\text{m/m}$$
Locate the resultant on the base and compare with the middle third. $$\bar{x}_R = \frac{M_r - M_o}{W} = \frac{1083.0-353.2}{517.97} = \boxed{1.41\ \text{m from the right edge}}$$ Eccentricity from the base centreline: $e = 3.0 - 1.41 = 1.59$ m, versus the kern limit $B/6 = 1.00$ m. Since $e = 1.59 > 1.00$ m (equivalently, 1.41 m < 2 m from the edge), the resultant falls outside the middle third.

Quantity	Result
Total self-weight W	517.97 kN/m
Horizontal thrust F_h	176.58 kN/m
Resultant location from right edge	1.41 m (middle third is 2–4 m)
Eccentricity e	1.59 m > B/6 = 1.00 m
Entire base under compression?	No — a heel zone will see tension/lift-off

Topic: Middle-third (kern) rule for gravity structures on soil.

Key relations: base pressure p = (W/B)(1 ± 6e/B); no-tension condition e ≤ B/6; hydrostatic thrust F_h = ½ρgH² acting at H/3.

Why this works. Soil (and most footing interfaces) cannot sustain tension, so the resultant of all vertical loads must stay within the kern (middle third) of the base or the pressure distribution predicted by beam theory goes negative at one edge, which is physically impossible and instead means partial lift-off/cracking there.

Common pitfall. Using only the stem's projected height for the hydrostatic force instead of the FULL submerged height (6 m); because the water is above the stepped toe as well, the horizontal thrust depends on the full vertical projection of the wetted surface, not on the shape of the step.

Source: White, Fluid Mechanics, Ch. 2 (hydrostatic forces); standard gravity-retaining-wall stability practice (middle-third/kern rule).

Question 3 of 13: Middle-Third (No-Tension) Check on a Concrete Canal Wall

Question 3: Middle-Third (No-Tension) Check on a Concrete Canal Wall (5 marks)