Gravity Dam Stability (Middle-Third Rule): 04-BS-7, December 2014, Q2

Question 2 — Gravity Dam Stability (Middle-Third Rule) (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
Water depth, H_w	18 m
Dam height	20 m
Base width	15 m
Density of concrete, ρ_c	2400 kg/m³

Find. The horizontal distance from toe O to where the resultant force crosses the base, and whether that point lies within the middle third (safety verdict), per metre of dam length.

Fig. Q2 — triangular gravity dam: vertical water-side face (18 m wetted, 20 m total), 15 m base, toe O at bottom-right. Moments are taken about O.

Approach. Compute the hydrostatic thrust (per metre of dam length) and its line of action, and the dam's self-weight and its centroid location; take moments about O to locate where the resultant crosses the base, then compare against the middle-third band [5 m, 10 m] from O.

Hydrostatic thrust on the vertical face. $$F_w = \tfrac{1}{2}\rho g H_w^2 = \tfrac{1}{2}(1000)(9.81)(18)^2 = \boxed{1589.2\text{ kN/m}}, \quad \text{acting at } H_w/3 = 6.00\text{ m above the base}$$
Dam self-weight. Triangular cross-section area = ½(15)(20) = 150 m², with centroid 10.00 m horizontally from O (toe): $$W = \rho_c g (\text{Area}) = 2400\times 9.81\times 150 = \boxed{3531.6\text{ kN/m}}$$
Location of the resultant on the base. Taking moments about O (overturning moment F_w·6.00, resisting moment W·10.00), the resultant crosses the base at a distance x̄ from O satisfying x̄ = 6F_w/W − 10: $$\bar{x} = \frac{6(1589.2)}{3531.6}-10 = 2.700-10 = \boxed{-7.30\text{ m (i.e. 7.30 m from O, toward the heel)}}$$
Middle-third check. The middle third of the 15 m base spans 5.00–10.00 m from O. Since 7.30 m falls inside this band, the resultant stays within the middle third: $$\boxed{\text{Dam is SAFE} - \text{compressive stress is maintained across the full base}}$$

Quantity	Result
Hydrostatic thrust, F_w	1589.2 kN/m
Dam weight, W	3531.6 kN/m
Resultant location from O	7.30 m
Middle-third band	5.00–10.00 m from O
Verdict	SAFE

Topic: Gravity dam stability — the middle-third rule.

Key relations: overturning moment M_ot = F_w(H_w/3); resisting moment M_r = W·(centroid arm); location of the resultant on the base, x̄ = (ΣM)/(ΣV), measured from a chosen reference point.

Why this works: the "middle third" rule keeps the resultant's eccentricity from the base centreline within b/6, which (for a trapezoidal contact-pressure distribution) guarantees the minimum edge pressure stays ≥ 0 — i.e. no tension or separation at the heel.

Common pitfall: forgetting the water force contributes only a MOMENT (no vertical component, since the wetted face is vertical) — only the dam's self-weight contributes to ΣV.

Source: White, Fluid Mechanics, 8th ed., §2.6 (hydrostatic force & centre of pressure), applied here to a structural stability check.

Question 2 of 13: Gravity Dam Stability (Middle-Third Rule)

Question 2 — Gravity Dam Stability (Middle-Third Rule) (5 marks)