--- title: "Middle third rule for gravity dam" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:middle-third-rule-derivation" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/middle-third-rule-derivation site: BCT Engineering --- # Middle third rule for gravity dam Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2075 Ashwin, 6 marks; 2070 Chaitra, 6 marks (asked 2 times) - Video: [Middle third rule gravity dam](https://www.youtube.com/watch?v=kha1MxG0kEI) — Dr.SUNIL KUTE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/middle-third-rule-derivation - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question Show that the resultant force in a concrete gravity dam should pass within the middle third of the base width in order to avoid tension in the heel (write about the "middle third rule" in the design of a concrete gravity dam with necessary derivation). ## Short answer For a gravity dam, base stress σ = (ΣV/B)(1 ± 6e/B). To avoid tension at the heel (σ_min ≥ 0), eccentricity e must satisfy e ≤ B/6, meaning the resultant lies within the middle third of the base. This keeps the entire base in compression, preventing cracking and uplift. ## Answer #### Statement For a gravity dam, the resultant of all forces on any horizontal section should cut the base within its middle third, i.e. the eccentricity from the centre must satisfy $e\le B/6$. Then there is no tension (at the heel when the reservoir is full, at the toe when it is empty). #### Derivation Consider a horizontal section of width $B$ and unit length. Let $\Sigma V$ be the total vertical load and $R$ the resultant, cutting the base at a distance $e$ from the centre of the base. The resultant is equivalent to a vertical force $\Sigma V$ at the centroid of the base plus a moment $M=\Sigma V\cdot e$. The stress at the base is (axial plus bending): $$ \sigma = \frac{\Sigma V}{A} \pm \frac{M\,y}{I} $$ with $A = B\times1$, $I = \dfrac{B^3}{12}$, and $y=\dfrac{B}{2}$ at the extreme edge: $$ \sigma = \frac{\Sigma V}{B} \pm \frac{\Sigma V\,e\,(B/2)}{B^3/12} = \frac{\Sigma V}{B}\left(1 \pm \frac{6e}{B}\right) $$ So $$ \sigma_{max} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right), \qquad \sigma_{min} = \frac{\Sigma V}{B}\left(1-\frac{6e}{B}\right) $$ The minimum stress is at the heel when the resultant lies towards the toe (reservoir full), and it becomes tensile if $$ 1-\frac{6e}{B} < 0 \;\Rightarrow\; e > \frac{B}{6} $$ To avoid tension, $\sigma_{min}\ge0$, hence $$ e \le \frac{B}{6} $$ An eccentricity of $e=B/6$ means the resultant passes through the limit of the middle third, a distance $B/3$ from the toe. Thus the resultant must lie within the middle third of the base ($B/3$ to $2B/3$ from the toe). At the limit $e=B/6$, the stress at the heel is zero and at the toe $\sigma_{max}=2\Sigma V/B$. #### Where $e$ comes from $$ \bar x = \frac{\Sigma M_R - \Sigma M_O}{\Sigma V}\ (\text{from the toe}),\qquad e = \frac{B}{2}-\bar x $$ #### Why it is used - It keeps the entire base in compression, so no crack opens at the heel to admit water (which would raise the uplift). - The dam is economical because the compressive strength of concrete is used well. - It also ensures an adequate margin of safety against overturning. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For resultant eccentricity e on a base of width B (unit length), the base stresses are: A. (ΣV/B)(1 ± 3e/B) B. (ΣV/B)(1 ± 6e/B) C. (ΣV/B)(1 ± e/B) D. (ΣV/B)(1 ± 12e/B) Answer: B. Axial plus bending stress with I = B^3/12 and y = B/2 gives (ΣV/B)(1 ± 6e/B). (Key point: Base stress formula) 2. What condition on e avoids tension at the heel (reservoir full)? A. e >= B/6 B. e <= B/6, i.e. the resultant lies within the middle third C. e = B/2 D. e <= B/3 Answer: B. sigma_min = (ΣV/B)(1 - 6e/B) >= 0 requires e <= B/6. (Key point: Condition for no tension) 3. When e = B/6 exactly, the stresses at the heel and toe are: A. Heel 2ΣV/B, toe zero B. Both ΣV/B C. Heel zero, toe 2ΣV/B D. Heel negative, toe ΣV/B Answer: C. At e = B/6, the minimum stress at the heel is zero and the maximum stress at the toe is 2ΣV/B. (Key point: Stresses at the limit) 4. Which is a reason for keeping the resultant within the middle third? A. It reduces concrete volume to nothing B. It eliminates the need for foundation checks C. It keeps the base in compression so no crack forms at the heel to admit water and raise uplift D. It makes the dam lighter than the water thrust Answer: C. A fully compressed base avoids heel cracking, which could let water in and increase uplift. 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