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Middle third rule for gravity dam

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.

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).

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≤B/6e\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 BB and unit length. Let ΣV\Sigma V be the total vertical load and RR the resultant, cutting the base at a distance ee from the centre of the base.

The resultant is equivalent to a vertical force ΣV\Sigma V at the centroid of the base plus a moment M=ΣV⋅eM=\Sigma V\cdot e. The stress at the base is (axial plus bending):

σ=ΣVA±M yI\sigma = \frac{\Sigma V}{A} \pm \frac{M\,y}{I}

with A=B×1A = B\times1, I=B312I = \dfrac{B^3}{12}, and y=B2y=\dfrac{B}{2} at the extreme edge:

σ=ΣVB±ΣV e (B/2)B3/12=ΣVB(1±6eB)\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

σmax=ΣVB(1+6eB),σmin=ΣVB(1−6eB)\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

1−6eB<0  ⇒  e>B61-\frac{6e}{B} < 0 \;\Rightarrow\; e > \frac{B}{6}

To avoid tension, σmin≥0\sigma_{min}\ge0, hence

e≤B6e \le \frac{B}{6}

An eccentricity of e=B/6e=B/6 means the resultant passes through the limit of the middle third, a distance B/3B/3 from the toe. Thus the resultant must lie within the middle third of the base (B/3B/3 to 2B/32B/3 from the toe). At the limit e=B/6e=B/6, the stress at the heel is zero and at the toe σmax=2ΣV/B\sigma_{max}=2\Sigma V/B.

Where ee comes from

xˉ=ΣMR−ΣMOΣV (from the toe),e=B2−xˉ\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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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