Elementary profile base width formula proof

Elementary profile

The elementary profile of a gravity dam is a right-angled triangle with a vertical upstream face, zero top width and base width B, with water up to its apex (height H). Water pressure is zero at the top and grows linearly with depth, so a triangle is the ideal shape: it uses the least material and, if the base is wide enough, has no tension under full or empty reservoir. Practical profiles are made by adding top width and freeboard to it.

        |\
        | \
  P --> |  \      H
  (H/3) | W \
        |    \
   heel |__.__\ toe
        ^  ^   ^
        U (KγwH at heel -> 0 at toe)
        <--B-->

Forces per metre length

With $G$ = specific gravity of dam material, $K$ = uplift intensity coefficient:

$$\begin{aligned} W &= \frac{1}{2}BH\,G\gamma_w \quad (\text{at } B/3 \text{ from heel}) \\ P &= \frac{1}{2}\gamma_w H^2 \quad (\text{at } H/3 \text{ above base}) \\ U &= \frac{1}{2}K\gamma_w H B \quad (\text{at } B/3 \text{ from heel}) \end{aligned}$$

Proof of $B = H/\sqrt{G-K}$ (no-tension condition)

For no tension at the heel with full reservoir, the resultant must pass through the outer middle-third point, i.e. at $2B/3$ from the heel ($B/3$ from the toe). Taking moments about this point:

• W and U act at $B/3$ from the heel, so their lever arm is $2B/3 - B/3 = B/3$.
• P has lever arm $H/3$.

For the resultant to pass through this point, the net moment must be zero:

$$\begin{aligned} (W - U)\frac{B}{3} &= P\frac{H}{3} \\ \left(\frac{1}{2}BHG\gamma_w - \frac{1}{2}K\gamma_w HB\right)B &= \frac{1}{2}\gamma_w H^2 \cdot H \\ \frac{1}{2}\gamma_w B^2 H (G - K) &= \frac{1}{2}\gamma_w H^3 \\ B^2 (G - K) &= H^2 \\ B &= \frac{H}{\sqrt{G - K}} \end{aligned}$$

Hence proved. Without uplift ($K = 0$), $B = H/\sqrt{G}$; for concrete ($G = 2.4$) this gives $B = 0.645H$, and with full uplift ($K = 1$) $B = 0.845H$.

No-sliding condition (for comparison)

$\mu(W - U) = P$ gives $B = \dfrac{H}{\mu(G - K)}$. The larger of the two widths is adopted.