Gravity dam equilibrium and overtopping discharge coefficient

Assumptions

The figure values are not given in the question, so these are assumed: crest level 720 masl (dam height $H_d=20$ m above the base at 700 masl), so HRWL $=719$ masl and the water depth $h=19$ m. Crest width 3 m, vertical u/s face. $\gamma_c=24$ kN/m³, $\gamma_w=9.81$ kN/m³, $\phi=43^\circ$. Uplift is taken as triangular and full (no drainage): $\gamma_wh$ at the heel and zero at the toe. Forces per metre length.

   3 m
  +---+ 720 crest
 ~|~~~| 719 HRWL
  |   \
  |    \ alpha
  |     \
 700 +----\
  heel <--B--> toe

a) Main forces at HRWL (in terms of B)

Split the section into a rectangle ($3\times20$) and a triangle (base $B-3$, height 20).

• $W_1=24\times3\times20=1440$ kN at $B-1.5$ from the toe
• $W_2=24\times\frac12(B-3)\times20=240(B-3)$ kN at $\frac23(B-3)$ from the toe
• Water thrust $P=\frac12\gamma_wh^2=\frac12\times9.81\times19^2=1,770.7$ kN acting at $h/3=6.33$ m above the base
• Uplift $U=\frac12\gamma_whB=\frac12\times9.81\times19\times B=93.20B$ kN acting upward at $\frac23B$ from the toe

b) Base width B and angle α for $FS_o=1.4$

Moments about the toe:

$$\Sigma M_R=1440(B-1.5)+240(B-3)\times\tfrac23(B-3)=1440(B-1.5)+160(B-3)^2$$ $$\Sigma M_O=P\times6.333+U\times\tfrac23B=11,214.5+62.13B^2$$

Condition $\Sigma M_R=1.4\,\Sigma M_O$ gives, after expanding, a quadratic in $B$:

$$160B^2-960B+1440+1440B-2160 = 1.4\,(11,214.5+62.13B^2)$$

Solving: $B=12.07$ m (check: $FS_o=1.40$).

Angle of the d/s face: $\tan\alpha=\dfrac{H_d}{B-3}=\dfrac{20}{9.07}$, so $\alpha=65.6^\circ$ with the horizontal (the face makes 24.4° with the vertical, slope $\approx0.45$ H : 1 V).

c) Tension and required shear resistance

$W_2=240\times(12.07-3)=2,176$ kN; $U=1,124$ kN. $\Sigma V=1440+2,176-1,124=2,491$ kN/m.

$\bar x=\dfrac{\Sigma M_R-\Sigma M_O}{\Sigma V}=3.25$ m from the toe, $e=B/2-\bar x=2.78$ m, and $B/6=2.01$ m.

Since $e>B/6$, the resultant is outside the middle third and the dam is not free from tension: $\sigma_{heel}=\dfrac{\Sigma V}{B}\left(1-\dfrac{6e}{B}\right)=-79.0$ kN/m² (tensile), and $\sigma_{toe}=491.9$ kN/m².

Shear safety factor $F_{SF}=\dfrac{cB+\Sigma V\tan\phi}{P}=2.5$:

$$c=\frac{2.5\,P-\Sigma V\tan43^\circ}{B}=\frac{2.5\times1,770.7-2,491\times0.9325}{12.07}=174.4\ \text{kN/m}^2$$

Required unit shear resistance ≈ 174 kN/m².

d) Discharge coefficient during overtopping

Head over the crest $H=722.5-720=2.5$ m, crest length $L=80$ m, $Q=320$ m³/s.

$$Q=C\,L\,H^{3/2}\Rightarrow C=\frac{320}{80\times2.5^{1.5}}=\frac{320}{80\times3.953}=1.01\ (\text{SI})$$

In the form $Q=\frac23C_d\sqrt{2g}\,L\,H^{3/2}$: $C_d=0.34$.

Comment: For a broad-crested or ogee crest, $C$ is about 1.7-2.2 ($C_d\approx0.5$-$0.75$). The value 1.01 is much lower, which means the estimated 320 m³/s is small for a head of 2.5 m. The reasons may be an over-estimated head, a submerged (tail water) flow, or a poor crest shape/rough overtopped surface, or the estimate of discharge is wrong. The check should be repeated with the measured crest level.

Answer: B ≈ 12.1 m, α ≈ 66°; tension occurs at the heel; required shear resistance ≈ 174 kN/m²; C ≈ 1.01 (C_d ≈ 0.34).