Gravity dam stability: flotation, overturning, sliding

Assumptions

• Height 60 m, crest width 6 m, base width $B=26$ m. Water level 5 m below the crest, so $H_w=55$ m. Tail water $H_t=40$ m. U/s face vertical; the d/s face is vertical for the top 15 m, then slopes to the toe (20 m run over 45 m).
• $\gamma_c=24$, $\gamma_w=9.81$ kN/m³. Uplift (no drainage): $\gamma_wH_w=539.6$ kN/m² at the heel and $\gamma_wH_t=392.4$ kN/m² at the toe (linear).
• Unit shear resistance $q=500$ kN/m², angle of shear resistance $\phi=35^\circ$.

   6 m
  +----+
  |    | 5 m freeboard
 ~|~~~~|  water 55 m
  |    |
  |    \   15 m vertical d/s
  |     \
  |      \  slope (20 m run)
  |       \ ~~ tail water 40 m
  +--------\
  B = 26 m

Forces and moments about the toe

Total resisting moment: $\Sigma M_R=363,390+104,640=468,030$ kN·m/m (weights plus the tail water thrust).

Checks

1. Flotation (uplift):

$$FS_f=\frac{\Sigma W}{U}=\frac{22,928.0}{12,115.4}=1.89>1$$

The dam does not float (its weight is 1.89 times the uplift).

2. Overturning:

$$FS_o=\frac{\Sigma M_R}{\Sigma M_O}=\frac{468,030}{437,812}=1.07$$

This is far below the required 1.5, so the dam is not safe against overturning.

3. Sliding (shear-friction): $\Sigma V=22,928.0-12,115.4=10,812.6$ kN/m and $\Sigma H=14,837.6-7,848.0=6,989.6$ kN/m.

$$SFF=\frac{qB+\Sigma V\tan\phi}{\Sigma H}=\frac{500\times26+10,812.6\times\tan35^\circ}{6,989.6}=\frac{13000+7,571.1}{6,989.6}=2.94$$

Friction alone gives $\tan35^\circ\,\Sigma V/\Sigma H=1.08$. With the shear resistance, $SFF=2.94$, which is below the usual 3-5, so sliding is also marginal.

4. Position of the resultant: $\bar x=30,218/10,812.6=2.79$ m from the toe, $e=10.21$ m $>B/6=4.33$ m, so tension develops at the heel ($\sigma_{heel}=-564$ kN/m², $\sigma_{toe}=1395$ kN/m²).

Conclusion: the profile is safe against flotation but unsafe against overturning, and it is marginal in sliding and has tension at the heel. The base must be widened or the uplift reduced by drainage.