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Sliding and overturning check of gravity dam

Short answer

Sliding FS 1.11, overturning FS 1.67, no tension. Safe, though sliding margin is narrow; recommend drainage gallery and foundation key to increase shear resistance.

Check safety against sliding and overturning for the gravity dam shown in figure. Assume, unit weight of water = 9.81 kN/m³, unit weight of concrete = 24 kN/m³, angle of internal friction ϕ = 37°, uplift coefficient = 80%. Consider self-weight, hydrostatic pressure and uplift pressure only. Mention your recommendations. [Figure: gravity dam with vertical upstream face and top width 2.0 m; the water surface is 1.0 m below the crest and the water depth is 30 m; the downstream face is vertical for the top 1.5 m and then slopes at 0.75 (H) : 1.0 (V) down to the toe.]

Geometry and data (per metre length)

  • Water depth 30 m, water surface 1 m below crest, so dam height =31= 31 m.
  • Top width 2 m; downstream face vertical for 1.5 m, then slopes 0.75H : 1V for the remaining 31−1.5=29.531 - 1.5 = 29.5 m, giving horizontal projection 0.75×29.5=22.1250.75 \times 29.5 = 22.125 m.
  • Base width B=2+22.125=24.125B = 2 + 22.125 = 24.125 m. No tailwater.
  • γw=9.81\gamma_w = 9.81, γc=24\gamma_c = 24 kN/m³; μ=tan⁡37°=0.754\mu = \tan37° = 0.754; uplift C = 0.8 (triangular, 0.8γwh0.8\gamma_w h at heel to zero at toe).
     2 m
    +--+    1 m above WL
    |  +    1.5 m vertical
 30 |   \
 m  |W1  \  0.75H:1V
 -> |  W2 \       29.5 m
heel+------\ toe
    <-24.125->

Forces and moments about the toe

ForceValue (kN)Arm (m)Moment (kN·m)
W1 = 2×31×241,488 ↓23.125+34,410
W2 = ½×22.125×29.5×247,832.3 ↓14.750+115,526
P = ½×9.81×30²4,414.5 →10.000−44,145
U = ½×0.8×9.81×30×24.1252,840.0 ↑16.083−45,677
∑V=1,488+7,832.3−2,840.0=6,480.3 kN∑H=4,414.5 kN∑MR=149,936,∑MO=89,822 kN⋅m\begin{aligned} \sum V &= 1{,}488 + 7{,}832.3 - 2{,}840.0 = 6{,}480.3\ \text{kN} \\ \sum H &= 4{,}414.5\ \text{kN} \\ \sum M_R &= 149{,}936, \quad \sum M_O = 89{,}822\ \text{kN·m} \end{aligned}

Safety against sliding

FSs=μ∑V∑H=0.754×6,480.34,414.5=1.11>1FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.754 \times 6{,}480.3}{4{,}414.5} = 1.11 > 1

Safety against overturning

FSo=149,93689,822=1.67>1.5FS_o = \frac{149{,}936}{89{,}822} = 1.67 > 1.5

Check of resultant: xˉ=(149,936−89,822)/6,480.3=9.28\bar{x} = (149{,}936 - 89{,}822)/6{,}480.3 = 9.28 m from toe, e=12.06−9.28=2.79e = 12.06 - 9.28 = 2.79 m <B/6=4.02< B/6 = 4.02 m, so there is no tension at the heel (toe stress 454.7 kN/m², heel stress 82.5 kN/m²).

Answer: FS against sliding = 1.11 and FS against overturning = 1.67; the dam is safe.

Recommendations: the sliding margin is small (just above 1), so:

  • provide foundation drainage holes/gallery to reduce uplift,
  • key the base into sound rock or give it a slight upstream slope to increase shear resistance,
  • check the shear friction factor using the shear strength of the rock–concrete joint.

✓ Answer checked by Claude Opus 5.5 (high effort)

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