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Stability of gravity dam (crest 1530 m)

Short answer

FS overturning 2.74, sliding 1.76, resultant 2.55 m from centre towards heel (no tension), maximum compressive stress 772.7 kN/m² at heel < 3000 kN/m². Dam is safe.

Check the stability of a gravity dam having trapezoidal section with following data: Crest level of dam = 1530 m, Maximum water level at upstream = 1525 m, Top width = 6 m, Base width = 60 m, Tail water depth = 6 m, Bed level = 1470 m. Drainage gallery provided at 10 m from the heel. Consider full uplift condition and permissible compressive strength of concrete 3000 kN/m². Take specific weight of concrete 24 kN/m³.

Data and assumptions (per metre length)

  • Dam height =1530−1470=60= 1530 - 1470 = 60 m; upstream water depth h1=1525−1470=55h_1 = 1525 - 1470 = 55 m; tailwater depth h2=6h_2 = 6 m.
  • Top width 6 m, base 60 m, upstream face vertical, downstream face straight from crest to toe: tan⁡α=54/60=0.9\tan\alpha = 54/60 = 0.9.
  • γw=9.81\gamma_w = 9.81, γc=24\gamma_c = 24 kN/m³; μ (not given) assumed 0.7.
  • "Full uplift": full hydrostatic uplift at the heel; with the gallery, IS 6512 rule: head at gallery =h2+13(h1−h2)=6+49/3=22.33= h_2 + \frac{1}{3}(h_1 - h_2) = 6 + 49/3 = 22.33 m.

Uplift intensities: heel 539.6, gallery 219.1, toe 58.9 kN/m².

      6
    +--+  1530
    |  |\
 55 |  | \     60 m
 m  |W1|W2\
    |  |   \  ~~ TW 6 m
    +--+----\ 1470
    <10>|<-50->
       gallery  B = 60 m

Forces and moments about the toe

ForceValue (kN)Arm (m)Moment (kN·m)
W1 = 6×60×248,640 ↓57.0+492,480
W2 = ½×54×60×2438,880 ↓36.0+1,399,680
Tailwater wedge = ½×5.4×6×9.81158.9 ↓1.8+286
P1 = ½×9.81×55²14,837.6 →18.333−272,023
P2 = ½×9.81×6²176.6 ←2.0+353
U1 = 219.1×102,190.9 ↑55.0−120,500
U2 = ½×320.5×101,602.3 ↑56.667−90,797
U3 = 58.9×502,943.0 ↑25.0−73,575
U4 = ½×160.2×504,005.8 ↑33.333−133,525
∑U=10,742 kN∑V=8,640+38,880+158.9−10,742=36,937 kN∑H=14,837.6−176.6=14,661 kN∑MR=1,892,799,∑MO=690,420 kN⋅m\begin{aligned} \sum U &= 10{,}742\ \text{kN} \\ \sum V &= 8{,}640 + 38{,}880 + 158.9 - 10{,}742 = 36{,}937\ \text{kN} \\ \sum H &= 14{,}837.6 - 176.6 = 14{,}661\ \text{kN} \\ \sum M_R &= 1{,}892{,}799, \quad \sum M_O = 690{,}420\ \text{kN·m} \end{aligned}

Overturning

FSo=1,892,799690,420=2.74>1.5(safe)FS_o = \frac{1{,}892{,}799}{690{,}420} = 2.74 > 1.5 \quad \text{(safe)}

Sliding

FSs=0.7×36,93714,661=1.76>1(safe)FS_s = \frac{0.7 \times 36{,}937}{14{,}661} = 1.76 > 1 \quad \text{(safe)}

Tension

xˉ=1,892,799−690,42036,937=32.55 m from toee=30−32.55=−2.55 m\begin{aligned} \bar{x} &= \frac{1{,}892{,}799 - 690{,}420}{36{,}937} = 32.55\ \text{m from toe} \\ e &= 30 - 32.55 = -2.55\ \text{m} \end{aligned}

The resultant is 2.55 m from the centre on the heel side; ∣e∣<B/6=10|e| < B/6 = 10 m, so no tension.

Compression

∑VB=615.6 kN/m2,6∣e∣B=0.255pheel=615.6(1+0.255)=772.7 kN/m2ptoe=615.6(1−0.255)=458.5 kN/m2\begin{aligned} \frac{\sum V}{B} &= 615.6\ \text{kN/m}^2, \quad \frac{6|e|}{B} = 0.255 \\ p_{heel} &= 615.6(1 + 0.255) = 772.7\ \text{kN/m}^2 \\ p_{toe} &= 615.6(1 - 0.255) = 458.5\ \text{kN/m}^2 \end{aligned}

Principal stress at toe =458.5(1.81)−58.9(0.81)=782.2= 458.5(1.81) - 58.9(0.81) = 782.2 kN/m². All stresses are well below 3000 kN/m² (safe).

Answer: FS overturning = 2.74, FS sliding = 1.76, e = 2.55 m (within middle third, no tension), maximum stress = 772.7 kN/m² at heel < 3000 kN/m². The dam is safe. (If drains are ignored and full uplift acts over the whole base, FS overturning = 1.98 and FS sliding = 1.42; still safe.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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