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Stability checks of 50 m trapezoidal dam

Short answer

FS overturning 2.23, sliding 1.60, eccentricity 0.26 m (no tension), maximum principal stress 829.6 kN/m² below permissible 981 kN/m². Safe; section is conservative.

A concrete dam of trapezoidal section is of height 50 m with upstream vertical face. Top width, base width and freeboard are 4.5 m, 42 m and 5 m respectively. Check the stability of the dam against (i) Overturning (ii) sliding (iii) Tension and (iv) Crushing. The unit weight of water is 10 kN/m³, the unit weight of concrete is 25 kN/m³ and Crushing strength of the material 10 kg/cm². Take angle of internal friction (ϕ = 37°), uplift coefficient 0.8. Consider self-weight, hydraulic pressure and uplift pressure only. Mention your recommendation.

Data (per metre length)

  • Dam height 50 m, freeboard 5 m, so water depth h=45h = 45 m (no tailwater).
  • Top width 4.5 m, base 42 m, upstream face vertical; downstream slope tan⁡α=37.5/50=0.75\tan\alpha = 37.5/50 = 0.75.
  • γw=10\gamma_w = 10, γc=25\gamma_c = 25 kN/m³; μ=tan⁡37°=0.754\mu = \tan 37° = 0.754; uplift coefficient C = 0.8.
  • Crushing strength =10= 10 kg/cm² =10×9.81×104= 10 \times 9.81 \times 10^4 N/m² =981= 981 kN/m².
     4.5
    +--+   ---
    |  |\   5 m FB
 45 |W1| \
 m  |  |W2\    50 m
 -> |  |   \
heel+--+----\ toe
    <-- 42 m -->

Forces and moments about the toe

ForceValue (kN)Arm (m)Moment (kN·m)
W1 = 4.5×50×255,625 ↓39.75+223,594
W2 = ½×37.5×50×2523,437.5 ↓25.00+585,938
P = ½×10×45²10,125 →15.00−151,875
U = ½×0.8×10×45×427,560 ↑28.00−211,680
∑V=5,625+23,437.5−7,560=21,502.5 kN∑MR=809,531,∑MO=363,555 kN⋅m\begin{aligned} \sum V &= 5{,}625 + 23{,}437.5 - 7{,}560 = 21{,}502.5\ \text{kN} \\ \sum M_R &= 809{,}531, \quad \sum M_O = 363{,}555\ \text{kN·m} \end{aligned}

(i) Overturning

FSo=809,531363,555=2.23>1.5(safe)FS_o = \frac{809{,}531}{363{,}555} = 2.23 > 1.5 \quad \text{(safe)}

(ii) Sliding

FSs=0.754×21,502.510,125=1.60>1(safe)FS_s = \frac{0.754 \times 21{,}502.5}{10{,}125} = 1.60 > 1 \quad \text{(safe)}

(iii) Tension

xˉ=809,531−363,55521,502.5=20.74 m from toee=21−20.74=0.26 m<B/6=7 m\begin{aligned} \bar{x} &= \frac{809{,}531 - 363{,}555}{21{,}502.5} = 20.74\ \text{m from toe} \\ e &= 21 - 20.74 = 0.26\ \text{m} < B/6 = 7\ \text{m} \end{aligned}

No tension (safe).

(iv) Crushing

ptoe=21,502.542(1+6×0.2642)=511.97×1.037=530.9 kN/m2pheel=511.97×0.963=493.0 kN/m2σ1=ptoesec⁡2α=530.9×1.5625=829.6 kN/m2\begin{aligned} p_{toe} &= \frac{21{,}502.5}{42}\left(1 + \frac{6\times0.26}{42}\right) = 511.97 \times 1.037 = 530.9\ \text{kN/m}^2 \\ p_{heel} &= 511.97 \times 0.963 = 493.0\ \text{kN/m}^2 \\ \sigma_1 &= p_{toe}\sec^2\alpha = 530.9 \times 1.5625 = 829.6\ \text{kN/m}^2 \end{aligned}

Both are below 981 kN/m² (safe).

CheckValueLimitResult
Overturning2.23≥ 1.5Safe
Sliding1.60≥ 1.0Safe
Tensione = 0.26 m≤ 7 mSafe
Crushing829.6 kN/m²≤ 981 kN/m²Safe

Recommendation: The dam is safe in all four checks. The resultant is almost at the centre of the base, so stresses are nearly uniform; the section is on the heavy side and the downstream slope could be steepened to save concrete, after re-checking with earthquake and silt loads.

✓ Answer checked by Claude Opus 5.5 (high effort)

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