Stability of trapezoidal gravity dam (RL 250 m)

The dam is checked for full reservoir against overturning, sliding, tension and compression (crushing), per metre length.

Data and assumptions

• Upstream water depth $h_1 = 250 - 200 = 50$ m; tailwater depth $h_2 = 210 - 200 = 10$ m.
• Dam height is not given, so the crest is taken at the maximum reservoir level: $H = 50$ m.
• Upstream face vertical; top width 6 m; downstream face slopes straight from the crest to the toe, so $\tan\alpha = (40-6)/50 = 0.68$.
• $\gamma_w = 9.81$ kN/m³, $\gamma_c = 24$ kN/m³. Coefficient of friction (not given) assumed $\mu = 0.7$.
• Uplift with drainage gallery (IS 6512 rule): heel $\gamma_w h_1$, at gallery $\gamma_w[h_2 + \frac{1}{3}(h_1 - h_2)]$, toe $\gamma_w h_2$.

Uplift intensities: heel $= 9.81 \times 50 = 490.5$ kN/m²; gallery $= 9.81 \times (10 + 40/3) = 9.81 \times 23.33 = 228.9$ kN/m²; toe $= 98.1$ kN/m².

     6 m
    +--+      El. 250 (crest = MRL)
    |  |\
 P1 |W1| \  W2         50 m
 -> |  |  \
    |  |   \  ~~ TW El. 210 (10 m)
heel+--+----\ toe   El. 200
    <10>|<- 30 ->
    gallery   B = 40 m

Forces and moments about the toe

$$\begin{aligned} \sum U &= 8{,}502\ \text{kN} \\ \sum V &= 7{,}200 + 20{,}400 + 333.5 - 8{,}502 = 19{,}431.5\ \text{kN} \\ \sum H &= 12{,}262.5 - 490.5 = 11{,}772\ \text{kN} \\ \sum M_R &= 731{,}191\ \text{kN·m}, \quad \sum M_O = 415{,}835\ \text{kN·m} \end{aligned}$$

(i) Overturning

$$FS_o = \frac{731{,}191}{415{,}835} = 1.76 > 1.5 \quad \text{(safe)}$$

(ii) Sliding

$$FS_s = \frac{\mu\sum V}{\sum H} = \frac{0.7 \times 19{,}431.5}{11{,}772} = 1.16 > 1 \quad \text{(safe)}$$

(iii) Tension

$$\begin{aligned} \bar{x} &= \frac{731{,}191 - 415{,}835}{19{,}431.5} = 16.23\ \text{m from toe} \\ e &= 20 - 16.23 = 3.77\ \text{m} < B/6 = 6.67\ \text{m} \end{aligned}$$

The resultant is within the middle third, so no tension.

(iv) Compression

$$\begin{aligned} \frac{\sum V}{B} &= 485.8\ \text{kN/m}^2, \quad \frac{6e}{B} = 0.566 \\ p_{toe} &= 485.8(1 + 0.566) = 760.6\ \text{kN/m}^2 \\ p_{heel} &= 485.8(1 - 0.566) = 211.0\ \text{kN/m}^2 \end{aligned}$$

Major principal stress at the toe ($p' = 98.1$ kN/m²):

$$\sigma_1 = p_v\sec^2\alpha - p'\tan^2\alpha = 760.6(1.4624) - 98.1(0.4624) = 1066.9\ \text{kN/m}^2$$

These exceed the permissible 300 kN/m², so the dam fails in crushing as the data stand.

Answer: FS overturning = 1.76 (safe); FS sliding = 1.16 (safe, μ = 0.7 assumed); e = 3.77 m < B/6 (no tension); toe stress 760.6 kN/m² and principal stress 1066.9 kN/m² > 300 kN/m² (unsafe in compression). A permissible strength of 300 kN/m² is very low for concrete; if 3000 kN/m² was intended, the dam is safe in all four checks.