Gravity dam forces and stability checks (2081 Baishakh)

Assumptions (figure not available in full)

• Dam height 50 m; crest width 5 m; water level 2 m below the crest, $H_w=48$ m; tail water $H_t=5$ m.
• U/s face vertical from the crest to 15 m above the base, then flared at 1 : 1 to the heel (15 m horizontal run). D/s face runs straight from the d/s edge of the crest to the toe; the total base width $B=57$ m gives a d/s run of $57-15-5=37$ m over 50 m ($\tan\phi_d=0.74$).
• $\gamma_c=24$, $\gamma_w=9.81$ kN/m³, $\mu=0.70$, uplift coefficient $K=0.45$: heel ordinate $=\gamma_wH_t+K\gamma_w(H_w-H_t)=238.9$ kN/m², toe ordinate $=\gamma_wH_t=49.1$ kN/m² (the gallery near the heel is the reason for this reduced uplift).
• Allowable stresses: foundation 50 kgf/cm² $=4905$ kN/m²; dam body 10.5 kgf/cm² $=1030$ kN/m².

        5 m
    +--------+
 ~~~|~~~~~~~~|  water (48 m)
    |        |\
    |        | \  d/s slope
    |        |  \
   /|        |   \ ~~ tail water 5 m
  /_|________|____\
  15 m flare      B = 57 m

Forces and moments about the toe

Checks

• $\Sigma V = \Sigma W - U = 36,950.3 - 8,205.8 = 28,744.5$ kN/m
• $\Sigma H = 11,301.1 - 122.6 = 11,178.5$ kN/m
• $\Sigma M_R = 1,209,574$ kN·m/m (tail water thrust moment included); $\Sigma M_O = 466,078$ kN·m/m

Overturning: $FS_o = \Sigma M_R/\Sigma M_O = 1,209,574/466,078 = 2.60$ — safe (> 1.5).

Sliding (friction): $FS_s = \mu\Sigma V/\Sigma H = 0.7\times28,744.5/11,178.5 = 1.80$ — safe (> 1).

Resultant: $\bar x = (\Sigma M_R-\Sigma M_O)/\Sigma V = 743,496/28,744.5 = 25.87$ m from the toe; $e = B/2-\bar x = 28.5-25.87 = 2.63$ m; $B/6 = 9.50$ m — inside the middle third (no tension).

$$\sigma_{toe} = \frac{\Sigma V}{B}\left(1+\frac{6e}{B}\right) = 504.3\,(1+0.277) = 644.1\ \text{kN/m}^2$$ $$\sigma_{heel} = 504.3\,(1-0.277) = 364.5\ \text{kN/m}^2$$

The toe stress of 644 kN/m² is below the dam-body allowable of 1030 kN/m². The heel stress of 364 kN/m² is compressive, so there is no tension.

Shear and principal stress at the toe ($\tan\phi_d=0.740$, $p'=49.1$ kN/m²):

$$\tau_{toe} = (\sigma_{toe}-p')\tan\phi_d = (644.1-49.1)\times0.740 = 440.4\ \text{kN/m}^2$$ $$\sigma_{1,toe} = \sigma_{toe}\sec^2\phi_d - p'\tan^2\phi_d = 644.1\,(1+0.548) - 49.1\times0.548 = 970.0\ \text{kN/m}^2$$

Shear and principal stress at the heel ($\tan\phi_u=1.000$, $p=470.9$ kN/m²):

$$\tau_{heel} = (p-\sigma_{heel})\tan\phi_u = (470.9-364.5)\times1.000 = 106.4\ \text{kN/m}^2$$ $$\sigma_{1,heel} = \sigma_{heel}\sec^2\phi_u - p\tan^2\phi_u = 258.0\ \text{kN/m}^2$$

• The maximum principal stress of 970 kN/m² (at the toe) is below the dam-body allowable of 1030 kN/m², and the toe stress is far below the foundation allowable of 4905 kN/m².

Conclusion: sliding ($FS_s=1.80$), overturning ($FS_o=2.60$), tension (none) and crushing (below 1030 and 4905 kN/m²) are all satisfied, so the dam section is safe.