Base width of 20 m high trapezoidal gravity dam

Data and assumptions

Height 20 m, vertical u/s face, top width 5 m, water depth $H_w=18$ m, no tail water. $\gamma_w=10$, $\gamma_c=24$ kN/m³. The resultant must act at the edge of the middle third, $e=B/6$ (so the heel stress is just zero). Uplift is assumed to act in full: triangular, $\gamma_wH_w=180$ kN/m² at the heel and zero at the toe (no drainage), which is the safe assumption.

Forces (per metre) for an unknown base width B

• $W_1=24\times5\times20=2400$ kN at $B-2.5$ from the toe
• $W_2=24\times\frac12(B-5)\times20=240(B-5)$ kN at $\frac23(B-5)$ from the toe
• $P=\frac12\times10\times18^2=1620$ kN at 6 m above the base
• $U=\frac12\times180\times B=90B$ kN at $\frac23B$ from the toe

$$\Sigma V=2400+240(B-5)-90B=1200+150B$$ $$\Sigma M_R=2400(B-2.5)+160(B-5)^2=160B^2+800B-2000,\qquad \Sigma M_O=1620\times6+90B\times\tfrac23B=9720+60B^2$$

Condition $e=B/6$

The resultant is at $\bar x=B/3$ from the toe, so $\Sigma M_R-\Sigma M_O=\Sigma V\cdot\dfrac{B}{3}$:

$$100B^2+800B-11720=\frac{(1200+150B)B}{3}=400B+50B^2$$ $$50B^2+400B-11720=0\ \Rightarrow\ B^2+8B-234.4=0$$

Solving the equation numerically gives $B=11.82$ m (with uplift). Without uplift the same condition gives $B=9.86$ m.

Check

$\Sigma V=1200+150\times11.82=2,974$ kN/m. At $e=B/6$ the stresses are $\sigma_{toe}=2\Sigma V/B=503.0$ kN/m² $=0.50$ MPa, and $\sigma_{heel}=0$. These are far below $\sigma_{concrete}=30$ MPa and $\sigma_{foundation}=80$ MPa, so the stress limits do not govern.

Sliding / shear: $\tau_s=6$ MPa $=6000$ kN/m², so $\dfrac{\tau_sB}{P}=\dfrac{6000\times11.82}{1620}=43.8\gg1$; the shear strength is more than sufficient.

Answer: base width ≈ 11.8 m (adopt 12 m); with a vertical u/s face this satisfies $e\le B/6$ and all stress limits.