--- title: "Base width of 20 m high trapezoidal gravity dam" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-base-width-trapezoidal-20m" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-base-width-trapezoidal-20m site: BCT Engineering --- # Base width of 20 m high trapezoidal gravity dam Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2078 Bhadra, 6 marks - Video: [Trapezoidal dam-Numerical](https://www.youtube.com/watch?v=8Bn-9tDWeG8) — The Motivated Mind - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-base-width-trapezoidal-20m - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question Determine the base width of a 20 m high trapezoidal concrete dam having a vertical upstream face and top width of 5 m. The design water depth is 18 m. There is no tail water. Ignore earthquake, silt and ice loads. Take e = B/6, $\sigma_{concrete}$ = 30 MPa, $\sigma_{foundation}$ = 80 MPa, $\tau_s$ = 6 MPa. Specific weights of water and concrete are 10 kN/m³ and 24 kN/m³ respectively. Assume suitable data, if necessary. ## Short answer Using no-tension condition e = B/6, force and moment equilibrium including full uplift gives quadratic 50B²+400B-11720=0, solving yields B=11.82 m, adopt 12 m. Stress and shear checks confirm safety. ## Answer #### 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the condition e = B/6 mean for the base? A. Resultant at the lower third point, B/3 from the toe, so the heel stress is just zero B. Resultant at the centre of the base C. Resultant at the toe, so the toe stress is zero D. Resultant outside the base Answer: A. At e = B/6 the resultant is B/3 from the toe, the limit of the middle third, so the heel stress is just zero (no tension). (Key point: Meaning of e = B/6) 2. How does including full triangular uplift change the required base width? A. It reduces B from 11.82 m to 9.86 m B. It increases B from 9.86 m (no uplift) to 11.82 m C. It has no effect D. It makes the base width independent of the water depth Answer: B. Uplift reduces the net vertical load and adds an overturning moment, so a wider base is needed: 11.82 m instead of 9.86 m. (Key point: Effect of uplift on B) 3. After finding B, why do the concrete (30 MPa) and foundation (80 MPa) stress limits not govern? A. Because toe stress is only about 0.50 MPa, far below the limits, and shear strength is ample B. Because they are exceeded but ignored C. Because the toe stress is 30 MPa D. Because uplift is assumed zero Answer: A. At e = B/6 the toe stress is 2 SumV / B = 503 kN/m^2, and tau_s B / P is about 43.8, so the stability condition e = B/6 controls. 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