Gravity dam forces
A concrete gravity dam holds back water by its weight. The design checks that, under every combination of loads, it will not slide on its base or on any weak plane, will not overturn, and will not overstress the concrete or the foundation. IS 6512 gives the Indian criteria.
The main forces are the water thrust on the upstream face, triangular in distribution and acting at one-third of the depth from the base; uplift, the water pressure under the base and in cracks, which reduces the effective weight; the self-weight of the dam; silt pressure behind the dam; wave pressure; and, in seismic zones, inertia forces on the dam and hydrodynamic pressure from the moving reservoir. Ice pressure applies in cold regions.
Uplift is the force that makes the difference. Without drains it is taken as varying from full reservoir head at the heel to tailwater head at the toe. A grout curtain and a line of drains near the heel cut it sharply, and design rules allow a reduced uplift diagram where drains are provided and maintained. That is why drain cleaning is part of dam maintenance.
Stability is checked by comparing the sum of horizontal forces with the frictional and cohesive resistance of the base (sliding), the moments about the toe (overturning), and the location of the resultant. For no tension at the heel, the resultant must lie within the middle third of the base.
Work forces per metre run of dam in a table with lever arms; it keeps the moments straight.
Water thrust: P = γ_w H² / 2 at H / 3 above the baseUplift without drains (no tailwater): U = γ_w H B / 2 at B / 3 from the heelSliding (friction only): FS = μ (W − U) / PMiddle third: eccentricity e ≤ B / 6
Ignoring uplift, which overstates the safety against sliding and overturning.
Problem. A triangular gravity dam 30 m high has a vertical upstream face and a 24 m base. Concrete weighs 24 kN/m³; water is at the crest; no tailwater, no drains. Take μ = 0.7. Check sliding by friction alone and the position of the resultant.
- W = 0.5 × 30 × 24 × 24 = 8 640 kN/m at 8 m from the heel; P = 9.81 × 30² / 2 = 4 414.5 kN/m at 10 m; U = 0.5 × 9.81 × 30 × 24 = 3 531.6 kN/m at 8 m from the heel.
- Sliding: FS = 0.7 × (8 640 − 3 531.6) / 4 414.5 = 0.7 × 5 108.4 / 4 414.5 = 0.81, less than 1.
- Moments about the toe: resisting 8 640 × 16 = 138 240; overturning 4 414.5 × 10 + 3 531.6 × 16 = 100 650.6 kNm/m. Resultant at (138 240 − 100 650.6) / 5 108.4 = 7.36 m from the toe.
- e = 12 − 7.36 = 4.64 m > B/6 = 4 m, so the heel goes into tension. This section needs a wider base, drains to cut uplift, or foundation shear strength (cohesion) to pass.
Answer. Friction-only sliding FS ≈ 0.81 and resultant outside the middle third: the trial section fails
Quick quiz
1Where does the water thrust act on a vertical face?
At one-third of the water depth above the base.
2How do drains help a gravity dam?
They relieve uplift pressure under the base.
3What is the middle-third rule?
For no tension, the resultant must fall within the middle third of the base.
- IS 6512 (India): Criteria for design of solid gravity dams.
- EN 1990 (Eurocode 0) (Europe): Basis of structural design: limit states, combinations and reliability.
Clauses and documents to read
- IS 6512:1984: criteria for design of solid gravity dams (load combinations, uplift, stability)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- P. Novak, A. I. B. Moffat, C. Nalluri, R. Narayanan, Hydraulic Structures, CRC Press, latest edition.
- U.S. Bureau of Reclamation, Design of Small Dams, USBR (official publication), 3rd edition, 1987.
- S. K. Garg, Irrigation Engineering and Hydraulic Structures, Khanna Publishers, latest edition.
Free and official sources
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