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Gravity dam concrete volume from hydrostatic thrust

Short answer

Net horizontal thrust 4.0×10^6 kN. Required dam weight is (2.5×4.0×10^6)/0.65 = 15.38×10^6 kN. Required concrete volume = 15.38×10^6 / 24 ≈ 6.41×10^5 m³.

A concrete gravity dam on a rocky foundation is acted on by an upstream horizontal hydrostatic force of 4.50 million kN and by a downstream one of 0.50 million kN. Determine the volume of concrete works (γcon\gamma_{con} = 24 kN/m³), neglecting bond stress and uplift force, and taking a factor of safety on the horizontal thrust of 2.5 and a friction coefficient between the concrete and rock of 0.65.

Principle

With no bond stress and no uplift, sliding resistance comes only from friction between the concrete and the rock, μΣV\mu\Sigma V. The factor of safety on the horizontal thrust is

FS=μ ΣVΣHFS=\frac{\mu\,\Sigma V}{\Sigma H}

Net horizontal thrust

ΣH=4.50×106−0.50×106=4.00×106 kN\Sigma H=4.50\times10^6-0.50\times10^6=4.00\times10^6\ \text{kN}

Required weight of the dam

ΣV=FS×ΣHμ=2.5×4.00×1060.65=15,384,615 kN=15.38×106 kN\Sigma V=\frac{FS\times\Sigma H}{\mu}=\frac{2.5\times4.00\times10^6}{0.65}=15,384,615\ \text{kN}=15.38\times10^6\ \text{kN}

The vertical load is taken as the weight of the concrete (no uplift and no water weight).

Volume of concrete

V=Wγcon=15,384,61524=641,026 m3≈6.41×105 m3V=\frac{W}{\gamma_{con}}=\frac{15,384,615}{24}=641,026\ \text{m}^3\approx6.41\times10^5\ \text{m}^3

Answer: volume of concrete ≈ 641,026 m³ (about 6.41 × 10⁵ m³).

✓ Answer checked by Claude Opus 5.5 (high effort)

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