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Uplift at inside corners, 50 m floor weir

Short answer

Khosla's theory: α=6.25, λ=3.665, basic pressures u/s inside corner 65.01%, d/s 34.99%. After interference (+2.43% for u/s,–2.43% d/s) and thickness corrections (+1.36%,–2.73%), final u/s corner uplift 68.8% (3.44m), d/s corner 29.8% (1.49m). Exit gradient 0.104.

A weir has a solid horizontal floor length of 50 m with two lines of cutoff of 8 m depth below the river bed at its two ends. The floor thickness is 1 m at upstream end and 2 m at downstream end, with its upper level being in flush with river bed. For an effective head of 5 m over the weir, calculate the uplift pressure at the two inside corner points and also the exit gradient.

Given: floor length b=50b = 50 m, cutoffs at both ends, depth d=8d = 8 m below the river bed (the floor top is flush with the bed), head H=5H = 5 m, floor thickness t1=1t_1 = 1 m (u/s end) and t2=2t_2 = 2 m (d/s end). Khosla's theory is used. The inside corners are the junctions of the floor with the u/s pile (C1C_1) and with the d/s pile (E2E_2).

Khosla's constants

α=bd=508=6.25,λ=1+1+α22=3.665\alpha = \frac{b}{d} = \frac{50}{8} = 6.25,\qquad \lambda = \frac{1 + \sqrt{1+\alpha^2}}{2} = 3.665 ϕE=1πcos⁡−1(λ−2λ)=34.99%ϕD=1πcos⁡−1(λ−1λ)=24.09%\begin{aligned} \phi_E &= \frac{1}{\pi}\cos^{-1}\left(\frac{\lambda-2}{\lambda}\right) = 34.99\% \\ \phi_D &= \frac{1}{\pi}\cos^{-1}\left(\frac{\lambda-1}{\lambda}\right) = 24.09\% \end{aligned}

Pressures from the simple profile (thin floor)

PointFormulaPressure (%)
E1E_1 (top of u/s pile)100100
D1D_1 (tip of u/s pile)100−ϕD100 - \phi_D75.91
C1C_1 (u/s inside corner)100−ϕE100 - \phi_E65.01
E2E_2 (d/s inside corner)ϕE\phi_E34.99
D2D_2 (tip of d/s pile)ϕD\phi_D24.09
C2C_2 (d/s end)00

Correction for mutual interference of piles

Both piles are 8 m deep (d=D=8d = D = 8 m), spaced b′=50b' = 50 m apart, total floor length b=50b = 50 m:

C=19Db′(d+Db)=19850×1650=2.43%C = 19\sqrt{\frac{D}{b'}}\left(\frac{d+D}{b}\right) = 19\sqrt{\frac{8}{50}}\times\frac{16}{50} = 2.43\%

It is positive at C1C_1 (upstream of the interfering pile) and negative at E2E_2 (downstream of it).

Correction for floor thickness

The uplift acts on the underside of the floor, a distance tt below the key point, so the pressure is interpolated along the pile:

at C1: (ϕD1−ϕC1)t1d=(75.91−65.01)×18=+1.36%at E2: (ϕE2−ϕD2)t2d=(34.99−24.09)×28=−2.73%\begin{aligned} \text{at } C_1: &\ (\phi_{D_1} - \phi_{C_1})\frac{t_1}{d} = (75.91 - 65.01)\times\frac{1}{8} = +1.36\% \\ \text{at } E_2: &\ (\phi_{E_2} - \phi_{D_2})\frac{t_2}{d} = (34.99 - 24.09)\times\frac{2}{8} = -2.73\% \end{aligned}

Corrected pressures and uplift heads

ϕC1′=65.01+2.43+1.36=68.80%,hC1=0.6880×5=3.44 mϕE2′=34.99−2.43−2.73=29.83%,hE2=0.2983×5=1.49 m\begin{aligned} \phi_{C_1}' &= 65.01 + 2.43 + 1.36 = 68.80\%,\qquad h_{C_1} = 0.6880\times 5 = 3.44\ \text{m} \\ \phi_{E_2}' &= 34.99 - 2.43 - 2.73 = 29.83\%,\qquad h_{E_2} = 0.2983\times 5 = 1.49\ \text{m} \end{aligned}

Exit gradient

GE=Hd⋅1πλ=58×1π3.665=0.1039 (≈1/9.6)G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{5}{8}\times\frac{1}{\pi\sqrt{3.665}} = 0.1039\ (\approx 1/9.6)

Answer: Uplift at the u/s inside corner ≈68.8%\approx 68.8\% of HH (3.44 m); at the d/s inside corner ≈29.8%\approx 29.8\% (1.49 m); exit gradient GE≈0.104G_E \approx 0.104, which is less than the safe gradient of about 1/6 (0.167) for fine sand, so the structure is safe against piping.

✓ Answer checked by Claude Opus 5.5 (high effort)

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