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Khosla: percentage pressure and piping, pond 503

Short answer

Percentage pressures: E1=100%, D1=81.1%, C1=77.2%, E2=31.1%, D2=21.6%, C2=0. Required thickness: A 1.17 m, B 2.57 m, C 1.95 m, D 1.33 m. Exit gradient 0.449 > 0.15, not safe against piping.

For the figure shown below ignoring floor thickness and slope corrections, find the percentage pressure at the key points of the piles using Khosla's theory. The floor thickness may not be less than 30 cm anywhere. If permissible exit gradient is 0.15, check the floor against piping failure. Also find the thickness of floor at A, B, C and D points.

[Figure: pond level 503 m; floor level 500 m (A); 497 m (B, C) and 497 m (D); pile depths 1.5 m and 2.0 m; horizontal distances 6 m and 10 m]

Data and assumptions

Reading of the figure: pond level 503 m; d/s floor and bed level 497 m, so H=6H = 6 m (no flow, no tail water). U/s pile 1.5 m deep at the u/s end (floor top 500 m, point A; pile bottom 498.5 m); d/s pile 2.0 m deep at the d/s end (floor top 497 m, point D; pile bottom 495 m); floor length b=6+10=16b = 6 + 10 = 16 m. B is at the foot of the drop (x=6x = 6 m) and C at x=11x = 11 m. Floor thickness and slope corrections are ignored (as stated), but the mutual interference correction still applies. G=2.4G = 2.4; minimum thickness 0.30 m; permissible exit gradient 0.15.

Khosla's constants

α=b/d\alpha = b/d, λ=1+1+α22\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}, ϕE=1πcos⁡−1λ−2λ\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}, ϕD=1πcos⁡−1λ−1λ\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda} (in percent), total floor length b=16b = 16 m. For the u/s pile, ϕD1=100−ϕD\phi_{D1} = 100-\phi_D and ϕC1=100−ϕE\phi_{C1} = 100-\phi_E.

Piled (m)αλφE (%)φD (%)
U/s pile1.5010.6675.85727.1218.88
D/s pile2.008.0004.53131.1321.56

Pressures at the key points

Pressure is expressed as a percentage of the head H=6.00H = 6.00 m (no flow).

PointSimple φ (%)Corrections (%)Corrected φ (%)Pressure head (m)
E1 (top of u/s pile)100.000100.006.00
D181.12-81.124.87
C172.88int +4.3177.204.63
E231.13-31.131.87
D221.56-21.561.29
C2 (d/s end)0000

Corrections: int = mutual interference at C1C_1 due to the d/s pile, 19D/b′ d+Db=195/16×1.5+516=4.31%19\sqrt{D/b'}\,\dfrac{d+D}{b} = 19\sqrt{5/16}\times\dfrac{1.5+5}{16} = 4.31\%, with D=500−495=5D = 500-495 = 5 m and d=1.5d = 1.5 m. At E2E_2 the u/s pile bottom (498.5 m) is above the d/s floor (497 m), so there is no interference correction.

Floor thickness

Unbalanced head hh = hydraulic-gradient level at the floor −- floor top level; required thickness t=hG−1t = \dfrac{h}{G-1} with G=2.40G = 2.40 (the HGL is taken as a straight line from C1C_1 to E2E_2 along the floor).

Locationx (m)φ (%)HG level (m)Floor top (m)Unbalanced head (m)Required t (m)Provided t (m)Check
A (x = 0, u/s)077.20501.63500.001.631.17-provide at least 1.17
B (x = 6)659.92500.60497.003.602.57-provide at least 2.57
C (x = 11)1145.53499.73497.002.731.95-provide at least 1.95
D (x = 16, d/s)1631.13498.87497.001.871.33-provide at least 1.33

All of these exceed the 0.30 m minimum, so they govern.

Exit gradient

GE=Hd⋅1πλ=6.002.00×1π4.531=0.4486 (≈1/2.2)G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{6.00}{2.00}\times\frac{1}{\pi\sqrt{4.531}} = 0.4486\ (\approx 1/2.2)

Permissible =0.1500= 0.1500. Not safe against piping; the d/s pile must be deepened or the floor lengthened.

Answer: Corrected pressures: E1=100%E_1 = 100\%, D1=81.1%D_1 = 81.1\%, C1=77.2%C_1 = 77.2\%, E2=31.1%E_2 = 31.1\%, D2=21.6%D_2 = 21.6\%, C2=0C_2 = 0. Floor thickness: A 1.17 m, B 2.57 m, C 1.95 m, D 1.33 m. The exit gradient (0.449) is about three times the permissible 0.15, so the floor is not safe against piping.

✓ Answer checked by Claude Opus 5.5 (high effort)

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