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Khosla: barrage HGL and floor thickness

Short answer

No‑flow head 11.25 m; required floor thickness up to 3.7 m exceeds provided 1.0 m; exit gradient 0.327 >0.200 (1/5) – unsafe against uplift and piping.

A barrage weir profile with an upstream pond level of 1135.0 m and downstream tailwater level of 1126.5 m is given below. Assume 1.0 m initial thickness of the floor throughout the length and the sheet pile is 0.5 m inside from the edge of the floor. Draw the hydraulic gradient line (HGL) and compute the designed thickness of the floor for no flow condition. Check the profile for piping failure. The safe exit gradient for the soil on which the weir is standing is 1:5.

[Figure: weir section; pond level 1135.0 m, crest 1127.0 m, upstream floor level 1126.5 m, upstream pile bottom 1120.5 m, downstream floor level 1123.75 m, downstream tailwater 1126.5 m, downstream pile bottom 1118.5 m; upstream glacis 1:3 and downstream 1:3 slopes; horizontal dimensions 13.5 m, 0.5 m, 6.25 m, 9.75 m and 40 m; end points E1, C1 (upstream) and E2, C2 (downstream)]

Data and assumptions

  • Pond level 1135.0 m. For the no-flow condition the d/s water level is taken at the d/s floor level 1123.75 m, so the head is H=1135.0−1123.75=11.25H = 1135.0 - 1123.75 = 11.25 m.
  • The u/s sheet pile (bottom 1120.5 m) is 6.0 m deep below the u/s floor (1126.5 m). The d/s pile (bottom 1118.5 m) is 5.25 m deep below the d/s floor (1123.75 m). Floor length between the piles b=40b = 40 m.
  • Floor thickness 1.0 m throughout (given); specific gravity of concrete G=2.4G = 2.4.
  • Slope correction for a 1:3 slope =4.5%= 4.5\%, multiplied by (slope length / bb): u/s slope length 1.51.5 m (up the slope, subtracted at C1C_1); d/s glacis length 9.759.75 m (down the slope, added at E2E_2).

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=40b = 40 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 pile6.006.6673.87133.9423.40
D/s pile5.257.6194.34231.8622.04

Pressures at the key points

Pressure is expressed as a percentage of the head H=11.25H = 11.25 m (no flow).

PointSimple φ (%)Corrections (%)Corrected φ (%)Pressure head (m)
E1 (top of u/s pile)100.000100.0011.25
D176.60-76.608.62
C166.06thk +1.76, int +2.97, slope -0.1770.627.94
E231.86thk -1.87, int -1.15, slope +1.1029.943.37
D222.04-22.042.48
C2 (d/s end)0000

Corrections: thk = floor thickness, +ϕD−ϕCdt+\dfrac{\phi_D-\phi_C}{d}t at C and −ϕE−ϕDdt-\dfrac{\phi_E-\phi_D}{d}t at E; int = mutual interference, 19D/b′ d+Db19\sqrt{D/b'}\,\dfrac{d+D}{b} (at C1C_1: D=1126.5−1118.5=8.0D = 1126.5-1118.5 = 8.0 m, d=6d = 6 m; at E2E_2: D=1123.75−1120.5=3.25D = 1123.75-1120.5 = 3.25 m, d=5.25d = 5.25 m; b′=40b' = 40 m), positive at C1C_1 and negative at E2E_2; slope = slope correction. The D points are not corrected.

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
C1 (u/s pile junction)070.621131.691126.505.193.711.00not safe
Toe of d/s glacis23.7546.471128.981123.755.233.731.00not safe
E2 (d/s pile junction)4029.941127.121123.753.372.411.00not safe

Exit gradient

GE=Hd⋅1πλ=11.255.25×1π4.342=0.3273 (≈1/3.1)G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{11.25}{5.25}\times\frac{1}{\pi\sqrt{4.342}} = 0.3273\ (\approx 1/3.1)

Permissible =1/5=0.2000= 1/5 = 0.2000. Not safe against piping; deepen the d/s pile or lengthen the floor.

Hydraulic gradient line

Bars show the corrected pressure (each # is 2.5%):

 E1  100.0% |########################################
 D1   76.6% |###############################
 C1   70.6% |############################
 E2   29.9% |############
 D2   22.0% |#########
 C2    0.0% |

The hydraulic gradient line (HGL) drops vertically across each pile (from E to D to C) and runs as a straight line along the floor between the pile junctions (C1C_1 to E2E_2).

Conclusion: For no flow, the floor needs about 3.7 m thickness at the u/s junction and at the toe of the glacis, reducing to about 2.4 m at the d/s pile. The 1.0 m provided is not enough anywhere, so the floor must be thickened (or the u/s floor lengthened and the u/s pile deepened to reduce the uplift). The exit gradient (1/3.1) exceeds the safe value of 1/5, so the profile is also unsafe against piping; the d/s pile must be deepened.

✓ Answer checked by Claude Opus 5.5 (high effort)

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