Skip to main content

Khosla: uplift at A, B, thickness, exit gradient 0.2

Short answer

Uplift pressures at key points: E1=6.00m, D1=4.36m, C1=3.87m, E2=1.56m, D2=1.37m, C2=0. Floor at B (1.0m provided) requires 2.77m — unsafe. Floor at A (2.5m provided) requires 1.11m — safe. Exit gradient = 0.190 < 0.200 (safe).

Calculate the uplift pressure at key points of piles in Fig 1. Also check the thickness of the floor at A, B location and exit gradient. The safe exit gradient is 0.2.

[Figure: levels 103 m (pond), 100 m, 99.5 m (point C), 97 m, 96 m, 94.5 m (point A), 95 m (point B), 92 m and 92 m (pile bottoms); horizontal distances 15 m and 20 m]

Data and assumptions

The figure is partly unclear. Reading adopted: pond level 103 m; two pile lines with floor length b=15+20=35b = 15 + 20 = 35 m. U/s pile: floor level (point C) 99.5 m, bottom 92 m (d=7.5d = 7.5 m). D/s pile: floor level 97 m, bottom 92 m (d=5d = 5 m). With no flow the d/s water level is the d/s floor level 97 m, so H=6H = 6 m. Point B is on the depressed floor at 15 m (floor top 96 m, bottom 95 m, thickness 1.0 m); point A is at the d/s end (floor top 97 m, bottom 94.5 m, thickness 2.5 m). Floor thickness 0.5 m at the u/s end. Safe exit gradient 0.2.

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=35b = 35 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 pile7.504.6672.88640.0727.33
D/s pile5.007.0004.03633.1722.90

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
D172.67-72.674.36
C159.93thk +0.85, int +3.7764.553.87
E233.17thk -5.14, int -2.0525.981.56
D222.90-22.901.37
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} with b′=35b' = 35 m (at C1C_1: D=99.5−92=7.5D = 99.5-92 = 7.5 m, d=7.5d = 7.5 m; at E2E_2: D=97−92=5D = 97-92 = 5 m, d=5d = 5 m), positive at C1C_1 and negative at E2E_2. 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
B (x = 15 m)1548.0299.8896.003.882.771.00not safe
A (d/s end)3525.9898.5697.001.561.112.50safe

Exit gradient

GE=Hd⋅1πλ=6.005.00×1π4.036=0.1901 (≈1/5.3)G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{6.00}{5.00}\times\frac{1}{\pi\sqrt{4.036}} = 0.1901\ (\approx 1/5.3)

Permissible =0.2000= 0.2000. Safe (less than the permissible value).

Hydraulic gradient line

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

 E1  100.0% |########################################
 D1   72.7% |#############################
 C1   64.6% |##########################
 E2   26.0% |##########
 D2   22.9% |#########
 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).

Answer: The floor at B (1.0 m provided, 2.77 m needed) is not safe; the floor at A (2.5 m provided, 1.11 m needed) is safe. The exit gradient of 0.190 is just within the safe value of 0.2.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Irrigation and Drainage Engineering chapter 5 questions →