Skip to main content

Khosla: uplift, HGL, thickness, exit gradient 1/5

Short answer

Using Khosla's theory with thickness and interference corrections, corrected pressure heads at E1, D1, C1, E2, D2, C2, E3, D3, C3 are 3.00, 2.52, 2.44, 1.81, 1.82, 1.84, 0.73, 0.58, 0 m. Required thicknesses exceed provided at C1, E2 and C2 but not E3. Exit gradient is 0.1354 (about 1/7.4), less than safe 1/5.

Calculate the uplift pressure at key points of the pile of the structure shown in figure below. Draw HGL and also check the thickness provided and safe exit gradient GE = 1/5.

[Figure: RL 103.0 m water level, 3 m head; floor RL 100.0 m; floor dimensions 1.0 m, 1.5 m, 1.25 m, 1.0 m; pile depths 2 m, 1.5 m and 3 m; 5.0 m; horizontal distances 10 m and 20 m]

Data and assumptions

Reading of the figure: water level RL 103.0 m, floor RL 100.0 m, so the head is H=3H = 3 m (d/s water at floor level). Pile depths 2 m (u/s, at 0 m), 1.5 m (intermediate, at 10 m) and 3 m (d/s, at 30 m); floor length b=10+20=30b = 10 + 20 = 30 m. Floor thickness provided: 1.0 m (u/s), 1.25 m (intermediate), 1.0 m (d/s). G=2.4G = 2.4. Safe exit gradient 1/51/5.

Khosla's constants

End piles: 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), b=30b = 30 m; for the u/s pile ϕD1=100−ϕD\phi_{D1} = 100-\phi_D, ϕC1=100−ϕE\phi_{C1} = 100-\phi_E.

Intermediate pile (b1b_1 from the u/s end, b2b_2 from the d/s end): λ1=1+α12−1+α222\lambda_1 = \dfrac{\sqrt{1+\alpha_1^2}-\sqrt{1+\alpha_2^2}}{2}, λ=1+α12+1+α222\lambda = \dfrac{\sqrt{1+\alpha_1^2}+\sqrt{1+\alpha_2^2}}{2}, ϕE=1πcos⁡−1λ1−1λ\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1-1}{\lambda}, ϕD=1πcos⁡−1λ1λ\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1}{\lambda}, ϕC=1πcos⁡−1λ1+1λ\phi_C = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1+1}{\lambda}.

Piled (m)αλφE (%)φD (%)
U/s pile2.0015.0008.01722.9816.07
D/s pile3.0010.0005.52527.9819.45
Intermediate piled (m)α1α2λ1λφE (%)φD (%)φC (%)
at 10 m1.506.66713.333-3.31510.05664.1260.6957.39

Pressures at the key points

Pressure is expressed as a percentage of the head H=3.00H = 3.00 m (no flow).

PointSimple φ (%)Corrections (%)Corrected φ (%)Pressure head (m)
E1 (top of u/s pile)100.000100.003.00
D183.93-83.932.52
C177.02thk +3.46, int +0.8681.332.44
E264.12thk -2.85, int -0.9960.271.81
D260.69-60.691.82
C257.39thk +2.75, int +1.1061.251.84
E327.98thk -2.84, int -0.7824.350.73
D319.45-19.450.58
C3 (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}, positive at C of the u/s pile and negative at E of the d/s pile of each pair (DD = depth of the neighbouring pile, b′b' = distance between the two piles). The D points are not corrected. (Around the shallow intermediate pile the corrected values come out nearly equal, about 60 to 61%.)

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.

Locationx (m)φ (%)HG level (m)Floor top (m)Unbalanced head (m)Required t (m)Provided t (m)Check
C1 (x = 0)081.33102.44100.002.441.741.00not safe
E2 (x = 10)1060.27101.81100.001.811.291.25not safe (marginal)
C2 (x = 10)1061.25101.84100.001.841.311.25not safe (marginal)
E3 (x = 30)3024.35100.73100.000.730.521.00safe

Exit gradient

GE=Hd⋅1πλ=3.003.00×1π5.525=0.1354 (≈1/7.4)G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{3.00}{3.00}\times\frac{1}{\pi\sqrt{5.525}} = 0.1354\ (\approx 1/7.4)

Permissible =1/5=0.2000= 1/5 = 0.2000. Safe (less than the permissible value).

Hydraulic gradient line

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

 E1  100.0% |########################################
 D1   83.9% |##################################
 C1   81.3% |#################################
 E2   60.3% |########################
 D2   60.7% |########################
 C2   61.2% |########################
 E3   24.4% |##########
 D3   19.5% |########
 C3    0.0% |

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

Answer: The u/s floor (1.74 m needed, 1.0 m provided) is not safe, the floor at the intermediate pile is marginally short (about 1.3 m needed, 1.25 m provided), and the d/s floor is safe. The exit gradient of 1/7.4 is within the safe value of 1/5.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Irrigation and Drainage Engineering chapter 5 questions →