Khosla's Theory Numerical

2. Example:
Determine the percentage pressure of the key points
and exit gradient. Check if the given structure is safe
against piping action or not. The permissible exit
gradient .
6
1
Ge 
204.0 m
4.5 m
205.0 m
204.0 m
199.0 m
203.5 m
198.5 m
18.4 m
4.2 m 2 m 2 m 6 m 4.2 m
19 m
Pile No.1 Pile No. 2

3. DD
EC
D
E
Cos
Cos
d
b




















 






 






100
100
11
21
2
11
1
1
1
1
2
1.
2.
3.
4.
5.
6.
Where,
b = Total floor length
d = Depth of pile

4. We have,
RL at water surface = 205.0 + 4.5 = 209.5 m
Total floor length, b = 19 m
Distance between two piles, b’ = 18.4 m
Depth of Pile no. 1 , d1 = 205.0 – 199.0 = 6.0 m
Depth of Pile no. 2, d2 = 204.0 – 198.5 = 5.5 m
Permissible exit gradient,
6
1
Ge 

5. For Pile No. 1
%6964.813066.18100100
%7074.722926.27100100
%100
%3066.18
1604.2
11604.2111
%2926.27
1604.2
21604.2121
1604.2
2
333.311
2
11
333.3
6
19
1
1
1
11
11
22








 





 






 





 









DD
EC
E
D
E
CosCos
CosCos
d
b















6. Correction for φC1
1. Correction due to thickness of floor
(+ve)
2. Correction due to mutual interference with Pile no. 2
(+ve)
D = Depth of pile no.2 from pile no.1 = 204 – 198.5 = 5.5 m
d = Length of pile no.1 = 204 – 199 = 5 m
3. Correction due to slope
The floor just above the pile is horizontal, so no correction is
needed.
Hence, corrected φC1 = 72.7074 + 1.4982 + 5.7400 = 79.9456 %
%4982.1)204205(
6
7074.726964.81
. 1
1
11




 xt
d
CD 
%7400.5
19
5.55
4.18
5.5
19
'
19 




 





 

b
Dd
b
D

7. For Pile No. 2
0
%7000.17
2982.2
12982.2111
%2748.26
2982.2
22982.2121
2982.2
2
4545.311
2
11
4545.3
5.5
19
2
11
2
11
2
22






 





 






 





 









C
DD
EE
CosCos
CosCos
d
b













8. Correction for φE2
1. Correction due to thickness of floor
(-ve)
2. Correction due to mutual interference with Pile no. 2
(-ve)
D = Depth of pile no.1 from pile no.2 = 203.5 – 199 = 4.5 m
d = Length of pile no.2 = 203.5 – 198.5 = 5 m
3. Correction due to slope
The floor just above the pile is horizontal, so no correction is
needed.
Hence, corrected φE2 = 26.2748 – 0.7795 – 4.6981 = 20.7972 %
%7795.0)5.203204(
5.5
7000.172748.26
. 2
2
22




 xt
d
DE 
%6981.4
19
5.45
4.18
5.4
19
'
19 




 





 

b
Dd
b
D

9. Pile No. 1 Pile No. 2
ΦE1 = 100 % ΦE2 = 20.7972 %
ΦD1 = 81.6964 % ΦD2 = 17.7000 %
ΦC1 = 79.9456 % ΦC2 = 0

10. For Exit Gradient
RL at surface of water = 209.5 m
Maximum seepage head, H = 209.5 – 204 = 5.5 m
Depth of downstream pile, d = 204 – 198.5 = 5.5 m
Total floor length, b = 19 m
For Exit Gradient,
= 1 in 4.762
Hence, floor is not safe against piping.
2982.2
2
4545.311
2
11
4545.3
5.5
19
22









d
b
2100.0
2982.2
1
.
5.5
5.51
. 
d
H
GE