Khosla's Theory: Problem Solution
Lecture-03
Example:
Determine the percentage of pressures at various key points in Figure below. Also determine the exit gradient.
Solution:
For upstream pile line 1:
Total
length of the floor = b = 57.0 m
Depth
of upstream pile line = d = 154.0 – 148.0 = 6.0 m
𝛼 = b/d = 57.0/6.0 = 9.5
1/ 𝛼 = 1/9.5 = 0.105
The
value of ØC1must be corrected for three corrections as below:
a)
Correction at C1 for mutual interference of piles ØC1 is effected by intermediate pile No. 2.
b′
= The distance between two pile lines =15.8 m
D
=
The depth of the neighboring pile = 153.0 - 148.0 =5.0 m
d
=
The depth of the pile on which the effect is considered = 153.0 - 148.0 =5.0 m
b
=
Total floor length = 57.0 m
Hence,
C= 1.88 % (+ve)
ØC1 = 71.33 + 1.88
= 73.21
b)
Correction at C1 Due to thickness of floor
c) Correction due to slope at C1 is nil as this point is neither situated
at the start nor at the end of the slope.
So, corrected,
ØC1 = 74.71%
ØD1 = 80 %
Intermediate pile line No. 2:
b1
= 0.6 + 15.8 = 16.4 m
b2
= 40 + 0.6 = 40.6 m
ØE2
should be corrected
a) Correction at E2 for mutual interference of piles
b′
= The distance between two pile lines =15.8 m
D
=
The depth of the neighboring pile = 153.0 - 148.0 =5.0 m
d
=
The depth of the pile on which the effect is considered = 153.0 - 148.0 =5.0 m
b
=
Total floor length = 57.0 m
Pile
No. (1) will affect the pressure at E2 and since E2 is in the forward direction
of flow and hence, this correction shall be –ve.
Hence,
C= 1.88 % (-ve)
b)
Correction at E2 Due to thickness of floor
since the pressure observed is at E2' and
not at E2 and by looking at the direction of flow, it can be stated easily that
the pressure at E2 shall be less than that at E2', hence, this correction is
negative.
ØE2 corr. = [(ØE2 - ØD2)/d1] * t1 = 1.17% (-ve)
c) Correction at E2 due to slope is nil as the point E2 is neither situated at the start of a slope nor at the end of a slope.
so, corrected,
ØE2 = 70.8% -1.88% -1.17% = 67.95 %
ØC2 should be corrected
a) Correction at C2 due to pile interference
b′
= The distance between two pile lines =40 m
D
=
The depth of the neighboring pile, depth
of pile No. 3, the effect of which is considered = 153.0
- 141.7 =11.3 m
d
=
The depth of the pile on which the effect is considered = 153.0 - 148.0 =5.0 m
b
=
Total floor length = 57.0 m
Pressure
at C2 is affected by pile No. 3 and since the point C2 is in the back water in
the direction of flow and hence, this correction is +ve.
Correction
= 2.89% (+ve)
b)
Correction at C2 due to floor thickness
The
pressures at C2 shall be more than that at C2', this correction shall be +ve
and its amount is the same as was calculated for the point E2 = 1.17%.
c)
Correction at C2 due to slope
bs = the horizontal length of the sloping
floor = 3m
b1 = the distance between the two piles =
40 m
C= the correction factor of slope taken
from table = 4.5
C2 is situated at the
start of a slope in the direction of flow, the correction is negative.
Cs = 0.34% (-ve)
Now, Corrected
ØC2 = 56.4+2.89+1.17-0.34 = 60.12%
Downstream pile line No.-3:
ØE3
should be corrected
a)
Correction due to mutual of piles
Correction at E3 due to pile interference
b′
= The distance between two pile lines =40 m
D
=
The depth of the neighboring pile, depth
of pile No. 2, the effect of which is considered = 150.7
- 148 =2.7 m
d
=
The depth of the pile on which the effect is considered = 150.7 – 141.7=9.0 m
b
=
Total floor length = 57.0 m
The
point E3 is affected by pile No. (2) and since E3 is in the forward direction
of flow from pile No. (3), the correction is negative
Correction
= 1.01% (-ve)
b)
Correction at E3 due to floor thickness
The
pressure at E3 shall be less than at E3' thus correction shall be negative
(-ve).
ØE3 corr. = [(ØE3-ØD3)/d3] * t2 , t2 = 1.3,
= 1.63% (-ve)
c)
Correction at E3 due to slope
Correction at E3 due to slope is nil
as the point E3 is neither situated at the start of a slope nor at
the end of a slope.
Now, Corrected
ØE3 = 40.6-1.01-1.63 = 37.96%
The corrected pressures at various key
points are tabulated in Table below:
|
Upstream pile No.1 |
Intermediate pile No.2 |
Downstream pile No.3 |
|
ØE1= 100 % |
ØE2= 67.95 % |
ØE3= 37.96 % |
|
ØD1= 80.0 % |
ØD2= 63.2 % |
ØD3=27.7 % |
|
ØC1= 74.71 % |
ØC2= 60.12 % |
ØC3=0.0 % |
Calculation of Exit Gradient
The maximum seepage head
= H = 158.0 – 152.0 = 6.0 m.
The depth of the d/s
cut-off = d = 152.0 – 141.70 = 10.3 m.
Total floor length = b =
57.0 m.
b/d
= 57.0/10.3 = 5.53
GE = KH/d = 0.102
Calculation of H.G. Grade lines
Figure: H.G. lines distribution