# Khosla theory in irrigation with Formula and Numerical Example

## Principles of Khosla’s theory

1) Seeping water below a hydraulic structure does not follow the bottom profile of the impervious floor as stated by Bligh but each particle traces its path along with a series of streamlines and equipotential lines as shown in 5.21.

2) The seepage water exerts force at each point in the direction of flow and tangential to the streamline as shown in the figure. This force has the maximum disturbing tendency at the exit end.

3) The steady seepage in a vertical plane for a homogeneous soil can be expressed by Laplacian equation, where, ∅

is the flow potential = kh.

k is the coefficient of permeability of soil.

h is the residual head at any point within the soil.

4) Undermining (piping) of the floor starts from the downstream end of the d/s pucca floor and if not checked it travels upstream towards the weir wall. It is, therefore, absolutely necessary to have a reasonably deep vertical cutoff at the d/s end of the floor to prevent undermining. Depth of d/s vertical cutoff is governed by;

a) Maximum depth of scour

5) The outer faces of the end sheet piles are much more effective than the inner ones and the horizontal length of floor.

6) The intermediate sheet piles, if smaller in length than the outer ones were ineffective except for local redistribution of pressure.

To find out the seepage flow by Khosla’s theory it is necessary to plot the flow net and also to solve the Laplacian equation mentioned in principal no (iii). But it is very difficult to solve the Laplacian equation mathematically.

So, for the design of hydraulic structures (such as weir, barrage, etc) and also to calculate the seepage flow and uplift pressure on pervious foundations, Khosla has developed simple profiles from practical complex profiles which is called the Method of independent variables.

These simple profiles can be solved easily unlike the Laplacian equation.

## Khosla’s Method of Independent variables (Calculation of percentage pressure, i.e., uplift pressure by Khosla’s Theory)

In this method, the complex weir or barrage section (profile) is broken into no. of simple section for which mathematical solutions have been obtained to find the seepage flow in terms of percentage pressure.

The percentage pressure later can be used to find uplift pressure due to seepage flow by directly multiplying with the total head available for the flow. The uplift pressure thus obtained is in terms of the head of water.

#### The simple sections (profiles) given by Khosla consists of:

1. A straight horizontal either floor of negligible thickness with a sheet pile either at the u/s end or d/s end. [Figure 5.22 (a) and 5.22 (b)]
2. A straight horizontal floor of negligible thickness with a sheet pile at some intermediate point. [Figure 5.22 (c)]
3. A straight horizontal floor depressed below the bed but no vertical cutoff. [Figure 5.22 (d)]

#### u/s pile [Figure (a)]

∅C1 = 100 – ∅E

∅D1 = 100 – ∅D #### d/s pile [Figure (b)] #### Intermediate pile [Figure (c)]

𝛼 1 = b1/d

𝛼 2 = b2/d #### Depressed floor [Figure (d)]

In above simple profiles, the key points where percentage pressures are to be calculated are:

• Junction of floor and pile lines on either sides [E and C]
• Bottom point of pile line [D]
• Bottom corners in case of depressed floor [D₁’ and D’].

The % pressure at these key points simple profiles will resemble the complex profiles only if the following corrections are applied.

• Correction for mutual interference of piles.
• Correction for thickness of floor.
• Correction for the slope of floor.

These corrections are described later.

The percentage pressure at key points can also be found out by using graph developed by Khosla which is known as Khosla’s pressure curves.

It is shown in figure 5.23 (end of chapter). The use of Khosla’s pressure curve will be clear from example 5.3.

## khosla’s theory corrections

#### Correction for percentage pressure

##### 1) Correction for the mutual interference of piles

where, b’ is the distance between two pile lines.

D is the depth of the pile line, the influence of which has to be determined on the neighboring pile of depth d.

D is to be measured below the level at which interference is desired.

d is the depth of the pile on which the effect is considered.

b is the total floor length.

Sign convention

• Positive for the points in rear or backwater
• Negative for the points on forward direction of flow
##### ii)Correction for thickness of floor

The percentage pressures calculated by Khosla’s equations or graphs, shall pertain to the top levels of the floor. While the actual junction points ‘E’ and ‘C’ are at the bottom of the floor.

In the figure 5.24;

Sign convention

• Positive for point in right side of pile
• Negative for point in left side of pile

In figure correction for C₁ is positive and E2 is negative.

##### iii)  Correction for slope

It is given by, [For E2 in  the figure 5.24]

where, bs and b1 are as shown in the figure 5.24.

Cs is the slope correction factor and is given in table.

Sign convention

• Positive for the downward slope
• Negative for upward slope

Exit gradient may be defined as the hydraulic or pressure gradient of subsoil flow or seepage flow at the d/s or the exit end of the floor.

#### Khosla theory exit gradient formula

It has been determined that for a standard form consisting of a floor length ‘b’ with a vertical cutoff of depth ‘d’, exit gradient at its downstream end is given by,

The value of exit gradient can also be found from figure 5.25 (end of chapter) developed by Khosla. In the figure (graph) for any value of the corresponding value of 𝜆

can be obtained directly. Knowing the value of H, d and 𝜆

the value of GE can be calculated using equation (5.3).

Table 5.3: Safe exit gradient for different types of soil

## Khosla’s theory problems

The given figure below shows the profile of a weir. Determine the uplift pressure at key points and the exit gradient and find whether the section provided is safe against piping if it is founded on fine sand with permissible exit gradient of 1/6. Also find uplift pressure at point X and check the safety against uplift.

Solution:

For u/s pile no. 1

Total length of floor (b) = 65 m

Depth of u/s pile line (d) = 100-91 = 9 m ∅ D = 23%

∅ E = 33%

.’.      ∅ D1= 100 – ∅ D = 77%

.’.      ∅ C1 = 100- ∅ E = 67%

Correction for ∅ C1’

i) Correction of C₁ for mutual interference of pile 2,

d = 98-91 = 7m

b’ = 48m

D = 98 – 89.5 = 8.5m

b = 65 m

(Positive since pile 1 is rear to pile 2)

ii) Correction for floor thickness,

= 2.2% (Positive since C, is in right of the pile 1)

.’. Corrected ∅

C1 = 67 +1.9 +22= 71.1%

For intermediated pile no. 2;

d = 96 – 89.5 = 6.5 m

b = 65 m

b₁ = 48.5 m

From the figure 5.23(b), for b1/b = 0.746 and 𝛼 = 10,

∅ E2 = 41%

∅ D2 = 34%

∅ C2 = 26%

Correction for ∅ E2 ,

d = 93 – 89.5 = 3.5m

b’ = 48 m

D = 93 – 91 = 2 m

b = 65 m

Correction of floor thickness, .

1. Slope correction,

Here, Cs=2(From the table 5.2)

bs = 32 m

b1 = 48 m

.’. Correction = 2 x (32/48) = 1.33 (Positive)

.’. Corrected ∅

E₂ = 41 -0.3-3.2+1.33= 38.83%.

Correction for ∅

c2

1.  Mutual interference due to pile 3,

d = 93-89.5 = 3.5 m; b’ = 16 m

D = 93-89.5 = 3.5 m; b= 65 m

1. Correction for floor thickness

No slope correction for ∅

1. C₂

.’. Corrected ∅

C2 = 26 + 1 + 3.7 = 30.7%

For d/s pile no 3,

d = 96 – 89.5 = 6.5m

b = 65 m

From figure 5.23 (a),

∅ E3 = 29%

∅ D3 = 20%

Correction of ∅ E3

1.  Mutual interference due to pile 2,

d=94-89.5 = 4.5 m

b’ = 16 m

D=94-89.5 = 4.5 m

b = 65 m

1. Correction for floor thickness

.’.  Corrected ∅

E3 = 29 – 1.4 – 2.8 = 24.8

The corrected % pressure at key point are tabulated as under,

Difference in head between pond level and d/s floor;

i.e., Maximum percolation head, H = 103-96 = 7 m

Uplift pressures @E₁ = 1×7=7m

@D₁ = 0.77 x 7 = 5.39 m

@C₁ = 0.711 x 7 = 4.977 m

@E2 = 0 388 x 7 = 2.716 m

@D₂ = 0.34 x 7 = 2.38 m

@C₂=0.307 x 7 = 2.149 m

@E3 = 0.248 x 7 = 1.736 m

@D3 = 0.20 x 7 = 1.4 m

@C3 = 0 m

H = 7 m

d = 96 – 89.5 = 6.5 m

From figure 5.25, for α = 10

Hence, safe.

Uplift pressure at X (Residual head at X) is,

.’. Floor thickness required at X,

Thickness provided = 96-93= 3 m > 2.8 m

Hence, safe against uplift.

Happy Learning – Civil Concept