**Eg:**Determine the lateral earth pressure at rest /unit length of the wall. Also, find the location of the resultant earth pressure on the retaining wall as shown below. Consider 𝛄w = 10KN/m³.

__Given data:__

Depth of 1st layer = H1 = 2.5m.

The angle of friction = 𝜱1 = 32°

Unit weight = 𝛄 = 16KN/m³

Depth of 2nd layer =H2 = 3m.

The angle of friction = 𝜱2 = 34°

Saturated unit weight = 𝛄sat = 19KN/m³.

Unit wt. of water 𝛄w = 10KN/m³.

**Calculation:**

**Coefficient of earth pressure at rest ****for the 1st layer**

Ka01 = [1-sin𝜱1]

= [1-sin32°]

= [1-0.5299]

**Ka01 = 0.47**

**Coefficient of earth pressure at rest **** for the 2nd layer**

Ka02 = [1-sin𝜱2]

= [1-sin34°]

= [1-0.5591]

**Ka02 = 0.4408**

**Intensity of pressure at layer B**

p01 = [ Ka01 x 𝛄 xH1]

= [ 0.47 x 16 x 2.5]

p01 = **18.80KN/m².**

**Intensity of pressure at layer C due to 1st layer**

p02 = [ Ka02 x 𝛄 xH1]

= [ 0.4408 x 16 x 2.5]

p02 =** 17.632KN/m².**

**The intensity of pressure at layer C due to 2nd layer**

p03 = [ Ka02 x 𝛄¹ xH2]

Where 𝛄¹ = submerged unit wt.

= [𝛄sat -𝛄w]

= [19 - 10]

= 9KN/m³.

p03 = [ 0.4408 x 9 x 3]

p03 =** 11.016KN/m².**

**Pressure due to water**

u = [𝛄w x H2]

= [10 x 3]

u = **30KN/m².**

**Note:** The pr. due to water in the 1st layer will be 0, as the water table is only in the 2nd layer.

**Now, let us draw the**

**pressure distribution diagram**from the above-calculated values.

Total pressure of triangle ABD

P1= [1/2 x base x height]

= [1/2 x 18.8 x 2.5]

P1= **23.5KN**.

Total pressure of rectangle BCFE

P2= [Base x height]

= [17.632 x 3]

P2= **52.896KN.**

The total pressure of triangle EFG

P3= [1/2 x 11.016 x 3]

P3= **16.524KN.**

The total pressure of the triangle EGH

P4= [1/2 x 30 x 3]

P4= **45KN.**

**The total active earth pressure **

P = [P1 + P2 + P3 + P4]

= [23.5 + 52.896 + 16.524 + 45]

**P = 137.92KN.**

Resultant active earth pressure:

The centroid of the total active earth pressure acts horizontally at a distance of Ȳ from the base of the pressure diagram.

For a triangular pressure diagram, the pressure acts at 1/3rd distance from the base of the diagram. For the rectangular diagram, the acting pressure will be at 1/2 the distance from the base.

( According to Rankine theory)

Therefore,

P**Ȳ = [P1****Ȳ1 + P2****Ȳ2 + P3****Ȳ3 + P4****Ȳ4]**

** = [{23.5 ****x (3m. + 1/3 ****x 2.5m.)} +{52.896 ****x (1/2 ****x****3m**** )} +{16.524 ****x (1/3 ****x 3m.)} +{30 ****x (1/3 ****x 3m.)}]**

** = [{23.5 ****x 3.833} + {52.896 ****x 1.5} + {16.524 ****x 1} + {30 ****x 1}]**

** = [90.076 + 79.344 + 16.524 + 30]**

**P Ȳ = 215.944**

**Ȳ = [****215.944 ➗ P]**

** = [215.944 ➗ 137.92]**

** Ȳ = 1.565m.**

**The total earth pressure of 137.92KN acts horizontally at a distance of 1.565m. from the base of the wall.**

**Thank you for going through these calculation steps❤. Have a good day 😄.**