Eg:
Determine active earth pressure on the retaining wall as shown in the below drawing.
Consider 𝛄w = 10KN/m³.
%20(1).png)
Given data:
Depth of 1st layer = H1 = 3m.
The angle of friction = 𝜱1 = 34°
Unit weight = 𝛄 = 16KN/m³
Depth of 2nd layer =H2 = 3m.
The angle of friction = 𝜱2 = 39°
Saturated unit weight = 𝛄sat = 18KN/m³.
Calculation:
Coefficient of earth pressure for 1st layer
Ka1 = [1-sin𝜱1 / 1+ sin𝜱1]
= [(1-sin34°) / (1+ sin34°)]
= [(1-0.5592) / (1+ 0.5592)]
= [0.4408/1.5592]
Ka1 = 0.283
Coefficient of earth pressure for 2nd layer
Ka2 = [1-sin𝜱2 / 1+ sin𝜱2]
= [(1-sin39°) / 1+ (sin39°)1]
= [(1-0.6293) / (1+ 0.6293)]
= [0.3706/1.6293]
Ka2 = 0.228
Intensity of pressure at layer B
p1 = [ Ka1 x 𝛄 xH1]
= [ 0.283 x 16 x 3]
p1 = 13.584KN/m².
Intensity of pressure at layer C due to 1st layer
p2 = [ Ka2 x 𝛄 xH1]
= [ 0.228 x 16 x 3]
p2 = 10.944KN/m².
The intensity of pressure at layer C due to 2nd layer
p3 = [ Ka2 x 𝛄¹ xH2]
Where 𝛄¹ = submerged unit wt.
= [𝛄sat -𝛄w]
= [18 - 10]
= 8KN/m³.
p3 = [ 0.228 x 8 x 3]
= 5.472KN/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.
.png)
Total pressure of triangle ABD
P1= [1/2 x base x height]
= [1/2 x 13.584 x 3]
P1= 20.376KN.
Total pressure of rectangle BCFE
P2= [Base x height]
= [10.944 x 3]
P2= 32.83KN.
The total pressure of triangle EFG
P3= [1/2 x 5.472 x 3]
P3= 8.208KN.
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]
= [20.376 + 32.83 + 8.208 + 45]
P = 106.414KN.
Thank you for going through these calculation steps❤. Have a good day 😄.
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