coulomb's theory of earth pressure coulomb's wedge theory of earth pressure coulomb's expression for active pressure coulomb's active earth pressure coefficient =Ka vedio link https://youtu.be/PSDwMjlTTGs for numerical problem https://youtu.be/ZPf3qAAtcpE

1. COULOMB’S WEDGE THEORY
SUBJECT :- FOUNDATION ENGINEERING
PREPARED
BY
ER. SANJEEV SINGH
SNJV432@GMAIL.COM
PART:- IV

2. CONTENT
 COULOMB’S WEDGE THEORY
 COULOMB’S ACTIVE PRESSURE IN COHESIONLESS SOILS.
 KA = COULOMB’S ACTIVE EARTH PRESSURE COEFFICIENT

3. COULOMB’S WEDGE THEORY
The following assumption are made:
 The soil is dry, isotropic and homogeneous.
 The surface of rupture is a plane.
 The failure wedge is a rigid body.
 There is friction between and the wall.
 The backfill can horizontal or inclined.
 Failure is two dimensional.
 The soil is cohesion-less.
 Coulomb’s equation of shear strength is valid.
 He assumed soil fails as a rigid wedge and stress is uniform distributed.

4. COULOMB’S ACTIVE PRESSURE
IN COHESIONLESS SOILS
 Expression for active case
From fig (b). Using the law of sine
𝑃𝑎
sin( - ϕ′
)
=
𝑊
sin(1800− β + δ − + ϕ′)
or
Where Pa= total active pressure force
The weight W of the wedge ABD can be determined from fig as
Taking AG = m, and BE = L we have W=
1
2
𝐿 ∗ 𝑚 ∗ γ
Or
𝑷𝒂=
𝑾𝒔𝒊𝒏( − 𝝓′)
𝒔𝒊𝒏(𝟏𝟖𝟎 𝟎−𝜷+𝜹− +𝝓′)
………….(a)
W =
𝟏
𝟐
(𝒎 ∗ 𝜸)(BG + GE) ………….. (b)

5. Now AB=H/sinβ
From triangle ABG,
𝑚
𝐴𝐵
=sin[1800 − (β + )] = sin(β + )
Therefore
From triangle ABG,
𝐵𝐺
sin(β+ -900)
=
𝑚
sin(β+ )
Or BG= m sin(β + −900
) / sin(β + ) …………
From triangle AGE,
𝐺𝐸
sin(900− +ί)
=
𝑚
sin( −ί)
GE = m
sin(900− +ί)
sin( −ί)
………….. (e)
m= 𝑨𝑩 sin(𝜷 + ) =
𝑯𝒔𝒊𝒏(𝜷+ )
sin 𝜷
…. (c)
BG= m sin(𝜷 + −𝟗𝟎 𝟎) / sin(𝜷 + ) …. (d)
GE = m
𝒔𝒊𝒏(𝟗𝟎 𝟎− +ί)
𝒔𝒊𝒏( −ί)
………….. (e)

6.  Substituting the values of m, BG and GE from Eqs.(c), (d) and (b),
W =
1
2
*
𝐻 𝑠𝑖𝑛(𝛽+ )
𝑠𝑖𝑛𝛽
* γ * m
sin(𝛽+ −900)
sin(𝛽+ )
+ m
𝑠𝑖𝑛(900− +ί)
𝑠𝑖𝑛( −ί)
Or W =
1
2
*
𝐻2γ 𝑠𝑖𝑛(𝛽+ )2
𝑠𝑖𝑛2 𝛽
sin(𝛽+ −900)
sin(𝛽+ )
+
𝑠𝑖𝑛(900− +ί)
𝑠𝑖𝑛( −ί)
Substituting the above value of W in Eqs. (a)
Pa =
𝟏
𝟐
𝐇 𝟐γ 𝐬𝐢𝐧(𝛃+ ) 𝟐
𝐬𝐢𝐧 𝟐 𝛃
*
𝐬𝐢𝐧( − 𝛟′)
𝐬𝐢𝐧(𝟏𝟖𝟎 𝟎−𝛃+𝛅− +𝛟′)
*
sin(𝛃+ − 𝟗𝟎 𝟎)
sin(𝛃+ )
+
𝐬𝐢𝐧(𝟗𝟎 𝟎− +ί)
𝐬𝐢𝐧( −ί)
The active pressure force Pa will be a maximum when the failure plane
makes an angle with the horizontal such that
Or W =
𝟏
𝟐
*
𝑯 𝟐γ 𝒔𝒊𝒏(𝜷+ ) 𝟐
𝒔𝒊𝒏 𝟐 𝜷
sin(𝜷+ − 𝟗𝟎 𝟎)
sin(𝜷+ )
+
𝒔𝒊𝒏(𝟗𝟎 𝟎− +ί)
𝒔𝒊𝒏( −ί)
𝛛𝐏𝐚
𝛛
= 0

7. The maximum value of Pa thus obtained is coulomb’s active force,
given by
Pa
1
2
𝐾𝑎γ𝐻2
Where Ka = Coulomb’s active earth pressure coefficient, given by
Where
 Φ’ is internal friction angle of the soil,
 β is the slope of the backfill
 is the angle of the back of retaining wall
 δ is friction angle between soil and back of retaining wall
Pa
𝟏
𝟐
𝑲𝒂γ𝑯 𝟐
Ka =
𝒔𝒊𝒏 𝟐(𝜷+𝝓′)
𝒔𝒊𝒏 𝟐 𝜷𝒔𝒊𝒏(𝜷−𝜹) 𝟏+
𝒔𝒊𝒏 𝝓′+𝜹 𝒔𝒊𝒏 𝝓′−ί
𝒔𝒊𝒏 𝜷−𝜹 𝒔𝒊𝒏(𝜷+ί)