7.5. Example Problems

Problem 7.1: Gravity Wall Rankine and Coulomb Methods

For the retaining wall shown in the figure below, calculate the active earth pressures and their application point and define the direction of the sliding surface. If the wall is characterized by a soil-wall friction angle of δ=30°, how much will the active and passive earth pressures be decreased (in %)? 

Solution

For a smooth wall (δ=0°) the factor of the active impulse will be given by the equation:

(Rankine)

For φ=30° we find that K_A=0.33.

The distribution of the active earth pressures on the wall is illustrated below.

For z=0:

For z=6.0m:

Therefore, we get:

The distance between the application point of P_α and the top surface of the wall will be: 

The sibling surface will form an angle with respect to the horizontal surface:  

If the angle of friction between the wall and the soil was 30°, the value of K_Α is calculated using Coulomb’s general relationship:

Because i=0°, β=90°, and for φ=30°, δ=30°, the equation above becomes:

consequently:

with an angle of 30° with respect to the horizontal.

Therefore, the active earth pressure is decreased by:

The horizontal component of the active earth pressure is:

and we have a decrease of:

Based on the above, it is concluded that that ignoring the friction between the wall - soil, is a conservative option.

Problem 7.2: Gravity wall - Culmann and Coulomb Methods

For the retaining wall below, calculate the active earth pressures of the soil:

1. using Culmann’s graphical method
2. using Coulomb’s general relationship

Solution

1) Culmann’s graphical method is shown in the figure below:

5 trial surfaces were considered BC₁, BC₂,...,BC₅. The surfaces were selected so that they cross the earthfill surface at points C₁, C₂,..., C₅ considering a horizontal distance between them of 2.0 m. In this way, the triangles ABC₁, C₁BC₂,...,C₄BC₅ are characterized by the same area. The weight of the triangle ABC₁ is equal to 2.0 x 6.0 x 0.5 x 1.8 = 10.8T, the weight of ABC₂ is twice the weight of ABC₁, (ABC₂) = 2 (ABC₁), etc.

The angle ψ is equal to: ψ = 180°-β-δ = 180°- 90°-20° = 70°.

The weights of the trial triangles were estimated assuming a scale on the line BF (points 10.8, 21.6, 32.4, 43.2, 54.0). From each point we draw a parallel line with regard to the BG and we define the crosspoints with the corresponding sides of the triangles. We draw the curve that passes through these points and find the point of the curve farther from line BF in the direction of BG. Thus, we define the direction of the slip surface (BC) and the value of the active earth pressure which is P_α = 8.8 T/m². 

2) Coulomb’s relationship for φ=38°, δ=20°, i=20°, β=90°: 

and the value of the active earth pressure will be:

The values of P_α that are estimated based on the graphical method, are less than 2% different than those resulting from the analytical solution. 

Problem 7.3: Anchored Wall

The vertical retaining wall of the figure below has a height of 2.7m, and it is pulled towards the sandy embankment through an anchor. For the embankment material, we have γ_t=1.9T/m² and φ=30°. The friction angle between the wall and the soil is δ= φ/2. Calculate the passive resistance exerted on the wall, as well as its horizontal component.

Solution

Since we have a vertical wall and a horizontal embankment the diagrams in Section 7.2 can be used. For φ=30° and δ=30°/2=15° we find K_P=4.83. Therefore, the passive resistance is: 

and it will form an angle 15° with respect to the vertical. As a result, the horizontal component of the passive resistance will be: P_PH=32.3 T/m².

Also, we could apply Coulomb’s general relationship for the passive earth pressure coefficient by using i= 0° and β = 90° (because δ):

We observe that Coulomb’s relationship, even though it is using a horizontal sliding surface, gives a value of Kₚ only 0.6% smaller than the value estimated by using fig. 7-2, in which curved sliding surfaces are considered.

Problem 7.4: Diaphragm Wall

For the diaphragm wall of the following figure, calculate the depth d so that the factor of safety against overturning is F=1.50. (The passive resistance is considered with a factor of safety F’=2). 

Solution

We assume that the wall will rotate around a point O near its bottom end. From point O and below, passive earth pressures are developed and their resultant is R, acting on point O. We  consider the moments of forces with regard to point O. The moment of the active earth pressures, P_α, is the overturning moment, M_A, while the moment corresponding to the passive resistance P_P/F' is the resisting moment, M_E. Accordingly, it must be M_E=1.5M_A.

The active earth pressure coefficient is: 

(assuming δ=0).

The active earth pressure is: 

and the overturning moment: 

The passive earth pressure coefficient is:

The passive resistance is:

And the resisting moment:

For F’=2 the above equation becomes: 

So, since M_E=1.5M_A we get:

Following common practice, the embedment depth is increased by about 20% so that the passive resistance is fully developed. In this way, the final value of d is d’=1.2xd=6.52 m.

Problem 7.6: Cantilever Wall

Check the stability of the cantilever wall of the figure below.

Given:

• unit weight of soil: γ_t=1.70 T/m³,
• angle of internal friction: φ=40°,
• angle of friction between wall base and soil: φ_B=30°
• unit weight of concrete: γ=2.35 Τ/m³

Solution

For this type of retaining wall, we assume that the soil between the wall and line AB is part of the retaining wall system. In this way, the active earth pressures are calculated over the imaginary surface ΑΒ by using Rankine’s relationship. The factor K_A is:

Active earth pressures

• From surcharge: 

• From the earthfill:

• Thus, in total:

Vertical forces

• wall weight, W_b1: 5.0 x 0.3 x 2.35 = 3.53 T

• foundation weight, W_b2: 5.0 x 0.4 x 2.35 = 2.82 T
• soil weight, W_b: 5.0 x 1.75 x 1.70 = 14.88 T
• surcharge, W: 1.75 x 4.0 = 7.00 T
• sum, N = 28.23 T

Check against slip

• Factor of Safety: 

Check against overturning

• Driving moment (with respect to point O):

• Resisting moment:

• Factor of Safety:

Problem 7.7: Gravity Wall - Surcharge Load

For the retaining wall of the figure below:

1. Calculate the active earth pressure of the soil acting on the wall
2. If a uniform surcharge of q_s = 10 T/m² is applied on the surface of the backfill, what will be the new active earth pressure?
3. What is the minimum value of q_s so that tensile stresses are not developed on the wall.

The soil and wall characteristics are shown in the figure.

Solution

1) As the wall is smooth (δ=0), we can use Rankine’s relationship. The distribution of the active stresses on the wall is given by the equation: 

Thus:

the depth in which σ_h=0 is:

The area of the triangle of the compressive stresses is:

This is the value of the active earth pressure on the wall. The point at which P_A is applied, is at:

above the base of the wall.

2) When the uniform load q_s is applied, we will have:

Thus:

The active earth pressure, now, is computed as:

The point of application of P_A is at distance z from the base: 

3) In this case, it is evident that if the value of σ_h is equal to zero at the top of the wall, then tensile stresses are not developed on the wall. The value of q_s is calculated by the relationship:

for z = 0: 

For this value of q_s, the horizontal stress at the base of the wall is:

The value of the active earth pressure in this case is P_A = 11.39 T/m² and its application point is located at distance from the base of the wall:

Problem 7.8: Gravity Wall - Active Earth Pressure

For the retaining wall of the following figure: calculate the active earth pressure acting on the wall. The characteristics of the soil and the wall are shown on the figure.

Solution

In this case, the active earth pressures acting on the wall are estimated considering first the effective stresses (for depth > 2.5 m) and then adding the pore water pressures. The active earth pressure coefficient, K_A, will be the same above and below the groundwater table, because φ and δ retain their values. Using the diagrams of Section 7.2, we get:

• φ = 30°
• δ =30°
• K_A = 0.29.

So the active earth pressure distribution on the wall will have the following form:

The active earth pressures are:

and will act at an angle of 30° with respect to the horizontal. The point of application of P_A is located at distance from the base of the wall:

The horizontal component of P_A is: 

The resultant of the hydrostatic pressures is: 

and it is applied at a distance from the base of the wall:

Therefore, we note that the presence of the groundwater table results in large pressures and thus water accumulation behind the wall should be avoided.