Earth retaining structures
Stability analysis
Learning objectives: 1. Basic Stability Analysis Background
2. Analysis: Gravity and Cantilever Retaining Walls
Lesson #1: Basic Stability Analysis Background:
When analyzing a retaining wall, the wall can fail in three different ways. Sliding, Toppling,
and soil bearing capacity failure. Doing this section we will look at the first two ways, and talk
discuss the bearing capacity in a different section.
1.
The retaining wall can slide. This is called a sliding failure. The example is
below, The retaining wall will fail if the Force force is less than the resultant
Vertical Force. The Factor of Safety (FoS) = F(floor) / F
Height (h)
Resultant
Vertical Force
F =.5 Ka * γ * h2
F(floor) = tan δ x W
σ’H = Ka * γ * h
Weight(W)
Where:
W = weight of the retaining wall
δ = friction angle between concrete and soil
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Earth retaining structures
Stability analysis
2.
The Resistance Against Overturning. The example is below, The retaining wall
will fail if the overturning moment (F x (h/3)) is greater than the resisting moment
(W x a). The Factor of Safety (FoS) against overturning = resisting moment
overturning moment = (3 x W x a) / (F x h)
Height (h)
PT. A
Resultant
Vertical Force
F = Ka * γ * h2/2
a
Weight(W)
σ’H = Ka * γ * h
Where:
a = distance from Pt. A to the center of
gravity of weight
The best way to explain these stability analysis problems are with examples;
Ex. 1: Gravity retaining wall with sand backfill with no groundwater. The height of
the retaining wall, h, is 25 ft, the weight of the retaining wall is 60 tons for a 1 ft
length of the wall and the weight acts at a distance of 12 ft from the tow of point
A. The friction angle of the soil backfill is 30o. The soil backfill mainly
consists of sandy soil. The density of the soils is 130 pcf. The friction angle
between the soil and earth at the bottom of the retaining wall was found to be
20o. Find the factor of safety for the retaining wall shown above.
Step 1: Calculate Ka.
Ka = tan2 (45 – φ/2) = tan2 (45 – 30/2) = .33
Step 2: Calculate the horizontal effective stress.
σ = Ka γ h = .33 x 130 lbs/cf x 25 ft = 1073 psf
Step 3: Find the resultant earth pressure
F = σ x (h/2) = 1073 psf x (25ft/2) = 13,413 lbs per foot of wall
Step 4: Find the resistance against sliding at the base.
FFloor = weight of the wall x tan (δ) = 60 tons 2000 lbs/ton x tan (20) = 43,676 lbs per
foot of wall
Step 5: Find the Factor of safety for Sliding.
The Factor of Safety (FoS) = F(floor) / F = 43,676 / 13,413 = 3.26 FS
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Earth retaining structures
Stability analysis
Ex. 1 continued:
Step 6: Find the resistance against overturning.
Resisting moment = W x a = 120,000 lbs x 12 ft = 1,440,000 lbs ft
Step 7: Find the overturning moment:
Overturning moment = F x H/3 = 13,413 lbs ft x (25/3) = 111,775 lb ft
Step 8: Find the FS.
Factor of Safety = 1,440,000 lbs ft / 111,775 lbs ft = 12.8 FS
So the concern limiting Factor of Safety is the sliding failure which is 3.26. This is still
acceptable.
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Earth retaining structures
Stability analysis
For a cantilever retaining wall system the concept is the same. The retaining can fail by
sliding and toppling. However, this system uses the earth weight to help with stability.
W3
W2
PT. A
W1
Height (h)
Passive Resultant
Vertical Force
F = Kp * γ * h2/2
a
F(floor) = tan δ x W
σ’H = Ka * γ * h
W4
Weight(W) = W1 +W2 +W3 + W4
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Active Resultant
Vertical Force
F = Ka * γ * h2/2
Weight from fill
Weight from wall
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Earth retaining structures
Stability analysis
For the cantilever retaining wall shown below, calculate the factor of factor of safety against
toppling? The concrete retaining walls density is 150 pcf.
1 ft
7 ft
3 ft
Sand
γ = 132 pcf
φ = 34°
c=0
13 ft
2 ft
PT. A
10 ft
Step 1: Find what you are trying to solve:
W3
W1
FS against toppling: Resisting Moment / Toppling moment
W2
Step 2: Find all the forces resisting:
All the weight is resisting toppling
Weight from fill
W1 = 132 pcf (2ft x 2ft) = 528 lbs per lf @ 1 ft
Weight from wall
W2 = 132 pcf (7ft x 12ft) = 11,088 lbs per lf @ 6.5 ft
W3 = 150 pcf (1ft x 12ft) = 1,800 lbs per lf @ 2.5 ft
W4 = 150 pcf (1ft x 10ft) = 1,500 lbs per lf @ 5 ft
So the moment force resisting due to the weight of the wall and soils
F = W1 x d1 + W2 x d2 + W3 x d3 + W4 x d4
= 528 x 1 + 11,088 x 6.5 + 1,800 x 2.5 + 1,500 x 5
= 528 + 72,072 + 4,500 + 7,500 = 84,600 lbs
-Also the passive resultant force due to the soil is a force resisting
Kp = tan2 (45 + φ/2) = tan2 (45 + 34/2) = 3.5
W4
The resultant force is F = Kp * γ * h2/2 = 3.5 x 132 x 32/2 = 2,079 lbs per lf @ 1 ft
So the resisting moment is 84,600 lbs + 2,079 lbs = 86,679 lbs
Step 3: Find the toppling moment:
Ka = tan2 (45 – φ/2) or 1/Kp = .285
F = Ka * γ * h2/2 = .285 x 132 pcf x 13ft2/2 = 3,179 lbs @ 4.33ft = 13,765 lbs
Step 4: Find the Factor of safety:
FS = Resisting moment / toppling moment = 86,679 / 13,765 =
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6.3
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Earth retaining structures
Stability analysis
Ex. 3: Below is a gravity retaining wall with sand backfill above the clay layer with no
groundwater. The top layer of the backfill is sand with a friction angle of 30 degrees. The
density of the sand layer is 130 pcf. The clay layer below the sand layer has a cohesion of
320 psf and a density of 120 pcf. The weight of wall is 40 tons per foot, and the center of
gravity is acting 6 ft from the toe of the wall. The friction angle between the concrete and
the soil at the bottom of the retaining wall was found to be 25o. Find the factor of
safety(tipping and sliding) for the retaining wall.
15 ft
Sand
γ = 130 pcf
φ = 30°
Height (h)
12 ft
PT. A
Clay
γ = 120 pcf
c = 320 psf
6ft
W = 40 tons per ft
Step 1: Calculate Ka.
Ka = tan2 (45 – φ/2) = tan2 (45 – 30/2) = .33
Step 2: Calculate the horizontal effective stress.
Sand layer  σ = Ka γ h = .33 x 130 pcf x 15 ft = 644 psf
Clay layer  @ depth 15 ft from ground level.
σ = γ h – 2c = 130 pcf x 15 ft – (2 x 320) = 1310 psf
Clay layer  @ depth 27 ft from ground level
σ = γ h + σ (cl at 15ft) = 120 pcf x 12 ft + 1310 psf = 1440 psf + 1310 psf = 2750 psf
Step 3: Find the resultant earth pressure
F1 from sand = σ x (h/2) = σ x (15/2) = 644 psf x (15ft/2) = 4,830 lbs per foot of wall
F2 from sand on clay = σ x h = 1310 psf x 12 = 15,720 lbs per foot of wall
F3 from clay = σ x (h/2) = 1440 psf x (12/2) = 8,640 lbs per foot of wall
Total Sliding force = 29,190 lbs per foot of wall
Step 4: Find the resistance against sliding at the base.
FFloor = weight of the wall x tan (δ) = 40 tons x 2000 lbs/ton x tan (25) = 37,304 lbs per
foot of wall
Step 5: Find the Factor of safety for Sliding.
The Factor of Safety (FoS) = F(floor) / F = 37,304 / 29,190 = 1.28 FS
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Earth retaining structures
Stability analysis
15 ft
F1 @ 1/3 from bottom
Height (h)
12 ft
F2 @ 1/2
from bottom
F3 @ 1/3
from bottom
6ft
PT. A
W = 40 tons per ft
Ex. 3 continued:
Step 6: Find the resistance against overturning.
Resisting moment = W x a = 80,000 lbs x 6 ft = 480,000 lbs ft
Step 7: Find the overturning moment:
Overturning moment = F1 x H/3 = 4,830 lbs ft x (15/3) = 24,150 lb ft
Overturning moment = F2 x H/2 = 15,720 lbs ft x (12/2) = 94,320 lb ft
Overturning moment = F3 x H/3 = 8,640 lbs ft x (12/3) = 34,560 lb ft
Step 8: Find the FS.
Factor of Safety = 480,000 lbs ft / (24,150 + 94,560 + 34,560) lbs ft
= 480,000 / 153,270 = 3.13 FS
So the limiting Factor of Safety of concern is the sliding failure which is 1.28. Which is
most likely unacceptable for your design so you would need to add more weight to the
gravity retaining wall.
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