Earth retaining structures

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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 ? h2
sH Ka ? h
F(floor) tan d x W
Weight(W)
Where W weight of the retaining wall d
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)
Resultant Vertical Force F Ka ? h2/2
sH Ka ? h
a
PT. A
Weight(W)
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 f/2) tan2 (45
30/2) .33 Step 2 Calculate the horizontal
effective stress. s Ka ? h .33 x 130 lbs/cf
x 25 ft 1073 psf Step 3 Find the resultant
earth pressure F s 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 (d) 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.
W2
W3
PT. A
W1
Height (h)
Active Resultant Vertical Force F Ka ?
h2/2
Passive Resultant Vertical Force F Kp ?
h2/2
sH Ka ? h
a
F(floor) tan d x W
Weight from fill
W4
Weight(W) W1 W2 W3 W4
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
Sand ? 132 pcf f 34 c 0
13 ft
7 ft
2 ft
3 ft
10 ft
W3
PT. A

• Step 1 Find what you are trying to solve
• FS against toppling Resisting Moment / Toppling
  moment
• Step 2 Find all the forces resisting
• All the weight is resisting toppling
• W1 132 pcf (2ft x 2ft) 528 lbs per lf _at_ 1
  ft
• W2 132 pcf (7ft x 12ft) 11,088 lbs per lf
  _at_ 6.5 ft
• W3 150 pcf (1ft x 12ft) 1,800 lbs per lf _at_
  2.5 ft
• W4 150 pcf (1ft x 10ft) 1,500 lbs per lf _at_
  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 f/2) tan2 (45 34/2)
  3.5
• The resultant force is F Kp ? h2/2 3.5 x
  132 x 32/2 2,079 lbs per lf _at_ 1 ft
• So the resisting moment is 84,600 lbs 2,079 lbs
  86,679 lbs
• Step 3 Find the toppling moment

W1
W2
Weight from fill
Weight from wall
W4
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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.
Sand ? 130 pcf f 30
15 ft
Height (h)
Clay ? 120 pcf c 320 psf
12 ft
6ft
PT. A
W 40 tons per ft
Step 1 Calculate Ka. Ka tan2 (45 f/2)
tan2 (45 30/2) .33 Step 2 Calculate the
horizontal effective stress. Sand layer ? s Ka
? h .33 x 130 pcf x 15 ft 644 psf Clay layer
? _at_ depth 15 ft from ground level. s ?
h 2c 130 pcf x 15 ft (2 x 320) 1310 psf
Clay layer ? _at_ depth 27 ft from ground level
s ? h s (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
s x (h/2) s x (15/2) 644 psf x (15ft/2)
4,830 lbs per foot of wall F2 from sand on clay
s x h 1310 psf x 12 15,720 lbs per foot of
wall F3 from clay s 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 (d) 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 _at_ 1/3 from bottom
Height (h)
F2 _at_ 1/2 from bottom
F3 _at_ 1/3 from bottom
12 ft
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.