Geosynthetics

y and
Reinforced Soil Structures
Reinforced Soil Embankments-III

Dr. K. Rajagopal
Professor of Civil Engineering
IIT Madras,, Chennai,, India
e-mail: gopalkr@iitm.ac.in
 Over View
Review of the design based on planar rupture
surface
Two‐part wedge analysis of reinforced soil
slopes
Design Example with two‐part wedge method of
analysis

Reinforced Soil Embankments - 3 2

Cross-section obtained from planar wedge
analysis
l i
Top reinforcement layer at
250 mm depth from surface

C=0
9m =30
 30
 = 20 kN/m3
16 llayers off reinforcement
i f t
Slope angle = 70
  TLTDS=25 kN/m
Length = 13 m
13m Spacing from 0.5 to 0.75 m

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Critical slip circle for the geometry that was designed – FS = 1.71
(toe circles were considered as foundation soil is strong)
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Bi-linear rupture surface approximation of the slip surface –
two planar surfaces
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Woods and Jewell ((1990)) two‐part
p wedge
g analysis
y
Considered the normal and shear components of
the reinforcement force on the rupture surface:
N
N

S
3 
PR

 PR sin PR PRcos
PR cos

PRsin

Triaxial compression test Direct Shear Test

Reinforced Soil Embankments - 3

6
The components of the reinforcement force PR help in the
following two ways:

1 Normal component of the force on the rupture surface

1.
increases the normal stress on the rupture surface
thereby generating higher resistance forces

2. Tangential component of the force on the rupture surface

directly opposes the shear force and adds to the shear
resistance.

 n  A N  PR cos 
1

  S  P R sin  
1
A

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 Planar Wedge Analysis by considering the
contribution of normal and shear components

shear resistance R v
cos tan +  Ti cos
FS = =
shear force R v
sin
sin    T i cos  T
R v
=1+ i

R sin 
v R tan
v
Considering the normal & shear component
shear resistance ( R v cos   T i sin  ) tan 
FS = =
shear force R v sin   Ti cos
sin    T i sin  tan 
R v

R sin    T cos
v i

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 FS  1 Rv sin 
T i

FS  cos   sin  tan 
For the previous design case of 70 and 9 m high slope,

Rv = (0.5*20*9*9+20*9)(cot(30)-cot(70)) = 1354.4 kN/m

T = (1.5-1)*1354.4
( )* * sin(30)/(1.5*cos30+sin30*tan30)
( )/( * * )=
213 kN/m

No. of reinforcement layers = 213/25  9

In the previous design analysis, the reinforcement

force required was 350 kN/m and the no. of
reinforcement
i f t layers
l was 16.
16
Reinforced Soil Embankments - 3 9
Let the
h reinforcement
i f llayers b
be provided
id d at
depths of:
9, 8.5, 8, 7.5, 7, 6.25, 5.5, 4.75, 4, 3, 2, 1, 0.25 m
from the surface
Total No. of layers = 13

The actual provided number of layers is more than the

theoretical required because of the limitation on
maximum allowable vertical spacing based on
strength and codal requirements.

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 Limit Equilibrium Analysis of Woods
and Jewell (1990)
Mobilised shear strength of soil

 FS
  max

c
  n
tan  
mob
s FS s FS s

max is the peak strength or the critical state strength as

applicable
PRmax is either the permissible
P reinforcement force at

P 
Rmax serviceability limit or the rupture
Rmob strength
FS R FSS and FSR are the factors of
safety on soil and reinforcement
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Actual forces generated in the reinforcement and the soil
should be based on strain compatibility.

• If the reinforcement is inextensible,

inextensible its peak force is
developed at low strain – mobilised soil strength
corresponding to that strain is used in calculations
• If the reinforcement is relatively extensible, its peak
force may develop at large strain corresponding to the
constant volume state of soil

Strain compatibility,
Woods and Jewell (1990)

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 Reinforcement

Inter‐wedge
boundary

Failure plane

Wedge “node” point

Two-part failure wedge considered by Woods and Jewell (1990)

Reinforcement force PR is the least of pullout capacity or rupture
strength

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 Pv
q
Ph P RV

Uw
P Rh
q=surcharge
W S Pv,Ph = external line loads
w N = normall force
f on rupture
t plane
l
U S = shear force on rupture plane
TW W = Weight
g of soil in wedgeg
N
N
Tw=inter‐wedge force
U = pore water force on rupture plane
 Uw = pore water force at inter‐wedge
inter wedge
w = inter‐wedge friction angle
Forces on a wedge in PR = reinforcement forces
the analysis  = wedge angle
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 VCF2
VCF1

TW1
UW2

UW1
TW2

2

Equilibrium of Wedges in the

1 two-part
p wedge g analysis
y
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For each pair of wedges (1, 2) a FS exists which keeps the
wedges in equilibrium i.e. TW1+TW2 = 0

There exists a pair of wedges for which the FS required to

maintain the equilibrium is the least.

Woods and Richards have developed a computer program

WAGGLE that searches for the minimum factor of safety of
reinforced soil slopes.

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Inter-wedge roughness coefficient f determines the
angle
l tto the
th h t l W
horizontal
i

 f  tan    yb = length
l th off th
the iinter-wedge
t d b boundary
d
 W
 tan 
1

 FS   = average friction angle of the soil

s
c=average cohesion along the inter-wedge
cy boundary
V cf
 f b

FS s

Results of a large number of parametric analyses

by WAGGLE computer program were compared
with those from other programs
programs.

f=0
0 was found to give conservative results.

Design charts for determining the quantity of

reinforcement and the length were developed
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 q
C1, C2 = cohesive forces on
th base
the b off wedges
d
W1, W2 = weight of wedges Wedge‐1
T1, T2 = reinforcement forces T1
= base sliding factor=0.8 C1
Ui = ru Wi Wedge‐2
1 c, , , ru
 T2 h
2 C2

 W1  Q1  tan 1  tan 1  U1 tan 1  C1  / cos 1 

T tot  T 1  T 2   (1  tan 1 tan 1) 

 W2  Q2  tan  2   tan 2   U 2 tan 2  C2  / cos  2 
 
 (1   tan  2 tan 2
 ) 
Effect of cohesion is to reduce the reinforcement
requirement Pore water forces increase the reinforcement
requirement.
requirement.
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Design charts for estimating the lateral earth pressures
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Design charts for estimating length of reinforcement layers

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• Solutions are given for different slope angles
g
and friction angles
• Designs are given for two pore pressure
coefficients of ru=0 and 0
0.25.
25 Linear
interpolation may be used for other values or
ru.

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Two-part
Two part wedge method of design for the earlier case

Slope angle,  = 70

Height of soil, H = 9 m
Friction angle of soil,  = 30
weight,  = 20 kN/m3
Unit weight
Uniform surcharge, q= 20 kPa
Modified height, HH = H + q/ = 9 + 20/20 = 10 m

Lateral earth pressure coefficient, K = 0.24

Design lateral force, P = 0.5 K H2

0 5*0 24*20*102= 240 kN/m
= 0.5*0.24*20*10
Minimum number of layers = 240/25 = 10
L/H = 0.66, L = 6.6 m  7m
L/H
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 Maximum permissible spacings
Depth Vertical pressure Permissible
from (kPa) Spacing = TLTDS/Ka*v Provided
surfac spacing
e (m)
1 1*20 + 20 = 40 25/(0.24*40)=2.6 m 1.0 m
2 2*20 + 20 = 60 25/(0.24*60)=1.7m 1.0 m
3 3*20 + 20 = 80 1.3 m 1.0 m
4 4*20 + 20 = 100 1 04
1.04 10m
1.0
5 5*20 + 20 = 120 0.87 0.75 m
6 6*20
6 20 + 20 = 140 0.74 0.50 m
7 7*20 + 20 = 160 0.65 m 0.50 m

8 8*20 + 20 = 180 0.58 m 0.50 m

9 9*20 + 20 = 200 0.52

0 52 m 0 50 m
0.50
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23
The reinforcement layers are provided at
the following depths:

9 , 8.5, 8.0, 7.5, 7, 6.5, 6, 5.25, 4.25,

3.25, 2.25, 1.25, 0.75, 0.25
Total No
No. of layers = 14
Reinforcement q quantityy = 14*25 = 350
kN/m
Length h off reinforcement
f layers
l = 7m
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24
Factor of safety against base sliding
Lateral force = 0.5
0.5*0.24*20*9*9
0.24 20 9 9 + 0.240.24*20*9
20 9
= 237 kN/m
Resistance force = 0.8*0.5*220*tan(30)*7 = 355
/ (base
kN/m ( slidingg factor of 0.8 is used to be
consistent with the design parameters)
Factor f t = 355/237  1.5
F t off safety 1 5 (safe)
( f )

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Optimisation of Reinforcement Layers

It is preferable to keep the vertical spacing as

low as possible.
p
It is possible to optimise the reinforcement
layout by using lower strength reinforcement
at the top of the embankment.

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 Geocell layer Geocell layer

Geocell layers used as anchor blocks to increase the

b i capacity
bearing i – concept is
i similar
i il to skirted
ki d
foundations

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Geocell layers used as thrust blocks – basal
reinforcement wrapped around the geocell blocks to
prevent pullout failure

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CONCLUSIONS

Consideration of the normal and shear

components
p of reinforcement forces
results in more economical deigns

Two-part wedge method off analysis is

more realistic as granular soils are used
for construction of most embankments.
embankments

Reinforced Soil Embankments - 3 29

Reinforced Soil Embankments - 3 30