Design and Analysis of MSE walls and

Design and Analysis of MSE
Walls and Embankments
Amit Prashant
Indian Institute of Technology Gandhinagar
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Dr. Amit Prashant
Reinforced Soil Concepts

Reinforcement placed parallel to principal strain direction
to compensate for lack of tensile resistance.

Mainly two mechanisms for ‘Stress Transfer’


Friction

Passive Resistance
The contribution of each transfer mechanism for a
particular reinforcement will depend on

Roughness of surface

Normal effective stress

Grid opening dimensions

Thickness of transverse members
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Dr. Amit Prashant
Stress transfer mechanisms
Frictional
Resistance
Passive
Resistance
Normal Pressure
Normal Pressure
Pullout Force
Pullout
Force
Frictional Resistance
Normal Pressure
Passive Resistance
Frictional Resistance
Pullout Force
Passive Resistance
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Dr. Amit Prashant
Evaluation of Pullout Performance

Pullout capacity


Allowable displacement


Pullout resist of each reinforcement should be adequate to resist
the design working tensile force with a specific factor of safety
Soil – reinforcement displacement required to mobilize the
design tensile force should be smaller than the allowable
displacement.
Long term displacement

Pull load < critical creep load
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Dr. Amit Prashant
Pullout Capacity
Pr  F *    v  Le  C
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Dr. Amit Prashant
Pullout Capacity

The F* can be obtained most accurately from laboratory
or field test. This can be derived from empirical or
theoretical relationship given below.
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Dr. Amit Prashant
Allowable Tensile Force/Width
The allowable tensile force per unit width of reinforcement, Ta
A- Metallic Reinforcement


Steel Strips
Ta  0.55
F y Ac
F.S = 1.8
b
Steel grids connected to concrete panels
Ta  0.48
b – Gross width of the strip, sheet of grid
Fy = Yield stress of steel
Ac =Design cross section area
F y Ac
F.S = 2.1
b
Greater potential for local overstress
due to load nonuniformities
B- Geosynthetic Reinforcement

The allowable design load considers all time dependent strength
losses
Ta 
TULT
T
 al
RF . FS FS
Ta – Design long-term reinforcement tension load
RF – Product of all applicable reduction factors
FS – Overall factor of safety
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Dr. Amit Prashant
Design using Limit Equilibrium
The overall stability of the structure needs analysis for
External, Internal and Combined stability

External stability involves the overall stability of the
stabilized soil mass considered as a whole and is
evaluated using slip surfaces outside the stabilized soil
mass

Internal stability analysis evaluates potential slip
surfaces within the reinforced soil mass

In some cases the slip surface is partly outside and
partly inside the reinforced zone. Hence: Combined
Analysis.
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Dr. Amit Prashant
External Stability
Sliding
Overturning
Deep
Seated
Stability
Bearing
Capacity
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Dr. Amit Prashant
External Stability Computational Steps
Define wall geometry and soil properties
Select performance criteria
Preliminary sizing
Evaluate external static stability
Sliding
Overturning
(eccentricity)
Bearing
capacity
Overall slope
stability
Settlement/lateral
deformation
Establish reinforcement length
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Dr. Amit Prashant
Performance Criteria and Preliminary Sizing
Performance Criteria

External stability FOS

Global stability FOS

Maximum differential settlement

Maximum horizontal displacement

Design life
Preliminary Sizing

A preliminary length of reinforcement is chosen should be greater of
0.7H and 2.5m

Structures with sloping surcharge fills or other concentrated loads
generally require longer reinforcements (0.8H to as much as 1.1H) for
stability
H: Design height of the structure
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Dr. Amit Prashant
Horizontal
back-face
with traffic
surcharge
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Dr. Amit Prashant
Sloping back-face
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Dr. Amit Prashant
Vertical Pressure Computations


Weight of any wall facing is typically neglected.
Calculation steps for determining vertical bearing stress are given in
the next slide

Compute eccentricity, e, of the resulting force on the base by moment
equilibrium about the center of reinforced base

Check if e < L/6 in soil and e < L/4 in rock. If not then increase
reinforcement length.
 Compute equivalent uniform vertical stress on the base, v:
Note this

Add the influence of surcharge and concentrated loads to v, where
applicable.
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Dr. Amit Prashant
Sliding Stability

The preliminary sizing should be checked w.r.t sliding at the base
layer
FSsliding 
 horizontal resisting forces
 horizontal driving forces

 PR  1.5
 Pd

Resisting force is the lesser of the shear resistance along the base
of the wall or of a weak layer near the base of the reinforced wall

Sliding force is the horizontal component of the thrust on the vertical
face at the back of the wall

Soil passive resistance at the toe due to embedment is ignored as
the soil may be removed
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Dr. Amit Prashant
Sliding Stability
Pd  FH  FT cos 

Calculate the driving force

Determine the most critical frictional properties fs of the base by choosing
minimum f for three possibilities below.




Sliding along the foundation soil if its shear strength (c f, ff) is smaller than that of
the backfill material.
Sliding along reinforced backfill (fr)
Sliding along weaker of the upper and lower soil-reinforcement interfaces.
Calculate the resisting force per unit length of wall:
PR  V1  V2  FT sin   .tan fs


Calculate the factor of safety w.r.t sliding and check if it is greater than 1.5
If NOT: Increase the reinforcement length, L, and repeat the calculations.
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Dr. Amit Prashant
Bearing Capacity Failure

General shear:

Vertical stress at the base should not exceed the allowable bearing
capacity of the foundation soil, determined considering a FOS of 2.5
qult
FS
qult  c f N c  0.5L f N f
 v  qa 


If NOT: Increase the reinforcement length, L, and repeat the calculations.
Local shear:

To prevent large horizontal movements of the structure on weak cohesive soils,
H  3c

If adequate support conditions cannot be achieved, ground improvement of
foundation soil is suggested
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Dr. Amit Prashant
Overall Stability

Overall stability is determined using rotational or wedge analyses
which can be performed by using a classical slope stability analysis
method

The reinforced soil wall is considered as a rigid body and only failure
surfaces completely outside a reinforced mass are considered

For simple structures (rectangular geometry, relatively uniform
reinforcement spacing and a near vertical face) compound failure is
normally not critical

For complex structures, compound failures must be considered

If FOS < 1.3, increase reinforcement length or improve foundation
soil
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Dr. Amit Prashant
Settlement Estimate

Conventional settlement analyses to ensure that immediate,
consolidation and secondary settlement of the wall satisfy the
performance requirements of the project

Significant total settlements at the end of construction indicate that
the planned top of wall elevations need to be adjusted

Significant differential settlements (greater than 1/100) indicate the
need of slip joints, which allow for independent vertical movement of
adjacent pre-cast panels

Where the differential settlement cannot be taken care of by these
measures, consideration should be given to ground improvement
techniques like wick drains, stone columns, dynamic compaction,
use of lightweight fill etc.
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Dr. Amit Prashant
Internal Failure of Reinforced Earth Walls

Internal failure of a reinforced earth wall can
occur in two different ways

Failure by elongation or breakage of reinforcement:
The tensile forces in the inclusions become so large that
the inclusion elongate excessively or break

Failure by pullout:
The tensile forces in the reinforcements become larger
than the pullout resistance which increases shear stresses
in the surrounding soil leading to large movements and
possible collapse.
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Dr. Amit Prashant
Internal Design Process

Select a reinforcement type

Select the location of critical failure surface

Select a reinforcement spacing

Calculate the maximum tensile force at each reinforcement level

Calculate the maximum tensile force at the connection to the facing

Calculate the pullout capacity at each reinforcement level
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Dr. Amit Prashant
A – Critical Slip Surface

The most critical slip surface in a simple reinforced soil wall is
assumed to coincide with the maximum tensile forces line

The shape and location of this line is assumed to be known from a
large number of previous experiments and theoretical studies

The maximum tensile force surface is assumed to be
approximately bilinear in the case of inextensible reinforcement,
approximately linear in the case of extensible reinforcement

Where the wall front batter is greater than 8 degrees the Coulomb
earth pressure relationship may be used to identify the failure
surface
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Dr. Amit Prashant
Potential Failure Surface For internal Stability
Inextensible Reinforcement
0.3H1*
Zone of maximum shear stress
or potential failure surface
H1  H 

La
Le
Active
Zone
Resistant
Zone
H1/2
H1
H
tan   0.3H
1  0.3 tan 
*If wall face is battered an offset
of 0.3H1 is still required, and the
upper portion of the zone of
maximum stress should be
parallel to the wall face.
Soil reinforcement
H1/2
L
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Dr. Amit Prashant
Potential Failure Surface For internal Stability
Extensible Reinforcement
Zone of maximum shear stress
or potential failure surface


For vertical wall
H
La
Le
Active
Zone
Resistant
Zone
  45 
fo
2
Soil
reinforcement
L
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Dr. Amit Prashant
B- Calculation of Maximum Tensile Forces in the
Reinforcement Layers

The maximum tensile force is primarily related to the
type of the reinforcement which is a function of the
modulus, extensibility and density of reinforcement

The resulting Kr/Ka for
inextensible
reinforcements ratio
decreases from the
top of the wall to a
constant value below
6m
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Dr. Amit Prashant
Maximum Tensile Forces (cont.)

The simplified coherent gravity method is used

Coeffcient of Lateral Earth Pressure is determined by
applying a multiplier to Ka.

For vertical walls use the active
earth pressure coefficient

For wall face batters equal
to or greater than 80 use
simplified form of Coulomb
equation
K a  tan 2 (45   ' )
2
Ka 
sin 2 (   ' )
 sin  ' 
sin  1 
 sin  
2
3
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Dr. Amit Prashant
Maximum Tensile Forces (cont.)

Calculation steps of
maximum tensile
forces
L/2

f, Kaf, ff
Retained Fill
L.tan 
2 
f
2
S 2
1. Calculate the horizontal
stress, H
Z
 H  K r v   h
Reinforced
soil mass
f, Kaf, Ko, ff
Zp =depth of soil
reinforcement layer
at beginning of
resistant zone, for
pull out calculations
where
 v   r Z   2  q   v
v – Increment of vertical stress due to
concentrated vertical loads
h – Increment of horizontal stress due to
horizontal concentrated surcharge
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Dr. Amit Prashant
Maximum Tensile Forces (cont.)
2. Calculate the maximum tension, Tmax
Tmax   H . Sv
-
For discrete reinforcements
Tmax   H . Sv .Sh
-
For discrete reinforcements and segmental
concrete facing
Tmax   H . At
At – area of (2 panel widths x the vertical spacing S v)
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Dr. Amit Prashant
Internal Stability with respect to breakage of the
reinforcement
3. Calculate internal stability with respect to breakage
of the reinforcement
Ta 
Tmax
Rc
The connection of the reinforcements with the facing,
shall be designed for Tmax for all loading conditions
Rc is the coverage ratio b/Sh
Ta - The allowable tension force per unit width of the reinforcement
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Dr. Amit Prashant
C - Internal Stability with Respect to Pullout
Tmax 
Where:
1
F *. .Z p .Le .C.Rc .
FS PO
FSPO
=
Safety factor against pullout ≥ 1.5
Tmax
=
Maximum reinforcement tension
C
=
2 for strip, grid, and sheet type reinforcement

=
Scale correction factor
F*
=
Pullout resistance factor
Rc
=
Coverage ratio
.Zp
=
The overburden pressure due to all permanent loads
Le
=
The length of embedment in the resisting zone
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Dr. Amit Prashant
Stability with Respect to Pullout (cont.)

The required embedment length in the resistance zone
Le 

1.5 Tmax
CF * Z p Rc 
 1m
The total length of reinforcement, L
L  La  Le
- For MSE walls with extensible reinforcement
La  ( H  Z ) tan (45  f ' )
2
- For wall with inextensible reinforcement
Base up to H/2
Upper half of the wall
La  0.6 ( H  Z )
La  0.3H
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Dr. Amit Prashant
Thank You
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